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English Pages 908 [910] Year 2011
OPERATIONS RESEARCH
OPERATIONS. RESEARCH (For B.Com., B.Tech., BBA, MBA, BCA, MCA and other Classes)
By
Col. D.S. CHEEMA (Retd.) Consultant Fashion Technology Park, Chandigarh Former Principal Bharatiya Vidya Bhavan's Dayanand College of Communication and Management, Chandigarh Former Directo,� Department of Professional Studies D.A. V. College Chandigarh
UNIVERSITY SCIENCE PRESS
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Dedicated to The Cherished Memory of My Father S. Santokh Singh Cheema Who was a great Educationist and a Great Human Being
CONTENTS Preface to the Second Edition Preface to the First Edition CHAPTER 1. Introduction to Operations Research Histmy and Background of Operations Research Why Study Operations Research ? Definition of Operations Research Operations Management, Production Management, System Management and Operations Research Management of Systems Salient Features of Operations Research Operation Research Models Methodology of Operation Research Tools of Operation Research Important Applications of Operation Research Pitfalls in the Use of Operation Research for Decisionmaking Limitations of Operations Research Tips on Formulating Linear Programming Models CHAPTER 2. Linear ProgrammingI Introduction Formulation of Linear Programming Problems Graphical Method of Solving Linear Programming Problems Corner Point Solutions Computer Solution Methods CHAPTER 3. Linear ProgrammingII Introduction Solving Operations Research Problem using Simplex Method Minimization Problems (All Constraints of the type�) Big 'M' Method ... Minimizing CaseConstraints of Mixed Type(:; and�) Limitations of LPP Simplex Method Sensitivity Analysis CHAPTER
4. Linear ProgrammingIII Concept of PrimalDual Relationship or Duality in Linear Programming Dual Problems when Primal is in the Standard Form Formulation of the Dual of the Primal Problem Interpreting PrimalDual Optimal Solutions
xi xii 1 1 1 2 3 3 6 6 7 8 10 11 12 12 17 17 19 35 76 79 97 97 97 117 122 141 142 158 158 159 160 172
VIII Contents
CHAPTER
5.
Dual Simplex Method Revised Simplex Method The Transportation Problems Introduction Terminology Used in Transportation Model Assumptions of Transportation Model Solution of the Transportation Model NorthWest Corner Rule RowMinima Method Column Minima Method Least Cost Method Vogel's Approximation Method (VAM) Performing Optimality Test Feasible Solution by VAM Optimality Test by MODI Method or UV Method
CHAPTER
6.
Assignment Model · Introduction Definition of Assignment Model Practical Steps Involved in Solving Minimization / Maximization Problems Unbalanced Problems
CHAPTER
CHAPTER
7.
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The Sequencing Problems Introduction Types of Sequencing Problems Replacement Theory Introduction Replacement Policy for Equipment/Policy that Breaks Down/Fails Suddenly
CHAPTER
9. Waiting Line (Queuing) Theory Objective and Models of the Theory Benefits and Limitations of Queuing Theory Important Terms Used in Queuing Theory MultiChannel Queuing Model (Arrival Poisson and Service Time Exponential) Poisson Arrival and Erlang Distribution for Service
CHAPTER
10. Decision Theory Introduction Decision Theory Approach Environment in, which Decisions are Made
175 186 198 198 '200 200 202 203 203 204 205 206 207 209 211 265 265 266 270 279 319 319 320 350 350 363 377 377 379 381 395 406 412 412 412 413
Contents CHAPTER
11. Theory of Games
Introduction Terms Used in Game Theory Limitations of Game Theory Situations of Twoperson Zerosum Pure Strategy Games Concept of Value of Game Concept of Saddle Point or Equilibrium Point Dominance Method or Principle of Dominance Approximation Method CHAPTER 12. Inventory Management Introduction Reasons for Carrying Inventories Classification of Inventories The Inventory Decision· Inventory Costs Objectives of Inventory Management Developing an Inventory Management Model Steps Involved in Developing an Inventory Model The Economic Order Quantity (EOQ) or Wilson's Lot Size Formula Graphic Method Algebraic Method Deterministic Inventory Model with Shortage (Back Order Model) Concept of Safety Stock or Buffer Stock Selective Inventory Management Objectives of ABC Analysis Some Limitations and Observations Probabilistic Inventory Models CHAPTER 13. Quality Management Introduction Historical Development of Quality Management Causes of Product Quality Variations Stable Variation Patterns and Control of Quality Objectives of Quality Control Traditionally used Quality Control Techniques Inspection and Quality Control Inspection Planning The Cost of Inspection and its Reduction CHAPTER
14. Investment Analysis and BreakEven Analysis Introduction Phases of Investment Decisions
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442 443 443 444 445 445 450 494 509 509 510 510 512 512 514 515 516 517 517 518 536 537 543 545 545 555 575 575 576 577 578 578 578 582 582 589 600 600 601
X Contents Factors Involved in Investment Decisions Compound Interest and Terminal Values Breakeven or Cost Volume Profit (CVP) Analysis
603 604 623
CHAPTER 15. Project Management PERT and CPM Introduction Project Management Network (Arrow Diagram) Steps in Project Crashing Probability and Project Planning CHAPTER 16. Simulation Introduction Scope of Simulation Applications Advantages of Simulation Teclmique Limitations or Demerits of Simulation Technique MonteCarlo Simulation Technique Random Number Generation CHAPTER 17. Work Study and Value Analysis Introduction Method Study Method Sh1dy Symbols Work Measurement Objectives of Work Measurement Techniques of Work Measurement Some Examples of Computation of Standard Time Work Sampling Value Analysis CHAPTER 18. Probability Theory and Markov Analysis Probability Theory Classical Definition of Probability The Normal Probability Distribution Markov Analysis CHAPTER 19. Concept of Goal, Integer and Dynamic Programming Concept of Goal Programming Concept of Integer Programming Limitations of Integer Linear Programming Methods of Integer Programming Integer Programming Formulation Concept of Dynamic Programming Formulation and Solution of Dynamic Programming Problems Appendices
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PREFACE TO THE SECOND EDITION Good, Better, Best, Never let it rest. 'Till your Good is Better And your Better is Best! Anonymous Although it is quite gratifying for me as an author to witness the success of this book, I do believe that a good book can be made better. It has been my effort to see the 'Good' becoming 'Better' in the second edition so that it is able to meet the needs of the shtdents much better than the first edition. Some of the chapters have been added, some sketches improved and printing mistakes corrected. In the process, I have hed my best to maintain the userfriendly style of the book. I truly hope that the students will get benefited by this revised edition. Through the medium of this book, I will continue to support the students all the way.
Author
PREFACE TO THE FIRST EDITION The discipline of Operations Res.earch is the gift of the Armed Forces to the modern society. This is a humble effort of an Army Officer who has spent years in teaching the subject at different levels. I have tried to make the book as studentfriendly as possible and it has been written in "Learn Yourself" style. The book adopts a fresh and novel approach to the learning of a subject which most of the students are scared of. I wish to express, through the medium of this book, my gratihtde to a large number of students, teachers, authors, publishers and institutions whose work I have consulted and got myself ben efited from. The teaching and shident community is requested to provide suggestions for improvement of the content and style of the present edition. In a subject like Operations Research certain mistakes in printing can l;Je expected, though the publishers have taken the pains to make the book error free. This book would not have been possible without the cooperation of my wife Mann and my two grandchildren Muskan and Ishaan who provided me the muchrequired destressing doses on daily basis during the days I was working on the book Author
Introduction to Operations Research 'LEARNING. OBiECTIVES . . ,
• • • • • • • •
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Operation Research (OR) as a Subject of Study and Understanding the History. Understanding Production Management, Operation Management, System Management and OR Use of OR in Different Fields as a Tool of Decisionmaking· Concept of Mathematical Modeling and Development of a Model Understanding the Steps in Application of OR. Classification of OR Techniques. Limitations in the Use of OR. Relationship of OR and Management.
HISTORY AND BACKGROUND OF OPERATIONS RESEARCH In the books of management one often finds a specific period of the development of management thought, called the Period of Scientific Management. It was in 1885 that Fredrick W. Taylor, "father of scientific management", developed the scientific management theories. It was also called the Modern era when rapid development of concepts, theories and techniques of management took place. During World War II, production bottlenecks forced the government of Great Britain to look up to scientists and engineers to help achieve maximum military production. These scientists and engineers created mathematical models to find the solution of the problems about increasing production of military equipments. This branch of study was called Operations Research (OR). Since it was used in the research in war operations of armed forces. These problems of the armed forces seemed to be similar to those that occurred in production systems. Because of the success of OR in military operations and approach to war problems it began to be used in industry as well.
WHY STUDY OPERATIONS RESEARCH ? We basically helps in determining the best (optimum) solution (course of action) to problems where decision has to be taken under the restriction of limited resources. It is possible to convert
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any real life problem into a mathematical model. The basic feature of OR is to formulate a real world problem as a .mathematical model. Since in the production industry, most of the manufacturers want to lower their labour or production costs to achieve higher profits, OR can be very usefully implied to real life production problems. OR should be seen as a problemsolving technique. Like management, OR is also a both Science and an Art. The Science part of OR is using mathematical techniques for solving decision problems. The Art part of OR is the ability of the OR team to develop good rapport with those supplying information and those who have to implement the recommended solutions. lt is important that both the Science and the Art parts of the OR are understood properly as a system of problemsolving. In lndia, OR society was formed in 1950's. The Journal of Operations Research has the mission to serve entire OR community including practitioners, researchers, educators and students. It celebrated its 50th Anniversary of Operations Research and published Anniversary issue in Jan.Feb. 2001. Industry has become quite aware of the potential of OR as a technique and many �strial and business houses have OR teams working to find solutions to their problems. Particularly, Railways, Indian Airlines, Defence Forces, Telco, DCM, etc., are using OR to their advantage. As a matter of fact, some techniques of OR like Programme Evaluation and Reviewing Techniques (PERT) and Critical Path Method (CPM) are frequently used by many organizations for effective planning and control of the construction projects. OR helps in taking decisions which optimize (maximize) the interest of the organization, it is a decisionmaking tool and should be seen as such. Many individuals and organizations see it as a management fad, which has limited use to them. At the sam'." time, tendency of some organizations to forcefit OR to prove that they use managerial techniques whether their functioning needs demand OR or not must be curbed. However, this is also true that complex real life problems can be solved for the advantage of the organizations by using OR techniques.
DEFINITION OF OPERATIONS RESEARCH Many authors have given different interpretation to the meaning of Operations Research as it is not possible to restrict the scope of Operations Research in a few sentences. Students must understand that there is no need to single definition of Operations Research which is acceptable to everyone. Two of the widely accepted definitions are provided below for understandmg the concept of Operations Research. "Operations Research is concerned with scientifically deciding how best to design and operate manmachine system usually under conditions requiring the allocation of scarce resources."  Operations Research Society of America
The salient features of the above definition are (a) (b) (c) (d)
It is a scientific decisionmaking technique. It deals with optimizing (maximizing) the results. It is concerned with manmachine systems. The resources are limited.
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!NT� ODUCTION TO OPERATIONS RESEARCH
"Operations Research is a scientific approach to problem solving for executive management."
 HM Wagner The above definition lays emphasis on
(a) OR being a scientific technique. (b) It is a problemsolving technique. (c) It is for the use of executives who have to take decisions for the organizations. A close observation of the essential aspects of the above two definitions will make it clear that both are in reality conveying the same meaning. Other definitions of OR also converge on these essential features. One need not remember the definitions word by word but understand the true meaning of the definition provided by different authors. The emphasis has to be on the application of technique so that organizations are benefitted. Hence, the real work of any managerial technique is the ability of the organizations to take advantage for meeting their objectives.
OPERATIONS MANAGEMENT, PRODUCTION MANAGEMENT, SYSTEM MAN AGEMENT AND OPERATIONS RESEARCH All the above subjects are interrelated and one must understand the fundamental concepts of these subjects before one is ready to study the details. Any production function brings together men, machines and materials. These are used to provide goods and services, which satisfy the needs, wants and desires of the people. For long, the term 'production' has been associated with factory like situations where goods are produced in the physical sense. In fact, a factory is defined as "a premises where people are employed for
making, altering, repairing, ornamenting, finishing, cleaning, working, breaking, demolishing or adopting any article for sale." For example, in a factory the mass production of any household product or goods may take place. The management of production of such goods is important but equally important is the management of the service part associated with it. Similarly, one can distinguish between the production of say McDonald's burger which is a product and its delivery, which is a service. If we generalize the concept of production as the 'process through which goods and services are created', both manufacturing of goods and service organization can be included in production management. Thus, nonmanufacturing processes like health, transport, banks, education etc., come tmder the scope of production management. It is because of this reason that term 'production' or 'operations management' has been suggested by many authors to include the application of teclmiques of management of men, machines and materials. So, general concept of 'Operations' and not production can include both manufacturing and service organizations. There is an operation function involved in all enterprises, big or small.
MANAGEMENT OF SYSTEMS Understanding system concept is important for understanding the process of operations manage ment. A system can be defined as a purposeful collection of men, objects and processes for operating within a particular environment. It includes two aspectsmanagement and systems. Management essentially is a social process involving coordination of human and material resources through
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the functions of forecasting, planning organizing and staffing, directing and controlling in order to accomplish the objectives of an organization in an effective manner. Effectiveness relates to achievements of right objectives by following the right methods. Efficiency relates to achievement of the objectives by using minimum amount of resources. For reaching any place one first decides the direction in which one should proceed and then runs very fast to reach the place as early as possible. A production system has the following five parts : (a) Inputs (b) Conversion process (c) Outputs (d) Feedback (e) Operating environment. A simple production system is explained in Figure. 1.1.
Inputs ❑ ❑ ❑ ❑
Land Labour Capital Management
Operating Environment Conversion Process
Outputs ❑ Goods ❑ Services
Manufacturing Assembling Delivery
Fig. 1.1 Production System The above basic process converts the inputs into outputs. Inputs are resources, which can be divided into: (a) Human resources,
(b) Material resources and (c) Time as a resource. It should be understood that all types of intangible resources, whether it is the management process, energy, knowledge, techniques etc., can be included in either the human or material resources. Output goods and services are obviously those, which are desired, and meet the specifications laid down. Operating environment in which the resources are converted into desirable results is of utmost importance. Environment provides different quality of manpower and other resources. For ease of understanding, let us take an example of Maruti Udyog Limited (MUL) production facilities established at Gurgaon. The type of workforce available and their work ethics as also the advantage of Delhi next door are the unique environment benefits as compared to locations in other parts of the country. Inputs to the operating system shown above are not only land, labour and capital but also patients in a hospital, vehicles in a repairing workshop, customers in a bank, passengers of road and railways, guests in a hotel and so on. All the inputs go through the conversion process and output may be in the form of finished or semifinished goods are: goods transported by the
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train/trucks plane, ship, messages delivered by a courier company post offices mobile phones or patients treated by a doctors of a hospital or a vehicle repaired by a mechanic of a workshop, etc. Another feature of the production system is the productivity of the conversion process. Productivity or the efficiency of conversion process can be found out by the ratio of output to input. Productivity (P) — Output (0) Input (I) It is clear from the above definition of productivity that it can be improved. (a) By increasing the output of specific and desirable standards with the same inputs. (b) By decreasing the level of inputs for same outputs. (c) By increasing outputs and decreasing inputs to change the productivity ratio favourably. The difference between the production and productivity must be understood. Whereas production is a process, productivity is a ratio as explained above. An essential part of any production system is the 'waste' or 'scrap' generated in the conversion process. The material may not be actually being processed (for example, a piece of iron not be loaded in the lathe and the tool of the lathe machine cannot cut it), it may only be waiting in store or at the workstation as either the machine or man is not available. Similarly, the vehicles may be waiting in a workshop because the numbers of repairing facilities are not adequate. In efficient production systems, this waste has to be reduced to increase the productivity of the system. If the outputs are not what is wanted, the system must be corrected through the process of feedback. Outputs must be monitored, a comparison between the actual outputs and desired outputs made and the difference fed to the conversion process for necessary correction. This is the feedback and is an important part of the production system. A conceptual model of a production/operation system is shown in Figure 1.2.
Subsystem to feed correction
Subsystem to maintain output
Inputs LI O O O
Land Labour Capital Management
0 Goods 0 Services
Process
_0.
t
Output
Conversion
t
Feedback Comparison between Desired and Actual
4—
t Operating Environment t Fig. 1.2 Conceptual model of a production/operation system
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SALIENT FEATURES OF OPERATIONS RESEARCH After having understood the basic concept of OR and the need, one can easily understand its salient features. 1. System Approach : OR is a systematic approach as is clear from the conceptual model of OR explained above. It encompasses all the subsystems and departments of an organization. Since it is a technique that effects the entire organization, optimizing results of one part of the organization is not the proper use of OR. Before applying OR techniques the management must understand its impact and implications·on the entire organization. 2. OR is both a Science and an Art : OR has the scientific orientation because of its inherent methodology and scientific methods are used for problemsolving. But its implementation needs the art of taking the entire organization along. OR does not perform experiment but helps in finding out solutions. OR must take into account the human factor which is the most· important factor in implementing any technique/methods of problemsolving. 3. Interdependency Approach : Problem of organizations could be related with economics, engineering, infrastructure related with markets, management of humanresources and so on. If OR has to find a solution to problems related to diverse fields, the OR team must be consti tuted of members with background disciplines of science, management and engineering etc. Only . then, practical solutions which can be implemented, can be found to the advantage of orgaruzations. 4. Management Decisionmaking : Management of any organization has to make decision, which has, impact on its profitability. All business organizations exist to make profits. Nonbusiness organizations like hospitals, educational institutions, NGOs etc., generate profits by reducing the inputs and increasing the outputs through effective and efficient manage ment. Decisionmaking involves generating different alternatives and selecting the best under · ,. the given situation. OR helps in making the right decisions. 5. Quantitative Technique: OR is a quantitative technique, which uses mathematical models and finds rational quantitative solutions to the managerial problems. The management may use the OR inputs and take into account the quantitative analysis of the problem in finding the solution in the best interest of the organization. 6. Use of Information Technology (IT) : OR extensively uses the IT for complex mathematical problems to its advc;\ntage, OR approach to decisionmaking depends heavily on the use of computers.
OPERATION RESEARCH MODELS What is a Model ? It is very difficult to represent the exact real life situations on a piece of paper. A model attempts to represent reality of the situation by identifying all factors of situation and by establishing some relationship between them. In real life situations, there are so many uncertainties and complexi ties, which cannot be exactly reproduced. Model helps in identifying such uncertainties and complexities in terms of different factors.
Types of Operation Research Models Following are some important types of operations research models:
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Symbolic or Mathematical Models This is the most important type of model. Mathematical modelling focuses on creating a math ematical representation of management problems in organisations. All the variables in a particu lar problem are expressed mathematically. The model then provides different outcomes, which will result from the different choices the management wishes to use. The best outcome in a particular situation will help the management in decisionmaking. These models use set of mathematical symbols and hence are also called symbolic models. The variables in many business and industry situations can be related together by mathemati cal equations. To understand the concepts of symbolic or mathematical model, visualise a balance sheet or profit and loss account as a symbolic representation of the budget. Similarly, the demand curve in economics can be seen as symbolic representation of the buyers' behaviour at varying price levels.
Simulation Models In simulation model, the behaviour of the system under study is 'initiated over a period of time'. Simulation models do not need mathematical variables to be related in the form of equations, normally, these models are used for solving such problems that cannot be solved mathematically. Simulation is a general technique, which helps us in developing dynamic models, which are similar to the real process. Developing good simulation models is difficult.because 'creating' a real life situation to perfection is extremely difficult.
Iconic Models .
.
These. models represent the physical simulations to the real life system under study. Physical dimensions are scaled up or down to simplify the actual characteristics and specifications of the . system. Preparation of prototype models for say, an automobile or 3D plant layout are some examples of iconic models. These are the physical replica of a system and are based on a smaller scale than the original. The models have all the operating features of the actual system. Flight simulators, missile firing simulators, etc., are also examples of iconic models.
Analog Models They are not the exact replica. Like the iconic models, these are smaller, simple physical systems as compared to the real life systems which are complex. These models are used to explain an actual system by analogy.
Deterministic Models When the change of one variable has a certain or definite change in the outcome, the model is called a Determmistic model. In fact, everything is absolutely clearly defined and the results are known. Economic Order Quantity (EOQ) is a deterministic model, as economic lot size can be exactly known, with change in one of the variables in the EOQ formula.
METHODOLOGY OF OPERATION RESEARCH There are many steps involved in application of OR. The methodology to be adopted involves the following:
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1. Observations of the Operating Environment: OR is a problemsolving technique. First step in solving the problem is the formulation of the problem. This is done through observation of the system and its environment. As much of information regarding the problem as possible is generated by using the researchers, observers etc. 2. Formulation of the Problem : Any problem has many interconnected factors related to the situation. The factors may or may not be in the control of the management. The factors which are relevant to the situation of the problem and are under the control of management must be identified. Once the problem area is known, different variables considered responsible for the problem are listed. Now, it is possible to define the problem in terms of the variables and their relationship. 3. Selecting and Developing a Suitable Model : At this stage, a suitable model which best represents the real life situation has to be selected. The model is developed to show the relations and interrelation between a cause and effect. Normally, the model is fully tested and modified to ensure that OR technique applied is able to solve the problem. 4. Collecting the Data: Next step is to collect the data required by the selected model. The process of the OR model in finding the solution depends to a large extent on the quantity and quality of data. More the data and lesser the errors in data, the quality of managerial decisions will be better. The required information can be obtained through observations or from recorded data or even based on experience and maturity of the OR team. 5. Finding the Solution : Once the model has been developed, it is possible to find the solution, OR solutions are under a particular situation and under certain assumptions. Many assumptions have to be made by the OR team to simplify the model. The solution is valid only under these assumptions. Once the solution by the OR techniques is found, certain input variables are changed to see the output. By this method the best possible solution can be found. 6. Presenting the Solution to the Management: The OR team has to present the solution to the management in a proper manner. The conditions under which the solution can be used and the conditions under which solution cannot work must be explained to the management. The assumptions made at arriving the solution and the weakness of the solutions should also be explained to the management. 7. Implementing the Solution: This is the last step in the OR application methodology. The solution provided by the OR technique is scientific but the application of this technique involves many behavioral aspects. This is the'art' part of OR and is of utmost importance. Any gap between the perception of the management and the approach of OR team must be removed.
TOOLS OF OPERATION RESEARCH Operation Research is a very versatile science and has many tools/techniques, which can be used for problem solving. However, it is not possible to list all these techniques as everyday new methods in the use of OR are being developed. Some of the tools of OR are discussed in the succeeding paragraphs 1. Linear Programming (LP): Most of the industrial and business organisations have the objec tives of minimizing costs and maximizing the profits. LP deals with maximizing a given objective. Since the objective function and boundry conditions are linear in nature, this math ematical model is called Linear Programming Model. It is a mathematical technique used to allocate limited resources amongst competing demands in an optimal manner. The applica tion of LP requires that there must be a welldefined objective function (like maximizing profits
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and minimizing costs) and there must be consfraints on the amount and extent of resources available for satisfying the objective ftmction. 2. Queuing Theory: In real life situations, the phenomenon of waiting is involved whether it is the people waiting to buy goods in a shop, patients waiting outside an Out Patient Depart ment (OPD), vehicles waiting to be serviced in a garage and so on. Because in general, customer's arrival and his service time is not known in advance; hence a queue is formed. Queuing or waiting line theory aims at minimizing the overall cost due to servicing and waiting. How many servicing facilities can be added at what cost to minimize the time in queue is the aim in the application of this theory. 3. Network Analysis Technique : A network can be used to present or depict the activities necessary to complete a project. This helps us in planning, scheduling, monitoring and con trol of large and complex projects. The project may be developing a new battle tank, construc tion of dam or a space flight. The project managers are interested in knowing the total project completion time, probability that a project can be completed by a particular time, and the least cost method of reducing the total project completion time. Techniques like Programme Evalu ation and Reviewing Technique (PERT) and Critical Path Method (CPM) are part of network analysis. These are popular techniques and widely used in project management. 4. Replacement Theory Model: All plants, machinery and equipment needs to be replaced at some point of time, either because there is deterioration in their efficiency or because new and better equipment is available and the old one has become obsolete. Sooner or later the equip ment needs to be replaced. The decision to be taken by the management involves consideration of the cost of new equipment which is to be purchased and what can be recovered from the old equipment tlU'ough its sale, or its scrap value, the residual life of the old equipment and many other related aspects. These are important decisions involving investment of capital and need to be taken very carefully. 5. Inventory Control : Inventory includes all the stocks of material, which an organization buys for production/manufacture of goods and services for sale. It will include raw material; semifinished and finished products, spare parts of machines, etc. Managers face the prob lems of how much of raw material should be purchased, when should it be purchased and how much should be kept in stock. Overstocking will result in locked capital not available for other purposes, whereas understocking will mean stockout and idle manpower and ma chine resulting in reduced output. It is desirable to have just the right amount of inventory at the right time. Inventory control models can help us in finding out the optimal order size, reorder level, etc., so that the capital resources are conserved and maximum output ensured. 6. Integer Programming: Integer programming deals with certain situations in which the vari able assumes nonnegative integer (complete or whole number) values only. In LP models the variable may take even a fraction value and the figures are rounded off to the nearest integer to get the solution, i.e., number of vehicles available in a problem cannot be in fractions. When such rounding off is done the solution does not remain an optimal solution. In integer pro gramming the solution containing unacceptable and fractional values are ruled out and the next best solution using whole numbers is obtained. An integer programming may be _called mixed or pure depending on whether some or all the variables are restricted to integer values. 7. Transportation Problems: Transportation problems are basically LP model problems. This model deals with finding out the minimum transportation cost for transporting the single commodity from a number of sources to number of destinations. Typical problem involves transportation of some manufactured products (say cars in 3 different plants) and these have
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to be sent to the warehouses of various dealers in different parts of country. This may be understood as a special case of simplex method developed for LP problems, allocating scare resources to competing demands. The main purpose of the transportation is to schedule the dispatch of the single product from different sources like factories to different destinations as total transportation cost is minimized. 8. Decision Theory and Games Theory: Information for making decisions is the most important factor. Many models of OR assume availability of perfect information which is called decision making under certainty. However, in real life situations, only partial or imperfect information is available. In such a situation we have two cases, either decision under risk or decision under uncertainty. Hence from the point of vie� of availability of information, there are three cases, certainty and uncertainty, the two extreme cases and risk is the "inbetween" case. Games theory is concerned with decisionmaking in a conflict situation where two or more intelligent opponents try to optimize their own decision. In Games theory, an opponent is referred to as a player and each player has a number of choices. The Games theory helps the decision maker to analyse the course of action available to his opponent. In decision theory, · we use decision tree which can be graphically represented to solve the decisionmaking problems. 9. Assignment Problems: We have the problem of assigning a number of tasks to a number of persons who may use machines. The objective is to assign the jobs to the machines in such_ a way that the cost is least. This may be considered a special case of LP transportation model. Here jobs may be treated as 'services' and machines may be considered the 'destinations'. Assignment of a particular job to a particular person so that ail the jobs can be completed in shortest possible time hence incurring the least cost, is the assignment problem. 10. Markov Analysis: Markov analysis is used to predict future conditions. It assumes that the occurrence of a future state depends upon the immediately preceding state and only on it. It is based on the probability theory and predicts the change in a system over a period of time if the present behaviour of the system is known. Predicting market share of the companies in future as also whether a machine will function properly or not in future, are examples of Markov analysis. 11. Simulation Techniques: Since all real life situations cannot be represented mathematically, certain assumptions are made and dynamic models which act like the real processes are developed. It is very difficult to develop simulation models which can give accurate solutions to the problems, but this is a good method of problem solving, when the problems are very complex and cannot be solved otherwise.
IMPORTANT APPLICATIONS OF OPERATION RESEARCH In today's world where decisionmaking does not depend on intuition; managerial techniques are widely used. All the applications of OR cannot be listed because OR as a tool finds . new application everyday. It finds typical applications in many activities related to work planning. Some important applications of OR are 1. Manufacturing/Production  Production planning and control  Inventory management. 2. Facilities Planning Design of logistic systems Factory+buildingJQgttion :?�s�ze decisions
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9.
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 . Transportation, loading and unloading ____::_ Planning warehouse locations. Accounting  Credit policy decisions  Cash flow and fund flow planning. · Construction Management Allocation of resources to different projects in hand  Workforce/labour planning  Project management (scheduling, monitoring and control). Financial Management  Investment decisions  Portfolio management. Marketing Management Productmix decisions  Advertisement/Promotion budget decisions  Launching new product decisions. Purchasing Decisions Inventory management (optimal level of purchase), Optimal reordering. Personnel Management Recruitment and selection of employees  Designing training and development programmes  Human Resources Planning (HRP). Research and Development . Planning and control of new research and development projects. Product launch planning.
PITFALLS IN THE USE OF OPERATION RESEARCH FOR DECISIONMAKING The first stage of OR application after collecting data/information through observation is the formulation of the problem. It is the most important and most difficult task in OR application. Have the OR team been able to identify the right problem for finding the solution ? Has the problem been accurately defined in unambiguous manner ? Selecting and developing a suitable model is not an easy task. The model must represent the real' life situation as far as possible. Collection of data needs a lot of time by a number of people. It is timeconsuming and expensive process. Collection of data is done either by observation or from the previous recorded data. When a system is being observed by the OR team, it effects the behaviour of the persons performing the task. The very fact that the workers know that they are being observed is likely to change their work behaviour. The second method of data collection, the records, are never reliable and do not provide sufficient information which is required. As OR problemsolving techniques is very timeconsuming, the quality of decisionmaking may become a causality. The management has to make a decision either way. Decision based on insufficient or incomplete informationwill not be the best decision. A reasonably good solution
12
OPERATIONS RESEARCH
without the use of OR may be preferred by the management as compared to a slightly better solution provided by the use of OR which is very expensive in time and money. Due to the above reasons, many OR specialists try and fit the solution they have, to the problem. This is dangerous and unethical and organizations must guard against this.
LIMITATIONS OF OPERATIONS RESEARCH Operation Research is an extremely powerful tool in the hands of a decisionmaker and to that extent the advantage of OR techniques are immense. Some of them are : (a) It helps in optimum use of resources. LP techniques suggest many methods of most effective and efficient ways of optimally using the production factors. (b) Quality of decision can be improved by suitable use of OR techniques. If a mathematical model representing the real life situation is wellformulated representing the real life situation, the computation tables give a clear picture of the happenings (changes in the various elements i.e., variables) in the model. The decisionmaker can use it to his advantage, specially if computer ised software can be used to make changes in variables as per requirement. The limitations of OR emerge only out of the time and cost involved as also the problem of formulating a suitable mathematical model, otherwise, as suggested above, it is a very powerful medium of getting the best out of limited resources. So, the problem is its application rather than its utility, which is beyond doubt. Some of the limitations are (a) Large number of cumbersome computations. Formulation of mathematical models which takes into account all possible factors which define a real life problem is difficult. Because of this, the computations involved in developing relationships in very large variables need the help of computers. This discourages small companies and other organisations from getting the best out of OR techniques. (b) Quantification of problems. All the problems cannot be qualified properly as there are a large number of intangible factors, such as human emotions, human relationship and so on. If these intangible elements/variables are excluded from the problem even though they may be more important than the tangible ones, the best solution cannot be determined. (c) Difficult to conceptualize and use by the managers. OR applications is a specialist's job, these persons may be mathematicians or statisticians who understand the formulation of models, finding solution and recommending the implementation. The managers really do not have the hang of it. Those who recommend a particular OR technique may not understand the problem well enough and those who have to use may not understand the 'why' of that recommenda tion. This creates a 'gap' between the two and the results may not be optimal.
TIPS ON FORMULATING LINEAR PROGRAMMING MODELS (a) Read the statement of the problem carefully. (b) IdentifiJ the decision variables. These are the decisions that are to be made. What set of variables has a direct impact on the level of achievement of the objectives and can be controlled by the decisionmaker? Once these variables are identified, list them providing a written definition (e.g., x1 = number of units produced and sold per week of product 1, x2 = number of units produced and sold per week of product 2).
INTRODUCTION TO OPERATIONS RESEARCH
13
(c) Identify the objective. What is to be maximised or minimised ? (e.g., maximize total weekly profit from producing product 1 and 2). (d) Identify the constraints. What conditions must be satisfied when we assign values to the decision variables? You may like to write a verbal description of the restriction before writing the mathematical representation (e.g., total production of product 1 > 100 units). (e) Write out the mathematical model. Depending on the problem, you might start by defining the objective function on the constraints. Do not forget to include the nonnegativity constraints. GRAPHICAL SOLUTION MI
Example : A firm manufactures two products. The products must be processed through one department. Product A requires 4 hours per unit and product B requires 2 hours per unit. Total production time available for the coming week is 60 hours. A restriction in planning the production schedule, therefore, is the total hours used in producing the two products cannot exceed. Also, since each variable represents a production quantity, neither variable can be negative. Determine the combination of products A and B that can be produced ? Solution. Let x1 represents the number of units produced of product A and x2 represents the number of units produced of B. Then the restriction is represented by 4 xi + 2 x2 60 The problem also implies that x1 0 and x2 0 In equation 4x1 + 2 x2 = 60 We can put different values of one variable to get the value of the other variable i.e., x1 = 0, x2 = 30 and x2 = 0, x1 = 15. Hence point A is (x1 = 0, x2 = 15) and Point B is (x1 = 15, x2 = 0). This is shown graphically here. (Units of product B)
30
25 20 15 10
(0, 0) 0 5
• (Units of product A) 10 15.. 20 25
Fig. 1.3 The shaded area represents the combination of products A and B which can be produced.
14
OPERATIONS RESEARCH
REVIEW AND DISCUSSION QUESTIONS What is the concept of Operation Research? Write a detailed note on its development. Discuss significance and scope of OR in business and industry. What are the different phases of OR? How is OR helpful in decisionmaking? Discuss briefly various steps involved in solving an OR problem. Illustrate with one example from the functional area of your choice. 5. Explain applications of Operations Research in business. 6. What is the significance and scope of Operation Research in the development of Indian Economy? 7. What is the role of OR in modern day business? Give examples in support of your answer. 8. Discuss the meaning, significance and scope of Operations Research. Describe some methods of OR. 9. Illustrate and explain various features of OR. 10. Define Operations Research in your own words and explain various tools of OR. 11. Give the role and significance of OR in business and industry for scientific decisionmaking. 12. "Operations Research is an aid for the executive in making his decisions by providing him with needed quantitative information based on the scientific method of analysis.'.' Discuss the statement and give examples to illustrate how OR is helpful in decisionmaking. 13. Briefly explain the technique of OR and its uses in Lndia. Which of the three techniques is most widely used in India and Why ? 14. Write a detailed essay on the scope and methodology of the use of OR. 15. "OR is useful only if applied with Information Technology." Comment. 16. Many believe that OR is a tecluuque which helps in resolving conflicts between production, finance, marketing and personnel functions of a manufacturing unit. Do you agree? Explain your answer giving examples. 17. What is a model? What are the types of models you are familiar with? What are the advantages and pitfalls of models? 18 . . Define an OR model. Give examples from industry and business to explain the use of models. 19. Define OR and discuss its scope. 20. Discuss the significance and scope of Operations Research ii1 modern management. 21. Write a detailed note on the use of models for decisionmaking. Your answer should specifically cover the following (i) Need for model building (it) Type of model appropriate to the situation (iii) Steps involved in the construction of a model (iv) Setting up criteria for evaluating different alternatives (v) Role of random numbers.. 22. "Modelbuilding is the essence of Operations Research approach." Discuss. 23. Give the essential characteristics of the following types of processes: . (a) Allocation (b) Competitive (c) Inventory (d) Waiting.line. 24. Comment on the following statements: (a) (i) OR is the art of winning wars without actually fighting them. 1. 2. 3. 4.
INTRODUCTION TO OPERA TIONS RESEARCH
15
(ii) OR is the art of finding bad answers where worse exist. (b) OR is not more than a quantitative analysis of the problems.
25.
26. 27. 28. 29. 30.
(c) OR advocates a system approach and is concerned with optimisation. It provides a quan titative analysis for decisionmaking. (d) OR replaces management by personality. Suggest a suitable OR model, giving reasons if any, for each of the following OR problems: (a) Stockpiling of crackers prior to Diwali. (b) Decision to replace the fleet of buses of a transport corporation after use for12 years even though some of them may be in working condition. (c) A book vendor deciding to place order for books before a new school session begins. (d) Modifications in the design of a new product due for launch in near future. (e) Statistical forecasting for sale of icecream. "OR replaces management by personality." Discuss. Write an essay on the scope and methodology of OR. Explain briefly the main phases of an OR study and techniques used in solving OR problems.. What are the steps involved in OR problems? What are the different types of models used in OR? Explain in detail. What are the situations when OR techniques will be applicable?
PUNJAB UNIVERSITY EXAMINATION PROBLEMS 1995SEP., 1998APR., 1999SEP., (1) Operations Research. 1998APR. (2) (i) Operations Research. (ii) Phases of solving a problem using Operation Research. 1995APR. (3) Discuss briefly the importance of Operation Research in decisionmaking. (Marks10) 1995 SEP. (4) Discuss briefly the limitations of Operations Research Techniques. (Marks10) 1998 SEP. (5) Discuss in brief of significance and scope of Operations Research in modern business management. (Marks 6) 1998APR. (6) Define Operations Research and explain its main characteristic features. Also give examples to highlight the scope of Operations Research from Industry and business view. (Marks 6) (7) Discuss significance and scope of Operations Research in business and 1998 SEP. industry. (Marks 6) ' . 1999APR. (8) Discuss significance and scope of Operations Research techniques. (Marks 6) 1999APR. (9) Discuss briefly various steps for solving an operations research problem. Illustrate with one example from the functional area of your choice. (Marks8) (10) Relate the methodology of operations research to that of scientific research and point out similarities and dissimilarities. (Marks 6) 2000SEP. (11) Explain the seven applications of operations research in business_. (Marks7) 2000SEP. (12) Discuss the phases of operations research. (Marks9) 2000SEP. (13) Illustrate the scope and limitation of operations research. (Marks7) 2001APR. (14) Explain application of OR in industries. (Marks9)
OPERATIONS RESEARCH
16
2001 APR. (15) What is significance and scope of OR in the development of Indian Economy ? (Marks 7) 2001 APR. (16) What is the role of OR in modern day business ? Give examples to support your answer. (Marks 9) 2002 APR. (17) Discuss the meaning, significance and scope of OR. Describe some methods of OR (Marks 18) (18) Explain : 2003 APR. 1. Methodology (Processes) of Operation Research. 2. Limitation of Operation Research. (Marks 10 + 8) 2004 APR. Explain the meaning, significance and scope of OR G.N.D.U. EXAMINATION PROBLEMS
1997 APR. 2000 APR. 2002 SEP. 2004 APR. 1996 SEP.
(1) (2) (3) (4) (5) (1)
1998 APR.
(2)
1998 SEP. 1999 APR. 1999 SEP.
(3) (4) (5)
2001 APR. (6) 2004 APR. (1) 2004 APR. (2) 1995 APR. 1995 SEP. 1999 APR.
(1) (2) (3)
2003 APR.
(1) (2)
2004 APR. (3)
Operations Research. Phases of solving a problem using Operation Research. Basic solution in Linear Programming Problems Discuss tools of Operation Research. O.R. is not more than a quantitative analysis of the problem. Discuss in brief the significance and scope of Operations Research in modern business management. Define Operations Research and explain its main characteristic features. Also give examples to highlight the scope of Operation Research from industry and business. Discuss significance and scope of Operations Research in business and industry. Discuss briefly the limitations of Operation Research techniques. Relate the methodology of Operations Research to that of scientific research and point out similarities and dissimilarities. Define Operations Research, give feature of OR. Briefly discuss the techniques or tools of OR. What are the situations where Operation Research techniques will be applicable ? Explain how and why Operations Research techniques have been valuable in aiding decisionmaking process. Give examples to support your views. Discuss briefly the importance of Operation Research in decisionmaking. Discuss briefly the limitations of Operations Research techniques. — Discuss briefly various steps for solving an Operation Research problem. Illustrate with one example from the functional area of your choice. What are the essential characteristics of Operations Research ? Mention different phases of an OR study. Point out its limitations, if any. Give application of OR in business and Industry. (B. Com. IIIrd Year Professional) Write an essay on the scope and methodology of Operations Research explaining briefly the main phases of OR study and techniques in solving OR problems.
Linear Program (Formulation of LPP and Graphical Solution Method)
LEARNING OBJECTIVES • • • • •
Understanding Linear Programming Problems. Understanding the Characteristics of an LPP. Application of LPP in Business and Industry. Demonstration of Use of LPP. Use of Graphic Solution in Solving LPP.
INTRODUCTION Linear Programming (LP) is a mathematical technique, which is used for allocating limited resources to a number of demands in an optimal manner. When a set of alternatives is available and one wants to select the best, this technique is very helpful. Management wants to make the best use of organizational resources. Human resources, which may be skilled, semiskilled or unskilled must be put to optimal use. Similarly, the material resources like machines must be used in an effective manner. Time is very important resource and any job must be completed in allotted time. Application of LP requires that the following conditions must be met: (a) There must be a welldefined objective of the organization such as: (i) Maximizing profit (ii) Minimizing cost. (b) This objective function must be expressed as a linear function of variables involved in decisionmaking. (c) There must be a constraint on availability of resources for the objective functions, i.e., for achieving maximum profit or for reducing the cost to a minimum. LP technique establishes a linear relationship between two or more variables involved in management decisions described above. Linear means it is directly proportional, i.e., if 5 per cent increase in manpower results in 5 per cent increase in output, it is a linear relationship.
OPERATIONS RWARCH
18
(d) Alternative course of action must be available to select the best, for example, if a company is producing four different types of products and wants to cut down one product, which one should stop manufacturing. The problem gives rise to a number of alternatives and so LP can be used. (e) Objective function must be expressed mathematically, i.e., we must be able to develop a linear mathematical relationship between the objective and its limitation. Linear equations are of first degree, i.e., if we want x and y as the variabiL, the equation 5x +.10y = 20 is a linear equation in which x and y can assume different values. However, an equation like 5x2 + 10y2 = 200 is not a linear equation, because of the variable x and y are squared, this is a typical second degree equation.
Formulation of Linear Programming Problem The formulation of linear programming requires the following steps: (a) (b) (c) (d)
Identifying/defining the decision variables. Specifying/defining the objective function to be maximized or minimized. Identifying the constraint equations, which have to be expressed as equalities or inequalities. Using the equation either in graphical or simplex method to find out the value of decision variables to optimize the objective function.
Assumptions for Solving a Linear Programming Problem The application of LP makes use of the following assumptions : (a) Linearity. The objective function and each constraint is linear. (b) Certain and Constant. It means that the number of resources available and production require
ments are known exactly and remain constant. (c) Nonnegative Variables. The values of decision variables are nonnegative and represent real
life solutions. Negative values of physical goods or products are impossible. Production of minus 10 refrigerators is meaningless.
Potential Applications of Linear Programming Some real life situations, where LP is very u ,eful are given below: 1. Product Mix Problem : Organizations often face the problems of making decision to manufac
ture different quantities of products, with the constraint of manpower, machines, availability of raw materials, etc. The idea is to minimize cost of production or maximization of profit under a given set of conditions. 2. Transportation Problems : LP finds typical use in finding solution to such problems. The problem is to transport products from a number of sources to a number of destinations with minimum cost. These are the real life situations where the goods have to be moved from the factory premises to warehouses in different parts of the country or from warehouses to Clearing and Forwarding (C ana 1) agents and so on. How many goods should be transported to meet the demands of different destinations so that the cost of transportation is minimum, can best be decided by use of LP. 3. Blending Problems : Large number of products use different types and quantities of raw materials. For example, in textiles industry a number of raw materials are used. The idea is to
LINEAR PROGRAMMINGI
19
make available different raw materials (with different specifications) in such quantities so that the product is manufactured at minimum raw material cost. Such problems are called blending problems. 4. The Diet Problem : This problem arises when one has to decide mixing of different type of foods to get a particular amount of nutritional values with minimizing cost of purchasing the diet. Hospitals can use LP methods for solving such problems. 5. Investment Decisions (Portfolio selection) Problem : This is a very common problem with those who want to make use of different investment opportunities. When the amount to be invested is fixed and different opportunities like investment in shares, bonds, mutual funds, postoffice schemes, banks, etc., are available, LP method can provide us the solution to get maximum returns. 6. Use of LP by Airlines : Operation of airlines routes is a very complex problem. With limited aircrafts and large number of destinations airlines would like to operate in the most economic routes at particular flight timings. LP is a very useful technique for solving such problems. FORMULATION OF LINEAR PROGRAMMING PROBLEMS Example 2.1. A manufacturing company is producing two products A and B. Each of the products A and B requires the use of two machines P and Q. A requires 4 hours of processing on machine P and 3 hours of processing on machine Q. Product B requires 3 hours of processing on machine P and 6 hours of processing on machine Q. The unit profits for products A and B are Rs. 20 and Rs. 30 respectively. The available time in a given quarter on machine P is 1000 hours and on machine Q is 1200 hours. The market survey has predicted that 250 units of products A and 300 units of product B can be consumed in a quarter. The company is interested in deciding the product mix to maximize the profits. Formulate this problem as LP model. Solution. Formulating the problem in mathematical equations Let XA = the quantity of product of type A manufactured in a quarter. XB = the quantity of products of type B manufactured in a quarter. Z = the profit earned in a quarter. (Objective function, which is to be maximized). Therefore, Z = 20 XA + 30 XB Z is to be maximized under the following conditions : 4XA + 3XB 1000 (Time constraint of machine P) 3XA + 6XB 1200 (Time constraint of machine Q) XA 5 250 (Selling constraint of product A) XB 300 (Selling constraint of product B) XA and XB ?_ 0 (Condition of nonnegativity). Example 2.2. M/s Steadfast Ltd. produces both the interior and exterior house paints for wholesale distribution. Two types of raw materials A and B are used to manufacture the paints. The maximum availability of A is 10 tons a day and that of B is 15 tons a day. Daily requirement of raw material per ton of interior and exterior paints are as follows :
20
OPERATIONS RESEARCH
Type of raw material
Requirement of Raw material per ton of paint Interior
Exterior
Max availability (tons)
Raw material A
3
2
10
Raw material B
2
3
15
The market survey indicates that daily demand of interior paints cannot exceed that of exterior paints by 2 tons. The survey also shows that maximum demand of interior paints is only 3 tons daily. The wholesale price per ton is Rs. 75000 for exterior paints and Rs. 50000 for interior paints. Problem. How much interior and exterior paints M/s Steadfast should produce to maximize its profits? Solution. Let XE — tons of external paints to be produced daily. XI — tons of internal paints to be produced daily. Z — Profit earned (objective function which is to be maximized). Therefore, Z = 75000 XE + 50000 X1 Z is to be maximized under the following constraints or conditions: 2XE + 3X1 5 10 (Availability constraint of raw material A) 3XE + 2X1 5_ 15 (Availability constraint of raw material B) XI —XE 2 (Demand constraint — Demand of interior paints daily cannot exceed more than 2 tons that of exterior paint) X/ 5. 3 (Demand of interior paint cannot exceed 3 tons everyday) Also, XI ?_ 0 (Nonnegativity constraint of interior paints) XE 0 (Nonnegativity constraint of exterior paints). The complete mathematical model for M/s Steadfast Ltd. problem may be written as given below: Determine the tons of interior and exterior paints, XI and XE to be produced in order to maximize Z = 75000 XE + 50000 X/ (objective function) under the constraints (conditions) of 2XE + 3)(1.5 10 3XE + 2)(1.5 15 — XE 2 XI _0 Example 2.4. M/s Plastics Ltd. produces two major products tables and chairs. Each table can be sold in the market at a profit of Rs. 50 and chair can be sold at a profit of Rs. 60. The requirement of manhours and machinehours for manufacturing of one table and one chair is as follows : $')
OPERATIONS RESEARCH
22 Product
Manhours (Hours per unit)
Table Chair
10 6
Machinehours (Hours per unit) 1.5 1.0
Total number of manhours available for this activity during the year is 1200 and machine is available only for 800 hours. The company carried out a survey to find out the maximum demand for their tables and chairs. It was revealed that they can sell only 100 tables and 160 chairs in one year. Problem. How many tables and chairs should M/s Plastics Ltd. produce to maximize their profit ? Solution. Let X be the number of tables to be produced. Y be the number of chair to be produced. Z = profit earned (Objective function which is to be maximized). Therefore, Z = 50X + 60Y Z is to be maximized under the following conditions: 10X + 6Y 5_ 1200 (Availability constraint of manhours) 1.5X + Y 800 (Availability constraint of machineshours) X 100 (Demand constraint) Y 5_ 160 (Demand constraint) X 0 (Nonnegativity constraint of tables) Y 0 (Nonnegativity constraint of chairs). Example 2.5. An oil refinery uses blending process to produce gasoline in a typical manufacturing process. Crude A and B are mixed to produce gasoline Gl and gasoline G2 . The inputs and outputs of the process are as follows : Input (tons) Output (tons) Process Crude A Crude B Gasoline (G1) Gasoline (G2) 1 1 2 6 8 2 8 5 7 6 Availability of crude A is only 200 tons and B 300 tons. Market demand of gasoline G1 is 150 tons and gasoline G2 is 120 tons. Profit by using process 1 is Rs. 200 per ton and by using process 2 is Rs. 250 per ton. What is the optimal mix of two blending processes so that the refinery can maximize its profits? Solution. Let X be the number of tons to be produced by process 1. Y be the number of tons to be produced by process 2. Z = Profit earned (Objective function which is to be maximized) . Therefore, Z = 200X + 250Y Z is to be maximized under the following conditions : X + 6Y 200 2X + 8Y 300 6X + 5Y 150 8X + 7Y 120 X>_0 Y>_0
LINEAR PROGRAMMINGI
23
Example 2.6. Vitamin C and Vitamin E are found in two different fruits F1 and F2 . One unit of fruit F1 contains 3 units of vitamin C and 2 units of vitamin E. Similarly, one unit of fruit F2 contains 2 units of vitamin C in it and 2 units of vitamin E in it. A patient needs minimum of 30 units of vitamin C and 20 units of vitamin E. Also one unit of fruit F1 costs Rs. 20 and one unit of fruit F2 costs Rs. 25. The problem, the hospitals faces is to find such units of fruit Fl and F2, which should be supplied to the patients at minimum cost. Solution. Let X be the number of units of fruit F1. Y be the number of units of fruit F2. Z = Minimum cost (Objective function which is to be minimized) . Therefore, Z = 20X + 25Y Z is to be minimized subject to the following constraints : 3X + 2Y 30 (Minimum requirement of vitamin C) 2X + 2Y 20 (Minimum requirement of vitamin E) X>_0 Y>_0 NOT II I I PP In this example, a unit means different measures like grams and calorific values, etc. Example 2.7. Manufacturing company XYZ Ltd. manufactures two different types of products, refrigerators and washing machines. Both these products have to be processed through two machines, Machine A and Machine B. Machine A is available for 200 hours and machine B is available for 100 hours. The requirement of time on these machines is as follows :
Machine A Machine B
Refrigerator 10 5
Washing machine 6 4
The company makes a profit of Rs. 800 on sale of one refrigerator and Rs. 500 on sale of one washing machine. What quantities of refrigerators and washing machines should company XYZ Ltd. produce to maximize its profits ? Solution. Let X be the number of refrigerators to be manufactured and Y be the number of washing machines to be manufactured. Z = Profit earned (Objective function which is to be maximized). Therefore, Z = 800X + 500Y Z is to be maximized subject to the following constraints : 10X + 6Y 200 (Availability of machine A) 5X + 4Y 100 (Availability of machine B) X>0 Y>0 Example 2.8. A manufacturer can manufacture two different types of products, FRP sheets and FRP bath tubs. Each unit of FRP sheets of a particular size needs 5 kg of raw material A and 2 kg of raw material B. Each unit of FRP bath tubs needs 7 kg of raw material A and 1 kg of raw material B. Availability of raw material A in the market is 500 kg and that of raw material B 100 kg. Each FRP sheet contributes profit of Rs. 100 and each FRP tub contributes profit of Rs. 400. What is the most suitable product mix for the manufacturer to maximize profits ?
OPERATIONS RESEARCH
24
Solution. Let X be the number of FRP sheets to be produced and Y be the number of FRP tubs to be produced. Z = Profit earned (Objective function which is to be maximized). Therefore, Z = 100X + 400Y Z is to be maximized subject to the following constraints: 5X + 2Y 5 500 7X + Y 5 100 X>_0 Y>0 Example 2.9. A company manufactures three types of electrical products, electric iron, fan and toaster. All the three products have to be processed on two machines A and B. The processing time required by each product on both the machines is as given below : Machine A Machine B
Electric Iron 2 1
Fan 3 2
Toaster 2 3
Machine A is available only for 200 hours and machine B is available for 160 hours. The firm should not manufacture more than 400 electric irons, more than 500 fans and more than 200 toasters. An electric iron gives a profit of Rs. 110, a fan of Rs. 150 and a toaster of Rs. 80. What product mix would you recommend to the company so that its profits are maximized ? Solution. Let X1 be the number of electric irons to be manufactured. X2 be the number of fans to be manufactured. X3 be the number of toasters to be manufactured. Z = profits generated (Objective functions which is to be maximized). Z is to be maximized under the following constraints : Z = 110X1 + 150X2 + 80X3 2X1 + 3X2 + 2X3 .5 200 + 2X2 + 3X3 5_ 160 Also, X1 , X2 , X3 0 Example 2.10. A manufacturer of furniture makes two products, chairs and tables. Processing of these products is done on two machines A and B. A chair requires two hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours of time per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and table is Rs. 2 and Rs. 10 respectively. What should be daily production of the two products ? Solution. Let X be the number of chairs to be manufactured and Y be the number of tables to be manufactured. Let Z = Profit earned (Objective function which is to be maximized). Z is to be maximized under the following constraints : Z = 2X + 10Y 2X + 5Y 5 16 (Availability constraint of machine A) 6X 5. 30 (Availability constraint of machine B) X>_0 Y>_0
LINEAR PROGRAMMINGI
25
Example 2.11. Sandeep Electric Company produces two products, motors and fans which are produced and sold on monthly basis. The monthly production of motors cannot exceed 100 and that of fans 150 because of the limitation of the facilities of production. Motor requires 12 manhours of labour per month and fan requires 12 manhours of labour per month. The company has a total of 50 employees. Profit margin on sale of motor is Rs. 110 and on sale of fan Rs. 90. Formulate a LP problem. Solution. Let X be the number of motors produced and Y be the number of fans produced. Z = Profit earned (Objective function) Z = 110X + 90Y X S 100 (Constraint of production of motors) Y 5_ 150 (Constraint of production of fans) 30X + 12Y 50 (Constraint of availability of manhours) X>0 Y>0 Example 2.12. A carpenter has 90, 80 and 50 running feet respectively of teak, plywood and rosewood. Product A requires 2, 1 and 1 running feet of teak, plywood and rosewood respectively. Product B requires 1, 2 and 1 running feet of teak, plywood and rosewood respectively. If A would sell for Rs. 48 and B would sell for Rs. 40 per unit, how much of each should he make and sell in order to obtain the maximum gross income out of his stock of wood ? Give a mathematical formulation to this linear programming problem. Solution. Let X be the number of products A produced and Y be the number of products B produced. Let Z = Gross income (Objective function) Z is to be maximized under the following constraints: Z = 48X + 40Y 2X + Y 90 (Constraint of availability of teak) X + 2Y 80 (Constraint of availability of plywood) X + Y 50 (Constraint of availability of rosewood) X>_0 Y>_0 Example 2.13. A retailer deals in two items only, item A and item B. He has Rs. 50000 to invest and a space to store at the most 60 pieces. An item of A costs him Rs. 2500 and B costs him 500. Net profit to him on items A is Rs. 500 and on item B is Rs. 150. If he can sell all the items that he purchases, how much should he invest his amount to have maximum profit ? Give a mathematical formulation to the linear programming problem. Solution. Let X be the number of items A the retailer should purchase and Y be the number of items B the retailer should purchase Z = Profit earned (Objective function) Z = 500X + 150Y X + Y 60 (Constraint of total capacity) 2500X + 500 Y 50000 (Constraint of total investment) X>_0 Y 0.
OPERATIONS RESEARCH
26
Example 2.14. A company produces two types of products A and B. Product B is superior quality and product B is lower quality. Profit on the two types of products are Rs. 30 and Rs. 40 respectively. The data about resources required and availability of resources are given below : Requirement of Resources Pro duct A Raw material (kg) Machine hours (per piece) Assembly
Product B
Capacity available (per month) 12000
8
120 5
3
4
500
60
630
How should the company manufacture the two types of products in order to get the maximum overall profits ? Formulate the problem as a LPP. Solution. Let X be the quantity of product A manufactured and Y be the quantity of product B manufactured. Z = Overall profits (Objective function) Z is to be maximized under the following conditions: Z = 30X + 40Y 60X + 120Y 12000 (Constraint of raw material) 8X + 5Y 630 (Constraint of machine hours) 3X + 4Y 500 (Constraint of assembly) )( 0;YO Example 2.15. A machine is producing either product A or B. It can produce product A by using 2 units of chemicals and 1 unit of a compound and can produce B by using 1 unit of chemical and 2 units of compound. Only 800 units of chemical and 100 units of compound are available. The profits available per unit of A and B are Rs. 30 and Rs. 20 respectively. Find the optimum allocation of units between A and B to maximize the total profit. Give a mathematical formulation to the LP problem. Solution. Let X be the quantity of product A produced and Y be the quantity of product B produced. Z = Total profits (Objective function) Z is to be maximized under the following constraints : Z = 30X + 20Y 2X + Y 800 (Constraint of availability of chemical) X + 2Y 1000 (Constraint of availability of compound) X>_0 Y>_0 Example 2.16. ABC Ltd. manufactures two products A and B. For manufacturing product A, a machine has to be used for 3 hours and the operator has to put in 4 hours. For manufacturing product B, the machine use is for 4 hours and the operator works only for 2 hours. In one week, the company can use a maximum of 60 hews of machine time and the worker is available for 80 hours. The contribution of profit by product A is Rs. 40 and by product B is Rs. 50. Formulate the problem as a LP model and find what quantities should ABC Ltd. manufacture of product A and B to be able to maximize their profits. Assume that all quantities of A and B produced are sold without any problem.
LINEAR PROGRAMMINGI
27
Solution. Let X and Y be the quantities of product A and B respectively to be manufactured. The data of the problem can be put in the form of a table as shown below: Decision Variable (Quantities to be manufactured)
Product
X Y Max time available (Hours)
A B
Time Required (Hours) Machine Operator 2 4 4 2 60 78
Profit/Unit (Rs.)
40 50
Objective function Maximise Z = 40X + 50Y Constraints (i) 2X + 4Y 60 (ii) 4X + 2Y 78 (iii) X , Y 0 Now, X and Y can be solved to have maximum of Z. X = 16 Y=7 Z = 40 x 16 + 50 x 7 = 640 + 350 = Rs. 990 ABC should manufacture 16 numbers of component A and 7 numbers of component B. Example 2.17. An ancillary unit of HMT tractors produces two types of tractor parts A and B. All the parts that are manufactured by the ancillary units are purchased by HMT. The manufacturing cost of part A is Rs. 2400 and of part B it is Rs. 3200. HMT purchases part A for Rs. 3200 and part B for Rs. 3600. The company's present production capacity is limited because of the following three constraints:
(a) Budget constraints—cash available at the beginning of year Rs. 1050000 (b) Machine time available 4200 hours/year (c) Manufactures available/year 5600. Part A needs 8 hours of machine time and part B needs only 4 hours. Also, the manhours required for part A are 8 hours and for part B 8 hours, find out what quantities of these should be produced by the ancillary unit to maximize its profits. Solution. The problem can be formulated as an LP problem with constraints as follows. The given data can be put in the form of a table for ease of understanding. Let X1 be the quantity of part A to be manufactured and X2 be the quantity of part B to be manufactured. Decision Variable (Quantities to be manufactured) X1
Part
A B
X2 Total availability Budget availability = Rs. 2560000
Machine Time (Hours) 8 4
4200
Operator Time (Hours)
Cost Price (Rs.)
Selling Price (Rs.)
20 8 5600
2400 3200
3200 3600
OPERATIONS RESEARCH
28
Objective Function Z = 800X1 + 400X2 Maximize 2400 X1 + 3200 X2 460000 (Constraints of Budget) 8X1 + 4X2 5. 4200 (Constraint of machine time) 20X1 + 8X2 5 5600 (Constraint of manhours) X1 , X2 0 (nonnegativity constraints) These inequalities can be converted into equations and solved. 3200 X1 + 3600 X2 = 2560000 8X1 + 4X2 = 4200 90X1 + 8X2 = 5600 X1 = 200 X2 = 650 It can be seen that the cost of quantity of parts manufactured remains within the total budget. Z = Rs. 420000. Also Example 2.18. A company supplying three types of parts to an automobile manufacturing company, purchases castings of three parts from a nearby foundry and performs three types of operations before selling these and cost per hour of these machines is given in the table below: Machine Cutting Drilling Polishing
A 20 40 50
Capacity/hour B 60 20 50
C 25 40 20
Costlhour (Rs.) 150 100 200
The cost of the castings for A, Rs. 120, for B Rs. 200, for C Rs. 400 and the selling price of these parts is Rs. 200, Rs. 350 and Rs. 500 respectively. All the parts that are processed by the company can be sold. What quantity of various parts should the company process for selling in order to maximize their profits? Solution. Let X1, X2 and X3 be the number of parts the company should process. 50 150 200 , B = —, C = The cutting cost per part A = 60 60 20 100 100 .pt, The drilling cost per part A = 40 =100 20 C  40 200 B = 200 = 200 The polishing cost per part A = 50 C 20 50 A = (200  120)  Cost of (cutting + drilling + polishing) Profit from part = 80  (
Similarly. profit for part Profit for part
150 100 200 + + ) = 80 14 = Rs. 66 20 40 60
100 2013) = 130  (2.5 + 5 + 4) = Rs. 118.5. B = (350  200)  (150 + 60 20 + 50 100 200 ) = 100  (6 + 2.5 + 10) = Rs. 81.5. C = (500  400) (150  25 + 40 + 20
LINEAR PROGRAMMINGI
29
Objective function, maximizing Z = 66 X1 + 118.5 X2 + 81.5 X3 with constraints: (a) Cutting machine constraint = X1 — + X2 + X3 < 1 20 60 25 (b) Drilling machine constraint = X X X 1 40 + 20 + 40 < X X X (c) Polishing machine constraint = 1 + 2 + 3 < 1 50 50 20 Nonnegativity constraint X1, X2, X3 > 0 These inequalities can be converted into equations and solved for X1 , X2 and X3 . Example 2.19. ABC Ltd. is assembling two products P1 and P2 . The cost of assembling one unit of products P1 and P2 is Rs. 200 and Rs. 240 respectively. The availability of work station for two products is limited to 60 hours and the two products spend 6 hours and 2 hours respectively on the work station. The products can be sold for Rs. 280 and Rs. 320 respectively. Total manhours available are 400 and P1 requires 2 manhours and P2 requires 4 manhours. Formulate the problem as a LPP. Solution. Let X1 and X2 be the number of products P1 and P2 to be assembled. Maximise Z = 80X1 + 80X2 with the following constraints: 6X1 + 2X2 .5 60 (Work station constraints) 2X1 + 4X2 5. 400 (manhours constraints) X1, X2 0 These inequalities can be converted into equations and solved for X1 and X2 . Example 2.20. A person has Rs. 1200000 which he wants to invest in different types of investment opportunities. He has consulted investment consultant to advise him on this, who has given him the following advice: (a) Government sector, Postoffice/NSC, etc., up to 40% (b) Share X and Share Y both put together Rs. 200000. (c) Mutual funds A and B. At least 25% The rate of return which he gets from these investments are: 0.1 (a) Government sectors 0.2 (b) Share X 0.25 (c) Share Y 0.15 (d) Mutual fund A 0.12 (e) Mutual fund B Formulate the above as a LP problem. X1 = Government sector investment Solution. Let X2 = Share X X3 = Share Y X4 = Mutual fund A X5 = Mutual fund B Objective function is to be maximized = Z = 0.1 X1 + 0.2 X2 + 0.25 X3 + 0.15 X4 + 0.12 X5 with the following constraints:
30
OPERATIONS RESEARCH
X1 + X2 + X3 + X4 + X5 1200000 (total investment constraint) 480000 X2 + X3 200000 X4 + X5 300000 Xi, X2, X3, X4, X5 0 These inequalities can be converted into equations and solved for X1, X2, X3, X4 and X5 and hence for objective function Z. Example 2.21. A firm is manufacturing three products A, B and C. Time to manufacture product A is twice that for B and thrice that for C and they are to be produced in the ratio 2 : 3 : 4. The relevant data is given below. If the whole labour is engaged in manufacturing product A, 2000 units of the product can be produced. There is a demand for at least 200, 300 and 400 units of the product A, B, C and the profit earned per unit is Rs. 100, Rs. 70 and Rs. 50 respectively. Formulate the problem as a linear programming problem. Requirement per unit of product (kg) A
B
C
Total availability (kg)
P
6
5
9
4000
Q
4
8
6
5000
Raw material
Solution. Maximize Z = 100A + 70B + 50C (Objective function) subject to the following constraints: 6A + 5B + 9C 4000 (Constraint of raw material P) 4A + 8B + 6C 5000 (Constraint of raw material Q) A+ 1B+1 2000 2 3 A 200 B 300 C 400 A:B:C::2:3:4
A 2
B i.e., 3A — 2B = 0 2 B C = — or 4B — 3C = 0
A, B 3C > 04 Example 2.22. A patient wants to decide the constituents of diet, which will fulfill his daily requirement of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from three different types of foods. The yields per unit of these foods are as follows : Food Type
Yield per unit Proteins
Fats
Carbohydrates
Cost per unit (Rs)
1
2
2
4
40
2 3
4
2
4
50
8
10
8
60
1200
400
800
Minimum requirement
Formulate linear programming model for the problem.
LINEAR PROGRAMMINGI
31
Solution. Let A, B and C be the number of units of food type 1, 2, and 3 respectively. Maximize Z = 40A + 50B + 60C (Objective function) subject to the following constraint: 2A + 4B + 8C 1200 (Constraint of Proteins) 2A + 2B + 10C 400 (Constraint of Fats) 4A + 4B + 8C 800 (Constraint of Carbohydrates) A>_0 B 0 (Nonnegativity constraints) C>_0 Example 2.23 A dietician is planning the menu of a meal to be provided in the hostel of a college. Three main items, having different nutritional content have to be served. The dietician wants to ensure at least the minimum daily requirement of each of the three vitamins is provided. The table below gives the details of vitamin content per gram of each type of food, cost per gram of the food and minimum daily requirement for the three vitamins. Any combination of the foods may be provided as long as the total serving is at least 10 grams. The problem is to determine the number of grams of each food to be included in the meal. The objective is to minimise the cost of each meal subject to satisfying daily minimum requirement of the three vitamins as well as the restriction of minimum serving size. Food 1 2 3 Minimum daily requirement
1 50 30 20 290
Vitamin (mg) 2 20 10 30 200
3 15 60 20 250
Cost per gram (Rupees) 0.20 0.40 0.25
Solution. To formulate the LP model for this problem. Let x1 , x2 and x3 be the number of grams in each type of food. The objective function should represent the total cost of the meal in rupees. The total cost equals the sum of the costs of three items. Objective function Z = 0.20 x1 + 0.40 x2 + 0.25 x3 Since we are interested in providing at least the minimum daily requirement for each of the three vitamins, there will be three greater than or equal to constraints. The constraint for each type of vitamin will have the form x1 gms of food 1, x2 gms of food 2, x3 gms of food 3 minimum daily requirement.
The constraints are : 50x1 + 30 x2 + 20 x3 290 20 + 10 x2 + 30 x3 200 15 xl + 60 x2 + 20 x3 250 The restriction that the serving size be at least 10 grams is stated as x1 + x2 + x3 10 The complete formulation of the problem is as follows : Minimize Z = 0.20 x1 + 0.40 x2 + 0.25 x3
OPERATIONS RESEARCH
32 subject to
50 x1 + 30 x2 + 20 x3 290 20 + 10 x2 + 30 x3 200 15 xi + 60 x2 + 20 x3 250 x1 + x2 + x3 10 x1, x2, x3 >_0. Note that nonnegativity constraint has been included to ensure that negative quantities of any of the foods will not be recommended. The above is a very simplified problem involving the planning of one meal, the use of first three food types, and consideration of three vitamins. In actual practice, models have been formulated, which consider (a) Menu planning over longer periods of time (daily, weekly, monthly, etc.) (b) The interrelationship of all the meals served during a particular day. (c) The interrelationship among meals served over the entire planning period. (d) Many food items and (e) Many nutritional requirements. The number of variables and number of constraints for such models can become extremely large. Example 2.24. A firm can produce three types of cloth A, B and C. Three kinds of wool are required for it, say, red wool, green and blue wool. One unit length of type A cloth needs 2 yards of red wool and 3 yards of blue wool, one unit length of type B cloth needs 3 yards of red wool, 2 yards of green wool and 2 yards of blue wool and one unit length of type C cloth needs 5 yards of green wool and 4 yards of blue wool. The firm has a stock of only 8 yards of red wool, 10 yards of green wool and 15 yards of blue wool. It is assumed that the income obtained from one unit of type A cloth is Rs. 3, of type B cloth is Rs. 5 and that of type C cloth is Rs. 4. Formulate the problem as linear programming problem. Solution. Maximize Z = 3A + 5B + 4C (Objective function) subject to the following constraints: 2A + 3B 8 (Constraints of availability of red wool) 2A + 5B S 10 (Constraints of availability of green wool) 3A + 2B + 4C 5_ 15 (Constraints of availability of blue wool) A, B, C 0 (Nonnegativity constraints) Example 2.25. Orient Paper Mills produces two grades of paper X and Y. Because of raw material restrictions not more than 400 tons of grade X and not more than 300 tons of grade Y can be produced in a week. There are 160 production hours in a week. It requires 0.2 and 0.4 hours to produce a ton of products X and Y respectively, with corresponding profit of Rs. 20 and Rs. 50 per ton. Formulate a linear programming problem to optimize the product mix for maximum profit. Solution.
Maximize Z = 20X + 50Y (Objective function) subject to the following constraints : 0 • 2X + 0 • 4Y 160 (Constraints of production hours) X 400 (Constraints of raw material) Y 300 (Constraints of raw material) X 0 (Nonnegativity constraints) Y>_0
33
LINEAR PROGRAMMINGI
Example 2.26. A town located at high altitude has two locations where kerosene and petrol is stored by Army for use in four different zones during winters when the highway is closed and no supplies of kerosene and petrol are possible to these locations. The table below provides the cost (Rs.) of supplying one kilolitres of kerosene and petrol from each stock location to each zone. In addition, the storing location capacity and normal level of demand for each zone are indicated in kilolitres. Formulate the LP problem. Zone Storage location 1 Storage location 2 Demand (k. litres)
1
2
3
4
4 5 300
6 2 500
2.50 3.50 400
3.00 4.50 350
Maximum Supply (k. Litre) 1000 800
Solution. In this problem, there are eight decisions to be madehow many k. litres should be transported from each storage location to each zone. In some cases the best decision may be not to transport any units from a particular location to a particular zone. Let x11 , x12 , X13 , X14, denote the number of k. litres supplied by location 1 to zone 1, 1 to 2, 1 to 3 and 1 to 4 respectively. Similarly, let x21 , x22 , x23 , x24 be the number of k. litres supplied by location 2 to zone 1, 2 to zone 2, 2 to zone 3 and 2 to zone 4. Total cost = 4 x11 + 6 x12 + 2.50 x13 + 3.50 x14 + 5 x21 + 2 x22 + 300 x23 + 4.50 x24. This function has to be minimised. The constraints are: x11 +x12 + x13 + x14 1000 (For location 1) x21 + x22 + x23 + x24 800 (For location 2) Also, the constraint of ensuring that each zone receives the quantity demanded. For zone 1, the sum of the transportation from location 1 and 2 should be 300 k. litres or xii + x21 = 300. Given that each origin can supply units to each destination in some measure x11 + x22 = 500 X13 + X23 = 400 x14 + x24 = 350 The complete formulation of LP model is as follows: Minimize Z = 4 x11 + 6 x12 + 2.5 x13 + 3 x14 + 5 x21 + 2 x22 + 3.5 x23 + 4.5 x24 Subject to x1i + x12 + X13 + X14 < 1000 X21 + X22 + X23 + X24 800 x11 + x21 = 300 x12 + x22 = 500 X13 + X23 = 400 x14 + x24 = 350 X11, X12, X13, X14, X21, X22, X23, X24 0.
34'
OPERATIONS RESEARCH
Example 2.27. A small refinery is about to blend four petroleum products into three final blends of gasoline. Although, the blending formulae are not precise, there are some restrictions which must be adhered to in blending process. These are : (a) Component 2 should constitute no more than 40 per cent of the volume of blend 1. (b) Component 3 should constitute at least 20 per cent of the value of blend 2. (c) Component 1 should be exactly 35 per cent of blend 3. (d) Components 2 and 4 together should constitute at least 60 per cent of the volume of blend 1. There is a limited availability of components 2 and 3 of 1600 k. litres and 1200 k. litres. The production manager wants to blend a total of 6000 k. litres. Of this total at least 2500 k. litres of final blend 1 should be produced. The wholesale price per litre from the sale of each final blend equals Rs. 20, Rs. 18 and Rs. 12 respectively. The input components cost Rs. 10, Rs. 15, Rs. 12 and Rs. 13 per litre respectively. Find out the number of litres of each component to be used in each final blend so as to maximise the total profit contribution from the production run. Solution. Let xi represent number of litres of component i used in the final blend j. Total profit contribution = Total revenue from all 3 blends — total cost of the 4 components = Profit blend 1 + Profit blend 2 + Profit blend 3 — (cost of components 1, 2, 3 and 4) = Rs (20 + 18 + 12) — Rs (10 + 15 + 12 + 13) Hence, the objective function is Z= [20 (x11 + X21 + X31 A X41) + 18 (x12 + X22 + X32 + X42) + 12 (X13 + X23 + X33 + X43)1 — [10 (X11 + X12 + X13) + 15 (X21 + X22 + X23) + 12 (X31 + X32 + X33) + 13 (x41 + x42 + x43)1 Z= 10 + 8 X12 ± 2 Xi3 I 5 X21 + 3 x22 — 3 X23 I 8 X31 ± 6 x32 + 0 x33 + 7 X4i + 5 X42 — X43. or The constraints are : (a) Total production run must be 6000 k. litres or xil + X12 + X13 + X21 + X22 + X23 + X31 + X32 + X33 + X41 + X42 + X43 = 6000000 litres (b) Recipe restrictions are: Amount of components 2 used 40% of amount of final blend 1 in blend 1 x21 < 0.4 (x11 + x21 + x31 + x41) — 0.4 xn + 0.6 x21 — 0.4 x31 — 0.4 x41 0 or Amount of component 3 used in blend 2 20% amount of final blend 2 x32 >_0.20 (x12 + x22 + X32 + X42) — 0.2 x12 — 2.0 x22 + 0.8 x32 — 0.20 x42 0. Or Component 1 should be exactly 35% of blend 3 X13 = 0.35 (x13 + x23 + X33 + X43) 0.65 x13 — 0.35 x23 — 0.35 x33 — 0.35 x43 = 0. Or Component 2 and 4 together should constitute at least 60 per cent of the value of blend 1 or x21 + x41 > 0.6 (x11 + x21 + x31 + x41) — 0.6 x11 + 0.4 x21 — 0.6 x31 + 0.4 x41 0 or Limitation of availability of components 2 and 3 can be represented as
LINEAR PROGRAMMINGI
35 x21 + x22 + X23 1600000 X31 + X32 + X33 < 1200000
Also, there is a constraint of minimum production requirement of final blend 1, that is x11 + x21 + x31 + x41 > 2500000 The complete LP model is Maximize Z = 10 + 8 xi2 + 2 X13 + 5 X21  3 x22 + 3 X23 I 8 x3i. + 6 X32  0.X33 4 7 X41 I 5 X42  X43 Subject to x11 + x12 + x13 + x21 + x22 + x23 + x31 4 x32 + x33 + x41 + X42 + x43 = 6000000 —0.4 xn + 0.6 x2i — 0.4 x31 — 0.4 x4i. 0 — 0.2 X12 — 0.20 x22 + 0.8 X32 — 0.2 x42 0 0.65 x13 — 0.35 x23 — 0.35 x33 — 0.35 x13 = 0 —0.6 x11 + 0.4 x21 — 0.6 x31 + 0.4 x41 0 x21 + X22 + X23 1600000 x31 + x32 + x33 < 1200000 x11 + x21 + x31 x41 2500000 xn X12 / X13 / X21 / X22 / X23, X31 / X32 / X33 / X41 / X42 / X43 0. GRAPHICAL METHOD OF SOLVING LINEAR PROGRAMMING PROBLEMS The various steps in solving the LP problem using the graphical method are as follows : 1. Formulate the problem with mathematical form by: (a) Specifying the decision variables, (b) Identifying the objective function and (c) Writing the constraint equations. 2. Plot the constraint equations on a graph 3. Identify the area of feasible solution 4. Locate the corner points of the feasible region 5. Plot the objective function 6. Choose the points where objective functions have optimal value. Example 2.28. A manufacturing company is producing two products A and B. Each requires processing on two machines 1 and 2. Product A requires 3 hours of processing on machine 1 and 2 hours on machine 2. Product B requires 2 hours of processing on machine 1 and 6 hours on machine 2. The unit profits for product A and B are Rs. 10 and Rs. 20 respectively. The available time in a given quarter on machine 1 and machine 2 are 1200 hours and 1500 hours respectively. The market survey has predicted that not more than 400 units of product A and not more than 250 units of product B can be sold in the given quarter. The company wants to determine the product mix to maximize the profits. Formulate the problem as linear programming mathematical model and solve it graphically. Solution. Step 1. Formulating the problem into a mathematical model. Let X be the number of product A manufactured in a quarter and Y be the number of product B manufactured in a quarter. Z = profit in a quarter (objective function)
36
OPERATIONS RESEARCH
Z is to be maximized under the following constraints: Z = 10X + 20Y 3X + 2Y 1200 (Time constraint of machine 1) 2X + 6Y 1500 (Time constraint of machine 2) X 400 (Selling constraint of product A) Y 5 250 (Selling constraint of product B) X > 0 (Nonnegativity constraint) Y>_0
Quarterly output product B
600
Time Constraint machine 1
A Feasible solution area
X 30 x1, x2 > 0.
39
LINEAR PROGRAMMINGI Solution. Converting the inequalities into equations 0.5x1 + x2 = 20 1.5x1 + 0.5x2 = 15 2x1 + 1.5x2 = 30 Equation (i) gives (0, 20) and (40, 0)
(i) (iii)
Equation (ii) gives (0, 30) and (10, 0) Equation (iii) gives (0, 20) and (15, 0) Points A, B, C, D can be found out by solving the above equations. The corner points are as shown in the table below : Corner A B C D The graph is shown in Figure 2.3.
Z = 10x1 + 5x2 150 70 + 60 = 130 40 + 90 = 130 400 + 0 = 400
Coordinates (15, 0) (7,12) (4,18) (40, 0)
40 30 20 10 D
7 10 A 20 30 Fig. 2.3
40
Hence, maximum, profit is 400. Example 2.31. A firm makes product X and Y and has a total production capacity of 9 tons per day, X and Y requiring the same production capacity. The firm has a permanent contract to supply at least 2 tons of X and at least 3 tons of Y per day to another company. Each ton of X requires 20 machine hours production time and each ton of Y requires 50 machine hours production time. The daily maximum possible number of machine hours is 360. All the firm's output can be sold and the profit made is Rs. 80 per tons of X and Rs. 120 per tons of Y. It is required to determine the production schedule for maximum profit and to calculate the profit. Solution. Step 1. Formulation of LP mathematical model Let X be the production of product X and Y be the production of product Y and Z be profit (objective function). Z is to be maximized under the following constraints: Z = 80X + 120Y X + Y 9 (Capacity constraint) X 2 (Demand constraint of product X) Y 3 (Demand constraint of product Y)
40
OPERATIONS RESEARCH
Product Y production
20X + 50Y 360 (Constraint of availability of machine hours) X>_0 Y>_0 Step 2. Plotting the constraint equations on the graph as shown in Figure 2.4. This can be done by converting the inequalities into equalities. i.e., X + Y = 9, X = 0, Y = 9, Y = 0, X = 9, X = 2, Y = 3 20X + 50Y = 360 X = 0, Y = 7.2 (0,7.2) Y = 0, X = 18 (18,0) 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
X =2
X+Y= 9
A Feasible solution area Isoprofit line
Y3 4 20X
CBI
D
I
+ 5Y = 360
I
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Product X production Fig. 2.4 Step 3. Identify area of feasible solution. It can be seen from the figure that RQPS is the area of feasible solution as indicated by shaded area. Step 4. Plot the objective function. It is clear that solution is either at R, Q, P or S. By inspection from the graph R= (2,3) Q = (6, 3) P = value of point P can be found out by solving the two equations which intersect at P, i.e., 20X + 50Y = 360
LINEAR PROGRAMMINGI
41
X+Y=9 Solution of these two equations can be found out by multiplying equation (ii) by 20 ...(iii) 20X + 20Y = 180 Subtract (ii) from (i) 30Y =180..Y = 6 If Y = 6 putting this value of Y in equation (i) X=3 Point P (3, 6) Similarly, we can find the value of point S by solving the two equations 20X + 50Y = 360 and X = 2 because S is the intersection of these two lines. If X=2 Y—
320 =6.6 50
S (2, 6.4) Let us now calculate the value of all the corner points to find out which point gives the maximum value. Coordinates X, Y
Value of objective function 80X + 120 Y
R
(2,3)
520
Q
(6, 3)
840
P
(3,6)
960
S
(2, 6.4)
928
Corner Point
Maximum profit is at point P Hence Z maximum = Rs. 960 Alternative method from step 4 onwards
Step 4. Plotting the objective function. The aim here is to find out the combination of products X and Y, which maximizes the objective function of profit Objective function 80X + 120Y can be plotted on the graph by selecting an arbitrary profit which can be achieved by production of some quantity of product X and some of product Y, within the feasible area. For example, let us consider a profit of Rs 600, it can be obtained by production of 5 tons of product X or 7.5 tons of product Y. The line joining these two points i.e., (0, 5) and (7.5, 0) is called the isoprofit line. A number of points on this line will give different quantities of products X and Y to give the same (iso) profit. Step 5. Determining the optimal solution. To determine the optimal solution, now draw a line parallel to the isoprofit line (AB) in such a way that it is the farthest away from the origin and also intersects the feasible area (shaded area) at some point. This is line CD in Figure 2.4. In this problem the point is P and its coordinates are P (3, 6) as can be seen from the figure. Z = 80X + 120Y Hence, Z = Rs. 960 Same as was found out by the other method.
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OPERATIONS RESEARCH
Example 2.32. Solve the following inequalities graphically: Maximize Z = 3X + 2Y X+Y.5 35 Subject to X—Y 0 Xs20 Y__>5 X> 0 Y__ >0 Solution. Convert the above constraints inequalities into equalities. X+Y= 35 Y=5 Y= 15 X—Y=0 X = 20 From (i) When X = 0, Y = 35 (0, 35) When X = 35, Y = 0 (35, 0) From (iv) X=0 Y= 0 X = 20
+ Y = 35 35
X—Y=0 30 25 Feasible solution area 20 15
Y = 15
10 Y=5
5
10
15
20
425 Fig. 2.5
I 30 35 40
X
The feasible solution region is shown with shaded area. Let us find out which point gives the maximum value of objective function Z.
43
LINEAR PROGRAMMINGI
Point P Q R S
Coordinates 
Max value 3X + 2Y
(5, 5) (20, 5) (20, 15) (15, 15)
25 70 90 75
Hence maximum value Z = Rs. 90 at point R. Example 2.33. Solve graphically the following linear programming problem Minimize Z = 3X1 + 5X2 Subject to the conditions —3X1 + 4X2 /2 2X1 + 3X2 _s" /2 2X1 — X2 = — 2 X2 2 whereas X1, X2 > —0 Solution. To solve the problem graphically, we have to convert the inequalities into equalities. — 3X1 + 4X2 = 12 2X1 + 3X2 = 12 2X1 — X2 = — 2 ...(iii) X1= 4 ...(iv) X2 = 2 ...(v) whereas XI , X2 From (i) X1 = 0 (0,3) X2 = 3 X2 = 0 (— 4, 0) X1 = — 4 From (ii) X1 = 0 (0, 4) X2 = 4 X2 = 0
X1= 6
From (iii)
(6, 0)
X1 = 0 (0, 2) X2 = 2 X2 = 0 X1 = — 1 (— 1, 0) From (iv) X1 = 4 (4, 0) From (v) X2 = 2 (0, 2) Now, let us plot the above lines on the graph. The values of points A, B, C, D and E can be found out graphically by reading the value on the graph or by solving the equations. For example, coordinates of point A can be found out by solving equations 2X1 — X2 = — 2 and 2X1 + 3X2 = 12 as lines representing these equations intersect at point A.
OPERATIONS RESEARCH
44
Fig. 2.6 Point
Coordinates
Z = 3X1 + 5X2
A
(0.75, 3.4)
19.25
B
(0 . 8, 3 . 7)
20.90
C
(4, 6)
42.00
D
(4, 2)
22.00
E
(3, 2)
19.00
It may be seen Z is minimum (19.00) at point E. Note that slight difference in the values of points A, B; etc., found by solving the equations and by reading from the graph can be attributed to graphical error and are negligible, so can be ignored. Example 2.34. Solve graphically. Z = 9X + 10Y Maximize Subject to 11X + 9Y 9900 7X + 12Y 8400 6X + 6Y 9600 where X >0 Y >0 __ Solution. Convert the inequalities into equalities. 11X + 9Y = 9900 7X + 12Y = 8400 6X + 6Y = 9600 X= 0
...(ii) ...(iii)
45
LINEAR PROGRAMMINGI Y=0 X=0 Y = 1100 Y=0 X.= 900 X=0 Y = 700 Y=0 X = 1200 X=0 Y = 1600 Y=0 X = 1600
From (i)
From (II)
From (iii)
(0, 1100) (900, 0) (0, 700) (1200, 0) (0, 1600) (1600, 0)
Plot the above lines on the graph PQRST (shaded area) is the feasible solution area coordinates and values of points PQRST are as follows : Y
1600
R 6X + 6Y = 9600
1400
>13 0 2
1200 1100 1000 800 700 600 A 400 11X +12Y = 9900
200 Isoprofit line 0
7X + 12Y = 8400
D T 200 400 600 800 1000 1200 1400 1600 Product X Fig. 2.7
X
OPERATIONS RESEARCH
46 Point
Coordinates
Value of Z = 9X + 10Y
P
(626, 335)
8984
Q
(0,1100)
12000
R
(0, 1600)
14400
S
(1600, 0)
14400
T
(1200, 0)
10800
Z maximizes at P as all other points are on the axis and not within the feasible area. Also, assume profit of Rs. 4500 and solve objective function Z = 9X + 10Y. If X = 0 then Y = 450 If Y = 0 then X = 500 Draw this isoprofit line AB on the graph. Now, draw a line parallel to AB farthest from origin which intersects the feasible area (shaded area) at any point. The line is CD and the point is P, where coordinates graphically are 627 and 335. Example 2.35. Solve the following linear programming problem by graphic method: Z = 2X1 — 3X2 Maximize 4X1 + 5X2 ..s 40 Subject to 2X1 + 6X2 "24 3X1 3X2 _26 XI ?. 4 X1 , X2 2 0 Solution. Convert the inequalities into equalities. 4X1 + 5X2 = 40 2X1 + 6X2 = 24 3X1  3X2 = 6 =4 X1 = 0 From (i) (0, 8) X2 = 8 X2 = 0 (10, 0) X1 = 10 X1 = 0 From (ii) (0, 4) X2 — 4 X2 = 0 (12, 0) X1 = 12 (0, 2) = 0, X2 =  2 From (iii) (2, 0) X1 = 2, X2 = 0
•••(i) ...(ii) ...(iii) ...(iv)
LINEAR PROGRAMMINGI
47
7 6 5 4 3 2
X2
Ilik ip 4h,, A
p
x1— 3 x2 = 6
_4;k+ 5 x2 =40
iik. Q
P
2 3 4 5 6 7 8 9 10 11 12 (6, 0) X1 Fig. 2.8
Coordinates and values of points PQRST (the shaded area representing the feasible solution area) are as follows. Point
Maximize
Coordinates
P
(4,0)
Q R
(10, 0)
20
(8.6,1.14)
13.78
S
(5.5,3.5)
0.5
T
(4,2)
2
8
Z = 20
Example 2.36. Maximize Z = 6X1 + 7X2 Subject to 2X1 + 3X2 12 2X1 + X2 X1 „..? 0 X2 > 0 Solution. Convert inequalities into equalities 2X1 + 3X2 = 12 2X1 + X2 = 8 From (i) j X1 = 0 (0,4) X1 = 6
From (ii)
Value of Z = 2X1  3X2
X2 = 0 X1 = 0 X2 = 8 X1 = 4 X2 = 0
(6, 0) (0,8) (4,0)
48
OPERATIONS RESEARCH
2X1 + X2 = 8
Feasible solution area
Q(3,2) 2X1 + 3X2 = 12
2
4R 5
6
7
8
Fig. 2.9 Coordinates and values of point OPQR (the shaded area corresponding the feasible solution area) are as follows. "Point 0 P. Q R
Coordinates (0, 0) (0, 4) (3, 2) (4, 0)
Value of Z = 6X1 + 7X2 0 28 32 24
Maximum Z = 32 at point Q (3, 2) Coordinates of point Q have been found out by solving the equation 2X1 + 3X2 = 12 and 2X1 + X2 = 8 which intersect at Q. 2X1 + 3X2 = 12 ...(iii) 2X1 + X2 = 8 ...(iv) Subtract (iv) from (iii)we get . 2X2 = 4 X2 = 2 Now substitute X2 = 2 in equation (iii) or (iv) X1 = 3 Example 2.37. Solve graphically: Maximize Z = 5X1 + 4X2 Subject to 4X1 + X2 "40 2X2 + 3X2 590
49
LINEAR PROGRAMMING1
whereas X1 , X2 Solution. Convert the inequalities into equalities 4X1 + X2 = 40 2X1 + 3X2 = 90 From (i) (0, 40) X1 = 0, X2 = 40 X2 = 0,
From (it)
= 10
(10,0)
X1 = 0, X2 = 30
(0, 30)
...(i) ...(ii)
X2 = 0, X1= 45
(45, 0) Coordinates and values of points OPQR (the shaded area representing the feasible solution area) are as follows. Point 0 P Q R
Coordinates (0, 0) (0, 30) (3, 28) (10, 0)
Value of Z = 5X1 + 4X2 0 120 127 50
At point Q (3, 28) Z is maximum 127. Xi 40
30
Q(3,28) 4x1 + x2 = 40 2x, + 3x2= 40
A 20
10
0 (0, 0)
10
20
30 Fig. 2.10
40
45
50
Example 2.38. A company produces two types of products say type A and type B. Product B is superior quality and product A is a lower quality. Profits on the two types of products are Rs. 30 and Rs. 40 respectively. The data about resources required and availability of resources are as follows.
OPERATIONS BfSEARCH
50 Requirement
Capacity available
Product A
Product B
60
120
12000
Machine hours (per piece)
8
5
630
Assembly
3
4
500
Raw material (kg)
How should the company manufacture the two types of products in order to get the maximum overall profits? Solution. Step 1. Formulate the linear programming problem. Z = 30A + 40B Subject to constraint of 60A + 120B 12000 (Constraint of raw material) 8A + 5B 630 (Constraint of machine hours) 3A + 4B S 500 (Constraint of assembly) A>_0 B>_0 Convert the inequalities into equalities 60A + 120B = 12000 8A + 5B = 630 3A + 4B = 500 (0, 100) A = 0, B = 100 From (i) B = 0, A = 200 (200, 0) x2 150
126 125 8A + 5B = 630 m A 100
3A + 4B = 500
z 2 50 ''414
60A + 120B = 12000 Xi
44thihi,. 50
B 78 150167 Product type A
Fig. 2.11
200
A
51
LINEAR PROGRAMMINGI From (ii) From (iii)
A = 0, B = 126 B = 0, A = 78.6 A = 0, B =125
(0,126,0) (78.6,0) (0, 125)
500 B = 0, A = 3
( 500 0) 3 )
OAPB is the area of feasible solution. Coordinates of points OAPB and values of Z are as follows : Point
Coordinates
Value of Z = 30A + 40B
0
(0, 0)
0
A
(0, 100)
4000
( 260 9701
46600
P B
n ' 11) ( 630 n) 8 )
11 18900 8
60 It can be easily seen that maximum Z = 46600so the company should manufacture = (24) 11 11 numbers type A product and 1 70= (89) numbers of type B product. 11, Example 2.39. A retailer deals in two items only, item A and item B. He has Rs. 50000 to invest and a space to store at the most 60 pieces. An item A costs him Rs. 2500 and B costs Rs. 500. A net profit to him on item A is Rs. 500. and item B is Rs. 150. If he can sell all the items he purchases, how should he invest his amount to have maximum profit ? (a) Give a mathematical formulation to the linear programming problem. (b) Use graphical method to solve the problem. (c) Indicate the feasible region on the graph. Solution. Z = 500 A + 150B is to be maximized (objective function) Subject to 2500A + 500B 5 50,00 (Constraint of amount to be invested) A + B 60 (Constraint of space) A>_0 B>_0 To solve it graphically, convert inequalities into equalities 2500A + 500B = 50,00 •••(i) A + B = 60 ...(ii) From (i) A = 0, B = 100 (0, 100) B = 0, A = 20 (20,0) From (it) A = 0, B = 60 (0, 60) B = 0, A = 60 (60,0) Shaded area represents the feasible area of the solution. Point B can be found out by solving equations (i) and (ii). Multiply (ii) by 50 and subtract from (ii).
OPERATIONS RESEARCH
52 A = 10 B= 50
2500 A + 500 B = 50,000
/ A
Feasible solution area
Q(3,2)
(0, 0) 20
40
60
80
100 120
A
Fig. 2.12 Graphically point B can be read as (10, 50) Max Z = 10 x 500 + 50 x 150 = 5000 + 7500 = 12,500 Example 2.40. (a) Maximize 6A + 5B 3A + 4B 5:120 Subject to 3A + 1B 540 A+B= 30 .A>0 B>0 using graphical method. (b)Three products are produced in three different operations. The limits on the available time for the three operations are respectively 430, 460 and 420 minutes and the profit per unit for three products are Rs. 3, Rs. 2 and Rs. 5 respectively. Time in minutes per unit on each machine operation is as follows :Operation
Product I
II
III
1
1
2
3
2
3
0
2
3
1
4
0
Write LP model for the problem Solution. Maximize Z = 6A + 4B Convert inequalities into equalities. 3A + 4B = 120 3A + 1B = 40 A + B = 30
53
LINEAR PROGRAMMINGI From (i) When
A = 0 B = 30 B = 0 A = 40 A = 0 B = 40
From (ii) When
(0, 30) (40, 0) (0, 40)
40 ( 40 0) , 3 3 From (iii) When A = 0 B = 30 (0, 30) B = 0 A = 30 (30, 0) Drawing the equations as straight lines on graph. B=0 A=
B
50 40 30 25 20
3A + B = 40 A + B = 30 3A + 4B = 120
10 0 (0, 0) 5
10 40 — 3 20
30
40 50 A
Fig. 2.13 Maximize Subject to
Z = 3A + 2B + 5C A + 2B + 3C 430 3A + 2C 460 A + 4B 420 A>_0 B>_0 C>_0
Example 2.41. A landlord has set up a stud farm where he wants to breed the best horses. His veterinary doctor has advised him to use two special diets, say A and B, for the horses. The nutrition value of these diets and minimum requirement of these nutrients is as follows: Availability of nutrients in the products Nutrients
A
B
Minimum requirement
1
40
12
120
2
20
10
60
3
8
36
108
54
OPERATIONS RESEARCH
If special diet A costs Rs. 60 per unit and diet B costs Rs. 80 per unit, using LP graphics method, determine how much of products A and B must be purchased by the landlord so as to provide the horses not less than_ the minimum required as per the advise of the vet. Solution. Mathematical formulation of the problem Let X1 and X2 be the number of units of diet A and diet B to be purchased. The other data given in the problem can be summarized in the table below. Decision Variable (units) ' X1 X2
Product A B
Minimum requirement
Nutrient available 20 8 40 12 10 36 120
60
108
Objective function Minimize Z = 60X1 + 80X2 with the following constraints: 40X1 + 12X2 � 120 20X1 + 10X2 � 60 8X1 + 36X2 � 108 X1 I Xi > 0 Plotting the constraint equations in the graph First, the inequalities have to be converted into equalities. 40X1 + 12X2 = 120 20X1 + 10X2 = 60 8X1 + 36X2 = 108 From (i) X1 = 0, X2::::; 10 X2 = 0, X1 =3 (3, 10) X1 = 0, X2 =6 From(it) X2 = 0, X1 =3 (3, 6) X1 = 0, X2 =3 From (iii) X2 = 0, X1 = 13.5 (13.5, 3) X2
14 13 12 11 10 9
✓
40X1 + 12X2 = 120
8 7 6 5 4 83 2
6
7
8
Cost (Rs) '60 80
9 10 11 12 13
Fig. 2.14
...(i) ... (ii) ...(iit)
55
LINEARRROGRAMMINGI These can be plotted in Figure 2.14 . Coordinates of P can be found out by solving the equations 20X1 + 10X2 = 60 and 8X1 + 30X2 = 108 X1 = 1.7 X2 = 2.6 Point
Coordinates
Objective function 60X1 + 80X2
B
(0, 3)
240
P
(1 . 7, 2 . 6)
102 + 208 = 310
Q
(3, 0)
180
Since the minimum cost is at point Q, the landlord should purchase 1.7 units of product X1 and 2.6 units of product X2 . Example 2.42. An advertising agency has been hired by a client launching a new product which can be used by middle income group and higher income group. The entrepreneur wants to reach two types of potential customers, one's with income bracket more than Rs. 200000 per annum and the other which has lesser income than this. The total money he wants to spend on this exercise is Rs, 400000. TV and Radio advertising is being considered. Each TV telecast of advertisement for 10 seconds costs Rs. 60000 and not more than 4 must be taken. Each Radio broadcast for 1 minute costs Rs. 40000 and at least 6 must be taken. Viewer and listener's survey carried out by advertising agency indicates that TV advertisement reaches 1200000 potential customers with income bracket higher than Rs. 200000 per annum and reaches 50000 potential customers in the lower bracket of income. Radio programme reaches 50000 customer in higher income group bracket and 400000 in lower category. Determine the media mix by using LP graphic method. Solution. Mathematical formulation of the problem Let X1 and X2 be the number of programmes on TV and Radio respectively. Maximize the objective function Z for maximum total TV and Radio audience. Z = X1 (1200000 + 60000) + X2 (50000 + 400000) Z = 1260000X1 + 450000X2 with the following constraints (Maximize): or 60000X1 + 40000X2 400000 3X1 + 2X2 20 (Budget constraint) or 4 X2 5_ 6 Convert inequalities into equations and plot as follows. X1 = 0, X2 = 10, (0, 10) X2 = 0, =
20 3
\3 0)
3X1 + 2X2 = 20, X1 = 4, X2 = 6
OPERATIONS RESEARCH
56 x2 •
10
A
X, = 4
60000X, + 40000X2 = 400000
X2 = 6
3X, + 2X2 = 20
0
2.7 4 6.67 Fig. 2.15
I
I
8
12
Point B can be determined by solving equations which intersect at B X1 = 2.7, X2 = 6. Point
Coordinates
Objective function
A
(10, 0)
12600000
B
(2 . 7, 6)
3402000 + 2700000 = 6102000
C
(4, 6)
5040000 + 2700000 = 7740000
Since maximum value occurs at point C, to maximise number of audience, the advertising agency should go for at least 4 programmes on TV and 6 on radio. Example 2.43. (a) Solve graphically: Z = 80X1 + 100X2 Maximize X1 + 2X2 5720 Subject to 5X1 + 4X2 _5 1800 3X1 + X2 5900 Xi,X2 ,> 0 (b) Formulate following LPP. One unit of product P1 contributes Rs. 70 and requires 30 units of raw material and 20 hours of labour. One unit of P2 contributes Rs. 50 and requires 10 units of raw material and 10 hours of labour. Availability of raw material is 480 units and time available is 400 hours. Solution. (a) Convert inequalities into equalities + 2X2 = 720 5X1 + 4X2 = 1800 3X1 + X2 = 900
•••(i) ...(ii) ...(iii)
LINEAR PROGRAMMINGI
57
From (i)
X1 = 0 X2 = 360 (0, 360) X2 = 0 X1 = 720 (720, 0) From (ii) X1 = 0 X2 = 360 (0, 450) X2 = 0 X1 = 450 (360, 0) From (iii) X1 = 0 X2 = 900 (0, 900) X2 = 0 X1 = 300 (300, 0) Now these points and lines connecting these points can be joined to determine the feasible solution area as follows. Point P can be found out by solving equations (i) and (ii) and point C can be found out by solving equations (ii) and (iii). Point P (120, 300) and C (257, 129)
(b)
Point
Coordinates
Z = 80X1 + 100X2
A
(0, 360)
360000
P
(120, 300)
9600 + 30000 = 39600
C
(257, 129)
20560 + 12900 = 33460
D
(300, 0)
24000
Maximize Z = 70P1 + 50P2 Subject to the constraints 30P1 + 10P2 5_ 480 (Constraint of raw material for product P1) 20P1 + 10P2 400 (Constraint of raw material for product P2) P1 > 0
P2 >0. 10.00 — 900 800
•100 200 300 400 500 600 700 800 900 1000
OPERATIONS RESEARCH
58
Example 2.44. Orient Paper Mills produces two grades of papers X and Y. Because of raw material restrictions not more than 400 tons of grade X and not more than 300 tons of grade Y can be produced in a week. There are 160 production hours in a week. It requires 0.2 and 0.4 hours to produce a ton of product X and Y respectively, into corresponding profit of Rs. 20 and Rs. 50 per ton. Formulate a linear programming problem to optimize the product mix for maximum profit. Solution. Maximize Z = 20X + 50Y (Objective function) Subject to constraint 0.2X + 0.4Y 160 (Constraint of production hours per week) X 400 (Constraint of raw material) Y 300 (Constraint of raw material) X>0 Y>_0 Example 2.45. A firm manufactures three products A, B and C. Time to manufacture product A is twice that for B and thrice that for C and they are to be produced in the ratio 2 : 3 : 4. The relevant data is given below. If the whole labour is engaged in manufacturing product A, 2000 units of the product can be produced. There is demand for at least 200, 300 and 400 units of product A, B and C and the profit earned per unit is Rs. 100, Rs. 70 and Rs. 50 respectively. Formulate the problem as a linear programming problem. Raw Material
Requirement per unit of product (Kg) Total availability (Kg) A
B
C
P
6
5
9
4000
Q
4
8
6
5000
Solution Maximize Z = 100A + 70B (Objective function) Subject to the following constraints: 6A + 5B + 9C 5_ 4000 (Constraint of raw material P) 4A + 8B + 6C 5_ 5000 (Constraint of raw material Q) A+ 1 B + 1 C 5. 2000 2 3 A 200 B 300 400 A: B:C:: 2 :3 :4 A_B 3 2
i.e., 3A — 2B = 0
B C i.e., 5B — 3C = 0 3 5 Example 2.46. A firm uses lathes, milling machines and grinding machines to produce two machine parts A and B. The machining time required for each part, the machining time available on different machines and the profit on each machine part are as follows :
LINEAR PROGRAMMINGI
59 Machining time required for the machine part (minutes) — A
B
Maximum time . available per month (hours)
Lathe
15
9
600
Milling Machine
8
10
400
Grinding Machine
2
3
150
Profit per unit
50
120
Type of machine
Find the number of parts A and B to be manufactured per month to maximize the profit. Solution. Maximize
Z = 50A + 120B (Objective function) 15A + 9B 5 600 8A + 10B 400 2A + 3B 160 A>_0 B>_0 Convert the inequalities into equalities. 15A + 8B = 600 8A + 10B = 400 2A + 3B = 160 From (i) A = 0 B = 75 (0, 75) B = 0 A = 40 (40, 0) From (ii) A = 0 B = 45 (0, 45) B = 0 A = 50 (50, 0) From (iii) A = 0 B = 50 (0, 50) B = 0 A = 75 (75, 0) Plotting these equations as a straight line. Point Q can be determined by solving equations (i) and (ii). Point Q (32.5, 14) Shaded area is the feasible solution area as shown in Figure 2.17 Coordinates and values of OPQR:
Subject to
.(0 ...(ii)
...(iii)
Point
Coordinates
Value of Z = 50A + 120B
0
(0, 0)
0
P
(0, 40)
4800
Q
(32, 20)
4840
R
(40, 0)
4800
Point R gives the maximum value of Z. Note that equation 2A + 3B = 150 is a redundant constraint as it does not effect the feasible solution area.
60
OPERATIONS RESEARCH
80 70 60 50
Part B
40 30 20
10
10 20 30 40 50 60 70 80 Part A Point Q can be determined by solving equations (i) and (ii) Point Q (32.5, 14)
Fig. 2.17 Example 2.47. A firm manufactures two products P1 and P2 on which the profits earned are Rs. 5 and Rs. 8 respectively. Each product is prepared on two machines M1 and M2 , the machine time required for these products on the two machines and their availability is as shown below.
Machine M1 Machine M2
Product P1
Product P2
Availability of machine (minutes) per day
2 4
1 1
400 600
Find the number of units of products P1 and P2 to be manufactured per day to get maximum profits. Solution. Maximize Z = 5P1 + 8P2 (Objective function) Subject to the following constraints: 2P1 + P2 5 400 •••(i) ...(ii) 4P1 + P2 600 P1 P2 0 Convert the inequalities into equalities. 2P1 + P2 = 400 •••(i) ...(ii) 4P1 + P2 = 600 (0, 400) From (i) P1 = 0 P2 = 400 (200, 0) P2 = 0 P1 = 200 (0, 600) P1 = 0 P2 = 600 From (ii) P2 = 0 _P1 7 150 (150, 0)
61
LINEAR PROGRAMMINGI Plotting the equation on the graph
600 500 — 400 Product P2
2P1 + P2 = 400 A
4Pi + P2 = 600
Feasible solution area
300 — 200 — 100 —
100
1
200 300 400 500 600 Product P1
Fig. 2.18 Shaded area is the feasible solution area. Coordinates of OABC and value of Z. Point
Coordinates
Value of Z = 5P1 + 8P2
0
(0, 0)
0
A
(0, 400)
3200
B
(100, 200)
2100
C
(150, 0)
750
Maximum value of Z = 3200 at point A. Example 2.48. A firm manufactures three products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respectively. The firm has two machines and the required processing time in number for each machine on each product is given below.
Machine
C D
A 4 2
Product B 3 2
C 5 4
Machines C and D have 2000 and 2500 machineminutes respectively. The firm must manufacture 100A's, 200B's and 50C's but no more than 150A's. Setup an LP model to maximize the profit. Solution. Maximize Z = 3A + 2B + 4C Subject to 4A + 3B + 5C 5 2000 2A + 2B + 4C 2500 A 200 B 200
62
OPERATIONS RESEARCH A 150 C 50 A, B, C > 0
Example 2.49. A plant manufactures two products A and B. The profit contribution of each product has been estimated as Rs. 20 for product A and Rs. 24 for product B. Each product passes through three departments of the plant. The time required for each product and total time available in each department are as follows : Department 1 2 3
Hours Required Product A Product B 2 3 3 2 1 1
Available hours during the month 1500 1500 600
The company has a contract to supply at least 250 units of product B per month. Formulate the problem as a linear programming model and solve by graphical method. Solution. Maximize Z = 20A + 24B (Objective function) Subject to the constraints 2A + 3B 1500 (Constraint of time in department 1) •••(i) ...(ii) 3A + 2B 1500 (Constraint of time in department 2) ...(iii) A + B 600 (Constraint of time in department 3) B 250 (Constraint of minimum production of product B) A>0 B>0
11111‘.
Product B
2A + 3B = 1500
3A + 2B = 1500
A + B = 600
100 200 300 400 500 600 700 800 Product A
Fig. 2.19
63
LINEAR PROGRAMMINGI
To solve this problem graphically, convert the inequalities into equalities. 2A + 3B = 1500 ...(ii) 3A + 2B = 1500 ...(iii) A + B = 600 B = 250 (0, 500) From (i) A = 0 B = 500 B = 0 A = 750 (750, 0) A = 0 B = 750 (0, 750) From (ii) Points P and Q can be determined by solving equations (i) and (ii), and (ii) and (iii). In fact, it is the same point with coordinates (300, 300). B = 0 A = 500 (500, 0) A = 0 B = 600 (0, 600) From (iii) B = 0 A = 600 (600, 0) Shaded area is the area of feasible solution. Point
Coordinates
Value of Z = 20A + 24B
A
(0, 500)
12000
P (Q) R
(300, 300)
13200
(500, 0)
10000
Maximum profit at point P (300, 300) equal to 13200. Example 2.50. The ABC electric company produces two products, refrigerators and cooking ranges. Production takes place in two separate departments, refrigerators are produced in department I and cooking ranges are produced in department II. The company's two products are produced and sold on a weekly basis. The weekly production cannot exceed 25 refrigerators in department I and 35 cooking ranges in department II. Because of limited available facilities in the two departments, the company regularly employs a total of 60 workers in the two departments. A refrigerator requires two menweeks of labour, while a range requires 1 manweek of labour. A refrigerator contributes a profit of Rs. 60 and a cooking range contributes a profit of Rs. 40. Formulate the problem as a LP problem. How many units of refrigerators and ranges should the company produce to get a maximum profit ? Solution. Let X and Y be the number of refrigerators and cooking ranges to be produced. Z = 60X + 40Y (Objective function) Maximize Subject to the following constraints: X 25 (Constraint on the production of refrigerators) Y 35 (Constraint on the production of cooking ranges) 2X + 1Y 60 (Constraint of availability of total labour) X0 Now, convert the inequalities into equalities. 400X + 600Y = 6000 ...(ii) 400X + 200Y = 4000 From (i) X = 0 Y = 10 (0, 10) Y = 0 X = 15 (15, 0) X = 0 Y = 20 (0, 20) From (ii) Y = 0 X = 10 (10, 0) Plotting these equations as straight lines on the graph to find the feasible solution area. Y
20 —
400X + 200Y = 400
15 —
Feasible solution area
10 — 400X + 600Y = 6000 C
5 10 15 20 X Fig. 2.21 OABC is the feasible solution area. Coordinates of these points and value of Z is as follows: Point
Coordinates
Value of Z = 2000 X + 2500Y
0
(0, 0)
0
A
(0, 10)
25000
B
(7.5,5)
27500
C
(10, 0)
25000
Coordinates of point B can also be found out by solving equations (i) and (ii) subtract (ii) from (i). 400Y = 2000 Y=5 Putting Y = 5 in (ii) gives X = 7.5.
66
OPERATIONS RESEARCH
The company should produce 7.5 pages in magazine I and 5 pages in magazine II to maximize its reach, i.e., to 27500 people. Exceptional Cases. We have so far discussed the linear programming problems where there is a unique optimal solution always available. However, it may not be the case for all the problems. In general, a LPP should have the following: (a) A definite and unique optimal solution. (b) An unbounded solution. (c) No solution. Let us discuss the case of an unbounded solution. Example 2.52. Maximize Z = 10X + 5Y Subject to 2X4Y s10 X+2Y6 X>0 Y> 0 Solution. Convert the inequalities into equalities and plot them on the graph. 2X — 4Y = 10 X + 2Y = 6 From (i) X = 0 Y = — 2.5 (0, — 2.5) Y=0X=5 (5,0) From (ii) X=0Y=—3 (0, — 3) Y=0X=6 (6,0)
r
Feasible solution area Isoprofit line
Fig. 2.22 The feasible solution area is shown shaded in the figure. Let us plot Z =10X + 5Y = 0 X=0X=1X=2 Y=0Y=—2Y=—4
...(i) ...(ii)
67
LINEAR PROGRAMMINGI
Draw PQ as isoprofit line. If we move the isoprofit line towards the right as Z increases, we still get Z maximum at a point (corner) farthest from the origin. It can be seen that the farthest point from the region will occurs at infinity. This is the case of an unbounded solution. Example 2.53. Maximize Z = 4X + 8Y Subject to  3X + 6Y 2X  6Y 6 X, Y > 0 Solution. Converting inequalities into equalities. ...(i)  3X + 6Y = 18 ...(ii) 2X  6Y  6 From (i) X=0Y=3 (0, 3) Y=0X=6 ( 6, 0) From (it) X=0Y=1 (0,1) Y=0X=3 ( 3, 0) Plotting these equations as straight lines on the graph. Y
6 5 4 3X + 6Y = 18 2X  6Y = 6
As%
Nivt.....1
0
2 1 Fig. 2.23 Figure 2.23 indicates two shaded regions as feasible solution areas since these two shaded regions do not overlap, there is no point common to the shaded regions. It means no solution exists and the problem cannot be solved either graphically or by any other method. Example 2.54. Maximize Z = 2x1 + 3x2 Subject to xi + x2 1 3 x1 + x2 .5" 4 x1, x2 2. 0 Solution. Converting the constraints into equalities. X1 4 X2 = 1
4 x1 = 0 x2 = 1 The two points are (0, 1) and (1, 0).
3x1 + x2 =
From equation (i)
OPERATIONS RESEARCH
68 and From equation (ii)
x2 = 0 x1 = 1 x1 = 0, X2 = 4
4 , 0) The two points are (0, 4) and (3
4 and x2 = 0 and x1 = — 3 X2 5— 4—c
3x1 + x2 4
2— (0,1), Feasible solution area 1 A
x1 + x2 0 Solution. The above constraints are plotted as shown in Figure 2.37.
X1 5. 10 20
4x, + 3x2 5_ 60
15 x2 5_12
B 10 Alternative optimal solution along line DE
5
10
15
20
Fig. 2.37 The corner points and their respective value for Z are summarized in the following table.
79
LINEAR PROGRAMMINGI Corner Point A B C D E F
Coordinates (0, 0) (0, 12) (4, 12) ( 60 60 ) 7'7
Value of Objective function Z = 20 xi + 15 x2 0 180 260
(10, 2°) 3 (10, 0)
300
300
200
It may be noted that there is a tie for the highest value of Z between points D and E. The slope of the objective function is the same as the constraint 4x1 + 3x2 60. In Figure 2.37, there are infinite alternative optimal solution along line DE. COMPUTER SOLUTION METHODS The graphical solutions discussed earlier are obviously limited to two variables problems. The other examples discussed do involve more than two variables but are relatively simple and compared to the real life problems encountered by the organisations of today. These problems could be solved manually by the Simplex Method but the method is complicated and involves laborious calculations. In actual applications, LP problems are solved by computer methods. Today, there are many efficient computer softwares available. While using the software for solving LP models, one need not always be concerned about the internal details of the solution method, though there is no doubt that one can become a better user of the software if one knows its details. Effective use of these software models can be made if one: (a) fully understands the LP model and its assumptions (b) is skillful in recognising an LP problem (c) can formulate the problems well (d) can arrange for solution using a computer package and is (e) is capable of interpreting the output from such packages. Before the computer solution methods are discussed, we need to understand certain basics. Most computer methods are based upon algebraic solution procedure, like of which are used in Simplex Method. Let us recapitulate the following three requirements in solving a LPP by Simplex Method. 1. All constraints must be stated as equations. 2. The right hand side of a constraint cannot be negative. 3. All variables are restricted to nonnegative values. It can be shown that the optimal solution to a LPP is included in the set of basic feasible solutions of the simplex procedure. So, the optimal solution can be found by performing a search of the set of basic feasible solutions. This is what the simplex method accomplishes. It begins with a basic feasible solution consisting of two pools of variables—basic variables and nonbasic variables. The Simplex method determines whether the objective function can be improved by exchanging a basic variable and a nonbasic variable. If an exchange will result in an improvement, an existing basic variable is set equal to 0 (becoming a nonbasic variable) an existing nonbasic variable is included in the pool of basic variables and the system of equations
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OPERATIONS RESEARCH
is resolved with the new set of basic variables to form a new basic feasible solution. A determination is made again regarding whether a better solution exists. If yes, another exchange takes place and the process repeats itself. The simplex method is termed as iterative process because a specified set of solution steps are repeated until an optimal solution is identified.
■ REVIEW AND DISCUSSION QUESTIONS ■ 1. What are the essential characteristics of a Linear Programming model ? 2. Explain the terms: Key decision, objective, alternative and restrictions in the context of linear optimisation models by assuming a suitable industrial situation. 3. Discuss the application of LP in any functional area of management. Use suitable example from business or industry. 4. "Linear Programming is the most widely used and most successfully used mathematical approach to decisionmaking". Comment. 5. Explain the advantages and major limitations of LP model. Illustrate your answer with suitable examples. 6. What are the major allocation problems that can be solved by using LP model ? Briefly discuss each one of them. 7. "Each LP problem with a feasible solution area offers infinite number of solutions". Do you agree with this statement ? Justify your answer. 8. What are the basic characteristics and assumptions of a LP model ? 9. Explain important characteristics of the industrial situations to which linear programming  method can be successfully applied. Illustrate applications of this technique with a suitable example. 10. Write at least five application areas of Linear Programming. 11. Explain the graphical method of solving an LPP involving two variables. 12. Write the general form of LPP and prove that the feasible region of an LPP is a convex set. 13. Illustrate graphically the following special cases of LP problems: (i) Multiple optimal solutions (it) Nonfeasible solution (iii) Unbounded problem. 14. Write the standard form of LPP in matrix form. 15. Write a short note on the limitations of Linear Programming. 16. Discuss briefly the steps to formulate a Linear Programming Problem. Give suitable examples. 17. Explain the following terms: (i) Linearity (it) Feasible solution (iii) Objective function (iv) Unbounded solution (v) Optimal solution. 18. Orient Paper Mills produces two grades of paper X and Y. Because of raw material availability limitations, not more than 500 tons of grade X and 250 tons of grade Y can be produced in a month. Total production hours in a month are 400 product X needs 15 minutes and product Y needs 25 minutes for production of one ton of product each. Profit margin of Xis Rs. 200 and Y is Rs. 400 per ton. Formulate a LPP to optimise the product mix for maximum profit.
81
LINEAR PROGRAMMINGI
1 19. Prime Enterprises manufactures three types of dolls. The boy requires — meter of red cloth, 2 1—
1 meters of green cloth and 1 — meters of black cloth and 5 kg of fibre. The girl requires
2 2 meter of red cloth, 2 meters of green cloth, one meter of black cloth and 6 kg of fibre. The dog
1 1 requires — meter of red, one meter of green, — meter of black and 2 kg of fibre. The profits on 2 4 the three are respectively Rs. 3, 5 and 2. The firm has 1000 meters of red, 1500 meters of green, 2000 meters of black and 6000 kg of fibre. Find the number of dolls of each type to be manufactured, setup LP for maximum profit. 20. Maximize P = 1.4 X1 + X2 Subject to X1 3 2X1 + X2 8 3X1 + 4X2 24 X1 0, X2 0 Using graphic method. 21. Three products are produced on three different operations. The limit of available time for the three operations are respectively 430, 460 and 420 minutes and profit per unit for the three products are Rs. 3, 2 and 5 respectively. Times in minutes per unit on each machine operation is as follows : Operations 1 2 3
I 1 3 1
Product II 2 0 4
III 1 2 0
Write LP model for this problem. 22. Use graphic method to solve the following LPP. Z = 3X1 + 4X2 Maximize Subject to 6X1 + 12X2 1200 0.8 X1 + 0.5 X2 60 0.3 X1 + 0.4 X2 50 X1 > 0,X2 ?_0 23. Use graphical method to solve the following: Maximize Z = 5X1 + 3X2 3X1 + 5X2 = 15 Subject to 5X1 + 2X2 = 10 0, X2 24. Solve the LPP given below by graphical method and shade the region representing the feasible solution. Z = 2X1 + 10 X2 Minimize X1 — X2 ?_ 0
82
OPERATIONS RENTONRCH X1  5X2  5
Xi 0, X, C 25. Solve graphically the following LPP : Minimize Z = 3X1 + 5X2 Subject to — 3X1 + 4X2 5.. 12 2X1  X2  2 2X1 + 3X 12 X1 _0, X2 >_0
26. Use graphical method to solve: Maximize Z = 2X1 + 3X2 Subject to X1 + X2 1 5X1  X2 0 X1 + X2 6 X1  5X2 0 X2  X1  1 X2 _0, X2 >_0
27. Solve graphically the LPP: Maximize Z = 6X1 + 4X2 2X1 + 3X2 30 Subject to 3X1 + 2X2 5. 24 X1 + X2 3 Xi , X2 0
28. Old hens can be bought at Rs. 2 each and young ones at Rs. 5 each. The old hen lays 3 eggs per week and young hen lays 5 eggs per week, each egg being worth 30 paise. A hen costs Rs. 1 per week to feed. Mr A has only Rs. 80 to spend for hens. How many of each kind should Mr A buy to give a profit of at least Rs. 6 per week, 8 assuming that Mr. A cannot house more than 20 hens? 29. Solve the LPP graphically: Maximize Z = 80X1 + 120 X2 Subject to X1 + X2 3 when X1 and X2 ?_ 0 30. A small manufacturer employs 5 skilled men and 10 semiskilled men and makes an article in two qualities, a deluxe model and an ordinary model. Making of a deluxe model requires 2 hours work by a skilled man and 2 hours work by a semiskilled man. The ordinary model requires one hour work by a skilled man and 3 hours work by a semiskilled man. By union rules no man can work for more than 8 hours a day. The manufacturer's clear profit of the deluxe model is Rs. 10 and of the ordinary model Rs. 8. Formulate the model of the problem.
LINEAR PROGRAMMINGI
83
31. A firm manufactures three products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respec
tively. The firm has two machines and the processing time in minutes for each machine on each product is given below. Products
Machine C
D
A
B
C
4 2
3 2
5 4
Machines C and D have 2,000 and 2,500 machineminutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's setup an LP model to maximize the profit. 32. A farmer has 100 acres farm. He can sell all the tomatoes, lettuce or reddish he can raise. The price he can obtain is Rs. 1 per kg for tomatoes, Rs. 0 • 75 head for lettuce and Rs. 2 per kg reddish. The average yield per acre is 2000 kg of tomatoes, 3000 heads of lettuce and 1000 kg of reddish. Fertiliser is available at Rs. 0.50 per kg and the amount required per acre is 100 kg each for tomatoes and lettuce and 50 kg for reddish. Labour required for sowing, cultivating and harvesting per acre is 5 mandays for tomatoes and reddish and 6 mandays for lettuce. A total of 400 mandays of labour are available at Rs. 20 per manday. Formulate an LP model for this problem in order to maximize the farmer's total profit. 33. A manufacturer of metal office equipment makes desks, chairs, cabinets and book cases. The work is carried out in three major departments—metal stamping, assembly and finishing. The exhibits A, B and C give requisite data of the problem. EXHIBIT—A Time required in per unit of product Products
Departments
Stamping Assembly Finishing
Desk
Chair
Cabinet
Bookcases
Hours available/week
4 10
2 6
3 8
3 7
800 1200
10
8
8
8
800
EXHIBIT—B Cost (Rs) of operation per unit of product Departments
Products Desk
Chair
Cabinet
Bookcases
Stamping Assembly
15 30
8 18
24 24
21 21
Finishing
35
28
25
21
EXHIBIT—C
Selling price (Rs) per unit of product Desk 175 Chair 55 Cabinet 145 Bookcase 130
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OPERATIONS RESEARCH
In order to maximize weekly profits, what should be the production programmes ? Assume that the items produced can be sold. Which df_ya, iment needs to be expanded for increasing profits ? 34. A used car dealer wishes to stock up his lot to maximize his profit. He can select cars A, B and C which are valued wholesale at Rs. 5000, Rs. 7000 and Rs. 8000 respectively. These can be sold at Rs. 6000, Rs. 8500 and Rs. 10500 respectively. For each car type, the probabilities of sale are: Type of Car A B C
Probability of sale in 90 days 0.7 0.8 0.6
For every two cars of type B, he should buy a one car of type A or C. If he has Rs 100000 to invest, what should he buy to maximize his expected gain ? Formulate the linear programming problem. 35. A certain farming organisation operates three farms of comparable productivity. The output of each farm is limited both by useable acreage and by the amount of water available for irrigation. Following is the data for upcoming season: Farm
Useable Acreage
1 2 3
400 600 300
Water available in acre feet 1500 2000 900
The organisation is considering three crops for planting which differ primarily in their expected profit per acre and their consumption of water. Furthermore, the total average that can be devoted to each of the crops is limited by the amount 8 appropriate harvesting equipment available. Crop A B C
Minimum Acreage 700 800 300
Water consumption in acre feet per acre 5 4 3
Expected profit per acre Rs 400 Rs 300 Rs 100
In order to maintain a uniform workload among the farms, it is the policy of the organisation that the percentage of useable acreage planted must be the same at each farm. However, any combination of the crops may be grown at any of the farms. The organisation wishes to know, how much of each crop should be planted at the respective farms in order to maximize expected profits. Formulate thus as a linear programming problem. ILLUSTRATED CASE STUDIES
Case Study No 1 ABC Ltd engages Quality Control inspectors from a consultancy company which has a pool of such manpower as it does not want to have inspector on its own pay roll. It uses 40 inspectors of Grade I level and 60 inspectors of Grade II level. The company expects the following standards:
LINE/kit PROGRAMMINGI Grade of Inspectors I II
85 No of pieces to be inspected in an 8 hours day 30 20
Wages per hour (Rs.) 25 40
Accuracy to be achieved 98% 96%
At least 1600 pieces must be inspected in a day of 8 hours. If an error in inspection is made, it will cost the company Rs 50. The company is interested in knowing the optimal assignment of inspectors so that the inspection costs are minimized. Hint. Two types of costs are incurred by each grade of inspector (a) Wages to be paid to the inspector. (b) Cost of inspection error. Cost of one grade I inspector per hour Rs. (25 + 50 x 0.02 x 3° —) 8 Similarly, cost of one grade II inspector/hour can be determined. Minimize Z = x1 x Cost of grade I inspector + x2 x cost of grade II inspector With the constraints x1 40 x2 60 (Constraint of number of inspectors) 30x1 + 20 x2 200 Case Study No. 2. A company produces special alloys used by the defence forces in manufacture of certain components of an air defence gun. The alloy specifications provided by the defence forces are: (a) Chromium 10% (b) Melting point 600°C (c) Specific gravity 0.99 The company uses three types of raw materials to produce the alloy of above specifications. The properties of the raw material required for this purpose are as given below. Parameter of Specification Chromium (%) Melting point (°C) Specific Gravity
Raw material P 8 700°C 0.98
Property Raw material Q 12 650°C 1.02
Raw material R 16 600°C 0.96
The cost of the raw materials per unit in open market are: P = Rs. 10000 Q = Rs. 25000 R = Rs. 8000 The company wants stock to the specifications so that they can continue getting the orders from defence department, which is their mainstay for profits and also give them the credibility and
86
OPERATIONS RESEARCH
•
goodwill in the market. At the same time, they are interested in reducing the cost of raw materials to the minimum. Advise the company as to what percentage of raw materials P, Q and R are to be used for making the alloy. [Hint. Minimize Z = 10000 x1 + 25000 x2 + 8000 x3 Constraints are : 8x1 + 12x2 + 16x 10 700x1 + 650 x2 + 600 x3 > 600 0.981 xi + 1.02 x2 + 0.96 x3 5 0.99 + x2 + x3 = 100] Case Study No. 3 ABC Ltd is engaged in assembling different types of washing machines. The balance sheet of the company as on 31 March 2003 is as follows: Liabilities Equity share capital Capital Reserves General Reserves Profit & Loss A/c LongTerm Loan Loans (a) Bank X (b) Bank Y
Assets 175000 20000 140000 20000 150000 75000 75000 75000 Total 730000
Land Building Plant & Machinery Furniture Transport Inventory Receivable Cash
200000 50000 120000 20000 25000 10000 70000 235000 Total 730000
The company is able to sell all the types of washing machine it manufactures but gets the payment only on the first of next month of the sales month. The company pays the labour, materials and other expenses in cash. The company had taken a loan of Rs. 75000 from bank X and this has to be paid back on 01.4.2003. However, the company has tiedup with bank Y to take a loan of Rs. 75000. The other overheads of the company are: (a) Salaries Rs. 30000 for a month. (b) Interest on longterm loan 8% (c) Interest on loan from Bank X and Bank Y — Rs 2000/month. ABC Ltd is manufacturing the following types of washing machines. These have to be finished and assembled in any assembly line. The following data is available for the production of the washing machines: Type of Washing Machine Mannual Semiautomatic Fully automatic Total available time
Polishing Time (Hours) 10 12 15 2000
Assembling Time (Hours) 4 6 8 1200
Variable Cost ' (Rs.) 8500 9000 12000.
Selling Price (Rs.) 10000 12000 16000
LINEMi. yRpGRAMMING I
87
The company has an order of 4 semiautomatic and 10 fully automatic machines for the next month. The management of the company is interested in knowing that how many units of each type of washing machines should they assemble next month so that they can earn maximum profit. The company is very careful regarding their functional matters and wants to be sure that the cash available next month is sufficient to meet all the operations of the next month. What advice will you give to the company management ? [Hint. Let x1, x2 and x3 the number of washing machines to be assembled. Cash requirement 8500 x1 + 9000 x2 + 12000 x3 From balance sheet and other data given the cash availability in the next month is = Cash balance + Receivables — (loan of bank x to be paid back + salaries + interest on longterm loan + interest on other loan) = (235000 + 70000) — (75000 + 30000 +
8 x 150000 ) + 2000 100x12
= (305000 — 108000) = Rs. 197000 Hence 8500 x1 + 9000 x2 + 12000 x3 197000 Also x2 > 4 x3 > 8 Maximize Z = 1500 x1 + 3000 x2 + 4000 x3 with the following constraints: 10x1 + 12x2 + 15x3 2000 (polishing machine constraint) 4x1 + 6x2 + 8x3 51200 (Assembly time constraint) Case Study No. 4. A corporate house has made huge profits from their cashcow product which is very wellaccepted in the market. It is planning to launch a new product after one year and R & D department is in the advance stages of the development of the product. The corporate house has consulted its investment research department to park Rs. 10 crore which has advised them as follows: Parameter for decisionmaking Risk Expected returns %
A High 40
Type of Investment B C High 30
Medium 10
D Low 6
The management in its Board meeting has decided the following guidelines for ensuring the safety of the funds. (a) Total amount invested in high risk funds must be less than 50%. (b) Total amount invested in medium risk funds must be less than 30%. (c) Total amount invested in low risk funds must be less than 10%. (d) Total amount invested in high risk funds A and B should be in the ratio of 2 : 3 of the total amount.
OPERATIONS RESEARCH
88
What should be the portfolio of the corporate house so that they are able to maximise their returns ? Formulate the problem as an LPP. Hint. Let x1 , x2 , x3 and x4 be the parts of the portfolio of different funds Maximize I = 0.40 x1 + 0.30 x2 + 0.10 x3 + 0.06 x4 Subject to the following constraints xi + x2 0.5 x3 0.30 x4 0.10 Also xi : x2 = 2 : 3 or 3x1  2x2 = 0. Case Study No. 5 A Company is manufacturing three products in its factory using three different routes by using Lathes, drills and shearing machines. The details of these are as follows : Type of Machines Lathes Drills Shearing Machine
Product A 1 0.4 0.6 0.3
2 0.5 0.2 0.6
Product B 3 0.4 0.6
1 0.6 0.4 0.2
2 0.2 0.8
Product C 1 0.4 0.2 0.5
2 0.6 0.2
Availability of Machines (Hours) 300 400 500
All the product manufactured can be sold in the market at sale price of product A Rs. 200, product B Rs. 260 and product C Rs. 100. The company has to meet the market demand of 200, 220 and 180 units of products A, B and C respectively. Formulate the problem as a LPP. Hint. Let the number of units of product A, B and C manufactured through the three routes by X Ai , XA2 , X A3 for product A, X , X for product B and Xc1, XC2 for product C. BI B2 Maximize objective function Z = 200 (X A3 + X A2 X A3 + 260 (X 53 + X52 )+100 (X c1 + )(C2 Constraint for Lathe machines 0.4 X A1 + 0.5 XA2 + 0.6 XBI +0.4 Xci < 300 Drills Constraint 0.6 XA3 + 0.2 X A2 + 0.4 XA3 + 0.4 X$1 +0.2X52 +0.2Xc1 + 0.6 Xe2 < 400 Shearing Machine Constraints 0.3 XA3 +0.6XA2 + 0.6XA3 +0.2)(51 +0.8X$2 +0.5Xc1 + 0.2Xc2 500 X AI
X A2 _> 200
XBI > 220 Xci > 180. I G.N.D.U. EXAMINATION PROBLEMS I
1995 APR. (1) Infeasible and unbounded solution (2) Linearity and certainty in LP problem.
LINEAR PROGRAMMINGI 1995 SEP.
89
(3) Basic feasible solution. (4) Objective functions of LPP 1996 APR. (5) Linearity and divisibility. (6) Multiple optimum solution. 1996 SEP. (7) Basic feasible solution. 1997 APR. (8) Linearity and divisibility. 1998 APR. (9) Multiple optimum solution. (10) Nonnegativity constraint in a LPP (11) Linear programming problem. (12) Graphic method. (13) Optimal solution of a LPP 1999 APR. (14) Initial basic Solution of a LPP 1999 SEP. (15) Assumptions underlying LPP (16) Objective function of LPP 2000 APR. (17) Basic solution in LPP (Professional) 2003 APR. (18) O.R. (Professional) 2003 APR. (19) Feasible solution v/s Optimum solution of LPP 2004 APR. (20) Objective function of a LPP (Professional) 2004 APR. (21) Feasible solution v/s Optimum solution. (Professional) 1996 APR. (22) Discuss briefly the application of Linear Programming in various functional areas of management. 1996 SEP. (23) Discuss briefly the application of Linear Programming in decisionmaking. 1997 SEP. (24) Discuss briefly the steps to formulate a Linear Programming problem. Explain with an example. 1995 SEP. (25) Discuss the formulation of a minimization linear programming problem. 1996 APR. (26) Solve graphically : Minimize : Z = 2500x1 + 3500x2 Constraints : 50 x1 + 60x2 2500 100 x1 + 60x2 3000 100 x1 + 200x2 7000 x1, x2 >_0. 1996 APR. (27) Vitamin A and B found is in foods F1 and F2. One unit of food F1 contains 3 units of vitamin A and 4 units of vitamin B and that of F2 contains 6 units of vitamin A and 3 units of vitamin B. One unit of food F1 and F2 costs Rs. 4 and Rs. 5 respectively. The minimum daily need per person of vitamin A and B is 80 and 100 units respectively. Assuming that anything in excess of daily minimum requirement is harmful, find out the optimum mixture of F1 and F2 at the minimum cost which meets the minimum requirement of vitamin A and B. Formulate this as a Linear Programming problem and solve. Minimize : Z = 4x1 + 5x2 Subject : 3x1 + 6x2 80 4x1 + 3x2 100 x2 ?_ 0
90
OPERATIONS RESEARCH
1996 SEP. (28) A company sells two products A and B and makes a profit of Rs. 40 and Rs. 30 per unit respectively. The two products are produced in a common production process and sold in different markets: Production process has a total capacity of 30000 manhours. It takes 3 hours to produce a unit of A and one hour to produce a unit of B. Market survey shows that the maximum units of A that can be sold is 8000 and that of B is 12000. Subject to these limitations, the product can be sold in any combination. Formulate the above as a Linear Programming problem and solve it. Maximum value of Z = 40 A + 30 B Subject to : 3 A + B 5. 30000 A 8000 B 12000; A, B 0 1997 APR. (29) Orient Paper Mills produces two grades of paper X and Y. Because of raw material restrictions not more than 400 tons of grade X and 300 tons of grade Y can be produced in a week. There are 160 production hours in a week. It requires 0.2 and 0.4 hours to produce a ton of product X and Y respectively with corresponding profits of Rs. 20 and Rs. 50 per ton. Formulate a Linear Programming problem to optimise the product mix for maximum profit and solve it. Maximum value of Z = 20 X + 50Y Subject to: 0.2 + 0.4 Y 160 X 400 Y__ 0 1998 APR. (30) Maximize : P = 1. 4X1 + X2 Subject to : 3 2 Xi + X2 8 3X1 + 4X2 5.. 24 X1 >_0, X2 >_0 Use graphic method. 1998 APR. (31) Prima Enterprises manufactures three type of dolls. The boy requires half metre of red cloth, 11/2 metres of green cloth, 11/2 metres of black cloth and 5 kg of fibre. The girl requires lh metre of red cloth, 2 metre of green cloth, 1 metre of black and 6 kg of fibre. The dog requires of 1/2 metre of red, 1 metre of green, 1/4 metre of black and 2 kg of fibre. The profits on the three are respectively 3, 5 and 2 rupees. The firm has 1000 metres of red, 1500 metres of green, 2000 metres of black and 6000 kg of fibre. To find the number of dolls of each type to be manufactured, setup a LP for maximum profit. Maximum value of Z = 3X1 + 5X2 + 2X3 Subject to 0.5 X1 + 0.5 X2 + 0.5 X3 1000 1.5 Xi + 2 X2 +X3 1500 0.5 Xi I X2 + 0.25 X3 2000 5 Xi + 6 X2 + 2 X3 5_ 6000 X1, X2, X3 0
LINEAR PROGRAMMINGI
91
1998 SEP. (32) Maximize : 6A + 5B Subject to
3 A+ 4135120 3A + 1B 5. 40 A+B= 30,A,B
Using graphic method. 1998 SEP. (33) Three products are produced on three different operations. The limits on the available times for the three operations are respectively 430, 460 and 420 minutes and the profit per unit for the three products are Rs. 3, Rs. 2. and Rs. 5 respectively. Time in minutes per unit on each machine operation is as follows : Operations 1 2 3
I 1 3 1
II 2 0 4
III 1 2 0
1999 APR. (34) Use Graphic method to solve the following LPP Maximize : Z = 3X1 + 4X2 Subject to : 6X1 + 12 X2 5 1200 0.8 X1 + 0.5 X2 5. 60 0.3X1 + 0.4 X2 5_ 50 X1 0, X2 >_ 0 2000 APR. (35) Solve Graphically : Minimize : Z = 2X + 4Y Subject to X + Y 14 2X + 2Y 30 2X + 18 X, Y>_ 0 2000 SEP. (36) Minimize Graphically: Z = 6x + 14y Subject to 5x + 4y 60 3x+ 84 x+2:>_ 18 2001 APR. (37) (a) Solve graphically: Maximize Subject to
Z = 80X1 + 100X2 X1 + 2X2 720 5 Xi + 4X2 1800 3X1 + X2 900 Xi, X2 0 (b) Formulate following LPP. One unit of product P1 contributes Rs. 70 and requires 30 units of raw material and 20 hours of labour. One unit of P2 contributes Rs. 50 and requires 10 units of raw material and 10 hours of labour. Availability of raw material is 480 units and time available is 400 hours.
OPERATIONS RESEARCH
92 Subject to :
2001 SEP.
2002 APR.
2002 SEP.
2003 APR.
2003 APR.
30 P1 + 10P2 480 20 P1 + 10P2 400
Pi, P2 0 (38) Solve graphically the following LPP: Minimize Z = 3X + 7Y Subject to: 5X+ 411 . 60 3X + 7Y 5_ 8 4 X+ 21/ 20 X, Y>_ 0 (39) Graphically solve the following LPP: Maximize Z = 100X + 120 Y Subject to 6X + 3Y 18 3 X+ 5Y 4X + 5Y 30 2X + 5Y 25 X, Y 0. (40) Graphically solve the following LPP: Maximize Z = 40X = 50 Y Subject to 2X + Y 5. 800 X + 2Y 5. 700 3X + 5Y S 900 x, y 0 (41) Maximum value of Z = 300X1 + 400X2 Subject to 5X1 + 4X2 200 3X1 + 4X2 150 5X1 + 4X2 100 8X1 + 4X2 80 X1 X2 0 (42) An advertising company wishes to plan an advertising compaign in three different media. T.V., radio and magazine. The purpose of advertising is to reach as many potential customers as possible. Result of market survey are as follows : TV
Cost of an Advertising unit Number of potential customers reached/Unit Number of women customers reached/Unit
Prime day (Rs.) 40000
Prime Time (Rs.) 75000
Radio
Magazine
(Rs.) 30000
(Rs.) 15000
400000
900000
500000
200000
300000
400000
200000
100000
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93
The company does not want to spend more than Rs. 800000 on advertising. It is further required that : (i) At least 2 million exposures take place among women. (ii) Advertising on T.V. be limited to Rs. 500000 (iii) At least 3 advertising units be bought on prime day and two units during prime time and (iv) Number of advertising units on Radio and Magazine should each be between 5 and 10. 2003 SEP. (43) Solve Graphically : Maximum value of Z = 5X1 + 4X2 Subject to X1 — 2X2 1 X1 + 2X2 ?_ 3 Xi , X2 ?_ 0 2004 APR. Minimize Z = 4X1 + 2X2 Subject to X1 + 2X2 2 3X1 + X2 3 4 X1  3X2 6 X1, X2 0 using graphic method. It PUNJAB UNIVERSITY EXAMINATION PROBLEMS LU
1995 APR. 1995 SEP. 1996 APR. 1996 APR. 1997 APR. 1998 APR. 1998 SEP.
(1) (i) Infeasible and unbounded solution (ii) Linearity and certainty in LP problem (2) (i) Basic feasible solution (ii) Objective functions of LPP (3) (i) Linearly and divisibility (ii) Multiple optimum solution (4) Basic feasible solution (5) (i) Linearity and divisibility (ii) Multiple optimum solution (6) Nonnegativity constraint in an LPP (7) (i) Linear Programming problem (ii) Graphical method (iii) Optimal solution of an LPP 1999 APR. (8) Initial basic solution of an LPP 1999 SEP. (9) (a) Assumption underlying LPP (b) Objective function of LPP 1995 SEP. (10) Discuss the formulation of a minimization linear programming problem (Marks 10) 1996 APR. (11) Discuss briefly the application of Linear Programming in various functional areas of management. (Marks 6) 1996 SEP. (12) Discuss briefly the application of Linear Programming in decisionmaking. (Marks 6) 1997 APR. (13) Discuss briefly the steps to formulate Linear Programming problem. Explain with an example. (Marks 6) 2000 SEP. (14) Define : (i) Feasible solution (ii) Basic solution (iii) Basic feasible solution (iv) Nondegenerate BFS (v) Degenerate BFS (vi) Optimum BFS (vii) Unbounded solution. (Marks 7)
94
OPERATIONS RESEARCH
1996 APR.
(15) (a) Solve Graphically : Z = 2500X1 + 3500 X2 Minimize : Constraints : 50X1 + 60X2 2500 100X1 + 60X2 3000 100X1 + 200X2 7,000 Xi, X2
0
Vitamin A and B is found in foods F1 and F2. One unit of food F1 contains 3 units of vitamin A and 4 units of vitamin B and that of F2 contains 6 units of vitamin A and 3 units of vitamin B. One unit of food F1 and F, cost Rs. 4 and Rs. 5 respectively. The minimum daily need per person of vitamin A and B is 80 and 100 units respectively. Assuming that anything in excess of daily minimum requirement is harmful, find out the optimum mixture of F1 and F2 at the minimum cost which meets the minimum requirement of vitamin A and B. (Marks 6) Formulate this as a Linear Programming Problem. 1996 SEP. (16) A company sells two products A and B and makes a profit of Rs. 40 and Rs. 30 per unit respectively. Two products are produced in a common production process and sold in different markets. Production process has a total capacity of 30000 manhours. It takes 3 hours to produce a unit of A and one hour to produce a unit of B. Market survey shows that the maximum units of A that can be sold 8000 and that of B 12000. Subject to these limitations, the product can be sold in any combination. Formulate the above as a Linear Programming problem. 3A + B 5_ 30000 Subject to : A 8000 B 12000 A, B 0 1997 APR. (17) Orient Paper Mills produces two grades of paper X and Y. Because of raw material restrictions not more than 400 tons of grade X and 300 tons of grade Y can be produced in a week. There are 160 production hours in a week. It requires 0.2 and 0.4 hours to produce a ton of products X and Y respectively, with corresponding profits of Rs. 20 and Rs. 50 per ton. Formulate a Linear Programming problem to optimise the product mix for maximum profit. P = 1.4 X1 + X2 1998 APR (18) (a) Maximise : Subject to : 3 2X1 + X2 8 3X1 + 4X2 24 0, X2 0 Using graphic method. (b) Prima Enterprises manufactures three types of dolls. The boy requires half metre of red cloth, 11/2 metres of green cloth, 2 metre of black cloth and 5 kg of fibre. The girls requires % metre of red cloth, 2 metre of green cloth, 1 metre of black and 6 kg of fibre. The dog requires lh metre of red, 1 metre of green, 1/4 metre of black and 2 kg of fibre. The profit on the three are respectively 3, 5 and 2 rupees. The firm has 1000 metres of red, 1500 metres of (b)
95
LINEAR PROGRAMMINGI
1998 SEP.
(19)
green, 2000 metres of black and 6000 kg of fibre. To find the number of dolls of each type to be manufactured, setup an LP for maximum profit. Subject to : 0.5X1 + 0.5X2 + 0.5X3 1000 1.5X1 + 2X2 + X3 5_ 1500 0.5 Xi + X2 + 0.25 X3 5_ 2000 5X1 + 6X2 + 2X3 6000 Xi, X2, X3>_0 (a) Maximize : 6A + 5B subject to 3A + 4B 120 3A+1B 40 A+B= 0,B?_Q Using graphic method. (b) Three products are produced on three different operations. The limits on the available time for the three operations are respectively 430, 460 and 420 minutes and the profit per unit for the three products are Rs. 3, Rs. 2 and Rs. 5 respectively. Time in minutes per unit on each machine operation is as follows : Operations
Product I
II
III
1
1
2
1
2
3 1
0
2
4
0
3
Write, LP model for this problem. 1999 APR. (20) Use Graphic Method to solve the following LPP Maximize : Z = 3X1 + 4X2 Subject to : 6X1 + 12X2 1200 0.8X1 + 0.5X2 60 0.3 X1 + 0.4 X2 50 0,X2 0 2000 APR. (21) Solve the following LPP using graphical method : Subject to : 6x + 3y 18 Maximize : Z = — 150X1 — 100X2 + 280000 Subject to : 20 5..X1. 60 70 5_ X2 140 120 + X2 140 and X1 0,X2 0 2001 APR. (22) Use graphic method to solve the following : Maximize : Z = 5X1 + 3X2 Subject to : 3X1 + 5X2 = 15 5X1 + 2X2 = 10
96
OPERATIONS RESEARCH X1 >_0, X2 >_0
2001 SEP.
(23) (a) Solve the LPP given below by graphical method and shade the region representing the feasible solution : Minimum value of Z = 2X1 —10 X2 X1 X2 0 X15X2 >_5
X1 >_0, X2 >_0 (b) Use graphical method to solve :
Maximize Subject to :
Z = 6X1 + 4X2 2X1 + 3X2 30 3X1 + 2X2 24 + X2 3 Xp X2 0 2002 APR. (24) Solve graphically the LPP : Maximize Z = 6X1 + 4X2 2X1 + 3X2 30 Subject to 3X1 + 2X2 24 X1+ X2 3 Xp X2 0. 2003 APR. Old hens can be bought at Rs. 2 each and young ones at Rs. 5 each. The old hens lay 3 eggs per week and the young ones lay 5 eggs per week, each egg being worth 30 paise. A hen costs Rs. 1 per week to feed. Mr Amit has only Rs. 80 to spend for hens. How many of each kind should Mr. Amit buy to give a profit of at least Rs. 6 per week, assuming that Mr. Amit cannot have more than 20 hens? Give the mathematical formulation of LPP only. (Marks 7) 2003 APR. Solve the LPP Graphically: Maximum value of Z = 80X1 + 120X2 X1 + X2 _< 9 Subject to 20X1 + 50X2 360 When X1 and X2 0 2, X2 > 3 2004 APR. A firm makes product A and B and has a total production of a capacity of 9 ton per day. A and B requiring the same production capacity. The firm has a permanent contract to supply al teast 2 tons of A and 3 tons of B per day to another Co. Each ton of A requires 20 machine hour production time and each ton of B requires 50 machine hour production time, the daily maximum possible of machine hours is 360. Profit per unit of product A is Rs. 80 and that of product B is Rs. 120. Formulate LPP.
Linear PrcgrarnrniF (Simplex Method)
LEARNING OBJECTIVES • • • • • •
Preparation of LPP before use of Simplex method Understanding of the need of use of Simplex method Steps involved in using a Simplex method Preparing a Simplex Table and understanding its various components Demonstrating the use of Simplex method for solving an LPP Solving LPP using maximization and minimizations problems
INTRODUCTION We have seen in the previous chapter that one can conveniently solve problems with two variables. If we have more than two variables, the solution becomes very cumbersome and complicated. Thus, there is a limitation of LPP in solving all types of complex real life problems, where the variables are always more than two. For such linear programming problems, solutions can be found with the help of simplex method. Simplex method is an algebraic procedure in which a series of repetitive operations are_ usedand we progressively approach the optimal solution. Thus, this procedure has a number of steps to find the solution to any problem, consisting of any number of variables and constraints, i however problems with more than 4 variables cannot be solved manually and require the use of computer for solving them. This method developed by the American mathematician G. B. Dantizg, can be used to solve any problem, which has a solution. The process of reaching the optimal solution through different stages is also called iterative, because the same computational steps are repeated a number of times before the optimum solution is reached. SOLVING OPERATIONS RESEARCH PROBLEM USING SIMPLEX METHOD Following are various steps for solving OR problem using simplex method.
98
OPERATIONS RESEARCH
Step I. Formulate the problem. The problem must be put in the form of a mathematical model. The standard form of the LP model has the following properties: (a) An objective function, which has to be maximized or minimized. (b) All the constraints can be put in the form of equations. (c) All the variables are nonnegative. Step II. Setup the initial simplex table with slack variable or surplus variables in the solution. A constraint of type or can be converted into an equation by adding a slack variable or subtracting a surplus variable on the left hand side of the constraint. For example, in the constraint XI + 3X2 15 we add a slack Si 0 to the left side to obtain an equation. Xi + 3X2 + S1 = 15, Si 0 Now, consider the constraint 2X1 + 3X2 — X3 4, since the left side is not smaller than the right side, we can subtract a surplus variable S2 > 0 from the left side to obtain the equation. 2X1 + 3X2 — X3 — S2 = 4 S2 > 0 The use of the slack variable or surplus variable will become clear in the actual example as we proceed. Step III. Determine the decision variables which are to be brought in the solution. Step IV. Determine which variables to replace. Step V. Calculate new row values for entering variables. Step VI. Revise remaining rows. Repeat step III to VI till an optimal solution is obtained. This procedure can best be explained with the help of a suitable example. Example 3.1. Solve the following linear programming problem by simplex method: Maximize
Z = 10X1 + 20X2
Subject to the following constraints: 3X1 + 2X2 — 0 Solution. Step I. Formulate the problem. Problem is already stated ir the mathematical model. Step II. Setup the initial simplex ,able with the slack variables in solution. By introducing the slack variables, the equations in step I, i.e., the mathematical model can be rewritten as follows: 3X1 + 2X2 + Si = 1200 2X1 + 6X2 + S2 = 1500 + S3 = 350
99
LINEAR PROGRAMMINGII X2 + 54 = 200
where SI, S2, S3 and S4 are the slack variables. Let us rewrite the above equation in a symmetrical manner so that all the four slacks S1, S2, S3 and S4 appear in all the equations. 3X1 + 2X2 +_151 + 0S2 + 0S3 + 0S4 = 1200 2X1 + 6X2 + OS1 + 1S2 + 0S3 + 0S4 = 1500 1X1 + OX2 + OSi + 052 +153 + 0S4 = 350 OXi + 1X2 + OSi + 0S2 + 053 + 1S4 = 200 Let us write the objective function Z by introducing the slacks in it. Z = 10X1 + 20X2 + OS1 + 0S2 + 0S3 + 0S4 The first simplex table can now be written as. TABLE 3.1 C. 0 0 0 0
Solution Mix S1
Rs. 10
Rs. 20
X1
X2
S3
3 2 1
2 6 0
S4
0
S2
4 (c,  4)
0 10
© Key element 0 20
0 S1 1 0 0
0
0
0
52
S3
S4
0 1 0
0 0 1
0 0 0
1200 1500 350
0
0
0
1
200
0 0
0 0
0 0
0 0
0
Contribution unit quantity
key TOW
key column The first simplex table is shown in Table 3.1. The table is explained as below: 1. Row 1 contains C1 or the contribution to total profit with the production of one unit of each product X1 and X2. This row gives the coefficients of the variables in the objective function which will remain the same. Under column 1(C1) is provided profit per unit of 4 variables S1, S2, S3, S4 which is zero. 2. All the variables S2, S3, S4 are listed under solution mix. Their profit is zero and written under column 1(C1) as explained above. 3. The constraint variables are written to the right of solution mix. These are X1, X2, Si, S2, S3 and S4. Under these are written coefficient of variables and under each are written the coefficients of particular variable as they appear in the constraint equations. For example, the coefficients X1, X2, Si, S2, S3 and S4 in first constraint equations are 3, 2, 1, 0, 0 and 0, respectively which are written under these variables in the first level. Similarly, the remaining 3 rows represent the coefficients of the variables as they appear in the other 3 constraint equations. The entries in the contribution unit quantity column represent the right hand side of each constraint equation. These values are 1200, 1500, 350 and 200 respectively, for the given problem. 4. The Z. values in the second row from the bottom refer to the amount of gross profit that is given up by the introducing one unit of that variable into the solution. The subscript j refers to the specific variable being considered. The Z1 value under the quantity column is the total profit
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100
for the solution. In the first column all the Z1 values will be zero because no real prOduct is being manufactured and hence there is no gross profit to be lost if they are replaced. 5. The bottom row of the table contains net profit per unit obtained by introducing one unit of a given variable into the solution. This row is designated as the Cj Z1 row. The procedure for calculating Zi and C1 Zi values is given below : Calculation of Z. C I X X2 C1 XX 1 Cj X S1 0x2=0 0x1=0 0X3=0
Similarly,
0x2=0
0x6=0
Ox 0 =0
0x1=0
Ox 0 =0
Ox 0 =0
Ox 0 =0
0x1=0
Ox 0 =0
Zx = 0
Zx2 = 0
Zs1 = 0
Zs2 , Zs3
and Zs4 , can be calculated as 0 each.
Calculation of C1 Cx2
1 = 10  0 = 10  Zx2 = 20  0 = 20
Csl  Zsl = 0  0 = 0 Cs2  Zs2 = 0  0 = 0 Cs,  Zs, = 0  0 = 0 Cs,  Zs, = 0  0 = 0 The total profit for this solution is Rs. zero. Step 3. Determine the variable to be brought into the solution. An improved solution is possible if there is a positive value in Ci  Z1 row. The variable with the largest positive value in the Ci= Zj row is subjected as the objective is to maximize the profit. The column associated with this variable is referred to as 'key column' and is designated by a small arrow beneath this column. In the given example, 20 is the largest possible value corresponding to X2 which is selected as the key column., Step 4. Determine which variable is to be replaced. To make this determination, divide each amount in the contribution quantity column by the amount in the comparable row of key column, X2 and choose the variable associated with smallest quotient as the one to be replaced. In the given example, these values are calculaitcl. as: for the S1 row, 1200/2 = 600 for the S2 row, 1500/6 = 250 for the S3 row, 350/0 = for the S4 row, 200/1 = 200
101
LINEAR PROGRAMMINGII
Since the smallest quotient is 200 corresponding to S4, S4 will be replaced, and its row is identified by the small arrow to the right of the table as shown. The quotient represents the maximum value of X which could be brought into the solution. Step 5. Calculate the new row values for entering the variable. The introduction of X2 into the solution requires that the entire S4 row be replaced. The values of X2, the replacing row, are obtained by dividing each value presently in the S4 row by the value in column X2 in the same row. This value is termed as the key or the pivotal element since it occurs at the intersection of key row and key column. X2 4 key column 2 6
key element
0 S4 0
0 0 0 1 200 key row
1
0
350
350.
0
1
200
0 0 20 4000 0
0
0
 20
key column The new variable entering the solution would be X1. It will replace the S2 row which can be shown as follows : For the Si row, 800/3 = 266.7 For the S2 row, 300/2 = 150 For the S3 row, 350/1 = 350 200/0 = 00 For the S4 row, As the quotient 150 corresponding to S2 row is the minimum, it will be replaced by X1 in the new solution. The corresponding elements of S2 row can be calculated as follows : 3 S4
Key element 0 0 1 0  6 300 key row
1 0 0 10 Key column New elements of S2 row to be replaced by X1 are: 2/2 = 1; 0/2 = 0; 0/2 = 0; 1/2 = 1/2; 0/2 = 0;  6/2 =  3; 300/2 = 150; The updated elements of Si and 53 rows can be calculated as follows : Elements of updated Si row Elements of updated S3 row Elements of updated X2 row 1  [1 x 1] = 0 3 [3 x1]= 0 0  [0 x 1] = 0 0  [1 x 0] = 0 0 [3x0]=0 1  [0 x 0] = 1 1  [3 x 0]= 1 0  [1 x 0] = 0 0[0x0]= 0 0  [1 x 1/2] =  1/2 0  [3 x 1/2] =  3/2 0  [0 x 1/2] = 0 0  [3 x 0]= 0 1  [1 x 0] = 1 0 [0x0]=0  2  [3 x  3] =  7 0  [1 x  3] = 3 1  [0 x 3]= 1 800  [3 x 150] = 350 350  [1 x 150] = 200 200  [0 x 150] = 200
103
LINEAR PROGRAMMINGII Revised simplex table can now be written as shown in Table 3.3 below : TABLE 3.3 Solution
Rs 10
Rs 20
0
0
0
0
C,
Mix
X1
X2
S1
S2
S3
S4
0 10
S1 X1
0
S3
0 1 0
0 0 0
1 0 0
 3/2 1/2 1/2
0 0 1
7 3 3
350 150 200
50 50 66.7
20
X2 Z,
0
1
0
0
0
1
200
200
10
20
0
5
0
 10
5500
(C1  Z1)
0
0
0
5
0
10
Contribution per unit Quantity
Minimum Ratio 4. key row
key column Now, the new entering variable will be S4 and it will replace S1 as shown below : 350/7 = 50 150/ 3 =  50 200/3 = 66.7 200/1 = 200 In these figures, 50 represent the minimum quotient which corresponds to row S1. The negative sign is not considered. The new elements of S1 row to be replaced by S4 can be calculated as follows : S4 Si 0 0 1 3/2 0 0 350 4—Key row 3
Key element
3 1 10 10 key ct olumn The new elements of S4 row would be 0/7 = 0; 0/7 = 0; 1/7 = 1/7; ( 3/2) x (1/7) =  3/14; 0/7 = 0; 7/7 = 1; 350/7 = 50
OPERATIONS RESEARCH
104
The updated elements of the other rows can be calculated as follows : Elements of updated X2 row Elements of updated X1 row Elements of updated S3 row 0  [1 x 0] = 0 1 [3 x 0]= 1 0 [3 x 0] =0 1  [1 x 0] = 1 0 [3 x 0]= 0  [3 x 0] =0  [1 x 1/7]  1/7 0 [3x1/7]=3/7 0  [3 x 1/7] =  3/7 1/2[3x3/14}= 1/7 0  [1 x  3/14] = 3/14 1/2  [ 3 x 3/14] =  1/7 0  [1 x 0] = 0 1  [3 x 0] = 1 0 [3 x 0]:= 1  [1 x 1] = 0 3 [3 x1]= 3  [3 x 1] = 0 200  [1 x 50] = 150 •150  [ 3 x 50] = 300 200  [3 x 50] = 50 The new simplex table can now be written as shown in Table 3.4. TABLE 3.4 C, 0 10 0 20
Solution Rs 10 Mix X, 0 S, X, 1 S3 0 X2 0 Z, 10 0 (C,  Zi)
Rs 20 X, 0 0 0 1 20 0
0 S, 1/7 3/7 3/7 1/7 10/7 10/7
0 S, 3/14 1/7 1/7 3/14 40/14 40/14
0 S, 0 0 1 0 0 0
0 S,i
Contribution Quantity
1 0 0 0 0 0
50 300 50 150 6000
As there is no positive value in CC  Z1 row it represents the optimal solution, which is given as : X1 = 300 units : X2 = 150 units And the maximum profit Z = Rs 6000 Minimization Problems
An identical procedure is followed for solving the minimization problems. Since the objective is to minimize rather than maximize, a negative (C1  Z1) value indicates potential improvement. Therefore, the variable associated with largest negative (CI  Z1) value would be brought into the solution first. Additional variables are brought into setup such problems. However, such problems involve greater than or equal to constraints, which need to be treated separately from less than or equal to constraints, which are typical of maximization problems. In order to convert such inequalities, the following procedure may be adopted: For example, if the constraint equation is represented as : 3X1 + 2X2 > 1200 To convert this into equality, it would be written as : 3X1 + 2X2  Si = 1200 where Si is a slack variable. However, this will create a difficulty in the simplex method because of the fact that the initial simplex solution starts with slack variables and a negative value (1S1) would be in the solution, a condition which is not permitted in linear programming. To overcome this problem, the simplex procedure requires that another variable known as artificial variable be
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105
added to each equation in which a slack variable is subtracted. An artificial variable may be thought of as representing a fictitious product having very high cost which though permitted in the initial solution to a simplex problem, would never appear in the final solution. Defining A as an artificial variable, the constraint equation can be written as : 3X1 + 2X2 —1S1 + 1A1 = 1200 Assuming the objective function is to minimize cost, it would be written as : 10X1 + 20X2 + OSi + MAi to be minimized. where M is assumed to be very large cost (say, 1 million). Also SI is added to the objective function even though it is negative in constraint equation. An artificial variable is also added to constraint equations with equality signs, e.g., if the constraint equation is 3X1 + 2X2 = 1200 then, in simplex it would change to 3X1 + 2X2 + 1A1 = 1200 to satisfy simplex requirement and would be reflected as MA in the objective function. Example 3.2. ABC company manufactures and sells two products P1 and P2. Each unit of P1 requires 2 hours of machining and 1 hour of skilled labour. Each unit of P2 requires 1 hour of machining and 2 hours of labour. The machine capacity is limited to 600 machine hours and skilled labour is limited to 650 man hours. Only 300 units of product P2 can be sold in the market. You are required to: (i) develop a suitable model to determine the optimal product mix; (ii) find out the optimal product mix and the maximum contribution. Unit contribution from product Pl is Rs. 8 and from product P2 is Rs. 12; (iii) determine the incremental contribution/unit of machinehours, per unit of labour and per unit of product P1. Solution. Step 1. Formulation of LP model. Let X1 and X2 be the number of units to be manufactured of the two products P1 and P2 respectively. We are required to find out the number of units of the two products to be manufactured to maximize contribution, i.e., profits when individual contribution of the two products are given. LP model can be formulated as follows : Maximize Z = 8X1 + 12X2 Subject to conditions/constraints 2X1 + X2 600 (Machine time constraint) X1 + 2X2 5.. 650 (Labourtime constraint) X2 S 300 (Marketing constraint of product P2) Step 2. Converting constraints into equations. LP problem has to be written in a standard form, for which the inequalities of the constraints have to be converted into equations. For this purpose, we add a slack variable to each constraint equation. Slack is the unused or spare capacity for the constraints to which it is added. In less than (5) type of constraint, the slack variable denoted by S, is added to convert inequalities into equations. S is always a nonnegative figure or 0. If S is negative, it may be seen that the capacity utilised will exceed the total capacity, which is absurd. The above inequalities of this problem can be rewritten by adding suitable slack variables and converted into equations as follows:
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OPERATIONS RESEARCH
2X1 + X2 + Si = 600 Xi+ 2X2 + S2 = 650 X2 + S3 = 300 Xi, X2, Si, S2, S3 > 0 Slack variables Si, S2 and S3 contribute zero to the objective function since they represent only unused resources. Let us include these slack variables in the objective function. Then maximize Z = 8X1 + 12X2 + 0S1 + 0S2 + 0S3 Step 3. Setup the initial solution. Let us recollect that the computational procedure in the simplex method is based on the following fundamental property: "The optimal solution to a Linear Programming problem always occurs at one of three corner points of the feasible solution space". It means that the corner points of the feasible solution region can provide the optimal solution. Let the search start with the origin which means nothing is produced at origin (0, 0) and the value of decision variable X1 and X2 is zero. In such a case, Si = 600, S2 = 650, S3 = 300 are the spare capacities as nothing (0) is being produced. In the solution at origin we have two variables X1 and X2 with zero value and three variables (S1, S2 and S3) with specific values of 600, 650 and 300. The variables with nonzero values, i.e., Si, S2 and S3 are called the basic variables whereas the other variables with zero values i.e. Xi, X2 and X3 are called nonbasic variables. It can be seen that the number of basic variables is the same as the number of constraints equations (three in the present problem). The solution with basic variables is called basic solution which can be further divided into Basic Feasible Solution and Basic Infeasible Solution. The first type of solutions are those which satisfy all the constraints. In Simplex Method, we search for basic feasible solution only. Step 4. Developing initial simplex table. The initial decision can be put in the form of a table which is called a Simplex Table or Simplex Matrix. The details of the matrix are as follows: 1. Row 1 contains Cj or the contribution to total profit with the production of one unit of each product Pi and P2. Under column 1 (C1) are listed the profit coefficients of the basic variables. In the present problem, the profit coefficients of Si, S2 and S3 are zero. 2. In the column labelled Solution Mix or Product Mix are listed the variables Si, S2 and S3, their profits are zero and written under column 1 (C1) as explained above. 3. In the column labelled 'contribution unit quantity' are listed the values of basic variables included in the solution. We have seen in the initial solution Si = 600, S2 = 650 and S3 = 300. These values are listed under this column on the right side as shown in Table 3.5. Any variables not listed under the solutionmix column are the nonbasic variables and their values are zero. 4. The total profit contribution can be calculated by multiplying the entries in column Ci and column 'contribution per unit quantity' and adding them up. The total profit contribution in the present case is 600 x 0 + 650 x 0 + 300 x 0 = 0. 5. Numbers under X1 and X2 are the physical ratio of substitution. For example, number 2 under X1, gives the ratio of substitution between X1 and Si. In simple words, if we wish to produce 2
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107
units of product P1, i.e., X1, 2 units of Si must be sacrificed. Other numbers have similar interpretation. Similarly, the number in the 'identity matrix' columns S1, S2 and S3 can be interpreted as ratios of exchange. Hence the numbers under the column Sl, represents the ratio of exchange between S1 and the basic variables S1, S2 and S3. 6. Z1 and C1— Z1 are the two final rows. These two rows provide us the total profit and help us in finding out whether the solution is optimal or not Z1 and C1 — Z1 can be found out in the following manner: (a) Z1 = Cj of S1 (0) x coefficients of X1 in S1 row (2) + C1 of S2 (0) x coefficients of X1 in S2 row (1) + Cj of S3 (0) x coefficient Xi in S3 row (1) = 0 x 2 + 0 x 1 + 0 x 1 = 0 Using the same procedure Z1 for all the other variable columns can be worked out as shown in the completed first Simplex table given in Table 3.5. (b) The number in the (C1— Z1) row represent the net profit that will result from introducing 1 unit of each product or variable into the solution. This can be worked out by subtracting, Z1 total for each column from the Cl values at the top of that variable's column. For example, C1 — Z1 number in the X1 column will 8 — 0 = 8, in the X2 column it will be 12 — 0 = 12, etc. 7. The value of the objective function can be obtained by multiplying the elements in Cj column with the corresponding elements in the C1 rows, i.e., in the present case Z = 8 x 0 + 12 x 0 = 0 TABLE 3.5 12 X2 1 2
0 S1 1 0
0
0
S2
S3
0 1
0 0
0
Solution mix S1
0
S2
8 X1 2 1
S3
0
1
0
0
1
0 8
0 12
0 0
0 0
0 0
C. I
0
4 (C1— 4 )
Contribution unit quantity (Solution values) 600 650 300
8. By examining the number in the (C1 — Z1) row, we can see that total profit can be increased by Rs 8 for each unit of product X1 added to the product mix or by Rs 12 for each unit of product X2 added to the product mix. A positive (C1 — Z1) indicates that profits can still be improved. A negative number of (C1— Z1) would indicate the amount by which the profits would decrease, if one unit of the variable was added to the solution. Hence, optimal solution is reached only when there are no positive numbers in (C1— Z1) row. Step 5. Test for optimality. Now, we must test whether the results obtained are optimal or it is possible to carryout any improvements. It can be done in the following manner: 1. Selecting the entering variable. We have to select which of the variables, out of the two nonbasic variables X1 and X2, will enter the solution. We select the one with maximum value of C1 — Z1. Variable X1 has a (C1 — Z1) value of 8 and X2 has a (C1 — Z1) value of 12. Hence, we select variable X2 as the variable to enter the solution mix and identify the column in which it occurs as the key column with the help of a small arrow.
OPERATIONS RESEARCH
108
2. Selecting the variable that leaves the solution. As a variable is entering the solution, we have to select a variable which will leave the solution. This can be done as follows : (a) Divide each number in the solution value or contribution unit quantity column by corresponding number in the key column, i.e., divide 600, 650 and 300 by 1, 2, 0. TABLE 3.6 C•
Solution mix
8 xi
12 x2
0 si
0 s2
0 s3
Solution values
Minimum ratio
0
S1
2
1
1
0
0
600
600
0
S2 S3
1
0
0
1
0
650
325
0
1
0
0
1
300
300
I
0
Zi 0 (C1  Zi )
8
key row
0 0 0 12 0
0
0
key column (b) Select the row with smallest nonnegative ratio as the row to be replaced, in present example the ratios are: For Si row, 600/1 = 600 unit of X2 For S2 row, 650/2 = 325 units of X2 For S3 row, 300/1 = 300 units of X2 It is clear that S2 (with minimum ratio) is the departing variable. This row is called the key row. (c) The number at the intersection of key row and key column is called the key number which is 2 in the present case and is circled in the table. Step 6. Developing second simplex table. Now, we can develop the second simplex table. TABLE 3.7 Second Simplex Table Solution mix
8
12
0
0
0
X,
X,
S,
S,
0
S,
2
0
1
0
S2
1
0
12
X,
0
Z, (CI  Zi )
c
'
S,
Solution value
Minimum ratio
0
0
300
150
0
1
0
50
1
0
0
1
300
50 co
0
12
0
0
0
3600
8
0
0
0
0
key column Now, X1 will be replaced.
—110. key row
109
LINEAR PROGRAMMINGII TABLE 3.8 Third Simplex Table c
C1
Solution mix
8 xi
12 X2
s,
2 1 0 8 0
0 0 1 12 0
0 8 12
X, X2 Z. CI Z1
0 Si 1 0 0 0 0
0
0
S2
S3
0 1 0 0 0
0 0 1 0 0
Solution Value 200 50 300 4000
As R2 will remain same because pivot element is already unity. Maximum Profit = 4000 We find that the value of objective function has been improved from 0 to oo. But the correct solution is not optimal as there are positive values (12) and (8) in the (C1  ZI) row. Also, since minimum ratio is 325, we select X2 row to leave the solution as X2 (key column) will enter the solution. The new X2 (key) row will remain same as its elements 1/2, 1, 0, 1/2, 0 and 325 have to be divided by key element, i.e., (shown circled in the above Table). However, row S1 and S3 elements will undergo change. Row S1 = old row number  [corresponding number in key row] x [corresponding fixed ratio] Fixed ratio = old row number in key column/key number = 0 It can be concluded that this problem does not have a optimal solution as X2 row is to be replaced by X2 row. Example 3.3. Solve the following problem using Simplex method 2x2 3x3
Z max = 15X1 Subject to
2x1 + 2x2  4x3 5 18 8x1 + 2x, + 2x3 _s36 xl, x2, x3 > 0. Solution. Converting inequalities into equalities condition are 2x1 + 2x2  4x3 + OS1 + 0S2 = 18 8x1 + 2x2 + 2x3 + OSi + 0S2 = 36 x1, x2, x3, Si, S2 = 0 Constructing the first simplex table TABLE 3.9 Solution mix
5 X1
2
3
0
0
x2
x3
S1
0
Si
2
2
4
0
S2
8
2
4 q  z,
0  15
Ci
S2
Solution Value
Minimum Ratio
1
0
18
9
2
0
1
36
4.5
0
0
0
0
2
3
0
0
OPERATIONS RESEARCH
110 Since — 15 is least in Ci — Zi, S2 will be replaced by X1. TABLE 3.10 Solution mix
15
2
3
0
0
xi
x2
x3
S1
Si
Solution Value
0
Si
0
23
—9 2
1
—1 —T i
9
15
X1
1
71
74
0
1 8
4.5
4
15
15 4
15 4
0
15 8
67.5
C — Zi
0
7 i
3 71
0
5 2
Ci
1
1
Elements of second row (x1) Divide every element of second row by 8 of previous table Elements of second row = Elements of first row by previous table — Elements of second row in previous table x conversion factor 2 18 — 36 x — =9 8 2 2 — 8 x — =0 8 2 _3 2—2x— 8 — 2 2 9 —4—2x— =—— 2 8 2 1 — 0 x — =1 8 2 0 — 1 x — = —1 4 8 In Table 3.10 all elements in Z1 — Ci row are positive. So, optimum solution has been achieved. Zmax = 67.5 where x1 = 4.5, x2 = 0 Example 3.4. Following data is available for a manufacturing coinpany engaged in production of three items X, Y and Z. Product X Y Z Company's capacity
Time required in hours Machining Finishing 12 3 6 8 8 6 3000 1500
Total contribution (Rs) 1000 800 400
LINEAR PROGRAMMINGII
111
You are required to present the above data in the form of LPP to maximize the profit from the production and solve the problem using simplex method. Solution. Mathematical formulation of the problem is Maximize Z = 1000 Xi + 800 X2 + 400 X3 Subject to constraints 12X1 + 6X2 + 8X3 3000 (Machining capacity constraint) 3X1 + 8X2 + 6X3 1500 (Finishing capacity constraint) Xi, X2, X3 > 0 where Xi, X2 and X3 are the number of products X, Y and Z to be manufactured respectively. Let us now introduce the nonnegative stocks Si, S2 as variables. Then, maximize Z = 12X1 + 6X2 + 8X3 + Si = 3000 3X1 + 8X2 + 6X3 + S2 = 1500 Xi > 0,. X2 > 0, X3 > 0, Si > 0, S2 > 0 Initial basic solution as discussed earlier is X1 =X2 =X3 =0andS1 =3000, S2 = 1500 Now let us construct the first Simplex table as below : TABLE 3.11 Solution mix
1000
800
400
0
0
X1
X2
X3
S1
Si S2
© 3 0 1000
6 8 0 800
8 6
1 0 0
4
(c,  4 )
0 400
S2
Solution values
Minimum ratio
0 1 0
3000. 1500 0
250 500
key row
key column X1 is the key column, Si is the key row and 12 is the key element (circled in the table). Also, it can be seen X1 is the entering variable and Si is the variable leaving the solution. Row (1) new = 1/12 Row (1) old, i.e., 1, 1/2, 2/3, 1/12, 0, 250 Row (2) new = Row (2) old — 3 row (1) new First element =33x1=0 Second element = 8 — 3 x 1/2 = 13/2 Third element = 6 — 3 x 2/3 = 4 Fourth element = 0 — 3 x 1/12 =— 1/4 Fifth element = 13x0=1 Sixth element = 1500 — 3 x 250 = 750 Hence, row 1 (S1) will be replaced by 1, 1/2, 2/3, 1/12, 0, 250 and row 2 (S2) will be replaced by 0, 13/2, 4, 1/4, 1, 750
OPERATIONS RESEARCH
112
Second simplex table can now be constructed. TABLE 3.12 Second Simplex Table C,
Solution mix
1000
800
400
0
0
X,
X,
X,
S,
S,
1000
xi
1
1 2
2 3
1 12
0
0
S2
0
0
4
1 71
Z.
1000
500
667
(C,  Z,)
0
300
 267
Solution values
Minimum ratio
250
500
1
750
115.4
83.3
0
250000
 83.3
0
key row
key column It may be seen that the value of objective function has increased from 0 to 250000. But there is a positive entry (300) in the (CI  Z1) hence this solution is not optimal. It can also be seen that entering variable is X2 and departing variable is S2.
Row  2 (new)
2 
13
Row1(new)
1 › row 1 (old)   row 2 (new) 2
Row  2 (new)
2 8 1 2 1 2 13 2 4 — x0=0, x =1,4x =, x =, 13 2 13 13 13 4 13 26 1x
i.e., Row1(new) or
row 2 (old)
2 2 1500 = — 750 x = 13 13 13 13 2
8 1 2 1500 0, 1„ „ 13 26 13 13 1 1 1 2 1 8 52  24 4 1  x 0 = 1,  x 1 = 0, x = 2 2 2 3 2 13 78 14 1 39 12 250
1 1 13  3 10 5 x = = 2 26 156 156 78
1 1500 2500 x = 2 13 13
14 5 1 2500 1, 0„ , 39 8 13 13
28 78
1 2 1 x =2 13 13
113
LINEAR PROGRAMMINGII
Third simplex table can now be constructed as follows : TABLE 3.13 Third Simplex Table (optimal solution) Ci
Solution mix
1000
800
400
0
0
X,
X,
X,
S,
1000
x,
1
0
14 39
S, 5 78
Solution value
1  13
2500 (192) 13
800
x,
0
1
8 13
1 26
2 13
1500(115) 13
Z.
284615
1000
800
33200 39
7400 78
600 13
0
0
(C)  Z) ,
17600 39
7400 78
 600 13
Zj (Yi) = 1000 x 1 + 800 x 0 = 1000 Zi (x2) = 0 x 1000 + 1 x 800 = 800 8 14 33 200 Zi (x3) = — X1000 + — x 800 _ 39 13 13 Zj (Si) = 1000 x 78 +800 x
1 6
=
7400 78
600 1 2 13 x 800 = Zi (S2) = 31x1000 + — 13 2500 1500 3700000 x1000 + 13 x800 .=  284615 13 13 Since all the entries in (C1  Z1) row are either zero or negative, this is the optimal solution. z.
Maximum value of Z, 284615 occurs at x1 = 192 x2 = 115 and x3 = 0. Example 3.5. ABC Ltd produces four products P1, P2, P3 and P4. Each one of these products has to be processed on three machines X, Y, Z. The capacity of the machines and the time required to manufacture one of each type of products are shown in the table below : Product P1 P2 P3 P4
Capacity (hours)
Processing time for production Machine X Machine Y Machine Z 2 4 3 3 2 2 4 1 2 3 1 1 800 600 420
The profit contribution/unit of products P1, P2, P3 and P4 are Rs. 8, 6, 4 and 2 respectively. You are required to formulate the above as a LPP and determine the optimal product mix by using simplex method.
OPERATIONS RESEARCH
114
Solution. Let x1, x2, x3 and x4 be the number of units of product 131, P2, P3 and P4 respectively. The mathematical model is as follows : Maximize Z = 8x1 + 6x2 + 4x3 + 2x4 Subject to the following constraints : 2x1 + 3x2 + 4x3 + 3x4 5_ 800 (capacity of machine X) 4x1 + 2x2 + 1x3 + 2x4 5_ 600 (capacity of machine Y) 3x1 + x2 + 2x3 + x4 5 420 (capacity of machine Z) x1, x2, x3, x4 0 After introducing slack variables Si, S2 and S3 the problem can be rewritten as Maximize Z = 8x1 + 6x2 + 4x3 + 2x4 + OS1 + 0S2 + 0S3 Subject to the constraints 2x1 + 3x2 + 4x3 + 3x4 + Si = 800 4x1 2X2 + X3 I 2X4 ± S2 = 600 3x1 + x2 + 2x3 + X4 + S3 = 420 xl, x2, x3, x4, S1, S2, S3 0 Initial feasible solution can be obtained by putting = x2 = x3 = x4 = 0 so that S1 = 800, S2 = 600, S3 = 420 and Z = 0. Now, first simplex table can be constructed. TABLE 3.14 CI 0 0 0
8 x, 2 4 0 0 8
Solution mix
S1 S2 S3 Z, (C,  Z, )
6 x2 3 2 1 0 6
4 x3 4 1 2 0 4
2 x4 3 2 1 0 2
0
0
Si
Si
1 0 0 0 0
0 1 0 0 0
0 S, 0 0 1 0 0
Solution Value
Minimum ratio
800 600 420
400 150 140
► key TOW
key column x1 is the key column. S3 is the key row. and 3 is the key number (circled in the table) Also xi is the entering variable and S3 is the outgoing variable. We use the following row operations to get second simplex table by entering x1 into the solution and removing S3 variable. 1 3 (old) 3R R1 (new) = R1 (old)  3R3 (new) R2 (new) = R2 (old)  2R3 (new) R3 (new) =
115
LINEAR PROGRAMMINGII TABLE 3.15 Second Simplex Table 8
6
4
2
x,
x2
x,
x4
1
3
2
2
1
3 0,
3 2 i
3 1 3
1120
8
8 3
16
(c,  z, )
0
10 —

