# Advanced optimization and operations research 9789813299665, 9789813299672

560 31 3MB

English Pages 626 Year 2019

Preface......Page 7
Contents......Page 9
1.2 Introduction......Page 17
1.3 Definitions of Operations Research......Page 18
1.4 Applications of OR......Page 19
1.5 Main Phases of OR......Page 21
1.6 Scope of OR......Page 22
1.7 Modelling in OR......Page 23
1.8 Development of OR in India......Page 26
1.9 Role of Computers in OR......Page 27
2.3 Convex and Concave Functions......Page 28
2.4 Geometrical Interpretation......Page 30
2.5 Properties of Convex and Concave Functions......Page 33
2.6 Differentiable Convex and Concave Functions......Page 35
2.7 Differentiable Strictly Convex and Concave Functions......Page 39
3.2 General Linear Programming Problem......Page 43
3.3 Slack and Surplus Variables......Page 44
3.4 Canonical and Standard Forms of LPP......Page 45
3.5 Fundamental Theorem of LPP......Page 47
3.6 Replacement of a Basis Vector......Page 51
3.7 Improved Basic Feasible Solution......Page 53
3.8 Optimality Condition......Page 54
3.9 Unbounded Solution......Page 56
3.10 Simplex Algorithm......Page 59
3.11 Simplex Table......Page 60
3.12 Use of Artificial Variables......Page 67
3.12.1 Big M Method......Page 68
3.12.2 Solution of Simultaneous Linear Equations by Simplex Method......Page 69
3.12.3 Inversion of a Matrix by Simplex Method......Page 70
3.12.4 Two-Phase Method......Page 80
3.13 Exercises......Page 90
4.2 Introduction......Page 93
4.4 Standard Form I of Revised Simplex Method......Page 94
4.5 Computational Procedure of Standard Form I of Revised Simplex Method......Page 97
4.6 Standard Form II of Revised Simplex Method......Page 103
4.6.1 Charne’s Big M Revised Simplex Method......Page 104
4.6.2 Two-Phase Revised Simplex Method......Page 109
4.7 Advantages of Revised Simplex Method......Page 120
4.9 Exercises......Page 121
5.2 Introduction......Page 123
5.3 Dual Simplex Algorithm......Page 128
5.5 Modified Dual Simplex Method......Page 134
5.6 Exercises......Page 138
6.2 Introduction......Page 140
6.3 The Computational Procedure......Page 141
6.4 Exercises......Page 148
7.2 Introduction......Page 150
7.3 Discrete Changes in the Profit Vector......Page 151
7.4 Discrete Changes in the Requirement Vector......Page 158
7.5 Discrete Changes in the Coefficient Matrix......Page 164
7.6 Addition of a Variable......Page 169
7.7 Deletion of a Variable......Page 171
7.8 Addition of a New Constraint......Page 172
7.9 Exercises......Page 179
8.2 Introduction......Page 183
8.3 Gomory’s All-Integer Cutting-Plane Method......Page 184
8.4 Construction of Gomory’s Constraint......Page 185
8.6 Construction of Additional Constraint for Mixed-Integer Programming Problems......Page 194
8.7 Branch and Bound Method......Page 199
8.8 Exercises......Page 205
9.2 Optimization of Functions of a Single Variable......Page 208
9.3 Optimization of Functions of Several Variables......Page 212
9.4 Exercises......Page 220
10.2 Introduction......Page 222
10.3 Direct Substitution Method......Page 223
10.4 Method of Constrained Variation......Page 226
10.4.2 Necessary Conditions for Extremum of a General Problem for Constrained Variation Method......Page 228
10.4.3 Sufficient Conditions for a General Problem in Constrained Variation Method......Page 231
10.5 Lagrange Multiplier Method......Page 236
10.5.1 Necessary Conditions for Optimality......Page 237
10.5.2 Sufficient Conditions for Optimality......Page 239
10.5.3 Alternative Approach for Testing Optimality......Page 241
10.6 Interpretation of Lagrange Multipliers......Page 252
10.7 Applications of Lagrange Multiplier Method......Page 256
10.8 Exercises......Page 257
11.2 Introduction......Page 259
11.3 Feasible Direction......Page 261
11.4.1 Kuhn-Tucker Necessary Conditions......Page 263
11.4.2 Kuhn-Tucker Sufficient Conditions......Page 265
11.5 Exercises......Page 277
12.2 Introduction......Page 278
12.3 Solution Procedure of QPP......Page 280
12.4 Wolfe’s Modified Simplex Method......Page 284
12.6 Beale’s Method......Page 301
12.7 Beale’s Algorithm for QPPs......Page 302
12.8 Exercises......Page 310
13.2 Introduction......Page 312
13.3 Cooperative and Non-cooperative Games......Page 313
13.4 Antagonistic Game......Page 316
13.5 Matrix Games or Rectangular Games......Page 317
13.6 Dominance......Page 333
13.7 Graphical Solution of 2 × n or m × 2 Games......Page 339
13.8 Matrix Games and Linear Programming Problems (LPPs)......Page 345
13.9 Infinite Antagonistic Game......Page 350
13.10 Continuous Game......Page 356
13.11 Separable Game......Page 357
13.12 Exercises......Page 373
14.2 Introduction......Page 376
14.3 Phases of Project Management......Page 377
14.5 Basic Components......Page 378
14.6 Common Errors......Page 381
14.7 Rules for Network Construction......Page 382
14.8 Numbering of Events......Page 383
14.9 Critical Path Analysis......Page 384
14.10 PERT Analysis......Page 389
14.11 Difference Between PERT and CPM......Page 395
14.13 Exercises......Page 408
15.2.1 Introduction......Page 412
15.2.2 Reason Behind the Queue......Page 413
15.2.3 Characteristics of Queueing Systems......Page 414
15.2.5 The State of the System......Page 418
15.2.6 Probability Distribution in a Queueing System......Page 419
15.2.7 Classification of Queueing Models......Page 424
15.3.1 Introduction......Page 425
15.3.2 Model (M/M/1):(∞/FCFS/∞)......Page 426
15.3.3 Model (M/M/1):(N/FCFS/∞)......Page 437
15.3.4 Model (M/M/C):(∞/FCFS/∞)......Page 443
15.3.5 Model (M/M/C):(N/FCFS/∞)......Page 451
15.3.6 Model (M/M/R):(K/GD) K  greaterthan  R......Page 457
15.3.7 Power Supply Model......Page 465
15.4.1 Introduction......Page 468
15.4.2 Model (M/Ek/1):(∞/FCFS/∞)......Page 469
15.4.3 Model (M/G/1):(\infty/GD)......Page 477
15.5.2 Mixed Queueing Model......Page 482
15.6 Cost Models in Queueing System......Page 486
15.7 Exercises......Page 488
16.2 Introduction......Page 496
16.3 Graphs......Page 497
16.4 Minimum Path Problem......Page 500
16.5 Problem of Potential Difference......Page 503
16.6 Network Flow Problem......Page 504
16.7 Generalized Problem of Maximum Flow......Page 508
16.8 Formulation of an LPP of a Flow Network......Page 509
16.9 Exercises......Page 526
17.2 Introduction......Page 529
17.2.1 Types of Inventories......Page 531
17.3 Basic Terminologies in Inventory......Page 532
17.4 Classification of Inventory Models......Page 534
17.5 Purchasing Inventory Model with No Shortage (Model 1)......Page 535
17.6 Manufacturing Model with No Shortage (Model 2)......Page 538
17.7 Purchasing Inventory Model with Shortage (Model 3)......Page 541
17.8 Manufacturing Model with Shortage......Page 546
17.9 Multi-item Inventory Model......Page 551
17.9.1 Limitation on Inventories......Page 552
17.9.2 Limitation of Floor Space (or Warehouse Capacity)......Page 554
17.9.3 Limitation on Investment......Page 555
17.10 Inventory Models with Price Breaks......Page 560
17.10.1 Purchasing Inventory Model with Single Price Break......Page 561
17.10.2 Purchase Inventory Model with Two Price Breaks......Page 563
17.11.1 Single Period Inventory Model with Continuous Demand (No Replenishment Cost Model)......Page 565
17.11.2 Single Period Inventory Model with Instantaneous Demand (no Replenishment Cost Model)......Page 572
17.12 Fuzzy Inventory Model......Page 578
17.13 Exercises......Page 584
18.2 Matrices......Page 588
18.3 Determinant......Page 594
18.4 Rank of a Matrix......Page 596
18.5 Solution of a System of Linear Equations......Page 597
18.6 Vectors......Page 598
18.7 Probability......Page 603
18.8 Markov Process......Page 617
Bibliography......Page 621
Index......Page 624
##### Citation preview

Springer Optimization and Its Applications 153

Asoke Kumar Bhunia Laxminarayan Sahoo Ali Akbar Shaikh

Springer Optimization and Its Applications Volume 153 Series Editor Panos M. Pardalos

, University of Florida, Gainesville, FL, USA

Honorary Editor Ding-Zhu Du, University of Texas at Dallas, Richardson, TX, USA Advisory Editors J. Birge, University of Chicago, Chicago, IL, USA S. Butenko, Texas A&M University, College Station, TX, USA F. Giannessi, University of Pisa, Pisa, Italy S. Rebennack, Karlsruhe Institute of Technology, Karlsruhe, Baden-Württemberg, Germany T. Terlaky, Lehigh University, Bethlehem, PA, USA Y. Ye, Stanford University, Stanford, CA, USA

Aims and Scope Optimization has continued to expand in all directions at an astonishing rate. New algorithmic and theoretical techniques are continually developing and the diffusion into other disciplines is proceeding at a rapid pace, with a spot light on machine learning, artiﬁcial intelligence, and quantum computing. Our knowledge of all aspects of the ﬁeld has grown even more profound. At the same time, one of the most striking trends in optimization is the constantly increasing emphasis on the interdisciplinary nature of the ﬁeld. Optimization has been a basic tool in areas not limited to applied mathematics, engineering, medicine, economics, computer science, operations research, and other sciences. The series Springer Optimization and Its Applications (SOIA) aims to publish state-of-the-art expository works (monographs, contributed volumes, textbooks, handbooks) that focus on theory, methods, and applications of optimization. Topics covered include, but are not limited to, non-linear optimization, combinatorial optimization, continuous optimization, stochastic optimization, Bayesian optimization, optimal control, discrete optimization, multi-objective optimization, and more. New to the series portfolio include Works at the intersection of optimization and machine learning, artiﬁcial intelligence, and quantum computing. Volumes from this series are indexed by Web of Science, zbMATH, Mathematical Reviews, and SCOPUS.

Asoke Kumar Bhunia Laxminarayan Sahoo Ali Akbar Shaikh •

123

Asoke Kumar Bhunia Department of Mathematics The University of Burdwan Burdwan, West Bengal, India

Laxminarayan Sahoo Department of Mathematics Raniganj Girls’ College Raniganj, West Bengal, India

Ali Akbar Shaikh Department of Mathematics The University of Burdwan Burdwan, West Bengal, India

Dedicated to our respected teachers Prof. M. Maiti and Prof. T. K. Pal Vidyasagar University West Bengal, India and Prof. M. Basu University of Kalyani West Bengal, India

Preface

Due to the gradual complexity of day-to-day decision-making problems, engineers, system analysts and managers face challenges in offering the best solutions for these problems. Most of these real-life decision-making problems can be modelled as non-linear constrained or unconstrained optimization problems. Optimization is the art of ﬁnding the best alternative(s) among a given set of options. The process of ﬁnding the maximum or minimum possible value which a given function can attain in its domain of deﬁnition is known as optimization. Operations research, on the other hand, a relatively new discipline, is an interdisciplinary subject. Mainly concerned with different techniques, it provides new insights and capabilities for determining better solutions of decision-making problems with higher accuracy, competence and conﬁdence. This book, Advanced Optimization and Operations Research, is intended to introduce the fundamental and advanced concepts of optimization and operations research. It has been written in simple and lucid language, according to the syllabi of the all major universities which include optimization and operations research in their graduate and post-graduate courses in areas of science, arts, commerce, engineering and management. Maximum care has been taken in preparing the chapters to include as many different topics as possible. There are several textbooks available on the market on optimization and operations research as separate topics. Our primary motivation to write this book was the absence of one book covering these two areas together. Existing books have been written in a conventional way, whereas in this book, all 18 chapters are written in details with equal emphasis. Each chapter states its objective in the beginning and includes a sufﬁcient number of solved examples and exercises which will be of much help for aspirants preparing for competitive examinations like NET, SET, GATE, CAT and MAT. Chapter 1 highlights the historical perspective, various deﬁnitions, characteristic scope and purpose of operations research. Chapter 2 discusses convex and concave functions and their properties. Chapter 3 introduces the simplex method. In Chaps. 4–6, advanced linear programming techniques, like revised simplex method, dual simplex method and bounded variable technique, are discussed. Chapter 7 vii

viii

Preface

introduces the post-optimality analysis of linear programming problems (LPPs), and Chap. 8 deals with techniques for solving LPPs with integer values. Chapters 9 through 11 deal with the basics of unconstrained and constrained non-linear optimization problems with equality and inequality constraints. In Chap. 12, quadratic programming problems and their solution by Wolfe’s modiﬁed simplex method and Beale’s method are explained in details. Chapter 13 talks about game theory, and Chap. 14 discusses project management, which includes the topics of critical path analysis, PERT analysis and time-cost trade-off. Chapter 15 presents the concepts of queueing theory, including Poisson and non-Poisson queues. Chapter 16 presents ideas on flow and potential in networks, including the maximum flow problem and its solution procedure. Chapter 17 introduces deterministic, probabilistic and fuzzy inventory control models. Chapter 18 discusses the preliminaries of several topics, like matrices, determinant, vectors, probability and Markov process, which will be essential for studying the different chapters of this book. Salient features of this book are as follows: • It is written in a self-instructional mode so that the readers can easily understand the subject matter without any help from other sources. • Each topic has been discussed in detail with numerous worked-out examples to help students properly understand the underlying theories. • Advanced-level (research-level) topics have also been included. • A set of exercises has been included at the end of each chapter. • It is suitable to help students to prepare for different competitive examinations. We would like to express our sincere gratitude to all the faculty members of the Department of Mathematics at the University of Burdwan, who have contributed greatly to the success of this project. We also express our heartfelt thanks to our research scholars—Avijit Duary, Subash Chandra Das, Tanmoy Banerjee, Rajan Mondal, Md. Akhtar, Md. Sadikur Rahaman and Nirmal Kumar—for their constant help in preparing the manuscript. Finally, we wish to thank Shamim Ahmad, Tooba Shaﬁque and the other publishing teams of Springer Nature for producing the book in its present form. West Bengal, India

Asoke Kumar Bhunia Laxminarayan Sahoo Ali Akbar Shaikh

Contents

1

Introduction to Operations Research . . . 1.1 Objectives . . . . . . . . . . . . . . . . . . . 1.2 Introduction . . . . . . . . . . . . . . . . . 1.3 Deﬁnitions of Operations Research . 1.4 Applications of OR . . . . . . . . . . . . 1.5 Main Phases of OR . . . . . . . . . . . . 1.6 Scope of OR . . . . . . . . . . . . . . . . . 1.7 Modelling in OR . . . . . . . . . . . . . . 1.8 Development of OR in India . . . . . 1.9 Role of Computers in OR . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

1 1 1 2 3 5 6 7 10 11

2

Convex and Concave Functions . . . . . . . . . . . . . . . . . . . . . 2.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Convex and Concave Functions . . . . . . . . . . . . . . . . . 2.4 Geometrical Interpretation . . . . . . . . . . . . . . . . . . . . . 2.5 Properties of Convex and Concave Functions . . . . . . . 2.6 Differentiable Convex and Concave Functions . . . . . . 2.7 Differentiable Strictly Convex and Concave Functions

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

13 13 13 13 15 18 20 24

3

Simplex Method . . . . . . . . . . . . . . . . . . . . . . 3.1 Objective . . . . . . . . . . . . . . . . . . . . . . 3.2 General Linear Programming Problem . 3.3 Slack and Surplus Variables . . . . . . . . 3.4 Canonical and Standard Forms of LPP . 3.5 Fundamental Theorem of LPP . . . . . . . 3.6 Replacement of a Basis Vector . . . . . . 3.7 Improved Basic Feasible Solution . . . . 3.8 Optimality Condition . . . . . . . . . . . . . . 3.9 Unbounded Solution . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

29 29 29 30 31 33 37 39 40 42

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

ix

x

Contents

3.10 3.11 3.12

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

45 46 53 54

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

55 56 66 76

Revised Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Standard Forms of Revised Simplex Method . . . . . . . . . . 4.4 Standard Form I of Revised Simplex Method . . . . . . . . . . 4.5 Computational Procedure of Standard Form I of Revised Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Standard Form II of Revised Simplex Method . . . . . . . . . 4.6.1 Charne’s Big M Revised Simplex Method . . . . . 4.6.2 Two-Phase Revised Simplex Method . . . . . . . . . 4.7 Advantages of Revised Simplex Method . . . . . . . . . . . . . 4.8 Difference Between Revised Simplex Method and Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

79 79 79 80 80

. . . . .

. . . . .

. 83 . 89 . 90 . 95 . 106

3.13 4

5

Simplex Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . Simplex Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Use of Artiﬁcial Variables . . . . . . . . . . . . . . . . . . . . 3.12.1 Big M Method . . . . . . . . . . . . . . . . . . . . . 3.12.2 Solution of Simultaneous Linear Equations by Simplex Method . . . . . . . . . . . . . . . . . 3.12.3 Inversion of a Matrix by Simplex Method . 3.12.4 Two-Phase Method . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . 107 . . . 107

109 109 109 114

Dual Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Dual Simplex Algorithm . . . . . . . . . . . . . . . . . . . . . . . 5.4 Difference Between Simplex Method and Dual Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Modiﬁed Dual Simplex Method . . . . . . . . . . . . . . . . . . 5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . 120 . . . . . 120 . . . . . 124

6

Bounded Variable Technique . . . . . 6.1 Objective . . . . . . . . . . . . . . . 6.2 Introduction . . . . . . . . . . . . . 6.3 The Computational Procedure 6.4 Exercises . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

127 127 127 128 135

7

Post-optimality Analysis in LPPs . . . . . . . . . . . . . 7.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Discrete Changes in the Proﬁt Vector . . . . . . 7.4 Discrete Changes in the Requirement Vector

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

137 137 137 138 145

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

Contents

7.5 7.6 7.7 7.8 7.9 8

xi

Discrete Changes in the Coefﬁcient Matrix . Addition of a Variable . . . . . . . . . . . . . . . . Deletion of a Variable . . . . . . . . . . . . . . . . Addition of a New Constraint . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

151 156 158 159 166

Integer Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Gomory’s All-Integer Cutting-Plane Method . . . . . . . . . 8.4 Construction of Gomory’s Constraint . . . . . . . . . . . . . . 8.5 Gomory’s Mixed-Integer Cutting-Plane Method . . . . . . 8.6 Construction of Additional Constraint for Mixed-Integer Programming Problems . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Branch and Bound Method . . . . . . . . . . . . . . . . . . . . . 8.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

171 171 171 172 173 182

Basics 9.1 9.2 9.3 9.4

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . . 182 . . . . . 187 . . . . . 193 . . . . .

. . . . .

. . . . .

197 197 197 201 209

10 Constrained Optimization with Equality Constraints . . . . . . . . . 10.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Direct Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Method of Constrained Variation . . . . . . . . . . . . . . . . . . . . 10.4.1 Geometrical Interpretation of Admissible Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.2 Necessary Conditions for Extremum of a General Problem for Constrained Variation Method . . . . . 10.4.3 Sufﬁcient Conditions for a General Problem in Constrained Variation Method . . . . . . . . . . . . . 10.5 Lagrange Multiplier Method . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Necessary Conditions for Optimality . . . . . . . . . . 10.5.2 Sufﬁcient Conditions for Optimality . . . . . . . . . . . 10.5.3 Alternative Approach for Testing Optimality . . . . 10.6 Interpretation of Lagrange Multipliers . . . . . . . . . . . . . . . . 10.7 Applications of Lagrange Multiplier Method . . . . . . . . . . . . 10.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

211 211 211 212 215

9

of Unconstrained Optimization . . . . . . . . . . . . Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . Optimization of Functions of a Single Variable . Optimization of Functions of Several Variables . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . 217 . . 217 . . . . . . . .

. . . . . . . .

220 225 226 228 230 241 245 246

11 Constrained Optimization with Inequality Constraints . . . . . . . . . . 249 11.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 11.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

xii

Contents

11.3 11.4

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

251 253 253 255 267

12 Quadratic Programming . . . . . . . . . . . . . . . . . . . . . . . 12.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Solution Procedure of QPP . . . . . . . . . . . . . . . . 12.4 Wolfe’s Modiﬁed Simplex Method . . . . . . . . . . 12.5 Advantage and Disadvantage of Wolfe’s Method 12.6 Beale’s Method . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Beale’s Algorithm for QPPs . . . . . . . . . . . . . . . 12.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

269 269 269 271 275 276 292 293 301

Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cooperative and Non-cooperative Games . . . . . . . . . . . . Antagonistic Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . Matrix Games or Rectangular Games . . . . . . . . . . . . . . . Dominance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphical Solution of 2  n or m  2 Games . . . . . . . . Matrix Games and Linear Programming Problems (LPPs) Inﬁnite Antagonistic Game . . . . . . . . . . . . . . . . . . . . . . Continuous Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Separable Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

303 303 303 304 307 308 324 330 336 341 347 348 364

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

367 367 367 368 369 369 372 373 374 375 380 386 387 399

11.5

13 Game 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12

Feasible Direction . . . . . . . . . . . . . . . . . . . . Kuhn-Tucker Conditions . . . . . . . . . . . . . . . 11.4.1 Kuhn-Tucker Necessary Conditions 11.4.2 Kuhn-Tucker Sufﬁcient Conditions Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

14 Project Management . . . . . . . . . . . . . . . . . 14.1 Objectives . . . . . . . . . . . . . . . . . . . . 14.2 Introduction . . . . . . . . . . . . . . . . . . 14.3 Phases of Project Management . . . . . 14.4 Advantages of Network Analysis . . . 14.5 Basic Components . . . . . . . . . . . . . 14.6 Common Errors . . . . . . . . . . . . . . . 14.7 Rules for Network Construction . . . . 14.8 Numbering of Events . . . . . . . . . . . 14.9 Critical Path Analysis . . . . . . . . . . . 14.10 PERT Analysis . . . . . . . . . . . . . . . . 14.11 Difference Between PERT and CPM 14.12 Project Time-Cost Trade-Off . . . . . . 14.13 Exercises . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . .

. . . . . . . . . . . . . .

. . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

Contents

xiii

15 Queueing Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Basics of Queueing Theory . . . . . . . . . . . . . . . . . . . . . 15.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.2 Reason Behind the Queue . . . . . . . . . . . . . . . 15.2.3 Characteristics of Queueing Systems . . . . . . . 15.2.4 Important Deﬁnitions . . . . . . . . . . . . . . . . . . 15.2.5 The State of the System . . . . . . . . . . . . . . . . 15.2.6 Probability Distribution in a Queueing System 15.2.7 Classiﬁcation of Queueing Models . . . . . . . . . 15.3 Poisson Queueing Models . . . . . . . . . . . . . . . . . . . . . . 15.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.2 Model ðM=M=1Þ:ð1=FCFS=1Þ . . . . . . . . . . 15.3.3 Model ðM=M=1Þ:ðN=FCFS=1Þ . . . . . . . . . . 15.3.4 Model ðM=M=CÞ:ð1=FCFS=1Þ . . . . . . . . . . 15.3.5 Model ðM=M=C Þ:ðN=FCFS=1Þ . . . . . . . . . . 15.3.6 Model ðM=M=RÞ:ðK=GDÞK [ R . . . . . . . . . . 15.3.7 Power Supply Model . . . . . . . . . . . . . . . . . . 15.4 Non-poisson Queueing Models . . . . . . . . . . . . . . . . . . 15.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 15.4.2 Model ðM=Ek =1Þ:ð1=FCFS=1Þ . . . . . . . . . . 15.4.3 Model ðM=G=1Þ:ð1=GDÞ . . . . . . . . . . . . . . . 15.5 Mixed Queueing Model and Cost . . . . . . . . . . . . . . . . . 15.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.2 Mixed Queueing Model . . . . . . . . . . . . . . . . 15.6 Cost Models in Queueing System . . . . . . . . . . . . . . . . 15.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

403 403 403 403 404 405 409 409 410 415 416 416 417 428 434 442 448 456 459 459 460 468 473 473 473 477 479

16 Flow in Networks . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . 16.3 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Minimum Path Problem . . . . . . . . . . . . . . . 16.5 Problem of Potential Difference . . . . . . . . . 16.6 Network Flow Problem . . . . . . . . . . . . . . . 16.7 Generalized Problem of Maximum Flow . . 16.8 Formulation of an LPP of a Flow Network . 16.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

487 487 487 488 491 494 495 499 500 517

17 Inventory Control Theory . . . . . . . . . . . 17.1 Objectives . . . . . . . . . . . . . . . . . . 17.2 Introduction . . . . . . . . . . . . . . . . 17.2.1 Types of Inventories . . . 17.3 Basic Terminologies in Inventory .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

521 521 521 523 524

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

xiv

Contents

17.4 17.5 17.6 17.7 17.8 17.9

17.10

17.11

17.12 17.13

Classiﬁcation of Inventory Models . . . . . . . . . . . . . . . . . . . Purchasing Inventory Model with No Shortage (Model 1) . . Manufacturing Model with No Shortage (Model 2) . . . . . . . Purchasing Inventory Model with Shortage (Model 3) . . . . . Manufacturing Model with Shortage . . . . . . . . . . . . . . . . . Multi-item Inventory Model . . . . . . . . . . . . . . . . . . . . . . . . 17.9.1 Limitation on Inventories . . . . . . . . . . . . . . . . . . 17.9.2 Limitation of Floor Space (or Warehouse Capacity) . . . . . . . . . . . . . . . . . . . 17.9.3 Limitation on Investment . . . . . . . . . . . . . . . . . . Inventory Models with Price Breaks . . . . . . . . . . . . . . . . . . 17.10.1 Purchasing Inventory Model with Single Price Break . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.10.2 Purchase Inventory Model with Two Price Breaks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Probabilistic Inventory Model . . . . . . . . . . . . . . . . . . . . . . 17.11.1 Single Period Inventory Model with Continuous Demand (No Replenishment Cost Model) . . . . . . 17.11.2 Single Period Inventory Model with Instantaneous Demand (no Replenishment Cost Model) . . . . . . . Fuzzy Inventory Model . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18 Mathematical Preliminaries . . . . . . . . . . . . . . . 18.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Determinant . . . . . . . . . . . . . . . . . . . . . . 18.4 Rank of a Matrix . . . . . . . . . . . . . . . . . . 18.5 Solution of a System of Linear Equations . 18.6 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 18.7 Probability . . . . . . . . . . . . . . . . . . . . . . . 18.8 Markov Process . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . .

. . . . . . .

526 527 530 533 538 543 544

. . 546 . . 547 . . 552 . . 553 . . 555 . . 557 . . 557 . . 564 . . 570 . . 576 . . . . . . . . .

. . . . . . . . .

581 581 581 587 589 590 591 596 610

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619

Asoke Kumar Bhunia is Professor at the Department of Mathematics, The University of Burdwan, West Bengal, India. He obtained his Ph.D. degree in Mathematics and M.Sc. in Applied Mathematics from Vidyasagar University, India. His research interests include computational optimization, soft computing, interval mathematics and interval ranking. He has published more than 125 research papers in various national and international journals of repute. He is a reviewer of several Science Citation Index (SCI) journals. He has guided 13 Ph.D. and two M.Phil. students. An author of four research monographs and six book chapters, he is an Indian National Science Academy (INSA) Visiting Fellow and former Associate Editor of the Springer journal OPSEARCH. Laxminarayan Sahoo is Assistant Professor at the Department of Mathematics, Raniganj Girls’ College, West Bengal, India. He obtained his Ph.D. in Mathematics from the University of Burdwan and his M.Sc. in Applied Mathematics from Vidyasagar University, India. His research interests include reliability optimization, computational optimization, artiﬁcial intelligence, soft computing, interval mathematics, interval ranking and fuzzy mathematics. He has published several research papers in various national and international journals of repute. He has received a Ministry of Human Resource Development (MHRD) fellowship from the Government of India and the Professor M.N. Gopalan Award for the best Ph.D. thesis in operations research from the Operational Research Society of India. Ali Akbar Shaikh is Assistant Professor at the Department of Mathematics in The University of Burdwan, West Bengal, India. Earlier, he was a postdoctoral fellow at the School of Engineering and Sciences, Tecnológico de Monterrey, Mexico. He received the award SNI of level 1 (out of 0–3) presented by the National System of Researchers of Mexico from the Government of Mexico in the year 2017. He earned his Ph.D. and M.Phil. in Mathematics from the University of Burdwan, and his

xv

xvi

M.Sc. in Applied Mathematics from the University of Kalyani, India. Dr. Shaikh has published more than 37 research papers in different international journals of repute. His research interests include inventory control, interval optimization and particle swarm optimization.

Chapter 1

Introduction to Operations Research

1.1

Objectives

The objectives of this chapter are to discuss: • The needs of using Operations Research (OR) • The historical perspective of the approach of OR • Several deﬁnitions of OR in different contexts and various phases of scientiﬁc study • The classiﬁcation and usage of various models for solving a problem under consideration • Applications of OR in different areas.

1.2

Introduction

The term ‘operations research (OR)’ was ﬁrst introduced by McClosky and Trefthen in the year 1940. This subject came into existence in a military context. The main origin of the development of OR was the Second World War. During that war, the military services of England gathered some scientists/researchers from various disciplines and formed a team to assist them in solving strategic and tactical problems relating to air and land defence of the country. As they had very limited military resources, it was necessary to decide upon their most effective utilization, e.g. efﬁcient ocean transport, effective bombing, etc. The mission of the team was to formulate speciﬁc proposals and plans to help the military services make decisions on optimal utilization of military resources and efforts and also to implement the decisions effectively. The members of this team were not actually engaged in the military operations. However, they worked as advisors in winning the war to the extent that the scientiﬁc and systematic approaches involved in OR provided good © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_1

1

2

1

Introduction to Operations Research

intellectual support to the strategic initiatives of the military services. In this context, Arthur C. Clarke gave an appropriate deﬁnition of OR, saying ‘OR is an art of winning the war without actually ﬁghting it’. After seeing the encouraging results by the British OR team, the US military department was quickly motivated to start similar activities in their country. The major developments of the US teams were the invention of new ﬁghting patterns, planning, sea mining and effective utilization of electronic equipment and their successful application in war. In USA, the work of the OR team was given different names, like Operational Analysis, Operational Evaluations, Operations Research, System Analysis, System Evaluation, etc. Among these, the name Operations Research became widely accepted. After the Second World War, industrial managers/executives who were seeking better solutions to their complex problems were attracted by the success of the military teams. The most common problem arising in industry was which method(s) to adopt so that the total cost/proﬁt is optimum. The ﬁrst mathematical technique in this ﬁeld, called the simplex method of linear programming problems (LPPs), was developed in 1947 by George B. Dantzig of the US Air Force. Since then, new techniques and applications have been developed through the efforts and cooperation of interested researchers in both academic institutions and industry. The impact of OR can now be felt in different areas of government as well as in the private sector. A number of consulting ﬁrms are currently engaged in OR activities. Besides military and business applications, OR activities include solving the problems of transportation systems, libraries, hospitals, city planning, ﬁnancial decision-making, etc. Most of the Indian industries, like Indian Railways, Indian airlines, defence organizations, Hindustan Unilever Ltd., TISCO, FCI, Telecom, Union Carbide and Hindustan Steel are using OR activities to obtain their best alternatives.

1.3

Deﬁnitions of Operations Research

OR has been deﬁned in various ways. It is not possible to provide uniformly acceptable deﬁnitions of OR due to its wide scope of applications. A few opinions about the deﬁnition of OR are given as follows: 1. ‘OR is the art of winning the war without actually ﬁghting it’. [Arthur C. Clarke] 2. ‘OR is a scientiﬁc method of providing executive departments with a quantitative basis for decisions regarding the operations under their control’. [P. M. Morse and G. E. Kimbol (1946)] 3. ‘OR is the scientiﬁc method of providing the executive with an analytical and objective basis for decisions’. [P. M. S. Blackett (1948)] 4. ‘OR is the art of giving bad answers to problems to which otherwise, worse answers are given’. [T. L. Saaty]

1.3 Deﬁnitions of Operations Research

3

5. ‘OR is a scientiﬁc approach to problem solving for executive management’. [H. M. Wagner] 6. ‘OR is an experimental and applied science devoted to observing, understanding and predicting the behaviour of purposeful man-machine systems and OR workers are actively engaged in applying this knowledge to practical problems in business, government and society’. [OR Society of America] 7. ‘OR is the application of scientiﬁc methods, techniques and tools to problems involving the operations of a system so as to provide those in control of the system with optimum solutions to the problem’. [Churchman, Ackoff and Arnoff (1957)] 8. ‘OR is applied decision theory. It uses many scientiﬁc, mathematical or logical means to attempt to cope with the problems that confront the executive when he tries to achieve a thorough going rationality in dealing with decision problems’. [D. W. Miller and M. K. Starr] 9. ‘OR has been characterized by the use of scientiﬁc knowledge through interdisciplinary team effort for the purpose of determining the best utilization of limited resources’. [H. A. Taha] 10. ‘The term “OR” has hitherto-fore been used to connate various attempts to study operations of war by scientiﬁc methods. From a more general point of view, OR can be considered to be an attempt to study those operations of modern society which involved organizations of men or of men and machines’. [P. M. Morse (1948)] 11. ‘OR is a management activity pursued in two complementary ways—one half by the free and bold exercise of commonsense untrammeled by any routine, and other half by the application of a repertoire of well established precreated methods and techniques’. [Jagjit Singh (1968)] 12. ‘OR is the application of the methods of science to complex problems in the direction and management of large systems of men, machines, materials and money in industry, business and defence. The distinctive approach is to develop a scientiﬁc model of the system, incorporating measurements of factors such as chance and risk with which to predict and compare the outcomes of alternative decisions, strategies or controls. The purpose is to help management to determine its policy and actions scientiﬁcally’. [Operational Research Society, UK (1971)]

1.4

Applications of OR

OR is mainly concerned with different techniques. It provides new insights and capabilities to determine better solutions to decision-making problems with higher speed, competence and conﬁdence. During the last few decades, OR has been applied successfully in various areas such as defence, government sectors, service organizations and industry. Some areas of management decision-making where the tools and techniques of OR are applied are outlined as follows:

4

1

Introduction to Operations Research

1. Budgeting and investments (i) Cash flow analysis, long-range capital requirements, dividend policies, investment portfolios (ii) Credit policies, credit risk, etc. 2. Purchasing (i) Rules for buying, supplies with stable or varying prices (ii) Determination of quantities and timing of purchases (iii) Replacement policies, etc. 3. Production management (i) Physical distribution (a) Location and size of warehouses, distribution centres and retail outlets (b) Distribution policy (ii) Facilities planning (a) Numbers and location of factories, warehouses, hospitals, etc. (b) Loading and unloading facilities for railways and trucks determining the transport schedule (iii) Manufacturing (a) Production scheduling and sequencing (b) Stabilization of production and employment training layoffs and optimum product mix (iv) Maintenance and project scheduling (a) Maintenance policies and preventive maintenance (b) Project scheduling and allocation of resources 4. Marketing (i) Product selection, timing, competitive actions (ii) Number of sales representatives (iii) Advertising media with respect to cost and time 5. Personnel management (i) Selection of suitable personnel on minimum salary (ii) Mixes of age and skills (iii) Recruitment policies and assignment of jobs 6. Research and development (i) Determination of the areas of concentration of research and development

1.4 Applications of OR

5

(ii) Project selection (iii) Reliability and alternative design (iv) Determination of time-cost trade-off and control of development projects.

1.5

Main Phases of OR

In the ﬁrst half of the twentieth century, it was difﬁcult to obtain an OR method to ﬁnd a procedure for conducting an OR project. The procedure for an OR study generally involves the following major phases: Phase I: Formulation of the Problem To ﬁnd the solution of a problem, ﬁrst it has to be formulated. Formulation of a problem for research consists of identifying, deﬁning and specifying the measure of the computation of a decision-making problem. To formulate a problem, the following information is required: (i) What are the objectives? (ii) Controlled variables and their ranges (iii) Uncontrolled variables that may affect the outcomes of the available choices. Since it is difﬁcult to extract a right answer from the wrong problem, this phase should be executed with considerable care. Phase II: Construction of a Mathematical Model After formulation of the problem, the next phase is to reformulate this problem into a speciﬁc form which is most useful for the analysis. The most convenient approach is to construct a mathematical model which represents the system. Three types of models are commonly used in OR and other sciences. These are (i) iconic, (ii) analogue and (iii) symbolic/mathematical. Phase III: Deriving a Solution from the Model After formulating the mathematical model for the problem under consideration, the next phase is to ﬁnd a solution from this model. In OR, our objective is to search for an optimal solution which optimizes (maximizes or minimizes) the objective function of the model. In addition to solving the model, sometimes it is also essential to perform a sensitivity analysis for studying the effect of changes of different system parameters. Phase IV: Testing the Model and its Solution (Updating the Model) After deriving a solution from the model of the problem, the entire solution should be tested again to determine whether or not there is any error. This may be done by reexamining the formulation of the problem and comparing it with that model, which may help to reveal any mistakes.

6

1

Introduction to Operations Research

The solution of an OR problem should yield a better performance than a solution obtained by some alternative procedure. A solution derived from the model of the problem should be tested further for its optimality. Phase V: Implementing the Solution The ﬁnal phase of an OR study is to implement the optimal solution derived by the OR team. Due to the constantly changing scenarios in the world, the model and its solution may not remain valid for a long period of time. With changes of scenario, the model, its solution and the resulting course of action can be modiﬁed accordingly.

1.6

Scope of OR

OR has considerable scope in the different sectors of our society. In general, we can say that wherever there is a problem, there is an OR technique for solving that problem. In addition to military applications, OR is widely used in many business and industry organizations among others. Here, we shall discuss the scope of OR in some particular areas of the government sector. (i) Defence: During World War II, the OR teams of Britain and America developed different techniques and strategies which helped them to win different battles. In modern warfare, military operations are carried out by the air force, army and navy. Therefore, it is necessary to formulate optimum strategies which will give maximum beneﬁts. Because OR helps military personnel to select the best possible strategy to win battles, OR has extensive applications in defence. (ii) Industry: After observing successful applications of OR in the military, industry personnel became interested in this ﬁeld. When a small organization expands, and in larger organizations in general, it is not possible to manage everything with a single department. As a result, the administration of an organization can establish the following departments with different objectives: (a) Production department The objective of this department is to produce an item with better quality by minimizing the manufacturing cost/time of the produced item. (b) Marketing department The objectives of this department are to: • Promote the item/product to the customers by attractive advertisements with the help of electronic and print media and also through sales representatives • Maximize the amount of proﬁt • Minimize the cost of sales.

1.6 Scope of OR

7

(c) Finance department The objective of this department is to minimize the capital required to maintain any level of business. These departments can come into conflict with each other as their policies may conflict. This difﬁculty can be overcome by the application of OR techniques. (iii) Life Insurence Company: OR techniques can be applied to determine the premium rates of different policies for the better interest of the organization. (iv) Agriculture: With an increasing population and consequent shortages of food, there is a need to increase the agricultural output of a country. However, many problems arise, e.g. climatic conditions and the problem of optimal distribution of water from resources, etc., for the agricultural department of a country. Thus, there is a need for best policies under the given restrictions. OR techniques may be beneﬁcial in determining the best policies. (v) Planning: Careful planning plays a crucial role in the economic development of a country, and OR techniques may be used in such planning. Thus, OR has extensive applications in the planning of different sectors of a country.

1.7

Modelling in OR

A model in OR may be deﬁned as an idealized representation of a real-life system in which only the basic aspects or the most important features of a typical problem under investigation are considered. The objective of a model is to analyse the behaviour of the system for the purpose of improving its performance. Advantages of a Model A model has many advantages over a verbal description of a problem. Some of them are as follows: (i) It describes a problem much more concisely. (ii) It provides a logical and systematic approach to the problem. (iii) It enables the use of high-powered mathematical techniques to analyse the problem. (iv) It indicates the limitations and scope of the problem. (v) It tends to make the overall structure of the problem more comprehensible. Disadvantages of a Model Models also have a few disadvantages; these are as follows: 1. Models are only an attempt to understand an operation and should never be considered absolute in any sense. 2. The validity of any model with regard to the corresponding operation can only be veriﬁed by experimentation and the use of relevant data characteristics.

8

1

Introduction to Operations Research

Types of Models Three types of models are commonly used in OR: (i) iconic models, (ii) analogue models, (iii) mathematical models. Iconic Models These models represent the system as it is, but in a different size. Thus, iconic models are obtained by enlarging or reducing the size of the system. An iconic model is a pictorial representation of the system. Some common examples are photographs, drawings, maps and globes. Iconic models of the Sun and its planets are scaled down, while the model of an atom is scaled up so as to make it visible. These models have some advantages as well as disadvantages. The advantages are that they are speciﬁc, concrete and easy to construct. They can be studied more easily than the system itself. The disadvantages are that these models are difﬁcult to manipulate for experimental purposes. To make any modiﬁcation or improvement in these models is not easy. Adjustments with changing situations cannot be performed with these models. Analogue Models These models are more abstract than iconic models; thus, there is no ‘look-alike’ correspondence between these models and real-life items. In analogue models, one set of properties is used to represent another set of properties. For example, graphs are analogues, as distance is used to represent a wide variety of variables such as time, percentage, age, weight, etc. These models are easier to manipulate than iconic models. However, they are less speciﬁc and less concrete. Mathematical (Symbolic) Models Mathematical models are more abstract and more general in nature. They employ a set of mathematical symbols to represent the components of a real system. Thus, these models are mathematical equations or inequalities reflecting the structure of the system they represent. Inventory models and queueing models are some examples of symbolic models. These models can be classiﬁed into two categories: (i) deterministic (crisp) models and (ii) non-deterministic models. In deterministic models, all the parameters and functional relationships are assumed to be known and precise. Linear programming and classical inventory models are examples of deterministic models. On the other hand, models in which at least one parameter or decision variable is a random variable or a fuzzy number or an interval-valued number are called non-deterministic models. These models reflect the complexity of the real world and the uncertainty surrounding it. In the context of generality, models can also be categorized as (i) speciﬁc models and (ii) general models. When a model presents a system at some speciﬁc time, it is known as a speciﬁc model. In these models, if a time factor is considered, they are termed dynamic models; otherwise, they are called static models. An inventory problem of determining the economic order quantity for the next period with a uniform demand rate is an example of a static model, whereas the same problem with a time-dependent demand is an example of a dynamic model. On the other

1.7 Modelling in OR

9

hand, simulation and heuristic models are called general models. These models do not yield an optimum solution to the problem, but they give a solution near to the optimal one. General Solution Methods for OR Models Generally, OR models are solved by the following three methods. (i) Analytical methods (ii) Numerical methods (iii) Monte Carlo technique. Analytical Methods In these methods, all the tools of basic mathematics (such as differential calculus, integral calculus, differential equations, difference equations, etc.) are used for solving an OR model. Numerical Methods These methods are concerned with iterative or trial and error methods. These methods are primarily used when analytical methods fail to derive a solution. The analytical methods may fail because of the complexity of the constraints and/or number of variables. In this procedure, the algorithm is started with an initial approximate (trial) solution and continued with a set of given rules for improving it towards optimality. Then the trial solution is replaced by the improved one, and the process is repeated for a ﬁxed number of times or until no further improvement is possible. Monte Carlo Technique This technique is based on the random sampling of a variable’s possible values. For this technique, some random numbers are required which may be converted into random variates whose behaviour is known from past experience. Darker and Kac developed this technique, combining probability and a sampling technique to provide the solution of partial or integral differential equations. Hence, this technique is concerned with experiments on random numbers, and it provides solutions to complicated OR problems. The Monte Carlo technique is useful in the following situations: (i) When the mathematical and statistical problems are too complicated and alternative methods are needed (ii) When it is not possible to gain any information from past experience for a particular problem (iii) To estimate the parameters of a model. Algorithm for Monte Carlo Technique The following steps are associated with the Monte Carlo technique: Step 1. To illustrate the general idea of the system, draw a flow diagram. Step 2. Make sample observations to select a suitable model for the system and determine the probability distribution for the variables of interest. Step 3. Convert the probability distribution to a cumulative distribution.

10

1

Introduction to Operations Research

Step 4. Select a sequence of random numbers with the help of random number tables. Step 5. Determine the sequence of values of the variables of interest with the sequence of random numbers obtained in Step 4. Step 6. Fit an appropriate standard mathematical function to the sequence of values obtained in Step 5. Advantages and Disadvantages of Monte Carlo Techniques The advantages of Monte Carlo techniques are as follows: (i) They are very helpful in ﬁnding solutions of complicated mathematical problems which cannot be solved otherwise. (ii) The difﬁculties of trial and error experimentation are avoided. Monte Carlo techniques have the following disadvantages: (i) These methods are time-consuming and costly ways of getting a solution of any problem. (ii) These techniques do not provide an optimal solution to a problem. The solution is nearly optimal only when the size of the sample is sufﬁciently large.

1.8

Development of OR in India

In India, OR started to appear when an OR unit was established at the Regional Research Laboratory in Hyderabad in the year 1949. At the same time, Professor R. S. Verma of Delhi University set up an OR team in the Defence Science Laboratory to solve different problems involving storing, purchasing and planning. In 1953, Professor P. C. Mahalanobis established an OR team at the Indian Statistical Institute in Kolkata for solving the problem of national planning and survey. In 1957, the OR Society of India (ORSI) was formed. India now publishes a number of research journals, namely, OPSEARCH (published by ORSI), Industrial Engineering and Management, Journal of Engineering Production, Defence Science Journal, SCIMA, IJOMAS, IAPQR, etc. Regarding OR education in India, the University of Delhi ﬁrst introduced a complete M.Sc. course in 1963. Simultaneously, the Institute of Management at Kolkata and Ahmedabad started teaching OR in their MBA courses. Nowadays, OR has become such a popular subject that it has been introduced in almost all institutes and universities in various disciplines, like Mathematics, Statistics, Commerce, Economics, Computer Science, Engineering, Business Administration, etc. Also, because of the importance of this subject, it has been introduced in different competitive examinations, like IAS, CA, ICWA and NET. Profesor P. C. Mahalanobis ﬁrst applied OR techniques in the formulation of a second ﬁve-year plan. The Planning Commission made use of OR techniques for planning the optimum size of the Caravelle fleet of Indian airlines. Some industries,

1.8 Development of OR in India

11

such as HLL, Union Carbide, TELCO, Hindustan Steel, TISCO, FCI, etc., have used these techniques, as well as other Indian organizations, like CSIR, TIFR, the Indian Institute of Science, the State Trading Corporation, and Indian Railways.

1.9

Role of Computers in OR

In the development of OR, computers play a signiﬁcant role. Without computers it would not be possible to achieve OR’s current position. In most OR techniques, the computations are very complicated and require a large number of iterations. Without a computer, these techniques have no practical use. Many large-scale applications of OR techniques, which require only a few seconds/minutes on the computer, would take a very long time (maybe weeks/months/years) to obtain the same results manually. In this context, the computer has become an essential and integral part of OR.

Chapter 2

Convex and Concave Functions

2.1

Objective

The objective of this chapter is to study convex & concave functions and their properties.

2.2

Introduction

Convex and concave functions are the key concepts of mathematical analysis and have interesting consequences in the areas of optimization theory, statistical estimation, inequalities and applied probability. These functions have some properties which can be used in developing suitable optimality conditions and computational schemes for optimization problems.

2.3

Convex and Concave Functions

A function f : SRn ! R (where S is a non-empty convex set) is said to be convex at x 2 S, if f ðkx þ ð1  kÞxÞ  kf ðxÞ þ ð1  kÞf ðxÞ for all x 2 S; 0  k  1 and kx þ ð1  kÞx 2 S:

The function f ðxÞ is said to be convex on S if it is convex at each x 2 S. The general deﬁnition of a convex function is as follows: A function f : S ! R is said to be convex on S, if x1 ; x2 2 S and 0  k  1 implies

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_2

13

14

2 Convex and Concave Functions

f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞf ðx2 Þ: By induction, this inequality can be extended as follows: f

n X

! 

ki x i

i¼1

n X

ki f ðxi Þwhere

i¼1

n X

ki ¼ 1

i¼1

for any convex combination of points from S. If there is a constant c > 0 such that for any x1 ; x2 2 S, 1 f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞf ðx2 Þ  ckð1  kÞkx1  x2 k2 ; 2 then f is called a uniformly convex or strongly convex function on S. A function f deﬁned on a convex set S  Rn is said to be strictly convex at x 2 S if f ðð1  kÞx þ kxÞ\ð1  kÞf ðxÞ þ kf ðxÞ for x 2 S; 0\k\1 and ð1  kÞx þ kx 2 S. f ðxÞ is said to be strictly convex on S if it is strictly convex at each x 2 S. A function f : SRn ! R (where S is a non-empty convex set) is said to be a concave function at x 2 S if f ðkx þ ð1  kÞxÞ  kf ðxÞ þ ð1  kÞf ðxÞ for all x 2 S; 0  k  1 and kx þ ð1  kÞx 2 S:

The function f ðxÞ is said to be concave on S if it is concave at each x 2 S. Then the general deﬁnition is as follows: A function f : S ! R is said to be concave on S if x1 ; x2 2 S and 0  k  1 implies f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞf ðx2 Þ: By induction, the preceding inequality can be extended as follows: f

n X i¼1

! ki x i



n X i¼1

ki f ðxi Þwhere

n X

ki ¼ 1

i¼1

for any convex combination of points from S. If there is a constant c > 0 such that for any x1 ; x2 2 S,

2.3 Convex and Concave Functions

15

1 f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞf ðx2 Þ  ckð1  kÞkx1  x2 k2 ; 2 then f is called a uniformly convex or strongly convex function on S. A function f deﬁned on a convex set S  Rn is said to be strictly concave at x 2 S if f ðð1  kÞx þ kxÞ [ ð1  kÞf ðxÞ þ kf ðxÞ for x 2 S; 0\k\1 and ð1  kÞx þ kx 2 S. f ðxÞ is said to be strictly concave on S if it is strictly concave at each x 2 S. Note that a strictly convex (strictly concave) function on a convex set SRn is convex (concave) on S, but the converse is not always true. For example, a constant function on Rn is both convex and concave on Rn , but neither strictly convex nor strictly concave on Rn .

2.4

Geometrical Interpretation

The geometrical interpretation of the convexity of a single variable function is as follows. For a convex function, the function values are below the corresponding chord (see Fig. 2.1); i.e. the values of a convex function at any point in the interval x1  x  x2 are less than or equal to the height of the chord joining the points ðx1 ; f ðx1 ÞÞ and ðx2 ; f ðx2 ÞÞ. This equality holds only for linear functions (See Fig. 2.1 and Fig. 2.3a, c).

Fig. 2.1 Convexity of single variable function f ðxÞ

16

2 Convex and Concave Functions

For a concave function the function values are above the corresponding chord; i.e. the values of a concave function at any point in the interval x1  x  x2 are greater than or equal to the height of the chord joining the points ðx1 ; f ðx1 ÞÞ and ðx2 ; f ðx2 ÞÞ. (See Figs. 2.2 and 2.3b, c). Examples of Convex Functions The following are some examples of convex functions: f ðxÞ ¼ eax is convex on R, for any a 2 R. f ðxÞ ¼ xa is convex on R þ , when a  1 or a  0. f ðxÞ ¼ j xjp for p  1 is convex on R. The function f with domain [0, 1] deﬁned by f ð0Þ ¼ f ð1Þ ¼ 1 f ðxÞ ¼ 0 for 0\x\1 is convex. It is continuous on the open interval (0, 1) but not continuous at x = 0 and x = 1. (v) Every afﬁne function of the form f ðxÞ ¼ ha; xi þ b in R is convex. Here instead of inequality equality holds, i.e.

(i) (ii) (iii) (iv)

f ðax þ ð1  aÞyÞ ¼ af ðxÞ þ ð1  aÞ f ðyÞ; a 2 ð0; 1Þ; x; y 2 S: ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ pP n 2 (vi) Every Euclidean norm in R, i.e. f ðxÞ ¼ kxk ¼ i¼1 xi , is convex. (vii) The distance to a convex set, i.e. the function d ðx; SÞ from points x 2 Rn to a convex set S, is convex. Since for any convex combination ax þ ð1  aÞy, if we take sequences uðkÞ and vðkÞ from S such that

Fig. 2.2 Concavity of single variable function f ðxÞ

2.4 Geometrical Interpretation

17

Fig. 2.3 a Convex function of two variables. b Concave function of two variables. c Function (of single variable) that is convex over a certain interval and concave over another certain interval

  d ðx; SÞ ¼ lim x  uðkÞ  k!1  d ðy; SÞ ¼ lim y  vðkÞ ; k!1

the points a uðkÞ þ ð1  aÞvðkÞ lie in S, and taking limits in the inequality,   d ½a x þ ð1  aÞy; S  a x þ ð1  aÞy  a uðkÞ  ð1  aÞvðkÞ       ax  uðkÞ  þ ð1  aÞy  vðkÞ ; i:e: d ½ax þ ð1  aÞy; S  a d ðx; SÞ þ ð1  aÞd ðy; SÞ: (viii) The function f ðxÞ ¼ Maxfx1 ; x2 ; . . .; xn g is convex on Rn . The function f ðxÞ ¼ Maxi xi satisﬁes

18

2 Convex and Concave Functions

f ðhx þ ð1  hÞyÞ ¼ Maxi ðhxi þ ð1  hÞyi Þ ½0  h  1  h Maxi xi þ ð1  hÞ Maxi yi ¼ hf ðxÞ þ ð1  hÞf ðyÞ (ix) The function f ðxÞ ¼ logðex1 þ    þ exn Þ is convex on Rn . Examples of Concave Functions Some examples of concave functions are as follows: (i) f ðxÞ ¼ log x on R þ is a concave function. (ii) f ðxÞ ¼ xa for 0  a  1 is concave. Qn 1n (iii) The geometric mean f ðxÞ ¼ on Rn is a concave function. i¼1 xi (iv) f ðxÞ ¼ 8x2 is a strictly concave function. (v) The function f ðxÞ ¼ sin x is concave on the interval ½0; p.

2.5

Properties of Convex and Concave Functions

The properties of convex and concave functions are described as follows: (i) A function f is said to be concave if f is convex. (ii) Every linear function is both concave and convex on Rn . Proof Let f ðxÞ ¼ ha; xi þ bða 2 Rn Þ be a linear function on Rn . Let x1 2 Rn ; x2 2 Rn and 0  k  1. So, kx1 þ ð1  kÞx2 2 Rn . Then f ðkx1 þ ð1  kÞx2 Þ ¼ ha; kx1 þ ð1  kÞx2 i þ b; 8k ¼ kha; x1 i þ ð1  kÞha; x2 i þ b ¼ kfha; x1 i þ bg þ ð1  kÞfha; x2 i þ bg ¼ kf ðx1 Þ þ ð1  kÞf ðx2 Þ: Hence, f ðkx1 þ ð1  kÞx2 Þ ¼ kf ðx1 Þ þ ð1  kÞf ðx2 Þ; 0  k  1, i.e. f ðxÞ is both convex and concave on Rn . (iii) If f : S  Rn ! R is concave, then S is a convex set on Rn . (iv) If f is linear and g is convex, then the composite function g f is convex. (v) If f is convex and g is convex and non-decreasing, then g f is convex. Proof Let Sð Rn Þ be a convex set and 0  k  1. Since f ðxÞ is a convex function on a convex set S, then

2.5 Properties of Convex and Concave Functions

19

f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞf ðx2 Þ: Since g is non-decreasing, we have gðf ðkx1 þ ð1  kÞx2 ÞÞ  gðkf ðx1 Þ þ ð1  kÞf ðx2 ÞÞ:

ð2:1Þ

Again as g is convex, then gðkf ðx1 Þ þ ð1  kÞf ðx2 ÞÞ  kgðf ðx1 ÞÞ þ ð1  kÞgðf ðx2 ÞÞ:

ð2:2Þ

From (2.1) and (2.2) we can write gðf ðkx1 þ ð1  kÞx2 ÞÞ  kgðf ðx1 ÞÞ þ ð1  kÞgðf ðx2 ÞÞ; which shows that g f is convex on S. (vi) The sum of convex (concave) functions is convex (concave). Let f1 ; f2 ; . . .; fk : Rn ! R be convex functions. Then we have the following: P (a) f ðxÞ ¼ kj¼1 aj fj ðxÞ, where aj [ 0 for j ¼ 1; 2; . . .; k, is convex, (b) f ðxÞ ¼ Maxff1 ðxÞ; f2 ðxÞ; . . .; fk ðxÞg is convex. (vii) Let us suppose that g : Rn ! R is a concave function. Let S ¼ 1 . Then f is convex over S. fx : gðxÞ [ 0g and deﬁne f : S ! R as f ðxÞ ¼ gðxÞ (viii) An n-dimensional vector function f deﬁned on a set S in Rn is convex (concave) at x 2 S, convex (concave) on S, if each of its components fi ði ¼ 1; 2; . . .; nÞ is convex (concave) at x 2 S, convex (concave) on S. Theorem Let f ðxÞ be a function deﬁned on a convex set S  Rn . If f ðxÞ is convex on S, then the set Aa ¼ fx : x 2 S; f ðxÞ  ag  S is convex for each real number a. Proof Let x1 ; x2 2 Aa ) f ðx1 Þ  a and f ðx2 Þ  a:

ð2:3Þ

Since f ðxÞ is convex on S, f ðkx1 þ ð1  kÞx2 Þ  kf ðx1 Þ þ ð1  kÞ f ðx2 Þ  ka þ ð1  kÞa ¼ a; i:e: f ðkx1 þ ð1  kÞx2 Þ  a. Hence, kx1 þ ð1  kÞx2 2 Aa and Aa is a convex set. Note that the condition of the preceding theorem is not sufﬁcient; i.e. if Aa is convex for each real number a, it does not follow that f ðxÞ is a convex function on S. Let us consider the function f ðxÞ ¼ x3 which is not convex on R. However, the     set Aa ¼ x : x 2 R; x3  a ¼ x : x 2 R; x  a1=3 is obviously convex for any a.

20

2 Convex and Concave Functions

Corollary Let f ðxÞ be a function deﬁned on a convex set S  Rn . If f ðxÞ is concave on S, then the set Ba ¼ fx : x 2 S; f ðxÞ  ag  S is a convex set for each real number a. Theorem Let f ðxÞ ¼ ðf1 ; f2 ; . . .; fm Þ be an m-dimensional vector function deﬁned on S  Rn . If f is convex (concave) at x 2 S, then each non-negative linear combination of its components fi ; /ðxÞ ¼ hc; f ðxÞi; c ¼ ðc1 ; c2 ; . . .; cm Þ  0 is convex (concave) at x. Proof Here we shall prove the theorem for a convex function. Let x 2 S; 0  k  1 and let kx þ ð1  kÞx 2 S. Then /½kx þ ð1  kÞ x ¼ \c; f ½kx þ ð1  kÞx [  \c; ½k f ðxÞ þ ð1  kÞf ðxÞ [ ¼ k\c; f ðxÞ [ þ ð1  kÞ\c; f ðxÞ [ ½By the convexity of f ðxÞ at x and c  0 ¼ k/ðxÞ þ ð1  kÞ/ðxÞ:

This proves that /ðxÞ is convex at x. Example Let the function f ðxÞ be convex on a convex set S  Rn . Show that for every x1 2 S; x2 2 S, the one-dimensional function gðyÞ ¼ f ðyx1 þ ð1  yÞx2 Þ is convex for all 0  y  1. Solution Let y1 ; y2 2 R. Hence, 0  y1 ; y2  1. So, ky1 þ ð1  kÞy2 2 R for 0  k  1. Now, gðky1 þ ð1  kÞy2 Þ ¼ f ½fky1 þ ð1  kÞy2 gx1 þ f1  ky1  ð1  kÞy2 gx2   As fky1 þ ð1  kÞy2 gx1 þ f1  ky1  ð1  kÞy2 gx2 ¼ kfy1 x1 þ ð1  y1 Þx2 g þ ð1  kÞfy2 x1 þ ð1  y2 Þx2 g ¼ f ½kfy1 x1 þ ð1  y1 Þx2 g þ ð1  kÞfy2 x1 þ ð1  y2 Þx2 g  kf ½y1 x1 þ ð1  y1 Þx2  þ ð1  kÞ f ½y2 x1 þ ð1  y2 Þx2  ¼ kgðy1 Þ þ ð1  kÞgðy2 Þ; which implies that gðyÞ is convex for all 0  y  1.

2.6

Differentiable Convex and Concave Functions

Here we shall discuss some of the properties of differentiable convex and concave functions. Let f be a function deﬁned on an open set S in Rn. If f is differentiable at x 2 S, then for x 2 Rn and x þ x 2 S; f ðx þ xÞ ¼ f ðxÞ þ hrf ðxÞ; xi þ aðx; xÞkxk and limx!0 aðx; xÞ ¼ 0, where rf ðxÞ is the n-dimensional gradient vector of f at x whose n components are the partial derivatives of f with respect to x1 ; x2 ; . . .; xn evaluated at x, and a is a function of x.

2.6 Differentiable Convex and Concave Functions

21

Theorem Let f be a function deﬁned on an open set S  Rn and let f be differentiable at x 2 S. If f is convex at x 2 S, then f ðxÞ  f ðxÞ  hrf ðxÞ; ðx  xÞi for each x 2 S: If f is concave at x 2 S, then f ðxÞ  f ðxÞ  hrf ðxÞ; ðx  xÞi for each x 2 S. Proof Let f be convex at x 2 S. Since S is open, then there exists an open ball Bd ðxÞ around x, which is contained in S. Let x 2 S and x 6¼ x. Then for some l such that 0\l\1 and l\d=kx  xk, we have ^x ¼ x þ lðx  xÞ ¼ ð1  lÞx þ lx 2 Bd ðxÞ  S: Since f is convex at x, from the convexity of Bd ðxÞ and the fact that ^x 2 Bd ðxÞ and 0\k\1, it follows that: ð1  kÞ f ðxÞ þ kf ð^xÞ  f ½ð1  kÞx þ k^x or; f ðxÞ  kf ðxÞ þ kf ð^xÞ  f ½ð1  kÞx þ k^x or; k½ f ð^xÞ  f ðxÞ  f ½ð1  kÞx þ k^x  f ðxÞ xÞf ðxÞ or; f ð^xÞ  f ðxÞ  f ½x þ kð^x k ¼ khrf ðxÞ;ð^xxÞi þ ka½x;kð^xxÞkk^xxk ½As f is differentiable at x 2 S ¼ hrf ðxÞ; ð^x  xÞi þ a½x; kð^x  xÞk^x  xk ) f ð^xÞ  f ðxÞ  hrf ðxÞ; ð^x  xÞi þ a½x; kð^x  xÞk^x  xk: Now taking the limit of both sides as k ! 0, we have f ð^xÞ  f ðxÞ  hrf ðxÞ; ð^x  xÞi; since lim

k!0

a½x; kð^x  xÞ ¼ 0:

Since f is convex at x; ^x 2 S and ^x ¼ ð1  lÞx þ lx;

ð2:4Þ ð2:5Þ

we have ð1  lÞf ðxÞ þ lf ðxÞ  f ½ð1  lÞx þ lx or l½ f ðxÞ  f ðxÞ  f ð^xÞ  f ðxÞ ð2:6Þ " # Since from ð2:4Þ or l½f ðxÞ  f ðxÞ  hrf ðxÞ; ð^x  xÞi f ð^xÞ  f ðxÞ  hrf ðxÞ; ð^x  xÞi " # From ð2:5Þ; since ^x ¼ ð1  lÞx þ lx or l½f ðxÞ  f ðxÞ  hrf ðxÞ; lðx  xÞi ) ^x ¼ x  lx þ lx or; ^x  x ¼ lðx  xÞ or f ðxÞ  f ðxÞ  hrf ðxÞ; ðx  xÞi

½*l [ 0:

22

2 Convex and Concave Functions

The second part of the theorem can be proved similarly. Theorem Let f be a differential function on an open convex set S  Rn . f is convex on S if and only if f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi for each x1 ; x2 2 S: f is concave on S if and only if f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi for each x1 ; x2 2 S: Proof Necessary part: This part of the proof follows from the previous theorem. Sufﬁcient part: Let f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi for each x1 ; x2 2 S:

ð2:7Þ

Since S is convex and x1 ; x2 2 S, )ð1  kÞx1 þ kx2 2 S for 0  k  1. Now from (2.4), we have f ðx1 Þ  f ½ð1  kÞx1 þ kx2   hrf ½ð1  kÞx1 þ kx2 ; ½x1  fð1  kÞx1 þ kx2 gi for x1 2 S and ð1  kÞx1 þ kx2 2 S or f ðx1 Þ  f ½ð1  kÞx1 þ kx2   hrf ½ð1  kÞx1 þ kx2 ; fx1  x1 þ kx1  kx2 gi or f ðx1 Þ  f ½ð1  kÞx1 þ kx2   khrf ½ð1  kÞx1 þ kx2 ; ðx1  x2 Þi: ð2:8Þ Again for x2 2 S; ð1  kÞx1 þ kx2 2 S, we have from (2.7) f ðx2 Þ  f ½ð1  kÞx1 þ kx2   hrf ½ð1  kÞx1 þ kx2 ; ½x2  fð1  kÞx1 þ kx2 gi or f ðx2 Þ  f ½ð1  kÞx1 þ kx2   hrf ½ð1  kÞx1 þ kx2 ; ½ð1  kÞx1 þ ð1  kÞx2 i or f ðx2 Þ  f ½ð1  kÞx1 þ kx2    ð1  kÞhrf ½ð1  kÞx1 þ kx2 ; ðx1  x2 Þi: ð2:9Þ Now multiply (2.8) by ð1  kÞ and (2.9) by k; then adding, we have ð1  kÞf ðx1 Þ  ð1  kÞf ½ð1  kÞx1 þ kx2  þ kf ðx2 Þ  kf ½ð1  kÞx1 þ kx2   kð1  kÞhrf ½ð1  kÞx1 þ kx2 ; ðx1  x2 Þi  kð1  kÞhrf ½ð1  kÞx1 þ kx2 ; ðx1  x2 Þi or ð1  kÞf ðx1 Þ þ kf ðx2 Þ  ð1  k þ kÞf ½ð1  kÞx1 þ kx2   0 or ð1  kÞf ðx1 Þ þ kf ðx2 Þ  f ½ð1  kÞx1 þ kx2 :

Hence, f is convex on S. Similarly, we can prove that f is concave on S.

2.6 Differentiable Convex and Concave Functions

23

Theorem Let f be a differentiable function on an open convex set S  Rn . A necessary and sufﬁcient condition that f be convex (concave) on S is that for each x1 ; x2 2 S h½rf ðx2 Þ  rf ðx1 Þ; ðx2  x1 Þi  0 ð  0Þ: Proof We shall establish the theorem for the convex case only. The concave case is similar. The condition is necessary. Let f be convex on S and let x1 ; x2 2 S. Then we have f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi, i:e: f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi  0 and

ð2:10Þ

f ðx1 Þ  f ðx2 Þ  hrf ðx2 Þ; ðx1  x2 Þi;

i:e: f ðx1 Þ  f ðx2 Þ   hrf ðx2 Þ; ðx2  x1 Þi or

f ðx1 Þ  f ðx2 Þ þ hrf ðx2 Þ; ðx2  x1 Þi  0:

ð2:11Þ

Now adding these two inequalities (2.10) and (2.11), we have f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi þ f ðx1 Þ  f ðx2 Þ þ hrf ðx2 Þ; ðx2  x1 Þi  0 or hrf ðx2 Þ; ðx2  x1 Þi  hrf ðx1 Þ; ðx2  x1 Þi  0 or h½rf ðx2 Þ  rf ðx1 Þ; ðx2  x1 Þi  0: The condition is sufﬁcient. Let h½rf ðx2 Þ  rf ðx1 Þ; ðx2  x1 Þi  0 for each x1 ; x2 2 S:

ð2:12Þ

We shall prove that f is convex on S. Since S is a convex set and x1 ; x2 2 S, then for 0  k  1; ð1  kÞx1 þ kx2 2 S. Now by the mean value theorem, we have for some k; 0\k\1,

f ðx2 Þ  f ðx1 Þ ¼ rf x1 þ kðx2  x1 Þ ; ðx2  x1 Þ : But by inequality (2.12), 



rf x1 þ kðx2  x1 Þ  rf ðx1 Þ ; kðx2  x1 Þ  0 



or rf x1 þ kðx2  x1 Þ  rf ðx1 Þ ; ðx2  x1 Þ  0

or rf x1 þ kðx2  x1 Þ ; ðx2  x1 Þ  hrf ðx1 Þ; ðx2  x1 Þi or f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi ½using ð2:4Þ:

ð2:13Þ

24

2 Convex and Concave Functions

Hence, f is convex on S [By the theorem ‘if f is a differentiable function on an open convex set S  Rn , f is convex on S and if f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi then f is convex on S.’] Note If f is an n-dimensional function on S  Rn and h½f ðx2 Þ  f ðx1 Þ; ðx2  x1 Þi  0 for all x1 ; x2 2 S, then f is said to be monotonic on S. So from the previous theorem, a differentiable function on the open convex set S  Rn is convex if and only if rf is monotonic on S.

2.7

Differentiable Strictly Convex and Concave Functions

Theorem Let f be a function deﬁned on an open set S  Rn and let f be differentiable at x 2 S. If f is strictly convex at x 2 S, then f ðxÞ  f ðxÞ [ hrf ðxÞ; ðx  xÞi for each x 2 S; x 6¼ x: If f is strictly concave at x 2 S, then f ð xÞ  f ðxÞ\hrf ðxÞ; ðx  xÞi for each x 2 S; x 6¼ x: Proof Let f be strictly convex at x. Then f ½ð1  kÞx þ kx\ð1  kÞf ðxÞ þ kf ðxÞ for all x 2 S; x 6¼ x

ð2:14Þ

0\k\1; and ð1  kÞx þ kx 2 S: Since f is convex and differentiable at x 2 S, then f ðxÞ  f ðxÞ  hrf ðxÞ; ðx  xÞi for all x 2 S:

ð2:15Þ

We shall now show that if equality holds in (2.15) for some x in S which is distinct from x, then a contradiction arises. Let the equality hold in (2.15) for x ¼ ~x; ~x 2 S and ~x 6¼ x. Then f ð~xÞ  f ðxÞ ¼ hrf ðxÞ; ð~x  xÞi or f ð~xÞ ¼ f ðxÞ þ hrf ðxÞ; ð~x  xÞi:

ð2:16Þ

Since ~x 2 S, from (2.14) we have f ½ð1  kÞx þ k~x\ð1  kÞ f ðxÞ þ k f ð~xÞ ½putting x ¼ ~x in ð2:14Þ or f ½ð1  kÞx þ k~x\ð1  kÞ f ðxÞ þ k½ f ðxÞ þ hrf ðxÞ; ð~x  xÞi ½using ð2:16Þ for 0\k\1 and ð1  kÞx þ k~x 2 S or f ½ð1  kÞx þ k~x\ f ðxÞ þ khr f ðxÞ; ð~x  xÞi:

ð2:17Þ

2.7 Differentiable Strictly Convex and Concave Functions

25

Again for the points x ¼ ð1  kÞx þ k~x 2 S and x, from (2.15) we have f ½ð1  kÞx þ k~x  f ðxÞ  hrf ðxÞ; kð~x  xÞi or f ½ð1  kÞx þ k~x  krf ðxÞð~x  xÞ þ f ðxÞ:

ð2:18Þ

Relation (2.18) contradicts relation (2.13). For small k; ð1  kÞx þ k~x 2 S is open. Hence, equality cannot hold in (2.15) for any x 2 S distinct. Theorem Let f ðxÞ be a twice differentiable function on an open convex set S  Rn and HðxÞ be the Hessian matrix of f ðxÞ. Then f ðxÞ is convex on S if and only if HðxÞ is positive semi-deﬁnite for each x 2 S. Proof Let x 2 S and y 2 Rn . Since S is convex and open, there exists a k [ 0 such that x þ ky 2 S for 0\k\k. Since f ðxÞ is a twice differentiable function on S, then f ðx þ kyÞ ¼ f ðxÞ þ khrf ðxÞ; yi þ

k2 hHðxÞy; yi þ k2 bðx; kyÞkyk2 8 k ð0\k\kÞ: 2

When b ! 0 as k ! 0, f ðx þ kyÞ  f ðxÞ  khrf ðxÞ; yi ¼

i k2 h hHðxÞy; yi þ 2bkyk2 : 2

For a differentiable convex function, we know that f ðx þ hkyÞ  f ðxÞ  khrf ðxÞ; i yi  0 2 k2 or 2 hHðxÞy; yi þ 2bk yk  0 or hHðxÞy; yi þ 2bkyk2  0 for 0\k\k: Letting k ! 0, we have hHðxÞy; yi  0 for all y 2 Rn , which means that HðxÞ is positive semi-deﬁnite for all x 2 S. Let the Hessian matrix HðxÞ be positive semi-deﬁnite for each x 2 S. Hence, hHðxÞy; yi  0 for all y 2 Rn . Now by Taylor’s theorem, we have for x1 ; x2 2 S 1 f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi ¼ hH ðx1 þ hðx2  x1 ÞÞðx2  x1 Þ; ðx2  x1 Þi 2 for some 0\h\1. Since x1 þ hðx2  x1 Þ 2 S, the right-hand side of the preceding equation is non-negative as hHðxÞy; yi  0 for all y 2 Rn . Hence, f ðx2 Þ  f ðx1 Þ  hrf ðx1 Þ; ðx2  x1 Þi  0 for x1 ; x2 2 S, i.e. f ðxÞ is convex on S.

26

2 Convex and Concave Functions

Theorem Let f ðxÞ be a twice differentiable function on an open convex set Sð Rn Þ and HðxÞ be the Hessian matrix of f ðxÞ. Then f ðxÞ is strictly convex on S if HðxÞ is positive deﬁnite for each x 2 S. Remark (i) The converse of the previous theorem is not true. To prove this, let us consider the function y ¼ x4 ; x 2 R. Hence, HðxÞ ¼ 12x2 , which is not positive deﬁnite since Hð0Þ ¼ 0, but f ðxÞ is strictly convex on R because f 0 ðxÞ ¼ 4x3 is strictly increasing on R. (ii) Let f ðxÞ be a twice differentiable function of a single variable x on an open set Sð Rn Þ. Then f ðxÞ is strictly convex if f 00 ðxÞ is positive for each x 2 S. Example Determine whether the following functions are convex or concave: (a) (b) (c) (d)

f ðxÞ ¼ ex f ðxÞ ¼ 8x2 f ðx1 ; x2 Þ ¼ 2x31  6x22 f ðx1 ; x2 ; x3 Þ ¼ 4x21 þ 3x22 þ 5x23 þ 6x1 x2 þ x1 x3  3x1  2x2 þ 15

Solution (a) f ðxÞ ¼ ex ) f 00 ðxÞ ¼ ex [ 0 for all real values of x. Hence, f ðxÞ is strictly convex. (b) f ðxÞ ¼ 8x2 ; f 00 ðxÞ ¼ 16\0 for all real values of x. Hence, f ðxÞ is strictly concave. (c) f ðx1 ; x2 Þ ¼ 2x31  6x22 2 HðxÞ ¼

@2 f 2 4 @x2 1 @ f @x1 @x2

Since

@2 f @x1 @x2 @2 f @x22

3

 5 ¼ 12x1 0

0 : 12

@2f ¼ 12x1  0 for x1  0 and @x21  0 for x1  0

and jHðxÞj ¼ 144x1  0 for x1  0  0 for x1  0; then f ðxÞ will be negative semi-deﬁnite for x1  0, i.e. f ðxÞ is convex for x1  0. (d) f ðx1 ; x2 ; x3 Þ ¼ 4x21 þ 3x22 þ 5x23 þ 6x1 x2 þ x1 x3  3x1  2x2 þ 15

2.7 Differentiable Strictly Convex and Concave Functions

2 HðxÞ ¼

@2 f 2 @x 6 21 6 @f 6 @x1 @x2 4 2 @ f @x1 @x3

@2 f @x1 @x2 @2 f @x22 @2 f @x2 @x3

@2 f @x1 @x3 @2 f @x2 @x3 @2 f @x23

3

2 8 7 7 4 ¼ 6 7 5 1

27

3 6 1 6 0 5: 0 10

Here the leading principal minors are given by j8j ¼ 8 [ 0:   8 6 1      8 6      6 6  ¼ 12 [ 0 and  6 6 0  ¼ 114 [ 0:  1 0 10  Hence, the matrix HðxÞ is positive deﬁnite for all real values of x1 ; x2 and x3 . Therefore, f ðxÞ is a strictly convex function.

Chapter 3

Simplex Method

3.1

Objective

The objective of this chapter is to discuss: • The concept of a linear programming problem (LPP) • The concepts of slack/surplus variables and canonical and standard forms of LPPs • The theories and algorithm of the simplex method • The Big M and two-phase methods • Finding the solution of linear simultaneous equations and inverse of a non-singular matrix by the simplex method.

3.2

General Linear Programming Problem

The general linear programming problem (LPP) is a problem of determining the value of n decision variables x1 ; x2 ; . . .; xn which optimizes (maximizes or minimizes) the linear function z ¼ c1 x1 þ c2 x2 þ    þ cn xn

ð3:1Þ

9 a11 x1 þ a12 x2 þ    þ a1n xn  ; ¼;  b1 > > = a21 x1 þ a22 x2 þ    þ a2n xn  ; ¼;  b2       > > ; am1 x1 þ am2 x2 þ    þ amn xn  ; ¼;  bm

ð3:2Þ

subject to the inequalities

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_3

29

30

3 Simplex Method

and xj  0; j ¼ 1; 2; . . .; n:

ð3:3Þ

The linear function z ¼ c1 x1 þ c2 x2 þ    þ cn xn which is to be maximized (or minimized) is called the objective function of the general LPP. The inequalities (3.2) are called the constraints of the general LPP, whereas the set of inequalities (3.3) is usually known as the non-negativity conditions of the general LPP. The right-hand quantities bi ði ¼ 1; 2; . . .; mÞ of the constraints (3.2) are known as the requirement parameters of the problem. In a compact form, the general LPP can be written as follows: Optimize z ¼

n P

cj xj

j¼1

subject to the constraints n P aij xj  ; ¼;  bi ; i ¼ 1; 2; . . .; m j¼1

and xj  0; j ¼ 1; 2; . . .; n: In matrix notation, the general LPP can be written as follows: Optimize z ¼ hcT ; xi subject to the constraints Ax  ; ¼;  b and x  0; where c; xT 2 Rn ; bT 2 Rm , and A is an m  n real matrix. Note The notation hcT ; xi represents the inner product of two vectors. In some books, it is denoted by either cT x or cx. In a mathematical sense, hcT ; xi is the appropriate representation as cT x or cx is a matrix.

3.3

Slack and Surplus Variables

In the formulation of a general LPP, if the constraint is in the form of a  type of inequality, such as ai1 x1 þ ai2 x2 þ    þ ain xn  bi ; then to convert this inequality into an equation, a non-negative quantity xn þ i is added on the left-hand side of the constraint. The converted equation is as follows:

3.3 Slack and Surplus Variables

31

ai1 x1 þ ai2 x2 þ    þ ain xn þ xn þ i ¼ bi : Hence, the non-negative quantity xn þ i is known as a slack variable. Similarly, if a constraint is in the form of a  type of inequality, such as ai1 x1 þ ai2 x2 þ    þ ain xn  bi ; then to transform this inequality into an equation, a non-negative quantity xn þ i is subtracted from the left-hand side of the constraint. The converted equation is as follows: ai1 x1 þ ai2 x2 þ    þ ain xn  xn þ i ¼ bi : Hence, the non-negative quantity xn þ i is known as a surplus variable. Note that the coefﬁcients of slack and/or surplus variables in the objective function are always assumed to be zero. This means that the conversion of the constraints to a system of simultaneous linear equations does not change the objective function.

3.4

Canonical and Standard Forms of LPP

After the formulation of the LPP, the next step is to determine its solution. But, to solve the LPP, the problem must be expressed in the following two forms: (i) Canonical form (ii) Standard form. Canonical form The formulation of a general LPP can always be expressed in the following form: n P Maximize z ¼ cj xj j¼1

subject to the constraints n P aij xj  bi ; i ¼ 1; 2; . . .; m j¼1

and xj  0;

j ¼ 1; 2; . . .; n

This form of LPP is called the canonical form of the LPP. The characteristics of this form are as follows:

32

3 Simplex Method

1. The objective function is of the maximization type. If the objective function is of the minimization type, it can be expressed as a maximization type by using the following mini-max property: Minimize f ð xÞ ¼ Maximize ff ð xÞg For example, the objective function Minimize z ¼ 3x1 þ x2  2x3 is equivalent to Maximize z0 ¼ z ¼ 3x1  x2 þ 2x3 : 2. All the constraints are of  type, except the non-negativity restrictions. An inequality of  type can changed to an inequality of  type by multiplying both sides of the inequality by ð1Þ. For example, the constraint 3x1 þ 2x2  x3  6 is equivalent to 3x1  2x2 þ x3   6: On the other hand, an equation can be replaced by two weak inequalities. For example, the constraint 5x1  x2 þ 7x3 ¼ 15 is equivalent to 5x1  x2 þ 7x3  15 and 5x1  x2 þ 7x3  15; i.e. 5x1  x2 þ 7x3  15 5x1 þ x2  7x3   15: 3. All the variables are non-negative. If a variable is unrestricted in sign (i.e. the value of that variable may be positive, negative or zero), it can be replaced by the difference between two non-negative variables. For example, if a variable x1 is unrestricted in sign, we can replace this variable by x01  x001 where both x01 and x001 are non-negative, i.e. x1 ¼ x01  x001 where x01  0 and x001  0.

3.4 Canonical and Standard Forms of LPP

33

Standard form A general LPP in the form Maximize or Minimize z ¼

n P

cj xj

j¼1

subject to the constraints n P aij xj ¼ bi ; i ¼ 1; 2; . . .; m j¼1

and xj  0; j ¼ 1; 2; . . .; n is known as the standard form of the LPP. The characteristics of this form are as follows: 1. All the constraints are expressed in the form of equations, except the non-negativity restrictions. 2. The right-hand side of each constraint equation is non-negative. The inequality constraint can be converted to an equation by introducing slack/ surplus variables on the left-hand side of that constraint.

3.5

Fundamental Theorem of LPP

Theorem If the LPP admits an optimal solution, then the optimal solution will coincide with at least one basic feasible solution of the problem. Proof Here we shall prove the theorem for the maximization problem. Let us assume that x is an optimal solution of the following LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b and x  0:

ð3:4Þ

Here A is an m  n (m\n) matrix and is given by A ¼ ½a1 ; a2 ; . . .; an , where aj is an m-component . . .; amj T ; j ¼ 1; 2; . . .; n.

column

vector

given

by

aj ¼ ½a1j ; a2j ;

Without any loss of generality, let us assume that the ﬁrst k components of x are non-zero and the remaining ðn  kÞ components of x are zero. Thus, x ¼ ½x1 ; x2 ; . . .; xk ; 0; 0; . . .; 0. Then, from (3.4), Ax ¼ b gives

34

3 Simplex Method k X

aij xj ¼ bi ; i ¼ 1; 2; . . .; m

j¼1

in which A ¼ ½a1 ; a2 ; . . .; ak ; ak þ 1 ; . . .; an  so that a1 x1 þ a2 x2 þ   ; ak xk ¼ b Also z ¼ zMax ¼

k P

ð3:5Þ

cj xj .

j¼1

Now, if the vectors a1 ; a2 ; . . .; ak corresponding to the non-zero components of x are linearly independent, then by deﬁnition x is a basic feasible solution; hence, the theorem holds in this case. If k ¼ m, then the basic feasible solution will be non-degenerate. Again, if k\m, then it will form a degenerate basic feasible solution with m  k of the basic variables equal to zero. But if the vectors a1 ; a2 ; . . .; ak are not linearly independent, then they must be linearly dependent and there exist scalars kj ; j ¼ 1; 2; . . .; k, with at least one non-zero kj such that k1 a1 þ k2 a2 þ   ; kk ak ¼ 0

ð3:6Þ

Let us suppose at least one kj [ 0; if the non-zero kj is not positive, then by multiplying (3.6) by (−1), one can get a positive kj .   kj : 1  j  k xj

Let l ¼ Max

ð3:7Þ

This l is positive as xj [ 0 for all j ¼ 1; 2; . . .; k and at least one kj is positive. Dividing (3.6) by l and subtracting it from (3.5), we have 

     k1 k2 kk x1  a1 þ x 2  a2 þ    þ x k  ak ¼ b l l l

and hence X1 ¼

     k1 kk x1  ; . . .; xk  ; 0; 0; . . .; 0 l l

is a solution of the system of equations Ax ¼ b.

ð3:8Þ

3.5 Fundamental Theorem of LPP

35

Again, from (3.7), we have k

l  xjj , for k

or xj  lj , for

j ¼ 1; 2; . . .; k j ¼ 1; 2; . . .; k

k

or xj  lj  0, for j ¼ 1; 2; . . .; k. This means that all the components of X1 are non-negative and hence X1 is a feasible solution of Ax ¼ b, x  0. Again, for at least one value of j, we have, from (3.7), l ¼

kj xj

k

or, in other words, xj  lj ¼ 0, for at least one value of j. Hence, it is seen that the feasible solution X1 will contain one more zero; i.e. the feasible solution X1 cannot contain more than ðk  1Þ non-zero variables. Therefore, the number of positive variables in an optimal solution can be reduced. Now we have to show that X1 is optimal. Let z0 be the value of the objective function for x ¼ X1 . Then z0 ¼

  X k k k X X kj kj c j xj  cj xj  cj ¼ l l j¼1 j¼1 j¼1

k 1X ¼z  c j kj l j¼1 



½* z ¼

k X

ð3:9Þ

cj xj :

j¼1

If possible, let us suppose that k X

cj kj 6¼ 0:

j¼1

Then for a suitable real number b, we can write bðc1 k1 þ c2 k2 þ   ; þ ck kk Þ [ 0; That is, c1 ðbk1 Þ þ c2 ðbk2 Þ þ    ; þ ck ðbkk Þ [ 0 Adding ðc1 x1 þ c2 x2 þ    ; þ ck xk Þ to both sides, we get

ð3:10Þ

36

3 Simplex Method

c1 ðx1 þ bk1 Þ þ c2 ðx2 þ bk2 Þ þ    ; þ ck ðxk þ bkk Þ [ c1 x1 þ c2 x2 þ   ; þ ck xk ¼ z :

ð3:11Þ

Again, multiplying (3.6) by b and adding it with (3.5), we get ðx1 þ bk1 Þa1 þ ðx2 þ bk2 Þa2 þ   ; þ ðxk þ bkk Þak ¼ b This relation implies that ½ðx1 þ bk1 Þ; . . .; ðxk þ bkk Þ; 0; 0; . . .; 0T

ð3:12Þ

is also the solution of the system Ax ¼ b. Now we choose b in such a way that xj þ bkj  0 for all j ¼ 1; 2; . . .; k or bkj   xj x or b   kjj , if kj [ 0, x

b   kjj ; if kj \0 and b is unrestricted, if kj ¼ 0. Then

    xj xj Max  ; kj [ 0  b  Min  ; kj \0 : j j kj kj

For this b, (3.12) becomes a feasible solution of Ax ¼ b; x  0. Hence, from (3.11) it is clear that the feasible solution (3.12) gives a greater value of the objective function than z for x . But this contradicts our assumption that z is the optimal value, and thus we must have k X

cj kj ¼ 0:

j¼1

Hence, X1 is also an optimal solution. Thus, it is seen that from the given optimal solution, we can construct a new optimal solution, whose number of non-zero variables is less than that of the given solution. If the vectors associated with the new non-zero variables are linearly independent, then the new solution will be a basic feasible solution, and hence it proves the theorem.