1 0
Solution Mix SI
0
S,
2
8
lc 1
Z,
c,
T 3
0 S, 1
0 S, 0
0
1
0 S3 1 2 
0
0
1 3
8 3
o
0
8

0
0
3
3
Solution Value
Minimum ratio
380
126.7
320
240
140
46.7
key column 1 1 1 1 2 1 R3 (new) =  x 3 =1, x1=,  x 2=  , , 0, 0, 1,140 3 3 33 33 3
1 2 1 1 i.e., 1, , — ,0,0,,140 33 3
3
RI (new)=23x1=1,33x 1  = 2,4 3 x 3 =2,33x 1 =2 3 3 1 13x0= 1,03 x0=0,03 x  =1,8003x140=380
i,e.,  1,2, 2,2,1,0, 1,380. R2(new)=42x1=2,22x 1 4
2
1
1 4
1 2 02x0= 0,12x 0=1,02x  =' 6002x140=320 3 3 4 3 3 3 Calculation of Z •
i.e., 2,
,320 3 Z1 (xi)=1x0+2x0+1x8=8 1 8 4 Z1 (x2)= 2 x +  x 0 + x 8 =3 3 3 16 Zi (x3) = 3 Zi (x4) = 8Z =380x 0+320 x0+140 x8=1120 Z (Si) = 0
Z1 (S2) = 0 Zi (S3) =
key row
OPERATIONS PR_ESEARCH
116
It can be seen that Z has improved from 0 to 1120 but since there is still a positive value in (CJ  Zj) it is not optimal solution. It is now clear that x2 is the entering variable and x1 the departing variable. Now, the third simplex table is to be constructed. We now use the following row operations to get a new solution by entering variable x2 and • removing x1 variable. R1 (new) = R1 (old) 2 R2 (new) = R2 (old)   RI (new) 3 2 R3 (new) = R3 (old)   RI (new) 3 R1 (new) =  1, 2, 2, 2, 1, 0,  1, 380 1 2 2 = 5 2 2   x 2 = 0,   5 x R2 (new) = 2 x  1 = 8 4 3 3 2_2 4 2 2 x1= ,u i x 0=0  i x 2 = 0,03 3 3 2 1 2  x  1 = 1, 320   x 380 = 200 3 3 3 3 8 5  2 200 i.e., , 0, , 0, , 01., 3 3 3 3 2 2 x  1= 53 1 2 x 2 1 2 2 = 1 '3 3 x 2 = 3 '3 3 3 2 1. _ x 2 = 2 1,0 3x1= 2,0 x0=0 3 3 i 3 340 1 2   i x  1 = 1, 140 2 x 380 3 3 3 2  340 5 2 1 , 1„0, 1, 3 3 3' ' 3 TABLE 3.16 Third Simplex Table R3 (new)
i.e.,
C, "
Solution mix
8
6
4
2
0
0
0
Solution
Xi
x2
x3
X4
S1
S2
S3
mix
0
S1
1
2
2
1
S2
0
1 2 y
0
0
2 5 5
0
1
6
X2
5 3
1
'i
0
1
380 200 3  3340
ZJ
10
6
(Cj  Z) )
2
0
8
2
1 32 8
0 1
2
6
4
0
6
8
4
0
6
LINEAR PROGRAMMINGII
117
The student should further attempt this problem to get the optimal solution. The present solution is not the optimal solution as positive values exist in Ci Z1
MINIMIZATION PROBLEMS (ALL CONSTRAINTS OF THE TYPE BIG 'M' METHOD We have till now seen in this chapter, the type of problems where profit had to be maximized and the constraints were of the type However, there could be problems where the objective function has to be minimized (like the availability of funds, raw material or the costs of operations have to be minimized) and the constraints involved may be of the type ?_ or = . In such cases, the simplex method is somewhat different and is discussed in the form of following steps: Step
1. Formulation of mathematical model.
Minimize
Z=
C„ x„
+ C2 X2 + C3 X3 +
 Subject to the constraints an
al „ x„
+ X2 + an x3 +
a2 „ X3 b2
a21 x1 + a22 x2 + a23 X3 + .
.
•
a„,, xi + a,,,2 x2 + a,,,3 x3 + where
a„, „ x„
xi., x2, x3
X„ 0
S,,, etc., to convert the inequalities into
Now, we subtract the surplus variables Si, S2, equations.
i.e.,
Minimize
Z = C1 xi + C2 X2 + C3 X3 +
+ C„
X„ + OSi + 0S2 +
+ OS„
Subject to the constraints an x1 + a12 X2 + a13 X3 +
„ x„ — =
a21 x1 + a22 x2 + a23 x3 +
a2 „ X3 — S2 = b2 •
a„,, xi + a„,2 X2 + a„, X3 + 3 where
•
+ an, n xn — S
xi
0 (i =1, 2,
Si
0 (j = 1, 2,
•
= b,,,
n) m)
As in the maximization problem, initial basic solution is obtained by putting x1 = x2 = =
x„ = 0 — S1= b1 or S1= — b1
So,
— S2 = b2 or S2 = — b2
—
= b„, or 5,„ — b,„
It may be seen that Si, S2
5„, being negative violate the nonnegativity constraint and hence
is not feasible. Hence, in the system of constraints we introduce
m new variables Ai, A2
known as artificial variable. By introducing these variables the equations are: an x1 + a12 x2 + a13 x3 +
+ai„x„—Si+A1= b1
+ a22 x2 + a23 x3 +
+ a2 „ x„ — S2 + A2 = b2
a21 •
•
a„,, x1 + a„,2 x2 + a„,3 X3 +
+
n xn — Sm + = b
A,,,
118
OPERATIONS RESEARCH
where
x1• > 0 (i =1, 2, 3, n) m) Si 0 (j = 1, 2, 3, A.> 0 (j = 1, 2, 3, m) As we have introduced artificial variables A1, A2 A,,, this has to be taken oirit, of the solution. For this purpose, we introduce a very large value (M) assigned to each of artificial variable and zero to each of the surplus variables as the coefficient values in the objective function. The problem now becomes Minimize Z = Cl xl + C2 x2 + C3 x3 + + C„ x„ + OSi + 0S2 + + OS. + MAI + MA2 + MAin Subject to the constraints xl + a12 X2 + al3 X3 4n x„Si +A1 =b1 a2,,x3 S2 + A2 = b2 a21 ± a22 + a23 x3 ± • • • am,x1 + an,2 x2 + a,„3 x3 + am „ xnStn + btn Step 2. Setting up of initial simplex table Here, we allot 0 values to variables x1 = x2 = x3 = = x„ = 0 so that Al = b1, A2 = b2 A,„ = b,„. TABLE 3.17 Initial Simplex Table
CBI C132
Solution mix Al A2
C1 Solution values b1 b2
CB„
A,,,
b,,,
CB
Z1 (C1  Z1)
Ci C2 C3 xi. x2 X3
Cn OOM
M
Xn Si Si S,„ Al A2
Am
au aiz an a22
al n 1 0 0 1 0 a2„ 0 1 0 0 1
0 0
a., a„,2
a„,„ 0 0 1 0
0
1
0 0 C1 C2
0 C„
0 0 0 0 0 00 0 M M
Minimum ratio
0 M
Step 3. Test for optimality Calculate the elements of (Cj  Z1) row. (a) If all (Cj  Z1) 0 then the basic feasible solution is optimal. (b) If any one (C1  Z1) 0 then pick up the largest negative number in this row. This is the key column and determines the variable entering the solution. Now, the second simplex table can be constructed. Step 4. Test for feasibility Determine the key row and key number (element) in the same manner as is done in the maximization problem.
X.,11sTkil. PROGRAMMING11 •
119
Example 3.6. A special diet for a patient in the hospital must have at least 8000 units of vitamins, 100 units of minerals and 2800 units of calories. Two types of foods X and Y are available in the market at the cost of Rs. 8 and Rs. 6 respectively. One unit of X contains 400 units of vitamins, 2 units of minerals and 80 units of calories. One unit of food B contains 200 units of vitamins, 4 units of minerals and 80 units of calories. What combination of foods X and Y be used so that the minimum requirement of vitamins, minerals and calories is maintained and the cost incurred by the hospital is minimized ? Use simplex method. Solution. Mathematical model of the problem is as follows : Z = 8x1 + 6k2 Minimize Subject to the constraints 400x1 + 200x2 8000 (Constraint of minimum vitamins) 2x1 + 4x2 100 (Constraint of minimum minerals) 80x1 + 80x2 2800 (Constraint of minimum calories) x1, x2 0 (Nonnegativity constraint) where x1 and x2 are the number of units of food X and food Y. Now, the constraint inequalities can be converted into equations. Here, we take an initial solution with very high cost, as opposed to the maximization problem where we had started with an initial solution with no profit. We subtract surplus variables S1, S2 and S3. 400x1 + 200X2 — S1 = 8000 2x1 + 4x2 — S2 = 100 80x1 + 80X2 — S3 = 2800 The surplus variables Si, S2 and S3 introduced in these equations represent the extra unit of vitamins, minerals and calories over 8000 units, 100 units and 2800 units in the least cost combination. Let xi, x2 be zero in the initial solution. S1 = — 8000 Hence S2 = —100 S3 = — 2800 This is not feasible as SI, S2 and S3 0 and cannot be negative. We have to see that Si, S2 and S3 do not appear (as they are — ye) in the initial solution. So, if xi, x2 and Si, S2, 53 are all zero, new foods which can substitute food X and Y must be introduced. A1, A2 and A3 are the artificial variables to be introduced. Let the artificial variables (foods) be of are very large price, M per unit 400x1 + 200x2 — Sl + A1 = 8000 2x1 + 4X2 — S2 + A2 = 100 80x1 + 80x2 — S3 + A3 = 2800 and Z objective function Z = 8x1 + 6x2 + OSi + 0S2 + 0S3 + MAi + MA2 + MA3 Minimize wherex1, x2, Si, S2, S3, A1, A2, A3 0 Now, it is possible to setup initial solution by putting x1 = x2 = Si = S2 = S3 = 0 in such a manner that Al = 8000, A2 = 100 and A3 = 2800.
OPERATIONS RESEARCH
120 TABLE 3.18 First Simplex Table
CB
Ci B b (= XB) Solution mix Solution values variables
8
6
0 0 0 M M M
Xi
x2
Sl
S2
S3
Al
A2
A3
Minimum ratio
200
1
0
0
1
0
0
20
0 1 0 0 4 80 0 0 1 0 284M M M M M 6M M M 0 284M
1 0 M
0 1 M
50 35
0
0
M
Al
8000
400
M
A2
M
A3
100 2800
2 80 482M 8482M
Zi (Ci  )
► key row
key column x1 is the key column entering the solution, A is the departing row and 400 (circled) in the table is the key number (element). Now, apply the row operations. (i) R 1 (new) >
R 1 (old) 400 (ii) R  2 (new) R  2 (old)  2R 1 (new) (iii) R  3 (new) R  3 (old)  80 R 1 (new) TABLE 3. 19 Second Simplex Table C1 8 6
CB
Solution Solution mix values variable b (= XB) (=B)
x1
8
xi
20
1
M
A2
60
0
M
A3
1200
0
Zi
8
(C  j)
0
x2
0
0 0 M M M
Si
S2
S3
0
1 400 1 0 — 200 1 40 5 4 +43 4+41 M M/200 243 441 M M/200 1
2
A2
A3
Minimum ratio
0
0
0
40
1
0
1
0
20
0
1
0
1
30
M
M
M
M
M
M
0
0
key column Value of Z calculated as follows: Zi (x1)= 8 x 1 +Mx 0= 8 Z1 (x2)= 2 x 8+3 xM+40M=4 +43M
Al
key TOW
LINEAR PROGRAMMINGII
121
s 1 m 1 m  4+ 41M Zi (Si) = 400 1 x + 200 — + 5 — 200 Zj (S2) =  M Z1 (S3) =  M Zi (A2) = M; Zi (A3) = M It is clear from the above table, that x2 enters the solution and A2 departs, using the following row operations: We introduce x2 and remove (i) R 2 (new)
A2.
1
 R 2 (old) 3
1 (ii) R 1 (new) R 1 (old)   R  2 (new) 2 (iii) R  3 (new) > R  3 (old)  40 R  2 (new) 1 1 1 R  2 (new) = 20, 0,1, 0, , 0. 600 3„ 3 1 1 n R  1 (new) = 10,1, 0, 300 6 2 40 R  3 (new) = 400, 0, 0, 15 , 1. Now, the third simplex table can be drawn. TABLE 3.20 Third Simplex Table
C.
6
C B
Solution Mix variables (= B)
8
xi
10
1
0
6
X2
20
0
1
M
A,
400
0
0
Z
8
6
(C,  Z1 )
0
0
Solution values
0
0
M M
0
Mini
x,
S
x2
S2
S,
A,
A
A,
2
b (= x6)
mum ratio
300
1 6
0


0
60
1 600
1 3
0


0
60 30

1
2 5 1+8M 60
t 3
1


1
2+4M 3
M


M
18M 60
2  4M 3
M


0
› key row
key column has to depart. This procedure can be adopted It can be seen S2 has to be introduced and for further improving the solution by constructing fourth simplex table and so on. A3
OPERATIONS RESEARCH
122 MINIMIZING CASE — CONSTRAINTS OF MIXED TYPE AND
We have seen the examples earlier where the constraints were either type or type. Both there are problems where the constraint equation could contain both types of constraints. This type of problem is illustrated with the help of an example. Example 3.7. A metal alloy used in manufacture of rifles uses two ingredients A and B. A total of 120 units of alloy is used for production. Not more than 60 units of A can be used and at least 40 units of ingredient B must be used in the alloy. Ingredient A costs Rs. 4 per unit and ingredient B costs Rs. 6 per unit. The company manufacturing rifles is keen to minimize its costs. Determine how much of A and B should be used. Solution. Mathematical formulation of the problem is Minimize cost Z = 4x1 + 6x2 Subject to constraints (Total units of alloy) x1 + x2 = 120 (Ingredient A constraint) x1 5. 60 (Ingredient B constraint) x2 40 (Nonnegativity constraint) xi, x2 0 where xi and x2 number of units of ingredient A and B respectively. Let xi and x2 = 0 and let us introduce an artificial variable which represents a new ingredient with very high cost M. xi + x2 + Ai= 120 Also xi + Si= 60 Third constraint x2  S2 + A2= 40 Now, the standard form of the problem is Minimize Z = 4x1 + 6x2 + MA1 + 0S1 + 0S2 + MA2 Subject to the constraints + x2 + = 120 xi + Si = 60 X2  S2 + A2 = 40 Xi, X2, Si, S2, Ap A2 Initial basic solution is obtained by putting xi = x2 = 0 and S1 = S2 = 0 so that Al = 100, S1 = 60, A2 = 40. TABLE 3.21 First Simplex Table Ci CB Solution mix Solution values M 120 Al 60 0 S1 M
A2
40 Z) (C,  Z,)
4 x1 1 1
6 x2 1 0
M 0 Al S1 1 0 0 1
0
0,
0
M 2M M 4M 6  2M 0 key column
0
M Minimum ratio
S2
A2
0 0
0 0
120 
0
1
0
40
0 0
M M M 0
key row
123
LINEAR PROGRAMMINGII
6  2M is the largest negative number hence, x2 will enter the solution and since 40 is the minimum ratio A2 will depart. R  3 (New) R  3 (old) as key element is 1. R  1 (New) > R  1 (old)  R  3 (New) TABLE 3.22 Second Simplex Table 4
6
M
0
0
M
x1
x2
Al
SI
S,
A2
1 ® 0 M 4M
0 0 1 6 0
1 0 0 M 0
0 1 0 0 0
1 0 1 M6 M+6
C, CB
Solution mix
M 0 6
A, SI x2
Solution values 80 60 40
Z, (C,  Z, )
Minimum ratio
80 60
E key row
key column R  1 (new) = 1  0 = 1 ; 1  1 = 0, 1  0 = 1, 0  0 = 0, 0  ( 1) = 1 0, 1,1, 0, 1, 100  40 = 60 i.e., x1 will be introduced and S1 will depart. Use the following row operations : (i) R  2 (new) > R2 (old) (ii) R 1 (new) R1 (old)  R2 (new) R  2 (new) = 1, 0, 0, 1, 0 R  1 (new) = 1  1 = 0, 0  0 = 0, 1  0 = 1, 0  1 =  1, 1  0 = 1 i.e., 0, 0, 1,  1, 1 TABLE 3.23 Third Simplex Table C, CB
Solution Solution values mix
4
6
M
0
0
M
x,
x2
A,
S,
S2
A2
40
0
0
1
1
T
4
A x,
60
1
0
0
1
0
6
x2
40
0
1
0
0
1
Z.
4
6
M
M+4
M6
 )
0 0 0 M  4 M+6
M
Minimum ratio
40  40
key column We now introduce S2 and take out Al using following row operations: R 1 (new) › R 1 (old) R  3 (new) > R  3 (old) + R 1 (new)
‹ key row
OPERATIONS RESEARCH
124
TABLE 3.24 Fourth Simplex Table C,
4
6
M Al
0
0
M
S1 S2 —1 1 1 0 —1 0
A2
CB
Solution mix
Solution values
x,
0 4 6
S2 x1
40 60 80
0 1 0
x2 0 0 1
Z,
4
6
—
—2
0
(C, — Z1 )
0
0
—
2
0
x2
Since all the numbers in (C1 — Z1) are either zero or positive, this is the optimal solution. Xi = 60, x2 = 80 and Z = 40 x 60 + 6 x 80 = Rs. 720 Maximization Caseconstraints of Mixed Type
A problem involving mixed type of constraints in which =, and are involved and the objective function is to be maximized. Example 3.8. Maximize Z = 2x1 + 4x2 — 3x3 Subject to the constraints x1 + x2 + x3 8 — x2 1 3x1 + 4x2 + x3 40 Solution. The problem can be formulated in the standard form. Maximize Z = 2x1 + 4x2 — 3x3 + 0S1 + 0S2 — MA1 — MA2 Subject to constraints + x2 + x3 + = 8 — x2 — Si + A2 = 1 3x1 + 4X2 4 X3 + S2 = 40 0, X2 0, Si 0, S2 0, Ai 0, A2 0 TABLE 3.25 First Simplex Table 2
4
—3
0 0 —M —M
CB
Solution mix variables(B)
Solution values b (= xB )
x1
x2
x3
S1
S2
A,
A,
Minimum ratio
—M —M
A, A2
8 1
1 ED
1 —1
1 0
0 —1
0 0
1 0
0 1
8 1
0
C2
40
3
4
1
0
1
0
0
40 3
Z,
—2M
0
—M
M
0
(C, — Z, )
2 + 2M
4
key column
—3 +M —M
0
—M —M 0
0
key row
LINEAR PROGRAMMINGII
125
where Al and A2 are the artificial constraints, S1 is the surplus variable, S2 is the slack variable and M is a very large quantity. For initial basic solution A1 = 8, A2 = 1, S2 = 40 This is a problem of maximization, hence we select 2 + 2M, the largest positive number in (C1  Z1) x1 will enter and A2 will depart. Use the following row operations: R  2 (New) R  2 (old) R  1 (New) R 1 (old)  R2 (new) R  3 (New) R  3 (old)  3 R2 (new) TABLE 3.26 Second Simplex Table
C1
2
4
3
0
0 M M
Solution mix variables (B)
Solution values
xl
x2
X3
S1
S2
M
Al
7
0
0
1
1
2
x1
1
1
1
0
0
S2
37
0
7
0
CB
ZI (Cj Z1 )
A2
Minimum ratio
0
1
7 2
1
0
1
3
1
—3
0 0
M+2 2
Al
b (= xB)
2 0
— 2M— 2 —M — M— 2 6+2M 3+M M+2
key TOW
37 7
key column R  2 (new) = R  2 (old) R 1 (new) = R 1 (old)  R  2 (new) R3(new)= 403 X 1 =37,33x1=0,43x1=7 03 x 0=0,03 x1 =3,13 x 0=1,03 x 1 =3 Now, x2 will enter as new variable and Al will depart as shown. Third Simplex table can be prepared by using the following row operations : R  1 (new) =
1
R  1 (old)
R  2 (new) = R  2 (old) + R 1 (new) R  3 (new) = R  3 (old)  7 R 1 (new) R  1 (new) ,
7
9
0,1,
1
2,
1
20
1 1 2, 0 2, 25 R3 (new)= 37 7 x7= , 07x0= 0,77x1=0 2 2 7 1 25 1 07x1=7,37x1 = ,1 7x0=1= 200, 2 2 2 2 2
R  2 (new) =
, 1, 0,
OPERATIONS RESEARCH
126 TABLE 3.27 Third Simplex Table
Solution mix variables (B)
CB 4
X2
2
x,
0
S2
C.
2
4
3
0
0
M
M
Solution values b (= xd
xl
x2
x3
S,
S,
A,
A,
0
1

1 2

1 2
0
1
0
1 2
1 2
0
0
0
2
1 2
1
2 0
4 0
3 6
1 1
0 0
2 2 25 2 Z, (C,  Z, )
Since all the entries in Ci  Zj are either 0 or negative, optimal solution has been obtained with 9 7 x1 = 2, x2 =  , x3 = 0, S2 = 11 and Z = 2x1 + 4x2  3x3 + OSi + 0S2 2 2 = 9 + 140 + 0 + 0 =Rs. 23. Example 3.9. A fertiliser manufacturing company produces three basic types of fertilisers A, B and C. It uses nitrate, phosphate and potash for this purpose. The following data is provided: Ingredients Nitrate Phosphate Potash Inert
% in type A 5 15 10
% in type B % in type C 10 15 15 10 10 10
Cost/ton (Rs.) 10000 4000 6000
Availability (tons) 1200 1600 1400
The selling price of three fertilisers/ton is Rs 3000, Rs 4000 and Rs 5000. The company must produce at least 6000 tons of type A fertiliser. The company wants to Maximize its profits. Advise the company how much quantity of three fertilisers should they produce. Solution. Let x1, x2 and x3 be the quantity of fertilisers A, B and C respectively the company should produce. The data can be put in the form of following table: TABLE 3.28 Fertilizers A B C Cost per ton (Rs.) Availability (tons)
Ingredients (%) Nitrate Phosphate Potash 10 5 15 15 10 10 10 15 10 10000 4000 6000 1200 1600 1400
Selling price/ton (Rs.) Inert 70 65 65 500
3000 4000 5000
127
LINEAR PROGRAMMINGII
Cost of A = 5% of 10000 + 15% of 4, 000 + 10% x6000 + 70% x 500 = 500 + 600 + 600 + 350 = 2050 Cost of B = 1000 + 600 + 600 + 325 = 2525 Cost of C = 1500 + 400 + 600 + 325 = 2825 Problem can be put in the mathematical formula. Maximize Z = (selling price — cost price) x1 + (SP — CP) x2 + (SP — CP) x3 = 950 x1 + 2375 x2 + 2175 x3 Subject to the constraints 0 x 05 + 0 x 10 x2 + 0 x 15x3 _1200 0 x 15 xi + 0 x 15 x2 + 0 x 10 x3 1600 0 x 10 + 0 x 10 x2 + 0 x 10 x3 1400 6000 xl, x2, x3 0 Or 5x1 10X2 15x3 120000 15x1 + 15x2 + 10x3 160000 10x1 + 10x2 + 10x3 140000 By introducing slack, surplus and artificial variables in the inequalities of the constraints, the LP problem in standard form becomes Maximize Z =950 x1 + 2375 x2 + 2175 x3 + 0S1 + 0S2 + 0S3 + 0S4 — MA1 Subject to the constraints 5x1 + 10x2 + 15x3 + S1 = 120000 15x1 + 15x2 + 10x3 + S2 = 160000 10x1 + 10x2 + 10x3 + S3 = 140000 x1 — S4 + A1 = 6000 and x1, x2, S1, S2, S3, S4, Al 0 An initial basic feasible solution is obtained by setting x1 = x2 = x3 = S4 = 0 so that S1 120000, S2 = 160000, S3 = 120000, Al = 6000 and maximum Z = — 6000 M Now, the first simplex table can be written. TABLE 3.29 First Simplex Table C
Solution mix Solution values variables C. b (= xd
950 2375 2175 0 0 0 xi
x3
x3
S1
S2
S2
5,
Minimum ratio Ai
5 15 10
10 15 10
15 10 10
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
T M
0 0
0 0
0 0
o 0
0 1 1 0 NI M
2175
0
0
0 —M
(B)
0 0 0 —M
S, S2 S3 A,
120000 160000 140000 600
Z (C,  Z. )
950 + M 2375 key column
0
24000 10667 14000 6000
key TOW
128
OPERATIONS RESEARCH
950 + M is the largest positive value in Ci  Zj row and 6000 is the minimum ratio so x1 is the entring variable and Al is the departing variable and 1 is the key number (circled in the table). Following elementary operations are applied to prepare the Second Simplex table: (i) R  4 (new) R  4 (old) (ii) R 1 (new) R 1 (old)  5 R  4 (new) (iii) R  2 (new) > R  2 (old) 15 R  4 (new) (iv) R  3 (new) > R  3 (old) 10 R  4 (new) R 1 (new) = 120000  5 x 6000 = 90000 5 5 x 5 = 0,105 x 0 = 10,15 5 x 0 = 15, 1 5 x 1, 05 x = 0,05 x = 0,0 5 x1 = 5, i.e., 90000, 0, 10, 15, 1, 0, 0, 5 R  2 (new) = 160000 x 15 x 6000 = 70000 15  15 x 1 = 0, 15  15 x 0 = 15, 10  15 x 0 = 10, 0  15 x 0 = 0 115 x =1,015 x 0 =0,015 x = 0,015 x1 =15 i.e., 700000, 15, 10, 0, 1, 0, 0, 15. R  3 (new) = 140000  10 x 6000 = 80000 10  10 x 1 = 0, 10  10 x 0 = 10, 10, 0, 0, 1  10 x 0 = 1, 0 10 x1 = 10 i.e., 80000, 0, 10, 10, 0, 0, 1, 10 TABLE 3.30 Second Simplex Table C.
0
S,
Solution Values b (= xd 90000
0
S,
70000
0
0
10
0
1
0
15
4666
0 950
S,
1:• 1
80000 6000
0 1 950 0
10 0 0 2375
0 0 0 2175
0 0 0 0
0 0 0 0
1 0 0 0
10 1  950 950
8000 6000
CB
'Basic variables
Minimum ratio
0
950 2375 2175
(B)
Z. (C,  Z,)
x,
x,
x,
S,
S,
S,
S,
A,
0
10
15
1
0
0
5
9000 key row
key column
Since 2375 is the largest positive number in (C1  Zi) row and 4666 is the minimum ratio, enter variable x2 and replace S2. Following row operations will be applied for writing the third simplex table: 1 R  2 (new)  R  2 (old) 15 R 1 (new) › R 1 (old)  10 R  2 (new) R  3 (new) = R  3 (old)  10 R  2 (new)
LINEAR PROGRAMMINGII
129
1 2 R  2 (new) = 4666, 0, 1, , 0, , 0, 1 R 1 (new) = 90000 10 x 4666 = 43334 2 25 010 x0 = 0,10 10 x 1 =0,1510 x  = , 3 3 1 2 110 x0 =1,0 10 x 15   , 0 10 x 0= 0 3 5 10 x 1 = 5. 25 i.e., 43334, 0, 0, — , 1,  , 0,  5. 3 3 R  3 (new) = 80000 10 x 4666 = 33340 0 0 10 x 0 = 0,1010 x 1 = 0,010 x 3 3 010 x 0 = 0,010 x
15
3
1 10 x 0 =1,
10  10 x 1 = 0. 20 2 = 33340, 0, 0,  — , 0, , 1, 0 3 3 TABLE 3.31 Third Simplex Table 950 2375 2175 0 Basic variables Solution values C ,, 1" 3 S1 i:1 T2 (B) b (=xd 0
S,
43334
0
0
75
x,
46666
0
1
0
S3
33340
950
x,
6000 Z, (C1 Z )
e 1 2
0
20 0 3 1 0 0 0 4750 950 2375 0 3 0
0
0
0
1 775 0 3
M
0
0
0
S2
S3
S4
2 3
0
5
5200
3 1
0
1
69999
Minimum ratio A, key row

1 0 3 0 0 1 475 0 1425 3 475 0 1425 3
key column 1775
Since is the largest positive value in (CI  Z1) row and 5200 is the minimum ratio, enter 3 x3 and replace S1. Following row operations will be applied for writing Third Simplex Table. 3 2 R 1 (new) 2 R 1 (old), R  2 (new) = R  2 (old)   R 1 (new) 5 3
130
OPERATIONS RESEARCH 20 R  3 (new) = R  3 (old) + — R 1 (new), 3 3 0 25 R 1 (new) = 5200, 0, 0, 1, —  , 0, , R  2 (new) 3 9 5 2 = 46666   x 5200 = 43199 3 2 2 2 0 x0=0,1  x0=1,  x1=0 3 3 3 3 2 25 50 1 2 50 100 2 0 = 0 x0=0 3x3 9 , 15 3 x 9 27 , 3 2 3 2 1   x — = . 3 5 5 50 100 43199, 0, 1, 0,  , , 0, .3. 9 27 5 20 20 20 R 3 (new) = 33340 + — x 5200 = 6807, 0 + — x 0 = 0, 0 + — x 0 = 0. 3 3 3 20 20 20 25 500 2 20 50 3006 x = + x = + x 1 = 0, 0 + 3 3 3 9 3 3 9 9 3 20 20x 3 1+ T x 0=1,0+ —  =4. 3 5 500 3000 9 ,1, 4. 6807, 0, 0, 0, 9
or
i.e.,
TABLE 3.32 Fourth Simplex Table C, Solution Basic variables C,3 Values (B) b(=xB )
950 2375 2175
0
0 0 0 M
x1
x2
x3
S,
S,
S,
S,
25 3  50 9 500
950
0
n
3 5
2175
x3
5200
0
0
1
2375
x,
43199
0
1
0
0
S3
6807
0
0
0
950
x,
6000
1
0
0
Z,
950
(C, Z,)
0
2375 2175 0
0
0 44375 9  44375 9
2 100 0 27 5  3000 1 4 9 0 0 1  88750 0  1305 27 88750 27
0
key column
1305
A,
Minimum ratio  ve 103202  ye  ve
LINEAR PROGRAMMINGII
131
The solution cannot be improved further and optimal solution is to produce 6000 tons of x1, 43199 of x2 and 5200 of x3. Example 3.10. Maximize Z = 3x1 + 5x2 + 4x3 Subject to 2x1 + 3x2 8 2x2 + 5x3 se 10 3x1 + 2x2 + 4x3 5 15 x1, x2, x3 0 using Simplex method. Solution. Introducing slack variables, the problem becomes: Z = 3x1 + 5x2 + 4x3 + 0S1 + 0S2 + 053 Three slack variables have been introduced. Subject to 2x1 + 3x2 + Ox3 + Si + 0S2 + 0S3 = 8 Oxi + 2x2 + 5x3 + 0S1 + S2 + 0S3 = 10 3X1 + 2X2 + 4X3 + OSi + 0S2 + S3 =15 Drawing the first simplex table TABLE 3.33 First Simplex Table C,
3
5
4
0
0
0
Basic Solution cB variables values
x,
x2
x3
Si
S2
S,
Minimum ratio
0
S,
8
2
®
0
1
0
0
2.6
0 0
S2 S3 Z, (C,  Z,)
10 15 0
0 3 0 3
2 2 0 5
5 4 0 4
0 0 0 0
1 0 0 0
0 1 0 0
5 7.5
key row
key column Since 5 is the maximum positive value of (Cj — Zj) in the maximizing problem. 5 is the key column and Si row has the minimum value, Si will be replaced by x2. 8 1 Dividing R — 1 row by 3 we get, — , , 1, 0, — , 0, 0 3 3 82 1 So, new x2 row is 3, — , 1, 0, — , 0, 0. 3 3 To get new R — 2 and R — 3, we use the following relationship: New row number = Old row number — [corresponding number of pivot row] x [corresponding element in replacing row] New
4 R2 = 102 x 8= 14,02x = , 2 2 x1 =0, 3 3 3
OPERATIONS RESEARCH
132
5  2 x 0 =5,0 2 x
2 1 33
= 1,
2 x
02x0=0. R2= New
4
14
0,5,
2 ,1, O. 3
2 5 29 R 3 = 152 x 13=,32x =, 22 x 1 =0, 3 3 3 3 1 4  2 x 0 =4,02 x  =3,02 x0 =0,12 x =1. 3
2 9 0, 4, ,0, 1. 3 3 3 Now, we can write the Second Simplex table as follows: =
TABLE 3.34 Second Simplex Table C
3 5 4 0 0 0 x2 x3 S2 S3 x1 S1
Basic variables
Solution values
5
X2
8 3
2 5
0
S,
14 3
4 5
S,
29 3
5 i
0
4
Z,
40 3
10 i
5
0
1 T
0
4
0
(C,z,)
1
0
0 @
Minimum ratio
1 3
0
0
00
2 3
, J.
,„ i.,
14 15
0
1
21 12
0
0
0
0
2 5 5 •5
key TOW
key column Since 4 is the maximum positive value, x3 is the key column. Minimum ratio is obtained by dividing each element of the three rows by their respective elements in the key column, i.e., by 0, 5 and 4. Since minimum ratio is for S2 it will be replaced by X3. New X3 row is obtained by dividing it by key element, i.e., 5. 2 1 14 4 0 It is —  0 1 15 5 * 15 15 New row R  3 is obtained by using the above relationship. New
4 41 29 14 89 5 , 0  4 x 0 = 0. = R3=34x 15 153  4 x 15 15 2 4x2 44x 1 =0, T , 04 x = 4 15 15 5 5 1 4x 0 = 1
133
LINEAR PROGRAMMINGII 2 4 1 89 41 , 0, 0 15 15 15 ' 5 • Now the Third Simplex table can be written as: R3=
TABLE 3.35 Third Simplex Table C.
Cl Basic variables Solution values 5
X2
4
x3
0
S3 Z, (C,  Z,)
8 3 14 15 89 15 256 15
3 5 4 0
0 0
x,
x2
x3
S,
S2
S,
Minimum Ratio
2 3
1
0
1 3
0
0
4
4 15
0
1
2 7 15
1 5
0
14 4
0
0
89 41
4
0
0
4 5 4 3 4 g
1
5
2 15 17 15 17 — 15
34 15 11 1—5
key row
0
0
key column 11 Since most positive value is — . It represents the key column. Minimum ratio is for R  3 it 15 41 . will be replaced by x1 and key element is — 15 New row R  3 is obtained by dividing old row by
41 . i
2 12 15 89 New row R  3 = — , 1, 0, 0, , , . 41 41 41 41 4) 89 62 4 4 14  4 x = , )x1=0 New row R2 = 15 15 41 41 15  ( 15 2 4 2 6 4 4 = 0 + x 0 = 0, 1 + x 0 = 1, 15 +15 x 15 41 41' 15 4 15 4 4 12 5 = , 0+ x = 1+ x 5 15 41 41 15 41 41 2 89 New row R  1 = 82 x = 50 2 2 x 1 = 0, 1   x 0 = 1 3 3 41 41 3 3 3 1 2 2 15 2 12 8 2 x = x = ,0 0  x 0 = 0, 3 3 41 41 3 3 41 41 2 15 10 0 x = 3 41 41
134
OPERATIONS RESVARCH
Now the fourth simplex table can be written as shown below. TABLE 3.36 Fourth Simplex Table C, —> Basic variables 5
Solution values 50 41 62 41 89 41 765 41
X2
4
Xi
3
X, Z
3
5
4
0
0
0
x,
x2
x3
S,
S2
S,
0
1
o
15 41 6 F ,1 2 —
8
 10
41
41
5
4
0
0
1
1
0
3
5
'
0
0
0
0 a
(C, — Z,)
41
41
12
15
41
41
41
45 41  45 41
24 41 24 41
11 41  11 41
Since all the values in CJ — Z are either 0 or negative, optimum solution has been achieved. 765 Zmax = 41 89 50 62 at Xi = — 41 41 3 41 Example 3.11. Minimize Z = 5yi + 6y2 Subject to constraints 2y1 + 5y2 >15 3y1 + y2 12 and y2 0 using simplex method. Solution. Since the problem is of minimization nature, artificial variables are also introduced. Z= 5y1 + 6y2 + OSI + 0S2 + MAi + MA2 Subject to 2y1 + 5y2 — 51 + 0S2 + Ai + 0A2 = 15 3y1 + y2 + OS1 — S2 I 0A1 0A2 = 12 y2, S1, S2, A1, A2 ?_ 0 Using these the first Simplex table can be written as follows: TABLE 3.37 First Simplex Table 5 C.
Basic variables
M M
Al A, Z, (C, — Z,)
6
0 0 M M S,
S,
A,
A,
—1 0 —M M
0 —1 —M M
1 0 M 0
0 1 M 0
Solution values
y,
y2
15 12 27M
2 3 5M 5— 5M
CI 1 6M 6— 6M
key column
Minimum ratio 3 12
key row
LINEAR PROGRAMMINGIT
135
Since M is a very large quantity and the problem is of minimization, most negative value will be taken as key column y2 will replace Al row and key element is 5. New R  1 is obtained by dividing the entire row by key element, i.e., 5 2 1 3,  , 1, , 0,  , 0 5 5 5 New R  2 is obtained by using the relationship already known. 1 2 13 12  1 x 3 = 9, 3  1 x  = — , 1  1 x 1 = 0, 0  1 x = 1. 5 5 5 5  1 + 1 x 0 =  1, 1  1 x 0 = 1. Second simplex table can be written as follows: TABLE 3.38 Second Simplex Table C. > Basic Solution variables values 6
Y2
3
M
A2
9
Z1
18 + 9M
5
6
0
0
,
.1/2
S,
S,
AI
A,
Minimum ratio
2 — 5
1
0
—
0
7.5
—1
—
1
45 13
—M
—
M
M
—
0
()1 12.53M 13  13M 5
(C,—Z.)
0 6 0
1 5 1 — 5 6. M 5 6M 5
key row
key column
13 —53M
is the most negative value, it is the key column. Minimum ratio is
45 13
13 key row and key element is — . A2 is the outgoing row which will be replaced by 5 13 Now, R  2 row will be obtained by dividing all elements by — . 5 45 1 5 i.e., — , 1, 0, —, 13 13 13 New row R  1 will be obtained by using the relationship already known. 2 45 15 2 2 2 x = , x 1 = 0, 1   x 0 = 1 5 13 13 5 5 5 3 0 2 5 2 _1 _ 2 x 1 5 5 13 13 5 13 13 3
so itis the
OPERATIONS RESEARCH
136 The third simplex table can be written as follows: TABLE 3.39 Third Simplex Table 5 6 Basic variables Solution values yl y2 15 6 0 1 Y2 13 45 5 1 0 Yi 13 135 Z, 5 6 13 (C,  Z,) 0 0
0 S, 3
0 S2 2
13
13
15 13
13
1
1
1
1
Since all the elements in Ci — Z1 are positive, the optimal solution is obtained —15 135 , 45 Minimum Z = — ror yi = — 13and Y2 — 13 13 5 x 45 6 x 15 225 — 90 _ 135 Z = 5yi + 6y2 = 13 13 — 13 13 Example 3.12. A firm has an advertising budget of Rs. 720000. It wishes to allocate this budget tb. two media, magazine or TV so that total exposure is Maximized. Each page of magazine advertising is estimated to result in 60000 exposures whereas each spot on TV is estimated to result in 120000 exposures. Each page of magazine advertising costs Rs. 900 and each spot on TV costs Rs.12000. An additional condition that the firm has specified is that at least two pages of magazines advertising be used and at least 3 spots on TV. Determine the optimal media mix for this firm. Solution. Let
x1 = Number of pages of magazine x2 = Number of spots on TV Z = 60000 x1 + 120000 x2
Maximize Subject to 9000 x1 + 12000 x2 5 720000 3x1 + 4x2 5_ 240 2 x2 3 x1,x2 — >0 Introducing slack and artificial variables, we have Maximize Z = 6000 x1 + 120000 x2 + OS1 + 0S2 — MA1 — MA2 Subject to
3x1 + 4x2 + S1 = 240 x1 —S2 +A1 =2 X2 — S3 + A2 = 3 x1, x2, 51, 52, A1, A2 0
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137
Now, the first simplex table can be constructed as follows: TABLE 3.40 First Simplex Table Cl > Basic Solution variables values 0 S1 240 Al M 2 3 A2 M —5M —, (CI — Z)
60000
120000
0
0
0—M—M
x,
x,
S
S,
S,
3 1 0 —M 60000 + M
4 0
1 0 0 0 0
0 —1 0 M —M
0 0 —1 M —M
0 —M 120000 + M
Al
A,
Minimum ratio
0 0 1 0 0 1 —M —M 0 0
60 00 3
► key row
key column Since x2 column has the largest positive value and minimum ratio is that of A — 2 row and ® is the key element. x2 will replace row A2. New elements of row A2 are obtained by dividing by the key element, i.e., 1 i.e., 3, 0, 1, 0, 0, — 1, 0, 1 New row — 1 is obtained by the relationship already known i.e., 240 — 4 x 3 = 228, 3 — 4 x 0 = 3, 4 — 4 x 1 = 0, 1 — 4 x 0 = 1, 0 — 4 x 0 = 0 0 — 4 x — 1 = 4, 0 — 4 x 0 = 0, New row 2 will remain the same as old row 2 as 0 is to be multiplied with new elements of row 3. Now the second simplex table can be constructed as follows: TABLE 3.41 Second Simplex Table Cl
0 —M 12000
60000 120000 0
Basic variables
Solution values
Si Al :r2 Z1 (Ci —Z)
228 2 3 360000 — 2M
Xi
X2
S1
S2
S3
Al
3
0 0 1 120000 0
1 0 0 0 0
0 —1 0 —M M
4 0 —1 120000 120000
0 1 0 —M 0
0 0 —M 60000 + M
Minimum ratio 76 2 co
key row
key column Since 60000 + M is the largest value of CI — ZI, x1 is the key column and since minimum ratio is that of A1, row Al will be replaced by column x1. New row Al is obtained by dividing all the elements of the row by key element, i.e., 1. New row Si is obtained by the relationship already known. 228 — 3 x 2 = 222, 3 — 3 X 1 = 0, 0 — 3 X 0 = 0, 1 — 3 x 0 = 1, 0 — 3 x — 1 = 3 4—3X0=4
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OPERATIONS RESEARCH
New row x2 elements are as follows: 3 — 0 x elements of departing row, so all elements remain the same. Third Simplex table can be constructed as follows: TABLE 3.42 Third Simplex Table
C, —› 1
Basic variables
Solution values
0
S,
222
60000 120000
x1
2 3 X2 Z, 480000 (C,  Z,)
60000
120000
0
0
t 1
X2
S,
S2
0
0
1
3
1 0 60000 0
0 1 120000 0
0 0 0 0
1 0  60000 60000
S3
Minimum Ratio
key row
55.5 0 1  120000 120000
00
 ye
t
key column S3 has the largest positive value and S1 has the minimum ratio so S1 will be replaced by S3. Key element is 4. New row S1 is obtained by dividing all its elements by key element, i.e., 4, 55.5, 0, 0, New row x1 element will be the same as for this row 0 occurs in the key column. New row x2 is obtained by using the relationship already known. 3—(1)x 55.5 =58.5,0—(1)x0=0,1—(1)x 0=1 1 0—(1)1=,0—(1)x 3 = 3 1—(1)x1 =0 4 4 4 4, The Fourth Simplex table can be constructed as follows: TABLE 3.43 Fourth Simplex Table
Cl —>
60000 120000
0
0
0
S,
S,
S,
1 4 0
3— 4 1
1
Basic variables
Solution values
x,
x,
0
S,
55.5
0
0
60000
xi
2
1
0
120000
x,
58.5
0
1
1 4
3— 4
0
Z
7140000
60000
120000
30000
30000
0
0
0
 30000
 30000
0
(C,  Z,)
0
1 3
1
139
LINEAR PROGRAMMINGII Since all the elements of Cl  Zi are 5 0 The optimal solution has been arrived. xi = 2 Z = 7140000 X2 = 58.5 Example 3.13. Minimize Z = xl + 2x2 + x3 Subject to 1
x1+ + 2
1 21 2 +  X3 2
3 — xi + 2x, + x3 8 2 xl, x2, x3 > 0. Solution. Introducing slack variables and artificial variables. Minimum value of Z = xi + 2x2 + x3 + OSi + 0S2 + MAi Subject to constraints 1 1 2 x3 + Si = 1 xi + — 2 x2 + — 3 — xi + 2x2 + x3  S2 + Ai =8 2 Writing the First Simplex table as follows.
and
TABLE 3.44 First Simplex Table Cj
*
0 M
1 Basic variables
Solution values
X1
S1
1
1
Al
8
3 2
Z)
8M
(C) — )
2
1
0
0
0
X2
x3
Si
S2
Al
Minimum ratio
1
0
0
2
0
1 2
2
1
0
—1
1
4
M
0
—M
M
1—M
0
M
0
3 2M 2M 3 1— M 2 — 2M 2
key TOW
key column Since M is a very large quantity and it is a minimizing problem, key column is x2 and key row or the departing row is Si, key element is . New x2 row (old Si) is obtained by dividing the 1 i.e., values are 2, 2, 1, 1, 2, 0, 0. 2, New values of row Al are obtained by the relationship we already know.
present elements by the key element, i.e., by
i.e., 8 — 2 x 2 = 4, 3 — 2 x 2 = —5 2 — 2 x 1 = 0, 1 — 2 x 1 = — 1 2 2 0 — 2 x 2 = — 4, — 1 — 2 x 0 = — 1, 1 — 2 x 0 = 1.
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140
Now, second simplex table can be constructed as follows: TABLE 3.45 Second Simplex Table
2 M
C,, Basic variables Solution values x2 2 A,
4
Z.
4 + 4M
(C,  Z)
1
2
1
0
0
xi
x2
x3
Si
S2
2 1 5 0 2 5 4— M 2 2 5 3+ — M 0 2
1
2
0
M Al 0
1
4
1
1
2M
4  4M
1+M 4+4M
M M M
0
Since all the values in Ci  Z1 are either 0 or positive, there is no feasible solution. x1 = 0, x2 = 2, Si = 0, S2 = 0 x3 =0,A1 =4 Zin• • um = 4 + 4M Two Phase Simplex Method
Example 3.14. Maximize Z = 5x1 + 3x2 Subject to constraints 2x1 + x2 5:1 + 4x2 6 xl, x2 0. Solution. Phase I. It consists of the following steps: Step 1. Adding slack variables, the problem becomes 2x1 + x2 + Si = 1 x1 + 4x2  S2 = 6
Step 2. Putting x1 = 0 and x2 = 0 Si = 1 S2 =  6. This gives the initial basic solution. However, it is not a basic feasible solution since S2 is negative. So, we will introduce artificial variable Al and the above constraint can be written as 2x1 + x2 + Si = 1 ...(i) + 4x2  S2 + Al = 6 ...(ii) Step 3. Substituting x1 = x2 = S2 = 0 in the constraint equation we get, Si = 1 and Al = 6. as the initial basic solution. This can be put in the form of a Simplex table as follows:
LINEAR PROGRAMMINGII
141 TABLE 3.46
0
C.› Basic variables
Solution values
x,
x2
S,
S2
A,
Minimum ratio
0
S,
1
2
@
1
0
0
1
1
A,
6
1
4
0
1
1
3 2
Z,
6
1
4 4
0 0
1
1 0
(C,  Z,)
1
1
key row
key column As (C1  Z1) is negative under some columns, the current basic feasible solution can 12? improved. Si will be replaced by x2 as x2 is the key column, and Si is the key row, also is the key elemeiit. New row x2 (old Si) will be obtained by dividing all the elements by 1. New row Al can be obtained by using the relationship already known. 64x1=2,14x2=7,44x1=0,04x1=4 14x0=1,14x0=1 New Simplex table can be constructed as follows: TABLE 3.47
C, 0 1
0 Basic variables
Solution values
x2 Al Z, (C,  Z,)
1 2 2
x, 2 7 7 7
0
0
0
1
x2 S, S, A, 1 1 0 0 0 4 1 1 0 4 1 1 0 4 1 0
Since all the elements are either positive or zero, an optimal basic solution has been arrived. However, Al = 2 which is > 0, the given LPP does not possess any feasible solution and the procedure stops. LIMITATIONS OF LPP SIMPLEX METHOD
Following are some limitations of LPP Simplex Method: 1. Simplex method involves understanding of many conceptual technical aspects. These cannot be understood by any manager not conversant with the subject. 2. Linear programming problems need lot of expertise, time and are cumbersome. A number of steps have to be adopted to proceed in a systematic manner before one can arrive at the solution.
142
OPERATIONS RESEARCH
3. Graphic solution method has lot of applications and is relatively short and simple. However, it has limitations and cannot be applied to problems with more than two variables in the objective function. 4. Simplex method of LPP can be applied to problems with more than two variables in the objective function, the procedure adopted is complicated and long. It may need computation of 4 to 5 simplex tables and can test the patience of the problem solver. Computers are of course helpful in such cases. 5. LPP does not lead to 'a unique' optimal solution. It can provide different types of solutions like feasible solution, infeasible solution, unbounded solution, degenerate solution, etc. 6. It gives absurd or impractical results in many solutions. The solution may ask for providing 24.59 men or 3.89 machines which is not possible. 7. LPP model makes many assumptions in the values of objective function and constraint variables, like the rate of profit. In fact, such assumptions may not be right. 8. The whole approach to the solution is based on the linearity of the functions, i.e., all the variables involved in the problem increase or decrease in a linear manner. This assumption does not hold good in all cases. In many cases, the objective function may assume the form of a quadratic equation. 9. LPP method cannot be used where a number of objectives are required to be fulfilled. It deals with either maximizing of profits or minimization of costs, etc. SENSITIVITY ANALYSIS The solution to LPP is based on a number of deterministic assumptions like the prices are known exactly and are fixed, resources are known with certainty and time needed to manufacture/ assemble/produce a product is fixed. In real life situations, which are dynamic and changing, the effect of variation of these variables must be studied and understood. This process of knowing the impact of variables on the outcome of optimal result is known as sensitivity analysis of linear programming problems. Let us say, for example, that if originally we had assumed the cost per unit to be Rs 10 but it turns out be Rs 11, how will the final profit and solution mix vary. Also, if we start with the assumption of certain fixed resources like man hours or machine hours and as we proceed we realise the availability can be improved, how will this change our optimal solution. Sensitivity analysis can be used to study the impact of changes in: (a) Addition or deletion of variables initially selected. (b) Change in the cost or price of the product under consideration. (c) Increase or decrease in the resources. Sensitivity analysis uses the following two approaches : (a) It involves solving the entire problem by trial and error approach and involves very cumbersome calculations. Every time data of a variable is changed, it becomes another set of the problem and has to be solved independently. (b) The last simplex table may be investigated. This reduces completion and computations considerably.
143
LINEAR PROGRAMMINGII Limitations of Sensitivity Analysis
Sensitivity analysis does take into account the uncertainty element, yet, it suffers from the following limitations: (a) Only one variable can be taken into account at one time. Hence, the impact of many variables changing cannot be considered simultaneously. (b) It suffers from the linearity limitations as only linear relationship between the variable is considered. (c) The extent of uncertainty cannot be studied. (d) As the result can be judged by individual analysts depending upon their skills and experience, it is to that extent subjective in nature. RI REVIEW AND DISCUSSION QUESTIONS uI 1. Explain the following terms : (a) Basic feasible solution. (b) Optimal solution. 2. Explain step by step the method used in solving LPP using simplex method. 3. Explain the use of slack, surplus and artificial variables when are these used and why. 4. How are the key column, key row and key element (number) selected ? 5. Explain the use of simplex method in solving the maximization and minimization problems. What are the differences in the approach ? 6. What do you understand by a redundant constraint ? Do these constraints influence analysis and final solution of a LPP ? 7. What are the limitations of LPP ? Give examples to support your argument. 8. What do the coefficient in a simplex table represent ? Why is it necessary to compute a new set of coefficients for each table in the analysis ? 9. Explain the terms decision variables, basic variables, entering and departing variables. 10. Write a detailed note on the sensitivity analysis. 11. Maximize x1 + 2x2 + 3x3 — x 4 Subject to x1 + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 xi + 2x2 + x3 + x4 = 10 Using simplex method. 12. Minimize Z = 8x1 + 4x2 + 2x3 Subject to 4x1 + 2x2 + x3 8 3x1 + 2x3 10 x1 + x2 + x3 = 4 xi, x2, x3 0 Using simplex method. 13. Maximize Subject to
Z = xi + 2x2 + 3x3 — x4 xi + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20
OPERATIONS RESEARCH
144 xi + 2x2 + x3 + x4 = 10 Xi, X2, X3, X4 0
Using simplex method. 14. Use Simplex Method to Maximize p = 5x  2y + 3z Subject to 2x + 2y  z ?. 2 3x  2y 3 y  3z 5 x, y, z 0 15. Solve the following LPP using simple method: Z = 25x1 + 80x2 Subject to 5x1 + 6x2 15 9x1 + 9x2 27 xi 0, x2 0. 16. The ABC company makes two products P1 and P2 with contribution per unit of Rs. 15 and Rs. 11 respectively. Each of the products is made from two raw materials A and B. P1 and P2 require the raw material in the following amounts: kgs
Product P1 P2
Availability in kgs
B 3 1 500
A 4 2 400
Find the optimum product mix for maximum profit. 17. ABC maufacturing company makes three products x1, x2 and x3 with contribution per unit to profit Rs. 2, Rs. 4 and Rs. 3 respectively. Each of the three product passes through three centres as part production process. Time required in each centre to procedure one unit of each product is as given below : Product X1 X2 X3 Time available (Hours)
Centre 1 3 4 2 60
Hours per unit Centre 2 2 1 2 40
Determine the optimal mix for next week production. 18. Solve the following LPP using simplex mt Maximize Z = 6x1 + 4x2 2x1 + 3x2 30 Subject to 3x1 + 2x2 24 x1 + x3 3 x1, x2 0 Is the solution uniqe ? If not give two different solutions.
Centre 3 1 3 2 80
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LINEAR PROGRAMMINGII
19. Explain the simplex method by carrying out the iteration in the following problem : Maximize + x5. Subject to xi + 2x2 + 3x3 + x4 = 8 3x1 + 4x2 + X3 4 X5 = 7 x1 to x5 0. 20. Maximize Z = 4x1 + 5x2 — 3x3 + 50 Subject to x1 + x2 + x3 = 10 x1 — x2 1 2x1 + 3x2 + x3 40 x1, x2, x3 0 21. A firm manufactures three products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respectively. The firm has two machines and below is the required processing time in minutes for each machine on each product. Z = 5X1 2X2 3X3
Machine
Product B 3 2
A 4 3
G H
C 5 4
Machines G and H have 2000 and 2500 machine minutes respectively. The firm must manufacture 100A's, 200B's and 50C's but no more than 150A's. Setup an LP problem to maximize profit. Find solution. 22. Solve the following problem by simplex method Minimize Z = 2x1 + x2 Subject to 3x1 + x2 3 4x1 + 3x2 6 + 2x2 3 0, x2 0. Explain your each step properly. 23. For the following production given in the table, formulate the problem as linear programming and solve. Machine Time (hours) Profit per product Product (Rs) A B C 20 P 8 4 2 6 2 3 0 Q 8 R 3 0 1 Available Machine 250 150 50 hours per week 24. Use simplex method to solve : Maximize Z = 6x1 + 4x2 Subject to 2x1 + 3x2 30 3x1 + 2x2 24 x1 + x2 >3 Xi,
0.
OPERATIONS RESEARCH
146 25. Solve the following problem : Maximize Subject to
Z = 2x1 + x2 x1 + 2x2 10 xi + x2 6 — x2 2 xi — 2x2 1 0, x2 0
26. Use simplex method to solve : Maximize Z = 3x1 + 2x2 + 5x3 xi + 2x2 + x3 430 Subject to 3x1 + 2x3 460 xl+ 4x2 420 x1 >_0, x2 >_0, x3 0. 27. Solve the following LPP by simplex method : Z = 60x1 + 80x2 Minimize x2 > Subject to  200 x1 5_ 400 x1 + x2 = 500 where x1 and x2 0 28. A plant makes two products A and B which are routed through four process centres 1, 2, 3, 4 as shown by the solid lines in figure shown below.
Centre 2
A Centre 4
Centre 1
B
B
Centre 3
♦A
If there is spare capacity in centre 3, it is possible to route product A through 3 instead of going through centre two twice, but this is more expensive. — Given the information below, how should production be scheduled so as to maximize profits.  Centres 1 and 4 run up to 16 hours a day, centres 2 and 3 run up to 12 hours a day. The shipping facility limit the daily output of A and B to a total of 2500 litres.
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147
Product
Centre
Input (Litres/hour)
% Recovery
Running cost per year
A
1 2 (First pass) 4 2 (Second pass) 3 1 3 4
300 450 250 400 350 500 480 400
90 95 85 80 75 90 85 80
150 200 180 220 250 300 250 240
B
Product
Rate material cost/litre
Sale price per finished litre
Maximum daily sales litres of finished product
A B
5 6
20 18
1700 1500
Formulate the problem as an LP problem. Do not solve. 29. Solve the following LP problem by revised simplex method :
Minimize Subject to
Z = — 3x1 + x2 + x3 — 2x2 + x3 11 — 4x1 + x.2 + 2x3 3 2x1 — x3 = — 1 Xlf x2, x3 >
30. Solve the following problem : Minimize
Z = 60 x1 + 80 x2
Subject to
20 xi + 30 x2 900 40x1 + 30 x2 1200 xi, x2 ?_ 0.
31. A firm makes two types of furniture chairs and tables. The contributions for each product as
calculated by accounting department are Rs 20 per chair unit and Rs 30 per table. Both products are processed on three machines M1, M2 and M3. The time required by each product and total time available per week on each machine is as follows : Machine
Mi M2
M3
Chair
3 5 2
Table
3 2 6
Available Time (hours)
36 50 60
How should the manufacture schedule his production in order to maximize contribution ? [CA (Final) May, 1979] 32. The products A, B and C are produced on three machines centres X, Y and Z. Each product involves operations on each of the machine centres. The time required for each operation for unit amount of each product is as follows :
OPERATIONS RESEARCH
148 Machine centres
Products A
B C
X
Y
Z
10 2 1
7 3 2
2 4 1
(Time in hours) There are 100, 77 and 80 hours available at machine centres X, Y and Z respectively. The profit per unit of A, B and C is Rs 12, Rs 3 and Re 1 respectively. Formulate the problem as LPP (Linear Programming Problem) and find the profit maximization product mix. [IGNOU MBA, Dec 2000] A company is engaged in producing three products viz A, B and C. The following data are 33. available: Products
Sale price (per unit) Cost (per unit)
A
B
C
10 6
12 9
15 10
The wholesaler who is responsible for selling to the customer is to be paid Rs 150 per day irrespective of the quantities sold of each of the products. The products are processed in three different operations. The time (hours) required to produce one product in each of the operations and the daily capacity (hours) available for each operation centre are given below: Operations 1
2 3
Products A
B
C
2 3 1
3 2 4
2 2 2
Daily capacity
400 350 300
What product would yield maximum profit and how much ? 34. A pharmaceutical company has 100 kgs of A, 180 kgs of B, and 120 kgs of C available per month. They can use these materials to make three basic pharmaceutical products, namely 5 — 10 — 5, 5 — 5 — 10 and 2 — 5 — 10 where the numbers in each case represent the percentage by weight of A, B and C respectively in each of the products. The costs of these raw materials are given below : Ingredient
A B C Inert ingredients
Cost per kg (Rs.)
80 20 50 20
Selling prices of these products are Rs. 40.50, Rs. 43 and Rs. 45 per kg respectively. There is a capacity restriction of the company for product 5 —10 — 5, so as they cannot produce more than 30 kg per month. Determine how much of each of the product they should produce in order to maximize their monthly profit.
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149
35. Noap's Boats makes three different kinds of boats, all can be made profitably in this company. But the company monthly production is constrained by the limited amount of labour, wood and screws available each month. The director will choose the combination of boats that Maximizes his revenue in view of the information given in the following table : Input Labour (Hours) Wood (Board feet) Screws (kg) Selling price (Rs.)
Row boat 12 22 2 4000
Camoe 7 18 4 2000
Kayak 9 16 3 5000
Monthly availability 1260 hours 19008 board feet 396 kg
(a) Formulate the above as a LPP. (b) Solve it by the simplex method from the optimal table of the solved LPP, answer the following questions : (i) How many boats of each type will be produced and what will be the resulting revenue ? (ii) Which, if any of the resources are not fully utilised ? If so how much of spore capacity is left ? (iii) How much of wood will be used to make all of the boats given in the optimal solution ? 36. An agriculturist has a farm with 125 acres. He produces Radish, or pea and Potato. Whatever he raises is fully sold in the market. He gets Rs. 5 for radish per kg, Rs. 4 for pea per kg, and Rs. 5 for potatoes per kg. The average yield is 1500 kg of radish per acre, 1800 kg of pea per acre and 1200 kg of potato per acre. To produce each 100 kg of radish and pea and to produce each 80 kg of potatoes, a sum of Rs. 12.50 has to be used for manure. Labour required for each acre to raise the crop is 6 man days for radish and potatoes each and 5 man days for pea. A total of 500 man days of labour at rate of Rs. 40 per man day are available. Formulate items as a LP model to Maximize the agriculturist's total profit. 37. The cost of production per unit of products A, B and Care Rs. 10, Rs. 15 and Rs. 18 respectively and their selling prices per unit are Rs. 16, Rs. 24 and Rs. 28 respectively but the agency commission of 25% on selling price should be borne by the company. These products are produced in 3 operations and their installed capacity with effective utilisation and working hours are given below: Operations
No. of machines installed
I II III
3 4 2
No. of working hours per day 15 16 24
Effective utilisation of machines 80% 75% 75%
The effective time required to produce one product in each of these operations are given below: Operations I II III
Product A 4 1 2
Product B 3 5 4
Product C 2 3 5
OPERATIONS RESEARCH
150
Demand is no constraint for all the three products. Find the optimum product mix per day that gives the maximum profit. 38. A factory works 8 hours a day, producing three products, viz A, B and C. Each of these products is processed in three different operations viz 1, 2 and 3. The processing time in minutes for each of these products in each of the operations are given below along with utilisation of the process and the cost and price in rupees for each of these three products which have unlimited demand.
Product
A B C Utilisation
Processing time (minutes) 1 2 3
4 2 3 80%
3 1 4 70%
1 4 5 90%
Cost per unit
Price per unit
(Rs.) 10 8 5
(Rs.) 16 12 10
(0 Determine the optimal product mix using simplex method. (ii) Give interpretation for the value obtained in the final simplex table. 39. A resourceful home decorator manufactures two types of lamps say A and B. Both the lamps
go through two technicals, first a cutter and second, a finisher. Lamp A requires 2 hours of cutter's time and 1 hour of finisher's time. Lamp B requires 1 hour of cutter's and 2 hours of finisher's time. The cutter has 104 hours and finisher 76 hours of available time each month. Profit on one lamp A is Rs 6.00 and on one lamp B is Rs. 11.00. Assuming that he can sell all that he produces, how many of each type lamp should be manufactured to obtain the best return ? 40. Using surplus and artificial variables, solve the following: Minimize Z = 5x1 + 6x2 Subject to
2x1 + 5x2 1500 3x1 + x2 1200 x1, x2 0
and
41. A product is manufactured by blending three different raw materials. The finished product
should meet certain requirements. Given the following data, what is your recommendation with regard to quantity for raw materials to be blended, which will meet the quality requirement with minimum cost ? Quality characteristics
1 2 3 Cost of raw materials per unit in Rs.
Contribution to quality by each unit of raw materials A
B
C
Minimum Quality Requirement
3 5 1
0 1 2
1 2 0
10 15 8
2
5
3
42. A TV company operates two assembly lines. Line I and Line II. Each line is used to assemble
the components of three types of TV's, colour, standard, economy. The expected daily production on each line is as follows:
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151
TV model Line —I Line — II 3 1 Colour 1 1 Standard 2 6 Economy The daily running cost for two lines average Rs. 6000 for line I and Rs. 4000 for line II. It is given that the company must produce at least 24 colour, 16 standard and 48 economy TV sets for which an order is pending. You are required to formulate the above problem as LPP model taking the objective function a minimization of total cost. Also, determine the number of days that the two lines should be seen to meet the requirement. 43. Maximize Z = 6x1 + 4x2 Subject to 2x1 + 3x2 30 3x1 + 2x2 24 x1 +x2 > 3 where x1, x2 0 44. Solve the following problem using simplex method : Maximize Z = 21x1 + 15x2 Subject to   2x2 6 4x1 + 3x2 5_ 12 where x1, x2 0 45. Maximize Z = 3x1 + 8x2 Subject to x1 + x2 = 200 x1 80 x2 60 where x1, x2 5 0. 46. The owner of fancy goods shop is interested to determine how many advertisements to release in selected three magazines A, B and C. His main purpose is to advertise in such a way that the total exposure to principal buyers of his goods is maximized. Percentage of readers for each magazine is known. Exposure in any particular magazine is the number of advertisements released multiplied by the number of principal buyers. The following data are available: Particulars Readers Principal buyers Cost per advertisement
A 1.0 lakh 20% Rs. 8000
Magazines B 0.6 lakhs 15% Rs. 6000
C 0.4 lakhs 8% Rs. 5000
The budgeted amount is at the most Rs. 1.0 lakh for the advertisement. The owner has already decided that magazine A should have no more than 15 advertisement and B and C each gets at least 8 advertisement. Formulate the Linear Programming Problem model and solve it. Z =  5x2 47. Maximize xl + x2 1 Subject to  0.5 x1  5x2 5_  10 xi, x2 0
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48. Maximize Subject to
Z = 3x1 + 2x2 2x1 + x2 5_ 2 + 4x2 12 0, x2 0 49. Solve the following problem by simplex method Maximize Z = 3x + 2y Subject to — x + 2y 5_ 4 3x + 2y 5_ 14 x _y< 3 x,:tR O. 50. A firm manufactures two qualities of tweed. A yard of quality A tweed requires 8 ozs of grey wool, 2.5 ozs of red wool and 2 ozs of green wool, one yard of tweed B is madeup of 10 ozs of grey wool, 1 oz of red wool and 4 ozs of green wool. The availability of wool in given periods is 5, 000 lbs of grey, 1250 lbs of red and 1875 lbs of green. Both tweeds can be produced on the same machines and both can be woven at the rate of 12 yards per hour. A total of 750 machine hours are available in given period. The contribution towards profit is Rs. 2 per yard of tweed. A and Rs. 4 for tweed B. Given that the company has firm order for and is obliged to produce, at least 300 yards of tweed; what is the optimal production policy for the firm? 11 G.N.D.U. EXAMINATION PROBLEMS RI 1995 APR. (1) 1996 APR. (2) 1998 APR. (3) 1999 APR. (4) 2000 APR. (5) 2002 APR. (6) 2004 APR. (7) 2004 APR. (8) 2004 APR. (9) 1997 APR. (10) 2003 APR. (11) (Professional) 2004 APR. (12) 1995 APR. (13)
2003 APR. (14) 1993 APR. (15)
Degeneracy in L.P. problem [April. 1995; Sept. 1995, April 1999] Slack and surplus variables Convex set Convex and Concave sets Slack variables Explain Degeneracy in L.P.P. Define Degeneracy Degenerate solution of L.P.P. (i) Unbounded solution (ii) Key element Explain briefly simplex method of solving a Linear Programming problem. Why is simplex method considered superior to graphic method ? Define slack, surplus and artificial variables in a LPP and explain how they help in finding a basic feasible solution to LPP using Simplex Method. Give sequence of steps in Simplex Method for solving a Linear Programming Problem. Why is the simplex method a better technique than a graphical approach for most real cases ? Discuss the advantages and limitations of Linear Programming. Briefly describe the graphic and simplex methods of solving a linear programming problem. Why is simplex method considered superior to graphic method ? A farm is engaged in breeding pigs. These pigs are fed in various products grown on the farm. Because of the need to ensure certain nutrient constituents,
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it is necessary to buy additional one or two products A and B. The nutrient constituents (vitamin and proteins) in each unit of the products are given: Nutrient
Nutrient Constituents in the products A B
1 2 3
36 3 20
Minimum Account on Nutrient
6 12 10
108 36 100
Product A costs Rs. 20 per unit and product B costs Rs. 40 per unit. How much of products A and B should be purchased at lowest possible cost as to provide the pigs' nutrients not less than given in the table ? 1993 APR. (16) Maximum : Z = 3a + 5b + 4c Subject to the constraints : 2a + 3b 8 2b+ 3c10 3a + 2b + 4c 5 where 1993 SEP. (17) The HandyDandy company wishes to schedule the production of kitchen appliances which requires two resources labour and material. The company is considering three different models and its production engineering department has furnished the following data: A
B
C
Profit (Rs./ unit)
4
3 4 2
6
Material (kg/unit)
7 4
Labour (Hours/unit)
5
3
The supply of the raw material is restricted to 200 kg per day. The daily availability of the manpower is 150 hours. Formulate the Linear Programming model to determine the daily production rate of the various models in order to maximize the total profit. 1993 SEP. (18) Solve the following Linear Programming problem: Minimize : Z = — 3a + b + c. Subject to: 9a — 2b + c 1 — 4a +b + 2c ?_. 3 2a — c = —1 (19) Use simplex method to solve LPP : Maximize : Z = 107x1 + x2 + 2x3 Subject : 6x1 + x2 — 2x3 3 16x1 + 3x2 — 6x3 5 5 3x1 — x2 — x3 0 xi, x2, x3 >_ 0 1995 APR. (20) A company produces the products P, Q and R from three raw materials A, B and C. One unit of product P requires two units of A and three units of B. One unit of
1994 SEP.
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Q requires two units of B and five units of C and one unit of product R needs three units of A, 2 units of B and 4 units of C. The company has 8 units of A, 10 units of B and 15 units of C available. Profit per unit of products P, Q and R are Rs. 3, Rs. 5 and Rs. 4 respectively. How many units of each product should be produced to maximize profits? 1995 SEP. (21) Use simplex method to verify that following problem has no optimal solution: Maximize : Z = xi + 2x3 Subject : 2x1 + X2 + X3 < 2 — x1 + x2 — x3 1 x1, x2, x3 >— 0 1996 APR. (22) Solve the following L.P. Problem using simplex method: Maximize: Z = 30x1 + 16x2 + 25x3 Subject to constraints : 0.8 x1 + 0.4x2 + 0.5x3 100 5x1 + 3x2 + 3x3 650 9x1 + 6x2 + 9x3 1260 x1, x2, x3 >— 0 1996 APR. (23) Solve the following L.P. problem using simplex method : Minimize : Z = 80x1 + 100x3 80x1 + 60x2 .?. 1500 20x1 + 90x2 1200 xi, x2 0 1997 APR. (24) Maximize : Z = xi + 2x2 + 3x3 — x4 and SEP. 1998 Subject to : xi + 2x2 + 3x3 = 15 2x1 + x2 + 3x3 = 20 x1 + 2x2 + x3 + x4 = 10 x1 + x2 + x3 + x4 0 Using Simplex Method: 1998 APR. (25) Minimize : x = 8x1 + 4x2 + 2x3 Subject to : 4x1 + 2x2 + x3 5 8 3x1 + 2x2 _. 10 xi + x2 + x3 = 4 xi, x2, x3 0 Using simplex method: 1999 SEP. (26) Solve the following L.P.P. Minimize : Z = 25x1 + 80x2 Subject : 5x1 + 6x2 15 9x1 + 4x2 27 x1 > 0, x2 > 0 (27) ABC manufacturing company makes three products X1, X2 and X3 with contribution per unit to profit Rs. 2, Rs. 4 and Rs. 3 respectively. Each of these products
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passes through three centres as part of production process. Time required in each centre to produce one unit of each product is as given below: Product
X, X2 X3 Time Available (Hours)
Centre 1
Hours per unit Centre 2
Centre 3
3 4 2 60
2 1 2 40
1 3 2 80
Find the optimum product mix. 2002 APR. (28) Solve the following maximization LPP using simplex method. Is your result a unique solution ? Discuss: Maximize Z = 8x + 16y Subject to x + y 200 y 5. 125 3x + 6y 900 x, y 0 2002 SEP.
(29) Solve by the simplex method. Minimize Z = 40x + 24y Subject to 50x + 20y 4800 50x = 80y 7200 x, y 0
2004 APR. (30) Solve the following Linear Programming (L.P. P.)
Minimize Subject to
Z = xi + x2 + x3 — 3x2 + 4x3 = 5 x1 2x2 5_3 2x1 — x3 4 x1, x2 ?_ 0 and x3 unrestricted in sign.
2004 APR. (31) A firm has 240, 370 and 280 kg of wood, plastic and steel respectively. The firm
2004 APR. 2004 APR.
produces two products A and B. Each unit of A requires 1, 3 and 2 kg of wood, plastic and Steel respectively. The corresponding requirement for each unit of B is 3, 4 and 1 kg respectively. If A sells for Rs. 4 and B sells for Rs. 6 per unit, then what product mix should the firm produce in order to have maximum gross income ? Formulate this as a LPP and solve. (a) Discuss the assumption of proportionality, additively, continuity, certainty and finite choices in the context of LPPs. (b) Use Simplex Method to maximize Z = 28x1 + 30x2 Subject to : 6x1 + 3x2 18 3x1 + x2 8 4x2 + 5x2 30 0, x2 O. and
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111 PUNJAB UNIVERSITY EXAMINATION PROBLEMS 11
1995 SEP. (1) Degeneracy in LPP. 1999 APR. (2) (i) Convex and concave sets (Marks 2 each) (ii)Degeneracy in LPP 1995 APR. (3) Why is the simplex method a better technique than the graphical approach for most real cases ? Discuss the advantage and limitations of Linear Programming. (Marks 10) (Marks 6) 1998 APR. (4) Explain briefly two phase simplex method. 1999 APR. (5) Explain briefly simplex method of solving a Linear Programming problem. How it is better than graphic method ? 1999 SEP. (6) Explain briefly two phase method of solving a Linear Programming program. (Marks 6) 2001 APR. (7) Illustrate the following and their importance in dealing with Linear Programming problem. (ii) Artificial Variables (i) Slack/Surplus variables (iii)Basic Variables and (iv) Nonbasic variables. (Marks 7) 1999 SEP. (8) Solve the following using two phase simplex method : Minimize : 8x1 + 4x2 + 2x3 Subject to : 4x1 + 2x2 + x35 8 3x1 + 2x2 510 x1+ x2 + x3 4 x1, x2, x3>_ 0 1999 SEP. (9) Use simplex method to verify that following problem has no optimal solution : Maximize : Z = x1 + 2x2 Subject to : — 2x1 + X2 + X3 2 — + x2 — x3 5. 1 x1, x2, x3 0. 1996 SEP. (10) Solve the following L.P. problem using simplex method : Minimize : Z = 80x1 + 100x2 Constraints : 80x1 + 60x2 1500 20x1 + 90x2 1200 x x2 > —0 1997 APR. (11) Solve the following L.P. problem using simplex method : Maximize : x1+ 2x2 + 3x3  x4 Subject to : x1 + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 X1
2X2 + X3 + X4 = 10 x1, x2, x3, x4 0
1998 APR. (12) Solve the following L.P. problem using simplex method Minimize : Z = 8x1 + 4x2 + 2x3 Subject to : 4x1 + 2x2 + x3 5 8 11%4 _ 0 1999 SEP.
(13) Solve the following L.P.P using simplex method: Minimize : Z = 25x1 + 80x2 Subject to : 5x1 + 6x2 ... 15 9x1 + 4x2 27 xi 0, x2 .?.. 0
1999 SEP.
(14) ABC manufacturing company makes three products X1, X2 and X3 with contribution per unit to profit Rs. 2, Rs. 3 and Rs. 3 respectively. Each of these products passes through three centres as part of production process. Time required in each centre to produce one unit of each product is as given below: Product
X, X2 X3 Time available (hours)
Centre 1
Hours per unit Centre 2
Centre 3
3 4 2 60
2 1 2 40
1 3 2 80
Determine the optimal product mix for next week production. 2000 APR. (15) Solve the following LPP using simplex method : Maximize : Z = 6x1 + 4x2 2x1 + 2x2 5. 30 Subject to : 3x1 + 2x2 5. 24 + x2 3 xi, x2 0 Is the solution unique? If not, give two different solutions : 2000 SEP. (16) (a) Explain the simplex method by carrying out one iteration in the following problem : Maximize : Z = 5x1 + 2x2 + 3x3 — x4 + x5 Subject to : x1 + 2x2 + 3x3 + X4 = 8 3x1 + 4x2 + x3 + X5 = 7 x1 to x5 — (b) Maximize : Z = 4x1 + 5x2 3x2 + 50 x1 + x2 + x3 =10 Subject to : — x2 ?_ 1 2x1 + 3x2 + x3 5 40 xi, x2, x3 5. 0 2003 APR. (17) Solve the following LPP using simplex method Minimize : Z = 60 x1 + 80x2 x2 2000 Subject to : xi 5. 400 + x2 = 500 xi and x2 ?. 0.
4 Linear ProgrammingIll (Duality In Linear Programming)
I LEA RIsTING OBJECTIVES • • • •
Understand optimal—dual concept. Understand dual formulation procedure. Interpret dual programming model. Solve LP problems using duality.
CONCEPT OF PRIMALDUAL RELATIONSHIP OR DUALITY IN LINEAR PROGRAMMING The original LPP as we have studied is called the Primal. For every LP problem there exists another related unique LP problem involving the same data which also describes the original problem. The original or primal programme can be solved by transposing or reversing the rows and columns of the statement of the problem. Reversing the rows and columns in this way gives us the dual programme. Solution to dual programme problem can be found out in a similar manner as we use for solving the primal problem. Each LP maximising problem has its corresponding dual, a minimising problem. Also, each LP minimising problem has its corresponding dual, a maximising problem. This duality is an extremely important and interesting feature of Linear Programming Problems (LPP). Important facts of this property are : (a) The optimal solution of the dual gives complete information about the optimal solution of the rimal and viceversa. (b) !ometimes converting the LPP into dual and then solving it gives many advantages, for example, if the primal problem contains a large number of constraints in the form of rows and comparatively a lesser number of variables in the form of columns, the solution can be considerably simplified by converting the original problem into dual and then solving it. (c) Duality can provide us economic information useful to the management. Hence, it has certain far reaching consequences of economic nature, since it helps managers in decisionmaking.
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(d) It provides us information as to how the optimal solution changes due to the results of the changes in coefficient and formulation of the problem. This can be used for sensitivity analysis after optimality tests are carried out. (e) Duality indicates that there is a fairly close relationship between LP and Games Theory as it shows each LPP is equivalent to a twoperson zerosum game. (f) Dual of the dual is a primal. DUAL PROBLEMS WHEN PRIMAL IS IN THE STANDARD FORM We have already seen the characteristics of the standard form of LPP, let us recall them once again. These are : (a) All constraints are expressed in the form of equation, only the nonnegativity constraint is expressed as > = 0. (b) The right hand side of each constraint equation is nonnegative. (c) All the decision variables are nonnegative. (d) The objective function Z, is either to be maximised or minimized. Let us consider a general problem. The primal problem can be expressed as Maximize Z = C1 X1 + C2 X2 + Subject to an + X2 + an + a22 X2 +
+ C„ X„ + al n Xn < bi a2 n Xti < b2
a„, 1 X1 + a,„ 2 X2 +
+ a „,,, X„ < = b,„
=0 X2, ,X„= The dual can be expressed as follows: Minimize Subject to
Z* = an an
+ B2 Y2 + + an Y2 + + a22 Y2 +
ant 1 Yin > Cl
+ am 2 Yin > = C2
al „ Y1 + a2„ Y2 +
where Y1, Y2,
, + B,„ Y„
+ a„,„ Y„, > = C,„
Y2, Yin > 0 , Y,,, are the dual decision variables.
In general, standard form of the primal is defined as Maximize or Minimize Z =
Ci xi j=1,
Subject to
,17;1 xi =b1 1 j=i
i = 1,
Xi > = 0
j = 1, 2,
, nt
,n
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160
For constructing a dual of this standard form, let us arrange the coefficient of primal as X2
Xi
x„
C2
Cl
C„
a,1
au au
a,„
b,
az Left side coefficients { of dual constraints a.
 a22 azi
a,„,
b,
— ain2a,q an,
b,,
x,
Right side of dual constraints
Cl
.
*— Primal variables
4—
Y2
i
Dual variables
I
.th
j
I— A,/
dual constraint dual objective
It may be noted that dual is obtained symmetrically from the primal using the following rules: X„ are the primal (a) For every primal constraint, there is a dual variable, here X1, X2, constraints and Y1, Y2, Y„, are the dual variables. X„ are the primal variables. (b) For every primal variable, there is a dual constraint X1, X2, (c) The constraint coefficients of a primal variable form, the left side coefficients of the corresponding dual constraints, and the objective coefficient of the same variable becomes the right hand side of the dual constraint as shown above. The above rules indicate that the dual problem will have m variables (Y1, Y2, Y,„ ) and n X„ ). The sense of optimisation, type of constraints and the constraints (related with X1, X2, sign of dual variables, for the maximisation and minimisation types of standard form are given below. Standard Primal
Dual
Objective
Constraints
Variables
Objective
Maximization
Equations with
All Nonnegative
Minimization
>=
Unrestricted
Maximization
0, x2 unrestricted Formulate its corresponding dual. Solution. Dual Problem
Primal Problem Minimize
Maximize
Z= 10x1 + 20x2 + 15x3 + 12x4
Z= 100y, + 140y, + 50y3
Subject to
Subject to
x, + x2 + x, + x4
100
2x,— X3 + 3x4
140
4X2  2X4
50
Xi
Yi
Y2
1
2
Y3
•
10 20 15 12
a. 0 x2 unrestricted XI , X3, X4
Dual is Maximize Z = 100 yi + 140 y2 + 50 y3 Subject to yl + 2y2 + y3 5 10 yl + 4y3 = 20 yl — Y2 ~ 15
yl + 3y2 — 2y3 512 yi 0, y2 5.. 0, y3 unrestricted It has been seen earlier in the table comparing the primal and the dual that an equality constraint in one problem corresponds to an unrestricted variable in the other problem. An unrestricted variable can assume a value which is positive, negative or 0. Similarly, a problem may have nonpositive variables. (xi 5_ 0) Example 4.2. The following primal is given Maximize Z = 5x1 + 6x2 Subject to 3x1 + 2x2 _5120 4x1 + 6x2 .5260, x1, x2 >0
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Find the corresponding dual of the primal. Solution. Primal Problem
Dual Problem
Maximize
Minimize
z =I 5x + 6x2 Subject to
Z= 120y1 + 260y2
► 3x, + 2x2
♦ 120
► 4x, + 6x
260
Subject to 3 2 •
Yi
4 Y2
Yi
6 Y2
Yi
Y2
Hence the dual is Z = 120 yi + 260 y2 Subject to
3:1/1 4Y2 5 2y1 + 6y2 6
Yv Y2 Example 4.3. Convert the following primal into its corresponding dual. Maximize Z = 5x1 + 12x2 + 4x3 Subject to xi + 2x2 + x3 S10 2x1 — x2 + 3x3 = 8 x1, x2, x3 0 Solution. Primal Problem
Dual Problem
Maximize
• Z=
Minimize
5x, + 12x2 + 4x3 Z= 10y, + 8y2 Subject to
Subject to
• 2x2 + x
2x, — x2 + 3x3
10
1
•
Yi
2 Y2
2 3'1
—1 Y2
1 Yi
3 Y2
I Yi
Y2
12 4
OPERATIONS RTSEARCH
164 The dual is Minimize Z = 10 y1 + 8y2 subject to constraints + 2y2 ?_ 5, 2yi — y2 12, yi + 3y2 4.
Example 4.4. Given the following primal problem, formulate the corresponding dual problem. Minimize Z = 8x1 + 5x2 + 6x3 xi + x2 + X3= 25 Subject to 4x1 5x2 >10 x1 — x2 + 2x3 _s48 x2 .s 12 x1, x2 > 0, x3 — unrestricted Solution. Dual Problem
Primal Problem Minimize
Maximize
•
z= 8x1 + 5x2 + 6x3
=
Subject to
4, 25y, + 10y2 + 48y3
Subject to x, + x2 + x,
25
4x, — 5x2
10
4 —5 0
X1  x2 + 2x3
48
Yi
X1, X2 > 0 x3 unrestricted
Corresponding dual is Maximize Z = 25 yl + 10 y2 + 48 y3 Subject to 4Y2+ Y3 8 4y1 — 5y2 5_ 5 — Y2+ 21/3 6 yv y2 > 0 y3 unrestricted Example 4.5. Obtain the dual problem of the following LPP: f(x) = 2x1 + 5x2 + 6x3 Maximize Subject to the constraints 5x1 + 6x2 — x3 5 3 —2x1 + x2 + 4x3 5 4 x1 — 5x2 + 3x3 51 — 3x1 — 3x2 + 7x3 56 x1, x2, x3 .?0. and
•
2 2
y3
165
LINEAR PROGRAMMINGIII Solution. It may be seen that the given problem can be rewritten as Maximize F (x) = CX where X = (X1, X2, X3) have to be determined. If we denote 5 6 1 2 1 4 A= 1 5 3 3 3 7 3
b=
4 1 6
C= [2 5 6] Then the constraints are AX b;X?_0 Dual Problem
Let Y = [yi, y2, y3, y4 I be the dual variables, then the dual problem is to determine Y so as to Minimize f (Y) = (3, 4, 1, 6) [yi, y2, y3, y41 Subject to the constraints 5 6 1
2 1 4
1 5 3
3 3 7
YI
Y2
2
5 Y1,1/2, Y3, Y4 ° Y36 y4
or Minimize f(Y)= 3Yi Subject to the constraints
4Y2+ Y3 + 6Y4
3Y4 2 5Y1 2Y2 61/1 + Y2  5Y3 5  + 4y2 3y3 + 7y4 6 and Yi, Y2f Y3, Y4 ° Example 4.6. Obtain the dual problem of the following LPP: Maximize Z = x1  2x2 + 3x3 subject to the constraints  2x1 + x2 + 3x3 = 2 2x1 + 3x2 + 4x3 = 1 x1, x2, x3 _>0 Solution. This can be done conveniently by the following method: Z = 2:th + y3 subject to Qualls, Minimize
OPERATIONS RESEARCH
166 — 2yi + 2y2
1
3Y2 — 2 3yi + 4y3 ?.. 3 yv y2 are unrestricted in sign. Dual Problem
Primal Problem
Maximize z=
Minimize x1 — 2x2 + 3x3
Z= 2y, +y3 Subject to
Subject to
—2
2x1 + x2 + 3x3
1
3
3
4
—2 3
2x, + 3x2 + 4x3
1 Yi
As for equal to (=) constraint, the variable is unrestricted. Example 4.7. Write the dual of LPP. Z = 4x1 + 6x2 + 18x3 Minimize subject to the constraints x1 + 3x2 3 X2 + 2X3 5 >0( j = 1, 2, 3) x.— Solution. Dual Problem
Primal Problem
Minimize
Maximize
Z= 4x1 + 6x2 + 18x3 Z= 3y1
5y2
,
Subject to
Subject to x, + 3x
0
x2 + 2x,
2
—A—
Y2
The dual is, Maximize Z = 3y1 + 5y2 subject to the constraints yl < 4
3Y3 + Y2 6 2y2 18
Y2
0
18
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LINEAR PROGRAMMINGIII Example 4.8. Obtain the dual problem of the following primal problem.
Minimize Z = x1 — 3x2 — 2x3 Subject to the constraints 3x1 — x2 + 2x3 s. 7 2x1 — 4x2 212 — 4x1 + 3x2 + 8x3 = 10 x1, x2 2 0 and x3 is unrestricted Solution. Dual Problem
Primal Problem Minimize Maximize Z=
xi  3x2  2x3 Z= 25y, + 10y2 + 48y3
Subject to Subject to x, + x2 + x3 I
2
1
4
3
12 0
2  4x, + 3x2 + 8x3
10
Yi
A, 72
x„ x, 0 x3 unrestricted
Dual is Maximize
Z = 7yi + 12y2 + 10y3
Subject to the constraints 3yi + 2y2  4y3 1 
 4yi + 3y3  3 or yi + 4y2  3y3 3 2y, + 8y3 = 2
Example 4.9. Obtain the dual problem of the following primal problem:
Minimize Z = 600 x1 + 500 x2 Subject to the constraints 3x1 + x2 .?10 8x1 + x2 .,_>18 6x1 + 4x2 20 10x1 + 20x2 ..?30 x1, x2 .? 0
8 3'3
2
OPERATIONS RESEARCH
168 Solution.
Dual Problem
Primal Problem Minimize
Maximize
z= 600x1 + 500x2
Z= 10y, + 18y2 + 20y3 + 30y4
Subject to
Subject to 3x, + x2
10
8x, + x2
18
3
8
6
10
600
1 A
1 A
4
20
500
•
Y2
6x, + 4x2
20
10x, + 20x2
30
y3
34
1
Dual problem is Maximize Z =10y1 + 18y2 + 20y3 + 30y4 Subject to 3y1 + 8y2 + 6y3 + 10y4 5 600 + y2 + 4y3 20y4 5. 500 Ylf Y2/ 1/3, 1:/,1•• 0 Example 4.10. Write the dual of the following LPP: Minimize Z = 3X1 — 6x2 + 4x3 Subject to constraints 4x1 + 3x2 + 6r3 29 x1 + 2x2 + 3x3 26 6x1 — 2x2 — 2x3 xi — 2x2 + 6x3 24 2x1 + 5x2 3x3 26 xi, x2, x3 2 O.
Solution. Since objective function of the given problem is minimization, any 5. type constraints has to be changed. Hence rewriting the above problem. Minimize Z = 3x1 — 6x2 + 4x3 Subject to constraints 4x1 + 3x2 + 6x3 9 xl + 2x2 + 3x3 6 — 6xi + 2x2 + 2x3 — 10 xi — 2x2 + 6x3 4 2x1 + 5x2 — 3x3 6 x1, x2, x3 0
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169
Dual can now be formulated in the usual manner. Primal Problem
Minimize
Dual Problem
Maximize
Z, = 3x, — 6x, + 4x3
Zy = 9y, + 6y, — Oy, + 4y4 + 6y5
Subject to constraints 4x + 3x, + 6x
Subject to constraints 4
3
—6
2x2 3x3 6x, + 2x2 + 2x
—10
—2
2
6
♦
A Y2
x, — 2x
2
Y3
Y4
+ 3x
Dual can now be written as Maximize Z, = 9yi + 6y2 —10y3 + 4y4 + 6y5 Subject to the constraints 4yi + y2 — 6y3 + y4 + 2y5 3 3Yi
2Y2 2Y3 2Y4 5Ys 6 6y1 + 3y2 + 2y3 + 6y4 — 3y5 4
:1/2, 1./3f Y4f Y5 Example 4.11. Find the dual problem of the following primal LP problem: Z = 30x1 + 100x2 Minimize Subject to the constraints 2x1 + 3x2 59 3x1 + 6x2 = 120 10x1 + 20x2 800 x1, x2 _> 0 type of constraint has to be converted into 1?_.' type. Also = equal to type of constraint Solution. should be written as equivalent to two constraints of the type and The primal can be rewritten as Minimize Z, = 30x1 + 100x2 Subject to the constraints — 2x1 — 3x2 — 9
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OPERATIONS RESEARCH
and or
Dual is, Maximize or if we put with constraint
or and
3x1 +6x2 2:90 3x1 +6x 2 :::;90 3x1 6x2 2: 90 10x1  20x2 2: 800 X1, Xz 2: 0 Zy=  9y1 + 90y2 90y3  800y4 Yzy3 =y Zy =  9y1 + 90y  800y4
2y1 + 3y2 3y3  10y4 ::;; 30  3y1 +6y 2 6y320y4:::;100 2y1 + 3y10tj4:::;30 3y1 + 6y 20tj4:::;100
y1, y4 2:0 and y is unrestricted because in the second constraint of the primal there is a= sign, hence the second dual variable, i.e., y2 should also be unrestricted. Since we have used y = y2  y3 let y be unrestricted.
Example 4.12. Write the dual programme of the following LPP: Maximize Z= 2x 1 + 4x2 + 6x3 Subject to constraints x1 + x2 + 4x3 5'12 4x1  x2 + 3x3 cl5 x1 , x2 cO x3 is unrestricted variable. Solution. Here x� is unrestricted variable, it can assume positive, negative or 0 value x3 needs to be replaced J:,y difference of two nonnegative variables say (x31 x32 ) where x31 and x32 2: 0. Let the problem ,be re·stated as _ ·
LINEAR PROGRAMMINGIII
171
Maximize Z = 2x1 + 4x2 + 6 (x31 X32 ) Subject to constraints X1 + X2 + 4 (X31  X32 ) 5. 12 —4x1+x2 3 (x31  x32) 2 + Y2 4 4y1— 3y2 6, yi, y2 0 Dual Problem
Primal Problem Minimize
Maximize
z= 2x1 + 4x2 + 6(x3i — .X32)
Z= 12y, — 15y2
Subject to constraints + x2 + 4(x3 i — — 4x
x2 — 3
Subject to constraints 12 —15
1 1 4
—4 1 —3
Example 4.13. Write the dual of the following LP problem: Maximize Z = 20x1 + 12x2 + 16x3 + 10; Subject to 3x1 — 4x2 + 10x3 + 6x4 5'90 x1 + x2 + x3 = 36 — 2x2 + 4x3 + 6x4 >50 and x4 unrestricted in sign. Solution. When ever an unrestricted variable is provided in primal, it must be converted or expressed as difference of two nonnegative variables. i.e., x4 = X41  X42 where x41 and x42 0 The given problem can be rewritten as Maximize Z = 20x1 + 12x2 + 16x3 + 10 (x41 — x42 ) Subject to constraints 3x1 — 4x2 + 10x3 + 6 (x41 — x42) 5. 90 Xi + X2 + X3 36 or — — x2 — x3 5 — 36 + X2 + X3 36 — 2x2 + 4x3 + 6 (x41 — X42 ) ?_ 50 or 2x2 — 4x3 — 6 (x41 — x42) — 50 Now, the dual can be formulated as 3y1 + y2 + y3 + 2y4 ?_ 20 Dual is
172
OPERATIONS RESEARCH 4Y1  Y2  Y3 12 10yi — y2 + y3  4y4 6y1 — 6y4 ?. 10 Yi' y4 42, y3 unrestricted Primal Problem
Dual Problerr7
Maximize
Minimize
= 20x, + 12x2 + 16x3 + 10 (x41 —x42)
Z,, =I 90y, — 36y2 + 36Y3 50Y4
Subject to constraints 3x, 4x2 + 10x3 + 6(x4, — x42)
V
► x1
—36
►
x, +
x2
x3
+ x,
,[2,2 _4x3 + 6(x41  x42)
90
Subject to constraints
V
3
1
1
2
20
—4
—1
1
0
12
10
—1
1
—4
16
6
0
0
—6
10
36 —50
INTERPRETING PRIMALDUAL OPTIMAL SOLUTIONS As has been said earlier, the solution values of the primal can be read directly from the optimal solution table of the dual. The reverse of this also is true. The following two properties of primaldual should be understood. PrimalDual Property 1 If feasible solution exists for both primal and dual the problems, then both the problems have an optimal solution for which the objective function values are equal. A peripheral relationship is that, if one problem has an unbounded solution, its dual has no feasible solution. PrimalDual Property 2 The optimal values for decision variables in one problem are read from row (0) of the optimal table for the other problem. The following steps are involved in reading the solution values for the primal from the optimal solution table of the dual: Step I. The slacksurplus variables in the dual problem are associated with the basic variables of the primal in the optimal solution. Hence, these slacksurplus variables have to be identified in the dual problem. Step II. Optimal value of basic primal variables can be directly read from the elements in the index row corresponding to the columns of the slacksurplus variables with changed signs. Step III. Values of the slack variables of the primal can be read from the index row under the nonbasic variables of the dual solution with changed signs. Step IV. Value of the objective function is same for primal and dual problems.
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173
Example 4.14. Solve the following LPP by using its dual. Maximize Subject to
Z = 5x1 — 2x2 + 3x3 2x1 + 2x2 — x3 ..?2 3x1 — 4x2 _s"3 x2 + 3x3 —.0
Solution. The problem can be rewritten as
Maximize Subject to
Z = 5x1  2x2 + 3x3  2x1  2x2 + x3  2 (converting sign into by multiplying both sides of the equation by  1) 3x1  4x2 3 x2 + 3x3 5 xl, x2, x3 0 Primal Problem Minimize
Dual Problem
Maximize
5x1  2x2 + 3x3
Z=
5Y3
Subject to
Subject to
2x  2x2
Y1 + 31'2
0
5
4
1
2
0
3
3
X3
3x1  4x2 X2 + 3x2
The dual is: Minimize Subject to the constraints
Z =  2y1 + 3y2 + 5y3  2yi + 3y2 5  2y1  4y2 + y3  2 yi+ 3y3 >3 Yi, Y2/ Y3 °
Step I. Convert the minimisation into a maximisation problem. Maximize Z* = 2yi  3y2 + 5y3 Step II. Make RHS of constraints positive.
 2y1  4y2 + y3 2 is rewritten as 2y1 + 4y2  y3 < 2
174
OPERATIONS RESEARCH
Step III. Make the problem as N + S coordinates problem Maximize Z* = 2y1  3y2 + 5y3 + 0S1 + 0S2 + 0S3  MAi  MA3 Subject to  2y1 + 3y2  Si + A1 = 5 2y1 + 4y2  y3 + S2 = 2 yi + 3y3  S3 + A3 = 3 Yi, Y2f Y3, S1, S2, S3, A1, A3 0 Step IV. Make N coordinates assume 0 values. Putting y1= Y2 = Y3 = S1= S3 = 0. we get Al = 5, S2 = 2, A3 = 3 is the basic feasible solution. This can be represented in the table as follows: Initial Solution
C
%
2
3
5
0
0
0
M
M
S1
S2
S,
A,
A,
Minimum ratio
CB
Basic variables
Solution variables
yr
y2
y3
M
A,
5
2
3
0
1
0
0
1
0
0
S2
2
2
®
1
0
1
0
0
0
1 .>
M
A,
3
1
0 3  3M 3M
0 M
0 0
1 0 M M
1
co
0
M
Z,
M
(C,  Z)
2M
+3M +3M M
5
M
0
0
Step V. C.  Z1 is positive under some columns, it is not the optimal solution. Perform the optimality test. Step VI. Write second, third or fourth Simplex table unless you come to the optimal solution. This has been provided in the table below: Optimal Solution
2
3
5
0
0
0
M
M
Solution variable
yi
Y2
y3
SI
S2
S3
Al
A3
S3
11
15
0
0
4
3
1
4
1
3
Y2
3
1
0
1 3
0
0
5
y3
0
1
3
1
0
3
5
5
0
0
0
5
0
CI Basic variables
0
CB
5
14
3 4
(CI  ZI)

2
14
T 76 — 3  70
4
23 — 3 23
T
1
3 4

3 23 — 3 23  m T
0 0 0
M
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175
Since all the values in (C1  Z1 ) are negative, this is the optimal solution. = (:// Y2 =
5
14
3, Y3 = 3
Z*min = — Zmax = —
85 ) 85 =— 3 3
DUAL SIMPLEX METHOD The basic difference between the regular Simplex Method and the Dual Simplex Method is that whereas the regular Simplex Method starts with basic feasible solution, which is not optimal and it works towards optimality, the dual Simplex Method starts with an infeasible solution which is optimal and works towards feasibility. The following steps are involved in this method: Step I. Convert the problem into a maximization problem, if initially it is a minimization problem. Step II. If there are any ?_ type constraints, these must be converted into type constraints by multiplying both sides by  1. Step III. Obtain the initial basic solution  For this, the inequality constraints have to be converted into equality by adding slack variables. Make a dual Simplex table by putting this information in the form of a table. Step IV. Compute Ci  Z1 for each column. (a) If all Ci  Z1 are negative or zero and all solution values are nonnegative, the solution found above is the optimum basic feasible solution. (b) If all Ci  Zj are negative and zero and at least one value of 'solution value' is negative then proceed to next step, i . e . , step V. (c) If any CI  Z1 is positive, this method cannot be applied. Step V. Select the row that contains the most negative 'solution value'. This row is called the key row or the pivot row. The corresponding basic variable leaves the current solution. Step VI. Scrutinise the elements of the key row. (a) If all elements are nonnegative, the problem does not have a feasible solution. (b) If at least one element is negative, find the ratios of corresponding elements of CI  Z1 row to these elements, ignoring the ratios associated with positive or zero elements of the key row. Select the smallest of these rows. The corresponding column is the key column and the associated variable is the entering variable. Mark (circle) the key element or pivot element. Step VII. Make the key element unity (1). Carryout the row operations as is done in the regular Simplex Method and repeat until. (a) An optimal feasible solution is obtained in a finite number of steps or, (b) An indication of nonexistence of feasible solution is found.
OPERATIONS RESEARCH
176
Example 4.15. Use dual simplex method to: Maximize Z = 3x1  2x2 Subject to x1 + x2 >1 x1 + x2 10 x2 x p x2 ,>0 Solution. The given problem may be put in the form as Z =  3x1  2x2 Maximize Subject to   x2 1 x1 + x2 7   2x2  10 x2 Basic variables
Solution variable
3
2
xi 1
2 1 2 1 2
(C)  ZI) CI  Z I Corresponding element in Key row
x2
0 Si
S2
0
1
0
0
0
1
1
0
0
0
0 S4
0
0 0 0
key row 0 0
key column Key row is marked with the arrow on the right side >
Selecting the smaller of ratio value =
C./  Z./ Corresponding element in Key row
Key column is marked is in the table. 1 Key element  is marked with a circle 2
in the table.
1 Row S4 is to be replaced by x1. All elements of S4 are divided by the key element (  out 2 these are the elements of new x1 row, i.e., 4, 1, 0, 0, 0,  1,  2. New values of Si row 1 Solution variable = 4  ( x 4) = 6 2 Element x1 = 2 ( 1 x 1) = 0 2 x2 = 0 
1 x=0 2
LINEAR PROGRAMMINGIII
179 = 1 — (1 x01 =1 S2 = 0 — ( X 0) = 0 2 1 2
1 2 x —1) — — 1
S4 = 0 — (— — 1 X —2) = — 1 2 The values are: 6, 0, 0, 1, 0, — 1, — 1. New values of S2 row ll Solution variable = 2 — (1 x 4) = 0 2 Element x1 = — (21 X 1) = 0 X 2 = 0 — 121 X 0) = 0 Si = 0 — (1 X 0) = 0 2 S2 = 1 — (2 X 0) = 1 S3 =
— (1 X —1) = 1 2
S4 = 0 — (1 X —2) =1 2 So, the new values are: 0, 0, 0, 0, 1, 1, 1. New values of x2 row are: 1 Solution variable = 5— 1 — x 4) = 3 2 Element xi
— (1 x 1) = 0 = 2 2
1 x2 = 1 — ( x = 1 2 S1 = 0 — (1 x 0) =0
2
S2 = 0 — (1 X 0) =0 2
180
_
OPERATIONS RESEARCH S3 =
 (1 x 1) = 0 2 1 S4 = 0  „, X  2) =1 The new values are 3, 0, 1, 0, 0, 0, 1 These values can be placed in the table to determine whether it turns out to be a feasible solution or not. C,
0 0 2 3
C3 ›
3
2
0
0
0
0
Basic Solution variables variable S, 6 S2 0 x2 3 x, 4 Zi 18 (C,  Z1)
x,
x2
S,
S2
S3
S4
0 0 0 1 3 0
0 0 1 0 2 0
1 0 0 0 0 0
0 1 0 0 0 0
1 1 0 1 3 3
1 1 1 2 4 4
Since all the Ci  Zi values are either zero or negative, this is the optimal feasible solution. x1 = 4, X2 = 3 Zmax =  3 x 4  2 x 3 =  18 Example 4.16. Solve by dual simplex method the following problem: Minimize Z = 2x1 + 2x2 + 4x3 Subject to 2x1 + 3x2 + 5x3 3x1 + x2 + 7x3 s:"3 xi + 4x2 + 6x3 x1, x2, x3 _?.0 Solution. Step I. Convert the minimizing problem into maximising problem Maximize  Z* = K =  2x1  2x2  4x3 Step II. The, constraint must be converted into 5_ by multiplying both sides of the first constraint by  1.  2x1  3x2  5x3 2 Step III. Add slack variables S1, S2, S3 in the constraints. The problem can now be rewritten as Maximize K=2x1 2x2  4x3 + S1 + 0S2 + 0S3 Subject to the constraints  2x1  3x2  5x3 + 51 =  2 3x1 + X2 1 7X3 + S2 = 3 + 4X2 6X3 + S3 = 5 where x1, x2, x3, Si, S2 and S3 0
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181
Putting x1 = x2 = x3 = 0 the vertical basic infeasible solution is Si =  2, S2 = 3, S3 = 5. This information is written in the following table: First Infeasible Solution
0
2 2 4
C, >
Cf
0
Basic variables Solution variable xi x2 x3 Si Si 2 2 0 5 1
0 0
3 5
S2 S3 4
3 1 0
0 (c;— 4)
1 4 0
7 6 o
—2 2 4
0
0
S2 S3 0 0
0 0 o
1 0 o
0 1 o
0
0
0
key row
key column Step IV. Since C1  Z1 are either 0 or negative and solution variable is negative the solution is optimal but infeasible. Hence we proceed to step V. Step V. The row containing the most negative solution variable is S1, hence it is the key row and S1 is the row to be replaced. Step VI. To find out the key element. Ratio of
Cl Zi Corresponding element in Key row
These ratios =
2
2 4 0 0 0 24 5 2' 3' 5' 1' 0' 0 or 1, 3'
Since we have to neglect element corresponding to 0 or positive values. 24 2 We have 1,  ,  ratios only since 3is the smallest ratio x2 is the incoming variable 3 5 replacing Si. and key element is  3 circled in the table. Step VII. Make the key element unity and find the new values of S2 and S3 by using the following relationship: New element = Old row element  [(Intersectional element in old row) x (Corresponding element in replacing row)] By making key element unity the new values of row S1 (to be replaced by x2) are 22 5 1 , 3,1,  , 0, 0. 3 3, 3 New values of row S2 New solution value = 3 (1 x 2)  = 73 3 Element x1 = 3  (1 x 2)  = 73 3
OPERATIONS RESEARCH
182 Element x2 = 1  (1 x 1) = 0 5) Element x3 = 7  (1 x 3 3 S1 = 0  11 X 72 = 3)3 S2 = 1  (1 X 0) = 1 S3 = 0  (1 X 0) = 0 7 7 16 1 So, new values are:  ,  , 0, — ,  ,1, 0 3 3 3 3 New values of row S3 New solution value = 5  (4 x 2)  = 73 3 Element x1 = 1  14 x
3) 3 x2 = 4  (4 x 1) = 0  =  2x3 = 6  (4 x 5) 3 3
4 = 0  (4 x )= 3 3 S2 = 0  (4 x 0) = 0 S3 = 1  (4 x 0) = 1 7 5 2 4 So the new values are 0— 0,1 3 3 3 3 Step VIII. Putting these values in the table below, we get C,
2
C.> Solution Basic variables variable 2 x2
0
S2
0
S, Z.
(C, Z)
2
2
4
0
0
0
x,
x2
x,
S1
S2
S,
0
0
1
0
0
1
0
0
0
0
2
3
3
7 i 7 34 3
7 5 3 4 3 2 3
1 0 0
2
0
5

1
16
1
2
4 3 2 32 3
3 10 3 2 3
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183
Step IX. Scrutinise CI — Z1 values and solution values. Since all C. — Z1 values or either negative or zero and all solution values are positive, this is the feasible optimal solution. x1 = 0 2 = — x2 3 X3 = 0
4 2 K=—2x0—2x— — 4 x 0 = 3 3 4 Z* = — 3
Maximum or Minimum
Example 4.17. Solve the following LPP by using its dual: Maximize
Z = 5x1 — 2x2 + 3x3
Subject to
2x1 + 2x2 — x3 22 3x1 — 4x2 5:3 x1 + x2 + 3x3 5:5 x1, x2, x3 20
Solution. constraints must be converted into 5 by multiplying both sides by — 1, the problem can be rewritten as Maximize Z = 5xi — 2x2 + 3x3 Subject to — 2x1 — 2x2 + x3 2 3x1 — 4x2 3 x2 +3x3 — 0 Primal Problem Maximize • 5x, — 2x2 + 3x3
Z=
Dual Problem Minimize Z* = —2y, + 3y2 + 5y3 Subject to
Subject to
• 2x„ — 2x2 + x, •
—2
3
0
5
—2
—4
1
—2
1
0
3
3
3x, — 4x2
—A— —A— Yi
x2 + 3x3
The dual is Minimize Subject to —
Z** = — 2y1 + 42 + 5y3 — 2y1 + 3y2 5 4y2 + y3 — 2
Y2
Y3
OPERATIONS RESEARCH
184
3Y3 3 Y1, Y2, Y3 ° Step I. Convert the minimisation problem into a maximization problem Minimize Z* =  2y1 + 3y2 + 5y3 or Maximize Z* = 2y1  3y2  5y3 Step II. Make RHS of the constraints positive The second constraint becomes 2:th + 4y2  y3 5 2 Step III. Make the problem as N + S coordinate problem. Introducing slack and artificial variables Z** = 2y1  3y2  5y3 + 0S1 + 0S2 + 0S3  MAi  MA3 Maximize Subject to  2y1  3y2  S1 + A1 = 5 2y1 + 4y2  y3 + S2 = 2 YI 3Y3 S3 + A3 = 3 Y2, Y3, Si, S2, S3, Ai, A3 Step IV. Let N coordinate assume 0 values
yl = Y2 = y3 = SI = S3 = 0; Al = 5, S2 = 2, A3 = 3 is the Basic Feasible Solution. This can be written as shown in the table below: 2 Basic Solution variables values
5 I 0
Minimum ratio
0
YI
Y2
Y3
S
S2
S3
A1
A3
M
Al
5
2
3
0
1
0
0
1
0
0
S2
2
2
®
1
0
1
0
0
0
M
A3
3
1
0
3
0
0
1
0
1
Z1
8M
M
M
0
M
M
0
M
(C)  Z1)
3M 3M
2M
+3M +3M
5 3 2
key row
00
M M 0
0
key column S2 is the key row, y2 is the key column and (4) is the key element S2 is the departing row and y2 is the entering variable. Step V. Optimality Test: Ci  Zi is positive in same column, the solution is not optimal. New elements of y2 are obtained by dividing the whole row by 4.
1 1 . 1 1 i. e., , , 1,   , 0, ,0, 0,0 4 4 2 2 New elements of A, row are obtained using the relationship
185
LINEAR PROGRAMMINGIII
New element = Old row element  [(Intersectional element in old row) x (Corresponding element in replacing row)] Solution value = 5 I3 x 11= — 13 2 2 Element y1 =  2  I3 x 11=   12 2 y2 =  3  [ 3 x 1] = 0 Element
y3 = 0  [3x1=34 4 =  1  [ 3 x 0] =  1
] =3 S2 = 0  [3 x I
4 4 S3 = 0  [ 3 x 0] = 0 = 1  [ 3 x 0] = 1 A3 = 0  [ 3 x 0] = 0 13 1 3 3 Al row new elements are 2,  0, , 1, .0,1, 0 4 New elements of A3 row are 1 Solution value = 3  [0 x 1 = 3 2 Element yi  1, y2  0, y3  3, Si  0, S2 = 0, S3 =  1, Iterating thrice we get
= 0, A3 =
1•
Final Simplex Table C1 *
2
3
5
0
0
0
M
M
Basic variables
Solution values
yi
Y2
y3
Si
S2
53
Al
A3
0
S3
1
 15
0
0
4
3
1
4
1
3
Y2
5
y3
5
1
2 14 3
Z1
(c,  z,)
1
_ 41
_ 14 3
0
1
76 — 3
3
5
_ 70 3
0
0
1 00 0 4 _ 4 — 3
23 23 3
1
0
5
0
5
0
4 3 23 3
_ 23 3
_ M
0 0 0 M
Optimality Test
The above is the optimal basic feasible solution, since all values of Ci  Zi are either 0 or negative. 14 The optimal solution is yi = 0, y2 = 3 , y3 = — . Substituting Zmin. =  2yi + 3y2 + 5y3 3 85 ) 85 Z*nn„. = = —3 3
OPERATIONS RESEARCH
186 REVISED SIMPLEX METHOD
We have seen how complex and cumbersome the iterating process in the Simplex Method can be. Every time a variable departs and new variable enters, complicated calculations requiring large space (and computer memory) have to be performed to get the new element values. Successive iteration are done by using number of row operations as has been discussed earlier. If a LPP with large number of variables and constraints (which is in real life problem the solution will be) has to be solved by the Simplex Method, it will need a lot of time and computer space as all the tables and data will have to be stored and retrieved again and again. However, there are methods available where the entire table does not have to be calculated during every iteration. The information required while constructing a new table is (a) C.  Zj value, the key column and the current basic variables and their values (or the solution value constants). This information can be directly obtained by using certain properties of the matrix. The following methods have been developed which make the computations much easier and simpler: (a) The revised Simplex Method (b) The bounded variables method, and (c) The decomposition method. The student is advised to go through the basics of matrix operation as given in the appendix of this book. Example 4.18. Maximize Z = 5x1 + 2x2 + 3x3  2x4 + x5 Subject to 3x1 + 2x2 + 3x3 + x4 =12 x1 + 3x2 + x3 + x5 = 10 x1, x2, x3, x4, x5 20 Solution. To find the basic feasible solution (nonoptimal), put x1 = x2 = x3 = 0. x4 = 12, x5 = 10. This is shown in the Initial Basic Feasible Solution below:
5 2 3 2 1
2 1
Basic variables
Solution values
Xi
X2
X3
X4
X5
Minimum ratio
x4
12
3
2
3
1
0
4
x5
5
®
3
1
0
1
5 — 3
Z1
19
5
1
5
2
1
19 5
10
3
8
0
0
(CI  ZI)
key row
key column x5 is the key row, x1 is the key column and (1) is the key element. New value of x5 element which is to be replaced by x1 variables are: 5, 1, 3, 1, 0, 1. New values of x4 element are: Solution value = 12,13 x,51 =  3
LINEAR PROGRAMMINGIII
187
element xi = 3  [3 x 1] = 0 element x2 = 2  [3 x 3] =  7 element x3 = 3  [3 x 1] = 0 element x4 = 1 [3 x 0] = 1 element x5 =0  [3 x 1] =  3 i.e.,  3, 0,  7, 0, 1,  3 Second feasible solution can be written as follows:
C,
5 2 3 2 1 Basic variables
Solution values
X1
X2
x3
x4
x5
2
x4
3
0
7
0
1
3
5
xi
5
1
3
1
0
1
Z1
31
5 0
(C1 Z)
29 5 2 27 2 0
Minimum ratio 3 7 5 3
11 10
This is also the optimal solution as all values of CI  Zi are either 0 or negative. = 5, X2 = X3 = X5 = 0, X4 = — 2, Zmax. = 31 The revised simplex method uses matrixvector operations to directly obtain the basic feasible solution from the original equations. We denote the column vector P1, P2, P3, P4, P5 to represent x1, x2, x3, x4 and x5 and let the column vector b represent the right hand side constraints i.e., solution values. From the original equation of constraints. P1= [3] = [2] = [2] 411 , r2 3 , r3 , r4 0 r5 The table giving the basic feasible solution (nonoptimal) can be generated directly as follows: The initial basic matrix is [1 01 iP4, Ps = 0 1 = I (Identity matrix) It is possible to calculate the new column coefficient corresponding to x4 and x5. MI SOME MORE DUALITY PROBLEMS
Example 4.19. Solve the following LPP with the help of Duality: Minimize Z = 2x1 + 2x2 Subject to 2x1 + 4x2 21 xi + 2x2 2x1 + x2 21 Solution. The dual of the given problem is Maximize Z=n+b+c 2a b 2 Subject to 4a 2b c 2.
OPERATIONS RESEARCH
188
As the problem is now of maximisation, we will introduce slack variables. Z=a +b+c+ 0S1 +0S2 + 0S3 Maximise 2a + b + c + S1 + 0S2 = 2 Subject to 4a + b + c + OSi + S2 = 2 The Simplex table can be written as follows: Cl 1
1
› Basic variables
Solution values
0
Si
0
52
2 2 0
4 (Ci — 4)
a
b
c
Si
S2
0 4 0 1
1 2 0 1
2 1 0 1
1 0 0 0
0 1 0 0
Minimum ratio
key row
1
key column Since all the values of (C1  Z1) are equal, i.e.,1,1, 1 we can take any one as the entering variable is the key element. by taking it as key column. Let us take column C. Si is the key row and New value of C row (old Si ) will be obtained by dividing by key element, i.e., 2. 1 1 So, the new elements are: 1, 1,  , 1,  , 0. 2 2 New value of S2 are obtained by the relationship already known. 1 3 21x1=1,41x1=3,21x  =,11x1=0 2 1 1 01x =2,11x0=1 2 Now the new Simplex table can be written Cl 4, 1
Basic variables
Solution values
a
b
c
S1
Sz
Minimum ratio
C
1
1
1 2
1
1 —
0
2
Sz
1
3
@
0
4
1
1
(CI — ZO
0
12 1
2
—
2
1
1
1 2
0
0
0
0
key column
3
key row
189
LINEAR PROGRAMMINGIII Row S2 will be replaced by column b, the key element is
. New elements of row b (S2 )
3 values by  . These are 3, 2, 1, 0, —, 2 3 3 New value of row C is obtained by the relationship already known.
will be obtained by dividing old
S2
1 2 2 1 1 1 x = 11x2=0,   x1=0, 1 x0=1 2 3 3 2 2 2 2 1 _ 1 _ 1 _ 2 0 _ 1 2 .1 x 2 2 3 3 2x3 3 Now the new Simplex table can be written as
1
Cl
Basic variables
1
C
1
B
1
1
1
0
0
a
b
c
Si
S2
0
0
1
2
1 3
2 — 3
2
1
0
_ — 31
2 3
4
2
1
1
1
1
—1
0
0
1 3
1

6
9

—
Solution values 2 —3
Z1
(c,  4) C1 _ Z1 S2
3
All the elements of Cl  Zi row are either 0 or negative. Optimum solution is achieved. 2 2 Solution is a = 0, b = , c = 3 3 4 Maximize Z =  . 3 Example 4.20. Minimize Z = 4x1 + 6x2 + 18x3 Subject to constraints x l + 3x3 >3 x2 + 2x3 x1, x2, x3 ,>0. by dual Simplex Method. Solution. Multiply the objective function (Z) and the constraints by  1 and introduce slack variables in order to convert it into equalities. As there are two constraints, two slack variables will be introduced.
OPERATIONS RESEARCH
190 Maximize Z =  4x1  6x2  18x3 + OS1 + 0S2 Subject to constraints  x1 + 0x2  3x3 + S1 + 0S2 =  3 0x1  x2  2x3 + OSi + S2 =  5 x1, X2, X3, S1, S2 The first simplex table can be constructed as follows. >
C;
4
6
18
0
0
Basic variables
Solution values
X1
X2
X3
S1
S2
0
Si
3
1
0
3
1
0
0
S2
5
0
ci),
_2
0
1
Zi
0
0
0
0
0
0
4
6
18
0
0

6
2


(C)  Z1) Ratio
C1 Z / / S2
key row
key column • C.Z. Since most negative value is in row S2 and the minimum ratio of " is for x2, it will replace row S2 . New elements of x2 (old row S2 ) will be obtained by dividing its existing elements by key element, i.e.,  1. New elements are: 5, 0, 1, 2, 0,  1 New elements of row S1 are obtained by the relationship already known. These are  3  0 x 5 =  3. All these elements will remain the same. Now new Simplex table can be written as follows: 4
6
18
0
0
x,
S,
S2
1 0
0 1
1
Basic variables
Solution values
x,
x,
0
S, x,
3 5
1 0
0O3 1 2
Z.
30
0
6
12
0
6
4
0
6
0
6
4