3.5 Fundamental Theorem of LPP

37

If the new solution is not a basic feasible solution, then we can further diminish the number of non-zero variables as above to get a new set of optimal solutions. We may continue the process until the optimal solution obtained is a basic feasible solution. This proves the theorem. For minimization problems, the theorem can be proved in a similar way.

3.6

Replacement of a Basis Vector

Theorem Let an LPP have a basic feasible solution. If one of the basis vectors is removed from the basis and a non-basic vector is introduced in a basis, then the new solution obtained is also a basic feasible solution. Proof Let us consider the LPP  Maximize z ¼ cT ; x subject to Ax ¼ b and x  0 where c; xT 2 Rn ; bT 2 Rm and A is an m  n real matrix: Also, let the rank of A be m. Let xB be a basic feasible solution to the LPP. ) BxB ¼ b

ð3:13Þ

where B is the basis matrix. Let aj be any column vector of A. Then aj ¼ y1j B1 þ y2j B2 þ    þ yrj Br þ    þ ymj Bm

ð3:14Þ

where Bi 2 B and yij are suitable scalars. Now, we know that if a basis vector Br with non-zero coefﬁcient is replaced by aj 2 A, then the new set of vectors also forms a basis. Now, for yrj 6¼ 0, from (3.14) we can write Br ¼

m X aj yij  Bi : yrj yrj i¼1 i 6¼ r

ð3:15Þ

38

3 Simplex Method

From (3.13), BxB ¼ b which implies

m P

xBi Bi þ xBr Br ¼ b

i¼1 i6¼r

or

m P

i¼1 i6¼r

or

2

3

m 6a P

xBi Bi þ xBr 4y j  rj

h m P

yij xB xBr y i rj

i

i¼1 i6¼r

yij yrj Bi

7 5¼b

xB Bi þ y r aj ¼b rj

i¼1 i6¼r

From the preceding relation, it is clear that the components of the new basic solution ^xB are given by ^xBi ¼ xBi  xBr

yij ; i ¼ 1; 2; . . .; m; i 6¼ r yrj

and ^xBr ¼ xyBrjr . Now we shall show that ^xB is also feasible; i.e. all the components of ^xB are non-negative. According to the value of xBr , two cases may arise. Case 1 xBr ¼ 0. In this case, ^xBi ¼ xBi  0; i ¼ 1; 2; . . .; m; i 6¼ r and xBr ¼ 0 i.e. ^xB  0. Case 2 xBr 6¼ 0. In this case, we must have yrj [ 0. For the remaining yij ði 6¼ rÞ, either yij ¼ 0 or yij 6¼ 0 for i 6¼ r. Now for yij 6¼ 0 for i 6¼ r, we have xBi xB  r for yij [ 0 and i 6¼ r yij yrj xBi xB  r for yij \0 and i 6¼ r: and yij yrj Therefore, if we select rðyrj 6¼ 0Þ in such a way that xyBrjr ¼ Min i

n

xBi yij

o ; yij [ 0; i 6¼ r ,

then the new set of basic variables is non-negative; hence, the basic solution ^xB is feasible.

3.7 Improved Basic Feasible Solution

3.7

39

Improved Basic Feasible Solution

Theorem Let xB be a basic feasible solution to the LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b and x  0; where c; xT 2 Rn , bT 2 Rm and A is an m  n real matrix. If ^xB is another basic feasible solution obtained by introducing a non-basic column vector aj in the basis, for which zj  cj is negative, then ^xB is an improved basic feasible solution to the problem, i.e.  T  T ^cB ; ^xB [ cB ; xB : Proof It is given that xB is a basic feasible solution.  Let z0 ¼ cTB ; xB . Let ^ aj be the column vector introduced in the basis such that zj  cj \0. Let Br be the vector removed from the basis and let ^xB be the new basic feasible solution. Then yij xB ^xBi ¼ xBi  xBr ði 6¼ rÞ and ^xBr ¼ r : yrj yrj Hence, the new value of the objective function is given by ^z ¼

m P

^cBi ^xBi ¼

i¼1 ¼

m P

m P

y

cBi ðxBi xBr y ij Þ þ ^cBr rj

i¼1 i6¼r ¼

m P

yij xB cB ðxB xBr y Þ þ cj y r i i rj rj

i¼1 i6¼r

¼

m P

i¼1

cBi ^xBi þ ^cBr ^xBr

i¼1 i6¼r

cB xB  i i

m P

i¼1 i6¼r

xBr yrj

½* ^cBr ¼cj 

yij xB cB xBr y cBr xBr þ cj y r i rj rj

40

3 Simplex Method m X

xB r xB þ cj r yrj yrj i¼1 ! m X xB r ¼ z0  cBi yij  cj yrj i¼1

xBr ¼ z 0  z j  cj yrj ¼ z0 

cBi yij

[ z0 : Hence, the new basic feasible solution ^xB gives an improved value of the objective function. Corollary If zj  cj ¼ 0 for at least one j corresponding to the non-basis vector for which yij [ 0, i ¼ 1; 2; . . .; m, then a new basic feasible solution is obtained which gives an unchanged value of the objective function. Proof From the preceding theorem,

xB r ^z ¼ z0  zj  cj yrj ¼ z0

; yrj [ 0

½* zj  cj ¼ 0:

Note According to the preceding theorem, for the maximum improvement, determine the entering vector corresponding to the net evolution as follows:

zk  ck ¼ Min zj  cj : zj  cj \0 : j

3.8

Optimality Condition

Theorem For a basic feasible solution xB of an LPP  Maximize z ¼ cT ; x subject to Ax ¼ b and x  0 where c; xT 2 Rn and bT 2 Rm ; if zj  cj  0 for every column vector aj of A, then xB is an optimal solution. Proof Let B ¼ ðB1 ; B2 ; . . .; Bm Þ be the basis matrix corresponding to the basic feasible solution xB so that BxB ¼ b.

3.8 Optimality Condition

41

Also, let zB be the value of the objective function corresponding to xB . m  P Hence, zB ¼ cTB ; xB ¼ cBi xBi . i¼1

 T Let X1 ¼ x01 ; x02 ; . . .; x0n be any other feasible solution of the LPP and let z0 be the value of the objective function corresponding to the solution. Then we have BxB ¼ b ¼ AX1 so that xB ¼ B1 ðAX1 Þ ¼ ðB1 AÞX1 ¼ yX1 ;

ð3:16Þ

where B1 A ¼ y ¼ ½y1 ; y2 ; . . .; yn . Now equating the ith component of (3.16) from both sides, we have xB i ¼

n X

yij x0j :

ð3:17Þ

j¼1

It is given that for every column vector aj of A, zj  cj  0; i:e: zj  cj ; j ¼ 1; 2; . . .; n:

ð3:18Þ

Hence, for all j ¼ 1; 2; . . .; n, zj x0j  cj x0j ; as x0j  0 n n P P or zj x0j  cj x0j j¼1 n P

j¼1

x0j ðcTB yj Þ  z0 m  P or x0j cBi yij  z0 j¼1 i¼1 ! m n P P 0 or cB i xj yij  z0 i¼1 j¼1 " # m n P P 0 0 or cBi xBi  z , From ð3:17Þ xBi ¼ yij xj or

j¼1 n P

i¼1

j¼1

or zB  z0 . This shows that the value of zB is greater than the objective function for any other feasible solution, and hence zB is the maximum value of the objective function. For a minimization problem, the optimality condition is as follows.

42

3 Simplex Method

Theorem For a basic feasible solution xB of an LPP  Minimize z ¼ cT ; x subject to Ax ¼ b and x  0; where c; xT 2 Rn ; bT 2 Rm and A is an m  n matrix; if zj  cj  0 for every column vector aj of A, then xB is an optimal solution.

3.9

Unbounded Solution

If the objective function has no ﬁnite optimum value and it can be increased or decreased arbitrarily, then we say that the problem has an unbounded solution. Theorem If at any iteration of the simplex algorithm, zj  cj \0 for at least one j, and for this j, yij \0 for all i ¼ 1; 2; . . .; m, then the LPP admits an unbounded solution in a maximization problem. Proof Let xB be the basic feasible solution of an LPP with basis matrix B at any iteration in which zj  cj \0 and yij \0, for all i ¼ 1; 2; . . .; m. If zB is the value of the objective function for this basic feasible solution, then  zB ¼ cTB ; xB and BxB ¼ b:

ð3:19Þ

Again, BxB ¼ b implies that m X

xBi Bi ¼ b;

ð3:20Þ

i¼1

where B ¼ ðB1 ; B2 ; . . .; Bm Þ, i.e. the Bi ’s are column vectors of B. In (3.20), adding and subtracting caj , where c is a positive scalar, we get m X

xBi Bi þ caj  caj ¼ b:

i¼1

Again, we know that aj ¼

m P

yij Bj ,

i¼1

i.e. caj ¼ c

m P

yij Bi .

i¼1

Using this result in (3.21), we have

ð3:21Þ

3.9 Unbounded Solution

43 m X

xBi Bi þ caj  c

i¼1

or

m X

yij Bi ¼ b

i¼1 m X

ð3:22Þ

ðxBi cyij ÞBi þ caj ¼ b:

i¼1

If now c is assumed to be positive and since yij \0; for i ¼ 1; 2; . . .; m; then we have xBi  cyij  0:

ð3:23Þ

Now, from (3.22) and (3.23), it is seen that we have obtained a new set of feasible solutions in which the ðm þ 1Þ variables xBi  cyij , together with c, can be different from zero and the remaining components are zero. This solution in general is feasible but not a basic solution. Then z0 ¼ ¼

m X i¼1 m X i¼1

cBi ðxBi cyij Þ þ cj c cBi xBi c

m X

cBi yij þ cj c

ð3:24Þ

i¼1

¼ zB  cðzj  cj Þ: From (3.24), it is observed that z0 will always be greater than zB as c [ 0 and ðzj  cj Þ\0. As c increases, z0 also goes on increasing, and by making c sufﬁciently large, we can make z0 as large as desired. Hence, we conclude that under such circumstances there is no upper bound of the objective function, although the solution is feasible. Hence, the problem admits an unbounded solution. For a minimization problem, the corresponding property is as follows: If at any iteration of the simplex algorithm we get ðzj  cj Þ [ 0 for at least one j, and for this j; yij \0, for all i ¼ 1; 2; . . .; m, then the LPP admits an unbounded solution in a minimization problem. Theorem Any convex combination of k different optimal solutions to an LPP is again an optimum solution to the problem. Proof Let X1 ; X2 ; . . .; Xk be k different optimum solutions of the following LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b and x  0; where c; xT 2 Rn ; bT 2 Rm and A is an m  n real matrix:

44

3 Simplex Method

Let z0 be the optimum value of z. Hence, z0 ¼ hc; X1 i ¼ hc; X2 i ¼    ¼ hc; Xk i

ð3:25Þ

AX1 ¼ AX2 ¼    ¼ AXk ¼ b;

ð3:26Þ

and

where X1  0; X2  0; . . .; Xk  0: Now, any combination of X1 ; X2 ; . . .; Xk can be written as x¼

k P

ai Xi , where the ai ’s are positive scalars such that

i¼1

k P

ai ¼ 1.

i¼1

Now k X

Ax ¼ A

! ai Xi

¼

i¼1

¼

k X

ai b

k X

ai ðAXi Þ

i¼1

½using ð3:26Þ

i¼1

¼b

k X

ai ¼ b;

since

i¼1

k X

ai ¼ 1:

i¼1

Again, since Xi  0 and ai  0 for all i ¼ 1; 2; . . .; k, then every component of x will be non-negative. Thus, x is a feasible solution to the LPP. Again, 

*

c ; x ¼ T

c ; T

k X i¼1

+ ai X i

¼

k  X

cT ; ai Xi

i¼1

k k X  X ¼ ai c T ; X i ¼ ai z 0 ¼ z 0 i¼1

i¼1

" *

k X

# ai ¼ 1 :

i¼1

Hence, a convex combination of k different optimum solutions to an LPP is also an optimal solution to the LPP.

3.10

3.10

Simplex Algorithm

45

Simplex Algorithm

The iterative procedure for ﬁnding an optimal basic feasible solution (BFS) to an LPP is as follows: Step 1: Check whether the given LPP is a maximization or minimization problem. If it is a minimization problem, then convert it into a problem of maximization, using Minimize z ¼ Maximize ðzÞ; where z is the objective function of the given LPP. Step 2: Check whether all bi ði ¼ 1; 2; . . .; mÞ are non-negative or not. If any one of bi is negative, then multiply the corresponding inequality/equation of the constraint by (−1) so that all bi ði ¼ 1; 2; . . .; mÞ will be positive. Step 3: Convert all the inequality constraints into equations by introducing slack and/or surplus variables in the constraints. Put the coefﬁcients of these variables in the objective function as zero. Step 4: Obtain an initial BFS xB using xB ¼ B1 b, where B is the initial unit basis matrix. Alternatively, by taking non-basic variables as zero, obtain the initial BFS. Step 5: Construct the simplex table. Step 6: Compute the net evaluations zj  cj ðj ¼ 1; 2; . . .; nÞ by using the for mula zj  cj ¼ cB ; yj  cj and examine their sign. Three cases may arise: (i) If all zj  cj  0 and either no artiﬁcial variable is in the basis or artiﬁcial variables appear in the basis at zero level, then the current BFS is optimal. Stop the process. (ii) If all zj  cj  0 but artiﬁcial variable(s) are in the basis at a positive level, then the given LPP has no feasible solution. (iii) If at least one zj  cj \0, go to the next step. Step 7: If at least one zj  cj \0, choose the most negative of zj  cj . Let it be zr  cr for some j ¼ r. Two cases may arise: (i) If all yir  0 ði ¼ 1; 2; . . .; mÞ, then there is an unbounded solution to the given LPP. Stop the process. (ii) If at least one yir [ 0, then the corresponding vector yr enters the basis. This vector is known as an incoming or entering vector. n o x Step 8: To select the departing or outgoing vector, compute Min yBiri ; yir [ 0 . x

i

Let the minimum value be yBkrk . Then the vector yk will leave the basis. This vector is known as a departing or outgoing vector. The common

46

3 Simplex Method

element ykr which is in the kth row and the rth column is known as a key or leading element. Step 9: Convert the key element to unity by dividing all the elements of its row by the key element and all other elements in its column to zeros using the following operations: ^yij ¼ yij 

ykj yir ; i ¼ 1; 2; . . .; m; i 6¼ k; j ¼ 1; 2; . . .; n ykr

y

and ^ykj ¼ ykrkj ; j ¼ 1; 2; . . .; n. Then modify the components of xB by using the relations ^xBi ¼ xBi 

ykj xB xB ; i ¼ 1; 2; . . .; m ði 6¼ kÞ and ^xBk ¼ k : ykr k ykr

Using these values, construct the next simplex table and go to Step 6.

3.11

Simplex Table

Let us consider the following standard LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b and x  0 where c ¼ ðc1 ; c2 ; . . .; cn Þ ¼ the profit vector x ¼ ðx1 ; x2 ; . . .; xn ÞT ¼ the decision vector A ¼ ða1 ; a2 ; . . .; an Þ ¼ the coefficient matrix b ¼ ðb1 ; b2 ; . . .; bm ÞT ¼ the requirement vector cB ¼ ðcB1 ; cB2 ; . . .; cBm ÞT ¼ the m-component column vector with elements corresponding to the basic variables B ¼ ðB1 ; B2 ; . . .; Bm Þ ¼ the basis matrix xB ¼ B1 b ¼ the basic feasible solution yj ¼ B1 aj ; j ¼ 1; 2; . . .; n: cj cB cB1 ... cBm

yB

yB1 ... yBm  zB ¼ cTB ; xB

xB xB1 ... xBm

c1 y1

c2 y2

... ...

cj yj

... ...

cn yn

y11 ... ym1 z1  c 1

y12 ... ym2 z2  c 2

... ... ... ...

y1j ... ymj zj  cj

... ... ... ...

y1n ... ymn zn  c n

Initially, B ¼ Im = a unit matrix of order m

3.11

Simplex Table

47

For the initial simplex table, xB ¼ B1 b ¼ Im b ¼ b [* B ¼ Im )B1 ¼ Im ] yj ¼ B1 aj ¼ aj : Example 1 Solve the following LPP: Maximize z ¼ 5x1 þ 4x2 subject to 3x1 þ 4x2  24 3x1 þ 2x2  18 x2  5 and x1 ; x2  0: Solution: Here the given problem is: Maximize z ¼ 5x1 þ 4x2 subject to 3x1 þ 4x2  24 3x1 þ 2x2  18 x2  5 and x1 ; x2  0: This is a maximization problem. All the bi ’s are positive. Now introducing the slack variables x3 ; x4 ; x5 , one to each constraint, we get the system of equations as follows: 3x1 þ 4x2 þ x3 ¼ 24 3x1 þ 2x2 þ x4 ¼ 18 x2 þ x5 ¼ 5: Then the new objective function is z ¼ 5x1 þ 4x2 þ 0  x3 þ 0  x4 þ 0  x5 : The set of above equations (1) can be written as 0

3 @3 0

4 2 1

1 0 0

1 x1 0 1 C 24 0 0 B B x2 C C ¼ @ 18 A 1 0 AB x 3 B C 5 0 1 @ x4 A x5 1

0

or A x ¼ b; 0

3 where A ¼ @ 3 0

4 2 1

1 0 0 1 0 0

1 0 0 0 0A @3 1 0

0 2 1

1 0 0

1 0 0 0 0 1 0A @0 0 1 0

0 0 0

1 0 0

1 0 0 1 0 A. 0 1

48

3 Simplex Method

0 1 0 1 0 1 1 0 0 Hence, the rank of A is 3. Then @ 0 A; @ 1 A and @ 0 A are three linearly 0 0 1 0 1 1 0 0 independent column vectors of A. Therefore, the submatrix B ¼ @ 0 1 0 A is a 0 0 1 non-singular basis submatrix of A. The basic variables are therefore x3 ; x4 ; x5 and an obvious initial basic feasible solution is xB ¼ B1 b 0 1 0 B ¼ @0 1

0

0 0

1

1 0 1 24 x3 i.e. @ x4 A ¼ @ 18 A 5 x5

10

24

1

0

24

1

CB C B C 0 A@ 18 A ¼ @ 18 A 5

5

0

and cB ¼ ð c3 ; c4 ; c5 Þ ¼ ð 0

0

) x3 ¼ 24; x4 ¼ 18; x5 ¼ 5

0 Þ.

Now we construct the starting simplex table. cj cB 0 0 0 zB ¼ 0

yB

xB

y3 y4 y5

x3 ¼ 24 x4 ¼ 18 x5 ¼ 5

5 y1

4 y2

0 y3

0 y4

0 y5

3 3 0 5

4 2 1 4

1 0 0 0

0 1 0 0

0 0 1 0

zj  cj

Here all zj  cj are not greater than or equal to zero. Hence, the current solution is

not optimal. Now, Minimum zj  cj ; zj  cj \0 ¼ Minimum f5; 4g ¼ 5 which is z1  c1 . Hence, y1 is the incoming vector. n o

x Again, Minimum yBi1i ; yi1 [ 0 ¼ Minimum 24 3 ;

18 3

¼ 6,

which occurs for the element y21 ð¼ 3Þ. Hence, y21 ¼ 3 is the key element and y4 is the outgoing vector. In the key row, we then replace 0; y4 ; x4 under cB ; yB ; xB by 5; y1 ; x1 respectively, the corresponding elements of the key column.

3.11

Simplex Table

49

Now converting the key element to unity and all other elements of this column to zero by suitable operations, we get the new simplex table as follows: cj cB

yB

0 y3 5 y1 0 y5 zB ¼ 30

xB x3 ¼ 6 x1 ¼ 6 x5 ¼ 5

5 y1

4 y2

0 y3

0 y4

0 y5

0 1 0 0

2 2/3 1 2=3

1 0 0 0

1 1/3 0 5/3

0 0 1 0

zj  c j

Here z2  c2 \0. Hence, the current solution is not optimal.

Now Minimum zj  cj ; zj  cj \0 ¼ Minimum f2=3g which is z2  c2 . Hence, y2 is the incoming vector. n o   x 6 ; 51 ¼ 3, which occurs for Again, Minimum yBi2i ; yi2 [ 0 ¼ Minimum 62 ; 2=3 the element y12 ð¼ 2Þ. Hence, y12 ¼ 2 is the key element and y3 is the outgoing vector. In the key row, we then replace 0; y3 ; x3 under cB ; yB ; xB by 4; y2 ; x2 respectively, the corresponding elements of the key column. Now converting the key element to unity and all other elements of this column to zero by suitable operations, we construct the new simplex table as follows: cj cB

yB

4 y2 5 y1 0 y5 zB ¼ 32

xB x2 ¼ 3 x1 ¼ 4 x5 ¼ 2

5 y1

4 y2

0 y3

0 y4

0 y5

0 1 0 0

1 0 0 0

1/2 1=3 1=2 1/3

1=2 2/3 1/2 4/3

0 0 1 0

zj  cj

Here all the zj  cj  0. Hence, the current BFS is optimal, and the optimal solution is x1 ¼ 4; x2 ¼ 3 and Max z ¼ 32: Example 2 Solve the following LPP: Maximize z ¼ 10x1 þ x2 þ 2x3 subject to x1 þ x2  2x3  10 4x1 þ x2 þ x3  20 and x1 ; x2 ; x3  0:

50

3 Simplex Method

Solution: This is a maximization problem. All the bi ’s are positive. Now introducing the slack variables x4 ; x5 , one to each constraint, we get the problem in the standard form as follows: Maximize z ¼ 10x1 þ x2 þ 2x3 þ 0x4 þ 0x5 subject to x1 þ x2  2x3 þ x4 ¼ 10 4x1 þ x2 þ x3 þ x5 ¼ 20 x1 ; x2 ; x3 ; x4 ; x5  0: Clearly, we may consider the variables x4 and x5 as basic variables. Considering the other variables as non-basic, we get x4 ¼ 10 and x5 ¼ 20, which is the initial basic feasible solution. Now we construct the initial simplex table. cj cB 0 0 zB ¼ 0

yB

xB

y4 y5

x4 = 10 x5 = 20

10 y1

1 y2

2 y3

0 y4

0 y5

Mini ratio

1 4 −10

1 1 −1

−2 1 −2

1 0 0

0 1 0

10/1 = 10 20/4 = 5 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y5 is the outgoing vector and y21 ¼ 4 is the key element. Now we construct the next simplex table. cj cB

yB

0 y4 10 y1 zB ¼ 50

xB x4 = 5 x1 = 5

10 y1

1 y2

2 y3

0 y4

0 y5

0 1 0

3/4 1/4 6/4

−9/4 1/4 2/4

1 0 0

−1/4 1/4 10/4

zj  c j

Here all zj  cj  0. Hence, an optimum solution has been attained, and this solution is given by x1 ¼ 5; x2 ¼ 0; x3 ¼ 0 and Max Z ¼ 50. Example 3 Maximize z ¼ 3x1 þ x2 þ 3x3 subject to 2x1 þ x2 þ x3  2 x1 þ 2x2 þ 3x3  5 2x1 þ 2x2 þ x3  6 and x1 ; x2 ; x3  0:

3.11

Simplex Table

51

Solution: This is a maximization problem. All the bi ’s are positive. Now introducing the slack variables x4 ; x5 ; x6 , one to each constraint, we get the problem in the standard form as follows: Maximize z ¼ 3x1 þ x2 þ 3x3 þ 0x4 þ 0x5 þ 0x6 subject to 2x1 þ x2 þ x3 þ x4 ¼ 2 x1 þ 2x2 þ 3x3 þ x5 ¼ 5 2x1 þ 2x2 þ x3 þ x6 ¼ 6 and xi  0i ¼ 1; 2; . . .; 6: Let us take the variables x4 ; x5 and x6 as basic variables. Considering the other variables as non-basic, we get x4 ¼ 2; x5 ¼ 5 and x6 ¼ 6, which is the initial basic feasible solution. Now we construct the initial simplex table. cj cB

yB

0 y4 0 y5 0 y6 zB ¼ 0

xB x4 = 2 x5 = 5 x6 = 6

3 y1

1 y2

3 y3

0 y4

0 y5

0 y6

Mini ratio

2 1 2 −3

1 2 2 −1

1 3 1 −3

1 0 0 0

0 1 0 0

0 0 1 0

2/2 = 1 5/1 = 5 6/2 = 3 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y4 is the outgoing vector and y11 ¼ 2 is the key element. Now we construct the next simplex table. cj cB

yB

3 y1 0 y5 0 y6 zB ¼ 3

xB x1 = 1 x5 = 4 x6 = 4

3 y1

1 y2

3 y3

0 y4

0 y5

0 y6

Mini ratio

1 0 0 0

1/2 3/2 1 1/2

1/2 5=2 0 −3/2

1/2 −1/2 −1 3/2

0 1 0 0

0 0 1 0

1 8/5 – zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y3 is the incoming vector, y5 is the outgoing vector and y23 ¼ 5=2 is the key element. Now we construct the next simplex table.

52 cj cB

3 Simplex Method

yB

3 y1 3 y3 0 y6 zB ¼ 27=5

xB x1 = 1/5 x3 = 8/5 x6 = 4

3 y1

1 y2

3 y3

0 y4

0 y5

0 y6

1 0 0 0

1/5 3/5 1 7/5

0 1 0 0

3/5 −1/5 −1 6/5

1/5 2/5 0 9/5

0 0 1 0

zj  c j

Here all zj  cj  0. Hence, the optimality condition has been satisﬁed, and the optimal solution is x1 ¼ 1=5; x2 ¼ 0; x3 ¼ 8=5 and Max z ¼

27 : 5

Example 4 Use simplex method to solve the following LPP: Minimize subject to

z ¼ x1  3x2 þ 2x3 3x1 þ x2 þ 2x3  7 2x1 þ 4x2  12 4x1 þ 3x2 þ 8x3  10 and x1 ; x2 ; x3  0:

Solution: This is a minimization problem. All the bi ’s are positive. Let z0 ¼ z ¼ x1 þ 3x2  2x3 . Hence, Minimize z ¼ Maximize z0 : Now introducing the slack variables x4 ; x5 ; x6 , one to each constraint, we get the reduced problem in the standard form as follows: Maximize z0 ¼ x1 þ 3x2  2x3 þ 0x4 þ 0x5 þ 0x6 subject to 3x1 þ x2 þ 2x3 þ x4 ¼ 7 2x1 þ 4x2 þ x5 ¼ 12 4x1 þ 3x2 þ 8x3 þ x6 ¼ 10 and x1 ; x2 ; x3 ; . . .; x6  0: Clearly, we can select the variables x4 ; x5 and x6 as the basic variables. Considering the other variables as non-basic, we get x4 ¼ 7; x5 ¼ 12 and x6 ¼ 10, which is the initial basic feasible solution. Now we construct the initial simplex table.

3.11 cj cB

Simplex Table

yB

0 y4 0 y5 0 y6 z0B ¼ 0

xB x4 = 7 x5 = 12 x6 = 10

53 −1 y1

3 y2

−2 y3

0 y4

0 y5

0 y6

Mini ratio

3 −2 −4 1

−1 4 3 −3

2 0 8 2

1 0 0 0

0 1 0 0

0 0 1 0

– 12/4 = 3 10/3 = 10/3 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y2 is the incoming vector, y5 is the outgoing vector and y22 ¼ 4 is the key element. Now we construct the next simplex table. cj cB

yB

0 y4 3 y2 0 y6 z0B ¼ 9

xB x4 = 10 x2 ¼ 3 x6 ¼ 1

−1 y1

3 y2

−2 y3

0 y4

0 y5

0 y6

Mini ratio

5=2 −1/2 −5/2 −1/2

0 1 0 0

2 0 8 0

1 0 0 0

1/4 1/4 −1/4 3/4

0 0 1 0

10/5/2 = 4 – – zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y4 is the outgoing vector and y11 ¼ 5=2 is the key element. Now we construct the next simplex table. cj cB

yB

−1 y1 3 y2 0 y6 z0B ¼ 11

xB x1 ¼ 4 x2 ¼ 5 x6 ¼ 11

−1 y1

3 y2

−2 y3

0 y4

0 y5

0 y6

1 0 0 0

0 1 0 0

4/5 2/5 10 11/5

2/5 1/5 1 1/5

1/10 3/10 −1/2 4/5

0 0 1 0

zj  cj

Here all zj  cj  0. Hence, optimality has been reached, and the optimal solution is x1 ¼ 4; x2 ¼ 5; x3 ¼ 0 and Min z ¼ Max z0 ¼ 11:

3.12

Use of Artiﬁcial Variables

In the standard form of an LPP, if the coefﬁcient matrix does not contain a unit basis matrix, then a non-negative variable is added to the left side of each equation which lacks the starting basic variables. This variable is known as an artiﬁcial variable. This variable plays the same role as a slack variable in providing the initial basic

54

3 Simplex Method

feasible solution. However, since such artiﬁcial variables have no physical meaning from the standpoint of the original problem, the method will be valid only if these variables are driven out or at zero level when the optimum solution is attained. In other words, to return to the original problem, artiﬁcial variables must be driven out or at zero level in the ﬁnal solution. Otherwise, the resulting solution may be infeasible. For solving an LPP involving artiﬁcial variables, the following two methods are generally employed: (i) Big M method (also known as Charne’s Big M method or the method of penalties) (ii) Two-phase method.

3.12.1 Big M Method The Big M method is an alternative method for solving an LPP having artiﬁcial variables. This method is applied only when the coefﬁcient matrix of standard form of an LPP does not give the unit basis matrix. In this method, a minimum number of artiﬁcial variables are introduced, and a very large negative price, say −M (M is positive and very large), is assigned in the objective function corresponding to the artiﬁcial variable. Due to the very large negative price, the objective function cannot be improved if one or more artiﬁcial variables appear in the basis. The corresponding problem can be solved in the usual simplex algorithm, and ﬁnally the following cases may arise: (i) At any iteration, if the optimality conditions are not satisﬁed, we have to proceed further to get an optimal solution by omitting the vectors corresponding to the artiﬁcial variables. Here the artiﬁcial variables are used as agents to get a unit basis. (ii) At any iteration, if the optimality conditions are satisﬁed and all the vectors corresponding to artiﬁcial variables are driven out from the basis, the given LPP has no feasible solution. (iii) At any iteration, if the optimality conditions are satisﬁed and all the artiﬁcial variables are at zero level but at least one artiﬁcial vector is present in the basis, the solution obtained is an optimal solution of the given LPP. (iv) At any iteration, if the optimality conditions are satisﬁed and some of the artiﬁcial vectors are present in the basis, i.e. some of the vectors corresponding to artiﬁcial variables are at a positive level, then the given LPP has no feasible solution.

3.12

Use of Artiﬁcial Variables

55

3.12.2 Solution of Simultaneous Linear Equations by Simplex Method Let us consider the system of m simultaneous linear equations in n unknowns in the form Ax ¼ b;

ð3:27Þ

where A is an m  n real matrix and b is an m  1 real matrix. The variables to be determined are given by the vector x. Now we shall solve this by the simplex method. For this purpose, we have to form an LPP with a dummy objective function which is to be optimized. To solve the given system of equations, we introduce artiﬁcial variables in the given system of equations, and the corresponding price in the dummy objective function z will be −M for each artiﬁcial variable. Thus, the given problem of solving the system of linear equations can be reduced to an LPP in which the objective function will be z ¼ M

m X

xa i ;

i¼1

where xai ði ¼ 1; 2; . . .; mÞ is the artiﬁcial variable. Since there is no non-negativity restriction for the required variables x, these variables are considered to be unrestricted in sign. Now to introduce the non-negativity restrictions, let us assume that x ¼ x0  x00 where x0 ; x00  0. Therefore, the transformed LPP becomes Maximize z ¼ M

m P

xa i

i¼1

subject to Aðx0  x00 Þ þ xa ¼ b and x0 ; x00 ; xa  0; where xa is the vector of artiﬁcial variables. Solving this LPP by the Big M method: (i) If we get Max z = 0 for which xai ði ¼ 1; 2; . . .; mÞ becomes zero, then a basic feasible solution of the system of equations will be obtained. (ii) At any iteration, if the optimality conditions are satisﬁed and some of the artiﬁcial vectors are present in the basis, then the given system of equations has no solution.

56

3 Simplex Method

3.12.3 Inversion of a Matrix by Simplex Method To ﬁnd the inverse of a given non-singular square matrix, an identity matrix of the same order is considered. Now, performing the same sequence of elementary row transformations on the given matrix and the identity matrix to reduce the given matrix to an identity matrix, we get the inverse matrix. Let us consider a real non-singular square matrix of order n. Let x be an ncomponent column vector and b be an n-component dummy column vector. Then let us consider the system of linear equations as follows: Ax ¼ b; x  0:

ð3:28Þ

Now we introduce the artiﬁcial vectors xa ð  0Þ and construct the dummy objective function as z ¼ M

m X

xa i ;

i¼1

where xai ð  0Þ is the ith component of the artiﬁcial vector xa . Hence, our problem is Maximize z ¼ M

m P

xa i

i¼1

subject to the constraints Ax þ Ixa ¼ b; x; xa  0: Solving this LPP, we get the optimal solution. If the basis of the optimum simplex table contains all the variables of vector x, then the inverse of A is directly obtained from the optimum simplex table. It consists of those column vectors in the last iteration of the simplex method which are present in the initial basis. If the optimum (ﬁnal) simplex table does not contain all the variables of vector x in the basis, the simplex method is continued until all the variables of the vector are in the basis. In this situation, the solution may be either feasible or infeasible but optimum. Example 5 Solve the following LPP by the Big M method: Maximize subject to

z ¼ 2x1  3x2  2x3 2x1 þ x2  2x3  3 x1 þ x2  x3  1 x1 ; x2 ; x3  0:

3.12

Use of Artiﬁcial Variables

57

Solution: The given problem is a maximization problem. Here all bi ’s are positive. Now, introducing a slack variable x4 and a surplus variable x5 to the ﬁrst and second constraints respectively, we get the following converted equations: 2x1 þ x2  2x3 þ x4 ¼ 3 x1 þ x2  x3  x5 ¼ 1; where xi  0; i ¼ 1; 2; . . .; 5. The coefﬁcient matrix of this system of equations does not contain a unit matrix. To get a unit matrix, an artiﬁcial variable x6 is added to the second equation; then the transformed equations are as follows: 2x1 þ x2  2x3 þ x4 ¼ 3 x1 þ x2  x3  x5 þ x6 ¼ 1 where xi  0; i ¼ 1; 2; . . .; 6. Then the new objective function is given by z ¼ 2x1  3x2  2x3 þ 0  x4 þ 0  x5  Mx6 [assign a very large negative price to the artiﬁcial variable]. cj cB

yB

0 y4 −M y6 zB ¼ M

xB x4 = 3 x6 = 1

2 y1

−3 y2

−2 y3

0 y4

0 y5

−M y6

Mini ratio

2 1 −M – 2

1 1 −M + 3

−2 −1 M+2

1 0 0

0 −1 M

0 1 0

3/2 = 3/2 1/1 = 1 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y6 is the outgoing vector and y21 ¼ 1 is the key element. Now we construct the next simplex table. cj cB

B

0 y4 2 y1 zB ¼ 2

xB x4 = 1 x1 = 1

2 y1

−3 y2

−2 y3

0 y4

0 y5

−M y6

Mini ratio

0 1 0

−1 1 5

0 −1 0

1 0 0

2 −1 −2

−2 1 2+M

1/2 – zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y5 is the incoming vector, y4 is the outgoing vector and y15 ¼ 1 is the key element.

58

3 Simplex Method

Now we construct the next simplex table. cj cB

B

0 y5 2 y1 zB ¼ 3

xB x5 = 1/2 x1 = 3/2

2 y1

−3 y2

−2 y3

0 y4

0 y5

−M y6

0 1 0

−1/2 1/2 4

0 −1 0

1/2 1/2 1

1 0 0

−1 0 M

zj  c j

Here all zj  cj  0. Hence, optimality has been reached, and the optimal solution is x1 ¼ 3=2; x2 ¼ 0; x3 ¼ 0 with Max z = 3. Example 6 Use Charne’s Big M method to solve the following LPP: Maximize subject to

z ¼ x1 þ 2x2 x1  5x2  10 2x1  x2  2 x1 þ x2 ¼ 0 and x1 ; x2  0:

Solution: The given problem is a maximization problem. Here all bi ’s are positive. Now, introducing a slack variable x3 and a surplus variable x4 to the ﬁrst and second constraints respectively, we get the following converted equations: x1  5x2 þ x3 ¼ 10 2x1  x2  x4 ¼ 2 x1 þ x2 ¼0 where xi  0; i ¼ 1; 2; . . .; 4. The coefﬁcient matrix of this system of equations does not contain a unit matrix. To get a unit matrix, two artiﬁcial variables x5 and x6 are added to second and third equations respectively, and the transformed equations are as follows: x1  5x2 þ x3

¼ 10

2x1  x2  x4 þ x5 ¼ 2 x1 þ x2 þ x 6 ¼0 where xi  0; i ¼ 1; 2; . . .; 6. Then the new objective function is given by z ¼ x1 þ 2x2 þ 0x3 þ 0x4  Mx5  Mx6 [assign a very large negative price to the artiﬁcial variable].

3.12

Use of Artiﬁcial Variables

59

Clearly, the variables x3 ; x5 and x6 are basic. Considering the other variables as non-basic, we get x3 ¼ 10; x5 ¼ 2 and x6 ¼ 0 which is the initial basic feasible solution. Now we construct the initial simplex table. cj cB

B

0 y3 −M y5 −M y6 zB ¼ 2M

xB x3 = 10 x5 = 2 x6 = 0

1 y1

2 y2

0 y3

0 y4

−M y5

−M y6

Mini ratio

1 2 1 −3M − 2

−5 −1 1 −2

1 0 0 0

0 −1 0 M

0 1 0 0

0 0 1 0

10/1 = 10 2/2 = 1 0/1 = 0 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y6 is the outgoing vector and y31 ¼ 1 is the key element. Now we construct the next simplex table. cj cB

B

0 y3 −M y5 1 y1 zB ¼ 2M

xB x3 = 10 x5 = 2 x1 = 0

1 y1

2 y2

0 y3

0 y4

−M y5

0 0 1 0

−6 −3 1 3M − 1

1 0 0 0

0 −1 0 M

0 1 0 0

zj  cj

Here all zj  cj  0 for all j; i.e. the optimality condition has been satisﬁed. But the artiﬁcial vector y5 appears in the basis at a positive level, and the value of the corresponding artiﬁcial variable x5 is 2. Hence, there is no feasible solution to the given LPP. Example 7 Solve the LPP Maximize z ¼ 3x1  x2 subject to x1 þ x2  2 5x1  2x2  2 and x1 ; x2  0: Solution: The given problem is a maximization problem. Here all bi ’s are positive. Now, introducing surplus and artiﬁcial variables, the given problem can be written in the standard form as follows:

60

3 Simplex Method

Maximize z ¼ 3x1  x2 þ 0  x3 þ 0  x4  Mx5  Mx6 subject to x1 þ x2  x3 þ x5 ¼ 2 5x1  2x2  x4 þ x6 ¼ 2 and xi  0; i ¼ 1; 2; . . .; 6: Clearly, the variables x5 and x6 are basic. Considering the other variables as non-basic, we get x5 ¼ 2 and x6 ¼ 2, which is the initial basic feasible solution. Now we construct the initial simplex table. cj cB

B

xB

−M y5 −M y6 zB ¼ 4M

x5 = 2 x6 = 2

3 y1

−1 y2

0 y3

0 y4

−M y5

−M y6

Mini ratio

−1 5 −4M − 3

1 −2 M+1

−1 0 M

0 −1 M

1 0 0

0 1 0

– 2/5 = 2/5 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y6 is the outgoing vector and y21 ¼ 5 is the key element. Now we construct the next simplex table. cj cB

B

xB

b

−M

y5

x5

12/ 5 2/5

x1 3 y1 zB ¼ 12M=5 þ 6=5

3 y1

−1 y2

0 y3

0 y4

−M y5

Mini ratio

0

3=5

−1

−1/5

1

1 0

−2/5 −3/5M − 1/5

0 M

−1/5 M/5 − 3/5

0 0

12/5/3/ 5=4 – zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y2 is the incoming vector, y5 is the outgoing vector and y12 ¼ 3=5 is the key element. Now we construct the next simplex table. cj cB −1 3 zB ¼ 2

B

xB

b

y2 y1

x2 x1

4 2

3 y1

−1 y2

0 y3

0 y4

0 1 0

1 0 0

−5/3 −2/3 −1/3

−1/3 −1/3 −2/3

zj  c j

Since all zj  cj l0, the current BFS is not optimal. The value of z4  c4 is mostly negative, with yi4 \0; i ¼ 1; 2. Hence, the outgoing vector cannot be selected because there is an unbounded solution to the given problem.

3.12

Use of Artiﬁcial Variables

61

Example 8 Solve the following LPP using the Big M method: Maximize z ¼ 3x1  x2 subject to the constraints 2x1 þ x2  2; x1 þ 3x2  3; x2  4 and x1 ; x2  0: Solution: The given problem is a maximization problem. Here all bi ’s are positive. Now introducing surplus variable x3 and slack variables x4 and x5 in the given constraints, we get the following converted equations: 2x1 þ x2  x3 ¼ 2 x1 þ 3x2 þ x4 ¼ 3 x2 þ x5 ¼ 4 where xi  0; i ¼ 1; 2; . . .; 5. The coefﬁcient matrix of the preceding system of equations does not contain a unit matrix. To get a unit matrix, an artiﬁcial variable x6 is added to the ﬁrst equation, and the transformed equations are as follows: 2x1 þ x2  x3 þ x6 ¼ 2 x1 þ 3x2 þ x4

¼3

x2 þ x5

¼4

where xi  0; i ¼ 1; 2; . . .; 6 . Then the new objective function is given by z ¼ 3x1  x2 þ 0  x3 þ 0  x4 þ 0  x5  Mx6 [assign a very large negative price to the artiﬁcial variable]. Clearly, the variables x4 ; x5 and x6 are basic. Considering the other variables as non-basic, we get x6 ¼ 2; x4 ¼ 3 and x5 ¼ 4, which is the initial basic feasible solution. Now we construct the initial simplex table. cj cB

yB

−M y6 0 y4 0 y5 zB ¼ 2M

xB x6 ¼ 2 x4 ¼ 3 x5 ¼ 4

3 y1

−1 y2

0 y3

0 y4

0 y5

−M y6

Mini ratio

2 1 0 2M  3

1 3 1 M þ 1

−1 0 0 M

0 1 0 0

0 0 1 0

1 0 0 0

2/2 = 1 3/1 = 3 – zj  cj

62

3 Simplex Method

Here all zj  cj l0. Hence, the current BFS is not optimal. Here y1 is the incoming vector, y11 ¼ 4 is the key element and y6 is the outgoing vector. Now we construct the next simplex table. cj cB 3 0 0 zB ¼ 3

yB

xB

y1 y4 y5

x1 ¼ 1 x4 ¼ 2 x5 ¼ 4

3 y1

−1 y2

0 y3

0 y4

0 y5

Mini ratio

1 0 0 0

1/2 5/2 1 5/2

−1/2 1=2 0 −3/2

0 1 0 0

0 0 1 0

– 2/½ = 4 – zj  c j

Here all zj  cj l0. Hence, the current BFS is not optimal. Here y3 is the incoming vector, y23 ¼ 1=2 is the key element and y4 is the outgoing vector. Now we construct the next simplex table. cj cB 3 0 0 zB ¼ 9

yB

xB

y1 y3 y5

x1 ¼ 3 x3 ¼ 4 x5 ¼ 4

3 y1

−1 y2

0 y3

0 y4

0 y5

1 0 0 0

3 5 1 10

0 1 0 0

1 2 0 3

0 0 1 0

zj  c j

Here all zj  cj  0 for all j. Hence, the current BFS is optimal, and the optimal solution is given by x1 ¼ 3; x2 ¼ 0 with Max z ¼ 9: Example 9 Using the simplex method, solve the following system of linear equations: x1 þ x2 ¼ 1; 2x1 þ x2 ¼ 3: Solution: Since the non-negativity restrictions of x1 and x2 are not given, these variables are unrestricted in sign. Now to introduce the non-negativity restrictions, let us assume that x1 ¼ x01  x001 and x2 ¼ x02  x002 ; where x01 ; x001 ; x02 ; x002  0.

3.12

Use of Artiﬁcial Variables

63

To solve the given system of equations, we introduce two artiﬁcial variables x3  0 and x4  0 in the ﬁrst and second equations respectively and a dummy objective function z ¼ Mx3  Mx4 (assigning a large negative price −M corresponding to each artiﬁcial variable) which is to be maximized subject to the constraints as follows: x01  x001 þ x02  x002 þ x3 ¼ 1 2x01  2x001 þ x02  x002 þ x4 ¼ 3 and x01 ; x001 ; x02 ; x002 ; x3 ; x4  0. Let us take x3 and x4 as the basic variables. Considering other variables as non-basic variables, we get x3 ¼ 1; x4 ¼ 3, which is the initial BFS. Now we construct the initial simplex table. cj cB

yB

−M y3 −M y4 zB ¼ 4M

xB x3 ¼ 1 x4 ¼ 3

0 y01

0 y001

0 y02

0 y002

−M y3

−M y4

Mini ratio

1 2 3M

−1 −2 −3M

1 1 2M

−1 −1 2M

1 0 0

0 1 0

1 3/2 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y01 is the incoming vector, y3 is the outgoing vector and y011 ¼ 1 is the key element. Now we construct the next simplex table. cj cB

yB

0 y01 −M y4 z ¼ M

xB x01 ¼ 1 x4 ¼ 1

0 y01

0 y001

0 y02

0 y002

−M y4

Mini ratio

1 0 0

−1 0 0

1 −1 M

−1 1 −M

0 1 0

– 1 zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y002 is the incoming vector, y4 is the outgoing vector and y0022 ¼ 1 is the key element. Now we construct the next simplex table.

64 cj cB 0 0 zB ¼ 0

3 Simplex Method

yB

xB

y01 y002

x01 ¼ 2 x002 ¼ 1

0 y01

0 y001

0 y02

0 y002

1 0 0

−1 0 0

0 −1 0

0 1 0

zj  cj

Since all zj  cj are non-negative, an optimum solution has been attained. Thus, the solution of the LPP is

i:e: and

x01 ¼ 2; x001 ¼ 0; x02 ¼ 0; x002 ¼ 1; x1 ¼ x01  x001 ¼ 2  0 ¼2 x2 ¼ x02  x002 ¼ 0  1 ¼ 1;

which is the solution of the given system of equations. Remark Note that the method is applied in solving an LPP with one or more unrestricted variables with the given objective function. Example 10 Using the simplex method, ﬁnd the inverse of the following matrix:  A¼

4 1

 2 : 5

Solution: Considering the given matrix as a coefﬁcient matrix, let us take the following system of linear equations: 

4 1

2 5



x1 x2



  6 ¼ ; 4

x1 ; x2  0;

where ð6; 4ÞT is a dummy column vector, i:e: and

4x1 þ 2x2 ¼ 6 x1 þ 5x2 ¼ 4 x1 ; x2  0:

To solve the system of equations, we introduce two artiﬁcial variables x3  0 and x4  0 in the ﬁrst and second equations respectively and a dummy objective function z ¼ Mx3  Mx4 (assigning a large negative price −M corresponding to each artiﬁcial variable)

3.12

Use of Artiﬁcial Variables

65

which is to be maximized subject to the constraints as follows: 4x1 þ 2x2 þ x3 ¼ 6 x1 þ 5x2 þ x4 ¼ 4 and x1 ; x2 ; x3 ; x4  0: Let us take x3 and x4 as the basic variables. Considering other variables as non-basic variables, we get x3 ¼ 6; x4 ¼ 4, which is the initial BFS. Now we construct the initial simplex table. cj cB

yB

−M y3 −M y4 zB ¼ 10M

xB x3 ¼ 6 x4 ¼ 4

0 y1

0 y2

−M y3

−M y4

Mini ratio

4 1 5M

2 5 7M

1 0 0

0 1 0

2 6/5 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y2 is the incoming vector, y4 is the outgoing vector and y22 ¼ 5 is the key element. Now we construct the next simplex table. cj cB

yB

−M y3 0 y2 zB ¼ 22=5M

xB x3 ¼ 22=5 x2 ¼ 4=5

0 y1

0 y2

−M y3

−M y4

Mini ratio

18=5 1/5 18=5M

0 1 0

1 0 0

−2/5 1/5 7/5M

11/9 4 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y3 is the outgoing vector and y11 ¼ 18=5 is the key element. Now we construct the next simplex table. cj cB 0 0 zB ¼ 0

yB

xB

y1 y2

x1 ¼ 11=9 x2 ¼ 5=9

0 y1

0 y2

−M y3

−M y4

1 0 0

0 1 0

5/18 −1/18 M

1=9 2/9 M

zj  cj

Since all zj  cj are non-negative, an optimum solution has been attained. From the preceding simplex table, it is seen that

66

3 Simplex Method

y3 ¼

5 18

1  18

T

and y4 ¼  19

2 T : 9

Since the initial basis consists of the column vectors y3 and y4 , and the column vectors y1 ; y2 corresponding to columns of matrix A reduce to unit vectors, then the inverse of matrix A is given by A1 ¼



5=18 1=18

1=9 2=9

 ¼

1 18



 5 2 : 1 4

3.12.4 Two-Phase Method In an earlier section, we discussed the Big M method for solving a particular type of LPP. To solve the same problem, Dantzig, Orden and Wolfe developed an alternative method known as the two-phase method. In this method, the problems are solved in two different phases, Phase I and Phase II. Phase I The objective of Phase I is either to remove all the artiﬁcial variables from the basis or to set some of the said variables (which are present in the basis) at a zero level. For this purpose, an auxiliary LPP is constructed with a new objective function X z0 ¼  xai ½xai is an artificial variable i

which is to be maximized. From the mathematical point of view, all artiﬁcial variables must be at a zero level at the optimal stage. Hence, the maximum value of z0 should be equal to zero. However, this will not be true for all problems. Thus, to ﬁnd the maximum value of the new objective function z0 , the following auxiliary LPP is to be solved ﬁrst: Maximize z0 ¼ 

X

xa i

i

subject to all the constraints: In solving this LPP, when optimality conditions are satisﬁed in an iteration, the following three cases may arise: Case I: Max z0 ¼ 0 and all artiﬁcial vectors are not present in the basis. Case II: Max z0 ¼ 0, but some of the artiﬁcial vectors are present in the basis at a zero level. In this case, some constraints may be redundant.

3.12

Use of Artiﬁcial Variables

67

Case III: Max z0 \0. This means that one or more artiﬁcial variables are present in the basis at a positive level. In this case, the original LPP has no feasible solution. Phase II If Phase I is terminated with either Case I or Case II, then we have to start Phase II to ﬁnd the optimal solution of the original problem. In this phase, the ﬁrst simplex table is almost identical with the last table of Phase I except that the rows for cj and zj  cj as the values of cj of the original LPP will be used in the table. In Case II, Max z0 ¼ 0 and one or more artiﬁcial vectors appear in the basis at zero level. This means that one or more artiﬁcial variables are basic, but their values are zero. In this case, we must take care that these artiﬁcial variables never become positive at any iteration of Phase II. Example 11 Use the two-phase simplex method to solve the following problem: Maximize z ¼ 3x1  x2 subject to the constraints 2x1 þ x2  2 x1 þ 3x2  2 x1  4 x1 ; x2  0: Solution: This is a maximization problem. All bi ’s are positive. Now introducing the surplus variable x3 and slack variables x4 ; x5 , the given problem can be rewritten as follows: Maximize z ¼ 3x1  x2 þ 0  x3 þ 0  x4 þ 0  x5 subject to the constraints 2x1 þ x2  x3 ¼ 2 x1 þ 3x2 þ x4 ¼ 2 x1 þ x5 ¼ 4 and xj  0; j ¼ 1; 2; . . .; 5:

The coefﬁcient matrix of the above constraints does not contain a unit matrix. To get a unit matrix, an artiﬁcial variable x6 is added to the ﬁrst equation, and the transformed equations are as follows:

68

3 Simplex Method

2x1 þ x2  x3 þ x6 ¼ 2 x1 þ 3x2 þ x4 ¼ 2 x1 þ x5 ¼ 4 where xj  0; j ¼ 1; 2; . . .; 6. Phase I: Assigning a proﬁt coefﬁcient −1 to the artiﬁcial variable and 0 to all other variables, the objective function of the auxiliary LPP to be maximized is given by z 0 ¼ 0  x1 þ 0  x2 þ 0  x3 þ 0  x4 þ 0  x5  x6 subject to the preceding system of equations. Let us take x4 ; x5 and x6 as the basic variables. Considering other variables as non-basic variables, we get x6 ¼ 2; x4 ¼ 2; x5 ¼ 4, which is the initial BFS. Now we construct the initial simplex table. cj cB

yB

−1 y6 0 y4 0 y5 0 z ¼ 2

xB x6 ¼ 2 x4 ¼ 2 x5 ¼ 4

0 y1

0 y2

0 y3

0 y4

0 y5

−1 y6

Mini ratio

2 1 1 −2

1 3 0 −1

−1 0 0 1

0 1 0 0

0 0 1 0

1 0 0 0

2/2 = 2 2/1 = 2 4/1 = 4 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y6 is the outgoing vector and y11 ¼ 2 is the key element. Now we construct the next simplex table. cj cB 0 0 0 z0B ¼ 0

yB y1 y4 y5

xB x1 ¼ 1 x4 ¼ 1 x5 ¼ 3

0

0

0

0

0

y1 1 0 0 0

y2 1/2 5/2 −1/2 0

y3 −1/2 1/2 1/2 0

y4 0 1 0 0

y5 0 0 1 0

zj  cj

Here all zj  cj are non-negative and Max z0 ¼ 0. Again no artiﬁcial variable appears in the ﬁnal table of Phase I. We then pass on to Phase II where the objective function of the original problem to be maximized is z ¼ 3x1  x2 þ 0  x3 þ 0  x4 þ 0  x5 :

3.12

Use of Artiﬁcial Variables

69

Now we construct the initial simplex table of Phase II. cj cB

yB

3 y1 0 y4 0 y5 zB ¼ 3

xB x1 ¼ 1 x4 ¼ 1 x5 ¼ 3

3 y1

−1 y2

0 y3

0 y4

0 y5

Mini ratio

1 0 0 0

1/2 5/2 −1/2 5/2

−1/2 1=2 1/2 −3/2

0 1 0 0

0 0 1 0

– 2 6 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y3 is the incoming vector, y4 is the outgoing vector and y32 ¼ 1=2 is the key element. Now we construct the next simplex table. cj cB 3 0 0 zB ¼ 6

yB

xB

y1 y3 y5

x1 ¼ 2 x3 ¼ 2 x5 ¼ 2

3 y1

−1 y2

0 y3

0 y4

0 y5

1 0 0 0

3 5 −3 10

0 1 0 0

1 2 −1 3

0 0 1 0

zj  c j

Here all zj  cj are non-negative. Hence, an optimum solution has been attained, and this solution is given by x1 ¼ 2; x2 ¼ 0 and Max z = 6. Example 12 Use the two-phase simplex method to solve the following problem: Maximize z ¼ 3x1 þ 2x2 subject to the constraints 2x1 þ x2  2 3x1 þ 4x2  12 and x1 ; x2  0:

Solution: This is a maximization problem. All bi ’s are positive. Now introducing the slack variable x3 and surplus variable x4 , the given problem can be rewritten as follows: Maximize z ¼ 3x1 þ 2x2 þ 0  x3 þ 0  x4 subject to 2x1 þ x2 þ x3 ¼ 2 3x1 þ 4x2  x4 ¼ 12 and x1 ; x2 ; x3 ; x4  0:

70

3 Simplex Method

The coefﬁcient matrix of these constraints does not contain a unit matrix. To get a unit matrix, an artiﬁcial variable x5 is added to the ﬁrst equation; the transformed equations are as follows: 2x1 þ x2 þ x3 ¼ 2 3x1 þ 4x2  x4 þ x5 ¼ 12 where xj  0; j ¼ 1; 2; 3; 4; 5. Phase I: Assigning a proﬁt coefﬁcient −1 to the artiﬁcial variable and 0 to all other variables, the objective function of the auxiliary LPP to be maximized is given by z 0 ¼ 0  x1 þ 0  x 2 þ 0  x3 þ 0  x4  x5 subject to the preceding system of equations. cj cB

yB

0 y3 −1 y5 z0B ¼ 12

xB x3 ¼ 2 x5 ¼ 12

0 y1

0 y2

0 y3

0 y4

−1 y5

Mini ratio

2 3 −3

1 4 −4

1 0 0

0 −1 1

0 1 0

2/1 = 2 12/4 = 3 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y2 is the incoming vector, y3 is the outgoing vector and y21 ¼ 1 is the key element. Now we construct the next simplex table. cj cB

yB

0 y2 −1 y5 z0B ¼ 4

xB x2 ¼ 2 x5 ¼ 4

0 y1

0 y2

0 y3

0 y4

−1 y5

2 −5 5

1 0 0

1 −4 4

0 −1 1

0 1 0

zj  c j

Here all zj  cj  0. Hence, an optimum basic feasible solution to the auxiliary LPP has been attained. Again, Max z0 \0 and the artiﬁcial variable x5 is basic with a positive value. Hence, the original LPP does not possess any feasible solution.

3.12

Use of Artiﬁcial Variables

71

Example 13 Use the two-phase simplex method to solve the following problem: Maximize z ¼ 5x1  4x2 þ 3x3 subject to the constraints 2x1 þ x2  6x3 ¼ 20 6x1 þ 5x2 þ 10x3  76 8x1  3x2 þ 6x3  50 and x1 ; x2 ; x3  0: Solution: This is a maximization problem. All bi ’s are positive. Now introducing the slack variables x3 and x4 , the given problem can be rewritten as follows: Maximize z ¼ 5x1  4x2 þ 3x3 þ 0  x4 þ 0  x5 subject to the constraints 2x1 þ x2  6x3 ¼ 20 6x1 þ 5x2 þ 10x3 þ x4 ¼ 76 8x1  3x2 þ 6x3 þ x5 ¼ 50 and xj  0; j ¼ 1; 2; 3; 4; 5: The coefﬁcient matrix of these constraints does not contain a unit matrix. To get a unit matrix, an artiﬁcial variable x6 is added to the ﬁrst equation, and the transformed equations are as follows: 2x1 þ x2  6x3 þ x6 ¼ 20 6x1 þ 5x2 þ 10x3 þ x4 ¼ 76 8x1  3x2 þ 6x3 þ x5 ¼ 50; where xj  0; j ¼ 1; 2; . . .; 6. Phase I: Assigning a proﬁt coefﬁcient −1 to the artiﬁcial variable and 0 to all other variables, the objective function of the auxiliary LPP to be maximized is given by z 0 ¼ 0  x1 þ 0  x2 þ 0  x3 þ 0  x4 þ 0  x5  x6 subject to the preceding system of equations. cj cB

xB

0 y1

0 y2

0 y3

0 y4

0 y5

−1 y6

Mini ratio

yB

−1 0

y6 y4

x6 ¼ 20 x4 ¼ 76

2 6

1 5

−6 10

0 1

0 0

1 0

10 38/3 (continued)

72

3 Simplex Method

(continued) cj cB

yB

0 y5 z0B ¼ 20

xB x5 ¼ 50

0 y1

0 y2

0 y3

0 y4

0 y5

−1 y6

Mini ratio

8 −2

−3 −1

6 6

0 0

1 0

0 0

25/4 zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y5 is the outgoing vector and y31 ¼ 8 is the key element. Now we construct the next simplex table. cj cB

yB

−1

y6

0

y4

0 y1 z0B ¼ 15=2

0 y2

0 y3

0 y4

0 y5

−1 y6

Mini ratio

xB

0 y1

x6 ¼ 15=2

0

7=4

−15/2

0

−1/4

1

x4 ¼ 77=2

0

29/4

11/2

1

−3/4

0

30 7 154 29

x1 ¼ 25=4

1 0

−3/8 −7/4

3/4 15/2

0 0

1/8 1/4

0 0

– zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y2 is the incoming vector, y6 is the outgoing vector and y12 ¼ 7=4 is the key element. Now we construct the next simplex table. cj cB

yB

0 y2 0 y4 0 y1 z0B ¼ 0

xB x2 ¼ 30=7 x4 ¼ 52=7 x1 ¼ 55=7

0 y1

0 y2

0 y3

0 y4

0 y5

0 0 1 0

1 0 0 0

−30/7 256/7 −6/7 0

0 1 0 0

−1/7 2/7 1/14 0

zj  cj

Here all zj  cj  0. Hence, an optimum solution to the auxiliary LPP has been attained. Furthermore, from this table, it is observed that no artiﬁcial vector appears in the basis. We then pass on to Phase II, where the objective function of the original problem to be maximized is z ¼ 5x1  4x2 þ 3x3 þ 0  x4 þ 0  x5 : Phase II: Now we construct the initial simplex table of Phase II.

3.12 cj cB

Use of Artiﬁcial Variables

yB

−4 y2 0 y4 0 y1 zB ¼ 155=7

xB x2 ¼ 30=7 x4 ¼ 52=7 x1 ¼ 55=7

73 5 y1

−4 y2

3 y3

0 y4

0 y5

0 0 1 0

1 0 0 0

−30/7 256/7 −6/7 69/7

0 1 0 0

−1/7 2/7 1/14 13/14

zj  cj

Here all zj  cj  0. Hence, an optimum basic feasible solution to the given LPP has been attained, and the optimal solution is given by x1 ¼ 55=7; x2 ¼ 30=7; x3 ¼ 0 with maximumz ¼ 155=7: Example 14 Using the simplex method, solve the following system of linear simultaneous equations: x1 þ x2 ¼ 1 2x1 þ x2 ¼ 3:

Solution: Since the non-negativity restrictions of x1 and x2 are not given, these variables are unrestricted in sign. Now to introduce the non-negativity restrictions, let us assume that x1 ¼ x01  x001 and x2 ¼ x02  x002 , where x01 ; x001 ; x02 ; x002  0. Now we introduce the artiﬁcial variables x3  0; x4  0 by assigning proﬁt −1 for each. Then the problem of solving the given system of equations reduces to the following LPP: Maximize z ¼ x3  x4 subject to the constraints x01  x001 þ x02  x002 þ x3 ¼ 1 2x01  2x001 þ x02  x00 þ x4 ¼ 3 and x01 ; x001 ; x02 ; x002 ; x3 ; x4  0: cj cB

yB

−1 y3 −1 y4 zB ¼ 4

xB x3 ¼ 1 x4 ¼ 3

0 y01

0 y001

0 y02

0 y002

−1 y3

−1 y4

Mini ratio

1 2 −3

−1 −2 3

1 1 −2

−1 −1 2

1 0 0

0 1 0

1 3/2 zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y01 is the incoming vector, y3 is the outgoing vector and y011 ¼ 1 is the key element. Now we construct the next simplex table.

74 cj cB

3 Simplex Method

yB

0 y01 −1 y4 zB ¼ 1

xB x01 ¼ 1 x4 ¼ 1

0 y01

0 y001

0 y02

0 y002

−1 y4

Mini ratio

1 0 0

−1 0 0

1 −1 1

−1 1 −1

0 1 0

– 1 zj  c j

Since all zj  cj l0, the current BFS is not optimal. Here y002 is the incoming vector, y4 is the outgoing vector and y0022 ¼ 1 is the key element. Now we construct the next simplex table. cj cB 0 0 zB ¼ 0

yB

xB

y01 y002

x001 ¼ 2 x002 ¼ 1

0 y01

0 y001

0 y02

0 y002

1 0 0

−1 0 0

0 −1 0

0 1 0

zj  cj

Since all zj  cj are non-negative, an optimum solution has been attained. Thus, the solution of the LPP is x01 ¼ 2; x002 ¼ 1; x001 ¼ 0; x02 ¼ 0;

i:e:x1 ¼ x01  x001 ¼ 2  0 x2 ¼ x02  x00 ¼ 0  1

¼2 ¼ 1;

which is the solution of the given system of equations. Remark Note that the method is applied in solving an LPP with one or more unrestricted variables with the given objective function. Example 15 Using the simplex method, ﬁnd the inverse of the following matrix:  A¼

3 4

 2 : 1

Solution: Taking the given matrix as a coefﬁcient matrix, let us consider the following system of linear equations: 

3 4

2 1



x1 x2



  4 ¼ ; 6

where ð4; 6ÞT is a dummy column vector,

x1 ; x2  0;

3.12

Use of Artiﬁcial Variables

75

i:e: and

3x1 þ 2x2 ¼ 4 4x1  x2 ¼ 6 x1 ; x2  0:

To solve the system of equations, we introduce two artiﬁcial variables x3  0 and x4  0 in the ﬁrst and second equations respectively and a dummy objective function z ¼ x3  x4 which is to be maximized subject to the constraints 3x1 þ 2x2 þ x3 ¼ 4 4x1  x2 þ x4 ¼ 6 and x1 ; x2 ; x3 ; x4  0: cj cB

yB

xB

−1 y3 −1 y4 zB ¼ 10

x3 ¼ 4 x4 ¼ 6

0 y1

0 y2

−1 y3

−1 y4

Mini ratio

3 4 −7

2 −1 −1

1 0 0

0 1 0

4/3 3/2 zj  cj

Since all zj  cj l0, the current BFS is not optimal. Here y1 is the incoming vector, y3 is the outgoing vector and y11 ¼ 3 is the key element. Now we construct the next simplex table. cj cB

yB

0 y1 −1 y4 zB ¼ 2=3

xB x1 ¼ 4=3 x4 ¼ 2=3

0 y1

0 y2

−1 y3

−1 y4

1 0 0

2/3 −11/3 11/3

1/3 −4/3 7/3

0 1 0

zj  c j

Since all zj  cj are non-negative, an optimal solution has been reached. But to ﬁnd the inverse of the matrix A, we have to convert the vector y2 as a unit vector. For this purpose, we consider y2 as the incoming vector and y4 as the outgoing vector and y22 ¼ 11=3 as the key element. Now we construct the next simplex table. cj cB 0 0 zB ¼ 0

yB

xB

y1 y2

x1 ¼ 16=11 x2 ¼ 2=11

0 y1

0 y2

−1 y3

−1 y4

1 0 0

0 1 0

1/11 4/11 1

2/11 −3/11 1

zj  cj

76

3 Simplex Method

1 4 T

2 3 T From this simplex table, it is seen that y3 ¼ 11 ; 11 and y4 ¼ 11 ; 11 , since the initial basis consists of the column vectors y3 and y4 and the column vectors y1 ; y2 corresponding to the columns of matrix A; therefore, the inverse of matrix A is given by A1 ¼

3.13



1 11 4 11

2 11 3 11

 ¼

1 11



1 4

 2 : 3

Exercises

1. Solve the following LPPs: (i) Maximize subject to (ii) Minimize subject to (iii) Maximize subject to

(iv) Maximize subject to

(v) Maximize subject to (vi) Maximize subject to

(vii) Maximize subject to (viii) Minimize subject to

(ix) Minimize subject to

z ¼ 10x1 þ x2 þ 2x3 x1 þ x2  2x3  10 4x1 þ x2 þ x3  20 and x1 ; x2  0: z ¼ 3x1  2x2 4x1 þ x2  8 2x1 þ 4x2  20 and x1 ; x2  0: z ¼ 2x1 þ 5x2 4x1  5x2   20 x1 þ x2  10 x2  2 and x1 ; x2  0: z ¼ 5x1 þ 11x2 2x1 þ x2  4 3x1 þ 4x2  24 2x1  3x2  6 and x1 ; x2  0: z ¼ 2x1 þ x2 þ 3x3 x1 þ x2 þ 2x3  5 2x1 þ 3x2 þ 4x3 ¼ 12 and x1 ; x2 ; x3  0: z ¼ 4x1 þ x2 3x1 þ x2 ¼ 3 4x1 þ 3x2  6 x1 þ 2x2  3 and x1 ; x2  0: z ¼ 2x1 þ x2  x3 þ 4x4 3x1  5x2 þ x3  2x4 ¼ 7 6x1  10x2  x3 þ 5x4 ¼ 11 and xi  0; i ¼ 1; 2; . . .; 4: z ¼ 2x1 þ x2 3x1 þ x2  3 4x1 þ 3x2  6 x1 þ 2x2  2 and x1 ; x2  0: z ¼ 4x1 þ 2x2 3x1 þ x2  27 x1 þ x2  21 x1 þ 2x2  30 and x1 ; x2  0:

3.13

Exercises

77

2. Solve the following LPPs: (i) Maximize subject to (ii) Maximize subject to (iii) Maximize subject to

(iv) Maximize subject to

(v) Maximize subject to (vi) Maximize subject to

(vii) Minimize subject to (viii) Maximize subject to

z ¼ x1 þ 5x2 3x1 þ 4x2  6 x1 þ 3x2  3 and x1 ; x2  0: z ¼ 2x1 þ x2 þ 3x3 x1 þ x2 þ 2x3  5 2x1 þ 3x2 þ 4x3 ¼ 12 and x1 ; x2 ; x3  0: z ¼ 4x1 þ x2 3x1 þ x2 ¼ 3 4x1 þ 3x2  6 x1 þ 2x2  3 and x1 ; x2  0: z ¼ 2x1 þ x2 4x1 þ 3x2  12 4x1 þ x2  8 4x1  x2  8 and x1 ; x2  0: z ¼ 2x1 þ x2 þ 10x3 x1  2x3 ¼ 0 x2 þ x3 ¼ 1 and x1 ; x2 ; x3  0: z ¼ 5x1  2x2 þ 3x3 2x1 þ 2x2  x3  2 3x1  4x2  3 x2 þ 3x3  5 and x1 ; x2 ; x3  0: z ¼ 4x1 þ 8x2 þ 3x3 x1 þ x2  2 2x1 þ x3  5 and x1 ; x2 ; x3  0: z ¼ 5x1 þ 3x2 x1 þ x2  2 5x1 þ 2x2  10 5x1 þ 2x2  10 and x1 ; x2  0:

3. Solve the following LPPs using the two-phase simplex method: (i)

(ii)

(iii)

(iv)

z ¼ x1 þ x2 2x1 þ x2  4 x1 þ 7x2  7 and x1 ; x2  0: Maximize z ¼ 3x1 þ 2x2 subject to 2x1 þ x2  2 3x1 þ 4x2  12 and x1 ; x2  0: Maximize z ¼ 5x1 þ 8x2 subject to 3x1 þ 2x2  3 x1 þ 4x2  4 x1 þ x2  5 and x1 ; x2  0: Minimize z ¼ 4x1 þ x2 subject to x1 þ 2x2  3 4x1 þ 3x2  6 3x1 þ x2 ¼ 3 and x1 ; x2  0: Minimize subject to

78

3 Simplex Method

(v)

Minimize subject to

z ¼ 40x1 þ 24x2 2x1 þ 25x2  240 8x1 þ 5x2  720 and x1 ; x2  0:

4. Solve the following systems of linear simultaneous equations using the simplex method: (i)

x1 þ x2 ¼ 1

(ii)

2x1 þ x2 ¼ 4: 3x1 þ x2 ¼ 7

(iii)

(iv)

x1 þ x2 ¼ 3: 7x1  x2 ¼ 1 11x1 þ 2x2 ¼ 23: x1 þ 2x2 þ 3x3 ¼ 14 2x1  x2 þ 5x3 ¼ 15  3x1 þ 2x2 þ 4x3 ¼ 13:

5. Using the simplex method, ﬁnd the inverse of the following matrices: 0 1      1 2 3 2 1 3 2 3 (i) (ii) (iii) @ 3 2 1 A (iv) 1 2 4 1 1 4 2 1   3 4 (v) 1 2

1 1



Chapter 4

Revised Simplex Method

4.1

Objective

The objective of this chapter is to discuss: • The revised simplex methods and their computational procedures • The relevant information required at each iteration of the revised simplex method • The use of the revised simplex in comparison to the usual simplex method.

4.2

Introduction

The revised simplex method is a modiﬁed form of the ordinary simplex method. It is also an efﬁcient method for solving linear programming problems (LPPs). It is efﬁcient in the sense that we need not recompute all the values of yj ; zj  cj ; xB at each iteration. Here, the word ‘revised’ refers to the procedure of changing or updating the ordinary simplex method. This method is economical on the computer, as it computes and stores only the relevant information required for testing the optimality condition and for updating the current solution. The basic information required at each iteration in the ordinary simplex method includes: (i) (ii) (iii) (iv)

Net evaluations corresponding to the non-basic variables The column vector corresponding to the most negative net evaluation The values of the current basic variables Selection of an outgoing vector from the basis.

This information can be computed directly from its deﬁnitions provided the inverse of the basis matrix is known. To get the inverse of the basis matrix at any © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_4

79

80

4 Revised Simplex Method

iteration, we need to compute the same from the previous inverse matrix. For this purpose, the relevant information we must know at each iteration is: (i) The coefﬁcients of non-basic variables in the objective function (ii) The components of the vectors corresponding to the non-basic variables. Another feature of the revised simplex method is to treat the objective function z ¼ hcT ; xi as a constraint considering z as an unrestricted variable.

4.3

Standard Forms of Revised Simplex Method

There are two standard forms of the revised simplex method. 1. Standard form I: In this form, an initial basis matrix is obtained by introducing slack variables and/or surplus variables. Basically, an identity is considered as an initial basis matrix. 2. Standard form II: In this form, artiﬁcial variables are also required for an identity matrix which is considered as the initial basis matrix. Thus, a two-phase method is used to handle the artiﬁcial variables. The method for solving standard form I may also be applied to solve these types of problems.

4.4

Standard Form I of Revised Simplex Method

Let us consider the following LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b; x  0, where c; xT 2 Rn ; bT 2 Rm and A is an m  n real matrix of rank m. In the revised simplex method we consider the objective function z ¼ hcT ; xi as one of the constraints and z as a variable. Thus, the problem is now to solve the new system of ðm þ 1Þ simultaneous linear equations in ðn þ 1Þ variables x1 ; x2 ; . . .; xn ; z, for which z is unrestricted in sign and may be as large as possible. The set of constraints can thus be represented as Ax þ O  z ¼ b    cT ; x þ z ¼ 0 x  0; z is unrestricted in sign  or

A O c 1



x z

! ¼

b 0

! ;

x  0:

ð4:1Þ

Let B be an initial basis submatrix of A and xB ¼ B1 b be an initial basic feasible solution (BFS) of the original LPP. Then the initial BFS of the reformulated problem (4.1) is given by

4.4 Standard Form I of Revised Simplex Method

81

BxB þ O  z ¼ b    cTB ; x þ z ¼ 0  or or or

B

O

cB

1



xB



  b

¼ 0  B b xB ¼ b b¼ Bb b where B cB z

     O xB b ; bx B ¼ ;b b¼ 1 z 0

b 1 b bx B ¼ B b;

which is the initial BFS of (4.1). We note that bx B is called the BFS whenever xB is the BFS of the original LPP, even when z is unrestricted in sign.   A1 A2 1 b Let B ¼ A3 A4   I O 1 b B b¼ ) B O 1      A1 A2 B O I O or ¼ cB 1 A3 A4 O 1     A1 B  A2 cB A2 I O or ¼ O 1 A3 B  A4 cB A4 ) A1 B  A2 cB ¼ I: A2 ¼ O A3 B  A4 cB ¼ O A4 ¼ 1: From (4.2), we have A1 B  A2 cB ¼ I or

A1 B ¼ I ½* A2 ¼ O or

A1 ¼ B1 :

From (4.3), A3 B  A4 cB ¼ O or

A3 B  cB ¼ O ½* A4 ¼ 1 or

A3 B ¼ cB

ð4:2Þ

ð4:3Þ

82

4 Revised Simplex Method

A3 ¼ cB B1   1 A2 B ¼ A4 cB B1

or b 1 ¼ )B



A1 A3

 O : 1

Let us deﬁne a new ðm þ 1Þ  n matrix b by ¼ B

1

b where A b¼ A,  )^y ¼  ¼

A c

!

B1 cB B1 y

O 1 

cB y  c



A

! ¼

c 



B1 A cB B1 A  c



 *B1 A ¼ y :

Now, cB y  c 1 y1n C y2n C C    C C  ðc1 ; c2 ; . . .; cn Þ C    A ym1 ym2    ymn ! m m m X X X cBi yi1  c1 ; cBi yi2  c2 ; . . .; cBi yin  cn 0

y11 B B y21 B ¼ ðcB1 ; cB2 ; . . .; cBm ÞB B  B @ 

¼

i¼1

y12 y22

 

i¼1

i¼1

¼ ðz1  c1 ; z2  c2 ; . . .; zn  cn Þ   yj ) by j ¼ ; Þ j ¼ 1; 2; . . .; n z j  cj  or ðby 1 ; by 2 ; . . .; by n Þ ¼

y1 z1  c 1

y2 z 2  c2

  yn :    z n  cn

ð4:4Þ

From these expressions, it is clear that the ﬁrst m components (i.e. the ﬁrst m rows) of the jth column constitute the vector yj and the (m + 1)th row is zj  cj (the net evaluation). Hence, the net evaluations are the components of 

cB B1 A  c ¼ cB B1 1

! A : c

4.4 Standard Form I of Revised Simplex Method

 Also; bx B ¼

xB z



83

b 1 b b ¼B

h

i b 1 b * bx B ¼ B b ;

where the ﬁrst m components of bx B are xB and the (m + 1)th component is the corresponding value of the objective function.

4.5

Computational Procedure of Standard Form I of Revised Simplex Method

Step 1 Introduce slack and surplus variables, if needed, and reformulate the given LPP in maximization standard form. Step 2 Obtain an initial BFS with an initial basis B ¼ Im and obtain the inverse of the new basis matrix    1  1 B O B O b b B¼ by the formula B ¼ : cB 1 cB B1 1 b and b Step 3 Find A b by the formula A

b¼ A

!

c

;

b b¼

b

!

0

:

Step 4 Compute the net evaluations zj  cj ð j ¼ 1; 2; . . .; nÞ by using the formula 



z j  cj ¼ cB B

1

1

A c

! :

(a) If all zj  cj are non-negative, the current BFS is an optimum one. (b) If at least one zj  cj is negative, determine the most negative of them, say zk  ck ; then the corresponding vector by k enters the basis. Go to Step 5. b 1 b Step 5 Compute by k ¼ B a k . In that case, (a) If all by ik  0, there exists an unbounded optimum solution to the given problem. (b) If at least one by ik [ 0, ﬁnd the minimum ratios of the components of bx B and the corresponding positive by ik and determine the outgoing vector and key or leading element. Step 6 Write down the results obtained in Step 2 through Step 5 in a tabular form known as a revised simplex table.

84

4 Revised Simplex Method

Step 7 Update the B1 matrix by suitable row operations to convert the key element to unity and all other elements of the entering column to zero, and b 1 b then update the current BFS using the formula bx B ¼ B b. Step 8 Go to Step 4 and repeat the procedure until an optimum BFS is obtained or there is an indication of an unbounded solution. Example 1 Using the revised simplex method, solve the following LPP: Maximize subject to

z ¼ 6x1  2x2 þ 3x3 2x1  x2 þ 2x3  2 x1 þ 4x3  4 and x1 ; x2 ; x3  0:

Solution The given LPP is a maximization problem. Here all bi ’s (i =1, 2) are positive. Now introducing the slack variables x4  0 and x5  0 in the ﬁrst and second constraints respectively, the given LPP is rewritten in standard form as follows: Maximize z ¼ 6x1  2x2 þ 3x3 þ 0  x4 þ 0  x5 subject to 2x1  x2 þ 2x3 þ x4 ¼ 2 x1 þ 4x3 þ x5   or Maximize z ¼ cT ; x subject to Ax ¼ b; x  0; where c ¼ ð 6 2

3

¼ 4 and xj  0;

0 0 Þ;  T

x ¼ ðx1 x2 x3 x4 x5 Þ ;

2 1

1 0

2 4

1 0 

The initial BFS is x4 ¼ 2; x5 ¼ 4 with B ¼ I2 ¼ Here cB ¼ ð0; 0Þ: b 1 ¼ )B



1

B

cB B1

0 1



j ¼ 1; 2; . . .; 5

0

1 0 B ¼ @0 1 0 0

 0 ; 1 1 0 0 1

! 2 : 4

b¼ 

as the initial basis.

1 0

 1 C 0 A * cB B1 ¼ ð0; 0Þ 0 1

0 1



¼ ð0; 0Þ :

4.5 Computational Procedure of Standard Form I of Revised Simplex Method

85

! A Since the net evaluations zj  cj are the components of ðcB B1 1Þ , the c net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z2  c2 z3  c3 Þ ¼ cB B1 1 ðb a1 b a2 b a3Þ 0 1 2 1 2 B C ¼ ð 0 0 1 ÞB 0 4 C @ 1 A 6 2 3 0 2 0 1 2 1 2 A 6 B 6* A A¼B 1 b¼@ 0 4 4 @ c 6 2 3 ¼ ð 6 2 3 Þ:

1

0

13

0

C7 7 1C A5

0

0

These calculations indicate that z1  c1 is the most negative, and hence by 1 enters the basis. 0

10

1

0

1

  y1 CB C B C 1 0 A@ 1 A ¼ @ 1 A ¼ : z 1  c1 0 0 1 6 6 0 10 1 0 12 0 13 1 0 0 2 2 2   b B CB C B C6 B C7 b 1 b b ¼ @ 0 1 0 A@ 4 A ¼ @ 4 A4 * b b¼ ¼ @ 4 A5 : Also; bx B ¼ B 0 0 0 1 0 0 0   2 ) xB ¼ ; zB ¼ 0: 4 Now;

1

B b 1 b by 1 ¼ B a1 ¼ @ 0

0 0

2

2

Now we construct the initial revised simplex table. by B

bx B

b 1 B

by 4 by 5

2 4 0

1 0 0

0 1 0

0 0 1

by 1

Mini ratio

2 1 6

1 4

The minimum ratio corresponds to the ﬁrst basic variable, i.e. x4 ; hence, by 4 leaves the basis and by 11 ¼ 2 is the key element. First iteration: We introduce by 1 and drop by 4 . In this iteration,

86

4 Revised Simplex Method

0

b B

1

1 0 0A 1

1=2 0 ¼ @ 1=2 1 3 0

by

R01 ¼ 12 R1 R02 ¼ R2  R01 : R03 ¼ R3 þ 6R01

Here cB ¼ ð 6 0 Þ. The net evaluations zj  cj corresponding to non-basic variables are given by  ðz2  c2 z3  c3 z4  c4 Þ ¼ cB B1 1 ðb a3 b a4Þ a2 b 0 1 1 2 1 B C ¼ ð 3 0 1 Þ@ 0 4 0 A ½ðcB B1 1Þ ¼ ð 3 0 2 3 0 ¼ ð 1 3 3 Þ:

Clearly; Now;

1 Þ

z2  c2 is negative: Hence; by 2 enters the basis: 0 10 1 0 1 1=2 0 0 1 1=2 B CB C B C b 1 b by 2 ¼ B a 2 ¼ @ 1=2 1 0 A@ 0 A ¼ @ 1=2 A 3

1=2 1 B b b and bx B ¼ B b ¼ @ 1=2

0 1

1 2 10 1 0 1 0 2 1 CB C B C 0 A@ 4 A ¼ @ 3 A

3 )xB ¼ ð 1 3 Þ ; zB ¼ 6:

0

1

0

0

0

1

6

T

Now we construct the next revised simplex table. by B

bx B

b 1 B next

by 1 by 5

1 3 6

1/2 1=2 3

0 1 0

0 0 1

by 2

Mini ratio

1=2 1/2 1

– 6

Second iteration: We introduce by 2 and drop by 5 . In this iteration, 0

b B

1

0 ¼ @ 1 2

1 2 2

Here

1 0 0A 1

by

R02 ¼ 2R2 R01 ¼ R1 þ 12 R02 : R03 ¼ R3 þ R02

cB ¼ ð 6 2 Þ

The net evaluations zj  cj corresponding to the non-basic variables are given by

4.5 Computational Procedure of Standard Form I of Revised Simplex Method

 ðz3  c3 z4  c4 z5  c5 Þ ¼ cB B1 1 ð b a3 0 ¼ ð2 ¼ ð9

b a4

2 B 2 1 Þ@ 4 2 2 Þ: 3

87

b a5 Þ 1 1 0 C 0 1A 0 0

Since all zj  cj [ 0 correspond to non-basic variables, an optimum basic feasible solution is obtained. The optimal solution is given by 0

b bx B ¼ B

1

0 B b b ¼ @ 1 2 6 ÞT and

) xB ¼ ð 4

10 1 0 1 0 2 4 CB C B C 0 A@ 4 A ¼ @ 6 A 1 0 12

1 2 2

zB ¼ 12:

Hence, the optimal solution of the given LPP is x1 ¼ 4; x2 ¼ 6; x3 ¼ 0 and zmax ¼ 12. Example 2 Using the revised simplex method, solve the following LPP: Maximize subject to

z ¼ x1 þ x2 þ 3x3 3x1 þ 2x2 þ x3  3 2x1 þ x2 þ 2x3  2 and

x1 ; x2 ; x3  0:

Solution This is a maximization problem. Here all bi ’s are positive. Now introducing the slack variables x4  0 and x5  0 in the ﬁrst and second constraints respectively, the given LPP is rewritten in the standard form as follows: Maximize subject to

z ¼ x1 þ x2 þ 3x3 þ 0  x4 þ 0  x5 3x1 þ 2x2 þ x3 þ x4 ¼ 3 2x1 þ x2 þ 2x3 þ x5 ¼ 2 and xj  0; j ¼ 1; 2; . . .; 5   or Maximize z ¼ cT ; x subject to Ax ¼ b; x  0; where

c ¼ ð1 

T

x ¼ ð x1 x2 x3 x4 x5 Þ ; A ¼

1

3

0 0 Þ;

3 2 2 1

1 2

1 0

0 1

 and

! 3 : 2

88

4 Revised Simplex Method

 Here the initial BFS is x4 ¼ 3; x5 ¼ 2 with B ¼ I2 ¼ Here A

b¼ Now A

b 1 ¼ )B

!

c 

B1 cB B1

0

3

2

1 0 0 1

 as the initial basis.

cB ¼ ð 0 0 Þ:

1

1

0

1

!

0 1 3 B C B ¼ @2C A 0

b B C b¼ ¼@ 2 1 2 0 1 A and b 0 1 1 3 0 0 0 1 1 0 0

   1 0 0 B C ) cB B1 ¼ ð 0 0 Þ ¼ ð0 0Þ : ¼ @0 1 0A 0 1 1 0 0 1

The net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z2  c2 z3  c3 Þ ¼ cB B1 1 ð b a1 0 ¼ ð0

This result indicates that z3  c3 0 1 0 b 1 b Now by 3 ¼ B a3 ¼ @ 0 1 0 0

1

b a3 Þ 2 1

1 1 3 Þ:

1

1

C 2 A 3

is the most negative. Hence, by 3 enter the basis. 10 1 0 1 2 0 13 0 1 1 1 0 A@ 2 A ¼ @ 2 A 4 * b a 3 ¼ @ 2 A5 : 1 3 3 3 0

1

B b 1 b Also; bx B ¼ B b ¼ @0   3 ) xB ¼ 2

3

B 1 Þ@ 2

0

¼ ð 1

b a2

0 and

10 1 0 3 1 3 C CB C B C 1 0 A@ 2 A ¼ B @ 2 A: 0 1 0 0 0 0

zB ¼ 0:

Now we construct the initial revised simplex table. by B

bx B

b 1 B

by 4 by 5

3 2 0

1 0 0

0 1 0

0 0 1

by 3

Mini ratio

1 2 3

3 1

The minimum ratio corresponds to the second basic variable, i.e. x5 ; hence, by 5 leaves the basis.