2


6
(C  Z,) CZ Ratio ) I Si
key column
key row
LINEAR PROGRAMMINGIll
191
C•  Z• " ratio is for column x3 hence row Si S1 will be replaced by column x3 and  3 is the key element. Most negative values occur in S1 row and minimum
1 1 New row x3 (old S1 ) will be obtained by dividing old S1 row by  3, i.e., 1,  ,0,1, 3 , 0. 3 New x2 row will be obtained by the relationship already known 2 ,1 _ 2 x 0 = 1, 2  2 x 1 = 0, 5  2 (1) = 3, 0  2 x 1 =3 3 1 2 0  2 x  = ,  1  2 x 0 =  1. 3  3Now the new simplex table can be written as follows:
c,
›
4
6
18
0
0
Xi
X2
X3
S1
S2
0
1
31
0
Basic variables
Solution values
18
X3
1
6
X2
3
— 32
1
0
4
36
2 2
6 0
18 0
(C,  4)
1
3 2 2
1 6 6
There is no negative value in the solution value, the optimal solution has been obtained. x1 = 0, x2 = 3, x3 = 1, Zi =  36. Example 4.21. Solve the following LPP by Dual Simplex: Z = 6x1 + 7x2 + 3x3 + 5x4 Minimize Subject to 5x1 + 6x2  3x3 + 4x4 x2 +5x3 6x4 10 2x1 + 5x2 + x3 + x4 .?8 x1, x2, x3, x4 Solution. To convert it into a maximization problem, the objective function and constraints will be multiplied by  1. Maximize Z =  6x1  7x2  3x3  5x4 12 Subject to  5x1  6x2 + 3x3  4x4  x2  5x3 + 6x4  10  2x1  5x2  X3  X4  8 x2, x3, x4 >_0 Introducing slack variables, the objective function and constraints can be written as Maximize Z =  6X1  7X 2  3X 3  5X 4 + OSi 0S2 + 053  5X1  6X2 + 3X3  4X4 + + 0S2 + 0S3 =  12 Subject to
OPERATIONS RESEARCH s.
192 — X2 — 5X3 + 6X4 + OSi + S2 + 0S3 = —10 — 2X1 — 5X2 — X3 — X4 + OSi + 0S2 + S3 = — 8 X1, X2, X3, X4, S1, S2, S3 0
and The first Simplex table can be written as follows. —>
C,
6
7
3
5
0
0
0
Basic variables
Solution values
X,
X,
X3
X,
S,
S,
S3
0
S,
 12
5
0
3
4
1
0
0
0 0
S2 S3 Z,
10 8 0
0 2 0 6
1 5 0 7
5 1 0 3
+6 1 0 5
0 0 0 0
1 0 0 0
0 1 0 0
6 5
7 6
1
—5 4
0
0
0
(C,  Z) C.,  Z I Si
key column Since least positive value is for column X3 and most negative value is for Si ( 12) X2 will replace Si and  6 is the key element. Row Si old or new row X2 is obtained by dividing all 1 2 1 5 elements by  6, i.e., the new elements are: 2,  , 1, —  — , 0, 0. 6 2 3 6 New row S2 is obtained by the relationship we know 10(1)x2=8,0(1)x
6
6' 11 1(1)x1=0,5(1)x 1 = 2 2 2 20 1 1 6(1)x  =,0(1)x = 3 3 6 6' 1(1)x 0 =1,0(1)x0= 0 11 20 1 i.e., the new row S2 values are  8, 5, 0, , , 1, 0. 6 2 3 6 New row S3 values are: 13  8  ( 5) x 2 = 2,  2  ( 5) x 5 = —  5  ( 5) x 1 = 0,  1  ( 5) x 1= 6 6 2 2 2 1 =5 , 0 _ (_ 5) x 0 = 0, 1  ( 5) x 0 = 1.  1  ( 5) x  = 3 , 0  ( 5) x 3 3 3 7 7 5 i.e., the new row S3 values are 2, L , 0, , 0, 1. 6 2 3 6
LINEAR PROGRAMMINGIII
193
The new Simplex table can be written as follows. >
6
7
3
5
0
0
0
X,
X,
X,
X4
S,
S,
S3
1
1 2
2
7
0
0
0
ei)
T
20
7
1
0
0
_7 2
7 3
5 6
0
1
0
7 2
 14 3
7
0
0
0
0
0
0
Basic variables
Solution values
7
X,
2
0
S,
8
0
S
2
13 6
Z
14
 35 6
u 5
1
_1
(C,  Z)
1 6
n 0
 13 2
1 T
_7
C1 Z1
_1
0
S2
5
13 11
1 20
7
key row
key column Since least positive value is for column X3, it will replace the most negative value row i.e. S2
(11) . 2
is the key element. 11 New element of S2 or replacing column Nfareobtained by dividing old elements by 2 , the 40 1 2 16 5 0, 1, „ , O. 11 33 33 33 11 New values of row X1 can be obtained by using the relationship we already know
new elements are :
2
1)x0=1 , 2)
16 = 30 5 (_ 1.)x 5 .25 1 2 ) 11 11 ' 6 2) 33 33' x
1 2_ _1 )x 40 2 3 ( 2) 33 33' 6
1) „ 2) 33
0
33
1) x 2) 11
1
1)1_0
2
2)
11
0 ()x 0=0 2 2 5 1 30 25 So the new values are — , — ,1, 0, — , , — , 0 11 33 33 33 11 New values of Row S3 are also obtained in the similar manner.
OPERATIONS RESEARCH
194
7) „16 _ 78 13
2
7
2 ) 11 11
7 10 2 ) 33
7 3
5 _ 18 nx 1=0 0 (7 )x 0 =0, 2 2 2 33 11
21 5 (7) 1 x 11 6 2 ) 33
8
(7)> S1 i.e., the demand is more than supply, then set X11 = Sl and proceed vertically. Step II. Proceed in this manner, step by step till a value is allotted to SouthEast right bottom corner. The NorthWest corner rule can be best demonstrated by the example in hand. 1. Set X11 = 1000, i.e., the smaller of the amount available at S1(1000) and that needed at D1 (2300). 2. Proceed to cell (BX) as per rule (c) above which demands that you should proceed vertically. If Di > S1 compare the quantity available at S2 (1500) with the amount required. Quantity available at D1(2300 — 1000 = 1300) and set X21 = 1300. 3. Proceed to cell BY (rule above) as now D < S. Here S2 is 1500 and the demand is 1400. So, set X22 =1400. We are required to proceed horizontally to next cell. Since there is no other horizontal cell, the allocation ends here. The transportation cost associated with the solution is Z = 2000 x 1000 + 2500 x 1300 + 2700 x 1400. = 2000000 + 3250000 + 3780000 = 9030000. ROWMINIMA METHOD In this method, we allocate maximum possible in the lowest cost cell of the first row. The idea is to exhaust either the capacity of the first source or the demand at destination centre is satisfied or both. Continue the process for the other reduced transportation costs until all the supply and demand conditions are satisfied. In the above problem, we first allot in cell AX of first row as it has the lowest cost of Rs. 2000. So, we allocate minimum out of (1000, 2200), i.e, 1000. This exhausts the supply capacity of plant A and thus the first row is crossed off. The next allocation is in cell BX as the minimum cost in row 2 is in this cell. We allocate minimum of (1500, 1300), i.e., 1300 in this cell. This exhausts the demand requirements of destination centre X and so column 1 is crossed off.
204
OPERATIONS RESEARCH
C
Distribution Centres X Y Rs. 5380 Rs. 2000 1000 Rs. 2700 Rs. 2500 1300 Rs. 1700 Rs. 2550
Demand
2300
A Plants
B
Supply 1000 200
1500
1200 1400
1200 3700
Now, we proceed to rote No. 3 in which the minimum cost Rs. 1700 is in cell CY. Here we allot minimum out of 1400 and 1200.Since the demand of distribution centres is 1400 and we have allotted only 1200 we allot 200 in cell BY. Now, column Y is satisfied and we cross out column Y. Also since in row two the complete supply of 1500 is satisfied (1300 + 200 = 1500) row two is also satisfied and can be crossed out. Similarly, row three is also satisfied and can be crossed out. Hence Z =Rs. (2000 x 1000 + 2500 x 1300 + 2700 x 200 + 1700 x 1200) = Rs. 7830000. COLUMN MINIMA METHOD In this method, we start with the first column and allocate as much as possible in the lowest cost cell of this column, so that either the demand of the first destination centre is satisfied or the cape ity of the second plant is exhausted or both. There are three cases : (a) If the demand of first distribution centre is satisfied, cross off the first column and move to the second column on the right. (b) If the supply (capacity) of the ith plant is satisfied, cross off the ith row and reconsider, the first column with the remaining demand. (c) If the demand (requirement) of the first distribution centre as also the capacity of ith plant are completely satisfied, make a zero allotment in the second lowest cost cell of the first column. Cross off the column as well as the ith row and move to the second column. Continue the process for the resulting reduced transportation table till all the conditions are satisfied. The matrix below shows the solution with this method which is sinfilar to Row Minima method
A Plants
Distribution Centres X Y Rs. 2000 Rs. 5380 1000 Rs. 2500 Rs. 2700 1300 Rs. 2550 Rs. 1700 _ _.
Supply 1000 200
1500
1200
1200 3700
205
THE TRANSPORTATION PROBLEMS
Let us solve the given problem with the help of this method. Lowest cost cell in the column is AX. We allocate minimum, i.e., 1000 out of 2300, 1000. With this the capacity of plant A is exhausted and thus row one is crossed off. The next allocation is made in cell BX as it now has the minimum, cost of Rs. 2500 in the first column. We allocate minimum 1300 in this cell. Now, the demand of distribution centre X is satisfied we can cross the first column. Now, we move to the second column in this minimum cost cell is CY. Allocate 1200 in this cell out of 1400 and 1200. Consider the next least cost cell in this column which is BY in which we can allot only 200. Now all the conditions are satisfied. Transporation cost associated with this solution is Z = Rs. (2000 x 1000 + 2500 x 1300 + 1700 x 1200 + 2700 x 200) = Rs. 7830000 which is same as obtained with solution by rowminima method. LEAST COST METHOD In this method, we allocate as much as possible in the lowest cost cell or cells and then move to the next lowest cost cell/cells and so on. Let us solve the above problem using the least cost method. Distribution Centres Supply X Y Rs. 2000 1000 Rs. 5380 X11 = 1000 X12 = 0 B Rs. 2500 Rs. 2700 1500 Plants X21 = 1300 Xn = 200 Rs. 2550 Rs. 1700 C X31 = 0 1200 X32 = 1200 Demand 2300 1400 Here the lowest cost cell is CY (Rs. 1700) and maximum possible allocation, meeting supply and demand requirement is made here, i.e., 1200. This meets the supply position of row 3 and hence it is crossed out. The next least cost cell is AX (2000). Maximum possible allocation of 1000 is made here and row one is crossed out. Next lowest cost cell is BX (2500) and maximum possible allocation of 1300 is made here as the total demand in column X is 2300 and we have already allocated 1000 in cell AX. Next lowest cost cell is CX (2550) only 0 can be allocated here to meet the demand (2300) and supply (1200) position. Next cell with lowest cost is BY (2700). Here allocation of 200 is possible. The next lowest cost cell is AY where only 0 allocation is possible. Hence, Z = Rs. (2000 x 1000 + 2500 x 1300 + 2700 x 200 + 1700 x 1200) = Rs. 7830000. A
OPERATIONS RESEARCH
206 VOGEL'S APPROXIMATION METHOD (VAM)
This method usually provides a better initial (starting) solution than the methods described already. In fact, VAM generally, yields an optimum or very close to optimum starting solution. This method takes into account not only the lest cost Ci) but also the costs that just exceeds cif . The following steps are involved in this method. Step I. Write down the cost matrix as shown below.
Plants
Distribution centres Rs. 2000 X12 Rs. 2500 X22 Rs. 2550 X32
X11 X21 X23
A B C
Rs. 5380 Rs. 2700 Rs. 1700
Supply 1000 (3380) 1500 (200) 1200 (850)
1400 (1000)
Demand 2300 (500)
Find out the difference between the smallest and second smallest cost elements in each column and write it below the column in brackets, i.e., in column X the difference is 500 and in the second column it is 1000. Find out the difference between the smallest and second smallest cost elements in each row and write it on the right side of each row in.brackets, i.e., in row A 3380, in row B 200 and in row C 850. It may be noted that the 'difference,' which is indicated under columns or rows actually indicates the unit penalty incurred by failing to make an allocation to the least cost cell in the row or column. Step II. Select the row or column with the maximum difference and allocate as much as possible (keeping the restrictions of supply and demand in mind) to the least cost cell in the row or column selected. In case of a tie, take up any one. Now, in this example, since 3380 is the greatest difference, we choose row A and allocate 1000 to least cost cell, i.e., AX. Step III. Cross out the row or column which satisfies the condition by allocation just made. So, row A is crossed out. The matrix without row A is as shown below. X B C
Rs. 2500 X21 Rs. 2550 X„
Rs. 2700 X22 Rs. 1700 X„ = 1200
1500 (200) 1200 (850)
Repeat steps I to II till all the allocations have been made. Now, column Y shows maximum difference, so we allocate to the least cost cell in Y column, i.e., CY an amount of 1200 but this does not satisfy column Y completely.
THE TRANSPORTATION PROBLEMS
207
Also, row C shows maximum difference (850) out of the two rows. We allocate 1200 to cell CY which is the least cost cell in row C. Since this allocation completely satisfies row C, we cross row C and the shrunken matrix is shown below.
B
X Rs. 2500 X2, = 1300
Y
Rs. 2700 X22 = 200
Since cell BX has the least cost, maximum possible allocation of 1300 is made here. In cell BY, we allocate 200. All the above allocations made can now be shown in one single matrix as below. Rs. 2000
Rs. 5380 X„
X„ 1000
Rs. 2700
Rs. 2500 X21
X22
1300
200 Rs. 2550
X23
Rs. 1700 X32
1200 The cost of transportation associated with this solution is Z = Rs. (2000 x 1000 + 2500 x 1300 + 1700 x 1200 + 2700 x 200) = Rs. 7830000. PERFORMING OPTIMALITY TEST We have found out a feasible solution. Now, we must find out whether this feasible solution is optimal or not. Such an optimality test can be performed only on such feasible solutions where 1. The number of allocation is m + n 1 m = number of rows and n = number of columns. where In given problem m = 3 and n = 2 so number of allocations is 4 which is the actual case hence optimality test can be applied. Also, all the allocations are independent of each other. We can test the optimality of a feasible solution by carrying out an examination of each vacant cell to find out whether or not an allocation in that cell reduces the total transportation cost. This can be done by the use of the following two methods: The SteppingStone Method
Let us consider the matrix of the above problem where we have already found out the feasible solution.
208
OPERATIONS RESEARCH Distribution centres X
Y
Rs. 2000 A
Rs. 5380
X11
X12
1000 Rs. 2700
Rs. 2500 Plants B
X21
X11
1300.
—100
2000
+100
Rs. 2550 C
X23
Rs. 1700 X11
1200
+100
—100
Let us take up any arbitrary empty cell, i.e., CX and allocate +100 units to this cell. Now in order to maintain the restrictions of column X, we must allocate —100 to cell BX and to maintain the row B restriction we must allocate +100 to cell BY. This will result in unbalance of column Y conditions and so we must allot —100 to cell CY. Now, let us work out the net change in the transportation cost by the changes we have made in allocations. = Rs. (2550 x 100 — 2500 x 100 + 2700 x 100 — 1700 x 100) Evaluation of cell CX = 255000 — 250000 + 270000 — 170000 = Rs. 105000 As the evaluation of the empty cell CX results in a positive value the total transportation cost cannot be reduced. The feasible solution is an optimal solution already. We must carryout evaluation of all the empty cells to be sure that optimal solution has been arrived. The total number of empty cells are in x n — (m + n — 1) = (m — 1) (n — 1). Hence (m — 1) (n — 1) cells must be evaluated. In the present problem m = 3 and n = 2, so only two empty cells are there but in other problems, the number of empty cells could be much more and this procedure becomes very lengthy and cumbersome. The Modified Distribution (MODI) Method or UV Method The problem encountered in the stepping stone method of optimality test can be overcome by MODI method because we don't have to evaluate the empty cells one by one, all of them can be evaluated simultaneously. This is considerably time saving. The method has the following steps: Step I. Setup the cost matrix of the problem only with the costs in those cells in which allocations have been made. Y
A B C
L_
Rs. 2000 Rs. 2500
Rs. 2700 Rs. 1700
I Step II. Let there be set of number Vi (V1, V2) across the top of the matrix and a set of number U; (U1, U2, U3) across the left side so that their sums equal the costs entered in the matrix shown above.
209
THE, TRANSPORTATION PROBLEMS V, = 0 2000 2500 1500
U1 U, U3
200
V, = 200
Rs. 2000 Rs. 2500
Rs. 2700 Rs. 1700
U1 + V1 = 2000 U2 + V2 = 2700 U2 + Vl = 2500 U3 + V2 = 1700 Let Vl = 0 then Ui = 2000, U2 = 2500 V2 = 2700  2500 = 200 U3 = 1500 Step III. Leave the already filled cells vacant and fill the vacant cells with sums of U1 and V1. This is shown in the matrix below. 0 2000 2500 1500
200 2200
V2
(V1 + V2)
U2
1500
U3
(U3 + V1)
Step IV. Subtract the vacant values now filled in step III from the original cost matrix. This will result in cell evaluation matrix and is shown below for the example in hand. ... ... 2550  1500 = 1050
5380  2200 = 3180 ... ...
Step V. If any of the cell evaluation turns out to be negative, then the feasible solution is not optimal. If the values are positive the solution is optimal. In the present example, since both the cell evaluation values are positive, the feasible solution is optimal. Let us take another example where some of the evolutions turns out to be negative to explain the entire procedure. Let us assume the following transportation model for this purpose:
A Plants C
Demand
P 200 100
Distribution centres Q R S 300 1100 700 0 600 100 @
500 7 (100)
800 5 (300)
1500 3 (500)
900 2 (600)
Supply 6 (100) 1 (100) 10 (300) 17 (Total)
FEASIBLE SOLUTION BY VAM Step I. In column P, the difference between the two lowest cost elements is 100 which is entered as (100) below column P. Similarly, the two smallest elements in row Q are 0 and 300. The
OPERATIONS SEARCH
210
difference is 300. We write their difference as (300) below column Q. Under column R, we write (500) and under column S we write (600). Similarly, against row A, we write (100), against row B (100) and against row C (300). Step II. We choose column S (having largest difference 600). In this column, cell BS has the lowest cost, i.e., 100 and we allot 1 as maximum possible allocation of only 1 is possible. Step III. Cross out row B as the supply 1 is completely satisfied by the allocation made, 1, in step H. Step IV. Write down the shrunken matrix after crossing out row B as follows: P 200 500 7 (300)
A C
Demand
R 1100 1500 3 (400)
Q
300 800 5 (500)
S 700 900 2 200
Supply 6 (100) 10 (300)
We repeat step I and write the difference in rows and columns as shown above. In column Q least cost is AQ, we make allocation of 5. Since this satisfies the condition of column Q completely, we cross out column Q and shrunken matrix is written as follows : P R S Supply 200 0 1100 700 1 (500) 500 1500 900 10 (400) 7 3 1 (300) (400) (200)
A C
Demand
Once again step I is repeated and the difference in rows and columns are written as shown above. We now make allocations in cell AP as this is the least cost cell. Only 1 can be allotted in this cell since this satisfies row A, it is crossed off and the shrunken matrix is rewritten as follows: C
P 500
R 1500
S 900
Supply
10 (400) ® 0 © 1 Demand 6 3 In this, as cell CP has the lowest cost, maximum possible allocation of 6 is made here. Next lowest least cost cell is CS and 1 is allotted here and 3 in the cell CR. The allocations made above are shown in the allocation matrix given below. P 200
A 10 Plant:
500 ®
Demand
R 1100
S 700
0 800
600
100
1500 ®
7
Supply 6
CD 100
C
Q 300
900 ©
3
1 10
211
Tilt TRANSPORTATION PROBLEMS
Z=Rs.(200 x 1 + 300 x 5 + 100 x 1 + 500 x 900 x 1 = 6 + 1500 x 3) = Rs. 10200 OPTIMALITY TEST BY MODI METHOD OR UV METHOD Step I. Setup cost matrix only for cells in which allocation have been made.
v, uti
P 200
A B
S
1500
100 900
300
500
C
R
Q
Step II. Enter a set of numbers V, across the top of the matrix and a set of numbers U, across the left side so that their sum is equal to the cost entered in step I. Ui+ V1 = 200 U3 + V4 = 900 + V2 = 300 U2 + V4 = 100 U3 +V1 =500 U3 + V3 = 1500 If V1 = 0,U1 = 200 V2 = 300, U3 = 500, V4 = 400 U2 = — 300, V3 = 1000 So, the matrix may be rewritten as =0 200 —300 500
200
100
1000
400
1500
100 900
300
500
Step III. Let us fill the vacant cells with the sums of U, and V,. This is shown below. Vi Ui
0
100
1000
400
200 300 500
... 300 ...
... 200 600
1200 700 ...
600 ... ...
Step IV. Now let us subtract the cell values of the matrix of step 3 from the original cost matrix. ... 100 + 300 ...
A B C
... 0 + 200 800  600 P ... 400 ...
This is called the cell evaluation matrix.
1100 —1200 600 — 700 ... Q
... 200 200
R 100 100 ...
1700 — 600 ... ... S
100 ... ...
OPERATI9NS RESEARCH
212
Step V. Now since two of the cell evaluations are negative, it means the basic feasible solution is not optimal. Hence, we will take steps to find an optimal solution. Step VI. Identify in the evaluation cell, the cell with most negative entry. In the present example there are two cells, i.e., AR and BR cells with same negative values of —100. So, let us take cell AR. Step VII. Write the initial feasible solution in the matrix. The cell value with most negative value is called the identified cell and is marked (✓). Step VIII. Trace a path in this matrix consisting of a series of alternatively horizontal and vertical lines. The path begins and terminates in the identified cell. All corners of the path lie in the cells for which allocation have already been made. As the path has to begin and end at the identified cell, it may skip over any number of occupied or vacant cells. This is shown in the table below. P Q R S Supply
1 0
A
+1 6
®
©
B
1
CD
C
CD
0
+1 —1 10 7 5 3 2 Step IX. Mark the identified cell (AR) as positive and each occupied cell at the corners of the path alternatively positive and negative and so on. Step X. Make a new allocation in the identified cell (AR) by entering the minimum allocation on the path that has been assigned a negative sign. Now, add or subtract as the case may be, these new allocation from the original values of the cells on the corners of the path traced, keeping the row and column requirement at the back of the mind. This makes one basic cell as zero and the other cells become nonnegative. That basic cell whose allocation has become zero (AP in this case) leaves the solution. The matrix of the second feasible solution can be rewritten as:
B C
P Q R S Supply 300 1200 6 ® CD 100 1 0 1500 500 900 10 CD 0 7 3 2 5
The total cost for this feasible solution is =Rs.(300 x 5 + 1200 x 1 + 100 x 1 + 500 x 7 + 1500 x 2 + 900 x 1) = Rs. 10200 Which is less than the cost found in the original feasible solution.
213
THE TRANSPORTATION PROBLEMS
Example 5.2. Find the feasible solution of the following transportation problem using NorthWest corner method. F, F2 F3 Demand
W, 14 65 35 4
W3 45 35 65 6
W2
25 25 3 7
Supply 6 8 16
W4 5 55 15 13
Solution. Initial feasible solution WI F,
14
25
®F2 F3
Demand
W3
W2
65 35
45
5
35
55
25
0 1
3
0 * 65
4
f:: O
7
6
Supply
W4
6 8
® 15 \
@ 13
16 '',,,,..,,3(1,1 30
Since requirement (4 + 7 + 6 + 13) is equal to the supply (6 + 8 + 16) it is a balanced problem. Step I. Set F1W1 (i.e., NorthWest corner cell) = 4, the smaller amount, Here S1 = 6 and D1 = 4 and so proceed to cell F1 W2 as D < S. Step II. Compare the number of units available in F1 and row (2) and the demand in column W2 (7) and so set 2 in row F1 W2. Since D > S in this case, we have to proceed vertically, so move to cell F2W2. Step III. Here supply is 8 and demand is 5. So, we set 5 in cell F2W, and proceed horizontally (D < S) to cell F2 W3. Step IV. In cell F2W3, supply is 3 and demand is 6 so we set 3 in cell F2W3 and proceed vertically (D > S) to cell F3W3. Step V. In cell F3W3 the demand is 16 and requirement is 3 so we set 3 in F3W3. Step VI. Allocate 13 in cell F3W4. NorthWest Corner Method F1W1 14 x 4 = 56 F1W2 25 x 2 = 50 F2W2 25 x 5 = 125 F2 W3 35 x 3 = 105 F3 W3 65 X 3 = 195 F3 W4 15 x 13 = 195 Total cost = 726
OPERATIONS RESEARCH
214
Example 5.3. Determine an initial basic feasible solution to the following transportation problem using NorthWest corner rule. P
Q
3
A
To : R 8
6
4
B
10
2
Availability
9 20
1
0 From :
T
S
5
1
30
CD
®
la
8
C
7
11
20
40
D
2
1
9
14
3 15
© 16
13
0 Demand
6
40
8
18
6
Solution. Since demand and availability both are equal, it is a balanced problem. Step I.
Set the NorthWest corner AP = 20, the lesser out of demand (40) and availability (20) as D > S, we proceed vertically to cell BP.
Step H.
Here in cell BP availability is 30 and demand is 20 so we set 20 in this cell and proceed horizontally to cell BQ as D < S.
Step III.
In cell BQ availability is 10 and demand is 6. So, we set 6 and proceed horizontally to cell BR as D < S.
Step IV.
In cell BR availability is 4 and demand is 8 so we set 4 in this cell. We proceed vertically down to cell CR as n > S.
Step V.
In cell CR availability is 15 and demand is 4 so we set 4 in this cell. We proceed horizontally to cell CS as D < S.
Step VI.
In cell CS availability is 11 and demand is 18 so we set 11 in this cell. We proceed vertically down to cell DS as D > S.
Step VII. In cell DS availability is 13 and demand is 7 so we set 7 in this cell. We proceed horizontally to cell DT as D < S. Step VIII. We allot 6 in cell DT. Example 5.4. Determine an initial basic feasible solution to the following transportation problem using row minima method: P
Q
To : R
S
Availability
A
22
B
15
From : 8 Demand
2
17
THE TRANSPORTATION PROBLEMS
215
Solution. Let us allot as much as possible in row A to the cell in which there is minimum cost, i.e., a AQ in which cost is 2 we allocate 12 in this cell. As the demand of the column Q is
completely satisfied, we cross off the column. Then next lowest cost cell in row A is AS. We allot 9 to this cell and hence S column can be crossed off as demand of column S is fully satisfied. Out of the rest matrix, the lowest cost in row A is in column AR. We can allot 1 in this cell. Now, row A is completely satisfied so row A is struck off. Let us now take row B in which minimum cost is in cell BR and we allot 15 in this cell so row B is completely satisfied and row B can be struck off. Now, let us take row C in which we have to allocate resources in cell CP as it is the minimum cost cell. We allot 7 in this cell. Which strikes off column P : only CR has been left in which only 1 can be allotted. Hence X12 =12, X13 = 1, X14 = 9, X23 = 15, X31 = 7, X33 = 1 Example 5.5. Find the initial basic feasible solution to the following transportation problem by (a) Minimum cost method (b) Northwest corner rule. State which of the methods is better. To :
P 2 3 5 1
A
B
From :
C
D Demand
R 4 1 7 2 18
Q
7 3 4 6
7
Supply 5 8 7 14 34
Solution. Initial basic feasible solution is shown below.
A
X
Y
2
7
4 0
B
3
® 1
3
5
8 0
C
5
4
7
7
0 D
1
Demand
7
0
6 9
2 18
0
14 34
(a) Minimum cost method
The lowest cost cells are BR and DP let us allot 7 in cell DP and 8 in cell BR. Now, we move to cells AP and DR as both have the next lowest cost i.e., 2. In cell AP only 0 can be allotted. In cell
216
OPERATIONS RESEARCH DR we can allot 7. The next minimum cost cells are BP and BQ in BP we can allot only 0 similarly in BQ we can allot only 0. The next minimum cost cells are CQ and AR. In cell CQ we can allot 7 and in cell AR we can allot 3. The next minimum cost cell is CP with cost 5. In this cell we allocate 0. In next lowest cost cell DQ, we can allot 0. The next lowest cost cells are AQ and CR. In AQ we allot 2 and in CR we allot 0. The cost of transportation associated with this solution is Z=Rs.(7 x2+4x3+1x 8+4x 7+1x7+2 x 7) = Rs. (14 + 12 + 8 + 28 + 7 + 14) = Rs. 83
(b) NorthWest corner rule Step I. We start with cell AP (top left) so in this cell we allot 5 since in this case D > S, we proceed vertically to cell BP. Step II. In cell BP we allot 2 and since D < S we proceed horizontally to cell BQ. To : A
2
Q
R
7
4
Supply 5
0 B
3
3
1
CD From :
5
8
CD 4
7 ®
D
1
6
0 2 14
Demand
7
7
9
14
18
Step III. In cell BQ we can allot only 6 since in this case D > S, proceed vertically to cell CQ. Step IV. In cell CQ we can allot only 3 and since here D < S, we proceed horizontally to cell CR. Step V. In cell CR we can allot only 4 since in this cell D < S, proceed vertically to cell DR. Step VI. In cell DR we can allot 14. Now, for this solution for transportation cost is Z=Rs.(2 x5+3x2+ 6 x3 +3x4+7x4 +2x 14) = Rs. (10 + 6 + 18 + 12 + 28 + 28) = Rs. 102. It is clear that minimum cost method gives a better solution. Example 5.6. A manufacturer wants to ship 8 loads of his product from production centres X, Y and Z to distribution centres A, B and C. Mileage from origin 0 to destination D is given in the following matrix:
THE TRANSPORTATION PROBLEMS
217 Distribution Centres
Production Centres
0
A
B
C
Available
X Y Z
50 90 250 3
30 45 200 3
220 170 50 2
1 3 4 8
Required
If the shipping cost is Rs. 10 per load per mile, what shipping schedule should be used ? Solution. Let us use VAM Method Step I.
Compute penalty for each row and column as shown below. A X Y Z Required Penalty
B
50 90 250
C
30 45 200 3 40
D 1 3 4 8
220 170 50 3 15
2 120
Penalty 20 45 150
This is obtained by computing the difference between smallest cost and the next smallest cost element in a particular row or column. Step II. Row with largest penalty is Z with 150 penalty and cell with smallest cost is ZC. Allocate maximum possible to this cell, i.e., 2. Hence column C can be eliminated. Reduced matrix is shown below. A X Y Z
50 90 250
B
Availability 1 3 2
30 45 200
Penalty 20 45 50
0 Required Penalty
3 40
3 15
Step III. Now the row with largest penalty is Z. We can allocate 2 to ZB and eliminate row, Z. Reduced matrix is as shown below. A X Y
50 90
B 30 45
Availability 1
Penalty 20
3
45
0 Required Penalty
3 40
1 15
OPERATIONS RESEARCH
218
We can allocate 1 to cell BY and eliminate column B. Reduced matrix is: A 50 90 3
X Y
Required
Availability 1 2 3
Allocate 1 to XA and 2 to YA Initial feasible solution is A X
501
Y
90 1
Z
250
B
C
30
2201
Availability 1
451
1701
3
200
50
4
0
T 0 Required
3
3
2
Totalcost= 1 x50 + 2 x 90 +1 x 45 + 2 x 200 + 2 x50 = 775 milesx 10 =Rs.7750 Example 5.7. Find the initial basic feasible solution of the following transportation problem by Vogel's Approximation Method. Werehouses
Factory
F2 F3
Requirement
WI
W2
W3
W4
Capacity
19 70
30 30 8
50 40
10 60
7 9
70 7
20 14
18
40 5
8
(34 Total)
Solution. By Vogel's Approximation Method (VAM) Step I.
Write down the cost matrix and enter the difference between the lowest and the second lowest element in columns and rows under the respective columns and on the right side of the respective rows. Warehouses F1 F2 F3
Requirement
Wi 19 70
W2
W3
W4
30 30
50 40
10 60
40 5 (21)
® 8 (22)
80 7 (10)
20 14 (34 Total) (10)
Capacity 7 (9) 9 (10) 18 (12)
Column W2 is the column with greatest difference. The lowest cost cell in this is F3W2, so we allot 8 in this cell.
THE TRANSPORTATION PROBLEMS
219
Step II. Strike out the column W2 since it completely satisfied the requirements. The shrunken matrix is shown below. To: Wi
F1
19
W3
W4
50
10
Capacity 7 (9)
70
F2
40
60
From :
9 (10) 40
F3
70
20 18 (12)
Requirement
5 (21)
7 (10)
14 (10)
Step III. Column W1 has the greatest difference so we make as much as possible allocation in the minimum cost cell of this column, i.e., F1 Wi (19) only 5 can be allotted in this cell since column W1 is completely satisfied, it is crossed out and the resulting matrix is shown below. W3
W4
F1
50
10
Capacity 7 (9)
F2
40
60
9 (10)
F3
70
20 10
Requirement
14 (10)
18 (12) (34 Total) 10
Step IV. Now row F3 has the greatest difference (12) and the least cost cell in this row F3W4 hence we allot 10 in this cell and resultant matrix is as shown below.
F1 F2
W3
W4
50 40
10 60 ®
(10)
7 (9)
® 9 (10) (10)
In row F2 we can allot to cell F2W3. Now, since column W3 is fully satisfied we can cross it out. In cell F3 W4 we can allot 2 to satisfy row F1 and in cell F2 W4 we can allot 2 to satisfy column 4.
OPERATIONS RESEARCH
220
All allocation made can now be written in a single matrix as shown below. F1
Capacity
W3
Wi
19
10
50
30 ®
F2
70
7
© 30
40
60 9
0 40
8
20
70
18
® Requirement
5
8
© 14
7
The cost associated with this solution is Z = Rs. (5x 19 + 2 x 10 + 7 x 40 + 2 x 60 + 8 x 8 + 10 x 20) = Rs. (95 + 20 + 280 + 120 + 64 + 200) = Rs. 779 Unbalanced Transportation Problems
Example 5.8. A departmental store wishes to purchase the following quantity of ladies dresses: Dress type Quantity
D
A 150
75
100
250
Tenders are submitted by three different manufacturers who undertake to supply not more than the quantity given below (all types of dresses combined) Manufacturer Total quantity
W 350
X 250
150
The store estimates that profit per dress will vary with the manufacturers as shown in the matrix below. How should orders be placed ?
Manufacturer
W X Y
A 2.75 3.00 2.50
Dress B 3.50 3.25 3.50
C 4.25 4.50 4.75
D 2.25 1.75 2.00
Solution. The problem can be written in the form of the following matrix: Step I.
Matrix W
Manufacturer
X Y
Demand
A B 2.75 3.50 3.00 3.25 2.50 3.50 150 100
C
4.25 4.50 4.75 75
D 2.25 1.75 2.00 250
Supply 300 250 150 (Total 700) (Total 575)
Since the supply and demand are not equal, it is not a balanced problem. Here total supply is 700 and total demand is 575, so surplus supplies are 125.
221
THE TRANSPORTATION PROBLEMS
We have to create dummy destination (store). The cost associated with store will be taken zero as the surplus quantity manufactured remains in the factory and is not transported at all, so the new matrix is: A B 2.75 3.00
Manufacturer
3.50 3.25
Supply
D E
C
4.25 4.50
2.25 1.75
300 (0.25) 250 (1.25)
0 0 125
X Y
Demand
2.50 3.50 4.75 2.00 0 150 100 75 250 (0.25) (0.25) (0.25) (0.25) (0)
150 (0.50)
Step II. Using Vogel's Approximation Method (VAM). Let us write the difference between the smallest and second cost in each column and each row and write it below the column or on right side of the rows respectively. Row with greatest difference is row X as indicated with an arrow. In this row the least cost cell is XE. In this we can allot 125 since column E is fully satisfied this column is crossed out. Now the shrunken matrix is shown below. A 2.75 3.00
Manufacturer
B 3.50 3.25
4.25 4.50
D 2.25 1.75
4.50 75 (0.25)
GO 2.00 250 (0.25)
C
X Y
Demand
2.50 150 (0.25)
3.50 100 (0.25)
300 (0.25) 250 (1.25) 150 (0.50)
Step III. In this matrix maximum difference is in row X and the least cost cell is XD. We can allot 125 units to this cell and since this row is fully satisfied it is crossed out. The new matrix is as follows: A 2.75 2.50
Manufacturer
B 3.50 3.50
4.25 4.75
D 2.25 2.00
150 100 75 (0.25) (0.25) (0.25)
250 (0.25)
C
Supply 300 (0.25) 150 (0.50)
Y
Step IV. In the above matrix maximum difference is in row Y which is shown with an 2
4
•.• The value of water square 23 is most negative, we draw closed loop through this cell and the new table will be =4
( 4)
® X
16 +
24
16
g ,
0 R3 =
4
Plant capacity 56
8
R, = 0 R2 =12
K, = 20
K2 = 12
Project requirement
77
124 16 0 +0
Y
72
102
82
41
215
218
OPERATIONS RESEARCH IS R, To From Ri = 0
W
Ki = 4
K2 = 12
K3 =20
Project A
Project B
Project C
4
8 ®
R2 = 12
X
16
R.3 = 4
Y
8
Project requirement
72
8 0
Cjil 24
®
16
e
13 16
®
24 41
R1 +K1 =4 0+K1 =4 K1 =4 R2 +Ki =16 R2 + 4 = 16 R2 = 12 2.—)2 R2 +K2 =24 12 + K2 = 24 K2 = 12 243 R2 +K3 =16 12 + K3 = 16 K3 =4 3 —> 2 R3 +K2 =16 R3 + 12 = 16 R3 = 4 Calculation of opportunity cost of water squares is as given
1>3
343 The opportunity cost of cell 1
C,) — KS C12— R1— K2 8 —0 —12 C13 — R1 — K3 8 —0 —4 C31 — R3 — K1 8 —4 — 4 C33 — R3 — K3 24 —4 —4 2 is negative
77 215
Stone square 1 > 1
Unused squares 1 2
82
CI
(t 102
Plant capacity 56
Improvement index —4 +4 0 + 16
229
THE TRANSPORTATION PROBLEMS
Third approved solution KJ
K, = 4
K, = 8
K, = 4
Project A
Project B
Project C
R.
To From R,= 0
W
4
R, = 12
X
16
R, = 8
Y
8
8
a
0 24
8
8
® 24
82 77
8
© 72
56
o e
16,
16
Plant capacity
102
215
41
The value of cell 3 > 1 is negative in the improved solution. K, R, To From R1 =0
W
4
R2 = 16
X
16
R3 = 8
Y
8
Project requirement
72
Ki = 0
K2 = 8
K3 = 0
A
B
C
8
Plant capacity
8
56
16
82
CD
@ 24
©
®
@ 16 g
77
24 CD
0 102
215
41
Total cost of optimal solution Shipping assignment
Quantities shipped
Limit cost
Total cost
WB XA XC YA YB
56 41 41 31 46
8 16 16 8 16
448 656 656 248 736 2744
230
OPERATIONS RESEARCH
Example 5.12. A company has 4 different locations in the country and 4 sales agencies in 4 other locations in the country. The cost of production, the sale price, shipping cost in the cells of the matrix, monthly capacities and monthly requirement are given below. Factory
1
2
3
4
Capacity
Cost of production
A
7
5
6
2
10
10
B C D Monthly requirement
3
5
4
2
15
15
4
6
4
5
20
16
8
7
6
5
15
15
8
12
18
22
Selling price
20
22
25
18
Find the monthly production and distribution schedule which will maximize profits. Solution. By adding cost of production and shipping cost of each cell, the total cost of matrix is I A B C D
' II
III
IV
17
15
16
14
18 20 23
20 22 22
19 20 21
17 21 20
By deducting cost of production from sales price of each agency, of we get profit matrix. Profit matrix I
II
III
IV
A B
3 2
7 2
C
0
0
9 6 5
4 1 3
D
3
0
4
2
Let us convert the matrix into a minimization matrix by deducting each element from the maximum profit i.e., Rs. 9. Sales Factory
SII
STH
SW
Capacity 10 15
FA FB
6
2
0
5
7
Fc FD
9 12
7 9 9
3 4
8 12
5
11
15
Required
8
12
18
22
60
20
THE TRANSPORTATION PROBLEMS
231 Initial Feasible Solution
Sales Factory
S1
S2
S3
S4
Capacity
UPI
6
2
0
5
10
2
3
8
15
12
5
FA
UP2
UP3
4
4
0
20
5
5
0
0
11
15
4
4
2
2
18
0 22
60
3
3
1
3
UP4
10
7
FB
7
15
Fc
9
9
FD
12
9
4 CD
Req.
® ® 8 12 5
UPI
1
UP2
2
2
UP3
2
2
UP4
3t
0
3
The stone squares are equal to rim requirement. The stone squares are 7 = Rim requirement 7. The above solution is nondegenerate, we proceed to test its optimality by MODI method. The optimality by MODI method
K. R,
K, = 12
K, = 9
K, = 7
K, = 11
S,
S,
S3
S,
Ag Fact K, =  7
FA
6
2
0
ri 8 K, =  3
FE,
7
K, =  3
Fc
9*
9
Fr,
12
9 0
Req
10
0 15
8
0
0 ►/1
®
® K, = 0
3
0
2
5
0
7
Capacity
12
20
11
15
0 5
1Y()T + 0•
0
OPERATIONS RESEARCH
232
We have selected FD S3 having negative value and draw a closed loop. K.
K1 =10
K2 =9
K3 =5
2
3
K4 = 11
Ki Ag
1
Fact K, =  7
A
2
6
0
® B
K2 =3
3
1C3 =  1
C
9
D
K, . 0
12
4
0 20
® 15
11
0
©
0 12
15
CD 5
0
10
12
© 9
8
10
O1
9
®
5
8
i
0
Capacity
2
0 7
7
4
22
18
60
The final improved matrix is Si
S3
S2
FA
3
7
FB
2
Fc FD
Capacity
S4
9
4
10
2
6
1
15
0
0
5
3
20
3
0
4
2
15
0 0 ©
0
©
0
The opportunity cost of all water squares is positive. So, the solution is optimum Sales Agency = Total profit Factory FA S2 = 7 x 10 = 70 FB S4 = 1 x 15 = 15 Fc S1 =0 x 8=0 Fe—> S3 = 5 x 12 = 60 FD S2 —0 x2 —0 FD —* 53 = 4 x 6 = 24 FD —> S4 = — 2 x 7 = — 14 Total profit = 155/
THE TRANSPORTATION PROBLEMS
233
Example 5.13. A transport corporation has trucks at 3 garages ABC in a city. The number of trucks available in each garage, the number of trucks required by each customer and the distance from garage to customer's locations are given below : Customers Garages A B C
Requirement
1
2
3
4
Availabilities
7 8 10 4
6 6 7 3
9 7 8 5
12 10 12 8
12 8 10
Just before sending the trucks, it is known that road from B to customers 2 is closed for traffic due to road block. How should the trucks be assigned to the customers in order to minimize the total distance to be covered due to road block ? Solution.The problem is unbalanced, we add dummy with 0 transportation cost and assign M prohibited cell To
1
From A
7
B
8
C
10
2
3
6
4
Dummy
Capacity
Upi
Up2
Up3
Up4
9
12
0
12
6
1
1
2
M
7
10
0
8
1
1
1
1
7
8
12
0
10
7
®
®
10
Req. Upi Up2 Up3 Up4
4 1 1 1 1
3 1 M7T
5 1 1 1 1
8 2 2
10 0
30
The stone square = 5 Rim requirement = m + n —1 =3+5—1=7 The above problem is of degeneracy. We introduce e at two cells and list its optimality by MODI method.
OPERATIONS RESEARCH
234
R;
K, = 7
Ki
R, = 0
K, = 6
1 7
A
K, = 9
2 6
3 9
B
8
R, = 0
C
10 4
4 12
K, = 0 0 0
Capacity 12 E
CO)
04 M 7 10 @ v 0) V7 12 8 ED M (1) 3 5 8 ®
R2 = 2
K, = 12
0
0
8 0) t. 0
0)
10
0 10
30
Optimal solution R,
K1 = 7
K2 = 6
K3 = 9
K4 = 12
K5 = 0
1
2
3
4
Dummy 0
R1 =.0
A
7
R2 = 1
B
8
R3 = 0
C
10
6 0
12
7
10
0 M
®
®
4310 7
® 4
9
12
0 3
0 5
0 0
®
0 8
8 0
0
0 8
Capacity 12
10
0 10
30
A1 = 28 A 2 = 18 B > 4 = 80 C —> 3 = 40 = 166 Example 5.14. A company manufacturing air coolers has two plants located at Bombay and Calcutta with weekly capacity of 200 units and 100 units respectively. The company supplies air coolers to its 4 showrooms situated at Ranchi, Delhi, Lucknow and Kanpur which have a demand of 75, 100, 100 and 30 units, respectively. The cost per unit (in Rs.) is shown in the following table. Bombay Calcutta
Ranchi 90 50
Delhi 90 70
Lucknow 100 130
Kanpur 100 85
Plan iite production programme so as to minimize the total cost of transportation. Solution. Here, the total demand is greater than the total capacity hence the problem is unbalanced. So we have to have a dummy plant with additional required capacity of 5 units. Transportation cost from the dummy plant will be zero.
235
THE TRANSPORTATION PROBLEMS
Let us use VAM's method to find out the Initial Feasible Solution. Constructing the cost, requirement and availability matrix as shown. To Ranchi
Delhi
Lucknow
Capacity
Kanpur
From Bombay
90
90
100
100
200
Calcutta
50
70
130
85
100
Dummy
0 I
0 I
0
0
5
Demand
75
100
100
305
30
Now, penalty for each row and column can be worked out as follows : To Ranchi
Delhi
Kanpur
Lucknow
Penalty
From Bombay
901
901
1001
1001
10
Calcutta
501
701
1301
85 1
20
Dummy
0 1
(:1
0
0
0
1
1
Demand
75
100
100
30
Penalty
40
20
30
15
Following the steps, we get the following Initial Feasible Solution : To Ranchi
Delhi
Kanpur
Lucknow
Capacity
From Bombay
901
100
901 0
Calcutta
501
Dummy
i::
70 1
200
100 0
0
1301
85 1
0
0
100
0
0 0 1
5
0 Demand
75
100
100
30
305
OPERATIONS RESEARCH
236
The transportation cost associated with this solution is = 90 x 75 + 100 x 95 + 100 x 30 + 50 x 75 + 25 x 70 + 0 x 5 Total cost = Rs. 24750 Example 5.15. A company has three factories A, .B and C which supply to four warehouses situated at P, Q, R, S. The monthly production capacity (tons) of A, B and C are 120, 80 and 200 respecitvely. Monthly requirement (tons) for the warehouses P, Q, R and S are 60, 50, 140 and 50 respectively. The transportation cost (Rs. per ton) matrix is as follows: Warehouses
Factories B 3 8 4 8
A 4 5 2 5
P Q
R S
C 7 4 7 4
Using Vogel's method, determine the optimum transportation distribution of products to warehouses to minimise the total transportation cost. Solution. Using the steps involved in VAM method, the student can easily determine the Initial Feasible Solution as given in the table below : B
A P
4
3
7
Q
5
8
4
R
2
4
7
40
Demand
C
60
50
140
qi)
S
5
8
4
Dummy
0
0
0
50
100 11
Supply
120
80
200
400 400
THE TRANSPORTATION PROBLEMS
237
The above solution is degenerate as Rim requirement = m + n 1 = 5 + 3  1 = 7 No of stone squares in present solution = 6 Now let E be allotted to the RC cell. Initial Feasible Solution is as shown in the matrix below : A
B 60
(6R
2
Demand I 60
(7
140
20 0 8
5
(7
01
(3
TO0
Dummy Supply
50
50
4 120
S
C 7 ; —
120
80
50 100 400
200
400
The above is a optimal solution as the opportunity costs are positive. Total cost= 3 x 60 + 20 x 4 + 50 x 4 + 120 x 2 + 50 x 4 + 0 x 100 = Rs 900 Example 5.16. ABC tool company has a sales force of 25 men who work out from three regional centres. The company produces four basic product lines of hand tools. Mr. Jain, Sales Manager feels that b salesmen are needed to distribute product line 1, 10 salesmen to distribute product line 2, 4 salemen to product line 3 and 5 salesmen to product line 4. The cost (in Rs.) per day of assigning salesmen for each of the office for selling each of the product lines are as follows : Product Lines Regional Office
1
2
3
4
A
20
21
16
18
B
17
28
14
16
C
29
23
19
20
At the present time, 10 salesmen are allowed to office A, 9 salesmen to office B and 7 salesmen to office C. How many salesmen should be assigned from each office to each product line in order to minimize costs ? Solution. Initial Feasible Solution is as follows
OPERATIONS RESEARCH
238 1
To .,,,„
From
20 1
21
B
17
28
C
29 1
A
3
2
18 1
16 1 14
Availability 10
®
0
0
0
5 Dummy ±:21_1
4
16 1
(:1
9
20 1
1) j
7
® 23
19 1
0
0 Requirement
6
10
4
5
1
26
Let us now apply MODI method to test the optimality of the above solution : K1 K3 K5 K2 K6 K1 2 19 21 16 18 R, 3 4 5 Availability To 1 2 Dummy From 16 10 0 A 20 21 18 0 RI O 1 5 (: +2 1 4 14 16 2 B 17 28 0 9 R2 ® r0 0 0 23 2 C 29 19 20 1 0 7 R3 17 (I) (+8 46 10 4 5 1 26 Requirement 6 The above is the optimum solution, since all optimality costs are positive. Totalcost=21 x 4 + 16x 1 +18 x5 + 17x 6 + 14 x 3 + 23 x 6 + 0 x 1 =Rs. 472 Example 5.17. A manufacturer wants to ship 8 loads of his products as shown below. The matrix gives the kilometres from origin to the destination. Destination
X Origin Requirement
A
B
C
Availability
50
30
220
1
Y
90
45
170
3
Z
50
200
50
4
3
3
2
8
Shipping costs are Rs. 10 per load per kilometre. What shipping schedule should be used ?
Solution. Step I: The student can easily find the initial feasible solution using VAM method by following the steps already explained. Initial solution is shown below :
239
THE TRANSPORTATION PROBLEMS
To j From i R,
X
K,
K2
K,
A
B
C
50
Availability
30
220
1
45
170
3
50
4
1 R2
Y
90
3 R3
Z
50
200 2
Requirement
3
2 3
2
8
It can be seen that this solution is degenerated as Required number of stone squares = m + n 1 = 3 + 3 1 = 5 But actually there are 4 stone squares. Total cost for the above solution = [50 x 1 + 45 x 3 + 50 x 2 + 50 x 2] x Rs. 10 = Rs. 3850 Step II. Assign e (epsilon) to that unoccupied cell which has minimum transportation cost, i.e., X B (30). It will now be considered as an occupied cell. Step III. Calculate the opportunity cost of all the unoccupied cells water square by using the relationship dij = cq  (Ri + KJ) where. di1 = Opportunity cost of square if in general c,1 =. Cost in square if, i.e., cost of square of intersection of row i and column j Ri = Value assigned to row i K1 = Value assigned to column j ' . Evaluation of di1is shoWn in the table below Unoccupied cell
Opportunity Cost
XC
d12= c„ (R, + K,) = 220  (0 + 50) = 170
YA
d„. c„  (R2 + Ki) = 90  (15 + 50) = 25
YC
d„= c„— (R, + K,) = 170 — (15 + 50) = 105
ZB
d„= c„— (R, + K,) = 200 — (0 + 30) = 170
Step IV. Alloting the opportunity costs, the matrix becomes:
OPERATIONS RESEARCH
240
IS
K,
K,
50
30
50
A
B
C
R. To From R,
0
X
50
CD 15
Y
90
Z
170
45
50
3
3
(iO
®
4
50
200 0
r 170
ED
25
o
1
220
30
r70 3
® 2
8
It can be seen that the opportunity cost of all unoccupied cells is a positive number. Hence, this solution is the optimal solution. The total transportation cost is XA + YB + ZA + ZC = 50 x 1 + 45 x 3 + 50 x 2 + 50 x 2 = 385 x Rs. 10 = Rs. 3850. Degeneracy in the Transportation Problem
We have seen that an initial feasible solution to an in resources/origins and 11 destination problem consists of (m + n — 1) basic variables which is the same as the number of occupied cells. However, if the number of occupied cells is less than (m + 11 — 1) at any stage of the solution, then the transportation problem is said to have a degenerate solution. Degeneracy as it is called can occur at two stages, i.e., at the initial solution or during the testing of the optimal solution. Let us discuss both the cases. Degeneracy at the Initial Solution Stage
If degeneracy occurs at the initial solution stage, we introduce a very small quantity e (Greek letter pronounced as epsilon) in one or more of the unoccupied cells to make the number of occupied cells equal to (ni + n — 1). e is so small a quantity that its introduction does not change the supply (sources) and demand (destinations) constraints or the rim conditions. E is placed in the unoccupied cell which has the least transportation cost and once e is allotted to it, it is supposed to have been occupied. e stays in the solution till degeneracy is removed or the final solution is achieved. The value of e is zero when used in the problems with movement of goods from one cell to another. The use of E and degeneracy can be explained with the help of examples. Here degeneracy occurs at the initial solution.
THE TRANSPORTATION PROBLEMS
241
LI REVIEW AND DISCUSSION QUESTIONS !I 1. What is a transportation problem? How is it useful in business and industry? 2. Explain the use of transportation problem in business and industry giving suitable examples. 3. What do you understand by (a) Feasible solution; (b) NorthWest solution; (c) Vogel's Approximation Method (VAM)? 4. Discuss various steps involved in finding initial feasible solution of a transportation problem. 5. Discuss any two methods of solving a transportation problem. State the advantages and disadvantages of these methods. 6. How can an unbalanced transportation problem be balanced ? How do you interpret the optimal solution of an unbalanced transportation problem ? 7. Explain the differences and similarities between the MODI method and stepping stone method used for solving transportation problems. 8. What is a transportation method? Explain its objectives. How can we used this model for solving a multiplesite facility beaten problem ? 9. Which method of solving transportation problems gives a more optimal solution ? How will you know when you have achieve the least cost allocation of products between origins and destinations? Explain with examples. 10. Formulate a cost minimization model for the allocation of facilities to locations in the following problem: ABC Ltd. is considering the layout of one of its plants divided into three different working areas. There are three different production facilities and each one has one of them. Assume the data not available. 11. An automobile manufacturing company has three factors say F1, F2 and F3 which are feeding 5 different zonesNorth, South, Eastern, Western and Central. The monthly demands of these zones thousand are 25, 40, 30, 25, 20 thousand units respectively. The cost of transporting one unit from the factories F1, F2 and F3 to each of the zones and factory capacity as shown in the table below. Zone (Transportation costs)
Factory
Capacity (units)
Northern
Southern
Eastern
Western
Central
F,
50000 40000 30000
8 15 12
4
6 6 8
10 8
12 8 10
F2 F3
10 8
10
Perpetrate a transportation model and a LP model. 12. XYZ Ltd. has three manufacturing plants Pp P2 and P3 which are stepping the production to three warehouses W1, W2 and W3. The following data is available: Plant P1 P2
Production (units) 150000 120000
P3
130000
Warehouse W1 W2 W3
Requirement (units) 160000 130000 8000
OPERATIONS RESEARCH
242 •
The rate of freight charges/unit is as shown below. To: W1 1.50 2.00 1.60
P1 From :
P2 P3
W3 1.80 2.50 3.00
W2 1.60 1.80 1.40
Determine the initial basic feasible solution using NorthWest Corner Method and VAM. 13. Plant location of a firm manufacturing a single product has three plants located at A, B and C. Their production during week has been 60, 40 and 50 units respectively. The company has firm commitment orders for 25, 40, 20, 20 and 30 units of the product to customers C1, C2, C3, C4 and C5 respectively. Unit cost of transporting from the three plants to the five customers is given in the table below. A Plant location C
C1 6 4 2
C2 1 4 6
C3 3 3 4
C4 4 3 4
C5 6 2 6
Use VAM to determine the cost of shipping the product from plant locations to the customers. 14. Solve the following transportation problem. Availability at each plant, requirements at each warehouse and the cost matrix is as shown below. Warehouse Pi Plant
P2 P3
Requirement
Wi 200 800 400 60
W2
W3
400 400 200 80
600 400 500 80
W4 200 500 400 120
Availability 80 100 190
15. There are four supply points P1, P2, P3 and P4 5 / destination A, B, C, D and E. The following table gives in cost of transportation of materials from supply points to demand statements in rupees. To:
From :
P1 P2 P3 P4
A 10 12 15 12
B 12 10 20 18
C 15 12 6 10
D 16 10 12 12
E 18 10 16 12
The present allocation is as follows: P1 to A 100, P1 to B20, P2 to B160, P3 to B10, P3 to C60, P3 to E120, P4 to D200, P4 to E100. Find an optimal solution for allocations. If we reduce the cost from any supply point to any destination, what do you think will be the impact. Select any case and discuss the outcome.
THE TRANSPORTATION PROBLEMS
243
16. Anand Lamps India Limited (ALIL) operates three manufacturing plants and four warehouses. Capacity of the plants and forecast demand of warehouses is as follows : Factory At P1
Capacity (tons) 16
P2 P3
8
Warehouse At A B
Forecast demand (tons) 6 7
8
C
6
D
12
The transportation cost per ton in Rs. is as given below. From / To P1 P2 P3
A 100 120 160
B 120 200 200
C 80 60 100
D 20 60 80
ALIL wishes to minimize its transport costs. 17. The recruitment policy of the personnel department of a company is such that three types of workers managers, supervisors and workers are required 20, 40, 150 from four placement service centres A, B, C, and D. The hourly rates in Rs. are given by the matrix below :
Category
Managers Supervisors Workers
Availability
Placement service centre A B C 16 10 8 20 24 30 200 180 60 50 200 60
D 12 20 160 80
Requirement 20 40 150 390
Determine the requirement pattern at the lowest cost. 18. Find the initial basic feasible solution to the following transportation problem by (a) Least cost method; (b) NorthWest Corner Rule. State which of the methods is better.
From
Demand
2 3 5 1 7
Table To 7 3 4 6 9
4 1 7 2 18
Supply 5 8 7 14
244
OPERATIONS RESEARCH
19. Solve the following transportation problem by VAM: Consumers A
B
C
Available
Suppliers I
6
7
4
14
II
4
3
1
12
III
1
4
7
5
Required
6
10
15
31
Use VAM to find an initial BFS. 20. The apex company is the distributor for television receivers. It owns three warehouses with stocking capacity as follows : Warehouse location
Sets in stock
A
100
B
25
C
75
It has the following order for set deliveries : Market location
Orders
X
80
Y
30
Z
90
Delivery costs for warehouses to each customers are largely a function of mileage or distance. The per unit costs have been determined to be : X
Y
Z
A
5
10
2
B
3
7
5
C
5
8
4
The deliveries could be made in many ways but the distributor would like to deliver the TV sets in a way that would minimize the delivery cost. Give the distribution schedule. Use VAM only. 21. A company has four factories situated in different locations and five warehouses in different cities. The matrix of transportation cost is as follows.
245
THE TRANSPORTATION PROBLEMS
Warehouse A B C D E Capacity (units)
I 4 9 6 5 7
II 8 5 5 8 6
Factories III 7 8 8 6 5
100
80
120
Iv 6 8 7 3 8 100
Requirement 150 50 40 60 200
Find the optimum transportation schedule using VAM only: 22. Obtain the initial BFS to the following transportation problem by VAM:
A, A2 A3 B.
D,
D2
D3
D,
D5
ai
5 8 10 3
7 6 9 3
10 9 8 10
5 12 10 5
3 14 15 4
5 10 10 25
23. A steel company has three furnaces and five rolling mills. Transportation cost (rupees per quintal) for sending steel from furnaces to rolling mills are given in the following table : Furnaces
M1
M2
M3
M4
M5
A B C 1 Requirement (Quintal)
4 5 6
2 4 5
3 5 4
2 2 7
6 1 3
4
4
10
8
8
Availability (Q)
8 12 14
How should they meet the requirement ? Use VAM. 24. A cement factory manager is considering the best way to transport cement from his three and manufacturing centres P, Q and R to depot A, B, C, D and E. The weekly production and demand along with transportation costs per ton are given below. P Q R Requirement (Quintal)
A 4 2 3
B 1 3 5
C 3 2 2
D 4 4 2
E 4 3 4
Tons 60 35 40
22
45
20
18
30
135
What should be the distribution programme ?
OPERATIONS RESEARCH
246 25. Solve the following problem and test its optimality: Project A
Project B
Project C
Plant
4
8
8
56
16
24
16
82
Pant Y
8
16
24
77
Project requirement
72
92
41
215
Pant W Pant X
•
26. The relevant data on demand, supply and profit per unit of a product manufactured and sold by a company are given below : 2 8 6 4 10
1 5 2 3 4 45
Factory P Q R Demand
3 14 7 5 7 70
65
4 7 8 9 8 35
5 8 7 8 6
Supply 100 20 60 20
15
135
Given the transportation from A to 3, and D to 2 are not allowed due to certain reasons. Find out using VAM, optimal method of transportation from factories to marketing centres. 27. Solve the following problem in which cell entries represent unit costs : DI 2 3 5 1 7
Q1 Q2 Q3
Q4 Required
D2
D3
7 3 4 6 9
4 1 7 2 18
Available 5 8 7 14 44
Apply MODI method to test optimality. 28. Solve the following transportation problem From To S, S, S3 Requirement
D,
D,
D3
D4
Available
4 5 3 50
3 2 5 60
1 3 6 20
2 4 3 50
80 60 40 180 Total
Apply MODI method to test its optimality.
THE TRANSPORTATION PROBLEMS
247
29. Solve the following transportation problem: F A C T 0 R Y Demand
1
1 7
2 5
3 7
4 7
5 5
6 3
Stock available 60
2
9
11
6
11

5
20
3
11
10
6
2
2
8
90
4
9 60
10 20
9 40
6 20
9 40
12 40
50
30. A manufacturer of jeans is interested in developing an advertising campaign that will reach four different age groups. Advertising compaigns can be conducted through TV, radio and magazines. The following table gives the estimated cost in paise or exposure for each age group according to the medium employed. In addition, maximum exposure levels possible in each of the media, namely TV, radio and magazines are 40,30 and 20 millions respectively. Also the minimum desired exposures with each age group, namely 1318,1925,2635,36 and older, are 30,25,15 and 10 millions. The objective is to minimize the cost of attaining the minimum exposure level in each age group. Media TV Radio Magazines
Age groups 1925 2635 7 10 9 12 12 9
1318 12 10 14
36 an older 10 10 12
Formulate the above as a transportation problem and find the optimal solution. Solve the problem if the policy is to provide at least 4 million exposures through TV in the 1318 age groups and at least 8 million exposures through TV in the age group 1925. 31. STRONGHOLD Construction Company is interested in taking loans from banks for some of its projects P, Q, R, S, T. The rates of interest and the lending capacity differ from bank to bank. All these projects are to be completed. The relevant details are provided in the following table. Assuming the role of a consultant, advise this company as to how it should take the loans so that the total interest payable may be the least. Are there alternative optimum solutions ? If so, indicate one such solution. (i) (it)
Bank Private Bank Nationalized Bank Cooperative Bank Amount required (in thousands)
Interest rate in percentage for project P R S T Q 20 18 18 17 17 16 16 16 16 16 15 15 15 14 14 200
150
200
75
75
Maximum credit (in thousands) Any amount 400 250
OPERATIONS RESEARCH
248
32. A company has four terminals it, v, w and x. At the start of particular day 10, 4, 6 and 5 trailers respectively are available at these terminals. During the previous night 13,10, 6 and 6 trailers respectively were loaded at plants A, B, C and D. The company dispatcher has come up with the costs between the terminals and plants as follows : Plants A 20 40 75 30
Terminals w x
D 28 20 50 20
10 45 45 40
36 20 35 35
Find the allocation of loaded trailer from plants to terminals in order to minimize transportation cost. 33. The cost conscious company requires for the next month 300, 260 and 180 tones of stone chips for its three constructions, C1, C2 and C3 respectively. Stone chips are produced by the company at three mineral fields taken on short lease. All the available boulders must be curshed into chips. Any excess chips over the demands at sites C1, C2 and C3 will be sold exfields. The fields M1, M2 and M3 will yield 250, 320 and 280 tones chips respectively. Transportation costs from mineral fields to construction sites vary according to distances, which are given below in monetary unit (MU). To From M1 M2 M3
Ci
C2
8 5 7
7 4 5
C3 6 9 5
(i) Determine the optimal economic transportation plan for the company and the overall transportation cost in MU. (it) What are the quantities to be sold from M1, M2 and M3 respectively ? 34. A company has 4 different factories in 4 different locations in the country and four sales agencies in four other locations in the country. The cost of productions, the sale price, shipping cost in the cell of matrix, monthly capacities and monthly requirements are given below. Sales Agency Factory
1
2
3
4
Capacity
A B C D Monthly requirements Selling price
7 3 4 8
5 5 6 7
6 4 4 6
4 2 5 5
10 15 20 15
8
12
18
22
20
22
25
18
Cost of production 10 15 16 15
Find the monthly production and distribution schedule, which will maximize profits.
THE TRANSPORTATION PROBLEMS
249
35. A company wishes to determine an investment strategy for each of the next four years. Five investment types have been selected, investment capital levels have been established for each investment type. An assumption is that amounts invested in any year will remain invested until the end of the planning horizon of four years. The following table summarises the data for this problem. The values in the body of the table represent net return on investment of one rupee upto the end of the planning horizon. For example, a rupee invested in type B at the beginning of year 1 will grow to Rs. 1.90 by the end of the fourth year, yielding a net return of 0.90. Investment made at the beginning of year 1 2 3 4 Maximum rupees investment (in 000's)
Investment type A
BCD
E
Rupees available in (000's)
0.90 0.55 0.30 0.15
NET RETURN DATA 0.90 0.60 0.75 0.65 0.40 0.60 0.25 0.30 0.50 0.12 0.25 0.35
1.00 0.50 0.20 0.10
500 600 750 800
750
600
500
1,000
The object in this problem is to determine the amount to be invested at the beginning of each year in an investment type so as to maximize the net rupee return for the fouryear period. Solve the above transportation problem and get an optimal solution. Also calculate the net return on investment for the planning horizon of fouryear period. 36. Investment made
Net return rate P
Available
R
S
70
60
70
1
95
Q 80
2
75
65
60
50
40
3
70
45
50
40
90
4
60
40
40
30
30
Maximum investment (in lacs)
40
50
60
60
Solve the given problem so as to give maximum return. 37. Following is the profit matrix based on four factories and three sales depots of the company. Sales depots
Factories
Requirements
Availability
F1
6
S2 6
F2 F3
2
2
4
150
3
2
2
50
F4
8
5
3
100
80
120
150
Si
1
10
OPERATIONS RESEARCH
250
Determine the most profitable distribution schedule and the corresponding profit, assuming [CA (Final) Nov., 2000] no profit in case of surplus production. 38. Determine the optimal solution to the problem given below and find the minimum cost of transportation. 1'13 From A B C D B.
E
F
G
H
I
a
4 1 7 4 8
7 4 2 8 3
3 7 4 2 7
8 3 7 4 2
2 8 7 2 2
4 7 9 2
39. A transport corporation has trucks at 3 garages A, B and C in a city. The number of trucks available in each garage, the number of trucks required by each customer and the distance (km) from garage to customer's locations are given below.
Garages A B C Requirement
1
2
3
4
Availability
7 8 10 4
6 6 7 3
9 7 8 5
12 10 12 8
12 8 10
Just before sending the trucks, it is known that road from B to customers 2 is closed for traffic due to road block. How should the trucks be assigned to the customer is order to minimize the total distance to be covered to the road block. 40. A leading firm has three auditors. Each auditor can work upto 160 hours during the next month, during which time three projects must be completed. Project 1 will take 130 hours. 41. Determine an initial basic feasible solution to the following transportation problem using NorthWest Corner Rule:
From :
Demand
3 2 7 2 40
4 10 11 1 6
To: 6 1 20 9 8
8 5 40 14 18
9 8 3 16 6
Available 20 30 15 13
42. Determine an initial basic feasible solution to the following transportation problem using row minima method:
251
THE TRANSPORTATION PROBLEMS
5 4 4 7
From : Demand
To : 4 1 7 17
2 8 6 12
Available 22 15 8
3 6 5 9
43. Find the initial basic feasible solution of the following transportation problem by Vogel's Approximation Method: F1 Factory
F2 F3
Requirement
WI
W2 W3 W4
19 70 40 5
30 30 8 8
50 40 70 7
Capacity 7 9 18 34 (Total)
10 60 20 14
44. Determine an initial basic feasible solution to the following L.P. using : (a) NorthWest corner Rule, (b) Vogel's Approximation Method. Supply B1 C1 2 4 A 11 10 3 7 1 4 7 2 1 8 Origin C 3 9 4 12 9 8 Demand 3 4 6 3 5 45. Solve the transportation problem for which the cost, origin availabilities and destination requirements are given below: Al
02 03 0, b1
1 3 4 3 20
D2
D3
2 3 2 1 40
1 2 5 7 30
D, 4 1 9 3 10
D5 5 4 6 4 50
D6
a,
2 3 2 6 25
30 50 75 20 175 (Total)
46. Give a mathematical formulation of the transportation and simplex methods. What are the differences in the nature of problems that can be solved by these methods? 47. Given below is the unit costs array with supplies a, = 1, 2, 3 and demands 1)1 = 1, 2, 3 and 4.
Source
1 2 3 b
1 8 12 9 25
2 10 9 11 32
3 7 4 10 40
Find the optimal solution to the above Hitchcock problem.
4 6 7 8 23
al 50 40 30 120 (Total)
252
OPERATIONS RESEARCH
48. Consider four bases of operations B and three tartets T. The tons of bombs per aircraft from any base that can be delivered to any target are given in the following table : 1
2
3
1
8
6
5
Base (B)
2
6
6
6
3
10
8
4
4
8
6
4
The daily sortie capability of each of the four bases is 150 sorties per day. The daily requirement in sorties over each individual target is 200. Find the allocation of sorties from each base to each target which maximizes the total tonnage over all the three targets explaining each step. 49. General Electrodes is a big electrode manufacturing company. It has two factories and three main distribution centres in three cities. The supply and demand transportation. How should the trips be scheduled so that the cost of transportation is minimized ? The present cost of transportation is around Rs. 3100 per month. What can be the maximum savings by proper scheduling ? Centres
A
Requirement
50
50
150
Cost per trip from X plant
25
35
10
Cost per trip from Y plant
20
5
80
Capacity of plant X
15 units of electrodes
Capacities of plant Y
100 units of electrodes
50. A company has decided to manufacture some or all of five new products at three of its plants. The production capacity of each of these three plants is as follows : Plant No.
Production capacity in total number of units
1
40
2
60
3
90
Sales potential of the five products is as follows : Product No.
1
2
3
4
5
Market potential in, units
30
40
70
40
60
Plant No. cannot produce product No. 5. The variable cost per unit for the respective plant and product combination is given below : Product No.
1
2
3
4
5
Plant No. 1
20
19
14
21
16
Plant No. 2
15
20
13
9
16
Plant No. 3
18
15
18
20
Based on above data, determine the optimum product to plant combination by using linear programming.
253
THE TRANSPORTATION PROBLEMS
51. A fertilizer company has three plants A, B and C which supply to six major distribution centres 1, 2, 3, 4, 5 and 6. The table below gives the transportation costs per case, the plant annual capacities and predicted annual demands at different centres in terms of thousands of cases. The variable production costs per case Rs. 8.50, Rs. 9.40 and Rs. 7 respectively at plants A, B and C. Determine the minimum cost production and transportation allocation.
Plants Annual demand in thousand of cases
Transportation cost (Rs.)per case Major distribution centres 2 5 1 3 • 4 A 2.50 3.50 5.50 4.50 1.50 B 4.60 3.60 2.60 5.10 3.10 C 5.30 4.30 4.80 2.30 3.30 850
750
420
580
6 4.00 4.10 2.80
Annual production in thousands of cases 2,200 3,400 1,800
1020920
Prove that if the variable production costs are the same at every plant, once can obtain an optimal allocation by using transportations costs only. 52. Describe the transportation problem. Give method of finding an initial feasible solution. Explain what is meant by an optimality test. Give the method of improving over the initial solution to reach the optimal feasible solution. 53. The unit costs of transportation from site i to site j are given below. At site i = 1, 2, 3 stocks of 150, 200 and 170 units respectively are available. 300 units are to be sent to site 4 and the rest to site 5. Find the cheapest way of doing this.
From :
1 2 3 4 5
1 1 7 8 2
To: 3 4 2 9 7
2 3 4 3 1
4 10 16 12 5
5 7 6 13 5 
[Hint. In accordance with the restrictions of supply and demand, above table reduces to the following table: To:
From :
1 2 3 Required
4 10 16 12 300
5 7 6 13 220
Available 150 200 170
254
OPERATIONS RESEARCH
54. Consider the following unbalanced problem : To From 1
1
2
3
Supply
5
1
7
10
2
6
4
6
80
3
3
2
5
15
Demand
75
20
50
Since there is not enough supply, some of the demands at these destinations may not be satisfied. Suppose that there are penalty costs for every unsatisfied demand unit which are given by 5, 3 and 2 for destinations 1, 2 and 3 respectively. Find the optimal solution. [Hints. The balanced transportation table with dummy source and associated penalty costs are shown below. To From 1
1
2
3
Supply
5
1
7
10
2
6
4
6
80
3
3
2
5
15
Dummy
5
3
2
40
Demand
75
20
50
This table can now be solved by the usual MODI method.] 55. A production control superintendent finds the following information on his desk. In departments A, B and C the number of surplus pallets is 18, 27 and 21 respectively. In departments G, H, I and J the number of pallets required is 14, 12, 23 and 17 respectively. The time in minutes to move a pallet from one department to another is given below. To
G
H
I
J
A
13
25
12
21
B
18
23
14
9
C
23
15
12
16
From
What is the optimal distribution plan to minimize the moving time ? 56. The following table gives the cost of transporting material from supply points A, B, C and D to demand points E, F, G, H and J. To
E
F
G
H
J
A
st
10
12
17
15
B
15
13
18
11
C
14
20
6
10
13
D
13
19
7
6
12
From
The present allocation is as follows: AtoE90,AtoF10,CtoF 16,CtoG 50,CtoJ120,DtoH 210,DtoJ 70.
255
THE TRANSPORTATION PROBLEMS
(a) Check if this allocation is optimum. If not, find an optimum schedule. (b) If in the above problem the transportation cost from A to G is reduced to 10. What will be the new optimum schedule ? 57. The following table shows all the necessary information on the available supply to each warehouse, the requirement of each market and the unit transportation cost in rupees from each warehouse to each market. Market A Warehouse C
Requirement
I
II
III
IV
Supply
5
2
4
3
22
4
8
1
6
15
4
6
7
5
8
7
12
17
9
The shipping clerk has working out the following schedule from experience : 12 units from A to II, 1 unit from A to III, 9 units from A to IV, 15 units from B to III, 7 units from C to Ito 1 unit from C to III. (a) Check and see if the clerk has made the optimal schedule. (b) Find the optimum schedule and minimum total shipping cost. (c) If the clerk is approached by a carrier of route C to II who offers to reduce his rate in the hope of getting some business, by how much must the rate be reduced before the clerk should consider giving him an order ? 58. A company has factories at A, B and C which supply warehouses at D, E, F and G. Monthly factory capacities are 250, 300 and 400 units respectively for regular production. If overtime production is utilised, factories A and B can produce 50 and 75 additional units respectively at overtime incremental costs of Rs. 4 and Rs. 5 respectively. The current warehouse requirements are 200, 225, 275 and 300 units respectively. Unit transportation costs in rupees from factories to warehouses are as follows : To From , A B C
D
E
F
G
11 16 21
13 18 24
17 14 13
14 10 10
Determine the optimum distribution for this company to minimize costs. [Hint. First table is made which takes into account the overtime production and the corresponding production costs. A B C
Demand
D 11 15 16 21 21 200
E 13 17 18 23 24 225
F 17 21 14 19 13 275
G
14 18 10 15 10 300
Supply 250 50 300 75 400
256
OPERATIONS RESEARCH In the above table, total supply = 1075 units Total demand = 1000 units Therefore, we add a dummy warehouse with demand of 75 units and cost coefficients zero in each of its cell D E F G Dummy A B C
Supply
11
13
17
14
0
250
15
17
21
18
0
50
16
18
14
10
0
300
21
23
19
15
0
75
21
24
13
10
0
400
225 Demand 200 275 300 75 The initial feasible solution can now be obtained and can be optimized using MODI method]. 59. A company has plants at A, B and C which have capacities to produce 300 kg, 200 kg and 500 kg respectively of a particular chemical per day. The production costs per kg in these plants are Re. 0.70, Re. 0.60 and Re. 66 respectively. Four bulk consumers have placed orders for the product on the following basis : kg required per day
Price offered Rs. 1 kg I 400 1.00 II 250 1.00 III Consumer 350 1.02 IV 150 1.03 Shipping costs (in paise per kg) from plants to consumers are given in the table below. IV 5 4 A 3 6 8 11 9 12 From B 4 2 C 6 8 Work out an optimal schedule for the above situation. Under what conditions would you change the schedule ? 60. ABC manufacturing company wishes to develop a monthly production schedule for the next months. Depending upon the sales commitments, the company can either keep the production constant, allowing fluctuations in inventory or inventories can be maintained at a constant level, with fluctuating production. Fluctuating production in working overtime, the cost of which is estimated to be double the normal production cost of Rs. 12 per unit. Fluctuating inventories result in inventory carrying cost of Rs. per unit. If the company fails to fulfill its sales commitment, it incurs a shortage cost of Rs. 4 per unit per month. The production capacities for the next three months are as follows.
257
THE TRANSPORTATION PROBLEMS
Production capacity Sales Overtime Regular Month 60 30 50 1 120 0 50 2 40 50 60 3 Determine the optimal production schedule. Hint. Here regular and overtime production capacity is the source and the sales is the destination. The costs for different cells may be computed as follows : (i) For items produced and sold in the same month, there will be no inventory carrying cost. Thus, the costs for cells (1, 1), (2, 2), (3, 3) are Rs. 12 each and for cells (11,1), (31, 3) are Rs. 24 each. (ii) For items produced in a particular month and sold in subsequent months, additional inventory cost of Rs. 2 per month will be incurred. Thus, cells (1, 2), (1, 3), (2, 3), (11, 2) and (11, 3) will have costs of Rs. 14, Rs. 16, Rs. 14, Rs. 26 and Rs. 28 respectively. (iii) For items produced in a particular month to meet the backlog of sales during previous months, in addition to the production costs (normal or overtime), shortage costs of Rs. 4 per month will be incurred. Therefore, for cells (2,1), (3, 2), (3,1), (31, 2) and (31,1) the costs will be Rs. 16, Rs. 16, Rs. 20, Rs. 28 and Rs. 32 respectively. Thus, the equivalent transportation table for the given problem will be one shown below :
Months
Dummy Production capacity
1
2
3
1
12
14
16
0
50
2
16
12
14
0
50
3
20
16
12
0
60
11
24
26
28
0
30
31
32
28
24
0
50
Sales
60
120
40
20
PUNJAB UNIVERSITY EXAMINATION PROBLEMS U
1995 APR.
(1)
(i) Initial basic feasible solution (ii) NorthWest Corner Rule (iii) Unbalanced
1995 SEP.
(2)
(i) Initial basic feasible solution (ii) Transportation problem (iii) NorthWest
1996 SEP. 1999 APR. 2002 APR. 1995 APR.
(3) (4)
(i) Degenerate solution (ii) Prohibited routes in transportation problem.
transportation problem. Corner rule. (Marks 2) Balanced transportation problem. Initial Basic Feasible Solution of a transportation problem. (ii) (5) Describe transportation problem with its general mathematical formation. (Marks 10) 1995 SEP. (6) Explain the various steps involved in solving a transportation problem by (Marks 10) applying the NorthWest Corner rule. (i)
OPERATIONS RESEARCH
258
1996 APR. (7) Describe transportation problem with its general mathematical formulation. (Marks 6) 1996 SEP. (8) Explain various steps in Vogel's Approximation Method for finding initial ba(Marks 6) sic feasible solution of the transportation problem. 1997 APR. (9) Explain various steps involved in solving a transportation problem by any one (Marks 6) of the methods to solve it. (Marks 2) 1998 APR. (10) Transportation Problem 1998 SEP. (11) Write a brief note on Vogel's Approximation Method to solving transportation 6) problem. 1999 APR. (12) Explain the various steps involved in solving a transportation probl.fnt by applying the NorthWest Corner rule. (Marks 6) 1999 SEP. (13) Describe sequences of steps in MODI Method of solving a transportation prob(Marks 6) lem. (Marks 7) 2000 APR. (14) Explain the transportation problem giving examples. 2004 APR. (15) Explain (i) NWCM, (ii) LCEM, (iii) VAM and Test of optimality by (i) Stepping stone method and (ii) MODI method. Take suitable examples. (16) Find the initial basic feasible solution of the following transportation problem 1996 APR. with the help of NorthWest Corner rule.
Plant
X Y
A 11 07 11 18
C
21 17 23 28
16 13 21 25
14 26 36
Available at plant
Market requirement. 1996 SEP. (17) Find the initial basic feasible solution of the following transportation problem using least cost method : F1 F2 F3
Warehouse requirement
Wi
w2
w3
30 29 31 90
25 26 33 160
40 35 37 200
20 40 30 50
Factory Capacity 100 250 150
1996 SEP. (18) A construction company needs 3, 3, 4 and 5 million cubic feet of fill at 4 dam sites. It can transfer the fill from three mounds A, B and C where 2, 6 and 7 million cubic feet of fill is available. Costs (in lakhs of Rs.) of transporting one million cubic feet of fill from the mounds to 4 dam sites are : To \ From A B C
I 15 16 12
II 10 13 17
III 17 12 20
IV 18 13 11
THE TRANSPORTATION PROBLEMS
259
Determine the optimum distribution for this company to minimize the total cost. 1998 APR. (19) Suggest an optimal transportation plan with a view to minimize cost from the following information: Source Destination W2 W3 Units Available
Rs. Rs. Rs.
Cost of shipping per unit F1 F3 F2 0.9 1 1 1 1.4 0.8 1 1.3 0.8 20 15 10
Unit Demand 5 20 20 45
1999 APR. (20) A distribution system has the following constraints : Factory A B
Capacity (units) 45 15 40
C
Warehouse I II III
Demand (units) 25 55 20
The transportation cost per unit (Rs.) associated with each route is as follows: From A
I 10 15 7
B
C
1999 SEP.
II
III
7 12 8
8 9 12
Find the optimum transportation schedule and the minimum total cost of transportation. (21) Given the following information, compute optimal transportation cost using any method. Requirement per week, Truck Available per week Project A B C
Cost information From Plant X Plant Y Plant Z
Loads 45 50 20
Plants X
35 40 40
To Project A 5 20 5
C
10 30 8
10 20 12
260
OPERATIONS RESEARCH
2000 APR. (22) Determine the optimum solution to the following problem :
T1 10 13 4 14 3 60
F1 F2
From :
F3 F4 F5
Demand 2000 SEP.
Cost Matrix To : T2 20 9 15 7 12 60
Availability T3 5 12 7 9 6 20
T4 7 8 9 10 19 10
10 20 30 40 50
(23) Solve the following transportation problem as : (a) maximization problem and; (b) minimization problem.
Q1 Q2 Q3 Q4
Required
D1 2 3 3 4 30
D2
D3
D4
D5
D6
Available
1 2 5 2 50
3 2 4 2
3 4 2 1 40
2 3 4 2 30
5 4 1 2 10
60 40 60 30 (Marks 18)
E 8 7 8 10 1
F 5 11 2 9 2
Available 3 4 2 8
20
2001 APR. (24) Solve the following transportation cost problem :
Q R S T Required
A 5 5 2 6 3
B 3 6 1 10 3
C 7 12 2 9 6
D 3 5 4 5 2
2001 SEP. (25) Find the initial basic feasible solution by at least three different methods for the following transportation problem: From \ To F1 F2 F3 Demand
D1 10 1 7 3
D2 7 6 4 2
D3 3 7 5 6
D4
6 3 6 4
Available 3 5 7
THE TRANSPORTATION PROBLEMS
261
2001 SEP. (26) Solve the following cost minimizing transportation problem: 95 115 195 5
Factory Demand
Warehouses 105 180 180 4
80 40 95 4
15 30 70 11
Capacity 12 7 1
G.N.D.U. EXAMINATION PROBLEMS I
1996 APR. (1) Describe transportation problem with its general mathematical formulation. 1996 SEP. (2) Explain various steps in Vogel's Approximation Method for finding initial basic feasible solution of the transportation problem. 1997 ARP. (3) Explain various steps involved in solving a transportation problem by any one of the method to solve it. 1998 SEP. (4) Write a brief note on Vogel's Approximation Method to solving transportation problem. 1999 APR. (5) Explain the various steps involved in solving a transportation problem by applying the NorthWest Corner rule. 1999 SEP. (6) Describe sequence of steps in MODI Method of solving a transportation problem. 2003 APR. (7) Discuss various methods of getting basic feasible solution of transportations problem. Which one would you prefer and why ? 1995 APR. (8) Describe transportation problem with its general mathematical formulation. 1995 SEP. (9) Explain the various steps involved in solving a transportation problem by applying the NorthWest Corner rule. 1996 APR. (10) Find the initial basic feasible solution of the following transportation problem with the help of NorthWest Corner rule.
Plant
X Y
z
Market B 11 21 07 17 11 23 18 28 Market requirement
C
16 13 21 25
14 26 36
Available at plant
1996 SEP. (11) Find the initial basic feasible solution of the following transportation problem using least cost method : F1
F2 F3
Warehouse requirement
W1 30 29 31 90
W2
W3
W4
25 26 33 160
40 35 37 200
20 40 30 50
Factory capacity 100 250 150
OPERATIONS RESEARCH
262
1999 APR. (12) Suggest an optimal transportation plan with view to minimize cost from the following information: Cost of shipping per unit Source Destination
1996 SEP.
Units Demanded 5 1 1 0.9 WI 1 1.4 20 0.8 W2 1 20 W3 1.3 0.8 Unit Available 20 10 45 15 (13) A construction company needs 3, 3, 4 and 5 million cubic feet of fill at 4 dam sites. It can transfer the fill from three mounds A, B and C where 2, 6 and 7 million cubic feet of fill is available. Costs (in lakhs of Rs.) of transporting one million cubic of fill from the mounds to 4 dam sites are: From \ To A B C
F1
I 15 16 12
F3
F2
II 10 13 17
III 17 12 20
IV 18 13 11
Determine the optimum distribution for this company to minimize the total cost. 1999 APR. (14) A distribution system has the following constraints: Factory A B C
Capacity (units) 45 15 40
Warehouse II III
Demand (units) 25 55 20
The transportation cost per unit (Rs.) associated with each route are as follows: From \ To A B
1999 SEP.
I 10 15 7
Warehouse II 7 12 8
III 8 9 12
Find the optimum transportation schedule and the minimum total cost of transportation. (15) Given the following information, compute optimal transportation cost using any method: Project A B C
Requirement per week, truck loads 45 50 20
Plants X Y Z
Available per week, truck loads 35 40 40
THE TRANSPORTATION PROBLEMS
263
Cost information From Plant X Plant Y Plant Z
To project B 10 30 8
A 5 20 5
C 10 20 12
2001 APR. (16) Solve the following transportation problem : D1 D2 D3 D4
2001 SEP.
Si
S2
S3
S4
5 10 12 5 120
7 12 10 7 130
8 15 7 6 140
9 18 9 9 150
125 125 125 125
(17) Solve the following transportation problem. The cost matrix is given below :
Requirement
Source 2 10 12 9 15
1 7 9 12 25
A B C
3 12 10 14 30
4 10 10 12 10
40 30 20
2002 APR. (18) Find the optimum cost of transportation for the following problem : W
X
Y
I II III IV
30 40 40 50
30 50 40 20
Requirement
350
450
30 30 40 30 200
Availability 200 400 300 200
60 50 60 70 100
Is your solution unique ? Discuss. 2003 APR. (19) Find the optimum solution to the following transportation problem in which the cells contain the transportation cost in rupees : W2
W3
W4
W5
Available
6 5
4
5
9
F3
WI 7 8 6
8
6 9
7 6
8 5
40 30 20
F4
5
7
7
8
Required
30
30
15
20
6 5
F1 F2
10 100
2003 APR. (20) The following table gives the information regarding the quantity required by 4 markets and supply capacity of 3 warehouses. The unit transportation cost
264
OPERATIONS RESEARCH from warehouse to market is also given below. Find the optimum allocation that minimizes the total shipping cost. Warehouse
1
2
Market 3
4
Supply
A B
5 4 4 7
2 8 6 12
4 1 7 17
3 6 5 9
22 15 8
C
Requirement 2003 SEP.
(21) Find the basic feasible solution of the following transportation problem by VAM. Also find the optimal transportation plan. 1
2
3
4
5
Available
D
4 5 3 2
3 2 5 4
1 3 6 4
2 4 3 5
6 5 2 3
80 60 40 20
Required
60
60
30
40
10
200 (Total)
A B C
2004 APR. (22) Solve the following transportation problem: From
A
B
C
Available
P1 P3
4 16 8
8 24 16
8 16 24
56 82 77
Required
72
102
41
P2
[B.Com. III, Prof. Apr. 2004] 2004 APR. (23) Solve the following transportation problem for maximum profit Market Warehouse X Y
12 8 14
B
C
D
18 7 3
6 10 11
25 18 20
Availability at Warehouse
Demand in Markets
X : 200 Units Y : 500 Units Z : 300 Units
A : 180 Units B : 320 Units C : 100 Units D : 400 Units [B.B.A III, Apr. 2004]
6 Assignment M oeei m BIVIIP101101,114r21 La .I nriblI II nL •••0.2•• MI MOM mura• IP .16.•
• • • • • • • •
•••••••,/
•
Knowing the type of assignment problem and use of assignment models in industry and business Formulating an assignment problem Understanding the methods of solving assignment problems Using Hungarian Method Solving Unbalanced Problems Understanding solving of maximization assignment problems Understanding various types of assignment problems Explaining with illustration the travelling salesman problems
INTRODUCTION In real life situations, problems arise where a number of resources have to be allotted to a number of activities. In a sense, a special case of the transportation model is the Assignment Model. This model is used when the resources, have to be assigned to the tasks, i.e., assign n persons to n different type of jobs. Since different types of resources whether human, i.e., men or material, machines, etc., have different efficiency of performing different types of jobs and it involves different costs, the problem is how to assign such resources to jobs so that total cost is minimized or given objective is optimized. A plant may have 10 persons and 10 different types of job, the plant manager would like to know which person should be allotted which job so that all the jobs can be completed in least time (and hence least cost). Similarly, if a transporter has six trucks available for loading in each of the cities A, B, C, D, E and F and it actually needs these trucks in six locations 1, 2, 3, 4, 5 and 6, obviously the trucker would like to know which truck should be assigned to which location so that the transportation costs are minimized. In the same manner if a sales agency has say four sales man available (with different abilities and perhaps different capacities) and there are four territories where the agency wants to assign these sales man, the problem is which salesman should be allotted to which territory so as overall sales can be maximized. An assignment problem is, in fact, a completely degenerate form of a transportation problem. In this case, the units (resources) available at each origin and units demanded at each destination
OPERATIONS RESEARCH
266
are all equal to one, i.e., exactly one occupied cell in each row and each column of the transportation table. DEFINITION OF ASSIGNMENT MODEL Let us consider an assignment problem involving n resources (origins) to n destinations. The objective in making the assignment can be one of minimization or maximization (i.e., minimization of total time required to complete n tasks or maximization of total profit from assigning salespersons to sales territories). The following assumptions have to be made while formulating assignment models: Assumptions 1. Each resource is assigned exclusively to one task. 2. Each task is assigned exactly to one resource. 3. For purposes of solution, the number of resources available for assignment must equal the number of tasks to be performed. Let xij be the variable in such a way that if 1 if resource i is assigned to task j x• • = 'I 0 if resource i is not assigned to task j Cii = objective function contribution if resource i is assigned to task j. n= number of resources and number of tasks. Clearly, since only one job is to be assigned to each resource. =1
and
I =
And the total assignment cost will be given by n z=
n
yxi; ci;
j=1 f=1
Hence, the mathematical formulation assumes the following form: Determine xi
0
(i, j =1, 2, ......... n) so as to minimize Z =
Subject to the following constraints :
I x;;
j = 1, 2, 3,
,n
n
xi, .1 y j=i and
i = 1, 2, 3,
xij = 0 or 1
,n
267
ASSIGNMENT MODEL The general assignment model can be written as Maximize (or minimize) x11 + C12 C„ „ x„„ subject to x11+ x12 + • • •• • •• • • + xl n Z = C11
x12 + • • • •• • " • •+ C1n
n
=1 =1
xi1+ xi 2 +............+ x,, „ x21 + • • •• • • + xn 1 x12 + X22 + ......+ x 2
=1 =1 =1
+ + x„ „ xii = 0 or 1 for all values of i and j.
=1
x21 + x22 + • • •• • •• • • + x2
x11+
„
X2 „
x1n +
C21 x21 +.........+
The student should notice that for this model the variables are restricted to the two values i.e., 0 for non —assignment of the resources or 1 for assignment of the resource. This restriction is quite different from the other Linear. Programming models we have seen so far.
Theorem 1. The optimum assignment schedule remains unaltered if we add or subtract a constant to/from all elements of the row or column of the assignment cost matrix. Theorem 2. If for an assignment problem all C,10 then an assignment schedule (xi) which satisfies E E xij Cij = 0 must be optimal. These two theorems are the basis of the assignment algorithm. We add or subtract suitable constant to/from the elements of cost matrix in such a way that new C,1 0 and can produce at least one new C,1 = 0 in each row and each column and try and make assignments from among these 0 positions. The assignment schedule will be optimal if there is exactly one assignment in each row and each column (i.e., exactly one assigned 0). Solution of Assignment Problems 1. Complete Enumeration Method In this method costs for all possible assignments are worked out and the one having the minimum cost is termed as the optimal solution. This method, for obvious reasons, can only be used for small problems. As the problems become complex, it is impractical to workout a very large number of alternatives and then pick up the optimal solution.
2. Simplex Method In this method the simplex algorithm is used and we n n Cif x11 Minimize or Maximize Z = 1=1 j=1 Subject to constraints (i) xi 1 + Xi 2 + ......:.. + Xi „ = 1 (ii) x11 + x21+ .........+ x„ = 1 (iii) xi = 0 or 1 for all values of i and j.
i=1, 2, ...... , n j=1, 2, ...... n
OPERATIONS RESEARCH
268
It can be seen there are n x n decision variables and n + n = 2n equalities. It means that for a problem involving 8 workers/jobs there will be 64 decision variables and 16 equalities to be solved. It is an extremely cumbersome method. 3. Transportation Method
We have earlier mentioned that assignment model is a special case of transportation model, so it should be possible to solve it by transportation method. However, we know that optimality test in the transportation method requires that there should be n + n —1 = 2n — 1 basic variables, the solution obtained by this method would be severally degenerate. For an assignment made there would be only n basic variables in the solution, hence to proceed for solving an assignment model by using transportation model, a very large number of dummy allocations will have to be made, which will make this method very inefficient to compute. 4. Hungarian Assignment Method or HAM (Minimization case)
This method was developed by Hungarian mathematician D Koning and is also known as the Flood's Technique or the reduced matrix method. It is a simpler and more efficient method of solving the assignment problems. The following steps are involved while using the Hungarian method: Step I. Formulate the opportunity cost table by the following method : (a) Subtract the smallest number in each row of the original cost matrix from every number in that row. (b)Subtract the smallest number in each column of the table obtained at (a) above from every number in that column. Step II. Make assignments in the following manner : (a) Examine all the rows looking for a row with exactly one unmarked zero in a square ) as assignment has to made there. Cross (x) all other zeros in the column as these will not get an assignment in future. Proceed in this manner for all the rows. (b)Now examine all the columns one by one until we find a column with exactly one marked zero is located. Make an assignment to this single zero and put a square ) around it. Also crossout (x) all other zeroes appearing in the corresponding row as no assignment will be made in that row. Proceed in this manner for all the columns. (c)Operations (a) and (b) are repeated till. (i)All the zeros in rows/columns are either put in the square ( ) or are crossed out (x) and exactly one assignment is in each row and in each column. This is the optimal solution. (ii)Some row or column may be left without assignment, if so proceed to step III. Step III. Revise the opportunity cost matrix by (i) Marking (4) all rows that have no assignment. (ii) Marking (1) all columns which have zeros but have not been marked earlier. (iii) Marking (I) all rows that have assignments and have not been marked earlier. (iv) Repeat step III (i) and (ii) until no more rows and columns can be marked. (v) Draw straight lines through each unmarked row and each marked column.
269
ASSIGNMENT MODEL
Now check the total assignments indicated by number of lines drawn is equal to the number of rows or columns, the optimal solution has been reached. Otherwise proceed to step IV. Step IV. Write the new revised opportunity cost matrix. Initial opportunity cost matrix may never give the optimal solution, we are normally required to revise this table in order to move one or more zero costs from present location to new uncovered locations. This is done by subtracting the smallest number not covered by a line from all numbers not covered by a straight line. This number (smallest) is added to every number, including zeros available at the intersection of any two lines. Step V. Repeat steps II to IV until an optimal solution is achieved. The above steps are shown in the form of a flow chart in the following figure: Prepare the Assignment Cost table for the Problem
Yes
Determine if a Maximization Problem
Convert is into a Minimization Problem. Subtract all elements from the largest element,
4
• Establish if it is a balanced problem
No
Convert into a balanced problem
Yes
• Subtract the smallest element in each row from all the elements of that row
Subtract the smallest element in each column from all the elements of that column
• Draw minimum number of lines to cover all zeros in each row and column
• Each of the number of lines drawn is equal to the order of matrix •
No
Yes
Subtract the smallest element of uncovered line from other elements and add to element lying at intersection Determine the total cost
Optimal solution obtained
Fig. 6.1
270
OPERATIONS RESEARCH
PRACTICAL STEPS INVOLVED IN SOLVING MINIMIZATION / MAXIMIZATION PROBLEMS Minimization Problems
Step I. Check if the number of rows is equal to number of columns. If it is equal, the problem is a balanced one. If not, add a dummy column or row to make it a balanced one, by allotting zero values to each element (cell) of dummy row or column as the case is. Step II. Row subtraction. Subtract the minimum or least value element of each row from all elements of that row. • Step III. Column subtraction. Subtract the minimum or least value element of each column from all elements of that column. Step IV. Draw minimum number of horizontal and /or vertical lines in such a manner that all zeros are covered. For doing this, follow the procedure as follows : (a)Select a row containing exactly one uncovered zero and draw a vertical line through the column containing this zero and repeat the process till no such row is left. (b)Select a column containing exactly one uncovered zero and draw a horizontal line through the row containing the zero and repeat the process till no such column is left. 7 Step V. If the total number of minimum lines covering all zeros is equal to the order of the matrix, then we have got the optimal solution and there is no need to proceed further. Step VI. If not, subtract the minimum uncovered element from all the uncovered elements and add this element to all the elements at the intersection points of the lines covering zeros. Step VII. Repeat steps IV, V and VI till minimum number of lines covering all zeros is equal to the order of the matrix. Step VIII. Now make assignments by selecting a row containing exactly one unmarked zero and put a square around it. Also, draw a vertical line through the column containing this zero. Repeat this process till no such row is left. Move to the column containing exactly one unmarked zero and put a square around it, also, draw a horizontal line through the row containing this zero and repeat this process till no such column is left. If there are more than one zeros in any one row or column, select any one arbitrarily and pass two lines horizontally and vertically. Step IX. Find the optimum value by adding up the values of the final assignments. Maximization Problems
Step I. Rewrite the problem as a minimization problem by subtracting all elements from the largest element. Step II. Follow the same steps as above from I to IX. Unbalanced Problems
Establish whether the problem is a balanced one, i.e., the number of rows = number of columns. If yes proceed as discussed earlier. If not, then add a dummy row or column to make the problem
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271
a balanced one by allotting zero values to each cell of the dummy row or column, as the case may be. Example 6.1. A department head has four subordinates and four tasks to be performed. Subordinates differ in efficiency and the tasks differ in their intrinsic difficulty, His estimate of time each man would take to perform each task is given in the matrix below. Men Tasks
E
F
G
H
A
18
26
17
11
B
13
28
14
26
C
38
19
18
15
D
19
26
24
10
How should the tasks be allotted, one to a man, so as to minimize the total man hours ? Solution. Step I. Subtract the smallest element of each row from every element of the corresponding
row, the reduced matrix is as follows (subtract 11 from row one, 13 from row 2, 15' from row 3 and 10 from row 4) 7 15 6 0 0 15 1 3 23 4 3 0 9 16 14 0 Step II. Subtract the smallest element of each column of the abovereduced matrix from every
element of the corresponding column, we get the following reduced matrix (subtract 0 from column 1,4 from column 2,1 from column 3 and 0 from column 4), 7 11 5 0 0 11 0 13 23 0 2 0 9 12 13 0 Step III. Starting with row 1, we make assignment to a single zero and put a square (
around it and cross out all other zeros in the column so marked. By doing so, we get 7
11
0
11
23
0 12
9
5
0 13
2 13
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272
N 0 711111V that row 2 had two zeros, we have arbitrarily made assignment to zero in column 1 and put a square around it. Also, note that column 3 and row 4 do not have any assignment.
Step IV. (i) Tick row 4 as it does not have any assignment. In fourth column of the ticked row (row 4), there is a zero, so we tick fourth column. (ii) In the first row of ticked column (column 4) there is an assignment made, so first row is ticked. (iii) Draw straight lines through all unmarked rows and marked columns. The following matrix shows the above operations: 7 11 5
0
11
X
13
2
X
23
9 12 13 X Step V. Is the present solution an optimal solution ? To find that we know, there are 3 lines drawn which is less than the order of the cost matrix (4). Hence it is not an optimal solution we have to generate new zeros in the matrix to increase the minimum number of lines. Step VI. Let us find out the smallest element not covered by the lines. It is 5. Subtract 5 from all the uncovered elements and add 5 to all the elements lying at the intersection of lines, we get the following reduced matrix. 2 6 0 0 0 11 0 18 23 0 2 5 4 7 8 0 Step VII. Repeat step III on the reduced matrix, i.e., scrutinise all the elements row wise, say with row one, make assignment to a single zero and crossout all other zeros in the column so marked, we get
A B C
D
E F G 6 0
H_ )3:
0 11 X 18 23
0
2
5
4
7
8
0
It can be seen now each row and each column has only one assignment, an optimal solution has been obtained. The optimum assignment is A G, B —> E, C F, DH. The minimum total time for this assignment would be obtained by adding the relevant times, i.e., 17 + 13 + 19 + 10 = 59 manhours.
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273
Example 6.2. A marketing manager has 5 salesmen and 5 districts. Considering the capabilities of the salesman and the nature of districts, the marketing manager estimates that sales figures per month (in hundred rupees) for each salesman in each district would be as follows: Salesman 1 2 3 4 5
A 30 40 41 22 29
Districts C 40 28 33 41 40
B 38 24 27 38 33
D 28 21 30 36 35
E 40 36 37 36 39
Find the assignment of salesman to districts that will result in maximum sales.
Solution. This is a maximization problem and has to be converted into a minimization problem by subtracting all the elements from the largest element of the sales table. Here largest element is 41. Hence, the equivalent sales table for this problem would be obtained by reducing all the elements from 41 and rewriting it as follows : 11 1 0 19 12
3 17 14 3 8
1 13 8 0 1
13 20 11 5 6
1 5 4 5 2
Step I. Subtract the smallest element of each row from every element of the corresponding row, we get 10 0 0 19 11
2 16 14 3 7
0 12 8 0 0
12 19 11 5 5
0 4 4 5 1
Step II. Subtract the smallest element of each column from every element of the corresponding column, we get the following reduced matrix : 10 0 0 19 11
0 14 12 1 5
0 12 8 0 0
7 14 6 0 0
0 4 4 5 1
Step III. Starting with row one, we make assignment to a single zero and cross out all other zeros in the column so marked, we get
OPERATIONS RESEARCH
274 10
0
X
7
0
0
14
12
14
4
12
8
6
4
19
1
0
11
5
5 0
1
Here row three and column 4 do not have any assignment. Step IV. Draw the minimum number of horizontal and vertical lines which cover all the zeros as follows. X
7
0
14
12
14
4
12
8
6
4
14
4
5
0
1 1 .5 X  i Since the number of lines (4) is less than the order of the matrix (5) the solution is not optimal. Step V. The least uncovered element 4 is subtracted from all the uncovered elements and added to the intersection of elements, we get the following reduced matrix. A B.  : 1 14 .0 11 2 41 4
23
5
15
C D E
'0
4 .4
>4
2 5
Optimum assignment is 1
B, 2 —> A, 3 —> E, 4 —> C, 5 —> D.
i.e., 38 + 40 + 37 + 41 + 35 = 191 Maximum sales would be Rs 19100. The Travelling Salesman Problem
One of the major applications of the assignment models is in the travelling salesman problem. Let us say that a salesman has to visit n destinations. He startsfrom a particular city, visits each destination once and then comes back to the city from where he started. Here the objective _would be to minimise the time this salesman takes to visit all the destinations. The problem is to select that sequence in which all the destinations are visited in such a manner that the time
275
ASSIGNMENT MODEL
taken is minimised. This problem is similar to the assignment problems already seen except that there is an additional restriction that the salesman starting from a particular city, visits each destinations only once and returns to his city from where he started. Formulation of Travelling Salesman Problem
Let us define the variables x, k as (notice constraint k has been added in addition to i and j already there)
Xi i k =
[1 if kth destination from i to j 0 if kth destination is not from city i to city j
where i, j and k are integers which vary between 1 and n. The following constraints can be put in the mathematical form as shown. k= 1, 2, 3, ...... n and i # j
(a) j
k
(b) As only one city can be reached from a specific city, say i, n
I j
II
I xiik = 1 where i = 1, 2, 3, k
And only from one city the salesman can go to a specific city say j 1 where j = 1, 2, 3, (d) Given that kth visit ends at some specific city j, (k + 1) th visit must start at the same city j, thus
we have xiik = iii
xii(k+i) for all values of j and k. imj
Now the objective function nnn xiik dij Minimise Z= j
i j
k
where dii is the distance from city i to city j. Example 6.3. Solve the following travelling salesman problem so as to minimize the cost per cycle: From A
B
C
D
E
3
6
2
3
5
2
3
6
4
B
3
C
6
5
D
2 3
2 3
6 4
6 6
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276
Solution. Step I. Row Reduction : Let us subtract the smallest element of each row from other elements of the same row. In prohibited cell 0. will be assigned as shown below : To 4
D 0
E 1
3
0
1
1
.0
2
()
0
0
4
00
4
0
0
1
3
00
C
1
B 1 00
C
2
D E
From A
A 00
B
Step II. Column Reduction : Subtract the smallest element of each column from other elements of the same column. Assignments will be made in rows and columns having single zero. To C
D
3
0
E 1
0
1 0
B
1
B 1 00
C
2
1
2 00
D
0
0
3
2 00
E
0
0
0
3
From A
4 00
Step III. Draw the minimum number of lines to cover all the zeros after marking rows and columns and after drawing lines through unmarked rows and marked columns. This is shown below. To C
1
B 1 00
C
2
1
D
0 X
From A
00
B
E 1
3
0
2
1
—
X 2
X
3
,,.,
4
X
0
3
—
0
Step IV. The above table can be modified by subtracting the lowest element, i.e., 1, from all the elements not yet covered by the lines and adding the same in the intersection of the two lines. Assignments can now be made. This gives the table as follows.
ASSIGNMENT MODEL
277 To
From A
A
B
X 2
2
D
0 ❑
E
X
00
1 00
1
X X
3
❑ 0
X
X
E
x
3
F01
00
4
4
00
Step V. The above table gives an infeasible solution as the optimum assignment is A —) B, D, D > A and the salesman does not visit C and E and returns home to A. Let us try and find the next best solution in which C and E cities are also visited by the salesman. Step VI. Let us make assignment at (2, 3) instead of 0 zero at (2, 4), which is the next higher value, 1 in the matrix. Assign 1 The table is as shown below. B
To From A B
00
C
X 2
D
X
E
C 2
D 0
E
x x 3
3 00
4
X
4
00
It may be seen optimum assignment is Minimum cost per cycle = 2 + 2 + 5 + 4 + 3 = 16. Example 6.4. A salesman has to visit five cities A, B, C, D and E. The intercity distances are tabulated below. To A B C D E From A — 12 25 16 26 B 7 — 17 19 8 C 10 11 — 12 19 D 15 18 23 — 17 14 E 12 24 26 — The above distances in km between two cities need not be same both ways. Which route would you advise the salesman to take so that the total distance travelled by him is minimum, if the salesman starts from city A and has to come back to city 'A' ?
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278
Solution. Step I. Select the smallest element in each row and subtract it from every element of that row. Row reduction results in the following matrix: To A B D E C From A 14 4 0 13 B 0 12 1 10 C 0 1 9 2 D 0 3 8 2 E 0 2 12 14 Step II. Select the minimum element in each column and subtract this element from every element in that column, we get A To B C From A 0 5 B 0 2 C 1 0 D 0 3 0 E 2 4 0 Step III. Draw minimum number of lines to cover all the zeros.
D
E
5 3 0 5
3 0 1 1
To A B C D E From A 0 5 5 3 B 0 2 & 3 C 1 1 0 0 D 0 0 3 E 0 2 4 5 Since the number of lines = order of matrix (5), this gives the optimal solution. Step IV. Making assignment in zero positions does not satisfy the constraints of the problem. Hence the following assignment are made by hit and trial. To From A
A
B
0
C
0
1
D
0
3
0
2
4
E
B
C
D
E
0
5
5
3
2
3
0
0
1 1
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279
Optimal schedule is A —> B, B C, C D, D E, E —> A Involving distance 12 + 17 + 19 + 17 + 12 = 77 km. This route satisfies the travelling salesman problem. UNBALANCED PROBLEMS Example 6.5. A transport corporation has three vehicles in three cities. Each of vehicles can be assigned to any of the four other cities. The distance differs from one city to the other as under.
A B C
1 33 42 40
2 40 30 31
3 43 31 37
4 32 24 31
You are required (a) To assign a vehicle to a city in such a way that the total distance travelled is minimized. (b) Formulate a mathematical model of the problem. Solution. (a) Step I. Introduce a dummy vehicle in city D as the problem is not balanced and take 0 distances as follows: 1 2 3 4 A 33 40 43 32 B 42 30 31 24 40 31 C 37 31 D 0 0 0 0 Step II. Subtract the minimum element of each row from all the elements of that row. There is no need to subtract 0 from each column. The matrix is 1 11 8 0 18 6 7 0 9 0 6 0 0 0 0 0 Step III. Draw minimum number of lines to cover all zeros. It can be seen that number of lines = 3, order of matrix = 4. Hence this is not an optimal solution and we have to take steps to increase number of zeros. 1 18
8 6 0
11 7 6
0 0
Step IV. Subtract the minimum uncovered element (1) from all the uncovered elements and adding it to the elements at the intersection point. Draw minimum number of lines to cover all the zeros, we get
OPERATIONS RESEARCH
280 7 5
17 9
6 6
0 Now, number of lines = 4, and is equal to the order of the matrix hence, this matrix will provide an optimal solution. Step V. Assignments can be made as shown below. 7
0
10 6
0
6
0
0
1.
Step VI. Minimum distance can be worked out as follows : Vehicle 1 4 2 3
City A B C D
Distance 33 24 31 0 Total = 88
(b) Formulation of LPP.
33
40 1x21
42
1x22
1x41
32
I X34
Ix33 37
31 I X43
I x42 0
I x24 24
31
31
lx14
1 X23
Ix32
I x31
0
43
30
40
1x13
lx12
1 xil
0
I x44 0
Minimize Z = 33 x11 + 40 x12 + 43 x13 + 32 x14 + 42 x21 + 30 x22 + 31 x23 + 24 x24 + 40 x31 Subject to
+ 31 x32 + 37 X33 + 31 X34 + 0 x41 + 0 X42 + 0 X43 + 0 X44 X11 + X12 + X13 + X14 = x21 + x22 + x23 + x24 = X31 + X32 + X33 + X34 = 1 X41 + x42 + x43 + x44 = xi •j >— 0•
ASSIGNMENT MODEL
281
Example 6.6. Solve the following unbalanced assignment problem of minimizing total time for doing all the jobs: Jobs Operator 1 2 3 4 5 A 8 3 6 3 7 B 3 6 9 8 8 C 9 9 7 9 9 D 7 2 3 5 6 E 10 3 8 9 7 F 5 7 4 7 8 Solution. Step I. Introducing dummy job 6 to make the problem balanced allotting 0 timing. Jobs Operator
1
2 4 5 3 6 A 8 3 6 3 7 0 B 3 6 9 8 8 0 C 9 9 7 9 9 0 D 7 2 3 5 6 0 E 10 3 8 9 7 0 F 5 7 4 7 0 8 Step II. Since 0 is in each row, there is no need of row deduction, column deduction is done by subtracting the minimum element of each column from all the elements of that column, we have Jobs 1 2 3 4 5 5. 1 3 0 1 B. 0 4 6 2 5 0' 6 7 4 C 6 3 0 D 4 0 2 0 0 *0 E 7 1 5 6 1 0 2 5 1 F 4 2 6 Draw minimum number of lines to cover all the zeros, since lines = 4 is less than the order of the matrix (6) this is not an optimal solution. Step III. Subtracting the minimum uncovered element (1) from all uncovered elements and adding it to the elements of intersection points of the lines, we have Operator A
• • • • •
Operator A B C
D
1
2
Jobs 3
3
4
4
3
Q
4
4
• • • • •
• • • • •
4 6 1
1
4 $ 3
Q 0
i
2
I
1 1 6 0 3
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282
Since number of lines drawn (6) = order of matrix, this is the optimal solution. Step IV. Assignments are made as shown below. Jobs Operator
6
4
A B
1 LI
4
6
2 5
C
2
D E
4
6
0
0
4 Step V. Computing the minimum time Operator A B
Job 4 1 6 5 2 3
Time 3 3 C 0 D 6 E 3 F 4 Total = 19 Example 6.7. Four operators A, B, C and D are available to a manager who has to get four jobs J —1, J — 2, J — 3 and J — 4, done by assigning one job to each operator. Given the time needed by different operators for different jobs as per the following matrix: J—1 J—2 J—4 J— 3 A 10 8 10 9 B 12 10 14 10 C 6 8 16 6 D 8 9 7 7 How should the manager assign the jobs so that the total time needed for all four jobs is minimum? Solution. Subtract the minimum element of each row from all the elements of that row, we have Step I. J—1 J—2 J—4 J— 3 A 2 0 2 1 B 2 0 4 0 C 0 2 10 0 D 1 2 0 0
ASSIGNMENT MODEL
283
Step II. Subtract the minimum element of each column from all elements of that column. Since all columns have zero, there will be no change in the matrix. Step III. Draw minimum number of lines to cover all the zeros, the matrix is as follows: J—1
J—2
2
0 0
A
B
J— 3
J4
2 4 0 2 10 D 1 2 0 As the number of lines (4) is equal to the order of matrix, this is the optimal solution. Step IV. Making the assignments
A
2
B
J —2
J3
0
2
0
4 10
C
0
2
D
1
2
J—4
0
0
Step V. Computing the minimum time Operator A
Job J—2
B
J—4 J—1
C D
J— 3
Time 8 10 6 7
Total = 31 Example 6.8. A manufacturing company has four zones Z — 1, Z — 2, Z — 3 and Z — 4 and four sales engineers A, B, C and D respectively for assignments. The zones are not equally rich in sales potential and it is estimated that a particular engineer operating in a particular zone will yield the following sales: Z —1 : 450000 Z — 2 : 340000 Z — 3 : 300000 Z — 4 : 460000 The engineers have different sales ability. Working under the same conditions their yearly sales are proportional to 15, 10, 12, 8 respectively. The criteria of maximum expected total sales is to be met by assigning the best engineer to the richest zone, the next best to the second richest zone and so on. Find the optimal assignment and the maximum sales.
OPERATIONS RESEARCH
284 Solution.Step I. The revenue matrix is as shown below. Zones (sales in thousands of Rs) Z 3 Z 2 15 15 340 x — 300 x — 45 45 B 10 340 x — 300 x 10 45 45 12 C 12 340 x — 300 x — 45 45 D 8 8 340 x — 300 x — 450 x$ 45 45 45 The matrix, after multiplications is reduced to Territory Engineers A
Z 1 15 450 x — 45 10 450 x — 45 12 450 x — 45
Z 4 15 460 x — 45 10 460 x 45 12 460 x — 45 8 460 x — 45
Z 4 Z2 Z3 154 114 100 A 102 B 67 75.5 (76) 123 80 120 C 90.5 (91) 82 D 80 53.5 60.5 (54) (61) Step II. Now the problem can be reduced to minimisation i.e., loss matrix can be formulated by subtracting all elements from the largest value i.e., 154. Z1 150 100
Z4 Z1 Z2 Z3 0 4 40 54 A 52 78 87 B 54 31 63 74 C 34 72 D 74 100 93 Step III. Subtract the minimum elements of each row from all elements of that row, we have Z4 Z1 Z2 Z3 A 4 40 54 0 B 2 26 35 0 32 43 0 C 3 D 2 21 28 0 Step IV. Subtract the minimum element of each column from all elements of that column and draw minimum number of lines to cover all zeros, we get
Z• A B C
D
1
Z 2 19 5 11
Z 3 26 7 15
ASSIGNMENT MODEL
285
Since there are 3 lines and order of matrix is four, this matrix does not provide the optimal solution. Step V. Subtract the minimum uncovered element (1) from all uncovered elements and add to the elements at the intersection points of the lines and again draw minimum number of lines to cover all zeros. Z1 A
1
Z2 18
B
0
5
Z3 25
Z€4 0
7
1
14 C 9 10 1/ } I3 .o 0 o i Again there are 3 lines hence it is not optimal solution, so subtract the minimum uncovered element (5) from all elements which are uncovered and add to the elements at the intersection points, we get A
Z1 1
Z2
Z3
Z4
13
20 2
0 1
B C
0 0 0 5 9 0 D 5 6 0 0 Now number of lines = order of matrix, so this will give us the optimal solution. Step VI. Making the assignments in the usual manner, we have z1
B
Z.2
Z3
Z4
13
2Q
4
0
1
Q
C
5
z
D Step VII. Computing the maximum sides Engineer
Zone
A
Z4
B
Z2
C
Z 1
D
Z3
Sales (Rs) 15 460000 x — = 460000 15 10 340000 x — = 226666 15 2 450000 x L = 360000 15 8 300000 x — = 160000 15 Total = 1206666
286
OPERATIONS RESEARCH
It can be seen from above assignments that the best engineer A has been assigned the richest zone Z — 4, the next best engineer C has been assigned the next richest zone Z — 1, and so on. Example 6.9. M/s Steadfast Enterprises has four plants each of which can manufacture any one of the four products. Product costs differ from one plant to another as follows: Product Plant 1 2
P 33 45 42 27
3 4
Q 40 28
R 43 30
29 42
35
S 32 23 29
45 38 Find out which product each plant should produce to minimize cost. Also, formulate a LPP. Solution.Step I. Subtracting the minimum element of each row from all the elements of that row. 1 12
8 5
13 0
0 15
1
8 5
5 1
15
12
11
0 0
7
6 0 18 11 Step II. Subtracting the minimum elements of each column of the above matrix from all the elements of that column.
12 13 0
11
Step III. Draw minimum number of lines (horizontal/vertical) to cover all the zeros. 1 12
8 5
1
13
0 15
0 12
0
5
(? (i)
11 Since the number of lines, drawn is (3) less than the order of the matrix (4), we have to take steps to increase zeros and present solution is not optimal solution. Step IV. Subtract the minimum uncovered element (1) from all the uncovered elements and add it to the elements at intersection point, we get the matrix. 0
11 13 0
7 4 0 15
4 0 12
1 12
287
ASSIGNMENT MODEL Step V. Draw the minimum number of lines to cover all the zeros, we get 4 0 .0. 12
4 fl
fl
0
12 6 15 Since the number of lines drawn (4) is equal to the order of the matrix (4), the above matrix gives the optimal solution. around it and Step VI. Select a row containing exactly one unmarked zero and put a draw a vertical line through the column containing this zero. Repeat this till no such row is left. around it and draw Then select a column containing exactly one unmarked zero and put a a horizontal line through the row containing this zero. Repeat this process till no such column is left. I"
R
1
4
2
11
3
13
4
"
4
Step VII. Optimal assignment is 1 S = 32 2>S=30 3 >S= 29 4>S=27 Minimum cost = 118 Formulation of LPP
I Xi 1 33
40
42
I X23
I x4 I
23
35 Ix42
42
I x24
I X33
I x32 29
lx14 32
30
28 1X31
27
43
Ix22
lx21 45
I X13
lx12
1X34 29
1x43 45
Maximize Z = 33 xii + 40 x12 + 43 x13 + 32 x14 + 45 x21 + 28 x22 + 30 x23 + 23 x23 +
Ix44 38
288
OPERATIONS RESEARCH ' 42 X31 + 29 X32 + 35 X33 + 29 X34 + 27 x41 + 42 X42 + 45 x43 + 38 x44
Subject to
+ x12 + x13 + x14 = 1 x21 + x22 + x23 + x24 = 1 X31 + X32 + X33 + X34 = 1
X41 + X42 + x43 + x44 = 1 and x11 ?_. O. Example 6.10. A Company is faced with assigning 5 jobs to 5 operators. Each job must be performed only by one operator. The cost of processing of each job by each operator is given below in Rs. Operators
A B
Jobs
C
D E
P
Q
R
S
T
7 9 8 7 5
5 12 5 3 6
9 6 4 6 7
8 11 6 8 5
11 10 8 5 11
Determine the assignment of jobs to the operators so that it will result in minimum cost. Solution.Step I. Select the minimum element in each row and subtract this element from every other element in that row.
A B C
D E
P
Q
R
S
2
0 6 1 0 1
4
3 5 2 5 0
3 4 4 0
0 0 3 2
T 6 4 4 2 6
Step II. Now select the minimum element in each column and subtract this element from every element of that column, we get the following matrix :
A B C
D E
P
Q
2
0 6 1 0 1
3 4 4 0
R 4 0 0 3 2
S 3 5 2 5 0
T 4 2 2 0 4
Step III. In row A, there is a single zero so assignment is made in cell AQ. In row B, there is a single zero, so assignment is made in cell BR. While making assignment in row AQ, the other zero appearing in the column Q, i.e., in element DQ is crossed out. Similarly, when assignment
289
ASSIGNMENT MODEL P A
2
Q
S
1
4
3
i
t
5
IF
,
B
3
C
4 4
E
0
1 x
1
I
)k si,
2 5
10
i
While inspecting rows, row D has a single zero, assignment can be made in cell DT. No assignment can be made in row E, since it has two zeros. Now inspect columns, column P has single zero, so assignment can be made in cell EP and other zero in row E, i.e., in cell ES can be crossed out. Since it is possible only to make 4 assignments against 5, so optimum solution is not reached. Number of lines drawn is equal to the number of assignments made. Step IV. Examine the elements not covered by the lines and select the smallest element, in this case 2 and subtract this from all the uncovered elements and add this to the elements lying at the intersection of the two lines. This gives us the following matrix : P A
Q
R
S
T
0
4
1
4
0
3
2
0
2
3
0
0
6
B
1
6
C
2
1
D
2
E
0
3 3
Step V. Make assignment in row C where there is a single zero. Hence assignment in cell CS is made. Step VI. Optimum solution A>Q=5 B R=6 C —> S = 6 D > T = 5 E P=5 Minimum Total Cost = Rs 27. Example 6.11. The following is the cost matrix of assigning 4 professors to 4 key courses. Class preparation time in hours for different topics varies from professor to professor and is given in the table below. Each professor is assigned only one course so as to minimize the total preparation time for all courses.
290
OPERATIONS RESEARCH Courses Professor
C1 2 15 13 4
A B C D
C2
C3
C4
10 4 14 15
9 14 16 13
7 8 11 9
Solution. Step I. Select minimum element from each row and subtract it from other elements of that row, we get Courses CI 0 11 2 0
A B Professors
C
D
C2
C3
8 0 3 11
7 10 5 9
5 4 0 5
Step II. Select minimum element from each column and subtract it from other elements of that column. Courses
Professors
A B C
D
C1 0 11 2 0
C2
C3
C4
8 0 3 11
2 5 0 4
5 4 0 5
Step III. In first row, as there is a single zero, assignment can be made in cell AC1 and crossing out other zeros in column C1. In second row assignment can be made in cell BC2, as there is a single zero. Similarly, assignment can be made in cell CC3 as follows: Courses C2
C3
C4
2
5
B
5
4
C
4".1 .46 •
A Professors
D
0
8
11
4
5
ASSIGNMENT MODEL
29,1
As the number of assignment is less than the order of matrix, we have to proceed further as it is not the optimum solution. Also, draw minimum lines to cover all the zeros, we have Courses C2
C3
C4
A
0
8
2
5
B
11
0
5
4
C
2
3
0
0
D
8
11
4
5
Professors
Step IV. Select the minimum number out of the uncovered elements (2) subtract it from the uncovered elements and add to the elements at the intersection of lines, we get Courses
Professors
C1
C2
C3
C4
A
0
6
0
3
B
13
0
5
4
C
4
3
0
0
D
0
9
2
3
C3
C4
6
0
3
5
4
Step V. Make fresh assignments Courses C2 A Professors B
13
0
C
4
3
D
0
9
0 2
3
Now the number of assignments is equal to the order of the matrix hence, this is the optimal solution. Professor
Course
A
C3
9
B
C2
4
C
C4
11
D
C1
4
Time
Total minimum course duration = 28 minutes
OPERATIONS R,ESEARCH
292
Example 6.12. In a modification of plant layout of a factory four new machines are to be installed in a machine shop. There are five vacant sheds A, B, C, D and E available. Because of limited space machine 2 cannot be placed at C and M3 cannot be placed at A. The cost is shown below. Find optimum assignment schedule. M1 M2
B 11 9 11 8
A 9 12
M3 M4
14
D 10 10 11 7
C
15 — 14 12
E 11 9 7 8
Solution. Since there are four rows and five columns, we introduce dummy row M5 to above matrix and assign 00 to the prohibited cell and we get
M2
A 9 12
B 11 9
M3
co
11
M4
14
M5
0
Ml
8
C 15 0. 14 12
D 10 10 11 7
E 11 9 7 8
0
0
0
0
Row Reduction Take least elements from all rows and deduct it from all other elements. A M,
0
M2
3
M3
00
M4
M5
7 0
B 2 0 4 1 0
C
6 7 5 0
D 1 1 4 0 0
E 2 0 0
1 0
Since every column contains zero, there is no need to follow column reduction. A M1
0
B 2
M2
3
0
1
M3
00
4
4
0
M4
7
1
0
1
M5
C
6
5 0
D 1
E 2
293
ASSIGNMENT MODEL Cost. M1 is given A = 9 M2 is given B = 9 M3 is given E = 7 M4 is given D = 7 M5 is given C = 0 32 As the number of lines equals the assigned zero, optimal solution has been obtained.
Example 6.13. Consider a problem of assigning four clerks to four tasks. The time required is given below. A Clerks
1 2 3 4
7 8 — 6
4
— 3 6
5 7 5 4
6 4 3 3
Clerk 2 cannot be assigned to task A and clerk 3 cannot be assigned to task B. Find all the optimum assigned schedules. Solution. The procedure is same as of earlier example. Subtracting the minimum element of each row from all its elements, we get A 0 co 0 3
1 2 3 4
1 3 2 1
3 4 oo 3
2 0 0 0
Column Reduction 00
1
0
00
0 2 1
3 there are 4 lines, which is equal to number of tasks, optimal solution has been found. A 1
B
C
D 2 0
0
2
00
1
2
3
0
00
1
4
3
4
0
OPERATIONS RESEARCH
294 1—B
2—D
3—A 4—C
7 + 4 + 3 + 4 = 18. 3 Example 6.14. A national car rental service has a surplus of one car in each of the cities 1 2 4 5 and 6 and a deficit of one car in each of the cities 7 8 9 10 11 12. The distance in miles between cities with a surplus and cities with a deficit are displayed in the table below. How should the cars be dispatched so as to minimize total mileage travelling ? 7
8
9
10
11
12
1
41
72
39
52
25
51
2
22
29
49
65
81
50
3
27
39
60
51
32
32
4
45
50
48
52
37
43
5
29
40
39
26
30
33
6
82
40
40
60
51
70
7
8
9
10
11
12
1
16
47
14
27
0
26
2
0
7
27
43
59
28
3
0
12
33
24
5
5
4
8
13
11
15
0
6
5
3
14
13
0
4
7
6
42
0
0
20
11
30
11
12
Solution. Row Reduction
Column Reduction and making assignments
1
4
7
8
9
10
16
47
14
27
21
7
27
4.3
26
j
12
39
24
6
1.3
11
1 .4
1`t
13
0
)3(
5 6
42
r).
20
1
25
295
ASSIGNMENT MODEL • the number of lines is less than the number of assignment, • further reduction will take place.
The smallest element will be subtracted from all the values from that particular element and added where two lines intersect and elements on line will remain same.
1
8 46
9 13
2
1
7
27
3
x
12
33
24
4
12
10
13
5
14
13
.0
)s(
4;2 6
1J 2;
11 (1;)
2 2b
6P
213
I )1(
Still the number of lines is 5 as against required 6. Second reduction will take place. 12
17 1 2 3 4 5 6
The minimum number of lines is 6 optimal solution has been achieved. 15
39
6
26
0
20
0
20
23
60
23
0
5
26
24
6
)3;
7
5
3
14
)3:
0
3
7
6
0
5
2
49
10
0
27
19
32
OPERATIONS RESEARCH
296 Row Reduction 1 2 3 4 5 6
7 16 0 0 8 3 49
9 14 27 33 11 13 0
8 47 7 12 13 14 0
10 27 43 24 15 0 20
11 0 59 5 0 4 11
12 26 28 5 6 7 30
Column Reduction and making assignments
1
7
8
9
10
11
12
16
47
14
27
0
21
27
43
59
3
12
33
24
5
.0
0
4
8
13
11
15
1
5
3
4
13
0
4
6
49
0
20
11
25
Now mark the unmarked row which is Row IVth. See the marked zero in that column. Draw lines on rest of the rows. 1 — 11 = 25
2 —8=
2
3 — 7 = 27
4 — 12 = 43
5 — 10 = 26
6 — 9 = 40
Total = 190 Example 6.15. Five lathes are to be allotted to five operators, the following table gives weekly output figures:
Operator
L1
L2
L3
L4
L5
P
20
22
27
32
36
Q R
19
23
29
34
40
23
28
35
34
S
21
24
31
39 37
T
24
28
31
36
41
42
Profit per piece is 25%. Find the maximum profit per week. Solution. As the given problem is a maximization problem, we convert it into an opportunity loss matrix by subtracting all the elements of the given table from the highest element of table, i.e., 42. Opportunity loss matrix is as follows :
ASSIGNMENT MODEL
297
P Operator
Q
R S
T
Li
L2
L3
L4
L5
22 23 19 21 18
20 19 14 18 14
15 13 7 1 11
10 8 3 5 6
6 2 8 0 1
L1
L2
L3
L4
L5
16 21 16 21 17
14 17 11 18 13
9 11 4 11 10
4 6 0 5 5
0 0 5 0 0
L3
L4
L5
5 7 0 7 6
4
0 0 5 0 0
Row Reduction
P Operator
Q
R S
T Column Reduction
P Operator
Q
R S
T
0 5 0 5 1
3 6 0 7 2
6 0 5 5
The minimum number of lines to cover all zeros is 3 which is less than 5, the above matrix will not give optimal solution. Subtract the least uncovered element which is 1 from all uncovered elements and add it to all the elements lying at the intersection of two lines, we get the following matrix :
Operator
P
0
Q
4 D 4
R S
T
L2
L3
L4
L5
3 5 0 6 1
5 6 0 6
4 5 0 4 4
R b 15
5
ib
The minimum lines drawn to cover all zero are 3. Repeating the above procedure, the matrix is
OPERATIONS RESEARCH
298
Operator
L1
L2
L3
L4
15
P
0
2
4
3
Q
4
4
5
4
(:)
R
1
0
U
S
4
5
5
3
(:)
0
0
4
3
The minimum number of lines to cover all zeros is 4 which is less than 5 the resultant matrix is L5 L3 L1 L2 L4 P
0
2
4
3
4
0 0
10
Operator ........ S
1
T
2 0
2 4
0
The minimum number of lines to cover all zeros is 5. Hence the above matrix will give optimal solution. L, P
Operator
L2
L3
L4
L5
2
4
3
4
1
0
Q
1
1
2
R
1
)3;
0
S
1
2
2
0
0
4
3
T
10
3
And the assignment of operator setting lathe time is given by P —>L1 = 20 Q > L2 = 40 R —> L3 = 35 S41,4 = 37 L L5 = 28 160 The maximum profit per week 25 x 160 = 4000. Example 6.16. The Captain of a cricket team has to allot five middle batting positions to five batsmen. The average runs scored by each batsman at these positions are as follows:
299
ASSIGNMENT MODEL
Batting Positions P Q
Batsman
R
I
II
III
IV
V
40 42
40 30 48 19 60
35 16
25 25 60 18 55
50 27
50 20 58
S
T
40 20 59
50 25 53
(i) Find the assignment of batsman to positions, which would give maximum number of runs. (ii) If another batsman U with the following average runs in batting positions as given below. Batting Positions Average Runs
I 45
II 52
III 38
IV 50
V 49
is added to the team, should be included to play in the team ? If so who will be replaced by whom ? Solution. (i) Convert the profit matrix into opportunity cost matrix by subtracting all the elements from the highest element 60 of given matrix, we obtain the following opportunity loss matrix. P Q
R S
T
I 20 18 10 40 2
II 20 30 12 41 0
III 25 44 20 40 1
IV 35 35 0 42 5
V 10 33 10 35 7
I 10 0 10 5 2
II 10 12 12 6 0
III 15 26 20 5 1
N 25 17 0
V 0 15 10
Row Reduction P R S
T
0
7
Column Reduction and Making Assignment I
II
III
IV
V
P
10
10
14
25
0
Q
1"'0
T2
25
1.7
15
I
R
14
1.2
19
S
5
6
4
.0.
1.0 7
5
7
Subtract the least uncovered element which is 4 from all uncovered elements, and add it to all the elements lying on the intersection of two lines and making assignment, we get
300
OPERATIONS RESEARCH I
II
III
IV
V
P
6
6
10
21
0
Q
0
12
25
17
19
R
10
12
19
0
14
S
1
2
0
3
T
2
0
)2(
5
11
The number of assignment is 5, the solution is optimum. P > V 4 50 Q —> I —> 42 R —> IV —> 60 S —> —> 20 L *60 232 (ii) If we introduce a new batsman U then number of batsmen is not equal to number of batting positions. So, a dummy batting position is created and the matrix will becomes. I P
40
40
35
25
Q
42 50 20 58 45
30 48 19 60 52
16 40 20 59 38
25 60 18 55 50
R S
T U
V
Dummy
50 27
0
50 25 53 49
0 0 0 0 0
Convert the profit matrix into opportunity cost matrix by subtracting all the elements from the highest element, i.e., 60 of the given matrix. I
II
III
IV
V
Dummy
P
20
20
25
35
10
60
Q
18
44
35
33
60
R
10
30 12
20
0
10
60
S
40
41
40
42
35
60
T
2
0
1
5
7
60
U
15
8
22
10
11
60
. 301
ASSIGNMENT MODEL Row Reduction I II III IV V Dummy P
10
10
15
25
0
50
Q R
0 10
12 12
26 20
17 0
15 10
42
S
5 2
6 00
7 5 2
0 7
7
5 1 14
T U
60 25 60 52
3
Column Reduction and Making Assignment I II III IV V Dummy P
10
10
14
25
0
25
Q
0
12
25
17
15
17
R
10
12
19
0
10
35
S
5
6
4
7
T
2
x
0
5
7
35
U
7
0
13
2
3
27
0
P —> V 4 50 Q > I 4 42 R —> IV 4 60 S Dummy 0 T —> —> 59 If 4 52 U 263 Example 6.17. A small school has five teachers teaching five different subjects. All the five teachers are capable of teaching all the five subjects. The output per day of the teacher and course coverage (%) for each subjects are given below. Teachers A B C
D E (Course Coverage %)
1 7 4 8 6 7 2
2 9 9 5 5 8 3
3 4 5 7 8 10 2
4 8 7 9 10 9 3
5 6 8 8 10 9 4
.
302
OPERATIONS RESEARCH
If teacher D is not available, will your answer be different ? Solution. We can multiply the output of teachers for different subjects with coverage of course (%), the resultant matrix is shown as :
Teachers
1 14 8 16 12 14
A B C
D E
2 27 27 15 15 24
4 24 21 27 30 27
3 8 10 14 16 20
5 24 32 32 40 36
The problem is of maximization. We will deduct the maximum element, i.e., 40. The resultant loss matrix will be Subjects Teachers
1
2
3
4
5
A
26
13
32
16
16
B C
32 24
13 25
30 26
19 13
8 8
D
28
25
24
10
0
E
26
16
20
13
4
1
2
3
4
5
13 24
0 5
19 22
3
3
16 28 22
17 25 12
18 24 16
11 5 10 9
0 0 0 0
Row Reduction
A B C
D E Column Reduction
Now subtracting the main elements of each column from all its elements.
A B C
D E
0 11 3 15 9
2 0 5 17 25 t2
3 3 6 2 8 '0
4 0 8 2 7 6
0 0
ASSIGNMENT MODEL
303
There are 3 lines, this is not the optimal solution. So, we will follow the deduction method. 2 A B D
9 1 13
3 15 23
E
9
12
3
4
5
3 4 0
i1 6
5 0
a
a
6 0
5 6
0 2
Still the number of lines are less. Again the same procedure will be followed. 1 0 6 1 10 9
A B C
D E
4 3 0 2 6
1 0
15 0 2
1
1 0 5
Optimum solution is obtained. 1 A
2
3
0
3 1
B
6
0
C
1
15
D
10
20
E
9
12
4
5 8
3 0
3
3
2
0
0
6
3
A —> 1 = 14 B —> 2 = 27 C —> 4 = 27 5 =40 E —> 3 = 20 128 IN REVIEW AND DISCUSSION QUESTIONS 1
1. (a) Show that assignment model is a special case of transportation model. (b) Consider the problem of assigning five operators to five machines. The assignment costs are as follows.
OPERATIONS RESEARCH
304
I 10 3 10 5 7
A B Machines
C
D E
II 5 9 7 11
Operators III 13 18 2 7
9
IV 15 3 2 7
4
V 16 6 2 12 12
4
Assign the operators to different machines so that total cost is minimized. 2. (a) If in an assignment problem we add a constant to every element of a row (or column) in the effectiveness matrix, prove that an assignment that minimizes the total effectiveness in one matrix also minimizes the total effectiveness in the other matrix. (b) A national carrental service has a surplus of one car in each of the cities 1, 2, 3, 4, 5, 6 and a deficit of one car in each of the cities 7, 8, 9, 10, 11, 12. The distance in miles between cities with a surplus and cities with a deficit are displayed in matrix. How should the cars be despatched so as to minimize the total mileage travelled. To 7 41 22 27 45 29 82
1 2 3 4 5 6
9 39 49 60 48 39 40
8 72 29 39 50 40 40
10 52 65 51 52 26 60
12 51 50 32 43 33 30
11 25 81 32 37 30 51
3. (a) Distinguish between transportation model and assignment model. (b) Four different jobs are to be done on four different machines. The setup and production times are prohibitively high for change over. Following table indicates the cost of producing job i on machine j in rupees. Machines
Jobs
1 2 3 4
1 5 8 4 10
2 7 5 7 4
3 11 9 10 8
4 6 6 7 3
Assign jobs to different machines so that the total cost is minimized. 4. Six machines M1, M2, M3, M4, M5 and M6 are to be located in six places P1, P2, P3, P4, P5 and P6. Cij the cost of locating machine M, at place P1 is given in the following matrix.
305
ASSIGNMENT MODEL P1
P2
P3
Mt
20
23
M2
50
M3
P5
P6
18
P4 10
16
20
20
17
16
15
11
60
30
40
55
8
7
M4
6
7
10
20
25
9
M5
18
19
28
17
60
70
M6
9
10
20
30
40
55
Formulate an LP model to determine an optimal assignment. Write the objective function and the constraints in detail. Define any symbol used. Find an optimal layout by assignment techniques of linear programming. 5. (a) Discuss assignment model. Indicate a method of solving an assignment problem. (b) A Company is faced with the problem of assigning six different machines to five different jobs. The costs estimated in hundreds of rupees are given in the table below. 1 2.5 2 3 3.5 4 6
1 2 3 4 5 6
2 5 5 6.5 7 7 9
3 1 1.5 2 2 3 5
4 6 7 8 9 9 10
5 2 3 3 4.5 6 6
Solve the problem assuming that the objective is to minimize the total cost. 6. Five new machines are to be located in machine shop. There are five possible locations in which machines can be located. Co the cost of placing machine i in place j is given in the table below.
Machine
1 2 3 4 5
1 15 1 8 14 10
2 10 8 9 10 8
Jobs 3 25 10 17 25 25
4 25 20 20 27 27
5 10 2 10 15 12
It is required to place the machines at suitable places so as to minimize the total cost. (i) Formulate an LP model to find an optimal assignment. (it) Solve the problem by assignment technique of LP.
OPERATIONS RESEARCH
306 7. Solve the following assignment problem : 1 2 3 4 5
I 11 9 13 21 14
II 17 7 16 24 10
IV 16 6 12 28 11
III 8 12 15 17 12
V 20 15 16 26 15
8. A team of 5 horses and 5 riders has entered a jumping show contest. The number of penalty points to be expected when each rider rides any horse is shown below. R, 5 2 4 6 4
Ell H2 Horse
H3 H4
H5
R2
R3
3 3 1 8 2
4 7 5 1 5
R4 7 6 2 2 7
R5
2 5 4 3 1
How should the horses be allotted to the riders so as to minimise the expected loss of the team? 9. Find the minimum cost solution for the 5 x 5 assignment problem whose cost coefficients are given below. 1
2
3
4
5
1
—2
—4
—8
—6
—1
2
0
—9
—5
—5
—4
3
—3
—8
—9
—2
—6
4
—4
—3
—1
0
—3
5
—9
—5
—8
—9
—5
10. A company has four machines on which to do three jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following table : Machines
Job
W
X
Y
Z
A
18
24
28
32
B
8
13
17
19
C
10
15
19
22
What are the job assignments which will minimize the cost ? 11. Assign four trucks 1, 2, 3 and 4 to vacant spaces 7, 8, 9, 10, 11 and 12 so that the distance travelled is minimized. The following matrix shows the distance.
ASSIGNMENT MODEL
307
7 8 9 10 11 12
1 4 8 4 7 6 6
2 7 2 9 5 3 8
3 3 5 6 4 5 7
4 7 5 9 8 4 3
12. A Company has five jobs to be done. The following matrix shows the return in rupees of assigning ith machine (i = 1, 2„ 5) to the job (j = 1, 2„ 5). Assign the five jobs to the five machines so as to maximize the total expected profit. Job
Machine
1 2 3 4 5
1 5 2 3 6 7
2 11 4 12 14 9
3 10 6 5 4 8
4 12 3 14 11 12
5 4 5 6 7 5
13. The owner of a small machine shop has four machinists available to assign to jobs for the day. Five jobs are offered with the expected profit in rupees for each machinist on each job being as follows. Find the assignment of machinists to jobs that will return in a maximum profit. Which job should be declined ? Job
Machinist
1 2 3 4
A 6.20 7.10 8.70 4.80
7.80 8.40 9.20 6.40
5.00 6.10 11.10 8.70
10.00 7.30 7.10 7.70
8.20 5.90 8.10 8.00
14. An airline that operates seven days a week has timetable as shown below. Crew must have a minimum layover of 5 hours between flights. Obtain the pairing of flights that minimizes layover time away from home. For any given pairing, the crew will be based at the city that results in smaller layover. Flight No. 1 2 3 4
DelhiJaipur Depart 7.00 A.M. 8.00 A.M. 1.30 A.M. 6.30 P.M.
Arrive 8.00 A.M. 9.00 A.M. 2.30 A.M. 7.30 P.M.
Flight No. 101 102 103 104
JaipurDelhi Depart 8.00 A.M. 8.30 A.M. 12.00 Noon 8.00 P.M.
For each pair also mention the town where the crew should be based.
Arrive 9.15 A.M. 9.45 A.M. 1.150 P.M. 6.45 P.M.
OPERATIONS RESEARcH
308
15. A company has four territories open and four salesmen available for assignment. The territories are not equally rich in their sales potential; it is estimated that a typical salesman operating in each territory would bring in the following annual sales : Territory
Annual Sales (Rs.)
I
60000
II
50000
III
40000
IV
30000
The four salesmen are also considered to differ in ability, it is estimated that working under the same conditions their yearly sales will be proportionately as follows : Salesman
Proportion
A
7
B
5
C
5
D
4
If the criterion is maximum expected total sales, the intuitive answer is to assign the best salesman to the richest territory, the next best salesman to the second richest and so on. Verify this answer by the assignment technique. 16. (a) State mathematical model of assignment problem. (b) The personnel manager of a medium size company decided to recruit two employees D and E in a particular section of the organisation. The section has five fairly defined tasks 1, 2, 3, 4 and 5 and three employees A, B and C are already employed in the section. Looking at the rather specialised nature of task 3 and the special qualifications of the recruit D for task 3, the manager decided to assign task 3 to employee D and then assign the ramaining tasks to remaining employees so as to maximize the total effectiveness. The index of effectiveness of each employee for different task is as under : Tasks
Employee
1
2
3
4
5
A
25
55
60
45
30
B
45
65
55
35
40
C
10
35
45
55
65
D
40
30
70
40
60
E
55
45
40
55
10
Assign the tasks for maximizing total effectiveness. Critically examine whether the decision of the manager to assign task 3 to employee D was correct.
ASSIGNMENT MODEL
309
111 PUNJAB UNIVERSITY EXAMINATION PROBLEMS
1995 SEP; 1998 SEP; 1999 SEP. (1) Assignment problem. (Marks 2) 1995 SEP. (2) Write a method of solving an assignment problem. Illustrate with an example. 1996 APR. (3) Enumerate various steps involved in Hungarian method of solving assignment problem. (Marks 6) 1998 APR. (4) Discuss briefly the Hungarian method of solving an assignment problem. 1998 SEP. (5) Write a method of solving an assignment problem. Illustrate with an example. (Marks. 6) 1995 APR. (6) A departmental head has 4 subordinates and 4 tasks to be performed, the subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimate of the times each man would take to perform each task is given below in the matrix : Tasks/Subordinates
1996 APR.
A
8
26
17
11
B C
13 38
28 19
4 18
26 15
D
19
26
24
10
How should the tasks to be allocated to subordinates so as to minimize the total manhours ? (7) A car hire company has one car at each of 5 depots a, b, c, d and e. A customer in each of five cities A, B, C, D and E require a car. The distance (in km.) between depots and the cities are as follows : Cities Depots
1996 SEP.
b
d
e
A
160
130
175
190
200
B
135
120
130
160
175
C
140
110
155
170
185
D
50
50
80
80
110
E
55
35
70
80
105
How should the cars be assigned to the customers so as to minimize the distance travelled ? (8) Find the assignment of men to jobs that will minimize the total time : JOBS
MEN
J,
Ji
J2
mi M2
8 13
26 28
17 4
11 26 .
M3
38
19
18
15
M4
19
26
24
10,
OPERATIONS RESEARCH
310 1997 APR.
(9) Three customers in a certain sales territory have requested technical assistance from available 3 technicians. Distance between the technicians and customers is reported below. Using assignment problems techniques assign the technicians to the customers for minimum cost of travel per km is Rs. 2 Customers A Technicians C
1998 SEP.
I
II
III
37
58
41
38
92
74
88
55
43
(10) Given the following cost matrix (in '000 Rs.) contract Bidder
Different
Subsystems
II
Weapon HI
IV
A
17
20
13
21
B
15
21
18
C
17 14
18 22
14 17 12
D
21 22
Assign the 4 different weapon subsystems, one each to bidder so as to minimize cost. 1992 APR. (11) Using following cost matrix, assign the different contractors to different jobs so as to minimize the total cost :
X JOBS L 1999 SEP.
B
C
D
17
20
13
21
15
21
14
18
17
18 22
17
21
12
22
14
(12) Using following cost matrix of cost of three jobs on three machines, assign one job to each machine so that the total cost is minimum : Machines
Jobs
X
Y
A
25
31
35
B
15
20
24
C
22
19
17
ASSIGNMENT MODEL
311
2000 ARP. (13) Solve the following assignment problem of maximization : 5 3 3 9 3
6 4 4 6 1
8 5 9 5 2
2 5 8 2
9 6 8 4 2
3
2000 APR. (14) Determine the optimum solution to each of the following problem : Jobs
2000 SEP.
Operators
1
2
3
4
5
1
6
2
5
2
6
2
2
5
8
7
7
3
7
8
6
9
8
4
6
2
3
4
5
4
9
3
8
9
7
6
4
7
4
6
8
(15) Assign the operators to machines to minimize costs for the following : Machines Operator
I
II
III
IV
V
A
5
5

2
6
B
7
4
2
3
4
C
9
3
5

3
D
7
2
6
7
2
E
6
5
7
9
1
2001 ARP. (16) Find the optimal solution for the assignment problem : I
II
III
IV
V
A
11
17
8
16
20
B
9
7
12
6
15
C
13
16
15
12
16
D
21
24
17
26
E
14
10
12
28 11
15
OPERATIONS RESEARCH
312 2001 APR. (17) For a given profit matrix, find the optimal assignments : Machines
2001 SEP.
4
5
8
16
20
7
12
6
15
13
16
15
12
16
D
21
24
17
28
26
E
14
10
12
11
15
Jobs
1
2
A
11
17
B
9
C
(18) Assigns five operators to five machines to minimize total cost : Machines
Operators
I
II
III
IV
V
A
5
5
—
2
6
B
7
4
2
3
4
C
9
3
5
—
3
D
7
2
6
7
—
E
6
5
7
9
1
2002 APR. (19) Solve the assignment problem : From \ To
A
A
B
C
D
E
7
6
8
4
8
5
6
9
7
B
7
C
6
8
—
D
8
5
9
E
4
6
7
8 8
(Marks 7) 2003 APR. (20) What is unbalance assignment problem ? How it is solved by HAM ? (Take any imaginary problem). HAM—Hungarian Assignment Method. (Marks 18) 2004 APR. (21) Suggest optimum assignment to sales territories, where the estimates of sales to be made by each salesman in different territories are given below : Territories
Salesmen
I
II
III
IV
V
A
10
15
17
14
14
B
6
18
10
12
16
C
12
5
13
13
6
D
8
11
16
10
12
ASSIGNMENT MODEL
313
If salesman B cannot be assigned to territory II for certain reasons, will the optimum assignment change ? If so, what will be the new assignment schedule and the total sales ? G.N.D.U. EXAMINATION PROBLEMS
2004 APR. (1) Unbalanced assignment problem. 2004 APR. (2) Distinguish between Assignment and Transportation problem. April 96, 98. (3) Enumerate various steps involved in Hungarian method of solving assignment problem. 1998 APR. (4) Write a method of solving a assignment problem. Illustrate with an example. 1995 SEP. (5) Write a method of solving a assignment problem. Illustrate with an example. 1996 SEP. (6) Find the assignment of men to jobs that will minimize the total time : Jobs J2 26 28 19 26
JI
Men
8 13 38 19
M2 M3 M4
1997 APR.
C
17 4 18 24
11 26 15 10
I 37 38 38
Customers II 58 92 55
III 41 74 43
(8) Given the following cost matrix (in '000 Rs.) contract : Bidder A B C
D
1999 APR.
J4
(7) Three customers in a certain sales territory has requested technical assistance from available 3 technicians. Distance between the technicians and customers is reported below. Using assignment problems techniques assign the technicians to the customers for minimum cost of travel per km is Rs. 2.
A TechniciansB
1998 SEP.
J3
I 17 15 17 14
Different Weapons subsystems II III IV 20 13 21 21 14 18 18 17 21 22 12 22
Assign the 4 different weapon subsystems, one each to each bidder so as to minimize cost. (9) Using following cost matrix, assign the different contractors to different jobs so as to minimize the total cost :
OPERATIONS RESEARCH
314 Contractors A Jobs
D
X
17
20
13
21
Y
15
21
14
18
17
18
21
14
22
17 12
L 1999 SEP.
C
22
(10) Using following cost matrix of cost of three jobs on three machines, assign one job to each machine so that total cost is minimum : Machines X Jobs
A
25
31
35
B
15
20
24
C
22
19
17
•
1993 APR. (11) A process can be carried out on any of the 4 machines by one of the operators. The average time taken by any operator on any specific machine is tabulated below. A proposal to replace one of the existing machines by a new machine is under consideration. The estimated timing to carryout the process on the new machine are also included in the table. Is it worthwhile to use new machinery ? Machines I
II
III
IV
New
A
12
14
8
12
11
11
12
12
9
10
C
10
9
9
8
7
D
14
15
8
7
6
Operators
1993 SEP.
(12) Consider the problem of assigning the five jobs to five persons. The assignment costs are given as follows :
A Persons C
D E
8 0 3 4 9
2 4 9 8 3 5
Jobs 3 2 5 9 1 8
4 6 5 2 0 9
5 1 4 6 3 5
ASSIGNMENT MODEL
315
1994 APR. (13) Assign the jobs to machines for maximizing profits given in table : Machines A Jobs
1994 SEP.
32
II
E 40 36
40
40
38 24
28
28 21
III
41
27
33
30
37
IV
22
41
36
36
V
39
38 33
40
35
39
(14) Assign machine to give jobs, you are given costs (in Rs.)
Machines
1994 SEP.
I
D
1
2
Jobs 3
1
2.5
5
1
6
1
2 3 4 5 6
2 3 3.5 4 6
5 6.5 7 7 9
1.5 2 2 3 5
7 8 9 9 10
3 3 4.5 6 6
4
5
(15) A firm wants to purchase three different types of equipments and the five manufactures have come forward to supply one or all the three machines. However, the firm's policy is to purchase one machine from one manufacturer only. The data relating to price (in lacs of Rs.) quoted by the different manufactures is given below.
A B Manufactures
C
D E
2.99 2.78 2.92 2.82 3.11
Machines 2 3.11 2.87 3.05 3.10 2.90
3 2.68 2.57 2.80 2.74 2.62
Determine how best the firm can purchase three machines, assuming that the machines of different manufactures have the same capacity. 1995 APR. (16) A departmental head has 4 subordinates and 4 tasks to be performed, the subordinates differ in efficiency and the task differ in their intrinsic difficulty. His estimate of the times each man would take to perform each task is given is the following matrix :
OPERATIONS RESEARCH
316 Tasks /Subordinates
I
II
III
IV
A
8
26
17
11
B C
13 38
28 19
4 18
26 15
D
19
26
24
10
How should the tasks to be allocated to subordinates so as to minimize the total manhours ? 1996 APR. (17) A car hire company has one car at each of 5 depots a, b, c, d and e. A customer in each of five cities, A, B, C, D and E require a car. The distance (in km) between depots and the cities are as follows Cities \ Depots A B C
D E
b 130 120 110 50 35
160 135 140 50 55
d 190 160 170 80 80
175 130 155 80 70
e
200 175 185 110 105
How should the cars be assigned to the customers so as to minimize the distance travelled ? 2001 APR. (18) Solve the folll owing assignment problem :
P1 P2 P3 P4
2001 SEP.
Jl
12
27 31 20 20
18 24 17 28
J4
X 21 20 20
20 12 X 16
Ts 21 21 16 27
(19) Solve the following assignment (minimization) problem : Machines
Jobs
1 2 3 4 5
A 12 12 8 8 10
7 5 4 6 11
9 10 9 7 9
D 10 11 7 5 7
ASSIGNMENT MODEL
317
2002 APR. (20) Solve the following Assignment problem : Sales in Rs. 000'S Salesmen Territories ,i,
2002 SEP.
Si
S2
S3
S4
A
50
48
47
45
B
70
72
70
60
C
90
80
85
82
D
60
48
52
55
(21) Solve the following assignment problem. The district wise sales of '5 salesmen is given as under : Districts 1
2
3
4
5
A
38 44 47 35 50
B
49 33 37 30 42
Salesmen
50
36
40
39
Sales is in Rs.000's
46
D
28 44 48 45 43
E
39 40 48 45 46
2003 APR. (22) A company has a team of four salesmen and there are four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts the company estimates that the profit per day in rupees for each salesman in each district is as below. 20 Districts 2
3
4
16
10
14
11
14
11
15
15
C
15
15
13
12
D
13
12
14
15
A Salesmen
2003 APR. (23) Following table gives the weekly output of five operators working on 5 lathes. Do the assignment of operators on the lathes so as to maximize total production: Lathes Weekly Output Operators
L,
L2
L3
L4
L5
P
20
22
27
32
36
Q
19
23
29
34
40
R
23
28
35
39
34
S
21
24
31
37
42
318 2004 APR. (22) B.B.A. III
OPERATIONS RESEARCH
Assignment problem is a special case of transportation problem discuss. (ii) Using following cost matrix assign jobs to different machines so as to minimize total cost : (i)
A Jobs B
10 5 12 8
C
D
2 12 10 14 15
Machines 3 19 7 13 11
4 11 8 11 9
2004 APR. (23) A company is faced with the problem of assigning 4 machines to 6 different jobs. (One machine to one job only). The profits are estimated as follows : Machines
Jobs
1 2 3 4 5 6
A 3 7 3 6 5 5
6 1 8 4 2 7
Solve the problem to maximize the profit.
2 4 5 3 4 6
D 6 4 8 7 3 4
7 The Sequencing Problems LEARNING OBIECTIVES • • • • •
The meaning and concept of sequencing Assumptions made in the sequencing problem Terminology used Processing of jobs through different number of machines Processing of two jobs through m machines and n jobs through m machines.
INTRODUCTION Every organization wants to utilize its productive systems effectively and efficiently and wants to maximize its profit by meeting the delivery deadlines. A number of jobs involving many operations have to be performed and there are limited resources in terms of plant and machinery on which the jobs have to be performed. It is necessary that available facilities are optimally utilized and they are loaded, scheduled and sequenced properly. A sequence is the order in which different jobs are to be performed. When there is a choice that a number of tasks can be performed in different orders, then the problem of sequencing arises. Such situations are very often encountered by manufacturing units, overhauling of equipments or aircraft engines, maintenance schedule of a large variety of equipment used in a factory, customers in a bank or car servicing garage and so on. The basic concept behind sequencing is to use the available facilities in such a manner that the cost (and time) is minimized. The sequencing theory has been developed to solve difficult problems of using limited number of facilities in an optimal manner to get the best production and minimum costs. Terms Commonly used 1. Job: These have to be sequenced, hence there should be a particular number of jobs (groups of tasks to be performed) say n to be processed. 2. Machine: Jobs have to be performed or processed on machines. It is a facility which has some processing capability.
OPERATIONS RESEARCH
320
3. Loading: Assigning of jobs to facilities and committing of facilities to jobs without specifying the time and sequence. 4. Scheduling: When the time and sequence of performing the job is specified, it is called scheduling.
5. Sequencing: Sequencing of operations refers to a systematic procedure of determining the order in which a series of jobs will be processed in a definite number, say k, facilities or machines. 6. Processing Time: Every operation that is required to be performed requires definite amount of time at each facility or machine when processing time is definite and certain, scheduling is easier as compared to the situation in which it is not known. 7. Total Elapsed Time: It is the time that lapses between the starting of first job and the completion of the last one. 8. Idle Time: The time for which the facilities or machine are not utilized during the total elapsed time. 9. Technological Order: It is the order which must be followed for completing a job. The requirement of the job dictates in which order various operations have to be performed, for example, painting cannot be done before welding. 10. Passing not allowed: If 'n' jobs have to be processed through 'm' machines in a particular order of M1 , M2 , M3 then each job will go to machine M1 first and then to M2 and finally to M3. This order cannot be passed. 11. Static arrival Pattern: If all the jobs to be done are received at the facilities simultaneously. 12. Dynamic arrival Pattern: Here the jobs keep arriving continuously. Assumptions
In sequencing problems, the following assumptions are made : (i) All machines can process only one job at a time. (ii) No time is wasted in shifting a job from one machine to other. (iii) Processing time of job on a machine has no relation with the order in which the job is processed. (iv) All machines have different capability and capacity. (v) All jobs are ready for processing. (vi) Each job when put on the machine is completed. (vii) All jobs are processed in specified order as soon as possible. TYPES OF SEQUENCING PROBLEMS The following types of sequencing problems will be discussed in this chapter: (a) n jobs one machine case (b) n jobs two machines case (c) n jobs im' machine case (d) Two jobs 'in' machines case.
321
THE SEQUENCING PROBLEMS The solution of these problems depends on many factors such as :
(a) The number of jobs to be scheduled (b) The number of machines in the machine shop (c) Type of manufacturing facility (slow shop or fast shop) (d) Manner in which jobs arrive at the facility (static or dynamic) (e) Criterion by which scheduling alternatives are to be evaluated. As the number of jobs (n) and the number of machines (in) increases, the sequencing problems become more complex. In fact, no exact or optimal solutions exist for sequencing problems with large n and in. Simulation seems to be a better solution technique for real life scheduling problems. nJobs One Machine Case This case of a number of jobs to be processed on one facility is very common in real life situations. The number of cars to be serviced in a garage, number of engines to be overhauled in one workshop, number of patients to be treated by one doctor, number of different jobs to be machined on a lathe, etc, are the cases which can be solved by using the method under study. In all such cases we are all used to 'first come first served' principle to give sense of satisfaction and justice to the waiting jobs. But if this is not the consideration, it is possible to get more favourable results in the interest of effectiveness and efficiency. The following assumptions are applicable : (a) The job shop is static. (b) Processing time of the job is known. The implication of the above assumption that job shop is static will mean that new job arrivals do not disturb the processing of n jobs already being processed and the new job arrivals wait to be attended to in next batch. Shortest Processing Time (SPT) Rule This rule says that jobs are sequenced in such a way that the job with least processing time is picked up first, followed by the job with the next Smallest Processing Time (SPT) and so on. This is referred to as shortest processing time sequencing. However, when the importance of the jobs to be performed varies, a different rule called WeightScheduling (Weight Scheduling Process Time) rule is used. Weights are allotted to jobs, greater weight meaning more important job. Let Wi be the weight allotted. By dividing the processing time by the weight factor, the tendency to move important job to an earlier position in the order is achieved. Wifi Weighted Mean Flow Time, (WMFT) — Wi i=i where fi = flow time of job i = Wi + ti ti = processing time of job i WSPT rule for minimizing Weighted MeanFlow Time (WMFT) puts n jobs in a sequence such that
322
OPERATIONS RESEARCH
t[1] < t[2] t [n] < < W[1] — W[2] — ••• W[n] The numbers in brackets above define the position of the jobs in the optimal sequence. Example 7.1. Consider the 8 jobs with processing times, due dates and importance weights as shown below. 8 jobs one machine case data Task (i)
Processing Time (t1 )
Due Date (di )
Importance Weight (WI )
Wi
1
5
15
1
5.0
2
8
10
2
4.0
3
6
15
3
2.0
4
3
25
1
3.0
5
10
20
2
5.0
6
14
40
3
4.7
7
7
45
2
3.5
8
3
50
1
3.0
ti
From processing time ti in the table the SPT sequence is 48 1 3 7 2 5 6 resulting in completion of these jobs at times 3, 6, 14, 20, 27, 36, 46, 60 respectively. 3+6+14+20+27+36+46+60 = 26.5 hours 8
No. of jobs working as inprocessing inventory
WMFT =
3 6
14
20
27
Flow time of Jobs
Fig. 7.1
36
46
60
THE SEQUENCING PROBLEMS
323
The sequence is shown graphically above from which the number of tasks waiting as inprocess inventory is seen to be 8 during 03, 7 during 36, 6 during 614, 5 during 1420, 4 during 2027, 3 during 2736, 2 during 3646 and one during 4660. Thus, the average inventory is given by Average inventory= 8x3+7x3+6x8+5x6+4x7+3x9+2x10+1x14 60 24+21+48+30+28+27+20+14 212 = = 3.53 jobs. 60 60 Weight Scheduling Process Time
If the important weights W, were to be considered the WSPT could be used to minimize the Weighted Mean Flow Time (WMFT) to yield the sequence 34827651. This results by first ti choosing job with minimum — in the table. The respective flow time of jobs in this sequence W, are 6, 9, 12, 21, 28, 42, 52, 58. Mean flow time is hours 6x3+9x1+12x1+21x3+28x2+42x3+52x2+58x1 3+1+1+3+2+3+2+1 18+9+12+63+56+126+104+58 446 = = 27.85 hours 16 16 Example 7.2. Eight jobs A, B, C, D, E, F, G and H arrive at one time to be processed on a single machine. Find out the optimal job sequence, when their operation time is given in the table below. WMFT
Job (n) A B C D E F G H
Operation time in minutes 16 12 10 8 7 4 2 1
Solution. For determining the optimal sequence, the jobs are selected in a nondescending operation time as follows : Nondecreasing operation time sequence is H > G F E —> D C > B A. Total processing time H=1 G=1+2 =3 F 1+2 +4 =7 E = 1 + 2 + 4 + 7 = 14
324
OPERATIONS RESEARCH
D = 1 + 2 + 4 + 7 + 8 = 22 C = 1 + 2 + 4 + 7 + 8 + 10 = 32 B = 1 + 2 + 4 + 7 + 8 + 10 + 12 = 44 A = 1 + 2 + 4 + 7 + 8 + 10 + 12 + 16 = 60 Average processing time = Total time /number of jobs = 183/8 = 23 minutes In case the jobs are processed in the order of their arrival, i.e., A > B > C > D E F G > H the total processing time would have been as follows : A = 16 B = 16 + 12 = 28 C = 16 + 12 + 10 = 38 D = 16 + 12 + 10 + 8 = 46 E = 16 + 12 + 10 + 8 + 7 = 53 F = 16 + 12 + 10 + 8 + 7 + 4 = 57 G = 16 + 12 + 10 + 8 + 7 + 4 + 2 = 59 H = 16 + 12 + 10 + 8 + 7 + 4 + 2 + 1 = 60 Average processing time = 357/8 = 44.6, which is much more than the previous time. Priority Sequencing Rules
The following priority sequencing rules are generally followed in production/service system: 1. First Come First Served(FCFS): As explained earlier, it is followed to avoid any heart burns and avoidable controversies. 2. Earliest Due Date (EDD): In this rule, top priority is allotted to the waiting job, which has the earliest due/delivery date. In this case the order of arrival of the job and processing time it takes is ignored. 3. Least Slack Rule (LS): It gives top priority to the waiting job whose slack time is the least. Slack time is the difference between the length of time remaining until the job is due and the length of its operation time. 4. Average Number of Jobs in the System: It is defined as the average number of jobs remaining in the system (waiting or being processed) from the beginning of sequence through the time when the last job is finished. 5. Average Job Lateness: Jobs lateness is defined as the difference between the actual completion time of the job and its due date. Average job lateness is sum of lateness of all jobs divided by the number of jobs in the system. This is also called Average Job Tardiness. 6. Average Earliness of Jobs: If a job is completed before its due date, the lateness value is negative and the magnitude is referred as earliness of job. Mean earliness of the job is the sum of earliness of all jobs divided by the number of jobs in the system. 7. Number of Tardy Jobs: It is the number of jobs which are completed after the due date. Sequencing n Jobs Through Two Machines
The sequencing algorithm for this case was developed by Johnson and is called Johnson's Algorithm. In this situation n jobs must be processed through machines M1 and M2. The processing time of all the n jobs on M1 and M2 is known and it is required to find the sequence, which minimizes the time to complete all the jobs.
THE SEQUENCING PROBLEMS
325
Johnson's algorithm is based on the following assumptions: (i) There are only two machines and the processing of all the jobs is done on both the Machinesin the same order, i.e., first on M1 and then on M2. (ii) All jobs arrive at the same time (static arrival pattern) have no priority for job completion. Johnson's algorithm involves following steps : 1. List operation time for each job on machine M1 and M2 . 2. Select the shortest operation or processing time in the above list. 3. If minimumprocessing time is on M1, place the corresponding job first in the sequence. If it is on M2, place the corresponding job last in the sequence. In case of tie in shortest processing time, it can be broken arbitrarily. 4. Eliminate the jobs which have already been sequenced as result of step 3. 5. Repeat steps 2 and 3 until all the jobs are sequenced. Example 7.3. Six jobs are to be sequenced, which require processing on two machines M1 and M2. The processing time in minutes for each of the six jobs on machines M1 and M2 is given below. All the jobs have to be processed in sequence M1, M2. Determine the optimum sequence for processing the jobs so that the total time of all the jobs is minimum. Use Johnson's algorithm. Jobs Processing Time
1
2
3
4
5
6
Machine M1
30
30
60
20
35
45
Machine M2
45
15
40
25
30
70
Solution. Step I. Operation time or processing time for each jobs on M1 and M2 is provided in the question. Step II. The shortest processing time is 15 for job 2 on M2. Step III. As the minimum processing time is on M2, job 2 has to be kept last as follows: 2 Step IV. We ignore job 2 and find out the shortest processing time of rest of jobs. Now the least processing time is 20 minutes on machine M1 for job 4. Since it is on M1, it is to be placed first as follows: 4
2
The next minimum processing time is 30 minutes for job 5 on M2 and Job 1 on M1. So, job 5 will be placed at the end. Job 1 will be sequenced earlier as shown below. 4
1
5
2
The next minimum processing time is 40 minutes for job 3 on M2, hence it is sequenced \ as follows : 4
1
3
5
2
OPERATIONS RESEARCH
326
Job 6 has to be sequenced in the gap or vacant space. The complete sequencing of the jobs is as follows. 4
1
6
2
5
3
The minimum time for six jobs on machine M1 and M2 can be shown with the help of a Gantt chart as shown below.
'MS F///80/§§08/5///.
20
45
90 95
165
205
235
67§§§§§ 8 /§ 1,§ §PF/3/ /// 6$ 0 95 15 5
250
220
250
I I I I I I 1 1 I 1 I I I 20 40 60 80 100 120 140 160 180 200 220 240 260 Time (Minutes)
7////
Machine working Machine idle
Fig. 7.2 The above figure shows idle time for M1 (30 minutes) after the last job (2) has been processed. Idle time for M2 is 20 minutes before job 4 is started and 5 minutes before processing 6 and finishing job 1. The percentage utilization of M1 = 250 — 30/250 = 88% and M2 = 250 — 25/250 = 90%. Example 7.4. A bookbinder has one printing press, one binding machine and manuscript of seven different books. The time required for performing printing and binding operations for different books are shown below. Book 1 2 3 4 5 6 7
Printing Time (days) 20 90 80 20 120 15 65
Binding Time (days) 25 60 75 30 90 35 50
THE SESUENCING PROBLEMS
327
Decide the optimum sequence of processing of books in order to minimize total time. Solution. Step I. Minimum time is 15 days on printing press (M1) for job 6 so it will be sequenced earlier as shown. 6 Step II. Now book 1 and 4 have the least time of 20 days on printing press (M1) so these two books will be sequenced as 6
4
Step III. After eliminating jobs 6, 1 and 4, the least time is for job 7 on binding machine (M2) so it will be placed last in the sequence. 6
1
7
4
Step IV. Now book 2 has least time of 60 on M2, hence it will be placed at the end. 1
6
4
7
2
Step' V. Book 3 has the least time of 75 days on M2 so it will be placed as below. 1
6
4
3
2
7
3
2
7
Step VI. Job 5 will be placed in the vacant place. 1
6
4
5
Step VII. Total processing time can be calculated as follows : Optimum sequence of jobs (books) 6 1 4 5 3 2 7
Printing In 0 15 35 55 175 255 345
Out 15 35 55 175 255 345 410
Binding In 15 50 75 175 265 345 410
Out 50 75 105 265 340 405 460
Idle time for Binding 15
70 5 5
Total idle time Printing = (460 — 410) = 50 days as the printing last job (7) is finished on 410 days but binding finishes only after 460 days, so printing machine is idle for 50 days. Binding = 15 + 70 + 5 + 5 = 95 days.
OPERATIONS RESEARCH
328
Example 7.5. A manufacturing company has 5 different jobs on two machines Ml and M2. The processing time for each of the jobs on M1 and M2 is given below. Decide the optimal sequence of processing of the jobs in order to minimize total time. Job. No.
Processing Time M1
1 2 3
M2
8 12
6 7 11
4
5 3
5
6
9 14
Solution. The shortest processing time is 3 on MI for 'ob 4 so it will be sequenced as follows: 4 Next is job 3 with time 5 and M1 , hence job 3 will be sequenced as 4
3
Next minimum time is for jobs 1 on M2 this will be sequenced last. 4
3
1
After eliminating jobs 4, 3 and 1, the next with minimum time is job 5 on M1 so it will be placed as 4
3
5
1
Now, job 2 will be sequenced in the vacant space. 4
3
5
2
1
n Jobs 3 Machines Case
Johnson's algorithm which we have just applied can be extended and made use of in n jobs 3 machine case, if the following conditions hold good : (a) Maximum processing time for a job on machine M1 is greater than or equal to maximum processing time for the same job. or (b) Minimum processing time for a job on machine M3 is greater than or equal to maximum processing time for a job on machine M2. The following assumptions are made : (a) Every job is processed on all the three machines M1, M2 and M3 in the same order, i.e., the job is first processed on M1 then on M2 and then on M3.
THE SEQUENCING PROBLEMS
329
(b) The passing of jobs is not permitted. (c) Processing time for each job on the machine M1, M2 and M3 are known. In this procedure two dummy machines M1' and M2 are assumed in such a manner that the processing time of jobs on these machines can be calculated as Processing time of jobs on M1' = Processing time (M1 + M2) Processing time of a job on M2 = Processing time (M2 + M3) After this Johnson's algorithm is applied on of jobs.
and M2' to find oust the optimal sequencing
Example 7.6. In a manufacturing process three operations have to be performed on machines M1, M2 and M3 in order M1, M2 and M3 . Find out the optimum sequencing when the processing time for four jobs on three machines is as follows: Job 1 2 3 4
M1 3 12 5 2
M2
8 6 4 6
M3
13 14 9 12
Solution. Step I. As the minimum processing time for job 2 on M1 > maximum processing time for job 2 on M2 , Johnson's algorithm can be applied to this problem. Step II. Let us combine the processing time of M1 and M2 and M3 to form two dummy machines M1' and M2'. This is shown in the matrix below. Job 1 2 3 4
M'1 11 (3 + 8) 18 (12 + 6) 9 (5 + 4) 8 (2+ 6)
M'2 21 (8 + 13) 20 (6 + 14) 13 (4 + 9) 18 (6 + 12)
Step III. Apply Johnson's algorithm. Minimum time of 8 occurs for job 4 on WI hence it is sequenced first. 4
3
1
The next minimum time is for job 3 on M 1 so it is sequenced next to job 4. Next is job 1 and so on. So the optimal sequencing is 4
3
1
2
Example 7.7. Six jobs have to be processed on machines M1 , M2 and M3 in order M1, M2 and M3. Time taken by each job on these machines is given below. Determine the sequence so as to minimize the processing time.
OPERATIONS RESEARCH
330 Job . 1 2 3 4 5 6
M1 12 8 7 11 10 5
M2 7 10 9 6 10 5
M, 3 4 2 5 3 4
Solution. Step I. As the minimum processing time for jobs 1 and 4 on M1 is greater than on M2, Johnson's algorithm can be applied. Step II. Combine the processing time of M1 and M2 and M3 and develop new matrix for machine M1' and M2 as follows: Job 1 2 3 4 5 6
M'1
M'2
19 (12 + 7) 18 (8 + 10) 16 (7 + 9) 17 (11 + 6) 20 (10 + 10) 10 (5 + 5)
10 (7 + 3) 14 (10 + 4) 11 (9 + 2) 11 (6 5) 13 (10 + 3) 9 (5 +4)
Step III. Use Johnson's algorithm and sequence the jobs Minimum processing time of 9 occurs for job 6 on M'2, so it will be sequenced the last. 6 Next minimum processing time of 10 occurs for job 1 on M2 so, job 1 will be sequenced next to job 6. 1
6
Next minimum processing time is 11 for jobs 3 and 4 on machine M'2 so, these will ,be sequenced as shown. 3
4
1
6
Next minimum is 13 for jobs 5 on machine M2 and after that job 2 has minimum processing time of 14 on M2', hence the sequencing is as follows : 2
5
3
4
1
6
'n' Jobs 'm' Machine Case m. The order of processing is n and 'm' machine M1, M2, M3 Let there be 'n' jobs 1, 2, 3, m and no passing is permitted. The processing time for the machine is shown M1, M2, M3 below.
THE SEQUENCING PROBLEMS
331
Job ' 1 2 3
M1 al a2 a3
M2
M3
m
b1 b2 b3
c,
Mi
c2
M2
c3
M3
• • •
• ••
• ••
• • •
•
n
a„
b„ .
c„
• • Mil
If the following conditions are used, we can replace machines by an equivalent of two machines problem: (a) Minimum ai maximum of M2, M3 ......... (m  1) (b) Minimum m maximum M2, M3  1) when M11 = a+ b. + ci + .........+(m  1)i M2 = bi + ci +............+(m 1); + mi Example 7.8. Determine the optimal sequence of performing 5 jobs on 4 machines. The machines are used in the order Ml, M2, M3 and M4 and the processing time is given below. Job 1 2 3 4 5
M1
8 9 .10 12 7'
M2
M3
M4
3 2 6 4 5
4 6 6 1. 2
7 5 8 9 3
Solution. Step I. Let us find out if any of the conditions stipulated is satisfied or not. Condition 1. Minimum ai maximum of M2 and M3. Minimum ai = 7 Maximum b, = 6 Maximum ci = 6 Hence, the condition is satisfied. Step II. Let us form the matrix of new processing time by creating two fictitious machines M1' and M'2. Job
Mi=ai+bi+ci 15(8+3+4)
M2'=bi +ci +d, 14(3+4+7)
2 3
17(9+2+6)
13(2+6+5)
22(10+6+6)
4
17(12+4+1)
20(6+6+8) 14 (4+1+9)
5
14 (7+5+ 2)
10(5+2+3)
1
332
OPERATIONS RESEARCH
Step III. Now solve 5 jobs 2 machine problem. Minimum time of processing is for job 5 on machine M2 so it will be sequenced last. 5 Next minimum time is 13 for jobs 2 on machlne 1\4'2 so it will be sequenced as shown. 2
5
Next minimum time is for jobs 1 and 4 on machine M'2 so it will be sequenced as shown. 1
4
2
5
Next minimum time is 20 for job 3 on machine M2'. 3
1
4
2
5
Example 7.9. Solve the following sequencing problem when passing off is not allowed : Job
Machine Processmg time in hours A
B
I
15
5
4
15
II
12
2
10
12
III
16
3
5
16
IV
17
3
4
17
D
C
Solution. Let us find out if one of the two conditions is satisfied. Step I. Minimum ai ?_ maximum bi and ci Or minimum di maximum bi and ci Here both the conditions are satisfied. Step II. The problem can be converted into 4 jobs 2 machines problem by introducing two fictitious machines M1' and M2 as follows: Job I II III IV where
M1 24 (15+5+4) 24 (12 + 2 + 10) 24 (16+3 +5) 24 (17+3 +4)
Mi = a; + bi + ci M2' = bi + ci + di
M2 24 (5+4+15) 24 (2 + 10 + 12) 24 (3+5 +16) 24(3+4+17)
THE SEQUENCING PROBLEMS
333
Since all the processing times are equal, the jobs can be sequenced in any manner and all sequences are optimal and will give the same minimum time. Total time can be worked out from the table below. Job
Machine A In Out 0 15 15 27 27 43 43 60
I II III IV
Machine B In Out 15 20 27 29 43 46 63 60
Machine C In  Out 20 24 39 29 46 51 63 67
Machine D In Out 24 39 39 51 51 67 67 84
Total time = 84 hours Idle time machine A = 84 — 60 = 24 hours Machine B = 15 + 7 + 14 + 14 + (84 — 63) = 15 + 7 + 14 + 14 + 21 = 71 hours Machine C = 20 + 5 + 7 + 12 + (84 — 67) = 61 hours Machine D = 24 hours. Example 7.10. Four Jobs 1, 2, 3 and 4 are to be processed on each of the five machines M1, M2, M3, M4 and M5 in the order M1, M2, M3, M4, and M5. Determine total minimum elapsed time if no passing off is allowed. Also find out the idle time of each of the machines. Processing time are given in the matrix below. Job 1 2 3 4
M1 8 7 6 9
M2 4 6 5 2
Machines M3 6 4 3 1
M4 3 5 2 4
M5 9 10 8 6
Solution. Step I. Find out if the condition minimum ei maximum bi, ci and di is satisfied. Job 1 2 3 4
M1 8 7 6 9 Minimum 6
M2 4 6 5 2 Maximum 6
Machines M3 6 4 3 1 Maximum 6
M4 3 5 2' 4 Maximum 5
M5 9 10 8 6 Minimum 6
This condition is satisfied hence we can convert the problem into four jobs and two ficticious machines M1' and M2'. Mi/ = ai bi + ci + di, M2' = bi + ci + di + ei
OPERATIONS RESEARCH
334 Step II. Job 1 2 3 4
M1' 21 (8 + 4 + 6 + 3) 22 (7 + 6 + 4 +5) 16 (6 + 5 + 3 + 2) 16 (9 + 2 + 1 + 4)
M,' 22 (4 + 6+ 3 + 9) 25 (6 + 4 4 5 +10) 18 (5 + 3 + 2 + 8) 13 (2 + 1 + 4 + 6)
Step III. The optimal sequence can be determined as minimum of processing time of 13 occurs on M2 for job 4 it will be processed last. Next minimum time is for job 3 on machine Mi so it will be processed earliest. Next shortest time is for machine 1 on Mi.', so it will be sequenced next to job 3 and so on. 3
1
2
4
Step IV. Total time can be calculated with the help of the matrix shown below. Job 1 2 3 4
In 0 8 15 21
M1 Out 8 15 21 30
M3
M2
In 8 15 21 30
Out 12 21 26 32
In 12 21 26 30
M5
M4
Out 18 25 29 31
In 18 25 29 32
Out 21 30 32 36
In 21 30 40 48
Out 30 40 48 54
Hence total minimum elapsed time is 51. Idle time for machines MI = 24 hours M2= 3 + 4 + 22 = 29 M3 = 3 + 1 + 1 + 23 = 28 M4 = 4 + 18 =22 Two Jobs 'in' Machines Case
1. Two axis to represent job 1 and 2 are drawn at right angles to each other. Same scale is used for X and Yaxes. Xaxis represents the processing time and sequence of job 1 and Yaxis represents the processing time and sequence of job 2. The processing time on machines are laid out in the technological order of the problem. 2. The area which represents processing times of jobs 1 and 2 and is common to both the jobs is shaded. As processing of both the jobs on it machine is not feasible, the shaded area represents the unfeasible region in the graph. 3. The processing of both the jobs 1 and 2 are represented by a continued path which consists of horizontal, vertical and 45 degree diagonal region. The path starts at the lower left corner and stops at upper right corner and the shaded area is avoided. The path is not allowed to pass through shaded area which as brought out in step II represents both the jobs being processed simultaneously on the same machine. Any vertical movement represents that job 2 is in progress and job 1 is waiting to be processed. Horizontal movement along the path indicates that job 1 is in progress and job 2 is idle waiting to be processed. The diagonal movement of the path indicates that both the jobs are being processed on different machines simultaneously.
THE SEQUENCING PROBLEMS
335
4. A feasible path maximizes the diagonal movement minimizes the total processing time. 5. Minimum elapsed time for any job = processing time of the job + idle time of the same job. Example 7.11. The operation time of two jobs 1 and 2 on 5 machines M1, M2, M3, M4 and M5 is given in the following table. Find out the optimum sequence in which the jobs should be processed so that the total time used is minimum. The technological order of use of machine for job 1 is M1, M2, M3, M4 and M5 for job 2 is M3, M1, M4, M5 and M2. Time Hours Job 1
M1 1
Job 2
M3 3
M2
M3
M4
M5
2
3
5
1
M1
M4
M5
M2
4
2
1
5
Job 1 preceeds job 2 on machine M1, job 1 preceeds job 2 on machine M2, job 2 preceeds job 1 on. machine M3, job 1 preceeds job 2 on M4 and job 2 preceeds job 1 on M5. The minimum processing time for jobs 1 and 2, total processing time for job 1 + idle time for Job 1 = 12 + 3 = 15 hours Total processing time for job 2 + idle time for job 2 = 15 + 0 = 15 hours.
15 14 
11__ 11 i i . Optithal Prc)cessing Path i
13 M6
12 
I
11 
M, M4
10
9 8 
I 7 M,
1
4 3 M3
2 1 0
1 014
k Mi
2
3 014
M2
7 M3
8
9 M4
Fig. 7.3
10
11
12 M5
OPERATIONS RESEARCH
336
Example 7.12. Two parts A and B for a product need processing of their operations through six machines at stations S1, S2, S3, S4, S5 and S6. The technological order of these parts and the manufacturing time on the machines are as given below. Part A Part B
Technological order Time (hours) Technological order Time (hours)
S3
S1
S5
S6
S4
S2
2
3
4
5
6
1
S2
S1
S5
S6
S3
S4
3
2
5
3
2
3
Determine the optimal sequencing order to minimize the total processing time for part A and B. Solution. Let us construct the twodimensional graph let Xaxis represent job A and Yaxis represent Job B. Total elapsed time = 23 Part A = 21 + 2 (idle time) = 23 Part B = 18 + 2 + 2 + 1 (idle time) = 23 Optimal processing path
19 18 S4
17 — 16 — 15
S3
14 13
S6
12 — 11 — 10 9—
S5
8— 7— 6—
S1
S2 I
I
I
I I I I I I I I I I I I I I I 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1+1
14 S3
S1
14
S6
S5
Fig. 7.4
S4
S2
THE SEQUENCING PROBLEMS
337
Example 7.13. There are two jobs to be performed on 5 machines. The following data is available. Find out the minimum total time sequence. Job 1 Sequence Time (Hours) Job 2 Sequence Time (Hours)
M1 2 M3 4
M2 3
M3 4
M1 5
M4 3
M5 2 M2 6
M4 6 M5 2
Solution. Let us draw Xaxis representing job1 and Yaxis representing job 2. Horizontal lines represent job 1 being processed when job 2 is waiting to be processed. Horizontal line indicates job 2 is being processed and job 1 is idle.
1 — 20 18 M2
16 4,_,_,,,6h4 ; < _._ ....41 .
14 45°
M5
il
12 M4
,._..
10
1
8
O
mi
i MI :
6 4
M,
.
ii
2
.I .
4i° 2
6
M1 M2
M3
, 10
I 12
I 14
, 16
18
, 20
14 M4
M5
Job 1 Fig. 7.5 Idle time for job 1 = 3 Idle time for job 2 = 0 Total elapsed time = 20 hours. Example 7.14. Sequencing problem as assignment problem Special case of travelling salesman problem The basic difference between the assignment and sequencing problem is that schedule in case of sequencing problem, cannot return to its original point until all the jobs have been completed.
OPERATIONS RESEARCH
338
Therefore, solving sequencing problem is more difficult than the assignment problem. Since both types of problems have certain common features, sequencing problems can be solved as assignment problem with some modifications. Travelling salesman problem is a typical sequencing problem where the salesman is expected to visit all the places before coming back to the starting point and complete the journey in minimum time. The following steps explain the procedure adopted to solve this type of sequencing problem. Step I. (i) Reduce the problem first by rows and then by columns. (ii) If zero cells in the matrix are so placed that they represent a feasible sequence for the jobs, this is taken as optimal solution of the sequencing problem. Step II. If the zero cells in the matrix are not placed in such a manner that feasible solution can be determined, then the solution is further improved till a lower cost schedule is found. If the zero cells do not give the optimal sequence then next higher cells may have to be considered for getting optimal sequence. Example 7.15. A travelling salesman has planned to visit to 5 cities. He would like to start from a particular city, visit each city once and then return to the city from where he had started. The travelling cost (Rs) for each city, is given below.
From
A . . — 7 7 3 5
A B C D E
To B 7
D 3 4 6
C 5 8
8 4 3
6 2
E 5 3 2 2
2
In what sequence the salesman should.visit all the cities in order to minimize the total travelling cost? Solution. Step I. Row reduction reduced matrix is as follows. 4
4 3 1 3
6 2 1
2 5 4 0
0 1 4 0
Total reduction in travel cost = R1 + R2 + R3 + R4 + R5 And lower bound = 3 + 3 + 2 + 2 + 2 = 12
2 0 0 0
339
THE SEQUENCING PROBLEMS
Step II. By column Reduction reduced method we get the following matrix: 3 3
2
0
2
5
1
0
4
0
2
5
00
0
2
4
2
0
0
0 0
Total reduction in travel cost = C1 + C2 = 1 + 1 = 2 and the lower bound = 12 + 2 = 14. Step III. By moving zero assignments we get
As the schedule is not completed, we deduce minimum uncovered value 1 out of all uncovered values and adding it at the intersection of lines we get. Making assignments and again moving same column and drawing parallel lines through unmarked rows and marked columns, we have (Lower bound is 14 + 1 = 15).
We can see the schedule is still not complete, let us deduct the minimum uncovered value 1 out of all uncovered values and add it at intersection of lines and lower bound is 15 + 1 = 16. Now the assignment becomes : 00
2
1
1
00
3
0
3
co
3
1.
4
55
2
0
3 0
2 .00
OPERATIONS RESEARCH
340
The schedule is still not complete. Again let us deduct the minimum value 1 out of uncovered elements and add it at the intersection of lines we get the improved matrix and the lower bound is 16 + 1 = 17. 00 1 0 0 2
A B C D E
1 00 2 0 0
0 2 co 3 0
0 0 3 00 2
3 0 0 2 Co
From the above, we get two alternatives of optimal sequence (a) A C —> E > B —› D —> A Travelling cost = 5 + 2 + 3 + 4 + 3 = 17 (b) 3 + 4 + 3 + 2 + 5 = 17 Example 7.16. Use geographical method to minimize time added to process the following jobs on the machines shown, i.e., for each machine find the job which should be done first. Also calculate the total time elapsed to complete both the jobs. D E A 2 4 2 6 3 E A D B C Sequence Time Job 2 3 2 6 4 5 Solution. The information provided in the problem can be used to draw the following diagram. The shaded area is of the overlap and need to be avoided. Sequence Time
Job 1
K 20 18 l kE 16 v.
ir 14
011011111112A
D v
N
A
12
A
10 9 8
rgA ,—,
C V A
B
3 A
6 7 8
9
10 11 12 13 14 15
16 17 18
Ofi
014
D Job 1
Fig. 7.6
E
01
341
THE SEQUENCING PROBLEMS
The path that minimizes the idle time for Job 1 is an optimal path. Also the ideal (optimal) path should minimize the idle time for Job 2. For working out the elapsed time, we have to add the idle time for either of the two jobs to that time which is taken for processing of that job. It can be seen that idle time for the chosen path for Job 1 is 5 hours and for Job 2 it is 2 hours, the elapsed time can be calculated as Processing time for Job 1 + idle time for Job 1 = 17 + (2 + 3) = 22 hours Processing time for Job 2 + idle time for Job 2 = 20 + 2 = 22 hours. II REVIEW AND DISCUSSION QUESTIONS FM 1. 2. 3. 4. 5.
What is no passing rule in a sequencing algorithm ? Explain the four elements that characterize a sequencing problem. Explain the principal assumptions made while dealing with sequencing problems. Describe the method of processing 'n' jobs through two machines. Give Johnson's procedure for determining an optimal sequence for processing n items on two machines. Give justification of the rules used in the procedure. 6. Explain the method of processing 'm' jobs on three machines A, B, C in the order ABC. 7. Explain the graphical method to solve the two jobs mmachines sequencing problem with given technological ordering for each job. What are the limitations of the method ? 8. A Company has 8 large machines, which receive preventive maintenance. The maintenance team is divided into two crews A and B. Crew A takes the machine 'Power' and replaces parts according to a given maintenance schedule. The second crew resets the machine and puts it back into operation. At all times 'no passing' rule is considered to be in effect. The servicing times for each machine are given below. Machine
a
b
c
d
e
g 9
h1
15
f 11
Crew A
5
4
22
16
Crew B
6
10
12
8
20
7
2
21
4
Determine the optimal sequence of scheduling the factory maintenance crew to minimize their idle time and represent it on a chart. 9. Use graphical method to find the minimum elapsed total time sequence of 2 jobs and 5 machines, when we are given the following information : Job 1 Job 2
Sequence
A
B
C
D
E
Time (hours)
2
3
4
6
2
Sequence
C
A
D
E
B
Time (hours) 4 5 3 2 6 10. Two jobs are to be processed on four machines a, b, c and d. The technological order for these jobs on machines is as follows : Job 1 Job 2
a d
b
a
OPERATIONS RESEARCH
342
Processing times are given in the following table : Machines Job
a
b
c
d
1
4 4
6 7
7 5
3 8
2
Find the optimal sequence of jobs on each of machines. 11. A machine shop has four machines A, B, C and D. Two jobs must be processed through each of these machines . The time (in hours) taken on each of these machines and the necessary sequence of jobs through the shop are given below. Job 1 Job 2
Sequence Time (hours) Sequence Time (hours)
A 2 D 6
B 4 B 4
C 5 C 2
D 1 A 3
Use graphic method to obtain the total minimum elapsed time. ■ G.N.D.U. EXAMINATION PROBLEMS ■
1995 APR. 1995 SEP. 1996 APR. 1996 SEP. 1998 SEP. 1997 APR. 1998 APR. 1998 SEP. 1999 SEP. 1999 SEP. 2000 SEP. 2004 APR. 2004 APR. 1995 APR.
(1) Job arrival pattern in sequencing (2) Processing time (3) Job arrival pattern (4) Processing time (5) Sequencing problem (6) What is sequencing problem ? Explain and illustrate with examples. (7) Explain sequencing problem and give its uses. Also point out the assumptions generally made, in case of simple sequencing problems. (8) Explain briefly : (i) Total elapsed time. (ii) No passing rule (iii) Processing orders in the context of sequencing problems. (9) Discuss briefly the importance of sequencing technique in decisionmaking. Also give assumptions generally made in case of simple sequencing problem. (10) What is the sequencing problem ? Explain with illustrations. (11) What general assumptions are made in solving "sequencing" problems ? List the steps of solving sequencing problem. (12) What is sequencing Problem ? How will you assign n jobs on 2 machines such that to minimize the total elapsed time ? (13) What is sequencing ? What are its assumptions ? Explain the procedure solving 'n' jobs and 3 machine sequencing problem. (14) What is a sequencing problem ? Explain the following terms in context of sequencing problems : (i) Total elapsed time and idle time, (ii) No passing rule, (iii) Processing order.
THE §kQUENCING PROBLEMS
343
1996 APR. (15) There are five jobs each of which must go through the two machines A and B in the order AB. Processing times are given below. Determine the optimum sequence for 5 jobs. Job :
1996 SEP.
J1 2
J2 4
J3 5
J4
J5
Machine A:
7
1
Machine B:
3
6
1
4
8
(16) A readymade garment manufacturer has to process 5 items through 2 stages of production, viz., cutting and sewing. The time taken from each at different stages in given below : Item :
1
2
3
4
5
Cutting
5
7
3
4
6
Sewing
2
6
7
5
9
Processing Times (Hours)
Find the order, in which these items should be processed so as to minimize the total processing time. 1998 APR. (17) Eight jobs are to be processed through 2 machines each day with no passing allowed between machines. The processing times are as follows : Job :
1
2
3
4
5
6
7
8
Processing time (hours) on :
1998 SEP.
Machine I:
4
8
7
8
2
1
3
9
Machine II:
6
3
6
4
6
5
7
2
Determine an optimum sequence to process the above jobs so as to minimize the total elapsed time. (18) Find the sequence that minimizes the total elapsed time required to complete the following jobs : ABC D E F G H I J Job Time (Hours) of M, 2 5 4 9 6 3 8 7 5 4 Time (Hours) on M2 6 8 7
4
3
11
9 8 5
2
1999 APR. (19) A certain manufacture has to process 6 items through two stages of production viz., assembling and polishing. The time taken for each of these items at different stages are given below. Find the optimal sequence so as to minimize the total processing time. Item :
1
2
3
4
5
6
Assembling :
8
10
6
7
9
14
Polishing :
5
9
10
8
12
8
OPERATIONS RESEARCH
344
1999 SEP. (20) Find the sequence that minimize the total elapsed time required to complete the following jobs : A B C DE F G H 6 3 8 7 2 5 4 9 4 3 11 9 8 6 8 7
Jobs Time (Hours) on M1 Time (Hours) on M2
2000 APR. (21) Graphically find the minimum elapsed time to perform jobs J1 and J2 on machines A through F. The sequence and process timing are given as under : E A B C D 3 4 4 5 6 B Job 2 Sequence A C E D 4 4 8 3 6 2000 APR. (22) Solve the following sequencing problem and find total elapsed time : Job 1 Sequence
F 2 F 4
4 5 6 Job : 1 2 3 9 8 M/c I: 7 4 2 5 4 1 M/c II : 3 8 6 6 2000 SEP. (23) Using graphical method find the minimum elapsed time to perform jobs J1 and J2 on machines M1 through M4. The sequence and process timing are given as follows: M3 Jobi Sequence M1 M2 M4 2 4 1 5 M3 Jobe Sequence M2 M4 M1 6 5 2 3 2000 SEP. (24) Determine the optimal sequencing of the following jobs and obtain the total elapsed time : Job Machine Mi Machine M2
1 7 8
2 3 8
3 6 2
4 8 4
5 9 7
6 5 5
7 4 6
8 3 8
5 16 13
6 20 9
2001 APR. (25) Solve the following sequencing problem :
Machine X Machine Y
1 8 7
2 10 15
Jobs 3 11 10
4 12 14
Find the total elapsed time. (Marks 6) 2001 APR. (26) The maintenance groups are G1 and G2, G1 is responsible for changing parts G2 Oils and Resets the machines. The time required by G1 and G2 are given as under. In what order should the machines be serviced so that total time taken is minimized ? Also find the waiting period of G1 and G2 :
THE SEQUENCING PROBLEMS
345
Machine
2001 SEP.
M2
Group 1
M1 8
Group 2
5
M3
M4
M5
M6
M7
6
10
11
10
14
4
3
7
12
8
6
7
(Marks 16) (27) Solve the following sequencing problems and find total elapsed time also : Job
1
2
3
4
5
6
7
Machine A
7
25
31
13
20
23
19
Machine B
17
21
21
13
24
3
7 (Marks 6)
2002 SEP.
Solve the following sequencing problem : 1
2
3
4
5
6
7
8
9
M/c X
2002 SEP.
5 6 12 15 7 8 9 4 5 M/c Y 8 10 6 9 15 3 6 5 9 Two maintenance teams T1 and T2 are responsible for maintaining seven machines. The time required is given as under. In what order the machines be serviced so that total time taken is minimized : M/cs
M1
M2
M3
M4
M
M6
M7
T1
18 15
16 13
20 17
21 22
20 18
24 16
14 17 (Marks 16)
T2
2004 APR. B.B.A. III
Script of a book has to pass through two stages—printing and binding before completion of the job. The time required during each of the above two processes in respect of six different books is given below Book No.
Printing Time (Hours)
1 2 3 4 5 6
30 120 50 20 90 110
Binding Time (Hours) 80 100 90 60 30 10
Determine the Optimal sequence and Idle time and Completion time for the Jobs. VI PUNJAB UNIVERSITY EXAMINATION PROBLEMS 11
1995 APR. 1995 SEP.
(1) Job arrival pattern in sequencing (2) Processing time
346
OPERATIONS RESEARCH
(3) Job arrival pattern (4) Processing time (5) Sequencing problem (6) What is sequencing problem? Explain the following terms in context of sequencing problems : (i)Total elapsed time and idle time, (ii)No passing rule, (Marks 10) (iii)Processing order. 1997 APR. (7) What is sequencing problem ? Explain and illustrate with examples. (Marks 6) (8) Explain sequencing problem and give its uses. Also point out the assumptions 1998 APR. (Marks 6) generally made in case of simple sequencing problems. 1998 SEP. (9) Explain briefly: (i)Total elapsed time, (ii)No passing rule, and (iii)processing orders in the context of sequencing problems. 1999 APR. (10) Discuss briefly the importance of sequencing technique in decisionmaking. Also give assumptions generally made in case of simple sequencing problem. (Marks 6)1999 SEP. (11) What is the sequencing problem ? Explain with illustrations. (Marks 6) 2000 SEP. (12) Give three different examples of sequencing problems from your daily life. What is the process of solving the sequencing problem ? (Marks 9) 2001 APR. (13) Give four different examples of sequencing problem from your daily life. (Marks 9) 2001 SEP. (14) Give an illustrative classification of sequencing problem and explain with examples. (Marks 9) 2002 APR. Explain the sequencing problem in proper details. (Marks 7) 1996 APR. (15) There are five jobs, each of which must go through the two machines A and B in the order AB. Processing times are given below. Determine the optimum sequence for jobs.
1996 APR. 1996 SEP. 1998 SEP. 1995 APR.
Job : b J2 J4 J3 J5 Machine A: 2 4 5 1 7 Machine B: 3 6 1 4 8 1996 SEP. (16) A readymade garment manufacturer has to process 5 items through 2 stages of production, viz., cutting and sewing. The time taken for each at different stages is as follows :
THE SEQUENCING PROBLEMS
347
Item :
1
2
3
4
5
Cutting
5
7
3
4
6
Sewing
2
6
7
5
9 I
Processing Times (Hours)
Find an order, in which these items should be processed so as to minimize the total processing time. 1998 APR. (17) Eight jobs are to be processed through 2 machines each day with no passing allowed between machines. The processing times are as follows : Job : Processing time (hours on) : Machine I: Machine II :
1
2
3
4
5
6
7
8
6
4 3
8 6
7 4
8 6
2 5
1 7
9 2
Determine an optimum sequence to process the above jobs so as to minimize the total elapsed time. 1998 SEP. (18) Find the sequence that minimizes the total elapsed time required to complete the following jobs : Jobs Time (Hours) on M, Time (Hours) on M2
A 2 6
B 5 8
C 4 7
D 9 4
E 6 3
F 3 11
G 8 9
H 7 8
I 5 5
J 4 2
1999 APR. (19) A certain manufacturer has to process 6 items through stages of production viz., assembling and polishing. The time taken for each of these items at different stages are given below. Find the optimal sequence so as to minimize the total processing time. Items : Assembling : Polishing:
1 8 5
2 10 9
3 6 10
4 7 8
5 9 12
6 14 8
1999 SEP. (20) Find the sequence that minimizes the total elapsed time required to complete the following jobs : Items : Assembling : Polishing :
1 8 5
2 10 9
3 6 10
4 7 8
5 9 12
6 14 8
2000 APR. (21) Convert the following three machine problems to two machine problems and hence solve for the optimal result : Tasks Time on MachineI Machine II Machine III
A 3 4 6
B 8 3 7
C 7 2 5
D 4 5 11
E 9 1 5
F 8 4 6
G 7 3 12
348 2000 SEP.
OPERATIONS RESEARCH (22) Find the optimal sequence and total elapsed time required to complete the following tasks : Time on Machine I Machine II
A _2 6
B 5 8
C 4 7
D 9 4
E 6 3
F 8 9
G 7 3
H 5 8
I 4 1
2001 APR. (23) Find the sequence that minimizes the total elapsed time required to complete the following jobs in the order AB : No. of Jobs Machine A Machine B 2001 SEP.
1 4 6
Processing times in hours 2 3 4 8 3 6 3 7 2
5 7 8
6 4 4
(24) Solve the sequencing problem AB and BA
Machines
A B
1 12 7
2 6 8
Jobs 3 5 9
4 11 4
5 5 7
6 7 8
7 6 3
2002 ARP. (25) Find the sequence that minimizes the total elapsed time required complete the following tasks. Each job is processed in any two of the machines A, B and C in any order : Jobs
Machines
A B C
1 12 7 3
2 6 8 4
3 5 9 11
4 3 8 5
5 5 7 2
6 7 8 8
7 6 3 4
2003 APR. (26) 7 Jobs, each of which has to go through the machines M1 and M2 in the order M2M1. Processing time (in Hour) are given : Job Machine Machine
MI M2
AB 3 12 8 10
C 15 10
Jobs D 6 6
E 10 12
F 11 1
G
6 3
(i)Determine the optimum sequence that will minimize the total elapsed time, with idle time of each machine. (ii)If the order is reversed to M2M1, what difference will it make to the calculated results ? (Marks 18) 2004 APR. (27) A manufacturing company processes 6 different jobs on two machines A and B. Number of units of each job and its processing time on A and B are given in table. Find the optimal sequence, the total minimum elapsed time and idle time for each machine.
349
THE SEQUENCING PROBLEMS
Table Job no.
No. of units of each job
1
3 4 2 5 2 3
2 3 4 5 6
Processing time Machine A Machine B (minutes) (minutes) 5 16 6 3 9 6
8 7 11 5 7 14
Replacement Theory I m Dm ma ming MI DIU IV • Understanding that depreciable assets have to be replaced • Formulation of assumptions of Replacement Theory • Understanding replacement policy for such items whose maintenances cost increases with time and also the money value changes with time (at constant rate) • Understand the mechanism of systems failing suddenly • Discuss group replacement policy. INTRODUCTION Replacement of old plant and equipment and items of use like bulbs/tubelights, refrigerators/ heating, tools/gadgets, etc, is a necessity. All these items are designed for performance up to the desired level for a particular time (years /hours) or particular number of operations. For example, when a refrigerator is given the warranty for 7 years the manufacturer knows that the design of the refrigerator is such that it will perform up to desired level of efficiency without breakdown for that period. Similarly, bulbs/tubelights may have been designed for say 10,000 onoff operations. But all these need to be replaced after a particular period/number of operations. The equipment is generally replaced because of the following reasons : (i) When the item/equipment fails and does not perform its function it is meant for. (ii) Item/equipment has been in use for sometimes and is expected to fail soon. (iii) The item/equipment in use has deteriorated in performance and needs expensive repairs i.e., it has gone beyond the economic repair situations. The cost of maintenance and repair of equipment keeps increasing with the age of the equipment. When it becomes uneconomical to continue with an old equipment, it must be replaced by a new equipment. (iv) Improved technology has given access to much better (convenient to use) and technically superior (using less power) products. This is the case of obsolescence. The equipment needs to be replaced not because it does not perform up to the standards it is designed for but because new equipment is capable of performance of much higher standards. It should be understood that all replacement decisions involve high financial costs. The financial decisions of such nature will depend upon a large number of factors, like the cost of
REPLACEMENT THEORY
351
new equipment, value of scrap, availability of funds, cost of funds that have to be arranged, tax benefits, government policy, etc. When making replacement decisions, the management has to make certain assumptions, these are : (i) The quality of the output remains unchanged. (ii) There is no change in the maintenance costs. (iii) Equipments perform to the same standards. Let us discuss some of the common replacement problems. Replacement of items, which deteriorate with time without considering the change in money value Most of the machinery and equipment having moving parts deteriorate in their performance with passage of time. The cost of maintenance and repair keeps increasing with passage of time and a stage may reach when it is more economical (in overall analysis) to replace the item with a new one. For example, a passenger car is bound to wear out with time and its repair and maintenance cost may go to such level that the owner has to replace it with a new one. Let C = Capital cost of the item, S (t) = Scrap value of the item after t years of use, 0 (t) = Operating and maintenance cost of the equipment at time t, n = number of years the item can be used, TC (n) = Total cost of using the equipment for n years, O (t)
TC (n) = C — S (t) + t=i Average TC (n) =
1 n [C — S (t) +
O(t)
Time 't' in this case is a discrete variable. In this case as long as the average TC (n) is minimum, the equipment can remain in use for that number of years. If average total cost keeps decreasing up to ith year and starts increasing from (i + 1)th year then ith year may be considered as most economic year for replacement of the equipment. The concept of depreciation cost also must be understood here. As the years pass by, the cost of the equipment or items keeps decreasing. How much the cost keeps decreasing can be calculated by two methods commonly used, i.e., straight line depreciation method and the diminishing value method. Example 8.1. A JCB excavator operator purchases the machine for Rs 1500000. The operating cost and the resale value of the machine is given below.
OPERATIONS RESEARCH
352 Year Operating
1
3
4
5
6
7
8
36000
40000
45000
52000
60000
70000
8
5
4.5
4
3
2
2
30000 32000
Cost (in Rs) Resale value
12
10
(in lakhs of Rs.) When should the machine be replaced ? C = 1500000 0 (t) = Operating cost S (t) = Resale value t = Time n = Number of years after when the machine is to be replaced. Let us draw a table showing the various variables required to make decision. This is shown in the table below.
Solution.
Year
1 2 3 4 5 6 7 8
Cumulative Resale value Depreciation Total cost 0 (t) TC (n) (in C — S (t) (in S (t) (in 0 (t) (in thousand thousands of thousands of thousands of Rupees) Rupees) Rupees) of Rupees) 300 330 30 30 1200 62 1000 500 562 32 800 700 798 36 98 500 1000 1138 40 138 450 1050 1233 45 183 400 1100 1335 52 235 300 1200 1495 60 295 200 1300 1665 70 365
Average TC (n) (in thousands of Rupees) 330 281 266 284.5 246.6 222.5 213.6 208
In third year the minimum average cost is 266000 as shown in the table above. So, replacement should take place at the end of three year. Example 8.2. A truck owner finds from his past experience that the maintenance costs Rs. 200 for the first year and then increases by Rs. 2000 every year. The cost of the truck type A is Rs. 9000. Determine the best age at which to replace the truck. Truck B type costs Rs. 10000. Annual maintenance costs are Rs. 400 and increases by Rs. 800 every year. The truck owner now has truck type A, which is one year old. Should it be replaced by type B and if so, when ? Solution. Let us work out the average total cost per year for the trucks, type A and type B.
353
REPLACEMENT THEORY Truck A Year
1 2 3 4 5 6 7
Operating Cost 0 (t)
Cumulative Operating Cost 0(t)
200 2200 4200 6200 8200 10200 12200
Depreciation C — S (t)
200 2400 6600 12800 21000 31200 43400
9000 9000 9000 9000 9000 9000 9000
Total Cost TC (n) 9200 11400 15600 21800 30000 40200 52400
Average TC (n) 9200 5700 5200 4550 6000 6700 7485
Truck B Year
Operating Cost 0 (t)
1 2 3 4 5 6 7
400 1200 2000 2800 3600 4400 5200
Cumulative Operating Cost 0 (t) 400 1600 3600 6400 10000 14400 19600
Depreciation C — S (t) 10000 10000 10000 10000 10000 10000 10000
Total Cost TC (n) 10400 11600 13600 16400 20000 24400 29600
Average TC (n) 10400 5800 4533 4100 4000 4067 4229
Truck A should be replaced by truck B, at the end of fourth year. Example 8.3. A taxi owner estimates from his past records that the cost per year for operating a taxi whose purchase price when new is Rs 60000 are as given below. Age Operating Cost (Rs.)
1
2
3
4
5
10000
12000
15000
18000
20000
After 5 years, the operating cost is Rs. 6000k where k = 6, 7, 8, 9, 10 (k denoting age in years). If the resale value decreases 10% of purchase price per year, what is the best replacement policy ? Cost of money is zero. Solution. Since the depreciation of the taxi is 10% of Rs. 60000, it means the resale value of the taxi decreases by Rs. 6000 every year. Average annual cost can be calculated as shown in the following table.
354
OPERATIONS RESEARCH
Year of Service
Operating Cost 0 (t)
1 2 3 4 5 6 7 8 9 10
10000 12000 15000 18000 20000 36000 42000 48000 54000 60000
Cumulative Operating Cost 0 (t) 10000 22000 37000 55000 75000 111000 153000 201000 255000 315000
Resale value S (t)
Depreciation C — S (t)
54000 48000 42000 36000 30000 24000 18000 12000 6000 0
6000 12000 18000 24000 30000 36000 42000 48000 54000 60000
Total Cost TC (n)
Average TC (n)
16000 34000 55000 79000 105000 147000 195000 249000 309000 375000
16000 17000 18333 19750 21000 24500 27857 31125 34333 37500
Total Cost TC (n) = Operating Cost 0 (t) + Depreciation C — S (t) It can be seen that average Total cost is minimum after one year. Hence the taxi should be replaced after one year. Example 8.4. A truck owner finds from his past records that the maintenance cost of a truck (whose purchase is Rs. 300000) during the first 8 years of its life and the resale price at the end of each year is as follows Year 1 2 3 4 5 6 7 8
Maintenance Cost (Rs.) 36000 48000 60000 72000 84000 96000 108000 120000
Resale Price (Rs.) 200000 150000 100000 80000 70000 60000 50000 40000
When should the truck be replaced? Solution. The average cost per year is calculated in the following table. Year of Service
Operating Cost 0 (t)
1 2 3 4 5 6 7 8
36000 48000 60000 72000 84000 96000 108000 120000
Cumulative Operating Cost 0 (t) 36000 84000 144000 216000 300000 396000 504000 624000
Resale Value S (t) 200000 150000 100000 80000 70000 60000 50000 40000
Depreciation C — S (t) 100000 150000 200000 220000 230000 240000 250000 260000
Total Cost TC (n)
Average TC (n)
136000 234000 344000 436000 530000 636000 754000 884000
136000 117000 114667 109000 106000 106000 107714 110500
355
REPLACEMENT THEORY
Since average total cost is minimum in the 5th and 6th year, the truck should be replaced at the end of six years. He gets no advantage by replacing it at the end of 5 years. Example 8.5. A new tempo costs Rs. 100000 and may be sold at the end of year at the following prices: 1 60000
Year Selling Price (Rs.)
2 45000
3 32000
5 10000
4 22000
6 2000
The corresponding annual operat'ng costs are: Year Cost/Year (Rs.)
1 10000
2 12000
3 15000
4 20000
5 30000
6 45000
It is not only possible to sell the tempo after use but also to buy a second hand tempo. It may be cheaper to do so than to buy a new tempo. Age of tempo Purchase Price (Rs.)
0 100000
1 60000
2 45000
3 33000
4 20000
5 10000
What is the age to buy and to sell so as to minimize average annual cost ? Solution. Cost of new tempo = Rs. 100000
Let us find out the average cost per year of the new tempo. Year of Service (1)
Operating Cost 0 (t) (2)
Cumulative Operating Cost 0 (t) (3)
1
10000 12000 15000 20000 30000 45000
10000 22000 37000 57000 87000 132000
2 3 4 5 6
Resale Value S (t) (4)
60000 45000 32000 22000 10000 2000
Depreciation C — S (t) (5)
40000 55000 68000 78000 90000 98000
Total Cost TC (n) (6) = (3) + (5)
50000 77000 105000 135000 177000 230000
Average TC (n) (7) = (6) + (1)
50000 38500 35000 33750
35400 35000
Average cost is minimum at the end of fourth year; hence the new tempo should be replaced after 4 years. Let us now find out the average total cost of second hand tempo. Year of Service (1)
Operating Cost 0 (t) (2)
Cumulative Operating Cost 0 (t) (3)
0
1 2 3 4 5
10000 12000 15000 20000 30000
10000 22000 37000 57000 87000
Resale Value S (t) (4)
100000 60000 45000 32000 20000 10000
Depreciation C — S (t) (5)
100000 40000 55000 68000 78000 90000
Total Cost TC (n) 6 =3 +5
Average TC (n) 7=6 ÷ 1
100000 50000 77000 105000 137000 177000
50000 38500 35000 33750 35400
OPERATIONS RfSEARCH
356
The tempo may be replaced by second hand tempo at the end of third year and the owner can save Rs. (3500034666), i.e., Rs. 334 instead of buying a new one. Example 8.6. A machine type A costs Rs. 50000 and the operating costs are estimated at Rs. 1200 for the first year and increased by Rs. 8000 in second and subsequent years. Another machine type B costs Rs. 60000 and operating costs are Rs. 1500 for the first year and increasing by Rs. 5000 per year. Should machine A be replaced by machine B assuming that both machines have no resale value and cost of money does not change with time ? Solution. Machine A Year
1 2 3 4 5 6
Operating Cost 0 (t) 1 1200 9200 17200 25200 33200 41200
Cumulative Operating Cost 0 (t) 2 1200 10400 27600 52800 86000 127200
Depreciation C — S (t) 3 50000 50000 50000 50000 50000 50000
Total Cost TC (n) 4=2+3 51200 60400 77600 102800 136000 177000
Average TC (n)
51200 30200 25866 25700 27200 29500
Machine B Year
Operating Cost 0 (t) 1
1
1500 6500 11500 16500 21500 26500
2 3 4 5 6
Cumulative Operating Cost 0 (t) 2
Depreciation C — S (t) 3
Total cost TC (n) 4 =2 +3
Average TC (n)
60000 60000 60000 60000 60000 60000
61500 68100 79600 96100 117600 144100
61500 34050 26533 24025
1500 8100 19600 36100 57600 84100
23520
24016
It can be exactly seen that the lowest average cost of machine B (Rs. 23520) is lesser than that of machine A (Rs. 25700), machine A should be replaced by machine B at the end of fourth year. Example 8.7. A firm is considering replacement of a machine, whose cost price is Rs. 12200 and the scrap value Rs. 200. The running (maintenance and operation) costs, in Rs. are found from experience to be as follows:
Year Running Cost
1 200
2 500
3 800
When should the machine be replaced ?
4 1200
5 1800
6 2500
7 3200
8 4000
REPLACEMENT THEORY
357
Solution. We are given the following data: C = Rs. 12200 S (t) = Rs. 200 0 (t) is also given. Let n be the number of years the item can be used. Average total cost per year during the life of the machine is shown in the table below. Year of Service n (1)
Running Cost (Rs) 0 (t) (2)
Cumulative Operating Cost Rs. X 0 (t) (3) 200
Depreciation Cost Rs. C — S (t) (4)
Total cost (Rs.) (3 + 4 =5)
Average Cost ( Total Cost) n )
12000 for all time
12200
12200
1
200
2
500
700
12700
6350
3
800
1500
13500
4500
4
1200
2700
14700
3675
5
1800
4500
16500
3300
6
2500
7000
19000
7
3200
10200
22200
3171
8
4000
14200
26200
3275
_
3167
The average total cost per year is minimum in sixth year i.e., Rs 3167 And the average cost in seventh year is Rs 3171 which is more than the cost in sixth year. Hence, the machine should be replaced after 6 years. Example 8.8. (a) Machine A costs Rs. 9000. Annual operating costs are Rs. 200 for the first year and then increases Rs. 2000 every year. Determine the best age at which to replace the machine. If the optimum replacement policy is followed, what will be the average yearly cost of owning and operating the machine? (b) Machine B costs Rs. 10000. Annual operating cost are Rs. 400 for the first year and then increases by Rs 800 every year. You now have a machine of type A which is one year old. Should you replace it with B, if so, when ? Solution. (a) Let us assume that there is no scrap value of the machine Average total cost can be computed as
OPERATIONS RESEARCH
358 Year (n)
Operating Cost 0 (t)
1 2 3 4 5
200 2200 4200 6200 8200
Cumulative Operating Cost I 0 (t) 200 2400 6600 12800 21000
Depreciation C — S (t)
Total Cost
9000 for all years
9200 11400 15600 21800 30000
Average Cost 9200 5700 5200 5450 6000
It can be seen that the best age for replacement is third year. (b) For machine B, the average cost can be calculated as follows: Year (n)
Operating Cost 0 (t)
1 2 3 4 5 6
400 1200 2000 2800 3600 4400
Cumulative Operating Cost 1 0 (t) 400 1600 3600 6400 10000 14400
Depreciation C — S (t) 10000 for all year
Total Cost
10400 11600 13600 16400 20000 24400
Average Cost 10400 5800 4533 4100 4000 4066
Since the minimum average cost for machine B is lower than for machine A, machine B should be replaced by machine A. Minimum average cost is (Rs. 4000), it should be replaced when it exceeds Rs. 4000. In case of one year old machine Rs. 2200/ will be spent next year and Rs. 4200 the following year. We should keep machine A for one year. Replacement policy of an equipment/item whose operating cost increases with time and money value also changes with time In previous examples, we assumed that the money value does not change and remains constant but it is wellknown that as the equipment deteriorates and operating costs keep increasing, the money value keeps decreasing with time. Hence we must calculate the Net Present Value (NPV) of the money to be spent a few years hence. Otherwise the resale value, the operating costs, which are to take place in future, will not be realistic and management will not be able to take optimal decisions. C = Initial cost of item/equipment Let O C = Operating cost R = Rate of interest A rupee invested at present will be equivalent to (1 + r) a year after (1 + r)2 two years hence and (1 + r)n in n years time. It means that making a payment of one rupee after n years is equivalent to paying (1 + r)" now. The quantity (1 + r)" is called the present worth or present value of one rupee spent n years from now.
REILACEMENT THEORY
359
Present value of a rupee V = (1 + r) I = 1/1 + r is called discount rate and is always less than 1. Then, year wise present value of expenditure in future years can be calculated as Present value (n) = (c + oc1) + oc2 v + oc3 v2+. .+ oc„ v" 1 + (c + oc1) v" + oc2 e+1 oc3 v"  2 +.........+0C„ y2" 1 (C OCi) 02"1 0C2 02 1+.........+0C„ V3 "1 Steps Involved in Calculation of Replacement Policy When Time Value Changes
Step I. Find out the present value factor at the given rate and multiply it with the operating/ maintenance cost of the equipment/items for different years. Step II. Work out the total cost by adding the cumulative present value to the original cost for all the years. Step III. Cumulate the discount factors. Step IV. Divide the total cost by corresponding value of the cumulated discount factor for every year. Step V. Find out the value of last column that exceeds the total cost. Equipment/item will be replaced in the latest year. These steps will be explained with the help of an example.
Example 8.9. The yearly cost of two machines X and Y, when money value is neglected is shown below. Find which machine is more economical if money value is 10% per year. Year Machine X (Rs.) Machine Y (Rs.)
1 2400 3200
2 1600 800
3 1800 1800
Solution. It may be seen that the total cost for each machine X and Y is Rs. 5800 (2400 + 1600 + 1800) or (3200 + 800 + 1800). When the money value is not discounted the machines are equally good, total cost wise, when money value is not changed with time, with money value 10% per year, the discount rate, it changes as follows : V—
1 1 1 = =0.9091 = 1+r 1 + 0.10 1.1
Discounted costs are obtained by multiplying the original costs with 0.9091 after one year. Total costs of machines X and Y are calculated as shown below. Year Machine X
1 2400
Machine Y
3200
2 1600 x 0.9091 = 1440 800 x 0.9091 = 720
3 1800 x 0.9081 = 1620 1620
Total cost (Rs.) 5460 5540
The total cost of machine X is less than that of machine Y, machine X is more economical.
Example 8.10. The cost of a new machine is Rs. 5000 the maintenance cost during nth year is given by M„ = Rs. 500 (n — 1) where n = 1, 2, 3, If the discount rate per year is 0.05, after how many years will it be economical to replace the machine by a new one ? Solution. The discounted rate is given as 0.05, i.e., 5% then the present value.
OPERATIONS RESEATCH
360
1 1 1 + r 1 + 0.05 = 0.9523 after one year. After 2 years it will be (.9523)2 and so on.
V
Year
Maintenance Cost
1 2 3 4 5 6
0 500 1000 1500 2000 2500
Discounted Maintenance Cost
Discounted Cost
Average Total Cost
5000 2738 2127 1919
5000 5476 6383 7679 9324 11283
0 467 907 1296 1645 1959
1.0 0.9523 0.9070 0.8638 0.8227 0.7835
Cumulative Total Discounted Cost
1865
1880
From above it is clear that it will be economical to replace the machine after fifth year. Example 8.11. The original cost of the machine is Rs. 10000. Maintenance costs vary as given below.
1 500
Year Maintenance Cost (Rs.)
2 800
3 1200
4 1500
5 2000
6 2500
7 3000
If the money is discounted at 10% per year, what is the optimum replacement policy ? Solution. Year Maintenance Cost OC 1
1 2 3 4 5 6 7
500 800 1200 1500 2000 2500 3000
V at 10% 2
PV of OC 3
1 .9523 .9070 .8638 .8227 .7835 0.7325
500 761.6 1088.4 1294.5 164.5 195.8 2197.5
Cumulative CS (t) TC (n) Cumulative Weighted PV of OC 5 6 = 4 + 5 Value of V Average 4
500 1261.6 2345.6 3640 5258 7243 9440.5
10000 10000 10000 10000 10000 10000 10000
10500 11261.6 123456 13640 15285 17243 19440
1 1.25 2.85 3.71 4.53 5.35 6.05
10500
3223 3213
Since the average cost is less in the seventh year, optimum replacement policy is at the end of seven year. Example 8.12. An engineering company is offered a material handling equipment A. A is purchased for
Rs. 60000 originally and maintenance costs are estimated to be Rs. 10000 for each of the first five years and increasing every year by Rs. 3000 from the sixth and subsequent years. The company expects a return of 10% on all its investments. What is the optimum replacement period ? Assume that maintenance cost is increased at the end of the year. Solution. Here C = 60000
r = 10%
REPLACEMENT THEORY Present value —
361
1 n = number of years (1+ r)"
Let us work out the weighted cost with the help of following table: Year
1 2 3 4 5 6 7 8 9 10
Operating Cost OC 1 10000 10000 10000 10000 10000 13000 16000 19000 22000 25000
Discounted Factor 2 0.909 0.826 0751 0683 0.621 0.564 0.513 0466 0.424 0.385
PV of OC 3 90090 8260 7510 6830 6210 7332 8208 8854 9328 9625
Total Cost Cumulative 4 69090 77350 84860 91690 97900 105232 113440 122294 131622 141247
Cumulative PV Factor 5 1.91 2.73 3.48 4.16 4.79 5.35 5.86 6.33 6.75 7.14
Weighted Average Cost 6 = 4+5 36192 28282 24343 21993 20438 19655 19335 19311 19479 19777
It can be seen above that the weighted average cost is minimum at the end of 8 years. Hence optimum replacement period is 8 years. Example 8.13. A manufacturer is offered two machines X and Y. Machine X is priced at Rs. 10000 with running cost of Rs. 1000 for first four years and increasing by Rs. 400 in fifth year and subsequent years. Machine Y which has the same capacity and performance as X costs Rs. 8000 but has maintenance cost of Rs. 1200 per year for first five years increasing by Rs. 400 in the sixth and subsequent years. If cost of money is 10% per year, which is a more economical machine ? Assume running cost is incurred at the beginning of the year. 1 = 0.909 1 + 0.10 Machine X, C = 10000.
Solution. PV =
Year
OC
1 2 3 4 5 6 7 8 9 10
1000 1000 1000 1000 1400 1800 2200 2600 3000 3400
PV factors 1.00 0909 0826 0751 0683 0.621, 0.564 0.513 0.466 0.424
PV of C + Cumulative OC PV of OC 1000 11000 909 11909 826 12735 751 13486 14442 956 1116 15558 1240 16798 1326 18124 1398 19532 1429 21951
Cumulative PV Factor 1.00 1.909 2.735 3.486 4.169 4.790 5.355 5868 6.334 6.759
Weighted Average Cost 11000 5984.5 4656.30 3868.6 3464 3248 3137 30886 3084 3247
OPERATIONS RESEARCH
362 Machine Y, C = 8000. Year
OC
1 2 3 4 5 6 7 8 9 10
1200 1200 1200 1200 1200 1600 2000 2400 2800 3200
PV factors
1.00 0.909 0.826 0.751 0.683 0.621 0.564 0.513 0.466 0.424
PV of OC
1200 1090.8 991.2 901.2 819.6 993.6 1128 1231.8 1304.8 1356.8
C+ Cumulative PV of OC 9200 10290.8 11282 12183.2 13002.8 13996.4 15124.4 16355.6 17660.4 19017.2
Cumulative PV Factor 1.00 1.909 2.735 3.486 4.169 4.790 5.355 5.868 6.334 6.759
Weighted Average Cost 9200 5390.67 4125 3494.8 3119 2922 2824.35 2787.25 2788.2 2813.6
It can be seen that weighted average cost of machine X is minimum, i.e., Rs. 3084 in ninth year. Where as the weighted average cost of machine Y is minimum in 8th year, i.e., 2787.25 so it is advisable to purchase machine Y. Example 8.14. An executive has the option to buy car A or car B. Car A costs Rs. 650000 and its running and maintenance cost is Rs. 60000 for each of the first five years increasing by Rs. 20000 per year from sixth year. Car B is considered as good as car A but costs only 585000. However, its running and maintenance cost is Rs. 100000 for each of the first 5 years and increases by Rs. 20000 per year thereafter. If the money is worth 9% per year which car should be purchased ? What is the optimal replacement period for each car assuming that there is no salvage value for either of the cars ? Solution. We have to make choice between car A and car B hence we should, work out the average weighted cost for both cars PV = 1 = 1 = 0.917 after one year. 1 + r 1+•09 Car A, cost C = 650000 Year
OC
1'V factors
PV of OC
1 2 3 4 5 6 7 8 9 10
60000 60000 60000 60000 60000 80000 100000 120000 140000 160000
1.000 0.917 0.841 0.772 0.708 0.650 0.596 0.546 0.501 0.459
60000 55020 50460 46320 42480 52000 59600 65520 70140 73440
C + Cumulative PV of OC 710000 765020 815480 861800 904280 956280 1015880 1081400 1151540 1224980
Cumulative PV Factor 1 1.917 2.758 3.530 4.238 4.888 5.484 6.03 6.531 6.99
Weighted Average Cost 710000 399071.46 295678 244136 213374 1956388 185244.3 179336.6 176319 175475
REPLACEMENT THEORY
363
Car B, cost C = 585000 Year
OC
PV factors
1 2 3 4 5 6 7 8 9 10
100000 100000 100000 100000 100000 120000 140000 160000 180000 200000
1.000 0.917 0.841 0.772 0.708 0.650 0.596 0.546 0.501 0.459
PV of OC C + Cumulative PV of OC 100000 685000 91700 776700 84100 860800 77200 938000 70800 1008800 184615.3 1193415.3 234899.3 1428314.6 293040.3 1721354.9 359281.4 2080636.3 435729.6 2516366
Cumulative PV Factor 1 1.917 2.758 3.530 4.238 4.888 5.484 6.03 6.531 6.99
Weighted Average Cost 685000 405164.3 312110.2 265722.3 238036.8 244152 260451.2 285465.1 318578.5 359995.1
It is obvious by comparing the weighted costs of two cars that car A is far better. Car A should be replaced after 10 years and car B should be replaced after 9 years. REPLACEMENT POLICY FOR EQUIPMENT/POLICY THAT BREAKS DOWN/ FAILS SUDDENLY As an equipment or item, which is made of a number of components ages with time, it deteriorates in its functional efficiency and the performance standard are reduced. However, in real life situation there are many such items whose performance does not deteriorate with time but fail suddenly without any warning. This can cause immense damage to the system or equipment and inconvenience to the user. When the item deteriorates with time, one is expecting reduced performance but other items, which may fail without being expected to stop performing, can create a lot of problems. A minor component in an electronic device or equipment like TV, fridge or washing machine, costs very little and may be replaced in no time but the entire equipment fails suddenly if the component fails. Hence the cost of failure in terms of the damage to the equipment and the inconvenience to the user is much more than the cost of the item. If it is possible to know exactly the life of the component, it is possible to predict that the component and hence equipment is likely to fail after performance of so many hours or miles, etc. This is the concept of preventive maintenance and preventive replacement. If the equipment is inspected at laid down intervals to know its conditions, it may not be possible to expect the failure of the item. The cost of failure must be brought down to minimum, preventive maintenance is cheap but avoids lots of problems. In many cases, it may not be possible to know the time of failure by direct inspection. In such cases the probability of failure can be determined from the past experience. Finding the Mean Time Between Failure (MTBF) of the equipment in past is one good way of finding this probability. It is possible by using the probabilities to find the number of items surviving up to certain time period or the number of items failing in a particular time period. In situations, when equipment/item fails without any notice, two types of situations arise. (a) Individual Replacement Policy. In this case an item is replaced immediately when it fails. (b) Group Replacement Policy. In this policy all the items are replaced irrespective of the fact whether the items have failed or not, of course, any item failing before the time fixed for group replacement is also replaced.
OPERATIONS RESEARCH
• 364 Individual Replacement Policy
In this policy, a particular time 't' is fixed to replace the item whether it has failed or not. It can be done when one knows that an item has been in service for a particular period of time and has been used for that time period. In case of moving parts like bearings, this policy is very useful to know when the bearing should be replaced whether it fails or not. Failure of a bearing can cause a lot of damage to the equipment in which it is fitted and the cost of repairing the equipment is much more than the cost of bearing if it had been replaced well in time. If it is possible to find out the optimum service life it'the sudden failure and hence loss to the equipment and production loss, etc, can be avoided. However, when we replace items on a fixed interval of preventive maintenance period certain items may be left with residual useful life which goes waste. Such items could still perform for another period of time (not known) and so the utility of items has been reduced. Consider the case of a city corporation wanting to replace its street lights. If individual replacement policy is adopted then replacement can be done simultaneously at every point of failure. If group replacement policy is adopted then many lights with residual life will be replaced incurring unnecessary costs. Analysis of the cost of replacement in case of items/equipments that fail without warning is similar to finding out the probability of human deaths or finding out the liability of claims of Life Insurance Company on the death of a policy holder. The probability of failure or survival at different times can be found out by using mortality tables or life tables. The problem of human births and deaths as also individual problems where death is equivalent to failure and birth is equivalent to replacement can also be studied as part of the replacement policy. For solving such problems, we make the following assumptions: (a) All deaths or part failures are immediately replaced by births or part replacements and (b) There are no other, except the ones under consideration, entries or exits. Let us find out the rate of deaths that occur during a particular time period assuming that each item in a system fails just before a particular time period. The aim is to find out the optimum period of time during which an item can be replaced so that the costs incurred are minimum. . Mortality or life tables are used to find out the probability destination of lifespan of items in the system. Let f (t)  number of items surviving at time (t 1) n = Total number of items with system under consideration. The probability of failure of items between 't' and (t  1) can be found out by P
((t 1)  f (t) )
Replacement Policy
Let the service life time of an item be T and try number of items in a system which need to be replaced whenever any of these fails or reaches T. F (t) = number of items surviving at T F'(t) = 1 f (t) number of items that have failed O (t) = Total operating time Cf = Cost of replacement after failure of item CPM = Cost of preventive maintenance.,
365
REPLACEMENT THEORY Cost of replacement after failure of service time T =n x f '(t) X C1 Also cost of replacement for item replaced before failure = n [1 —1(T)] Cpm
= n + f '(T) c f + n [1 — f ' (T) Cpm Hence we can replace an item when the total replacement cost given above is minimum where 0 (t) = f (t) dt
Group Replacement Policy Under this policy, all items are replaced at a fixed interval 't' irrespective of the fact they have failed or not and at the same time keep replacing the items as and when they fail. This policy is applicable to a case where a large number of identical low cost items which are more and more likely to fail at a time. In such cases, i.e., like the case of replacement of street lights, bulbs, it may be economical to replace all items at fixed intervals. Let
n = total number of items in the system N (t) = number of items that fail during time t C (t) = Cost of group replacement after time t C (t) / t = average cost per unit time C g = Cost of group replacement Cf = Cost of replacing one item on failure C (t) =nCg F (t) = Average cost per unit time =C(t)/ t=nCg +Cf(ni +n2+.........+nt _ 1 ) / t
We have to minimize average cost per unit time, so optimum group replacement time would be that period which minimize this time. It can be concluded that the best group replacement policy is that which makes replacement at the end of 't'th period if the cost of individual replacement for the same period is more than the average cost per unit time. Example 8.15. The following mortality rates have been observed for certain type of light bulbs:
End of week
1
2
3
4
5
Percentage Failing
10
20
50
70
100
There are 1000 bulbs in use and it costs Rs. 10 to replace an individual bulb which has burnt out. If all the bulbs are replaced simultaneously, it would cost Rs. 5 per bulb. It is proposed to replace all the bulbs at fixed intervals whether they have fixed or not and to continue replacing fused bulbs as and when they fail. At what intervals should all the. bulbs be replaced so that the proposal is economical ? Solution. Average life of a bulbi,n weeks = Probability of failure at the end of week x number of 4 r • bulbs = (1 X 10/100 4 21X a0/100 + 3 x 30/100 + 4 x 20/100 + 5 x 30/100) = 0.10 + 0 .2 FOA 9)+ . 8 + 1 =3 .5 number of bulbs 1000 t = 285 Average number., of replacement p r week = average life 3.5 Cost per week @ Rs 10 per bulb = 285 x 10 = Rs.2850
OPERATIONS RESEARCH
366
Let n1, n2, n3, n4 and 125 be the number of bulbs being replaced at the end of first,second, third, fourth and fifth week respectively then ni = number of bulbs in the beginning of the first week x probability of the bulbs failing during first week =1000 x 10/100 = 100 n2 = (number of bulbs in the beginning x probability of the bulbs failing during second week) + number of bulbs replaced in first week x probability of these replaced bulbs failing in second week. = 1000 x (20 — 10) /100 + 100 x 10/100 = 100 + 10 = 110 123 = (number of bulbs in the beginning x probability of the bulbs failing during third week) + number of bulbs being replaced in first week x probability of these replaced bulbs failing in second week) + number of bulbs being replaced in second week x probability of those failing in third week) = 100 x (50 — 20) /100 + 100 x (20 — 10)/100 + 110 + 10/100 = 300 + 10 + 11 = 321 124 = 1000 x (70 — 50) /100 + 100 x (50 — 20) /100 + 110 + 20 — 10/100 + 321 x 10/100 = 200 + 30 + 11 + 32 = 273 n5 = 100 x 30/100 + 100 x 20/100 + 110 x 30/100 + 321 x 10/100 + 273 x 10/100 = 300 + 20 + 33 + 32 + 28 = 413 The economics of individual or group replacement can be worked out as shown in the table below. End of week
No. of bulbs failing
Cumulative No. of failed bulbs
Cost of individual replacement
Cost of group replacement
Total cost
Average Total Cost
1 2 3 4 5
100 110 321 273 413
100 220 541 814 1227
1000 2200 5410 8140 1227n
5000 5000 5000 5000 5000
6000 7200 10410 13140 17270
6000 3600 3470 3285 3454
Individual replacement cost was worked out to be Rs. 2850. Minimum average cost per week corresponding to 4th week is Rs. 3285, it is more than individual replacement cost. So it will be economical to follow individual replacement policy. Example 8.16. An automatic machine uses 250 moving parts as part of its assembly. The average cost of a failed moving part is Rs. 200. Removing the failed part and replacing it is time consuming and disrupts manufacture. Due to this problem, the management is considering group replacement policy, i.e., replacing all the moving parts at a specific interval. What replacement policy should the manufacture adopt? The information regarding the machine breakdown and the cost is as given below.
Use time in months Probability failure
1 0.05
2 0.05
3 0.10
4 0.15
5 0.25
6 0.40
REPLACEMENT THEORY
367 Replacement Cost Total 700 350
Individual Replacement
Purchase Installation 200 500 Group Replacement 150 _ 200 Solution. Let us find out the average life of a moving part. Month 1 2 3 4 5 6
Prob. of failure 0.05 0.05 0.10 0.15 0.25 0.40
Months x Probability 0.05 0.10 0.30 0.60 0.75 2.40 Total = 4.20
Average number of replacement per month = Number of moving parts = 250 = 60 4.2 average life Average cost per month when the moving part is individually replaced 60 x 700 = 42000 Now, we must find out the failure of the moving parts per month sixth Let n1, n2, .......,n6 be the number of moving parts failing at the end of first, second, month = p1 = number of parts in the first month x probability of a part failing in first month = 250 x 0.05 = 12.5 = 13 n2 = no + pi = (250 x 0.05) + (13 x 0.05) = 12.5 + .65 = 13.15 = 14 n3 = no p3 +n1 p2 + n2 = 250 x 0.10 + 13 x 0.05 + 14 x 0.05 = 26.35 = 27 = no /34 +n1 p3 + n2 p2 + n3 = 250 x 0.15 + 13 x 0.10 + 14 x 0.05 + 27 x 0 05 = 37.5 + 1.3 + 0.7 + 1.35 = 40.85 = 41 n5 = no p5 + + n2 p3 + n3 /22 + = 250 x 0.25 + 13 x 0.15 + 14 x 0.10 + 27 x 0.05 = 62.5 + 1.95 + 1.4 + 1.35 = 67.2 = 68 n6 =no p6 +ni p5 +n2 p4 +n3 p3 +n4 p2 +n5 p1 = 250 x 0.40 + 13 x0.25 + 14 x 0.15 + 27x 0.10 + 41 x 0.05 + 68 x 0.05 = 100 + 3.25 + 2.1 + 2.7 + 2.05 + 3.4 = 113.5 = 114 Total Average Cost can be found out with the help of following table. Cumulative failure
Individual replacement
Group replacement
Total cost
Average Total Cost
1
No. of moving parts failed 13
13
9100
350 x 250 = Rs 87500
96600
96600
2 3 4 5 6
14 27 41 68 114
27 54 95 16.3 227
18900 37800 66500 114100 193900
106400 125300 154000 201600 281400
53200 41767 38500 40320 46900
Month
OPERATIONS RESEARCH
368
It can be seen that the average total cost in third month, i.e., Rs. 38500 is the minimum, the optimum group replacement period is 3 months. Also the individual replacement cost of Rs. 42000 is more than the minimum group replacement cost of Rs. 38500, group replacement is a better replacement policy. Example 8.17. A machine system contains 4000 ICs and the present policy is to replace an IC as and when it fails. The average cost of replacing one IC is Rs. 100. If all the ICs are replaced under a preventive maintenance policy, the average cost of IC comes down to Rs. 50. Existing number of ICs at the end of the year and the probability of failing during the year is shown below. Year Present Functional ICs Probability of failure During the year
0 1000
1 800 0.04
2 700 0.06
3 500 0.25
4 300 0.30
5 100 0.15
6 0 0.20
Compute the associated costs, if individual or group replacement policy is followed. Which policy should be adopted and why ?
Solution. Let us first find out the average failure of ICs Average failure = Number of ICs / Average failure of ICs =4000 / (1 x 0.04 + 2 x 0.06 + 3 x 0.25 + 4 x 0.30 + 5 x 0.15 + 6 x 0.20) = 4000/ (0.04 + 0.12 + 0.75 + 1.20 + 0.75 + 1.20) = 4000/4.06 = 986 ICs. Cost if individual replacement policy is adopted = Rs 986 x 100 = Rs 98600. Now, we must find out the failure of ICs per month ni = no pi = 4000 x 0.04 = 160 112 = no Pz ni = 4000 x 0.06 + 160 x 0.04 = 240 + 6.4 247 173 = no p3 + n1 p2 + n2 pi = 4000 x 0.25 + 160 x 0.06 + 247 x 0.04 = 1000 + 9.6 + 9.88  1020 11 4 = nO P4 + n1P3 nz P2 ± n3 Pi = 4000 x 0.30 + 160 x 0.25 + 247 x 0.06 + 1020 x 0.04 = 1200 + 40 + 14.82 + 40.8 = 1296 n5 = no es n1 p4 + n2 p3 + n3 p2 + 114 PI = 4000 x 0.15 + 160 x 0.30 + 247 x 0.25 + 1020 x 0.06 + 1296 x 0.04 = 600 + 48 + 61.75 + 61.2 + 51.84 823 116 = no P6 + n1 Ps + /12 P4 + n3133+ n4P2 ns = 4000 x 0.20 + 160 x 0.15 + 247 x 0.30 + 1020 x 0.25 + 1296 x 0.06 + 823 x 0.04 = 800 + 24 + 74.1 + 255 + 77.76 + 32.92 1264 Now average cost of group replacement can be worked out with the help of the following table. Month
Failure of ICs during month
Cumulative failure
1
160
160
Individual replacement cost @ 100 16000
2 3 4 5 6
247 1020 1296 823 1264
407 1427 2723 3546 4810
40700 142700 272300 354600 481000
Group replacement cost @ 50 4000 X 50 = 200000
Total cost 216000
Average Total Cost 216000
240700 342700 472300 554600 68100
120350 114233 118072 110920 113517
369
REPLACEMENT THEORY
The group replacement cost is minimum at the end of five months, i.e., Rs. 110920. The individual replacement cost is Rs. 98600. Hence, it is better to adopt individual replacement policy. Example 8.18. The computer system has a large number of transistors. These are subject to a mortality rate given below : Period 1 2 3 4
Age of failure in hours 0 — 400 400 — 800 800 —1200 1200 —1600
Probability of failure 0.05 0.15 0.35 0.45
If the transistors are replaced individually the cost per transistor is Rs. 20. But if it can be done as a group at a specific interval determined by the preventive maintenance policy of the user, then the cost per transistor comes down to Rs. 10. Should the transistor be replaced individually or as a group ? Solution. Let us assume that a block of 400 hours is the one period and total number of transistors in the system are 1600. Find out the average failure of transistors Average failure = Number of transistors /Average mean life = 1600 / (0.05 x 1 + 0.15 x 2 + 0.35 x 3 + 4 x 0.45) = 1600 / (0.05 + 0.30 + 1.05 + 1.80) = 1600/3.2 = 500 If cost of individual replacement policy is adopted, Cost = Rs. 500 x 20 = Rs. 10000, now we must find out the failure of transistors per period of block of 400 hours. 111 = no pi =1600 x 0.05 = 80 122 = no p2 + 111 pi = 1600 x 0.15 + 80 x 0.05 = 240 + 4 = 244 n3 = n3 = no P3+ P2+ n2 Pi = 1600 x 0.35 + 80 x 0.15 + 244 x 0.05 = 560 + 12 + 12.2 = 585 114 = no P4 + 111 P3 4. 112 P2 + n3 P = 1600 x 0.45 + 80 x 0.35 + 244 x 0.15 + 585 x 0.05 = 720 + 28 + 36.6 + 29.25 = 814 Now, average cost of group replacement must be found. Period 400 hours block 1 2 3 4
Failure of ICs during month 80 244 585 814
Cumulative failure 80
Individual replacement cost @ 20 1600
324 909 1723
6480 18180 34460
Group replacement cost @ 10 1600 x 10 = 16000
Total cost 17600
Average Total Cost 17600
222480 34180 50460
11240 11393 12615
OPERATIONS RESE/ARCH
370
The minimum cost of group replacement is Rs. 11240 for an interval of 400 hours. Individual replacement is optimal policy since the cost is Rs. 10000, which is less than the group replacement cost. Manpower replacement policy (Staffing policy)
All organizations face the problem of initial recruitment and filling up of vacancies caused by promotions, transfer, employee quitting their jobs or retirement and deaths. The principle of replacement used in industry for replacement of parts, etc, can also be used for recruitment and promotion policies, which are laid down as personnel policy of an organization. The assumption made in such case is that the destination of manpower is already decided. Few examples will illustrate this point. Example 8.19. An army unit requires 200 men, 20 Junior Commissioned Officers (JCOs) and 10 officers. Men are recruited at the age of 18 and JCOs and officers are selected out of these. If they continue in service, they retire at the age of 40. At present there are 800 jawans and every year 20 of them retire. How many jawans should be recruited every year and at what age promotions should take place ? Solution. If 800 jawans had been recruited for the past 22 years (age of recruitment 40 years — age of entry 18 years), the total number of them serving up to age of 39 years = 20 x 22 = 440 Total number of jawans required = 200 + 20 + 10 = 230 Total number of jawans to be recruited every year in order to maintain a strength of 230 = 800/440 x 230 = 418 Let a jawan be promoted at age of X, then up to X —1 year, number of jawans recruited is 200 out of 230. Hence out of 800, jawans required = 200/230 x 800 = 696. 696 will be available up to 5 years as 20 retire every year and (800 — 20 x 5) = 700. Hence promotion of jawans is due in sixth year. Out of 230 jawans required, 20 are JCOs, therefore if recruited 800, number of JCOs = 20/230 x 800 = 70 approximately. In a recruitment of 800, total number of men and JCOs = 697 + 70 = 766 Number of officers required = 800 — 766 = 34 This number will be available in 20 years of service, so promotion of JCOs to officers is due in 21 year of service. Example 8.20. College X plans to raise the strength of its faculty to 150 and then keep it at that level. The wastage of faculty due to retirement, quitting, deaths, etc, based on the length of service of the faculty member is as given below. Block years % of teachers Those who live up to end of year
1 0 0
2 5 5
3 10 35
4 15 60
5 20 65
6 30 70
7 35 85
8
9
10
100
(i) Find the number of faculty members to be recruited every year. (ii) If there are 10 posts of Head of Departments (HODs) for which length of service is the only criterion of promotion, what is the average length of service after which a new faculty member should expect promotion ?
371
REPLApEMENT THEORY
Solution. Let us assume that the recruitment per year is 100. These 100 teachers join initially in the block of 0 — 5 years, will serve for 35 years and will become 0 in 7th block of 5 years, i.e., at the service of 35 years. Those 100 who join between the block of year 5 — 10 will serve for 30 years and become 15, the third set of teachers will become 30 after 25 years of service and soon. Year 0 5 10 15 20 25 30 35
No. of faculty members 100 95 65 40 35 30 15 0
Hence, if 100 faculty members are required every year, the total number of staff members after 35 years (7 block of 5 years) of service = 380 To maintain staff strength of 150, the number to be recruited every year = 100 /380 x 150 = 40 If the college recruits 40 every year, then they want 10 as HODs. Hence if the college recruits 100 every year then they will need HODs = 10/40 x 100 = 25. It can be seen from the above table that 0 + 15 + 30 25, i.e., the promotion of the newly recruited faculty member as HODs can be done after 20 years of service. IN REVIEW AND DISCUSSION QUESTIONS 1. 2. 3. 4. 5.
ri
What is replacement problem ? When does it arise ? Describe various types of replacement situations. Enumerate various replacement problems. What are the situations which make the replacement of items necessary ? Give a brief account of situations of which the replacement problems arise. What does the theory of replacement establish ? 6. Discuss in brief replacement procedure for the items that deteriorate with time. 7. The cost of maintenance of a machine is given as a function increasing with time and its scrap value is constant. Show that the average annual cost will be minimized by replacing the machine when the average cost to date becomes equal to the current maintenance cost. 8. Discuss the replacement problem where items are such that maintenance costs increase with time and the value of money also changes with time. 9. Find the optimum replacement policy which minimizes the total of all future discounted costs for an equipment which costs Rs. A and which needs maintenance costs of Rs. C1 , C2, ...,C„ etc. (C„ +1 > C„ ) during the first year, second year etc., and further D is the depreciation value per unit of money during a year. 10. State some of the simple replacement policies.
OPERATIONS RESEARCH
372
11. Construct the cost equation reflecting the discounted value of all future costs for a policy of replacing equipment after every n periods. Hence establish the following : (i) Replace if the next period cost is greater than the weighted average of previous costs. (it) Do not replace if the next period's cost is less than the weighted average of previous costs. 12. What is "group replacement"? Give an example. 13. Write a short note on group replacement and individual replacement policies. 14. The cost per item for the individual replacement is C1 and the cost per item of group replacement is C2 . If only individual replacement is more economical than the group replacement along with the individual, find relation between C1 and C2 . 15. Consider a group replacement model. There are N items in the group. Replacement is made after every t periods. Assume that all the failures in a group are replaced at the end of the period. Further C1 is the cost of replacing a unit in the group C2 is the cost of replacing a failure (C2 > C1 ). Find an expression for the least cost associated with group replacement. 16. A large population is subject to a given mortality curve for a long period of time. All deaths are immediately replaced by births and there are no other entries or exists. Show that the age distribution becomes stable and that the number of deaths per unit time becomes constant. 17. In a machine shop, a particular cutting tool costs Rs. 6 to replace. If a tool breaks on the job, the production disruption and associate costs amount to Rs. 30. The past life of a tool is given as follows: Job No.: Proportion of broken tools on Job:
1
2
3
4
5
6
7
0.01
0.03
0.09
0.13
0.25
0.55
0.95
After how many jobs should the shop replace a tool before it breaks down ? 18. It has been suggested by a data processing firm that they adopt a policy of periodically replacing all the tubes in a certain piece of equipment. A given type of tube is known to have the mortality distribution shown in the table : Tube failures/week : Probability of broken tools on Job :
1
2
3
4
5
0.01
0.03
0.09
043
0.25
There are approximately 1000 tubes of this type in all the combined equipment. The cost of replacing the tubes on an individual basis is estimated to be Re. 1.00 per tube and the cost of a group replacement policy average Rs. 0.30 per tube. Compare the cost of preventive replacement with that of remedial replacement. 19. A truck has been purchased at a cost of Rs. 160000. The value of the truck is depreciated in the first three years by Rs. 20000 each year and Rs. 16000 per year thereafter. Its maintenance and operating costs for the first three years are Rs. 16000, Rs. 18000 and Rs. 20000 in that order and increase by Rs. 4000 every year. assuming an interest rate of 10% find the economic life of the truck. 20. A manual stamper currently valued at Rs. 10000 is expected to last 2 years and costs Rs. 4000 per year to operate. An automatic stamper which can be purchased for Rs. 3000 will last 4 years and can be operated at an annual cost of Rs. 3000. If money carries the rate of interest 10% per annum, determine which stamper should be purchased.,
REPACEMENT THEORY
373
21. The cost of a new machine is Rs. 5000. The maintenance cost of nth year is given by R„ = 500 (n — 1); n =1,2, ....... Suppose that the discount rate per year is 0.05. After how many years will it be economical to replace the machine by new one ? 22. A machine costs Rs. 10000 operating costs are Rs. 500 per year for the first five years. Operating costs increase by Rs. 100 per year in the sixth and succeeding years. Assuming a 10 per cent discount rate of money per year, find the optimum length of time to hold the machine before it is replaced. State clearly the assumptions made. 23. A pipeline is due for repairs. It will costs Rs. 10000 and last for 3 years. Alternatively, a new pipeline can be laid at a cost of Rs. 30000 and lasts for 10 years. Assuming cost of capital to be 10% and ignoring salvage value, which alternative should be chosen ? 24. The cost pattern for two machines M1 and M2 when money value is not considered is given below : Cost at the beginning of the year (in Rs.) Year M1 M2 1 900 1400 2 600 100 3 700 700 Find the cost pattern for each machine when money is worth 10% per year and hence find the machine which is less costly. 25. An engineering company is offered two types of material handling equipment A and B. A is priced at Rs. 60000 including cost of installation, and the costs for operation and maintenance are estimated to be Rs. 10000 for each of the first five years, increasing by Rs. 3000 per year in the sixth and subsequent year. quipment B with a rated capacity same as A, requires an initial investment of Rs. 30000 but in terms of operation and maintenance costs more than A. These costs for B are estimated to be Rs. 13000 per year for the first six years, increasing by Rs. 4000 per year for each year from the seventh year onwards. The company expects a return of 10 per cent on all its investments. Neglecting the scrap value of the equipment at the end of its economic life, determine which equipment the company should buy. 26. An individual is planning to putrchase a car. A new car will cost Rs. 120000. The reasale value of the car at the end of the year is 85% of the previous year value. Maintenance and operation costs during the first year are Rs. 20000 and they increase by 15% every year. The minimum resale value of the car can be Rs. 40000. (i) When should the car be replaced to minimum average annual cost (ignore interest) ? (ii) If interest of 12% is assumed, when should the car be replaced ? 27. A large computer installation contains 2000 components of identical nature which are subject to failure as per probability distribution given below : Weekend 1 2 3 4 5 Percentage failure to date : 10 25 50 80 100 Components which fail have to be replaced for efficient functioning of the system. If they are replaced as an when failure occur, the cost of replacement per unit is Rs. 3. Alternatively, if all components are replaced in one lot at periodical intervals and individually replaced only as such failures occur between group replacement, the cost of component replaced is Re. 1.
OPERATIONS RESEARCH
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(a) Access which policy of replacement would be economical. (b) If group replacement is economical at current costs, then assess at what cost of individual replacement would group replacement be uneconomical. (c) How high can the cost per unit in group replacement be to make a preference for individual replacement policy ? 28. Let p (t) be the probability that a machine in a group of 30 machines would breakdown in period t. The cost of repairing a broken machine is Rs. 200. Preventive maintenance is performed by servicing all the 30 machines at the end of T unit of time. Preventive maintenance cost is Rs. 15 per machine. Find optimum T which will minimize the expected total cost per period of servicing, given that 0.03
for t=1 for t= 2, 3, ... , 10 for t =11, 12, 13, ...
p (t —1) + 0.01
p(t) =
0.13
29. An electric company which generates and distributes electricity conducted a study on the life of poles. The appropriate life data are given in the following table : Years after installation Percentage poles failing
1 1
2 2
3 3
4 5
6 12
5 7
7 20
8 30
9 16
10 4
(i) If the company now instals 5000 poles and follows a policy of replacing poles only when they fail, how many poles are expected to be replaced each year during the next ten years? To simplify the computation assume that failures occur and replacements are made only at the end of a year. (ii) If the cost of replacing individually is Rs. 160 per pole and if we have a common group replacement policy, it costs Rs. 80 per pole, find out the optimal period for group replacement. 30. A computer contains 10000 resistors. When any resistor fails, it is replaced. The cost of replacing a resistor individually is Re. 1. If all the resistors are replaced at the same time, the cost per resistor would be reduced to 35 paise. The percentage surviving say S (t) at the end of month t and P (t) the probability of failure during the month t, are t
S (t) P (t)
0
1
2
3
4
5
6
100
97 0.03
90
70
30
0.07
0.20
0.40
15 0.15
0 0.15
—
What is the optimum replacement plan ? 31. Given that cost of failure replacement is 3 times compared to the cost of common preventive replacement of Rs. 100 per item in a system of 1000 items and the following data, suggest the strategy between failure replacement and common preventive replacement that should be followed for cost reduction. Week % of item surviving at the end of the week
0
1
2
3
4
5
100
90
75
40
15
6
REPLACEMEr THEORY
375
32. A manufacturer wants to replace a machine. The purchase price of the machine is Rs. 10000. Following other details are available: Year 1 2 3 4 5 6 7 8
Maintenance (Rs.) 1200 1300 1500 2000 2200 3000 3200 3800
Resale Price (Rs.) 6000 3000 2000 1000 800 500 400 300
Suggest in which year the machine may be replaced, if the supplier of the machine is prepared to prove 3 years insitu maintenance free of cost. G.N.D.U. EXAMINATION PROBLEMS II
1995 APR. 1995 APR. 1996 APR. 1996 SEP. 1997 APR. 1996 APR. 1997 SEP. 1998 SEP. 1995 APR. 1995 SEP. 2003 APR.
1997 APR.
(1) Preventive replacement (2) Present value of investment (3) Group replacement (4) Group replacement (5) Individual replacement (6) Present value of investment (7) What is replacement ? Write a note on preventive replacement. (8) Describe briefly the model to explain the importance of Replacement Policy in an organisation. (9) Describe briefly the replacement model to explain the importance of Replacement Policy in an organisation. (10) Write note on group replacement policy. (11) What is replacement ? Write a note on preventive replacement ? (12) Explain how the theory of replacement is used in the following problems : (i)Replacement of items that fail completely. (ii)Replacement of items whose maintenance cost varies with time. (13) A certain transport company operating in Rajasthan has a fleet of 20 trucks. A truck costs Rs. 175000. From last experience, the transporting company has following information : Age of Truck in years Operating cost (Rs.) Salvage value (Rs.)
1 15600 110000
2 3 20000 27000 95000 88000
What is the optimum replacement period ?
4 45000 70000
5 85000 60000
6 98000 45000
OPERATIONS RESEARCH
376
2001 APR. (14) The original cost of a machine is Rs. 45000. Operating cost varies as follows : Year OpCost
1 3400
2 4500
3 7500
4 10000
5 12500
6 18000
7 21000
If 10% is the discount rate of money what should be the optimum replacement (Marks 16) interval ? 2001 SEP. (15) Determine the optimal replacement period for the following. The maintenance cost and resale value of an equipment whose purchase price is 300000 is as follows : 6 7 8 Year 1 2 3 4 5 Maintenance 0.36 0.48 0.60 0.72 0.84 0.96 1.08 1.20 Cost Rs. in lacs, 2.00 1.50 1.00 0.80 0.70 0.60 0.50 0.40 Resale price Rs. in lacs 2002 APR. (15) Cost price of a machine is Rs. 50000 the maintenance cost and scrap values are given in the following table. Calculate the economic life and minimum average cost : Year Maintenance cost Scrap value
1 2 3 4 5 6 7 2500 2800 3500 5000 6000 7500 10000 30000 25000 22000 18000 12000 10000 10000
2002 APR. (16) Determine the Optimum Replacement Period for a machine costing Rs. 6000. Maintenance cost is Rs. 10000 in each of the first 4 years and increases by Rs. 200 every year thereafter. Salvage is NIL. The time value of money is 10 %. 2002 SEP. (17) Obtain the economic life of the machine and minimum average cost from the following data. Purchase price is Rs. 20000/1 2 3 4 5 Year 6 Maintenance cost 1500 1700 2000 2500 3500 5500 Resale price 17000 15300 14000 12000 8000 3000 (18) Original cost of equipment is Rs. 120000. The year wise maintenance cost is given as under. Find the optimum replacement period if discount rate of money is 10 %, scrap value is Rs. 20000. 1 Year 2 3 4 5 6 7 8 Maintenance 10000 13000 15000 20000 25000 32000 40000 50000 cost 2004 APR. (19) The probability (P„) of failure just before age (n) are shown below. In individual replacement costs Rs. 1.25 and group replacement costs Rs. 0.50 per item, find the optimal group replacement policy (assuming that there are 1000 bulbs in use). 2 3 4 5 6 7 8 9 10 11 .01 .03 .05 .07 .10 .15 .20 .15 .11 .08 .06
(n) 1
Ating Line (Queuing) Theory LEARNING' OBJECTIVES • • • • • •
Appreciate situations in which queuing problems are generated. Learning to set objectives for the operation of queuing. Understand standard queuing language and symbols. Understanding operating characteristics of queuing. Solving queuing problems in singlechannel and multiplechannel situations. Understand the advantages and limitations of queuing systems.
OBJECTIVE AND MODELS OF THE THEORY Queues of customers arriving for service of one kind or another arise in many different fields of activity. Businesses of all types, government, industry, telephone exchanges, and airports, large and small, all have queuing problems. Many of these congestion situations could benefit from OR analysis, which employs to this purpose a variety of queuing models, referred to as queuing systems or simply queues. A queuing system involves a number of servers (or serving facilities) which we will also call service channels (in deference to the source field of the theory telephone communication system). The serving channels can be communications links, workstations, check out counters, retailers, elevators, buses, to mention but a few. According to the number of servers, queuing systems can be of single and multichannel type. Customers arriving at a queuing system at random intervals of time are serviced generally for random times too. When a service is completed, the customer leaves the servicing facility rendering it empty and ready and gets next arrival. The random nature of arrival and service times may be a cause of congestion at the input to the system at some periods when the incoming customers either queue up for service or leave the system unserviced; in other periods the system might not be completely busy because of the lack of customers, or even be idle altogether. A queuing system operation is a random process with discrete states and continuous time. The state changes jumpwise at the instant some events occurs : an arrival occurs, a service is completed, or a customer unable to wait any longer leaves the queue.
378
OPERATIONS RESEARCH
The subject matter of queuing theory is to build mathematical models, which relate the specified operating conditions for the system (number of channels, their servicing mechanism, distribution of arrivals) to the concerned characteristics of valuemeasures of effectiveness describing the ability of the system to handle the incoming demands. Depending on the circumstances and the objective of the study, such measures may be : the expected (mean) number of arrivals served per unit time, the expected number of busy channels, the expected number of customers in the queue and the mean waiting time for service, the probability that the number in queue is above some limit, and so on. We do not single out purposely among intended for the given operating conditions, those intended for decision variables, since they may be either of these characteristics, for example, the number of channels, their capacity, service mechanism, etc. The most important part of a study is model establishment (primal problems) while its optimisation (inverse problem) is indeed depending on which parameters are selected to work with or to change. We are not going to consider optimization of queuing models in this text with the exception made only for the simplest queuing situations. The mathematical analysis of a queuing system simplifies considerably when the process concerned is Markovian. Markov process may be defined as , "A random process is referred to as Markov, if for any moment of time, its probability characteristics in the future, depend only on its state at time and are independent of when and how this state was acquired." As we already know a sufficient condition for this is that all the process changing system's states (arrival intervals, service intervals) are Poisson. If this property does not hold, the mathematical description of the process complicates substantially and acquires an explicit analytical form only in seldom cases. However, the simplest mathematics of Markov queues may prove of value for approximate handling even of those queuing problems whose arrivals are distributed not in a Poisson process. In many situations a reasonable decision on queuing system organization suffices with an approximate model. All of the queuing systems have certain common basic characteristics. They are: (a) input process (arrival pattern) which may be specified by the source of arrivals, type of arrivals and the interarrival times, (b) service mechanism which is the duration and mode of service and may be characterized by the servicetime distribution, capacity of the system, and service availability, and (c) queue discipline which includes all other factors regarding the rules of conduct of the queue. We start illustrating the classification breakdown with a loss and delay system. In a purely loss system, customers arriving when all the servers are busy are denied service and are lost to the system. Examples of the loss system may be met in telephony : an incoming call arrived at an instant when all the channels are busy cannot be placed and leaves the exchange unserved. In a delay system an arrival incoming when all the channels are busy does not leave the system but joins the queue and waits (if there is enough waiting room) until a server is free. These latter situations more often occur in applications and are of great importance, which can be readily inferred from the name of the theory. According to the type of the sourcesupplying customers to the system, the models are divided into those of a finite population size, when the customers are only few, and the infinitepopulation
WAITING LINE (QUEUING) THEORY
379
systems. The length of the queue is subject to further limitations imposed by allowable waiting time or handling of impatient customers which are liable to be lost to the system. The queue discipline, that is the rule followed by the server in taking the customers in service, may be according to such selfexplanatory principles as "firstcome, firstserved", "lastcome, firstserved", or chain "random selection for serve". In some situations priority disciplines need be introduced to allow for realistic queues with high priority arrivals. To illustrate, in extreme cases, the server may stop the service of a customer of lower priority in order to deal with a customer of high priority; this is called preemptive priority. For example, a gantry crane working on a container ship may stop the unloading halfway and shift to another load to unload perishable goods of a later arrived ship. The situation when a service of a low priority customer started prior to the arrival of a high priority customer is completed and the high priority customer receives only a better position in the queue is called nonpreemptive priority. This situation can be exemplified by an aeroplane which enters a queue of a few other aircraft circling around an airport and asks a permission for an emergency landing; the ground control issues the permission on the condition that it lands next to the aeroplane being on a runway at the moment. Turning over to the service mechanism, we may find systems whose servicing channels are placed in parallel or in series. When in series, a customer leaving a previous server enters a queue for the next channel in the sequence. For example, a work piece being through the operation with one robot on a conveyor is stacked to wait when the next robot in the process is free to handle it. These operation stages of a serieschannel queuing system are called phases. The arrival pattern may and may not correlate with the other aspects of the system. Accordingly, the system can be loosely divided into "Open" and "Close." In an open system, the distribution of arrivals does not depend on the status of the system, say on how many channel are busy. To contrast, in a close system, in does. For example, if a single operator tends a few similar machines each of which has a chance of stopping, i.e., arriving for serve, at random, then the arrival rate of stopping depends on how many mechanics have been already adjusted and put on the yet served. An optimisation of a queuing system may be attempted from either of two standpoints, the first in favour of owners of the queue, the second to favour the "queuers", i.e., the customers. The first stand makes a point of the efficiency of the system and would tend to load all the channels as high as possible, i.e., to cut down idle times. The customers on the contrary would like to cut down waiting time in a queue. Therefore, any optimisation of congestion necessitates a "system approach" with the intrinsic complex evaluation and assessment of all consequences for each possible decision. The need for optimality over conflicting requirements may be illustrated with the viewpoint of the customer wishing to increase the number of channels, which, however, would increase the total servicing cost. The development of a reasonable model may help solving the optimisation problem by choosing the number of channels which account for all pros and cons. This is the reason why we do not suggest a single measure of effectiveness for all queuing problems, formulating them instead as multiple objective problems. BENEFITS AND LIMITATIONS OF QUEUING THEORY Queuing theory has been used for many real life applications to a great advantage. It is so because many problems of business and industry can be assumed/simulated to be arrivaldeparture or
380
OPERATIONS RESEARCH
queuing problems. In any practical life situations, it is not possible to accurately determine the arrival and departure of customers when the number and types of facilities as also the requirements of the customers are not known. Queuing theory techniques, in particular, can help us to determine suitable number and type of service facilities to be provided to different types of customers. Queuing theory techniques can be applied to problems such as: (a) Planning, scheduling and sequencing of parts and components to assembly lines in a mass production system. (b) Scheduling of workstations and machines performing different operations in mass production. (c) Scheduling and dispatch of war material of special nature based on operational needs. (d) Scheduling of service facilities in a repair and maintenance workshop. (e) Scheduling of overhaul of used engines and other assemblies of aircrafts, missile systems, transport fleet, etc. (f) Scheduling of limited transport fleet to a large number of users. (g) Scheduling of landing and takeoff from airports with heavy duty of air traffic and limited facilities. (h) Decision of replacement of plant, machinery, special maintenance tools and other equipment based on different criteria. Special benefit which this technique enjoys in solving problems such as above are: (i) Queuing theory attempts to solve problems based on a scientific understanding of the problems and solving them in optimal manner so that facilities are fully utilised and waiting time is reduced to minimum possible. (ii) Waiting time (or queuing) theory models can recommend arrival of customers to be serviced, setting up of workstations, requirement of manpower, etc., based on probability theory. Limitation of Queuing theory
Though queuing theory provides us a scientific method of understanding the queues and solving such problems, the theory has certain limitations which must be understood while using the technique, some of these are : (a) Mathematical distributions, which we assume while solving queuing theory problems, are only a close approximation of the behaviour of customers, time between their arrival and service time required by each customer. (b) Most of the real life queuing problems are complex situation and are very difficult to use the queuing theory technique, even then uncertainty will remain. (c) Many situations in industry and service are multichannel queuing problems. When a customer has been attended to and the service provided, it may still have to get some other service from another service point and may have to fall in queue once again. Here the departure of one channel queue becomes the arrival of the other channel queue. In such situations, the problem becomes still more difficult to analyse. (d) Queuing model may not be the ideal method to solve certain very difficult and complex problems and one may have to resort to other techniques like MonteCarlo simulation method.
WAITING LINE (QUEUING) THEORY
381
IMPORTANT TERMS USED IN QUEUING THEORY Following are some important terms used in queuing theory: 1. Arrival Pattern: It is the pattern of the arrival of a customer to be serviced. The pattern may be regular or at random. Regular interval arrival patterns are rare, in most of the cases, arrival of the customers cannot be predicted. Regular pattern of arrival of customers follows Poisson's distribution. 2. Poisson's Distribution: It is discrete probability distribution which is used to determine the number of customers in a particular time. It involves allotting probability of occurrence of the arrival of a customer. Greek letter X, (lamda) is used to denote mean arrival rate. A special feature of the Poisson's distribution is that its mean is equal to the variance. It can be represented with the notation as explained below. P(n) = Probability of n arrivals (customers) = Mean arrival rate e = Costant = 2.71828 P (n) —
ex (X)" L
where n = 0, 1, 2,
Notation L or ! is called the factorial and it means that ILt or n! = n(n1) (n2) (n3)
2,1
Poisson's distribution tables for different values of n is available and can be used for solving problems where Poisson's distribution is used. However, It has certain limitations because of which its use is restricted. It assumes that arrivals are random and independent of all other variables or parameters. Such can never be the case. 3. Exponential Distribution: This is based on the probability of completion of a service and is the most commonly used distribution in queuing theory. In queuing theory, our effort is to minimize the total cost of queue and it includes cost of waiting and cost of providing service. A queue model is prepared by taking different variables into consideration. In this distribution system, no maximization or minimization is attempted. Queue models with different alternatives are considered and the most suitable for a particular is attempted. Queue models with different alternatives are considered and the most suitable for a particular situation is selected. 4. Service Pattern: We have seen that arrival pattern is random and poissons distribution can be used for use in queue model. Service pattern are assumed to be exponential for purpose of avoiding complex mathematical problem. 5. Channels: A service system has a number of facilities positioned in a suitable manner. These could be (a) Single Channel Single Phase System. This is very simple system where all the customers wait in a single line in front of a single service facility and depart after service is provided. In a shop if there is only one person to attend to a customer, is an example of the system. Input
Service Facility
Arriving Customers (units etc.) Queue or Waiting Line
Servr"Cr
Fig. 9.1
Served Customer Output
OPERATIONS RESEARCH
.382
(b) Service in series: Here the input gets serviced at one service station and then moves to
second and or third and so on before going out. This is the case when a raw material input has to undergo a number of operations like cutting, turning drilling etc. Queue
1===>
1
Served Customer
Facility 2
Facility Queue
Fig. 9.2 (c) Multiparallel facility with a single queue: Here the service can be provided at a number of
points to one queue. This happens when in a grocery store, there are 3 persons servicing the same queue or a service station having more than one facility of washing cars. This is shown in Figure 9.3 Facility 1
Facility 2 Queue
c> E
Served Customers Departs
Facility 3
Fig. 9.3 (d) Multiple parallel facilities with multiple queue: Here there are a number of queues and sepa
rate facility to service each queue. Booking of tickets at railway stations, bus stands, etc., is a good example of this. This is shown in Figure 9.4. 1
2
3
Facility 1
Facility 2
Departure of Served Customers
Facility 3
Fig. 9.4 6. Service Time: Service time, i.e., the time taken by the customer when the facility is dedicated to it for serving depends upon the requirement of the customer and what needs to be done as assessed by the facility provider. The arrival pattern is random so also is the service time required by different customers. For the sake of simplicity the time required by all the customers is considered constant under the distribution. If the assumption of exponential distribution is not valid, Erlang Distribution is applied to the queuing model. 7. Erlang Distribution: It has been assumed in the queuing process we have seen that service is either constant or it follows negative exponential distribution in which case the standard deviation a (sigma) is equal to its mean. This assumption makes the use of the exponential distribution simple. However, in cases where a and mean are not equal, Erlang distribution developed by AK Erlang is used. In this method, the service time is divided into number of
.
WAITING LINE (QUEUING) THEORY
.383„
phases assuming that total service can be provided by different phases of service. It is assumed that service time of each phase follows the exponential distribution, i.e., r = mean. 8. Traffic Intensity or Utilisation Rate: This is the rate of at which the service facility is utilised by the components. If A. = mean arrival rate and (Mue) 1.t = Mean service rate, then utilisation rate (p) = X / Ix it can be easily seen from the equation that p > 1 when arrival rate is more than the service rate and new arrivals will keep increasing the queue. p < 1 means that service rate is more than the arrival rate and the waiting time will keep reducing as µ keeps increasing. This is true from the commonsense. 9. Idle Rate. This is the rate at which the service facility remains unutilised and is lying idle. Idle rate = 1 — utilisation rate =1—p = (1—
x total service facility = 1— A= x . IL 11 11 10. Expected number of customers in the system. This is the number of customers in queue plus the number of customers being serviced and is denoted by E„ =
. (µ — X) 11. Expected number of customers in queue (Average queue length). This is the number of expected customers minus the number being serviced and is denoted by Eq. X, X A.2 (11 — X) 11 11 (11 —20 12. Expected time spent by customer in system. It is the time that a customer spends waiting in queue plus the time it takes for servicing the customer and is denoted by Et Eq = E„ — p =
En
—
1 . — 13. Expected waiting time in queue. It is known that Et = expected waiting time in queue 1 + expected service time, therefore expected waiting time in queue (En,) = Et — Et =
=
=,
1 X = 1 = (11, — X) (II — X) P. k(11— X) 15. Probability that customer wait is zero. It means that the customer is attended to for servicing at the point of arrival and the customer does not wait at all. This depends upon the utilization A, rate of the service or idle rate of the system, Po = 0 persons waiting in the queue = 1— — and the 11, probability of 1, 2, 3 .. , n persons waiting in the queue will be given by 2 Pi= Po(11) P2 = 1/1(" ) /Pn = Po( ) • 1t I1 16. Queuing Discipline. All the customers get into a queue on arrival and are then serviced. The order in which the customer is selected for servicing is known as queuing discipline. A number of systems are used to select the customer to be served. Some of these are : (a) First in First Served (F1FS): This is the most commonly used method and the customers are served in the order of their arrival. 14. Average length of nonempty queue. Et =
II
OPERATIONS RESEARCH
384
(b) Last in First Served (LIFS): This is rarely used as it will create controversies and ego problems amongst the customers. Any one who comes first expects to be served first. It is used in store management, where it is convenient to issue the store last received and is called Last In First Out (LIFO). (c) Service in Priority (SIP): The priority in servicing is allotted based on the special requirement of a customer like a doctor may attend to a serious patient out of turn, so may be the case with a vital machine which has broken down. In such cases the customer being serviced may be put on hold and the priority customer attended to or the priority may be on hold and the priority customer waits till the servicing of the customer already being serviced is over. 17. Customer Behaviour: Different types of customers behave in different manner while they are waiting in queue, some of the patterns of behaviour are : (a) Collusion: Some customers who do not want to wait they make one customer as their representative and he represents a group of customers. Now only the representative waits in queue and not all members of the group. (b) Balking: When a customer does not wait to join the queue at the correct place which he warrants because of his arrival. They want to jump the queue and move ahead of others to reduce their waiting time in the queue. This behaviour is called balking. (c) Jockeying: This is the process of a customer leaving the queue which he had joined and goes and joins another queue to get advantage of being served earlier because the new queue has lesser customers ahead of him. (d) Reneging: Some customers either do not have time to wait in queue for a long time or they do not have the patience to wait, they leave the queue without being served. 18. Queuing Cost Behaviour. The total cost a service provider system incurs is the sum of cost of providing the services and the cost of waiting of the customers. Suppose the garage owner wants to install another car washing facility so that the waiting time of the customer is reduced. He has to manage a suitable compromise in his best interest. If the cost of adding another facility is more than offset by reducing the customer waiting time and hence getting more customers, it is definitely worth it. The relationship between these two costs is shown below.
Total Cost of Service
Cost of Providing Service
Cost
Cost of Waiting Optivial Service Level Waiting Tithe/Level of Service
Vig. 9.5
WAITING LINE (QUEUING) THEORY
385
TYPES OF QUEUING MODELS
Different types of models are in use. The three possible types of categories are : (a) Deterministic model: Where the arrival and service rates are known. This is rarely used as it is not a practical model. (b) Probabilistic model: Here both the parameters, i.e., the arrival rate as also the service rate are unknown and are assumed random in nature. Probability distribution, i.e., Poissons, Exponential or Erlang distributions are used. (c) Mixed model: Where one of the parameters out of the two is known and the other is unknown. SINGLE CHANNEL QUEUING MODEL
(Arrival — Poisson and Service time Exponential) This is the simplest queuing model and is commonly used. It makes the following assumptions: (a) Arriving customers are served on First Come First Serve (FCFS) basis. (b) There is no Balking or Reneging. All the customers wait the queue to be served, no one jumps the queue and no one leaves it. (c) Arrival rate is constant and does not change with time. (d) New customers arrival is independent of the earlier arrivals. (e) Arrivals are not of infinite population and follow Poisson's distribution. (f) Rate of serving is known. (g) All customers have different service time requirements and are independent of each other. (h) Service time can be described by negative exponential probability distribution. •! (i) Average service rate is higher than the average arrival rate and over a period of time the queue : keeps reducing. Exmple 9.1. Assume a single channel service system of a library in a school. From past experiences it is known that on an average every hour 8 students come for issue of the books at an average rate of 10 per hour. Determine the following: (a) Probability of the assistant librarian being idle. (b) Probability that there are at least 3 students in system. (c) Expected time that a student is in queue. Solution. (a) Probability that server is idle = (
1— 2' in this example X, = 8, µ= 10 N, C II
(1_ 8 _ 1670 = 016. 10) — (b) Probability that at least 3 students are in the system 3+1 En = ( 1 —04 1.1 — 10 — * Po — 10
386
OPERATIONS RESEARCH
(c) Expected time that a students is in queue 21 /4,2 64 = 3.2 hours. µ(µ — X) (10 x 2) Example 9.2. At a garage, car owners arrive at the rate of 6 per hour and are served at the rate of 8 per hour. It is assumed that the arrival follows Poisson's distribution and the service pattern is exponentially distributed. Determine (a) Average queue length, (b) Average waiting time. Solution. Average arrival (mean arrival rate) = 6 per hour. Average (mean) service rate µ = 8 per hour. Utilisation rate (traffic intensity) p = — = II
(a) Average queue length Et =
X1 —
8
= 0.75
= 36 = 2.25 cars. [8(8 — 6)]
1 0.1 —
1 = = 30 minutes. 2 Example 9.3. Customers arrive at a sales counter managed by a single person, according to a Poisson's process with a mean rate of 20 per hour. The time required to serve a customer has an exponential distribution with mean of 100 seconds. Find the average waiting time of a customer. Solution. Mean arrival rate = 20 per hour (b) Average weighting time Et =
Average or mean service rate = 36 per hour as in 100 seconds one customer is served in 1 hour = 60 x 60 = 3600 seconds, 36 will be served Average waiting time of a customer in queue Er„ =
µ
20 5 5 hours — — x 60 —— 2.08 minutes. =— — 2%) 36 (36 — 20) 144 144 =
Average weighting time in the system Et =
1— X) = (36120) —— —16 1
hours = 16 x 60 = 3.75 minutes.
Example 9.4. Selfhelp canteen employs one cashier at its counter, 8 customers arrive every 10 minutes on an average. The cashier can serve at the rate of one customer per minute. Assume Poisson's distribution for arrival and exponential distribution for service patterns. Determine (a) Average number of customers in the system; (b) Average queue length; (c) Average time a customer spends in the system. 8 Solution. Arrival rate = — customers/minute 10
WAITING LINE (QUEUING) THEORY
387
Service rate p. = 1 customer/minute (a) Average number of customers in the system A, 0.8 = E= = 4 n 1121. 10.8
(b) Average queue length E—
X2 (0.8)2 — 3 x 2. 1.1. at— X) 1x0.2
(c) Average time a customer spends in the queue 0.8 E21 = = = 4 minutes. X) 1 x 0.2 Example 9.5. A branch library of State Library has only one clerk. The clerk is expected to perform venous duties of issuing the books which are randomly distributed and can be approximated with Poisson's distribution. He is able to issue 12 books per hour. The readers arrive at the rate of 10 per hour during the 10 hours the library is open. Determine (a) Idle rate of the clerk; (b) % time that the student has to wait; (c) Average system time; Solution. Arrival rate A, = 10 per hours Service rate p. = 12 per hour X, (a) Idle rate p = — = 0.83 p. =1 0 2 (b) % time a student has to wait = % time the clerk is busy = 83% 1 1 (c) Average system time = Et = = = 30 minutes. (p, — X) 12 10 Example 9.6. An electrician repairs geysers, presses, etc. He finds that the time he spends on repair of a geyser is exponentially distributed with mean 20 minutes. The geysers are repaired in the order in which these are received and their arrival approximates Poisson's distribution with an average rate 16 per 8 hours day. Determine (a) Electrician's idle time each day; (b) How many geysers are ahead of the geyser just brought for repairs ? Solution. Arrival rate ? = 2/hour Service rate p. = 3/hour (One geyser is repaired in 20 minutes three will be repaired in one hour). (a) Electrician idle time = 8 — utilisation time. X =8——x 8 2 8 = 8 — — x 8 = — hours = 2 hours 40 minutes. 3
OPERATIONS RESEARCH
388
(b) Number of geysers ahead of the geyser just brought in = Average number of geysers in the' system. 2 X, E„ = = =2 —X 3—2 Example 9.7. Arrival rate of telephone calls at telephone booth are according to Poisson distribution, with an average time of 12 minutes between two consecutive calls arrival. The length of telephone calls is assumed to be exponentially distributed with mean 4 minutes. (a) Determine the probability that person arriving at the booth will have to wait. (b) Find the average queue length that is formed from time to time. (c) The telephone company will install second booth when convinced that an arrival would expect to have to wait at least 5 minutes for the phone. Find the increase inflows of arrivals which will justify a second booth. (d) What is the probability that an arrival will have to wait for more than 15 minutes before the phone is free ? (e) Find the fraction of a day that the phone will be in use. Solution. Arrival rate X = 1/12 minutes Service rate p, = 1/4 minutes. 1 1 1x4=— 3 = 0 33 (a) Probability that a person will have to wait = = 1212 1 4 1 2 /4 = 1 x41 x =1 person. = (b) Average queue length = E = q µ(µ—) 1 1 1 14 2 4 (4 12) (c) Average waiting time in the queue E„, =
R(I — Xi)
=
Xi 1 µ 4 — x1)
+1) X1 1(1 16 4 ) 4l 4 1) 5 4 = — x — = 5 arrivals/minute 16 9 36 5 1 1 Increase in flow of arrivals = — — — = — minutes 36 12 18 (d) Probability of waiting time > 15 minutes. 1 1 _2 5 „, (X015 12 (1f4)15 —_ —3e =1 —3e . = v 5=
e
4 (e) Fraction of a day that phone will be in use = — = 0.33 .
is9
WAITING LINE (QUEUING) THEORY
Example 9.8. An emergency facility in a hospital where only one patient can be attended to at any one time receives 96 patients in 24 hours. Based on past experiences, the hospital knows that one such patient, on an average needs 10 minutes of attention and this time would cost Rs. 20 per patient treated. The hospital wants to reduce the queue of patients from the present number to — patients. How much will it 2 cost the hospital ? 96 Solution. using the usual notations X = — 4 patients/hour. 24 1.t = 10 x 60 = 6 patients/hour. 16 16 4 Average expected number of patients in the queue = E = = = = patients q µ(µ — X) 6(6 — 4) 12 3 1 16 This number is to reduced to — . Therefore, 1 = or 11.12 —4µi = 32 2 2 Ili (Ili —4) or 1.42  4111  32 = 0 or (.4 — 8) (1.11 + 4) = 0 or 1.ti = 8 patient per hours. 1 15 For Ri = 8 Average time required to attend to a patient = — x 60 = — minutes 2 8 15 5 Decrease in time = 10 — = — minutes. 2 2 5 Budget required for each patient = 100 + — x 20 = Rs. 150 2 4 1 Thus, to decrease the queue from — to — , the budget per patient will have to be increased from 3 2 Rs. 100 to Rs. 150. Example 9.9. Customers arrive at the executive class air ticketing at the rate of 10 per hour. There is only one airlines clerk serving the customer at the rate of 20 per hour. If the conditions of single channel queuing model apply to this problem, i.e., arrival rate and service rate probability distribution are approximated to Poisson's and Exponential respectively; determine (a) System being idle probability; (b) The probability that there are three customers at the counter; (c) The probability that there is no customer waiting to buy the ticket; (d) The probability that the customer is being served and nobody is waiting. Solution. — = 10 per hour p. = 20 per hour p„= Probability that there are n customer in the system 4 A, jn =1
) (m,
=