4.5 Computational Procedure of Standard Form I of Revised Simplex Method

89

First iteration: In this iteration, 0

b B

1

1 ¼ @0 0

1=2 1=2 3=2

1 0 0A 1

by

R02 ¼ R22 R01 ¼ R1  R02 : R03 ¼ R3 þ 3R02

The net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z2  c2 z5  c5 Þ ¼ cB B1 1 ð b a1 b a2 b a5 Þ 0 1 3 2 0  B C 1 1A ¼ 0 32 1 @ 2 1 1 0 ! " 1  1  2 cB B1 ¼ ð 0 3 Þ ¼ 0 1 0 2  1 3 ¼ 2 2 2 :

3 2

Since all zj  cj  0, an optimum basic feasible solution is obtained. The optimal solution is given by 0

1 B b 1 b bx B ¼ B b ¼ @0 ) xB ¼ ð 2

1=2 1=2

10 1 0 2 1 0 3 CB C B C C 0 A@ 2 A ¼ B @1A

0 3=2 1 1 Þ and zB ¼ 3: T

0

3

Hence, the optimal solution is x1 ¼ 0; x2 ¼ 0; x3 ¼ 1, and the maximum value of z is 3.

4.6

Standard Form II of Revised Simplex Method

In this form, if artiﬁcial variables appear in an LPP, then the initial basis matrix is obtained by using the vectors’ corresponding to slack and artiﬁcial variables. In that case, to solve the problem, either Charne’s Big M method or the two-phase method may be used. We denote these methods as Charne’s Big M revised simplex method and the two-phase revised simplex method. We shall discuss both methods separately.

90

4.6.1

4 Revised Simplex Method

Charne’s Big M Revised Simplex Method

In this case, the auxiliary basis matrix b¼ B



B cB

0 1



is not a unit basis, and the associated proﬁt vector cB is not a null vector. Thus, b 1 ¼ B



B1 cB B1

0 1



will not be an identity matrix. This is the only difference between the standard forms I and II of the revised simplex method. All other theoretical developments and computational procedures are the same as in standard form I. Example 3: Using the revised simplex method, solve the following LPP: Maximize z ¼ x1 þ 2x2 subject to 2x1 þ 5x2  6 x1 þ x2  2; x1 ; x2  0: Solution This is a maximization problem. All the bi’s (i = 1, 2) are positive. Now introducing the surplus variables x3  0 and x4  0 in the ﬁrst and second constraints respectively, the given LPP is written as follows: Maximize z ¼ x1 þ 2x2 þ 0  x3 þ 0  x4 subject to 2x1 þ 5x2  x3 ¼ 6 x1 þ x2  x4 ¼2; xj  0; j ¼ 1; 2; 3; 4: The coefﬁcient matrix of the constraints does not contain a unit basis matrix. To obtain a unit matrix, two artiﬁcial variables x5 ; x6 are added to the left-hand side of the ﬁrst and second constraints. Then the LPP becomes: Maximize

z ¼ x1 þ 2x2 þ 0  x3 þ 0  x4  Mx5  Mx6

subject to 2x1 þ 5x2  x3 þ x5 ¼ 6 x1 þ x2  x4 þ x6 ¼ 2; xj  0; j ¼ 1; 2; . . .; 6   or Maximize z ¼ cT ; x subject to

Ax ¼ b; x  0;

where c ¼ ð 1 2 0 0  M  M Þ; x ¼ ðx1 x2 . . . x6 ÞT     6 2 5 1 0 1 0 b¼ A¼ : 2 1 1 0 1 0 1

4.6 Standard Form II of Revised Simplex Method

91

The coefﬁcients of the variables constitute the column vectors y1 ¼

! 2 ; 1

y2 ¼

!   5 1 ; y3 ¼ ; 0 1

and the requirement vector b ¼

6

 y4 ¼

 0 ; 1

y5 ¼

! 1 ; 0

y6 ¼

  0 ; 1

!

. 2 Here the artiﬁcial vectors y5 and y6 constitute the initial basis matrix B ¼ I2 , and taking this as the initial basis, we have the initial BFS as xB ¼ B1 b ¼ I2 b ¼ b  0: Therefore xB ¼ ð xB1 xB2 ÞT ¼ ð x5 x6 ÞT ¼ ð 6 2 ÞT ) x5 ¼ 6; x6 ¼ 2:   1 0 Here B ¼ ; cB ¼ ð M M Þ: 0 1 0 1 2 5 1 0 1 0   A B C b¼ Now; A ¼@ 1 1 0 1 0 1 A c 1 2 0 0 M M 0 1 1 0 0  1  0 B B C b 1 ¼ )B ¼@ 0 1 0A 1 cB B 1 M M 1

  1 0 ¼ ð M M Þ: * cB B1 ¼ ð M M Þ 0 1 Now, the net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z2  c2 z3  c3 z4  c4 Þ ¼ cB B1 1 ðb a1 b a2 b a3 b a4Þ 0 2 5 B ¼ ð M M 1 Þ@ 1 1

1 0

1 2 0 ¼ ð 3M  1 6M  2 M M Þ:

1 0 C 1 A 0

The result indicates that z2  c2 is the most negative, and hence by 2 enters the basis.

92

4 Revised Simplex Method

1 1 0 10 5 5 1 0 0 C C B CB B b 1 b 1 a2 ¼ @ 0 Now; by 2 ¼ B 1 0 A@ 1 A ¼ @ A 6M  2 2 M M 1 1 10 1 0 0 6 6 1 0 0 C CB C B B b 1 b Also; bx B ¼ B b¼@ 0 1 0 A@ 2 A ¼ @ 2 A: 8M 0 M M 1   6 ) xB ¼ and zB ¼ 8M: 2 0

13 5 6 C7 B a 2 ¼ @ 1 A5 : 4* b 2 2

0

Now, we construct the initial revised simplex table. by B

bx B

b 1 B

by 5 by 6

6 2

1 0 M

0 1 M

0 0 1

by 2

Mini ratio

5 1 6M  2

6/5 2

The minimum ratio corresponds to the ﬁrst basic variable, i.e.x5 ; hence, by 5 leaves the basis. First iteration: We introduce by 2 and drop by 5 .In this case, 0

b B

1

1=5 ¼ @ 1=5 ðM þ 2Þ=5

0 1 M

1 0 0A 1

by

R01 ¼ R51 R02 ¼ R2  R01 R03 ¼ R3 þ ð6M þ 2ÞR01 :

Now, the net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z3  c3 z4  c4 z5  c5 Þ ¼ cB B1 1 ðb a1 b a3 b a4 b a5Þ 0 2 1 0  M þ2 B M 1 @ 1 0 1 ¼ 5 1  ¼  3M5þ 1  M 5þ 2 M

0 6M þ 2 5

0 :

1 1 C 0A M

Since z1  c1 is the most negative, then by 1 enters the basis. 0

1=5 b 1 b a 1 ¼ @ 1=5 Now; by 1 ¼ B ðM þ 2Þ=5

0 1 M

1 10 1 0 2=5 0 2 0 A@ 1 A ¼ @ 3=5 A:  3M5þ 1 1 1

4.6 Standard Form II of Revised Simplex Method

93

10 1 2 6 1=5 0 0 1 B CB C 6 b b b Again; bx B ¼ B b ¼ @ 1=5 1 0 A@ 2 A 6 4* b ¼ 0 ðM þ 2Þ=5 M 1  T 6 4 4M þ 12 ¼ 5 5 5  T 6 4 4M þ 12 ) xB ¼ and zB ¼ : 5 5 5 0

b

!

0

0 13 6 B C7 C7 ¼B @ 2 A5 0

The next revised simplex table is as follows: by B

bx B

b 1 B

by 2 by 6

6/5 4/5

1/5 1=5

4M þ 12 5

M þ2 5

0 1 M

0 0 1

by 1

Mini ratio

2/5 3/5

3 4/3

 3M5þ 1

Second iteration: We introduce by 1 and drop by 6 . In this case, 0

b B

1

1 1=3 2=3 0 ¼ @ 1=3 5=3 0 A by 1=3 1=3 1

R02 ¼ 53 R2 ; R01 ¼ R1  25 R02 ; R03 ¼ R3 þ

3M þ 1 0 5 R2

:

Now, the net evaluations zj  cj corresponding to non-basic variables are given by  ðz3  c3 z4  c4 z5  c5 z6  c6 Þ ¼ cB B1 1 ð b a4 b a5 b a6 Þ a3 b 0 1 1 0 1 0 1 1 B C ¼ 3 3 1 @ 0 1 0 1 A 0 0 M M  ¼  13  13 M þ 13 M þ 13 : Clearly, z3  c3 and z4  c4 are most negative. Any one of them may enter the basis. Arbitrarily, we take by 3 as the entering vector. 0

b Now; by 3 ¼ B

1

1=3 b a 3 ¼ @ 1=3 1=3

2=3 5=3 1=3

10 1 0 1 0 1 1=3 0 A@ 0 A ¼ @ 1=3 A: 0 1=3 1

94

4 Revised Simplex Method

0

1=3

2=3

10 1 6 CB C 0 A@ 2 A 1 0

2

0

B b 1 b Again; bx 1 5=3 B ¼ B b ¼ @ 1=3 1=3 1=3  T 248 ¼ 333  T 24 8 ) xB ¼ and zB ¼ : 33 3

6 b 4* b ¼

b

!

0

0 13 6 B C7 ¼ @ 2 A5 0

The next revised simplex table is as follows: by B

bx B

b 1 B next

by 2 by 1

2/3 4/3 8/3

1/3 1=3 1/3

2=3 5/3 1/3

0 0 1

by 3

Mini ratio

1=3 1/3 1=3

– 4

The minimum ratio corresponds to the second basic variable, i.e. x1 ; hence, by 1 leaves the basis. Third iteration: In this case, 0

b B

1

0 ¼ @ 1 0

1 1 0 5 0 A by 2 1

R02 ¼ 3R1 ; R01 ¼ R1 þ

1 0 0 1 R ; R ¼ R3 þ R02 : 3 2 3 3

The net evaluations zj  cj corresponding to non-basic variables are given by  ðz1  c1 z4  c4 z5  c5 z6  c6 Þ ¼ cB B1 1 ð b a4 b a5 b a6 Þ a1 b 0 1 2 0 1 0 B C ¼ ð 0 2 1 Þ@ 1 1 0 1 A ¼ ð 1 2 M M Þ: 1 0 M M Clearly, z4  c4 is negative. ) by 4 enters the basis. 0

b Now; by 4 ¼ B

1

0 b a 4 ¼ @ 1 0

1 5 2

10 1 0 1 0 0 1 0 A@ 1 A ¼ @ 5 A: 1 0 2

Since the ﬁrst two components of the entering vector by 4 are negative, there is an indication of an unbounded solution. Hence, the given LPP has an unbounded solution.

4.6 Standard Form II of Revised Simplex Method

4.6.2

95

Two-Phase Revised Simplex Method

In this method, besides the original objective function, an auxiliary objective function z is formed by considering the coefﬁcients of artiﬁcial variables as −1 and the coefﬁcients of the other variables as zero. Then both the objective functions are converted to constraints, considering z and z as unrestricted variables. Example 4 Use the two-phase revised simplex method to solve the following LPP: Maximize subject to

z ¼ 3x1  x2 2x1 þ x2  2 x1 þ 3x2  2 and x1 ; x2  0:

Solution This is a maximization problem. Here all bi ’s are positive. Now introducing the slack variable x3 and surplus variable x4 in the second and ﬁrst constraints respectively, and then considering the given objective function as a constraint, with z as a variable, we get the reduced set of constraints as follows: 2x1 þ x2  x4

¼2

x1 þ 3x2 þ x3 ¼ 2  3x1 þ x2 þ z ¼ 0 and xj  0; j ¼ 1; 2; 3; 4: To get the initial basis matrix B as an identity matrix, introducing one artiﬁcial variable x5 ð  0Þ in the ﬁrst constraint, we get the reduced constraints as 2x1 þ x2  x4 þ x5 ¼ 2 x1 þ 3x2 þ x3 ¼2  3x1 þ x2 þ z

¼ 0:

We shall solve this problem using the two-phase revised simplex method. In Phase I, the auxiliary objective function is Maximize

z ¼ x5

i.e. Maximize z ¼ 2x1 þ x2  x4  2 [substituting x5 from the0ﬁrst constraint]. 1 1 0 0 Here the initial BFS is x3 ¼ 2; x5 ¼ 2; z ¼ 0 with I3 ¼ @ 0 1 0 A as the 0 0 1 initial basis,

96

4 Revised Simplex Method

0

1 0 0 1 0 A. Here cB ¼ ð 0 0 0 1 0 1 1 0 0 ¼ @ 0 1 0 A. 0 0 1

1 i.e. B ¼ @ 0 0 Now, B1

0 Þ.

0 1 1 2 1 0 0 0 B C B0 1 0 0C B1 0 C. Here b ¼ B 2 C. ) B1 ¼ ¼B 1 @ @ A 0 A 0 0 1 0 cB B 1 2 0 0 0 1 The net evaluations zj  cj corresponding to non-basic variables are given by 

ð z 1  c1



z 2  c2

¼ ð0 0

0

¼ ð 2

1

0

z 4  c4 Þ

1 Þð a1

a2

1 Þ:

2

0

a1

6 B 6 B 2 6 B  6  a4 Þ 6 * A ¼ B B 1 6 B 4 @ 3 2

a2

a3

a4

1

0

1

3 1

1 0

0 0

1

0

1

a5

13

C7 1 C7 C7 7 0C C7 C7 0 A5 0

This indicates that z1  c1 is the most negative. Hence, y1 enters the basis. 0 Now;

1

B0 B y1 ¼ B1 a1 ¼ B @0 0

Also;

xB ¼ B1 b ¼ ð 2

0

0

1

0

0 0

1 0

2 0

0

10

2

1

0

1

2

C B C B 0C CB 1 C B 1 C C: C¼B CB 0 A@ 3 A @ 3 A 1

2

2

2 ÞT :

Now we construct the initial revised simplex table. yB y5 y3 z z

xB 2 2 0 −2

y1

B1 1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1



2 1 −3 −2

Mini ratio 2/2 = 1 2/1 = 2

According to the minimum ratio, y5 leaves the basis and y11 ¼ 1 is the key element. First iteration: We introduce y1 and drop y5 . In this iteration, the reduced basis inverse is given by

4.6 Standard Form II of Revised Simplex Method

0

B1

1=2 B 1=2 ¼B @ 3=2 1

97

0 1 0 0

1 0 0C C: 0A 1

0 0 1 0

The net evaluations zj  cj corresponding to non-basic variables are given by ð z 2  c2 ¼ ð1

z5  c5 Þ ¼ ð 1 0 0 1 Þð a2 a4 0 1 1 1 1 B 3 0 0C B C 1 ÞB C ¼ ð 0 0 1 Þ: @ 1 0 0A

z 4  c4

0

0

1

1

a5 Þ

0

This indicates that all zj  cj  0 correspond to non-basic variables. Hence, the current solution is an optimum for Phase I. The optimum solution for this phase is 0

1=2 B 1=2 B xB ¼ B1 b ¼ B @ 3=2 1 ) xB ¼ ð 1

1

3 ÞT

0 1

0 0

0 0

1 0

and

1 0 1 10 1 2 0 B B C C 0 CB 2 C B 1 C C C¼B C CB @ @ A A 3A 0 0 0 2 1

zB ¼ 0;

i.e. x1 ¼ 1; x3 ¼ 1; z ¼ 3 and Maximize z ¼ 0. Since there is no artiﬁcial variable in the current basis, we shall apply Phase II. Phase II: From the 0 basis inverse 1 B1 of the optimal obtained in Phase I, we 1=2 0 0 b 1 ¼ @ 1=2 1 0 A by deleting the fourth row and the fourth column. have B 3=2 0 1 0

Here

a1 B 2 b¼B A @ 1 3

a2 1 3 1

a3 0 1 0

1 a4 1 C C: 0 A 0

Now the net evaluations zj  cj corresponding to non-basic variables are given by 0

ð z2  c2

z4  c4 Þ ¼ ð 3=2 0

1 Þð b a2

b a 4 Þ ¼ ð 3=2

0

1 1 Þ@ 3 1

1 1 0 A ¼ ð 5=2 0

This indicates that z4  c4 is negative. Hence, by 4 enters the basis.

3=2 Þ:

98

4 Revised Simplex Method

0

b ) by 4 ¼ B

1

1=2 b a 4 ¼ @ 1=2 3=2

0 1 0

10 1 0 1 0 1 1=2 0 A@ 0 A ¼ @ 1=2 A 1 0 3=2

0 1 0 2 1=2 0 b b and bx B ¼ B b, where b b ¼ @ 2 A ¼ @ 1=2 1 0 3=2 0 Now we construct the revised simplex table.

10 1 0 1 0 2 1 0 A @ 2 A ¼ @ 1 A. 1 0 3

1

by B

bx B

b 1 B

by 1 by 3 z

1 1 3

1/2 −1/2 3/2

0 1 0

0 0 1

by 4

Mini ratio

−1/2 1/2 −3/2

– 1/(½) = 2

According to the minimum ratio, by 3 leaves the basis and by 24 ¼ 1=2 is the key element. First iteration: In this iteration, the reduced basis inverse matrix is given by 0

b 1 B

0 1 ¼ @ 1 2 0 3

1 0 0 A: 1

Now the net evaluations zj  cj corresponding to non-basic variables are given by ð z 2  c2 ¼ ð0

z 3  c3 Þ ¼ ð 0 3 1 Þ ð b a2 0 1 1 0 B C 3 1 Þ@ 3 1 A ¼ ð 10 3 Þ: 1 0

b a3 Þ

Since all zj  cj [ 0 correspond to non-basic variables, an optimum basic feasible solution has been obtained. The optimal solution is given by 0

0 1 B b b bx B ¼ B b ¼ @ 1

) bx B ¼

2

!

0 and

0

i:e: x1 ¼ 2; x2 ¼ 0

1 2

10 1 0 1 0 2 2 CB C B C 0 A@ 2 A ¼ @ 2 A

3

1

0

zB ¼ 6; and zMax ¼ 6:

6

4.6 Standard Form II of Revised Simplex Method

99

Example 5 Solve the following LPP by the two-phase revised simplex method: Maximize subject to

z ¼ x1 þ 2x2 þ 3x3 þ 4x4 3x1 þ 2x2 þ 3x3  x4  25 2x1 þ x2  2x3 þ x4  5 2x1 þ 2x2 þ x3 þ x4 ¼ 20 and x1 ; x2 x3 ; x4  0:

Solution This is a maximization problem. Here all bi ’s are positive. Now introducing the slack variable x5 ð  0Þ and the surplus variable x6 ð  0Þ in the ﬁrst and second constraints respectively, the given LPP is rewritten in the standard form as follows: Maximize subject to

z ¼ x1 þ 2x2 þ 3x3 þ 4x4 þ 0  x5 þ 0  x6 3x1 þ 2x2 þ 3x3  x4 þ x5 ¼ 25 2x1 þ x2  2x3 þ x4  x6 ¼ 5 2x1 þ 2x2 þ x3 þ x4 ¼ 20 and xi  0; i ¼ 1; 2; . . .; 6:

Considering the objective function as a constraint and z as a variable, we get the reduced set of constraints as 3x1 þ 2x2 þ 3x3  x4 þ x5

¼ 25

 2x1 þ x2  2x3 þ x4  x6 ¼ 5 2x1 þ 2x2 þ x3 þ x4 ¼ 20  x1  2x2  3x3  4x4 þ z ¼ 0: To get the initial unit basis matrix, introducing two artiﬁcial variables x7 ð  0Þ and x8 ð  0Þ in the second and third constraints respectively, we get the reduced constraints as 3x1 þ 2x2 þ 3x3  x4 þ x5 ¼ 25  2x1 þ x2  2x3 þ x4  x6 þ x7 ¼ 5 2x1 þ 2x2 þ x3 þ x4 þ x8

¼ 20

 x1  2x2  3x3  4x4 þ z

¼ 0:

We shall solve this problem by the two-phase method. In Phase I, the auxiliary objective function is to Maximize z ¼ x7  x8 ; i.e. Maximize z ¼ 3x2  x3 þ 2x4  x6  25 [substituting x7 and x8 from the second and third constraints].

100

4 Revised Simplex Method

Here the initial BFS is x5 ¼ 25; x7 ¼ 5; x8 ¼ 20; z ¼ 0 with B ¼ I4 as the initial basis. cB ¼ ð 0 0

Here

) B1 ¼



0 Þ:

0

1

0 1

B cB B1



0

1

B B0 B ¼B B0 B @0 0

0

0

1

0

0 0

1 0

0

0

0 0

1

C 0 0C C 0 0C C: C 1 0A 0 1

The net evaluations zj  cj corresponding to non-basic variables are ð z 1  c1

z 2  c2

z 3  c3

0 0 1 Þ½ a1 0 3 2 6 B 6 B 2 1 6 B 6* A ¼ B 2 2 6 B 6 B 4 @ 1 2 0 3

¼ ð0 2

¼ ð0

0

3 1

2

z 4  c4 z 6  c6 Þ a2 a3 a4 a6 

3

1

1

2 1

1 1

0 0

3

4

0

0

0

0

1

2

0

1

0

0

0

0

0

1 1 0 0

0 1

00

13

C7 0 0 C7 C7 7 0 0C C7 C7 1 0 A5 01

1 Þ:

This indicates that z2  c2 is the most negative. Hence, y2 enters the basis. Now; Also;

y2 ¼ B1 a2 ¼ ð 2 1 2 2 3 ÞT : xB ¼ B1 b ¼ ð 25 5 20 0 25 ÞT :

) xB ¼ ð 25

5

20 0 ÞT and zB ¼ 25:

Now we construct the initial revised simplex table. yB

xB

B1

y5 y7 y8

25 5 20 0 25

1 0 0 0 0

z z

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

y2

Mini ratio

2 1 2 2 3

25/2 5 10 – –

4.6 Standard Form II of Revised Simplex Method

101

According to the minimum ratio, y7 leaves the basis and y22 ¼ 1 is the key element. First iteration: We introduce y2 and drop y7 . In this iteration, the reduced basis inverse is given by 0

B1

1 B0 B ¼B B0 @0 0

2 1 2 2 3

0 0 1 0 0

1 0 0C C 0C C 0A 1

0 0 0 1 0

by

R01 R03 R04 R05

¼ R1  2R02 ¼ R3  2R02 : ¼ R4 þ 2R02 0 ¼ R5 þ 3R2

Again, the net evaluations zj  cj corresponding to non-basic variables are given by ð z1  c1

¼ ð0

z4  c4 z6  c6 Þ ¼ ð 0 3 0 0 1 Þ½ a1 a3 a4 1 0 3 3 1 0 C B B 2 2 1 1 C C B 1 ÞB 1 1 0 C C ¼ ð 6 5 1 2 Þ: B 2 C B @ 1 3 4 0 A

z 3  c3

3 0

0

0

1

2

a6 

1

This indicates that z1  c1 is the most negative. Hence, y1 enters the basis. 0

1 B0 B ) y1 ¼ B1 a1 ¼ B B0 @0 0 0

1 B0 B and xB ¼ B1 b ¼ B B0 @0 0

2 1 2 2 3

0 0 1 0 0

0 0 0 1 0

10 1 0 1 0 3 7 B C B C 0C CB 2 C B 2 C B 2 C¼B 6 C 0C CB C B C 0 A@ 1 A @ 5 A 1 0 6

2 1 2 2 3

0 0 1 0 0

0 0 0 1 0

1 1 0 10 15 25 0 C C B B 0C CB 5 C B 5 C B C B C 0 CB 20 C ¼ B 10 C C: 0 A@ 0 A @ 10 A 10 25 1

Hence, the next revised simplex table is as follows: yB

xB

B1

y5 y2

15 5

1 0

2 1

0 0

0 0

0 0

y1

Mini ratio

7 2

15/7 – (continued)

102

4 Revised Simplex Method

(continued) yB

xB

B1

y8

10 10 10

0 0 0

z z

2 2 3

1 0 0

0 1 0

0 0 1

y1

Mini ratio

6 5 6

5/3 – –

According to the minimum ratio, y8 leaves the basis and y13 ¼ 6 is the key element. Second iteration: We introduce y1 and drop y8 . In this iteration, the reduced basis inverse is given by 0

B1

1 B0 B ¼B B0 @0 0

1=3 1=3 1=3 1=3 1

7=6 1=3 1=6 5=6 1

0 0 0 1 0

1 0 0C C 0C C: 0A 1

Now, the net evaluations zj  cj corresponding to non-basic variables are ð z 3  c3

¼ ð0

z 4  c4

1

1

z6  c6 Þ ¼ ð 0 0 3 1 B B 2 1 B 0 1 ÞB 1 B 1 B @ 3 4 1

2

1 1 0 1 Þð a3 a4 1 0 C 1 C C 0 C C ¼ ð 0 0 0 Þ: C 0 A 1

a6 Þ

Since all zj  cj corresponding to non-basic variables are zero, the current solution is an optimum one for Phase I. Therefore, the optimum solution for this phase is 0

1

B B0 B  1  xB ¼ B b ¼ B B0 B @0 0

1=3

7=6

0

1=3 1=3

1=3 1=6

0 0

1=3

5=6

1

1

1

0

0

10

25

1

0

10=3

1

C C B CB 0 CB 5 C B 25=3 C C C B CB C C B B 0C CB 20 C ¼ B 5=3 C: C C B CB 0 A@ 0 A @ 55=3 A 0 25 1

5 25 ) x1 ¼ ; x2 ¼ ; x3 ¼ 0; x4 ¼ 0 and the Max z ¼ 0: 3 3 Since there is no artiﬁcial variable in the current basis, we shall apply Phase II.

4.6 Standard Form II of Revised Simplex Method

103

Phase II: From the basis inverse B1 of the optimal solution obtained in Phase I, we have 0

b 1 B

Here

1 B0 B ¼B @0

1=3 0 1=3 0 3 B 2 b¼B A B @ 2 1

1 0 0C C C by deleting the fifth row and the fifth column: 0A

7=6 1=3

1=3 1=3

1=6 5=6

1

2

3

1

1

1

2

1

0

2 2

1 3

1 4

0 0

0

1

1 C C C: 0 A 0

Now the net evaluations zj  cj corresponding to non-basic variables are ð z3  c 3

z 4  c4

z 6  c6 Þ

 a4 ¼ 0 13 56 1 ð b a3 b  17 1  ¼  6  17 6 3 :

0

 b a6 Þ ¼ 0

1 3

5 6

3 B B 2 1 B @ 1

1 1

3

4

1

1 0 1 C C C 0 A 0

This result indicates that either by 3 or by 4 can enter the basis. Arbitrarily, we choose by 3 as the entering vector. 0

1 B0 1 b b Now; by 3 ¼ B a3 ¼ B @0 0

1=3 7=6 1=3 1=3 1=3 1=6 1=3 5=6

b 1 b and bx B ¼ B b ¼ ð 10=3

10 1 0 1 0 3 7=6 B C B C 0C CB 2 C ¼ B 1=3 C A @ A @ 0 1 5=6 A 1 3 17=6

25=3 5=3

55=3 ÞT ) zB ¼

55 : 3

Hence, the next revised simplex table is as follows: by B

bx B

b 1 B

by 5 by 2 by 1 z

10/3 25/3 5/3 55/3

1 0 0 0

1/3 1/3 −1/3 1/3

−7/6 1/3 1/6 5/6

0 0 0 1

by 3

Mini ratio

7/6 −1/3 5/6* −17/6

20/7 – 2 –

According to the minimum ratio, by 1 leaves the basis and by 33 ¼ 5=6 is the key element.

104

4 Revised Simplex Method

First iteration: In this iteration, the reduced basis matrix is given by 0

b 1 B

1 B0 ¼B @0 0

4=5 7=5 1=5 2=5 2=5 1=5 4=5 7=5

1 0 0C C: 0A 1

Now, the net evaluations zj  cj corresponding to non-basic variables are ð z1  c1  ¼ 0

 z6  c6 Þ ¼ 0  45 75 1 ð b a1 0 1 3 1 0 C B B 2 1 1 C  17 1 B C ¼ 5  17 5 @ 2 1 0 A 1 4 0

z 4  c4  45

7 5

b a4

4 5

b a6 Þ :

This indicates that by 4 enters the basis. 0

1 B0 1 b b ) by 4 ¼ B a4 ¼ B @0 0 0

4=5 1=5 2=5 4=5

1 B0 1 b b b¼B and bx B ¼ B @0 0

7=5 2=5 1=5 7=5

10 1 0 1 0 1 8=5 B C B C 0C CB 1 C ¼ B 3=5 C 0 A@ 1 A @ 1=5 A 1 4 17=5 10 1 0 1 0 25 1 B 5 C B 9 C 0C CB C ¼ B C: 0 A@ 20 A @ 2 A 1 0 24

4=5 7=5 1=5 2=5 2=5 1=5 4=5 7=5

) zB ¼ ð 1

9

2 ÞT ; zB ¼ 24:

Hence, the next revised simplex table is as follows: by B

bx B

b 1 B

by 5 by 2 by 3 z

1 9 2 24

1 0 0 0

4/5 1/5 −2/5 −4/5

−7/5 2/5 1/5 7/5

0 0 0 1

by 4

Mini ratio

– 8/5 3/5* −1/5 −17/5

– 15 –

According to the minimum ratio, by 2 leaves the basis and by 24 ¼ 3=5 is the key element.

4.6 Standard Form II of Revised Simplex Method

105

Second iteration: In this iteration, the reduced basis matrix is given by 0

b 1 B

1 B0 ¼B @0 0

4=3 1=3 1=3 2=3 1=3 1=3 1=3 11=3

1 0 0C C: 0A 1

Now the net evaluations zj  cj corresponding to non-basic variables are given by ð z1  c1 z2  c2 z6  c6 Þ  1 ðb ¼ 0 13 11 a2 b a6 Þ a1 b 3 0 1 3 2 0 C  B B 2 1 1 C  17 1 ¼ 0 13 11 B C¼ 3 3 @ 2 2 0 A 1

2

17 3

 13 :

0

This indicates that by 6 enters the basis. 0

1 B0 1 b b a6 ¼ B ) by 6 ¼ B @0 0 0

1 B0 1 b B b¼B and bx B ¼ B @0 0

4=3 1=3 1=3 1=3

1=3 2=3 1=3 11=3

4=3 1=3 1=3 2=3 1=3 1=3 1=3 11=3

10 1 0 1 0 0 4=3 B C B C 0C CB 1 C ¼ B 1=3 C 0 A@ 0 A @ 1=3 A 1 0 1=3 10 1 0 1 0 25 25 B 5 C B 15 C 0C CB C ¼ B C: 0 A@ 20 A @ 5 A 1 0 75

The next revised simplex table is as follows: by B

bx B

by 5 by 4 by 3 z

25

b 1 B 1

4/3

−1/3

15 5 75

0 0 0

1/3 −1/3 1/3

2/3 1/3 11/3

by 6

Mini ratio

0

−4/3

0 0 1

−1/3 1=3 −1/3

– 15

According to the minimum ratio, by 3 leaves the basis and by 36 ¼ 13 is the key element.

106

4 Revised Simplex Method

Third iteration: In this iteration, the reduced basis matrix is given by 0

b 1 B

1 B0 ¼B @0 0

0 0 1 0

1 1 1 4

1 0 0C C: 0A 1

Now the net evaluations zj  cj corresponding to non-basic variables are given by ð z1  c1 ¼ ð0

z 3  c3 Þ ¼ ð 0 0 4 1 Þ ð b a1 b a2 0 1 3 2 3 B 2 1 2 C B C 1 ÞB C ¼ ð 7 6 1 Þ: @ 2 2 1 A

z 2  c2

0 4

1

2

b a3 Þ

3

Since the net evaluations zj  cj corresponding to non-basic variables are greater than zero, an optimum basic feasible solution has been obtained. The optimal solution is given by 0 0

1 1

1

1

10 1 0 1 45 25 0 B B C C 0 CB 5 C B 20 C C CB C ¼ B C: 0 A@ 20 A @ 15 A

0

4

1

0

1 B B0 b 1 b bx B ¼ B b¼B @0 0

45

1

B C )bx B ¼ @ 20 A

0 and

0

80

zB ¼ 80;

15 i:e: x1 ¼ 0; x2 ¼ 0; x3 ¼ 0; x4 ¼ 20 and zMax ¼ 80:

4.7

The revised simplex method has the following advantages: (i) This method automatically generates the inverse of the current basis matrix and the new basic feasible solution as well. (ii) It provides more information with less computational effort. (iii) It requires fewer computations than that of the ordinary simplex method. (iv) Fewer entries need to be made in each table of the revised simplex method.

4.8 Difference Between Revised Simplex Method and Simplex Method

4.8

107

Difference Between Revised Simplex Method and Simplex Method

Although the simplex method is a straightforward algebraic procedure, it is not an efﬁcient computational procedure for solving LPPs on a digital computer. The ordinary simplex method requires the storing of all the elements of the entire table in the memory of the computer, which may not be feasible for a very large problem. Also, at each iteration the ordinary simplex method requires many computations that are not needed in the subsequent iteration. In fact, it is not necessary to compute the entire table during each iteration. The only pieces of information needed at each iteration are: (i) net evaluations zj  cj , (ii) entering vector, (iii) current basic variables and their values. In the revised simplex method the preceding information is directly obtained very efﬁciently, avoiding unnecessary calculations as compared to the ordinary simplex method. Thus, the revised simplex method computes and stores only those pieces of information which are needed in the current iteration.

4.9

Exercises

Solve the following LPPs using the revised simplex method: 1. Maximize such that 2. Maximize such that 3. Maximize such that 4. Maximize such that 5.

Maximize such that

6. Maximize such that 7. Maximize such that

z ¼ x1 þ x2 x1 þ 2x2  2 4x1 þ x2  4 and x1 ; x2  0: z ¼ 5x1 þ 3x2 3x1 þ 5x2  15 3x1 þ 2x2  10 and x1 ; x2  0: z ¼ x1 þ x2 þ 3x3 3x1 þ 2x2 þ x3  3 2x1 þ x2 þ 2x3  2 and x1 ; x2 ; x3  0: z ¼ x1 þ x2 3x1 þ 3x2  6 x1 þ 4x2  4 and x1 ; x2  0: z ¼ 3x1 þ 2x2 þ 5x3 x1 þ 2x2 þ x3  430 3x1 þ 2x3  460 x1 þ 4x2  420 and x1 ; x2 ; x3  0: z ¼ x1 þ 2x2 x1 þ 2x2  3 x1 þ 3x2  1 and x1 ; x2  0: z ¼ x1 þ 2x2 x1 þ x2  3 x1 þ 2x2  5 3x1 þ x2  6 and x1 ; x2  0:

108

4 Revised Simplex Method

z ¼ 5x1 þ 3x2 4x1 þ 5x2  10 5x1 þ 2x2  10 3x1 þ 8x2  12 and x1 ; x2  0: 9. Minimize z ¼ 3x1 þ x2 such that x1 þ x2  1 2x1 þ x2  0 and x1 ; x2  0: 10. Minimize z ¼ 4x1 þ 2x2 þ 3x3 such that 2x1 þ 4x3  5 2x1 þ 4x2 þ x3  4 and x1 ; x2 ; x3  0: Maximize z ¼ 4x1 þ 3x2 11. such that 3x1 þ 4x2  12 3x1 þ 3x2  10 2x1 þ x2  4 x1 ; x2  0: 12. Maximize z ¼ 2x1 þ 3x2 such that x1 þ x2  0 x1  4 and x1 ; x2  0: Minimize z ¼ 2x1 þ x2 13. such that 3x1 þ x2  3 4x1 þ 3x2  6 x1 þ 2x2  3 and x1 ; x2  0: 14. Maximize z ¼ x1 þ x2 such that 2x1 þ 5x2  6 x1 þ x2  2 and x1 ; x2  0: 8. Maximize such that

Chapter 5

Dual Simplex Method

5.1

Objective

The objective of this chapter is to discuss an advanced technique, called the dual simplex method, for solving linear programming problems with  type constraints.

5.2

Introduction

In Chap. 3, the simplex method was discussed for solving linear programming problems (LPPs). In this method, a negative net evaluation (for a maximization primal problem) indicates that the current solution is not optimum. Consider the jth dual constraint of a primal maximization problem, namely m X

aij wi  cj ; j ¼ 1; 2; . . .; n

i¼1

where aij is the constraint coefﬁcient of the jth primal variable xj in the ith primal constraint, and cj is the proﬁt/cost associated with xj in the primal objective funcm P tion. Then clearly the net evaluation zj  cj corresponding to xj is aij wi  cj . i¼1

Thus, a negative value of zj  cj indicates that the corresponding dual constraint is not satisﬁed. This leads to the conclusion that the dual is infeasible when the primal is non-optimum and vice versa. This gives the following result. While the primal problem starts as feasible but non-optimum and continues to be feasible until the optimum is reached, the dual problem starts as infeasible but better than optimum and continues to be infeasible until the true optimal solution is reached. In other words, while the primal problem is seeking optimality, the dual problem is automatically seeking feasibility. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_5

109

110

5 Dual Simplex Method

Hence, the solution will start optimum (or actually better than optimum) and infeasible and will remain infeasible until the true optimum is reached, at which the solution becomes feasible. Such a procedure is called the dual simplex method. The dual simplex method is very similar to the ordinary simplex method. In these methods, the only difference is in the criterion used for selecting the entering and outgoing vectors. In this context, we note that in the dual simplex method the outgoing vector is determined ﬁrst and then the entering vector. This is just the reverse of what is done in the ordinary simplex method. The dual simplex method gives an algorithm in which we start with a basic optimal solution of the primal problem with all zj  cj  0, but it is not a feasible solution, as the values of some basic variables are negative. In each iteration, the numbers of negative basic variables are decreased while maintaining the optimality. An optimal solution is reached in a ﬁnite number of steps. In this method, inclusions of artiﬁcial variables are not required. Hence, this method greatly reduces the amount of computation. The dual simplex method has a wide range of applications. Some are as follows: (i) Parametric programming problem (ii) Integer linear programming problem (iii) Post-optimality analysis when the requirement vector is changed or when new constraints are added. Theorem For a basic solution (not necessarily feasible) with all zj  cj  0 8j; the value of the objective function corresponding to this basic solution is optimal or better than optimal. Proof Let the primal LPP be as follows:   Maximize z ¼ cT ; x subject toAx ¼ b; x  0; where c; xT 2 Rn ; bT 2 Rm and A is an m  n real matrix: Let B be the current basis of the primal LPP and xB be the corresponding basic solution. ) BxB ¼ b

or

xB ¼ B1 b:

Let zj  cj  0; 8j ¼ 1; 2; . . .; n and one or more basic variables has negative value. The dual of the preceding LPP is given by the following: Minimize subject to

z0 ¼ hbT ; wi AT w  cT ; w is unrestricted:

The value of z corresponding to the basic solution xB is given by

5.2 Introduction

ðzB Þ ¼ cB xB ¼ ½cB xB T ¼

111

½* cB xB is (1,1) matrix as cB is 1  m and xB is m  1 matrix

xTB cTB

 T   ¼ B1 b cTB * xB ¼ B1 b  T  T ¼ bT B1 cTB ¼ bT cB B1  T ¼ bT w ; where w ¼ cB B1 :

Therefore zB ¼ hbT ; w i: It should be noted that w 2 Rm : Since all zj  cj  0, we have    A 1 0 1 cB B c   or cB B1 A  c  0   T  T or cB B1 A  cT ; or AT cB B1  cT or AT w0  cT This implies that w0 is a basic feasible solution of the dual LPP. Now hbT ; w0 i is the value of the dual objective function for the solution w0 :   z0Min  bT ; w0 : But we have hbT ; w0 i ¼ zB : Therefore, z0Min  zB : According to the duality theorem, we have ) z0Min  zMax : Hence, zMax  zB or zB  zMax ; i.e. the value of the objective function corresponding to the basic solution xB ¼ B1 b is equal to or greater than the optimal value of the objective function. This proves the theorem. Theorem Let xBr be a negative basic variable in a dual simplex table, let all net evaluations zj  cj be non-negative and let the primal LPP be of maximization type. If yrj  0 for all non-basic variables xj , then there does not exist any feasible solution to the primal LPP. Proof Let the primal LPP be as follows:   Maximize z ¼ cT ; x subject to Ax ¼ b; x  0; where c; xT 2 Rn ; bT 2 Rm and A is an m  n real matrix:

112

5 Dual Simplex Method

Let xB ¼ B1 b be a basic solution of this primal LPP. The dual of this LPP is given by

  Minimize z0 ¼ bT ; w subject to AT w  cT ; where w is unrestricted and wT 2 Rm :

Since all zj  cj  0; we have  cB B1

1



A c

! 0

T

or AT w0  cT , where w0 ¼ ðcB B1 Þ . Hence, w0 is a feasible solution of the dual LPP. Let g be any positive real number and w1 ¼ w0 þ gbr , where br is the rth column of T ðB1 Þ : Now AT w1 ¼ AT ðw0 þ gbr Þ ¼ AT w0 þ gAT br : ð5:1Þ We know that yj ¼ B1 aj for all j. T

Hence, Y ¼ B1 A or Y T ¼ ðB1 AÞ ¼ AT ðB1 Þ

T

  ) Y T ¼ AT ½b1 b2 . . . bm  ¼ AT b1 AT b2 . . . AT bm 2

y11 6 y21 6 . 6 or6 .. 6 . 4 .. ym1 2

y11 6 y12 6 . 6 or6 .. 6 . 4 .. y1n

3 y1n y2n 7 7   7 7 ¼ AT b1 AT b2 . . . AT bm 7 5

y12 y22

... ...

ym2

. . . ymn

y21 y22

y2n

3 . . . ym1 . . . ym2 7 7   7 7 ¼ AT b1 AT b2 . . . AT bm 7 5 . . . ymn

3 yr1 6 yr2 7 6 . 7 6 7 ) 6 .. 7 ¼ AT br for r ¼ 1; 2; . . .; m 6 . 7 4 .. 5 yrn 2

5.2 Introduction

113

Since all yrj  0; we have AT br  0: As g  0 we have gAT br  0. Hence from (5.1), we have AT w0  AT w1 or AT w0  cT as AT w1  cT : Hence, w0 is a feasible solution to the dual LPP for all a: The value of the dual objective function z0 for the feasible solution w0 is given by         z0 jw0 ¼ bT ; w0 ¼ bT ; w1 þ gbr ¼ bT ; w1 þ g bT ; br : T

ð5:2Þ

T

As, xB ¼ B1 b; xTB ¼ ðB1 bÞ or xTB ¼ bT ðB1 Þ or ½ xB1 xB2 . . . xBr . . . xBm  ¼ bT ½ b1 b2

. . . br

. . . bm :

Hence, xBr ¼ hbT ; br ifor all r: Thus, from (5.2), z0 jw0 ¼ hbT ; w1 i þ gxBr : Letting g ! 1, we have z0 jw0 ! 1 ½* xBr \0 and g [ 0. This means that the dual LPP has an unbounded solution. Therefore, the primal LPP has no feasible solution. Criteria for incoming and outgoing vectors in dual simplex method In the simplex method, if the basic variable xBr is replaced by the non-basic variable xk , then we have ^z ¼ z 

xBr ð z k  ck Þ yrk

and ^zj  ^cj ¼ zj  cj 

yrj ðzk  ck Þ: yrk

ð5:3Þ ð5:4Þ

Now to remove negative basic variables, ﬁrst the most negative basic variable xBr is selected as the outgoing basic variable from all basic variables. So the corresponding vector yBr is the outgoing vector. We know that if for some basic solution (not necessarily feasible) all components of net evaluations are non-negative, the value of the objective function to this basic solution is either optimal or better than optimal. So, we have to decrease the value of the objective function to obtain the value of zMax : Since xBr \0 and zk  ck  0, in order to have ^z  z, we must choose an entering variable xk such that yrk \0: In this case, the new values of the net evaluations are given by

114

5 Dual Simplex Method

^zj  ^cj ¼ zj  cj  Again;

yrj ðzk  ck Þ: yrk

^zj  ^cj  0 8 j:

Hence, we can write yrj ðzk  ck Þ; yrk z j  cj z k  ck i:e:  for yrj \0 yrj yrk

z k  ck z j  cj or ¼ Max ; yrj \0 : j yrk yrj z j  cj 

From this criterion, the incoming vector and key element can be found easily. Here the incoming vector is yk and yrk is the key element.

5.3

Dual Simplex Algorithm

The iterative procedures for the dual simplex method are as follows: Step 1 If the given LPP is in the minimization form, then convert it into a maximization problem using the mini-max principle. Step 2 Convert  type constraints of the given LPP, if any, into  type, multiplying both sides of the corresponding constraints by ð1Þ. Step 3 Introduce slack variables in the constraints of the given LPP and obtain an initial basic solution. Put this solution in the starting dual simplex table. Step 4 Test the nature of the net evaluations zj  cj in the starting simplex table. (i) If all zj  cj and xBj are non-negative for all i and j, then an optimum basic feasible solution has been obtained. (ii) If all zj  cj are non-negative and at least one basic variable, say xBi , is negative, then go to Step 5. (iii) If at least one zj  cj is negative, then the dual simplex method is not applicable for the given problem. Step 5 Select the most negative of xBi , say xBr . The corresponding basis vector yBr leaves the basis. Step 6 Test the nature of all yrj ; j ¼ 1; 2; . . .; n, i.e. the elements of the rth row. (i) If all yrj are non-negative, there does not exist any feasible solution to the given LPP.

5.3 Dual Simplex Algorithm

115

(ii) If at least one yrj is negative, then compute

Max j

z j  cj : yrj \0 ; yrj

j ¼ 1; 2; . . .; n;

k If this is zkyc , then yk enters the basis in place of yBr and yrk is the rk key element.

Step 7 With yrk as the key element, form the next table. Using elementary row operations, convert the key element to unity and all other elements of the key column to zero to get the improved solution. Step 8 Repeat Steps 4 through 7 until either an optimum basic feasible solution is obtained or there is an indication of no feasible solution. Example 1 Using the dual simplex method, solve the following LPP: Minimize z ¼ x1 þ x2 subject to 2x1 þ x2  2 x1  x2  1 and x1 ; x2  0: Solution The given LPP can be written as follows: Minimize subject to

z ¼ x1 þ x2 2x1  x2   2 x1 þ x2   1 and x1 ; x2  0:

Here the problem is a minimization problem. Let z0 ¼ z. Then Min z ¼ MaxðzÞ ¼ Max z0 using the mini-max principle. Hence, the problem is to maximize z0 ¼ x1  x2 . Now introducing two slack variables x3 and x4 in the ﬁrst and second constraints respectively in the given LPP, we get the following converted equations: 2x1  x2 þ x3 ¼ 2 x1 þ x2 þ x4 ¼ 1 and x1 ; x2 ; x3 ; x4  0 Then the readjusted objective function z0 is given by z0 ¼ x1  x2 þ 0  x3 þ 0  x4 :

116

5 Dual Simplex Method

Now taking x1 ¼ x2 ¼ 0, we get x3 ¼ 2; x4 ¼ 1, which is an initial basic but infeasible solution to the given problem. Now we construct the starting dual simplex table as follows: cj cB 0 0

yB

xB

y3 y4 z0B ¼ 0

x3 ¼ 2 x4 ¼ 1

1 y1

1 y2

0 y3

0 y4

2 1 1

1 1 1

1 0 0

0 1 0

zj  c j

Since all zj  cj  0 and xB1 ¼ 2; xB2 ¼ 1, the current basic solution is optimum or more than optimum but infeasible. Now xBr ¼ MinfxBi ; xBi \0g ¼ Minf2; 1g ¼ 2 ¼ xB1 )r¼1 Hence, the vector yB1 , i.e. y3 , leaves the basis.

z j  cj z j  cj Again; Max ; yrj \0 ¼ Max ; y1j \0 yrj y1j

z 1  c1 z 2  c2 1 1 1 ; ¼ Max ; ¼ ; ¼ Max 2 1 2 y11 y12 which occurs for j =1. Then the vector y1 enters the basis, and yrj i.e. y11 ¼ 2 is the key element. Now we construct the next table.

cB

cj yB

xB

1 y1

1 y2

0 y3

0 y4

1

y1

x1 = 1

1

 12

0

y4

x4 ¼ 2

1 2 1 2 1 2

1 2 1 2

1

0

z0B ¼ 1

0 0

0

zj  c j

Since all zj  cj  0 and xB2 ¼ 2 is negative, the current basic solution is an optimum or more than optimum but infeasible solution. Since only xB2 ¼ 2 is negative, then r =2 and the corresponding basis vector yB2 i.e. y4 leaves the basis. But since all y2j ðj ¼ 1; 2; 3; 4Þ are non-negative, no other vector enters the basis. Thus, there does not exist any feasible solution to the given LPP. Example 2 Using the dual simplex method, solve the following LPP:

5.3 Dual Simplex Algorithm

Minimize subject to

117

z ¼ 2x1 þ x2 þ x3 4x1 þ 6x2 þ 3x3  8 x1  9x2 þ x3   3 2x1  3x2 þ 5x3   4 and x1 ; x2 ; x3  0:

Solution Here the problem is a minimization problem. Let z0 ¼ z. Then Min z ¼ MaxðzÞ ¼ Max z0 using the mini-max principle. Hence, the problem is to maximize z0 ¼ 2x1  x2  x3 . Now introducing three slack variables x4  0; x5  0; x6  0, one to each constraint respectively, we get the following converted equations: 4x1 þ 6x2 þ 3x3 þ x4 ¼ 8 x1  9x2 þ x3 þ x5 ¼ 3 2x1  3x2 þ 5x3 þ x6 ¼ 4 and xj  0; j ¼ 1; 2; . . .; 6 Then the readjusted objective function z0 is given by z0 ¼ 2x1  x2  x3 þ 0  x4 þ 0  x5 þ 0  x6 : Now taking x1 ¼ x2 ¼ x3 ¼ 0, we get x8 ¼ 8; x5 ¼ 3; x6 ¼ 4, which is an initial basic but infeasible solution to the given problem. Now we construct the starting dual simplex table.

cB

cj yB

0 y4 0 y5 0 y6 z0B ¼ 0

xB x4 ¼ 8 x5 ¼ 3 x6 ¼ 4

2 y1

1 y2

1 y3

0 y4

0 y5

0 y6

4 1 2 2

6 9 3 1

3 1 5 1

1 0 0 0

0 1 0 0

0 0 1 0

zj  c j

Since all zj  cj  0 and xB2 ¼ 3, xB3 ¼ 4, the current solution is optimum or more than optimum but infeasible. Since xBr ¼ Min½xBi ; xBi \0 ¼ Min½xB2 ; xB3  ¼ Min½3; 4 ¼ 4 ¼ xB3 ; ) r ¼ 3:

118

5 Dual Simplex Method

Hence, the vector yB3 , i.e. y6 , leaves the basis. Again;

zj  cj zj  cj z1  c1 z2  c2 Max ; yrj \0 ¼ Max ; y3j \0 ¼ Max ; yrj y3j y31 y32

2 1 1 ; ¼  ; which occurs for j ¼ 2: ¼ Max 2 3 3

Then the vector y2 enters the basis, and yrj , i.e. y32 ¼ 3, is the key element. Now we construct the next table.

cB

cj yB

0 0 −1

y4 y5 y2

z0B

¼

xB

2 y1

1 y2

1 y3

0 y4

0 y5

0 y6

x4 ¼ 0 x5 ¼ 9

0 7

13 14

2 3 4 3

1 0 0

0 1 0

2 3

x2 ¼ 43

0 0 1

0

0

1 3

 43

0

 53 8 3

 13 zj  cj

Since all zj  cj  0 8 j ¼ 1; 2; . . .; 6 and all xBi  0, an optimal basic feasible solution has been obtained. Hence, the required optimal basic feasible  4solution  4 to the 4 0 given LPP is x1 ¼ 0; x2 ¼ 3 ; x3 ¼ 0 with Min z ¼ Max z ¼   3 ¼ 3. Example 3 Use the dual simplex method to solve the following LPP: Maximize z ¼ 3x1  x2 subject to x1 þ x2  1 2x1 þ 3x2  2 and x1 ; x2  0: Solution The given LPP can be written as follows: Maximize z ¼ 3x1  x2 subject to x1  x2   1 2x1  3x2   2 and x1 ; x2  0: Now introducing the slack variables x3  0 and x4  0 in the constraint of the given LPP, we get the following reduced LPP: Maximize subject to

z ¼ 3x1  x2 þ 0  x3 þ 0  x4 x1  x2 þ x3 ¼ 1 2x1  3x2 þ x4 ¼ 2

and

x1 ; x2 ; x3 ; x4  0:

5.3 Dual Simplex Algorithm

119

Now setting x1 ¼ x2 ¼ 0, we get x3 ¼ 1; x4 ¼ 2, which is an initial basic but non-feasible solution to the given problem. The starting dual simplex table is as follows:

cB 0 0 zB ¼ 0

cj yB

xB

y3 y4

x3 ¼ 1 x4 ¼ 2

3 y1

1 y2

0 y3

0 y4

1 2 3

1 3 1

1 0 0

0 1 0

zj  c j

Since all zj  cj  0 and xB1 ¼ 1; xB2 ¼ 2, the current basic solution is optimum or more than optimum but infeasible. Here the most negative basic variable is xB2 ¼ 2. Hence r = 2, and the corresponding basis vector y4 leaves the basis. For the vector which enters the basis we select

z j  cj z 1  c 1 z 2  c2 Max ; yij \0 ¼ Max ; y2j y21 y22

3 1 3 1 1 ; ¼ Max ¼ Max  ;  ¼  ; which occurs for j ¼ 2: 2 3 2 3 3 Then y2 enters the basis, and y2j , i.e. y22 ¼ 3, is the key element. Now we construct the next table. 3 y1

1 y2

0 y3

0 y4

x3 ¼  13

 13

0

1

 13

x2 ¼

2 3 7 3

1

0

 13

0

0

1 3

cB

cj yB

xB

0

y3 y2

1 zB ¼

 23

2 3

zj  cj

Since all zj  cj  0 and xB1 ¼  13, the current basic solution is optimum or more than optimum but infeasible. Since xB1 ¼  13 is negative, hence r ¼ 1 and the corresponding basis vector y3 leaves the basis.

zj  cj ; y1j \0 y1j

z 1  c1 z 4  c4 7=3 1=3 ¼ Max ; ; ¼ Max ¼ Maxf7; 1g ¼ 1; which occurs for j ¼ 4: y11 y14 1=3 1=3

Now; Max

120

5 Dual Simplex Method

Then y4 enters the basis and y14 ¼  13 is the key element. Now we construct the third table.

cB 0 1 zB ¼ 1

cj yB

xB

y4 y2

x4 ¼ 1 x2 ¼ 1

3 y1

1 y2

0 y3

0 y4

1 1 2

0 1 0

3 1 1

1 0 0

zj  cj

Here all zj  cj  0 8 j ¼ 1; 2; 3; 4 and all xBi ’s are non-negative. This shows that an optimal basic feasible solution has been obtained. Thus, the required optimal basic feasible solution is x1 ¼ 0; x2 ¼ 1 and zMax ¼ 1.

5.4

Difference Between Simplex Method and Dual Simplex Method

The dual simplex method is similar to the regular simplex method, except that in the latter the starting initial basic solution is feasible but not optimum, while in the former it is infeasible but optimum or better than optimum. The simplex method works towards optimality, while the dual simplex method works towards feasibility. In the consecutive iterations of the simplex method for solving a maximization problem, the value of the objective function gradually increases and ﬁnally reaches to its optimal value. In each iteration, the solution is basic feasible but non-optimal. On the other hand, in the consecutive iterations of the dual simplex method for solving the same problem, the value of the objective function gradually decreases and ﬁnally reaches to its optimal value. In each iteration, the solution is basic non-feasible but optimal (or better than optimal).

5.5

Modiﬁed Dual Simplex Method

If the initial table of the dual simplex method contains some negative basic variables and some of the net evaluations are negative, then the dual simplex method is not applicable. In such situations, the dual simplex method is to be modiﬁed to form an equivalent LPP in which some basic variables are negative but all net evaluations are non-negative. Hence, the standard dual simplex method can be applied to that equivalent LPP. One of the modiﬁed methods is the artiﬁcial constraint method. In this method, the variables corresponding to the negative net evaluations and the variable

5.5 Modiﬁed Dual Simplex Method

121

corresponding to the most negative component of net evaluations are considered. Let zp  cp be the most negative net evaluation corresponding to the variable xp : In this method we have to consider the artiﬁcial constraint X xj  M; P where is extended over all j’s for which zj  cj \0, and M is a sufﬁciently large positive number. Adding the slack variable xM to this constraint, we have X xj þ xM ¼ M: From this equation, we obtain xp ¼ M 

xM þ

P j6¼p

! xj .

Then substituting this xp in the original objective function and all the constraints, we get a reduced problem. This reduced problem together with the artiﬁcial constraint is equivalent to the given problem. For this equivalent LPP, all zj  cj  0: Thus, to solve this problem, the dual simplex method can now be applied. Example 4 Solve the following problem by the modiﬁed dual simplex method: Minimize z ¼ 2x1  x2  x3 subject to 4x1 þ 6x2 þ 3x3  8 x1 þ 9x2  x3  3 2x1 þ 3x2  5x3  4 and x1 ; x2 ; x3  0: Solution The given minimization problem is written as a maximization problem as follows: Maximize z0 ¼ 2x1 þ x2 þ x3 þ 0  x4 þ 0  x5 þ 0  x6 subject to 4x1 þ 6x2 þ 3x3 þ x4 ¼ 8 x1  9x2 þ x3 þ x5 ¼ 3 2x1  3x2 þ 5x3 þ x6 ¼ 4 and x1 ; x2 ; x3 ; x4 ; x5 ; x6  0: Now we construct the initial dual simplex table.

cB 0 0 0 z0B ¼ 0

yB y4 y5 y6

cj

2

1

1

0

0

0

xB x4 ¼ 8 x5 ¼ 3 x6 ¼ 4

y1 4 1 −2 −2

y2 6 −9 −3 −1

y3 3 1 5 −1

y4 1 0 0 0

y5 0 1 0 0

y6 0 0 1 0

zj  c j

122

5 Dual Simplex Method

Here there are negative net evaluations, so the standard dual simplex method is not applicable. From this simplex table, it is seen that the most negative evaluation is z1  c1 ¼ 2 corresponding to the variablex x1 : Hence, the artiﬁcial constraint is x1 þ x2 þ x3  M, where M is a very large positive number. Adding the slack variable xM ; we have x1 þ x2 þ x3 þ xM ¼ M; from which we have x1 ¼ M  x2  x3  xM : Substituting the expression for x1 in the given LPP, we have: Maximize subject to

z00 ¼ 2ðM  x2  x3  xM Þ þ x2 þ x3 þ 0  x4 þ 0  x5 þ 0  x6 4ðM  x2  x3  xM Þ þ 6x2 þ 3x3 þ x4 ¼ 8 M  x2  x3  xM  9x2 þ x3 þ ¼ 3 2ðM  x2  x3  xM Þ  3x2 þ 5x3 þ x6 ¼ 4 x1 þ x2 þ x3 þ xM ¼M and x1 ; x2 ; x3 ; x4 ; x5 ; x6 ; xM  0

or Maximize subject to

z00 ¼ 2xM  x2  x3 þ 0  x4 þ 0  x5 þ 0  x6 þ 2M 4xM þ 2x2  x3 þ x4 ¼ 8  4M xM  10x2 þ x5 ¼ 3  M 2xM  x2 þ 7x3 þ x6 ¼ 4 þ 2M xM þ x1 þ x2 þ x3 ¼M and x1 ; x2 ; x3 ; x4 ; x5 ; x6 ; xM  0

Now we construct the dual simplex table.

cB yB 0 y4 0 y5 0 y6 0 y1 z00B ¼ 2M

Now,

cj

−2

0

−1

−1

0

0

0

xB x4 ¼ 8  4M x5 ¼ 3  M x6 ¼ 4 þ 2M x1 ¼ M

yM −4 −1 2 1 2

y1 0 0 0 1 1

y2

y3 −1 0 7 1 1

y4 1 0 0 0 0

y5 0 1 0 0 0

y6 0 0 1 0 0

2 −10 −1 1 1

zj  cj

xBr ¼ MinfxBi ; xBi \0g ¼ Minf8  4M; 3  M g ¼ 8  4M ¼ xB1 :

Hence, r ¼ 1 and the corresponding vector yB1 ; i:e: y4 leaves the basis.

5.5 Modiﬁed Dual Simplex Method

n Again,

Max

zj cj yrj

123

o n o 2 1 z c 2 ; yrj \0 ¼ Max jy1j j ; y1j \0 ¼ Max 4 ; 1 ¼ 4 ,

which

occurs for the vector yM : Then the vector yM enters the basis and yM1 ¼ 4 is the key element. Now we construct the next dual simplex table.

yB cB −2 yM 0 y5 0 y6 0 y1 z00B ¼ 4

cj

−2

0

−1

−1

0

0

0

xB xM ¼ 2 þ M x5 ¼ 5 x6 ¼ 0 x1 ¼ 2

yM 1 0 0 0 0

y1 0 0 0 1 0

y2 −1/2 −21/2 0 3/2 2

y3 1/4 1/4 13/2 3/4 1/2

y4 −1/4 −1/4 1/2 ¼ 1/2

y5 0 1 0 0 0

y6 0 0 1 0 0

zj  cj

Since only xB2 is negative hence r ¼ 2 and the corresponding basis vector y5 leaves the basis. n o n o z c 2 4 ; 1=2 Again, Max jy2j j ; y2j \0 ¼ Max 21=2 1=4 ¼ 21, which occurs for the vector y2 : Then the vector y2 enters the basis and y22 ¼ 21=2 is the key element. Now we construct the next dual simplex table.

yB cB −2 yM −1 y2 0 y6 0 y1 z00B ¼ 64=21

cj

−2

0

−1

−1

0

0

0

xB xM ¼ M  37=21 x2 ¼ 10=21 x6 ¼ 0 x1 ¼ 9=7

yM 1 0 0 0 0

y1 0 0 0 1 0

y2 0 1 0 0 0

y3 5/21 −1/42 13/2 11/14 13/42

y4 −5/21 1/42 1/2 3/14 19/42

y5 −1/21 −2/21 0 1/7 4/21

y6 0 0 1 0 0

zj  cj

Here all the basic variables are non-negative. So an optimal basic feasible solution has been obtained. Therefore, the required optimal basic feasible solution is given by x1 ¼ 9=7; x2 ¼ 10=21; x3 ¼ 0 and z0Max ¼ ð2M þ 64=21Þ þ 2M ¼ 64=21: Therefore, zMin ¼ z0Max ¼ 64=21:

124

5.6

5 Dual Simplex Method

Exercises

Use the dual simplex method to solve the following LPPs: 1. Minimize subject to

z ¼ 3x1  x2 x1 þ x2  1 2x1 þ 3x2  2 and x1 ; x2  0:

2. Minimize subject to

z ¼ 2x1 þ x2 3x1 þ x2  3 4x1 þ 3x2  6 x1 þ 2x2  3 and x1 ; x2  0:

3. Maximize subject to

z ¼ 2x1  3x2  x3 2x1 þ x2 þ 2x3  3 3x1 þ 2x2 þ x3  4 and x1 ; x2 ; x3  0:

4. Minimize subject to

z ¼ x1 þ 2x2 þ 3x3 2x1  x2 þ x3  4 x1 þ x2 þ 2x3  8 x2  x3  2 and x1 ; x2 ; x3  0:

5. Minimize subject to

z ¼ 2x1  x2  x3 4x1 þ 6x2 þ 3x3  8 x1  9x2 þ x3   3 2x1  3x2 þ 5x3   4 and x1 ; x2 ; x3  0:

6. Minimize subject to

z ¼ 10x1 þ 6x2 þ 2x3 x1 þ x2 þ x3  1 3x1 þ x2  x3  2 and x1 ; x2 ; x3  0:

7. Minimize subject to

z ¼ x1 þ x2 2x1 þ x2  2 x1  x2  1 and x1 ; x2  0:

5.6 Exercises

125

8. Maximize subject to

z ¼ 3x1  2x2 x1 þ x2  1 x1 þ x2  7 x1 þ 2x2  10 x2  3 and x1 ; x2  0:

9. Maximize subject to

z ¼ 3x1 þ 2x2 2x1 þ x2  5 x1 þ x2  3 and x1 ; x2  0:

10. Maximize subject to

z ¼ 2x1  3x2  2x3 x1  2x2  3x3 ¼ 8 2x2 þ x3  10 x2  2x3  4 and x1 ; x2 ; x3  0:

11. Minimize subject to

z ¼ 3x1 þ x2 x1 þ x2  1 2x1 þ 3x2  2 and x1 ; x2  0:

12. Minimize subject to

z ¼ 6x1 þ 7x2 þ 3x3 þ 5x4 5x1 þ 6x2  3x3 þ 4x4  12 x2 þ 5x3  6x4  10 2x1 þ 5x2 þ x3 þ x4  8 and x1 ; x2 ; x3 ; x4  0:

13. Minimize subject to

z ¼ 5x1 þ 6x2 x1 þ x2  2 4x1 þ x2  4 and x1 ; x2  0:

14. Minimize subject to

z ¼ 4x1 þ 2x2 3x1 þ x2  27 x1 þ x2  21 x1 þ 2x2  30 and x1 ; x2  0:

Chapter 6

Bounded Variable Technique

6.1

Objective

The objective of this chapter is to discuss the modiﬁed simplex method to solve LPPs whose variables are restricted with lower and upper bounds.

6.2

Introduction

In the theory of linear programming, it is well known that the decision variables of a linear programming problem (LPP) satisfy non-negativity conditions/restrictions only. However, in real-life situations, some (or all) of the variables are bounded by lower and upper limits. These bounds of the variables are expressed as bound constraints. This means that, in an LPP, the bound constraints of some (or all) variables are given in addition to the general constraints. In such cases, the standard form of the LPP will be as follows: Maximize z ¼

n P

cj xj

j¼1

subject to the constraints Ax = b, l  x  u, where

x ¼ ½x1 ; x2 ; . . .; xn T ; b ¼ ½b1 ; b2 ; . . .; bn T ; l ¼ ½l1 ; l2 ; . . .; ln T ; u ¼ ½u1 ; u2 ; . . .; un T ; A ¼ aij mxn ; i ¼ 1; 2; . . .; m; j ¼ 1; 2; . . .; n:

Here A is an m  n real matrix, and l and u denote the lower and upper bounds of x respectively. In cases of unbounded variables, these limits are considered as 0 and 1 respectively.

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_6

127

128

6 Bounded Variable Technique

The bound constraints l  x  u can be written separately as x  l and x  u. Now introducing surplus variables s0 and slack variables s00 in the preceding bound constraints, we get the equality constraints as follows: x  s0 ¼ l and x þ s00 ¼ u;

where

x; s0 ; s00  0:

The lower bound constraints can be written as x ¼ l þ s0 . Thus, x can be eliminated from all the remaining constraints by substituting x ¼ l þ s0 . On the other hand, the upper bound constraints can be written x ¼ u  s00 . This does not serve the purpose, since there is no guarantee that x will be non-negative. This difﬁculty is overcome by using a special technique called the bounded variable simplex method. In this method, the optimality condition for a solution will be the same, as the constraints x þ s00 ¼ u require modiﬁcation in the feasibility condition of decision variables due to the following reasons: (i) A basic variable should become a non-basic variable at its upper bound (in the usual simplex method, all non-basic variables are at zero level). (ii) When a non-basic variable becomes a basic variable, its value should not exceed its upper bound without affecting the non-negativity and upper bound conditions of all existing basic variables.

6.3

The Computational Procedure

The iterative procedure for ﬁnding an optimum basic feasible solution (BFS) to an LPP with bounded variables is as follows: Step 1: Check whether the objective function of the given LPP is to be maximized or minimized. If it is to be minimized, then convert it into a problem of maximizing by using Minimum z ¼ Maximum(  zÞ: Step 2: Check whether all bi (i = 1, 2,…, m) are non-negative. If any one of bi is negative, then multiply the corresponding inequality of the constraints by (−1) so as to get all bi (i = 1, 2,…, m) as non-negative. Step 3: Convert all the inequality constraints (except bound constraints) into equations by introducing slack and/or surplus variables in the constraints. Set the coefﬁcients of these variables in the objective function equal to zero. We note that the lower and upper bounds of slack and surplus variables are assumed as 0 and 1 respectively. Step 4: If the lower bound lj (j = 1, 2, …, n) of any variable xj is at a positive level (i.e. if lj  xj ; lj> 0), set xj= lj+ xj, xj  0 in the reduced problem obtained from Step 3.

6.3 The Computational Procedure

129

Step 5: Obtain an initial BFS. Step 6: Compute the net evaluations zj − cj (j = 1, 2,…, n) and examine their signs. Two cases may arise: (i) If all zj − cj  0, then an optimum BFS has been attained. (ii) If at least one zj − cj< 0, choose the most negative of zj − cj. Let it be zr − cr for some j = r. Step 7: Compute h ¼ Minfh1 ; h2 ; ur g, where h1 ¼ Min i

h2 ¼ Min i

n

xBi yir

n

o ; yir [ 0 ;

uðxBi ÞxBi yir

o ; yir \0 ;

where the xBi’s correspond to the current BFS, and ur is the upper bound for the variable xr. Note that h2 ¼ 1 if all yir > 0 and h1 ¼ 1 if all yir < 0. In ﬁnding the value of h, three cases may arise: (i) If h ¼ h1 which corresponds to xBk, then yr enters the basis and yk leaves it. (ii) If h ¼ h2 which corresponds to xBk, then yr enters the basis and yk leaves it. Due to this, the value of one or more basic variable will be negative in the next iteration. In that case, the values of basic variables are to be updated. This can be done by putting the non-basic variables xr (corresponding to the most negative zj − cj) at zero level at the upper bound by substituting xr ¼ ur  x0r ð0  x0r  ur Þ, as the upper bound of xr is ur. In this case, ðxBi Þnew ¼ ðxBi Þold  yir ur (iii) If h ¼ ur ; xr is substituted at its upper bound and it remains non-basic; i.e. it can be put at zero level by using xr ¼ ur  x0r ; 0  x0r  ur : In this situation, the basic variables are to be updated by the formula ðxBi Þnew ¼ ðxBi Þold  yir ur : Then Go to Step 6.

130

6 Bounded Variable Technique

Example 1 Using the bounded variable technique, solve the following LPP: Maximize z ¼ 3x1 þ 5x2 þ 2x3 subject to x1 þ 2x2 þ 2x3  14; 2x1 þ 4x2 þ 3x3  23; 0 0  x3  3:

 x1  4; 2  x2  5;

Solution By introducing the stack variables x4 ð  0Þ and x5 ð  0Þ, in the ﬁrst and second constraints respectively, we get the reduced problem as Maximize z ¼ 3x1 þ 5x2 þ 2x3 þ 0:x4 þ 0:x5 x1 þ 2x2 þ 2x3 þ x4 ¼ 14 2x1 þ 4x2 þ 3x3 þ x5 ¼ 23 and 0  x1  4; 2  x2  5; 0  x3  3; x4 ; x5  0: Since there are no upper bounds speciﬁed for the variables x4 and x5 , we arbitrarily assume that all have an upper bound at /, i.e. u4 ¼ u5 ¼/. Since x2 has a positive lower bound, let us take x2 ¼ x02 þ 2. Then 2  x2  5 implies 0  x02  3. Then the given LPP reduces to Maximize z ¼ 3x1 þ 5ðx02 þ 2Þ þ 2x3 þ 0:x4 þ 0:x5 x1 þ 2ðx02 þ 2Þ þ 2x3 þ x4 ¼ 14 or, x1 þ 2x02 þ 2x3 þ x4 2x1 þ 4ðx02

þ 2Þ þ 3x3 þ x5 ¼ 23

and 0  x1  4;

0  x02

or, 2x1 þ 4x02

¼ 10

þ 3x3 þ x5 ¼ 15

 3; 0  x3  3; x4 ; x5  0

Here the initial BFS is x4 ¼ 10; x5 ¼ 15 with B ¼ I2 as initial basis. Now we construct the starting simplex table:

cB yB 0 y4 0 y5 zB ¼ 10 uj

cj

3

5

2

0

0

xB x4 ¼ 10 x5 ¼ 15

y1 1 2 –3 4

y02 2 4 –5 3

y3 2 3 –2 3

y4 1 0 0 1

y5 0 1 0 1

Since D2 is most negative, y02 enters the basis. n o  15 15 x Now, h1 ¼ Min yB0 i ; y0i2 [ 0 ¼ Min 10 2 ; 4 ¼ 4 (corresponds to x5 ) i

i2

Dj

6.3 The Computational Procedure



131

uðxBi Þ  xBi 0 h2 ¼ Min ; yi2 \0 i y0i2



¼ 1 ½As y0i2 [ 0

i ¼ 1; 2

for

and u02 ¼ 3     0 Hence, h ¼ Min h1 ; h2 ; u02 ¼ Min 15 4 ; 1; 3 ¼ 3, which implies that h ¼ u2 . Therefore, x02 is substituted at its upper bound and it remains non-basic. Thus, x02 ¼ u02  x002 ¼ 3  x002 , where 0  x002  3. Due to this substitution, the values of the basic variables will be updated, and these updated values are ðxB1 Þnew ¼ ðxB1 Þold y012 u02 ¼ 10  2  3 ¼ 4 ðxB2 Þnew ¼ ðxB2 Þold y022 u02 ¼ 15  4  3 ¼ 3: Therefore, by using the updated values of the basic and non-basic variables, the initial simplex table becomes cj cB yB xB 0 y4 x4 ¼ 4 x5 ¼ 3 0 y5 zB ¼ 15 þ 10 ¼ 25 uj

3

–5

2

0

0

y1 1 2 −3 4

y002 −2 −4 5 3

y3 2 3 −2 3

y4 1 0 0 1

y5 0 1 0 1

Dj

Since D1 , i.e. z1  c1 , is most negative; hence y1 enters the basis.   Now, h1 ¼ Min 41 ; 32 ¼ 32 (corresponds to x5 ) h2 ¼ 1; since all the elements of y1 are positive and u1 ¼ 4   3 3 ) h ¼ Minfh1 ; h2 ; u1 g ¼ Min ; 1; 4 ¼ ; which implies h ¼ h1 : 2 2 Hence y12 ¼ 2 is the key element and y5 leaves the basis. Now, we construct the next table. cj cB yB xB 0 y4 x4 ¼ 5=2 3 y1 x1 ¼ 3=2 zB ¼ ð25 þ 3  3=2Þ ¼ 59=2 uj

3

−5

2

0

0

y1 0 1 0 4

y002 0 −2 −1 3

y3 1/2 3=2 5=2 3

y4 1 0 0 1

y5 1=2 1=2 3=2 1

Dj

132

6 Bounded Variable Technique

Since Dj corresponding to y002 is negative, then y002 enters the basis. Now, h1 ¼ 1, since all the elements of y002 are either negative or zero.  h2 ¼ Min

4  32 ð2Þ



5 ¼ ; which corresponds to x1 : 4

u002 ¼ 3:   Hence, h ¼ Min h1 ; h2 ; u002 ¼ 54, which implies h ¼ h2 . Hence, y0022 ¼ 2 is the key element and y1 leaves the basis. Now we construct the next table. cj yB xB cB 0 y4 x4 ¼ 5=2 x002 ¼ 3=4 –5 y002 zB ¼ ð25 þ 15=4Þ ¼ 115=4 uj

3

−5

2

0

0

y1 0 1=2 1=2 4

y002

y3 1/2 −3/4 7=4 3

y4 1 0 0 1

y5 1=2 1=4 5=4 1

0 1 0 3

Dj

Since D1 is only negative, then y1 enters the basis. But from the preceding table, it is seen that the value of one basic variable is negative. So, we have to update the basic variables. This can be done by putting the non-basic variable x1 at zero level at its upper bound by substituting x1 ¼ 4  x01 ð0  x01  4Þ, as the upper bound of x1 is 4. Due to this substitution, the values of the basic variables will be updated, and these updated values are 5 5 04¼ 2 2 3 1 5 ¼ ðxB2 Þold y21 u1 ¼   ð Þ  4 ¼ : 4 2 4

ðxB1 Þnew ¼ ðxB1 Þold y11 u1 ¼ ðxB2 Þnew

Thus using the updated values of the basic variables, we construct the simplex table. cj

−3

−5

2

0

0

cB 0

yB y4

xB

y01 0

y002 0

y3

y4 1

y5

–5

y002

x002

1

 34 7 4

zB ¼ 25 þ 3  4 

x4 ¼ 52 ¼ 54 25 123 4 ¼ 4

1 2 1 2

0

1 2

 12

0

 14

0

5 4

Dj

Since all Dj  0, an optimal solution has been attained. The solution is given by x01 ¼ 0; x002 ¼ 54 ; x3 ¼ 0 with Maximum z ¼ 123 4 .

6.3 The Computational Procedure

133

Now, since x1 ¼ 4  x01 ; x01 ¼ 0 implies x1 ¼ 4. Again, since x02 ¼ 3  x002 ; x002 ¼ 54 implies x02 ¼ 3  54 ¼ 74. Also, x2 ¼ x02 þ 2 ¼ 74 þ 2 ¼ 15 4. 123 Hence, the optimal BFS is x1 ¼ 4; x2 ¼ 15 4 ; x3 ¼ 0 and Maximum z ¼ 4 :

Example 2 Solve the LPP by using the bounded variable method. Minimize z ¼ 6x1  2x2  3x3 subject to 2x1 þ 4x2 þ 2x3  8 x1  2x2 þ 3x3  7 0  x1  2; 0  x2  2 and 0  x3  1: Solution The given LPP is a minimization problem. Let z0 ¼ z; then Min z = −Max (−z) = −Max z*. Hence, the problem is to Maximize z* = 6x1 þ 2x2 þ 3x3 subject to the given constraints. Introducing slack variables x4 and x5 , the given LPP reduces to Maximize z ¼ 6x1 þ 2x2 þ 3x3 subject to 2x1 þ 4x2 þ 2x3 þ x4 ¼ 8 x1  2x2 þ 3x3 þ x5 ¼ 7 0  x1 ; x2  2; 0  x3  1 and x4 ; x5  0: Since there are no speciﬁed upper bounds for the variables x4, x5, we may arbitrarily assume that u4 ¼ 1; u5 ¼ 1. Here the initial BFS is x4 = 8, x5 = 7, with B = I2 as the initial basis. Now, we construct the initial simplex table as follows:

cB 0 0 zB ¼ 0

yB y4 y5

cj

−6

2

3

0

0

xB 8 7

y1 2 1 6 2

y2 4 −2 −2 2

y3 2 3 −3 1

y4 1 0 0 1

y5 0 1 0 1

uj

Dj

134

6 Bounded Variable Technique

Since D3 is most negative,   y3 enters the basis. Now h1 ¼ Min 82 ; 73 ¼ 73, which corresponds to y5, h2 ¼ 1 since all yi3 [ 0 for i = 1,2: u3 ¼ 1:   h ¼ Minfh1 ; h2 ; u3 g ¼ Min 73 ; 1; 1 ¼ 1, which corresponds to h ¼ u3 . Therefore, x3 is substituted at its upper bound and it remains non-basic, thus x3 ¼ u3  x03 ¼ 1  x03 ; where 0  x03  1:

Due to this substitution, the basic variables will be updated, and these updated values are ðxB1 Þnew ¼ ðxB1 Þold y13 u3 ¼ 8  2  1 ¼ 6 ðxB2 Þnew ¼ ðxB2 Þold y23 u3 ¼ 7  3  1 ¼ 4: The initial table is as follows:

cB 0 0 zB ¼ 3

yB y4 y5

cj

−6

2

-3

0

0

xB 6 4

y1 2 1 6 2

y2 4 −2 −2 2

y03 −2 −3 3 1

y4 1 0 0 1

y5 0 1 0 1

uj

Dj

Since, D2 is negative  only then y2 enters the basis. Now, h1 ¼ Min 64 ¼ 32, which corresponds to y4, h2 ¼ Minf1g ¼ 1 and u2 ¼ 2: ) h ¼ Min

3

2 ; 1; 2



¼ 32, which corresponds to h ¼ h1 .

Hence, y12 = 4 is the key element, and y4 leaves the basis. Now we construct the next table: cj

−6

cB 2

yB y2

xB

y1

3 2

1 2

0

y5

7

2

2

−3

0

0

y2 1

y03

y4

 12

1 4

y5 0

0

−4

1 (continued)

6.3 The Computational Procedure

135

(continued) −6

cj zB ¼ 6

2

7

0

−3 2

0 1 2 1 2

0

0

Dj

Since all Dj  0, the optimal solution has been reached and the optimal solution is given by 3 and x03 ¼ 0; 2 3 i:e: x1 ¼ 0; x2 ¼ and x3 ¼ 1 as x3 ¼ 1  x03 2 3 and Minimum z ¼ 6:0  2:  3:1 ¼ 6: 2

x1 ¼ 0; x2 ¼

6.4

Exercises

Solve the following LPP: 1. Maximize z ¼ 3x1 þ 2x2 subject to the constraints x1  3x2  3; x1  2x2  4; 2x1 þ x2  20 x1 þ 3x2  30; x1 þ x2  6 and 0  x1  8; 0  x2  6: 2. Maximize z ¼ 3x1 þ 5x2 þ 2x3 subject to the constraints x1 þ 2x2 þ 2x3  14; 2x1 þ 4x2 þ 3x3  23 0  x1  4; 2  x2  5; 0  x3  3:

and 3. Maximize z ¼ x2 þ 3x3 subject to the constraints

x1 þ x2 þ x3  10; x1  2x3  0; 2x2  x3  10 and

0  x1  8; 0  x2  4; x3  0:

136

6 Bounded Variable Technique

4. Maximize z ¼ 4x1 þ 4x2 þ 3x3 subject to the constraints  x1 þ 2x2 þ 3x3  15; x2 þ x3  4; 2x1 þ x2  x3  6; x1  x2 þ 2x3  10 0  x1  8; x2  0; 0  x3  4:

and

5. Maximize z ¼ 3x1 þ x2 þ x3 þ 7x4 subject to the constraints 2x1 þ 3x2  x3 þ 4x4  40; 2x1 þ 2x2 þ 5x3  x4  35 x1 þ x2  2x3 þ 3x4  100 and

x1  2; x2  1; x3  3; x4  4:

6. Maximize z ¼ 4x1 þ 10x2 þ 9x3 þ 11x4 subject to the constraints 2x1 þ 2x2 þ 2x3 þ 2x4  5; 48x1 þ 80x2 þ 160x3 þ 240x4  257 and

0  xj  1; j ¼ 1; 2; 3; 4:

7. Maximize z ¼ 6x1  2x2  3x3 subject to the constraints 2x1 þ 4x2 þ 2x3  8; x1  2x2 þ 3x3  7 and 0  x1  2; 0  x2  2; 0  x3  1:

Chapter 7

Post-optimality Analysis in LPPs

7.1

Objectives

The objectives of this chapter are: • To discuss the effect of discrete changes of the cost/proﬁt vector and requirement vector on the optimal solution of a linear programming problem (LPP) • To ﬁnd the feasibility and optimality conditions for the discrete changes of the elements of the coefﬁcient matrix of the LPP • To study the effect of structural changes (i.e. inclusion and deletion of a decision variable and constraint) in an LPP.

7.2

Introduction

When solving a real-life linear programming problem (LPP), after formulation and obtaining an optimal solution of the said LPP, sometimes we are interested in studying the effect of changes (discrete as well as continuous) of different parameters of the problem on the current optimal solution. This situation arises in the following two cases: Case I: An error (wrong value used for cost/proﬁt coefﬁcient, omission of variables or constraints, etc.) has been observed in the problem formulation after obtaining the optimal solution. Case II: The system is analysed to study the effect of changes in different parameters. For Case I, the error can be removed in two different ways: (i) Solve the new LPP after replacing the correct values of the components which are used wrongly. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_7

137

138

7 Post-optimality Analysis in LPPs

(ii) Develop some procedure so that the current optimum solution can be used to ﬁnd the correct solution of the problem. For Case I, to reduce the computational effort, one should choose option (ii). On the other hand, for Case II, option (ii) of Case I is used, as there is no other option. An analysis of such a post-optimal problem can thus be termed a post-optimality analysis. The changing parameters may be either discrete or continuous. The study of the effect of discrete parameter changes on the optimal solution is called sensitivity analysis, and that of continuous changes is termed parametric programming. The changes or variations in an LPP which are studied in the post-optimality analysis are: (i) Changes in the cost coefﬁcients cj of the objective function, i.e. a change in the cost or proﬁt vector (ii) Changes in the requirement vector b (iii) Changes in the coefﬁcient of the constraints (iv) Structural changes due to addition or deletion of some variables (v) Structural changes due to addition or deletion of some linear constraints. After these changes have been effected in an LPP, we may have any of the following cases: (i) The optimum solution remains unaltered; i.e. the basic variables and their values are essentially unchanged. (ii) The basic variables remain the same, but their values may change. (iii) The basic solution is changed entirely.

7.3

Discrete Changes in the Proﬁt Vector

Let xB be the optimal basic feasible solution of the following LPP:   Maximize z ¼ cT ; x subject to Ax = b and x  0, where c, xT 2 Rn ; bT 2 Rm and A is an m  n real matrix of rank m. Let Dck be the amount which is added to the kth component ck of c = (c1, c2, …, cn), so that the new value of the kth component becomes ck ¼ ck þ Dck . Since x B ¼ B  1 b is independent of c, then for any change of ck ; k ¼ 1; 2; . . .; n, the value of x B will not be changed. Hence, the solution x B remains basic feasible for the current problem. Here, two cases may arise: Case I: ck is not in cB . Case II: ck is in cB .