20)(20) = 0.5 (0.5)" for values of n 1, 2, 3 ... (a) System being idle probability or 0 customer at the counter = Po = 1 — p„ = 1 — (0.5) x (0.5)° = 1 — 0.5 = 0.5 II
OPERATIONS RESEARCH
390
(b) Probability that there are more than 3 customers at the counter 3+1 ?k.) 10 p (>3) (_ = (05)4 = 0.06 20 11 (c) Probability that there is no customer waiting = Probability that at the most 1 customer is waiting = pc, + pi = 0.5 + 0.5 x 0.5 = 0.5 + 0.25 = 0.75. (d) Probability of customers being served and nobody is waiting pi = 0.5 x 0.5 = 0.25. Example 9.10 An electricity bill receiving window in a small town has only one cashier who handles and issues receipts to the customers. He takes on an average 5 minutes per customer. It has been estimated that the persons coming for bill payment have no set pattern but on an average 8 persons come per hour. The management receives a lot of complaints regarding customers waiting for long in queue and so decided to find out. (a) What is the average length of queue ? (b) What time on an average, the cashier is idle ? (c) What is the average time for which a person has to wait to pay his bill ? (d) What is the probability that a person would have to wait for at least 10 minutes ? Solution. Making use of the usual notations X, = 8 persons/hour = 10 persons/hour (a) Average queue length =
2k,2
64 = = 3.2 persons p.(11  X) 10 (10  8)
8 (b) Probability that cashier is idle = Po = 1 — = 1 10 = 0.2, i.e., the cashier would be idle for, p. 20 % of his time. (c) Average length of time that a person is expected to wait in queue. 8 = 24 minutes Ew = p.(p.  X) 10 (10  8) (d) Probability that a customer will have to wait for at least 10 minutes. _33, t = 1 hours. X, x0100p(8) = — e 6 =8 e 10
Example 9.11. A small town has only one bus stand where the bus comes every 10 minutes. The commuters arrive in a random manner to use the bus facility. The commuters have complained that they have to wait for a longtimes in a queue to board the bus. Average rate of arrival of commuters is 4 per hours. Calculate (a) The probability that a commuter has to wait; (b) The waiting time of the commuter. Solution. X. = 4 per hour = 6 per hour
WAITING LINE (QUEUING) THEORY
391
(a) Probability that a commuter has to wait Po = 1—(1 — = = — 4 = 0.66 µ 6 There is a 66 per cent probability that a commuter has to wait. 1 (b) Expected time spent by the commuter in a queue Et =
X)
1 = 1 = 0.5 hours = 2 6—4
30 minutes Example 9.12. A bank plans to open a single server derivein banking facility at a particular centre. It is estimated that 28 customers will arrive each hour on an average. If on an average, it requires 2 minutes to process a customer transaction, determine (a) The probability of time that the system will be idle. (b) On the average how long the customer will have to wait before receiving the server. (c) The length of the drive way required for accommodating all the arrivals. On the average 20 feet of derive way is required for each car that is waiting for service. [B.B.A. III P.U. April, 20021 Solution. = 28 per hour 60 = — = 30 per hour 2 28 = = 0.93 µ 30 (a) System idle Po = 1 — P = 1 — 0.93 = 0.07 7% of the time the system will be idle. Traffic intensity
p=
(b) Average time a customer is waiting in the queue Et =
µ(µ  21/4,)
=
28 = 28 minutes. 60
28 (c) Average number of customers waiting E = = 28 x — =13 60 q µ(µ—?) Length of drive way =13 x 20 = 260 feet. Example 9.13. A factory manufacturing tanks for military use has a separate tool room where Special Maintenance Tools (SMTs) are stored. The average time between requirements of a tool from tool room is 10 minutes and this follow the Poisson's distribution. Average service time of the storekeeper is 9 minutes. Determine (a) Average queue length. (b) Average length of nonempty queues. (c) Average number of mechanics in the system including one who is being attended to. (d) Mean waiting time of a mechanic. (e) Average waiting time of mechanic who waits and ( f ) Whether there is a need of employing another storekeeper so that cost of storekeeper idle time and mechanics waiting is reduced to the minimum. Assuming that a skilled mechanics cost Rs. 10 per hour and the storekeeper cost Rs. 1 per hour.
OPERATIONS RESEARCH
392 Solution. Using the usual notations 1 = — x 60 = 6 per hour. 10 1 15 = — x 60= per hour 2 8 36
36 x 4
— 3.2 mechanics 15 (15 A) 45 goi 2 2 (b) Average number of workers/mechanics in the system. 6 = 4 mechanics. En = — 15 —6 2 Mean waiting time of a mechanic in the system (d) 1 2 E 1 = = 40 minutes Er "= p — A 15 _ 6 3 2 (e) Average waiting time of a mechanic in queue (or average time of a mechanic in queue)
(a) Average queue length Eq =
=
Er = µ I = 2 hours = 40 minutes. 3 (f) Probability that the store keeper remains idle = Po =
1 — X, 3 = = 0.2 . Idle time cost of one store 15
2 keeper = — x 8 x 1= Rs.1.6 / day.(assuming 8 hours working day) 10 E Waiting time cost of mechanics = W x 8 x 10 = 0.53 x 8 x 10 = Rs. 42.4 . hour It can be seen that the time cost of mechanics is much higher than the idle time cost, it is reasonable to use another storekeeper. Example 9.14. A large transport fleet employs vehicle repairmen on daily basis. The vehicles break down at an average rate of 4 per hour and the breakdown follows Poisson's distribution. Idle time of the vehicle cost Rs. 20/hour . Transport manager has the choice of selecting one out of two mechanics, one is a very efficient mechanic and for repairing 6 vehicles per hour demands 25 rupees per hour. The other charges only Rs. 15 per hour but repairs 5 vehicles per hour. Assuming a working day of 8 hours, which mechanic should the transport manager hire? Solution. Using the usual notations = 4 hours Idle time cost of a repairable vehicle = Rs. 20/hour Efficient mechanic case: = 4/hour = 6/hour Average number of vehicles in the system = E„ = in 8 hours = 8 x 2 = 16 hours.
4 = 2 vehicles. Vehicle hours lost 1.t — X 6 — 4
WAITING LINE (QUEUING) THEORY
393
Total cost per day = Cost of an idlerepairable vehicle + charges of mechanic = (20 x 16) + (25 x 8) = 320 + 200 = Rs. 520. Inefficient mechanic case: X = 4/hour = 5 /hour E„ =
= 4 = 4 vehicles. —X 5—4 Vehicle hours lost in 8 hours = 8 x 4 = 32 Total cost per day = cost of repairable vehicles + charges of vehicles mechanics (20 x 32) + (15 x 8) = 640 + 120 = Rs. 760. Since the cost of engaging an inefficient mechanic is more than that of an efficient mechanic, the efficient mechanic should be hired. Example 9.15. Customer arrive at a onewindowsdrivein bank according to Poisson's distribution with mean 10 per hour. Service time per customer is exponential with mean 5 minutes. The space in front of the window, including for the service car accommodates a maximum of three cars. Other cars can wait outside the space. (a) What is the probability that an arriving customer can drive directly to the space in front of the window ? (b) What is the probability that an arriving customer will have to wait outside the indicated space ? (c) How long is an arriving customer expected to wait before starting service ? Solution. Using the usual notations Here 2.= 10/hour 60 =— 5 = 12/hour (a) Probability that an arriving customer can directly drive to the space in front of the window. Since a maximum of three cars can be accommodated, we must determine the total probability, i.e., of po, pi and p2. (Ix — X) 2 Po = 12 10 2 20 — = Pi = 12x12=144 \2 100 2 200 P2 = (112011.= 144 x 12 12x144 11 2 20 + 200 = 728= 0.42 Total probability = — + 12 144 12 x 144 144 x 12 (b) Probability that an arriving customer has to wait = 1 — 0.42 = 0.58
OPERATIONS RESEARCH
394 (c) Average waiting time of a customer in the queue
10 = 0.417 hours — X) 12(12 —10) = 25 minutes. Example 9.16. ABC Diesel engineering works gets on an average 40 engines for overhaul per week, the need of getting a diesel engine overhauled is almost constant and the arrival of the repairable engines follows Poissons's distribution. However, the repair or overhaul time is exponentially distributed. An engine not available for use costs Rs. 500 per day. There are six working days and the company works for 52 weeks per year. At the moment the company has established the following overhaul facilities. Facilities Installation Charges Operating Expenses /year Economic life (years) Service Rate/Week
1 1200000 200000 8 50
2 1600000 350000 10 80
The facilities scrap value may be assumed to be nil. Determine wh .ch facility should be preferred by the company, assuming time value of money is zero ? Solution. Let us work out the total cost of using both the facilitates. Facility 1: = 40/week, [1. = 50/week Total annual cost = Annual capital cost + Annual operating cost + Annual cost of lost time of overhaul able engines. Expected annual lost time = (Expected time spent by repairable engines in system) x (Expected number of arrivals in a year). Et = 1 (X x number of weeks) =
1 x 40 x 52 = 208 weeks. (50 — 40)
Cost of the lost time = Rs. 208 x 6 x 500 = 624000 Total annual cost 1200000 + 200000 + 624000 =150000 + 200000 + 624000 8 = Rs. 974000 Facility 2: Annual capital cost 1600000 + 350000 + cost of lost engine availability time 10 Cost of lost availability time = Et x (X x number of weeks) =
1 x (X x number of weeks) (P, —
WAITING LINE (QUEUING) THEORY Here
395
= 40 = 80
Hence, cost of lost availability time =
2080 1 x (40 x 52) = = 52 weeks /years. 80 — 40 40
Cost of lost time = 52 x 6 x 52 = Rs. 162245 Total cost =
1600000+
350000 + 162245 = Rs. 672245 10 Hence, facility No. 2 should be preferred to facility number one.
MULTICHANNEL QUEUING MODEL (ARRIVAL POISSON AND SERVICE TIME EXPONENTIAL) This is a common facilities system used in hospitals or banks where there are more than one service facilities and the customers arriving for service are attended to by these facilities on first come first serve basis. It amounts to parallel service points in front of which there is a queue. This shortens the length of the queue if there was only one service station. The customer has the advantage of shifting from a longer queue where he has to spend more time to shorter queue and can be serviced in lesser time. Following assumptions are made in this model: (a) The input population is infinite, i.e., the customers arrive out of a large number and follow Poisson's distribution. (b) Arriving customers form one queue. (c) Customer are served on First come First served (FCFS) basis. (d) Service time follows an exponential distribution. (e) There are a number of service station (K) and each one provides exactly the same service. (f) The service rate of all the service stations put together is more than arrival rate. In this analysis we will use the following notations. = Average rate of arrival = Average rate of service of each of the service stations K = Number of service stations = Mean combined service rate of all the service stations. Hence p(rho) the utilisation factor for the system =
Kg 1
(a) Probability that system will be idle Po =
+ „_,) [Li
(b) Probability of n customers in the system. p„ —
x pon k
p) ]
OPERATIONS RESEARCH
396
p„
()" K"—k xpo n > k
(c) Expected number of customers in queues or queue length k ) E= q L(102 x Po • (d) Expected number of customers in the system = E„ = Eq + (e) Average time a customer spends in queue = (f) Average time a customer spends in waiting line 1 = E,, + — . Example 9.17. A workshop engaged in the repair of cars has two separate repair lines assembled and there are two tools stores one for each repair line. Both the stores keep in identical type of tools. Arrival of vehicle mechanics has a mean of 16 per hour and follows a Poisson distribution. Service time has a mean of 3 minutes per machine and follows an exponential distribution. Is it desirable to combine both the tool stores in the interest of reducing waiting time of the machine and improving the efficiency? Solution.
= 16/hour = 1 x 60 = 20 hours 3
Expected waiting time in queue, E,0 =
16 = = 0 . 2 hour = 12 minutes. If we (u. — X) 20 (20 — 16)
combine the two tools stores. X = Mean arrival rate = 16 + 16 = 32 / hour K = 2, n =1. = Mean service rate = 20/hour 0 k2 x En x po • Expected waiting time in queue, E, , — ' = X k — 1(1
where
Po =
1,=0
= 0.182
fi
2x3224 }
WAITING LINE (QUEUING) THEORY E10 =
E
q x po
Eq A.
Hence
397
32(f)
= 32
121(4032)2 25
Eu, = —x0.182=14 25 minutes.
Since the waiting time in queue has increased, it is not desirable to combine both the tools stores. Present system is more efficient. Example 9.18. XYZ is a large corporate house having two independent plants A and B working next to
each other. Its production manager is concerned with increasing the overall output and so has suggested the two plants being combined with facilities in both the plants. The maintenance manager has indicated that at least 6 breakdowns occur in plants A and B each in 12 hours shift and it follows the Poisson's distribution. He feels that when both the plants are combined on an average 8 breakdowns per shift will take place following Poisson's distribution. The existing service rate per shift is 9 and follows exponential distribution. The company management is considering two options, one combining the two plants. This will increase the average service rate to 12, second retaining the two plants A and B and the capacity of serving in this will be 10 servicing per shift in each of the plants. Servicing/repair time follows exponential distribution. Which alternative will reduce the customer waiting time? Solution. (First alternative) combining two plants X=8 µ=12
Po = 1—
Ix
8 1 = 1— — = — = 0.33 12 3
Expected number of machines waiting for service (in queue) E =
p.(µ — X)
=
64 =1.33 48
Expected time before a machine is repaired or (Expected time spent by machine in a system)
1 1 = = 0.25 hour = 15 minutes (II — X) 128
Second alternative (two channels)/having two plants: 1
k_1(,) Po =
(jk
in n=o
Here
)1.
k= 2, X = 6 = 10 6
Po = 1+10+ 11
36 100 2 (1— 6 ) 20
1
OPERATIONS RESEARCH
398
Po
=
[i+( ),( 36 x 2014 0) 100 28)
(1, 6 10
[140+84+3611 _I 17 26f 140 4 E Ew = q
x po lk1(1cµ 202
=
36 )140)
0 4
26 = .5
36 100 x0.54 196
10x
36 x0.54 19600 10(t)2
Or
14
+
12 1(2 x 10  6).
X 0.54 =
36 x0.54 100x196
Expected number of machines waiting for services =
36
19600
x0.54 = 0.0018
Expected time before a machine is repaired = 0.0018 hour + 1= 0.108 hours In 8 hours = 8 x 0.108 hours = 52 minutes Single channel or combined facility has less waiting time as compared to having two plants, hence, combining the two plants is preferable. Example 9.19. A bank has three different single window service counters. Any customer can get any service from any of the three counters. Average time of arrival of customer is 12 per hour and it follows Poisson's distribution. Also, on average the bank officer at the counter takes 4 minutes for servicing the customer. The bank is considering the option of installing ATM, which is expected to be more efficient and service the customer twice as the bank officers do at present. If the only consideration of the bank is to reduce the waiting time of the customer, which system is better ? Solution. The existing system is a multichannel system, using the normal notations here 60 X = 12 / hour = — = 15 / hour 4 Average time a customer spends in the queue waiting to be served. Eq = Average number of customer in the queue waiting to be served. ki.t(
E=
Or
X,)
2 xPo
E
q=
2 xPo lk1(kµk)
399
WAITING LINE (QUEUING) THEORY 1
k [
where
Here
=
X)
X
k=1
n=o
ILc
X}
k= (16)3 12 25) Po = 1+ 12} 15x6+ 6 1{ 45
1
Po = [1 + 0.133 + 0.06]1 = [1.193]1 = 0.83 02)3 Ew =
L2(18)2
xpt,=15x
= 15 x 64 x 0.83
64 xp3 (125 x 2 x 324)
(250 x 324)
= 0.009 hour
= 0.33 seconds. Proposed System
12 12 Ez = = x 60 here = 12/hour, p, = 15/hour, = 15(1512) 45 v il(11— X) = 16 minutes Hence, it is better to continue with the present system rather than installing ATM purely on the consideration of customer waiting time. Example 9.20. At a polyclinic three facilities of clinical laboratories have been provided for blood testing. Three lab technicians attend to the patients. The technicians are equally qualified and experienced and they take 30 minutes to serve a patient. This average tiny follows exponential distribution. The patients arrive at an average rate of 4 per hour and this follows Poisson's distribution. The management is interested in finding out the following : (a) Expected number of patients waiting in the queue. (b) Average time that a patient spends at the polyclinic. (c) Probability that a patient must wait before being served. (d) Average percentage idle time for each of the lab technicians. Solution. In this example X= 4/hour 11. =
30 K=3
x 60 = 2 /hour
400
OPERATIONS RESEARCH
po = Probability that there is no patient in the system. 71k 1 k1 n=1
la Lti. 41. ) kg) 1
1 2 22
1
21 22 (2)3 13 x —1 4] = 1+ + + 12 + — I +1— 6 16 12: 12 2 x 2 = [111 + — 6_ 1
x6 1 = [1+1+ 213] = (26) 1 = 0.038 4 (a) Expected number of patients waiting in the queue 1
qk
E 2 X po I I k 1 Cµ) (kg  k) 1 8 x 8 x  x 0.038 = 8 x 0.038 = 0.304 or one patient 2 4 =[ (b) Average time a patient spends in the system 41 1 0' + = 0.076 + 0.5 = 0576 hours = 35 minutes 1.t 44 2 (c) Probability that a patient must wait k 1 x po p(n~k)= ik 1 (X) =
E
q+
1 =  x8x8 x0.038 6 = 0.40 3 2 1 (d) p (idle technician) =  po + 3pi_ +  p2 when p„ =11 21n Po 3 3 11 po = when all the 3 technician are idle (no one is busy) pi = when only one technician is idle (two are busy) p2 = when two technicians are idle (only one busy) 3 1 1 p (idle technician) =  x 0.038 +2 x (4 x 0.038 +  x  (2)2 x 0.038 3 2 3 3 g = 0.038 + 0.05 + 0.025 = 0.113 Example 9.21. A telephone exchange has made special arrangement for ISD long distance calls and placed two operators for handling these calls. The calls arrive at an average rate of 12 per hour and follow Poisson's distribution. Service time for such calls is on an average 6 minutes per call and it follows
WAITING LINE (QUEUING) THEORY
401
exponential distribution. What is the probability that a subscriber will have to wait for his ISD call ? What is expected waiting time ? Assume that the policy of First Come First Served (FCFS) is followed. Solution. Using usual notations A, = 12 calls per hour 60 = — =10 calls per hour 6 K=2 (a) Probability that a subscriber has to wait p (n 2) = (Po + pi) as there are two operators, a subscriber will have to wait only if there are either 2 or more than 2 calls. 4A1" 1" + 1 ()k 1
Lykit/
k Icy
n=0
11
1 1 (12 )+ 1 (12)2 + 1 (12Y 1 + [0 ) gOo) p0o) (1;:)12)11
[
1+ [
6 1 1 (12)2 1 0.2 )3 i 1 +0.72+ x + 6 [._ io ) Lq. i.o) (v) 5
6 +0.72+ 61x 1000 1728 = [1+5
= 11+1.2+0.72+6]