7.3 Discrete Changes in the Proﬁt Vector

139

Case I: When ck is not in cB , ck is not the coefﬁcient of the basic variable in the objective function. In that case, the net evaluation corresponding to the kth proﬁt coefﬁcient is given by zk  ck ¼ zk  ðck þ Dck Þ;

8ck 62 cB

since zk is not affected for any change in ck . Also, we note that the net evaluations corresponding to the proﬁt coefﬁcients other than ck remain unchanged. Thus, the current solution xB remains optimum for the new problem if zk  ck  Dck  0 i:e: if Dck  zk  ck

8ck 2 6 cB ; 8ck 2 6 cB :

Case II: When ck is in cB, ck is the coefﬁcient of the basic variable in the objective function. In this case, the net evaluations are given by z j  cj ¼

m X

¼

cBi yij  cj

i¼1 m X

cBi yij þ cBk ykj  cj ; where ck ¼ cBk :

i¼1

i 6¼ k If ck is replaced by ck þ Dck ¼ ck , then zj  cj ¼

m X

cBi yij þ ðcBk þ DcBk Þykj  cj ðj 6¼ kÞ

i¼1

i 6¼ k m X ¼ cBi yij þ DcBk ykj  cj i¼1

  ¼ zj  cj þ DcBk ykj : Thus, the new basic feasible solution xB will remain optimum so long as Dck satisﬁes or When ykj [ 0, DcBk  

zj  cj þ DcBkykj  0  DcBk ykj   zj  cj :

zj cj ykj ,

140

7 Post-optimality Analysis in LPPs

     z j  cj i:e: DcBk  Max ; ykj [ 0 : ykj   ðzj cj Þ ðzj cj Þ ; y \0 . Again, when ykj \0, DcBk  ykj , i.e. DcBk  Min kj ykj     ðzj cj Þ ðzj cj Þ Hence, Max ; ykj [ 0  DcBk ¼Dck  Min ; ykj \0 ð j 6¼ kÞ. ykj ykj Example 1 We are given the following LPP: z ¼ x1 þ 2x2  x3

Max

subject to 3x1 þ x2  x3  10  x1 þ 4x2 þ x3  6 and

x2 þ x3  4 xj  0; j ¼ 1; 2; 3:

(a) Determine the effect of discrete changes of those components of the proﬁt vector which correspond to the basic variables. (b) Discuss the effect of discrete changes in those components of the proﬁt vector which do not correspond to basic variables. Solution Introducing the slack variables x4  0, x6  0, surplus variable x5  0 and an artiﬁcial variable x7  0 in the constraints of the given LPP and then solving the resulting problem by the simplex method, the following optimum simplex table is obtained:

cj cB

yB

0 y4 2 y2 0 y5 zB ¼ 8

xB x4 = 6 x2 = 4 x5 = 10

−1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

3 0 1 1

0 1 0 0

−2 1 3 3

1 0 0 0

0 0 1 0

−1 1 4 2

0 0 −1 M

Dj

(a) Here the basic variables are x4 ; x2 ; x5 . The corresponding proﬁt coefﬁcients are c4 ; c2 ; c5 . We know that an optimum basic feasible solution will maintain its optimality if a change Dck of ck 2 cB satisﬁes the following inequalities:

7.3 Discrete Changes in the Proﬁt Vector

141

         z j  cj  zj  cj Max ; ykj [ 0  DcBk ¼ Dck  Min ; ykj \0 ðj 6¼ k Þ: ykj ykj For k ¼ 2, i.e. k ¼ 2, we have          z j  cj  z j  cj Max ; y2j [ 0  DcB2 ¼ Dc2  Min ; y2j \0 ; y2j y2j  2 i.e. Max 3 1 ; 1  Dc2 , i.e. 2  Dc2 or Dc2   2. (b) Here the non-basic variables are x1 ; x3 ; x6 . Among these variables, only x1 and x3 are the decision variables of the given LPP. The corresponding proﬁt coefﬁcients are c1 and c3 : We know that the change Dck in ck 2 cB must satisfy the upper limit Dck  ðzk  ck Þ in order to maintain the optimality of the optimum basic feasible solution. Thus, for k ¼ 1; Dc1  z1  c1 , i.e. Dc1  1. For k ¼ 3; Dc3  z3  c3 , i.e. Dc3  3. Alternative method (a) Here the basic variables are x4 , x2 , x5 . Among these variables, only x2 is the given decision variable of the given LPP. The corresponding proﬁt coefﬁcient is c2 ¼ 2. Let Dc2 be the amount by which c2 is changed. Hence, the new value of c2 is c2 ¼ c2 þ Dc2 ¼ 2 þ Dc2 . Now for the change of c2 ¼ 2 to c2 , we have the following corresponding changed simplex table: cj cB

yB

xB

0 2 þ Dc2 0

y4 y2 y5

x4 ¼ 6 x2 ¼ 4 x5 ¼ 10

−1 y1

2 þ Dc2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

3 0 1 1

0 1 0 0

−2 1 3 3 þ Dc2

1 0 0 0

0 1 0 0

−1 1 4 2 þ Dc2

0 0 −1 M

The current solution xB remains optimum for the new problem if 3 þ Dc2  0 and 2 þ Dc2  0, i.e. if Dc2   3 and Dc2   2, i.e. if Dc2  Maxf3; 2g, i.e. if Dc2   2.

Dj

142

7 Post-optimality Analysis in LPPs

(b) Here the non-basic variables are x1 , x3 , x6 . Among these, only x1 and x3 are the given decision variables. The corresponding proﬁt coefﬁcients are c1 and c3 . Let Dc1 be the amount by which c1 is changed. Hence, the new value of c1 is c1 ¼ c1 þ Dc1 ¼ 1 þ Dc1 . Now for the change of c1 ¼ 1 to c1 , we have the following corresponding changed simplex table: cj cB

yB

xB

0 2 0

y4 y2 y5

x4 ¼ 6 x2 ¼ 4 x5 ¼ 10

1 þ Dc1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

3 0 1 1  Dc1

0 1 0 0

−2 1 3 3

1 0 0 0

0 0 1 0

−1 1 4 2

0 0 −1 M

Dj

The current solution remains optimum for the new problem if 1  Dc1  0, i.e. if Dc1  1: Similarly, the range of Dc3 can be obtained. Example 2 Given the LPP z ¼ 15x1 þ 45x2

Max

subject to x1 þ 16x2  240 5x1 þ 2x2  162 x2  50 and x1 ; x2  0; if Maximize z ¼

P

cj xj ; j ¼ 1; 2 and c 2 is kept ﬁxed at 45, determine how much c 1

j

can be changed without affecting the optimal solution. Solution By introducing the slack variables x3  0; x4  0 and x5  0 in the constraints of the given LPP and then solving the resulting problem by the simplex method, the following optimum simplex table is obtained: cj cB

yB

45 y2 15 y1 0 y5 zB ¼ 1005

xB x2 ¼ 173=13 x1 ¼ 352=13 x5 ¼ 477=13

15 y1

45 y2

0 y3

0 y4

0 y5

0 1 0 0

1 0 0 0

5/78 −1/39 -5/78 5/2

+1/78 8/39 1/78 5/2

0 0 1 0

Dj

From this optimum table, it is seen that y 1 ; y 2 ; y 5 are the basic vectors. Let c 1 be changed to c 1 þ Dc 1 .

7.3 Discrete Changes in the Proﬁt Vector

143

Since c 2 is kept ﬁxed and c 1 can vary, an optimum basic feasible solution will maintain its optimality if Dc 1 in c1 2 cB satisﬁes          z j  cj  z j  cj Max ; y2j [ 0  DcB2 ¼ Dc1  Min ; y2j \0 ; y2j y2j     ðz4  c4 Þ ðz3  c3 Þ  Dc1  Min or Max y24 y23     5=2 5=2  Dc1  Min or Max 8=39 1=39 195 195  Dc1  : or  16 2

here k ¼ 2

Example 3 We are given the following LPP: Max z ¼ 3x1 þ 5x2 subject to 3 x 1 þ 2 x 2  18 x1  2 x2  6 and x1 ; x2  0: Discuss the effect on the optimality of the solution, where the objective function is changed to 3x1 þ x2 . Solution After introducing the slack variables x3  0; x4  0 and x5  0 in the constraints of the preceding LPP and then solving the reformulated LPP by the simplex method, the optimum simplex table is as follows: cj cB

yB

3 y1 0 y4 5 y2 zB ¼ 36

xB x1 ¼ 2 x4 ¼ 0 x2 ¼ 6

3 y1

5 y2

0 y3

0 y4

0 y5

1 0 0 0

0 0 1 0

1/3 −2/3 0 1

0 1 0 0

−2/3 4/3 1 3

Dj

Since the objective function is now changed to z ¼ 3x1 þ x2 , the new cB becomes (3 0 1). Hence, the new net evaluations corresponding to the variables x3 ; x4 ; x5 become 1 2 þ0  þ1  0  0 ¼1 3 3 D4 ¼ 3  0 þ 0  1 þ 1  0  0 ¼0

2 4 þ 0  þ 1  1  0 ¼ 1: D5 ¼ 3   3 3 D3 ¼ 3 

144

7 Post-optimality Analysis in LPPs

This indicates that y5 enters the basis in the next iteration. The new simplex table is the following: cj cB

yB

3 y1 0 y4 1 y2 zB ¼ 12

xB x1 ¼ 2 x4 ¼ 0 x2 ¼ 6

3 y1

1 y2

0 y3

0 y4

0 y5

Mini ratio

1 0 0 0

0 0 1 0

1/3 −2/3 0 1

0 1 0 0

−2/3 4/3 1 −1

– 0/(4/3) = 0 6/1 = 6 Dj

Here y 4 is the outgoing vector, and y 25 ¼ Now we construct the next table: cj cB

yB

3 y1 0 y5 1 y2 zB ¼ 12

xB x1 ¼ 2 x5 ¼ 0 x2 ¼ 6

4 3

is the key element.

3 y1

1 y2

0 y3

0 y4

0 y5

1 0 0 0

0 0 1 0

0 −1/2 1/2 1/2

1/2 3/4 −3/4 3/4

0 1 0 0

Dj

Here all D j  0. Hence, the solution is optimal, and it is x1 ¼ 2; x2 ¼ 6 and Max z ¼ 12. Example 4 We are given the LPP Maximize z ¼ 6x1  2x2 þ 3x3 subject to 2x1  x2 þ 2x3  2 x1 þ 4x3  4 and x1 ; x2 ; x3  0: Find the ranges of the proﬁt components when (i) c1 and c2 are the two proﬁt components changed at the same time, (ii) all three proﬁt components c1 , c2 and c3 are changed at a time to keep the optimal solution the same. Solution Introducing the slack variables x4  0 and x5  0 in the constraints of the given LPP and then solving the resulting problem by the simplex method, the following optimal simplex table is obtained:

cB

cj yB

xB

6 -2

y1 y2

x1 ¼ 4 x2 ¼ 6

6 y1

−2 y2

3 y3

0 y4

0 y5

1 0 0

0 1 0

4 6 9

0 −1 2

1 2 2

Dj

7.3 Discrete Changes in the Proﬁt Vector

145

(i) When two proﬁt components c1 and c2 are changed at a time, for the change of c1 ¼ 6 and c2 ¼ 2 to c1 and c2 , the modiﬁed simplex table for the new problem is as follows:

cj cB

yB

xB

c1 c2

y1 y2

x1 ¼ 4 x2 ¼ 6

c2 y1

c2 y2

3 y3

0 y4

0 y5

1 0 0

0 1 0

4 6 4c1 þ 6c2  3

0 −1 c2

1 2 c1 þ 2c2

Dj

The current solution remains optimum for the new problem if zj  cj  0 for all j, i.e. if 4c1 þ 6c2  3  0, c2  0 and c1 þ 2c2  0, c

1 , c  0 and c2   21 , i.e. if c2  34c n6  2 o 34c1 c1 i.e. if Max 6 ; 2  c2  0 and c1 is any real number. 

(ii) When all the three components c1 , c2 and c3 are changed at a time, for the change of c1 ¼ 6, c2 ¼ 2 and c3 ¼ 3 to c1 , c2 and c3 , the modiﬁed simplex table for the new problem is as follows:

cj cB

yB

xB

c1 c2

y1 y2

x1 ¼ 4 x2 ¼ 6

c2 y1

c2 y2

c3 y3

0 y4

0 y5

1 0 0

0 1 0

4 6 4c1 þ 6c2  c3

0 −1 c2

1 2 c1 þ 2c2

Dj

The current solution remains optimum for the new problem if zj  cj  0 for all j, i.e. if 4c1 þ 6c2  c3  0, c2  0 and c1 þ 2c2  0, c

1 , c  0 and c2   21 , i.e. if c2  c3 4c n 6  2  o c3 4c1 c1 i.e. if Max  c2  0 and c1 , c3 are any real numbers. 6 ; 2 

7.4



Discrete Changes in the Requirement Vector

Let xB be the optimum basic feasible solution of the following LPP: Maximize z ¼ hcT ; xi subject to Ax ¼ b and x  0

146

7 Post-optimality Analysis in LPPs

where c, xT 2 Rn ; bT 2 Rm and A is an m  n real matrix of rank m. Let Db k be the amount which is added to b k , and the new value of b k is b0k ¼ bk þ Dbk ; k ¼ 1; 2; . . .; m. Let the new basic solution be x0B . 3 b1 7 6 6 b2 7 7 6 6 ...7 7 6 0 1 0 0 Then xB ¼ B b , where b ¼ 6 7 6 bk þ Dbk 7 7 6 6 ...7 5 4 bm 2

2

2

3

0

3

b1 7 6 6 07 6 7 7 6 6 b2 7 7 6 6 7 6 ...7 6 7 7 1 6 1 6 7 ¼ B 6...7þB 6 7 6 Dbk 7 6 7 7 6 6...7 7 6 4 5 6 ...7 5 4 bm 0 2

b11 6b 6 21 6 6 ... ¼ xB þ 6 6 ... 6 6 4 ...

b12 b22

... ...

b1k b2k

... ...

...

...

...

...

... ...

... ...

... ...

... ...

bm1

bm2

...

bmk

...

3

2

3

2

or

x0B1

xB 1

2

3

2

0

3

b1m 6 7 6 07 7 b2m 7 76 7 76 7 . . . . . . 76 7 6 76 7 7 . . . 76 Dbk 7 7 6 7 7 . . . 56 6 ...7 5 4 bmm 0

b1k Dbk

3

7 6 7 6 0 7 6 6 xB2 7 6 xB2 7 6 b2k Dbk 7 7 6 7 6 7 6 6 ...7 6 ...7 6 ...7 7 6 7 6 7 6 7þ6 7: 6 0 7¼6 6 xB 7 6 xBi 7 6 bik Dbk 7 7 6 7 6 i7 6 7 6 7 6 6 ...7 5 4 ...5 4 ...5 4 0 xB m bmm Dbk xB m

) x0Bi ¼ xBi þ bik Dbk , where bik is the (i, k)th element.

7.4 Discrete Changes in the Requirement Vector

147

For maintaining the feasibility of the basic solution, we must have x0B  0 i:e: x0Bi  0 8i ¼ 1; 2; . . .; m i:e: xBi þ bik Dbk  0 8i ¼ 1; 2; . . .; m i:e: bik Dbk   xBi : x

When bik [ 0; Dbk   bBi ki ; 8i ¼ 1; 2; . . .; m, n o x i.e. Max bi kBi ; bik [ 0  Dbk . n o x Similarly, for bi k \0; Dbk  Min bi kBi ; bik \0 8i ¼ 1; 2; . . .; m. n o n o x x Hence, Max bi kBi ; bik [ 0  Dbk  Min bi kBi ; bik [ 0 ; 8i ¼ 1; 2; . . .; m. We observe that x0B remains optimal since the optimality condition Dj  0 remains unaffected by the change of bk. Example 5 We are given the following LPP: Maximize z ¼ x1 þ 2x2 x3 subject to 3x1 þ x2  x3  10  x1 þ 4x2 þ x3  6 x2 þ x3  4 and xj  0;

j ¼ 1; 2; 3:

(a) Determine the ranges for discrete changes in the components b2 and b3 of the requirement vector so as to maintain the optimality of the current optimum solution. Solution Introducing slack variables x4 and x6 , surplus variable x5 and artiﬁcial variable x7 , the given LPP can be written as Max z ¼ x1 þ 2x2  x3 þ 0:x4 þ 0:x5 þ 0:x6  Mx7 subject to 3x1 þ x2  x3 þ x4 ¼ 10  x1 þ 4x2 þ x3  x5 þ x7 ¼ 6 x2 þ x3 þ x6 ¼ 4: The initial simplex table is the following: cj cB

yB

xB

−1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

0 −M

y4 y7

x4 = 10 x7 = 6

3 −1

1 4

-1 1

1 0

0 −1

0 0

0 1 (continued)

148

7 Post-optimality Analysis in LPPs

(continued) cj cB

yB

0 y6 zB ¼ 6M

xB x6 = 4

−1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

0 M−1

1 −4 M−2

1 −M + 1

0 0

0 M

1 0

0 0

Dj

The optimum table is as follows: cj cB

yB

0 y4 2 y2 0 y5 zB ¼ 8

xB x4 = 6 x2 = 4 x5 = 10

−1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

3 0 1 1

0 1 0 0

−2 1 3 3

1 0 0 0

0 0 1 0

−1 1 4 2

0 0 −1 M

Dj

From the optimum simplex table, we have xB ¼ ½6; 4; 10; B1

b ¼ ½10; 6; 4 0 1 1 0 1 B C ¼ ½ y4 ; y 7 ; y6  ¼ @ 0 0 1 A: 0 1 4

The individual effects of changes in b2 and b3 where b = [b1, b2, b3] without violating the optimality of the basic feasible solution are given by  Max

   xBi xBi ; bik [ 0  Dbk  Min ; bik \0 ; k ¼ 2; 3: bik bik

When k = 2, then 

   xBi xBi ; bi2 [ 0  Db2  Min ; bi2 \0 bi2 bi 2   xB 3 Db2  Min  b32   10 Db2  Min or Db2  10: 1

Max or or

7.4 Discrete Changes in the Requirement Vector

149

Again, when k = 3, then 

   xBi xBi ; bi3 [ 0  Db3  Min ; bi3 \0 bi 3 bi 3     xB2 xB3 xBi Max ;  Db3  Min b23 b33 b13     4 10 6 ; Max  Db3  Min 1 4 1

Max or or or

Maxf4; 5=2g  Db3  6

or

 5=2  Db3  6:

Example 6 Consider the LPP Maximize z ¼ 2x1 þ x2 þ 4x3  x4 subject to x1 þ 2x2 þ x3  3x4  8  x2 þ x3 þ 2x4  0 2x1 þ 7x2  5x3  10x4  21 x1 ; x2 ; x3 ; x4  0 The optimal simplex table of this LPP is

cB 2 1 0 zB ¼ 16

yB y1 y2 y7

xB 8 0 5

2

1

4

−1

0

0

0

y1 1 0 0 0

y2 0 1 0 0

y3 3 −1 −4 1

y4 1 −2 2 1

y5 1 0 −2 2

y6 2 −1 3 3

y7 0 0 1 0

When b is changed to ½ 3 2 4 T , make the necessary corrections in the optimal simplex table and solve the resulting problem. Solution From the given optimal simplex table, we have xB ¼ ½ 8 B1 ¼ ½ y5

5 T ;

0

y6

When b is changed to ½ 3 2 ables are given by

b ¼ ½ 8 0 21 T 2 3 1 2 0 6 7 y7  ¼ 4 0 1 0 5: 2 3 1

4 T , the new values of the current basic vari-

150

7 Post-optimality Analysis in LPPs

2

3 2 x1 1 2 4 x2 5 ¼ 4 0 1 2 3 x7

32 3 2 3 0 3 1 0 54 2 5 ¼ 4 2 5: 1 4 8

Since the values of x1 and x7 are negative, the current basic solution becomes infeasible. Hence, we have to apply the dual simplex method to obtain the basic feasible solution. Now we construct the next dual simplex table.

cj cB 2 1 0 zB ¼ 0

yB

xB

y1 y2 y7

−1 2 −8

2 y1

1 y2

4 y3

−1 y4

0 y5

0 y6

0 y7

1 0 0 0

0 1 0 0

3 −1 −4 1

1 −2 2 1

1 0 −2 2

2 −1 3 3

0 0 1 0

Now xBr ¼ minfxBi ; xBi \0g ¼ minf1; 8g ¼ 8 ¼ xB3 . ) r ¼ 3. Hence, the vector yB3 , i.e. y7 , has the basis.   z j  cj Again; Max ; yrj \0 yrj     z j  cj z3  c3 z5  c5 ¼ Max ; y3j \0 ¼ Max ; y3j y33 y35   1 2 1 ; ¼ Max ¼  ; which occurs for j ¼ 3: 4 2 4 Hence, the vector y3 enters the basis, and yrj , i.e. y33 ¼ 4, is the key element. Now we construct the next dual simplex table.

cj cB

yB

xB

2 y1

1 y2

4 y3

−1 y4

0 y5

0 y6

0 y7

2

y1

−7

1

0

0

5 2

 12

17 4

3 4

1 2 1 2

 74  34

 14

3/2

15/4

1/4

1

y2

4

0

1

0

4

y3

2

0

0

1

 52  12

0

0

0

3/2

zB ¼ 2

 14

7.4 Discrete Changes in the Requirement Vector

151

cj cB

yB

xB

2 y1

1 y2

4 y3

−1 y4

0 y5

0 y6

0 y7

0

y5

14

−2

0

0

−5

1

 17 2

 32

1

y2

−3

1

1

0

0

0

4

y3

−5

1

0

1

2

0

3

0

0

9

0

5 2 7 2 33 2

1 2 1 2 5 2

zB ¼ 23

This table shows that y3 leaves the basis, but since all the elements of the departing row are non-negative, by the dual simplex method, the given problem does not possess any feasible solution.

7.5

Discrete Changes in the Coefﬁcient Matrix

Here, we shall discuss the post-optimality problems which involve discrete changes in the element of the coefﬁcient matrix A. Let us consider the following LPP:   Maximize z ¼ cT ; x subject to Ax ¼ b and x  0; where c, xT 2 Rn ; bT 2 Rm and A is an m  n real matrix of rank m. Let xB be the optimum basic feasible solution of that LPP. Again, let us assume that the element ark in the rth row and kth column of matrix A is changed to a0rk ð¼ ark þ Dark Þ: Two possibilities may arise: Case I: ak 62 B Case II: ak 2 B, where B is the basis matrix. Case I: In this case, the post-optimal change does not have any effect on the feasibility of the current optimal solution xB = B−1b. Thus, the only effect, if any, may be on the optimality of the current solution. Let B1 ¼ ðb1 ; b2 ; . . .; bm Þ and ak ¼ ða1k ; a2k ; . . .; amk ÞT . Therefore, a0k ¼ ða1k ; a2k ; . . .; ark þ Dark ; . . .; amk ÞT ¼ ða1k ; a2k ; . . .; ark ; . . .; amk ÞT þ ð0; 0; . . .; Dark ; . . .; 0ÞT ¼ ak þ ð0; 0; . . .; Dark ; . . .; 0ÞT Now, to preserve the optimality of the optimal solution of the original LPP, we must have zk  ck  0.

152

7 Post-optimality Analysis in LPPs

D E D E where zk ¼ cTB ; yk ¼ cTB ; B1 ak D h iE ¼ cTB ; B1 ak þ ð0; 0; . . .; Dark ; . . .; 0ÞT D E D E ¼ cTB ; B1 ak þ cTB ; B1 ð0; 0; . . .; Dark ; . . .; 0ÞT D E ¼ zk þ cTB ; ðb1 ; b2 ; . . .br ; . . .; bm Þð0; 0; . . .; Dark ; . . .; 0ÞT D E ¼ zk þ cTB ; br Dark D E D E ) zk  ck ¼ zk  ck þ cTB ; br Dark ¼ zk  ck þ cTB ; br Dark : Now, from the optimality condition, we have zk  ck  0

D E ) zk  ck þ cTB ; br Dark  0 D E ) cTB ; br Dark   ðzk  ck Þ D E 2 ck Þ Dark  ðzcTk;b for cTB ; br [ 0 h B ri 6 6 D E 6 ðzk ck Þ ) 6 Dark  cT ;b for cTB ; br \0 h B ri 6 4 D E Dark is unrestrieted for cTB ; br ¼ 0

Case II: In this case, ak 2 B, i.e. ak is a basis vector. So, the feasibility of the current optimum solution may also be destroyed. Therefore, in this case, we have to ﬁnd a range for discrete changes Dark in ark so as to maintain the feasibility as well as the optimality of the solution. Now, we shall ﬁnd the conditions for maintaining feasibility of the solution. Let B = (B1, B2, …, Bm), Bj 2 A; j ¼ 1; 2; . . .; m and B1 ¼ ðb1 ; b2 ; . . .; bm Þ, where bj is the jth column vector of B−1. Since ak 2 B; any change in ark will affect only the pth column of B. Then the basis matrix B* becomes B ¼ B1 ; B2 ; . . .; Bp ; . . .; Bm ,

where Bp ¼ ðB1p; B2p; . . .; Brp; þ Darp ; . . .; Bmp ÞT . Since Bp 2 B , it can be expressed as the linear combination for vectors B1 ; B2 ; . . .; Bp ; . . .; Bm in B; therefore, Bp ¼ k1 B1 þ k2 B2 þ    þ km Bm ¼ Bk where k ¼ ðk1 ; k2 ; . . .; km ÞT .

7.5 Discrete Changes in the Coefﬁcient Matrix

153

) k ¼ B1 Bp ¼ B1 ½Bp þ ð0; 0; . . .; Darp ; . . .; 0ÞT  ¼ B1 Bpþ br Dark ¼ ep þ br Dark

"

# since B1 Bp ¼ ep   ; as Bep ¼ B1 ; B2 ; . . .; Bp ; . . .; Bm ð0; 0; . . .; 1; . . .; 0Þ ¼ Bp

i.e. ðk1 ; k2 ; . . .; kp ; . . .; km ÞT ¼ ð0; 0; . . .; 1; . . .; 0ÞT ðb1r ; b2r ; . . .; bpr ; . . .; bmr ÞT Dark . Now, equating the pth element and ith (i 6¼ p) element of both sides, we have kp ¼ 1 þ bpr Dark and ki ¼ bir Dark for i 6¼ p. Now, B*−1 exists only if B* is the non-singular ði:e:; jB j 6¼ 0Þ. Therefore, we must have kp 6¼ 0; i.e. 1 þ bpr Dark 6¼ 0. Now, we shall introduce a new matrix E which differs from an identity matrix Im in the pth column only. The pth column of matrix E can be obtained as follows: Bp ¼ k1 B1 þ k2 B2 þ    þ kp Bp þ    þ km Bm or

k1 k2 1 km B1  B2  . . . þ Bp . . .  Bm kp kp kp kp

  k1 k2 1 km T  ¼ B1 ; B2 ; . . .; BP ; . . .; Bm ; ; . . . þ ; . . .; ¼ B g; kp kp kp kp

Bp ¼ 

 T where g ¼ k1 =kp ; k2 =kp ; . . .; 1=kp ; km =kp = 0 1 0 ::: ::: ::: k1 =kp ::: B 0 1 ::: ::: ::: k2 =kp ::: B B ::: ::: ::: ::: ::: ::: ::: B B ::: ::: ::: ::: ::: ::: ::: Hence, E ¼ B B 0 0 ::: ::: ::: 1=k ::: p B B ::: ::: ::: ::: ::: ::: ::: B @ ::: ::: ::: ::: ::: ::: ::: 0 0 ::: ::: ::: km =kp :::

the pth column of E. 1 ::: ::: 0 ::: ::: 0 C C ::: ::: ::: C C ::: ::: ::: C C. ::: ::: 0 C C ::: ::: ::: C C ::: ::: ::: A ::: ::: 1

Therefore; B E ¼ B ðe1 ; e2 ; . . .; g; . . .; em Þ ¼ ðB e1 ; B e2 ; . . .; B g; . . .; B em Þ   ¼ B1 ; B2 ; . . .; Bp ; . . .; Bm h

as B e1 ¼ B1 ; B2 ; Bp ; . . .; Bm ð1; 0; . . .; 0ÞT ¼ B1  and B g ¼ Bp :

154

7 Post-optimality Analysis in LPPs

) B E ¼ B ) ðB EÞB1 ¼ BB1   ) B EB1 ¼ I ) ðB Þ1 ¼ EB1: Hence, the new solution becomes xB ¼ ðB Þ1 b ¼ ðEB1 Þb ¼ EðB1 bÞ   ¼ ExB *xB ¼ B1 b ) xB ¼ ExB

 T ) x0B1 ; x0B2 ; . . .; x0Bp ; . . .; x0Bm ¼ E xB1 ; xB2 ; . . .; xBp ; . . .; xBm ( xBi  kkpi xBp for i 6¼ p 0 ) xBi ¼ xBp for i ¼ p kp 8 < xB i  bi r Dar k xB p for i 6¼ p; i ¼ 1; 2; . . .; m 1 þ bp r Dar k or x0Bi ¼ : xB p for i ¼ p 1 þ b Dar k pr

Thus, to maintain the feasibility of x0B , we must have x0Bi  0 for i 6¼p and x0Bp  0. Now; x0Bp  0 )

xBp  1 þ bpr Dark

) 1 þ Dbp r Dar k [ 0 if xBp 6¼ 0:

Again; x0B i  0ði 6¼ pÞ ) xB i 

bi r Dar k xB p  0 1 þ bp r Dar k

) xB i þ Dar k ðbp r xB i  bi r xB p Þ  0 ) Dar k ðbp r xB i  bi r xB p Þ   xB i 8 xB i > ; for bp r xB i  bi r xB p [ 0 Da   > < rk bp r xB i  bi r xB p ) xB i > > ; for bp r xB i  bi r xB p \0 : Dark   b p r x B i  bi r x B p

7.5 Discrete Changes in the Coefﬁcient Matrix

155

Hence, in order to maintain the feasibility, combining these two inequalities, we have  Max

   xBi xBi ; Qi [ 0  Dark  Min ; Qi \ 0 ; Qi Qi

where Qi ¼ bpr xBi  bir xBp : The preceding inequalities give a range for discrete changes Dark in the coefﬁcient matrix A to maintain the feasibility of the solution. Condition for maintaining the optimality of the solution To maintain the optimality of the new solution, we must have zj  cj  0

for all j ¼ 1; 2; . . .; n: D E Now; zj  cj ¼ cTB ; B1 aj  cj D E   ¼ cTB ; ðEB1 Þaj  cj *B1 ¼ EB1 D E ¼ cTB ; ðEB1 aj Þ  cj D E h i 1 ¼ cTB ; ðEyj Þ  cj )yj ¼ B aj ¼

m X

cBi yij  cj ;

i¼1

where

yi j

¼

) zj  cj ¼ ¼

8 < yi j  :

Dar k bi r 1 þ bp r Dar k yp j ;

yp j 1 þ bp r Dar k

m X i¼0 m X

cB i



1 þ bp r Dar k

condition 0

or zj  cj þ Dark

of

i¼p ! Dar k bi r yp j yi j   cj 1 þ bp r Dar k m X

cB i

i¼1

¼ z j  cj 

Thus, the yp j Dar k hcTB ;br i

;

cB i yi j  cj 

i¼1

i 6¼ p

D yp j Dar k cTB ; br

Dar k bi r yp j 1 þ bp r Dar k E

1 þ bp r Dar k

optimality

requires

n D Eo  zj  cj bp r  yp j cTB ; br  0.

:

zj  cj  0;

i.e.

D E ðzj cj Þ T ; if ðz  c Þb  y c ; b j j pr pj r [0 B ðzj cj Þbpr ypj hcTB ;br i D E ðzj cj Þ and D ar k   z c b y cT ;b ; if ðzj  cj Þbpr  ypj cTB ; br \0. ð j j Þ pr pj h B r i

Hence, D ar k  

z j  cj

156

7 Post-optimality Analysis in LPPs

Hence, in order to maintain the optimality of the new solution, Dark must satisfy the following inequalities (which are the combination of the preceding two inequalities):     ðzj  cj Þ ðzj  cj Þ Max  ; Pj [ 0  Dark  Min ; Pj \0 ; Pj Pj D E   where Pj ¼ zj  cj bpr  ypj cTB ; br .

7.6

Addition of a new variable in an LPP plays an important role in post-optimality analysis. Due to this change, structural variation in the simplex table takes place. Let a new variable xn+1 be introduced in the following LPP whose optimal solution is known:   Maximize z ¼ cT ; x subject to Ax ¼ b and x  0: Also, let an+1 be the coefﬁcient vector associated with xn+1 and let the proﬁt coefﬁcient for xn+1 be cn+1. Since the requirement vector b is not changed, the old optimal solution will be a feasible solution of the new LPP, but it may not be optimal. Now, we have to compute   yn þ 1 ¼ B1 an þ 1 and zn þ 1  cn þ 1 ¼ cTB ; yn þ 1  cn þ 1 : If zn þ 1  cn þ 1  0; then xn+1 = 0, and the current optimal solution remains optimal for the new problem. If zn þ 1  cn þ 1 \0; yn þ 1 enters the basis, and the simplex method is applied to obtain the optimal solution of the resulting new problem. Example 7 Consider the LPP Maximize z ¼ x1 þ 2x2 þ x3 subject to 2x1 þ x2  x3  2 2x1  x2 þ 5x3  6 4x1 þ x2 þ x3  6 and x1 ; x2 ; x3  0:

157

The optimal simplex table is as follows:

cB 2 1 2 zB ¼ 10

yB y2 y3 y6

cj

1

2

1

0

0

0

xB x2 ¼ 4 x3 ¼ 2 x6 ¼ 0

y1 3 1 0 6

y2 1 0 0 0

y3 0 1 0 0

y4 5/4 1/4 −3/2 1/4

y5 1/4 1/4 −1/2 3/4

y6 0 0 1 0

Dj

If a new variable x7 ð  0Þ is introduced with proﬁt (i) 4, (ii) 6 and a7 ¼ ½ 2 1 4 T , discuss the effect of the introduction of the new variable x7 . Solution From the given optimal simplex table, we get the inverse of the basis matrix as 2

B1 ¼ ½ y4

y5

5=4 y6  ¼ 4 1=4 3=2

1=4 1=4 1=2

3 0 0 5: 1

Since a7 ¼ ½ 2 1 4 T , the corresponding column vector to be introduced in the optimal simplex table is given by 2

5=4 y7 ¼ B1 a7 ¼ 4 1=4 3=2

1=4 1=4 1=2

32 3 2 3 0 2 9=4 0 54 1 5 ¼ 4 1=4 5: 1 4 3=2

(i) When c7 ¼ 4,

z 7  c7 ¼ 2 

9 1 3 3 þ 1  þ 0   4 ¼ [ 0: 4 4 2 4

Hence, the optimality condition is satisﬁed for the modiﬁed problem also. Therefore, the optimal solution for the modiﬁed problem is given by x1 ¼ 0; x2 ¼ 4; x3 ¼ 2; x7 ¼ 0 with zMax ¼ 10: (ii) When c7 ¼ 6,

z7  c7 ¼ 2 

9 1 3 5 þ 1  þ 0   6 ¼  \0: 4 4 2 4

158

7 Post-optimality Analysis in LPPs

Hence, the optimality condition is not satisﬁed. We have to modify the optimal simplex table of the old problem with the added column vector y7 ¼ ½ 9=4 1=4 3=2T ; z7  c7 ¼  54 and c7 ¼ 6. Now, to get the optimal solution, we shall apply the simplex method. The consecutive tables obtained are as follows:

cj cB

yB

2 y2 1 y3 0 y6 zB ¼ 10

cj cB

yB

2 y2 1 y3 6 y7 zB ¼ 10

xB x2 ¼ 4 x3 ¼ 2 x6 ¼ 0

xB x2 ¼ 4 x3 ¼ 2 x7 ¼ 0

1 y1

2 y2

1 y3

0 y4

0 y5

0 y6

6 y7

Min ratio

3 1 0 6

1 0 0 0

0 1 0 0

5/4 1/4 −3/2 11/4

1/4 1/4 −1/2 3/4

0 0 1 0

9/4 1/4 3/2 −5/4

16/9 8 0 Dj

1 y1

2 y2

1 y3

0 y4

0 y5

0 y6

6 y7

3 1 0 6

1 0 0 0

0 1 0 0

7/2 1/2 −1 3/2

1 1/3 −1/3 1/3

−3/2 −1/6 2/3 5/6

0 0 1 0

Dj

From this table, it is seen that all xBi  0 and all Dj  0, and so the optimality conditions are satisﬁed. Hence, the optimal solution is given by x1 ¼ 0; x2 ¼ 4; x3 ¼ 2; x7 ¼ 0 and zMax ¼ 10.

7.7

Deletion of a Variable

Like addition of a new variable, deletion of a variable (basic or non-basic) also changes the structure of the problem. According to the type of deleted variables, two cases may arise. Case 1: If the variable to be deleted is non-basic, then the feasibility and optimality conditions are not affected. This means that the current optimal solution is the optimal solution of the new problem. Case 2: If the deleted variable is basic, then the conditions of optimality may be affected, and a new solution must be obtained. To ﬁnd this new optimal solution, we have to assign a proﬁt −M corresponding to the basic variable to be deleted and apply the simplex method, taking the modiﬁed optimal simplex table as the starting simplex table.

7.8 Addition of a New Constraint

7.8

159

Let us suppose that a new constraint, i.e. the (m + 1)th constraint am þ 1; 1 x1 þ am þ 1; 2 x2 þ    þ am þ 1; n xn  bm þ 1 ;

i:e: ux  bm þ 1 ;

is added to the system of original constraints Ax = b, x  0, where u ¼   am þ 1;1 ; am þ 1;2 ; . . .; am þ 1;n and bm+1 is positive or 0 or negative. Then two cases may arise. (i) The current optimal solution xB of the original problem satisﬁes the new constraint. If so, the solution remains feasible as well as optimal. (ii) If the current optimal solution xB does not satisfy ! the new constraint, then the xB new basic feasible solution will be xB ¼ , where xs is the slack variable xS introduced in the additional constraint. In that case the new basis matrix B* of

B O , where a is a submatrix of u [the order m + 1 is given by B ¼ a 1 components are the coefﬁcients of basic variables].

)B1 ¼

B1 aB1

O : 1

b xB B1 O ¼ . bm þ 1 bm þ 1  axB aB1 1 If xB  0, the current solution remains basic feasible for the new problem also; otherwise, we apply the dual simplex method to obtain the basic feasible solution. In this case, note that the following three situations may arise depending on the nature of the solution to the original LPP. Now, xB ¼ B1 b ¼

(i) If the original LPP has an optimal solution, then the modiﬁed LPP may have an optimal solution or it will have no feasible solution. (ii) If the original LPP has an unbounded solution, then the modiﬁed LPP may have an optimal solution or it will have no feasible solution or it will have an unbounded solution. (iii) If the original LPP has no feasible solution, then the modiﬁed LPP will also have no feasible solution.

160

7 Post-optimality Analysis in LPPs

Example 8 We are given the following LPP: Maximize z ¼ 3x1 þ 4x2 þ x3 þ 7x4 subject to the constraints 8x1 þ 3x2 þ 4x3 þ x4  7 2x1 þ 6x2 þ x3 þ 5x4  3 x1 þ 4x2 þ 5x3 þ 2x4  8 and x1 ; x2 ; x3 ; x4  0: (i) Discuss the effect of discrete changes in b1. (ii) Let the variable x4 be deleted from the LPP part of the problem. Obtain an optimum solution to the resulting LPP. (iii) Let a linear constraint 2x1 þ 3x2 þ x3 þ 5x4  4 be the addition to the constraints of the problem. Check whether there is any change in the optimum solution of the original problem. (iv) Also discuss the case where the upper limit of the preceding constraint is reduced to 2. Solution For the given LPP, the optimum table is the following:

cj cB

yB

xB

3 y1 x1 ¼ 16/19 x4 ¼ 5/19 7 y4 x7 ¼ 126/19 0 y7 zB ¼ 83=19

3 y1

4 y2

1 y3

7 y4

0 y5

0 y6

0 y7

1 0 0 0

9/38 21/19 59/38 169/38

1/2 0 9/2 1/2

0 1 0 0

5/38 −1/19 −1/38 1/38

−1/38 8/38 −15/38 53/38

0 0 1 0

Dj

(i) From this optimum table, we have xB = [16/19, 5/19, 126/19]T, b = [7, 3, 8]T 0 and B1 ¼ ðy5 y6 y7 Þ ¼

5 38 B 1 B @ 19 1 38

1 38 8 38 15 38

0

1

C 0C A: 1

Let the requirement vector component b1 be changed to b1+Δb1. Then the range of Δb1 for such a change, maintaining the optimality to the basic feasible solution, is given by

7.8 Addition of a New Constraint

161

    xBi xBi Max ; bi1 [ 0  Db1  Min ; bi1 \0 bi1 bi1 or or

Maxfð16=19Þ=ð5=38Þg  Db1  Minfð5=19Þ=ð1=19Þ; ð126=19Þ=ð1=38Þg  32=5  Db1  Minf5; 252g or; 32=5  Db1  5:

(ii) Since the variable x4 to be deleted is a basic variable, we assign a cost—M to x4 and take the current optimum simplex table as the initial simplex table for the new LPP. 3

4

1

−M

0

0

0

cB

yB

xB

y1

y2

y3

y4

y5

y6

y7

Mini ratio

3 −M 0

y1 y4 y7

16/19 5/19 126/19

1 0 0 0

9/38 21/19 59/38

1/2 0 9/2

5/38 −1/19 −1/38

−1/38 4/19 −15/38

125  21M 19  38

1 2

0 1 0 0

0 0 1 0

32/9 5/21 252/59 Dj

cj

zB ¼ 5M=19 þ 48=19

M 19

þ

15 38

3  4M 19  38

Here y2 is the incoming vector, y4 is the outgoing vector and y22 = 21/19 is the key element. Now introducing y2 and dropping y4, we develop the next table.

cj cB

xB

3 y1

4 y2

1 y3

0 y5

0 y6

0 y7

3 y1 x1 ¼ 209=266 x2 ¼ 5=21 4 y2 x7 ¼ 13997=798 0 y7 zB ¼ 139=42

1 0 0 0

0 1 0 0

1/2 0 9/2 1/2

1/7 −1/21 1/21 5/21

−1/14 4/21 −187/266 23/42

0 0 1 0

yB

Dj

This shows that the new optimum solution to the LPP when the variable x4 has been deleted is given by x1 = 209/266, x2 = 9/2, x3 = 0 and x4 = 0, Max z = 139/ 42. (iii) From the optimum simplex table of the given LPP, we have x1 = 16/19, x2 = 0, x3 = 0, x4 = 5/19. For this solution, 2x1 þ 3x2 þ x3 þ 5x4 ¼ 2:16=19 þ 5:5=19 ¼ ð32 þ 25Þ=19 ¼ 3\4. Hence, the current optimal solution also satisﬁes the additional constraint. Thus, the additional constraint is redundant, and the optimum solution of the given LPP is also the optimum solution of the modiﬁed LPP (the original LPP including the additional constraint).

162

7 Post-optimality Analysis in LPPs

(iv) When the upper limit of the additional constraint is reduced to 2, the additional constraint will be 2x1 þ 3x2 þ x3 þ 5x4  2. The current optimum solution does not satisfy the additional constraint, so by introducing a slack variable x8 ð  0Þ in the additional constraint, we have 2x1 þ 3x2 þ x3 þ 5x4 þ x8 ¼ 2: Now appending this constraint in the optimum table, we have the following: cj cB

yB

xB

3 y1

4 y2

1 y3

7 y4

0 y5

0 y6

0 y7

0 y8

3 7 0 0

y1 y4 y7 y8

16/19 5/19 126/19 2

1 0 0 2

9/38 21/19 59/38 3

1/2 0 9/2 1

0 1 0 5

5/38 −1/19 −1/38 0

−1/38 4/19 −15/38 0

0 0 1 0

0 0 0 1

The basis B = (y1, y2, y7, y8) is obtained by performing the operation R04 ¼ R4  2R1 and then R04 ¼ R4  5R2 , to obtain the following simplex table:

cj cB

yB

3 y1 7 y4 0 y7 0 y8 zB ¼ 83=19

xB 16/19 5/19 126/19 −1

3 y1

4 y2

1 y3

7 y4

0 y5

0 y6

0 y7

0 y8

1 0 0 0 0

9/38 21/19 59/38 −3 169/38

1/2 0 9/2 0 1/2

0 1 0 0 0

5/38 −1/19 −1/38 0 1/38

−1/38 4/19 −15/38 −1 53/38

0 0 1 0 0

0 0 0 1 0

Dj

Since all zj  cj  0 and xB4 ¼ 1, an optimum or more than optimum but infeasible solution has been obtained. Hence, we have to apply the dual simplex method to solve the problem. As xB4 ¼ 1, which is negative, the corresponding basis vector y8 leaves the basis. 

 z j  cj Now Max ; y4j \0 ; j ¼ 1; 2; . . .; 8 y4j     169=38 53=38 169 53 ; ; ¼ Max ¼ Max  3 1 114 38 53 ¼  ; which occurs for j ¼ 6: 38

7.8 Addition of a New Constraint

163

Therefore, y6 enters the basis, and y46 = −1 is the key element. Now we construct the next simplex table.

cj cB

yB

xB

3 y1 33/38 1/19 7 y4 267/38 0 y7 1 0 y6 zB ¼ 113=38

3 y1

4 y2

1 y3

7 y4

0 y5

0 y6

0 y7

0 y8

1 0 0 0 0

6/19 9/19 7/19 3 5/19

1/2 0 9/2 0 1/2

0 1 0 0 0

5/38 −1/19 −1/38 0 1/38

0 0 0 1 0

0 0 1 0 0

−1/38 4/19 −15/38 −1 53/38

Dj

Since all zj  cj  0, an optimum solution has been attained. From the preceding simplex table, we see that the additional constraint has decreased the optimum value of the objective function from 83/19 to 113/38. Example 9 We are given the following LPP: Maximize Z ¼ 3x1 þ 4x2 þ x3 þ 7x4 subject to the constraints 8x1 þ 3x2 þ 4x3 þ x4  7 2x1 þ 6x2 þ x3 þ 5x4  3 x1 þ 4x2 þ 5x3 þ 2x4  8 and

x1 ; x2 ; x3 ; x4  0:

The following table gives the optimum solution:

cj cB

yB

xB

3 y1 16/19 5/19 7 y4 126/19 0 y7 zB ¼ 83=19

3 y1

4 y2

1 y3

7 y4

0 y5

0 y6

0 y7

1 0 0 0

9/38 21/19 59/38 169/38

1/2 0 9/2 1/2

0 1 0 0

5/38 −1/19 −1/38 1/38

−1/38 8/38 −15/38 53/38

0 0 1 0

Dj

Discuss the effect of discrete changes in the activity coefﬁcients a12 and a24 of the coefﬁcient matrix A on the current optimum basic feasible solution to the given LPP.

164

7 Post-optimality Analysis in LPPs

Solution From the given optimum simplex table, we have b1 5 B 38 B B B 1 ¼B B 19 B @ 1 38

b2 b3 1 1 0C 38 C C 8 C 0 C; C 38 C A 15 1 38

0

B1

where B is the basis for the given problem. Let the element a12 of A be changed to a12 ¼ a12 þ Da12 . In the optimum simplex table, the element a12, i.e. the corresponding column vector y2, is not in the basis. We know that the range for discrete change Δark in the coefﬁcient matrix A for maintaining the optimality of the solution is given by D E cTB ; br [ 0 D E ck Þ for cTB ; br \0 Dark  ðzcTk;b h B ri   Dapk is unrestricted for cTs ; br ¼ 0: Dark 

ðzk ck Þ hcTB ;br i

for

Here r = 1, k = 2. Now cTB ; b1 ¼ 3ð5=38Þ þ 7ð1=19Þ þ 0 ¼ 1=38 [ 0. Therefore, the range for discrete change Δa12 in the coefﬁcient matrix A for maintaining the optimality of the solution is given by ðz2  c2 Þ E Da12  D cTB ; b1

½)cB b1 [ 0

or

Da12   ð169=38Þ=ð1=38Þ

or

Da12   169:

Let the element a24 of A be changed to a24 ¼ a24 þ Da24 . In the optimum simplex table, the element a24 corresponding to the column vector y4 is in the basis. Hence, the discrete change in a24 belonging to vector y4 may affect the feasibility as well as the optimality of the original optimum basic feasible solution. We know that the range for discrete change in the element a24 of the coefﬁcient matrix in order to maintain the optimality of the new solution is given by

7.8 Addition of a New Constraint

165



   ðzj  cj Þ ðzj  cj Þ Max  ; Pj [ 0  Dark  Min  ; Pj \0 ; Pj Pj D E   where Pj ¼ zj  cj bp r  yp j cTB ; br : Here rD= 2, k E= 4 and p = 2. Now, cTB ; b2 ¼ 3ð1=38Þ þ 7ð8=38Þ þ 0ð15=38Þ ¼ 53=38. D E Hence; P2 ¼ ðz2  c2 Þb22  y22 cTB ; b2 ¼ ð169=38Þð8=38Þ  ð21=19Þð53=38Þ ¼  23=38 D E P3 ¼ ðz3  c3 Þb22  y23 cTB ; b2 ¼ ð1=2Þð8=38Þ  0ð53=38Þ ¼ 2=19 D E P5 ¼ ðz5  c5 Þb22  y25 cTB ; b2 ¼ ð1=38Þð8=38Þ  ð1=19Þð53=38Þ ¼ 3=38 D E P6 ¼ ðz6  c6 Þb22  y26 cTB ; b2 ¼ ð53=8Þð8=38Þ  ð8=38Þð53=38Þ ¼ 0:

Therefore, the range for discrete change in the element a24 of the coefﬁcient matrix in order to maintain the optimality of the new solution is given by or or

n o n o Max  ðz3Pc3 3 Þ ;  ðz5Pc5 5 Þ  D a24  Min  ðz2Pc2 2 Þ 1=3  a24  169=23:

Again, we know that the range for discrete change in the element a24 in order to maintain the optimality of the new solution is given by     xB xB Max  i ; Qi [ 0  D ark  Min  i ; Qi \0 ; Qi Qi where Qi ¼ bpr xBi  bir xBp . Here r = 2, k = 4, p = 2.  1 5 8 7 Now Q1 ¼ b22 xB1  b12 xB2 ¼ 38  16 19   38  19 ¼ 38 Q3 ¼ b22 xB3  b32 xB2

8 126 15 5 3    ¼ : ¼  38 19 38 19 2

Therefore, the range for the discrete change in the element a24 in order to n xB1 maintain the optimality of the new solution is given by Max  Q1 ; x

 QB33 g  Da24 \1 n o 126=19 ;  or Max  16=19  Da24 or  84 19  Da24 \1. 7=38 3=2

166

7.9

7 Post-optimality Analysis in LPPs

Exercises

1. Solve the following LPP:

Maximize z ¼ x1 þ 3x2  2x3 subject to the constraints 3x1  x2 þ 2x3  7  2x1 þ 4x2  12  4x1 þ 3x2 þ 8x3  10: and x1 ; x2 ; x3  0: Determine the separate range for discrete changes in a13 ; a23 ; a21 consistent with the optimal solution of the given LPP. 2. The optimal solution to the problem

Maximize Z ¼ x1 þ 2x2  x3 subject to the constraints 3x1 þ x2  x3  10; x1 þ 4x2 þ x3  6; and x1 ; x2 ; x3  0

x2 þ x3  4

is given in the following table: cj ! yB

cB

y4 0 2 y2 0 y6 zB ¼ 8

xB 6 4 10

−1 y1

2 y2

−1 y3

0 y4

0 y5

0 y6

−M y7

3 0 1 1

0 1 0 0

−2 1 3 3

1 0 0 0

−1 1 4 2

0 0 1 0

0 0 −1 M

Dj

Determine the ranges for discrete changes in the components b2 and b3 of the required vector so as to maintain the optimality of the current optimal solution. 3. Will the optimal solution to the following LPP remain optimal if the requirement vector (3,5,6)T changes to (3, 10, 6)T?

7.9 Exercises

167

Minimize z ¼ 8x1 þ 3x2 þ 6x3 þ 3x4 subject to x2 þ 4x3 þ 5x4  3 3x1 þ 2x3  5 x1 þ 2x2 þ x4  6 x 1 ; x2 ; x3 ; x4  0 4. Find the optimal solution of the LPP

Maximize z ¼ 4x1 þ 3x2 subject to x1 þ x2  5 3x1 þ x2  7 x1 þ 2x2  10; x1 ; x2  0: Show how to ﬁnd the optimal solution of the problem if: (i) The ﬁrst component of the original requirement vector is increased by one unit and the third component is decreased by one unit. (ii) The second component of the original requirement vector is decreased by two units. 5. For the LPP

Maximize z ¼ x1 þ 2x2 þ x3 subject to 2x1 þ x2  x3  2 2x1  x2 þ 5x3  6 4x1 þ x2 þ x3  6 x1 ; x2 ; x3  0; the optimal table is as follows: cB

yB

2 y2 1 y3 0 y6 zB ¼ 10

xB

y1

y2

y3

y4

y5

y6

4 2 0

3 1 0 6

1 0 0 0

0 1 0 0

5/4 1/4 −3/2 1/4

1/4 1/4 −1/2 3/4

0 0 1 0

Discuss the effect of deletion of the variable (i) x1, (ii) x2, (iii) x3.

Dj

168

7 Post-optimality Analysis in LPPs

6. Let us consider the ﬁnal table of an LPP. cB 2 4 1

yB y1 y2 y3

cj

2

4

1

8

2

0

0

0

xB 3 1 7

y1 1 0 0 0

y2 0 1 0 0

y3 0 0 1 0

y4 −1 2 1 −1

y5 0 1 −2 0

y6 1/2 −1 5 2

y7 1/5 0 −3/10 1/10

y8 −1 ½ 2 2

Dj

Here y6, y7 and y8 are slack variables. If the constraints ðiÞ 2x1 þ 3x2  x3 þ 2x4 þ 4x5  5 ðiiÞ 2x1 þ 3x2  x3 þ 2x4 þ 4x5  1 are added, then ﬁnd the solution of the changed LPP. 7. For the LPP

Maximize z ¼ 15x1 þ 45x2 subject to 5x1 þ 2x2  162 x1 þ 16x2  240 x2  50 and x1 ; x2  0; ﬁnd the optimal solution. Find the range of each cost coefﬁcient (changed one at a time) for which the current solution will remain optimal. 8. Find the optimal solution of the LPP and the separate ranges of variations of b2 and b3 consistent with the optimality of the solution

Minimize z ¼ x1 þ 2x2  x3 subject to 3x1 þ x2  x3  10  x1 þ 4x2 þ x3  6 x2 þ x3  4 and x1 ; x2 ; x3  0: Determine also the efﬁcient discrete changes in the components of the cost vector which correspond to the basic variables.

7.9 Exercises

169

9. Following is the optimal table for an LPP: cB 2 1

yB y1 y2

cj

2

1

1

2

0

xB 3 4

y1 1 0 0

y2 0 1 0

y3 −1 4 1

y4 3 −1 3

y5 2 −2 2

(i) Find the limitations of the values of c3 ; c4 ; c5 (taking one at a time) for which the current solution will remain optimal. (ii) Find the optimal solution to the problem if c3 is changed to 3. (iii) Find the limitations of the values of c1 for which the current solution remains optimal. (iv) Find the optimal solution to this problem if c1 is changed to 5. 10. Find the optimal solution of the LPP

Maximize z ¼ 4x1 þ 3x2 subject to x1 þ x2  5 3x1 þ x2  7 x1 þ 2x2  10 and x1 ; x2  0: Then ﬁnd the optimal solution of the problem, if: (i) The ﬁrst component of the original requirement vector is increased by one unit and the third component is decreased by one unit. (ii) The second component of the original requirement vector is decreased by two units.

Chapter 8

Integer Programming

8.1

Objectives

After studying this chapter, the readers are able to learn: • The limitations of the simplex method in deriving integer solutions to LPPs • Gomory’s all-integer and mixed-integer programming techniques • The branch and bound method to solve integer programming problems.

8.2

Introduction

Integer programming problems (IPPs) are a special class of linear programming problem (LPP) where all or some of the decision variables are restricted to the integers. If all the variables are restricted to take integer values, the corresponding IPP is called a pure IPP. On the other hand, if some of the variables are restricted to take integer values, then the problem is called a mixed-integer linear programming problem (mixed ILPP). In the year 1956, R. E. Gomory ﬁrst developed a method to solve pure IPPs. Later, he extended the method to solve mixed ILPPs. Another important approach, called the branch and bound method, was developed for solving both all-integer and mixed-integer programming problems. The integer solution of a given ILPP can be obtained by ﬁnding the non-integer optimal solution through the simplex method or the graphical method and then rounding off the fractional values of the variables. But, in some cases, the derivation from the exact optimal integer solution (obtained as a result of rounding) may become large enough to give an infeasible solution. Hence, it is necessary to develop a systematic procedure to determine the optimal integer solution to such problems. The following example will make the concept more understandable. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_8

171

172

8 Integer Programming

Let us consider a simple problem: Maximize z ¼ 10x1 þ 4x2 subject to 3x1 þ 4x2  8; x1  0; x2  0, where x1, x2 are integers. First ignoring the integer-valued restrictions, we get the optimal solution x1 ¼ 2 23 ; x2 ¼ 0 and Maximize z ¼ 26 23 using the graphical method. Then by rounding off the fractional value of x1 ¼ 2 23, the optimum solution becomes x1 ¼ 3; x2 ¼ 0 with Maximize z ¼ 30. But this solution does not satisfy the constraint 3x1 þ 4x2  8. Therefore, this solution is not feasible. Now, let us take x1 ¼ 2; x3 ¼ 0, which is feasible and integer-valued. But this solution gives z = 20, which is far away from the value z ¼ 26 23. So this is another disadvantage of getting an integer-valued solution by rounding down the solution. The result may be far away from the optimum solution. There are two methods used to solve IPPs: (i) The Gomory method or the Gomory cutting-plane method (ii) The branch and bound method.

8.3

Gomory’s All-Integer Cutting-Plane Method

A systematic procedure for solving all IPPs was ﬁrst developed by R. E. Gomory in 1956, Later, he extended the method to deal with the more complicated case of mixed-integer programming problems. In this method, ﬁrst of all, we have to ﬁnd the optimal solution of the given IPP by the simplex method, ignoring the restriction of integer values of the variables. (i) If all the variables in the optimal solution obtained have integer values, the current solution will be the required optimal integer solution. (ii) If not, the considered LPP requires modiﬁcations by introducing a secondary constraint (also called the Gomory constraint) that reduces some of the non-integer values of variables to integer ones but does not eliminate any feasible integers. (iii) Then, an optimal solution to this modiﬁed LPP is obtained by using the dual simplex method. If all the variables in this solution are integers, then the optimal integer solution is obtained. Otherwise, another constraint is added to the LPP, and the entire procedure is repeated. All Integer Programming Algorithm Gomory’s fractional cut method may be summarized in the following steps: Step 1 Express the given LPP in its standard forms by converting the minimization problem to a maximization problem (if necessary) and introducing slack and/or surplus variables with the readjustment of the objective function. Step 2 Determine the optimal solution by using the simplex method, ignoring the integer value restriction.

8.3 Gomory’s All-Integer Cutting-Plane Method

173

Step 3 Test the integrality of the optimum solution (a) If the optimum solution contains all integer values, an optimal integer basic feasible solution (BFS) has been achieved. Stop the process. (b) If not, go Step 4. Step 4 Examine the values of basic variables which give fractional values. Express them as the sum of an integer and a proper fraction. Then, select the largest fractional value of the basic variables, i.e. ﬁnd Maxf fBi g. Let i

it be fBk for i ¼ k. Step 5 Express each of the negative fractions, if any, in the kth row of the optimal simplex table as the sum of a negative integer and a non-negative proper fraction. Step 6 Construct the Gomorian constraint n X

fkj xj  fBk

j¼1

P or  nj¼1 fkj xj þ G1 ¼ fBk , where 0  fkj \1 and 0\ fBk \1 and G1 is the Gomorian slack variable. Step 7 Add the Gomorian constraint constructed in Step 6 at the bottom of the optimum simplex table. Use the dual simplex method to ﬁnd an improved optimal solution. Then go to Step 3.

8.4

Construction of Gomory’s Constraint

Let the optimal non-integer solution to the maximization LPP be obtained. In our usual notation this solution is shown by the following simplex table: cB

B

XB

y1 ðb1 Þ

y 2 ð b2 Þ



y i ð bi Þ

y m ðbm Þ

ymþ1



yn

cB1

b1

x B1

1

0



0



0

y1; m þ 1



y1; n

cB2

b2

x B2

0

1



0



0

y2; m þ 1



y2; n

.. .

.. .

.. .

.. .

.. .

.. .

.. .

cBi

Bi

x Bi

0

0

0

yi; m þ 1

.. .

.. .

.. .

.. .

.. .

.. .

.. .

cBm

bm

x Bm

0

0



0

1

ym; m þ 1

0

0



0

0

Dm þ 1

z ¼ \cTB ; xB [

.. . 

1



.. . 

.. . 

yi; n .. .



ym; n Dn

Dj

174

8 Integer Programming

In this table, the variables xBi ði ¼ 1; 2; . . .; mÞ are taken as basic variables and the remaining xm þ 1 ; xm þ 2 ; . . .; xn as the non-basic variables. (For simplicity, basic variables are taken consecutively.) Let the ith basic variable xBi possess a non-integer value which is given by

or

0:x1 þ 0:x2 þ    þ xi þ    þ 0:xm þ yi;m þ 1 xm þ 1 þ    þ yi;n xn ¼ xBi n X xi þ y ij x j ¼ xBi : j¼m þ 1

ð8:1Þ Now, let us assume that xBi ¼ IBi þ f Bi and yij ¼ Iij þ f_ij , where I Bi and Iij are the largest integral parts of xBi and yij respectively, such that IBi \xBi and Iij  yij . Hence, 0\ fBi \1 and 0  fij \1; i.e. fBi is a strictly positive proper fraction, while fij is a non-negative proper fraction or zero. Now substituting xBi ¼ IBi þ fBi and yij ¼ Iij þ fij in (8.1), we have n X 

xi þ or

j¼m þ 1 n X

xi þ

 Iij þ fij xj ¼ IBi þ fBi

I ij xj þ

j¼m þ 1

n X

ð8:2Þ f ij xj ¼ IBi þ fBi

j¼m þ 1

Now for all the variables xi (for i = 1, 2, …, m) and xj (for j = m + 1, m + 2, …, n) to be integer-valued, the sum of terms with fractional coefﬁcients on the left-hand side must be greater than or equal to the fraction on the right-hand side, n X

i:e: or or

fij xj  fBi

j¼m þ 1 n X

fij xj j¼m þ 1 n X

 fBi  0

fij xj þ fBi  0



j¼m þ 1

This constraint is known as the Gomorian constraint. Example 1 Find the optimal integer solution to the following LPP: Maximize z ¼ x1 þ x2 subject to 3x1 þ 2x2  5 x2  2; x1 ; x2  0 and are integers:

ð8:3Þ

8.4 Construction of Gomory’s Constraint

175

Solution: The given problem is a maximization problem. Now introducing two slack variables x3 ð  0Þ and x4 ð  0Þ, the standard form of given LPP becomes Maximize z ¼ x1 þ x2 þ 0:x3 þ 0:x4 3x1 þ 2x2 þ x3 ¼ 5 x2 þ x4 ¼ 2; x1 ; x2 ; x3 ; x4  0 and are integers:

subject to

Now, ignoring the integer condition, we shall solve the problem by the simplex method.   1 0 Here the initial BFS is x3 = 5, x4 = 2 with the initial basis as B ¼ . 0 1 Now we construct the initial simplex table. cj xB

1 y1

1 y2

0 y3

0 y4

Minimum ratio

x3 = 5 x4 = 2 zB ¼ 0

3 0 −1 "

2 1 −1

1 0 0 #

0 1 0

5/1 = 5 – Dj

1 y2

0 y3

0 y4

Minimum ratio

2/3 1 −1/3

1/3 0 1/3

0 1 0

(5/3)/(2/3) = 5/2 2/1 = 2 Dj

cB

yB

0 0

y3 y4

cB

yB

cj xB

1 y1

1 0

y1 y4

x1 = 5/3 x4 = 2

1 0 0

zB ¼ 53

"

cB 1 1

yB y1 y2

#

cj

1

1

0

0

xB x1 = 1/3 x2 = 2

y1 1 0 0

y2 0 1 0

y3 1/3 0 1/3

y4 −2/3 1 1/3

zB ¼ 73

Since all Dj  0, an optimal solution is obtained, given by x1 ¼ 13 ; x2 ¼ 2. The optimal solution is not an integer solution because x1 ¼ 13, so we have to add a fractional cut constraint, i.e. a Gomorian constraint, in the optimal simplex table.

176

8 Integer Programming

Since xB1 ¼ 13 ¼ 0 þ 13, we have fB1 ¼ 13. Hence, Maxð fB1 Þ ¼ 13 ¼ f B1 , which corresponds to the ﬁrst row. So, the ﬁrst row is the source row. Now, from the ﬁrst row of the optimal simplex table, we get 1 2 1 x3  x4 ¼ 3 3 3 1 1 1 x1 þ 0:x3 þ x3  x4 þ x4 ¼ : 3 3 3 x1 þ

Hence, the Gomorian constraint will be

or or

1 1 1 x3 þ x4  3 3 3 1 1 1  x3  x4   3 3 3 1 1 1  x3  x4 þ G1 ¼  ; 3 3 3

where G1 is the Gomorian slack variable. Adding this new constraint at the bottom of the preceding optimal simplex table, we have used the dual simplex method to solve the problem.

cB

yB

cj xB

1 y1

1 y2

0 y3

0 y4

0 g1

1 1 0

y1 y2 g1

x1 = 1/3 x2 = 2 G1 = −1/3

1 0 0 0

0 1 0 0

1/3 0 −1/3 1/3

−2/3 1 −1/3 1/3

0 0 1 0

zB ¼ 73

Dj

  We have xBr ¼ Min½xBi ; xBi \0 ¼ Min  13 ¼  13 ¼ xB3 . Hence, r ¼ 3. So yB3 , i.e. g1 , leaves the basis. Now, 

Dj Max ; yrj \0 yrj

(

¼ Max

1 3

1 3

)

;  13  13

¼ Maxf1; 1g ¼ 1; which corresponds to y3 and y4. Arbitrarily, we choose y3 as the entering vector, and y33 ¼  13 is the key element. Now we construct the next simplex table.

8.4 Construction of Gomory’s Constraint

cB

yB

1 1 0

y1 y2 y3

cj xB x1 x2 x3 zB

=0 =2 ¼1 ¼2

177

1 y1

1 y2

0 y3

0 y4

0 g1

1 0 0 0

0 1 0 0

0 0 1 0

−1 1 1 0

1 0 −3 1

Dj

Since all Dj  0 and all xBi  0, we obtain an optimal feasible integer solution. The optimal integer solution is x1 ¼ 0; x2 ¼ 2 and Max z ¼ 0 þ 2 ¼ 2. Example 2 Find the optimum integer solution to the following LPP: Maximize z ¼ 2x1 þ 1:7x2 subject to 4x1 þ 3x2  7 x1 þ x2  4 x1 ; x2  0 and are integers: Solution: The given problem is a maximization problem. Now introducing two slack variables x3 ð  0Þ and x4 ð  0Þ, the standard form of the given LPP becomes Maximize z ¼ 2x1 þ 1:7x2 þ 0:x3 þ 0:x4 subject to 4x1 þ 3x2 þ x3 ¼ 7 x1 þ x2 þ x4 ¼ 4; x1 ; x2 ; x3 ; x4  0 and are integers:

Now, ignoring the integer condition, we shall solve the problem by the simplex method.   1 0 Here the initial BFS is x3 = 7, x4 = 4 with the initial basis as B ¼ . 0 1 Now we construct the initial simplex table.

cB

yB

0 0

y3 y4

cj xB

2 y1

1.7 y2

0 y3

0 y4

Minimum ratio

x3 = 7 x4 = 4 zB ¼ 0

4 1 −2 "

3 1 −1.7

1 0 0 #

0 1 0

7/4 = 1.75 4 Dj

178

8 Integer Programming

yB

cj xB

2 y1

1.7 y2

0 y3

0 y4

Minimum ratio

cB 2 0

y1 y4

x1 = 7/4 x4 = 9/4

1 0 0

3/4 1/4 −1/5

1/4 −1/4 1/2

0 1 0

7/3 9

#

"

zB ¼ 72

Dj

cB

yB

cj xB

2 y1

1.7 y2

0 y3

0 y4

1.7 0

y2 y4

x2 = 7/3 x4 = 5/3

4/3 −1/3 0.267

1 0 0

1/3 −1/3 1.7/3

0 1 0

zB ¼ 11:9 3

Since all D j  0, an optimum solution is obtained, given by x2 ¼ 73 ; x4 ¼ 53. The optimal solution is not an integer solution because x2 ¼ 73 ; x4 ¼ 53, so we have to add a fractional cut constraint or Gomorian constraint in the optimal simplex table. Since xB1 ¼ 73 ¼ 2 þ 13 and xB2 ¼ 53 ¼ 1 þ 23, we have fB1 ¼ 13 and fB2 ¼ 23. Hence,   Max fB1 ; fB2 ¼ 23 ¼ fB2 , which corresponds to the second row. So, the second row is the source row. Now, from the second row of the optimal simplex table, we get x1 þ

2 2 2 x1  x3 þ x3 þ x4 ¼ 1 þ : 3 3 3

Hence, the Gomorian constraint will be

or or

2 2 2 x1 þ x3  3 3 3 2 2 2  x1  x3   3 3 3 2 2 2  x1  x3 þ G1 ¼  ; 3 3 3

where G1 is the Gomorian slack variable. Adding this new constraint at the bottom of the preceding optimal simplex table, we have used the dual simplex method to solve the problem.

8.4 Construction of Gomory’s Constraint

179

cB

yB

cj xB

2 y1

1.7 y2

0 y3

0 y4

0 g1

1.7 0 0

y2 y4 g1

x2 = 7/3 x4 = 5/3 G1 = −2/3

4/3 −1/3 −2/3 0.267

1 0 0 0

1/3 −1/3 −2/3 1.7/3

0 1 0 0

0 0 1 0

zB ¼ 11:9 3

Dj

  We have xBr ¼ Min ½xBi ; xBi \0 ¼ Min  23 ¼  23 ¼ xB3 . Hence r ¼ 3. So yB3 i.e. g1 leaves the basis. Now, 

Dj Max ; yrj \0 yrj

(

0:267 1:7 ¼ Max ; 3  23  23

)

¼ Maxf0:801=2; 1:7=2g ¼ 0:801=2; which corresponds to y1. Hence y1 is the entering vector, and y31 ¼  23 is the key element. Now we construct the next simplex table.

cB

yB

1.7 0 2

y2 y4 y1

cj xB x2 x4 x1 zB

=1 =2 =1 ¼ 3:7

2 y1

1.7 y2

0 y3

0 y4

0 g1

0 0 1 0

1 0 0 0

−1 0 1 0.3

0 1 0 0

2 −1/2 −3/2 0.4

Dj

Since all Dj  0 and all xBi  0, we obtain an optimal feasible integer solution. The optimal integer solution is x1 ¼ 1; x2 ¼ 1 and Max z ¼ 2 þ 1:7 ¼ 3:7. Example 3 Find the optimum integer solution to the following LPP: Maximize z ¼ 7x1 þ 6x2 subject to

2x1 þ 3x2  12 6x1 þ 5x2  30 x1 ; x2  0 and are integers:

Solution: The given problem is a maximization problem. Now introducing two slack variables x3 ð  0Þ and x4 ð  0Þ, the standard form of the given LPP becomes

180

8 Integer Programming

Maximize z ¼ 7x1 þ 6x2 þ 0:x3 þ 0:x4 2x1 þ 3x2 þ x3 ¼ 12 6x1 þ 5x2 þ x4 ¼ 30; x1 ; x2 ; x3 ; x4  0 and are integers:

subject to

Now, ignoring the integer condition, we shall solve the problem by the simplex method.   1 0 Here the initial BFS is x3 = 12, x4 = 30 with the initial basis as B ¼ . 0 1 Now we construct the initial simplex table.

cB

yB

0 0

y3 y4

cB

yB

0 7

y3 y1

cB

yB

6 7

y2 y1

cj xB

7 y1

6 y2

0 y3

0 y4

Minimum ratio

x3 = 12 x4 = 30 zB ¼ 0

2 6 −7 "

3 5 −6

1 0 0

0 1 0 #

12/2 = 6 30/6 = 5 Dj

cj xB

7 y1

6 y2

0 y3

0 y4

Minimum ratio

x3 = 2 x1 = 5 zB ¼ 35

0 1 0

4/3 5/6 −1/6 "

1 0 0 #

−1/3 1/6 7/6

3/2 6 Dj

cj xB

7 y1

6 y2

0 y3

0 y4

x2 = 3/2 x1 = 15/4 zB ¼ 141=4

0 1 0

1 0 0

3/4 −5/8 1/8

−1/4 3/8 9/8

Since all Dj  0, an optimum solution is obtained, given by x2 ¼ 32 ; x1 ¼ 15 4 . The optimal solution is not an integer solution because x2 ¼ 32 ; x1 ¼ 15 , so we have to 4 add a fractional cut constraint or Gomorian constraint in the optimal simplex table. 3 1 3 Since xB1 ¼ 32 ¼ 1 þ 12 and xB2 ¼ 15 4 ¼ 3 þ 4, we have fB1 ¼ 2 and fB2 ¼ 4. Hence   3 Max fB1 ; fB2 ¼ 4 ¼ fB2 , which corresponds to the second row. So, the second row is the source row.

8.4 Construction of Gomory’s Constraint

181

Now, from the second row of the optimal simplex table, we get x1  x3 þ

3 3 3 x3 þ x4 ¼ 3 þ : 8 8 4

Hence, the Gomorian constraint will be 3 3 3 x3 þ x4  8 8 4 3 3 3  x3  x4   8 8 4 3 3 3  x3  x4 þ G1 ¼  ; 8 8 4

or or

where G1 is the Gomorian slack variable. Adding this new constraint at the bottom of the preceding optimal simplex table, we have used the dual simplex method to solve the problem.

cB

yB

6 7 0

y2 y1 g1

cj xB

7 y1

6 y2

0 y3

0 y4

0 g1

x2 = 3/2 x1 = 15/4 G1 = −3/4 zB ¼ 141=4

0 1 0 0

1 0 0 0

3/4 −5/8 −3/8 1/8

−1/4 3/8 −3/8 9/8

0 0 1 0

Dj

  We have xBr ¼ Min½xBi ; xBi \0 ¼ Min  34 ¼  34 ¼ xB3 . Hence, r ¼ 3. So yB3 i.e. g1 leaves the basis. Now, 

Dj Max ; yrj \0 yrj

(

)

1=8 9=8 ¼ Max ;  38  38 ¼ Maxf1=3; 3g ¼ 1=3;

which corresponds to y3. Hence y3 is the entering vector, and y33 ¼  38 is the key element. Now we construct the next simplex table. cj

2

1.7

0

0

0

cB

yB

xB

y1

y2

y3

y4

g1

6

y2

x2 = 0

0

1

0

−1

2

7

y1

x1 = 5

1

0

0

1

−5/3

0

y3

x3 =4

0

0

1

1

−8/3

zB ¼ 35

0

0

0

1

1/3

Dj

182

8 Integer Programming

Since all D j  0 and all x Bi  0, we obtain an optimal feasible integer solution. The optimal integer solution is x1 ¼ 5; x2 ¼ 0 and Max z ¼ 5  7 þ 6  0 ¼ 35.

8.5

Gomory’s Mixed-Integer Cutting-Plane Method

In all-integer programming problems, the fractional cut or Gomory’s constraint is formed under the assumption that all the variables including slack and surplus variables are integers. This means that the application of the fractional cut will not yield a feasible integer solution unless all variables assume integer values. The importance of this assumption is discussed in the following constraint: x1 þ

1 13 x2  ; x1 ; x2  0 and are integers: 3 2

From the stand point of solving the associated integer linear programming program, the constraint is converted to an equation by introducing a non-negative slack variable G1, i.e. x1 þ

1 13 x2 þ G 1 ¼ : 3 2

From this equation, it is clear that it can have a feasible integer solution in x1 and x2 only if G1 is non-integer. Thus, the application of the fractional cut will yield no feasible integer solution because all the variables x1, x2 and G1 cannot be integers simultaneously. To avoid this situation, a special cut, called Gomory’s mixed-integer cut, is developed. For this cut, only a subset of variables is assumed to take integer values, and the remaining variables including slack and surplus variables are assumed as non-integer variables.

8.6

Construction of Additional Constraint for Mixed-Integer Programming Problems

Let us consider the following mixed-integer programming problem: Maximize z ¼ subject to

x X j¼1 n X

cj xj aij xj ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

and xj  0;

j ¼ 1; 2; . . .; n

where xj are integers for j ¼ 1; 2; . . .; kð\nÞ:

8.6 Construction of Additional Constraint for Mixed …

183

Let us suppose that the basic variable xr has the largest fractional value among the values of all other basic variables which are restricted to take integer values in the optimal simplex table (which is developed ignoring the integer restrictions of variables). Then, from the optimal simplex table, the rth constraint can be written as follows: X xBr ¼ xr þ arj xj ð8:4Þ j2R

where R is the set of indices corresponding to non-basic variables. Let xBr ¼ ½xBr  þ fr ; 0\fr \1

and j þ ¼ j : arj  0 ; j ¼ j : arj \0 . Then Eq. (8.4) can be rewritten as X X ½xBr  þ fr ¼ xr þ arj xj þ arj xj or

X

X

arj xj þ

j2j

j2j þ

arj xj ¼ ½xBr   xr þ fr ¼ I þ fr

ð8:5Þ

j2j

j2j þ

where 0\fr \1 and I is the integer value. Clearly, the left-hand side of (8.5) is either positive or negative according to whether I þ fr is positive or negative respectively. Case I: Let I þ fr be positive. In this case, I þ fr must be greater than or equal to fr, i:e: I þ fr  fr : Hence, from (8.5), we have

X

arj xj þ

X

arj xj  fr

ð8:6Þ

j2j

j2j þ

Since for j 2 j ; arj are non-positive and xj  0, then X j2j þ

i:e:

arj xj 

X

X j2j þ

arj xj  fr

arj xj þ

X j2j

½by (8:6Þ

arj xj ; ð8:7Þ

j2j þ

Case II: Let I + fr be negative. In this case, the value of I + fr will be either −1 + fr or −2 + fr or −3 + fr or …

184

8 Integer Programming

Therefore, X

arj xj 

j2j

or

X

X

arj xj þ

X

arj xj   1 þ fr

j2j

j2j þ

arj xj   1 þ fr

j2j

or or

1 X arj xj  1 ½Dividing both sides by  1 þ fr  1 þ fr j2j fr X arj xj  fr ½Multiplying both sides by fr  1 þ fr j2j

ð8:8Þ

In both cases, the left-hand sides of inequalities (8.7) and (8.8) are positive and greater than or equal to fr. Thus, any feasible solution of mixed-integer programming must satisfy the following inequality: X

arj xj þ

j2j þ

fr X arj xj  fr 1 þ fr j2j

ð8:9Þ

This inequality is not satisﬁed by the optimal solution of the LPP without the integer requirement, because by setting xj ¼ 0 8 j, the left-hand side becomes zero, whereas the right-hand side becomes positive. Now, introducing a non-negative slack variable in (8.9), we get the following equation: 

X j2j þ

arj xj 

fr X arj xj þ G1 ¼ fr 1 þ fr j2j

ð8:10Þ

This equation represents Gomory’s constraint or Gomory’s cut. Example 4 Solve the following mixed-integer programming problem: Maximize z ¼ 7x1 þ 9x2 subject to  x1 þ 3x2  6 7x1 þ x2  35; x1 ; x2  0 and x1 is an integer:

Solution: Ignoring the integer restriction, we get the optimal simplex table as follows:

8.6 Construction of Additional Constraint for Mixed …

185

cB

yB

cj xB

7 y1

9 y2

0 y3

0 y4

9 7 zB = 63

y2 y1

7/2 9/2

0 1 0

1 0 0

7/22 −1/22 28/11

1/22 3/22 15/11

From the preceding table, it is observed that the optimal value of x1 is not an integer. Hence from the second row, we have

or

9 1 3 ¼ x1  x3 þ x4 22 2  22 1 1 3 x4 : 4þ ¼ x1  x3 þ 2 22 22

Here r = 2, fBr ¼ 12. Hence, the corresponding fractional cut is X fBr X arj xj  arj xj þ G1 ¼ f Br ;  1 þ fBr j2j j2j þ

where j þ ¼ j : arj  0 and j ¼ j : arj \0 and xj are non-basic variables in the optimal simplex table   1 3 1 1 or  x4  2 1  x3 þ G1 ¼  22 22 2 1 þ 2 3 1 1 or  x4  x3 þ G1 ¼  : 22 22 2 Adding this new constraint at bottom of the preceding simplex table, we have used the dual simplex method to solve the problem.

cB

yB

9 7 0

y2 y1 g1 zB = 63

cj xB

7 y1

9 y2

0 y3

0 y4

0 g1

7/2 9/2 −1/2

0 1 0 0

1 0 0 0

7/22 −1/22 −1/22 28/11

1/22 3/22 −3/22 15/11

0 0 1 0

  Since xBr ¼ Min½xBi ; xBi \0 ¼ Min  12 ¼ xB3 , then B3 or g1 leaves the basis. n o n o D 28=11 15=11 Now Max y3jj ; y3j \0 ¼ Max 1=22 ; 3=22 ¼ Maxf56; 10g ¼ 10, which corresponds to y4.

186

8 Integer Programming

Hence, y4 enters the basis and g1 leaves the basis.

cB

yB

9 7 0

y2 y1 y4 zB = 58

cj xB

7 y1

9 y2

0 y3

0 y4

0 g1

10/3 4 11/3

0 1 0 0

1 0 0 0

10/33 −1/11 1/3 23/11

0 0 1 0

1/3 1 −22/3 10

The required optimum solution is x1 ¼ 4; x2 ¼ 10 3 and Max z ¼ 58 with x1 an integer. Example 5 Solve the following mixed-integer programming problem: Maximize z ¼ x1 þ x2 subject to 2x1 þ 5x2  16 6x1 þ 5x2  30; x2  0 and x1 is a non-negative integer: Solution: Ignoring the integer restriction, we get the optimal simplex table as follows: cB

yB

1 y2 1 y1 zB = 53/10

cj xB

1 y1

1 y2

0 y3

0 y4

9/5 7/2

0 1 0

1 0 0

3/10 −1/4 1/20

−1/10 1/4 3/20

From the preceding table, it is observed that the optimal value of x1 is not an integer. Hence from the second row, we have

or

7 1 1 ¼ x1 þ 0:x2  x3 þ x4 2 4 4   1 1 1 3þ ¼ x1  x3 þ x4 : 2 4 4

Here r = 2, fBr ¼ 12. Hence, the corresponding fractional cut is 

X j2j þ

arj xj 

fBr X arj xj þ G1 ¼ fBr ; 1 þ fBr j2j

8.6 Construction of Additional Constraint for Mixed …

187

where j þ ¼ j : arj  0 and j ¼ j : arj \0 and xj are non-basic variables in the optimal simplex table.   1 1 1 1 2  x4   x3 þ G 1 ¼  4 4 2 1 þ 12 1 1 1 or  x4  x3 þ G1 ¼  : 4 4 2 Adding this new constraint at bottom of the preceding simplex table,

or

we have used the dual simplex method to solve the problem. cj xB

cB

yB

1 1 0

y2 9/5 7/2 y1 −1/2 g1 zB = 53/10

1 y1

1 y2

0 y3

0 y4

0 g1

0 1 0 0

1 0 0 0

3/10 −1/4 −1/4 1/20

−1/10 1/4 −1/4 3/20

0 0 1 0

  Since xBr ¼ Min½xBi ; xBi \0 ¼ Min  12 ¼  12 ¼ xB3 then yB3 i.e. g1 leaves the basis. n o n o D 1=20 3=20 ; 1=4 ¼ Maxf1=5; 3=5g ¼ 1=5 which Now, Max y3jj ; y3j \0 ¼ Max 1=4 corresponds to y3. Hence, y3 enters the basis and g1 leaves the basis. cj xB

cB

yB

1 1 0

y2 6/5 4 y1 2 y3 zB = 26/5

1 y1

1 y2

0 y3

0 y4

0 g1

0 1 0 0

1 0 0 0

0 0 1 0

−4/10 1/2 1 1/10

6/5 −1 −4 1/5

The required optimum solution is x1 ¼ 4; x2 ¼ 65 and Max z ¼ 26=5 with x1 as integer.

8.7

Branch and Bound Method

This method was ﬁrst developed by A. H. Land and A. G. Doig, and it was further studied by J. D. C. Little and other researchers. This method can be used to solve all-integer, mixed-integer and zero-one LPPs. The concept behind this method is to divide the entire feasible region of the LPP into smaller parts by forming (or branching) subproblems and then examine each

188

8 Integer Programming

successively until a feasible solution which gives the optimal solution is obtained. Here, we shall discuss its application to all-integer programming problems. Branch and Bound (B & B) Algorithm The iterative procedure of this method is summarized here for solving an integer programming problem (maximization case): Let us consider the following all-integer programming problem (AIPP): Maximize z ¼

n X

cj xj

j¼1

subject to

n X

aij xj ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

xj  0 ðj ¼ 1; 2; . . .; nÞ are integers: Step 1 Ignoring the integer restrictions of the variables, obtain the optimal solution of the given LPP (we denote this LPP as LP-1). Step 2 Test the integrality of the optimum solution obtained in Step 1. There may arise different possibilities. (i) If the solution of LP-1 is infeasible or unbounded, then the solution to the given AIPP is also infeasible or unbounded respectively. Then stop the process. (ii) If the optimum solution of LP-1 satisﬁes the integer restrictions, then the current solution is optimum to the given AIPP. Stop the process. (iii) If one or more basic variables of LP-1 do not satisfy the integer restrictions, then go to Step 3. Let the optimal value of the objective function of LP-1 be z1. This value provides an initial upper bound. Step 3 Branching step (i) The value of z1 will be considered as the initial upper bound of the objective function value for the given AIPP. Let it be denoted by zU. The lower bound of the objective function of the given integer LPP can be obtained by rounding off to integer all values of the variables. (ii) Let the optimum value xk of the basic variable xk not be an integer. This selection can be done by considering the largest fractional value of the basic variables whose values are not integer, or it can be done arbitrarily. (iii) Branch or partition the problem LP-1 into two new subproblems (called nodes) based on the integer value of xk. This can be done by adding two mutually exclusive constraints

8.7 Branch and Bound Method

189

    xk  xk and xk  xk þ 1   to the LPP LP-1, where xk is the largest integer value not greater than xk . In this case, the two new LPPs (LP-2 and LP-3) are as follows: n X cj xj LP-2: Maximize z ¼ j¼1

subject to

n X

aij xj ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

  xk [ xk þ 1 and xj  0; j ¼ 1; 2; . . .; n: n X LP-3: Maximize z ¼ cj xj j¼1

subject to   xk  xk þ 1 and xj  0;

n X

aij xj ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

j ¼ 1; 2; . . .; n:

Step 4 Bound step Solve the two subproblems LP-2 and LP-3 obtained in Step 3. Let the optimal value of the objective function of LP-2 be z2 and that of LP-3 be z3. For the best integer solution (if it exists), the objective function value becomes the lower bound of the IPP objective function value (initially this is the rounding off value). Let the lower bound be denoted by zL. Step 5 Fathoming step Examine the solutions of both LP-2 and LP-3, which contain the optimal solution. There may arise different cases: (i) If the optimal solutions of the two subproblems are integral, then the required solution is the one that gives the larger value of z. (ii) If the optimal solution of one subproblem is integral and the other subproblem has no feasible solution, the required solution is the same as that of the subproblem having an integer-valued solution. (iii) If the optimal solution of one-subproblem is integral while that of the other is not integer-valued, then repeat Steps 3 and 4 for the non-integer-valued subproblem. When the solution of any subproblem is either integer-valued or infeasible, then the subproblem or node is said to be fathomed. If the solution of any subproblem is infeasible, then it is said to be fathomed by infeasibility. If zL  zU , it is said to be fathomed by bounding.

190

8 Integer Programming

If zL is obtained as an integer feasible solution in any subproblem and zL \zU , it is said to be fathomed by integrality. Step 6 Repeat Steps 3 through 5 until all integer-valued solutions are recorded. Step 7 Choose the solution among the recorded integer-valued solutions that yields an optimum value of z. Step 8 Stop. Now, we shall illustrate the B & B algorithm by the following numerical example. Example 6 Solve the following IPP by the branch and bound method: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1 ; x2  0 and they are integers:

Solution: Let us consider the following IPP: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1 ; x2  0 and are integers: Ignoring the integer restrictions of the variables x1 and x2, the associated problem of the given IPP is denoted by LP-1 as follows: Maximize z ¼ 5x1 þ 4x2 ðLP-1Þ subject to x1 þ x2  5; 10x1 þ 6x2  45; x1 ; x2  0 Its optimum solution is x1 ¼ 3:75; x2 ¼ 1:25 and z = 23.75 (see Fig. 8.1). The optimum solution of LP-1 does not satisfy the integer requirements. Hence, this solution is not optimal for the given IPP. Hence, the initial upper bound of the objective function to the given problem is given by zu ¼ 23:75. Again, the initial lower bound can also be obtained by rounding of the values of x1 and x2 to the nearest integer, i.e. x1 ¼ 4, x2 ¼ 1, which does not satisfy the second constraint. Now, we ﬁrst select one of the integer variables whose value at LP-1 is not integer.

8.7 Branch and Bound Method

191

We select x1 ð¼ 3:75Þ as it contains the largest fractional part. The region 3\x1 \4 of the LP-1 solution space does not contain any integer value. Hence, we can eliminate this region. So, we can write two new restrictions x1  3 and x1  4 which are mutually exclusive. As a result, we get two new LPPs, LP-2 and LP-3 as follows: LP-2: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1  3 and x1 ; x2  0 ½Here LP-2 solution space = LP-1 solution space + ðx  3Þ The optimal solution of LP-2 is given by x1 ¼ 3, x2 ¼ 2 and z ¼ 23 (Fig. 8.2). LP-3: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1  4 and x1 ; x2  0 x2

Fig. 8.1 Graphical representation of LP-1 8

C (0, 5)

6

4

2

B(3.75, 1.25)

O

x1 2

4

6

A (4.5, 0)

192

8 Integer Programming

½Here LP-3 solution space ¼ LP-1 solution space þ ðx1  4Þ The optimal solution of LP-3 is given by x1 ¼ 4, x2 ¼ 0:83 and z ¼ 23:33. Since this solution does not satisfy the integer restriction, we can write two new constraints, x2  ½0:83 and x2  ½0:83 þ 1, i.e. x2  0 and x2  1. These constraints are mutually exclusive. As a result, we get two new LPPs, LP-4 and LP-5 as follows: LP4: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1  4; x2  0: ½Here; LP-4 solution space ¼ LP-3 solution space þ ðx2  0Þ ¼ LP-1 solution space þ ðx1 [ 4Þ þ ðx2  0Þ The optimal solution of LP-4 is x1 ¼ 4:5, x2 ¼ 0 and z ¼ 22:5. LP5: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1  4; x2  1:

x2

Fig. 8.2 Graphical representation of LP-2 & LP-3

8

C(0, 5)

6

4

D 2

B

O

2

E

4

G

A

x1 6

8.7 Branch and Bound Method

193

½Here LP-5 solution space ¼ LP-3 solution space þ ðx2  1Þ ¼ LP-1 solution space þ ðx1  4Þ þ ðx2  1Þ LP-5 has no feasible solution. Since the solution of LP-4 does not satisfy the integer restriction, we can write two new mutually exclusive constraints x1=5. As a result, we get two new LPPs, LP-6 and LP-7 as follows: LP-6: Maximize z ¼ 5x1 þ 4x2 subject to x1 þ x2  5 10x1 þ 6x2  45 x1  4; x2  0 and x1 ; x2  0:

½Here LP-6 solution space ¼ LP-4 solution space þ ðx1  4Þ ¼ LP-1 solution space þ ðx1  4Þ þ ðx2  0Þ þ ðx1  4Þ The optimal solution of LP-6 is x1 ¼ 4, x2 ¼ 0 and z ¼ 20. LP-7: Maximize z ¼ 5x1 þ 4x2 subject to

x1 þ x2  5

10x1 þ 6x2  45 x1  5; x2  0 ½Here LP-7 solution space ¼ LP-4 solution space þ ðx  5Þ LP-7 has no feasible solution. Hence, the optimal solution of the given AIPP is x1 ¼ 3; x2 ¼ 2 and Max z ¼ 23.

8.8

Exercises

1. Find the optimum integer solution to the following all-integer programming problems: (i)

Maximize z ¼ x1 þ 2x2 subject to 2x2  7; x1 þ x2  7; 2x1  11 x1  0; x2  0 and they are integers:

194

8 Integer Programming

(ii)

Maximize z ¼ x1 þ 5x2 subject to x1 þ 5x2  20;

(iii)

x1  2 x1  0; x2  0 and they are integers: Maximize z ¼ x1 þ x2 subject to 3x1  2x2  5; x1  2 x1  0; x2  0 and they are integers: Maximize z ¼ 3x1 þ 4x2 subject to

(iv)

(v)

(vi)

3x1 þ 2x2  8;

x1 þ 4x2  10 x1  0; x2  0 and they are integers:

Answer: x1 ¼ 0; 2 ¼ 4; Max z ¼ 16 Maximize z ¼ 2x1 þ 2x2 subject to 5x1 þ 3x2  8 x1 þ 2x2  4 x1  0; x2  0 and they are integers: Maximize z ¼ x1  x2 subject to x1 þ 2x2  4; 6x1 þ 2x2  9 x1  0; x2  0 and they are integers:

2. Find the optimum integer solution to the following mixed-integer programming problems. (i)

Maximize z ¼ 1:5x1 þ 3x2 þ 4x3 subject to 2:5x1 þ 2x2 þ 4x3  12 2x1 þ 4x2  x3  7

(ii)

(iii)

x1 ; x2 ; x3  0 and x3 is an integer: Maximize z ¼ 2x1 þ 6x2 subject to x1 þ x2  5  x1 þ 2x2  6 x1 ; x2  0 and x2 is an integer: Maximize z ¼ x1 þ 6x2 subject to 3x1 þ 2x2  5 x2  2 x1 ; x2  0 and x1 is an integer:

8.8 Exercises

195

3. Use the branch and bound method to solve the following all-integer programming problems. (i)

(ii)

Maximize z ¼ x1 þ x2 subject to 4x1  x2  10 2x1 þ 5x2  10 x1  0; x2  0 and they are integers: Maximize z ¼ 7x1 þ 9x2 subject to

(iii)

 x1 þ 3x2  6

7x1 þ x2  35 0  x1 ; x2  7 and they are integers: Maximize z ¼ 21x1 þ 11x2 subject to 7x1 þ 4x2 þ x3 ¼ 13 x2  5

(iv)

x1 ; x2 ; x3  0 and they are integers: Maximize z ¼ 4x1 þ 3x2 subject to x1 þ 2x2  4 2x1 þ x2  6 x1  0; x2  0 and they are integers:

Chapter 9

Basics of Unconstrained Optimization

9.1

Objectives

The objectives of this chapter are to: • Discuss different techniques for solving unconstrained optimization problems of a single variable • Study the necessary and sufﬁcient conditions for optimizing a function of several variables.

9.2

Optimization of Functions of a Single Variable

Let I  R be any interval, ﬁnite or inﬁnite, and let f: I ! R be a real-valued function. Deﬁnition (i) x 2 I is said to be a local minimum or minimizer of f if 9 a d [ 0 such that f ðx Þ  f ð xÞ, 8x 2 ðx  d; x þ dÞ. (ii) x 2 I is said to be a global minimum or minimizer of f if f ðx Þ  f ð xÞ; 8x 2 I.

Necessary condition for optimality The condition for optimality of f (x) at the point x = x* is f 0 ðx Þ ¼ 0. This is a necessary condition but not sufﬁcient. (i) For example, if f ð xÞ ¼ x3 , then at x = 0, f 0 ð0Þ ¼ 0. But when x > 0, f (x) > f (0) and f (x) < f (0) for x < 0. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_9

197

198

9 Basics of Unconstrained Optimization

Here f (x) has no extreme value at x = 0. (ii) Even when f 0 ðx Þ does not exist, f ðx Þ may be optimum. For example, if f ð xÞ ¼ j xj, then obviously f (0) is a minimum value of f (x), but f 0 ð0Þ does not exist. Sufﬁcient condition for optimality The sufﬁcient condition for optimality of f(x) is as follows: If at x = x* for the function f (x): (i) (ii) (a) (b)

f 0 ðx Þ ¼ f 00 ðx Þ ¼    ¼ f n1 ðx Þ ¼ 0 f n ðx Þ exists and 6¼ 0 then at x = x*, f (x) has no extreme value if n is odd an extreme value if n is even, i.e. a maximum if f n ðx Þ\0, minimum if f n ð x Þ [ 0

Alternative approach Theorem If (i) f (x) is deﬁned at x = x*. (ii) f 0 ð xÞ exists in a suitably small neighbourhood of 0\jx  x j\d [ f 0 ð xÞ may or may not exist]. (iii) f 0 ð xÞ has a ﬁxed sign when x < x*, also a unique sign when x < x*. Then with the increase of x through x*, (a) f (x) has no extreme value at x = x* if the sign of f 0 ð xÞ does not change. (b) f (x) has a maximum value at x = x* if the sign of f 0 ð xÞ changes from positive to negative. (c) f (x) has a minimum value at x = x* if the sign of f 0 ð xÞ changes from negative to positive. Note: This method does not require higher order derivative information or the derivative information at the point where the optimality is tested. Example 1 Test for optimality at x = 0 for f ð xÞ ¼ sin x  x þ

x3 3!

Solution Here 3

5

f ð xÞ ¼ sin x  x þ x3!  x5! 2 4 f 0 ð xÞ ¼ cos x  1 þ x2!  x4! ; 3 f 00 ð xÞ ¼  sin x þ x  x3! ; 2 f 000 ð xÞ ¼  cos x þ 1  x2! ; f iv ð xÞ ¼ sin x  x; f v ð xÞ ¼ cos x  1; f vi ð xÞ ¼  sin x; f vii ð xÞ ¼  cos x;

5

 x5! .

) f 0 ð 0Þ ¼ 0 ) f 00 ð0Þ ¼ 0 ) f 000 ð0Þ ¼ 0 ) f iv ð0Þ ¼ 0 ) f v ð 0Þ ¼ 0 ) f vi ð0Þ ¼ 0 ) f vii ð0Þ ¼ 1 6¼ 0:

9.2 Optimization of Functions of a Single Variable

199

Since the order of the ﬁrst non-zero derivative is odd, then f (x) is neither maximum nor minimum at x = 0. Example 2 If f 0 ð xÞ ¼ ðx  aÞ2m ðx  bÞ2n þ 1 , where m and n are positive integers, show that x ¼ a gives neither a maximum nor a minimum value of f (x). But x = b gives a minimum. Solution Let h be any positive number, however small. Now; f 0 ða  hÞ ¼ ða  h  aÞ2m ða  h  bÞ2n þ 1 ¼ ðhÞ2m ða  b  hÞ2n þ 1 ¼ h2m ða  b  hÞ2n þ 1 and f 0 ða þ hÞ ¼ ða þ h  aÞ2m ða þ h  bÞ2n þ 1 ¼ h2m ða  b þ hÞ2n þ 1 : Since h is a very small quantity, ða  b  hÞ2n þ 1 and ða  b þ hÞ2n þ 1 have the same sign. ) f 0 ða  hÞ and f 0 ða þ hÞ have the same sign. ) f (x) is neither maximum nor minimum at x = a. Again, f 0 ðb  hÞ ¼ ðb  h  aÞ2m ðb  h  bÞ2n þ 1 ¼ ðb  a  hÞ2m ðhÞ2n þ 1 ¼ h2n þ 1 ðb  a  hÞ2m f 0 ðb þ hÞ ¼ ðb þ h  aÞ2m ðb þ h  bÞ2n þ 1 ¼ h2n þ 1 ðb  a þ hÞ2m : Since h is a very small quantity, ðb  a þ hÞ2m and ðb  a  hÞ2m have the same sign. ) f 0 ðb  hÞ\0 and f 0 ðb þ hÞ [ 0; i.e. f 0 ð xÞ changes its sign from negative to positive. Therefore, f (x) is minimum at x = b. Example 3 If f ð xÞ ¼ j xj, show that f (x) is minimum at x = 0. Solution Since f ð xÞ ¼ j xj

) f ð0Þ ¼ 0.

Let h be any positive number, however small.

200

9 Basics of Unconstrained Optimization

Now; f ð0  hÞ ¼ f ðhÞ ¼ jhj ¼ h [ 0 ¼ f ð0Þ ) f ð0  hÞ [ f ð0Þ: Again; f ð0 þ hÞ ¼ f ðhÞ ¼ jhj ¼ h [ 0 ¼ f ð0Þ ) f ð0 þ hÞ [ f ð0Þ: Hence, f (x) has a minimum at x = 0. 1

2

Example 4 Show that ðx  aÞ3 ð2x  aÞ3 is maximum for x ¼ a2 and neither maximum nor minimum for x = a (a > 0).  1 2 Solution Let f ð xÞ ¼ ðx  aÞ3 ð2x  aÞ3 ) f a2 ¼ 0 and f ðaÞ ¼ 0. Again, let h be any positive number, however small. 1

2

) f ða  hÞ ¼ ða  h  aÞ3 ð2a  2h  aÞ3 1

2

1

2

¼ ðhÞ3 ða  2hÞ3 ¼ h3 ða  2hÞ3 \0 ¼ f ðaÞ 1

2

f ða þ hÞ ¼ ða þ h  aÞ3 ð2a þ 2h  aÞ3 2

1

¼ h3 ða þ 2hÞ3 [ 0 ¼ f ðaÞ ) f ða  hÞ\f ðaÞ\f ða þ hÞ: Hence, f(x) is neither maximum nor minimum for x = a (a > 0). Again; f

Now; f

a

 a 13 h a  i23 þh ¼ þh  a 2 þh  a 2 2 2 n a o13 2 ¼  h ð2hÞ3 2 a 13  a 2 ¼   h ð2hÞ3 \0 ¼ f : 2 2

a  a 13 h a  i23 h ¼ ha 2 h a 2 2 2   2 1 2 a13 a13 ¼ h  ða  2h  aÞ3 ¼ ð1Þ3 h þ ð2hÞ3 2 2 a 13 2 ¼  þ h ð2hÞ3 2   13 2 a þ h [ 0 and ð2hÞ3 [ 0 as h [ 0 ¼ negative * 2 a   a ) f  h \0 ¼ f : 2 2

) f (x) is maximum at x ¼ a2.

9.3 Optimization of Functions of Several Variables

9.3

201

Optimization of Functions of Several Variables

Here we consider the optimization problem f ðxÞ x 2 X;

Minimize subject to

ð9:1Þ

where the function f : Rn ! R is a real-valued function, called the objective function. The vector x is an n-tuple vector of independent variables x1 ; x2 ; . . .; xn : The set X is a subset of Rn , called the feasible set. There are also optimization problems that require maximization of the objective function, in which case we seek maximizers. Maximization problems, however, can be represented equivalently in the minimization form above because maximizing f is equivalent to minimizing f . Therefore, we can conﬁne our attention to minimization problems without loss of generality. Deﬁnition Let f : X  Rn ! R be a real-valued function. A point x 2 X is said to be a: (i) Local minimizer of f if 9 a d [ 0 such that f ðx Þ  f ð xÞ; 8x 2 X \ Bðx ; dÞ, where Bðx ; dÞ is an open ball whose centre is at x and radius d (ii) Global minimizer of f if f ðx Þ  f ð xÞ; 8x 2 X. If strict inequality holds in the preceding deﬁnition, x is said to be a strict local minimizer/global minimizer. Necessary condition for optimality Theorem The necessary condition for the continuous function f (x) to have an extreme point at x ¼ x is that the gradient rf ðx Þ ¼ 0, i:e:

@f ðx Þ @f ðx Þ @f ðx Þ ¼ ¼  ¼ ¼0 @x1 @x2 @xn

ð9:2Þ

Proof From the Taylor series expansion, we have f ðx þ hÞ ¼ f ð xÞ þ hh; rf ð xÞi þ

1 hH ðx þ hhÞh; hi 2!

0\h\1

Setting x ¼ x in this expression, we have 1 hH ðx þ hhÞh; hi 2! n X @f ðx Þ 1 ) f ð x  þ hÞ  f ð x  Þ ¼ hi þ hH ðx þ hhÞh; hi: @x 2! i i¼1

f ðx þ hÞ ¼ f ðx Þ þ hh; rf ðx Þi þ

202

9 Basics of Unconstrained Optimization

Since the term hH ðx þ hhÞh; hi is of order h2i , so the term of order h only will dominate the higher order terms for small h.  Thus, the sign of f ðx þ hÞ  f ðx Þ is decided by the sign of hi @f@xðxi Þ. 

Let us suppose that @f@xðxi Þ [ 0; then the sign of f ðx þ hÞ  f ðx Þ will be positive for hi > 0 (i = 1, 2, …, n), i.e. f ðx þ hÞ [ f ðx Þ and negative for hi < 0 (i = 1, 2, …, n), i.e. f ðx þ hÞ\f ðx Þ.

This means that x cannot be an extreme point. The same conclusion can be  obtained if we assume that @f@xðxi Þ \0. These two conclusions contradict our original statement that x is an extreme point. Hence, we conclude that @f ðx Þ ¼ 0; i ¼ 1; 2; . . .; n; @xi i:e: rf ðx Þ ¼ 0:

Sufﬁcient condition for optimality Theorem A sufﬁcient condition for a stationary point x to be an extreme point is that the Hessian matrix of f ð xÞ evaluated at x is: (i) Positive deﬁnite when x is a local minimizer. (ii) Negative deﬁnite when x is a local maximizer. Proof From Taylor’s theorem we can write f ð x  þ hÞ ¼ f ð x  Þ þ

n X i¼1

hi

n X n @f ðx Þ 1X @ 2 f ðx þ hhÞ þ hi hj ; @xi 2! i¼1 j¼1 @xi @xj

0\h\1: ð9:3Þ

Since x is an extreme point, from the necessary conditions for f ð xÞ to have an  extreme point, we have @f@xðxi Þ ¼ 0; i ¼ 1; 2; . . .; n. Then Eq. (9.3) reduces to  n X n 1X @ 2 f ð xÞ f ð x þ hÞ  f ð x Þ ¼ hi hj : 2! i¼1 j¼1 @xi @xj x¼x þ hh 



9.3 Optimization of Functions of Several Variables

203

) The sign of f ðx þ hÞ  f ðx Þ will be the same as that of XX i

j

 @ 2 f ð xÞ hi hj  @xi @xj 

: x¼x þ hh

ð xÞ Since the second-order partial derivative @@xfi @x is continuous in the neighbourhood j   2 @ 2 f ð xÞ  ð xÞ  of x , then @xi @xj   will have the same sign as @@xfi @x   for sufﬁciently small j 2

x¼x þ hh

x¼x

components of h.   This implies that f ðx þ  hÞ  f ðx Þ will be positive and x will be a local minimum  n P n P 2 ð xÞ  if Q ¼ hi hj @@xfi @x is positive.  j i¼1 j¼1  x¼x

This Q is a quadratic form and can be written as Q ¼ hHh; hijx¼x ; where Hjx¼x is called a Hessian matrix of f ð xÞ. From linear algebra, we know that the quadratic form will be positive for all h if and only if H is positive deﬁnite at x ¼ x . This means that a sufﬁcient condition for the extreme point x to be a local minimum is that the Hessian matrix of f ð xÞ evaluated at x ¼ x is positive deﬁnite. This completes the proof for the minimization case. By proceeding in a similar manner, it can be proved that the Hessian matrix will be negative deﬁnite if x is a local maximum point. Remark For the semi-deﬁnite case of the Hessian matrix of f ð xÞ, no general conclusion can be drawn regarding the optimality of f ð xÞ. Example 5 Examine whether f ðx1 ; x2 Þ ¼ 12 k2 x21 þ 12 k3 ðx2  x1 Þ2 þ 12 k1 x22  Px2 (P being constant) is optimum or not. Solution It is given that f ðx1 ; x2 Þ ¼ 12 k2 x21 þ 12 k3 ðx2  x1 Þ2 þ 12 k1 x22  Px2 . Now for the extremum of f, we have @f ¼ 0 i:e:; k2 x1  k3 ðx2  x1 Þ ¼ 0 @x1

ð9:4Þ

204

9 Basics of Unconstrained Optimization

and

@f ¼0 @x2

i:e:; k3 ðx2  x1 Þ þ k1 x2  P ¼ 0:

ð9:5Þ

From (9.4), we have k2 x1  k3 x2 þ k3 x1 ¼ 0

ð9:6Þ

or ðk2 þ k3 Þx1  k3 x2 ¼ 0: From (9.5), we have k3 ð x2  x1 Þ þ k1 x2  P ¼ 0 or k3 x2  k3 x1 þ k1 x2 ¼ P

ð9:7Þ

or  k3 x1 þ ðk1 þ k3 Þx2 ¼ P: Now multiplying (9.6) and (9.7) by k3 and (k2 + k3) respectively and then adding, we have  2 k3 þ ðk1 þ k3 Þðk2 þ k3 Þðk2 þ k3 Þ x2 ¼ Pðk2 þ k3 Þ Pðk2 þ k3 Þ : or x2 ¼ k1 k2 þ k1 k3 þ k2 k3 Now multiplying (9.6) and (9.7) by (k1 + k3) and k3 respectively and then adding, we have   k1 k2 þ k1 k3 þ k2 k3 þ k32 x1 ¼ Pk3 Pk3 or x1 ¼ : k1 k2 þ k1 k3 þ k2 k3 Now the Hessian matrix of f evaluated at ðx1 ; x2 Þ is 2 Hjðx1 ;x2 Þ ¼

@2f 2 4 @x2 1 @ f @x1 @x2

@2f @x1 @x2 @2f @x22

3 5 ðx1 ;x2 Þ



k þ k3 ¼ 2 k3

k3 k1 þ k3



k þ k3 ¼ 2 k3 ðx1 ;x2 Þ

k3 : k1 þ k3

The leading principal minors of the Hessian matrix are

and

A1 ¼ jk2 þ k3 j ¼ k2 þ k3 [ 0 ½* k2 ; k3 [ 0  k þ k3 k3  A2 ¼  2 ¼ ðk2 þ k3 Þðk1 þ k3 Þ  k32 k3 k1 þ k3  ¼ k1 k2 þ k2 k3 þ k3 k1 þ k32  k32 ¼ k1 k2 þ k2 k3 þ k3 k1 [ 0 ½* k1 ; k2 ; k [ 0:

9.3 Optimization of Functions of Several Variables

205

  Therefore, the Hessian matrix of f evaluated at x1 ; x2 , i.e. Hjðx1 ;x2 Þ , is positive deﬁnite.   Hence, f is minimum at x1 ; x2 , which are given by Pk3 k1 k2 þ k2 k3 þ k2 k1 P ð k2 þ k3 Þ x2 ¼ : k1 k2 þ k2 k3 þ k3 k1

x1 ¼

Example 6 Find the extreme point of f ðx1 ; x2 Þ ¼ x31 þ x32 þ 2x21 þ 4x22 þ 6. Solution Here and

f ðx1 ; x2 Þ ¼ x31 þ x32 þ 2x21 þ 4x22 þ 6 @f ) @x ¼ 3x21 þ 4x1 ¼ x1 ð3x1 þ 4Þ 1 @f 2 @x2 ¼ 3x2 þ 8x2 ¼ x2 ð3x2 þ 8Þ:

The necessary conditions for the optimum of f ðx1 ; x2 Þ are given by @f ¼0 @x1

and

@f ¼ 0; @x2

i:e: x1 ð3x1 þ 4Þ ¼ 0

ð9:8Þ

and x2 ð3x2 þ 8Þ ¼ 0:

ð9:9Þ

From (9.8), we have x1 = 0 or x1 ¼  43. From (9.9), we have x2 = 0 or x2 ¼  83. Hence, Eqs. (9.8) and (9.9) are satisﬁed at the points [or the solution of (9.8) and (9.9)] 8 4 4 8 ð0; 0Þ; 0;  ;  ; 0 and  ;  : 3 3 3 3 Since f ðx1 ; x2 Þ ¼ x31 þ x32 þ 2x21 þ 4x22 þ 6, @2f ¼ 6x1 þ 4; @x21

@2f ¼ 6x2 þ 8; @x22

@2f ¼ 0: @x1 @x2

206

9 Basics of Unconstrained Optimization

The Hessian matrix of f ðx1 ; x2 Þ is given by  H¼

6x1 þ 4 0 : 0 6x2 þ 8

Now, the leading principal minors of the Hessian matrix are given by    6x1 þ 4 0   H1 ¼ j6x1 þ 4j and H2 ¼  : 0 6x2 þ 8  Now, we construct the following table: Point ðx1 ; x2 Þ

Value of H1

Value of H2

Nature of H

Nature of x

Value of f(x)

(0, 0)

4

32

Local minimum

6

4

−32

Positive deﬁnite Indeﬁnite

−4

−32

Indeﬁnite

−4

32

Negative

Local maximum

418 27 194 27 50 3



 0;  83  4   ;0  34   3 ;  83

Example 7 Examine the extreme values of f ðx; y; zÞ ¼ 2xyz  4zx  2yz þ x2 þ y2 þ z2  2x  4y þ 4z: Solution Here; fx ¼ 2yz  4z þ 2x  2 fy ¼ 2xz  2z þ 2y  4 fz ¼ 2xy  4x  2y þ 2z þ 4: For extreme values of f ðx; y; zÞ, the necessary conditions are fx ¼ 0;

fy ¼ 0;

fz ¼ 0;

i:e: yz  2z þ x  1 ¼ 0

ð9:10Þ

xz  z þ y  2 ¼ 0

ð9:11Þ

xy  2x  y þ z þ 2 ¼ 0:

ð9:12Þ

9.3 Optimization of Functions of Several Variables

207

From (9.10), we have x ¼ 1 þ 2z  yz:

ð9:13Þ

Substituting x ¼ 1 þ 2z  yz in (9.11) and then simplifying, we have ð2  yÞz2 þ y  2 ¼ 0:

ð9:14Þ

Again, substituting x ¼ 1 þ 2z  yz in (9.11) and then simplifying, we have zðy2  4y þ 3Þ ¼ 0; i:e: z ¼ 0 or y ¼ 3; 1: When z = 0, then from (9.13) and (9.14), we have x = 1, y = 2. Again, for y = 2, z = ± 1, x = 0, 2 and for y = 1, z = ± 1, x = 2, 0. Therefore, the solutions of (9.10) through (9.12) are ð1; 2; 0Þ; ð0; 3; 1Þ; ð2; 3; 1Þ; ð2; 1; 1Þ; ð0; 1; 1Þ: Now, fxx ¼ 2;

fyy ¼ 2;

fzz ¼ 2

fyx ¼ fxy ¼ 2z;

fyz ¼ fzy ¼ 2x  2;

fzx ¼ fxz ¼ 2y  4:

) The Hessian matrix of f ðx; y; zÞ is 2

fxx H ¼ 4 fyx fzx

fxy fyy fzy

3 2 fxz 2 fyz 5 ¼ 4 2z fzz 2y  4

2z 2 2x  2

3 2y  4 2x  2 5: 2

Now at (1, 2, 0), 2

2 H ¼ 40 0

3 0 0 2 0 5: 0 2

The leading principal minors of H are   2 0   ¼ 4 and H3 ¼ jH j ¼ 8: H1 ¼ j2j ¼ 2; H2 ¼  0 2 As all the leading principal minors are positive, the Hessian matrix at (1, 2, 0) is positive deﬁnite.

208

9 Basics of Unconstrained Optimization

Hence, the given function f ðx; y; zÞ is minimum at (1, 2, 0), and the minimum value of f ðx; y; zÞ is −5. In a similar way, test the optimality at other points. Example 8 Show that f ðx; y; zÞ ¼ ðx þ y þ zÞ3 3ðx þ y þ zÞ  24xyz þ a3 has a minimum at (1, 1, 1) and a maximum at (−1, −1, −1).

Solution Here f ðx; y; zÞ ¼ ðx þ y þ zÞ3 3ðx þ y þ zÞ  24xyz þ a3 Hence; fx ¼ 3ðx þ y þ zÞ2 3  24yz fy ¼ 3ðx þ y þ zÞ2 3  24zx fz ¼ 3ðx þ y þ zÞ2 3  24xy Again; fxx ¼ 6ðx þ y þ zÞ fyy ¼ 6ðx þ y þ zÞ fzz ¼ 6ðx þ y þ zÞ fxy ¼ fyx ¼ 6ðx þ y þ zÞ  24z fyz ¼ fzy ¼ 6ðx þ y þ zÞ  24x fxz ¼ fzx ¼ 6ðx þ y þ zÞ  24y: At (1, 1, 1), fx ¼ 3:32  3  24:1:1 ¼ 0 fy ¼ 3:32  3  24:1:1 ¼ 0 fz ¼ 3:32  3  24:1:1 ¼ 0: Again at (−1, −1, −1), fx ¼ 3ð3Þ2 3  24 ¼ 0 fy ¼ 3ð3Þ2 3  24 ¼ 0 fz ¼ 3ð3Þ2 3  24 ¼ 0: Hence, (1, 1, 1) and (−1, −1, −1) are the extreme points of f (x, y, z). The Hessian matrix for the function f (x, y, z) is 2

3 fxy fxz fyy fyz 5 fzy fzz 6ðx þ y þ zÞ ¼ 4 6ðx þ y þ zÞ  24z 6ðx þ y þ zÞ  24y

fxx 4 f H ¼ yx 2 fzx

6ðx þ y þ zÞ  24z 6ð x þ y þ z Þ 6ðx þ y þ zÞ  24x

3 6ðx þ y þ zÞ  24y 6ðx þ y þ zÞ  24x 5: 6ð x þ y þ z Þ

9.3 Optimization of Functions of Several Variables

209

At (1, 1, 1), 2

18 H ¼ 4 6 6

6 18 6

3 6 6 5: 18

The leading principal minors are    18 6    ¼ 288 and H3 ¼ jH j ¼ 18 288 ¼ 5184: H1 ¼ j18j ¼ 18; H2 ¼  6 18  As all the leading principal minors are positive, the Hessian matrix H is positive deﬁnite at (1, 1, 1), Therefore, f (x, y, z) has a minimum value at ð1; 1; 1Þ, and the minimum value of f (x, y, z) is a3−6. At (−1, −1, −1),   18  H ¼  6  6

 6 6  18 6 : 6 18 

The leading principal minors are   18 H1 ¼ j18j ¼ 18; H2 ¼  6

 6  ¼ ð18Þ2 62 ¼ 288 18 

and H3 ¼ jH j ¼ 18ð324  36Þ  6ð108  36Þ þ 6ð36 þ 108Þ ¼ 5184 þ 2 864 ¼ 3456: Hence, f (x, y, z) has a maximum value at (−1, −1, −1) [since the Hessian matrix is negative deﬁnite].

9.4

Exercises 1

2

1. Show that ðx  aÞ3 ð2x  aÞ3 is maximum for x ¼ a2 and neither maximum nor minimum for x = a (a > 0).

210

9 Basics of Unconstrained Optimization

2. Find the extreme points of the function f ðx1 ; x2 Þ ¼ 2x31 þ 3x32 þ 5x21 þ 4x22 þ 6. 3. Find the extreme points of the function f ðx1 ; x2 Þ ¼ 8x1 þ 4x2 þ x1 x2  x21  x22 . 4. Examine the extreme values of f ðx; y; zÞ ¼ 2xyz  4zx  2yz þ x2 þ y2 þ z2  2x  4y þ 4z:

Chapter 10

Constrained Optimization with Equality Constraints

10.1

Objective

The objective of this chapter is to introduce the most important methods, namely direct substitution, constrained variation and the Lagrange multiplier method, for solving non-linear constrained optimization problems with equality constraints.

10.2

Introduction

The general form of non-linear constrained optimization problems with equality constraints can be written as follows: Optimize z ¼ f ð xÞ subject to gi ð xÞ ¼ 0;

i ¼ 1; 2; . . .; m;

where x ¼ ðx1 ; x2 ; . . .; xn Þ: i ¼ 1; 2; . . .; m subject to gi ðxÞ ¼ 0; x 2 Rn ; f : Rn ! R; gi : Rn ! R and m  n: If m > n, the problem becomes overdeﬁned, and in general there will be no solution. For solving this problem, there are several methods, e.g. (i) Direct substitution method (ii) Constrained variation method (iii) Lagrange multiplier method. We shall discuss these methods in detail. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_10

211

212

10.3

10 Constrained Optimization with Equality Constraints

Direct Substitution Method

For an optimization problem with n variables and m equality constraints, it is possible (theoretically) to express any set of m variables in terms of the remaining (n − m) variables. When these expressions are substituted into the original objective function, then the reduced objective function involves only n − m variables. This reduced objective function is not subjected to any constraint, so its optimum value can be found by using the unconstrained optimization technique discussed in Chap. 9. Theoretically, the method of direct substitution is very simple. However, from a practical point of view, it is not convenient. In most of practical problems, the constraint equations are highly non-linear in nature. In those cases, it becomes impossible to solve them and express any m variables in terms of the remaining (n − m) variables from the given constraints. Example 1 Minimize

z ¼ 9  8x1  6x2  4x3 þ 2x21 þ 2x22 þ x23 þ 2x1 x2 þ 2x1 x3

subject to

x1 þ x2 þ 2x3 ¼ 3:

Solution From the given constraint, we have x2 ¼ 3  x1  2x3 , and substituting it in the objective function, we have z ¼ 9  8x1  6ð3  x1  2x3 Þ  4x3 þ 2x21 þ 2ð3  x1  2x3 Þ2 þ x23 þ 2x1 ð3  x1  2x3 Þ þ 2x1 x3 ¼ 2x21 þ 9x23 þ 6x1 x3  8x1  16x3 þ 9:

Now we have to optimize z with respect to the decision variables x1 and x3 . The necessary conditions for optimality of z are given by @z @z ¼ 0 and ¼ 0; @x1 @x3 i:e: 2x1 þ 3x3 ¼ 4

ð10:1Þ

3x1 þ 9x3 ¼ 8:

ð10:2Þ

Solving (10.1) and (10.2), we have x1 ¼ 43 ; x3 ¼ 49. Now x2 ¼ 3  x1  2x3 ¼ 3  43  2 49 ¼ 79. We have

@2 z @x21

¼ 4;

@2 z @x23

¼ 18;

@2 z @x1 @x3

¼ 6.

10.3

Direct Substitution Method

213

Hence, the Hessian matrix is given by 0 H¼

@2 z 2 @ @x2 1 @ z @x3 @x1

@2z @x1 @x3 @2z @x23

1 A¼



4 6

 6 : 18

 4 The leading principal minors are H1 ¼ j4j ¼ 4 and H2 ¼  6

 6  ¼ 36. 18 

Since H1 [ 0 and H2 [ 0, Z is minimum for x1 ¼ 43 ; x2 ¼ 79 ; x3 ¼ 49 and the minimum value of z is 19. Example 2 Find the dimension of a box of largest volume that can be inscribed in a sphere of unit radius. Solution Let the origin O of the Cartesian coordinate system OX1 X2 X3 be the centre of the sphere and the sides of the box be 2x1, 2x2 and 2x3. Obviously, x1 ; x2 ; x3 [ 0. The volume of the box is given by f ðx1 ; x2 ; x3 Þ ¼ 8x1 x2 x3 : Since the box is inscribed in a sphere of unit radius, the corners of the box lie on the surface of the sphere of unit radius; that is, x1, x2 and x3 have to satisfy the constraint equation x21 þ x22 þ x23 ¼ 1, as the equation of a sphere with unit radius and centre at the origin is x21 þ x22 þ x23 ¼ 1. Hence, our problem is Maximize f ðx1 ; x2 ; x3 Þ ¼ 8x1 x2 x3 subject to x21 þ x22 þ x23 ¼ 1 and x1 ; x2 ; x3 [ 0. This problem has three variables and one equality constraint. Hence, any one variable can be eliminated using the constraint. Now, from the constraint x21 þ x22 þ x23 ¼ 1, we have  1 x3 ¼ 1  x21  x22 2 as x3 [ 0 Substituting the expression of x3 in f ðx1 ; x2 ; x3 Þ, we get the reduced objective function as

214

10 Constrained Optimization with Equality Constraints

 1 f ¼ 8x1 x2 1  x21  x22 2 ; which can be maximized as an unconstrained function of two variables. The necessary conditions for the optimality of f are @f @f ¼ 0 and ¼ 0: @x1 @x2 2 3 2 1   @f x1 ¼ 0 implies 8x2 4 1  x21  x22 2   Now; 1 5 ¼ 0: @x1 1  x21  x22 2 2 3 2 1   @f x 2 Again; ¼ 0 implies 8x1 4 1  x21  x22 2   12 5 ¼ 0: @x2 2 2 1x x 1

ð10:3Þ

ð10:4Þ

2

After simplifying (10.3) and (10.4), we have 1  2x21  x22 ¼ 0 1  x21  2x22 ¼ 0

as x1 ; x2 [ 0:

Solving these equations, we have 1 x1 ¼ x2 ¼ pﬃﬃﬃ : 3  1 1 2 Hence; x3 ¼ 1  x1  x22 3 ¼ pﬃﬃﬃ : 3

2 3 3  12 @2f 8x1 x2 8x2 x 2 2 1 4 Now; ¼  1  1 þ 2x1 1  x1  x2 5 2 2  @x21 1  x21  x22 2 1  x1  x2 1  x21  x22 2 2 3 3  12 @2f 8x1 x2 8x1 x 2 2 2 4 ¼  1  1 þ 2x2 1  x1  x2 5 2 2  @x22 1  x21  x22 2 1  x1  x2 1  x21  x22 2 2 3  1  12 @2f 8x22 8x21 x22 2 2 2 2 2 4 ¼ 8 1  x1  x2   1  1  x2  x2  1  x1  x2 þ  1 5: @x1 @x2 1 2 1  x21  x22 2 1  x21  x22 2

At x1 ¼ p1ﬃﬃ3 ;

x2 ¼ p1ﬃﬃ3 ;

@2f @x21

¼  p32ﬃﬃ3 ;

@2 f @x22

¼  p32ﬃﬃ3 ;

@2 f @x1 @x2

¼  p16ﬃﬃ3 :

10.3

Direct Substitution Method

215

Hence, at x1 ¼ x2 ¼ p1ﬃﬃ3, the Hessian matrix H of f is given by " H¼

#  p32ﬃﬃ3  p16ﬃﬃ3 :  p16ﬃﬃ3  p32ﬃﬃ3

The leading principal minors are    32  32 H1 ¼  pﬃﬃﬃ ¼  pﬃﬃﬃ \0 3 3     p32ﬃﬃ  p16ﬃﬃ  1024 256 768  3  ¼ [ 0: and H2 ¼  163 ¼   pﬃﬃ3  p32ﬃﬃ3  9 9 9 Hence, f is maximum at x1 ¼ x2 ¼ p1ﬃﬃ3, and the maximum value of f is 8 Hence, the maximum volume of the box is

10.4

8 ﬃﬃ p 3 3

with dimensions p2ﬃﬃ3 ;

 3 p1ﬃﬃ ¼ 3

8 ﬃﬃ p . 3 3

p2ﬃﬃ ; p2ﬃﬃ. 3 3

Method of Constrained Variation

The basic idea used in the method of constrained variation is to ﬁnd a closed-form expression for the ﬁrst-order differential of f(x) (i.e. df) at all points at which the constraints gj ð xÞ ¼ 0; j ¼ 1; 2; . . .; m, are satisﬁed. Then, by setting df = 0, the desired optimum points are obtained. Before discussing the general method, we shall indicate the important features through the following simple problem with n = 2 and m = 1. f ðx1 ; x2 Þ subject to gðx1 ; x2 Þ ¼ 0: Optimize

Theorem 1 The necessary condition for the preceding problem to have an    extremum at x1 ; x2 is 

 @f @g @f @g   ¼ 0: @x1 @x2 @x2 @x1 ðx ;x Þ 1 2

  Proof The necessary condition for f to have an extremum at some point x1 ; x2 is    that the total differential of f ðx1 ; x2 Þ must be zero at x1 ; x2 ,

216

10 Constrained Optimization with Equality Constraints

i:e: df ¼ 0 @f @f or dx1 þ dx2 ¼ 0 @x1 @x2

ð10:5Þ

  Since g x1 ; x2 ¼ 0 at the extreme point, any inﬁnitesimal variations dx1 and dx2   taken about the point x1 ; x2 are called the admissible variations, provided that the new point lies on the constraint,   i:e: g x1 þ dx1 ; x2 þ dx2 ¼ 0:

ð10:6Þ

  The Taylor series expansion of the function g x1 þ dx1 ; x2 þ dx2 about the point    x1 ; x2 gives g



x1 ; x2



  @g  @g  dx1 þ dx2 ¼ 0: þ @x1 ðx ;x Þ @x2 ðx ;x Þ 1 2 1 2

ð10:7Þ

  Since g x1 ; x2 ¼ 0, Eq. (10.7) reduces to @g @g dx1 þ dx2 ¼ 0 @x1 @x2

  at x1 ; x2 :

ð10:8Þ

Thus, Eq. (10.8) has to be satisﬁed by all the admissible variations. Now, assuming that

@g @x2

6¼ 0, Eq. (10.8) can be rewritten as dx2 ¼ 

@g=@x1 dx1 @g=@x2

  at x1 ; x2 :

ð10:9Þ

This relation indicates that once the variation in x1 (i.e. dx1) is chosen arbitrarily, the variation in x2 (i.e. dx2) is obtained automatically in order to have dx1 and dx2 as a set of admissible variations. Now, by substituting the preceding expression for dx2 in (10.5), we have 

 @f @f @g=@x1   dx1 ¼ 0: @x1 @x2 @g=@x2 ðx ;x Þ 1 2

ð10:10Þ

The expression on the left-hand side of Eq. (10.10) is called the constrained variation of f. Since dx1 is chosen arbitrarily, the coefﬁcient of dx1 in (10.10) must be zero,

10.4

Method of Constrained Variation

217



 @f @f @g=@x1  i:e:  ¼0 @x1 @x2 @g=@x2 ðx ;x Þ 1 2   @f @g @f @g  i:e:  ¼ 0; @x1 @x2 @x2 @x1 ðx ;x Þ 1 2

ð10:11Þ

which is the necessary condition for the given problem to have an extremum at   x1 ; x2 .

10.4.1 Geometrical Interpretation of Admissible Variation We now illustrate the admissible variations geometrically in Fig. 10.1, where PQ indicates the curve  at each point of which gðx1 ; x2 Þ ¼ 0 is satisﬁed. If A is taken as the point x1 ; x2 , then the inﬁnitesimal variations in x1 and x2 leading to the points B and C are called admissible variations. Thus, by the coordinates of B and C, gðx1 ; x2 Þ ¼ 0 will be satisﬁed, as these points lie on the same curve. On the other hand, the inﬁnitesimal variations in x1 and x2 representing D are not admissible, since the point D does not lie on the constraint curve gðx1 ; x2 Þ ¼ 0. Thus, any set of variations ðdx1 ; dx2 Þ that does not satisfy Eq. (10.8) leads to points which do not satisfy the constraint gðx1 ; x2 Þ ¼ 0.

10.4.2 Necessary Conditions for Extremum of a General Problem for Constrained Variation Method Let the function to be optimized be f ðx1 ; x2 ; . . .; xn Þ subject to the equality constraints

x1 D

Q C

A B P

x2

218

10 Constrained Optimization with Equality Constraints

gi ðx1 ; x2 ; . . .; xn Þ ¼ 0;

i ¼ 1; 2; . . .; m:

ð10:12Þ

  Let x ¼ x1 ; x2 ; . . .; xn be an extreme point and (dx1, dx2, …, dxn) be the inﬁnitesimal admissible variations about the point x*.   Then, gi x1 ; x2 ; . . .; xn ¼ 0; i ¼ 1; 2; . . .; m.   n        P @gi    Now, gi x1 þ dx1 ; . . .; xn þ dxn ’ gi x1 ; x2 ; . . .; xn þ @xj  dxj .  j¼1 x

Since (dx1, dx2, …, dxn) are the admissible inﬁnitesimal variations, then  gi x1 þ dx1 ; . . .; xn þ dxn ¼ 0. 

 

) gi x1 ; x2 ; . . .; xn  n X @gi  or  dxj ¼ 0; @xj  j¼1

 n X @gi  þ  dxj ¼ 0 @xj  j¼1 x

  * gi x1 ; x2 ; . . .; xn ¼ 0 i ¼ 1; 2; . . .; m:

ð10:13Þ

x

There are m equations in (10.13). These equations can be solved to express any m variations, say, the ﬁrst m variations, in terms of the remaining variations as follows:  @gi i Writing @g @xj   simply as @xj , we get x

m X @gi j¼1

@xj

dxj ¼ 

n X @gi dxk ¼ hi ðsayÞ; @xk k¼m þ 1

i ¼ 1; 2; . . .; m:

By Cramer’s rule, we have dx1 ¼ D1 =D,   h1    h2 where D1 ¼  .  ..   hm

@g1 @x2 @g2 @x2

@g1 @x3 @g2 @x3

@gm @x2

@gm @x3

  @g   1   @x1   @g2    and D ¼  @x1   ..   .   @g m   m    @g @xm @x1  

@g1 @xm @g2 @xm

@g1 @x2 @g2 @x2

 

@gm @x2



     :    @gm  @g1 @xm @g2 @xm

@xm

10.4

Method of Constrained Variation

219

  n P  @g1 @g1 @g1    dx . . . @xm   k¼m þ 1 @xk k @x2     n P  @g2 @g2 @g2  dx     k @x2 @xm   k¼m þ 1 @xk   :     :       :   n   P @gm @gm @gm   . . . @xm   

@xk dxk @x2  g1 ; g2 ; . . .; gm k¼m þ 1  Hence; dx1 ¼ assuming J 6¼ 0 ...; gm x1 ; x2 ; . . .; xm J gx11;; gx22;; ...; xm  P ...; gm  nk¼m þ 1 dxk J gx1k ;; gx22;; ...; xm  ¼ : g1 ; g2 ; ...; gm J x1 ; x2 ; ...; xm

) dx1 ¼

In general; dxa ¼



Pn

k¼m þ 1



 g1 ; g2 ; ...; gm dx J k k¼m þ 1 xk ; x2 ; ...; xm  : g1 ; g2 ; ...; gm J x1 ; x2 ; ...; xm

Pn

 ga1 ; ga ; ga þ 1 ; ...; gm dxk J gx11;; gx22;; ...; ...; xa1 ; xk ; xa þ 1 ; ...; xm  ; g1 ; g2 ; ...; gm J x1 ; x2 ; ...; xm

a ¼ 1; 2; . . .; m:

At extreme point x*, we have df = 0 @f @f @f dx1 þ dx2 þ    þ dxn ¼ 0 @x1 @x2 @xn m n X X @f @f or dxa þ dxk ¼ 0 @x @x a k a¼1 k¼m þ 1

or

Denoting



@f  @f by @xj x @xj

Putting the expression dxa ; a ¼ 1; 2; . . .; m in above, we have "  J

1 g1 ; g2 ; ...; gn x1 ; x2 ; ...; xn



m X @f @xa a¼1

n X k¼m þ 1

 dxk J

g1 ; g2 ; . . .; ga1 ; ga ; ga þ 1 ; . . .; gm x1 ; x2 ; . . .; xa1 ; xk ; xa þ 1 ; . . .; xm #

)

 @f g1 ; g2 ; . . .; gm dxk ¼ 0 J @xk x1 ; x2 ; . . .; xn k¼m þ 1 "     n m X X @f g1 ; g2 ; . . .; ga1 ; ga ; ga þ 1 ; . . .; gm @f g1 ; g2 ; . . .; gm þ dxk ¼ 0:  J J or @xa @xk x1 ; x2 ; . . .; xa1 ; xk ; xa1 ; . . .; xm x1 ; x2 ; . . .; xm a¼1 k¼m þ 1 þ

n X



(

Since dxm þ 1 ; dxm þ 2 ; . . .; dxn are all independent variations, the coefﬁcient of each variation must be zero. Thus, we have

220

10 Constrained Optimization with Equality Constraints

    m X @f g1 ; g2 ; . . .; ga1 ; ga ; ga þ 1 ; . . .; gm @f g1 ; g2 ; . . .; gm þ ¼0  J J @xa x1 ; x2 ; . . .; xa1 ; xk ; xa þ 1 ; . . .; xm @xk x1 ; x2 ; . . .; xm a¼1 for k ¼ m þ 1; m þ 2; . . .; n     @f g1 ; g2 ; . . .; gm @f g1 ; g2 ; . . .; gm or    J J @x1 @x2 xk ; x2 ; . . .; xm x1 ; xk ; . . .; xm     @f g1 ; g2 ; . . .; gm1 ; gm @f g1 ; g2 ; . . .; gm þ ¼ 0:  J J @xm @xk x1 ; x2 ; . . .; xm1 ; xk x1 ; x2 ; . . .; xm

ð10:14Þ h By transforming the column

@g1 @xk

@g2 @xk

@g3 @xk

...

@gm @xk

iT in each determinant of

Eq. (10.14) to the ﬁrst column, we get       @f g1 ; g2 ; . . .; gm @f g1 ; g2 ;    ; gm @f g1 ; g2 ; . . .; gm  þ  J J J @xk @x1 @x2 xk ; x1 ; x3 ; . . .; xm x1 ; x2 ; . . .; xm xk ; x2 ; . . .; xm   @f g1 ; g2 ; . . .; gm þ ð1Þm J ¼ 0; k ¼ m þ 1; m þ 2; . . .; n @xm xk ; x1 ; . . .; xm1

  @f @f   @f  @xk @x1    @xm     @g1 @g1    @g1   @xk @x1 @xm  ¼0 or     ...     @gm @gm @g  @x @x    @xm  k 1 m   f ; g1 ; g2 ; . . .; gm or J ¼0 xk ; x1 ; x2 ; . . .; xm

ð10:15Þ

where k ¼ m þ 1; m þ 2; . . .; n:

Thus, the (n − m) equations represented by (10.15) give the necessary conditions for an extremum of f ðx1 ; x2 ; . . .; xn Þ under m equality constraints gi ðx1 ; x2 ; . . .; xn Þ ¼ 0; i ¼ 1; 2; . . .; m.

10.4.3 Sufﬁcient Conditions for a General Problem in Constrained Variation Method For the optimization problem with n variables and m equality constraints, m variables are dependent and the other n − m variables are independent. Let us assume that x1 ; x2 ; . . .; xm are dependent variables and the others xm þ 1 ; xm þ 2 ; . . .; xn are independent.

10.4

Method of Constrained Variation

221

Then the Taylor series expansion of f, in terms of independent variables, about the extreme point x* gives ð10:16Þ  where

@f @xi g

denotes the partial derivative of f with respect to xi (holding all the

other variables xm þ 1 ; xm þ 2 ; . . .; xj1 ; xj þ 1 ; . . .; xn as constant) when x1, x2, …,  xm are allowed to change so that the constraints  gi ðx þ dxÞ ¼ 0; i ¼ 1; 2; . . .; m @2f @xj @xk g

are satisﬁed. The second-order derivative

is used to denote a similar

meaning.



Since all the variations dxj appearing in (10.16) are independent,

@f @xj g

has to be

zero for j = m + 1, m + 2, …, n. Thus, the necessary conditions for the existence of a constrained optimum at x* can also be expressed as 

@f @xj

 ¼ 0;

j ¼ m þ 1; m þ 2; . . .; n:

ð10:17Þ

g

Hence, from (10.16), we have

Therefore, the sufﬁcient condition for x* to be a local optimum is that the quadratic form Q deﬁned by  2  n X @ f Q¼ dxj dxk @xj @xk g j¼m þ 1 k¼m þ 1 n X

is positive (or negative) for all non-vanishing variations dxj. That is, the matrix 2  6 6 6 4

@2f @x2m þ 1 g

.. . 2



@ f @xn @xm þ 1 g

@2f @xm þ 1 @xm þ 2 g



@2f @xn @xm þ 2 g

 



3

@2 f

@xm þ 1 @xn g



@2 f @x2n g

7 7 7 5

is positive (or negative) deﬁnite. Remark After a little manipulation, it can be shown that Eq. (10.17) and  f ; g1 ; g2 ; ...; gm J xk ; x1 ; x2 ; ...; xm ¼ 0 are the same.

222

10 Constrained Optimization with Equality Constraints

Disadvantages of the method In  this method, the computation of the constrained second-order partial derivative @2f is a formidable task for a problem with more than three constraints. Also, @xj @xk g

the necessary conditions involve the evaluation of (n − m) numbers of determinants of order m + 1. This is also a complicated task. Example 3 Using the method of constrained variation, Minimize f ðx1 ; x2 ; x3 Þ ¼ x21 þ 2x22 þ x23 subject to 2x1 þ 4x2 þ 3x3 ¼ 9 4x1 þ 8x2 þ 5x3 ¼ 17: Solution The given problem is f ¼ x21 þ 2x22 þ x23

ð10:18Þ

subject to g1 ¼ 2x1 þ 4x2 þ 3x3  9 ¼ 0

ð10:19Þ

g2 ¼ 4x1 þ 8x2 þ 5x3  17 ¼ 0:

ð10:20Þ

In this problem, m ¼ number of constraints ¼ 2 n ¼ number of variables ¼ 3: Hence (n − m) = 1 variable is independent and the other two are dependent. Let us choose x1 ; x2 as dependent variables.      @g1 @g1    @x1 @x2   2 4  g1 ; g2 ¼ 0. Then, J x1 ; x2 ¼  @g2 @g2  ¼  4 8  @x @x  1

2

So x1 and x2 cannot be chosen as dependent variables. Now, we consider x1 and x3 as dependent variables.     @g1 @g1   g1 ; g 2  @x1 @x3   2 3  ¼ 2 6¼ 0: J ¼  @g2 @g2  ¼  4 5  @x @x  x1 ; x 3 1

3

Thus, x1 ; x3 can be chosen as dependent variables. The necessary condition for optimality of f is given by 

f ; g1 ; g2 J x2 ; x1 ; x3

 ¼ 0;

10.4

Method of Constrained Variation

223

 @f   @x2  1 i:e: @g @x2  @g  2 @x2

  4x2  or;  4  8

2x1 2 4

 2x3  3  ¼ 0 5 

@f @x1 @g1 @x1 @g2 @x1

@f @x3 @g1 @x3 @g2 @x3

    ¼0  

or,  8x2 þ 8x1 ¼ 0

or, x1 ¼ x2 :

ð10:21Þ

Using (10.21) in (10.19) and (10.20), we have 6x2 þ 3x3 ¼ 9 12x2 þ 5x3 ¼ 17: Solving, we get x2 ¼ 1; x3 ¼ 1

) x1 ¼ x2 ¼ x3 ¼ 1.

So the required optimum solution is x1 ¼ x2 ¼ x3 ¼ 1 with fmin ¼ 6: Example 4 Find at which point f ð xÞ ¼ 2x21  x22 þ 4x23 þ x24 is optimum subject to g1 ð xÞ ¼ x1 þ x2  x3 þ x4  10 ¼ 0

ð10:22Þ

g2 ð xÞ ¼ x1 þ x2 þ 2x3 þ 3x4  12 ¼ 0:

ð10:23Þ

Solution Here m ¼ number of constraints ¼ 2 n ¼ number of variables ¼ 4: So (n − m) = 2 variables are independent and the other two are dependent. We ﬁrst take x1 and x2 as dependent variables depending on x3 ; x4 .    @g1 g1 ; g 2  @x1 Now; J ¼  @g  @x2 x1 ; x 2 1

@g1 @x2 @g2 @x2

     1 ¼ 1 

 1  ¼ 0: 1

So x1 , x2 cannot be chosen as dependent variables. 

g1 ; g2 Again; J x1 ; x3



  @g1  @x1 ¼  @g  @x2 1

@g1 @x3 @g2 @x3

     1 ¼ 1 

 1  ¼ 3 6¼ 0: 2 

224

10 Constrained Optimization with Equality Constraints

So x1 , x3 can be chosen as dependent variables. Now the necessary conditions for optimality are given by 

f ; g1 ; g2 J x2 ; x1 ; x3  @f   @x2  1 i:e:  @g @x2  @g  2 @x2

  2x2  i:e:;  1  1

@f @x1 @g1 @x1 @g2 @x1

4x1 1 1





f ; g1 ; g2 ¼ 0 and J x4 ; x1 ; x3

@f @x3 @g1 @x3 @g2 @x3

  @f     @x4   1  ¼ 0 and  @g @x4   @g   2 @x4

   2x4 8x3   1  ¼ 0 and  1  3 2 

@f @x1 @g1 @x1 @g2 @x1

4x1 1 1

 ¼0 @f @x3 @g1 @x3 @g2 @x3

    ¼0  

 8x3  1  ¼ 0: 2 

Hence; 6x2  12x1 ¼ 0 implies x2 ¼ 2x1

ð10:24Þ

and 6x4  20x1  16x3 ¼ 0 implies 3x4  10x1  8x3 ¼ 0:

ð10:25Þ

Now from (10.22) and (10.23) using (10.24), we have x1  x3 þ x4 ¼ 10

ð10:26Þ

and  x1 þ 2x3 þ 3x4 ¼ 12:

ð10:27Þ

From (10.26) and (10.27), we have 3x3 þ 2x4 ¼ 2;

ð10:28Þ

and from (10.25) and (10.26), we have 2x3  7x4 ¼ 100:

ð10:29Þ

304 Solving (10.28) and (10.29), we have x3 ¼  186 25 ; x4 ¼ 25 .

) x1 ¼ x4  x3  10 ¼

48 5

) x2 ¼ 2x1 ¼ 

96 : 5

96 186 304 So at the point x1 ¼ 48 5 ; x2 ¼  5 ; x3 ¼  25 ; x4 ¼ 25 , the given function f ð xÞ is optimum.

10.5

10.5

Lagrange Multiplier Method

225

Lagrange Multiplier Method

In the area of optimization, the method of Lagrange multipliers (named after Joseph-Louis Lagrange) provides a strategy for ﬁnding the optimum of a function subject to the given equality constraints. The Lagrange multiplier method is a very useful technique in multivariate calculus. One of the most common problems in calculus is ﬁnding the optimum of a function. However, it is difﬁcult to ﬁnd the closed-form solution for a problem to be optimized. Such difﬁculties often arise when one wishes to maximize or minimize a function subject to some equality constraints. The method of Lagrange multipliers is a very well-known method for solving non-linear constrained optimization problems. The method of Lagrange multipliers is a systematic process of generating the necessary conditions for a stationary point. Let us consider the following optimization problem: Optimize z ¼ f ð xÞ subject to gi ð xÞ ¼ 0;

i ¼ 1; 2; . . .; m

where x 2 Rn ; f : Rn ! R; gi : Rm ! R and m  n. Now we introduce a new variable for each of the given constraints. These new variables k1 ; k2 ; . . .; km are known as Lagrange multipliers. Using these multipliers, Lagrange deﬁned a function now called the Lagrange function. It is denoted by Lðx1 ; x2 ; . . .; xn ; k1 ; k2 ; . . .; km Þ and is deﬁned by Lðx1 ; x2 ; . . .; xn ; k1 ; k2 ; m P . . .; km Þ ¼ f ð xÞ þ ki gi ð xÞ (ki may be positive or negative). i¼1

Since the problem has n variables and m equality constraints, m additional variables are to be added so that the Lagrange function will have n + m variables. Stationary point The points where the partial derivations of the Lagrange function L are zero are called stationary points. If ðx1 ; x2 ; . . .; xn Þ is an optimum point for the original constrained optimization problem, then there exists a set of real numbers ðk1 ; k2 ; . . .; km Þ such that ðx1 ; x2 ; . . .; xn ; k1 ; k2 ; . . .; km Þ is a stationary point for the Lagrange function. However, not all stationary points yield a solution of the original problem. Thus, the method of Lagrange multipliers yields necessary conditions for optimality in constrained optimization problems. Now we shall prove the theorem regarding the necessary conditions for optimality.

226

10 Constrained Optimization with Equality Constraints

10.5.1 Necessary Conditions for Optimality Theorem 2 The necessary condition for a function f(x) subject to the constraints gi ð xÞ ¼ 0; i ¼ 1; 2; . . .; m to have a local optimum at a point x* is that the partial derivative of the Lagrange function deﬁned by L ¼ Lðx1 ; x2 ; Pm . . .; xn ; k1 ; k2 ; . . .; km Þ ¼ f ð xÞ þ i¼1 ki gi ð xÞ with respect to each of its arguments is zero at x*. Proof Since x* is a local extreme point, then at x*, df = 0  n X @f  or  @xj  j¼1

dxj ¼ 0;

ð10:30Þ

x¼x

where dxj (j = 1, 2, …, n) are the admissible variations about the point x*. Since x* is an extreme point and dxj (j = 1, 2, …, n) are the admissible variations about the point x*, then gi ðx Þ ¼ 0 and gi ðx þ dxÞ ¼ 0:

ð10:31Þ

From (10.31), we have gi ðx þ dxÞ ¼ 0

 n X @gi  dxj ¼ 0 or gi ðx Þ þ  @xj   j¼1 x¼x  n X @gi  dxj ¼ 0 i ¼ 1; 2; . . .; m: or  @xj  j¼1 

ð10:32Þ

x¼x

Multiplying each equation in (10.32) by a constant ki ði ¼ 1; 2; . . .; mÞ, we have  n X @gi  ki  @xj  j¼1

dxj ¼ 0 x¼x

ði ¼ 1; 2; . . .; mÞ:

ð10:33Þ

10.5

Lagrange Multiplier Method

227

From (10.30) and (10.33), we have   n m n X X X @f  @gi  dxj þ ki dxj ¼ 0   @xj   @xj   j¼1 i¼1 j¼1 x¼x x¼x  ! n n m X X X @f  @gi  dxj þ ki or dxj ¼ 0   @xj   @xj   j¼1 j¼1 i¼1 x¼x x¼x  "  # n m X X @f  @gi  ki or dxj ¼ 0:  x¼x þ @xj  @xj   j¼1 i¼1

ð10:34Þ

x¼x

Since there are m constraints, any n − m variables can be chosen independently. Let x1 ; x2 ; . . .; xm be a set of dependent variables and xm þ 1 ; xm þ 2 ; . . .; xn be the set of independent variables. We choose the values of ki ði ¼ 1; 2; . . .; mÞ in such a way that the coefﬁcients of the ﬁrst m variations dxi ði ¼ 1; 2; . . .; mÞ become zero. Thus, ki are deﬁned by   m X @f  @gi  þ ki  @xj x¼x @xj  i¼1

¼0

for j ¼ 1; 2; . . .; m:

ð10:35Þ

x¼x

Using (10.35), from (10.34) we have n X j¼m þ 1

"

  m X @f  @gi  þ ki  @xj x¼x @xj  i¼1

# dxj ¼ 0:

x¼x

Since dxm þ 1 ; dxm þ 2 ; . . .; dxn are all independent, we have   m X @f  @gi  þ ki ¼ 0 for j ¼ m þ 1; m þ 2; . . .; n:  @xj x¼x i¼1 @xj  

ð10:36Þ

x¼x

From (10.35) and (10.36), we have   m X @f  @gi  þ ki  @xj x¼x i¼1 @xj 

¼ 0 for j ¼ 1; 2; . . .; n:

ð10:37Þ

for i ¼ 1; 2; . . .; m:

ð10:38Þ

x¼x

Also; gi ðx Þ ¼ 0

228

Now L ¼ f ð xÞ þ

10 Constrained Optimization with Equality Constraints m X

ki gi ðxÞ:

i¼1

   m X @L  @f  @gi  ¼ þ ki ¼ 0 for j ¼ 1; 2; . . .; n ½using ð10:37Þ )  @xj x¼x @xj x¼x i¼1 @xj   x¼x  @L  and ¼ gk ðx Þ ¼ 0 for k ¼ 1; 2; . . .; m ½using ð10:38Þ @kk x¼x  @L  ¼ 0 for k ¼ 1; 2; . . .; m: or @kk x¼x

These are the necessary conditions.

10.5.2 Sufﬁcient Conditions for Optimality Theorem 3 A sufﬁcient condition for f(x) to have a local minimum (maximum) subject to the constraints gi ð xÞ ¼ 0; i ¼ 1; 2; . . .; m at x* is that the partial derivative of the Lagrange function Lðx1 ; x2 ; . . .; xn ; k1 ; k2 ; . . .; kn Þ at x* with respect to each of its arguments is zero, and the quadratic Q deﬁned by  n X n X @ 2 L  Q¼ dxj dxk  @xj @xk  j¼1 k¼1 x¼x

must be positive (negative) for all choices of the admissible variations dxj ðj ¼ 1; 2; . . .; nÞ. Proof The Lagrange function is deﬁned as Lðx; kÞ ¼ f ð xÞ þ

m X

ki gi ð xÞ:

i¼1

Let dx be the vector of admissible variations about the point x*. Then the point x* + dx must satisfy the given constraints. ) gi ðx þ dxÞ ¼ 0:  @L  Also; ¼ gi ðx Þ ¼ 0: @ki x

ð10:39Þ ð10:40Þ

10.5

Lagrange Multiplier Method

229

) Lðx þ dx; kÞ  Lðx ; kÞ ¼ f ðx þ dxÞ þ

m X

ki gi ðx þ dxÞ  f ðx Þ 

i¼1

m X

ki gi ðx Þ

i¼1

or Lðx þ dx; kÞ  Lðx ; kÞ ¼ f ðx þ dxÞ  f ðx Þ ½using ð10:39Þ and ð10:40Þ:

ð10:41Þ The Taylor series expansion of Lðx þ dx; kÞ about the point x* (keeping all the components of k as ﬁxed) gives  n X @L  Lðx þ dx; kÞ ¼ Lðx ; kÞ þ  @xj  j¼1

 n X n X @ 2 L  dxj þ dxj dxk ; 0\h\1  @xj @xk   j¼1 k¼1 x¼x x¼x þ hdx 3   n X n X @ 2 L  @L    dxj dxk * ¼ 05: or Lðx þ dx; kÞ  Lðx ; kÞ ¼ 0 þ  @xj @xk  @xj x¼x j¼1 k¼1 



x¼x þ hdx

ð10:42Þ L are continuous in If it is assumed that the second-order partial derivatives @x@i @x j *   the neighbourhood of x , then the sign of Lðx þ dx; kÞ  Lðx ; kÞ will be the same  n P n P @2 L  dxj dxk . as that of @xj @xk  j¼1 k¼1  2

x¼x

Therefore, from (10.41) and (10.42), it is clear that the signs of f ðx þ dxÞ   P P 2  f ðx Þ and Q ¼ nj¼1 nk¼1 @x@j @xL k   dxj dxk be the same for all admissible x¼x

variations. If Q > 0 for all admissible variations, then f ðx þ dxÞ  f ðx Þ [ 0 for all admissible variations. That is, f ðx þ dxÞ [ f ðx Þ. ) x is a local minimum. If Q < 0 for all admissible variations, then f ðx þ dxÞ  f ðx Þ\0 or f ðx þ dxÞ \ f ðx Þ for all admissible variations: ) x is a local maximum.

230

10 Constrained Optimization with Equality Constraints

10.5.3 Alternative Approach for Testing Optimality To test the optimality of a constrained optimization problem by the above-described sufﬁcient conditions is a formable task. Therefore, the following two methods are also used to test the optimality. Method-1: In this method, a square matrix HB of order (m + n)  (m + n), called a bordered Hessian matrix, is evaluated at the stationary points. This matrix is arranged as follows: HB ¼

where O is an m  m null matrix, U ¼

O UT h i

U ; V

@gi @xj mn ;

h

i

@2L @xi @xj nn :

Then the sufﬁcient conditions for maximum and minimum stationary points are given as follows. Let ðx ; k Þ be the stationary points for the Lagrangian function and HB be the value of the corresponding bordered Hessian matrix computed at this stationary point. Then: (i) x* is a local maximum if, starting with the leading principal minor of order (2 m + 1), the last (n − m) leading principal minors of HB have alternating sign starting with ð1Þm þ n . (ii) x* is a local minimum if, starting with the leading principal minor of order ð2m þ 1Þ, the last (n − m) leading principal minors of HB have the sign of ð1Þm . Method-2: In this method, optimality is tested by evaluating the roots of the polynomial in z, deﬁned by the following Hancock determinantal equation:    L11  z L12 L13    L1n g11 g21    gm1    L21 L22  z L23    L2n g12 g22    gm2     ..   .    Ln1 Ln2 Ln3    Lnn  z g1n g2n    gmn  ¼ 0   g11 g12 g13    g1n 0 0    0     ..   .    gm1 gm2 gn3    gmn 0 0  0  at x ¼ x and k ¼ k ;

where Ljk ¼ @x@j @xL k ðx ; k Þ; 2



i ðx Þ gij ¼ @g@x . j

10.5

Lagrange Multiplier Method

231

This equation leads to an (n − m)th-order polynomial in z. Then the sufﬁcient conditions for minimum and maximum are, respectively: (i) All positive roots of the polynomial (ii) All negative roots of the polynomial. If some of the roots of this polynomial are positive while the others are negative, the point x* is not an extreme point. Remark It may be observed that the conditions in both the methods are only sufﬁcient for identifying an extreme point but not necessary. That is, a stationary point may be an extreme point without satisfying the conditions described in Methods 1 and 2. Example 5 Solve the following problem by the Lagrange multiplier method: Minimize subject to

z ¼ x21  10x1 þ x22  6x2 þ x23  4x3 x1 þ x2 þ x3 ¼ 7:

Solution The given problem is Minimize subject to

z ¼ x21  10x1 þ x22  6x2 þ x23  4x3 gð xÞ ¼ x1 þ x2 þ x3  7 ¼ 0:

Now the Lagrange function is Lðx1 ; x2 ; x3 ; kÞ ¼ x21  10x1 þ x22  6x2 þ x23  4x3 þ kðx1 þ x2 þ x3  7Þ: ) The necessary conditions for the minimum of z are given by @L ¼ 0; @x1 Now;

@L ¼ 0; @x2

@L ¼ 0; @x3

@L ¼ 0: @k

@L 10  k ¼ 0 gives 2x1  10 þ k ¼ 0 or, x1 ¼ @x1 2

ð10:43Þ

@L ¼ 0 gives 2x2  6 þ k ¼ 0 @x2

or, x2 ¼

6k 2

ð10:44Þ

@L ¼ 0 gives 2x3  4 þ k ¼ 0 @x3

or, x3 ¼

4k 2

ð10:45Þ

@L ¼ 0 gives x1 þ x2 þ x3  7 ¼ 0 or, x1 þ x2 þ x3 ¼ 7 @k

ð10:46Þ

232

10 Constrained Optimization with Equality Constraints

From (10.43), (10.44), (10.45) and (10.46), we have 20  3k ¼ 7 or, k ¼ 2: 2 ) x1 ¼ 4; x2 ¼ 2; x3 ¼ 1: Now

@2L ¼ 2 for i ¼ 1; 2; 3 @x2i @2L ¼ 0 for j ¼ 2; 3 @x1 @xj @2L ¼0 @x2 @xj

for

j ¼ 1; 3

@2L ¼ 0 for j ¼ 1; 2 @x3 @xj @g and ¼ 1 for i ¼ 1; 2; 3: @xi Hence, the bordered Hessian matrix at x1 ¼ 4; x2 ¼ 2 and x3 ¼ 1 is given by HB ¼

O UT

U , where O is a 1  1 null matrix, V 2 @gi @ L U¼ and V ¼ ; @xi @xj 33 @xj 13 2

0 61 i:e: HB ¼ 6 41 1

1 2 0 0

1 0 2 0

3 1 07 7: 05 2

Since m = 1 and n = 3, then n − m = 3 − 1 = 2 and 2 m + 1 = 3. Hence, only two leading principal minors of HB of order 3 and 4 (i.e. D3 and D4 ) are to be evaluated, and both the minors must have negative sign, as ð1Þ1 = negative.   0 1 1   Now; D3 ¼  1 2 0  ¼ 2  2 ¼ 4\0 1 0 2      0 1 1 1 1 0 0 1      1 2 0 0  ¼  1 2 0  þ  1 and D4 ¼      1 0 2 1 1 0 2 0 1 0 0 2 ¼ 4 þ ð4Þ  4 ¼ 12\0

  2 0   1 2 0 0    1 0 0 2 1 0

 0  2  0

10.5

Lagrange Multiplier Method

233

Hence, z is minimum subject to gð xÞ ¼ 0 at x1 ¼ 4; x2 ¼ 2; x3 ¼ 1 and zmin ¼ 16  40 þ 4  12 þ 1  4 ¼ 35. Alternative test for optimality Again, the Hancock determinantal equation is  L13 g11  L23 g12  ¼ 0; L23  z g13  g13 0 

  L11  z L12   L21 L z 22   L31 L 32   g11 g12   2 @gi  L  where Lij ¼ @x@i @x   and gij ¼ @x  j x ¼ x j

x¼x

k ¼ k

or,

or,

 2  z 0 0   0 2z 0    0 0 2z   1 1 1  2  z 0   ð2  zÞ 0 2z   1 1

 1  1  ¼0 1  0   1   0   1  0   0 1

2z 0 1

 0   2z ¼ 0  1 

or ð2  zÞfð2  zÞð1Þ  ð2  zÞg  ð2  zÞ2 ¼ 0 or ð2  zÞð2  zÞð2Þ  ð2  zÞ2 ¼ 0 or  3ð2  zÞ2 ¼ 0 or z ¼ 2; 2: Hence, the extreme point (4, 2, 1) minimizes the objective function, and zmin ¼ 35. Example 6 Solve the following problem by the Lagrange multiplier method: Minimize z ¼ 4x21 þ 2x22 þ x23  4x1 x2 subject to the constraints x1 þ x2 þ x3 ¼ 15; 2x1  x2 þ 2x3 ¼ 20: Solution The given problem is Minimize z ¼ f ð xÞ ¼ 4x21 þ 2x22 þ x23  4x1 x2 subject to g1 ð xÞ ¼ x1 þ x2 þ x3  15 ¼ 0 and

g2 ð xÞ ¼ 2x1  x2 þ 2x3  20 ¼ 0:

234

10 Constrained Optimization with Equality Constraints

Now the Lagrange function is given by L ¼ f ð x Þ þ k1 g1 ð x Þ þ k2 g2 ð x Þ ¼ 4x21 þ 2x22 þ x23  4x1 x2 þ k1 ðx1 þ x2 þ x3  15Þ þ k2 ð2x1  x2 þ 2x3  20Þ: The necessary conditions for the minimum of f ð xÞ are given by @L ¼ 0; @x1

@L ¼ 0; @x2

Now;

@L ¼ 0; @x3

@L ¼ 0; @k1

@L ¼ 0: @k2

@L ¼ 0 gives 8x1  4x2 þ k1 þ 2k2 ¼ 0 @x1

@L ¼ 0 gives 4x2  4x1 þ k1  k2 ¼ 0 @x2

ð10:48Þ

@L ¼ 0 gives 2x3 þ k1 þ 2k2 ¼ 0 @x3

ð10:49Þ

@L ¼ 0 gives x1 þ x2 þ x3  15 ¼ 0 @k1

ð10:50Þ

@L ¼ 0 gives 2x1  x2 þ 2x3  20 ¼ 0: @k2

ð10:51Þ

From (10.47) and (10.48), we have 4x1 þ 2k1 þ k2 ¼ 0 or x1 ¼ 

2k1 þ k2 : 4

From (10.47), we have 4x2 ¼ 8

ð10:47Þ

2k1 þ k2 þ k1 þ 2k2 ¼ 4k1  2k2 þ k1 þ 2k2 ¼ 3k1 4 3 or x2 ¼  k1 4

From (10.49), we have x3 ¼ 

k1 þ 2k2 : 2

Now putting the values of x1 ; x2 ; x3 in (10.50), we get

10.5

Lagrange Multiplier Method



235

2k1 þ k2 3 k1 þ 2k2  k1  ¼ 15 4 4 2 or 7k1 þ 5k2 þ 60 ¼ 0:

ð10:52Þ

Putting the values of x1 ; x2 ; x3 in (10.51), we get 2k1 þ k2 3 k1 þ 2k2 þ k1  2 ¼ 20 4 4 2 or k1 þ 2k2 þ 16 ¼ 0: 2

ð10:53Þ

Now solving (10.52) and (10.53) by the cross-multiplication method, we have k1 k2 1 ¼ ¼ 80  120 60  112 14  5 40 52 and k2 ¼  : or k1 ¼  9 9  2k1 þ k2 1 40 52 11 2   ) x1 ¼  ¼ ¼ 4 9 9 3 4   3 3 40 10  x 2 ¼  k1 ¼  ¼ 4 4 9 3   k1 þ 2k2 1 40 52  2 and x3 ¼  ¼ ¼ 8: 2 9 9 2 @2L @2L @2L ¼8 ¼ 4 ¼0 Now 2 @x1 @x2 @x1 @x3 @x1 @2L ¼4 @x22

@2L ¼ 4 @x2 @x1

@2L ¼0 @x2 @x3

@2L @2L @2L ¼2 ¼0 ¼ 0: 2 @x3 @x1 @x3 @x2 @x3 @g1 @g1 @g1 @g2 ¼1 ¼1 ¼1 ¼2 Again; @x1 @x2 @x3 @x1

@g2 ¼ 1 @x2

@g2 ¼ 2: @x3

10 Hence, the bordered Hessian matrix at x1 ¼ 11 3 ; x2 ¼ 3 ; x3 ¼ 8 is given by

HB ¼

O UT

U ; V

where O is an m  m null matrix, U¼

@gi @xj

and

mn

@2L @xi @xj

; nn

236

10 Constrained Optimization with Equality Constraints

2

0 60 6 i:e: HB ¼ 6 61 41 1

0 0 2 1 2

1 2 8 4 0

1 1 4 4 0

3 1 27 7 07 7: 05 2

Since m = 2 and n = 3, then n − m = 1 and 2 m + 1 = 5. Thus, only one leading principal minor of HB of order 5 (i.e. D5 ) is to be evaluated, and the minor must have positive sign, as ð1Þm ¼ positive. Here D5 ¼ jHB j ¼ 90 [ 0. Hence, f ð xÞ is minimum subject to the constraints g1 ð xÞ ¼ 0 and g2 ð xÞ ¼ 0 at 112  2 10 10 820 x1 ¼ 11 þ 2 10 þ ð8Þ2 4  11 3 ; x2 ¼ 3 and x3 ¼ 8, and fmin ¼ 4 3 3 3  3 ¼ 9 . Example 7 Solve the following problem by using the Lagrange multiplier method: Minimize z ¼ 2x21  24x1 þ 2x22  8x2 þ 2x23  12x3 þ 200 subject to the constraints x1 þ x2 þ x3 ¼ 11 and x1 ; x2 ; x3  0: Solution The given problem can be written as Minimize z ¼ 2x21  24x1 þ 2x22  8x2 þ 2x23  12x3 þ 200 subject to gð xÞ ¼ x1 þ x2 þ x3  11 ¼ 0: Now the Lagrange function is Lðx1 ; x2 ; x3 ; kÞ ¼ 2x21  24x1 þ 2x22  8x2 þ 2x23  12x3 þ 200 þ kðx1 þ x2 þ x3  11Þ:

The necessary conditions for the minimum of Z are given by @L ¼ 0; @x1 Now;

@L ¼ 0; @x2

@L ¼ 0; @x3

@L ¼ 0: @k

@L 24  k ¼ 0 gives 4x1  24 þ k ¼ 0 ) x1 ¼ @x1 4

ð10:54Þ

@L 8k ¼ 0 gives 4x2  8 þ k ¼ 0 ) x2 ¼ @x2 4

ð10:55Þ

@L 12  k ¼ 0 gives 4x3  12 þ k ¼ 0 ) x3 ¼ @x3 4

ð10:56Þ

10.5

Lagrange Multiplier Method

237

@L ¼ 0 gives x1 þ x2 þ x3  11 ¼ 0: @k

ð10:57Þ

Substituting the values of x1 ; x2 and x3 [from (10.54), (10.55), (10.56)] in (10.57), we have 44  3k ¼ 11 ) k ¼ 0 4 ) x1 ¼ 6; x2 ¼ 2; x3 ¼ 3 Now;

@2L ¼ 4; @x21

@2L ¼ 0; @x1 @xj

j ¼ 2; 3

@2L ¼ 4; @x22

@2L ¼ 0; @x2 @xj

j ¼ 1; 3

@2L @2L ¼ 4; ¼ 0; j ¼ 1; 2: 2 @x3 @xj @x3 @g @g @g ¼ 1; ¼ 1; ¼ 1: Again; @x1 @x2 @x3 Now we shall test the optimality by the bordered Hessian matrix method. The bordered Hessian matrix at x1 ¼ 6; x2 ¼ 2; x3 ¼ 3 is given by

U ; V

O HB ¼ UT where 0 is an m  m matrix, U ¼

h i

@gi @xj mn

2

0 61 i:e: HB ¼ 6 41 1

and V ¼ 1 4 0 0

1 0 4 0

h

i

@2L @xi @xj nn ,

3 1 07 7: 05 4

Since m = 1, n = 3, n − m = 3 − 1 = 2 and 2m + 1 = 3, only two leading principal minors of HB of order 3 and 4 (i.e. D3 and D4 ) are to be evaluated, and both the minors must have negative sign as ð1Þ1 ¼ 1 = negative.   0 1 1   Now; H3 ¼  1 4 0  ¼ 4  4 ¼ 8\0 1 0 4

238

10 Constrained Optimization with Equality Constraints

 0  1  and H4 ¼ jHB j ¼  1  1

1 4 0 0

  1 1 1    0 0   ¼ 1 1  4 0  1  0 4

  0 0   1   4 0þ1   0 4 1

4 0 0

  0   1 4   0  1 0   4 1 0

 0   4  0

¼ 16 þ ð4Þð4  0Þ ¼ ð4Þð0  4Þ ¼ 16  16  16 ¼ 48\0: Hence, z is minimum subject to the given constraint at x1 ¼ 6; x2 ¼ 2; x3 ¼ 3 and Zmin ¼ 102. Alternative test for optimality For the given problem, the Hancock determinantal equation at x1 ¼ 6; x2 ¼ 2; x3 ¼ 3 is given by   L11  z L12   L21 L z 22   L31 L 32   g11 g12

 L13 g11  L23 g12  ¼ 0; L33  z g13   g13 0

  @ 2 L  @gi  where Lij ¼ gij ¼  ; @xi @xj  x ¼ x  @xj x¼x k¼k   4  z 0 0 1    0 4z 0 1  ¼0 or  0 0 4  z 1    1 1 1 0  4z   or ð4  zÞ 0   1

  1   0 4  z   0 4 z 1 0     1 1 1 0 0

 0   4  z ¼ 0  1 

or ð4  zÞ½ðz  4Þ  ð4  zÞ  ð4  zÞð4  zÞ ¼ 0 or ð4  zÞðz  4  4 þ zÞ  ð4  zÞ2 ¼ 0 or ð4  zÞ2 ¼ 0 or z ¼ 4; 4: Since all the values of z are positive, Z is minimum subject to the given constraint at x1 ¼ 6; x2 ¼ 2; x3 ¼ 3, and Zmin ¼ 102. Example 8 Find the dimension of a cylindrical tin (with top and bottom) made up of sheet metal to maximize its volume such that the total surface area is equal to 24p.

10.5

Lagrange Multiplier Method

239

Solution Let x1 be the radius of the top or bottom and x2 be the height of the tin. Then the volume of the tin ¼ px21 x2 and the total surface area ¼ 2px21 þ 2px1 x2 . Therefore, the problem is f ðx1 ; x2 Þ ¼ px21 x2 2px21 þ 2px1 x2 ¼ 24p px21 þ px1 x2  12p ¼ 0 g1 ðx1 ; x2 Þ ¼ px21 þ px1 x2  12p ¼ 0:

maximize subject to or or

Now, the Lagrange function is   Lðx1 ; x2 ; kÞ ¼ px21 x2 þ k px21 þ px1 x2  12p : The necessary conditions for the maximum of f are given by @L ¼ 0; @x1 Now;

@L ¼ 0; @x2

@L ¼ 0: @k

@L ¼ 0 implies 2px1 x2 þ kpð2x1 þ x2 Þ ¼ 0: @x1 Again;

@L ¼ 0 gives px21 þ kpx1 ¼ 0 @x2 or; px21 þ kpx1 ¼ 0

ð10:58Þ

ð10:59Þ

or; x1 þ k ¼ 0 ½* x1 6¼ 0 or; x1 ¼ k: Again;

@L ¼ 0 gives x21 þ x1 x2  12 ¼ 0: @k

Now setting x1 ¼ k in (10.58), we have 2pðkÞx2 þ kpð2k þ x2 Þ ¼ 0 or  pkx2 ¼ 2k2 p or x2 ¼ 2k ½* k 6¼ 0 or x2 ¼ 2x1 ½From ð10:59Þ; x1 ¼ k or, k ¼ x1 :

ð10:60Þ

240

10 Constrained Optimization with Equality Constraints

Setting x2 ¼ 2x1 in (10.60), we have x21 þ x1  2x1  12 ¼ 0 or, x21 ¼ 4 or x1 ¼ 2 ½As x1 is the radius of the top or bottom of the tin: ) x2 ¼ 2x1 ¼ 2:2 ¼ 4 ) x ¼ ð2; 4Þ: From (10.59), we have x1 = −k or, k = −2. Now the Hancock determinantal equation is  g11  g12  ¼ 0: 0 

  L11  z L12   L21 L 22  z   g11 g12

We have

 @ 2 L ¼ L11 @x21  L12

x¼x k¼2

 @ 2 L  ¼ @x1 @x2 

ð10:61Þ

¼ ½2px2 þ 2pkx0 ¼ 2p  4 þ 2p  ð2Þ ¼ 8p  4p ¼ 4p

x¼x k¼2

¼ ½2px1 þ 2px0 ¼ 2p  2 þ ð2Þp ¼ 2p:

Similarly, L21 = 2p.  2  Again; L22 ¼ @@xL2  2

x¼x k¼2

¼0

Therefore, (10.61) becomes    4p  z 2p 8p     0  z 2p  ¼ 0  2p    8p 2p 0        or; ð4p  zÞ 0  4p2  2p 0  16p2 þ 8p 4p2 þ 8pz ¼ 0 or; 68p2 z þ 48p3 ¼ 0 or; 68p2 z ¼ 48p3

or;

z¼

48 12 p ¼  p: 68 17

Since the value of z is negative, f is maximum for x ¼ ð2; 4Þ.

10.6

10.6

Interpretation of Lagrange Multipliers

241

Interpretation of Lagrange Multipliers

To ﬁnd the physical meaning of the Lagrange multipliers, let us consider the following optimization problem involving only one constraint: Minimize f ð xÞ subject to hð xÞ ¼ b; i:e: Minimize f ð xÞ

ð10:62Þ

subject to gð xÞ ¼ bhð xÞ ¼ 0;

ð10:63Þ

where b is a constant. The necessary conditions for the minimum of f (x) are given by @L @f @g ¼ þk ¼ 0; @xj @xj @xj and

j ¼ 1; 2; . . .; n

ð10:64Þ

@L ¼ g ¼ 0; @k

where Lðx; kÞ ¼ f ð xÞ þ kgð xÞ:

ð10:65Þ

Let the solution of (10.64) and (10.65) be given by x*, k* and f  ¼ f ðx Þ. Now, we want to ﬁnd the effect of a small relaxation or tightening of the constraint on the optimum value of the objective function (i.e. we want to ﬁnd the effect of a small change in b on f*). From (10.63), g(x) = 0, where g(x) = b – h(x). ) dg ¼ 0; which implies dbdh ¼ 0 or; db ¼ dh; i:e: db ¼

n X @h j¼1

@xj

dxj :

ð10:66Þ

Equation (10.64) can be rewritten as @f @h k ¼0 @xj @xj

or,

From (10.66) and (10.67), we have

@h 1 @f ¼ ; @xj k @xj

j ¼ 1; 2; . . .; n:

ð10:67Þ

242

10 Constrained Optimization with Equality Constraints

db ¼

"

n X 1 @f j¼1

or, k ¼

df dxj ¼ k @xj k

df db

or, k ¼

Since df ¼

n X @f

dxj

ð10:68Þ

or, df  ¼ k db:

ð10:69Þ

j¼1

df  db

#

@xj

Thus, k denotes the sensitivity (or rate of change) of f with respect to b or the marginal or incremental change in f  with respect to b at x*. In other words, k indicates how the constraint can be relaxed or tightly binding at the optimum point. Depending on the value of k , three cases may arise: Case I (k [ 0), Case II  (k \0), Case III (k ¼ 0). Case-I: k [ 0 In this case, for a unit decrease in b, the decrease in f  will be exactly equal to k as df ¼ k ð1Þ ¼ k . Hence, k may be interpreted as the marginal gain (or reduction) in f  due to the tightening of the constraint. On the other hand, if b is increased by one unit, f will be increased to a new optimum level by the amount df ¼ k ð þ 1Þ ¼ k . In this case, k may be interpreted as the marginal cost (increase) in f  due to relaxation of the constraint. Case-II: k \0 In this case, the optimum value of f will be decreased with the increase of b. Here, the marginal gain (reduction) in f* due to a relaxation of the constraint by one unit is determined by the value of k as df  ¼ k ð þ 1Þ\0. If b is decreased by one unit, the marginal cost (increase) in f* by the tightening of the constraint is df  ¼ k ð1Þ [ 0, since, in this case, the minimum value of the objective function increases. Case-III: k ¼ 0 In this case, any incremental change in b has absolutely no effect on the optimum value of f, and hence the constraint will not be binding. This means that the optimization of f subject to g = 0 leads to the same optimum point x* as with the unconstrained optimization of f. In economics and operations research, Lagrange multipliers are known as shadow prices of the constraints, since they indicate the changes in optimal value of the objective function per unit change in the right-hand side of the equality constraints. Procedure for ﬁnding the effect of changes of b on f* This can be done in two ways: (i) For ﬁnding the effect of changes of b (right-hand side of the constraints) on the optimal value f*, one can solve the problem with the new value of b. (ii) Another procedure is to use the relation df  ¼ k db, where k is the optimal value.