x 18 ]
(3.4)1
I
1 4 = 0.294 = 3.
1 VI" in 11
Pi= — — x Po

1 (12) x 0.294 = 0.352 10
p (n 2) = 1(p) +1)1)=1 (0.294 + 0.352) = 1 0.646 = 0.354. k (b) Expected waiting time for subscriber = 2 10(0)
[t(t)
1(141. X) Po•
,., X 0.294
LI (2O12)2
(14.4 x 0.294 x 60) = 3.97 minutes. 64
OPERATIONS RESEARCH
402
Example 9.22. A general insurance company handles the vehicle accident claims and employs three officers for this purpose. The policy holders make on an average 24 claims during 8 hours working day and it follows the Poisson's distribution. The officers attending the claims of policy holders spend on an average 30 minutes per claim and this follows the exponential distribution. Claims of the policy holders are processed on first come first served basis. How many hours does the claim officers spend with the policy holder per day ? 24 Solution. Arrival rate X = — = 3 claims/hour. 8 60 Service rate p. = — = 2 claims/hour. 30 Probability that no policy holder is with bank officer k1
Po=
[ n
+
1
k
1 ll
[I n=olq
1
1 + 1 3 + 1 (3)2 1(3Y 1 + IP a 2 12 2 ) 6 2) C13) 6 )_ = C1+ 3 + 8 + 8112 9 9
t8+12+9+9)) 1 =8 38 —0.21 8
Probability that one policy holder is with bank officer p1=
3 1X1 —(—) x Po =1x x 0.21 = 0.315 2 ill•
Probability that two policy holders are with bank officer 2
p2 = 1(12 Po = 1 x
2
x 0.21 =1.89 = 0.236 8
Expected number of bank officers being idle = All three idle + any two idle + one idle = 3 po + 2 pi + 1 p2 = 3 x 0.21 + 2 x 0.315 + 1 x 0.236 = 0.63 + 0.630 + 0.236 = 0.866 Probability of any bank officer not remaining idle = 1 — 0.21 = 0.79 Time bank officers will spend with the policy holder per day = 0.79 x 8 = 6.02 hours (Assuming 8 hours working day) Example 9.23. A new company, entering the business of repair and maintenance of small generators for household use, wants to decide about the number of mechanics and other related tool for repair of such generators. The company has no experience of its own but has carried out a survey and determined that such generators would need repair at the rate of one generator every 5 hours and this follows Poisson's distribution. If only one mechanic is used, his mean repair time is two per hour and it follows the
WAITING LINE (QUEUING) THEORY
403 •
exponential distribution. It is estimated that generator down time cost is Rs. 60 per hour and the generator repair costs Rs. 100 per day of 8 hours. Calculate the expected number of operating generators and expected down time work per day. Would you recommend the company to employ two mechanics each repairing two generators out of the total 4 rather than only one mechanic repairing all the four generators ? Solution. In usual notations, we have Arrival rate (X) = 1/hour Service rate (1) = 2/hour, n = 4 Probability of no generator in the repair shop [V1111 14414 Ili + po= 12 111 —_14
1
= [1+111+(1)214—+(1)314+(1)41=411 8 1_2 8 12 8 8 IQ 1+ 1 + 3 + 3 + 3 1 2 16 64 1024
=[
= [1+0.5+0.187+0.046+0] =(1.733)1 = 0.577 Expected number of generators in the repair shop = [(1xpi )+2xp2 +3xp3 +4xp4 ] where,
Pii =
1
" (II ) 'Po
1 (1)° _ 0.577 Pi= [ p_ 8 x po 1 = 0.577 8 P2 = 11 (8 ) x Po 1 ( 1)2x P3 — g 8) P2 .=
0.577 P° = 128
(1)3 x, = 0.577 8 vo 3072
= (0.577+2x
0.577 0.577) +3x 0.577 +4 x 8 128 3072 )
4 1 3 = 0.577(1+ + + ) = 0.577(1+0.25+0.046+0.005) 4 426 . 768 0.577 x 1.301 = 0.7501
404
OPERATIONS REMARCH
Or 1 generator approximately. Expected number of operating generators = 4 —1 = 3 generators. Expected down time cost per day = 8 x 60 = Rs. 480 per generator. Total breakdown cost of generator = Expected down time cost + cost of generator repair = 480 + 100 = Rs. 580. Second case. When two mechanics repair two generators each K = 2, n = 2 po = Probability of no generator in repairshop x_1
1
X „ 1 1
  ±
II )
„=0
ILCX
kg
xi _1)2 ± 1. 1 = 1+1.1+1 8 2 8)
(114 /4)]
= [1+ 1 + 1 + 1 x 4 11 L
8 128 2 15 j
= [1+ 0.125 +0.007 + 0.133] = (1.265) 1 = 0.79 Expected number of generators in the repair shop for each mechanic 1xpi F2p2 —1 1 1 " x — x po , where p„ = — x(—) x po =1 8 IL/ P2 =
11y 1 2 8) P° = 128 P°
Expected number of generators 1 1 1 1 —po + 2 x o = 0.79(8 + 128 p =8 64 = 0.79 x
9 64
=0.11
Total down time cost of generators = 0.11 x 2 x 60 = Rs.105 Total breakdown cost of generators = down time cost + cost of two mechanics = Rs. (105 + 2 x 100) = Rs. 305 It may be seen that the total breakdown cost is lesser when two mechanics are used for repairing two generators each, hence this option is preferable. Example 9.24. You have been asked to consider three systems of providing service when customers arrive with a mean arrival rate of 24 per hour. (i) Single channel with a mean service rate of 30 per hour at Rs. 5 per customer with a fixed cost of Rs. 50 per hour.
WAITING LINE (QUEUING) THEORY
405
t
(ii) 3 channels in parallel each with a mean service rate of 10 per hour at Rs. 3 per customer and fixed cost of Rs. 25 per hour per channel. It is confirmed that the systems are identical in all other aspects with a simple queue. Average time a customer is in the system is given by
( p)22
and
1 x po + — x (where symbols have usual meaning)
If.(1—p (1 Po = 0.2 when c = 1 po = 0./1/ when c = 2 po = 0.056 when c = 3
You are required to calculate: (i) The average time a customer is in the system when 1, 2, 3 channels are in use; (ii) The most economical system to adopt if the value of the customer's time is ignored and to state the total cost per hour of this system. Solution. Average time in system Arrival rate A, = 24/hour Service rate µ = 30/hour (c = 1) = 15/hour (c = 2) = 10/hour (c = 3) 11() x po +tt
(a) Average time in system =
1(11)(1—X)2
for c = 1 pc, = 1— traffic intensity =
Average time in system =
1 — c 1 24 = = 1.8 = 0.2 cp, 30
1 30 x 24x 0.2+3024 30 30
24 2 1 48 1 50 = — x —+ — = —+ =— = 50 minutes 6 10 30 60 30 60 ,m
2
14'1
15
(b) Average time in the system
x 0.111+1 15 for c = 2 36
1(24 x 24 x 0.11x i5 15 x 36 = 11.06x0.11+ 1 ) 15 1 = (0.118 + 0.06) = 0.184 x 60 =11 minutes (c) Average time in the system 101
10
3 x 0.056 +
10
OPERATIONS RESEpCH
406
p(20 24)2
for c=3
(24 x 24 x 24 x 0.056 +1) 0.34 hour 2 x16 x100 = 20.5 minutes Total cost/hour Po =
One channel = (24 x 5)+ (50 x 1) = Rs.170 As cost @ Rs.5 per hour for 24 customers + fixed cost. Two channels = (24 x 4) + (30 x 2) = Rs. 156 Three channels = (24 x 3) + (25 x 3) = Rs. 147 POISSON ARRIVAL AND ERLANG DISTRIBUTION FOR SERVICE We have assumed in our earlier problems that the two service pattern distributions follow exponential distribution in a manner that its standard deviation is equal to its mean. But there are many situations where these two will vary, we must use a model which is more relevant and applicable to real life situations. In this method the service is considered in a number of phases 1 each with a service time — and time taken in each phase is exponentially distributed. With same mean time of 1, with different channels we get different distribution. The method makes the follows assumptions: (a) The arrival pattern follows Poisson distribution. (b) One unit completes service in all the phases and only then the other unit is served. (c) In each phase the service follows exponential distribution.
Phases 1› * Queue Fig. 9.6 The following formulae are used in this method: 1. Expected number of customer in the system 1 x2 x + =E+ 2k µ(.t. — k) µ q 2. Expected number of customers in the queue (or Average queue length)
E„= k+
E= k+
2 1 x 2k ROI —20
3. Average waiting time of a customer in queue Et = k +
2 1 x 2k µ(µ X,)
Departure after Service
WAITING,i6INE (QUEUING) THEORY
• 407 .
4. Expected waiting time of a customer in the system k+1 1 x + 2k µ(µ,— X) Example 9.25. Maintenance of machine can be carried out in 5 operations which have to be performed in a sequence. Time taken for each of these operations has a mean time of 5 minutes and follows exponential distribution. The breakdown of machine follows Poisson distribution and the average rate of breakdown is 3 per hour. Assume that there is only one mechanic available, find out the average idle time for each machine breakdown. Solution. K = 3 3 Arrival = — =120 machines/hour 60 Total service time for one machine = 5 x 3 = 15 minutes Service rate 1.1. = 1/15 machines/hour X P = Utilisation rate/traffic intensity = w. .t V. 1 Expected idle time for machine = k + — k = 2 µ(µ — A.)
20 x 1
3) x 15 = 1 = 0.25 4
_ 4 1 1 x/5( 1 ) 1 x x 1+1 6 20 20 15 20 ) 15 = 1 +15 x 60 +15 = 1.5 +15 = 16.5 minutes. 600 Example 9.26. A servicing garage carries out the servicing in two stages. Service time at each stage is 40 minutes and follows exponential distribution. The arrival pattern is one car every 2 hours and it follows Poisson's distribution. Determine (a) Expected number of customer in the queue; (b) Expected number of vehicles in the system; (c) Expected time in the system. Solution. We have, = 1vehicles/hour 2 40 2 =— 60 vehicles/hour = — 3 vehicles/hour k=2 (a) Expected number of customers in a queue 1 262 3 1 3 (2 1_9 54 27 = xx E = k+ x — x6= — per hour. q 2k µ(1.t. — X) 4 4 2 3 2 ) 32 32 16
OPERATIONS RESEARCH
408
(b) Expected number of vehicles in the system A. 27 1 3 _ 27 3 (27 +12) 39 16 = 16 vehicles /hour. E„ = Eq + = 16 4 2 x 2 16 + 4 48 102 1 54 3 1 2k.,2 x = = 7, i — = 54 +— = — hours. 32 32 2k µ(µ — X) X, 32 2 Example 9.27. In a restaurant, the customers are required to collect the coupons after making the payment at one counter, after which he moves to the second counter where he collects the snacks and then to the third 1 counter, where he collects the cold drinks. At each counter he spends 1 x — minutes on an average and this 2 time of service at each counter is exponentially distributed. The arrival of customer is at the rate of 10 customers per hour and it follows Poisson's distribution. Determine (a) Average time a customer spends waiting in the restaurant; (b) Average time the customer is in queue. X. = 10 customer/hour Solution. = Total service time for one customer 9 3 = — x 3 = — customers 2 4 4 80 = — x 60 = — hours. 9 3 1 X, (a) Average time a customer spends waiting in the restaurant Et = k + x (c) Expected time in the system = Et = k +
2k
— X)
3 1 3 3 3 80 x 600 = 0.9 minutes. — = 10 x — x — — 10 = — x — = — minutes or 200 80 3 4 50 200 9 (b) Average time the customer is in queue 1 1— 3 x 60 = 4 minutes. = 80 80 3 II REVIEW AND DISCUSSION QUESTIONS El
1. What is a queue ? Give an example and explain the basic concept of queue. 2. Define a queue. Give a brief description of the type of queue discipline commonly faced. 3. (a) Explain the single channel and multichannel queuing models. (b) Draw a diagram showing the physical layout of a queuing system with a multi server, multichannel service facility. 4. (a) Give some applications of queuing theory. (b) State three applications of waiting line theory in business enterprises. 5. With respect to the queue system, explain the following : (i) Input process, (ii) Queue discipline, (iii) capaCity of the system, (iv) Holding time, (v) Balking and (vi) Jockeying. • 6. Briefly explain the important characteristic of queuing system.
WAITING LINE (QUEUING) THEORY
409
7. What do you understand by : (a) (i) queue length, (ii) traffic intensity, (iii) the service channels ? (b) (i) steady and transient state and (ii) utilization factor ? 8. Show that if the interarrival times are exponentially distributed, the number of arrivals in a period of time is a Poisson process and conversely. 9. Consider the pure birth process, where the system starts with K customers at t = 0. Derive the equation describing the system and then show that pn(t)=
e (At)" ; n=k,k+1,... (n _01
10. Consider the pure birth process, where the number of departures in some time interval follows a Poisson distribution. Show that the line between successive departures is exponential. 11. If kAt is the probability of a single arrival during a small interval of time At, and if the probability of more than one arrival is negligible, prove that the arrivals follows the Poisson's law. 12. (a) Derive Poisson's process assuming that the number of arrival, in nonoverlapping intervals, are statistically independent and then apply the binomial distribution. (b) What are the various queuing models available ? 14. Explain (i) Single queue, single server queuing system, and (ii) Single queue, multiple servers in series queues. [Hint. GD indicates that discipline is general, i.e., it may be FCFS or LCFS or SIRO] 15. For an M / M / 1 queuing system derive the set of steadystate differencedifferential equations. 16. For a single server, Poisson arrival and exponential service time queuing system, obtain the steady equations satisfied by the probability P„ of n customer in system and hence obtain P,,. 17. (a) Write the short note on M/M/1 queue and its applications. (b) Establish the system characteristic of a M/M/1 system. Also list of the measure effectiveness for the system. P„ (1— p) p", n 0 where p is the traffic intensity. Also find the expected number of units in the system. 18. Show that for (M/M/I) : (./FCFS) the distribution of waiting time in the queue is P(w) =
— p)e(11k)w w > 0 where p =
19. For a (M/M/1) : (oo/F/FO) queuing model, in the steadystate case, obtain expressions for the mean and variance of queue length in terms of relevant parameters : A, and [t. 20. For a (M/M/1) : (oo/F/FO) queuing model in the steadystate case, show that (a) The expected number of units in the system and in the queue is given by E(n) =
and E(m) = —
11(11— X)
(b) (i) Expected waiting time of an arrival in the queue is
P)
OPERATIONS RESEARCH
410
(ii) Expected waiting time the customer spends in the system (including services) is
1  A) 21. Define busy period of a queuing system. Obtain the busy period distribution for the simple (M/M/1) : (oo/FCFS) queue. What is the condition that the busy period will terminate eventually ? 22. Derive the differentialdifferential equations for the queuing model (M/M/1) : (N/FCFS) and solve the same. 23. For a (M/M/1) : (N/FIFO) queuing model : (i) find the expression for E(n), (ii) derive the formula for P„ and E(n) when p =1. 24. What is multicham,el queuing system ? Deduce the differencedifferential equation when there are K channels. Show that, for the steadystate case, the solution of the equations can be put in the form n _0
OPERATIONS RESEARCH
448,
Hence, if au represents, the (i, j) entry of the game matrix, xi and yi will appear as in the following matrix. Player B Probabilities
X3
yi all an a31
x,„
a,,, l
xi x2
Player A
Y2
Y3
a12
a13
a22
a23
a32
a33
a2 „ a3 n
am 2 am n a,,, 3 The solution of the mixed strategy problem is also based on the minimax criterion. However, if A selects xi that maximizes the lowest expected payoff in a column and if B selects yi, it minimizes the highest expected payoff in a row. This will be illustrated in the examples that follow : Odds Method
This method can be used only for 2 x 2matrix games. In this method we ensure that sum of column odds and row odds is equal. Finding out Odds
Step I. Take first row and find out the difference between values of cell (1, 1) and that of cell (1, 2) place this value in front of second row on the right side. Step II. Take second row, find out the difference between the value of cell (2, 1) and that of cell (2, 2). Place it in front of the first row on the right side. Step III. Take first column, find out the difference between the value of cell (1, 1) and that of value of cell (2, 1). Place it below the second column. Step IV. Take second column, find out the difference between the value of cell (1, 2) and that of the value of cell (2, 2). Place this value below the first column. Example 11.6. Consider a modified form of "matching biased coins" game problem. The matching player is paid Rs. 8 if two coins turn both heads and Rs. 1 if the coins turn both tail. The nonmatching player is paid Rs. 3 when the two coins do not match. Given the choice of being the matching or nonmatching players, which one would you choose and what would be your strategy ? Solution. Let us prepare the payoff matrix.
Player A
H T
Player B H 8
CD
CD
1
T
Let us see if the saddle point exists. Minimum of row one is — 3 and similarly minimum of row two is also — 3, a circle has been put around these figures. Maximum of column is 8 and that of
449
THEORY OF GAMES
column 2 is 1. A square has been put around these two figures. There is no value, which is the lowest in its row and maximum of its column. Hence no saddle point exists. So, both the player will use mixed strategy. Use of Odds Method
Player B B1
B2
Player A
Al 8 3 A2 3 1 4 Odds 11 (a) Take first row  difference between the cell Al B1 and Al B2 8  ( 3) = 8 + 3 = 11 place it in front of second row. (b) Take second row  difference between the cell A2 B1 and A2 B2  3 1 =  4 (ignore sign) (c) Take first column  3 1 =  4 (ignore sign) (d) Take second column 8  ( 3) = 11
Odds 4 11
Value of the game
For finding out the value of the game, following formula is used : Player B
Player A
Bl
B2
Odds
Al
al
A2
b1
a2 b2
(b1 b2 ) (al  a2 )
Odds
(a2  b2 )
Value V =
 )
al (bi  b2 ) + bl (al  a2 (b1  b2 ) + (a1  a2 )
Probability of A1 =
b1  b2 , (bi  b2 ) + (al  a2)
A2 =
Probability of B1 =
a2  b2 (a2  b2 ) +  b1)'
al  bl Bn = (a2  b2 ) + (al  bi )
8 x 4 3 x11 1 4 + 11 15 4 1 Probabilities of A1 = 15 A2 1 15 Game value =
Probabilities of B1 = E,
B2 = 11 15
al  a2 (bi  b2) (al  a2 )
OPERATIONS RESEARCH
450
Example 11.7. Two players A and B are involved in a game of matching coins. When there are both heads, player A wins 100 points and wins 0 when there are two tails. When there is one head and one tail A wins 50 points. Determine the payoff of the matrix, the best strategy for both players A and B. Find out the value of game to A. Solution. The payoff matrix for player A and B is shown below. Player B H
Player A
H
T
[no
c03 0
T
Let us find out if there is saddle point. There is no saddle point so we apply ODDS method. Player B H H Player A
100 (al) —50 (bi)
Odds — 50 (a2) 0 (b2)
50 150
Odds 50 150 First Row = difference between al and a2 = 100 — (— 50) = 150, place it in front of second row Second Row = difference between bl and b2 = — 50 — (— 0) = — 50, ignore sign and place it in front of first row First Column = al — bi =100 — (— 50) = 150, place it below second column Second Column = a2 — b2 = — 50 — (— 0) = — 50, ignore sign and place it below first column l (bi — b2 ) + bi (al — a2 ) 100(50)+ — 50 (150) Game value V = a — — 125 — 50 + 150 Oh — b2 ) + (al — a2) DOMINANCE METHOD OR PRINCIPLE OF DOMINANCE
jj
t•
This method basically states that if a particular strategy of a player dominates in values his other strategies then this strategy, which dominates, can be retained and what is dominated is deleted. The Dominance Rule for Column
Every value in the dominating column (s) must be equal to or less than the corresponding value of the dominated column. The Dominance Rule for Row
Every value in the dominating row (s) must be greater than or equal to the corresponding value of the dominated row. A given strategy can be dominated if on average its value is lesser than the average of two or more pure strategies. To illustrate this point, consider the following game:
451
THEORY OF GAMES
A, A
A2 A3
B,
B B,
8 2 3
3 9 4
B, 4 8 5
This game has no saddle point. Also none of the pure strategies of A (A1, A2, A3) is lesser in value to any of his other pure strategies. Let us find out the average of A's pure and second pure strategies (A1 and A2). ( (8 + 2) (3 + 9) (4 + 8)) = (5, 6, 6) 2 2 2 This is superior to A's third pure strategy (A3). Hence strategy A3 may be deleted and the matrix becomes B,
B, 8 2
A, A2
3
B3 4
9
8
Sometimes game, which is reduced by dominance method, shows a saddle point but in original matrix there was no saddle point. This saddle point must be ignored since it does not have the properties of a saddle point, i.e., least value in its row and the highest value in its column. Example 11.8. Reduce the following game by dominance and find the game value Player B II
III
Iv
2 4
4
0
II
3 3
2
4
III
4
2
4
0
Iv
0
4
0
8
I Player A
Solution. Let us find if there is a saddle point in the matrix. This matrix has no saddle point. From player A's point of view, row III dominates row I as every value of row IV is either equal to or greater than every value in row I. Hence, row I can be deleted. The reduced matrix is Player B
Player A
I
II
III
IV
II
3
4
2
4
III
4
2
4
0
Iv
0
4
0
8
From player B's point of view, column III dominates column I as every value of column III is equal to or lesser than the value of column I. Hence, column I can be deleted. The resulting matrix is
OPERATIOT,RESEARCH
452
II 4 2 4
II
Player A
III Iv
Player B III 2 4 0
IV 4 0 8
In the above matrix, no single row or column dominates another row or column. Let us find if average of any two rows dominates the pure strategy of the other. There is no such possibility. Now let us try if the average of any two columns dominates the third, i.e., if the average value of the two (2 + 4) (40 + 2) (0 + 8) columns is equal to or less than the third average of columns III and IV is 2 2 2 = (3, 2, 4). This value is equal to or lesser than value of column II, so column II can be deleted. The resulting matrix is
II
Player A
III IV
Player B N III 4 4 0 8 0
(4 + 0) (0 + 8)  (2, 4). Hence row II , 2 2 is deleted. This results in the reduced matrix shown below. Now, row II is dominated by average of row III and IV
III
Player A
Player B III 4 0 0 8
This is now a 2 x 2 matrix and can be solved by Odds method. Step I. Subtract the two digits of column III and write them under column IV (ignoring signs). Step II. Subtract the two digits of column III and write them under column IV (ignoring signs). Step III. Subtract two digits of row III and write in front of row II (ignoring signs). Step IV. Subtract two digits of row II and write it in front of row III (ignoring sign) The resulting matrix with odds is as follows : Player B III HI
4 (a1) 0 (a2) N 0 (b) 8 (b) Odds 8_ 2/3, 1/3 Probability of player A III 8/12, IV 4/12, i.e., Player A
Odds 8 4
THEORY, OF GAMES
453
Probability of player B III 8/12,14/12, i.e., 2/3, 1/3 i — b2 ) + — ) j4 x 8 + 0 x 4 32 8 Value of the game = (b (b1 — b2 ) + (ai —a2 ) 8+4 12 3 Example 11.9. Using the dominance probability, obtain the optimal strategies for both the players and determine the value of the game. The payoff matrix for player A is given. Player B
Player A
I
II
III
IV
V
2
4
3
8
4
II
5
6
3
7
8
III
6
7
9
8
7
IV
4
2
8
4
3
Solution. This problem has a saddle point entry of cell III — I (6) but since we have to use the dominance method according to the problem, we are using it. By inspection of rows, it is clear 'Tat row III dominates row IV because all elements of row III are equal to or higher than the value of row IV. Hence, row IV can be deleted. The resulting matrix is
Player A
II III
I
II
2
4
5 6
6 7
Player B III
IV
3 3
8 7
9
8
V 4 8 7
We can now see that column I dominates column IV as all the elements of column I are equal I° or lesser than those of column IV. Hence, column IV can be deleted. The resulting matrix is Player B
Player A
I
II
III
N
2
4
3
4
II
5
6
3
8
HI
6
7
9
7
It can be seen that row III dominates row I as all values of row III are higher in value than those values of row I. Hence delete row I. The resulting matrix is Player B
Player A
I
II
III
IV
5
6
3
8
6
7
' 9
7
Column I dominates column V as all elements of colgmnI are lesser than the elements of column I. Column I can be deleted. The resulting matrix is
OPERATIONS RESEARCH
454 Player B
Player A
I
II
III
5
6
3
6
7
9
Column I dominates column II, since all its values are lesser than the values of II. Hence delete column II. The resulting matrix is Player B
I
III 3
Player A
6
9
Row IV dominates row II, since all value of row II are lesser than the values of all elements of row II. So, row II can be deleted the resulting index is Player B
Player A
III
I
II
6
9
Value of game V = 6 Strategy for player A — III Strategy for player B — I Example 11.10. Solve the following game by using the principle of dominance
Player A
2
Player B 3 4
6
1
4
2
0
2
1
1
2 3
4 4
3 3
1 7
3 —5
2 1
2 2
4
4
3
4
—1
2
2
5
4
3
3
—2
2
2
Solution. Let us search for the saddle point. This problem has no saddle point, but even it had
one we have to solve the problem by using dominance method as per the requirements of the problem. Let us take the rows (player A's point of row). Row 2 dominates row 1 as all elements of row 2 are either equal or higher in value of all the elements of row 1. Hence row 1 can be deleted. Also row 4 dominates row 5 because all the elements values of row 4 are either equal or more than the value of row 5. Hence row 5 can also be deleted. The reduced matrix is as follows:
455
THEORY OF GAMES
2 Player A 3 4
4 4 4
Player B 3 4 1 3 7 5 4 1
2 3 3 3
5 2 1 2
6 2 2 2
Let us take the columns (player B's point of view) Column 1 is dominated by columns 4, 5 and 6. Column 2 is dominated by column 4, 5 and 6. Column 4 is dominated by column 5. This is so because the elements value of column 4, 5 and 6 are equal to or lesser than the element value of column 1 and 2. Also, column 6 is dominated by column 5 because the element values of column 5 are equal or less than that of the element values of column 6. The reduced matrix by deleting columns 1, 2 and 6 is as shown below.
2 Player A 3 4
3 1 7 4
Player B 4 3 5 1
5 2 1 2
At this stage none of the single rows or columns dominates another row or column. However, it can be seen that the average of column 3 and 4, dominates column 5.
(1+3) (7  5) (4 1) (
3 = 2,1, —), as values of (2, 1,1 2 are lesser than or equal to values of 2 2 2 3 column 5. Hence column 5 can be deleted. The resulting matrix is
Player A
2 3 4
Player B 3 4 1 3 7 5 4 1
Now, it can be seen that row 4 is dominated by the average values of row 2 and 3. (7 + 1) (3  5)  (4,  1). Hence row 4 is deleted, the 2 x 2 game is as shown below. 2 2
Player A
2 3
Player B 3 4 1 4 7 5
456
OPERATIONS RESEARCH
Solving the problem by methods of odds 3 Player A
2 3 Odds
Probability of player A
1 (a) 7 (b) 8
Player B 4 3 (a)  5 (b2) 6
Odds 12 2
Strategy 2 = 1— 2, 14 7 Strategy 3 = 4 ' 17)
Probability of player B
Strategy III = 8 , (4 ) 1. 4 Strategy IV = — 4 , (;) .14
Game value =
ai (bi b2 )+171 (ai  a2) 1 x12 +7 x 2 12 +14 26_13 (al  a2 ) + (bi  b2 ) 14 14 14 7
Optimal strategy for A
Strategy 1, 0 (0, 67 , 1  , 0, 0) 7 Strategy 2, 67 Strategy 3, 71
Optimal Strategy for B
Strategy 4, 0 Strategy 5, 0 Strategy I = 0 Strategy II = 0 (0, 0, 4 , o, o) 77 4 Strategy III = 7 Strategy IV = Strategy V =0 Strategy VI = 0
THEORY OF GAMES
457
Example 11.11. Solve the following game by using the dominance method : Player B
Player A
B,
B2
B,
A,
4
5
8
A2
6
4
6
A,
4
2
4
Solution. Let us find the column maxima and row minima Player B
Player A
B2
B3
Row Minimax
A,
4
5
8
A2
6
4
6
0
A,
4
2
4
2
6
CI
8
Column Maximin Since Maximin (5) * Minimax (4).
We can't use the method for the pure strategies and mixed strategies methods has to be used. Step I. It can be seen that row A2 dominates row A3 as every element of row A2 is of higher
value than that of row A3, this row (A3) can be deleted. The resulting matrix is as shown below. Player B
Player A
B,
B2
B3
AI
4
5
8
A2
6
4
6
Step II. Column B1, dominates column B3 as the element values of B1 are either equal to or
lower than the element values of column B3, this column (B3) can be deleted. The reduced matrix is as follows : Player B
Player A
B,
B2
A,
4 (a, )
5 (a, )
2
A2
6 (b1 )
4 (b2 )
1
1
2
Step III. Let us now use the odds method, take up row Al. Find difference of (a2  al) and write it in front of row A2. Take up row A2 and find the difference between B1 and B2 and place it in front of row Al.
Now take up column B1 and find the difference of al and b1 and write it below B2. Also, find out the difference in two values of column B2, i.e., a2  b2 and write it below column B1. It ms,, be noted that the signs are ignored in these calculations.
OPERATIONS RESEARCH
458
Step IV. Gain of the Game value =
ai (bi — b2) +bi (ai — a2) 4 x 2 + 6 x —1 2 = = 3 2 +1 (ai — a2 )+ (bi — b2)
Probabilities of selection of strategies Al B1
A2 B2
2 Player A — 3
3
1 Player B — 3
2 3
A3 B3
0 0
Example 11.12. In a game, player A has three choices X, Y and Z and player B has two choices P and Q. Payment have been agreed to be made as per the arrangement shown below solve the game. XP A pays B Rs. 5 XQ B pays A Rs. 5 YP A pays B Rs. 6 YQ B pays A Rs. 4 ZP B pays B Rs. 8 ZQ B pays A Rs. 4 Solution. Step I. Let us write the payoff matrix for player A using the above information. Player B
Player A
P
Q
CD
0 4
8 Step II. Find out if a saddle point exists. No saddle point exists so the two players are using mixed strategies. Step III. Take up row X since each element of row X is higher in value than the elements of row Y, row X dominates row Y hence, row Y is deleted. The resulting matrix is as below. Player B Player A
x z
P —5 8
Q
5 4
Step IV. It is a 2 x 2 matrix and can be solved by using odds method.
THEORY OF GAMES
459 Player B
Player A
P  5 (a)
5 (a)
Odds 4
8 (b1)
4 (b2)
10
1
13
Odds
Q
In row X difference of the two values  5 and 5 is 10 and it is placed in front of row Z. In row Z the difference is 4 and it is placed in front of row X. Similarly, difference of the two elements of column P is 13 and it is placed below column Q. Difference of the two elements of column Q is 1 and it is placed below column P. Step V. Value of game can be found out from the following formula : Game value =
al (b1  b2) + (al  a2) , where al and a2 are the two elements of row X and b1  a2 ) + (bi b2)
and b2 are the values of two elements of row Z. V=
 5 x 4 + 8 x 10 4 + 10
20+80 = 60 = 30 14 14 7
Step VI. Probabilities of selection of strategies X P Player A — 4 14
Q
0
0
10 14
13 1 Player B 14 14 SubGames Method for 2 x n or m x 2 Games
0
In this method we subdivide the given game (2 x n or m x 2) into a number of 2 x 2 games. Now, each of these 2 x 2 games can be solved and then optimal strategies are selected. These are games when one of the players has 2 alternatives; where as the other player has more than two alternatives. When there is no saddle point or the game cannot be solved by using the dominance method the subgames method is very useful. It is suitable when the number of alternatives is limited to 4. In case of large number of alternatives, the solution becomes lengthy and complicated. It follows the following procedure : Step I. Divide the 2 x n or m x 2 game matrix in 2 x 2 matrix subgames. Step II. Take up each game one by one and find out if a saddle point exists. Such a subgame has pure strategies. Step III. If the subgame has no saddle point, then use odds method to solve the subgame. Step IV. Select the best subgame out of all the subgames from the point of view of the player who has more than two alternatives. Step V. Find out the strategies of this selected subgame. This is applicable to both the players and for the entire game. Step VI. Find out the value of the selected subgame, this will be the value of the whole game.
460
OPERATIONS RESEARCH
Example 11.13. Two airlines A and B operate their flights to an island and are interested in increasing their market share. Airline A has two alternatives, it either advertises its special fare or it advertises its features unique to it. On the other hand, airlines B have three alternatives of doing nothing, advertising their special fares or advertising their own special features. The matrix showing gains and losses of the two airlines in lakhs of rupees is shown below. Positive values favour airline or A and negative values favour airline B. Find the value of the game and best strategy by both the airlines using subgame method. Solution. B, 350 200
A, A2
Airline A
Airline B B, 100 180
B3  75 175
Al  Advertising special fares A2  Advertise special features B1  Do nothing B2  Advertise special fares B3  Advertise special features This game has to be solved by subgames method as per the requirement of the question. Step I. Let us divide the complete game (2 x 3) game as (2 x 2) game Subgame I B B, B, A, 350  100 A A2 200 180 Subgame II B
A
A, A2
B,  75 175
35 200
Subgame III B
A
A, A3
B2 100 180
B3 75 175
THEORY OF GAMES
461
Step II. Solve all three subgames Solution to subgame I B Odds A
A,
350
A2
200
C:7115.0 CIRO
50 450
Odds 250 150 It has a saddle point, as minimin of row A2 is also the maximum of column B2. Value of game = 180 Solution to subgame II B A,
A
A2
B,
B,
35 200
al) 4100
As minimum of row A2 is the maximum of column B3, A2, B3 is the saddle point. Value of game = 175 Solution to subgame III B B2
100 180
A,
A
A2
75
ago
Value of the game = 175 Step II. Select the best subgame from the point of view of the player who has more alternatives, i.e., B. Subgame
Value 180
II
175
III 175 B will select that game which has minimum V. B will select either subgame II or III as both have equal V. Step IV. Now, we find out the probabilities for the player A and B using their strategies while using subgame II or III. B B Subgame II A Odds
B,
B3
Odds
A,
35
 75
25
A2
200
175
120
250
165
Subgame III A Odds
B2
B,
Odd:
A,
 100
 75
5
A2
180
175
175
250
280
OPERATIONS RESEARCH
462 Probability of A to select
Probability of A to select
Strategy A1 =
Strategy A1 = 5 530 175 Strategy A2 = 530 Probability of B to select
415 120 Strategy A2 = 415 Probability of B to select Strategy B1 =
250 415
Strategy B1 = 0 Strategy 2B= 250 530
Strategy B2 = 0 Strategy B = 3 165 415 Example 11.14. Solve the following game:
Strategy 3B= 280 530 B
A
B,
B2
A,
2
6
A,
4
5
A,
12
4
Solution. Step I. Check if this game has a saddle point: B
A
A,
B1
B2
O
[61
A, A3
5 12
This game does not have a saddle point so we have to solve it by using subgame method. Step II. Divide the above game into subgames: Subgame I B B, A
B,
A,
6
A2
5
It has a saddle point since minimum of row A2 is also the maximum of column B1 value of the game = 4
THEORY OF GAMES
463
Subgame II B B, A
A,
2
A2
12
4
This game does not have a saddle point so let us solve it by method of odds. B
A A2 Odds
B,
B2
Odds
2 (a,)
6 (a2)
8
12 (b, )
4 (b, )
4
2
10
Value of game can be found out from the formula V=
— b2) + (al — a2 ) , where al and a2 are two values of elements of row Al and b1 and (a1 — a2 ) + (bi —b2 )
b2 are the values of row A2 V =
2 x 8 + 12 x 4
12
=
8 3
Subgame III B
A
B,
B2
A2
®
5
A,
F.fi
This game does not have a saddle point. So, let us use the odds method. V, using the formula given above. (4 x 8 + 12 x — 1) (32 — 12) 20 9 9 9 Step III. Select the best subgame from the point of view of the player having more alternatives. V—
Subgame
Value
I
4
II
8 3
III
20 9
Player A will select the subgame with maximum payoff, i.e., I with payoff 4. Step IV. Probability of A and B selecting their strategies
464
OPERATIONS RESEARCH B
A
A, A, Odds
B,
B2
2 (a,)
6 (a2 ) 5 (b2)
4 (b1) 1
Odds 1 4
V 2x1+4x4 18 4+(1) 5
Strategy
Player A Probability 1 A1 5 4 A2 5
Player B Strategy
B1 B2
Probability 1 5 4 5
Solution of 2 x 2 matrix without saddle point using Equal Gains Method (Probability Method) We have already seen that in case of games, which do not have a saddle point, the players use mixed strategies. Both have the options of selecting their strategies in such a manner that net gains of one is not effected by combinations of strategies of other player or opponent. Each player selects a strategy with certain probability. Example 11.15. Player A has two options Al and A2. Let the probabilities of Al be p and that of A2 will be (1  p) and p + 1  p = 1. Similarly, player B has two options B1 and B2, B1 with probability q and B2 with probability (1  q) in such a manner that q + 1  q = 1 we can solve for p and q and the value of game can be calculated. Take the case of a payoff matrix as shown below and assign the probabilities as discussed above.
A, A, Probability
Player A
Player B B, B2 6 10 8 4 q
Probability P (1  p)
(1 q)
This matrix gives the following two equations : 6p+8(1p) =10p+4(1p) 6q+10(1q) =8q+4(1q) Solving equation (i) for p
P=
Solving equation (ii) for q
q
1
=3
Note that the logic shown above can be used only for square matrices, i.e., 2 x 2 or 3 x 3, etc.
THEORY OF GAMES
465
Example 11.16. Solve the following game by equal gains or probability method: Player B
Player A
B, 8 4
A, A2
B2 2 6
Solution. Let p be the probability of player A selecting strategy Al so (1  p) is the probability of A selecting strategy A2, Also, let q be the probability of player B selecting strategy B1 then (1  q) will be the probability of B selecting strategy B2. Redraw the matrix after introducing the probability. Player B A, A2 Probability
Player A
B, 6 8
B2
Probability
10 4
(1p)
p
(1 q) If player B selects strategy B1 then payoff to A will be 8p + (1 p). If player B selects strategy B2 payoff to player A will be 2p + 6 (1 p). Since payoff under both the situations must be equal. 8p+4(1p)=2p+6(1p) 8p+44p =2p+66p 8p = 2 1 3 P= — , and (1  p) = — 4 4 Similarly, we can work out the pay off to player B 8q+2(1q)=4q+6(1q) 8p + 2  2q = 4q + 6  6q 1 8q = 4, q = (1  q) = — 2 2 Value of game = (Expected pay off of player A when player B uses strategy B1) x (Probability of player B using strategy B1) + (Expected payoff player A when player B uses strategy B2) x (Probability of player B using strategy B2) = [{8p + 4 (1  p)} + q + 12p + 6 (1  p)) x (1  q)] = [(8p + 4  4p) q + (2p + 6  6p) (1  q)] = [(4p + 4) q +( 4p + 6) (1 q)] =[4pq + 4q + (6  6q  4p + 4pq)] = (4pq + 4q + 6  6q  4p + 4pq) =  4p  2q + 8pq + 6 Substituting the value of p and q, we get the value of game. q
OPERATIONS RESEARCH
466 1 1 1 I Value of game =  4 x 4  2 x  + 8 x  x + 6 2 4 =11+1+6=5 Probability of player a selecting strategy 1 3 Al = , A2 = Probability of player B selecting strategy 1 1 B1 = 2, B2 = 2 .
Example 11.17. Solve the following game with equal gain or probability method Player B
Player A
A, A,
B, 2 6
B2 8 4
Solution. Let p be the probability of player a selecting strategy Al then probability of a selecting strategy A2 will be (1  p). Also, let q be the probability of player B selecting strategy B1, then probability of B selecting strategy B2 will be (1  q). Reconstruct the matrix after introducing the probabilities. Player B B, B2 Probability 2 P A, 8 Player A A2 6 4 (1—p) Probability (1— q) If player B selects strategy B1 then payoff to player A will be 2p + 6 (1  p). If player B selects strategy B2, then payoff to player A will be 8p + 4 (1  p). Since payoff under both the conditions must be equal. 2p+6(1p)=8p+4(1p) 2p + 6  6p = 8p + 4  4p  4p  4p = 4  6, i.e.,  8p = 2, p = 4, (1  p) =
3
Similarly, we can work out the payoff to player B 2q +8 (1q) =6q+4(1q) 2p + 8  8q 6q + 4  4q  6q  2q = 4  8 1 1  8q  4, q =  (1  q) = 2' Value of the game = [2p + 6 (1  p)] q + [2p + 6 (1  p)] (1  q) = (2p + 6  6p) q + 2p + 6  6q) (1  q)
THEORY OF GAMES
467 = 2pq + 6q — 6pq + (6 — 4p) (1— q) = — 4pq + 6q + 6 — 6q — 4p + 4pq = 6 — 4p 1 = 6 — 4 x — =5 4
3 Probabilities are 4 — 4 player A selecting strategies Al and A2. 1 1 — Player B selecting strategies B1 and B2. 22 Example 11.18. Player X is paid Rs. 10 if two coins turn both Heads and Rs. 2 if both coins turn both Tails. Player Y is paid Rs. 4 when the two coins do not match. If you had the choice of becoming player X or player Y, which one would you like to be and what will be your strategy ? Solve the problem using equal gains or probability method. Solution. Let us construct the payoff matrix for the given problem. Player Y Y, Y2 Player X
X, X,
10 —4
—4 2
Let p be the probability of player X selecting strategy X1 so, (1 — p) is the probability of player X selecting strategy X2, similarly, let q be the probability of player Y selecting strategy Y1 then (1 — q) will be the probability of player Y selecting strategy Y2. If player Y selects strategy Y1 then payoff to player X is = 10 p — 4 (1 — p) If player Y selects strategy Y2 then payoff to player Xis — 4p + 2 (1 — p). Since payoff in both situations must be equal, i.e., 10p — 4 + 4p = — 4p + 2 — 2p 10p+4p+4p+2p =2+4 20p = 6 _6 3 P — 20 10 7 (1 —P) = 173 If player X selects strategy X1 then payoff to player Y is 10q — 4 (1 — q) If player X selects strategy X2 then payoff to player Y is —4q+2(1—q) These two payoffs must be equal, i.e., 10q+4q=4q+22q 10q+4q+4q+2q =2+4
468
OPERATIONS RESEARCH
3 q = 10 7 (1  q) = 10 Now let us calculate the value of the game. V = (Expected pay off of player X when player Y uses strategy Y1) x probability of player Y using strategy Y1) + (Expected pay off of player Y using strategy Y2) x (probability of player Y using strategy Y2) V = [(10p  4 (1  p)} q + { 4p + 2 (1  p)} x (1  q)] = [(10p  4 + 4p) q + ( 4p + 2  2p) (1  q)] = 10pq  4q + 4pq + (2  6p) (1 q) = 14pq  4q + 2  2q  6p + 6pq = 20pq  6p  6q + 2 Substituting the value of p and q. P=q=
3 10
20
33(6x3) (6 x 3) +2 10 10 10 10 = 1.8 1.8  1.8 + 2 = 02 =
Pay off Player X Uses X1 3 10p  4 (1  q) = 10 x —  4(1 3 ) 10 10 = 3  (4 x 7) /10 = 3  28/10 = 1/5 Player X uses X2 3 3 4p+ 2 (1 p)= 4 x .  4 2(1  ) 170 To
5 Pay off player Y uses
1 Yi = 10q  4 (1  q) = 5 1 Y2= 5
So, whether I am player X or Y, I have equal gains and equal probabilities.
THEORY OF GAMES
469
Example 11.19. Two firms are competing for business under the conditions so that one firm's gain is other firm's loss. Firm A's payoff matrix is given below. Firm B II — Medium advertising 5 12 14
I — No advertising Firm A
I No advertising II — Medium advertising III — Heavy advertising
10 13 16
III — Heavy advertising —2 15 10
Suggest optimal strategies for the two firms and the net outcome thereof Solution. Let us find the Row Minima and Column Maxima as shown.
Firm A
I II III '1. Column Maxima
I 10 13 16 16
Firm B III 2 15 10 15
II 5 12 14 14
Row Minima 2 12 10
Since Maximin # Minimax The players have to use equal gains method using dominance method to reduce the above game. Since column II elements are smaller than each element of column I, column II dominates column I. Hence column I is eliminated. Firm B
Firm A
II 5 12 14
II III
III 2 15 10
Now, elements of row III are bigger in value than corresponding each element of Row I. Hence, Row III dominates Row I, so Row I is eliminated. Resulting matrix is Firm B
Firm A
II III
II 12 14
III 15 10
Qi
(1  q1)
Let pi be the probability of firm A selecting strategy number II Hence (1  pi) is the probability of firm A selecting strategy number III. Similarly, let qi be the probability of firm B selecting strategy number II. Hence (1  q1) is the probability of firm B selecting strategy number III.
OPERATIONS RESEARCH
470
Let us assume firm B selects strategy II. Then, payoff to firm A = 12pi + 14(1 — pi) If firm B selects strategy III Then, payoff to firm A = 15pi + 10(1 — pi) As payoff has to be equal, we get 12p1 + 14(1 — pi) = 15pi + 10(1 — pi) Solving this equation
4 Pi = — 7
3 7 Now, if firm A selects strategy II. Then payoff to firm B = 12q1 + 15(1 — q1) And if firm A selects strategy III Then payoff to firm B = 14q1 + 10 (1— qi) As payoffs under both situations have to be equal 12q1 + 15(1 — qi) = 14q1 + 10 (1— q1) 1 — pi =
Solving this equation
5 qi = — 7 (1 — qi) = — 7 Expected payoff to firm A Expected payoff of firm A when selecting strategy II when selecting strategy III V= x Probability of firm B x Probability of firm B selecting strategy II selecting strategy III V = [12pi + 14(1 — pi)] qi + [15pi + 10(1 — Pi)] (1 — 4 )2 = (12 x — 4 + 14 x 7x) 7 — 5 + (15 x — + 10 x 2  . 7 7 77 = (48 + 4) 2 5 _ r 60 .30) 2 x F x 7 7 )7 7 +7 ) 7. =
450 180 + . 49 49 90 7
Graphical Solution of (2 x n) and (m x 2) Games
Such solutions are possible only to the games in which at least one of the players has only two strategies. Let us consider a (2 x n) game of player A and B. Player A has only two strategies with probabilities pi and p2 where P2 = 1 — p and player B has n strategies. The game does not have a saddle point.
THEORY OF GAMES
471
Probability
Player B B1
B2
B3
B„
Player A
p1
an
a12
al3
a,„
P2= 1 — 191
P2
a21
a22
a23
a2„
A's expected payoff corresponding to the pure strategies of B are as follows : B's pure strategyA's expected payoff B1 aii pi + a21 (1 — pi)  (a11 — 021) p1 a21 B2 a12 P1 + a22 (1 — pl) = (a12 a22) B3
a22
a13 pi + a23 (1 — pi) = (013 — 023) pl + a23 •
•
•
B,, a1„ pi + a2 „ (1 — Pi) = (al „ — 02 „) p1 + a2 „ A should selects such a value of pi in such a manner that it maximizes his minimum payoff. This can be done by plotting the payoff equations as straight line of functions of pi. The steps involved in this solutions are as follows : Step I. The game must be reduced to such a subgame that at least one of the players has only two strategies. Step II. Take the probability of two alternatives of a player (say A) having only two strategies as pi and (1 — pi). We formulate equations of net gain of A from different strategies of B. Step III. Two parallel lines are drawn on the graph to include the boundaries of two strategies
of first player say A. Step IV. Pay off equations as functions of probabilities of two alternatives of A for different strategies of player B are plotted on the graph as straightline functions of P. Step V. If player A is maximizing, the point is identified where minimum expected gain is maximized, on the other hand, in case of minimizing player B, the point as identified where maximum loss is minimized.
The method will be demonstrated with the help of an example. Example 11.19A. Solve the following game using the graphic method.
B
A
1
2
3
4
1
2
2
3
—1
2
4
3
2
6
Solution. The game does not have a saddle point. A's expected payoff according to the pure strategies of B is shown in the matrix below. p is the probability of A selecting strategy 1 and (1 — pi) is the probability of A selecting strategy 2.
OPERATIONS RESEARCH
472 B's Strategy
A's expected payoff
B,
2p+4(1—p)=2p+4
B2 B3
2p+3(1—p)=—p+3 3p+2(1—p)=p+2 —p+6(1—p)=7p+6
B4
——2 ——3 ——4 ——5 —6 —
——6
Fig. 11.1 Let us plot — 2p + 4 when p = 0 value = 4 when p = 1 value = 2 — p + 3 when p = 0 value = 3 when p = 1 value = 2 p + 2 when p = 0 value = 2, p = 1, value = 3 — 7p + 6 when p = 0 value = 6 p = 1 value = —1 1 It can be seen from the graph that maximin occurs at p = — . This is the point of inter2 1 1 section of any two of the lines. Hence A's optimal strategy is p = — , 1 — p = — . The value of 2 2 game can be found out by substituting the value of p in the equation of any of the lines passing through P.
THEORY OF GAMES
473 5.
1  — +3= 2 2 75 V= 12 + 2 7 5
P is the point of intersection of any three of the lines B2, B3 and B4. To find out the optimal strategies of B as three lines pass through P, it indicates that B can mix all the three strategies, i.e., B2, B3 and B4. The combination of B2  B3, B3  B4, B2  B4 and must be considered. Example 11.20. Solve the following game graphically:
A
1 5 8
1 2
B 3 0 1
2 5 4
4 1 6
5 8 5
[B.B.A. III (Punjab University.), April. 2001] Solution. Step I. We know if minimax = maximin, then pure strategies are used by the players. Let us find out the values of Minimix and Maximin in this example. B A
1 2
Column Maxima
Row Minima
1
2
3
4
5
5
5
0
1
8
5
8
4
1
6
5
5
8
5
0
6
8
Since minimax maximin, the two players will use mixed strategies. Step II. Let us find if there is a saddle point in the problem. It is obvious, (since the minimum of the rows are not maximum of the columns) there is no saddle point. Step III. Let p be the probability of player A selecting strategy Al and hence (1  p) is the probability of player A selecting strategy A2. B selects Strategy
A's expected payoff to A
B1
5p+8 (1p) =13p+8
B2
5p + 4 (1  p) =9p 4
B3
0 xp+1(1p)=p1
B4
1 xp+6(1p)=7p+6
B5
8p +  5 (1  p) =13p  5
OPERATIONS RESEARCH
474 These values can now be plotted on the graph given below.
Fig. 11.2 It can be seen that out of point P, Q and R in the lower envelope R is the maximin point where lines corresponding to strategies B1 and B3 intersect. Selecting these strategies, the matrix is as shown below.
A, A2 V— A
B,
B3
odds
— 5 (a1) 8 (b1)
0 (a)
9
—1(b2)
5
(bi — b2 ) + (ai — a2 ) — 5 x 9 + 8 x 5 — 85 (al — a2 ) +(bl — b2 ) 14 9+5
Probabilities of selecting strategies
Player Al
9 14
A2
5 14
475
THEORY OF GAMES B
Probabilities of selecting strategies 1 14 0
13 14 B2 B4 0 B5 0 Example 11.21. Solve the following game graphically where payoff matrix for player A has been prepared : Player B1
B3
Player A Player B
1
5
7
4
2
2
4
9
3
1
[B.B.A.  III Punjab University, April; 2002] Solution. Step I. Let us construct the matrix from the point of view of player A. Player B Strategies
B1
B2
A,
1
2
A2
5
4
A3
7
9
A4
4
3
A5
2
1
Player A
Step II. We know if minimax = maximin, the pure strategies are used by the players. Here it is not the case. Step III. Let us find out if there is a saddle point since no entry with both minimum of the row and maximum of the column, in this matrix no saddle point exists. Step IV. Let player B have mixed strategies of B1 and B2 with probabilities of q and (1  q) so that q +1— q =1. A selects Strategy
A's expected payoff to B
Al
q +2(1—q)=—q +2
A2
5 q+4 (1q)=q+4
A3
7q+9 (1 q)=16q+ 9
A4
 4 q + (3 (1q))=q3
A5
2q+1(1q)=q +1
Step V. These equations are plotted as functions of q as straight lines on the following graph :
OPERATIONS RESEARCH
476 10
 10
9A3 8
9
7
7
8
65
6 (tAistiotal)
5
A2
4
4
3
 3 A1
2
2
A5
1
 1 0
0 1 
 1
2 
 2 A4
3 
 3
4 
4
5 
 5
6 
 6
7 
7
8 
 8
9 
 9
10 
 10
Fig. 11.3 The point P represents the minimax point hence we are plotting B's expected payoff. This point is the intersection of straight lines representing strategies A2 and A3 of player A. The resulting matrix is : Player B
Player A
B,
B2
Odds
A2
5 (a)
4 (a2)
16
A3
 7 (b,)
9 (b2)
1
 b2 ) + b1 (al  a2 )  5 x  (16) + ( 7) x 1  87 Value of the game = al (b1 (al  a2 ) + (bi  b2 ) 17 17 Strategies Strategies
I
A
0
B
5 17
II 16 17 12 17
III 1 17
IV
V
0
0



477
THEORY OF GAMES
Example 11.22. Solve by algebraic method the game whose payoff matrix is
2 1 —1 1 —2 2. 3 4 —3
Solution. Let the strategies and probabilities for player A be A1, A2, A3 and pi, p2, p3 respectively and let the strategies and probabilities for player B be B1, B2, B3 and q1, q2, q3 respectively. It is obvious that pi + p2 + p3 = 1 and q1 + q2 + q3 = 1 1 A2 A3 A = [A P1 P2 P3
For player
B = 1131 B2 B31 L ql q2 q3 i Also let V be the value of the game for player A. For player
— Pi + P2 4 3P3 V 2p1 — 2p2 + 4p3 ?. V Pi + 2P2 — 3P3 V Similarly, for player B, we have — qi + 2q2 + q3 5.. V q1— 2q2 + 2q3 V 3q1 + 4q2 — 3q3 V Converting these inequalities into equalities we get —qi + 2q2 +q3 = V — P1 + P2 + 3P3 = V 2p1 — 2p2 + 4p3 =V and qi — 2q2 + 2p3 = V 3q2 + 4q2 — 3q3 = V Pi + 2P2 — 3P3 = V These equations can be solved in terms of V and, we get 17V Pi= 30
P2 =
2V
_ 3V P3 — 10
i
7V 2V q2 = 5 q1 = 15 Using these values and knowing pi + p2 + p3 =1 and q1 + q2 + q3 =1, we get
V=
15 23
7 q1= 23,
17 Pi= 46; ,
10 P2_23'
6 q2 = E ,
_ 10 (13 — 23
Optimal strategies are : l A2 A3 Player A [A = 17 10 9 46 23 46
2V q3= 3 9 P3 = 46
OPERATIONS RESEARCH
478 B1
Player B =
B2
B3 
7 6 10 23 23 23
. 23 Example 11.23. Reduce the following Game by using dominance and modified dominance property, and then solve the Game. B, B, B2 B, 2 A, 1 2 1 A2 3 1 2 3 A, 1 3 2 1 2 A, 2 0 3
and value of the game is
Solution. B1 1
A, A2 A,
B,
B,
2
O
O
2 2 0
3 2
2
A4
B, 2 3 1
Mixed strategy will be used as problem is of unique saddle point. R4 will be deleted. B1 1 3 1
Al A2 A3
B,
B2
1
2 1 3
All the elements of column B4 131 1 Al 3 A, 1 A, 2 Average A, + A3
B4
2 3 1
2 2
B1, B4 will be deleted.
B2 2 1 3 2
B, 1 2 2 2
R1 will be deleted. All its element is
the corresponding elements of row A2 & A3. B,
B,
Al
3
1
A2
1
3
B1 +B2
2
B,
2 2
1
THEORY OF GAMES
479
B3 will be deleted. Its elements are equal to average of elements B1 & B2 column. The reduced matrix is
B, 3 1 1 1 = 4
A, A2
B2 1 3 3 1 = 2
Odds 13 = 4 31 = 2 2
Value of Game V
3(4) 1(2) = 10 = 5 4+2 6 3
Probabilities of 'A' s selecting strategies A, = 0
Probabilities of 'B' s selecting strategies 2 1 B,= 2+4 3
A„
4 2 = 4+ 2 3
4 _2 B .= 2 + 4 3
Am =
2 1 2+2 3
Bm = 0
AN = 0 2 1 A) 0, , , 0 33
13,,, = 0 B > 1, 2,0,0 33
Example 11.24. A is paid Rs. 8 if coins turn both heads and Rs. 1 if two coins turn both tail B. wins Rs 3 when the two coins do not match give the choices to be a or B. Find the values of Game.
111 (i) A T 3
3 1 I
Solution. The above problem does not have saddle point, the players will use mixed strategy. Let pi be the probability of player A selecting strategies I (1 pi) is probability the player A will select strategy II Similarly, qi is the probability. of player B selecting strategy II (1  q1) is probability of player B selecting I. Then payoff of A is = 8pi  3 (1  pi). If B selects strategy II the payoff =  3pi + 1 (1  pi). Gains are equal Bpi  3 (1 pi) =  3pi + (1  pi)
OPERATIONS RESEARCH
480 8Pi
3 + 3Pi = 3Pi + 1 Pi 11pi = — 4pi + 1 15 pi = 4 4
1—
=
11 4 1— — 175 15
Probability of B 8q1 —3(1—q1) = —3q1 + 1(1—q1) 8q1 — 3 + 3q1 = — 3q1 + 1 — q1 11q1 — 3 = — 4q1 + 1 15q1 = 4 4 q1 = 15 1 — q1 = (
11 15
V = Expected payoff to player A When B uses strategy I X probability of player B selecting strategy I + Expected pay off is player A When B uses strategy II X probability of player B selecting strategy II, + [{(3P1) + 1 (1 — PO) (1— th)]
V = [{8Pi — 3(1— Putting the value of (pi) (q1) (1 — q1) (1 — Pi)
11(11)( 1 ) 4 + — 1( 11) _ 8(4 ) 3(11)( 45j+ ( 4 4 ) +41 1 15 15 15 15 5 5 1 5 5 15 —1 15 It is better to B as the value of Game is positive. Example 11.25.Determine the optimum strategies and the value of the Game following payoff matrix of 2 person 4 x 2 Game. Y
X
I II III IV
I r —6 —3 2 —7
II —4 —9 —1
THEORY OF GAMES
481
Solution. Y
X
I II III
I II 'CO  2 \ 3 C4)
111 C_D
It is n x 2 game the subgames method has been applied to solve it. We divide the given game into maximum number of sub games. SubGame I I II Odds I 6 2 3  ( 4) = 1 II 3 4 6  ( 2) = 4 odds 2  ( 4) = 2  6  ( 3) = 3  6 (1)  3(4) 18 V= 1+4 5 SubGame II I II Odds 2 6 11 2 9 4 7 8 6(11)+2 x 4 66 +8 58 V= 11+4 15 15 SubGame III I II I 2 IV Value of Game = SubGame IV I
II
3 2
CD
Value of Game =  4 SubGame V II
III Odds
I II 3 4 +7 1 4  (1) = 3,  3  (7) = 4
Odds +7(1)=6 3  ( 4) =1
OPERATIONS RESEARCH —3(6) + — 7(1) —187 —25 = ,V 6+1 7 7
V=
SubGame VI I
II
Odds
III IV
2 —7
9 —1
6 11
Odds
8
9
• 2(6)— 7(11) V= 6+11
1
2
—18 5
—58 15
—65 17
Value of subgames tabulated 3 4 —6
—4
5
6
—25 7
—65 17
X has more than two strategies the sub game V with highest value will be selected I
II
Odds
I
—3
—4
6
II
—7
—1
1
Odds
3
4
V=
—3(6)— 7(1) —25 = . 6+1 7
Probability of setting strategies I
II 6 7 4 7
0 Player
HI
3 7
IV
0
7
Example 11.26. Solve the following game. B I
11
—6
7
lI
+4
—5
III
—1
—2
IV
—2
5
V
7
—6
A
THEORY OF GAMES
483
Solution.
B 7
CU II III
A
4 1 5 6
IV
V
[7 '
The above game does not have a saddle point and no row or column is dominated. We will apply subgame method •'‘
Subgame I
I 6 4 7 (5) =12
I II Odds
A
V=
 6 (9) + 4(13) 9 + 13
II 7 5 6  4 = 10
Odds  4  (5) = 9 6  7 = 13
1 11
Subgame II
B I 6 1 7  (2) = 9
I II
A
II 7 2  6  (1) = 5
Odds 1  (2) = 1 6  (7) = 13
V = (6)(1) + (1)(13) 9 1 +13 14 Subgame III B A IV
I 6 2
II 7 5
At saddle point V = (2) SubGame IV B A
I V Odds
I II 6 7 7 6  7  ( 6) = 13  6  7 = 13 =13 =13
Odds 7  ( 6) = 13 6 7 = 13
484
OPERATIONS RESEARCH V
(6)(13) + 7(13) 13 1  13 + 13 26 2 SubGame V B I 4 1
A
II
At saddle point V = 2 SubGame VI B I 4 2 55 =10
II IV
A
II 5 5 4 (2) =6
25 = 7 4  (5) = 9
V  4(7) + (2) (9) 10 = =5 7+9 16 8 Subgame VII B A
I 4 7
II IV
II
6
At saddle point V = 5 Subgame VIII B A
I 1 2  2 5 =7
II IV Odds
V = (1) (7) + (2) (1) 7 +1 9 8
II 2 5 1 (2) =1
Odds 2 5 = 7 1  (  2) = 1
485
THEORY OF GAMES Subgame IX B II III
A
CD
V At saddle point V = 21 Subgame X B II
IV
I 2
5
Odds 7  ( 6) = 13
V
7
6
2  5 = 7
Odds
5  (  6) =11
2  7 =9
A
V= =
 2(13) + 7(7) = 23 13+7 20 Value of subgames
II III IV V VI VII VIII IX X 1 11
11
2
1 2
2
5 8
9 8
5
2
23 20
The game having maximum V, i.e., i23 has been selected. 0 23 V= 20 Optimum
II
III
IV
V
0
13 20
7 20
A
0
0
B
11 20
9 20
Example 11.27. Solve the following game by using graphic method : Y
X
I II III Iv
I 6 3 2 7
II 2 4 9 1
OPERATIONS RESEARCH•
486
Solution. Expected payoff of Y when X adopts strategies.
I=6q2(1q) =6q2+2q =  4q  2. II =  3q  4 (1  q) = 3q  4 + 4q = q 4 III =2q9(1 q)=2q 9+9q =11q9 IV = 7q  (1 q)= 7q 1+ q = 6q 1 We have, II IV Odds
I
II
Odds
3 7 41 (1) =3
4 1
7(1)=6 3  ( 4) =1
3 (7) =4
Value of game V=
—3(6) — 7(1) 6+1 —187 —25 7 7 Probability of selecting strategies
I
II
III
X
0
6 7
0
Y
3 7
4 7
Player
N 1 7
Algebraic Method without Saddle Point
When the given game is 3 x 3 and game does not have saddle point such games can be solved
by Algebraic method. Example 11.28. Solve the following game : B 1
A
II III
I
II
III
3 —3 —1.
4 0 4
—2 1 2
487
THEORY AF GAMES Solution. B
A
I II III Column
I 3
II 4
3 1 3
0 4 4
Maxima
III 2 1
Row 2 3
2 2
4
Minima Maxima
Minimax
... Maximin # Minimax Let pi the probability of player A selecting strategy I. Let P2 the probability of players A selecting strategy II. Let p3 be the probability of player A selecting strategy III. So that Pi+ P2 FP3 = 1 Similarly, qi be the probability of strategy I. q2 be the probability of strategy II. q3 be the probability of strategy III. 3Pi  3P 1P3 .. z) 4pi + Op2  4p3 ?_ v  2pi + p2 + 2p3 3q1 + 4q2  2q3 v  3q1 + 0q2 + q3 v  1q1  4q2 + 2q3 v Pi+ P2 +P3 = 1 qi + q2 + q3 = 1 Thus, there are in all seven unknown and eight known relations. Assuming all the inequalities as equations, we get 31/1  3P2  1 P3 = v 4pi + Op2  4p3 = V  2pi + 1P2 + 2/93 = v 3q1 + 4q2  2q3 = v  3q1 + 0q2 + 1q3 = v  1q1  4q2 + 2q3 = v Pi + P2 + 793 =1 q1 + q2 + q3 =1 Now, multiply equation (iii) by 2 and add the same with equations (ii), we get =V  4pi + 2792 + 4793 = 2v 4791 + 0P2  4P3
2p2 = 3z'
...(t) ...(ii) ...(iii) ...(iv) ...(v) ...(vi) ...(vii) ...(viii)
488
OPERATIONS RESEARCH 3 p2 = 2 v
or
Now, multiply equation (i) with 4, equation (ii) by 3 and subtract (ii) form (i) 12pi 12p2  4p3 = 4v 12pi + Op2 12p3 = 3v  + _ 12p2 + 8p3 = v Now, by substituting the value of p2, we get 3  (12 x  v) + 8p3 = v 18 v+ 8p3 =v 19 P3 = — v 8 Substituting the value of p3 in (ii), we get
9
4pi  Op2  (4 x L v) =v 8 4pi = v +
9 v 2
21 Pi = — 8 v• Now, by substituting the values of pi, P2, p3 in (vii), we get Pi + P2 + P3 = 1 21 3 19 —v+v+—v =1 8 2 8 21v + 12v + 19v = 8 8 2 v=—=— 52 13 Now,
21 21 2 _ 21 Pi = 8 V _8 x 13  52 3 3 2 3 _12 P2 2 v = 2 x 13 = 13 52
19 19 2 19 v P3 = 8 = 8 x 13 = 52 Now, by adding (iv) and (v), we get 4q2  q3 = 2v
489
THEORY OF GAMES By multiplying (vi) with 3 and subract from (v), we get  3q1 + 0 q2 + 1q3 = v 3q1 1282 + 6q2 = 3v + + 12q2  5q3 =  2v Multiply (6) with 3, we get 12q2  3q3 = 6v  12q2 + 4q3 =2v q3 = 4v Putting the value of q3 in (6), we get 4q2  4v = 2v 4q2 = 6v 3 92= Zv Putting the value of q3 in (v), we get 3q1 + 0q2 + 4v = v  3q1 =  3v qi = From (viii) q1 + q2 + q3 = 1 By putting the value of q1, q2, and q3 we get 3 V+  v + 4v = 1 2 13v = 2 2 v= . 13
v=
Hence
13
Probability of A selecting strategy
Probability of B selecting strategy
, 21 52
2 I=— 13
12 II = — 52
II =
9
III = 152 Same problem can also be solved by Simplex Method.
I I
1 2
2 4 2
13
8 III= — 13 I 4 2
490
OPERATIONS RESEARCH
Row Minima II III I 4 3 2 4 1 4 2 1 2* Maximin 2 2 6 4 6 3* Minimax Maximin * Minimax. The value of game lies between 2 and 3 we have not added any constant. Let pi is the probability of A selecting strategy = I P2 is the probability of A selecting strategy = II P3 is the probability of A strategy = III Pi + P2 + P3 = 1 From the point of view of A.
1 Max. V = Min. 
P1 + P2 + P3
V
V
3pi  1 p2 + 2p3 v 2pi + 4p2 + 2p3 V  4pi + 2p2 + 6p3 v Pi P2 P3 Now, in order to have numerical value in R.H.S. we divide the constraints by v P2 —— MM • . = PI
P3
V V V
Subject to 2p 1p2 2p3 + >1 v v v 
4p2 2p3 >1 ✓ v v  4p1 2p2 6p3 >1 ✓ v v P21 P3 ° PI =
Now, substituting
(i= 1, 2, 3), we get
Min. Pi + P2 + P3 subject to
3/31  1P2 + 2/33 1 2pi + 4p2 + 2p3 1  4p1 + 2p2 + 6p3 1 Pi, P2, P3 >0 Similarly, let qi q2 q3 are the probability of players B selecting strategy I, II and III so that q1 + q2 + q3 =1 For player B Min. v = Max. 1= Max.
+ q2 + q3
491
THEORY OF GAMES
3q1  2q2  4q3 v 1q1 + 4q2 + 2q3 v 2q1 + 2q2 + 6q3 v Q1,g2, g3 ?O.
In order to get numerical value on R.H.S we divide the constraints by v 91 — 92 93 Max, — V V V
3q1 2q2 4q3 < v v v 1q1 4q2 2q3 < 1 v v v 2q1 2q2 6q3 < v v v 91t 92, 93
Let (2; = (31: (i= 1,2,3) Max. Q1 + Q2 + Q3 Sub to 3Q1 2Q2  4Q3 1 1(21 + 4Q2 + 2Q3 _5 1 2Q1 + 2Q2 + 6Q3 5 1 (211 Q21 (23 ?•• °
Now, we can solve any set of equations form pt of new of any players Each is dual of other. Max Q1 + Q2 + Q3 +0 S1 +0 S2 +0 S3 subject to 3Q1  2Q2  4Q3 + + 0 S2 A 0 S3 = 1 1Q0 4Q2 2Q3 + 0 Si + S2 + 0S3 = 1 2Q1 2Q2 6Q3 OSi 0S2 + S3 =1 Y~ Q1Q2Q3S1 S3= 0
Ratio
1
1
1
0
0
0
Q,
Q,
Q3
s,
S2
S3
1
3*
2
4
1
0
0
— —> 3
0
1
1
4
2
0
1
0
ye
0
1
2
2
6
0
0
1
Y
0
0
0
0
0
0
1
1
0
0
By
C.
S,
0
S2 S,
V.
YY Now, Q1 will be placed as S1(i)
1
31
0
(ii) R2 + Ri (iii) R3 
Ri
1
1
2
492
OPERATIONS RESEARCH By
CB
Y
S2
0
S3
0
1
1
0
QI
Q2
Q3
SI
2 3 10 3
4 3 2 3 26 * 3 4 3 7 3
51
1
Q,
1
VB
1
4 3 1 3 1 3
0 0
2 3 5 3
1
Y.Y
0
3 1 3 2 3 1 3 1 3
0
0
0
0
Now, Q3 will be placed as S3. 3 26
(i) — R3 „I
2 , 26
(ii) .1‘.2  — Rs
3
x
2
4 26' (iii) R3 +— R3 + x _ 246 3 3 ,26 Y
CB 1 0 1
V, 5 13 5 13 1 26
1
1
1
QI
Q2
Q3
1
0
0
0
0
1
1 0
3 13 10 13
Now, Q2 will be replaced in place of Q3. 13 (i) — R3 5 (ii) R2  8R3.
40 5 pt 1313 x 
1 0
0 S1 3 13 5 13 3 13 2 13 2 13
0 S, 0 1 0 0 0
0
Ratio
S3
2 13 1 13 2 13 7 26 7 26
ye 17 40 1 10
493
THEORY OF GAMES
26 + — R3 15
(iii)
2 2 5 26   —X 3 13 ,157 3
Y 4 By CB VB
Ql
Q2
S2
2 5 8 13 5
Si 1 5 1 _1 5
Q3
Qi
1
2 5
1
0
Sz
0
1
0
0
0
1
0
1
3
0
0
0
0
3
0
0
Q2
1 :1/
1
10 1 2 0
S3
1 5 1 3 10 1 2 1 2
0 1 0
All the values in (c)  z1) are either 0 or negative 1 Max.  = 1 v 2 Min. v = 2 =
q1
4 ,., ch = Viz) = 2 x 2 = 5 5 1 1 q2 = Q2 v x= 10 5 q3 = Q3 7)=OX 2 = 0 Similarly, for players A D
1
I
V
=pi v=0x2=0 P2 = p2 = 0 x 2 = 0 1 P3 =p3 v=  x 2 = 1 2 v = 2. A
B
4
I=0
I=
II = 0
II= 
Strategy
III = 1
5
1 5 III = 0
Ratio
494
OPERATIONS RESEARCH
APPROXIMATION METHOD Example 11.29. Solve the following game approximately : Player B 1 1 1 Player A 1 1 3 —1 2 —1 Solution. Step I. Player A chooses any row arbitrarily and places that row below the matrix [1 —1 3]. Step II.Player B examines this row and chooses a column corresponding to the smallest no in that row. This column is placed to the right of the matrix. Here (1) is smallest number and 1 —1 he chooses 1st column, i.e.,[ —1 Step III. Player A examines this column and chooses a row corresponding to the smallest number in that column. 1 1 1 CI 1
00 0 0
0 1
1 3 1 2 3  4 5 2 1 2 1_ 1 0 0
O1 0
2
1
O2
@ 0Ol @ 0 1
Ci 1
0 2 1 0
@
1
(11
1
OO 1 0
4 10 3 0 0 ® 10 3 2 3  4 10
3 2 1 0 1
a 0 g
2
5 2 3 10 10 10 Step IV. In case of tie the player should choose the row or column preferably different from his last choice.
495
THEORY OF GAMES Step V. The procedure is repeated for a number of iteration.
Step VI. The upper limit is calculated by dividing the highest number in the last column by the total no. of iteration and lowest limit by dividing the lowest no. in the last now by iteration. The approximation for players are : I
Ill
III
A
4 10
3 10
3 10
B
5 10
2 10
3 10
Player
The highest no. In the last column are 2 and smallest element in last row is 2. 2 So upper limit is — and lower limit is 10
2 10
2 Or [— _0.20 0.20 0.90>_0.20 0.35 0.70 ?_0•20 0 25 0.35 ?_0•20 0.10 0.10 0.17 0.95 > 0.17 0.75 > 0.17 0.35 > 0.17 0.15 p (d) no setup cost, the optimum stock level z can be obtained by
E
•
d =0
Cl
+ c2
d =0
INVENTORY MANAGEMENT
567
.24.
Discuss the continuous case of a probabilistic inventory model with instantaneous demand and no setup cost. 25. The owner of a fleet of wagons has to determine the optimal size of his fleet so that the expected cost of maintaining the fleet and of hiring extra wagons in case the demand exceeds of his fleet, is minimized. Assuming that the cost of maintaining a wagon is 'a', the cost of hiring is 'b' (b > a) and pn is the probability on n wagons being demanded on any particular day, determine the optimal size of the fleet. 26. Construct the mathematical model for the following inventory problem. "Stock is reviewed continuously and an order of size y is placed every time the stock level reaches a certain reorder point R. The p.d.f. of demand during lead'time is given as f (x), p and h denote the penalty cost and the holding cost per unit time. K is the setup cost per order". 27. What is selective inventorycontrol? 28. Explain ABC/analysis. What are its advantages and limitations, if any? 29. What is ABC analysis? Why is it necessary? What are the basic steps in implementing it? 30. Ten items kept in inventory of school of Management studies of a state university are listed beloW. Which items should be classified as A items, B items and C items? What percentage of items is in each class? What percentage of total annual value is in each class? Item 1 2 3 4 5 6 7 8 9 10
Annual Usage 200 100 200 400 6000 1200 120 2000 1000 • 80
Value per unit (Rs.) 40.00 260.00 0.20 20.00 0.04 0.80 100.00 0.70 1.00 400.00
G.N.D.U. EXAMINATION PROBLEMS III
1. Lead time (April 95, April 96, Sept. 99) Economic lot size (April 95, Sept. 98, Sept. 99 Sept. 2000) Reorder point (April 95). 2. Lead time. 3. Quantity discounts Sept. 96. 4. Inventory holding cost April 97, Sept 99. 5. Economic lot size. 6. Inventory carrying cost Sept 98. 7. Lead time. 8. Inventory holding cost Sept. 99.
568
OPERATIONS RESEARCH
9. Inventory classification.E.O.Q. 10. Classification of inventorySept. 2000. (1) "Most of the businessmen view inventory as a necessary evil". Dou you agree 1996APR. with this statement? Explain. 1997APR. (2) Some businessman believes that inventories are necessary evils. Do you agree or disagree?Explain with examples. 1998APR. Narrate the benefits of effective inventory control system to business and indus (3) trial units. 1999APR. (4) Discuss briefly the impact of quantity discount on economic order quantity and hence on inventory control procedure. 2000APR. (5) List the assumptions ofE.O.Q. model and write the constituents of carrying cost and ordering costs. 2000APR. (6) Derive the formula forEconomic order quantity where total annual requirement is R and ordering cost is C0 per order. Carrying cost is C% of average inventory carried "P" is price per unit. 2000 SEP. (7) What constitutes the ordering and carrying costs? Hence deduce that the most economical order quantity is that at which carrying cost is equal to ordering cost. 2002APR. (8) DeriveEOQ formula and state the assumptions maqe by you while arriving at the formulae. 2002APR. (9) What constitutes the ordering cost and carrying cost in inventory management, discuss in detail? 1994APR. (1) (a) Explain the following: (i) Ordering cost (ii) Carrying costs (iii) Lead time. (b) What isEconomic Order Quantity? How it is calculated? Explain taking suitable example. (2) Write short notes on the following: 1994 SEP. (1) Economic order quantity (2) The setup cost (3) Lead time (4) Holding costs (5) Shortage costs. 1995APR. (3) (a)Discuss four costs associated in developing an inventory model. 2000APR. (4) Write assumptions ofE.O.Q. model. Discuss classicalE.O.Q. model and derive formula forE.O.Q. How is safety stock calculated? Discuss. (5) What are the assumptions ofE.O.Q� model of inventory management? Discuss 2000 SEP. the classical E.O.Q. model and derive the formula for E.O.Q. How you will determine the safety stock? Discuss. 1996APR. (1) Calculate economic order quantity from the following: Annual requirement = 100000 units = Rs. 160 per order Ordering cost = Rs. 20 Material cost/unit Carrying cost = 20% 1996 SEP.
(2) The annual requirement for a particular raw material is 2000 units costing Rs. 1 each. The ordering cost is Rs. 10 per order and carrying cost is 16 % per annum of the average inventory value. FindE.O.Q. and total inventory cost per annum.
INVENTORY MANAGEMENT 1997 APR.
569
(3) Calculate economic order quantity from the following : Annual requirement = 50 units per month Ordering cost/order = Rs. 10 Material cost /unit = Rs. 6 Inventory Carrying Cost = 20 % p.a. 1998 SEP. (4) A manufacturing company uses certain part at a constant rate of 4000 units per year. Each unit cost Rs. 2 and the company personal estimate that it will cost Rs. 50 to place an order and that carrying cost of inventory is 20 % per year. Find the optimum size of each order and the minimum yearly cost. 1999 A PR. (5) A company has found that its cost to purchase of component is Rs. 40 per order and the carrying cost is 10 % on the average inventory. The company currently purchases Rs. 20000 worth of component in a year. Assuming that same demand will be there in the next year, suggest a suitable policy of purchase in terms of number of orders in a year and quantity to be ordered for each year. 1999 SEP. (6) A manufacture of motors uses Rs. 50000 of values per year. The administrative cost per purchase in Rs. 50 and the carrying charge is 20 % of the average inventory. Find the economic order quantity and number of orders per year of optimum purchasing policy of the company. 2000 APR. (7) Determine the best order size from the following given data: Expected annual requirement = 8000 units Carrying cost = 10 % of the average inventory Ordering cost = Rs. 180 per order Lot Size Unit Prize Q < 1000 Rs. 22.00 1000 5. Q 5 1500 Rs. 20.00 1500 5. Q 5 2000 Rs. 19.00 2000 5_ Q Rs. 18.50 2000 SEP. (8) Determine the best order size from the data given below: Lost Size Unit Price Rs. 22.00 1 — 999 1000 — 1499 Rs. 20.00 1500 — 1999 Rs. 19.00 2000 and above Rs. 18.50 Annual requirement is 8000 units, carrying cost 10% of the average inventory and ordering cost of Rs. 180 per order. 1993 APR. (1) A company is requiring 10000 units of raw materials per annum. The cost per order is estimated to be Rs. 50. The storage cost is estimated to be Rs. 5 per unit of average inventory. What quantity should be ordered so that the total cost is minimum? Also find the total minimum cost. 1993 APR. (2) A contractor has to supply 10000 bearings per day to an automobiles manufacturer. He finds that when he starts a production run, he can produce 25000 bearings per day. The cost of holding bearing in stock for one year is 2 paise, and the setup cost of production run is Rs. 18. How frequently should production run be made ?
•570 1993 SEP.
1993 SEP.
1994 APR.
1994 SEP.
• 1995 APR.
1995 SEP.
1998 APR.
1998 SIP.
OPERATIONS RESEARCH (3) A company is requiring 1000 units of raw material per month. The ordering cost is Rs. 15 per order. The carrying cost in addition to Rs. 2 per unit is estimated to be 15 % of average inventory per unit per year. The purchase price of raw material is Rs. 100 per unit. Find the economic lot size and total cost. The company gets concession of 5% on purchase price if it orders 2000 units or more but less than 5000 units. On orders of 5000 units or above it gets 2% commission in addition to 5%. Which of the three ways of the orders the company should adopt ? (4) An item is produced at the rate of 50 items per day. The demand occurs at the rate of 26 items per day. If the setup cost is Rs. 100 per setup and holding cost is Re. 0.01 per unit per day. Find the economic lot size for one run, assuming that the shortages are not permitted. (5) The Himavan Manufacturing Company wishes to determine the most economic order quantity for one of its product. Manufacturing cost amounts to Rs. 15 per unit, the production is 5000 units per annum. Each new lot requires a setup cost of Rs. 25 and the inventory carrying cost is 25 % of inventory value. What is most economic size to manufacture ? What is corresponding total yearly cost ? (6) The demand for an item is uniform rate of 25 units per month. The fixed cost is Rs. 15 each time a production run is made. The production cost per item is Re. 1 and the inventory carrying costs per item per month is Re. 0.30. If the storage cost is Rs. 1.50 per item per month, determine the frequency and the lot size of production runs. (7) A company purchases raw material from outside supplier for its annual requirement. During the coming year, the company plans to manufacture at a constant rate, 1,00,000 units of its product. The cost of placing each order is Rs. 160. For any item in inventory, the company used an annual carrying cost to 20 `1/0 of the item's cost. The product cost Rs. 20 each. Answer the following questions : (a) What is the optimal size ? (b) What is the total inventory cost ? (c) How many orders will be placed in the next year ? (8) The demand for an item is uniform at the rate of 25 units a month. The fixed cost is Rs. 15 each time a production run is made. The production cost is Re. 1 per item and inventorycarrying cost is Re. 0.30 per item per•month. If the shortage cost is Rs. 1.50 per item per month, find out the frequency and size of production runs. (9) Given the following information : (1) Yearly use of grinding wheels = 16000 wheels (2) Ordering cost per order = Rs. 8 (3) Inventory storing cost = 10 % (4) Price per wheel = Re. 1 . • (5) Number of orders may be = 1, 5,10, 20,40 Find : (i) E.O.Q., (ii) Number of orders per year for most economical, and (iii) Inventory cost for the year. (10) The following information is available from the records of X manufacturing company : (1) Setup cost Rs. 80 per lot (2) Annual demand = 24000 units (2) Lot size =• 8000 units (4) Carrying cost = 20 % (5) Price product per unit = Rs. 20. Is the policy practised by the company
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good with regards to inventory management ? If not, why ? Can you suggest a more economical alternative plan to this company ? If yes, explain it. 2000 APR. (11) Describe types of classification of inventory management. Perform A, B, C analysis using the following data: Item rI NCO '''l LC) %.0
Units 950 2400 200 100 4000 4500
Price/Unit 5.00 3.00 10.00 25.00 2.00 0.60
Item 7 8 9 10 11 12
Units 5000 5.00 50 3000 1200 500
Price/Unit 0.20 500 8.00 0.40 7.00 8.00
2001 APR. (12) Annual demand is 48,000 units; cost per unit is Rs. 1.50. Ordering cost is Rs. 45, carrying cost is 12 %. Supposing the company is working for 3000 days in a year and safety stock is 500 units. If leadtime is 12 days, calculate E.O.Q., Reorder point, Maximum Inventory Level. (Marks 6) 2001 SEP. (13) Find the optimal order quantity using the following information : Annual demand = 1000000 units Ordering cost = Rs. 2880 per order Carrying cost = 20 % of unit cost Price Schedule Order size 10000 — 19999 20000 and above
Price unit 0.9999 1.60 1.40
2002 APR. (14) A company requires 16000 units of a part costing Rs. 2 per price. If carrying cost is 8 % of the inventory value and ordering cost is Rs. 40 per order; then calculate EOQ, optimum No. of order per annum, the total holding cost, the reorder level, the length of the inventory cycle. What would be the saving if ordering is done as per EOQ instead of thrice a year? 2004 APR. (a) Describe briefly the benefits of effective inventory control system to business B.Com. III and industry. Also explain briefly the method of selective inventory man(Prof.) agement. (b) The annual demand of an item is 3200 units. The unit cost is Rs. 61 an item and inventorycarrying cost is 25 % of cost per annum. If the cost of per order is Rs. 150, determine (i) EOQ (ii) No. of orders per year. 2004 APR. A manufacturer of engines is required to purchase 4800 castings per year. The requirements is assumed to be known and fixed. These casting are subject to quantity discounts. The price schedule is as follow: Quantity Cost Less than 500 Rs. 150 per unit 500 or more but not more than 750 unit Rs. 138.75 per unit 750 or more units Rs. 131.25 per unit
572
2004 APR. B.B.A. III
OPERATIONS RESEARCH Monthly holding cost expected as a decimal fraction of the units is 0.02. Setup cost associated with the procurement of purchased item is Rs. 750. per procurement. Find the optimum purchase quantity per procurement. A factory requires 15000 units of an item per month, each costing Rs. 27. The cost per order is Rs. 150 and the inventory carrying charges work out to be 20 % of the inventory cost. Find the Economic Order Quantity and the number of orders per year. Would you accept a 2 % price discount on a minimum supply quantity of 1200 numbers ? Compare total cost in both cases. I I PUNJAB UNIVERSITY EXAMINATION PROBLEMS II
1995 APR. 1996 APR. 1996 SEP. 1997 APR. 1998 APR. 1998 SEP. 1999 SEP. 2002 APR. 1995 APR. 1996 APR.
(1) (i) Lead time (ii) Economic lot size (iii) Recorder point (2) Lead time (3) Quantity discounts (4) Inventory holding cost (5) Economic lot size (6) Inventory carrying cost (7) (a) Lead Time (b) Economic lot size (c) Inventory holding cost. (Marks 2 Each) Inventory Classification (8) Discuss four costs associated in developing an inventory model. (9) "Most of the businessmen view inventory as a necessary evil". Do you agree with this statement ? Explain. (Marks 6) 1997 APR. (10) Some businessmen believe that Inventories are necessary evils. Do you agree or disagree ? Explain with examples. (Marks 6) 1998 APR. (11) Narrate the benefits of effective inventory control system to business and industrial units. 1999 APR. (12) Discuss briefly the impact of quantity discount on economic order quantity and hence on inventory control procedure. (Marks 6) 2000 SEP. (13) Illustrate the types, nature and classification of inventory. (Marks 7) 2001 APR. (14) Illustrate the types, nature and classification of inventory. (Marks 9) 2001 SEP. (15) Discuss the concept of economic lot size. How is it determined ? What happens when total carrying cost per year is not equal to the total ordering cost per year ? (Marks 7) 1995 APR. (16) A company purchases raw material from outside supplier for its annual requirement. During the coming year, company plans to manufacture at a constant rate, 1,00,000 units of its product. The cost of placing each order in Rs. 160. For any item in inventory, the company uses an annual carrying cost to 20 % of the item's cost. The product cost Rs. 20 each. Answer the following questions : (a) What is the optimal size ? (b) What is the total inventory cost ? (c) How many orders will be placed in the next year ? 1995 SEP. (17) The demand for an item is uniform at the rate of 25 units a month. The fixed cost is Rs. 15 each time a production run is made. The production cost is Re. 1 per item and inventorycarrying cost is Re. 0.30 per item per month. If the shortage cost is Rs. 1.50 per item per month, find out the frequency and size of production runs.
INVENTORY MANAGEMENT
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1996 APR. (18) Calculate economic order quantity from the following: Annual requirement = 1,00,000 units Ordering cost = Rs. 160 per order Material cost/unit = Rs. 20 Carrying cost = 20 % 1996 SEP. (19) The annual requirement for a particular raw material is 2,000 units costing Re. 1 each. The ordering cost is Rs. 10 per order and carrying cost is 16 % per annum of the average inventory value. Find E.O.Q. and total inventory cost per annum. 1997 APR. (20) Calculate economic order quantity from the following: Annual requirement = 50 units per month Ordering cost/order = Rs. 10 Material cost/unit = Rs. 6 = 20 % p .a Inventory carrying cost 1998 APR. (21) Given the following information: (1) Yearly use of grinding wheels = 16000 wheels (2) Ordering cost per order = Rs. 8 (3) Inventory storing cost = 10 "Yo (4) Price per wheel = Re. 1 (5) Number of orders may be =1, 5,10, 20, 40 Find : (i) E.O.Q., (ii) Number of orders per year for most economical, and (iii) Inventory cost for the year. 1998 SEP. (22) A manufacturing company uses certain part at a constant rate of 4,000 units per year. Each unit cost Rs. 2 and the company personal estimate that it will cost Rs. 50 to place an order and that carrying cost of inventory is 20% per year. Find the optimum size of each order and the minimum yearly cost. 1998 SEP. (23) The following information is available from the records of X manufacturing company : (1) Setup cost Rs. 80 per lot (2) Annual demand = 24,000 units (3) Lot size = 8000 units (4) Carrying cost = 20 % (5) Price of product per unit = Rs. 20. Is the policy practiced by the company good with regards to inventory management? If not, why? Can you suggest a more economical alternative plan to this company? If yes, explain it. 1999 APR. (24) A company has found that its cost to purchase a component is Rs. 40 per order and the carrying cost is 10 % on the average inventory. The company currently purchases Rs. 20,000 worth of component in a year. Assuring that same demand will be there in the next year, suggest a suitable policy of purchase in terms of number of orders in a year and quantity to be ordered for each year. 1999 SEP. (25) A manufacturer of motors uses Rs. 50000 of valves per year. The administrative cost per purchase in Rs. 50 and the carrying charge is 20 % of the average inventory. Find the economic order quantity and number of orders per year for optimum purchasing policy of the company. 2000 APR. (26) Find the optimal order quantity for which price breaks are as follows: Quantity : Unit cost (Rs.) :
0 5_ qi< 500
10.00
500 5_ q2 < 750 9.25
750 q3 8.75
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OPERATIONS RESEARCH
The monthly demand for the product is 200 imits,the cost of storage is 2% of the unit cost and the cost of ordering is (a) Rs. 350 and (b) Rs. 100. 2000 SEP. (27) A certain product has demand of 25 unit's per month and the items are withdrawn uniformly. Each time a production run is made the setup cost is Rs. 15. The per item per month production cost is Re. 1 per item and inventory holding cost is Re. 0.30 per item per month. (a) Assuming shortages are not permitted. Determine how often to make a production run and what size it should be ? (b) If shortage cost is Rs. 1.50 per item per month, determine how often to make a production run and what size it should be ? 2001 APR. (28) The annual demand of an item is 3200 units. The unit cost is Rs. 6 and inventory carrying charge is 25 % per annum. If the cost of procurement of one order is Rs. 150, determine (i) E.O.Q., (ii) Optimal cost. 2001 APR. (29) Demand for a particular item is 2000 units per year. Unit cost is Rs. 5, carrying charge is 12 % per year and ordering cost is Rs. 10. (i) Find E.O.Q., (ii) A 1 % discount is offered for this item, if ordered in quantity between 600 to 1,000. Should it be taken? If so, what should be the size of order? Explain each step. 2001 SEP. (30) Determine an optimal ordering rule for the following problem: Annual Demand = 6000 units; Ordering cost = Rs. 50 per order and inventory charge is 20 % per year, where price discounts are as follows: PI = Rs. 1.25 for 0 5_ q < 100; P2 = Rs. 1.20 for 100 5. q2 < 300 P2 = Rs. 1.15 for 300 5_ q3 < 500; P4 = Rs. 1.00 for 500 5_ q4 < 1,000 P5 = Rs. 0.95 for 1000 5. q5 < 2000; P6 = Rs. 0.90 for q6 2000.
2000 APR. (31) Determine the EOQ for the data as follows: Annual Demand Ordering Cost Carrying Cost
= Rs. 1,00,000 worth of an item = 1 % of the order value = 20 % of the average inventory value.
7
2002 APR. (32) A motor company purchases 9000 motor parts for its annual requirements, ordering onemonth usage at a time. Each spare parts costs Rs. 20. The ordering cost per order is Rs. 15 and the carrying charges are 15 % of the average inventory per year. You have been asked to suggest more economical purchasing policy for the company. What advice would you offer and how much would it save the company per year? 2004 APR. (33) A shopkeeper has a uniform demand of an item at the rate of 50 items per month. He buys from a supplier at a cost of Rs. 6 per item and the cost of ordering is Rs. 10 each time. If the stock holding costs are 20 % of stock value, how frequently should he replenish his stock ? Suppose the supplier offers a 5 `1/0 discount on orders between 200 and 999 items and 10 % discount on orders exceeding or equal to 1000, can shopkeeper reduce his cost by taking advantage of either of the discounts ?
Q(
" A [1 M
r c ge[cent
LEARNING OBJECTIVES • • • • • •
Understanding the Concept of Quality Learning Difference between Quality and Inspection Techniques of Quality Control Understanding the TQM concept SQC and control charts Advantages of SQC
INTRODUCTION In everyday life customersupplier chains exist wherein goods and services are produced by using raw materials and processes and these goods and services are used by the customers. The output from the processes must be of the highest standards, i.e., quality so that the customer is satisfied. Quality may be defined as the "Totality of features and characteristics of a product or service that bears on. its ability to satisfy the customer". [B. S. 4778 (1)1. It is being increasingly recognized that a high quality of product and service and their associated customer satisfaction are the key to survival for any enterprise. The nature of the current worldwide competition generally demands the following four types of ability characteristics : 1. To understand what the customer wants and to provide it, immediately on demand, at the lowest cost. 2. To provide product and services of high quality and reliability consistently. 3. To keep up with the pace of change, technological as well as political and social. 4. To be one step ahead of customer's needs, that is, to predict what the customer will want one year or ten years from now. Dr. W. Edwards Deming has been called the founder of The Third Wave of the Industrial Revolution. His name has become synonymous with the reason for Japanese industrial success in the secondhalf of the twentieth century. The successful implementation of his approach has greatly contributed to Japan's high reputation for quality and reliability and therefore to its
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OPERATIONS RESEARCH
industrial success. Japan's highest industrial award is named after him, the Deming prize is awarded annually to the one company or individual who has contributed most to the enhancement of statistical techniques and to their improved application of design, research and development, manufacturing or service. Deming advocates the implementation of a statistical quality management approach. J. M. Juran called the father of Quality Management, defines quality as "Quality is customer satisfaction achieved through product features and freedom from deficiencies". Product features imply "Quality of design" freedom from deficiencies means "quality of conformance". Quality of design is the quality specified by the designer on behalf of the customer. Quality of conformance measures the extent the product manufactured conforms to the laid down design, i.e., design specifications. Both quality of design and quality of conformance are important for total quality management. HISTORIC AL DEVELOPMENT OF QUALITY MANAGEMENT Quality has undergone many stages as under: Inspection Stage
During this phase, there were only simple inspection based systems where teams of inspectors were employed to examine, measure or test a product and compare it with product standard. Inspection was conducted at various stages of manufacture such as incoming materials, semifinished and finished goods. Often, lots of product were subjected to hundred per cent inspection and nonconfirming products i.e., products of poor quality were segregated from good ones. Such nonconforming products were subsequently scrapped, reworked or sold at lower price. However, it was realized that inspection based systems often failed to find poor quality items as 100% inspection was never fool proof and were costly. Quality Control Stage
Quality control (inspection based) stage has the following distinguishing features : 1. Acceptance of good quality products by rejecting and segregating the defective ones, is done to utilize results for prevention of defectives in future lots. 2. Data on defects is generated from the inspection results of products testing done at various stages of manufacturing. 3. Sampling inspection plans are adopted for product control thereby replacing 100% inspection by sampling inspection. 4. Statistical Quality Control is employed for process control. Quality Assurance Stage
This stage has the following distinguishing features : 1. Inspection based quality control is replaced by prevention based Quality Assurance. 2. Emphasis is placed on quality of processes. 3. Quality planning and quality manuals are prepared with the objective of building quality into the manufacturing process. However, quality assurance (prevention based) pays no attention to service industries and other soft areas of quality such as delivery, customer satisfaction.
QUALITY MANAGEMENT
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Quality, to most people, is a variable. It is subjectively judged because it deals with the relative goodness of a product. Buyers tend to boast that their new houses or cars are the best. The implication is clear to them "quality" means high quality. It is clear that consumers assume there is a measure of goodness applicable to any kind and all kinds of products, whether the product is a box of pepper, a new bestselling novel, a television set, or even an airline. To a manufacturer, in part, quality is similarly subjective and vaguely measurable. But his concept of quality imperatives are that the consumer's quality level of goodness is attainable at a satisfactory cost level. He does not aim for the best. He sets quality standards for his product, including small tolerable departures. This, permits evaluating quality in terms of precision measurement. Thus, subjectively, quality refers to degree of goodness of a product. Objectively, it consists of a set of measurable characteristic for which standard dimensions, together with small, allowable departures up and down, may be prescribed. CAUSES OF PRODUCT QUALITY VARIATIONS Products quality variation arises out of the variables constituting a given industrial process. Among these, the following are most important : 1. Basic raw materials : The raw materials entering a process themselves vary in form and composition. They are originally products of nature and as such, are never uniform. If manufacturing processes have performed conversion operations on them already, these tend to increase rather than decrease. In manufacturing plant even identical processing could not yield identical units of product. (a) Cost of defective inputs (i) Costs of unusable items, (it) Costs of not having a part when it is needed, (iii) Costs of disgruntled customers. One hundred `1/0 inspection Sampling inspection Proportional sampling. (b) Average outgoing quality limit (i) Output quality monitor, (ii) Inspection standards. (c) SQC Calculations (d) Control Chart 2. Machines and tooling : A given industrial process is composed of machine tools and ha r)7._ dling equipment, each setup for given production runs the requisite jigs, fixtures, forms, dies, or other machine accessories. Even machines of the same kind and make differ of capability due to age, variation and other factors. The tooling items used to setup these machines also vary in their impact on product quality because of age and wear differentials. Thus, in any industrial process, identical processing is impossible. 3. Setup men, machine operators and inspectors : Machine setup in any plant seldom, if ever are perfectly executed. The setup man may be careless in reading inspection devices used to
578
OPERATIONS RESEARCH
test and measure initial pieces of product; he may also use worn devices that prevent correct readings. He may simply lack the skill needed to setup the machine properly. 4. Working conditions : Working conditions also affect quality of end product. Operators work best with good lighting, humidity and temperature figures and an absence of noise or other distractions. They work with more confidence and better quality results if they know health and accident hazards have been minimized. STABLE VARIATION PATTERNS AND CONTROL OF QUALITY If and only if, all these variables are properly controlled, it can be demonstrated that the given process has a "stable variation pattern." This means it then has predictable, reasonably narrow limits of variation for future product. Accordingly, quality control is systematic control by management of the variables in the manufacturing process that affect goodness of end product. The statistical methods is very useful because it can show whether or not a given process has a stable variation pattern. In particular, since a normal universe of data is known to have a stable variation pattern, with known probabilities of item departures from the average, it is helpful to determine whenever. or not a process is normal. A normal process is not absolutely required, however, nonnormal processes may also have stable variation patterns. If a process variation pattern is stable the process quality level will tend to remain fixed in value over time as processing continues. Also, its average variation from the process average will be constant for the several units of product manufactured. OBJECTIVES OF QUALITY CONTROL One obvious objective of quality control.in a manufacturing plant is to eliminate defective pieces of product. This may be done by placing inspectors at the end of the process to look over and test the product and pass judgement on it, piece by piece. In this way all defectives will be sorted out before the lot is shipped to.a customer. This objective is remedial, but "is very important. Preventing creation of defectives during processing is even more important, however. This is never fully possible.. Accordingly, reduction of the number of defective created, must be sought. Responsibility (For Quality Mtainment)
The ultimate responsibility for product belongs inescapably to. top management. Nevertheless, this responsibility is shared by those in subordinate management and superVisory positions. In turn, it is delegated by them to production workers and inspectors. In fact, making an quality product is the.responsibility of every member of a manufacturing organization. • Quality .must be built in a product during its processing. This undeniable fact makes the production worker a key person in attaining quality. He cannot evade his responsibility. TRADITIONALLY USED QUALITY CONTROL TECHNIQUES Inspection : Inspection is the crutch on which manufacturers of yester years, primarily depended. So, much has been said about it already that further exposition here is unnecessary. Other Techniques : Noninspection techniques traditionally used include, in some industries, the use of a "Quality man" for spotchecking the quality of both product and process. To the
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QUALITY MANAGEMENT
uninitiated outsider, he would appear to wander aimlessly through the plant, but his contribution as a quality trouble shooter is often impressive. He applies pressure, where and as necessary, to foreman and operators. In addition, he gives quality talks to departmental groups, places quality posted on bulletin boards, and tries to promote qualitymindedness and good workmanship generally. Another generally used, timehonoured technique involves preparing and issuing scrap reporting showing number and causes of defectives found by inspectors. A third technique traditionally employed to help control product quality is using quality factors in the supervisor's bonus plan. This tends to keep the foremen alert and quality minded. The advantage is that the foremen, in turn, keep their workers alert to the importance of quality. That they will do, is obvious, since their own bonus pay depends on their department's quality standings. Management may also use yield and/or bonus factors in wage plans for determining worker's pay. This tends to equalize the workers interest in quantity and quality. This fourth noninspection technique, however is much less frequently used than the first three discussed. Charging workers for product damaged by them is the final traditional quality control technique requiring mention. This is seldom used now in industry. Deducting such charges from workers' next cheques was•once common practice. Later, only partial charges were made. To assure identification of those responsible, workers were required to enter their clock numbers on work ticket stubs. Prevention (Costs associated with design and planning of a quality control programme)  Design and systems planning (QC engineering work)  Incoming in processes final inspection  Special process planning  Quality data analysis  Vendor surveys .  Reliability studies and control equipment material
Appraisal . (Cost involved in the direct appraisal of quality both in the . plant and in the field).  Inspection .  Quality audits • laboratory acceptance  Checking labour  Laboratory or other measurement service  Setup for test and inspection  Test and inspection •  Outside endorsements  Maintenance and Product engineering review and shipping  Field testing
Internal Failure (Cost directly related to the occurrence of defective production within the plant)  Cost  Rework at shop due to fault of vendor  Material procurement  Factory contract engineering  QC investigation (of failures)  Material review activity  Repair and troubleshooting
External Failure (Costs associated with the failure of a product or service in the field)  Loss of customer goodwill and service  Returned material processing and repair  Replacement inventories  Strained distributor relations
SQC Tools and Techniques
It may be restated here profitably, that there are two kind of tools — control charts and sampling inspection plans. The hourly control chart samples permit rapid, regular evaluation of the process quality level and the spread of. the individual measurements around the level. Then, if quality lapses between two sampling,. managements learns this, investigates and corrects the process, so
OPERATIONS RESEARCH
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that relatively few defective ever are proceeded. And this means that final inspectors receive lots which are almost all very good, so that sampling inspection plan are safely used. That is quality assurance to customer while cost and amount of inspection drop drastically. Setup and Organizational Placement of Quality Control In large concerns, the top quality man would normally have an assistant director to help him with departmental administration. In medium sized and small factory organizations, there probably would not be such an assistant. Immediate subordinates would included the quality control statistician—who would design or select sampling inspection plans to be used, choose the type of quality control charts to be used, do analyses on quality data and help evaluate results of quality audits and the chief inspector. The latter would supervise all inspectors, whether those doing control chart sampling and plotting or those doing final inspection. Where to Inspect ? Perhaps the most important inspection problem facing the modern industrialist is : "Where to inspect?" This question has at least two sets of answers. The first of these considers essential types of inspection. Among these are the following: 1. 2. 3. 4.
Receiving inspection Inprocess inspection Clearance inspection (Departmental or Final) Equipment and tooling inspection.
Receiving Inspection Raw materials and component parts are commonly inspected in the receiving room of the using manufacturer. This inspection occurs immediately after the materials of parts reach the receiving clerks. It is done to ascertain the quality and condition of all items received. Such inspection could be done by representatives of the buyer prior to shipment of the goods, by the vendor in the later's final inspection room. Such a procedure would provide acceptance or rejection of the goods while they were still in the seller's plant, thus eliminating vendors quality levels; this information might be useful to the purchasing agent in deciding on suppliers to be utilized in the future. InProcess Inspection The inspection of work in process must be integrated into the production line, but the inspectors doing this work must also remain independent of the chief inspector, not reporting to the operating department foremen. The objective of inprocess inspection is to prevent the continuing production of an excessive amount of defective product. Traditional inprocess inspection consists of: (1) spotchecking product quality during processing, and (2) 100 per cent inspection of lots of production at critical inspection points, between two direct labour operations. Statistical quality control charting involves using inprocess inspection also to obtain the data for computing control chart entries. Inspecting product in process is necessary at the following critical points: 1. After machine operations typically causing a high ratio of defectives. 2. Before operations that would conceal defects caused prior operations such as painting, plastic costing or assembly operations.
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3. Before very expensive operations, so that further costly processing of defectives already created will be eliminated. 4. Before operations where defective product might cause an accident of a machine breakdown. 5. Before welding, painting and other similar operations that cannot be undone. Two other types of inpress inspections require mention. This first of these is the considerable amount of informal inspection of workinprocess done by machine tenders themselves. The second is inspection of first pieces of production. In the latter case, the purpose is to make sure that setup is right prior to turning the machine over to the machine tender for the lengthy production run. Clearance Inspection
The third class of industrial inspection mentioned above was clearance inspection. This may assume the following three forms: 1. Departmental clearance, prior to sending a lot on to the next producing department. 2. Storage clearance, prior to sending lots of fabricated parts of subassemblies to special stock rooms for storage, pending issue to the final assembly line at a later date. 3. Final inspection, prior to shipping product to customers. Of course, if the concern maintains a finished product stock room, the final inspection becomes a prestocking clearance inspection, with lots of finished product that pass inspection going into the stock room, where they are stored until required for shipping to a customer. Inspection of Equipment and Tooling
The fourth class of inspection stipulated is inspection of equipment, tooling and gauges. This is a function of plant maintenance. Inspection of processing and handling equipment is typically done on the factory floor. This is true because such equipment is not portable because of its bulk, weight and permanent mounting on the production floor. The inspection of jigs, fixtures, tools, dies, gauges and machine shop is done, immediately after their return to storage. It should be standard operating procedure to clean and inspect all tooling items thus, repairs or adjustments can be made immediately. The objective is to prevent machine breakdowns and production delays except for gauges and other inspection devices, which are inspected to make sure they function properly and yield accurate measurements when used by inspectors. How much to Inspect ?
The second significant inspection problems is how much to inspect. There are just two possible solutions to this problem: (1) Inspecting every item of product made, one by one (known as 100 per cent inspection, sorting detail inspection and screening inspection), and (2) Inspecting a representative portion of a lot known as sampling inspection. If detailing is done, theoretically every defective in every lot inspected will be found and eliminated. However, in practice, this result is not usually attained. Inspectors are human and subject to error in judgements. The monotony of repetition may induce error, especially in inspecting very large lots of small parts. Other sources of errors include: 1. Misreading measuring scales on inspecting instruments. 2. Using worn or outofadjustments gauges.
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3. Inspecting while fatigued and or physically handicapped. 4. Improper positioning of a place of product in the measuring instrument. 5. Misunderstanding or negligence in applying inspection standards. 6. Flinching that is passing all borderline items. This is probably psychological. 7. Conspiring with production operators to pass everything, regardless of how good or how bad. 8. Not knowing of a changing in inspecting standards. Many companies, in the belief that their inspectors can and do make errors at times have used the techniques of secret reinsertion of lots of products. This consists of reinspecting a lot by the same or a different inspector after it has once passed inspection so that, in theory, all defectives have been eliminated. The objective is to see if further defectives are found. It is also to see if inspection judgement made were all accurately made; hence, this techniques includes reinspecting both in items placed in scrap and replacements for the defectives found at the earlier inspection. Still another instance of sampling, already noted in another connection, must be recalled. This is the piece and occasional followup inspection of production coming off automatic or semiautomatic machines. Once such machines are properly setup as demonstrated by inspecting the first few pieces of products run off the sense that only a portion of the total product processed is inspected. INSPECTION AND QUALITY CONTROL Inspection and quality control ensures: 1. Adequate maintenance of quality necessary to satisfy customers and to meet competition of rival customers. 2. That parts are within the specified limits of accuracy and they fit properly during assembly. 3. That economy in production is achieved through reduction of a defective work and consequent increase in utilization of facilities and labour. 4. The labour and machine time is prevented from being spent on work already identified as defective. 5. That control on work produced by incentive workers is provided since they need to be paid only for the acceptable quantities. INSPECTION PLANNING Inspection is an important function though it does not add any value to the product but adds to its cost. Inspection function too must be planned. Too much as well as too little inspection, both are undesirable. Too much inspection is a needless expense and too little inspection may not provide desirable quality assurance level. Inspection planning, therefore must serve as an essential element in a quality assurance programme. It consists of six basic elements : 1. What to inspect ? The product should be inspected for important quality characteristics to achieve desired quality at optirnium cost. 2. When to inspect ? The stages at which product should be inspected must be fixed.
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3. Where to inspect ? The areas to be covered by inspection (e.g., incoming materials, manufactured items, functional tests, shelf tests, final inspection, customer's return, etc.,) should be specified. 4. How much to inspect ? Decision whether product should be subject to sampling inspection or cent per cent inspection must be taken. And in case of sampling inspection, the type of sampling plan (single or double sampling plan) must be specified. 5. Who should inspect ? The qualification and experience, type of training and traits required in the inspectors should be specified. 6. How to inspect ? Inspection should be conducted by trained inspectors using the right type of tools and equipment. What to Inspect ?
Specifications are the tools of inspection. These are the clear statements of requirements. In the absence of specifications, neither buying nor selling can be thought of blue prints, operation sheets, bill of materials, packaging instructions, assembly and testing instructions, etc., are the examples of specifications. Each and every quality characteristics of the product may not be checked. Only important quality characteristics affecting performance of the product may be checked to reduce the cost of inspection. When to Inspect ?
Stages of inspection identify within the conversion process the points at which inspection need to be carried out to identify defects and initiate corrective actions, thus ensuring production of goods of the right quality at minimum cost to the organization. The following principles give valuable guidelines. 1. "Inspection sampling or cent per cent inspection should be conducted on all including materials, semifinished parts, bought out components, subcontract items and other prior to any operation whatsoever, or prior to their use in assembly, or prior to their fitment on the machine". This saves money which otherwise will be wasted due to subsequent operations on the defective work. Inspection of raw materials (for cracks, etc.,) parts and supplies on their receipt or at vendor's plant belongs to this. 2. "Inspection should be conducted prior to long run so as to reduce possibility of defectives due to errors in setting, judgment or drawing." This kind of inspection is commonly known as First off Inspection or Pilot piece inspection. "First off inspection" is followed for a process layout while "Pilot piece inspection" is carried out for a product layout. (i) First off inspection : The first piece produced after a machine is setup is checked thoroughly against the prescribed quality standards. The operator is allowed to proceed for regular production if the first piece is approved by the inspection. If any discrepancy is found, necessary adjustments are made and the new piece is produced on the machine and is offered for inspection. Only after the satisfactory piece is produced, the operator is given "go ahead" signal for production. "First off inspection" reduces the possibility of defectives due to errors in calculations of gearings (differential gears, etc.,) errors in turning, wrong machine setups and mistakes in reading the blue prints. This type of inspection,
584
OPERATIONS RESEARCH
though is especially useful when semiautomatic or automatic machinescopying lathes, machines in gear shaping or gear hobbing, gear shaving, gear grinding hack saw machine, milling machine, thread milling machine and automates are involved, yet it pays to cover all operations, manual as well as machine operations. (ii) 'Pilot piece inspection' is employed for a product layout where a series of operations are to be employed on a job. All the operations are completed on a pilot piece and the piece is offered for inspection. The production is undertaken only if the pilot satisfies the prescribed quality standards. 3. "Inspection should be conducted as the production is coming off the line. Correction (if any) is made at the earliest". This kind of inspection which also goes by the name "patrolling" inspection, is an inspection at the place of a actual production. Such an inspection is performed by the stationary inspector on or near the production line if there is sufficient volume of work. Alternatively, the inspection is done by roaming or patrolling inspector who wanders around the section under his charge in a systematic manner and checks the parts. The inspector gives the necessary feedback to the worker and to the suprevisor to affect the adjustments, if necessary. Floor inspection limits the tendency of machines to drift out of limits and produce defectives. The defects, if any, are found at an early stage and rectified before large lots are produced and declared defective subsequently. The worker is saved the anxiety and avoidable firing since the defectives, if any due to operator related errors are made known to him as the work is progressing which we can correct. Occasionally, decision of rejection by the inspector on the spot is found to create friction between the workers and the inspectors. Floor inspection is used when inspection does not require use of delicate instruments. 4. "Inspection should be carried out on the components before they are transferred to other section, department or division". This kind of inspection is called "stage inspection". Stage inspection identifies sources of defects. 5. "Inspection should be conducted prior to an operation after whose completion no rework whatsoever is possible". This type of inspection is called critical operation inspection. A critical operation is one on whose completion, no rework is possible. A few illustrations of such inspection are heat treatment operation in gears and shafts requiring hardness above 40 RC. It forms a critical operation since no rework whatsoever can be done after heat treatment. 6. "Inspection should be conducted prior to an earlier operation". This kind of inspection is called key operation inspection. A key operation is one which is characterized by an excessive cost. Inspection should be carried out on components before they are sent for costlier operations such as heat treatment, jig boring, gear cutting, etc. 7. "Inspection should be carried out prior to an operation which tends to conceal the defects of the previous operations". Surface treatments such as painting and plating that covers surface defects, assembly that hides a defective part, machining that conceals forging or casting defects, are good points of inspection. This kind of inspection called functional inspection or vendor/inward inspection. 8. "Inspection should be conducted before an operation which is likely to endanger costly tooling".
QUALITY MANAGEMENT
585
This kind of inspection is commonly known as 'critical operation inspection'. all gear blanks must be inspected thoroughly at least for the outside diameter before they are sent for gear hobbing or gear shaping failing which a hob costing few thousand rupees, may break if any gear blank has over size diameter. Similarly, gear blanks must be checked thoroughly at least for inside diameter and undercut before they are sent for gear shaping of internal teeth or a gear shaper to prevent breaking of cutter during machining. 9. "Inspection should be conducted on first few pieces produced after replacement of form tools: broaches, hobs, shaping cutters, shaving cutter, etc". Inspection after regrinding the tool is especially important where the jobs run continuously but the cutting tools having a special firmgear hob, gear shaping cutter, form tool is taken out for grinding. This stage of inspection ensures that the tool after regrinding retains the form failing which defectives may be produced e.g., indexing errors in hobs and face runout in the shaping cutter can result in heavy rejection. 10. "Inspection should be conducted on the completed assemblies under conditions similar to field conditions". This kind of inspection is commonly known as Functional Inspection. Trial runs of engines, pumps, ships, valves, actuations, component assemblies in general are the example of this type of inspection. 11. "Inspection should be concluded prior to shipment of goods to the customers". This kind of inspection is referred to as Final Inspection. Although, this kind of inspection does not give information for immediate correctable action, but it ensures that only products of predetermined quality are allowed to leave the plant and are sent to the customers. Final inspection provides excellent extent of rejections and the reason there of which enables those concerned to reassess the inspection procedure along the line, to revise manufacturing process if possible, to identify defects in existing jigs and fixtures, to know inadequacies of inspection stages, etc., and thereby suggest corrective action to prevent defectives. Final inspection, however, is the costliest inspection as loss to the organization for pieces removed as defective is more than other stages of inspection because of high incidence of processing cost and overheads until last operation. Final inspection should be employed to reaffirm product quality, not to sort out good from bad. 12. "Inspection should be carried out on all jobs received back from customers as defectives". This kind of inspection is referred to as Appraisal Inspection and the purpose is to establish the causes of defectives inspite of the fact that goods were inspected prior to dispatch. 13. "Inspection should be concluded on all nonreworkable segregated at one or more of above mentioned stages prior to their movement to the scrap yard". Where to Inspect ?
Another important aspect of inspection is to decide the place where inspection shall be carried out so as to bring out the desired quality with minimum inspection and minimum cost. The place of inspection requires to be decided with respect to incoming materials, manufactured parts, finished products, jigs and fixture, tools, assemblies, fractional testing, shelflife and environmental testing. (a) Inspection of incoming materials : Inspection of incoming material may either be performed at the vendors place or at the home plant.
OPERATIONS RESEARCH
586
Inspection at the vendor's place is performed by the buyer's inspection team at the vendor's plant prior to shipment of goods to the buyer. This inspection has number of advantages. • Inspection workload at the home plant is reduced. • Unnecessary expenses of returning the rejected material to vender are saved. • Production stoppages and last minute rush due to rejection are avoided. • Papers are cleared faster. • It gives a clear picture to the supplier regarding inspection standards of the buyer and knowledge of the functionally critical dimensions. Inspection at source, at time, is employed while developing an item with the supplier. The presence of the buyer's inspection team helps supplier to know methods of inspection similar to those of buyer. (ii) Inspection at home plant is performed on the receipt of the shipment. Such an inspection goes by the name of 'Inward inspection' or 'vendor inspection' and is performed near the receiving department. It inspects goods for three conformity characteristics, conformity to dimensions, conformity to material specifications and conformity to performance. (b) Inspection during manufacturing : It maybe performed either at the place of production (Floor inspection) or in a separate room at special bays where material to the inspected is brought (centralized inspection). (c) Inspection of finished products : Inspection of finished product should be performed as near as possible to the wrapping or packing operation. This avoids the need to transport and the likelyhood of damage or deterioration in quality. Such an inspection moreover, does not disturb production schedule and is generally faster. (d) Inspection of jigs and fixtures : The inspection of jigs and fiXture should be carried out in a separate inspection room. The inspection department often requires few pieces to be made on a machine on whose result the approval is accorded. (e) Inspection of tools : The inspection of tools especially form tools, broaches, hobs, cutters, etc., is carried out after the tool is made, sharpened or repaired. Inspection department usually requires few pieces produced to be submitted to them. Inspection is performed on these tools before the operator is allowed to continue production. (f) Inspection of gauges : Inspection of gauges is carried out in the standards room where a permanent history card is prepared and maintained for each gauge. The card shows wear allowance based on tolerance, minimum clearance permitted and other information. It also requires frequency of inspection date. This gives complete history of gauge in terms of progressive wear and the exact dimensions. After every inspections the gauges are drawn from active service on the dates entered against their names for detailed inspection. (i)
How to Inspect ?
The accuracy of measurement, a prime factor in inspection, is achieved be use of various types of measuring devices such as: • Measuring instruments • Limit gauges • Air gauges • Multiple gauging
QUALITY MANAGEMENT • • • •
587
Optical comparators Automatic gauging Nondestructive testing Visual/ dimensional inspection.
How much to Inspect ?
Inspection may be done either on each piece, called cent per cent inspection or on samples, called sampling inspection. Sampling inspection is often more practical and economical than cent per cent inspection. The method lies in selecting random samples of the lots to be inspected, conducting inspection on the samples, determining number of defective in the sample and deciding the fate of lotaccepting the entire lot if defective found in the random sample are within the specified limit or either rejecting the lot or inspecting all the pieces in the lot if sizes, Sample size and number of defectives that are permissible in the sample so as to achieve desired degree of quality. The choice between cent per cent inspection and sampling inspection depends on the following factors: (i) Cost of inspection versus cost of failure (ii) Nature of inspection (iii) Lot size (iv) Manufacturing process (v) Stage of inspection (vi) Functional importance of the item (vii) Stage of development (viii) Results of sampling inspection (ix) Reinspection of segregated lots (x) Vendor's quality rating (xi) Inspection performed as a moral check on stage inspectors or line inspectors (xii) Natural tolerance versus design tolerance (xiii) Industry practices. Total Quality Management (TOM) Stage
We have earlier seen that any corporation has to have four types of ability characteristics. Deming says "you don't have to do this; survival is not compulsory !" But the fact is that any company that lags behind in terms of any of the characteristics given will inevitably be overtaken by its competitor. The attainment of these abilities requires an organized approach to management, an approach for managing for total quality, of managing for efficiencies and competitiveness, involving each and every activity and person at all levels in the organization. This is the Total Quality Management approach. It has the following distinguishing features: 1. TQM lays stress on all processes and all persons in an organization. 2. It covers all kind of industries, manufacturing as well as service. 3. TQM implementation requires unambiguous vision of top management and gradual removal of interdepartmental barriers.
OPERATIONS RESEARCH
588
4. TQM places great emphasis on customer's, (internal and external) focus. 5. TQM lays greater trust on business processes and advocates continuous improvement in every business activity. Whereas quality control involves the following aspects: 1. Design specifications 2. Conforming to specifications 3. Feedback about the performance of the product so that design can be further improved. Essential features of quality control are shown in Fig. 13.1
•
\c‘ka` OA ":01 ci.
tl
fool and Process Design Emerging Efforts
Market Research A
1.12 t2 a, E o
s
w
a.) "c o o co 4
= 15. Fig. 13.1 Features of quality control SQC defined "SQC is a simple statistical method for determining the extent to which quality goals are being met without necessary checking every item produced and for indicating whether or not the variations which occur are exceeding normal expectation. SQC enables us to decide whether to reject or accept a product". "SQC is it;:i effective system for coordinating the quality maintenance and quality improvement efforts of various groups inan organization so as to enable production at the most economical levels which allow for a customer satisfaction".
QUALITY MANAGEMENT
589
"SQC may be broadly defined as that industrial management technique by means of which products of uniform acceptable quality are manufactured. It is mainly concerned with making things right rather than discovering and rejecting those made wrong". THE COST OF INSPECTION AND ITS REDUCTION This brings us to a consideration of the problem of inspection cost. First the cost of inspection is a "deadweight" cost. Inspection cannot improve the quality of any product. Quality should be built into a product during processing. Inspection constitutes a postmortem judgement on goodness or defectiveness of product; it is indirect labour, so it is a proper overhead charge. This is so because it does not aid in converting of raw materials into final product. Since it is a deadweight cost to be born by the product, it follows that it should be reduced to the minimum necessary to provide the level of quality assurance required by customers. The proper amount of inspection, thus, is the least amount with which the manufacturer can "get by". Aid in minimizing inspection cost may be obtained by using statistical sampling inspection. This is true because quality assurance is protected while amount and cost of inspection both are reduced. But the amount of inspection must maintain a satisfactory level of quality assurance. 1. X and R chart for process control 2. P chart for analysis of fraction defectives 3. C chart for control of number of defects per unit. The X and R charts are used for the quality characteristics which are specified as variables on the basis of actual readings is taken. The purpose or objectives of these charts is : (a) To establish whether the process is in statistical control or not; (b) To determine whether the process capability is compatible with the specifications; (c) To detect trends in the process so as to assist in planning, adjustment and resetting of the process; (d) To show when the process is likely to be out of control. Control Charts for Variables A number of samples of components coming out of the process are taken over a period of time. Each sample must be taken at random and the size of sample is generally kept 5 but 1015 units can be taken for sensitive control charts. For each sample the average value X of all the measurements and the range R are calculated. The grand average X (X is equal to the average value of all the average X) and R (R is equal to the average of all the range) are then found and from these we can calculate the control limits for the X and R charts. Use of Quality Control Charts It was emphasized in the preceding section that effective use of SQC charts will enable management to deliver typically good lots to final inspection This implies first their successful inauguration in use and second their continuing use to preserve a highquality level and tight variation spread for the process. Hence, we must first look at the problem of initiating the use of SQC chart in a plant; after this we shall look at the continuing operation of such charts.
590
OPERATIONS RESEARCH
Reliability of Central Lines and Control limits on First Initiation of SOC Charts
X and R Charts : Since the probability of crossing an Xchart control limit, in random sampling of a normal universe is 0.000135 the probability of obtaining it falling on either control limit in n consecutive samples is given by the binomial, or repeated trials theorem of elementary probability. With n > 7 the lower limit of the Rchart cannot be crossed in such sampling; the probability of crossing the upper control limit, with n = 5, is 0.0049. Hence in the same way as for the Xchart case, the probability of obtaining an R on the upper control limit is obtainable for any n consecutive samples. These are respective probabilities of a point out of control. Sampling, for none on the Xchart and on the Rchart, out of control, the probability is 0.0560. For all other possible paired numbers of points out of control on X and Rcharts with N = 25, n = 5 the probabilities are respectively all less than 0.01 hence when initiating X and Rcharts for the first time in a plant, 25 samples of live items each sampled at random from a normal process find the process in control if, and only if, the maximum number of point of control on both charts combined is one. Even one point out of control on each chart is too many. This is important because the 25 samples of five each must find the process in control for the resulting calculating central line and control limit value to be reliable. Only when such sampling shows the process in control may one depend on the control limits computed in counting the use of the control charts for judging process quality. pCharts : A similar procedure should be used in initiating the use of pcharts in a plant or department. First, take 25 to 50 samples, figuring the ratio defective on each. The average of these sample ratios defective is the central line value for the pchart. Control limits are calculated also from above and below the central line. Again, the number of points on or outside the control limits should be counted. And the binomial probability of getting on the outside such limits for an item sample should be computedzero time in 25 samples, and so on, which will be the yardstick for judging whether the initial group of 25 samples had too many points on or outside control limits. If not, the trial control limits and central line may be safely projected for judgement of future sampling results. However, if too many points are on or outside the trial limits, judged at the 5 per cent level of significance, then investigation is needed to find the assignable cause of quality trouble. When found, these should be eliminated and strengthen the whole initiating procedure must be repeated which continues until a satisfactory initial set of 25 samples is obtained. cCharts : Similar reasoning and a similar procedure are applicable to the case of initiating the use of ccharts in a plant or department. However, each of the 25 samples will be measuring unit sample of product, and the variable to be controlled is the number of counted defects per measuring unit sample. Benefits realized by management from using control charts are as follows: The Variable Case: (Xand Rchart)
The principal benefit realized by management from variables quality control charts is that their use leads to such prompt detection of quality trouble that remedial action restores a satisfactory process before too many excess defectives are processed. A number of other managerial benefits result using quality control charts for variables. Among these are the following : 1. The amount and cost of rework and scarp are both greatly reduced from fewer total defective.
QUALITY MANAGEMENT
591
2. Delivery of typically good lot final inspection is the result of creating few defectives in processing. 3. The control charts define the range of tolerable product variation, which is that within control limits and caused by chance. 4. Knowing the range of tolerable product variable places great emphasis upon variation beyond this range. 5. Using quality control charts contributes greatly to elevating the quality consciousness and quality mindedness of the work force. 6. Using control charts contributes to great understanding and improvement of applied inspection standards. 7. Control charts identify the proper tolerance range, that is they tell what tolerance can be met using present machines. 8. Gradually reducing the total number of assignable causes that might adversely affect the process is made possible by using control charts. They point out the advent of assignable cause and call an investigation to identify and eliminate them from the process. 9. Product improvement deriving from control charts leads to actual elimination of selective mating of parts supposed to be interchangeable in assembly operations. 10. Individual machine capabilities to hold to tolerances in force may be successfully determined by maintaining separate variables control charts on each machine in a given group of machines. 11. Such machine capability data is useful in production scheduling since the production controll can then assign order of customers demanding tight tolerances of machines capable of holding such tolerances and assign orders of other, less demanding customers to machines incapable of holding tight tolerances. 12. This facilitates the work of the production control, as it releases more machines for regular production. It may also prelude the necessity of purchasing added machines if sales increase, or it may permit the sale of the machines not needed for rework. 13. Control charts constitute a permanent historical record of the process quality level and variation pattern, whether these are satisfactory or unsatisfactory. 14. Control chart identification and recording of assignable causes of variation assist in training new machine operators and designing equipment, tools and accessories. 15. Quality improvement from using control charts conserves and makes more effective use of manpower. 16. Substandard operators working on piece rates can be assisted by a special application of the control chart principle. Both X and R charts will be required to carryout this procedure for each operation element separately. The Attributes Case : pcharts
Benefits obtained from using pcharts also are such that management cannot afford to exclude their use. First, pcharts make possible direct control of the ratio defective. This makes maintenance of good process more likely. As long as a process is controlled with the charts, management knows that typically good lots are reaching final inspection. Placing pcharts at inspection points in the plant leads to the greatest in product quality on the chart of the machine operators concerned, if the charts are kept uptodate and commented
592
OPERATIONS RESEARCH
upon the workers concerned by their foreman and the inspector posting the chart. Several other favourable results derive from using pcharts. Among these are the following : 1. pcharts clearly point out quality trouble. If trouble cannot be located promptly, it becomes clear that management should change to variables charts for their greater diagnostics power. So, pcharts can signal the need for more effective tools. 2. pcharts provide a history of process and product quality, including process quality level and of variations from it. 3. pcharts reveal discrepancies in applying inspection procedures and standards. 4. Quality reports of definite trouble and significant quality improvement, specifying time and place provoke action to improve the process and further stimulate quality mindedness in all employees. 5. pcharts may also be used to provide the company with a historical record of incoming raw material and parts quality, by vendors that should be of material assistance to the purchasing agent. The Attributes Case : ccharts
First, using ccharts paves the way to attaining and maintaining a controlled process, otherwise unattainable, since variables charts and pcharts cannot be used when the quality control required is over the number of defects per unit. The cchart, if it results in process, is also an instrument for reducing the cost of scrap and rework. Moreover, the amount and cost of inspection necessary to identify the defects will be less when the process is in control. This assumes, of course, that management will be fully responsive to the process improvement and will reduce detail inspection to scientific sampling inspection. ccharts also inform management about the current quality level and changes in it. Total Quality Control
All of the following sections and individuals have definite quality responsibilities : 1. Quality control engineering 2. Statistician 3. Inspectors 4. Product design 5. Plant engineering 6. Industrial engineering 7. Plant maintenance 8. Purchasing 9. Shop supervision 10. Production workers 11. Company salesmen 12. Field servicemen 13. Cost accounting 14. Production control 15. Top management.
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Total quality control offers an underlying, unifying philosophy for the whole quality effort. With it a firm can minimize quality failures and overall cost. Total control contemplates a complete programme of quality cost accounting. It envisages three categories of costs. 1. Failure costs 2. Appraisal costs 3. Preventive costs. Management continually recall the importance of the human factor, however. One formal approach to this has been the inception of "zero defects". In such defects programmes employees can point out most of the potential causes of quality trouble and management must evoke their interest in doing so. Quality must be built into the product as it is made. And it is made by people, who use material and run machines, using prescribed work methods. The "people" part of quality is very important. Advantages Statistical Quality Control is one of the tools for scientific management and has following main advantages over 100 per cent inspection. (i) Reduction in costs: Since only a fractional production is inspected, hence cost of inspection is greatly reduced. (ii) Greater efficiency: It requires a lesser time and boredom as compared to the 100 per cent inspection and hence the efficiency increases. (iii) Easy to apply: Once the statistical quality control plan is established, it is easy to apply even by a man who does not have extensive specialized training. (iv) Accurate prediction: Specifications can easily be predicted for the future which is not possible even with 100 per cent inspection. (v) Can be used where inspection needs destruction of items: In case where destruction of product is necessary for inspecting it 100 per inspection is not possible (which will spoil all the products) sampling inspection is resorted to: (vi) Early detection of faults: The moment a sample point falls outside the control limits, it is taken as a danger signal and necessary corrective measures are taken. Whereas in 100 per cent inspection unwanted variations in quality may be detected after number of defective items have already been produced. Thus, by using the control charts we can know from graphic picture that how the production is proceeding and where corrective action is required and where it is not required. Types of Control Charts in S.Q.C. What the X and R charts tell ? Sometimes X chart does not give satisfactory results, this may occur due to old machine or worn out parts or misalignment or where processing is inherently quite variable. Here the range chart is used as an additional tool to control. The purpose of this chart is to have constant check over the variability of process. Process variability shows that though the means or average of the process may be perfectly centred about the specified dimension, excessive variable will result in poor quality products.
594
OPERATIONS RESEARCH
The use of Rchart is called for if after using the X chart for sometime, it is found that frequently fails to indicate trouble promptly. The Rchart does not replace the X chart but simply supplements with additional information about the production process. The Rchart is also used for high precision process whose variability must be carefully held within prescribed limits. Similarly, many electrochemical processes such as plating and microchemical biological production supplements with additional information about the production process. Inference : In the X charts, most of the time the plotted points representing averages are well within the control limits but in some samples plotted points fall outside the control limits. It means something has probably gone wrong or is about to go wrong with the process and check is needed to prevent the appearance of defective products. If the cause has been eliminated the following plotted points will stay well within the control limits but if more points fall outside the control limits, then a very thorough investigation should be made, even if it is necessary to shut down production temporarily until every thing is adjusted again and no more points fall outside. Major Parts of a Control Chart
A control chart has the following major parts : (a) The vertical scale, also called the quality scale. The scale represents the quality characteristics of different samples. (b) Quality of sample is plotted. Quality of different items of a sample are not plotted on the control charts. Value plotted on the chart which represents the quality of a sample is in the form of a dot, circle or cross. (c) Sample Numbers. The samples which are plotted on a control chart are numbered. The lines is normally placed at the end of the chart. These samples are also called subgroups in SQC. Normally 2025 subgroups are used to construct a control chart. (d) The Horizontal lines. A control line represents the average quality of the sample. Then there is a Upper Control Limit (UCL) which is above the average quality lines and is obtained by adding a to the average, i.e., Mean + 3 SD. Lower Control Line (LCL) is drawn by subtracting 3o from the average i.e., Mean —3 SD. Normally the central horizontal line representing the average is drawn as a thick line and the UCL and LCL are drawn as dotted lines. Figure 13.2 shows the theoretical or statistical basis for control charts. UCL = + 30
LCL = p.— 3a
Fig. 13.2
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QUALITY MANAGEMENT
The 36 limits were provided by Dr. W.A. Shewert. If a variable X is normally distributed, the probability that a random variable will lie betweenµ ± 36 is 0.9974 and the variable lying outside these limits is only 0.0026 which is very low. Hence, if we use 3a limits and the random variable is assumed to be normally distributed, the probability of a sample point falling outside these limits is extremely small if the process is in control.
Example 13.1. A weighing machine gives the packets of a given weight. Ten sample of size five each were checked and the weights were found to be as follows : Sample No. Mean Range
1 49 6
2 42 5
3 38 7
4 45 4
5 47 8
6 45 6
7 37 5
8 51 4
9 46 7
10 44 6
Calculate the values for the central lines (mean) and control limits for the mean chart and the range chart. For n = 5, d2 = 2.326, d3 = 0.8649. Solution. From the information which is given in the question X=
K