10.6

Interpretation of Lagrange Multipliers

243

Example 9 Find the maximum of the function f (x) = 2x1 + x2 + 10 subject to gð xÞ ¼ x1 þ 2x22 ¼ 3 using the Lagrange multiplier method. Also, ﬁnd the effect of changing the right-hand side of the constraint on the optimum value of f. Solution Here the Lagrange function is given by L ¼ f ð x Þ þ k gð x Þ

  ¼ 2x1 þ x2 þ 10 þ k 3  x1  2x22 :

ð10:70Þ

The necessary conditions for the maximum of f (x) are given by @L ¼ 0; @x1

@L ¼ 0; @x2

@L ¼ 0: @k

ð10:71Þ

@L ¼ 0 gives 2  k ¼ 0 or k ¼ 2 @x1 @L 1 1 ¼ ¼ 0:13 ¼ 0 gives 1  4k x2 ¼ 0 or x2 ¼ @x2 4k 8 @L 1 95 2 ¼ 0 gives x1 þ 2x2  3 ¼ 0 or x1 ¼ 3  2x22 ¼ 3  ¼ ¼ 2:97: @k 32 32

Now;

Hence, the solution of (10.71) is given by x1 ¼ 2:97;

x2 ¼ 0:13;

k ¼ 2:

To test the sufﬁcient condition, solve the Hancock determinantal equation for ﬁnding the value of z. The value of z will be negative. Hence, f will be maximum for x1 ¼ 2:97; x2 ¼ 0:13, and the maximum value of f will be f  ¼ f ðx Þ ¼ 16:07: Second part: When the original constraint is tightened by one unit, i.e. db = −1, df  ¼ k db ¼ 2ð1Þ ¼ 2:  Due to this change, the new value of f* is fold þ df  ¼ 16:07  2 ¼ 14:07.

On the other hand, if we relax the original constraint by one unit, i.e. db = + 1, df  ¼ k db ¼ 2ð1Þ ¼ 2:  Due to this change, the new value of f* is fold þ df  ¼ 16:07 þ 2 ¼ 18:07

244

10 Constrained Optimization with Equality Constraints

Example 10 Find the maximum of the function f ð xÞ ¼ 4x1 þ x2 þ 15 subject to gð xÞ ¼ 5  x1  4x22 ¼ 0, using the Lagrange multiplier method. Also, ﬁnd the effect of changing the right-hand side of the constraint on the optimum value of f. Solution The Lagrange function is given by L ¼ f ð x Þ þ k gð x Þ

  ¼ 4x1 þ x2 þ 15 þ k 5  x1  4x22 :

ð10:72Þ

The necessary conditions for the maximum of f ð xÞ are given by @L ¼ 0; @x1 Now;

@L ¼ 0; @x2

@L ¼ 0: @k

ð10:73Þ

@L ¼ 0 gives 4  k ¼ 0 ) k ¼ 4 @x1 @L 1 1 ¼ ¼ 0:03 ¼ 0 gives 1  8kx2 ¼ 0 ) x2 ¼ @x2 8k 32 @L 4 ¼ 0 gives 5  x1  4x22 ¼ 0 ) x1 ¼ 4x22 þ 5 ¼ 5  ¼ 4:99: @k ð32Þ2

Hence, the solution of (10.73) is given by x1 ¼ 4:99;

x2 ¼ 0:03; k ¼ 4:

To test the sufﬁcient condition, solve the Hancock determinantal equation for ﬁnding the value of z. Now;

@2L ¼ 0; @x21

@2L ¼0 @x1 @x2

@2L @2L ¼ 8k ¼ 32; ¼0 2 @x2 @x1 @x2 @g @g and ¼ 1; ¼ 8x2 ¼ 0:24: @x1 @x2 Now the Hancock determinantal equation at x1 ¼ 4:99; x2 ¼ 0:03 and k ¼ 4 is given by  0  z   0   1

0 32  z 0:24

 1  0:24  ¼ 0 0 

10.6

Interpretation of Lagrange Multipliers

or, z ¼ 

245

32 ¼ 30:26; 1:0576

since the value of z is negative. Hence, f will be maximum for x1 ¼ 4:99; x2 ¼ 0:03, and the maximum value of f will be f  ¼ f ðx Þ ¼ 34:99 Second part: When the original constraint is tightened by one unit, i.e. db ¼ 1, df  ¼ k db ¼ 4ð1Þ ¼ 4: Due to this change, the new value of f  is  þ df  ¼ 34:99  4 ¼ 30:99: fold

On the other hand, if we relax the original constraint by one unit, i.e. db ¼ þ 1, df  ¼ k db ¼ 4ð1Þ ¼ 4: Due to this change, the new value of f  is  þ df  ¼ 34:99 þ 4 ¼ 38:99: fold

10.7

Applications of Lagrange Multiplier Method

Financial mathematics: Constrained optimization plays an important role in ﬁnancial mathematics; e.g. the choice problem for a consumer is represented as one of maximizing a utility function subject to a budget constraint. The Lagrange multiplier has an economic interpretation as the shadow price associated with the constraint, since the Lagrange multipliers indicate the changes in the optimal value of the objective function per unit change on the right-hand side of the equality constraint. Other examples include proﬁt maximization for a ﬁrm, along with various applications in the branch of ﬁnancial mathematics dealing with performance, structure, behaviour and decision-making of the whole economy. Control theory: The Lagrange multiplier method is used in optimal control theory to ﬁnd the best possible control for taking a dynamical system from one state to another, especially in the presence of constraints for the state or input controls.

246

10 Constrained Optimization with Equality Constraints

10.8

Exercises

1. Solve the following problems by the Lagrange multiplier method: ðiÞ

Optimize z ¼ x21  10x1 þ x22  6x2 þ x23  4x3 subject to x1 þ x2 þ x3 ¼ 7:

ðiiÞ

Optimize z ¼ 3x21 þ x22 þ x23 subject to x1 þ x2 þ x3 ¼ 2:

2. Solve the problem by the Lagrange multiplier method: Minimize z ¼ 4x21 þ 2x22 þ x23  4x1 x2 subject to the constraints x1 þ x2 þ x3 ¼ 15 2x1  x2 þ 2x3 ¼ 20: 3. Solve the following problem by the Lagrange multiplier method: Minimize z ¼ x21 þ x22 þ x23 subject to the constraints x1 þ x2 þ 3x3 ¼ 2 5x1 þ 2x2 þ x3 ¼ 5 4. Solve the problem by the Lagrange multiplier method: Maximize z ¼ 6x1 þ 8x2 þ x21  x22 subject to the constraints 4x1 þ 3x2 ¼ 16 3x1 þ 5x2 ¼ 15: 5. Solve the following optimization problems: (i) Minimize z ¼ x1 x2 x3 subject to the constraints subject to x1 þ x2 þ x3 ¼ 1 and x1 ; x2 ; x3 [ 0:

10.8

Exercises

(ii) Minimize z ¼ x21 þ x22 subject to the constraint x1 þ x2 ¼ 2 : (iii) Maximize z ¼ x1 þ x2 subject to the constraint x21 þ x22 ¼ 1:  1 2 x1 þ x22 þ x23 2 subject to the constraint x1 þ x2 þ x3 ¼ 1:

(iv) Minimize z ¼

(v) Minimize z ¼ 2x21 þ 2x22 þ x23 þ 2x1 x2 þ 2x1 x3  8x1  6x2  4x3 þ 9 subject to the constraint x1 þ x2 þ 2x3 ¼ 3 :

247

Chapter 11

Constrained Optimization with Inequality Constraints

11.1

Objective

The objective of this chapter is to derive the Kuhn-Tucker necessary and sufﬁcient conditions to solve multivariate constrained optimization problems with inequality constraints.

11.2

Introduction

Let us consider a multivariate minimization problem as follows: Minimize f ð xÞ subject to gi ð xÞ  0;

i ¼ 1; 2; . . .; m: :

where x 2 Rn ; f ; gi : Rn ! R Now adding the non-negative slack variables, y2i in the given inequality constraints, we get the equality constraints as gi ð xÞ þ y2i ¼ 0;

i ¼ 1; 2; . . .; m:

Let Gi ðx; yÞ ¼ gi ð xÞ þ y2i ¼ 0; i ¼ 1; 2; . . .; m, where y ¼ ðy1 ; y2 ; . . .; ym ÞT is the vector of slack variables. Therefore, the reduced problem is Minimize f ð xÞ subject to Gi ðx; yÞ ¼ gi ð xÞ þ y2i ¼ 0;

i ¼ 1; 2; . . .; m:

This problem can be solved by the method of Lagrange multiplier. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_11

249

250

11 Constrained Optimization with Inequality Constraints

In this case, the Lagrange function is Lðx; y; kÞ ¼ f ð xÞ þ

m X

ki Gi ðx; yÞ;

i¼1

where k ¼ ðk1 ; k2 ;    ; km ÞT is the vector of Lagrange multiplier. Now, the necessary conditions for a stationary point to be a local minimum are given by @L ¼ 0; @xj @L ¼ 0; @ki @L and ¼ 0; @yi i:e:

j ¼ 1; 2; . . .; n i ¼ 1; 2; . . .; m i ¼ 1; 2; . . .; m;

m X @f @gi þ ki ¼ 0; @xj @xj i¼1

gi ð xÞ þ y2i ¼ 0;

j ¼ 1; 2; . . .; n

i ¼ 1; 2; . . .; m

ð11:1Þ ð11:2Þ

and 2ki yi ¼ 0;

i ¼ 1; 2; . . .; m:

ð11:3Þ

From (11.3), we have ki yi ¼ 0. Hence, either ki ¼ 0 or yi ¼ 0; i ¼ 1; 2; . . .; m. For any i, if ki ¼ 0, it means that the corresponding constraint is inactive and hence can be ignored. On the other hand, if yi ¼ 0, it means that the constraint is active at the optimum point. Now, we divide the indices of the constraints into two subsets J1 and J2. Let the set J1 indicate the indices of those constraints which are active at the optimum point, and let J2 include the indices of all the inactive constraints, i:e: J1 ¼ fi: yi ¼ 0; i ¼ 1; 2; . . .; mg J2 ¼ fi: ki ¼ 0; i ¼ 1; 2; . . .; mg; i.e. i 2 J1 when yi ¼ 0 and i 2 J2 when ki ¼ 0; i ¼ 1; 2; . . .; m. Hence, (11.1) becomes X @gi @f þ ki ¼ 0; j ¼ 1; 2; . . .; n: @xj @xj i2J1

ð11:4Þ

11.2

Introduction

251

Similarly, (11.2) becomes gi ð xÞ ¼ 0 for i 2 J1

ð11:5Þ

gi ð xÞ þ y2i ¼ 0 for i 2 J2 :

ð11:6Þ

and

If the number of active constraints is p, then the number of inactive constraints is m − p. Therefore, in (11.4), (11.5), and (11.6), the number of equations are n + p + m − p = m + n and the unknowns are x1, x2, …, xn, ki ði 2 J1 Þ and yi ði 2 J2 Þ. Therefore, the number of unknowns are n + p + m − p = m + n. Hence, we can solve the system (11.4), (11.5), and (11.6) easily. Let  rf ¼

@f @f @f  @x1 @x2 @xn

T

 and rgi ¼

@gi @gi @gi  @x1 @x2 @xn

T

be the gradients of the objective function and the ith constraint respectively. Then Eq. (11.4) may be written as rf ¼

X

ki rgi :

ð11:7Þ

i2J1

Thus, the negative of the gradient of the objective function can be expressed as a linear combination of the gradients of the active constraints at the optimum point. Now, it can be proved that: (a) (b) (c) (d)

For For For For

11.3

a a a a

minimization problem with gi ð xÞ  0, we have ki  0. maximization problem with gi ð xÞ  0, we have ki  0. minimization problem with gi ð xÞ  0, we have ki  0. maximization problem with gi ð xÞ  0, we have ki  0.

Feasible Direction

The direction of a vector S is called a feasible direction from a point x ¼ ðx1 ; x2 ; . . .; xn Þ if at least a small step can be taken along S that does not immediately leave the feasible region. Thus, for problems with sufﬁciently smooth constraint surfaces, any vector S satisfying the relation \ST ; rgi [ \0

252

11 Constrained Optimization with Inequality Constraints

g1 = 0

Fig. 11.1 Feasible direction S when both constraint functions are convex

g2 = 0 g10

g1

g2

g1 = 0

Fig. 11.2 Feasible direction S when one constraint function is linear

g10

0

90

g1

g2

Angle greater than 900

g1 = 0

Fig. 11.3 Feasible direction S when one constraint function is concave and other one is convex

g10

900 g1

Angle greater than 900

g2

11.3

Feasible Direction

253

can be called a feasible direction (see Fig. 11.1). On the other hand, if the constraint is either linear or concave (see Figs. 11.2 and 11.3), any vector satisfying the relation \ST ; rgi [\0 is called a feasible direction. Let us suppose that only two constraints are active at the optimum point, i.e. p = 2. Then Eq. (11.7) reduces to rf ¼ k1 rg1 þ k2 rg2 :

ð11:8Þ

Let S be a feasible direction at the optimum point. Now by premultiplying both sides of (11.8) by ST, we have \ST ; rf [ ¼ k1 \ST ; rg1 [ þ k2 \ST ; rg2 [ : Since S is a feasible direction, then we have \ST ; rg1 [\0

and \ST ; rg2 [\0:

ð11:9Þ

Thus, if k1 [ 0 and k2 [ 0, the quantity \ST ; rf[ is always positive. As rf indicates the gradient direction along which the values of the function f increase at the maximum rate, \ST ; rf[ represents the component of the increment of f along the direction S. If \ST ; rf [ [ 0, the function value increases as we move along the direction S. Hence, if k1 ; k2 are positive, we will not be able to ﬁnd any direction in the feasible region along which the function value can further be decreased. Since the point at which Eq. (11.9) is valid is assumed to be optimum, k1 and k2 have to be positive. This reason can be extended to cases where there are more than two active constraints.

11.4

Kuhn-Tucker Conditions

In this section, we shall derive the necessary and sufﬁcient conditions for a local optimum of the general non-linear programming problem.

11.4.1 Kuhn-Tucker Necessary Conditions Let us consider a minimization problem as follows: Minimize f ð xÞ subject to gi ð xÞ  0;

i ¼ 1; 2; . . .; m:

wherex 2 Rn ; f ; gi : Rn ! R

254

11 Constrained Optimization with Inequality Constraints

Now adding the non-negative slack variables y2i in the given inequality constraints, we get the equality constraints as gi ð xÞ þ y2i ¼ 0;

i ¼ 1; 2; . . .; m:

Therefore, the reduced problem is Minimize f ð xÞ subject to gi ð xÞ þ y2i ¼ 0;

i ¼ 1; 2; . . .; m:

This problem can be solved by the Lagrange multiplier method. In this case, the Lagrange function is Lðx; y; kÞ ¼ f ð xÞ þ

m X   ki gi ð xÞ þ y2i ; i¼1

where k ¼ ðk1 ; k2 ; . . .; km ÞT is the vector of Lagrange multipliers. Now the necessary conditions for a stationary point to be a local minimum are given by @L ¼ 0; @xj @L ¼ 0; @ki @L and ¼ 0; @yi i:e:

j ¼ 1; 2; . . .; n i ¼ 1; 2; . . .; m i ¼ 1; 2; . . .; m;

m X @L @gi þ ki ¼ 0; j ¼ 1; 2; . . .; n @xj @xj i¼1

ð11:10Þ

gi ð xÞ þ y2i ¼ 0;

ð11:11Þ

i ¼ 1; 2; . . .; m

and 2ki yi ¼ 0;

i ¼ 1; 2; . . .; m:

ð11:12Þ

If a constraint is satisﬁed with an equality sign gi ð xÞ ¼ 0 at the optimum point, then it is called the active constraint. Otherwise, it will be called inactive. Equation (11.12) implies that either ki ¼ 0 or yi ¼ 0. Now, if ki ¼ 0 and y2i ¼ 0, then the ith constraint is inactive and can be discarded. If yi ¼ 0; ki [ 0, then Eq. (11.11) implies gi ð xÞ ¼ 0 at the optimum point. This means that the constraint is active. Therefore, Eqs. (11.11) and (11.12) imply

11.4

Kuhn-Tucker Conditions

255

ki gi ð xÞ ¼ 0:

ð11:13Þ

Again, y2i  0. Then from (11.11), we have gi ð xÞ  0:

ð11:14Þ

Thus, (11.13) implies that if gi ð xÞ\0, then ki ¼ 0 and when gi ð xÞ ¼ 0; ki [ 0. Hence, the necessary conditions to be satisﬁed at a local minimum point are as follows: m X @f @gi þ ki ¼ 0; @xj @xj i¼1

j ¼ 1; 2; . . .; m

ki g i ð x Þ ¼ 0 gi ð x Þ  0 ki  0; i ¼ 1; 2; . . .; m: This set of necessary conditions is known as the Kuhn-Tucker necessary conditions. Note: The ﬁrst two conditions will be the same for both minimization as well as maximization problems. However, the last two conditions will be different, and these are dependent on the type of problem. The constraint conditions are given in the table. Type of problem

Type of constraint

Form of last two necessary conditions

Minimization Minimization Maximization Maximization

gi ð xÞ  0 gi ð xÞ  0 gi ð xÞ  0 gi ð xÞ  0

gi ð xÞ  0; ki  0 gi ð xÞ  0; ki  0 gi ð xÞ  0; ki  0 gi ð xÞ  0; ki  0

11.4.2 Kuhn-Tucker Sufﬁcient Conditions Theorem The Kuhn-Tucker necessary conditions for the optimization problem Minimize f ð xÞ subject to the constraints gi ð xÞ  0; i ¼ 1; 2; . . .; m are also sufﬁcient conditions if f(x) is convex and all gi(x) are convex functions of x. Proof The Kuhn-Tucker necessary conditions for the given problem

256

11 Constrained Optimization with Inequality Constraints

Minimize f ð xÞ subject to gi ð xÞ  0; i ¼ 1; 2; . . .; m m X @f @gi þ ki ¼ 0; j ¼ 1; 2; . . .; n are @xj @xj i¼1 ki gi ð xÞ ¼ 0; i ¼ 1; 2; . . .; m gi ð xÞ  0; i ¼ 1; 2; . . .; m ki  0;

i ¼ 1; 2; . . .; m:

Now, we have to prove that these conditions are sufﬁcient if f(x) is convex and all gi(x) are convex functions of x. The Lagrangian function of the given problem is Lðx; y; kÞ ¼ f ð xÞ þ

m X   ki gi ð xÞ þ y2i : i¼1

If ki  0, then ki gi ð xÞ is convex. Further, since ki yi ¼ 0, we have gi ð xÞ þ y2i ¼ 0. Thus, it follows that Lðx; y; kÞ is a convex function. Again, the necessary condition for f(x) to be minimum at x is that Lðx; y; kÞ has a stationary point. However, if Lðx; y; kÞ is convex, its partial derivatives must vanish at only one point. According to the convexity analysis, this point must be a global minimum point. Hence, the Kuhn-Tucker necessary conditions are also sufﬁcient conditions for a global minimum of f(x) at x if f(x) and all gi(x) are convex. Example 1 Determine x1, x2, x3 so as to maximize z ¼ x21  x22  x23 þ 4x1 þ 6x2 subject to the constraints x1 þ x2  2; 2x1 þ 3x2  12 and x1 ; x2 ; x3  0: Solution Let f ¼ x21  x22  x23 þ 4x1 þ 6x2 subject to g1 ¼ x1 þ x2  2  0; g2 ¼ 2x1 þ 3x2  12  0: Hence, the Kuhn-Tucker necessary conditions are

11.4

Kuhn-Tucker Conditions

257

2 X @f @gi þ ki ¼ 0; @xj @xj i¼1

ki gi ¼ 0;

j ¼ 1; 2; 3

i ¼ 1; 2

ð11:15Þ ð11:16Þ

gi  0;

i ¼ 1; 2

ð11:17Þ

k1  0;

k2  0:

ð11:18Þ

From (11.15), we have 2x1 þ 4 þ k1 þ 2k2 ¼ 0

ð11:19aÞ

2x2 þ 6 þ k1 þ 3k2 ¼ 0

ð11:19bÞ

2x3 ¼ 0:

ð11:19cÞ

From (11.16), we have ki gi ¼ 0; i ¼ 1; 2, i:e: k1 ðx1 þ x2  2Þ ¼ 0

ð11:20aÞ

k2 ð2x1 þ 3x2  12Þ ¼ 0:

ð11:20bÞ

From (11.17), we have gi  0; i ¼ 1; 2, i:e: x1 þ x2  2  0

ð11:21aÞ

2x1 þ 3x2  12  0:

ð11:21bÞ

and

According to (11.18), four different cases may arise: Case 1: k1 ¼ 0;

k2 ¼ 0

Case 2: k1 ¼ 0;

k2 6¼ 0

Case 3: k1 6¼ 0;

k2 ¼ 0

Case 4: k1 6¼ 0; k2 6¼ 0. Now we shall examine all the cases. Case 1: k1 ¼ 0; k2 ¼ 0 In this case, from (11.19a), (11.19b) and (11.19c), we have 2x1 þ 4 ¼ 0; 2x2 þ 6 ¼ 0 and x3 ¼ 0, i.e. x1 ¼ 2; x2 ¼ 3; x3 ¼ 0. However, this solution violates both constraints (11.21a) and (11.21b). So, this solution is rejected. Case 2: k1 ¼ 0; k2 6¼ 0

258

11 Constrained Optimization with Inequality Constraints

In this case, (11.20a) and (11.20b) give 2x1 þ 3x2  12 ¼ 0, and (11.19a) and (11.19b) give 2x1 þ 4 þ 2k2 ¼ 0, 2x2 þ 6 þ 3k2 ¼ 0, i.e. x1 ¼ k2 þ 2 and x2 ¼ 3 2 k2 þ 3. Substituting these results in 2x1 þ 3x2 ¼ 12, we have   3 2ðk2 þ 2Þ þ 3 k2 þ 3 ¼ 12 2 9 13 2 k2 ¼ 1 or; k2 ¼  \0: or 2k2 þ 4 þ k2 þ 9 ¼ 12 or 2 2 13   2 24 3 3 2 3 36 þ 2 ¼ ; x 2 ¼ k2 þ 3 ¼  þ3 ¼ : ) x1 ¼  þ3 ¼  13 13 2 2 13 13 13 Again, from (11.19c), x3 = 0 ) x1 ¼

24 36 ; x2 ¼ ; x3 ¼ 0: 13 13

This solution violates the constraint x1 þ x2  2  0. So this solution is rejected. Case 3: k1 6¼ 0; k2 ¼ 0 In this case, (11.20a) gives x1 þ x2 ¼ 2, and (11.19a) and (11.19b) give x1 ¼ 2 þ k21 ; x2 ¼ 3 þ k21 . k1 k1 þ3þ ¼ 2 or, k1 ¼ 3\0 2 2 3 1 3 3 ) x1 ¼ 2  ¼ ; x2 ¼ 3  ¼ : 2 2 2 2

) 2þ

Again from (11.19c), we have x3 = 0. Hence, the solution is x1 ¼ 12 ; x2 ¼ 32 ; x3 ¼ 0. This solution satisﬁes the constraint 2x1 þ 3x2  12. Case 4: k1 6¼ 0; k2 6¼ 0 In this case, (11.20a) and (11.20b) give x1 + x2 = 2, 2x1 + 3x2 = 12. Solving these equations, we have x1 = –6, x2 = 8. Now, from (11.19a) and (11.19b) we have k1 þ 2k2 ¼ 16; k1 þ 3k2 ¼ 10. Solving these equations, we have k2 ¼ 26; k1 ¼ 68. Since k2 ¼ 26 violates the condition k2  0, this solution is also rejected. Hence, the optimum (maximum) solution of the given problem is x1 ¼ 12 ; x2 ¼ 32 ; x3 ¼ 0 with k1 ¼ 3; k2 ¼ 0 and Max z ¼ 17 2. Example 2

11.4

Kuhn-Tucker Conditions

259

  Optimize z ¼ 2x1 þ 3x2  x21 þ x22 þ x23 subject to the constraints x1 þ x2  1 2x1 þ 3x2  6: Solution Here we have f ð xÞ ¼ 2x1 þ 3x2  x21  x22  x23 g1 ð x Þ ¼ x 1 þ x 2  1  0 g2 ð xÞ ¼ 2x1 þ 3x2  6  0: First of all, we have to determine whether the given problem is of minimization type or maximization type. For this purpose, we have to construct the bordered Hessian matrix as follows: 

O HB ¼ UT

U ; V

where O is an m  m null matrix, 

@gi U¼ @xj



mn

@2L and V ¼ @xi @xj

: nn

Here m = 2, n = 3. Hence,

0

0

jH B j ¼ 1

1

0 ¼ 2:

0

1

1

0 2

2 2

3 0

3

0

2

0

0

0

0

0

0

0

2

Since n − m = 1, 2 m + 1 = 5, ð1Þm þ n ¼ negative and jHB j\0, the solution point should maximize the objective function. Hence the Kuhn-Tucker conditions for the given problem are given by 2 X @f @gi þ ki ¼ 0; @xj @xj i¼1

j ¼ 1; 2; 3

ð11:22Þ

260

11 Constrained Optimization with Inequality Constraints

ki gi ¼ 0;

i ¼ 1; 2

ð11:23Þ

gi  0;

i ¼ 1; 2

ð11:24Þ

k1  0;

k2  0:

ð11:25Þ

From (11.22), we have 2  2x1 þ k1 þ 2k2 ¼ 0

ð11:26aÞ

3  2x2 þ k1 þ 3k2 ¼ 0

ð11:26bÞ

2x3 ¼ 0:

ð11:26cÞ

From (11.23), we have k1 ð x 1 þ x 2  1Þ ¼ 0 : k2 ð2x1 þ 3x2  6Þ ¼ 0

ð11:27a; bÞ

From (11.24), we have x1 þ x2  1  0

ð11:28aÞ

2x1 þ 3x2  6  0:

ð11:28bÞ

According to (11.25), four different cases may arise: Case 1: k1 ¼ 0; k2 ¼ 0 Case 2: k1 ¼ 0; k2 6¼ 0 Case 3: k1 6¼ 0; k2 ¼ 0 Case 4: k1 6¼ 0; k2 6¼ 0. Now we shall examine the different cases. Case 1: k1 ¼ 0; k2 ¼ 0 In this case, from (11.26a)–(11.26c), we have x1 ¼ 1; x2 ¼ 3=2; x3 ¼ 0. This solution does not satisfy the inequalities (11.28a) and (11.28b). So this solution is rejected. Case 2: k1 ¼ 0; k2 6¼ 0 In this case, solving (11.26a)–(11.26c), (11.27a) and (11.27b), we have

11.4

Kuhn-Tucker Conditions

x1 ¼

12 ; 13

261

x2 ¼

18 1 ; x3 ¼ 0 and k2 ¼ : 13 13

This solution does not satisfy inequality (11.28a) and is therefore rejected. Case 3: k1 6¼ 0; k2 ¼ 0 In this case, the solution of Eqs. (11.26a)–(11.26c) and (11.27a) yields x1 ¼ 1 3 3 4 ; x2 ¼ 4 ; x3 ¼ 0 and k1 ¼ 2. This solution satisﬁes all the conditions and is therefore accepted. Case 4: k1 6¼ 0; k2 6¼ 0 In this case, the solution of Eqs. (11.26a)–(11.26c), (11.27a) and (11.27b) gives x1 ¼ 3; x2 ¼ 4; x3 ¼ 0, k1 ¼ 34; k2 ¼ 13. This solution violates the conditions (11.25) and is therefore discarded. Since only one solution satisﬁes all the conditions, this one is thus the optimal solution, and it is given by x1 ¼ 14 ; x2 ¼ 34 ; x3 ¼ 0 and Max z ¼ 17 8. Example 3 Solve the following problem: Minimize f ¼ x21 þ x22 þ x23 þ 40x1 þ 20x2  3000 subject to g1 ¼ x1  50  0 g2 ¼ x1 þ x2  100  0 g3 ¼ x1 þ x2 þ x3  150  0: Solution The Kuhn-Tucker necessary conditions are 3 X @f @gi þ ki ¼ 0; @xj @xj i¼1

ki gi ¼ 0;

From (11.29), we have

j ¼ 1; 2; 3

i ¼ 1; 2; 3

ð11:29Þ ð11:30Þ

gi  0;

i ¼ 1; 2; 3

ð11:31Þ

ki  0;

i ¼ 1; 2; 3:

ð11:32Þ

262

11 Constrained Optimization with Inequality Constraints 3 X @f @gi þ ki ¼ 0; @xj @xj i¼1

i:e:

j ¼ 1; 2; 3;

@f @g1 @g2 @g3 þ k1 þ k2 þ k3 ¼ 0; j ¼ 1; 2; 3; @xj @xj @xj @xj i:e: 2x1 þ 40 þ k1 þ k2 þ k3 ¼ 0

ð11:33aÞ

2x2 þ 20 þ k2 þ k3 ¼ 0

ð11:33bÞ

2x3 þ k3 ¼ 0:

ð11:33cÞ

From (11.30), we have ki gi ¼ 0; i ¼ 1; 2; 3, i:e: k1 ðx1  50Þ ¼ 0

ð11:34aÞ

k2 ðx1 þ x2  100Þ ¼ 0

ð11:34bÞ

k3 ðx1 þ x2 þ x3  150Þ ¼ 0:

ð11:34cÞ

From (11.31), we have gi  0; i ¼ 1; 2; 3, i:e: g1  0 g2  0 g3  0; i:e: x1  50  0

ð11:35aÞ

x1 þ x2  100  0

ð11:35bÞ

x1 þ x2 þ x3  150  0:

ð11:35cÞ

From (11.32), we have ki  0; i ¼ 1; 2; 3, i:e: k1  0

ð11:36aÞ

k2  0

ð11:36bÞ

k3  0:

ð11:36cÞ

Now from (11.34a), we have k1 ðx1  50Þ ¼ 0. Hence, either k1 ¼ 0 or x1  50 ¼ 0, i.e. either k1 ¼ 0 or x1 ¼ 50. Case 1: When k1 ¼ 0 Now from Eqs. (11.33a)–(11.33c), we have

11.4

Kuhn-Tucker Conditions

263

2x1 ¼ 40  k1  k2  k3

or x1 ¼ 20 

k2 k3  2 2

2x2 ¼ 20  k2  k3 or; x2 ¼ 10  k22  k23 2x3 ¼ k3 or; x3 ¼  k23

½* k1 ¼ 0 ) :

ð11:37Þ

Substituting these values of x1, x2, x3 in Eqs. (11.34b) and (11.34c), we have   k2 k3 k2 k 3 k2 20    10    100 ¼ 0 2 2 2 2 and   k2 k3 k2 k3 k3 k3 20    10     150 ¼ 0; 2 2 2 2 2   3 i:e: k2 ð130  k2  k3 Þ ¼ 0 and k3 180  k2  k3 ¼ 0: 2

ð11:38Þ

From the ﬁrst equation of (11.38), we have either k2 ¼ 0 or  130  k2  k3 ¼ 0: From the second equation of (11.38), we have 3 either k3 ¼ 0 or  180  k2  k3 ¼ 0: 2 Hence, the four possible solutions of (11.38) are given by 3 k2 ¼ 0; 180  k2  k3 ¼ 0 2

ðAÞ

k3 ¼ 0; 130  k2  k3 ¼ 0

ðBÞ

k2 ¼ 0; 130  k2  k3 ¼ 0;

k3 ¼ 0 3 180  k2  k3 ¼ 0: 2

ðCÞ ðDÞ

From (A), k2 ¼ 0; 180  k2  32 k3 ¼ 0, i.e. k2 ¼ 0 and  32 k3 ¼ 180 or k2 ¼ 0 and k3 ¼ 120. Now substituting k2 ¼ 0 and k3 ¼ 120 in (11.37), we have

264

11 Constrained Optimization with Inequality Constraints

0 120 i:e:; x1 ¼ 40 x1 ¼ 20   2 2 0 120 i:e:; x2 ¼ 50 x2 ¼ 10   2 2 120 ¼ 60: x3 ¼  2 Hence, (A) gives the solution x1 = 40, x2 = 50, x3 = 60. This solution satisﬁes (11.35c), but it does not satisfy (11.35a) and (11.35b). So, we reject this solution. Now from (B), we have k3 ¼ 0; 130  k2  k3 ¼ 0, i:e: k3 ¼ 0 and k2 ¼ 130: Hence, from (11.37), we have 130 0  ¼ 20 þ 65 ¼ 45; 2 2 130 0  ¼ 55; x3 ¼ 0: x2 ¼ 10  2 2 ) x1 ¼ 45; x2 ¼ 55; x3 ¼ 0: x1 ¼ 20 

This solution satisﬁes (11.36), but it does not satisfy (11.35a) and (11.35c). Hence, we reject this solution. Now from (C), we have k2 ¼ 0; k3 ¼ 0. From (11.37), we have x1 ¼ 20;

x2 ¼ 10; x3 ¼ 0:

This solution satisﬁes (11.36), but it does not satisfy (11.35a)–(11.35c). Hence, we reject this solution. Now from (D), we have 3  130  k2  k3 ¼ 0; 180  k2  k3 ¼ 0; 2 3 i:e: k2 þ k3 ¼ 130 and k2 þ k3 ¼ 180: 2 Subtracting the ﬁrst equation from the second, we have 1 k3 ¼ 50 2 From the ﬁrst, k2  100 ¼ 130

or k3 ¼ 100: or; k2 ¼ 30.

11.4

Kuhn-Tucker Conditions

265

Hence, from (11.37), we have 30 100  ¼ 20 þ 15 þ 50 ¼ 45 2 2 30 100  ¼ 10 þ 15 þ 50 ¼ 55 x2 ¼ 10  2 2 100 ¼ 50: x3 ¼  2

x1 ¼ 20 

This solution satisﬁes (11.35b) and (11.35c) but does not satisfy (11.35a). Hence, we reject this solution. Case 2: When x1 = 50 Now from Eqs. (11.33a)–(11.33c), we have k3 ¼ 2x3 k2 ¼ 20  2x2  k3 ¼ 20  2x2 þ 2x3 and k1 ¼ 40  2x1  k2  k3 ¼ 40  2:50  ð20  2x2 þ 2x3 Þ þ 2x3 ¼ 40  100 þ 20 þ 2x2  2x3 þ 2x3 ¼ 120 þ 2x2 : 9 ) k1 ¼ 120 þ 2x2 = k2 ¼ 20  2x2 þ 2x3 : ; k3 ¼ 2x3

ð11:39Þ

Now substituting these values of k1 ; k2 ; k3 in (11.34b) and (11.34c), we have ð20  2x2 þ 2x3 Þðx1 þ x2  100Þ ¼ 0

ð11:40aÞ

2x3 ðx1 þ x2 þ x3  150Þ ¼ 0:

ð11:40bÞ

and

From (11.40a), we have either  20  2x2 þ 2x3 ¼ 0 or; x1 þ x2  100 ¼ 0: From (11.40b), we have either x3 ¼ 0 or x1 þ x2 þ x3 150 ¼ 0. Therefore, the four possible solutions of (11.40a) and (11.40b) are given by 202x2 þ 2x3 ¼ 0;

x1 þ x2 þ x3 150 ¼ 0

202x2 þ 2x3 ¼ 0; x1 þ x2 100 ¼ 0;

x3 ¼ 0 x3 ¼ 0

ð11:41aÞ ð11:41bÞ ð11:41cÞ

266

11 Constrained Optimization with Inequality Constraints

x1 þ x2 100 ¼ 0;

x1 þ x2 þ x3 150 ¼ 0:

ð11:41dÞ

From (11.41a), we have  202x2 þ 2x3 ¼ 0; x1 þ x2 þ x3 150 ¼ 0; i:e:

10x2 þ x3 ¼ 0; 50 þ x2 þ x3 150 ¼ 0

½* x1 ¼ 50:

Adding the preceding two equations, we have 2x3 ¼ 1050 þ 150

or 2x3 ¼ 110 or x3 ¼ 55:

Now substituting x3 = 55 in −10 − x2 + x3 = 0, we have 10x2 þ 55 ¼ 0 or x2 ¼ 45: )

x1 ¼ 50; x2 ¼ 45; x3 ¼ 55:

This solution does not satisfy (11.35b); hence, we reject it. From (11.41b), we have  20  2x2 þ 2x3 ¼ 0; x3 ¼ 0; i:e:  10  x2 þ x3 ¼ 0; x3 ¼ 0: ) x1 ¼ 50;

x2 ¼ 10;

x3 ¼ 0:

This solution does not satisfy the constraints (11.35b) and (11.35c). From (11.41c), we have

or

x1 þ x2  100 ¼ 0; 50 þ x2  100 ¼ 0;

or

x2 ¼ 50; x3 ¼ 0:

) x1 ¼ 50;

x2 ¼ 50;

x3 ¼ 0 x3 ¼ 0

½* x1 ¼ 50

x3 ¼ 0:

This solution does not satisfy the constraint (11.35c). From (11.41d), we have x1 þ x2  100 ¼ 0;

x1 þ x2 þ x3  150 ¼ 0:

From x1 þ x2  100 ¼ 0 we get 50 þ x2  100 ¼ 0 or x2 = 50. From x1 þ x2 þ x3  150 ¼ 0 we get 50 þ 50 þ x3  150 ¼ 0 or x3 = 50. ) x1 ¼ 50;

x2 ¼ 50;

x3 ¼ 50:

This solution satisﬁes all the constraints (11.35a)–(11.35c).

11.4

Kuhn-Tucker Conditions

267

Now substituting x1 ¼ 50; x2 ¼ 50; x3 ¼ 50 in (11.39), we have k1 ¼ 120 þ 2:50 ¼ 20 k2 ¼ 20  2:50 þ 2:50 ¼ 20 k3 ¼ 2:50 ¼ 100: Since all these values of ki satisfy the requirement ki  0; i ¼ 1; 2; 3, then the optimum solution is x1 ¼ 50; x2 ¼ 50; x3 ¼ 50, and the minimum value of f is ð50Þ2 þ ð50Þ2 þ ð50Þ2 þ 40  50 þ 20  50  3000 ¼ 7500:

11.5 ð1Þ

Exercises Maximize f ðx1 ; x2 Þ ¼ 2x1 þ bx2 subject to g1 ðx1 ; x2 Þ ¼ x21 þ x22  5

g2 ðx1 ; x2 Þ ¼ x1  x2  2: Using the Kuhn-Tucker conditions, ﬁnd the values of b for which x1 ¼ 1; x2 ¼ 2 will be the optimal solution. 2. Solve the following problems using the Kuhn-Tucker conditions: ðiÞ Minimize f ðx1 ; x2 Þ ¼ ðx1  1Þ2 þ ðx2  5Þ2 subject to  x21 þ x2  4; ðx1  2Þ2 þ x2  3: ðiiÞ Maximize Z ¼ 10x1  x21 þ 10x2  x22 subject to x1 þ x2  14; x1 þ x2  6: ðiiiÞ Maximize z ¼ 12x1 þ 21x2 þ 2x1 x2  2x21  2x22 subject to x2  8; x1 þ x2  10: ðivÞ Maximize Z ¼ 2x1 þ 3x2  2x21 subject to x1 þ 4x2  4; x1 þ x2  2: ðvÞ Minimize Z ¼ 6  6x1  2x21  2x1 x2 þ 2x22 subject to x1 þ x2  2: ðviÞ Maximize z ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to x1 þ 2x2  10; x1 þ x2  9:

Chapter 12

12.1

Objective

The main objective of this chapter is to discuss: • Solution procedures of quadratic programming problems • Wolfe’s modiﬁed simplex method • Beale’s method for solving quadratic programming problems.

12.2

Introduction

Quadratic programming deals with non-linear programming problems of optimizing (maximizing or minimizing) the quadratic function subject to a set of linear inequality constraints. The general form of a quadratic programming problem (QPP) may be written as follows: Optimize f ðxÞ ¼

n X

cj xj þ

j¼1

n X n 1X djk xj xk 2 j¼1 k¼1

subject to the constraints n X

aij xj  bi ;

i ¼ 1; 2; . . .; m

j¼1

and xj  0;

j ¼ 1; 2; . . .; n:

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_12

269

270

In matrix notation, it is written as   1 Optimize f ðxÞ ¼ cT ; x þ hQx; xi 2 subject to the constraints Ax  b and x  0;   where x ¼ ðx1 ; x2 ; . . .; xn ÞT ; c ¼ ðc1 ; c2 ; . . .; cn Þ b ¼ ðb1 ; b2 ; . . .; bn ÞT , A ¼ aij   is an m  n real matrix and Q ¼ djk is an n  n real symmetric matrix, i.e. djk ¼ dkj for all j and k. The matrix Q can be taken as non-null otherwise the QPP is a linear programing problem. The term hQx; xi deﬁnes a quadratic form where Q is a symmetric matrix. The quadratic form hQx; xi is said to be positive deﬁnite (negative deﬁnite) if hQx; xi [ 0ð\0Þ for x 6¼ 0 and positive semi-deﬁnite (negative semi-deﬁnite) if hQx; xi  0 ð  0Þ for all x such that there is one x 6¼ 0 satisfying hQx; xi ¼ 0. It may easily be veriﬁed that: (i) If hQx; xi is positive semi-deﬁnite (negative semi-deﬁnite), then it is convex (concave) in x over all of Rn (ii) If hQx; xi is positive deﬁnite (negative deﬁnite), then it is strictly convex (strictly concave) in x over all of Rn. These results help the students in determining whether the quadratic objective function f(x) is concave (convex) and the implication of the same on the sufﬁciency of the Kuhn-Tucker conditions for constrained maxima (minima) of f(x). The feasibility of the problem can be tested by Phase I of the simplex method. The solution of a general constrained optimization problem can be classiﬁed as follows: (i) No feasible solution (ii) An unbounded solution (iii) An optimal solution. Theorem 1 The quadratic programming problem cannot have an unbounded solution when the matrix Q is positive deﬁnite. Proof Let x 6¼ 0. The objective function can be written as   1  T  f ðxÞ ¼ hQx; xi þ c ; x hQx; xi : 2

ð12:1Þ

Let x be any point on the hypersphere j xj ¼ r, where j xj2 ¼ hx; xi. Then x ¼ r^x, where j^xj ¼ 1, and then hQx; xi ¼ r 2 hQ^x; ^xi.

12.2

Introduction

271

Let m be the minimum value of hQ^x; ^xi. Since Q is positive deﬁnite, hQx; xi  r 2 m [ 0 ) hQx; xi ! þ 1 as j xj ¼ r ! 1:

ð12:2Þ

Now, let M be the maximum value of hcT ; ^xi=hQ^x; ^xi. Then,

hcT ;xi hQx;xi

hcT ;^xi ¼ 1r hQ^x;^xi  Mr ,

and therefore h cT ; xi ! 0 as r ! 1: hQx; xi

ð12:3Þ

From (12.1), (12.2) and (12.3), we get f ðxÞ ! 1 and j xj ! 1: Hence, for ﬁnite x; f ð xÞ will be ﬁnite. This means that the QPP cannot have an unbounded solution. Now we shall show that the QPP has a unique optimal solution provided the problem has a feasible solution and the matrix Q is positive deﬁnite. If Q is positive semi-deﬁnite, the solution could be an unbounded one. But the difﬁculties in this case can be avoided if we consider the perturbed matrix Q þ eI, where ðe [ 0Þ is substantially less than the magnitude of any element of Q, since this matrix is positive deﬁnite when Q is positive semi-deﬁnite.

12.3

Solution Procedure of QPP

Let us consider a QPP in the following form: Maximize z ¼ f ðxÞ ¼

n X j¼1

cj xj þ

n X n 1X djk xj xk 2 j¼1 k¼1

subject to the constraints n X

aij xj  bi and xj  0;

i ¼ 1; 2; . . .; m; j ¼ 1; 2; . . .; n;

j¼1

  where djk ¼ dkj for all j, k, bi  0 for i ¼ 1; 2; . . .; m and the matrix Q ¼ djk nn is symmetric negative deﬁnite.

272

Now, introducing slack variables q2i in the ith constraint ði ¼ 1; 2; . . .; mÞ and rj2 in the jth non-negativity constraint ðj ¼ 1; 2; . . .; nÞ, the problem becomes Maximize z ¼ f ðxÞ ¼

n X

cj xj þ

j¼1

n X n 1X djk xj xk 2 j¼1 k¼1

subject to the constraints n X

aij xj þ q2i ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

 xj þ rj2 ¼ 0;

j ¼ 1; 2; . . .; n:

Now the Lagrangian function is given by Lðx1 ; x2 ; . . .; xn ; q1 ; q2 ; . . .; qm ; r1 ; r2 ; . . .; rn ; k1 ; k2 ; . . .; km ; l1 ; l2 ; . . .; ln Þ ¼ f ðxÞ ! m n n

X X X 2 ki aij xj þ qi  bi  lj xj þ rj2 ;  i¼1

j¼1

j¼1

where ki  0; i ¼ 1; 2; . . .; m and lj  0; j ¼ 1; 2; . . .; n. The necessary conditions for the maximum of L are given by m @L @f X ¼0)  ki aij þ lj ¼ 0; @xj @xj i¼1

j ¼ 1; 2; . . .; n

ð12:4Þ

@L ¼ 0 ) 2ki qi ¼ 0; @qi

i ¼ 1; 2; . . .; m

ð12:5Þ

@L ¼ 0 ) 2lj rj ¼ 0; @rj

j ¼ 1; 2; . . .; n

ð12:6Þ

n X @L ¼0) aij xj þ q2i  bi ¼ 0; @ki j¼1

@L ¼ 0 ) xj þ rj2 ¼ 0; @lj

i ¼ 1; 2; . . .; m

j ¼ 1; 2; . . .; n:

These conditions are known as Kuhn-Tucker conditions.

ð12:7Þ

ð12:8Þ

12.3

Solution Procedure of QPP

273

By deﬁning a set of new variables si as si ¼ q2i  0; i ¼ 1; 2; . . .; m, the equations in (12.5) become ki si ¼ 0;

i ¼ 1; 2; . . .; m:

Then the equations in (12.7) reduce to n X

aij xj þ si  bi ¼ 0;

i ¼ 1; 2; . . .; m:

j¼1

Now from (12.6) and (12.8), we have lj xj ¼ 0 and

xj  0;

j ¼ 1; 2; . . .; n:

The equations in (12.4) can be written as cj þ

n X

djk xk 

m X

ki aij þ lj ¼ 0;

j ¼ 1; 2; . . .; n:

i¼1

k¼1

Thus, the necessary conditions can be summarized as follows: cj þ

n X

djk xk 

m X

ki aij þ lj ¼ 0;

j ¼ 1; 2; . . .; n

ð12:9Þ

i¼1

k¼1 n X

aij xj þ si ¼ bi ;

i ¼ 1; 2; . . .; m

ð12:10Þ

j¼1

ki si ¼ 0;

i ¼ 1; 2; . . .; m

ð12:11Þ

lj xj ¼ 0;

j ¼ 1; 2; . . .; n

ð12:12Þ

xj ; lj  0;

j ¼ 1; 2; . . .; n

ð12:13Þ

si ; ki  0;

i ¼ 1; 2; . . .; m:

ð12:14Þ

These conditions, except ki si ¼ 0 and lj xj ¼ 0, are linear constraints involving 2 (n + m) variables. Thus, the solution of the original QPP can be obtained by ﬁnding a non-negative solution to the set of (n + m) linear equations given by the equations in (12.9) and (12.10), which also satisﬁes the (n + m) equations stated in (12.11) and (12.12).

274

Since Q is a negative deﬁnite matrix, f(x) is a strictly concave function and the feasible space is convex (because of linear equations), any local maximum of the problem will be the global maximum. Again, from (12.9) to (12.14), it is observed that there are 2ðn þ mÞ equations with 2ðn þ mÞ variables in the necessary conditions. Hence, the solution of the system (12.9)–(12.14) must be unique. Thus, the feasible solution of the system (12.9)–(12.14), if it exists, must give the optimum solution of the QPP directly. The solution of the system (12.9)–(12.14) can be obtained by using Phase I of the two-phase simplex method. The only restriction here is that the satisfaction of non-linear relations (12.11) and (12.12) has to be maintained all the time. Since our objective is just to ﬁnd a feasible solution to the system, there is no necessity for Phase II computations. For the application of Phase I of the two-phase method, to get the initial basis as an identity matrix, we have to introduce artiﬁcial variables Aj into Eq. (12.9) so that cj þ

n X

djk xk 

m X

ki aij þ lj þ Aj ¼ 0;

j ¼ 1; 2; . . .; n:

i¼1

k¼1

Hence, the equivalent optimization QPP becomes Maximize z0 ¼ 

n X

Aj

j¼1

subject to the constraints cj þ

n X

djk xk 

k¼1 n X

m X

ki aij þ lj þ Aj ¼ 0;

j ¼ 1; 2; . . .; n

i¼1

aij xj þ si ¼ bi ;

i ¼ 1; 2; . . .; m

j¼1

xj ; lj ; Aj  0;

j ¼ 1; 2; . . .; n si ; ki  0; i ¼ 1; 2; . . .; m:

While solving this problem, we have to consider the following additional conditions known as complementary slackness conditions: ki si ¼ 0;

i ¼ 1; 2; . . .; m

lj xj ¼ 0;

j ¼ 1; 2; . . .; n

Thus, when deciding whether to introduce si into the basic solution, we have to ensure ﬁrst that either ki is not in the basic solution or ki will be removed when si enters the basic solution. Similar care has to be taken regarding the variables lj and xj.

12.4

12.4

Wolfe’s Modiﬁed Simplex Method

275

Wolfe’s Modiﬁed Simplex Method

Let the QPP be the following: Maximize z ¼ f ðxÞ ¼

n X j¼1

cj xj þ

n X n 1X djk xj xk 2 j¼1 k¼1

subject to the constraints n X

aij xj  bi ;

i ¼ 1; 2; . . .; m and xj  0; j ¼ 1; 2; . . .; n;

j¼1

where djk ¼ dkj for all j and k, bi  0 for all i. P P Also, assume that the quadratic form nj¼1 nk¼1 djk xj xk is negative semi-deﬁnite. Then the iterative procedure for the solution of a modiﬁed simplex method is as follows: Step 1. Check whether the given QPP is in the maximization form or not. If it is in the minimization form, convert it into the maximization form by appropriate adjustment in f(x). Step 2. Convert the inequality constraints into equations by introducing the slack variables q2i in the ith constraint (i = 1, 2, …, m) and the slack variables rj2 in the jth non-negativity constraint (j = 1, 2, …, n). Step 3. Construct the Lagrangian function Lðx1 ; x2 ; . . .; xn ; q1 ; q2 ; . . .; qm ; r1 ; r2 ; . . .; rn ; k1 ; k2 ; . . .; km ; l1 ; l2 ; . . .; ln Þ ! m n n

X X X 2 ¼ f ðxÞ  ki aij xj þ qi  bi  lj xj þ rj2 ; i¼1

j¼1

j¼1

where ki  0; i ¼ 1; 2; . . .; m and lj  0; j ¼ 1; 2; . . .; n. Step 4. Derive the Kuhn-Tucker conditions by differentiating L partially with respect to x1 ; x2 ; . . .; xn ; q1 ; q2 ; . . .; qm ; r1 ; r2 ; . . .; rn ; k1 ; k2 ; . . .; km ; l1 ; l2 ; . . .; ln and then equating to zero. Step 5. Introduce the non-negative artiﬁcial variables Aj ; j ¼ 1; 2; . . .; n in the ﬁrst @L ¼ 0; j ¼ 1; 2; . . .; n and construct the new n equations obtained from @x j P n 0 objective function z ¼  j¼1 Aj , and form an LPP with this new objective function. Step 6. Use two-phase simplex method to obtain an optimum solution to the LPP formed in Step 5 satisfying the complementary slackness conditions.

276

Step 7. The optimal solution obtained in Step 6 is an optimal solution to the given QPP also. Step 8. Stop.

12.5

The advantage in using the Wolfe method is that only a slight modiﬁcation has to be made in the simplex method to take care of the restricted basis entry rule. However, the disadvantage is that the number of rows of the coefﬁcient matrix increases from m to m + n. Example 1 Solve the following problem by Wolfe’s method: Minimize z ¼ x21  x1 x2 þ 2x22  x1  x2 subject to 2x1 þ x2  1 x1  0; x2  0: Solution Maximize z0 ¼ x21 þ x1 x2  2x22 þ x1 þ x2 subject to 2x1 þ x2  1  x1  0; x2  0 where z ¼ z 0

Now introducing slack variables q21 ; r12 and r22 , we have the reduced problem as Maximize z0 ¼ x21 þ x1 x2  2x22 þ x1 þ x2 subject to 2x1 þ x2 þ q21 ¼ 1  x1 þ r12 ¼ 0; x2 þ r22 ¼ 0: Now the Lagrange function L is given by Lðx1 ; x2 ; q1 ; q2 ; r1 ; r2 ; k1 ; k2 ; l1 ; l2 ;Þ

  ¼ x21 þ x1 x2  2x22 þ x1 þ x2  k1 2x1 þ x2 þ q21  1      l1 x1 þ r12  l2 x2 þ r22 :

12.5

277

Clearly, x21 þ x1 x2  2x22 ¼  x21 þ x22  2x1 x2  x1 x2  x22 ¼ ðx1  x2 Þ2 x1 x2  x22 is a negative semi-deﬁnite quadratic form. Hence, z0 ¼ x21 þ x1 x2  2x22 þ x1 þ x2 is a concave function in x1 ; x2 . Hence, the maximum of L will be the maximum of z0 and vice versa. Now the necessary conditions for maxima of L (and hence of z0 ) are given by

@L ¼ 0 ) 2x1 þ x2 þ 1  2k1 þ l1 ¼ 0 @x1

ð12:15Þ

@L ¼ 0 ) x1  4x2 þ 1  k1 þ l2 ¼ 0 @x2

ð12:16Þ

@L ¼ 0 ) 2x1 þ x2 þ q21  1 ¼ 0 @k1

ð12:17Þ

@L ¼ 0 ) x1 þ r12 ¼ 0 @l1

ð12:18Þ

@L ¼ 0 ) x2 þ r22 ¼ 0 @l2

ð12:19Þ

@L ¼ 0 ) k1 q1 ¼ 0 @q1

ð12:20Þ

@L ¼ 0 or l1 r1 ¼ 0 or l1 x1 ¼ 0 as  x1 þ r12 ¼ 0 @r1

ð12:21Þ

@L ¼ 0 or l2 r2 ¼ 0 or l2 x2 ¼ 0 as  x2 þ r22 ¼ 0 @r2

ð12:22Þ

Now letting q21 ¼ s1  0, from (12.17) we get 2x1 þ x2 þ s1  1 ¼ 0;

ð12:23Þ

k1 s1 ¼ 0;

ð12:24Þ

and from (12.20), we get

where x1 ; x2 ; s1 ; l1 ; l2 ; k1  0.

278

Now the equivalent problem of the given QPP is as follows: 9 2x1  x2 þ 2k1  l1 ¼ 1 = x1 þ 4x2 þ k1  l2 ¼ 1 ; 2x1 þ x2 þ s1 ¼ 1

ð12:25Þ

x1 ; x2 ; s1 ; k1 ; l1 ; l2  0

ð12:26Þ

k1 s1 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0:

A solution xj ; j ¼ 1; 2 of (12.25) and satisfying (12.26) shall necessarily be an optimal one for maximizing L (i.e. maximizing z). Now to determine the solution to the simultaneous equation (12.25), we introduce two artiﬁcial variables A1 ð  0Þ and A2 ð  0Þ in the ﬁrst two equations of (12.25) and construct the dummy objective function z00 ¼ A1  A2 . Then the equivalent problem of the given QPP becomes Maximize z00 ¼ A1  A2 subject to 2x1  x2 þ 2k1  l1 þ A1 ¼ 1  x1 þ 4x2 þ k1  l2 þ A2 ¼ 1 2x1 þ x2 þ s1 ¼ 1 x1 ; x2  0; k1  0; l1 ; l2 ; s1 ; A1 ; A2  0; satisfying the complementary slackness conditions k1 s1 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0: To solve the problem, we shall apply two-phase method. Here the initial basic feasible solution (BFS) is A1 ¼ 1; A2 ¼ 1; s1 ¼ 1. Now we construct the initial table of Phase I.

Basic variables 1 A1 1 A2 0 s1 z0B ¼ 2 cB

cj !

0

0

0

0

0

1

1

0

xB

x01

x02

k01

l01

l02

A01

A02

s01

1 1 1

2 1 2 1

1 4 1 3

2 1 0 3

1 0 0 1

0 1 0 1

1 0 0 0

0 1 0 0

0 0 1 0

Mini ratio – 1/4 1

12.5

279

Here all Dj j0. Hence, the current BFS is not optimal. From this table, we observe that either x02 or k01 as ðD2 ¼ 3; D3 ¼ 3Þ can be introduced into the basic solution. Arbitrarily, we choose x02 as the entering vector, since by the complementary condition, s1 is already in the basis.

Basic variables 1 A1 0 x2 0 s1 z0B ¼ 5=4 cB

cj !

0

0

0

0

0

1

1

0

xB

x01

x02

k01

l01

l02

A01

A02

s01

5/4 1/4 3/4

7/4 1=4 9/4 7=4

0 1 0 0

9/4 1/4 1=4 9=4

1 0 0 1

1=4 1=4 1/4 1/4

1 0 0 0

1/4 1/4 1=4 3/4

0 0 1 0

Mini ratio 5/7 – 1/3

Here all Dj j0. Hence, the current BFS is not optimal. Dj corresponding to k01 is most negative, but it does not enter into the basis, since by the complementary condition, s1 is already in the basis.

cB

Basic variables

1 A1 0 x2 0 x1 z0B ¼ 2=3 cB

cj ! xB 2/3 1/3 1/3

Basic variables

0 k1 0 x2 0 x1 z0B ¼ 0

0 x01

0 x02

k01

0 l01

0 l02

1 A01

0 s01

0 0 1 0 cj ! xB

0 1 0 0 0 x01

22/9 2/9 1=9 22=9 0 0 x02 k01

1 0 0 1 0 l01

4=9 2=9 1/9 4/9

1 0 0 0 0 l02

7=9 1/9 4/9 7/9 1 A01

0 s01

3/4 3/11 4/11

0 0 1 0

0 1 0 0

2=11 2 2 0

9/22 1 1 0

7=22 2 9 0

0

1 0 0 0

9=22 1 1 0

The optimal solution is x1 ¼ 4=11; x2 ¼ 3=11 and Max z0 ¼ 5=11. Example 2 Solve the following problem by Wolfe’s method: Minimize z ¼ 10x1  25x2 þ 10x21 þ x22 þ 4x1 x2 subject to x1 þ 2x2  10 x1 þ x 2  9 and x1  0; x2  0:

Mini ratio 3/11 3/2 –

280

Solution This is a minimization problem. Now converting the problem into a maximization problem and then converting all the inequality constraints, including the non-negativity constraints, into  type, we get the reduced problem as follows: Maximize z0 ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to x1 þ 2x2  10 x1 þ x2  9  x1  0; x2  0 0

where z ¼ z: Now introducing slack variables q21 ; q22 ; r12 and r22 , we have the reduced problem as Maximize z0 ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to x1 þ 2x2 þ q21 ¼ 10 x1 þ x2 þ q22 ¼ 9  x1 þ r12 ¼ 0; x2 þ r22 ¼ 0: Now the Lagrange function L is given by Lðx1 ; x2 ; q1 ; q2 ; r1 ; r2 ; k1 ; k2 ; l1 ; l2 Þ

  ¼ 10x1 þ 25x2  10x21  x22  4x1 x2  k1 x1 þ 2x2 þ q21  10    k2 x1 þ x2 þ q22  9      l1 x1 þ r12  l2 x2 þ r22 :

Clearly, 10x21  x22  4x1 x2 is a negative semi-deﬁnite quadratic expression. Hence, z0 ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 is a concave function in x1 ; x2 . Hence, the maximum of L will be the maximum of z0 and vice versa. Now the necessary conditions for maxima of L (and hence of z0 ) are given by @L ¼ 0 or 10  20x1  4x2  k1  k2 þ l1 ¼ 0 @x1

ð12:27Þ

@L ¼ 0 or 25  2x2  4x1  2k1  k2 þ l2 ¼ 0 @x2

ð12:28Þ

12.5

@L ¼ 0 or x1 þ 2x2 þ q21  10 ¼ 0 @k1 @L ¼0 @k2

or x1 þ x2 þ q22  9 ¼ 0

281

ð12:29Þ ð12:30Þ

@L ¼ 0 or  x1 þ r12 ¼ 0 @l1

ð12:31Þ

@L ¼ 0 or  x1 þ r22 ¼ 0 @l2

ð12:32Þ

@L ¼0 @q1

or k1 q1 ¼ 0

ð12:33Þ

@L ¼0 @q2

or k2 q2 ¼ 0

ð12:34Þ

@L ¼0 @r1

or l1 r1 ¼ 0 or l1 x1 ¼ 0 as  x1 þ r12 ¼ 0

ð12:35Þ

@L ¼0 @r2

or l2 r2 ¼ 0 or l2 x2 ¼ 0 as  x2 þ r22 ¼ 0

ð12:36Þ

Now letting q21 ¼ s1  0, we have from (12.29) x1 þ 2x2 þ s1  10 ¼ 0:

ð12:37Þ

k1 s1 ¼ 0:

ð12:38Þ

From (12.33),

Let q22 ¼ s2  0; then we have from (12.30) and (12.34) x1 þ x2 þ s 2  9 ¼ 0

ð12:39Þ

k2 s 2 ¼ 0

ð12:40Þ

where x1 ; x2 ; s1 ; s2 ; l1 ; l2 ; k1 ; k2  0. Now the equivalent problem of the given QPP is as follows: 9 20x1 þ 4x2 þ k1 þ k2  l1 ¼ 10 > > = 4x1 þ 2x2 þ 2k1 þ k2  l2 ¼ 25 x1 þ 2x2 þ s1 ¼ 10 > > ; x1 þ x2 þ s2¼ 9

ð12:41Þ

282

x1 ; x2  0; s1 ; s2  0; k1 ; k2  0; l1 ; l2  0

ð12:42Þ

k1 s1 ¼ 0; k2 s2 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0:

A solution xj ; j ¼ 1; 2 of (12.41) satisfying (12.42) shall necessarily be an optimal one for maximizing L (i.e. maximizing z). Now to determine the solution to the simultaneous Eqs. (12.41), we introduce two artiﬁcial variables A1 ð  0Þ and A2 ð  0Þ in the ﬁrst two equations of (12.41) and construct the dummy objective function z00 ¼ A1  A2 . Then the equivalent problem of the given QPP becomes Maximize z00 ¼ A1  A2 subject to 20x1 þ 4x2 þ k1 þ k2  l1 þ A1 ¼ 10 4x1 þ 2x2 þ 2k1 þ k2  l2 þ A2 ¼ 25 x1 þ 2x2 þ s1 ¼ 10 x1 þ x2 þ s 2 ¼ 9 x1 ; x2  0; k1 ; k2  0; l1 ; l2  0; s1 ; s2  0; A1 ; A2  0; satisfying the complementary slackness conditions k1 s1 ¼ 0; k2 s2 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0: To solve the problem we shall apply the two-phase method. Here the initial BFS is A1 ¼ 10; A2 ¼ 25; s1 ¼ 10; s2 ¼ 9. Now we construct the initial table of Phase I.

Basic variables 1 A1 1 A2 0 s1 0 s2 z00B ¼ 35 cB

cj

0

0

0

0

0

0

1

1

0

0

xB

x01

x02

k01

k02

l01

l02

A01

A02

s01

s02

10 25 10 9

20 4 1 1 24

4 2 2 1 6

1 2 0 0 3

1 1 0 0 2

1 0 0 0 1

0 1 0 0 1

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

Mini ratio 1/2 25/4 10 9 Dj

Here all Dj l0. Hence, the current BFS is not optimal. Here x01 is introduced into the basis as l1 ¼ 0 with A01 as the outgoing (leaving) vector. Now we construct the next table, dropping the A01 column.

12.5

cB

283

cj

0

0

0

0

0

0

1

1

0

0

Basic variables

xB

x01

x02

k01

k02

l01

l02

A01

A02

s01

s02

Mini ratio

0

x1

1/2

1

1/5

1/20

1/20

1=20

0

1/20

0

0

0

5/2

1

A2

23

0

6/5

9/5

4/5

1/5

1

1=5

1

0

0

115/6

0

s1

19/ 2

0

9/5

1=20

1=20

1/20

0

1=20

0

1

0

95/18

0

s2

17/ 2

0

4/5

1=20

1=20

1/20

0

1=20

0

0

1

85/8

0

6=5

9=5

4=5

1=5

1

6/5

0

0

0

z00B ¼ 23

Dj

Here all Dj l0. Hence, the current BFS is not optimal. Again, Dj corresponding to k01 is most negative, but it does not enter into the basis, since by the complementary condition, s01 is already in the basis. So the entering vector is x02 , as l02 is not present in the basis and x01 is the leaving vector.

cB 0 1 0 0

Basic variables x2 A2 s1 s2

cj

0

0

0

0

0

0

1

1

0

0

xB

x01

x02

k01

k02

l01

l02

A01

A02

s01

s02

5/2 20 5 13/ 2

5 6 9 4

1 0 0 0

1/4 3/2 1=2 1=4

1/4 1/2 1=2 1=4

1=4 1/2 1/2 1/4

0 1 0 0

1/4 1=2 1=2 1=4

0 1 0 0

0 0 1 0

0 0 0 1

6

0

3=2

1=2

1=2

1

3/2

0

0

0

z00B ¼ 20

Mini ratio 40 10 26 Dj

Here all Dj l0. Hence, the current BFS is not optimal. Dj corresponding to k01 is most negative, but it does not enter into the basis, as by the complementary condition, s01 is already in the basis. Now k02 does not enter into the basis, as by the complementary condition, s02 is already in the basis. Now l01 is the entering vector, since by the complementary condition, l1 x1 ¼ 0 and s01 is the leaving vector.

cB 0

Basic variables x2

cj

0

0

0

0

0

0

1

1

0

0

xB

x01

x02

k01

k02

l01

l02

A01

A02

s01

s02

1/2

1

0

0

0

0

0

0

1/2

0

5

Mini ratio (continued)

284

(continued) 1 A2 0 l1 0 s2 z00B ¼ 15

cj

0

0

0

0

0

0

1

1

0

0

15 10 4

3 18 1/2 3

0 0 0 0

2 1 0 2

1 1 0 1

0 1 0 0

1 0 0 1

0 0 0

1 0 0 0

1 2 1=2 1

0 0 1 0

15/2

Dj

Here all Dj l0. Hence, the BFS is not optimal. From the table we observe that x01 is most negative, but it does not enter into the basis, since by the complementary condition, l01 is already in the basis. So by the complementary condition, k01 is entering the basis and A02 is the leaving vector.

cB

Basic variables

0 x2 0 k1 0 l1 0 s2 z00B ¼ 0

cj

0

0

0

0

0

0

0

0

xB

x01

x02

l02

s01

s02

1/2 3/2 33/2 1/2 0

1 0 0 0 0

k02 0 1/2 1=2 0 0

l01

5 15/2 35/2 4

k01 0 1 0 0 0

0 0 1 0 0

0 1=2 1=2 0 0

1/2 1=2 3/2 1=2 0

0 0 0 1 0

Mini ratio

Dj

Since Dj  0, the current BFS is optimal. The optimal solution is x1 ¼ 0; x2 ¼ 5, k1 ¼

15 ; 2

k2 ¼ 0; l1 ¼

35 ; l2 ¼ 0; s1 ¼ 0; s2 ¼ 4: 2

This solution satisﬁes the complementary slackness conditions, k1 s1 ¼ 0 ¼ k2 s2 and l1 x1 ¼ 0 ¼ l2 x2 . Since z00B ¼ 0, the current solution is also feasible, and

Minimize z ¼ 10x1  25x2 þ 10x21 þ x22 þ 4x1 x2 ¼ 10:0  25:5 þ 10:0 þ 52 þ 4:0 ¼ 125 þ 25 ¼ 100:

Example 3 Apply Wolfe’s method for solving the following QPP:

12.5

285

Maximize z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 subject to x1 þ 2x2  2 and x1 ; x2  0: Solution The given problem is a maximization problem. Now, converting all the inequality constraints including the non-negativity constraints, into  type, we have x1 þ 2x2  2; x1  0 and  x2  0: Now introducing slack variables q21 ; r12 and r22 , we have the reduced problem as Maximize z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 subject to x1 þ 2x2 þ q21 ¼ 2  x1 þ r12 ¼ 0  x2 þ r22 ¼ 0 Now, the Lagrangian function L is given by L ¼ Lðx1 ; x2 ; q1 ; l1 ; l2 ; k1 ; r1 ; r2 Þ ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22        k1 x1 þ 2x2 þ q21  2  l1 x1 þ r12  l2 x2 þ r22 : As 2x21  2x1 x2  2x22 represents a negative semi-deﬁnite quadratic form, then z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 is concave in x1, x2. Therefore, the maxima of L will be the maxima of z and vice versa. Now, the necessary conditions for the maxima of L (and hence of z) are @L ¼ 0 or 4  4x1  2x2  k1 þ l1 ¼ 0 @x1 @L ¼ 0 or 6  2x1  4x2  2k1 þ l2 ¼ 0 @x2 @L ¼ 0 or x1 þ 2x2 þ q21 ¼ 2 @k1 @L ¼ 0 or  x1 þ r12 ¼ 0 @l1

286

@L ¼ 0 or  x2 þ r22 ¼ 0 @l2 @L ¼ 0 or k1 q1 ¼ 0 ) k1 q22 ¼ 0 @q1 @L ¼ 0 or l1 r1 ¼ 0 or l1 r12 ¼ 0 or l1 x1 ¼ 0 as  x1 þ r12 ¼ 0 @r1 @L ¼ 0 or l2 r2 ¼ 0 or l2 r22 ¼ 0 or l2 x2 ¼ 0 as  x2 þ r22 ¼ 0: @r2 Now, letting q21 ¼ s1  0, we have k1 s1 ¼ 0 as k1 q22 ¼ 0. Also, x1 þ 2x2 þ s1 ¼ 2 and ﬁnally, x1 ; x2 ; s1 ; k1 ; l1 ; l2  0. Hence, to determine all the variables, the equivalent problem of the given QPP is as follows: 9 4x1 þ 2x2 þ k1  l1 ¼ 4 = 2x1 þ 4x2 þ 2k1  l2 ¼ 6 ; ; x1 þ 2x2 þ s1 ¼ 2

ð12:43Þ

where all variables are non-negative and k1 s1 ¼ 0; l1 x1 ¼ 0 l2 x2 ¼ 0:

ð12:44Þ

A solution xj ; j ¼ 1; 2; of (12.43) and satisfying (12.44) shall necessarily be an optimal one for maximizing L (i.e. maximizing z). Now to determine the solution to the simultaneous equations in (12.43), we introduce two artiﬁcial variables A1 ð  0Þ and A2 ð  0Þ in the ﬁrst two equations of (12.43) and construct the dummy objective function z0 ¼ A1  A2 . Then the equivalent problem of the given QPP becomes Maximize z0 ¼ A1  A2 subject to 4x1 þ 2x2 þ k1  l1 þ A1 ¼ 4 2x1 þ 4x2 þ 2k1  l2 þ A2 ¼ 6 x1 þ 2x2 þ s1 ¼ 2 x1 ; x2  0; k1  0; l1 ; l2 ; s1 ; A1 ; A2  0;

12.5

287

satisfying the complementary slackness conditions k1 s1 ¼ 0;

l1 x1 ¼ l2 x2 ¼ 0:

To solve the problem, we shall apply the two-phase method. Here the initial BFS is A1 ¼ 4; A2 ¼ 6; s1 ¼ 0. Now we construct the initial table of Phase I.

cB

Basic variables

−1 A1 −1 A2 0 s1 z0B ¼ 10

cj xB

0 x01

0 x02

k01

0 l01

0 l02

−1 A01

−1 A02

0 s01

4 6 2

4 2 1 −6

2 4 2 −6

1 2 0 −3

−1 0 0 1

0 −1 0 1

1 0 0 0

0 1 0 0

0 0 1 0

0

Mini ratio 4/4 = 1 6/2 = 3 2/1 = 2 Dj

Here all Dj l0. Hence, the current BFS is not optimal. From the preceding table, we observe that either x1 or x2 (as D1 ¼ 6; D2 ¼ 6) can be introduced into the basic solution. Arbitrarily, we choose x1 as the entering basic variable, as l1 ¼ 0 (for satisfying l1 x1 ¼ 0). Therefore, A1 is the leaving basic variable. Now we construct the next table, dropping the A1 column. cj

0

0

0

0

0

−1

0

x02

k01

l01

l02

A02

s01

½ 3 3/ 2 −3

¼ 3/2 −¼

−¼ ½ ¼

0 −1 0

0 1 0

0 0 1

Mini Ratio 2 4/3 2/3

−3/2

−½

1

0

0

Dj

cB

Basic variables

xB

x01

0 −1 0

x1 A2 s1

1 4 1

1 0 0

ZB0 ¼ 4

0

Since l2 ¼ 0; x2 is introduced as the basic variable with s1 as the leaving basic variable. Now, we construct the next table. cj

0

0

0

0

0

−1

0

cB

Basic variables

xB

x01

x02

k01

l01

l02

A02

s01

0 −1 0

x1 A2 s1

1 4 1

1 0 0

½ 3 3/ 2 −3

¼ 3/2 −¼

−¼ ½ ¼

0 −1 0

0 1 0

0 0 1

Mini Ratio 2 4/3 2/3

−3/2

−½

1

0

0

Dj

ZB0 ¼ 4

0

288

Since s1 ¼ 0, k1 is introduced as the basic variable. A2 is the leaving basic variable. Now we construct the next table, dropping the A2 column.

cB

Basic variables

0 x1 0 k1 0 x2 ZB0 ¼ 2

cj

0

0

0

0

0

0

xB

x01

x02

k01

l01

l02

s01

1/3 1 5/6

1 0 0 0

0 0 1 0

0 1 0 0

−1/3 0 1/6 0

1/6 −1/2 −1/12 0

0 −1 ½ 0

Dj

Here all Dj l0. Hence, the optimal solution for Phase I has been obtained, and the solution is x1 ¼ 13 ; x2 ¼ 56 ; k1 ¼ 1; l1 ¼ 0; l2 ¼ 0; s1 ¼ 0. Thus, the optimal solution of the original QPP is given by      2     2 1 5 1 5 1 1 5 5 and zmax ¼ 4 x1 ¼ ; x2 ¼ 2 þ 2 2 3 6 3 6 3 3 6 6 25 ¼ : 6 Example 4 Apply Wolfe’s method for solving the following problem: Maximize z ¼ 2x1 þ x2  x21 subject to 2x1 þ 3x2  6 2x1 þ x2  4 and x1 ; x2  0: Solution The given problem is a maximization problem. Now, converting all the inequality constraints, including the non-negativity constraints, into  type, we have 2x1 þ 3x2  6 2x1 þ x2  4  x1  0; x2  0: Now introducing slack variables q21 ; r12 and r22 , we have the reduced problem as

12.5

289

Maximize z ¼ 2x1 þ x2  x21 2x1 þ 3x2 þ q21 ¼ 6 2x1 þ x2 þ q22 ¼ 4 x1 þ r12 ¼ 0 x2 þ r22 ¼ 0 The Lagrangian function L is given by Lðx1 ; x2 ; q1 ; q2 ; r1 ; r2 ; k1 ; k2 ; l1 ; l2 Þ       ¼ 2x1 þ x2  x21  k1 2x1 þ 3x2 þ q21  6  k2 2x1 þ x2 þ q22  4      l1 x1 þ r12  l2 x2 þ r22 : As x21 represents a negative semi-deﬁnite quadratic form, z ¼ 2x1 þ x2  x21 is concave in x1, x2. Hence, the maxima of L will be the maxima of z and vice versa. Now, the necessary conditions for maxima of L (and hence of z) are @L ¼ 2  2x1  2k1  2k2 þ l1 ¼ 0 or 2x1 þ 2k1 þ 2k2  l1 ¼ 2 @x1 @L ¼ 1  3k1  k2 þ l2 ¼ 0 or 3k1 þ k2  l2 ¼ 1 @x2   @L ¼  2x1 þ 3x2 þ q21  6 ¼ 0 or 2x1 þ 3x2 þ s1 ¼ 6 putting q21 ¼ s1 @k1   @L ¼  2x1 þ x2 þ q22  4 ¼ 0 or 2x1 þ x2 þ s2 ¼ 4 putting q22 ¼ s2 @k2 @L ¼ 2k1 q1 ¼ 0 or k1 q21 ¼ 0 or k1 s1 ¼ 0 @q1 @L ¼ 2k2 q2 ¼ 0 or k2 q22 ¼ 0 or k2 s2 ¼ 0 @q2   @L ¼  x1 þ r12 ¼ 0 or x1 ¼ r12 @l1   @L ¼  x2 þ r22 ¼ 0 or x2 ¼ r22 @l2

290

@L ¼ 2l1 r1 ¼ 0 or l1 r1 ¼ 0 or l1 r12 ¼ 0 or l1 x1 ¼ 0 @r1 @L ¼ 2l2 r2 ¼ 0 or l2 r2 ¼ 0 or l2 r22 ¼ 0 or l2 x2 ¼ 0 @r2 and x1 ; x2  0; k1 ; k2  0; l1 ; l2 ; s1 ; s2  0: Hence, to determine all the variables, the equivalent problem of the given QPP is as follows: 9 2x1 þ 2k1 þ 2k2  l1 ¼ 2 > > = 3k1 þ k2  l2 ¼ 1 ; ð12:45Þ 2x1 þ 3x2 þ s1 ¼ 6 > > ; 2x1 þ x2 þ s2 ¼ 4 where all variables are non-negative and k1 s1 ¼ 0; k2 s2 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0:

ð12:46Þ

A solution xj ; j ¼ 1; 2; of (12.45) and satisfying (12.46) shall necessarily be an optimal one for maximizing L (i.e. maximizing z). Now to determine the solution to the simultaneous Eq. (12.45), we introduce two artiﬁcial variables A1 ð  0Þ and A2 ð  0Þ in the ﬁrst two equations of (12.45) and construct the dummy objective function z0 ¼ A1  A2 . Then the equivalent problem of the given QPP becomes Maximize z0 ¼ A1  A2 subject to 2x1 þ 2k1 þ 2k2  l1 þ A1 ¼ 2 3k1 þ k2  l2 þ A2 ¼ 1 2x1 þ 3x2 þ s1 ¼ 6 2x1 þ x2 þ s2 ¼ 4 and x1 ; x2  0; k1 ; k2  0; l1 ; l2 ; s1 ; s2 ; A1 ; A2  0; satisfying the complementary slackness conditions k1 s1 ¼ 0; k2 s2 ¼ 0; l1 x1 ¼ 0; l2 x2 ¼ 0: To solve the problem, we shall apply the two-phase method. Here the initial BFS is A1 ¼ 2; A2 ¼ 1; s1 ¼ 6; s2 ¼ 4.

12.5

291

Now we construct the initial table of Phase I. 0

0

0

0

0

0

0

0

−1

−1

xB

x01

x02

k01

k02

l01

l02

s01

s02

A01

A02

Mini ratio

2 1 6 4

2 0 2 2 −2

0 0 3 1 0

2 3 0 0 −5

2 1 0 0 −3

−1 0 0 0 1

0 −1 0 0 1

0 0 1 0 0

0 0 0 1 0

1 0 0 0 0

0 1 0 0 0

1 – 3 2

cj cB

Basic variables

−1 A1 −1 A2 0 s1 0 s2 z0B ¼ 3

Dj

Here all Dj j0. Hence, the current BFS is not optimal. Dj corresponding to k01 is most negative and corresponding to k02 is second most negative, but the corresponding variables k1 ; k2 will not be considered as basic variables, since s1 and s2 are basic variables. So the entering basic variable will be x1, as l1 is not a basic variable and A1 is the leaving basic variable. Now we construct the next table, dropping the A01 column.

cB

cj

0

0

0

0

0

0

0

0

−1

Basic variables

x01

x02

k01

k02

l01

l02

s01

s02

A02

Mini ratio

0

0

0

0

−1 0 0 1

0 1 0 0

0 0 1 0

1 0 0 0

– 4/3 2

xB

x1

1

1

0

1

1

−1 A2 0 s1 0 s2 z0B ¼ 1

1 4 2

0 0 0 0

0 3 1 0

3 −2 −2 −3

1 −2 −2 −1

0

 12 0 1 1 0

Dj

Here all Dj j0. Hence, the current BFS is not optimal. From this table, we observe that D1 corresponding to k1 is most negative and k2 is second most negative. However, the corresponding variables k1 ; k2 will not be considered as basic variables, as s1 and s2 are already basic variables. So the basic variable will be x2 (as l2 is not the basic variable) and s1 is the leaving basic variable. Now, we construct the next table.

cB

cj ! Basic variables

0

x1

−1

A2

xB

0 x01

0 x02

0

0

k01

k02

1

1

1

0

0 l01

0

1

1

 12

0

3

1

0

0 l02

0 s01

0 s02

−1 A02

Mini ratio

0

0

0

0

1

−1

0

0

1

1/3 (continued)

292

(continued) cj ! Basic variables

0 x02

0

xB

0 x01

0

cB

k01

k02

0 l01

0

x2

4/3

0

1

 23

 23

1/3

0

0

 43

 43

0

0

−3

−1

0

s2

2/3

z0B ¼ 1

0 s01

0 s02

−1 A02

Mini ratio

0

1/3

0

0

2/3

0

 13

1

0

0

1

0

0

0

0 l02

Dj

Here all Dj 0. Hence, the current solution is not optimal. From the preceding table, it is observed that Dj corresponding to k1 is most negative. Hence, k1 is the new basic variable, since s1 is not in the basis and A2 is the leaving basic variable. cj Basic variables

0 x02

0

xB

0 x01

0

cB

k01

k02

0 l01

0 l02

0 s01

0 s02

0

x1

2/3

1

0

0

2/3

 12

1/3

0

0

0

0

1/3

0

 13

1

0

0

0

k1

1/3

0

0

1

1/3

0

0

x2

14/9

0

1

0

 49

1/3

0

s2

10/9

0

0

0

 89

2/3

 13  29  49

0

0

0

0

0

0

z0B

¼0

Dj

From the table, it is clear that Dj  0. Hence, the current solution is optimal. The 1 10 optimal solution is x1 ¼ 23 ; x2 ¼ 14 9 , k1 ¼ 3 ; k2 ¼ 0; l1 ¼ 0; l2 ¼ 0; s1 ¼ 0; s2 ¼ 9 . This solution also satisﬁes the complementary slackness conditions k1 s1 ¼ 0 ¼ k2 s2 and l1 x1 ¼ 0 ¼ l2 x2 : Again, since zB ¼ 0, the current solution is also feasible. 14 4 12 þ 144 ¼ 22 Now, Max z ¼ 2:2 3 þ 9 9¼ 9 9. Therefore, the optimal solution of the original QPP is given by x1 ¼ 23 ; x2 ¼ 14 9 and 22 zmax ¼ 9 .

12.6

Beale’s Method

In the year 1959, Beale developed an alternative technique for solving QPPs without using Kuhn-Tucker conditions. In this technique, the classical calculus results have been used to partition the variables into basic and non-basic ones. In

12.6

Beale’s Method

293

each iteration, all the basic variables and the objective function are expressed in terms of non-basic variables.

12.7

Beale’s Algorithm for QPPs

Let the general QPP be the following: 1 Maximize z ¼ hc; xi þ hQx; xi 2 subject to the constraints Ax  or  or ¼ b and x  0; where xT 2 Rn , A is an m  n matrix, b is an m  1 matrix and Q is an n  n symmetric matrix. Beale’s iterative procedure for solving this problem can be summarized as follows: Step 1. Convert the minimization problem into a maximization problem, if required. Introduce slack and/or surplus variables in the inequality constraints, if any, and put the QPP in the standard form. Step 2. Select arbitrarily m number of variables as basic variables; the remaining variables are considered as non-basic variables. Let the basic variables be denoted by xBi (i = 1, 2, …, m) and the non-basic variables as xNBk ; k ¼ 1; 2; . . .; n  m. Step 3. Express each basic variable xBi entirely in terms of the non-basic variables, i.e. the xNBi ’s (and ui ’s, if any), using the given constraints (as well as any additional ones). Step 4. Express the objective function z, also in terms of the non-basic variable xNBi ’s (and ui ’s, if any). Step 5. Examine the partial derivatives of z with respect to the non-basic variables at the point xNB ¼ 0 (and u = 0):

(i) If @x@zNB x ¼ 0  0 for each k ¼ 1; 2; . . .; n  m k

NB

u¼0 @z and @ui x ¼ 0 ¼ 0 for each i = 1,2, …, m, NB

u¼0 the current basic solution is optimal. Go to Step 7.

@z (ii) If @xNB x ¼ 0 [ 0 for each least one k, NB k u¼0

294

choose the most positive one. The corresponding non-basic variables will

enter the basis. @z (iii) If @xNB x ¼ 0 \0 for each k ¼ 1; 2; . . .; n  m NB k

u¼0 @z 6¼ 0 for some i = r, but @u

i xNB ¼ 0 u¼0 @z , and treat introduce a new non-basic variable uj, deﬁned by uj ¼ @u r ur as a basic variable. Go to Step 3. Step 6. Let xNBi ¼ xk be the entering variable identiﬁed in Step 5 (ii). Compute the largest non-negative values of xk from the non-negativity conditions of @z each basic variable and the value of xk from @x ¼ 0 using other non-basic k variables as zero. Step 7. Find the minimum value of xk. (i) If the minimum value of xk occurs from the non-negative condition of a basic variable, then that variable will be the leaving variable from the basis and xk will be the new basic variable. @z (ii) If the minimum value of xk occurs from @x ¼ 0, introduce an addik tional non-basic variable, called a free variable, deﬁned by @z (ui is unrestricted). ui ¼ @x k This relation becomes an additional constraint equation. Step 8. Go to Step 3 and repeat the procedure until an optimal basic solution is reached. Step 9. Determine the optimal values of xB and z by setting xNB ¼ 0 in their expressions obtained in Step 3 and Step 4. Step 10. Stop. Example-5 Use Beale’s method to solve the following QPP: Maximize z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 subject to x1 þ 2x2  2 and x1 ; x2  0: Solution The given problem is Maximize z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 subject to x1 þ 2x2  2 and x1 ; x2  0:

12.7

Beale’s Algorithm for QPPs

295

Now introducing slack variable x3  0 in the given constraint, the given problem reduces to Maximize z ¼ 4x1 þ 6x2  2x21  2x1 x2  2x22 subject to x1 þ 2x2 þ x3 ¼ 2 and x1 ; x2 ; x3  0: Let us consider x2 as basic variables xB and the remaining two variables x1 ; x3 as non-basic variables xNB . Now expressing xB and z in terms of xNB , we have 1 x2 ¼ 1  ðx1 þ x3 Þ 2    2 x1 x3

1 1 and z ¼ 4x1 þ 6 1    2x21  2x1 1  ðx1 þ x3 Þ  2 1  ðx1 þ x3 Þ 2 2 2 2        @z 6 1 1 1 1  2 1  ðx1 þ x3 Þ  4 1  ðx1 þ x3 Þ  ) ¼ 4   4x1  2x1  @x1 2 2 2 2 2 ¼ 4  3  4x1 þ x1  2 þ x1 þ x3 þ 2  x1  x3 ¼ 1  3x1      @z 6 1 1 1  4 1  ðx1 þ x3 Þ  ¼ 0   2x1  @x3 2 2 2 2 ¼ 3 þ x1 þ 2  x1  x3 ¼ 1  x3

@z

@z

) ¼ 1  3:0 ¼ 1 and ¼ 1  0 ¼ 1: @x1 xNB ¼0 @x3 x ¼0 NB

The values of the partial derivatives of z with respect to x1 and x3 at xNB ¼ 0 indicate that z would be increased if x1 is increased. Therefore, x1 will be the new basic variable. Now we have to ﬁnd the value of x1 . Since x2 ¼ 1  12 ðx1 þ x3 Þ and x3 = 0, then x1 = 2 if x2 is selected as the non-basic variable. Again,

@z @x1

¼ 0 with x3 ¼ 0 implies 1 1  3x1 ¼ 0 or x1 ¼ : 3

Hence, the new value of x1 is given by   x1 ¼ Min 2; 13 ¼ 13, which corresponds to Hence, x2 cannot be removed from the basis.

@z @x1

¼ 0 with x3 ¼ 0.

296

Now, we introduce a non-basic free variable u1 and the new constraint u1 ¼

@z ; i:e: u1 ¼ 13x1 : @x1

Hence, the current basic variables will be x2, x1, and the remaining variables x3 and u1 will be non-basic variables.  ) xB ¼

x2 x1



 and xNB ¼

 x3 : u1

Now expressing xB and z in terms of xNB , we have x1 ¼ ð1  u1 Þ=3 1 1 5 u1 x3 x2 ¼ 1  ðx1 þ x3 Þ=2 ¼ 1  ð1  u1 Þ  x3 ¼ þ  6 2 6 6 2   5 u1 x3 2 þ  =3 and z ¼ 4ð1  u1 Þ=3 þ 5 þ u1  3x3  2ð1  u1 Þ =9  2ð1  u1 Þ 6 6 2   5 u1 x3 2 2 þ  6 6 2 @z 1 5 u 1 x3 ) ¼ 3 þ þ þ  @x3 3 6 6 2   @z 4 5 u1 x3 þ  ¼  þ 1 þ 4ð1  u1 Þ=9 þ 2 =3  2ð1  u1 Þ=9 @u1 3 6 6 2   5 u1 x3 þ  2 =3: 6 6 2

Now evaluating these partial derivatives at xNB ¼ 0, i.e. at x3 = 0 and u1 = 0, we have

@z

@z

¼ negative and ¼ 0; @x3 xNB ¼0 @u1 xNB ¼0 which indicate that the current xB provides the optimal solution. As x3 ¼ 0, u1 ¼ 0, x1 ¼ 13 ; x2 ¼ 56  2     2 and Max z ¼ 4  13 þ 6  56  2  13 2 13 56  2 56 ¼ 25 6.

12.7

Beale’s Algorithm for QPPs

297

Example 6 Use Beale’s method to solve the following QPP: Maximize z ¼ 2x1 þ 3x2  2x22 subject to the constraints x1 þ 4x2  4; x1 ; þ x2  2 and x1 ; x2  0: Solution Introducing slack variables x3 ð  0Þ and x4 ð  0Þ into the given constraints, we get the reduced problem as Maximize z ¼ 2x1 þ 3x2  2x22 subject to x1 þ 4x2 þ x3 ¼ 4 x1 þ x2 þ x4 ¼ 2 and x1 ; x2 ; x3 ; x4  0:

Let us take x1 and x2 as basic variables and the remaining variables x3 and x4 as non-basic variables.  ) xB ¼

x1 x2



 and xNB ¼

 x3 : x4

Now expressing xB and z in terms of xNB , we have x2 ¼ ð2  x3 þ x4 Þ=3 x1 ¼ ð4 þ x3  4x4 Þ=3 and z ¼ 2ð4 þ x3  4x4 Þ=3 þ ð2  x3 þ x4 Þ  2ð2  x3 þ x4 Þ2 =9 @z 2 4 1 4 ) ¼  1  ð2  x3 þ x4 Þð1Þ ¼  þ ð2  x3 þ x4 Þ @x3 3 9 3 9 @z 8 4 5 4 and ¼  þ 1  ð2  x3 þ x4 Þ ¼   ð2  x3 þ x4 Þ: @x4 3 9 3 9 Now evaluating these partial derivatives at xNB ¼ 0, i.e. at x3 = 0, x4 = 0, we have

@z

1 8 5 @z

23 þ ¼ and ¼  ¼ :

@x3 xNB ¼0 3 9 9 @x4 xNB ¼0 9 The values of the partial derivatives of z with respect to x3 and x4 at xNB ¼ 0 indicate that z would be increase if x3 is increased. Therefore, x3 will be the new basic variable. Now we have to ﬁnd the value of x3. Since x1 ¼ ð4 þ x3  4x4 Þ=3 and x4 = 0, then x3 = –4 if x1 is selected as the non-basic variable. This is impossible, as x3  0.

298

Again, since x2 ¼ ð2  x3 þ x4 Þ=3 with x4 = 0, then x3 = 2 if x2 is selected as the @z non-basic variable. On the other hand, @x ¼ 0 with x4 = 0 gives 3 1 4 4 1 5  þ ð 2  x3 Þ ¼ 0 ) ð 2  x3 Þ ¼ ) x3 ¼ : 3 9 9 3 4 Hence, the new value of x3 is given by   x3 ¼ min 2; 54 ¼ 54, which corresponds to

@z @x3

¼ 0 with x4 = 0.

Hence, we cannot remove x1 and x2 from the basis. Now, we introduce a non-basic free variable u1 and the new constraint u1 ¼

@z @x3

i:e:; u1 ¼ 

1 4 þ ð2  x3 þ x4 Þ: 3 9

Hence, the current basic variables will be x1, x2, x3, and the remaining variables x4 and u1 will be non-basic variables. 0 1   x1 x4 @ A ) xB ¼ x2 and xNB ¼ : u1 x3 Now expressing xB and z in terms of xNB, we have

  9 5 4 5 9  u1 þ x 4 ¼  u1 þ x 4 4 9 9 4 4     1 1 5 9 1 21 9 7 3 x1 ¼ ð4 þ x3  4x4 Þ ¼ 4 þ  u1 þ x4  4x4 ¼  u1  3x4 ¼  u1  x4 3 3 4 4 3 4 4 4 4     1 1 5 9 1 3 9 1 x2 ¼ ð 2  x3 þ x4 Þ ¼ 2  þ u1  x 4 þ x 4 ¼ þ u1 ¼ ð1 þ 3u1 Þ 3 3 4 4 3 4 4 4   7 3 3 1 2 and z ¼ 2  u1  x4 þ ð1 þ 3u1 Þ  ð1 þ 3u1 Þ : 4 4 4 8 x3 ¼

Now,

@z @x4

¼ 2 and

@z @u1

¼  32 þ 94  34 ð1 þ 3u1 Þ.

Now evaluating these partial derivatives at xNB ¼ 0, i.e. at x4 = 0, u1 = 0, we have

@z

@z

3 9 3 ¼ 2 and ¼  þ  ¼ 0:

@x4 xNB ¼0 @u1 xNB ¼0 2 4 4 These partial derivatives indicate that the current xB provides the optimal solution. Since x4 ¼ 0; u1 ¼ 0; then x1 ¼ 74 ; x2 ¼ 14 and z ¼ 72 þ 34  18 ¼ 33 8.

12.7

Beale’s Algorithm for QPPs

299

Hence, the optimal solution is x1 ¼ 74 ; x2 ¼ 14 and maxz ¼ 33 8. Example 7 Solve the following QPP by Beale’s method: Maximize z ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to

x1 þ 2x2  10 x1 þ x2  9 and x1 ; x2  0:

Solution Introducing slack variables x3 ð  0Þ and x4 ð  0Þ into the given problem, we convert it into Maximize z ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to x1 þ 2x2 þ x3 ¼ 10 x1 þ x2 þ x4 ¼ 9 and x1 ; x2 ; x3 ; x4  0 Let us consider x1 ; x2 as basic variables and x3 ; x4 as non-basic variables. Now expressing xB and z in terms of xNB ; we have x1 ¼ 8 þ x3  2x4 and x2 ¼ 1  x3 þ x4 and z ¼ 10ð8 þ x3  2x4 Þ þ 25ð1  x3 þ x4 Þ  10ð8 þ x3  2x4 Þ2 ð1  x3 þ x4 Þ2 4ð8 þ x3  2x4 Þð1  x3 þ x4 Þ: Now @z ¼ 10  25  20ð8 þ x3  2x4 Þ þ 2ð1  x3 þ x4 Þ @x3  4ð1  x3 þ x4 Þ þ 4ð1 þ x3  2x4 Þ and @z ¼ 20 þ 25  20ð2Þð8 þ x3  2x4 Þ  2ð1  x3 þ x4 Þ þ 8ð1  x3 þ x4 Þ @x4  4ð8 þ x3  2x4 Þ:

300

@z

@x

3 xNB ¼0

@z

¼ 145 and @x

¼ 299:

4 xNB ¼0

The values of the partial derivatives of z with respect to x3 and x4 at xNB ¼ 0 indicate that z will increase if x4 increases. Since 2x4 ¼ ð8 þ x3  x1 Þ; if x4 is increased to a value greater than 4, then x1 will become negative and x3 ¼ 0. Now @z ¼0 @x4 or  20 þ 25 þ 40ð8 þ x3  2x4 Þ  2ð1  x3 þ x4 Þ þ 8ð1  x3 þ x4 Þ  4ð8 þ x3  2x4 Þ ¼ 0:

Setting x3 ¼ 0; we get 5 þ 40ð8  2x4 Þ  2ð1 þ x4 Þ þ 8ð1 þ x4 Þ  4ð8  2x4 Þ ¼ 0 or; 5 þ 320  80x4  2  2x4 þ 8 þ 8x4  32 þ 8x4 ¼ 0 or; 66x4 ¼ 299 299 or; x4 ¼ 66 Hence, the new value of x4 is given by   @z Min 4; 299 66 ¼ 4 which corresponds to @x4 ¼ 0 with x3 ¼ 0. So, the new basic variables are x4 and x2 . Now, expressing xB and z in terms of xNB , we have 1 x2 ¼ 5  ðx1 þ x3 Þ 2 1 x4 ¼ 4 þ ð x 3  x 1 Þ 2    2   1 1 1 z ¼ 10x1 þ 25 5  ðx1 þ x3 Þ  10x21  5  ðx1 þ x3 Þ 4x1 5  ðx1 þ x3 Þ ; 2 2 2

where xB ¼ ðx2 ; x4 Þ; xNB ¼ ðx1 ; x3 Þ. Now,

12.7

Beale’s Algorithm for QPPs

301

     @z 25 1 1 1 ¼ 10   20x1  2 5  ðx1 þ x3 Þ   4 5  ð x1 þ x3 Þ @x1 2 2 2 2   1  4x1  2 and      @z 25 1 1 1 ¼   2 5  ð x1 þ x3 Þ   4x1  : @x3 2 2 2 2

@z

35 @z

15 and ¼ ¼ : @x1 xNB ¼0 2 @x3 xNB ¼0 2 Both partial derivatives are negative at xNB ; hence, neither x1 nor x3 can be introduced to increase z, and the current xB provides the optimal solution. Setting x1 ¼ 0; x3 ¼ 0, we have x2 ¼ 5 and x4 ¼ 4. Hence, the optimal solution is given by x2 ¼ 5 and x4 ¼ 4 and Max z ¼ 0 þ 25  5  0  25  0 ¼ 100. Note In this problem, if we initially consider x2 ; x4 as basic variables and x1 ; x3 as non-basic variables, then this problem can be solved very easily in one iteration.

12.8

Exercises

1. Use Wolfe’s method for solving the following problem: Maximize z ¼ 10x1  x21 þ 10x2  x22 , where x1 þ x2  14; x1 þ x2  6; x1 ; x2  0. 2. Use the Kuhn-Tucker conditions to solve the problem: Maximize z ¼ 12x1 þ 21x2 þ 2x1 x2  2x21  2x22 subject to x2  8; x1 þ x2  10: 3. Use Wolfe’s method for solving the following problem: Maximize Z ¼ 2x1 þ 3x2  2x21 subject to the constraints x1 þ 4x2  4; x1 þ x2  2; x1 ; x2  0:

302

4. Use Beale’s method to solve the following problem: Minimize Z ¼ 6  6x1  2x21  2x1 x2 þ 2x22 such that x1 þ x2  2 and x1 ; x2  0: 5. Use Wolfe’s method for solving the following problem: Maximize z ¼ 10x1 þ 25x2  10x21  x22  4x1 x2 subject to x1 þ 2x2  10; x1 þ x2  9 and x1 ; x2  0: 6. Use Beale’s method to solve the following problem: Minimize z ¼ x21  x1 x2 þ 2x22  x1  x2 subject to 2x1 þ x2  1 and x1  0; x2  0:

Chapter 13

Game Theory

13.1

Objectives

The objectives of this chapter are to discuss: • The process of ﬁnding the optimal strategies in conflict and competitive situations. • Different types of ﬁnite and inﬁnite games and their solution procedures.

13.2

Introduction

In most practical problems, there arises a situation where there are two or more opposite parties (particularly players) with conflicting of interests and actions. This type of situation is called a competitive situation. A large variety of competitive situations are commonly observed in daily life, e.g. in political campaigns, elections, advertisements, military battles, etc. A competitive situation is called a game if the following properties are satisﬁed: (i) (ii) (iii) (iv)

There is a conflict of interests between the participants. The number of competitors (participants), called players, is ﬁnite. Each of the participants has a ﬁnite/inﬁnite set of possible courses of action. The rules governing these choices are speciﬁed and known to the players; a play of the game results when each of the players chooses a single course of action from a list of courses available to him. (v) The outcome of the game is affected by the choices made by all the players. (vi) The outcome for all speciﬁc sets of choices by all the players is known in advance and deﬁned.

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_13

303

304

13

Game Theory

(vii) Every play, i.e. combination of courses of action, determines an outcome (which may be money or points) which determines a set of payments (positive, negative or zero), one to each player.

13.3

Cooperative and Non-cooperative Games

Games in which the objective of each participant (player) is to achieve the largest possible individual gain (the ‘payoff’) are called non-cooperative games. Games in which the actions of the players are directed to maximize the gain of ‘collectives’ without subsequent subdivision of the gain among the players within the condition are called cooperative games. Here we shall consider only non-cooperative games. Let I denote the set of all players. We shall assume that I is ﬁnite. Let I ¼ f1; 2; . . .; ng, where n denotes the number of players in the game. Let each player i 2 I have at his disposal a certain set Si of the available actions, which are called strategies. Each player has at least two distinct strategies. The process of the game consists of each one of the players choosing a certain strategy si 2 Si . Thus, as a result of each ‘round’ of the game, a system of strategies ðs1 ; . . .; sn Þ  s is put together. This system is called a situation. The set of all strategies is denoted by S ¼ S1  S2      Sn . In each situation the players attain certain gains (payoffs). The payoff of player i in situation s is denoted by Hi(s). The function Hi deﬁned on the set of all situations is called the payoff function of player i. Thus, Hi : S ! R, where R is a set of real numbers.   Deﬁnition A system denoted by C ¼ I; fSi gi2I ; fHi gi2I , where I and Si ði 2 I Þ are Qn sets and Hi’s are functions deﬁned on the set S ¼ i¼1 Si taking on real values, is called a non-cooperative game.   is called a Deﬁnition A non-cooperative game C ¼ I; fSi gi2I ; fHi gi2I P constant-sum game if there exists a constant C such that i2I Hi ðsÞ ¼ C for each and every situation s 2 S. Admissible situation and equilibrium situation Let s ¼ ðs1 ; . . .; si1 ; si ; si þ 1 ; . . .; sn Þ be an arbitrary situation in the game C, and let si be a strategy of player i. We form a new situation that differs from the situation sonly in that the strategy si for player i is now replaced by s0i . Denotethis as   ss0i ¼ s1 ; . . .; si1 ; s0i ; si þ 1 ; . . .; sn . If si and s0i coincide, then obviously ss0i ¼ s.

13.3

Cooperative and Non-cooperative Games

305

Admissible situation A situation s ¼ ðs1 ; . . .; sn Þ in a game C ¼ hI; fsi g; fHi gi is called an admissible situation for player i if for any other strategy s0i of this player we have   Hi ss0i  Hi ðsÞ: The term ‘admissible’ can thus be justiﬁed  by0 the fact that if in situation s there 0 exists a strategy si for player i such that Hi ssi [ Hi ðsÞ, then player i, knowing that the situation s will materialize, may choose the strategy s0i at the last moment and as a result of this choice end up with a large payoff. In this sense the situation may be viewed as admissible for player i. Equilibrium situation An admissible situation for all players is called an equilibrium situation; i.e.   Hi ss0i  Hi ðsÞ is satisﬁed for any player i and for any strategy s0i 2 Si . Equilibrium strategy An equilibrium strategy is one that appears in at least one equilibrium situation of the game. Solution of the game The process of determination of an equilibrium situation in a non-cooperative game is often referred to as the solution of the game. Strategic equivalence of games Let two non-cooperative games with the same set of players and the same set of strategies be given (i.e. the games differ only in their payoff functions) as D n o E ð1Þ C1 ¼ I; fSi gi2I ; Hi i2I

D n o E ð2Þ C2 ¼ I; fSi gi2I ; Hi : i2I

Deﬁnition The games C1 and C2 are called strategically equivalent if a positive number k and numbers Ci (for each of the players i 2 I) exist such that for any situations ð1Þ

ð2Þ

Hi ðsÞ ¼ k Hi ðsÞ þ Ci : We denote this fact by the symbol C1  C2 .

306

13

Game Theory

Properties of strategic equivalence (i) Reflexivity: C  C; this can easily be proved choosing k = 1 and Ci = 0. (ii) Symmetry: C1  C2 ) C2  C1 . ð1Þ

Proof Since C1  C2 , we can write Hi

ð2Þ

¼ k Hi ðsÞ þ Ci

k [ 0; 8 i 2 I

1 ð1Þ Ci ð2Þ or Hi ðsÞ ¼ Hi ðsÞ  k k 1 ð1Þ ¼ k 0 Hi ðsÞ þ Ci0 where k 0 ¼ ; k

Ci0 ¼ 

Ci : k

Hence, by deﬁnition, C2  C1 . (iii) Transitivity: If C1  C2 and C2  C3 , then C1  C3 . This can easily be proved using the deﬁnition. Note The basic difference between two strategically equivalent games lies in the amount of initial capital possessed by the players and in the relative units in which the payoffs are measured. This is indicated by the coefﬁcient k. It is therefore natural to suppose that the rational behaviour of players in distinct strategically equivalent games should be the same. Theorem 13.1 Strategically equivalent games possess the same equilibrium situation. D n o E ð1Þ Proof Let C1  C2 , where C1 ¼ I; fSi gi2I ; Hi i2I

D n o E ð2Þ C2 ¼ I; fSi gi2I ; Hi i2I

and let s be an equilibrium situation in C1 . This means that 8i 2 I and s0i 2 Si , we have the inequality  ð1Þ  ð1Þ Hi s k s0i  Hi ðsÞ, where s ¼ ðs1 ; . . .; si1 ; si ; si þ 1 ; . . .; sn Þ, s k s0i ¼ ðs1 ; . . .; si1 ; s0i ; si þ 1 ; . . .; sn Þ. ð1Þ

ð2Þ

Now since C1  C2 ; Hi ¼ k Hi þ Ci ðk [ 0Þ 8i 2 I and si 2 Si  ð1Þ  ð2Þ i:e: Hi s k s0i  k Hi þ Ci    ð2Þ  ð2Þ ð1Þ  ð2Þ  or k Hi s k s0i þ Ci  Hi ðsÞ þ Ci as Hi s k s0i ¼ k Hi s k s0i þ Ci  ð2Þ  ð2Þ or Hi s k s0i  Hi ðsÞ 8i 2 I and s0i 2 Si since k [ 0 which implies s is an equilibrium situation in C2 .

13.3

Cooperative and Non-cooperative Games

307

Zero-sum game AP non-cooperative game C is called a zero-sum game if for each situation s; i2I Hi ðsÞ ¼ 0. Theorem 13.2 Any non-cooperative constant-sum game is strategically equivalent to a certain zero-sum game.   P Proof Let C ¼ I; fSi gi2I ; fHi gi2I be a constant-sum game. Then i2I Hi ðsÞ ¼ C, D   E C being a constant. Now let us consider the game C0 ¼ I; fSi gi2I ; Hi0 i2I , where P Hi0 ¼ Hi  Ci and i2I Ci ¼ C. Then obviously C  C0 when k = 1. But

P i

Hi0 ðsÞ ¼

P i

Hi ðsÞ 

P i

Ci ¼ C  C ¼ 0,

So C′ is a zero-sum game.

13.4

Antagonistic Game

  The game C ¼ I; fSi gi2I ; fHi gi2I is called antagonistic if there are only two players in the game (i.e. I ¼ f1; 2g), and the values of the payoff function for these players in each situation are the same in absolute value but are of opposite sign. Thus, for an antagonistic game, I ¼ f1; 2g and H2 ðsÞ ¼ H1 ðsÞ. Hence, H1 ðsÞ þ H2 ðsÞ ¼ 0; s ¼ ðs1 ; s2 Þ 2 S1  S2 . Thus, an antagonistic game is also a two-person zero-sum game. Thus, to deﬁne an antagonistic game, it is sufﬁcient to stipulate the payoff function of one of the players only. It is usually denoted by the triplet hX; Y; M i, where X, Y are the sets of strategies for players 1 and 2 respectively and M is the payoff function for player 1 which is a real-valued function such that M : X  Y ! R. Note It is obvious that each non-cooperative two-person constant zero-sum game is strategically equivalent to a certain antagonistic game.

Equilibrium situation for an antagonistic game Let hX; Y; M i be an antagonistic game, H1 ¼ M; H2 ¼ M. Hence, s1  x 2 X; s2  y 2 Y; s  ðx; yÞ 2 X  Y. If (x, y) is an equilibrium situation, then H1 ðx; yÞ  H1 ðx0 ; yÞ8x0 2 X and H2 ðx; yÞ  H2 ðx; y0 Þ8y0 2 Y; i:e: M ðx; yÞ  M ðx0 ; yÞ8x0 2 X and  M ðx; yÞ   M ðx; y0 Þ8y0 2 Y; i:e: M ðx0 ; yÞ  M ðx; yÞ  M ðx0 ; yÞ8x0 2 X; y0 2 Y:

308

13

Game Theory

Changing the notation (x, y) to (x*, y*) and (x′, y′) to (x, y), we see that (x*, y*) will be an equilibrium situation for an antagonistic game hX; Y; M i if M ðx; y Þ  M ðx ; y Þ  M ðx ; yÞ 8x 2 X; y 2 Y: The point (x*, y*) is called the saddle point of M(x, y).

13.5

Matrix Games or Rectangular Games

Antagonistic games in which each player possesses a ﬁnite number of strategies are called matrix games or rectangular games. When there are two competitors playing a game, it is called a two-person game. In a game, if the number of competitors is more than two, say n, the game is referred to as an n-person game. If, in a game, the algebraic sum of the payments to all the competitors is zero for every possible outcome of the game, then the game is said to be a zero-sum game. Otherwise, it is said to be non-zero-sum game. A game with only two players in which the gains of one player are exactly equal to the losses of another player is called a two-person zero-sum game. It is also called a rectangular game because their payoff matrix is in a rectangular form. Some basic terms Player: The competitors in the game are known as players. A player may be an individual or a group of individuals or an organization. Strategy: A strategy for a player is deﬁned as a set of rules or alternative courses of action available to him/her in advance, by which the player decides the course of action that he should adopt. A strategy may be of two types: (i) pure strategy, (ii) mixed strategy. Pure strategy: If the players select the same strategy each time, then it is referred to as a pure strategy. In this case, each player knows exactly what the other player is going to do, and the objective of the players is to maximize the gains or to minimize the losses. Payoff matrix: The payoff is the outcome of playing the game. When the players select their particular strategies, the payoffs (gains or losses) can be represented in the form of a matrix called a payoff matrix. Since the game is zero sum, the gain of one player is exactly equal to the loss of the other and vice versa. In other words, one player’s payoff table would contain the same amounts as in the payoff table of the other player with the sign changed. Thus, it is sufﬁcient to construct payoff only for one of the players.

13.5

Matrix Games or Rectangular Games

309

Let player A have m strategies, say, A1, A2, …, Am and player B have n strategies, say, B1, B2, …, Bn. Here, it is assumed that each player has his choices from among the pure strategies. Also, it is assumed that player A is always the gainer and player B is always the loser. That is, all payoffs are assumed in terms of player A. Let aij be the payoff which is the gain of player A from player B if player A chooses strategy Ai whereas player B chooses Bj. Then the payoff matrix of player A is Player B 2

A1

B1 B2    a11 a12   

Bn 3 a1n a2n 7 7

.. 7 7i:e:; aij mxn . 5

A2 6 6 a21 Player A . 6 . .. 6 4 ..

a22 .. .



am1

am2

   amn

Am



The payoff matrix of player B is ½aij mn : Example 1 Consider a two-person coin tossing game. Each player tosses an unbiased coin simultaneously. Player B pays Rs. 7.00 to player A if the outcomes of both tosses are heads and Rs. 4.00 if both the outcomes are tails; otherwise, player A pays Rs. 3.00 to player B. This two-person game is a zero-sum game, since the winnings of one player are the losses of the other. Each player has choices from among two pure strategies H and T. In that case, A’s payoff matrix will be

Player A

Player B H T H 7 3 T

3

4

Maximin-minimax principle or maximin-minimax criteria of optimality This principle is used for the selection of optimal strategies by two players. It states that ‘If a player lists his worst possible outcomes of all potential strategies then he will choose that strategy which corresponds to the best of these worst outcomes’.

310

13

Game Theory

Let player A’s payoff matrix be

B1 2a

Player B    Bj

B2

11 A1 6 a A2 6 21 .. 6 6 . . 6 .. 6 Player A Ai 6 6 ai1 .. 6 6 . . 4 .. Am am1



a12



a1j



a22 .. .



a2j .. .



 

ai2 .. . am2

aij .    ..    amj

   

Bn

a1n 3 a2n 7 7 .. 7 7 . 7 7: ain 7 7 7 .. 7 . 5 amn

If player A takes the strategy Ai, then surely he will get at least aij ði ¼ 1; 2; . . .; m; j ¼ 1; 2; . . .; nÞ for taking any strategy by the opponent player B. Thus, by the maximin-minimax criteria of optimality, player A will choose that strategy which corresponds to the best of these worst outcomes: Min a1j ; Min a2j ; . . .; Min amj : j

j

j

Thus, the maximin value for player A is given by Maxi Minj aij . Similarly, player B will choose that strategy which corresponds to the best (minimum) of the worst outcomes (maximum losses): Max ai1 ; Max ai2 ; . . .; Max ain : i i i

Thus, the minimax value for player B is given by Minj Maxi aij .

 Let Max Min aij ¼ apq

ð13:1Þ

 and Min Max aij ¼ ars :

ð13:2Þ

i

j

i

i

From (13.1), it follows that apq is the minimum element in the pth row, i:e: apq  aps ;

ð13:3Þ

where aps is another element of the pth row. From (13.2), it follows that ars is the maximum element in the sth column, i:e: aps  ars ; where aps is another element of the sth column.

ð13:4Þ

13.5

Matrix Games or Rectangular Games

311

From (13.3) and (13.4), we have apq  ars .





Max Min aij  Min Max aij ; i

j

j

i

i.e. maximin for A  minimax for B.

Therefore, Maxi Minj aij ; i.e. maximin for A is called the lower value of the game

and is denoted by Minj Maxi aij , and minimax for B is called the upper value of the game and is denoted by v. If v is the value of the game, then it will always satisfy the inequality maximin for A  v  minimax for B: If for a game, v ¼ v ¼ alk , then the game possesses a solution given by: (i) The optimal strategy for player A is the strategy Al. (ii) The optimal strategy for player B is the strategy Bk. (iii) The value of the game is v = alk. For the case of pure strategy, such a game is called a game with a saddle point. Saddle point: A saddle point is a position in the payoff matrix where the maximum of row minima coincides with the minimum of the column maxima. The cell entry (or payoff) at the saddle point position is called the value of the game. A game for which maximin for A = minimax for B is called a game with a saddle point. Thus, in a game with a saddle point, the players use pure strategies; i.e. they choose the same course of action throughout the game. Note (i) A game is said to be fair if v ¼ v ¼ 0: (ii) A game is said to be strictly determinable if maximin value = minimax value = the value of the game 6¼ 0, i.e. v ¼ v ¼ vð6¼ 0Þ. (iii) Not all payoff matrices possess a saddle point, and so the determination of the optimal solution is not an easy task. Example 2 Solve the game whose payoff matrix is given by Player B

Player A

B1 B2 2 A1 1 3 6 A2 4 0 4 A3 1 5

B3 3

3

7 3 5 1

312

13

Game Theory

Solution To ﬁnd the saddle point, the row minima and column maxima are selected and displayed on the right side of the corresponding row and in the bottom of the corresponding column respectively.

2

Player A

Player B B1 B2 B3

A1 1 6 A2 4 0 A3 1

Column maxima

3

3 7 3 5 1

4 5

1

Row minima

3

5

1 4 1

3

Hence, the maximin value of the game is v = Max{1, −4, −1} = 1 whose location is in the cell (1, 1), and the minimax value of the payoff matrix is v ¼ Minf1; 5; 3g ¼ 1 whose locations are in the cells (1, 1) and (3, 1). Hence, the payoff matrix has a saddle point at the position (1, 1). The solution of the game is given by: (i) The optimal strategy for player A is A1. (ii) The optimal strategy for player B is B1. (iii) The value of the game is 1. Example 3 For what value of k is the game with the following payoff matrix strictly determinable? Player B

Player A

B1 B2 2 A1 k 6 6 A2 4 1 k A3 2 4

B3 2

3

7 7 5 k

Solution Ignoring the value of k, we shall ﬁnd the maximin and minimax values of the payoff. For this purpose, we have

Player A

Player B B1 B2 B3 Row minima 2 3 k 6 2 A1 2 6 7 A2 4 1 k 7 5 7 A3 2 4 k 2

Column maxima 1

6

2

13.5

Matrix Games or Rectangular Games

313

) v = Max{2, −7, −2} = 2 and Min{−1, 6, 2} = −1. We know that the game is strictly determinable if v ¼ v ¼ v 6¼ 0. Hence, 1  k  2ðk 6¼ 0Þ. Mixed strategy If the payoff matrix does not have a saddle point, then there exists no equilibrium situation of the form (i*, j*); i.e. an optimal strategy of the players cannot be a single (pure) strategy.

Let us consider the matrix game whose payoff matrix is A ¼ aij mn . By a mixed strategy for player 1, we shall mean an ordered m-tuple ðx1 ; . . .; xm Þ or the Pm vector n ¼ ðx1 ; . . .; xm Þ0 , where xi  0 i = 1, 2, …, m and i¼1 xi ¼ 1: The components x1 ; x2 ; . . .; xm may be thought of as the relative frequencies (or probabilities) with which player 1 chooses the strategies (pure) 1; 2; . . .; m. It should be noted that em . .; 0; 1; 0; . . .; 0Þ0 ; i ¼ 1; 2; . . .; m represents the pure i ¼ ð0; .P m strategy of player 1, and n ¼ m i¼1 ei xi . Similarly, a mixed strategy for player 2 is denoted by g ¼ ðy1 ; y2 ; . . .; yn Þ0 , where Pn 0 n yj  0; j ¼ 1; 2; . . .; n and j¼1 yj ¼ 1. Note that ej ¼ ð0; 0; . . .; 0; 1; 0; . . .; 0Þ Pn m represents the pure strategy of player 2, and g ¼ j¼1 ej yj .   P Let Sm ¼ n : xi  0; m i¼1 xi ¼ 1 , then Sm 2 Em . This Sm is the space of mixed strategies for player 1. Note that a pure strategy is a particular case of a mixed strategy. n o P Similarly, let Sn ¼ g : yj  0; nj¼1 yj ¼ 1 , then Sn 2 En . Thus, Sn is the space of mixed strategies for player 2. Mixed extension of a matrix game without a saddle point Let players 1 and 2 choose their mixed strategies as n ¼ ðx1 ; x2 ; . . .; xm Þ0 , g ¼ ðy1 ; y 2 ; . . .; yn Þ0 independently of one another in a matrix game with payoff matrix A ¼ aij mn . Deﬁnition A pair ðn; gÞ of mixed strategies for the players in a matrix game is called a situation in mixed strategies. In a situation ðn; gÞ of mixed strategies, each usual situation (i, j) in pure strategies becomes a random event occurring with probabilities xi, yj. Since in the situation (i, j) player 1 receives a payoff aij, the expected value of his payoff under ðn; gÞ is equal to Eðn; gÞ ¼

m X n X

aij xi yj :

i¼1 j¼1

We thus arrive at a new game which can be described as follows.

314

13

Game Theory

Deﬁnition A mixed extension of a matrix game is an antagonistic game, denoted by hSm ; Sn ; E i, in which the set of strategies for the players is the set of their mixed strategies in the original game, and the payoff function for player 1 is deﬁned by E ðn; gÞ ¼

m X n X

aij xi yj :

i¼1 j¼1

Recalling the deﬁnition of saddle point of an antagonistic game, we observe that the situation ðn ; g Þ in a mixed extension of a matrix game will be a saddle point (i.e. an equilibrium situation), provided that for any n 2 Sm and g 2 Sn the following double inequality is satisﬁed: E ðn; g Þ  E ðn ; g Þ  E ðn ; gÞ

8n 2 Sm ; g 2 Sn :

Theorem 13.3 Let E (n,η) be such that both Min Max Eðn; gÞ and Max Min g2Sn n2Sm

Eðn; gÞ exist; then

n2Sm g2Sn

Min Max Eðn; gÞ  Max Min Eðn; gÞ: g2Sn n2Sm

n2Sm g2Sn

Proof Let n and g be two arbitrarily chosen mixed strategies for players A and B respectively. Then, for every n 2 Sm , we have Max Eðn; g Þ  Eðn ; g Þ; n2Sm

ð13:5Þ

and for every g 2 Sn , we have Min Eðn ; gÞ  Eðn ; g Þ: g2Sn

ð13:6Þ

Hence, from (13.5) and (13.6), the inequality Min Eðn ; gÞ  Max Eðn; g Þ g2Sn

n2Sm

or Max Eðn; g Þ  Min Eðn ; gÞ n2Sm

g2Sn

holds for all n and η. Since g is an arbitrarily chosen mixed strategy, the preceding inequality holds for all values of η. Hence, if g is such a strategy for which Maxn Eðn; gÞ has the

13.5

Matrix Games or Rectangular Games

315

maximum value, the inequality remains true. Therefore, Min Max Eðn; gÞ  g

Min Eðn ; gÞ.

n

g

Again, since n is any strategy, the preceding inequality holds even if we select n which gives the maximum value of Min Eðn; gÞ. g

Therefore, Min Max Eðn; gÞ  Max Min Eðn; gÞ. This proves the theorem. g

n

g

n

Saddle point of a function Let E(n,η) be a function of two variables (vectors) n and η in Sm and Sn respectively. The point ðn ; g Þ; n 2 Sm ; g 2 Sn is said to be the saddle point of the function E(n,η) if Eðn; g Þ  Eðn ; g Þ  Eðn ; gÞ: Now we shall discuss a theorem regarding the existence of the saddle point of a function. Theorem 13.4 Let E(n,η) be a function of two variables n 2 Sm and g 2 Sn such that Max Min Eðn; gÞ and Min Max Eðn; gÞ exist. Then the necessary and sufﬁn

g

g

n

cient condition for the existence of a saddle point ðn ; g Þ of E(n,η) is that Eðn ; g Þ ¼ Max Min Eðn; gÞ ¼ Min Max Eðn; gÞ: g

g

n

n

Proof The condition is necessary; i.e. the point ðn ; g Þ is the saddle point of E(n,η). Hence, from the deﬁnition of saddle point, we have Eðn; g Þ  Eðn ; g Þ  Eðn ; gÞ

ð13:7Þ

for all n 2 Sm and g 2 Sn . From (13.7), clearly, Max Eðn; g Þ  Eðn ; g Þ holds for all n 2 Sm . n

Hence; Min Max Eðn; gÞ  Eðn ; g Þ: g

n

ð13:8Þ

Similarly, from (13.7), we have Eðn ; g Þ  Min Eðn ; gÞ; g

i:e: Eðn ; g Þ  Max Min Eðn; gÞ: n

g

ð13:9Þ

316

13

Game Theory

From (13.8) and (13.9), we have Min Max Eðn; gÞ  Max Min Eðn; gÞ:

ð13:10Þ

Min Max Eðn; gÞ  Max Min Eðn; gÞ:

ð13:11Þ

g

n

n

g

But, we know that g

n

n

g

Hence, from (13.10) and (13.11), we have Eðn ; g Þ ¼ Max Min Eðn; gÞ ¼ Min Max Eðn; gÞ: n

g

g

n

The condition is sufﬁcient. Let the point ðn ; g Þ be such that Max Min Eðn; gÞ ¼ Min Max Eðn; gÞ: g

n

g

n

ð13:12Þ

Also, let Max Min Eðn; gÞ ¼ Min Eðn ; gÞ n

g

g

and Min Max Eðn; gÞ ¼ Max Eðn; g Þ. g

n

n

Hence, from (13.12), we have Min Eðn ; gÞ ¼ Max Eðn; g Þ: g

n

ð13:13Þ

Now, from the deﬁnition of maximum and minimum, we have Min Eðn ; gÞ  Eðn ; g Þ

ð13:14Þ

and Eðn ; g Þ  Max Eðn; g Þ:

ð13:15Þ

g

n

From (13.13) and (13.14), we have Max Eðn; g Þ  Eðn ; g Þ; n

which implies E ðn; g Þ  Eðn ; g Þ for all n 2 Sm :

ð13:16Þ

From (13.13) and (13.15), we have Eðn; g Þ  Min Eðn ; gÞ; g

which implies Eðn ; g Þ  Eðn ; gÞ for all g 2 Sn : Combining (13.16) and (13.17), we have

ð13:17Þ

13.5

Matrix Games or Rectangular Games

317

Eðn; g Þ  Eðn ; g Þ  Eðn ; gÞ; which implies ðn ; g Þ is a saddle point of E(n, η). Fundamental theorem of a matrix game 







Theorem 13.5 The quantities Max Min E ðn; gÞ and Min Max E ðn; gÞ exist g2Sn

n2Sm

g2Sn

n2Sm

and are equal. Proof This theorem can easily be proved from Theorem 13.4. Value of a matrix game









The common value of Max Min E ðn; gÞ

and Min Max Eðn; gÞ is called the g n

value of the matrix game with payoff matrix A ¼ aij and denoted by v(A) or simply v. n

g

Deﬁnition Equilibrium strategies of players in a matrix game are called their optimal strategies. Deﬁnition The solution of a matrix game is the process of determining the value of the game and the pairs of optimal strategies. Note: Thus, if ðn ; g Þ is an equilibrium situation in mixed strategies of the game hSm ; Sn ; E i, then n ; g are the optimal strategies for players 1 and 2 respectively in the matrix game with payoff matrix A ¼ aij mn . Hence, n ; g are optimal strategies for players 1 and 2 respectively iff E ðn; g Þ  Eðn ; g Þ  Eðn ; gÞ 8n 2 Sm ; g 2 Sn : Note:   Min E ðn; gÞ ¼ Eðn; g Þ ) Max Min E ðn; gÞ ¼ Max Eðn; g Þ ¼ E ðn ; g Þ n

g

n

and Max Eðn; gÞ ¼ Eðn ; gÞ n

n

  ) Min Max Eðn; gÞ ¼ Min E ðn; gÞ ¼ Eðn ; g Þ: g

n

g

Notation The expected payoff to player 1 under ðn; gÞ is E ðn; gÞ ¼

P i; j

aij xi yj .

  The expected payoff to player 1 under n; enj or simply ðn; jÞ is

318

13

Game Theory

  X E n; enj ¼ E ðn; jÞ ¼ aij xi : i

  The expected payoff to player 1 under em i ; g or simply ði; gÞ is   X E em aij yj : j ; g ¼ E ði; gÞ ¼ j

Thus, E ðn; gÞ ¼

P

P

E ði; gÞ ¼

i

Eðn; jÞ.

j

  n Also, E em ; e ¼ Eði; jÞ ¼ aij . i j Some properties of the value of a matrix game

The matrix game is characterized by the payoff matrix A ¼ aij mn and v is its value.     Theorem 13.6 v ¼ Max Min E ðn; jÞ ¼ Min Max E ði; gÞ , and the outer g

j

n

i

extrema are attained at optimal strategies of the players.   Proof We have, by deﬁnition, v ¼ Max Min E ðn; gÞ . g

n

For a ﬁxed n, let Eðn; j0 Þ ¼ Min E ðn; jÞ. Then j

E ðn; j0 Þ  E ðn; jÞ; j ¼ 1; 2; . . .; n X X Eðn; j0 Þyj  E ðn; jÞyj ; or j

j

i:e: E ðn; j0 Þ  E ðn; gÞ for a fixed n and all g 2 Sn Hence, Eðn; j0 Þ  Min E ðn; gÞ: g

ð13:18Þ

Again since Min E ðn; gÞ  E ðn; gÞ 8g 2 Sn , then setting g ¼ enj0 , we have g

  Min E ðn; gÞ  E n; enj0 ¼ Eðn; j0 Þ: g

ð13:19Þ

By (13.18) and (13.19), we have Min E ðn; gÞ ¼ E ðn; j0 Þ ¼ Min Eðn; jÞ: g

j0

This is valid for any n 2 Sm ; hence, taking the maximum of both sides with respect to n, we have

13.5

Matrix Games or Rectangular Games



319



Max Min E ðn; gÞ



¼ Max Min E ðn; jÞ ;

g

n

 j

n

where the outer maxima on both sides are attained for the same value of n. It is known, however, that on the left side the outer extremum is attained at the optimal strategies of player 1. Therefore, the same strategies yield the maximum on the right-hand side. Thus, the ﬁrst part of the theorem is proved. The proof of the second part is similar. Theorem 13.7









Max Min aij  v  Min Max aij : i

j

j

i

  Proof By the preceding theorem, we have v ¼ Max Min E ðn; jÞ 8n 2 Sm . 

j

n



But Max Min E ðn; jÞ  Min E ðn; jÞ 8n 2 Sm j

n

j

) v  Min Eðn; jÞ j

Setting

n ¼ em i ,

we

8n 2 Sm : have

  v  Min E em i ; j ¼ Min E ði; jÞ ¼ Min aij , j

j

j

i.e.

v  Min aij . j

The left side v is independent of i, so taking the maximum with respect to i, we obtain 



v  Max Min aij : i

j

The proof of the second part is similar. Theorem 13.8 (i) If player 1 possesses a pure optimal strategy i*, then v ¼ Max Min aij ¼ Min ai j : i

j

j

(ii) If player 2 possesses a pure optimal strategy j*, then v ¼ Min Max aij ¼ Max aij j

Proof We know that

i

i

320

13

Game Theory

v ¼ Max Min E ðn; jÞ j n  m  ¼ Min E ei ; j as n ¼ em i is optimal: j

The proof of the second part is similar.

Solution of matrix game Let us

consider a 2  2 matrix game whose payoff matrix A is given by a12 a A ¼ 11 . a21 a22 If A has a saddle point, the solution is obvious. Let A have no saddle point. Let player 1 have the strategy n ¼ ðx1 ; x2 Þ0  ðx; 1  xÞð0  x  1Þ and player 2 have the strategy g ¼ ðy; 1  yÞ0 ð0  y  1Þ. Then Eðn; gÞ ¼

2 X 2 X

aij xi yj

i¼1 j¼1

¼ a11 xy þ a12 xð1  yÞ þ a21 ð1  xÞy þ a22 ð1  xÞð1  yÞ ¼ f ðx; yÞ; say: If n ¼ ðx ; 1  x Þ0 ; g ¼ ðy ; 1  y Þ are optimal strategies, then from E ðn; g Þ  E ðn ; g Þ  E ðn ; gÞ 8n 2 S1 ; g 2 S2 we have f ðx; y Þ  f ðx ; y Þ  f ðx ; yÞ 8x 2 ð0; 1Þ; y 2 ð0; 1Þ. From the ﬁrst part of the inequality, we set that f ðx; y Þ regarded as a function of x has a maximum at x , i.e.  @f  ¼ 0: @xðx ; y Þ This gives a11 y þ a12 ð1  y Þ  a21 y  a22 ð1  y Þ ¼ 0 or ða11  a12  a21 þ a22 Þy ¼ a22  a12 a22  a12 provided that a11 þ a22  ða12 þ a21 Þ 6¼ 0: or y ¼ a11 þ a22  ða12 þ a21 Þ Similarly, from the second part of the inequality, it is seen that f ðx ; yÞ regarded as a function of y has a minimum at y*, i.e.

13.5

Matrix Games or Rectangular Games

321

 @f  ¼ 0: @yðx ; y Þ This gives a11 x  a12 x þ a21 ð1  x Þ  a22 ð1  x Þ ¼ 0 a22  a21 provided that a11 þ a22  ða12 þ a21 Þ 6¼ 0: or x ¼ a11 þ a22  ða12 þ a21 Þ Now v ¼ f ðx ; y Þ ¼ a11 x y þ a12 x ð1  y Þ þ a21 ð1  x Þy þ a22 ð1  x Þð1  y Þ a11 a22  a12 a21 : ¼ a11 þ a22  ða12 þ a21 Þ It can be proved that a11 þ a22  ða12 þ a21 Þ ¼ 0 implies that A has a saddle point. Some properties of optimal strategies in a matrix game

Let A ¼ aij mn be a payoff matrix of a game whose value is v. Theorem 13.9 Let u be an arbitrary real number. Then (i) u  E ðn; jÞ; (ii) u  E ði; gÞ;

j ¼ 1; 2; . . .; n for some ﬁxed n implies u  v. i ¼ 1; 2; . . .; m for some ﬁxed η implies u  v.

Proof Let g be an optimal strategy for player 2. Now we have u  Eðn; jÞ j ¼ 1; . . .; n for some ﬁxed n. )

u

n X

y j 

n X

j¼1

E ðn; jÞy j ;

j¼1

i:e: u  E ðn; g Þ  Eðn ; g Þ; where n is an optimal strategy for player 1; i:e: u  v The proof of the second part is similar. Theorem 13.10 Let u be a real number such that Eði; g0 Þ  u  E ðn0 ; jÞ for i ¼ 1; 2; . . .; m, j ¼ 1; 2; . . .; n and some particular n0 ; g0 . Then u is the value of the game and n0 ; g0 are the optimal strategies for players 1 and 2 respectively. Proof Given that E ði; g0 Þ  u; ði ¼ 1; 2; . . .; mÞ and for some particular g0 , X X ) E ði; g0 Þx i  u x i i

i

or E ðn0 ; g0 Þ  u: Again u  E ðn0 ; jÞ;

j ¼ 1; 2; . . .; n gives similarly

ð13:20Þ

322

13

Game Theory

u  Eðn0 ; g0 Þ:

ð13:21Þ

Hence, (13.20) and (13.21) imply that u ¼ E ðn0 ; g0 Þ. Again we have Eði; g0 Þ  u ¼ E ðn0 ; g0 Þ; i ¼ 1; 2; . . .; m X X Eði; g0 Þxi  E ðn0 ; g0 Þ xi so that i

i

ð13:22Þ

or Eðn; g0 Þ  E ðn0 ; g0 Þ 8n 2 Sm : Again u ¼ E ðn0 ; g0 Þ  E ðn0 ; jÞ; j ¼ 1; 2; . . .; n X X yj  E ðn0 ; jÞyj so that E ðn0 ; g0 Þ j

j

or E ðn0 ; g0 Þ  Eðn0 ; gÞ

ð13:23Þ

8g 2 Sn :

Hence, from (13.22) and (13.23), we have E ðn; g0 Þ  Eðn0 ; g0 Þ  Eðn0 ; gÞ 8n 2 Sm ; g 2 Sn : Hence, by the deﬁnition of optimal strategies, n0 ; g0 are optimal strategies and E ðn0 ; g0 Þ ¼ u is the value of the game. Theorem 13.11 If Max E ði; g0 Þ  Min E ðn0 ; jÞ is valid for some ﬁxed n0 and g0 , i

j

then n0 ; g0 are optimal strategies and the inequality becomes equality. Proof We have E ði; g0 Þ  Max E ði; g0 Þ i

8i ¼ 1; 2; . . .; m [by the deﬁnition of

maximum]  Min Eðn0 ; jÞ [by the given condition] j

 E ðn0 ; jÞ 8j ¼ 1; 2; . . .; n [by the deﬁnition of minimum] 8i ¼ 1; 2; . . .; m and j ¼ 1; 2; . . .; n. Let u1 ¼ Maxi E ði; g0 Þ. Then E ði; g0 Þ  u1  Eðn0 ; jÞ 8i ¼ 1; 2; . . .; m and j ¼ 1; 2; . . .; n. Hence, by Theorem 13.10, u1 is the value of the game and n0 ; g0 are the optimal strategies. Again let u2 ¼ Min Eðn0 ; jÞ. j

Now by the deﬁnition of minima, we have Eðn0 ; jÞ  Min E ðn0 ; jÞ 8j ¼ 1; 2; . . .; n j

 Max E ði; g0 Þ ½by the given condition

i

 Eði; g0 Þ 8i ¼ 1; 2; . . .; m ½by the definition of maxima

13.5

Matrix Games or Rectangular Games

323

) Eðn0 ; jÞ  Min Eðn0 ; jÞ  Eði; g0 Þ: j

i:e: Eði; g0 Þ  Min E ðn0 ; jÞ  E ðn0 ; jÞ j

or Eði; g0 Þ  u2  E ðn0 ; jÞ Hence, by Theorem 13.10, u2 is the value of the game and n0 ; g0 are the optimal strategies. But the value of the game is unique. Hence, u1 = u2; i.e. the inequality becomes an equality. Theorem 13.12 If Eði; g0 Þ  Eðn0 ; jÞ n0 ; g0 are the optimal strategies.

8i ¼ 1; 2; . . .; m;

j ¼ 1; 2; . . .; n, then

Proof This theorem can be proved from Theorem 13.3. Theorem 13.13 (i) If for a strategy n0 of player 1, v  E ðn0 ; jÞ; optimal strategy for player 1. (ii) If for a strategy g0 of player 2, v  E ði; g0 Þ; optimal strategy for player 2.

8j ¼ 1; 2; . . .; n, then n0 is an 8i ¼ 1; 2; . . .; m, then g0 is an

Proof Let g be an optimal strategy for player 2, then E ðn; g Þ  v; 8n 2 Sm . Setting n ¼ em i , we have E ði; g Þ  v 8i ¼ 1; 2; . . .; m. But v  Eðn0 ; jÞ (given). Hence, Eði; g Þ  v  E ðn0 ; jÞ 8i ¼ 1; 2; . . .; m; 8j ¼ 1; 2; . . .; n. By Theorem 13.2, it follows that n0 is an optimal strategy for player 1. The proof of the second part is similar. Theorem 13.14 (i) If v  Min E ðn0 ; jÞ for a ﬁxed n0 , n0 is an optimal strategy for player 1. j

(ii) If v  Max E ði; g0 Þ for a ﬁxed g0 , g0 is an optimal strategy for player 2. i

Proof (i) v  Min Eðn0 ; jÞ  Eðn0 ; jÞ 8j ¼ 1; 2; . . .; n:: j

Then by Theorem 13.5, this theorem can be proved. Corollary (i) A necessary and sufﬁcient condition for the optimality of strategy n0 for player 1 is v ¼ Min E ðn0 ; jÞ. j

(ii) A similar condition for player 2 is v ¼ Max E ði; g0 Þ. i

324

13

13.6

Game Theory

Dominance

In a game, if one course of action (pure strategy) of one player is better than or as good as another, for all possible courses of action of the opponent, then the ﬁrst course of action is said to dominate the second. In this situation, the dominated (or inferior) course of action can simply be discarded from the payoff matrix, and a reduced matrix is obtained. In this case, the optimal strategies for the reduced matrix are also optimal for the original matrix with zero probability for discarded strategies. When there is no saddle point in a payoff matrix, then the size of the game can be reduced by this process, which is known as dominance. For example, let us consider a game with the following payoff matrix:

Player B B1

B2

B3

A1 4 Player A A2 7

9 4

4 8

A3 8

5

9

From this payoff matrix, it is clear that the second row is inferior to the third row for the maximizing player A, as every element of the third row is either greater than or equal to the corresponding element of the second row. Then by deleting the second row, we get the reduced payoff matrix as follows:

B1 A1 4 A3 8

B2

B3

7 5

4 9

Again, from player B’s point of view, the third strategy ðB3 Þ is dominated by B1 , as every element of the third column is either greater than or equal to the corresponding element of the ﬁrst column. Then by deleting the third column, we get the reduced payoff matrix as follows:

B1 A1 4 A3 8

B2 7 5

    Now, if x 1 ; x 3 and y 1 ; y 2 are the optimal strategies for this reduced matrix, then     x1 ; 0; x 3 and y 1 ; y 2 ; 0 are the optimal strategies for the original payoff matrix.

13.6

Dominance

325

Theorem 13.15 If the ith row of the payoff matrix of an m  n rectangular game is dominated by its rth row, then the deletion of the ith row from the matrix does not change the optimal strategies of the player whose objective is to maximize the gain. Proof Let A and B be the

players of a rectangular game whose payoff matrix is given by the matrix aij mn . According to the given condition of the theorem, aij  arj for j ¼ 1; 2; . . .; n

i 6¼ r;

and

ð13:24Þ

and for at least one j; aij 6¼ arj .   Let g ¼ y 1 ; y 2 ; . . .; y n be the optimal mixed strategy for player B. Then from (13.24), we have n n X X aij y j \ arj y j ; ð13:25Þ j¼1 j¼1 i:e: E ðAi ; g Þ\EðAr ; g Þ; where Ai and Ar are the ith and rth pure strategies of A. Let v be the value of the game. Hence, from (13.25), we have v  E ðAr ; g Þ [ E ðAi ; g Þ:

ð13:26Þ

  Let n ¼ x 1 ; x 2 ; . . .; x m be an optimal strategy for player A. If possible, let x i [ 0. Then, from (13.26), we have x i v [ x i EðAi ; g0 Þ: Since ðn0 ; g0 Þ is the optimal solution of the game problem, we have v ¼ E ðn ; g Þ ¼

m X

x k E ðAk ; g Þ

k¼1

¼ x i EðAi ; g Þ þ

m X k¼1 k6¼i

x k EðAk ; g Þ\x i v þ v

m X k¼1

x k ¼ v

m X

x k ¼ v

k¼1

k 6¼ i

i:e: v \ v: This is a contradiction to our assumption that x i [ 0 and hence x i ¼ 0. Hence, the deletion of the ith row does not alter the optimal solution.

326

13

Game Theory

Theorem 13.16 If the jth column of the payoff matrix of an m  n rectangular game dominates its kth column, then the deletion of the jth column from the matrix does not change the optimal strategy of the player whose objective is to minimize the loss. [The proof of this theorem is similar to that of the preceding theorem.] Theorem 13.17 If the ith row of the payoff matrix of an m  n rectangular game is strictly dominated by a convex combination of the other rows of the matrix, then the deletion of the ith row from the matrix does not affect the optimal strategy of the player whose objective is to maximize the gain. Proof Let A and B be the

players of a rectangular game whose payoff matrix is given by the matrix aij mn . Let ðx1 ; x2 ; . . .; xm Þ be the mixed strategy of player A where

P

k xk

¼ 1.

According to the given condition, xi ¼ 0 and aij 

m X

akj xk ;

j ¼ 1; 2; . . .; n

k¼1 k6¼i

or aij 

m P

ð13:27Þ

akj xk

k¼1

in which strict inequality holds for at least one j.   Let g ¼ y 1 ; y 2 ; . . .; y n be the optimal strategy for player B, whose objective is to minimize the loss. Hence, from (13.27), we have n X

aij y j \

n X m X

j¼1

or E ðAi ; g Þ\

akj xk y j

j¼1 k¼1 n X m X

ð13:28Þ akj xk y j  v;

j¼1 k¼1

where Ai is the of A and v is the value of the game.  ith pure strategy  Let n ¼ x1 ; x2 ; . . .; xm be the optimal strategy of player A, whose objective is to maximize the gain. If possible, let x i 6¼ 0. From (13.28), we have x i E ðAi ; g Þ\x i v;

since x i 6¼ 0:

13.6

Dominance

327

Hence, we have v ¼ E ð n ; g Þ ¼

n P m P j¼1 k¼1

¼ x i E ðAi ; g Þ þ ¼ x i E ðAi ; g Þ þ

m P k¼1 k6¼i m P k¼1 k6¼i

x k

akj x k y j n P j¼1

akj y j

x k EðAk ; g Þ\x i v þ v

m P k¼1 k6¼i

x k ¼ v

m P k¼1

x k ¼ v:

This is a contradiction to our assumption that x i 6¼ 0. Thus, x i ¼ 0. Hence, the deletion of the ith row does not affect the optimal solution. Theorem 13.18 If the jth column of the matrix strictly dominates a convex combination of other columns, then the deletion of the jth column of the matrix does not affect the optimal strategy of the player whose objective is to minimize the loss. [The proof of this theorem is similar to that of Theorem 13.3.] Theorem 13.19 R1 and R2 are the subsets of the rows and C1 and C2 are the subsets of the columns of the payoff matrix aij mn of a matrix game. (i) If a convex combination of the rows in R1 strictly dominates a convex combination of the rows in R2 , then there exists a row in R2 which, if discarded, does not change the optimal strategy of the player whose objective is to maximize the gain. (ii) If a convex combination of the columns in C1 dominates a convex combination of the columns in C2 , then there exists a column in C1 which, if discarded, does not change the optimal strategy of the player whose objective is to minimize the loss. The Rules of dominance Sometimes, in a rectangular game, it is seen that one or more pure strategies of a player are inferior to at least one of the remaining strategies. In such a case, this inferior strategy is never used. In other words, we can say that this inferior pure strategy is dominated by a superior pure strategy. In such cases of dominance, we can reduce the size of the payoff matrix by removing the pure strategies which are dominated by other strategies. Hence, the rules of dominance are used to reduce the size of the payoff matrix. These rules are especially used for solving a two-person zero-sum game without a saddle point. Rule 1: If each element in a row (say, the ith row, i.e. Ri) of a payoff matrix is either less than or equal to the corresponding element in another row (say, the jth row, i.e. Rj), then the ith strategy is dominated by the jth strategy, and that row (i.e. the ith row) can be deleted from the payoff matrix.

328

13

Game Theory

In other words, player A will never use that strategy, because if player A chooses that strategy, then he will gain less payoff. Rule 2: If each element in a column (say, the ith column, i.e. Ci) of a payoff matrix is either greater than or equal to the corresponding element in another column (say, the jth column, i.e. Cj), then the ith strategy is dominated by the jth strategy, and that column (i.e. the ith column) can be deleted from the payoff matrix. Rule 3: If the ith row is dominated by a convex combination of some/all other rows, then the ith row is deleted from the payoff matrix. If the jth column dominates a convex combination of some/all other columns, then the jth column is deleted from the payoff matrix. Rule 4: If R1 , R2 are the subsets of the rows of the payoff matrix, and if a convex combination of rows in R1 dominates a convex combination of the rows in R2 , then there exists a row in R2 which is deleted. If C1 ; C2 are the subsets of the columns of the payoff matrix, and if a convex combination of the columns in C1 dominates a convex combination of the columns in C2 , then there exists a column in C1 which is deleted. Rules 3 and 4 are called the modiﬁed dominance property. We note that the element or probability value corresponding to the deleted strategy by the rules of dominance is taken as zero. Example 4 Solve the game whose payoff matrix is given by

A1

Player A

2

B1

Player B B2 B3

B4

2

2

2 2

3 1

0

3

1

2

A2 6 6 3 1 6 A3 4 1 3 A4

2 2

3 7 7 7 5

Solution First of all, we shall compute the maximin and minimax values in the case of a pure strategy. 2

B1

B2

B3

2 1

2 2

2

3 2

2 0

3

3

3

2

3

A1 1 A2 6 6 3 6 A3 4 1 A4 Column maxima

B4

3

2 3 7 7 7 1 5

Row minima 2 1 1 3

In the case of a pure strategy, maximin value = 1 and minimax value = 2. As maximin value 6¼ minimax value, this game has no saddle point for a pure strategy.

13.6

Dominance

329

Hence, this game can be solved for the mixed strategy. For this purpose, we shall try to reduce the size of the payoff matrix by using the dominance rule. Since every element of the fourth row is less than the corresponding elements of the third row, then from player A’s point of view, the fourth strategy, i.e. A4, is dominated by the third strategy (A3); thus, we can delete the fourth row. In this situation, the optimal strategy will not be affected. Now deleting the fourth row, we get the reduced payoff matrix as follows: 2

B1

A1 1 6 A2 4 3 1

A3

B2

B3

B4 3 2 2 2 7 1 2 35

3

2

1

Again, from player B’s point of view, the fourth strategy (B4) is dominated by B1, as every element of the fourth column is either greater than or equal to the corresponding element of the ﬁrst column. Then by deleting the fourth column, we get the reduced payoff matrix as follows:

A1

2

B1 1

6 A2 4 3 A3 1

B2

B3

2 2 1 3

3

7 2 5 2

From the reduced payoff matrix, it is seen that none of the pure strategies of players A and B is inferior to any of their other strategies. However, the convex combination due to strategies A2 and A3 (i.e. the average of the second and third rows, 1, 2, 2) is superior to the payoff due to strategy A1. Thus, strategy A1 may be deleted, giving us the reduced payoff matrix as follows: B1 B2 B3

 A2 3 1 2 A3 1 3 2 In this reduced matrix, the convex combination due to strategies B1 and B2 is superior to the payoff due to strategy B3. Thus, strategy B3 may be deleted, and we will get the reduced 2  2 payoff matrix as follows: B1 B2

 A2 3 1 A3 1 3 Now, we have to solve this reduced game whose payoff matrix is of 2  2 order.

330

13

Game Theory

Clearly, this game has no saddle point and cannot be reduced further. Therefore, the optimal strategies will be mixed strategies. Let player A choose his strategies A2 and A3 with probabilities x2 and x3; those of B are y1 and y2. ) x2 þ x3 ¼ 1, y1 + y2 = 1 and x2, x3, y1, y2 >0 and v is the value of the game. To determine the optimal values of x2 and x3, we have 3x2  x3 ¼ x2 þ 3x3 ¼ v; which implies x2 ¼ 23 ) x1 ¼ 1  x2 ¼ 13. To determine the optimal values of y1 and y2, we have 3y1 þ y2 ¼ y1 þ 3y2 which implies y1 ¼ 13 ) y2 ¼ 23. The value of the game v ¼ 3x2  x3 ¼ 3  23  13 ¼ 53. Hence, the solution of the game is as follows:     Optimal strategies are n ¼ 0; 23 ; 13 ; 0 , g ¼ 13 ; 23 ; 0; 0 , and the value of the game is 53. An important result If a constant is added to all the elements of a matrix game, the optimal strategies remain unaltered, and the value of the game is increased by this constant. This result can easily be proved.

13.7

Graphical Solution of 2  n or m  2 Games

Any rectangular game of order 2  n or m  2 can be reduced to a 2  2 game by using a graphical method, and the reduced game of order 2  2 can be solved by an algebraic method. Let us consider a 2  n game without a saddle point. Its payoff matrix is as follows: Player B

Player A

A1

B1 B2    Bn  a11 a12    a1n

A2 a21

a22

   a2n

13.7

Graphical Solution of 2  n or m  2 Games

331

Let n = (x1, x2) and η = (y1, y2, …, yn) be the mixed strategies for players A and B respectively. When player B uses his pure strategy Bj, then the expected gain of player A is given by Eðn; Bj Þ ¼ a1j x1 þ a2j x2 ¼ a1j x1 þ a2j ð1  x1 Þ; j ¼ 1; 2; . . .; n

ð13:29Þ

as x1 + x2 = 1. Clearly, both x1 and x2 must lie in the open interval (0, 1) [because if either x1 = 1 or x2 = 1, the game reduces to a game of pure strategy, which is against our assumption]. Hence Ej (n) is a linear function of either x1 or x2. Considering Ej as a linear function of x1 (say), we have   Ej n2 ; Bj ¼ a2j for x1 ¼ 0 ¼ a1j for x1 ¼ 1 Hence, Ej ðn; Bj Þ represents a line segment joining the points (0, a2j) and (1, a1j). Now player A expects a least possible gain v. Therefore, Ej ðn; Bj Þ  v for all j. Now considering the strict equations for inequalities, and with the help of a graphical method, we shall ﬁnd out two particular moves or choices of B which will maximize the minimum gain of A. Let us draw two parallel lines x1 = 0 and x1 = 1 at unit distance apart. Now draw n line segments joining the points (0, a2j) and (1, a1j), j = 1, 2, …, n. The lower envelope (or lower boundary) of these line segments (indicated by a thick line segment) will give the minimum expected gain of A as a function of x1. Now the highest point of the lower envelope will give the maximum of minimum gain of A. The line segments passing through the point corresponding to B’s two pure moves, say, Bk and Bl, are the critical moves for B which will maximize the minimum expected gain of A. Now the 2  2 payoff matrix corresponding to A’s moves A1, A2 and B’s moves Bk , Bl will produce the required result. Thus, solving the 2  2 game algebraically, we can ﬁnd the value of the game. If there are more than two line segments passing through the highest (maximin) point, there are ties for the optimum mixed strategies for player B. Thus, any two such lines with opposite sign slopes will determine an alternative optimum for player B. Again, if there is more than one maximin point, alternative optima exist corresponding to these points. Example 5 Solve the following 2  4 game graphically: Player B

Player A

B1

A1 1

B2 B3 3 3

B4  7

A2 2

5

6

4

332

13

Game Theory

Solution Clearly, the given problem does not possess any saddle point in the case of pure strategy. Hence, this problem can be solved with a mixed strategy. Let player A play with mixed strategy n = (x1, x2), where x1+ x2 = 1 and both x1 and x2 lie in the open interval (0, 1). Then player A’s expected gains against B’s pure moves are given by B0 s Pure strategy B1 B2 B3 B4

A0 s expected gain Eðx1 ; B1 Þ ¼ x1 þ 2x2 ¼ x1 þ 2ð1  x1 Þ Eðx1 ; B2 Þ ¼ 3x1 þ 5ð1  x1 Þ Eðx1 ; B3 Þ ¼ 3x1 þ 4ð1  x1 Þ Eðx1 ; B4 Þ ¼ 7x1 þ 6ð1  x1 Þ

Draw two vertical lines x1 = 0 and x1 = 1 at unit distance apart. Mark the lines x1= 0 and x1 = 1 by using the same scale as given in Fig. 13.1. Now draw the line segments for the expected gain equations between the two vertical lines x1 = 0 and x1 = 1. These line segments represent A’s expected gain due to B’s pure move. We denote these line segments as B1, B2, B3, B4. Since player A wishes to maximize his minimum expected gain, the highest point of intersection H of two line segments B3 and B4 represents the maximin expected value of the game for player A. Hence, the solution to the original game reduces to that of a simpler game with the following 2  2 payoff matrix:

Player A

Fig. 13.1 Graphical representation of 2  4 game

Player B B3 B4

 A1 3 7 A2

4

6

7

7 B4

B2

4 B3

2 0

B1 H

-6

0 -2

-2 -4

2

Lower Envelop

-4 -6

x 1=1

x 1=0

4

13.7

Graphical Solution of 2  n or m  2 Games

333

Now, if n = (x1, x2) and η = (y3, y4) are the optimum strategies for players A and B, then we have a22  a21 6  4 10 1 ¼ ¼ ¼ ða11 þ a22 Þ  ða12 þ a21 Þ 3  6  ð7 þ 4Þ 20 2 1 ) x2 ¼ 1  x1 ¼ : 2 x1 ¼

67 13 7 Again, y3 ¼ 36ð7 þ 4Þ ¼ 20, y4 ¼ 1  y3 ¼ 20 and the value of the game is given by

a11 a22  a12 a21 18  28 1 ¼ : ¼ 20 2 ða11 þ a22 Þ  ða12 þ a21 Þ

Hence, the solution of the game is as follows:     7 (i) Optimal strategies n ¼ 12 ; 12 , g ¼ 0; 0; 13 20 ; 20 . (ii) The value of the game is 12. Example 6 By the graphical method, solve the game whose payoff matrix is given by Player B B1 B2 B3 B4

 A1 2 2 3 1 Player A A2 4 3 2 6 Solution In a similar way, draw the graph. Here three lines pass at the highest point of the lower envelope. Thus, accordingly we get three square matrices of order 2, and there are three optimal solutions. But actually we shall have to select a pair of lines which have the slope opposite in sign. Thus, we get two reduced games with a payoff matrix of order 2  2 (see Fig. 13.2)

Fig. 13.2 Graphical representation of 2  4 game

6 4 2 0

B1

6

B4 B3 H

B2

Lower Envelope

4 2 0

334

13

Player B

Player A

Player B

B2 B3 A1 2 3 3

A2

Game Theory

and Player A

2

B3 B4 A1 3 1 A2

2

6

Solving these, we get the value of the game as 52 and the optimal strategies as         n ¼ 12 ; 12 ; g ¼ 0; 12 ; 12 ; 0 for the ﬁrst case and n ¼ 12 ; 12 ; g ¼ 0; 0; 78 ; 18 for the second case. Any m  2 game problem can be solved by using B’s mixed strategy g ¼ ðy1 ; y2 Þ, where y1 þ y2 ¼ 1 and both lie in the open interval (0,1), and A’s particular critical moves can be determined graphically. In this case, the problem is to determine the minimum of the maximum expected loss of B. Thus, we shall have to select the lowest point (minimax) of the upper envelope, and A’s critical moves are those whose corresponding line segments pass through the minimax point. Now selecting a 2  2 payoff matrix, the value of the game can be determined easily. Example 7 Solve the game whose payoff matrix is as follows: Player B B1 B2 2 3 A1 2 3 6 7 A2 4 2 5 5

Player A

A3

1

0

Here H is the lowest point of the upper envelope. The point H is the point of intersection of the two line segments A1, A2 (see Fig. 13.3). Hence, the payoff matrix of the reduced game will be B1

B2

A1 2 A2 2

3 5



Now solving the reduced game by the algebraic method, we shall easily ﬁnd optimal strategies and the value of the game. For this problem, the optimal strategies will be

n ¼ and the value of the game is 13.

7 5 2 1 ; ;0 ;g ¼ ; 12 12 3 3

13.7

Graphical Solution of 2  n or m  2 Games

Fig. 13.3 Graphical representation of 3  2 game

335

5

Upper Envelope A2

3

0

A3

-3

A1

5 3 0

H -3

Symmetric game A game is said to be symmetric if the payoff matrix remains unchanged on interchanging the positions of two players. Hence, the payoff matrix of any player of a symmetric game is skew-symmetric; i.e. the payoff matrix of any player is a square matrix in which aij ¼ aji and aii ¼ 0, since the gain of one player is equal to the loss of the other. Theorem 13.20 In a symmetric game, the sets of all optimal strategies of players 1 and 2 are the same, and the value of the game is zero. Proof For a matrix game, the expected payoff for player 1 is given by Eðn; gÞ ¼

m X n X

aij xi yj ;

i¼1 j¼1

where n ¼ ðx1 ; x2 ; . . .; xm Þ and

g ¼ ðy1 ; y2 ; . . .; yn Þ are the mixed strategies of A and B respectively and aij mn is the payoff matrix. In a symmetric game, the payoff matrix is skew-symmetric, and Eðn; nÞ ¼ 0 and E ðg; gÞ ¼ 0. Let n and g be the optimal strategies for players 1 and 2 respectively. Then the value of the game is given by v ¼ E ðn ; g Þ: Again, we know that v ¼ Max Min Eðn; gÞ ¼ Minðn ; gÞ n

g

) E ðn ; gÞ  v: Setting g ¼ n above, we have E ðn ; n Þ  v.

g

336

13

But Eðn ; n Þ ¼ 0; therefore v  0:

Game Theory

ð13:30Þ

Again v ¼ Min Max E ðn; gÞ ¼ Max Eðn; g Þ g

n

n

) E ðn; g Þ  v: Setting n ¼ g , we have E ðg ; g Þ  v. But Eðg ; g Þ ¼ 0; therefore v  0:

ð13:31Þ

Hence, from (13.30) and (13.31), we have v¼0 and therefore n ¼ g . This completes the proof.

13.8

Matrix Games and Linear Programming Problems (LPPs)

Theorem 13.21 A matrix game is equivalent to a certain LPP.

Proof Let A ¼ aij mn be the payoff matrix and v be the value of the game. It is assumed that v > 0. If not, then a suitable constant may be added to each element of A so as to make v positive. This change does not affect the optimal strategies for two players; only the value of the game is increased by this constant. Let n ; g be the optimal strategies and v (> 0) be the value of the game. Then v  E ð n ; gÞ

8g 2 Sn ;

i:e: v  E ðn ; jÞ 8j ¼ 1; 2; . . .; n We can say that v is the largest number u (v  u) such that for some n 2 Sm u  E ðn; jÞ;

j ¼ 1; 2; . . .; n

ð13:32Þ

n is an optimal strategy for player 1 iff n satisﬁes (13.32) with u replaced by v. Now (13.32) can also be written as m X i¼1

aij xi  u j ¼ 1; 2; . . .; n:

ð13:33Þ

13.8

Matrix Games and Linear Programming Problems (LPPs)

Also, we have

m X

xi ¼ 1;

xi  0;

i ¼ 1; 2; . . .; m

337

ð13:34Þ

i¼1

Since v is assumed to be positive, u is also positive, so that from (13.33) and (13.34), we have m X

and

aij

i¼1 m X i¼1

xi  1; u

xi 1 ¼ ; u u

j ¼ 1; 2; . . .; n xi  0; u

i ¼ 1; 2; . . .; m:

Now substituting xi for xui , we get m X

and

aij xi  1;

i¼1 m X i¼1

1 xi ¼ ; u

j ¼ 1; 2; . . .; n xi  0;

i ¼ 1; 2; . . .; m:

For the optimal strategy, v ¼ Max u (we can regard u as a function of n) or 1 v

¼

Min 1u n

¼ Min

m P

n

xi .

i¼1

Thus, ﬁnding v and n is equivalent to solving the following LPP: Minimize z ¼

m X

xi

i¼1

subject to m P

aij xi  1;

i¼1

xi  0;

9 j ¼ 1; 2; . . .; n = : ; i ¼ 1; 2; . . .; m

ð13:35Þ

 0 If zmin is the optimal value and  n ¼ x 1 ; . . .; x m is an optimal solution of (13.35), 1 then the value of the game is v ¼ zMin , and an optimal strategy for player 1 is  n  n ¼ vn ¼ . zMin

Again, we have v  E ðn; g Þ 8n 2 Sm . Hence, v  Eði; g Þ;

i ¼ 1; 2; . . .; m.

Thus there exists w such that (v  w) for g 2 Sn ,

338

13

w  Eði; gÞ

i ¼ 1; . . .; m:

Game Theory

ð13:36Þ

η is an optimal strategy for player 2 iff η satisﬁes (13.36) with w replaced by v. Now (13.36) can be written as n X

aij yj  w

i ¼ 1; 2; . . .; m:

ð13:37Þ

j¼1

Also, we have

n X

yj ¼ 1;

yj  0;

j ¼ 1; 2; . . .; n:

ð13:38Þ

j¼1

Since v is assumed to be positive, we can assume w > 0, so that from (13.37) and (13.38), we have n X j¼1 n X

aij yj  1; yj ¼

j¼1

1 ; w

i ¼ 1; 2; . . .; m yj  0;

j ¼ 1; 2; . . .; n;

where yj ¼ yj =w. For the optimal strategy, v ¼ Min w ðw can be regarded as a function of gÞ g

or

n X 1 1 yj : ¼ Min ¼ Min v w j¼1

Thus, ﬁnding v and g is equivalent to solving the following LPP: Maximize z0 ¼

n X

yj

j¼1

subject to n P

aij yj  1;

j¼1

yj  0;

9 i ¼ 1; 2; . . .; m = : ; j ¼ 1; 2; . . .; n

ð13:39Þ

 is an optimal solution of (13.39), then v ¼ z0 1 If z0Max is the optimal value and g

and g ¼ v  g ¼ z0g . Max

Max

13.8

Matrix Games and Linear Programming Problems (LPPs)

339

Note that (13.39) is the dual of (13.35) and vice versa. Example 8 Solve the following game by reducing it to an LPP: 0 B1 B2 B3 1 1 1 3 @ 3 5 3 A 6 2 2

A1 A2 A3

Solution First we construct the table for row minimum and column maximum as follows: Row minima 2

1 1

6 6 63 4

5

6

2

Column maxima 6

5

3

1 7 7 3 7 3 5 3

2 2 3

Hence, maximin value = −1, minimax value = 3. Hence, the value of the game lies between −1 and 3, i.e. 1  v  3. This means that the value of the game may be negative or zero. Thus, we add a constant k = 4 to all the elements of the matrix so that all these elements of the payoff matrix became positive, assuming that the value of the game represented by this new matrix is non-negative and non-zero. The transformed (reduced) matrix is as follows: B B 2 1 B2 A1 5 3 A2 4 7 9 A3 10 6

B33 7 15 2

Let p ¼ ðp1 ; p2 ; p3 Þ and q ¼ ðq1 ; q2 ; q3 Þ be the mixed strategies of the two players respectively. Hence, the problem of player B is as follows: Maximize z0 ¼ x1 þ x2 þ x3 subject to 5x1 þ 3x2 þ 7x3  1 7x1 þ 9x2 þ x3  1 10x1 þ 6x2 þ 2x3  1 and x1 ; x2 ; x3  0: Now, if max z0 ¼ v1 , the value of the original game is v ¼ v  4, and the optimal solution is qj ¼ xj  v , j = 1, 2, 3.

340

13

Game Theory

Introducing the slack variables x4 ; x5 ; x6 ; we get 5x1 þ 3x2 þ 7x3 þ x4 ¼ 1 7x1 þ 9x2 þ x3 þ x5 ¼ 1 10x1 þ 6x2 þ 2x3 þ x6 ¼ 1:

cB

yB

0 y4 0 y5 0 y6 zB ¼ 0

cB

yB

1 y3 0 y5 0 y6 zB ¼ 1=7

yB cB 1 y3 1 y2 0 y6 zB ¼ 1=5

cj xB

1 y1

1 y2

1 y3

0 y4

0 y5

0 y6

Mini ratio

x4 ¼ 1 x5 ¼ 1 x6 ¼ 1

5 7 10 1

3 9 6 1

7 1 2 1 "

1 0 0 0 #

0 1 0 0

0 0 1 0

1/7 1/1 1/2 Dj

cj xB

1 y1

1 y2

1 y3

0 y4

0 y5

0 y6

Mini ratio

x3 ¼ 1=7 x5 ¼ 6=7 x6 ¼ 5=7

5/7 44/7 60/7 2=7

3/7 60=7 36/7 4=7 "

1 0 0 0

1/7 1=7 2=7 1=7

0 1 0 0 #

0 0 1 0

1/3 1/10 5/36 Dj

cj

1

1

1

0

0

0

xB x3 ¼ 1=10 x2 ¼ 1=10 x6 ¼ 1=5

y1 2/5 11/15 24/5 2=5

y2 0 1 0 0

y3 1 0 0 0

y4 3/20 1=60 1=5 2=15

y5 1=20 7/60 3=5 1/5

y6 0 0 1 0

Dj

Since all zj  cj  0, the optimal solution is attained, and the optimal solution is 1 1 x1 ¼ 0; x2 ¼ 10 ; x3 ¼ 10 and Max z0 ¼ 15. 0 1 From z ¼ v , we have v ¼ 5 ) q1 ¼ x1 v ¼ 0;

q2 ¼ x 2 v ¼

1 1 5¼ ; 10 2

and the value of the game v ¼ v  4 ¼ 5  4 ¼ 1.

q3 ¼ x 3 v ¼

1 1 5¼ 10 2

13.8

Matrix Games and Linear Programming Problems (LPPs)

341

Since A’s problem is the dual of B’s problem, then using duality theory, we have w1 ¼

2 ; 15

w2 ¼

1 ; 15

w3 ¼ 0

) p1 ¼ x 1  v ¼

values of D4 ; D5 ; D6 if Dj ¼ zj  cj

2 2 5¼ ; 15 3

1 p2 ¼ ; 3

p3 ¼ 0:

Hence, the optimal solution is the following:   (i) Best strategy for A is p ¼ 23 ; 13 ; 0 .   (ii) Best strategy for B is q ¼ 0; 12 ; 12 . (iii) Value of the game = 1.

13.9

Inﬁnite Antagonistic Game

Deﬁnition An inﬁnite antagonistic game is a game in which at least one of the players possesses an inﬁnite number of strategies. Let X and Y be the set of strategies for players 1 and 2 respectively and M(x, y) ðx 2 X; y 2 Y Þ be the payoff function of the antagonistic game hX; Y; M i. In the case of the matrix game, recall that X, Y are discrete sets of strategies (pure) for players 1 and 2 respectively. In that case, we have deﬁned a mixed strategy for player 1 as a vector n having m (the number of pure strategies) components, each component diminishing the probability of choosing a pure strategy. A mixed strategy g for player 2 was similarly deﬁned. We can extend this idea of mixed strategy when X and Y are inﬁnite. In this case, a mixed strategy for player  00 1 will be denoted by a probability 0 distribution function F(x), where F xi  F xi denotes the probability of choosing x when x00i \x\x0i . Similarly, the probability function G(y) will denote a mixed strategy for player 2. Note that F(x) is a non-negative, non-decreasingR function and R dFðxÞ ¼ 1. G(y) is also non-negative and non-decreasing, and Y dGðyÞ ¼ 1. X Here the integration is in the Stieltjes sense. Let DX ¼

DY ¼

8 < : 8 < :

Z FðxÞ : 0  FðxÞ  1;

dFðxÞ ¼ 1

9 =

; 9 Z = dGðyÞ ¼ 1 : ;

X

GðyÞ : 0  GðyÞ  1;

Y

For a ﬁxed y, let player 1 choose x 2 X with probability distribution F(x); i.e. the     probability of choosing x, where x00i \x\x0i , is F x0i  F x00i . Then the expected

342

13

Game Theory

     payoff to player 1 is M ðx; yÞ F x0i  F x00i . Hence, when x 2 X, the expected payoff to the player is (for a ﬁxed y) n X

M ðx; yÞfF ðxi Þ  F ðxi1 Þg;

i¼1

where X is divided into n intervals with the help of the points xi ; i ¼ 0; 1; . . .; n. If we now make n ! 1 in such a way that Max jxi  xi1 j ! 0, then this becomes i

the Stieltjes integral with respect to F(x) over the interval X; Z i:e: EðF; yÞ ¼ J ð yÞ ¼ M ðx; yÞ dF ð xÞ: X

In a similar way, when player 2 chooses y with probability distribution G(y),      P0 then the expected payoff to player j is nj¼1 J ð yÞ G yj  G yj1 , where Y is 0 0 divided into n0 intervals with  the help of the points y0 ; . . .; yn . Now making n !   1 in such a way that Max yj  yj1 ! 0, the expected payoff to player 1, when j

his strategy is F(x) and player 2’s strategy is G(y), is Z E ðF; GÞ ¼

0 @

Y

Z

1 M ðx; yÞ dF ð xÞA dGð yÞ:

X

Note (i) If M(x, y) is a continuous function of x, y for x 2 X; y 2 Y, then Z E ðF; GÞ ¼ Y

0 @

Z X

1

M ðx; yÞ dxA dy ¼

Z X

0 @

Z

1

M ðx; yÞ dyA dx ¼

Y

Z Z M ðx; yÞ dx dy: Y

(

X

0 for x\x0 (ii) When F(x) is the Heaviside function H ðx  x0 Þ ¼ , then 1 for x [ x0 R E(F, G) becomes Y M ðx0 ; yÞ dGð yÞ and we denote it by E ðx0 ; GÞ. Thus, E ðx0 ; GÞ is the expected payoff to player 1 when he chooses a simple strategy H ðRx  x0 Þ and player 2 chooses the strategy G. Similarly, E ðF; y0 Þ ¼ X M ðx; y0 Þ dGð xÞ is the expected payoff to player 1 when he chooses the strategy F(x) and player 2 chooses the simple strategy H ðy  y0 Þ. (iii) Let hX; Y; M i be an inﬁnite antagonistic game. Then this is strategically equivalent to the game hDX ; DY ; Ei, where E ¼ E ðF; GÞ; F 2 DX ; G 2 DY .

13.9

Inﬁnite Antagonistic Game

343

Equilibrium situation Deﬁnition Let ðF ; G Þ be a situation in mixed strategies for the inﬁnite antagonistic game hX; Y; M i or hDX ; DY ; E i. This situation is called an equilibrium situation in mixed strategies or a saddle point in mixed strategies if for any ﬁxed strategies F, G for player 1 and player 2 respectively, the following inequality is satisﬁed: E ðF; G Þ  EðF ; G Þ  E ðF ; GÞ 8F 2 DX ; G 2 DY : For mixed strategies in inﬁnite antagonistic games, we shall prove the following theorem. Theorem 13.22 If E ðx; GÞ  u 8x 2 X and a ﬁxed G, then E ðF; GÞ  u 8F 2 DX [for a matrix game: F ði; gÞ  u; i ¼ 1; 2; . . .; m for a ﬁxed g 2 Sn implies that E ðn; gÞ  u 8n 2 Sm ]. Proof We R integrate both sides R of E ðx; GÞ  u with respect to F(x) over X so as to obtain X Eðx; GÞ dF ð xÞ  u X dF ð xÞ, 2 i:e: E ðF; GÞ  u

4*

Z

3 dF ð xÞ ¼ 15:

X

Similarly; E ðx; GÞ  u 8x 2 X implies EðF; GÞ  u 8F 2 DX ; EðF; yÞ [ u; 8y 2 Y implies E ðF; GÞ [ u; 8G 2 DY and E ðF; yÞ\u; 8y 2 Y implies E ðF; GÞ\u; 8G 2 DY Theorem 13.23 In order for the situation ðF ; G Þ to be an equilibrium situation in an inﬁnite antagonistic game C ¼ hX; Y; M i, it is necessary and sufﬁcient that for all x 2 X and y 2 Y, the following inequalities are satisﬁed: Eðx; G Þ  E ðF ; G Þ  EðF ; yÞ: ½For a matrix game: Eði; g Þ  E ðn ; g Þ  E ðn ; jÞ

i ¼ 1; 2; . . .; m;

j ¼ 1; 2; . . .; n

This proof is left to the reader. Theorem 13.24 For any mixed strategy G for player 2, sup E ðx; GÞ ¼ sup E ðF; GÞ; x2X

F2DX

where the supremum on the left side is taken over all the pure strategies for player 1, and on the right side it is taken over all the mixed strategies.