=
10
(49 + 42 + 38 + 45 + 47 + 45 + 37+51 +46+44)
= 43.6 1 1 R = K K=— (6+5+ 7+4 +8 + 6+5+4+ 7+6) 10
= 5.8 The 36 limits can be computed as Central Line = X = 43.6 3 x 5.8 UCL = X + 3R = 43.6 + = 43.6 + 3.346 = 46.946 d2 n 2.326v 5 LCL = X
3R = 43.6 — 3.346 = 40.254 d2 VI/
For Rchart: Central Line = R = 5.8
•
UCL = R+
d2
=5.8+
2.326
= 5.8 + 6.463 = 12.263 LCL = R
3d3R = 5.8 — 6.463 d2
= —ve value Sample 1, 5, 8, 3 and 7 in the mean chart are beyond the control limits.
OPERATIONS RESEARCH
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Example 13.2. A bulb manufacturing company ABC samples the fused bulb, taking sample of 5 each every hour. These samples sets of five have been arranged in increasing orders as follows : 45
42
20
35
43
52
61
20
16
70
65
60
28
100
85
75
68
46
25
55
52
70
65
25
75
65
82
69
57
75
70
32
40
110
95
94
77
70
87
78
60
80
90
55
65
115
100
109
88
92
87
85
79
120
110
65
85
160
1120
140
Construct a control chart for mean and the range. Is the production under control ?
Solution. For finding the control limits, we have to determine the sample mean, i.e.,
= 1 RT K ' and sample Range
= R = K IR.
This can be done by tabulating all the 12 samples as follows : Sample No.
Sample observation
1 2 3 4 5 6 7 8 9 10 11 12
45 42 20 35 43 52 61 20 16 70 65 60
Total
68 46 25 55 52 70 65 25 28 100 85 75
75 65 82 69 57 75 70 32 40 110 95 94
77 70 87 78 60 80 90 55 65 115 100 109
8 92 87 85 79 120 110 65 85 160 110 140
IX
Sample Mean
Sample Range
353 315 301 322 291 397 396 197 234 555 455 478
70.6 63 60.2 64.4 58.2 79.2 79.2 39.4 46.8 111 91 95.6
43
IR = 858.6
= 1 X = — L Xi 858.6 = 71.55 K 12 R=
K
ERi = — 697 = 58.08 12
50 67 50 36 49 49 69 69 90 45 80 IR = 697
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MANAGEMENT
For X chart: Central line = 71.55 UCL = X + A2 R = 71.55 + 0.577 x 58.08, for n =5 = A2 = 0.577
= 71.55 + 33.51 = 105.06 LCL = X — A2 R = 71.55 — 33.51 = 38.04. It can be seen that sample number 10 falls outside the control limits. • 110 —
UCL
105.06
100 — 90 — Sample Mean
80 — 70 — 60 —
Central Line
50 — 40 —
38.04
LCL
30 — 20 — 10 — 1
2
3
4
5
6 7 Sample Number
8
Fig. 13.3 For Rchart: Central line = R = 58.08 UCL = D4 R = 2.115 x 58.08 = 122.83 LCL = D3 R = 0 The control chart for R is shown in Figure 13.4.
9
10
11
12
OPERATIONS RESEARCH
598
A 130 —
UCL
121.97
120 — 110 — 100 — 90 — Sample Mean
80 — 70 — Control Line
60 — 50 — 40 — 30 — 20 — 10 —
1
2
3
4
5
6
7
8
9
10
11
12
Sample Number
Fig. 13.4 Process variability is under control. WI REVIEW AND DISCUSSION QUESTIONS I
1. What do you understand by Quality ? What is the difference between inspection and quality ? 2. Briefly explain the historical development of quality. 3. What are the causes of product quality variations ? 4. What are the traditionally used quality control techniques ? Discuss some of these with examples from business and industry. 5. Describe various kinds of inspection. Discuss certain principles which can be used as guidelines for carrying out inspection. 6. What is the concept of Total Quality Management ? How is it different from isnpection and quality control ? 7. What do you understand by SQC ? What is the necessity of SQC in industry ? 8. What are the various charts which are used for process control in SQC ? Discuss the relative advantages and disadvantages of use of control charts. 9. What is a control chart ? Discuss the underlying principle in drawing of control charts. 10. Explain the examples how control charts help in controlling the quality of manufacturised products.
599
QUALITY MANAGEMENT
11. Explain the basis of drawing control charts for X and R. What are the basic assumptions for construction of such charts ? 12. What do you mean by pchart ? Discuss the construction of pchart with all samples are of same size. 13. What are the conditions in which pchart or ccharts are used ? Examine the basis and approximation involved in the calculations. 14. What do you understand by the sampling inspection plans ? How can you use these in controlling the quality of a manufactured product ? 15. Prepare an appropriate control chart from the following data : Date of Month (Nov. 2003)
No. of defects
Date of Month (Nov. 2003)
No. of defects
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
20 32 36 40 32 42 40 32 45 38 41 68 70 80 45
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
32 81 44 28 38 68 55 45 30 32 45 38 40 38 28
Assume that the subgroups from which the defectives were taken were of the same size, i.e, 100 items each. [Hint. p = 100 whre d is number of defectives in sample size 16. 20 samples each of size 10 were checked for specifications. The number of defective parts
found in each of the samples are given below : Sample No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
No. of defectives
1
0
1
3
8
0
3
7
1
0
0
3
1
2
0
0
2
1
0
1
Construct the number defective chart and interpret the chart.
Investment AnaFlysk and BreakEven Anallyshjs II WI!Kimmel tii
• • • • • • •
II VA
Understanding the concept of money value change with time Learning the present value and discounting needs Knowing the importance of investment decisions Investment analysis—understanding the mechanics of investment criteria Understanding risk and investment relationship Learning the decision tree approach Highlighting the concept and use of the process of BreakEven analysis
INTRODUCTION All important decisions of the enterprises, whether they are business entities or hospitals, educational institutions or government departments, are based on the investment of funds in longterm assets and other activities with the basic objective of making profits. Peter F. Drucker has given certain objectives of business organisation and one of the objectives he has listed is to make profit. As a matter of fact no organisation can survive or is viable unless it makes legitimate profits, unless it is charitable organisation for which funds come from elsewhere. Investment is acquiring of facilities (machines, buildings, furniture.) which can be used in production and future financial gains are expected by their use. Such investment may not always be tangible like equipment but could be in the form of training and development. The investments, generally very large amounts of money are made at one point of time and the gains that occur are at another point of time in future. Since, once the funds are committed it is not possible to put them to other use immediately, and the funds involved are large, investments have major impact on the functioning of the organization. Any organisation is scarce of funds as there are number of requirements which have to be met, efficient utilization of available financial resources is of utmost important. If the investment of funds is not done prudently and funds are committed which do not get good revenue, the organization will suffer a financial setback and may not be
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
601
able to continue its operations. Thus, investment decisions involve the most efficient investment of the financial resources in longterm activities in anticipation of the expected flow of future benefits over a long period of time. The ultimate aim of making decision is to increase the wealth of the company, thus investment decisions effect the value of the firm. These decisions are not easy to make as the funds are limited and the alternatives to which funds have to be allotted are many. Whether the firm should acquire new equipment which can improve its productivity or increase the number of production lines with the same type of additional equipment; how much should be spent on raw material, manpower cost, on repairs and maintenance of plant, research and development and so on. We shall be dealing with different problems relative to outcomes of financial decisions. All such problems need detailed analysis as uncertain future outcomes like profits over a long period of time depend upon a large number of variables like government policies, legal framework, availability of funds, technical requirement and constraints, social responsibility and so on. Organisation use two sources for funding, internal sources and external sources. Future benefits are considered in terms of cash flows, outflow and inflow. Internal sources, like retained earnings are not included in this. External sources are in the form of debt capital or equity capital. Investment analysis involves the following : (a) All investment proposals available to the company must be considered. A list of all the alternatives should be prepared and no proposal left out at this stage. (b) Costs and benefits of each of the above proposals must be analysed in detail by using a suitable investment analysis technique. Only that proposal can be selected which makes financial sense at this stage, other consideration like social responsibility, etc. may still force the management to weigh in favour of other proposals at a later stage. (c) Estimates of total amount of funds must be made in a realistic manner, some committed funds which are expected to be released and have uncertainty associated with it cannot be considered available funds for use in the present project. PHASES OF INVESTMENT DECISIONS Investment decisions are basically capital budgeting decisions. The process of capital budgeting involves the following steps: Step I. Identification of investment opportunities These could fall under two categories —investment opportunities for a new business activity or investment opportunities for an ongoing enterprise. Ideas for new proposals or projects can come from many sources, depends upon the entreprenuer's attitude, risk taking ability and operating environment. Brain storming sessions or other techniques of ideas generation could be used to identify investment opportunities. This is the fist phase before the analysis of investment opportunities is carried out. Step II. Collection of relevant information Any decision depends heavily on the information. More the information about the opportunities that have been identified, better the decision. A plan for the implementation of the project muss be formulated and the cash inflows and outflows estimated. All possible inputs required foi implementing a project must be taken into account and only relative outcomes projected failini which the information will be misguiding and the decision will not be a good decision.
602
OPERATIONS RESEARCH
Step III. Establishing the discount rate Cash flows will be meaningful only if they are discounted properly. Cost of capital must be selected carefully. Step IV. Cash flow analysis The cash flow which have been estimated in step II must be analysed. For this purpose, techniques of capital budgeting are used. Step V. Decision making After the above steps, it is possible to take a decision with the inputs provided in these steps. However, the final decision is not only based on the cash flow analysis, which in any case is a major factor for consideration. Other aspects like the impact of the decision on overall health of the company, social acceptability, government support, legal remifications, how this particular proposal fits in the overall vision, philosophy and mission of the company. That is why the investment decisions are not simple decisions, for large companies huge investment could have major impact on the profitability of overall organisation. Step VI. Implementation of the selected proposal A project on a drawing board and on papers looks entirely different than the same project on ground. Something which appears very simple on papers, may be extremely difficult to physically achieve. When feasibility study of the project is carried out, even the most experienced consultant will miss out on something or make very different estimates than what practically may turn out to be. After detailed implementation plans are worked out, responsibilities of implementation have to be allotted to individuals and groups. Time involved in implementing the project varies with the type of project, technological process involved and availability of resources. Setting up a new plant, expansion of existing plant and replacement of machinery, all obviously have different time frames for implementation. Step VII. Project evaluation After the project is made operational, it must be evaluated to assess whether what was planned is being achieved or not. Suitable monitoring system to correct any deviations and overall assessment of the investment decision is an important phase of the investment decision, though it is often neglected. As discussed earlier, investment decisions relate to the selection of assets in which funds will be invested by a firm. The assets which can be acquired fall into two broad groups. (a) Longterm assets which will yield a return over a period of time in future. (b) Shortterm or current assets, defined as those assets which in the normal course of business are convertible into cash usually within a year. Accordingly, the asset selection decision of a firm is of two types, the first category of assets is known in the financial management literature as capital budgeting, the second category is the one with reference to current assents or shortterm assets and is popularly designated as working capital management. (a) Capital budgeting: This long investment decision is the most crucial financial decision. (b) Working capital management: This is concerned with management of current assets. It is an important and integral part of financial management as shortterm survival is prerequisite to
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
603
longterm success. One aspect of the working capital management is the trade off between profitably and liquidity. (c) Financing decision: The investment decisions are basically concernd with the asset mix, or the composition of the assets of a firm. The concern of financing the decision is with the financing mix or capital structure or leverage. The term capital structure refers to the proportion of debt (fixed interest sources) and equity capital (variable — dividend securities/sources of funds). (d) Dividend policy decisions: The dividend decision should be analysed in relation to other financing decision of the firm. One significant element in the dividend decision is the dividend pay out ratio, i.e., what proportion of net profits should be paid out to the shareholders. FACTORS INVOLVED IN INVESTMENT DECISIONS Investment decisions, the most important a firm takes are influenced by a large number of parameters like the environment in which these decisions are made, the politico — legal, social and technological changes, which take place in the society. The following factors have impact direct or indirect on investment decisions: 1. Fast changing technology: This is one factor which has major impact on the capital investment decisions. Initially while setting up a new plant and processing new machinery, a dicision has to be taken to buy a machine which will not become obsolescent or obsolete for at least a particular period. Since the vary fast changes in technology supports innovative processes and operations, the likelyhood of the machines and process getting outdated very fast is very high. Suitable cost of new equipment as also the salvage value of the old equipment must be taken into account. 2. Competition: Staying ahead of the competitor in technology and other strategies adopted by them is the crux of survival of the firm in a highly competitive environment. A company will perish if it does not keep abreast with the latest techniques and technology adopted by its competitors. Hence, it is a race between the strategies of the companies which they adopt to outsmart each other. Investment decisions often get dictated by the competitors in the same industry/market. 3. Future market potential: Though demand forecasting is extremely difficult and at best it may give some approximations (estimate, even when obtained by techniques like OR/simulation etc.) of what to expect in reality at a future point of time. If there is a potential for a future demand of a product the investment decision will go in its favour since business organisations are dynamic in nature, they will have to take a decision even under condition of uncertainties. 4. Management policies: Ultimately, it is the management which has to take decision to invest funds even when the project has a great profit potential in future. Progressive and modern managements take risk and encourage R & D as also invest in innovate products as they can feel the pulse of the market for improved/convenient to use products. However, because of the market share they have, they may not like to take risk. Certain products also have limited scope of using latest technology whereas some other can adopt the latest technology with ease. Management in all such cases, means the top management which is involved in making decisions. 5. Government policies: Governments in different countries play the role of an entrepreneur or a promoter or a regulator. Depending on the fiscal policies of the government, the investment
604
OPERATIONS RESEARCH
decision will vary. Policy regarding tax holiday, rebates on setting up industry in backward rural areas, special concessions for export oriented units, rebate on new investments, etc. have direct or indirect favourable or adverse impact on investment decisions. 6. Cash flow Budget: Cash flow budgets are prepared to analyse the availability of funds for the purpose of investment. The budget indicates the cash inflows at a particular point of time, a factor that will help the company in deciding when to invest and how much. 7. Expected returns: No decision involving finances can be made without considering the returns expected. Unless the expected returns over a period of time override the investment, no prudent manager will commit the funds in that project. The company must evaluate the future returns before making an investment decision. 8. Noneconomic factors: There are certain factors which do not directly effect the return or other benefits which will accrue from the investment decision. However, they can have major impact. Improving the workplace in a factory or providing the doctor, or installing music or keeping the workplace neat and clean, can improve the motivation of workers, reduce their absentism and hence increase the productivity resulting in increased profit. Installing a new machine which reduces reworks and rejects will give more confidence to the worker in machine and he may be more committed to his work. Time Value of Money
This concept is drawn from the simple fact that money has time value. A rupee today has greater value than rupee tomorrow . It is utmost essential for a financial manager to understand this concept and apply the same very objectively in all the decisions he makes. It can be easily seen that all business decisions or decisions involving personal life involve time value of money consideration. When one puts money in the bank and gets a particular return in terms of interest he understands that the money when received will have lesser value than when it is put in the bank. The main objective of the financial management should be maximization of wealth of the shareholders and thus among other things depends on the timings of cash flow. The recognition of the time value of the money and risk is extremely vital in financial decisionmaking. If the timing and risk of cash flow is not considered then the firm may make decisions which will be maximized only when the net wealth or net present value is created from making a financial decision. Let us now look at one of the most important principles in all financial management, the relation between rupee one today and rupee one tomorrow. From common sense and practical point of view it is known that rupee one, two years from now is less valuable than rupee one, a year from now. COMPOUND INTEREST AND TERMINAL VALUES The notation of compound interest is central to understanding the mathematics of finance. The term itself merely implies that interest paid on loan or an investment is added to the principal. As a result interest is earned on interest. This concept can be used to solve a class of problems illustrated in the following examples. Let us consider the case of person who has Rs. 100 in a bank account. If the rate of interest is 5% compounded annually, how much Rs. 100 will be worth after one year. At the end of the year, future value of Rs. 100 would be
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
605
= Rs. 100 + Rs. 100 x 100 = Rs. 100 + 5 = 100 (1 + 0.05) = Rs. 105 For a deposit of two years the future value would be = Rs. (105 + 105 x
100
= Rs. (105 + 5.25) = 100 (1 + .05)2 = Rs. 110.25 For a deposit of 3 years the future value would be = Rs. 1110.25 + 110.25 x 100) = Rs. (110.25 + 5.51) = Rs. 100 (1 + .05)3 = Rs. 115.76 It can also be looked at a different way that Rs. 100 grows to Rs. 105 at the end of the first year, if the interest rate is 5 percent and when we multiply this amount (i.e., Rs. 105) by 1.05 we get Rs. 110.25 at the end of the second year. Multiplying this amount (Rs. 110.25) by 1.06 we get Rs. 115.76 at the end of the third year and so on Similarly, at the end of the year the future value is Terminal value = TV„ = X0 (1 + r") where X0 = Initial amount or capital r = rate of interest n = number of years at the end of which we want to know the future value. It is obvious from the above equation that greater the interest rate r and greater the value of n, the greater is the terminal value. For our example of Rs. 100 at the interest rate of 5% for a period of 5 years the terminal values are shown in the following table. Period n(years)
Capital (Rs.)
Interest earned (Rs. @ 5% at capital)
Terminal value (TV) Rs.
TV (Rs.)
1
100 105 110.25 115.76 121.54
5 5.25 5. 51 5.78 6.07
105.00 110.25 115.76 121.54 127.61
100 = (1 + .05)1 100 = (1 + .05)2 100 = (1 + .05)3 100 = (1 + .05)4 100 = (1 + .05)5
2 3 4 5
Using the above equation, we can derive the TV or future values. The table below shows interest rates of 5 to 10 per cent on principal value of Rs. one. Year 1 2 3 4 5
10/0 1.0100 1.0201 1.0306 1.0406 1.0510
2%
3%
1.0200 1.0404 1.0612 1.0824 1.1041
1.0300 0.0609 1.0927 1.1255 1.0593
4% 1.0400 1.0816 1.1249 1.1699 1.2167
5% 1.0500 1.1026 1.1576 1.2155 1.2763
OPERATIONS RESEARCH
606
If we look at the 5% interest rate column, note that the terminal values (TV) shown for Rs. 1 invested at the compound rate for 1 year, 2 years and 3 years corresponds to our earlier calculations, if we multiply it by 100. As the interest rate rises the proportional increase in TV becomes greater as the number of years increases say from 3 onwards. It can be shown that Rs. 1 deposited today will be worth only Rs. 2.7048 after 100 years at interest rate of 1 %. But if the interest rate is 15 % Rs. 1 at the end of 100 years will amount to Rs. 11,74,313.45. This shows the wonder of compound interest. In the above examples, we have compounded the interest annually as this is easy to calculate. Let us also understand the relationship between TV and interest rates if compounding is done half yearly or quarterly. Continuing with the same example of capital of Rs. 100 and the interest 5 rate of 5%. Suppose this interest is paid every six months i.e., per six months the interest is — % 2 The TV at the end of six months will be .05 TV1 = Rs.100 (1 + —= Rs. 102.525 2 2 at the end of one year it will be lz TV1 = Rs.100 I + *(15 = Rs. 105.062 2 J We can see that if the interest was paid annually as in our earlier example, after one year Rs. 100 had become Rs. 105 but if the interest is paid half yearly Rs. 100 becomes Rs. 105.062. This increasing of Rs. 0.062 can be attributed to the fact that Rs. 102.525 also earns interest at the rate 5 of — = 2.5% . Hence it can be concluded that more times the interest is paid in a year, greater TV 2 at the end of a given year. Suppose interest paid is in times a year, the general formula will be r TV„ = Xo I 1+— — Let us work out the TV if in the above example the interest rate was paid quarterly 05 4 xl TV1 = 100 (1 +
4
JI
= 105.094
which is higher than amount (Rs. 105) and semiannual (Rs. 105. 062) interest compounding. The TV at the end of years with quarterly payment of interest will be .05 r2
TV2 = 100 (1 +
4
= 121.55 05 2 x2
But if it was paid semiannually TV2 = 100 (1 + 2 These can be compared in the following table X0 = Rs. 100, r = 5%
= Rs 110.38
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS Period
607
Terminal value (TV) in Rs. Compounding yearly
Compounding 6 half yearly
Compounding quarterly
105 110.25 115.76
105.062 110.38 115.96
105.094 121.55 217.92
1 2 3
Greater the number of years, greater the difference in TVs obtained by different methods of compounding. As it approaches infinity mn r Lt (1+ — = e where e = 2.71828 approximately. in>co m or
TV„= xoem" For the above example, for 3 years period, TV3 = 100 e3x°.°5 = 100 (2.718)0.15
Present value of money concept
We have seen how much a particular amount of money will grow to at given rate of interest (and mode of interest payment) in given period of time. But we may like to know how much of money should be saved and put in bank at a given rate of interest (and mode of interest payment) so that we get a specific return after a specific period of time. Suppose one wants to use Rs. 1000 one year from now and wants to know how much of money should be put in the bank now at the interest rate of 5 per cent so that he gets Rs. 1000 after one year. Then PV = Present value S = Amount payable in n periods of time n = Number of years or time period r = Interest rate S1 =PV(1+r)orPV=
1 1+r
For our example
1000 = PV x 1.05 1000 PV = Rs. = Rs. 952.38 1.05 It means if he deposits Rs. 952.38 now at the rate of interest of 5% paid annually, he can get Rs. 1000 after one year. Similarly, PV of an amount to be received 2 years from now. PV 
2 (1+ 02 in our above example if we want Rs. 1000 at the end of two years 1 PV = 1000 = Rs. 907.44 (1.05) Thus, it can be seen Rs. 1000 two years from now has lower present value than Rs. 1000 one year from now.
OPERATIONS RESEARCH
608
Similarly, the PV of Rs. 1 to be received at the end of n years is 1 (1 + r)" • We can calculate the PV of say Rs. 1 at the end of 4 years at the interest rate of 10%. 1 1 = = 0.683 PV (1 + 040)4 (140)4 PV=
It means that one rupee which he gets after 5 years at interest rate of 10% is worth paise 68 today. Present value tables are available which avoid tedious calculations PV of Rs.1 is given in the tables for different rate of interest and for different periods of time. This is called the discount factor. Using PV tables, we can find out the present value for any series of future cash flow. Present Value of an Annuity
A series of even cash flows i.e., same amount being received say at the end of every four years is known as annuity. Suppose Rs. 1 is to be received at the end of each of the next three years, PV can be calculated using say 10% discount rate or taken from PV tables. PV to be received after one year = 0.90909 PV to be received after two years = 0.82645 PV to be received after three years = 0.75131 Discount factor for PV of the series of 3 years = 2.48685. If we have a series where the cash flows are every three years, it is not necessary to go through all the calculations. The discount factor of 2.48685 can be applied straight way. Present value tables for even series are also available and appropriate discount value can be picked up from the tables. This fact can be put in formulae which are given in the table below : R = the periodic payment to be received after a particular period. Let n = number of payments of the time period after which payment is to be received. r = rate of interest. It can be seen that after (n1) period the first payment would be received with accumulated interest, the second payment would carry interest for (n2) period, etc. The last payment is made at the end of time period and will not carry interest. Period / number of payments
I
R Amount invested (principal) (n1) Time period for which interest will be accrued R(1 + r)'" Amount at the end of period.
II
R (n2) R(1 + /4
III...
(n1)th
nth
(n3) ... R (1 + r)n3 ...
R 1 R(1 + r)
R 0 R
If we add up the amount in the last column, we will get amount of ordinary annuity of Rs. R per period for n time period at the rate of r per period. Hence, S=R(1 +r)nl +R(1 +r)i2 +R(1+r)i3 +...+R(1+r)+R. Multiply RHS and LHS of the above equation by (1 + r) S(1 +r)=R(1+r)n+R(1+0111+R(1+r)"2+...+R(1+02+R(1+r)
609
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS Subtracting the first equation from this equation, we get S (1+ r)  S = R (1 + r)" R S = R[(1 + r)"  1]
or Amortizing A Loan
Installment loans is very common feature of repayments in financing of car, and other consumer loans. What is the payment required to be paid under an installments type of loan is an important use of PV concept. The loan is repaid in equal installments monthly, quarterly or annually and the repayment includes both the principal amount as also the interest accrued on it. This can be illustrated with the following example. Suppose a person has borrowed Rs. 22000 loan at 12 per cent interest to be paid over the next 6 years. Equal installments payments are required annually, these payments should be such that the principal amount of Rs. 22000 and interest at the rate of 12 per cent is paid back. 6 Rs. 22000 = X (1.12)t Consulting the PV tables we can find the discount factor for 6 years annuity with a 12 interest rate, it is 4.1114. 22000 = 4.114 x or
x = 22000= 5250.9 or Rs. 5351 4.114 It means that repayment Rs. 5351 per year will completely amortize loan of Rs. 22000 in 6 years. Amortization schedule for the above example is shown below in the form of a table. Year end 0 1 2 3 4 5 6 Total
Installment amount (Rs.) — 5351 5351 5351 5351 5351 5351 32106
Principal amount to be paid at year end (Rs.) 22000 19289 16253 12853 9044 4778 0
Annual interest at the outstanding amount (Rs.) — 2640 2315 1951 1542 1085 573 10106
Principal amount paid — 2711 3036 3400 3809 4266 4778 22000
Internal Rate of Return (IRR) or Yield
IRR is that rate of interest that can equal the present value of the expected cash outflow with the present value of the expected inflow. Let the rate of interest be r then. [ At —0 tt (1 + r)"1
OPERATIONS RESEARCH
610
where At is the net cash flow whether outflow or inflow, n is the period when the cash flow is expected or
Ao =
A„ Al + A2 2 + + (1 + r) (1 r) (1 +
So, r is the discount rate that discounts the future cash flow, Al to A„ so that it is equal to the initial outlay, Ao at time to. Let us illustrate this point with an example. Suppose there is an investment opportunity involving Rs. 20000 which is expected to prove cash inflows of Rs. 6000 at the end of the year for next 4 years. The problem can be presented as Rs. 20000 =
6000 6000 6000 6000 + + (1 + r) (1 + r)2 (1 + r)3 (1+ r)4
We have to select such a discount rate or rate of interest for which Rs. 6000 will give the present value of Rs. 20000. This can best be done by using computer. However, manually it can be done by assuming some arbitrary discount rate of say, 16% and 14%, determine the discount factors and see how much is the present value of Rs. 6000. 16% Discount rate. 3.1272 discount factor. PV of Rs. 6000 = Rs. 1873.2, but we are interested in getting this value as close to Rs. 20000 as possible. Hence select discount rate 14%. Discount factor = 3.4331 PV of Rs. 6000 = 20598.6 Discount rate is more than 14% but less than 16%. How much, this can be found by determining the difference in two PV values and then by interpolating a close enough value. 14 % = Rs. 20598.6 16% = Rs. 18763.2 2% difference = 1835.4 1% difference = 917.7 0.5 differences = 458.8 14.5 % = Rs. 20598.6 — 458.6 = Rs. 20140. Example 14.1. An individual purchases furniture from XYZ company on installment basis. He pays Rs. 20000 which is to be paid at the time of taking delivery of the furniture and agrees to pay the balance amount in 3 yearly installments of Rs. 10000 each payable at the end of the year. If the interest rate is 7% compounded annually, what would he pay, if he had the ready cash with him ? Solution. Cash down price of furniture = cash down price at the time of delivery of furniture + PV of annuity of Rs. 1000 for 3 years at 7% annually compounded interest. = Rs. (20000 + PV) PV
R [1 — (1 — "
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
= 10000
611
1— (1 + .07)s 0
= 1000 [1
(1 — 16 ) 1 1 = 1000 0.07 ) (1.07)'
18.4 x 10000 = Rs. 26285.7 7 Hence required cash down price of furniture = 20000 + 26285.7 = Rs. 46285.7 Example 14.2. A marketing expert has recommended the following for increasing the sale of tractors among the rural population. The total cost of the tractor Rs. 1200000.
"No down payment, pay only Rs. 25000 per month for 48 months, interest rate of 9% per annum one of the lowest ; compounded monthly on balance amount". Calculate how much will a customer pay for the tractor and how much for the interest ? Solution. Let us assume that the cost of the tractor is s. If amount s was invested at 9% per annum compounded monthly to yield an annuity of Rs. 25000 per month for 48 months ; the purchaser of the tractor would have paid back the complete amount. It can be seen from this how the cost of the tractor is PV of such an annuity. PV
[1 — (1 + r) n r
R = Rs. 25000
n = 48 r=
PV = Let
0.9 = 0.075 12 25000 11 (1 + 0.75) 48 } 0.075
x = (1.075) 8 log x = — 48 log 1.075
Find the value of x from here and use this to find the PV. Evaluation of NPV method
NPV method provides the following major benefits. (a) It provides the interpretation of business activities if it is properly understood. If the investment proposal is acceptable, positive NPV can be interpreted at immediate increase in the firm's wealth and it is equivalent to unrealised capital gains. A second interpretations of NPV being positive is that it represents the maximum amount a firm would be ready to pay for purchasing the opportunity for making investment or the amount at which the firm would be willing to sell the right to invest without being financially worseoff. (b) The most important advantage of NPV is that it recognises the value of money (c) All cash flows over the entire period of the project are taken into account. (d) It is consistant with the major objectives of the firm of maximising the wealth of the owner.
612
OPERATIONS RESEARCH
The NPV method suffers from the following limitations: (a) The ranking of projects by this method is not independent of the discount rate chosen for the analysis. If we consider two projects costing a particular amount and if we calculate the NPV at different discount rates, it can be easily proved that the ranking of the projects by this methods is dependent on the discount rate adopted. (b) Though the method is simple in concept, it is difficult to be used as it involves the use of complex calculations which may need the help of computers or advanced calculaters. (c) NPV method assumes that the discount rate is known. Discount rate is a firm's cost of capital, it is difficult and complex to understand and workout. (d) Project ranking irrespective of NPV may depend on the investments to be made, project with higher NPV may not be desirable if it also requires a large investment. (e) NPV method may give misleading outcomes if the life of the project is also taken into account. The alternative with higher NPV may involve larger economic life to the point that it would be less desirable than an alternative having a shorter life. Example 14.3. A project costs Rs. 20000 and is expected to generate cash inflows of and espected to generate Rs. 10000, Rs. 8000 and Rs. 7000 over its life of three years. Calculate the IRR for the project. Solution. Let us assume 20% rate of interest and workout the NPV of cash inflows
Total
Years
Cash inflows (Rs.)
Discount factor at 20%
NPV (Rs.)
1 2 3
10000 8000 7000
0.833 0.694 0.579
8330 5552 4053
NPV = Rs. 17935 Less projects cost = Rs. 20000 NPV = — Rs. 2065
This indicates the rate of interest chosen i.e., 20% is very high, now let us work out NPV with 18%, 16% and 14% to increase the total NPV and bring it as close to Rs. 25000 as possible. Year
Cash inflow (Rs.)
Discount factor (DF) 18%
NPV (Rs.)
DF 16%
NPV (Rs.)
DF 14%
NPV (Rs.)
1000 8000 7000
0.847 0.718 0.609
8470 5744 4263 18477
0.862 0.743 0.641
8620 5744 4487 19051
0.877 0.769 0.675
8770 6152 4725 19647
1
2 3 Total
Less cost 2000 Total
NPV = —1523 NPV = — 0949 NPV = — 353
It means we have to choose a still lower rate of interest. IRR Method Acceptance/Rejection Rule
The project should be accepted if the IRR is higher than or equal to the minimum required rate of return k i.e., r k, k is to firm's cost of capital and is also known as cutoff rate or hurdle rate. And if the IRR is lower than its cost of capital i.e., r k.
INVESTMENT ANigrlYSIS AND BREAKEVEN ANALYSIS
613
IRR may be interpreted as that higher rate of interest an enterprise is prepared to pay on the funds that it borrows to finance the project without being adversely affected by paying back the total (interest & principal) out of the cash inflows generated by the project. IRR is also referred to as the 'Breakeven' rate of borrowing from the bank. Evaluation of IRR Method
IRR method has the following major advantages. (a) Like the NPV ; IRR method also considers the present value of money which is an important parameter for investment which give the results in future. (b) Cash inflows over the entire lifecycle of the project are evaluated. (c) The management understands the concepts of rate of return on capital much better and the figures in % are more meaningful and acceptable to them. (d) In NPV method the cost of capital or discount rate is expected to be known. However, in this method even if the cost of capital is not known, IRR suggests the maximum rate of return and gives a reasonably good idea regarding the possibility and probability of the project. (e) It is consistent with the owners objective of maximing his wealth. Though IRR method is considered theoretically superior to the NPV method, it suffers from the following limitations. (a) The method involves complicated calculations and is tedious to understand and use. (b) In this method, it is assumed that cash inflows during different time periods of the project are recommended at the internal rate of the project. In NPV method, it is assumed that the intermediate cash inflows are reinvested at the firm's cost of capital. NPV method of reinvestment is definitely more logical and hence more appropriate. (c) The two methods i.e., IRR and PV may give different outcomes depending upon the (ii) initial project cost and (z) expected lives of projects (iii) time periods of receipt of cash inflows. In (d) certain situations, the IRR method may provide absurd results. Profitability Index (PI) Method
This is another method of evaluating projects for investment. It is also called B/C (Benefit Cost) ratio. Here the ratio of present value of cash flows at required rate of rectums to the initial cash outflow of the investment is found out. The following relationship is used B/C ratio or PI = PV cash inflows Initial cash outflow
Vf=1(1 Pt+ r)t It can be easily seen from the above relationship that project should be accepted if its PI or B/C ratio is greater than 1. Such projects will have a positive NPV. Example 14.4. The initial cash outflow of a project is Rs. 200000 and it generates cash inflows Rs. 120 000, Rs. 80000, Rs. 50000 and Rs. 30000 in four years. Determine NPV and PI of the project at 10% discount.
OPERATIONS RESEARCH
614
Solution. NPV of the cash inflows for four different years at the discount rate of 10% are shown in the table below. Year 1 2 3 4
Cash Inflow (Rs.) 120000 80000 50000 30000
Discount Factor(DF) 0.909 0.8626 0.751 0.683
Present value (Rs.) 109080 66080 37550 20550
Total = Rs. 233200 Less outlay = Rs. 200000 NPV = Rs. 33200 233200 =1166 200000 PI(net) = 1.166 — 1.0 = 0.166 Example 14.5. XYZ has Rs. 1500000 to invest, the company is evaluating a number of projects but details of three of them are given below. The cost of capital for the company is 15 %. PI =
Project 1 2 3
Initial outlay (Rs.) 500000 200000 100000
Annual cash Inflow (Rs.) 150000 100000 40000
Life of the project (years) 8 10 20
You are required to rank the above projects on the basis of (a) NPV method (b) Payback method (c) PI method. Value of annuity of Rs. 1 received is discounted at the rate of 15% as 8 years = 4.6586 10 years = 5.1790 20 years = 6.3345. Solution. (a) NPV method. The projects are ranked in terms of higher value of NPV, highest as 1. Pro .j ect 1 2 3
Initial out— lay (Rs.) 500000 200000 100000
Annual cash inflow (Rs.) 150000 100000 40000
Life of project (years) 8 10 20
PV factor at 15% 4.6586 5.1790 6.3345
PV (Rs.) 698790 517900 253380
NPV 198790 317900 213380
(b) Pay Back Method. Project 1 2 3
Initial outlay (Rs.) 500000 200000 100000
Annual Cash inflow (Rs.) 150000 100000 40000
Payback period (Years) 3.33 2.00 2.5
Rank III I II
Rank III I II
INVESTIMONT ANALYSIS AND BREAKEVEN ANALYSIS
615
Lowest pay back period project will be ranked first. PI Method. The highest PI project will be ranked first.
1
Initial out— flow (Rs.) 500000
Annual cash inflow (Rs.) 150000
2 3
200000 100000
100000 40000
Project
Project life (years)
PV factor 15 %
PV (Rs.)
PI
Rank
8 10 20
4.6586 5.1790 6.3345
698790 51790 253380
1.397 2.589 2.533
III I II
Analysis of Project Rank
1. 2. 3.
Method
Project 1
Project 2
Project 3
NPV Pay Back PI
Third Third Third
First First First
Second Second Second
Example 14.6. ABC Ltd. is considering three capital projects for three years, 2004 to 2006. The company's cost of capital is 12%. The expected cash inflows of the projects are
Project X Y Z
2004 (80000) (50000) (40000)
2006 50000 25000 15000
2005 40000 20000 10000
Figures in brackets represent cash out flow. All projects are flexible enough and capital investment can be adjusted to suit the availability of funds. The capital available is limited to Rs. 220000 in years 2004. But there is no problem of funds beyond 2004. PV factors are as follows: Year PV factors
2004 1.00
2005 0.89
2006 0.80
Determine which project should the company select for implementat 'on. Solution. Let us work out the present value and PI of the these projects Project
X Y Z
2004 (80000) (50000) (40000)
2005 35600 44500 35600
2006 40000 20000 12000
NPV (Rs.) — 4400 14500 7600
PI
Example 14.7. A company XYZ Ltd. is considering two projects on the recommendations of a management consultant. It wants to evaluate the projects on the basis of NPV method and IRR method. The cost of capital is 10%. The cash inflows of the two projects are as given below . Year
Project A Project B
1 70 160
Cash inflows are in '000 of rupees.
_
2 80 20
3 90 10
4 100 5
OPERATIONS RESEARCH
616 The discount factors for 10 % and 20% over four years are as follows DF 10%
1
2
3
4
0.91
0.83
0.75
0.68
20%
0.83
0.69
0.58
0.48
Calculate NPV and IRR of the both the projects and justify which project should be selected by the company. Assume the initial out lay and life for both the projects to be 200000 and four years, respectively. Solution.
(a) NPV and IRR of Project A
Year
Cash Inflow (in'000 Rs.)
DF (10%)
PV (in '000 Rs.)
DF (20%)
PV (in '000 Rs.)
1
70
0.91
63.7
0.83
58.1
2
80
0.83
66.4
0.69
55.2
3
90
0.75
67.5
0.58
52.2
4
100
0.68
68.0
0.48
48.00
Total
265.6
213.5
NPV
200.0
200.0
+ 65.6
+13.5
IRR = 10+
65.6x 10 = 10 + 12.59 = 22.59% 65.6  13.5
(b) NPV and IRR of project B
Year
Cash Inflow (1000 Rs.)
DF (10%)
PV (in '000 Rs.)
DF (20%)
PV (In Rs.)
1
160
0.91
145.6
0.83
132.8
2
20
0.83
16.6
0.69
13.8
3
10
0.75
7.5
0.58
5.8
4
5
0.68
3.40
0.48
2.40
Total =
173.10
154.80
26.9
 45.2
NPV = 200.00 IRR = IRR = 10+
 26.9 x 10 = 10 + ( 14.69)  26.9 + 45.2
=  4.69 % It is obvious that project A should be selected.
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
617
Investments under Risk and Uncertainty
In earlier discussion, we have not taken risk and uncertainty into account as we have been discussing the cases with expected cash flows in future. The quantum of cash outflows and inflows at a future point of time and the intervals of time at which these expenditures will be incurred or incomes accrued are extremely important for the enterprise to do well to meet its objectives. However, in real life situations there are so many uncertainties involved because of number of factors not always within control of the management that if we assume certain cash inflows and outflows the results could be misleading. It is a well known fact that risk and returns have direct relationship. Risk must be taken into account while making decisions. The following techniques are used for risk adjustment for investment decisionmaking. Certainty Equivalent Approach
In this approach unlike in the IRR method, the risk element is compensated by making suitable adjustments in cash flows rather than in the discount rate. Certainty equivalent coefficient (ar) is calculated and the cash flows of each year are multiplied by this coefficient to adjust the risk of uncertain cash flows. This is an approximation method where the coefficient is determined based on the attitude of the management towards risk and uncertainly. If the management has two proposals and cash inflows of one are known with certainty, then the management compares the two, one without certainty of returns and another with uncertainty and hence risk of return. Take the case of a company considering two proposals where one with an investment of Rs. 10 lakhs does not have assured returns but another with investment of 8 lakhs has guaranteed returns. If these two proposals are ranked at par or equal, then the certainty coefficient is 0.8 for the period. a=
Assured/certain cash flow at time t Risky cash flow at time t
This coefficient is inversely proportional to the level of risk involved i.e., if the risk is more the coefficient is smaller value between 0 and 1. a's are determined for various time periods, and are then multiplied by the risky cash flows to get equivalent of assured or certain cash flow. After the risk element has been taken into account, NPV can be determined by using usual (risk free) rate of discount to make a more optimal decision. NPV = initial cash outflow c2, c3...c„ = future cash flow without risk adjustment a2, a3...a„ = Certainty equivalent coefficient for first year, second year and so on r = discount rate Example 14.8. A company wants to use the certainty equivalent approach to evaluate two proposals A and B for making capital investments. The risk free borrowing rate is 10%. Use the following information to decide which proposal should be selected. Year 0 1 2
3
Proposal A Cash flow (Rs.) (80000) 60000 30000 12000
a
t 1.30 0.85 065 0.65
Proposal B Cash flow (Rs.) (60000) 50000 20000 10000
at 1.0 0.95 0.65 0.70
OPERATIONS RESEARCH
618
Solution. Let us find out the NPV of two proposals. (a) Proposal A Years
Cash flow (Rs.)
at
(1) 0 1 2 3
(2) (80000) 60000 30000 20000
(3) 1.0 .085 0.70 0.65
Total (b)
Adjusted cash flow (Rs.) 4=2x3 (80000) 51000 21000 13000
PV Factor 10%
Total PV (Rs.)
(5) 1.0 0.909 0.8284 0.7513
6=4x5 (80000) 54540 17354.4 9767
NPV = (800000 + 801661.3) = + Rs. 1661.3
Proposal B 0 1 2 3
(60000) 20000 20000 10000
1.0 0.65 0.70 0.70
(60000) 13000 13000 7000
1.0 0.8264 0.8264 0.7513
(60000) 10743.2 10743.2 5259.1
Total
NPV = 63027.3 — 60000 = 3027.3 Since NPV of proposal B is higher, this proposal should be selected. Statistical Approach
This method is also named after Fredric Hillier. The degree of a risk of a proposal is measured by using standard deviation (a) of the NPV distribution. The mean of the present values cash flows and their standard deviation is calculated. Cash flow distribution is used to calculate the mean and a as illustrated below. (i) Let us consider an investment proposal where probability over the number of years of the project life is known, has mean values c1, c2, c3......c„ and standard deviations equal to ap 62, 63f• • vGii• Then
NPV = +
cl (1 + r)1 (1 + r)2 ". (1 +
ct = c1, E2 , , „ are the expected monitory values. or
= ci Pi + cz P24, • •• • • where c1, c2, etc., are the cash flows value after one year, two years and so on and pi, P2,...., p„ are the corresponding probabilities. (ii) Now the variance of NPV or 62 has to be determined. In case of nonrelated cash flows n.2 ,,2 6°2 62=6z+ 60 + + (1 + 1.)2 (1+ r)4 (1+02" where
Z't
at = [(c1 — •,)
2
2 (C2 — 4)
2 (c„ — 4) ]
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
619
In case of perfectly related cash flows an (1 + )"
02
60 + (1 + r) (1 + r)2
Example 14.9. A company manufacturing garments is considering to purchase an automatic sewing machine costing Rs. 25000. Probable cash flow during the three year life of the machine are given in the form of a table. Year 1 Cash flow 8000 10000 20000
Probability 0.2 0.5 0.3
Year 2 Cash flow Probability 8000 0.10 1000 0.8 20000 0.1
Year 3 Cash flow 8000 10000 20000
Probability 0.10 0.85 0.05
If the discount rate is 8% calculate the NPV and standard deviations of the investment opportunity, assuming that the cash flows of each year is independent of the other year. Solution. (a) To compute the NPV, the expected cash flows must be worked out first. Year I Cash Probability flow(Rs.) c, x p, PI Ci
Year II Cash flow (Rs.) c,
8000 0.2 1600 10000 0.5 5000 20000 0.3 6000 Expected cash flow el = 12600
8000 10000 20000
Year III
Probability p2
c2 X P2
Cash flow (Rs.) c3
800 8000 2000
8000 10000 20000
0.1 0.8 0.1 U2 = 10800
Probability 0.1 0.85 0.5 E3 = 10300
NPV Year 0 1 2 3
Expected cash flow (Rs.)  1., ?2 , — 25000 126600 10000 10300
Total expected
PV factor @ 8%
Present Value (Rs.)
1 0.926 0357 10.794
— 25000 11667.6 9255.6 8178.2
NPV = 29101.4 — 25000 = 4101.4.
Standard Deviation. 2 al =
(Ci — cl )2 Pi
Using the above formula let us calculate the variance a2
c3 x p3
P3
800 8500 1000
OPERATIONS RESEARCH
620 Year 1.
2.
3.
Cash flow (Rs.) ci 8000 10000 20000 8000 10000 20000 8000 10000 20000
Probability pi 0.20 0.50 0.30 0.80 0.80 0.10 0.10 0.85 0.05
(c1 — .e.1)2
(c1 — F1)2 pi
(8000 — 12600)2 = 2316 x 104 (10000 — 12600)2 = 676 x 104 (20000 — 12600)2 = 5476 x 102 (8000 — 10800)2 = 784 x 104 (8000 — 10800)2 = 64 x 104 (20000 — 10800)2 = 8464 x 104 (8000 — 10300)2 = 529 x 104 (10000 — 10300)2 = 9 x 104 (20000 — 10300)2 = 9409 x 104
231.6 x 104 388 x 104 1624.8 x 104 627.2 x 104 6.4 x 104 846.4 x 104 52.9 x 104 7.65 x 104 470.45 x 104
a? = (231.6 + 388 + 16428)104 = 2262.4 x 104 cyi = 1480 x 104 = 531 x 104 Year 1 2 3
Variance (a2) 2262.4 x 104 1480.0 x 104 531 x 104
PV Factor 0.926 0.857 0.794
Example 14.10. XYZ Ltd. is under lot of pressure from its competitor and is considering to add a new product by investing 600000. The management feels that it will be able to generate the following cash per year standard deviation is also known. Year 1 2 3 4
Expected cash flows (Rs.) 250000 200000 150600 150000
S.D. (Rs.) 12000 10000 8000 6000
Use the discount rate of 8%. Determine the mean of present value and standard deviation of its distribution. Solution. (a) Expected NPV Year 0 1 2 3 4 Expected NPV
Cash flow (Rs.) (600000) 250000 2500000 160000 150000 = 56190
PV factor at % 1.000 0.926 0.857 0.794 0.835
Present value (Rs.) (600000) 232500 171400 127040 125250
621
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS (b) SD Year 1 2 3 4
Variance o2 14400 x 104 1000 x 104 6400 x 104 3600 x 104
at 2.8. 4 12000 10000 8000 6000
PV factor 0.926 0.857 0.794 0.835
(i 1334.4 x 104 857 x 104 4841.6 x 104 3006 x 104
452 = (1334.4 + 857 + 4841.6 + 3006)10 = 10049 x 104 Standard Deviation
= a = V10049 x 104 =102.2 x 102
= 10220. Example 14.11. An NRI is keen to set up some projects for the benefit of his ancestral Village, he is considering investment of Rs. 5000000. He has consulted some management experts who have given him the following information for two mutually exclusive projects. Project A 3500000 2 Years 2500000
Initial cost Life of project Cash flow each year
Project B 1500000 2 years 1000000
Probability cash flow in each year for the two proposals are : Cash Flow 3000000 3500000 4000000
Probability Project A 0.1 0.7 0.2
Cash flow 1500000 1000000 1200000
Probability Project B 0.1 0.8 0.1
If the cost of capital is 12% which project should the NRI select ? Solution. We have to calculate the mean and variance of each proposal. This is calculated in the table below : Project A. Cash Flow ct
Probability pt,
Ct ph 2
(ct, — Et )
(c1 —et )2 x pt,
3000000 3500000 4000000
0.1 0.7 0.2
300000 2450000 8000000
500000 0 500000
250 x 108 0 500 x 108
_ 1 ct = — (3000000 + 3500000 + 4000000) 3 = 3500000 a? = 375 x 108 Also
= — (300000 + 2450000 + 800000) 3 = 1183333
622
OPERATIONS RESEARCH
Project B. 1500000 1000000 1200000
0.1 0.8 0.1
150000 800000 120000
266667 Assume 266000 233000 (233333) 33333 (33330)
70.75 x 108 734.3 x 108 1.11 x 108
ct = 1233333. Assumption has been made to simply the calculation
cl= —12 (150000 + 800000 + 120000) = 356667. Expected inflows and variances are calculated as given below: Year (t)
at
0 1 2
(3500000) 3500000 1233333
PV factor for 10% 1.00 0.893 0.797
PV of c, '
PV factor Project A (1 + r)2' variance
(3500000) 3125500 982966 GA =
1.00 0.797 0.635
pv. (Rs )
'
Project B variance
0 0 0 375 x 108 298.8 x 108 168.72 x 108 375 x 108 238 x 108 168.72 x 108
PV (Rs.) 0 134.5 x 108 107.1 x 108
(298.8 + 238)108
= 23.16 x 104 = 231600 GB =
J(134.5 + 107.1)108 =1554x104
= 155400. Example 14.12. A new manufacturer of automobile parts and accessories is considering manufacture of either car air conditioners or steering column. The returns are independent of each other. His experts have worked out the following expected returns and variances: Year 1 2 3 4
Expected returns (Rs.) 200000 150000 300000 260000
Variance of returns (Rs.) 40000 50000 20000 30000
On the other hand, returns, if he manufactures car steering columns have an expected value of 220000 and variance of 25000 each year. Also, returns are dependent on each as it will largely depend upon the success of the new car in the market. The manufacturer wants to invest only Rs. 600000 in each of the projects. Also the cost of the capital of the firm is expected to be 8%. You are required to determine the expected value and variance of NPV of both the projects. Solution. The return on car air conditioners is independent of time. The following table shows the expected NPV and the standard deviation.
623
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
Year (t)
c, (Rs.)
0 1 2 3 4
600 x 104 20 x 104 20 x 104 30 x 104 20 x 104 Total
PV factor 12 % 1.000 0.893 0.797 0.712 0.636
PV of c, (Rs.) 60 x 104 178600 119550 213360 165360 77110
ut2
(1 + 0.12r
0 40 x 103 50 x 103 20 x 103 30 x 103
1.000 0.797 0.635 0.506 0.404
(1 + 0.1212' x at2 0 31.88 x 103 31.75 x 103 10.12 x 103 12.12 x 103 85.87 x 103
Mean of PV distribution of return = Expected NPV = Rs. 77110. SD of the PV distribution, a= [E(1 + r)2t
= J8587x103 = 926
Project manufacture of steering column In this case the returns have interdependence over the period of the process SD is given by 6=
al 1+r
+ a2
(1 + 02
an
(1 + 0"
The calculations of expected NPV are as follows : Year (t)
4 (Rs.)
PV factors 12%
Present value of et (Rs.)
0 1 2 3 4
60 x 104 22 x 104 2 x 104 22 x 104 22 x 104 Expected NPV
1.000 0.893 0.797 0.712 0.636
 60 x 104 19.6 x 104 17.53 x 104 16.66 x 104 13.99 x 104 67890
PV of a, at
et (i. + o .12)t at
0 160 160 160 160
0 142.85 128 113.96 101.71 486.57
BREAKEVEN OR COST VOLUME PROFIT (CVP) ANALYSIS Costvolumeprofit analysis is a technique to give the management an overview of the volume of product which the company must sell so that no losses are incurred and the sale volume at which the profit objective of the firm can be achieved. An organisation would like to find the ideal combination of costs incurred and the volumes sold so that it can maximize its profits. CVP analysis can be a very useful technique for the management to understand the implication of fixed and variable cost, volume and related selling price for overall profit plan of the company. CVP analysis can answer the following questions any organisation is interested in finding the answer of : (a) What might be the maximum volume of sales so that company does not suffer any loss ? (b) What is level of sales or sale volume to achieve the target of profit the firm has set for itself ?
624
OPERATIONS RESEARCH
(c) How are the profits changed with change in price, cost and volume ? (d) Should the company wind up its operations ? If yes, at what point of time ? As it can be seen this tool can help the management in planning their profits for all range of products over a period of time. Breakeven analysis is one form of the CVP analysis and so widely used that both the terms CVP and Breakeven are used interchangeably by many. Breakeven is a special way of CVP analysis where the relationship between costs incurred, volumes of products to be produced and profits achieved can be presented in a simple and easy to understand manner. A breakeven analysis involves the study of following factors: 1. Fixed and variable costs: Fixed costs are such which do not change with the volume of production and volume of sales. Such costs include capital assets which the company uses for its operations. Fixed costs depreciate over a period of time. Variable costs are such costs which will change depending upon the volume of production. If the production increases the cost of raw material expenditure on electricity, repair and maintenance, taxes, etc., will increase and when the production decreases these will decrease. Sometimes, a category of semivariable costs is also talked about. These are such costs which can be broken down to either fixed or variable. 2. Sales price: This is important as change in sales price will change the profit. 3. Contribution Margin: It is defined as the excess of selling price over variable cost or contribution margin = Total sales — total variable cost. or Contribution margin = Fixed cost + profits. Breakeven analysis establishes a relationship between revenue generated and costs incurred with different volumes. A point at which revenues and costs are equal at a particular sales level is called breakeven point or in other words breakeven point may be defined as "point of sales volume at which total revenue generated is equal to total costs incurred". It implies that beyond this point, the company will make profit but below this point the company will suffer losses. Assumptions: Determining the breakeven point presupposes the following:. (a) Costs can be separated as fixed and variable costs. (b) Semivariable costs are ignored and clubbed into either the fixed costs or the variable costs. (c) Sales price of the product is constant and hence the curve of total revenue is a linear curve. (d) Increase in variable costs is at a constant rate, hence the total cost curve is linear in nature. (e) There is no change in technology during the time period under consideration. (f) There is no change in the labour efficiency during the time period under consideration . Determining the Breakeven Point
It must be clearly understood that the breakeven or no profit no loss point only gives an idea to the firm of how much must they produce and sell so that they do not fall into the loss trap. However, the significant aspect of the CVP analysis is to study the effects of change in costs, volume and prices on profits. Two approaches to determine breakeven point are adopted. Breakeven Formula F = Fixed cost/period
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
625
V = Variable cost per unit P = Selling price per unit Q = Number of units produced and sold. Then total revenue generated = Q x P Total cost = F + VQ Profit = QP — (F + VQ) At breakeven point profit is zero or total revenue is equal to the total cost Hence, QP = F + VQ QP — VQ = F or Q (P—V) = F or Q(units) =
F Fixed cost P — V Selling price per unitvariable cost / unit
It can be seen from the above formula that the selling price per unit should be greater than the variable cost per unit if the breakeven point has to be a positive or a rational number of units. The three situations are (a) P > V but P # V selling price should be greater than the variable cost for a positive break—even point. (b) If P < V will give negative break even point i.e., negative sales volume will result which has no meaning. (c) If P = V, no breakeven point is possible unless the firm has zero fixed cost F, under this situation, every sales volume will give a breakeven point because at this point revenue will be exactly equal to total sales cost at any volume of sale. Breakeven Point (rupees) =
F V 1—— P
Fixed costs Total variable costs 1 total revenue
Concept of Margin of safety Margin of safety is expressed in terms of sales. It is the excess of actual sales over the breakeven sales volume. Margin of safety = Actual sales — Breakeven sales. or
Margin of safety ratio =
Actual sales — Breakeven sales x 100 Actual sales
P/V or contribution ratio
V In the above formula of breakeven point in units and in rupees the denominator 1— — or 1
Variable cost . is called the P / V ratio or contribution ratio. sales (revenue) Breakeven point (rupees)
Fixed costs P/V ratio
OPERATIONS RESEARCH
626 Breakeven chart or costvolume graph
We have calculated the breakeven point using mathematical formula. It is also possible to depict it in the form of graph or chart. It can help us in visualizing the extent of profit or loss for different levels of sales at a glance. In other words, it is a pictorial view of the relationship between costs, volumes of sales and profits incurred or loss suffered. In graph the point at which total cost line and total sales line (representing revenue) intersect, is called the breakeven point. Drawing of Breakeven Graph Let horizontal axis indicate the output volumes and vertical axis the revenue generated. The changes in output can be shown as changes in costs, revenue and profit. Total cost and total revenue curves are drawn as linear straight profits. Total cost and total revenue curves are drawn as linear lines as per the assumptions made. Linear total revenue curve means as the volume of sale or output increases the revenue keeps increasing. Similarly, linear total cost curve is because of the assumption that variable costs change at constant rate. Let us take the following example to illustrate the drawing of the graph assuming the following data. Estimated sales (20,000 units @ Rs. 10) Variable material cost Direct material cost Factory overheads Miscellaneous over heads Contribution Fixed Costs Factory overheads Administrative overheads Net Profit
Rs. 200000 Rs. 30000 Rs. 50000 Rs. 20000 Rs. 20000
Rs. 30000 Rs. 20000
120000 Rs. 80000
Rs. 50000 Rs. 30000
The following steps are involved in preparation of breakeven graph. Step I. Draw the horizontal axis and mark sales (rupees) or units along the Xaxis. On Xaxis, rupees, units or percentage of capacity can be plotted. Equal distances may be marked to indicate different sales volumes at different levels. Step II. Draw the vertical or Yaxis and revenue and fixed costs and variable costs are represented on this axis. Step III. Total sales line can be drawn by joining the two points, zero sales, at zero volume i.e., origin. The other point is the estimated sales point. (in this case Rs. 200000 point on Yaxis.) Step IV. Total cost line is drawn from the point of fixed costs in years and the point of fixed cost Rs. 50000 point is joined with the total cost point in Rs. (50000 + 120000) point. It may be noted that if the vertical and horizontal axes are spaced equally with the same distance, the sales line will connect the two opposite corners of the graph at 45°. The point of intersection of the sales line and total cost line is the breakeven point.
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
627
20 18 1 17
Profit Zone
Variable Cost Line
170000
16 B/E Point a) 14 c a) i, )
CC
0
"c' 12 2
1 "/
ne
C CC C 1 0
Variable Cost
(I) c 0
C.)
Fixed Cost Line Margin of Safety Loss Zone
t11 /1//11
I
I
I
I
I
I
2
4
6
8
10
12
14
16
18
20
Sales (Units, Rupees) % of capacity Fig. 14.1
The angle formed by intersection of sales line and the total cost line is known as angle of incidence. It can be seen larger this angle, lower the breakeven point and viceversa. The area to the right of the B/E point is called the profit zone and to the left of B/E point is the loss zone. The variable cost is shown as the gap between the fixed cost line and the total cost line as marked in the graph. Margin of safety, as seen earlier is the excess of actual sales over the breakeven sales. Margin of safety is expressed as a ratio. Actual sales — B / E sales Actual sales The margin of safety indicates up to which level the sales can fall before the breakeven point is reached beyond which the firm will suffer loss. A firm will be more safe if the margin of safety is more. Margin of safety ratio =
Merits and Drawbacks of Breakeven Analysis Merits
1. Accounting data can be represented in the form that it can be understood and interpreted easily. 2. The management can use this analysis to understand how to reach a point beyond which they will generate profits and other safeguards against suffering losses. 3. Helps in deciding the products mix. 4. Increasing of capacity and shutting down of certain facilities can be planned.
OPERATIONS RESEARCH
628 Drawbacks
1. It is difficult in most of the situations to segregate fixed and variable costs, hence assumptions are made which result in approximation and not optimal results. 2. The analysis becomes complicated if it is to be used for multiproducts. 3. Takes into account only the production costs and doesn't take into account the selling costs. 4. Useful only in short run and not very long periods of time. Examples 14.13. A manufacturing firm produces only one product whose selling price is Rs.50 per unit and the variable cost per unit works out to be Rs. 10. The fixed costs of the company workout to be Rs. 200000. Find the breakeven quantity and the sales volume. Solution. B/E point (quantity) = B/E (sales) =
F = 200000 P — V 50 — 10
5500
F 200000 5 = = 220000 x — = Rs. 275000 1—P1 — 4
i.e., The B/E quantity 5500 at selling price of Rs. 50 per unit gives 55000 x 50 = Rs. 275000 sales. Example 14.14. The fixed cost of a firm is estimated as Rs. 100000. It manufactures one product whose variable cost is Rs. 30 per unit and the selling price is Rs. 50 per unit. Calculate the breakeven point of the firm in units, rupees and as percentage capacity. Solution. B/E point (numbers) = prupee) — B/Epoint (
F = 100000 P — V 50 — 30 F 1—
=
5000
100000 100000 x 50 = 250000 30 = 20 1 —( 50 )
% capacity if the firm is assured to have capacity of 10000 units. Then % capacity5000 = x 100 = 50%, i.e., the breakeven would be reached at 50% of the 10000 capacity of firm. Example 14.15. A manufacturing firm produces only one product whose selling price is Rs. 100 per unit and variable cost is Rs. 60 per unit. The annual fixed cost of the company works out to be Rs. 160000. Also, the company has estimated capacity of Rs. 8000 units. Find out the B/E point in units, rupees and as percentage of capacity. What is the margin ? Solution. F = Rs. 160000 P = Rs. 100 V = Rs. 60 160000 = 4000 units B/E (units) = F P — V 100 — 60 F = 160000 = Rs. 40000 B/E (rupees) — 1—P 1 — (21loo)
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
629
B/E sales 40000 _50% capacity 8000 Margin of safety = Capacity sales — B/E sales = 8000 — 4000 = 4000 units
B/E (% of capacity) =
Example 14.16. Three firms A, B C manufacture the same product. The fixed cost of the firms are Rs. 160000, Rs. 220000 and Rs. 300000 respectively and the variable cost of the firms are Rs. 20, Rs. 15 and Rs. 10 respectively. The selling price of the product is same for all the three firms i.e., Rs. 25. Find out the B/E point for the three firms. If each of these firm sells 50000 units, what is the profit earned by them ? Determine the impact on their profits if their sales decrease by 10%. Solution. Firm A: B/E (units) =
F = 160000= 32000 P — V 15 — 20
Firm B: B/E (units) =
F 220000 = =22000 P — V 25 — 15
Firm C: B/E (units) =
F — 300000 — 20000 P — V 25 — 10
Profits for sale of 20000 units. From A = (Selling price/unit — variable cost/unit) = 50000 — fixed cost. = (25 —20) 50000 — 160000 = 250000 — 160000 = Rs. 90000 From B = (25 — 15) x 50000 — 220000 = 500000 — 220000 = Rs. 280000 From C = (25 — 10) 50000 — 300000 = 750000 — 300000 = 450000 If the sales decrease by 20%, i.e., now the sales are 50000 x 0.8 = 40000 units Profit of firm A = (25 — 20) 40000 — 160000 = Rs. 40000 Firm B = (25 — 20) 40000 — 220000 = Rs. 180000 Firm C = (25 — 10) x 40000 — 300000 = Rs. 300000 % decrease in profit Firm A =
9000000 — 40000 = 55.55 90000
Firm B =
220000 — 180000 40000 = =18.18 220000 220000
450000 — 300000 150000 = 33.33 450000 450000 Example 14.17. Suppose a manufacturing company has the following budget for the particular year: Sales (50000 units @ Rs. 60) = Rs. 30000000 Variable cost (50000 units @ Rs. 20) = Rs. 10000000 Fixed costs = Rs. 1000000 Net profit = Rs. 1000000 Firm C =
What shall be the impact on the profits of the firm if the company is forced to make the following changes: Increase in price = 25% Decrease in sales volume = 20%
OPERATIONS RES/ARCH
630
Increase in variable cost = 15% Increase in fixed cost = 5% Solution. P is changed from Rs. 50 to 60 x 1.25 = Rs. 75 Q is decreased by 20% = 50000 x 0.8 = 40000 units V is increased by 15% = 20 x 1.015 = Rs. 23 F is increased by 5% = 10000000 x 1.05 = 1050000 Profit = (Selling price/unit— Variable cost/unit) x Number of units sold —Fixed cost = (75 — 23) x 40000 — 1050000 = 2080000 — 1050000 1030000 — 1000000 = 2.91% 1030000 Example 14.18. An automobile engine manufacturing plant is manufacturing 5000 engines for a particular passenger car. The plant is operating at 80% of its capacity. % change in profit =
The sales return per year is Rs. 25000,000. The fixed cost of the plant is Rs. 2000000 and variable cost is Rs. 45000 per unit. The manufacturer proposes to use the spare capacity the plant to manufacture an important version of the engine. This will increase the fixed cost by Rs. 1000000, but would reduce the variable cost by 20%. Do you think the proposal is viable ? Also, workout the viability of the proposal if selling price is reduced by Rs. 200 per unit and if the plant is run at 85% of its capacity. Solution. Present capacity of the plant = 5000 units B/E = F = 2000000 = 4000 units. P — V 5000 — 4500 Profit = (P — V) x units manufactured — Fixed cost = 500 x 5000 — 2000000 = 2500000 — 2000000 = Rs. 500000 00 Full capacity of the plant = 5 0 = 6250 units 08 New fixed cost = Rs. 3000000 New variable cost = 4500 — 20% of 4500 = Rs. 3600 250000 Rs. 4000 6250 = New profit = (4000 — 3600) x 6250 — 3000000 500000 This proposal is uneconomical as it gives a loss of Rs. —500000. Second Proposal New selling price = 4000 — 200 = Rs. 3800 85% of B/E capacity = 4000 x 0.85 = 3400 units Profit = (3800 — 4500) x 3400 — 2000000 = — 4380000 New sale price = Rs.
This proposal results in heavy loss and hence is not acceptable. Example 14.19. Breakeven production of XYZ Ltd. is 8000 units of a product. The fixed cost is Rs. 64000 and the variable cost is estimated to be Rs. 2/unit . What is the price of the product ? How much the company should produce to earn a minimum profit of Rs. 56000 ?
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS Solution.
B/E units = 8000 F = 64000 V = Rs. 2/unit 8000 P = 80000 P = Rs. 10/unit Sales (units) =
631 64000 8000 = F P—VP—2 8000 P — 16000 = 64000
Fixed cost + Profit (Sale price/unit + variable cost/unit)
64000 + 56000 (10 + 2)
120000 = 10000 units. 12 Example 14. 20. The following data is available for company ABC Ltd. for one year. Net sales Cost of goods produced and sold 800000 Fixed cost 120000 Gross Profit 100000 Selling costs Variable cost 20000 Fixed cost 12000 Net profit 68000 Determine: (a) B/E point. (b) What is the profit ? Solution. Total fixed cost = 132000 Total values cost = 600000 132000= Rs. 528000 (a) B/E point (rupees) — Fixed cost P / V ratio 1 6000000 800000 (b) Contribution = Net sales — Variable cost = 800000 — 600000 = Rs. 200000 Profit = 200000 — Fixed cost = 200000 — 132000 = Rs. 68000 Example 14.21. Following information is available for ABC Ltd: Total fixed costs = Rs . 8000 Total variable costs = Rs. 12000 Total sales = Rs. 25000 Units sold = Rs. 4000 Calculate the following: (a) Contribution (b) B/E in units (c) Profit
OPERATIONS RESEARCH
632 (d) Margin of safety (e) Volume of sales to earn a profit of Rs. 12000. Solution. = Total sales  Total variable costs (a) Contribution
= 25000  12000 = Rs. 13000 or contribution/units Fixed cost 8000 = 2461.5 units = Contribution per unit 3.25
(b) B/E (sales)
B/E (sales  money units) = Fixed costs P / V ratio P/V = 25000 12000 = 13000 = 0.52 25000 25000 = 0.52 x 100 .= 52% x 100 = Rs. 15384.6 B/E (sales)  F = 8000 52% 52 (c) Profit = Total sales  Total costs = 25000  (8000 + 12000) = Rs. 5000 (d) Margin of safety' = Total sales  B/E sales = 25000 15384.6 = Rs. 9615.4 F +profit 8000 + 12000 20000 x 100 = 38461.5 P/V Ratio 52% 52 Example 14.22. The following information is available for a company: Fixed costs = Rs. 48000 Variable costs = Rs. 4/unit Selling price = Rs. 12/unit Sales = Rs. 120000 Calculate: (a) Profit if expected sales are Rs. 160000 (b) Sales if profit target of Rs. 100000 has been planned (c) Selling price/unit if B/E point is 4000 units. Solution. (a) Profit if expected sales are Rs. 160000 (e) Volume of sales (units) =
12 4 x 100 = 66.67% 12 P (for sale of Rs. 160000) = (160000 x 66.67%  48000) Rs. = (106672  48000) = Rs. 58672 P/V ratio =
PV
x 100 =
13000= 3.25 4000
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS
633
(b) Sales to earn a profit of Rs. 100000 F + Profit _(48000 + 100000) x 100 P /Vratio o 52 = 284616 units
Sales (units) = (c) Selling price for B/E of 4000 units B/E (units) at selling price of Rs. 12 —
Fixed cost 48000 = 6000 units Contribution / unit 12 — 4
48000 + 4 +V= B/E (unit) 4000 = 12 + 4 = Rs. 16 Example 14.23. Under the corporate house 'A', two plants X and Y are functioning at two different locations. The corporate house has been planning to merge the two plants for a number of administrative reasons. However, the management is extremely careful and wants to know the following before taking a final decision. =
(i) Capacity of the plant after merger for purposes of breakeven. (ii) Probability if the merged plant works at 80% of the merged capacity. Following particulars of the two plants are known: Capacity of operations Sales Variable costs Fixed costs
Plant A 100% Rs. 100000000 Rs. 50000000 Rs. 12000000
Plant B 75% Rs. 40000000 Rs. 20000000 Rs. 8000000
Solution. (i) P/V ratio =
Contribution x 100 Sales
Contribution at 100% capacity = (100000000 — 50000000) + — 1°° x (40000000 — 20000000) 75 = 50000000 + 26666666 = 76666666 66666666x 100 Sales 100 Sales at 100 % capacity = 100000000 + x 40000000 5 = 100000000 + 53200000 = 153200000 P/V ratio
P/V ratio
B/E (rupees)
76666666x 100 153200000 = 50% F 12000000 + 800000 P/Vratio = 50%
20000000 = 40000000. 50%
634
OPERATIONS RESEARCH
40000000 = 26.1% 153200000 (ii) Calculation if the capacity of merged plant to B/E at 80% sales 80 % sales at 100% capacity sale = 0.8 x 153200000 = 122560000 % capacity
100 x 20000000 ) 80 = 40000000 + 25000000 = 65000000 Contribution = 122560000 — 65000000 = 57440000 = Contribution—Fixed cost Profit = 57440000 — (12000000 + 8000000) = Rs. 37440000 Example 14.24. A manufacturer of plastic product has collected the following data about the three products he manufactures: = (8050000000 +
Variable costs
Product A
Selling price per unit
Variable cost per unit
Sales volume
25 18 20
20 12 15
25 % Total Sales 40 % Total Sales 35 % Total Sales
B C
If the total sales is Rs. 2000000 and the annual fixed cost Rs. 420000 you are required to determine (a) Breakeven point in rupees (b) The profit or loss at usual capacity of 85 %. Solution. A
% Sales 25
Volume (Rs.) 2000000
= Rs. 500000
B C
40 35
2000000 2000000
= Rs. 800000 = Rs. 2000000
Product B (Rs.)
Product C (Rs.)
Total (Rs.)
500000
800000
700000
2000000
500000 = 20000 15
800000 = 44445 18
700000 = 35000 20
2000000
20000 x 20 = 400000
44445 x 12 = Rs. 533333
35000 x 15 = Rs. 525000
500000 — 400000 =100000
800000 — 533333 = 266667
700000 — 525000 = 175000
541667
100000 x 100 500000 = 20%
266667 x100 800000 = 33.33%
775000 x 100 700000 = 25%
26.11%
Product A (Rs.) 1. Sales Volume (Rs.) 2. Sales (units) sales volume/selling price 3. Total variable cost sales volume x variable cost/unit 4. Contribution (PV) 5. P/V ratio P — V ) x 100
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS (a) B/E point sale (rupees) =
420000 =R s.1608579 P/V Ratio 26.11%
(b) Profit or loss at 85% of capacity 85 % capacity = 2000000 x 0.85 = Rs. 1700000 F + Profit P/V ratio
P
Profit = P x P/V Ratio — F Loss = 170000 x 26.11 — 420000 = 443870 — 420000 = Rs. 23870 Example 14.25. The following data is available for ABC Ltd: Direct material cost/unit = Rs. 800 Direction labour cost/unit = Rs. 420 Fixed overheads = Rs. 60000 Variable overheads 40 % on direct labour @ selling price/unit = Rs. 2400 Trade discount 5% Calculate: (a) B/E point (b) Net sales volume (c) Net profit for sales 30% above B/E volume (d) Net profit for sales 20% below B/E volume. Solution. (a) B/E point Total cost per unit = Rs. (material cost + labour cost + variable overheads) x = Rs. (800 + 420 + 40100420) = Rs. 1388 Contribution per unit = Sales (P) — variable cost (V) Net receiveable/unit sold = Selling price per unit — discount = 2400 — 5% of 2400 — Rs. 2280 Contribution/unit = Rs. 2280 — 1388 = Rs. 892. B/E (sales units) =
F =60000= 68 units Contribution/unit 892
(b) Net sales value at B/E point Sales Value = Units sold x Sales price/unit = Rs. 682400 = Rs. 163200
635
OPERATIONS RESEARCH
636 (c) Net profit if sales are 30% above B/E point sales B/E point sales = 68 units 30% above this = 68 x 1.3 = 85 units Sales value = Rs. 85 x 2400 = Rs. 204000 P/V ratio =
Sales value =
Contribution/unit 892 x 100 = x 100 = 37.16% Selling price/unit 2400 F + Profit P / V ratio
Profit = Sales value x P/V ratio — F = 204000 x 37.16% — 60000 = (75820 — 60000) = Rs. 15820 (d) Net profit if sales are 20% below B/E point sales B/E point sales = 68 units 20 `1/0 below this = 54.4 = 55 units Sales value = 55 x 2400 = Rs. 1, 32, 000 P/V ratio =
Contribution / unit = 37.16% (already determined) Selling price / unit
Profit = Sales value x P /V ratio — F = Rs. (132000 x
37.16 00 60000) 1
= Rs. (49051 — 60000) = — Rs. 10949 (a loss of Rs. 10949) Example 14.26. XYZ Ltd. deals with consumer household electronic goods. It has at present three types of working machines in stock. A marketing expert has advised the company to drop a particular brand and two more brands to increase their profits . The marketing expert has provided the following data to the management of XYZ Ltd. Washing machines stocked at present (Type) A B C
Selling price (Rs.) 10000 12000 16000
Variable cost (Rs.) 6000 8000 10000
Fixed costs = Rs. 400000. Stocking of washing machines as per proposal of marketing expert.
Sales during this year 500 400 3000
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS A B D E
10000 12000 10800 12500
637
6000 8000 5000 7000
400 600 800 1000
Would you recommend the store to stock new line of products ? Solution. Let us workout the net profit in the two cases. Net profit on the existing stocking pattern is shown below. Type
Sales (Numbers)
(Selling price  Variable cost) (Rs.)
Contribution (Rs.) net realisation/unit x Number of units sold
500 400 300
(10000 — 6000) = 4000 (12000 — 8000) = 4000 (16000 — 10000) = 6000
4000 x 500 = 2000000 1600000 1800000
A B
C
Net profit = Contribution — Fixed cost
Total Rs. 5400000
= 5400000 — 400000 = Rs. 5000000 Net profit on the new proposed stocking pattern Type
Expected sales (Numbers)
A
500
B D
400
E
800 1000
Net realisation (Rs.)
Contribution (Rs.)
4000 400 (10800 — 5000) = 5800 5500 Total
2000000 1600000 5800 x 800 = 4640000 5500000 Rs. 13740000
Net profit = Contribution — Fixed cost = 13740000 — 400000 = Rs. 13340000 This proposal of the marketing expert yields an extra profit of Rs. 83,40,000 and should be accepted. REVIEW AND DISCUSSION QUESTIONS
1. A credit company has issued a loan of the Rs. 50,000. Simple interest is charged at a rate of 10 per cent per year. The principal plus interest is to be paid at end of the third year compute the interest for the three year period. What amount will be repaid at the end of third year ? 2. A person gives Rs. 100000 to a corporation by purchasing a bond from the corporation. Simple interest is computed quarterly at the rate of 3 per cent per quarter and is paid at the end of each quater. Bonds expire at the end of 5th year and the final payment includes the original principal plus interest earned during the last quarter. Compute the interest earned each quarter and the total interest, which will be earned over 5 year life of the bonds. 3. Suppose that Rs. 1000 is invested in saving bank, which earns interest at a rate of 8 per cent per
year compounded annually. If all interest is left in the account, what will the balance in account after 10 year ?
OPERATIONS RESEARCH
638
4. A longterm investment of Rs. 250000 has been made by a medium sized company. The interest rate is 8 per cent per year and interest is compounded annually if all interest is reinvested at the same rate of interest, what will the value of investment be after 8 years ? 5. A young man has recently received an inheritance amount of Rs. 2000000. He wants to take a position of this and invest it of later years. This goal is to accumulate Rs. 3000000 in 15 years. If the money earns 10 per cent per year compound annually how much of money should be invested ? How much of interest will be over 15 years. 6. A lumpsum of money is invested at a rate of 10 per cent per year compounded quarterly. How long will it take the investment to double ? To triple ? To increase by 50 per cent ? 7. A person wishes to invest Rs. 50000 and wants the investment to grow to Rs. 100000 over the next 10 years. At what annual rate of interest the amount will have to be invested for this growth to occur, assuming annual compounding ? 8. M/s Batra & Co. has 500000 to invest the following proposals are under consideration. The cost of capital for the company is estimated to be 12 per cent. Project A B C D E
Initial out lay (Rs.) 120000 60000 40000 50000 45000
Annual Cost Flow (Rs.) 28000 18000 7000 12000 10000
Life of Project (year) 10 7 18 10 20
Rank the above project on the basis of : (a) Payback method (b) NPV method PI (Profitability Index) method. Present value of annually of Rs. 1 receiving is discounted at the rate of 15% 5 year = 4.5587,10 year = 5.1790,20 year = 6.3345. 9. M/s ABC Ltd. is considering five capital projects for the year 1999 to 2002. The company is financed by equity entirely and its cost of capital is 10%. The expected cash flows of the project are as follows : Year and cash flow (Rs.'000) Project A B C D E
1999 (60) (50) (40) — 70
2000 40 (30) (65) (40) 25
2001 40 50 65 50 45
2002 30 50 85 70 55
Figures in brackets represent cash outflow. Size of investment in the project can be reduced if necessary in relation to funds. None of the project can be delayed or undertaken more than once. Find out which projects ABC Ltd. should undertake if the capital available for investment is limited to Rs. 150000 in 1999 and with no limitation in subsequent year. PV factors as given below may be used.
639
INVESTMENT ANALYSIS AND BREAKEVEN ANALYSIS Year Factor 1.0
1999 1.0
2000 0.89
2001 0.80
2002 0.71
10. XYZ corporation is considering three alternative investments characterized by the data shown in the table below. Notice that the three investments have equal initial outlays, equal lifetime and equal rupee returns. Patterns of rupee return are different for the three investments. Initial Investment Cash Inflows Total Cash Inflows
180000 80000 80000 80000
180000 120000 80000 40000
180000 40000 80000 120000
Assume that cash inflows occur at the end of each year. XYZ corporation has a minimum desired rate of return on investment of 15 per cent. Determine the NPV of each of this investment and determine which meet the rate of return criterion. 11. A Ltd. has Rs. 1500000 allocated for capital budgeting purposes. The following proposals and associated profitability indexes have been determined : Project 1 2 3 4 5 6
Amount (Rs.) 350000 175000 300000 460000 460000 420000
Profitability index 1.24 0.98 1.11 1.21 1.18 1.28
Which of the above investment should be undertaken. Assume that the projects are indivisible and there is not alternative use of the money allocated for capital budgeting. 12. Three firms A, B and C manufactures the same product. The selling price for the product per unit is Rs. 15 for all the companies. The fixed costs of the companies are Rs. 160000, Rs. 350000 and Rs. 520000 respectively and the variable cost per unit as Rs. 10, Rs. 7 and Rs. 6. (i) B/E point (physical units) for each company. (ii) Profit of the companies of each of the them sells 200000 units. (iii) Percentage impact on profits of the firms increases their sales by 20 %.
15 Projecl Management BERT and CPM LEARNING OBJECTIVES • • • • • • •
Understanding the concept of project and project management The use of Networks for project management Basic terms related with Network Analysis Developing Simple network diagrams Identification of critical path and project duration Understanding PERT and different time estimates Crashing of projects.
INTRODUCTION
Programming Evaluation and Review Technique (PERT) and Critical Path Method (CPM) are two techniques used in project management. Project management is necessary to ensure that a project is completed within the stipulated budget, within the allocated time and perform to satisfaction. PERT was developed by US Navy in 1958 for managing its Polaris Missile Project. It is very useful device for planning time and resources of a project. Polaris Missile project involved 3000 separate contracting organizations and was regarded as the most complex project experience till that time. Parallel efforts, at almost the same time, were undertaken by Du Pont Company, which developed Critical Path Method (CPM) to plan and control the maintenance of chemical plants. These methods were subsequently widely used by Du Pont for many engineering functions. PROJECT MANAGEMENT Definition of terms commonly used in PERT and CPM Activity
Activity is the smallest unit of productive efforts to be planned, scheduled and controlled. It is an identifiable part of the project, which consumes time and resources. In fact, a project is a
PROJECT MANAGEMENT PERT AND CPM
641
combination of interrelated activities, which must be performed in a certain order for its completion. The project is divided into different activities by the work breakdown into smaller work contents. In network (arrow diagram) an activity is represented by an arrow, the tail that represents the start and the head, the finish of the activity. The length, shape and direction of the arrow have no relation to the size of the activity.
Fig. 15.1 Event
An event is an instant of time at which an activity starts and finishes. An event is represented by a node, i.e., 0. The beginning of an activity is Tail Event and finishing of an event is Head Event. Path
An unbroken chain of activity arrows connecting the initial event to some other event is called a path. Predecessor Activity
This is an activity that must be completed immediately before the start of another activity.. Successor Activity
Activity, which cannot be started until one or more activities are completed but immediately succeeds them is called successor activity of a project. Dummy Activity
As seen in the definition of activities, all activities take some time and resources. A dummy activity is the one which is introduced in the network for communication when two or more activities have the same head and tail events. It means that two or more activities share the same start and finish nodes simultaneously. A dummy activity takes no time and requires no resources. It is shown as a dotted line in Figure 15.2. Figure 15.3 shows wrong representation
Fig. 15.2. Correct Representation a Tail Event or Start Node Head Event or Finish Node
Fig. 15.3. Wrong Representation
642
OPERATIONS RESEARCH
Let us assume that the start of activity C depends upon the completion of activity A and B and the start of activity D depends only on the finish of activity. For this situation, we draw wrong and right representation in Figure 15.4 to understand the introduction of dummy activity in network. C
Dummy Activity
(a) Wrong Representation
(b) Correct Representation
Fig 15.4 NETWORK (ARROW DIAGRAM) A network is the graphical representation of logically and sequentially connected arrows representing activities and nodes representing events of a project. Looping
Sometimes, due to errors in network logic, a situation of looping or cycling error occurs in which no activity can be completed as all the activities of the network are interlinked. In such situations, there is need to reexamine the network logic and redraw the network. To understand looping, see Figure 15.5.
Ci
A
E
Fig 15.5 Activity B cannot start until activity D is completed and activity D depends on the completion of activity C but C is dependent on the completion of activity B. Thus activities B, C and D form a loop and the network cannot proceed. Such condition can be avoided by checking the precedence relationship of the activities and numbering them in a logical sequence. Dangling
In a network all activities except the final activity has a successor activity. A situation may occur when an activity other than the final activity, does not have a successor activity. The situation is shown in Figure 15.6.
643
PROJECT MANAGEMENT PERT AND CPM
•
A 0 B P
•
C
Dangling Activity
Fig 15.6 In must be remembered that except the first node and last node, all nodes must have at least one activity entering it and one activity leaving it. Construction of Networks
Construction of a network is a simple procedure of putting all the events and activities in a logical and sequential manner to meet the requirement of a particular project/ problem. Difficulty occurs only when the basic rules are ignored. The following steps are helpful in constructing the network: (a) Divide the project into activities by following the procedure of Work Breakdown Structure (WBS). (b) Decide the start event, and the end event of project for all the activities. This is called establishing the precedence order and is the most important part of drawing the network. (c) The activities decided by the precedence order are put in a logical sequence by using the graphical representation notations. Logical sequence can be decided by asking the following questions : (i) What are the activities that must be completed before the start of a particular activity ? (Predecessor Activities) (ii) What activities must follow the activity already drawn ? (Successor Activities) (iii) Are there any activities which must be performed simultaneously with a particular activity ? Rules to Construct a Network
1. Activities are represented by arrows3 and events are represented by circles 0. 2. Each activity is represented by one and only one arrow. The tail of the arrow represents the start and head the end of the activity. 3. Each activity must start and end in a node. 4. Arrow representing activities must be kept straight and should not be shown curved or bent. 5. Angles between arrows should be as larges as possible to make the activities clearly distinguishable from each other. 6. Arrows should not cross each other. 7. Event Number 1 represents the start of the project. There will be no activities (arrows) entering this node. 8. All events (nodes) should be numbered in an ascending order. 9. No events numbers can be repeated. 10. Dangling is not permitted.
644
OPERATIONS RESEARCH
11. Dummy activities also must follow the above rules, even though they do not consume any resource or time. Numbering of events
For numbering of the events, Fulkerson's Rule is very helpful. (a) Initial or start event, having no preceeding event is numbered 1. (b) Numbering of other events is done from left to right or from top to bottom as 2, 3, 4, etc. (c) The events, which has been numbered are ignored or deleted. This will result in new initial events; these must be numbered in ascending order. (d) Continue numbering all the events till we reach the last event out of which no activity (arrow) will emerge. It will be allotted the highest number, as it is the end event. The numbering of activities is illustrated with the help of Figure 15.7
Fig. 15.7 Skipping of event numbers
In large projects in which the activities run into hundreds, it is not always possible to list all the activities at the initial stage and some additional activities may have to be added as the project progresses. Hence, while numbering the events continuously as 1, 2, 3, 4...and so on, the events are numbered in gaps of 5's or 10's so that other events can be inserted without causing any inconvenience to the logic of the network. The first event may be numbered 5 and subsequent events may be numbered as 10, 15, 20 and so on. Example 15.1. Let us use a simple example to illustrate the procedure we have just learnt. Listed below is the precedence chart showing the activities, their precedence (sequence), etc., for the project, ' Launching a new product' Sequencing is very important part of the construction of a network. The precedence given below must be carefully understood, as this example will be used to draw the network at a later stage. Activity A B C D
Description Arranging a sales office Hiring sales persons Training sales persons Selecting advertising Agency
Immediate predecessor activity A B A
Time (weeks) 6 4 7
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Plan advertising campaign Conduct advertising campaign Design packaging of product Establish packaging facility Package initial stocks Order stock from manufacturer Select distributors Sell to distributors Transport stock to distributors
D E — G H, J — A C, K I, L
4 10 2 10 6 13 9 3 5
The logic of the predecessor activities for each activity listed in the above table should be understood properly. The project 'Launching a new product' can be broken down into a number of activities. The set of activities given in the table are one perception based on simple logic. Other such logic could also be developed. The students are advised to carefully study the precedence of the activities. Solution. Network diagram for the activities listed in the table is shown in Figure 15.8.
Start (0)
Finish (0)
Fig 15.8 Please see each activity carefully to understand the logic. The network is listed with 0 event and the activity has 0 time and is written as start (0). Arranging a sales office does not have any immediate predecessor activity. This is written as activity A with its time (6 weeks) written in the brackets as A (6) on top of the arrow. From node 1 there are three activities, which do not have any immediate predecessor activity, i.e., A (6), G (2) and J (13). This may be verified from the precedence table. Activity B, hiring of salespersons can only commence after arranging sales office (activity A) so activity B (4) is shown as arrow coming out of node 2. Also activity D (2) and K (9) can also start only after activity A has been completed and they are shown with arrows moving out of node 2. There is only one activity C (7), which can start after completion of B and is shown as leaving node another node 11 has been created and the finish activity moving out of node 10. Finish activity has 0 times as shown in the network diagram.
OPERATIONS RESEARCH
646 Example 15.2. Develop a network with the following data : Activity
Preceded by initial activity
Activity Time
A
NIL
4
B
NIL
6
C
A
10
D
B
15
E
B
10
F
C, D
8
E G Solution. The network is shown in Figure 15.9
12
Fig 15.9 The nodes or events have been numbered according to Fulkerson's rule. Activity timings have been shown in the brackets. Example 15.3. An automobile company manufacturing scooters has decided to come up with a scooter specially designed for the women only. The project involves several activities listed in the following table: Activity
Description
A
Study design of scooters in the market
Predecessor activity
Design the new scoter
A
C
Design the marketing programme
A
D
Design new production system
E
Select advertising media
F
Test prototype
G
Release scooter in market
D, E
Draw a suitable network. Solution. The network is shown in Figure 15.10
Fig 15.10
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Example 15.4. The characteristics of a project schedule are given below : S. No.
Activity
Time
1. 3.
1—2 2—4
6 1
5. 7.
3—5 56
9. 11.
8—7 8—9
5 8 2 1
S. No. 2. 4.
Activity 13 3—4
Time 4 2
6. 8.
4 —7 6—8
7 4
10.
7 —9
2
Construct a suitable network. Solution. The network is shown in Figure 15.11 7
8
4
Fig 15.11 Example 15.5. Draw a network diagram based on the following project schedule information available: S. No. 1.
Activity A
Immediate Predecessor Activity —
Time 2
2.
B
—
3. 4. 5
C D
A B
4 6 5
E
C, D
8
6
F
3
7
G
E F
Solution. The network is shown in Figure 15.12
Fig 15.12
2
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/
Critical Path and Activity Times
As explained earlier PERT is a very useful technique for planning the time and resources of any project. It is an eventoriented approach as it is mainly concerned with various events in a project. PERT deals with probability of completion of a project in particular time, as the time of various activities involved cannot be known accurately. It is only the time an activity is expected to take for completion, which can best be calculated. Expected time of completion of each activity can be found out form the following three timings : (a) Optimistic Time (b) The most likely time (c) Pessimistic Time. These three timings are based on Beta Statistical Distribution. Beta distribution is used as it is extremely flexible and can take on any form of activity and times that are associated in a typical project. Four typical Beta curves are shown below.
a
a
E.T (A)
E.T in (C)
b
a
a
in
E.T (B)
E.T (D)
Fig 15.13 It can be seen that the Beta distribution has finite end points like (a), the optimistic time and the pessimistic time and the Expected Time (ET) of the activity is limited between these two ends. Curve (A) is a symmetrical curve and the difference between the most likely time (m) and Expected Time (ET) is very small. Had the curve been exactly symmetrical, the firm line (m) and dotted line (ET) would be exactly the same. Curve B indicates a high probability of finishing the activity m and ET indicates that if something goes wrong, the activity time can be greatly extended. Curve C is something like a rectangular distribution. Here the probability of finishing the activity early or late is almost equal. Similarly, curve D indicates very small probability of finishing the activity early but it is more probable that it will take an extended period of time. The Expected Time (ET) can be calculated from the following formula : (b)
PROJECT MANAGEMENT PERT AND CPM ET=
649
+ 4m + b 6
Activity Times— Estimated Time After constructing a network reflecting the precedence relationship, we have to ascertain the time estimate for each activity. We must calculate ET for each activity using the above formula: ET=
a + 4m + b 6
Now the variance of the activity time has to be calculated. v2
b — (2)2 6)
Earliest Start and Finish Times
Let us take zero as the start time for the project, then for each activity there is an Earliest Start Time (EST) relative to the project starting time. It is the earliest possible time that activity can start, assuming that all of the predecessor activities are also started at their EST. In that case for that particular activity, its Earliest Finish Time (EFT) is EST + activity time. Latest Start and Finish Times
If we assume that the effort is to complete the project in as soon as possible time, this is the Latest Finish Time (LFT) of the finish activity or of the project. The Latest Start Time (LST) is the latest time when an activity can start, if the project schedule is to be maintained LST = LFT — activity time Finish activity has zero time, hence LST = LFT Slack. Slack of an activity can be defined as the difference between the Latest Start Time (LST) and Earliest Start Time (EST) or the difference between the Latest Finish Time (LFT) and Earliest Finish Time (EFT). This is the significance of slack or Total Slack Time (TST), that the TST for any activity must be used up. Critical Path
If we observe the network, we can see that there are a number of paths that lead to the finish activity, i.e., completion of the project. But the longest path is the most limiting path. This path is called the Critical Path. It can be easily determined by adding the activity times of all the activities on the largest path from start to finish of the project. Calculation of EST and EFT
These calculations can best be described with the help of a network. Let us draw a network as shown in Figure 15.14:
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0
(4) 8E(4) 12
Start (0)dik
‘ 170:,
O0 0
C (7)
017 22 0
10 17
I (5)
Finish (0)
J (4) 0 K (10) 0 22 26
26 36
36 36
•
Fig 15.14 This is the same network as was drawn in example 15.1 In the above figure the name of the activities are written above the arrow and their timings are written in the brackets. The start activity and the finish activity with zero timing have only been listed for convenience. For calculations of EST and EFT let us proceed forward through the network as follows : (a)
Put the value of the project start time in both EST and EFT positions near the start activity arrow. So for start activity EST and EFT is zero, which is placed under the start activity as 0 0
(b) Consider activity A with activity time of 6. For this EST is zero and EFT is 6 because that is the minimum time the activity will take. It has been placed near activity A as
0 6
(c) All activities emanating from node 2 will have EST as 6 and EFT = EST + activity time, hence for activity B it is 6 10 because activity B has a timing of 4. Similarly, near activity D has been 6 8 as it has activity time of 2. All the timings have been written in this manner. (d) Continue through the entire network and mark the EST and EFT. The critical path is ABCIJK and is 36. Hence for the finish activity EST = EFT = 36. Calculation of LST and LFT For this purpose we work backward through the network. These timings have been listed in Figure 15.15.
Fig 15.15
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PROJECT MANAGEMENT PERT AND CPM
For activity K, EST was 26 and EFT was 36. LST for this activity is 36 — 10 = 26 and LFT is 36 so mark it next to activity K as shown. Similarly, let us take activity F. EST was 12 and EFT 22 as activity time is 10. LST can only be 26 — 10 = 16 and LFT is 26. Calculation of Float (Slack) and Crashing the Network Example 15.5. A project consists of the following activities. The Optimistic Time (OT), Pessimistic Time (PT) and Most Likely Time or the Expected Time for the activities is also listed in front of them. Predecessor Activity
Successor Activity
OT
Most Likely Time
PT
1— 2
2
2
3
4
2—3
3
3
6
9
2—4
4
3
4
5
3—5
5
2
4
6
3—6
6
0
—
46
6
—
0
—
47
7
4
5
6
5—7
7
4
6
8
6—7
7
6
7.5
12
Draw a network diagram of the above project and calculate associate timings of the project, i.e., Earliest and Latest Occurrence times of different events, slack, identify critical events and mark the Critical Path in the diagram. What is the total project duration ? Solution. The network diagram is as shown below.
Fig 15.16
(4, 6), (3, 6) —4 Dummy Activities Critical Path 1, —2, —3 —5 —7, Longest Path Event Predecessor
Successor
1 2
2 3
2 3
4 5
ET =
9 + 4m + 6 6 3 6 4 4
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3 4 4 5 6
6 6 7 7 7
0 0 5 6 8
EST Event 1= 0 Event 2 = 0 + 3 = 3 Event 3 = 3 + 6 = 9 Event 4 = 3 + 4 Event 5 = 9 + 4 = 13 Event 6 = 9 + 0 Event 7 = 13 + 6 = 19 Now the slack can be calculated. Event 1 2 3 4 5 6 7
LST Event 7 = 19 Event 6 = 19 — 8 = 11 Event 5 = 19 — 6 = 13 Event 4 = 11 — 0 = 11 Event 3 = 13 — 4 = 9 Event 2 = 9 — 6 = 3 Event 1 = 3 — 3 = 0 EST 0 3 9 7 . 13 9 19
LST 0 3 9 11 13 11 19
Slack 0 0 0
4 0 2 0
All the events having zero slack are the Critical Events, i.e., 1, 2, 3, 5 and 7. This is the Critical Path. The project duration is 19 (days/weeks). Crashing of Network
Most of the projects result into cost overruns because of the inability of the project management team to complete the project in minimum possible time frame. The crashing of network involves considering the cost incurred on different activities required for completing the project. Let us understand certain terminology associated with crashing of network. Normal Cost This is the cost of the project when all the normal activities are carried out, i.e., there is no overtime or there are no special resources for which extra payment has to be made. Normal Time It is that time in which project can be completed with the normal cost as defined above. Crash Cost
It is the minimum possible time, which is associated with the crash cost.
653
PROJECT MANAGEMENT PERT AND CPM The relationship between these costs can be expressed as shown in Figure 15.17. Crash Cost
D
E Normal Cost
A D, C F, DE, E G 5 8 0 E — H, F > H, G —> H
2003 APR.
There is a constraint that activity F —> H cannot start till the activity D > E is completed. Determine the critical path and tabulate the earliest start time, earliest finish time, latest start time, latest finish time, total float, free float. 20 A small project consists of T activities for which the relevant data are given below: Activity A B C D E F
Predecessor Activity — —
6 5
(i) Draw the network and find the project completion time. (ii) Calculate total float for each of the activity. 20 Draw the network diagram and find the critical path and duration of the following project. Also find the total free float and independent float for each activity :
Activity Duration 2004 APR.
6 5 7
A, B A, B C, D, E C, D, E
G
2003 SEP.
Duration (days) 4 7
a—b a—c a—d
b—f c—e
c—f d—e e— f f—g
13 15 24 29 16 15 17 12 12 10 A small project consists of seven activities, whose time estimates are given in the following table :
Activity 1—2 1—3 1—4 2—5 3—5 4—6 5—6
Optimistic time 1 1 2 1 2 2
Most likely time 1 4 2 1 5 5
3
6
Pessimistic time 7 7 8 1 14 8 15
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(a) Draw the network and determine project duration and critical path. (b) Determine floats and variance for each activity. (c) If the target time is 18 weeks what is the probability of not meeting the target. A project consists of 8 independent activities as shown below :
2004 APR.
2004 APR. B.B.A. III
Activity
Predecessor
A B C D E F G H
A C B, D, E A C
Time estimate (weeks) Optimal Most like Pessimistic 1 5 3 4 2 3. 5 4 3 10 9 2 6 4 5 13 5 6 6 2 4 9 6 3
(i) Draw the network diagram and identify the critical activities. (ii) What is expected line of Completion of the project ? (a) What is PERT ? Explain 'Optimistic Time', ' Pessimistic Time' and 'Most Likely time' in relation to PERT. (b) Time requirement of various activities of a project are as follows :
Activity A B C D E F G
Event 12 1 3 1 4 15 35 46 56
Most Likely Time 6 12 12 6 30 30 30
Pessimistic Time 24 18 30 6 48 42 54
Optimistic Time 6 6 12 6 12 12 18
Determine the Optimal sequence and Idle time and Completion time for tne jobs. Find : (i) The expected duration and variance of each activity. (ii) Expected Project Length. (iii) Variance and Standard Deviation of Project Length.
6 Simulation LEARNING OBJECTIVES • Learning the meaning of Simulation and relating it with real life problems • Learning why management problems use the Simulation technique • Discuss the situations where any mathematical formulation of the problem is extremely difficult and Simulation is the only means of solving the problem • Understanding the concept of Simulation • Understanding some of the basic techniques used in Simulation • Apply the technique of Simulation in queuing, inventory control problems etc • Use of table of random numbers. INTRODUCTION In previous chapters, we have considered problems that can be solved by formulating a mathematical model. All real life situations cannot be converted into exact mathematical models because of the complexity of the situations. In such cases, simulation is a very useful technique. To simulate is to try and 'duplicate' the behaviour of the system under study and analyse the interaction among its components. The effort to develop a 'duplicate' involves recreating the fe