344

13

Game Theory

Similarly, for a ﬁxed strategy F for player 1, inf E ðF; yÞ ¼ inf E ðF; GÞ:

y2Y

2 4

G2DY

For a matrix game: Max Eði; gÞ ¼ Max ðn; gÞ for some fixed g i

n

and Min E ðn; jÞ ¼ Min E ðn; gÞ for some fixed n

3 5

g

j

Proof Since the set of mixed strategies contains all the pure strategies ( e:g:; F ð xÞ ¼ H ðx  x0 Þ ¼

1

x [ x0

1

x \ x0

! ;

we have sup E ðx; GÞ  sup E ðF; GÞ. x2X

F2DX

Now, let if possible, sup E ðx; GÞ\ sup EðF; GÞ. x2X

F2DX

This implies that there exists F1 2 DX such that for some e [ 0 sup Eðx; GÞ\EðF1 ; GÞ  e; x2X

i:e: Eðx; GÞ\EðF1 ; GÞ  e 8x 2 X: Now integrating both sides with respect to F1(x), we have Z

Z Eðx; GÞ dF1 ð xÞ\fE ðF1 ; GÞ  eg

X

2

i:e: EðF1 ; GÞ\EðF1 ; GÞ  e

4*

dF1 ð xÞ; X

Z

3

dF1 ð xÞ ¼ 15;

X

which is a contradiction Hence, sup Eðx; GÞ ¼ sup E ðF; GÞ. x2X

F2DX

The proof of the second result is similar. Value of an inﬁnite antagonistic game Deﬁnition If sup inf EðF; GÞ ¼ inf sup E ðF; GÞ, then the common value of these F

G

G

F

mixed extrema is called the value of the game hX; Y; M i.

13.9

Inﬁnite Antagonistic Game

345

Some properties of the value of the game and optimal strategies Let v be the value of the game hX; Y; M i. Theorem 13.25 sup inf M ðx; yÞ  v  inf sup M ðx; yÞ y

x

y

x

 For a matrix game: Max Min aij  u  Min Max aij : i

j

j

i

Proof We know that inf E ðF; yÞ ¼ inf EðF; GÞ for any arbitrary mixed strategy y2Y

G2DY

F for player 1.

(

Let us choose a simple strategy F ¼ H ðx  x0 Þ ¼

0 for x\x0 . 1 for x [ x0

Then inf M ðx0 ; yÞ ¼ E ðx0 ; GÞ 8x0 2 X;

y2Y

i.e. changing x0 to x, inf M ðx; yÞ ¼ inf E ðx; GÞ. y2Y

G2DY

    But sup inf M ðx; yÞ  sup inf E ðF; yÞ x

y

y

F

½as the set of mixed strategies DX contains all pure strategies

  ¼ sup inf E ðF; GÞ ¼ v; G F   i:e: sup inf M ðx; yÞ  v: x

y

ð13:40Þ

Similarly, we can prove that v  inf sup M ðx; yÞ: y

ð13:41Þ

x

Hence, from (13.40) and (13.41), we have   sup inf M ðx; yÞ  v  inf sup M ðx; yÞ: x

y

y

x

Theorem 13.26 If player 1 possesses a pure optimal strategy x0 and player 2 possesses an arbitrary strategy (in general mixed), then

346

13





v ¼ Max Min M ðx; yÞ x

Game Theory

¼ Min M ðx0 ; yÞ:

y

y

Similarly, if player 2 possesses a pure strategy y0 and player 1 possesses an arbitrary strategy, then n o v ¼ Min Max M ðx; yÞ ¼ Max M ðx; y0 Þ: y

x

x

The reader may easily prove this theorem. Theorem 13.27 Let u be a real number. (i) If F0 is a mixed strategy for player 1, then u  E ðF0 ; yÞ 8y 2 Y ) u  v: (ii) If G0 is a mixed strategy for player 2, then u  E ðx; G0 Þ

8x 2 X ) u  v:

Proof We have u  EðF0 ; yÞ 8y 2 Y. Now integrating both sides with respect to G(y) over Y, we have Z Z u dGð yÞ  EðF0 ; yÞ dGð yÞ Y

Y

or u  E ðF0 ; GÞ

8G 2 DY   Hence u  inf E ðF0 ; GÞ  sup inf E ðF; GÞ ¼ v G

F

G

) u  v: The proof of (ii) is similar. Theorem 13.28 (i) In order for the strategy F0 for player 1 to be optimal, it is necessary and sufﬁcient that EðF0 ; yÞ  v 8y 2 Y:

13.9

Inﬁnite Antagonistic Game

347

(ii) In order for the strategy G0 for player 2 be optimal, it is necessary and sufﬁcient that E ðx; G0 Þ  v 8x 2 X: Proof of (i) Let F0 be an optimal strategy. This implies that inf E ðF0 ; GÞ ¼ sup inf E ðF; GÞ ¼ v; G

F

G

so that E ðF0 ; GÞ  v for all G 2 DY : Let us choose G to be a pure strategy H ðy  y0 Þ. Then we obtain EðF0 ; y0 Þ  v for all y0 2 Y; i:e: E ðF0 ; yÞ  v for all y 2 Y:

Sufficient: Let E ðF0 ; yÞ  v for all y 2 Y:

ð13:42Þ

Now, we have to show that F0 is an optimal strategy for player 1. Integrating both sides of (13.42) with respect to G(y) over Y, we have Z

Z EðF0 ; yÞ dGð yÞ  v

Y

dGð yÞ Y

or E ðF0 ; GÞ  v

8G 2 DY :   Hence; inf E ðF0 ; GÞ ¼ v ¼ sup inf E ðF; GÞ ; G

F

G

i.e. the supremum of the function inf EðF0 ; GÞ is attained at F0. This implies that G

F0 is optimal.

13.10

Continuous Game

Deﬁnition An antagonistic game C ¼ hX; Y; M i is called a game on the unit square if the sets of pure strategies for each of the players are the unit segments, i.e. X = Y = [0, 1]. Hence, the mixed strategies for the players in a game on the unit square are R1 probability distributions on the segment [0, 1] such that 0 dF ð xÞ ¼ 1 and R1 0 dGð yÞ ¼ 1.

348

13

Game Theory

Here DX ¼ DY ¼ D: Deﬁnition A game on the unit square is called continuous if the payoff function is continuous.

Fundamental theorem for continuous game Theorem 13.29 For any continuous game on the unit square with payoff function M(x, y), where both players possess optimal strategies, 



Max Min E ðF; GÞ F

G





and Min Max EðF; GÞ G

F

exist and are equal. The proof of the theorem is left to the reader.

13.11

Separable Game

Deﬁnition A function M(x, y) of two variables x, y is called separable if there exist m continuous functions ri (x), i = 1, 2, …, m, n continuous functions sj (y), j = 1, 2, …, n and mn constants aij i = 1, 2, …, m, j = 1, 2, …, n such that M ðx; yÞ ¼

m X n X

aij ri ð xÞ sj ð yÞ:

i¼1 j¼1

Example 9 M ðx; yÞ ¼ x sin y þ x cos y þ 2x2 is separable. We can take r 1 ð xÞ ¼ x r 2 ð xÞ ¼ x2

s1 ð yÞ ¼ sin y s2 ð yÞ ¼ cos y s 3 ð yÞ ¼ 1

a11 ¼ 1; a12 ¼ 1; a13 ¼ 0 a21 ¼ 0; a22 ¼ 0; a23 ¼ 2

We can also take r 1 ð xÞ ¼ x r2 ð xÞ ¼ 2x2

s1 ð yÞ ¼ sin y þ cos y s2 ð yÞ ¼ 1

a11 ¼ 1; a12 ¼ 0 a21 ¼ 0; a22 ¼ 1

13.11

Separable Game

349

Note

P Pn (i) The representation M ðx; yÞ ¼ m i¼1 j¼1 aij ri ð xÞ sj ð yÞ of a separable function is not unique. (ii) If M(x, y) is separable, we may also write M ðx; yÞ ¼

n m X X j¼1

! aij ri ð xÞ sj ð yÞ ¼

i¼1

n X

tj ð xÞ sj ð yÞ;

j¼1

where ti(x) and sj(y) are continuous functions of x and y respectively. Thus, a continuous game on the unit square whose payoff function is separable is called a separable game. Expectation function in a separable game A separable game is hX; Y; M i, where X = Y = [0, 1], and M(x, y) is separable. Hence, a mixed strategy for player 1 is a probability distribution function F(x), and a mixed strategy for player 2 is a probability distribution function G(y), where F 2 D; G 2 D and  D¼

Z

1

F : 0  F  1; 0

 or D ¼

Z

1

G : 0  G  1;

 dF ð xÞ ¼ 1  dGð yÞ ¼ 1 :

0

If player 1 chooses a mixed strategy F(x) and player 2, a mixed strategy G (y) ðF; G 2 DÞ, then the expectation of player 1 is Z E ðF; GÞ ¼

1

Z

1

M ðx; yÞ dF ð xÞ dGð yÞ ) Z 1 Z 1 (X m X n ¼ aij ri ð xÞ sj ð yÞ dF ð xÞ dGð yÞ

¼ ¼

0

0

0

0

i¼1 j¼1

m X n X i¼1 j¼1 m X n X

Z

1

aij

and vj ¼

R1 R01 0

1

ri ð xÞ dF ð xÞ

0

aij ui vj ;

i¼1 j¼1

where ui ¼

Z

ri ð xÞ dF ð xÞ;

i ¼ 1; 2; . . .; m

sj ð yÞ dGð yÞ;

j ¼ 1; 2; . . .; n.

0

 sj ð yÞ dGð yÞ

350

13

Game Theory

Note 1: There may exist two different strategies F1 ð xÞ; F2 ð xÞ such that Z1

Z1 ri ð xÞ dF1 ð xÞ ¼

0

ri ð xÞ dF2 ð xÞ

i ¼ 1; 2; . . .; m:

0

But E ðF1 ; GÞ ¼

m X n X i¼1 j¼1

aij ui vj ¼

m X n X i¼1 j¼1

0 aij @

Z1

1 ri ð xÞ dF2 ð xÞAvj

0

¼ EðF2 ; GÞ 8 G 2 D: Thus, it is immaterial whether player 1 uses strategy F1 ð xÞ or F2 ð xÞ. In this case we shall say that the two strategies F1 and F2 are equivalent (with respect to the game). P-space and Q-space Let a ¼ ðu1 ; . . .; um Þ0 ; b ¼ ðv1 ; . . .; vn Þ0 ; then a 2 E m ; b 2 En . n o R1 Let P ¼ a ¼ ðu1 ; u2 ; . . .; um Þ0 ; ui ¼ 0 ri ð xÞ dF ð xÞ; i ¼ 1; 2; . . .; m; F 2 D . Then obviously P E m . We also say that a and F correspond. Note that a given point a of the P-space, in general, corresponds to many different distribution functions. n R1 Similarly, let Q ¼ b ¼ ðv1 ; v2 ; . . .; vn Þ0 ; vj ¼ 0 sj ð yÞ dGð yÞ; j ¼ 1; 2; . . .; n; G 2 Dg. Then obviously Q E n . We also say that b and G correspond. Note that a given point b in the Q-space, in general, corresponds to many different distribution functions. Note If u corresponds to both F(x) and F1(x), and v corresponds to both G(y) and G1(y), then E ðF; GÞ ¼ E ðF1 ; GÞ ¼ E ðF; G1 Þ ¼ E ðF1 ; G1 Þ X aij ui vj ; and let us denote it by E ða; bÞ ¼ i; j

Geometrical properties of P- and Q-spaces Theorem 13.30 Let Fk ð xÞ ðk ¼ 1; . . .; pÞ be a distribution function (i.e. strategies P for player 1), and let F ð xÞ ¼ pk¼1 ck Fk ð xÞ, where ck  0; k ¼ 1; 2; . . .; p and

13.11 p P

Separable Game

351

ck ¼ 1. Then F 2 D. Let ak and Fk correspond ðk ¼ 1; . . .; pÞ. Then a ¼

k¼1 p P

ck ak corresponds to F so that a 2 P.

k¼1

A similar result holds for the Q-space. m P n P Proof We have M ðx; yÞ ¼ aij ri ð xÞ sj ð yÞ. i¼1 j¼1  0 ðk Þ Let ak ¼ u1 ; . . .; uðmkÞ u ¼ 1; . . .; p.

Since ak corresponds to Fk ð xÞ; k ¼ 1; . . .; p, we have Z1 ui ¼

ri ð xÞ dFk ð xÞ; i ¼ 1; . . .; m: 0

Let a ¼ ðu1 ; . . .; um Þ0 . Since a ¼

p X

ck ak ; then ui ¼

k¼1

¼

p X k¼1

Z

ðk Þ

ck ui ; i ¼ 1; 2; . . .; m

k¼1 1

ck

ri ð xÞ dFk ð xÞ

0

Z1 ¼

p X

r i ð xÞ d 0

p X k¼1

! ck Fk ð xÞ

Z1 ¼

ri ð xÞ dF ð xÞ 0

Since F 2 D, a corresponds to F, and hence a 2 P. The proof for the Q-space is similar. P0 - and Q0 -spaces

  Let us deﬁne P ¼ q ¼ ðr1 ;    ; rm Þ0 : ri ðtÞ; 0  t  1 ) P 2 E m   Q ¼ r ¼ ðs1 ; . . .; sn Þ0 : sj ðtÞ; 0  t  1 ) Q E n :

Properties of P0 - and Q0 -spaces Property-1: P P and Q Q.

352

13

Game Theory

Property-2: Since ri(x) and sj(y) are continuous functions deﬁned on a closed set, P* and Q* are bounded, closed and connected sets. Property-3: The P-space is the convex hull of P*. Similarly, the Q-space is the convex hull of Q*. Example 10 Let the payoff function be M ðx; yÞ ¼ cosð2pxÞ cosð2pyÞ þ x þ y; 0  x; y  1. Here we can write M ðx; yÞ ¼

3 P 3 P

aij ri ð xÞ sj ð yÞ,

i¼1 j¼1

where r1 ð xÞ ¼ x; r2 ð xÞ ¼ cos 2px; r3 ð xÞ ¼ 1;

s1 ð yÞ ¼ y; a11 ¼ 0; a12 ¼ 0; a13 ¼ 1 s2 ð yÞ ¼ cos 2py; a21 ¼ 0; a22 ¼ 1; a23 ¼ 0 s3 ð yÞ ¼ 1; a31 ¼ 1; a32 ¼ 0; a33 ¼ 0

  Thus; P ¼ q ¼ ðr1 ; r2 ; r3 Þ0 : r1 ðtÞ ¼ t; r2 ðtÞ ¼ cos 2pt; r3 ðtÞ ¼ 1; 0  t  1   Q ¼ r ¼ ðs1 ; s2 Þ0 : s1 ðtÞ ¼ t; s2 ðtÞ ¼ cos 2pt; s3 ðtÞ ¼ 1; 0  t  1 :

Image of a 2 P Deﬁnition The image of a 2 P is deﬁned as the set of points b 2 Q such that E ða; bÞ ¼ Min E ða; gÞ, and we denote this image of a by QðaÞ and QðaÞ Q. g2Q

Note that QðaÞ is in general a set of points 2 Q. Image of b 2 Q Deﬁnition Similarly, the image of b 2 Q is deﬁned as the set of points a 2 P such that Eða; bÞ ¼ Max E ðn; bÞ, and we denote this image of b by P(b) and PðbÞ P. n2P

Note that P(b) in general is a set of points 2 P. Fixed point Deﬁnition If b 2 Q is an image point of a 2 P and a 2 P is an image point of b 2 Q, then a is called a ﬁxed point of P-space and b is a ﬁxed point of Q-space. Theorem 13.31 If F1 is any distribution function and a 2 P corresponds to F1, then F1 is an optimal strategy for player 1 iff a is a ﬁxed point P. Similarly, G is an optimal strategy for player 2 iff the corresponding point b 2 Q is a ﬁxed point Q. Proof Let F1 be a mixed strategy for player 1 and let it correspond to a. Now, a is a ﬁxed point of P means that for some point b 2 Q, we have b 2 QðaÞ and a 2 PðbÞ. That is, b is an image point of a and a is an image point of b, i.e.

13.11

Separable Game

353

E ða; bÞ ¼ Min E ða; gÞ g2Q

and Eða; bÞ ¼ Max Eðn; bÞ: n2P

Let G1 be a strategy (mixed) for player 2, and let it correspond to b 2 Q. Then E ðF1 ; G1 Þ  E ða; bÞ ¼ Min E ða; gÞ ¼ Min E ðF1 ; GÞ g2Q

G2D

) E ðF1 ; G1 Þ  E ðF1 ; GÞ 8G 2 D: Again E ðF1 ; G1 Þ  E ða; bÞ ¼ Max E ðn; bÞ ¼ Max E ðF; G1 Þ  E ðF; G1 Þ 8F 2 D; F2D

n2P

i:e: E ðF; G1 Þ  E ðF1 ; G1 Þ 8F 2 D;

Hence, we obtain EðF; G1 Þ  EðF1 ; G1 Þ  E ðF1 ; GÞ 8F; G 2 D: Thus, F1 and G1 are optimal strategies. Theorem 13.32 If a is any ﬁxed point of P and b is any ﬁxed point of Q, then a is an image point of b, and b is an image point of a. Proof Since a is a ﬁxed point of P, then 9 a point b1 of Q such that b1 is an image point of a and a is an image point b1 , i.e. a 2 Pðb1 Þ and b1 2 QðaÞ, i.e. such that E ða; b1 Þ ¼ Max E ðn; b1 Þ

ð13:43Þ

and E ða; b1 Þ ¼ Min Eða; gÞ:

ð13:44Þ

n2P

g2Q

Similarly, since b is a ﬁxed point of Q, then there exists a point a1 2 P such that a1 is an image point of b and b is an image point of a1 , i:e: a1 2 PðbÞ and b 2 Qða1 Þ i:e: E ða1 ; bÞ ¼ Max E ðn; bÞ

ð13:45Þ

and Eða1 ; bÞ ¼ Min E ða1 ; gÞ:

ð13:46Þ

n2P

g2Q

Thus, we have

354

13

Game Theory

E ða; bÞ  Max E ðn; bÞ ½by the definition of maxima

n2P

¼ E ða1 ; bÞ ½by ð13:45Þ

¼ Min E ða1 ; gÞ ½by ð13:46Þ

g2Q

 E ða1 ; b1 Þ ½by the definition of minima

 Max E ðn; b1 Þ

½by the definition of maxima

n

¼ E ða; b1 Þ ½by ð13:43Þ

¼ Min E ða; gÞ ½by ð13:44Þ

g2Q

 E ða; bÞ

½by the definition of minima :

Since the ﬁrst and last terms of these inequalities are equal, we therefore conclude that all the quantities involved are equal, and thus in particular, E ða; bÞ ¼ Max E ðn; bÞ ¼ Min E ða; gÞ; g2Q

n2P

which means that a 2 PðbÞ and b 2 QðaÞ, i.e. a is an image point of b and b is an image point of a. General procedure to solve a separable game Step-1 Plot the curves P* and Q* and determine their convex hulls to give the Pspace and Q-space. Step-2 Find Q(a) for every a 2 P and P(b) for every b 2 Q; i.e. ﬁnd all the image points of the Pspace and Q-space. Step-3 Using the results of Step 2, ﬁnd the ﬁxed points of P and Q. Step-4 Express the ﬁxed points as convex linear combinations of points in P* and Q* respectively, then ﬁnd the distribution function in which these ﬁxed points corresponding to the distribution function will be the optimal strategies Example 11 Given that M ðx; yÞ ¼ 3 cos 4x cos 5y þ 5 cos 4x sin 5y þ sin 4x cos 5y þ sin 4x sin 5y: Find P and Q

Solution We can write M ðx; yÞ ¼

P2 P2 i¼1

j¼1

aij ri ð xÞ sj ð yÞ,

where r1 ð xÞ ¼ cos 4x s1 ð yÞ ¼ cos 5y r2 ð xÞ ¼ sin 4x

s2 ð yÞ ¼ sin 5y

and a11 ¼ 3; a12 ¼ 5; a21 ¼ 1; a22 ¼ 1.

0x1 0y1

13.11

Separable Game

355

  Thus, P ¼ q ¼ ðr1 ; r2 Þ0 r1 ¼ cos 4t; r2 ¼ sin 4t; 0  t  1   and Q ¼ r ¼ ðs1 ; s2 Þ0 ; s1 ¼ cos 5t; s2 ¼ sin 5t; 0  t  1 . Canonical representation of the payoff function in a separable game For a separable game, let the payoff function M(x, y) be represented by M ðx; yÞ ¼

n X n X

aij ri ð xÞ sj ð yÞ þ

i¼1 j¼1

n X

bi r i ð x Þ þ

i¼1

n X

cj sj ð yÞ þ d;

j¼1

where (i) r1 ð x Þ; r 2 ð xÞ; . . . rn ð xÞ and s1 ð yÞ; . . .; sn ð yÞ are continuous functions over [0, 1]. (ii) det aij 6¼ 0. Any representation of a separable function in this form is called a canonical representation of the function M(x, y). We note that every separable function has a canonical Pn form. For if M(x, y) is separable, then it can always be written in the form j¼1 rj ð xÞsj ð yÞ so that det

aij 6¼ 0. Example 12 Examine whether the representation M ðx; yÞ ¼ xy  xey þ 2x cos y þ 2ex y þ 3ex ey þ ex cos y þ 5 cos xey  3 cos x cos y is canonical or not: Let r1 ð xÞ ¼ x s1 ð yÞ ¼ y, r2 ð xÞ ¼ ex s2 ð yÞ ¼ ey , r3 ð xÞ ¼ cos x Then a11 ¼ 1; a21 ¼ 2; a31 ¼ 0;

a12 ¼ 1; a22 ¼ 3; a32 ¼ 5;

a13 ¼ 2;

s3 ð yÞ ¼ cos y.

b1 ¼ b2 ¼ b3 ¼ 0

a23 ¼ 1; c1 ¼ c2 ¼ c3 ¼ 0 a33 ¼ 3; d ¼ 0:

   2   1 1 3 1  ¼ 0. Now, det aij ¼  2 0 5 3  Now from M ðx; yÞ ¼ xy  xey þ 2x cos y þ 2ex y þ 3ex ey þ ex cos y þ 5 cos xey  3 cos x cos y we have

356

13

Game Theory

7 3 y M ðx; yÞ ¼ ðx  2 cos xÞ y þ cos y  ðx  2 cos xÞ e  cos y 5 5 3 3 x x y þ 2ðe þ cos xÞ y þ cos y þ 3ðe þ cos xÞ e  cos y ; 5 5 i:e: M ðx; yÞ ¼ r1 ð xÞs1 ð yÞ  r1 ð xÞs2 ð yÞ þ 2r2 ð xÞs1 ð yÞ þ 3r2 ð xÞs2 ð yÞ where r1 ð xÞ ¼ x  2 cos x r2 ð xÞ ¼ ex þ cos x

7 s1 ð yÞ ¼ y þ cos y 5 3 y s2 ð yÞ ¼ e  cos y: 5

Hence, a11 ¼ 1;

a22 ¼ 1;

b1 ¼ b2 ¼ 0

a21 ¼ 2;

a22 ¼ 3

c1 ¼ c2 ¼ 0

d ¼ 0:

   1 1   ¼ 5 6¼ 0. Therefore, det aij ¼  2 3 Thus, this representation is canonical.

First critical point and second critical point Let a separable function M(x, y) have the following canonical representation: M ðx; yÞ ¼

n X n X

aij ri ð xÞ sj ð yÞ þ

i¼1 j¼1

n X i¼1

bi r i ð x Þ þ

n X

cj sj ð xÞ þ d;

j¼1

where ri ð xÞ; sj ð yÞ ði; j ¼ 1; 2; . . .; nÞ are continuous functions on [0, 1] and

det aij 6¼ 0. In this case, let F, a mixed strategy for player 1, correspond to a ¼ ðu1 ; . . .; un Þ0 2 P, and let G, a mixed strategy for player 2, correspond to b ¼ ðv1 ; . . .; vn Þ0 2 Q (note that both P; Q E n ). Z1 Then ui ¼

ri ð xÞ dF ð xÞ;

i ¼ 1; 2; . . .; n

sj ð yÞ dGð yÞ;

j ¼ 1; 2; . . .; n

0

Z1 vj ¼ 0

13.11

Separable Game

357

and EðF; GÞ  E ða; bÞ ¼

n X n X i¼1 j¼1

¼

n n X X j¼1

aij ui vj þ

n X

bi ui þ

i¼1

!

aij ui þ cj vj þ

i¼1

n X

n X

cj vj þ d

j¼1

bi ui þ d:

i¼1

Alternatively, E ðF; GÞ ¼

n n X X i¼1

! aij vi þ bi ui þ

n X

j¼1

cj vj þ d:

j¼1

Since det aij 6¼ 0, then the system of equations n X

aij ui þ cj ¼ 0;

j ¼ 1; 2; . . .; n

ð13:47Þ

i¼1

has a unique solution, as does the system of equations n X

aij vj þ bi ¼ 0;

i ¼ 1; 2; . . .; n

ð13:48Þ

j¼1

Q The unique solution of (13.47), say ¼ ðp1 ; p2 ; . . .; pn Þ0 , is called the ﬁrst critical point of the game (with respect to the given representation), and the unique solution of (13.48), say v ¼ ðq1 ; q2 ; . . .; qn Þ0 , is called the second critical point of the game (with respect to the given representation). Q Note The point may notQbelong to the P-space, and the point v may not belong to the Q-space. However, if 2 P and v 2 Q, then the corresponding game problem can be solved easily by using the following two theorems. Theorem 13.33 Let the payoff function of a separable game have a given canonical form M ðx; yÞ ¼

n X n X i¼1 j¼1

aij ri ð xÞ sj ð yÞ þ

n X

b i r i ð xÞ þ

i¼1

n X

cj sj ð yÞ þ d; det aij 6¼ 0

j¼1

Q and also let ¼ ðp1 ; p2 ; . . .; pn Þ0 ; v ¼Qðq1 ; q2 ; . . .; qn Þ0 beQrespectively the ﬁrst and second critical points and suppose 2 P; v 2 Q. Then and v are the ﬁxed n n Q P P points, and the value of the game is given by E ð ; vÞ ¼ bi pi þ d ¼ cj qj þ d. i¼1

Proof By the deﬁnition of critical points,

j¼1

358

13 n X

and

aij pi þ cj ¼ 0;

i¼1 n X

aij qj þ bi ¼ 0;

Game Theory

j ¼ 1; 2; . . .; n i ¼ 1; 2; . . .; n:

j¼1

Now, for any point b 2 Q b ¼ ðv1 ; v2 ; . . .; vn Þ0 , ! n n n Y  X X X aij pi þ cj vj þ bi pi þ d E ;b ¼ j¼1

¼ 0þ

i¼1 n X

i¼1

bi pi þ d; which is independent of b:

i¼1

Q Hence, Qð Þ ¼ Q. Similarly, PðvÞ ¼ P. Q Q Q Hence, 2 PðvÞ and v 2 Qð Þ, so that and v are ﬁxed points, and the value of the game is n n Y  X X E ;v ¼ bi pi þ d ¼ cj qj þ d: i¼1

j¼1

Theorem 13.34 Let the interior of Q contain a ﬁxed point; then the ﬁrst critical point belongs to P and is the only ﬁxed point. Similarly, if the interior of P contains a ﬁxed point, then the second critical point belongs to Q and is the only ﬁxed point of Q. Proof Let b ¼ ðv1 ; . . .; vn Þ0 be a ﬁxed point of Q which lies in the interior of Q. Let Q 0 a ¼ ðu1 ; . . .; un Þ0 be a ﬁxed point of QP and let ¼ ðp1 ; . . .; pn ÞQbe the ﬁrst critical point. WeQwish to show that a ¼ . Suppose Pn if possible, a 6¼ . Since is the unique solution of j ¼ 1; 2; . . .; n, we i¼1 aij pi þ cj ¼ 0; have for some k  n, Pn i¼1 aik ui þ ck 6¼ 0 and is equal to g. Let h be a real number of opposite sign to g, and let it be small enough to ensure that the point  ¼ ðv1 ; . . .; vk1 ; vk þ h; vk þ 1 ; . . .; vn Þ 2 Q: b ! n n n X X X Now E ða; bÞ ¼ aij ui þ cj vj þ bi ui þ d j¼1



i¼1

  ¼ E ða; bÞ þ h E a; b

i¼1 n X

!

aik vk þ ck

i¼1

¼ E ða; bÞ þ hg \E ða; bÞ as hg\0:

13.11

Separable Game

359

This means that a 62 PðbÞ, which contradicts the assumption that a and Q b are both ﬁxed points of P and Q respectively. Hence, a must coincide with . Corollary 1 Let the payoff of a separable game be given in a canonical form and let the ﬁrst critical point 62 P. Then every ﬁxed point of Q is in the boundary of Q. Similarly, if the second critical point 62 Q, then every ﬁxed point of P is in the boundary of P. CorollaryQ2 Let the payoff function of a separable game be given in a canonical Q form and and v be the ﬁrst and second critical points respectively. Let belong Q to the interior of P and v belong to the interior of Q; then is the only ﬁxed point of P and v is the only ﬁxed point of Q. Example 13 Solve the separable game whose payoff function is M ðx; yÞ ¼ 3 cos 7x cos 8y þ 5 cos 7x sin 8y þ 2 sin 7x cos 8y þ sin 7x sin 8y: Solution Let r1 ð xÞ ¼ cos 7x;

r2 ð xÞ ¼ sin 7x

s1 ð yÞ ¼ cos 8y;

s2 ð yÞ ¼ sin 8y:

Then M ðx; yÞ ¼ 3r1 ð xÞ s1 ð yÞ þ 5r1 ð xÞs2 ð yÞ þ 2r2 ð xÞ s1 ð yÞ þ r2 ð xÞ s2 ð yÞ. Hence; a11 ¼ 3; a12 ¼ 5; b1 ¼ b2 ¼ 0 a21 ¼ 2; a22 ¼ 1; c1 ¼ c2 ¼ 0   3 5  ¼ 3  10 ¼ 7 6¼ 0: ) det aij ¼  2 1 Hence, this representation is canonical. P* and Q* spaces  Here P ¼ q ¼ ðr1 ; r2 Þ0 : r1 ¼ cos 7t; r2 ¼ sin 7t;  Q ¼ r ¼ ðs1 ; s2 Þ0 : s1 ¼ cos 8t; s2 ¼ sin 8t;

0t1 0t1

 

The P* space is the circle with its interior, and the Q* space is the circle with its interior. See Fig. 13.4a, b. First and second critical points Q ¼ ðp1 ; p2 Þ0 is the unique solution of

360

13

(a)

r2

Game Theory

s2

(b)

s1

r1

Q* space

P* space

Fig. 13.4 a Graphical representation of P* space. b Graphical representation of Q* space

n X

aij pi þ cj ¼ 0;

j ¼ 1; 2;

i¼1

i:e: 3p1 þ 2p2 ¼ 0 5p1 þ p2 ¼ 0 ) p1 ¼ 0; p2 ¼ 0 Q ) ¼ ð0; 0Þ0 2 interior of P-space. Now v ¼ ðq1 ; q2 Þ0 is the unique solution of q X

aij qj þ ci ¼ 0;

i ¼ 1; 2;

j¼1

i:e: 2q1 þ 5q2 ¼ 0 2q1 þ q2 ¼ 0 ) q1 ¼ 0; q2 ¼ 0 )v ¼ ð0; 0Þ0 2 interior of Q: Q Hence, is the only ﬁxed point of P and v is the only ﬁxed point of Q. Value of the game The value of the game is E

2 2 Y  X X ;v ¼ bi pi þ d ¼ cj qj þ d ¼ 0: i¼1

j¼1

13.11

Separable Game

361

P*

Fig. 13.5 Graphical representation of P* space

B (1, 0)

(-1, 0) A

fixed point

Optimal strategies Q We write the ﬁxed point of P, i.e. ¼ ð0; 0Þ0 , as a convex combination of the * points on P . See Fig. 13.5. We can obviously write 1 1 ð0; 0Þ0 ¼ ð1; 0Þ0 þ ð1; 0Þ0 : 2 2 Hence, an optimal strategy for player −1 is 1 1 F ð xÞ ¼ FA ð xÞ þ FB ð xÞ; 2 2 where FA ð xÞ is a strategy corresponding to A  ð1; 0Þ0 and FB ð xÞ is a strategy corresponding to B ¼ ð1; 0Þ0 . Now to ﬁnd FA ð xÞ, we note that it satisﬁes Z1

Z1 r1 ðtÞ dFA ðtÞ ¼

1 ¼

cos7t dFA ðtÞ

0

0

Z1

Z1

r2 ðtÞ dFA ðtÞ ¼ 0

sin7t dFA ðtÞ: 0

  These expressions are satisﬁed by FA ð xÞ ¼ H x  p7 . Similarly, FB ð xÞ satisﬁes

t1 ( cos 7t1 ,sin 7t1 )

t2 ( cos 7t2 ,sin 7t2 )

362

13

Z1 1¼

Z1 r1 ðtÞ dFB ðtÞ ¼

cos7t dFB ðtÞ

0

0

Z1

Z1

Game Theory

r2 ðtÞ dFB ðtÞ ¼ 0

sin7t dFB ðtÞ: 0

  These are satisﬁed by FB ð xÞ ¼ H ð xÞ or H x  2p 7 . Hence, an optimal strategy for player 1 is

t1 ( cos 7t1 ,sin 7t1 )

t2 ( cos 7t2 ,sin 7t2 ) 1  p either F ð xÞ ¼ H x  þ 2 7 1  p or F ð xÞ ¼ H x  þ 2 7

1 H ð xÞ 2 1 2p H x : 2 7

Hence, we can also write 1 1 p 0  t1 ; t2  1: ð0; 0Þ0 ¼ ðcos 7t1 ; sin 7t1 Þ0 þ ðcos 7t2 ; sin 7t2 Þ0 where t2  t1 ¼ 2 2 7 Hence, a most general optimal strategy for player 1 is 1 1 p 0  x1 ; x2  1: F ð xÞ ¼ H ðx  x1 Þ þ H ðx  x2 Þ where x2  x1 ¼ 2 2 7 Similarly, a most general optimal strategy for player 2 is

13.11

Separable Game

363

1 1 p G ð yÞ ¼ H ðy  y1 Þ þ H ðy  y2 Þ; y2  y1 ¼ 2 2 8

0  y1 ; x2  1:

Z1 Also, 0 ¼

cos7x dF ð xÞ 0

Z1 0¼

cos7x dF ð xÞ 0

are satisﬁed by choosing F ð xÞ ¼

 7x

2p ;

1 Z7

Z1

2p

cos7x dF ð xÞ ¼

In fact;

0  x  2p 7 . x1

2p 7

Z1 7x cos 7x d þ cos7x d ð1Þ 2p

0

0

2p 7 sin 7x 7 ¼0 ¼ 2p 7 0

2p 7

and

2p

Z7

Z1 sin7x dF ð xÞ ¼

sin7x d

Z1 7x þ sin7x dð1Þ 2p

0

0

¼

7 cos 7x  2p 7

2p7 0

2p 7

¼

7 ð1  1Þ ¼ 0: 2p

Similarly, it can easily be veriﬁed that 8 8y 2p > < 0y 2p 8 G ð yÞ ¼ 2p > :1 y1 8 is an optimal strategy for player 2. Note This is example of a game where there are an inﬁnite number of optimal strategies.

364

13

13.12

Game Theory

Exercises

1. Solve the game whose payoff matrix is given below: 2

2 6 3 6 4 4 5

0 2 3 3

0 1 0 4

5 2 2 2

3 3 2 7 7 6 5 6

2. Solve the following 4  2 game graphically: Player B 3 2 0 6 3 1 7 6 7 PlayerA6 7 4 3 2 5 2

5

4

3. Determine the optimum minimax strategies for each player in the following games:

ðiÞ

ðiiÞ

A1 A2 A3 A1 A2 A3

2 B1 B2 B3 B4 3 5 2 0 7 4 5 6 4 8 5 4 0 2 3 2B1 B2 2 15 4 5 6 5 20

B3 3 2 4 5 8

4. Use dominance to solve the following game: 2

1 62 6 43 4

2 2 1 3

3 1 0 2

3 1 5 7 7 2 5 6

5. Solve the following two-person zero-sum game to ﬁnd the value of the game: Company B 2 2 4 6 6 1 12 6 4 3 2 0 2 3 7 2 Company A

3 1 37 7 65 7

13.12

Exercises

365

6. Use the notion of dominance to simplify the rectangular game with the following payoff, and then solve it graphically: Player K I II III B 18 4 6 B B 6 2 13 B @ 11 5 17 7 6 12 0 Player L

1 2 3 4

1 IV 4 C C 7 C C 3 A 2

7. Convert the following game problem into a linear programming problem and solve:

2 Player 5 7 4 10 4 6 2

Player A

B 3 2 95 0

8. For the following two-person zero-sum game, ﬁnd the strategy of each player and the value of the game: 2 A

3 7 5 5 3

B 3 4 2

8 4 6 2

9. Solve the following 3  3 game: 2

7 49 5

1 1 7

3 7 15 6

10. Consider the following 2  2 game:

4 6

7 5

(i) Does it have a saddle point? (ii) Is it correct to state that the value of game G will satisfy 5\G\6? (iii) Determine the frequency of optimum strategies by the arithmetic method and ﬁnd the value of the game.

366

13

Game Theory

11. Solve the following game:

I II III IV

2I II III IV3 3 2 4 0 63 4 2 47 6 7 44 2 4 05 0 4 0 8

12. Solve the following game whose payoff matrix is given by the graphical method:

A1 A2 A3

2 B1 B2 B3 4 2 3 4 1 2 0 2 1 2

B4 3 1 1 5 0

13. Obtain the optimal strategies for both persons and the value of the game for a two-person zero-sum game whose payoff matrix is as follows: 2 Player B 3 1 3 6 3 5 7 7 6 6 1 6 7 7 6 6 4 1 7 7 6 4 2 2 5 5 0

Player A

14. Use the graphical method to solve the following game:

Player B

2 4

Player A  2 3 2 3 2 6

Chapter 14

Project Management

14.1

Objectives

The objectives of this chapter are to: • Discuss the importance of PERT and CPM techniques for project management. • Distinguish between PERT and CPM network techniques. • Discuss the different phases of any project and various activities to be done during these phases. • Draw the network diagrams with single and three-time estimates of activities involved in a project. • Determine the critical path and floats associated with non-critical activities and events along with the total project completion time. • Determine the probability of completing a project within the scheduled time. • Establish a time-cost trade-off for completion of a project.

14.2

Introduction

A project is a well-deﬁned set of activities, jobs or tasks, all of which must be completed to ﬁnish the project. Construction of a highway or power plant, production and marketing of a new product and research & development work are the examples of projects. Such projects involve a large number of inter-related activities (or tasks) which must be completed in a speciﬁed time and in a speciﬁed sequence (or order) and require resources such as personnel, money, materials, facilities and/ or space. The main objective before starting any project is to schedule the required activities in an efﬁcient manner so as to: (i) Complete it on or before a speciﬁed time limit (ii) Minimize the total time © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_14

367

368

(iii) (iv) (v) (vi)

14

Minimize Minimize Minimize Minimize

the the the the

Project Management

time for a prescribed cost cost for a speciﬁed time total cost idle resources.

Therefore, before starting any project, it is essential to prepare a plan for scheduling and controlling the various activities involved in the project. The techniques of operations research (OR) used for planning, scheduling and controlling large & complex projects are often referred to as network analysis, network planning or network scheduling techniques. In all these techniques, a project is broken down into various activities which are arranged in a logical sequence in the form of a network. This approach helps managers to visualize a project as a number of tasks which can easily be deﬁned in terms of duration, cost and starting time. The sequence of activities is also deﬁned. There are two basic planning and control techniques that utilize a network to complete a predetermined project or schedule. These are PERT (program evaluation and review technique) and CPM (critical path method). A PERT network was developed during 1956–1958 by a research team of the US Navy’s Polaris Nuclear Submarine Missile development project. Since 1958, this technique has been used to plan almost all types of projects. At the same time but independently, CPM was developed jointly by two companies: E. I. DuPont de Nemours and Company and Remington Rand Corporation. Other network techniques are PEP (performance evaluation programme), LCES (least cost estimating and scheduling), SCANS (scheduling and control by automated network system), etc.

14.3

Phases of Project Management

The work involved in a project can be divided into three phases corresponding to the management functions of planning, scheduling and control. Planning This phase involves setting the objectives of the project and the assumptions to be made. It also involves the listing of tasks or jobs that must be performed to complete a project under consideration. In this phase, the men, machines and materials required for the project in addition to the estimates of costs and duration of the various activities of the project are also determined. Scheduling This consists of organizing the activities according to the precedence order and to determine

14.3

Phases of Project Management

369

(i) The starting and ﬁnishing times for each activity (ii) The critical path for which the activities require special attention (iii) The slack and float for the non-critical paths. Control This phase is exercised after the planning and scheduling and involves the following: (i) (ii) (iii) (iv)

14.4

Making periodical progress reports Reviewing the progress Analysing the status of the project Management decisions regarding updating, crashing and resource allocation, etc.

The network analysis: (i) Shows the inter-relationships of all jobs in the project (ii) Gives a clear picture, other than a typical bar chart, of the relationship controlling the order of performance of various activities (iii) Helps in communication of ideas. The pictorial approach helps to clarify the verbal instructions (iv) Provides time schedules containing much more information than other methods such as bar charts (v) Identiﬁes jobs which are critical for a project completion date (vi) Permits an accurate forecast of resource requirements (vii) Provides a method of resource allocation to meet the limiting condition and to maintain or minimize the overall costs (viii) Integrates all elements of a programme to whatever detail is desired by management (ix) Relates time to costs, which allows a money value to be placed on proposed changes.

14.5

Basic Components

There are two basic components in a network. These are: (i) Event/node (ii) Activity.

370

14

Project Management

Event/node An event/node is a particular instant in time showing the end or beginning of one or more activities. It is a point of accomplishment or decision. The starting and end points of an activity are thus described by two events, usually called the tail event and head event. An event is generally represented by a circle, rectangle, hexagon or some other geometric shape. These geometric shapes are numbered for distinguishing one activity from another. The occurrence of an event indicates that the work has been accomplished up to that point (See Fig. 14.1). Merge and burst event It is necessary for an event to be the ending event of at least one activity, but it can also be the ending event of two or more activities. Then it is deﬁned as a merge event. If an event is the beginning event of two or more activities, it is deﬁned as a burst event. See Fig. 14.2. Activity An activity is a task or item of work to be done that consumes time, effort, money or other resources. Activities are represented by arrows. Activities are identiﬁed by the numbers of their starting (tail) event and ending (head) event. Generally, an ordered pair (i, j) represents an activity where events i and j represent the starting and ending of the activity respectively. It can also be denoted by i − j. Sometimes, activities are denoted by capital letters (Fig. 14.3). The activities can further be classiﬁed into different categories: (i) Predecessor activity: An activity which must be completed before one or more other activities start is known as a predecessor activity. (ii) Successor activity: An activity which started immediately after the completion of one or more other activities is known as a successor activity. (iii) Dummy activity: In connecting events by activities showing their interdependencies, very often a situation arises where a certain event j cannot occur until another event i has taken place, but the activity connecting i and j does not involve any time or expenditure of other resources. In such a case, the activity is called a dummy activity. It is depicted by a dotted line in the network diagram.

Activity Tail event / node Fig. 14.1 Head and trail events

14.5

Basic Components

371

(a)

(b)

Merge event

Burst event

Fig. 14.2 a Merge event. b Burst event

Activity

Starting event (tail event)

Fig. 14.3 Activity

Let us consider an example of a car taken to a garage for cleaning. Both the inside and the outside of the car are to be cleaned before it is taken away from the garage. The events can be stated as follows: Event Event Event Event Event Event

1: 2: 3: 4: 5: 6:

Starting of car from house Parking of car in garage Completion of outside cleaning Completion of inside cleaning Taking of car from garage Parking of car in house.

The network diagram for the problem is given in Fig. 14.4. It is assumed that the inside cleaning and outside cleaning can be done concurrently by two garage assistants. Activities B and C represent these cleaning operations. What do activities D and E stand for? Their time consumptions are zero, but they express the condition that events 3 and 4 must occur before event 5 can take place. Activities D and E are called dummy activities.

3

D

B

F

A 1

5

2 4

C Fig. 14.4 Project network

E

6

372

14

Project Management

Network A network is the graphical representation of logically and sequentially connected arrows and nodes representing the activities and events of a project. Networks are also called arrow diagrams. Path An unbroken chain of activity arrows connecting the initial event to some other event is called a path.

14.6

Common Errors

There are three common errors in a network construction: looping, dangling and redundancy. Looping (cycling) In a network diagram, a looping error is also known as a cycling error. The drawing of an endless loop in a network is known as an error of looping. A looping network is shown in Fig. 14.5. Dangling To disconnect an activity before the completion of all the activities in a network diagram is known as dangling. It should be avoided. In that case, a dummy activity is introduced in order to maintain the continuity of the system (see Fig. 14.6). Redundancy If a dummy activity is the only activity emanating from an event and it can be eliminated, this is known as redundancy (see Fig. 14.7).

1

2

3

4 Fig. 14.5 Looping in project network

5

14.7

Rules for Network Construction

373

5 2

1

4

7 6

3

Fig. 14.6 Dangling in project network

5 2

4

6

7

1 3 Fig. 14.7 Redundancy in project network

14.7

Rules for Network Construction

For the construction of a network, certain rules are generally followed: (i) Each activity is represented by one and only one arrow. (ii) Crossing of arrows and curved arrows should be avoided; only straight arrows are to be used. (iii) Each activity must be identiﬁed by its starting and ending nodes. (iv) No event can occur until every activity preceding it has been completed. (v) An event cannot occur twice; i.e. there must be no loops. (vi) An activity succeeding an event cannot be started until that event has occurred. (vii) Events are numbered to identify an activity uniquely. The number of the tail event (starting event) should be lower than that of the head (ending) event of an activity. (viii) Between any pair of nodes (events), there should be one and only one activity. However, more than one activity may emanate from a node or terminate at a node. (ix) Dummy activities should be introduced only if it is extremely necessary. (x) The network has only one entry point, called the starting event, and one terminal point, called the end or terminal event.

374

14

14.8

Project Management

Numbering of Events

After the network is drawn in a logical sequence, every event is assigned by a number. The number sequence must reflect the flow of the network. In numbering the events, Fulkerson’s rules (for D. R. Fulkerson) are used. These rules are as follows: (i) The event number should be unique. (ii) Event numbering should be carried out on a sequential basis from left to right. (iii) An initial event is one which has all outgoing arrows with no incoming arrow. In any network, there will be one such event. Number it as 1 (one). (iv) Delete all arrows emerging from event 1 (one). This will create at least one more initial event. (v) Number these initial events as 2, 3, …, etc. (vi) Delete all emerging arrows from these numbered events, which will create new initial events. (vii) Repeat steps (v) and (vi) until the last event is obtained which has no arrows emerging from it. Number the last event. Example 1 Construct a network of a project whose activities and their precedence relationships are given below. Then number the events (Fig. 14.8). Activity

A

B

C

D

E

F

G

H

I

Immediate predecessor

A

A

D

B, C, E

F

D

G, H

G

7

I

8

2

B

4

A C 1

F

5 D

E 3

H

Fig. 14.8 Construction of project network

6

14.9

14.9

Critical Path Analysis

375

Critical Path Analysis

Once the network of a project is constructed, the time analysis of the network becomes essential for planning various activities of the project. The main objective of the time analysis is to prepare a planning schedule of the project. The planning schedule should include the following factors: (i) Total completion time for the project (ii) Earliest time when each activity can start (iii) Latest time when each activity can be started without delaying the total project (iv) Float for each activity, i.e. the duration of time by which the completion of an activity can be delayed without delaying the total project completion (v) Identiﬁcation of critical activities and critical path. Notation The following notation is used in this analysis: Ei = Earliest occurrence time of event i, i.e. the earliest time at which the event i can occur without affecting the total project duration Li = Latest allowable occurrence time of event i. It is the latest allowable time at which an event can occur without affecting the total project duration tij = Duration of activity (i, j) ESij = Earliest starting time of activity (i, j) LSij = Latest starting time of activity (i, j) EFij = Earliest ﬁnishing time of activity (i, j) LFij = Latest ﬁnishing time of activity (i, j). The critical path calculations are done in the following two ways: (a) Forward pass calculations method (b) Backward pass calculations method. Forward pass calculations method In this method, calculation begins from the initial event, proceeds through the events in an increasing order of event numbers and ends at the ﬁnal event of the network. At each node (event), the earliest starting and ﬁnishing times are calculated for each activity. The method may be summarized as follows: Step 1 Set E1 = 0, i = 1. Step 2 Calculate the earliest starting time ESij for each activity that begins at event i, i.e. ESij = Ei for all activities (i, j) that start at node i. Step 3 Calculate the earliest ﬁnishing time EFij of each activity that begins at event i by adding the earliest starting time of the activity with the duration of the activity. Thus, EFij = ESij + tij = Ei + tij.

376

14

Project Management

Step 4 Go to the next event (node), say event j(j > i), and compute the earliest occurrence time for event j. This is the maximum of the earliest ﬁnishing times of all ending  activities   at that  event, i.e. Ej ¼ Max EFij ¼ Max Ei þ tij for all immediate predecessor activities. i

i

i

i

Step 5 If j = n (ﬁnal event number), then the earliest ﬁnishing time for the project is given  by   En ¼ Max EFij ¼ Max Ei þ tij for all terminal activities. Backward pass calculations method In this method, calculations begin from the terminal event, proceed through the events in a decreasing order of event numbers and end at the initial event of the network. At each node (event), the latest starting and ﬁnishing times are calculated for each activity. The method may be summarized as follows: Step 1 Set Ln = En, j = n. Step 2 Calculate the latest ﬁnishing time LFij for each activity that ends at event j, i.e. LFij= Lj for all activities (i, j) that end at node j. Step 3 Calculate the latest starting time LSij of each activity that ends at event j by subtracting the duration of each activity from the latest ﬁnishing time of the activity. Thus, LSij ¼ LFij  tij ¼ Lj  tij : Step 4 Proceed backward to the node in the sequence that decreases j by 1. Also compute the latest occurrence time of node i (i < j). This is the minimum of the latest times  starting   of allactivities starting from that event, i.e. Li ¼ Min LSij ¼ Min Lj  tij for all immediate successor activities. j

j

Step 5 If i = 1 (initial  node), then   L1 ¼ Min LSij ¼ Min Lj  tij for all initial activities. j

j

Determination of floats and slack times When the network is completely drawn and properly labelled, and the earliest and latest event times are computed, the next objective is to determine the floats of each activity and the slack time of each event. The float of an activity is the amount of time by which it is possible to delay its completion time without affecting the total project completion time. There are three types of activity floats: (i) Total float (ii) Free float (iii) Independent float.

14.9

Critical Path Analysis

377

Total float The total float of an activity represents the amount of time by which an activity can be delayed without a delay in the project completion time. Mathematically, the total float of an activity (i, j) is the difference between the latest start time and earliest start time of that activity (or the difference between the earliest ﬁnish time and latest ﬁnish time). Hence, the total float for an activity (i, j) is denoted by TFij and is computed by the following formula:   TFij ¼ LSij ESij or TFij ¼ LFij EFij or TFij ¼ Lj  Ei þ tij as Lj ¼ LFij and EFij ¼ Ei þ tij : Free float Sometimes we may need to know how much an activity’s completion time may be delayed without causing any delay in its immediate successor activities. This amount of float is called a free float. Mathematically, the free float for an activity (i, j) is denoted by FFij and is computed by FFij ¼ Ej Ei tij as TFij ¼ Li Ei tij and Li  Ej ) TFij  Ej Ei tij i:e: TFij  FFij : Hence, for all activities, free float can take values from zero up to total float, but it does not exceed total float. Again, free float is very useful for rescheduling activities with a minimum disruption of earlier plans. Independent float In some cases, the delay in the completion of an activity affects neither its predecessor nor its successor activities. This amount of delay is called an independent float. Mathematically, independence of an activity (i, j) denoted by IFij is computed by the following formula: IFij ¼ Ej Li  tij : A negative independent float is always taken as zero. Event slack or event float The slack of an event is the difference between its latest time and its earliest time. Hence, for an event i, slack ¼ Li  Ei :

378

14

Project Management

Critical event: An event is said to be critical if its slack is zero, i.e. Li= Ei for the ith event. Critical activity: An activity is critical if its total float is zero, i.e. LSij = ESij or LFij = EFij for an activity (i, j). Otherwise, an activity is called non-critical. Critical path The critical path is the continuous chain or sequence of critical activities in a network diagram path from starting event to ending event. It is shown by a dark line or double lines to make a distinction from other, non-critical paths. The length of the critical path is the sum of the individual times of all critical activities lying on it and deﬁnes the minimum time required to complete the project. The critical path on a network diagram can be identiﬁed as follows: (i) For all activities (i, j) lying on the critical path, the E values and L values for tail and head events are equal, i.e. Ej = Lj and Ei = Li. (ii) On the critical path, Ej − Ei = Lj − Li = tij. Main features of the critical path The critical path has two main features: (i) If the project has to be shortened, then some of the activities on that path must be shortened. The application of additional resources on other activities will not give the desired results unless that critical path is shortened ﬁrst. (ii) The variation in actual performance from the expected activity duration time will be completely reflected in one-to-one fashion in the anticipated completion of the whole project. Example 2 Determine the critical path and minimum time of completion of the project whose network diagram is shown in Fig. 14.9. Also ﬁnd the different floats.

2

G(19)

5 H(4)

C(20)

1

B(8)

6

4 F(18) D(16)

A(23)

E(24) 3

Fig. 14.9 Project network

I(10)

7

14.9

Critical Path Analysis

379

Solution Forward pass calculations At At At At

node 1: Set E1 = 0 node 2: E2 = E1 + t12 = 0 + 20 = 20 node 3: E3 = E1 + t13 = 0 + 23 = 23 node 4: E4 ¼ MaxfEi þ ti4 g ¼ MaxfE1 þ t14 ; E3 þ t34 g ¼ Maxf0 þ 8; 23 þ i¼1;3

16g ¼ 39 At node 5: E5 ¼ MaxfEi þ ti5 g ¼ MaxfE2 þ t25 ; E4 þ t45 g ¼ Maxf20 þ 19; 39 þ i¼2;4

0g ¼ 39 At node 6: E6 ¼ MaxfEi þ ti6 g ¼ MaxfE4 þ t46 ; E5 þ t56 g ¼ Maxf39 þ 18; 39 þ i¼4;5

0g ¼ 57 At node 7: E7 ¼ Max fEi þ ti7 g ¼ MaxfE3 þ t37 ; E5 þ t57 ; E6 þ t67 g i¼3;5;6

¼ Maxf23 þ 24; 39 þ 4; 57 þ 10g ¼ 67 Backward pass calculations At node 7: Set L7 = E7 = 67 At node 6: L6 = L7 − t67 = 67 − 10  = 57 At node 5: L5 ¼ Min Lj  t5j ¼ MinfL6  t56 ; t7  t57 g ¼ Minf57  0; 67 j¼6;7

4g ¼ 57   At node 4: L4 ¼ Min Lj  t4j ¼ MinfL5  t45 ; L6  t46 g ¼ Minf57  0; 57 j¼5;6

18g ¼ 39   At node 3: L3 ¼ Min Lj  t3j ¼ MinfL4  t34 ; L7  t37 g ¼ Minf39  16; 67 j¼4;7

24g ¼ 23 At node 2: L2 = L5 − t 25 = 57 −  19 = 38 At node 1: L1 ¼ Min Lj  t1j ¼ MinfL2  t12 ; L3  t13 ; L4  t14 g j¼2;3;4

¼ Minf38  20; 23  23; 39  8g ¼ 0 To ﬁnd the critical activities and different floats, we construct the table (cf. Table 14.1). Table 14.1 Computation of critical activity and different floats Activity

Duration of activity

Earliest time

Latest time

Start (Ei)

Finish (Ei+ tij)

Start (Lj − tij)

Float Finish (Lj)

Total Lj − Ei − tij

Free Ej − Ei − tij

Independent Ej − Li − tij

(1, 2)

20

0

20

18

38

18

0

0

(1, 3)

23

0

23

0

23

0

0

0

(1, 4)

8

0

8

31

39

31

31

31

(2, 5)

19

20

39

38

57

18

0

0

(3, 4)

16

23

39

23

39

0

0

0

(3, 7)

24

23

47

43

67

20

20

20

Critical activity

(1, 3)

(3, 4)

(continued)

380

14

Project Management

Table 14.1 (continued) Activity

Duration of activity

Earliest time

Latest time

Start (Ei)

Finish (Ei+ tij)

Start (Lj − tij)

Finish (Lj)

Total Lj − Ei − tij

Float Free Ej − Ei − tij

Independent Ej − Li − tij

(4, 5)

0

39

39

57

57

18

0

(4, 6)

18

39

57

39

57

0

0

0

(5, 6)

0

39

39

57

57

18

18

0

(5, 7)

4

39

43

63

67

24

24

46

(6, 7)

10

57

67

57

67

0

0

0

Critical activity

0 (4, 6)

(6, 7)

From the preceding table, it is clear that the critical activities (zero total float) are (1, 3), (3, 4), (4, 6) and (6, 7). Hence, the critical path is 1–3−4–6–7, and the duration of the project is 67 time units (as E7 = L7 = 67).

14.10

PERT Analysis

Time estimates It is very difﬁcult to estimate the time required for the execution of each activity or because of various uncertainties. Taking the uncertainties into account, three types of time estimates are generally obtained. The PERT system is based on these three time estimates of the performance time of an activity: (i) Optimistic time (to): This is the estimate of the shortest possible time in which an activity can be completed under ideal conditions. (ii) Pessimistic time (tp): This is the maximum time which is required to perform the activity under extremely bad conditions. However, such conditions do not include labour strikes or acts of nature (like floods, earthquakes, tornados, etc.). (iii) Most likely time (tm): This is the estimate of the normal time which an activity would require. This time estimate lies between the optimistic and pessimistic time estimates. Statistically, it is the modal value of the duration of the activity. The range speciﬁed by the optimistic time (to) and pessimistic time (tp) estimates supposedly must close every possible estimate of duration of the activity. The most likely time (tm) estimate may not coincide with the midpoint tmid = (to + tp)/2 and may occur to its left or right, as shown in Fig. 14.10. Keeping in mind the above-described time properties, it may be justiﬁed to assume that the duration of each activity may follow a beta (b)-distribution with its unimodal point occurring at tm and its end point at to and tp.

14.10

t0

PERT Analysis

te

tm

381

tp

t0

tm

tmid

tp

t0

tm

tp

Fig. 14.10 Time estimates of an activity

The expected or mean value of an activity duration can be approximated by a linear combination of the three time estimates or by the weighted average of three time estimates to, tp and tm, i.e. te = (to + 4tm + tp)/6. Again, to determine the activity duration variance in PERT, the unimodal property of the b-distribution is used. However, in PERT, the standard deviation is expressed as r¼

t  t 2  1 p o tp  to or variance r2 ¼ : 6 6

Note that in PERT analysis, the b-distribution is assumed as it is unimodal, has non-negative end points and is approximately symmetric. Probability of meeting the schedule time After identifying the critical path and the occurrence time of all activities, there arises a question: What is the probability that a particular event will occur on or before the schedule date? This particular event may be any event in the network. Let us recall that the expected time of an activity is the weighted average of the three time estimates to, tp and tm,   i:e: te ¼ to þ 4tm þ tp =6: The probability that the activity (i, j) will be completed in time te is 0.5; i.e. the chance of completion of that activity is 50%. In the frequency distribution curve, for the activity (i, j) the vertical line through te will divide the area under the curve into two equal parts, as shown in Fig. 14.11. For completing the activity in any other time tk, the probability will be p¼

Area under AEK : Area under AEB

A project consists of a number of activities. All activities, as we know, are independent random variables, and hence the length of the project up to a certain event through a certain path is also a random variable. But the point of difference is that

Probability density function

382

14

Project Management

E

K

A

B Activity duration

t0

te

tk

tp

Fig. 14.11 Time distribution curve of an activity

the expected project length Te does not have the same probability distribution as the expected activity time te. While a b-distribution curve approximately represents the activity time probability distribution, the project expected time Te follows approximately a standard normal distribution. This standard normal distribution curve has an area equal to unity and standard deviation 1 and is symmetrical about the mean, as shown in Fig. 14.12. The probability of completing a project in scheduling time Ts is given by probðTs Þ ¼

Area under ACS : Area under ACB

Probability prob(Ts) depends upon the location of Ts. Taking Te as a reference point, the distance Ts − Te can be expressed in terms of the standard deviation for a network, calculated as

Probability density

C

S

A

B Project duration

Te

Ts

Fig. 14.12 Time distribution of a project network

14.10

PERT Analysis

383

Standard deviation for a network ¼ re pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ sum of the variances along the critical path;

i.e. d for a network ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ t t 2 P 2 rij , where r2ij for an activity (i, j) ¼ p 6 o .

Since the standard deviation for a standard normal curve is unity, the standard deviation re calculated above is used as a scale factor for calculation of the normal deviate, The normal deviate D ¼

T s  Te : re

Hence, the probability of completing the project by scheduling time (Ts) is given by e prob(Z  D), where D ¼ TsrT and Z is the standard normal variate. e The values of the probabilities for a normal distribution curve corresponding to the different values of the normal deviate are available from the table of the standard normal curve. Example 3 A small project is composed of seven activities whose time estimates (in weeks) are listed in the Table 14.2. (a) Draw the project network. (b) Find the expected duration and variance of each activity. (c) Calculate the earliest and latest occurrence time for each event and the expected project length. (d) Calculate the variance and standard deviation of the project length. (e) What is the probability that the project will be completed: (i) At least 4 weeks earlier than expected? (ii) Not more than 4 weeks later than expected? (f) If the project due date is 19 weeks, what is the probability of meeting the due date? (g) Find also the schedule time at which the project will be completed with a probability 0.90. Table 14.2 Activities and their different time estimates Activity

(1, 2)

(1, 3)

(1, 4)

(2, 5)

(3, 5)

(4, 6)

(5, 6)

t0 tm tp

1 1 7

1 4 7

2 2 8

1 1 1

2 5 14

2 5 8

3 6 15

384

14

Project Management

Solution (a) The project network diagram is shown in Fig. 14.13. (b) The expected time and variance of each activity is computed and displayed in the Table 14.3. Table 14.3 Computation of expected time and variance of each activity Activity

to

tm

tp

te ¼

(1, (1, (1, (2, (3, (4, (5,

1 1 2 1 2 2 3

1 4 2 1 5 5 6

7 7 8 1 14 8 15

2 4 3 1 6 5 7

2) 3) 4) 5) 5) 6) 6)

to þ 4tm þ tp 6

r2 ¼

tp to 2 6

1 1 1 0 4 1 4

Let Ei be the earliest occurrence time of event i. Set E1 = 0 E2 = E1 + t12 = 0 + 2 = 2 E3 = E1 + t13 = 0 + 4 = 4 E4 = E1 + t14 = 0 + 3 = 3

E5 ¼ MaxfEi þ ti5 g ¼ MaxfE2 þ t25 ; E3 þ t35 g ¼ Maxf2 þ 1; 4 þ 6g ¼ 10 i¼2;3

E6 ¼ MaxfEi þ ti6 g ¼ MaxfE4 þ t46 ; E5 þ t56 g ¼ Maxf3 þ 5; 10 þ 7g ¼ 17 i¼4;5

2 5

1 3

6 4 Fig. 14.13 Project network diagram

14.10

PERT Analysis

385

(c) Backward pass calculations Let Lj be the latest Set L6 = E6 = 17 L5 = L6 − t56 = 17 L4 = L6 − t46 = 17 L3 = L5 − t35 = 10 L2 = L5 − t25 = 10

occurrence time of event j. − − − −

7 5 6 1

= = = =

10 12 4 9

  L1 ¼ Min Lj  t1j ¼ MinfL2  t12 ; L3  t13 ; L4  t14 g j¼2;3;4

¼ Minf9  2; 4  4; 12  3g ¼ 0: From these calculations, it is seen that E1 ¼ L1 ; E3 ¼ L3 ; E5 ¼ L5 ; E6 ¼ L6 : Hence, the critical events are 1, 3, 5, 6 and the critical path is 1–3–5–6. Also, the expected project length = E6 = L6 = 17 weeks. (d) The variance of the project length is given by r2e ¼ 1 þ 4 þ 4 ¼ 9

or; re ¼ 3:

(e) The standard normal deviate is given by D¼

schedule time  expected time of completion pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : variance or standard deviation

(i) Now, the probability that the project will be completed at least 4 weeks earlier than expected is given by probðZ  DÞ where D ¼

ð17  4Þ  17 4 ¼ ¼ 1:33 ðapprox:Þ 3 3

¼ probðZ   1:33Þ ¼ 0:5  Probð0\Z  1:33Þ ¼ 0:5  uð1:33Þ ¼ 0:5  0:4082 ¼ 0:0918:

½From the table of the area under the standard normal curve

386

14

Project Management

(ii) Again, the probability that the project will be completed not more than 4 weeks later than expected = the probability that the project will be completed within 17 + 4 weeks, i.e. within 21 weeks:

probðZ  DÞ; where D ¼

21  17 4 ¼ ¼ 1:33 ðapprox:Þ; 3 3

¼ probðZ  1:33Þ ¼ 0:5 þ uð1:33Þ ¼ 0:5 þ 0:4082 ¼ 0:9082: (f) When the due date is 19 weeks, D ¼ 1917 ¼ 23 ¼ 0:67. 3 Then the probability of meeting the due date is given by probðZ  0:67Þ ¼ 0:5 þ uð0:67Þ ¼ 0:5 þ 0:2514 ¼ 0:7514: (g) Since the probability for the completion of the project is 0.90, e prob(Z  D) = 0.90, where D ¼ Ts T and Ts is the schedule time. r

As prob(Z  1.29) = 0.90, ∴ D = 1.29, which implies

14.11

Ts 17 3

¼ 1:29, i.e. Ts = 17 + 3  1.29, i.e. Ts = 20.87.

Difference Between PERT and CPM

The differences between PERT and CPM are listed as follows: PERT

CPM

1. This technique was developed in connection with R&D (research and development) works; therefore, it had to cope with the uncertainty which is associated with R&D activities. In this case, the total project duration is regarded as a random variable. As a result, multiple time estimates are made to calculate the probability of completing the project within the schedule time. Therefore, it is a probabilistic model.

1. This technique was developed in connection with a construction and maintenance project which consists of routine tasks or jobs whose resource requirements and duration are known with certainty. Therefore, it is basically a deterministic model.

2. It is used for projects involving activities of a non-repetitive nature.

2. It is used for projects involving activities of a repetitive nature.

3. It is an event-oriented technique, because the results of analyses are expressed in terms of events or distinct points in time indicative of progress.

3. It is an activity oriented technique, as the results of calculations are considered in terms of activities.

4. It incorporates statistical analysis and thereby enables the determination of probabilities concerning the time by

4. It does not incorporate statistical analysis in determining the estimates because the time is precise and known.

(continued)

14.11

Difference Between PERT and CPM

387

(continued) PERT

CPM

which each activity and the entire project would be completed. 5. It serves as a useful control device, as it assists management in controlling a project by calling attention through constant review to delays in activities which might lead to a delay in the project completion date.

14.12

5. It is difﬁcult to use this technique as a controlling device, because one must repeat the entire evaluation of the project each time changes are introduced into the network.

In this section, the costs of resources consumed by activities are taken into consideration. The project completion time can be reduced (crashing) by reducing the normal completion time of critical activities. The reduction in normal time of completion will increase the total budget of the project. However, the decision-maker will always look for a trade-off between the total cost of the project and the total time required to complete it. Project cost In order to include the cost aspect in project scheduling we have to ﬁnd the cost duration relationships for various activities in the project. The total cost of any project comprises direct and indirect costs. Direct cost This cost is directly dependent upon the amount of resources in the execution of individual activities such as manpower loading, material consumed, etc. The direct cost increases if the activity duration is to be reduced. Indirect cost This cost is associated with expenditures which cannot be allocated to individual activities of the project. This cost may include managerial services, loss of revenue, ﬁxed overheads, etc. The indirect cost is computed on a per-day, per-week or per-month basis. This cost decreases if the activity duration is to be reduced. The network diagram can be used to identify the activities whose duration should be shortened so that the completion time of the project can be shortened in the most economic manner. The process of reducing the activity duration by putting in extra effort is called crashing the activity. The crash time (Tc) represents the minimum activity duration time that is possible; any attempts to further crash would only raise the activity cost without reducing the time. The activity cost corresponding to the crash time is called the crash cost (Cc), which is the minimum direct cost required to achieve the crash performance time.

388

14

Project Management

Fig. 14.14 Direct cost w.r.t. time

Direct Cost

A

B

Time The normal cost (Cn) is equal to the absolute minimum of the direct cost required to perform an activity. The corresponding time duration taken by an activity is known as the normal time (Tn). The direct cost curve (from the relationship of direct cost and time) is shown in Fig. 14.14. The point B denotes the normal time for completion of an activity, whereas point A denotes the crash time which indicates the least duration in which activity can be completed. The cost curve has a non-linear and asymptotic nature. But, for simplicity, it can be approximated by a straight line whose slope (in magnitude) is given by Cost slope =

Crash cost  Normal cost Cc  Cn ¼ : Normal time  Crash time Tn  Tc

It is also called the crash cost slope or crash rate of increase in the cost of performing the activity per unit reduction in time and is also known as the time-cost trade-off. It varies from activity to activity. After assessing the direct and indirect project costs, the total project cost, which is the sum of the direct and indirect costs, can be determined. Time-cost optimization algorithm/time-cost trade-off procedure The following are the steps involved in the project crashing. Step 1 Considering normal times of all activities, identify the critical activities and ﬁnd the critical path. Step 2 Calculate the cost slope for different activities and rank the activities in ascending order of cost slope.

14.12

389

Step 3 Crash the activities on the critical path as per ranking of cost slopes of different activities; i.e. an activity having a lower cost slope would be crashed ﬁrst to the maximum extent possible. (For the crashing of a lower cost slope, i.e. for the reduction of activity duration time, the direct cost of the project would be increased very slowly.) Step 4 Due to the reduction of the critical path duration by crashing in Step 3, other paths also become critical; i.e. we get parallel critical paths. In such cases, the project duration can be reduced by crashing of activities simultaneously in the parallel critical paths. Step 5 Repeat the process until all the critical activities are fully crashed or no further crashing is possible. In the case of indirect cost, the process of crashing is repeated until the total cost is minimum, beyond which it may increase. The minimum cost is called the optimum project cost and the corresponding time, the optimum project time. Example 4 The following table shows activities, their normal time, cost and crash time and cost for a project: Activity

Normal time (days)

Cost (Rs.)

Crash time (days)

Cost (Rs.)

(1, (1, (2, (2, (3, (3, (4, (5,

6 8 4 3 Dummy 6 10 3

1400 2000 1100 800 _ 900 2500 500

4 5 2 2 _ 3 6 2

1900 2800 1500 1400 _ 1600 3500 800

2) 3) 3) 4) 4) 5) 6) 6)

The indirect cost for the project is Rs. 300 per day. (i) (ii) (iii) (iv)

Draw the network of the project. What are the normal duration and associated cost of the project? What will be the least project duration and the corresponding cost? Find the optimum duration and minimum project cost.

Solution (i) The network of the project is shown in Fig. 14.15. (ii) Using the normal time duration of each activity, the earliest and latest occurrence times at various nodes are computed and displayed in Fig. 14.15 of the network.

390

14

L2 = 6 E2 = 6

2

6(4)

L4 = 10 E4 = 10

3(2)

• •

4

Crash time are displayed in bracket Double line shows the critical path

10(6)

4(2)

L1 = 0 1 E1 = 0

Project Management

6 8(5)

E6 = 20 L6 = 20

3(2)

5

3

6(3)

E5 = 16 L5 = 17

E3 = 10 L3 = 10

Fig. 14.15 Project network with time duration 20 days

L4 = 9 E4 = 9

3(2)

L2 = 6 E2 = 6

2

6(4)

10(6)

4

3(2)

1 L1 = 0 E1 = 0

6

E6 = 19 L6 = 19

6

E6 = 18 L6 = 18

3(2)

8(5) E3=9 L3=9

5

3

E5 = 15 L5 = 16

6(3)

Fig. 14.16 Project network with time duration 19 days

L2 = 5 E2 = 5

L4 = 8 E4 = 8

3(2)

2

4 10(6)

5(4)

3(2)

1 L1 = 0 E1 = 0

8(5)

3(2)

3 E3=8 L3=8

5 6(3)

Fig. 14.17 Project network with time duration 18 days

E5 = 14 L5 = 15

14.12

Project Time-Cost Trade-Off L1 = 5 E1 = 5

3(2)

2

4

5(4)

L1 = 0 E1 = 0

391

9(6) L1 = 8 E1 = 8

3(2)

1

6

E6 = 17 L6 = 17

6

E6 = 16 L6 = 16

3(2)

8(5)

5

3

L1 = 8 E1 = 8

L1 = 14 E1 = 14

6(3)

Fig. 14.18 Project network with time duration 17 days

L1 = 5 E1 = 5

L1 = 0 E1 = 0

L1 = 8 E1 = 8

3(2)

2

4

5(4)

8(6)

3(2)

1 8(5)

3(2) 5

3

L1 = 8 E1 = 8

5(3)

L1 = 13 E1 = 13

Fig. 14.19 Project network with time duration 16 days

L1 = 5 E1 = 5

L1 = 0 E1 = 0

3(2) 2

4

5(4)

7(6)

L1 = 8 E1 = 8

3(2)

1

6

E6 = 15 L6 = 15

3(2)

8(5) L1 = 8 E1 = 8

3

5 4(3)

L1 = 12 E1 = 12

Fig. 14.20 Project network with time duration 15 days

From the network, it is seen that the L values and E values at nodes 1, 2, 3, 4, 6 are same. This means that the critical path is 1–2–3–4–6 and the normal duration of the project is 20 days.

392

14

L1 = 5 E1 = 5

L1 = 8 E1 = 8

3(2)

2

Project Management

4 6(6)

5(4) L1 = 0 E1 = 0

3(2)

1

6 8(5)

E6 = 14 L6 = 14

3(2) 5

3

L1 = 8 E1 = 8

3(3)

L1 = 11 E1 = 11

Fig. 14.21 Project network with time duration 14 days L1 = 4 E1 = 4

L1 = 7 E1 = 7

3(2)

2

4 6(6)

4(4)

L1 = 0 E1 = 0

3(2)

1

6 7(5)

E6 = 13 L6 = 13

3(2) L1 = 7 E1 = 7

5

3

L1 = 10 E1 = 10

3(3)

Fig. 14.22 Project network with time duration 13 days

The associated cost of the project are as follows: ¼ Direct normal cost þ indirect cost for 20 days ¼ Rs: ½ð1400 þ 2000 þ 1100 þ 800 þ 900 þ 2500 þ 500Þ þ 20  300 ¼ Rs: ½9200 þ 6000 ¼ Rs: 15;200: (iii) The cost slopes of different activities are computed by using the formula Cost slope ¼

Crash cost  Normal cost Normal time  Crash time

and these are shown in the following table: Activity Slope

(1, 2) 250

(1, 3) 267

(2, 3) 200

(2, 4) 600

(3, 5) 233

(4, 6) 250

(5, 6) 300

(1, 2) (1)

(4, 6) (1)

Figure 14.15

Figure 14.16

Figure 14.17

Figure 14.18

Figure 14.19

Figure 14.20

Figure 14.21

Figure 14.22

1–2–3–4–6

1–2–3–4–6, 1–2–4–6

1–2–3–4–6, 1–2–4–6 1–3–4–6

1–2–3–4–6, 1–2–4–6, 1–3–4–6, 1–3–5–6, 1–2–3–5–6

1–2–3–4–6, 1–2–4–6, 1–3–4–6, 1–3–5–6, 1–2–3–5–6

1–2–3–4–6, 1–2–4–6, 1–3–4–6, 1–3– 5–6, 1–2–3–5–6,

1–2–3–4–6, 1–2–4–6, 1–3–4–6, 1–3–5–6, 1–2–3–5–6

1–2–3–4–6, 1–2–4–6, 1–3–4–6, 1–3–5–6, 1–2–3–5–6

(2,3) (1) (2,4) (1) (1,3) (1)

(1, 2) (1) (1, 3) (1)

(3, 5) (1) (4, 6) (1)

(3, 5)(1) (4, 6)(1)

(3, 5) (1) (4, 6) (1)

(2, 3) (1)

_

Figure 14.15

1–2–3–4–6

Activities crashed (time)

See ﬁgure (before crashing)

Critical path(s)

12

13

14

15

16

17

18

19

20

Project length after crashing (days)

9200

9200

9200

9200

9200

9200

9200

9200

15,200

300  19 300  18 300  17

300  16

300  15

300  14

300  13

300  12

200  1 = 200 200 + 250  1 = 450 450 + 250  1 = 700

700 + 233  1 + 250  1 = 1183

1183 + 233  1 + 250  1 = 1666

1666 + 233  1 + 250  1 = 2149

2149 + 250  1 + 267  1 = 2666

2666 + 200  1 + 600  1 + 267  1 = 3733

16,533

15,766

15,549

15,366

15,183

15,000

15,050

15,100

(A + B + C)

300  20

Total cost (Rs.)

(C)

Indirect cost (Rs. 300/ day)

_

(B)

(A) 9200

Crashing cost (Rs.)

Normal direct cost (Rs.)

Table 14.4 Computation of total cost with duration due to crashing for Example 4

394

14

L1 = 4 E1 = 4

4(4) L1 = 0 E1 = 0

L1 = 6 E1 = 6

2(2)

2

Project Management

4

6(6)

2(2)

1

6

L1=12 E1= 12

6(5)

L1 = 6 E1 = 6

5

3

3(3)

3(2)

L1 = 9 E1 = 9

Fig. 14.23 Project network with time duration 12 days

(iv) Now we construct the table for ﬁnding the optimal cost with duration and minimum project duration with cost (cf. Table 14.4). From Fig. 14.23, it is seen that no further crashing is possible beyond 12 days. Hence, the least project duration is 12 days, and the corresponding cost of the project will be Rs. 16,533.00. As the minimum cost occurs for a 17-day schedule, the optimum duration of the project is 17 days, and the minimum project cost is Rs. 15,000.00. Example 5 The following table gives data on normal time & cost and crashed time & cost for a project: Activity (1,2) (1,3) (2,3) (2,4) (2,5) (3,6) (4,5) (5,6)

Time (days) Normal

Crash

Cost (Rs.) Normal

Crash

9 15 7 7 12 12 6 9

4 13 4 3 6 11 2 6

1300 1000 700 1200 1700 600 1000 900

2400 1380 1540 1920 2240 700 1600 1200

The indirect cost for the project is Rs. 400 per day. (i) (ii) (iii) (iv)

Draw the network of the project. What are the normal duration and associated cost of the project? What are the optimum project duration and corresponding cost? Find the optimum cost and corresponding duration.

14.12

L2 = 9 E2 = 9

395

L4 = 16 E4 = 16

7(3)

2

4 12(6) 9(4) L1 = 0 E1 = 0

6(2) 9(6)

7(4)

L6 = 31 E6 = 31

6

5

1

E5 = 22 L5 = 22 15(13)

12(11) 3

L3 = 16 E3 = 19

Fig. 14.24 Project network with time duration 31 days L2 = 9 E2 = 9

(a)

L4 = 16 E4 = 16

7(3)

2

4 12(6) 9(4) L1 = 0 E1 = 0

6(2) 8(6)

7(4)

1

L6 = 30 E6 = 30

E5 = 22 L5 = 22 15(13)

12(11) 3

L3 = 16 E3 = 18 L2 = 9 E2 = 9

(b)

6

5

L4 = 16 E4 = 16

7(3)

2

4 12(6) 9(4) L1 = 0 E1 = 0

7(4)

1

6(2) 7(6) 6

5

L6 = 29 E6 = 29

E5 = 22 L5 = 22 15(13)

12(11) L3 = 16 E3 = 17

3

Fig. 14.25 a Project network with time duration 30 days. b Project network with time duration 29 days

396

14

L2 = 9 E2 = 9

L4 = 16 E4 = 16

7(3)

2

Project Management

4 12(6) 9(4) L1 = 0 E1 = 0

6(2) 6(6)

7(4)

5

1

6

L6 = 28 E6 = 28

6

L6 = 27 E6 = 27

E5 = 22 L5 = 22 15(13)

12(11) L3 = 16 E3 = 16

3

Fig. 14.26 Project network with time duration 28 days

L2 = 8 E2 = 8

L4 = 15 E4 = 15

7(3)

2

4 12(6) 8(4) L1 = 0 E1 = 0

7(4)

1

6(2) 6(6) 5 E5 = 21 L5 = 21

15(13)

12(11) L3 = 15 E3 = 15

3

Fig. 14.27 Project network with time duration 27 days

Solution (i) The network diagram of the project is shown in Fig. 14.24. (ii) The network of the project has been drawn using the normal time duration for each activity. Then for each event, the earliest and latest occurrence times for each event have been computed and displayed in the network. The normal and crash time durations are also shown in the network.

14.12

L2 = 8 E2 = 8

397

L4 = 15 E4 = 15

7(3)

2

4 12(6) 8(4) L1 = 0 E1 = 0

5(2) 6(6)

7(4)

5

1

6

L6 = 26 E6 = 26

6

L6 = 25 E6 = 25

E5 = 20 L5 = 20 15(13)

11(11) L3 = 15 E3 = 15

3

Fig. 14.28 Project network with time duration 26 days

L2 = 7 E2 = 7

L4 = 14 E4 = 14

7(3)

2

4 12(6) 7(4) L1 = 0 E1 = 0

7(4)

1

5(2) 6(6) 5 E5 = 19 L5 = 19

14(13)

11(11) L3 = 14 E3 = 14

3

Fig. 14.29 Project network with time duration 25 days

From the network, it is seen that L1 ¼ E1 ¼ 0; L2 ¼ E2 ¼ 9; L4 ¼ E4 ¼ 16; L5 ¼ E5 ¼ 22; L6 ¼ E6 ¼ 31: Hence, the critical path is 1–2–4–5–6. As E6 ¼ L6 ¼ 31, the normal duration of the project is 31 days. The associated cost of the project ¼ Direct normal cost þ indirect cost for 31 days ¼ Rs: ½ð1300 þ 1000 þ 700 þ 1200 þ 1700 þ 600 þ 1000 þ 900Þ þ 31  400 ¼ Rs: 20;800:

398

14

Project Management

(iii) The cost slopes of different activities are computed by using the formula Cost slope = (Crash Cost − Normal cost)/(Normal time − Crash time), and these values are shown in the following table: Activity Cost slope

(1,2) 220

(1,3) 190

(2,3) 280

(2,4) 180

(2,5) 90

(3,6) 100

(4,5) 150

(5,6) 100

Now we construct the table for ﬁnding the optimal cost with duration and minimum duration with cost (cf. Table 14.5). Table 14.5 Computation of total cost with duration due to crashing for Example 5 Critical path(s)

See ﬁgure (before crashing)

1–2–4– 5–6

Figure 14.24

1–2–4– 5–6

Figure 14.24

1–2–4– 5–6

Indirect cost (Rs. 400/day) (C)

Total cost (Rs.) (A + B + C)

400  31

20,800

100  1 = 100

400  30

20,500

8400

100 + 100  1 = 200

40ion ion 0  29

20,200

28

8400

200 + 100  1 = 300

400  28

19,900

(1,2) (1)

27

8400

300 + 220  1 = 520

400  27

19,720

Figure 14.27

(4,5) (1) (3,6) (1)

26

8400

520 + 150  1 + 100  1 = 770

400  26

19,570

1–2–4– 5–6 1–2–3– 6 1–3–6 1–2–5– 6

Figure 14.28

(1,2) (1) (1,3) (1)

25

8400

770 + 220  1 + 190  1 = 1180

400  25

19,580

Do

Figure 14.29

(1,2) (1) (1,3) (1)

24

8400

1180 + 220  1 + 190  1 = 1590

400  24

19,590

Project length after crashing (days)

Normal direct cost (Rs.) (A)

31

8400

(5,6) (1)

30

8400

Figure 14.25a

(5,6) (1)

29

1–2–4– 5–6 1–2–3– 6

Figure 14.25b

(5,6) (1)

1–2–4– 5–6 1–2–3– 6

Figure 14.26

1–2–4– 5–6 1–2–3– 6 1–3–6 1–2–5– 6

Activities crashed (time)

Crashing cost (Rs.) (B)

From Fig. 14.30, it is seen that no further crashing is possible beyond 24 days. Hence, the optimum project duration, i.e. the least project duration, is 24 days and the corresponding cost of the project will be Rs. 19,590.00.

14.12

L2 = 6 E2 = 6

399

L4 = 13 E4 = 13

7(3)

2

4 12(6) 6(4) L1 = 0 E1 = 0

5(2) 6(6)

7(4)

6

5

1

L6 = 24 E6 = 24

E5 = 18 L5 = 18 13(13)

11(11) L3 = 13 E3 = 13

3

Fig. 14.30 Project network with time duration 24 days

(iv) From Table 14.5, it is observed that the optimum cost is Rs. 19,570.00 and the corresponding project duration is 26 days.

14.13

Exercises

1. A project consists of a series of tasks labelled A, B, …, H, I with the following relationships (W < X, Y means X and Y cannot start until W is completed; X, Y < W means W cannot start until both X and Y are completed). With this notation, construct the network diagram having the following constraints: A\D; E; B; D\F; C\G; G\H; F; G\I: Find also the minimum time of completion of the project, when the time (in days) of completion of each task is as follows: Task Time

A 23

B 8

C 20

D 16

E 24

F 18

G 19

H 4

I 10

400

14

Project Management

2. The following are the details of the estimated times of activities of a certain project: Activity Immediate predecessor Estimated time (weeks)

A _ 2

B A 3

C A 4

D B, C 6

E _ 2

F E 8

(a) Find the critical path and the expected time of the project. (b) Calculate the earliest start time and earliest ﬁnish time for each activity. (c) Calculate the float for each activity. 3. A project consists of eight activities with the following relevant information: Activity

Immediate predecessor

Estimated duration (days) Optimistic Most likely

Pessimistic

A B C D E F G H

_ _ _ A B C D, E F, G

1 1 2 1 2 2 3 1

7 7 8 1 14 8 15 3

1 4 2 1 5 5 6 2

(ii) Draw the network and ﬁnd the expected project completion time. (iii) What duration will have 95% conﬁdence for project completion? (iv) If the average duration for activity F increases to 14 days, what will be its effect on the expected project completion time which will have 95% conﬁdence? 4. Draw the network for the following project and compute the earliest and latest times for each event and also ﬁnd the critical path: Activity

(1, 2)

(1, 3)

(2, 4)

(3, 4)

(4, 5)

(4,6)

(5, 7)

(6,7)

Immediate predecessor

_

_

(1, 2)

(1, 3)

(2, 4)

(2,4), (3,4)

(4, 5)

(4, 6)

(7, 8) (6, 7), (5, 7)

Time (days)

5

4

6

2

1

7

8

4

3

14.13

Exercises

401

5. A small project consists of seven activities, the details of which are given below: Activity A B C D E F G

Time estimates tm to

tp

3 2 2 2 1 4 1

9 8 6 10 11 8 15

6 5 4 3 3 6 5

Predecessor None None A B B C, D E

Find the critical path. What is the probability that the project will be completed by 18 weeks? 6. The following table gives data on normal time-cost and crash time-cost for a project: Activity (1, (1, (2, (2, (3, (4, (5, (6,

2) 3) 4) 5) 4) 6) 6) 7)

Normal Time (days)

Cost (Rs.)

Crash Time (days)

Cost (Rs.)

6 4 5 3 6 8 4 3

650 600 500 450 900 800 400 450

4 2 3 1 4 4 2 2

1000 2000 1500 650 2000 3000 1000 800

The indirect cost per day is Rs. 100. (i) Draw the network and identify the critical path. (ii) What are the normal project duration and associated cost? (iii) Crash the relevant activities systematically and determine the optimum project completion time and cost.

402

14

Project Management

7. The following table gives the activities in a construction project and other relevant information: Activity

Immediate predecessor

Time (days) Normal Crash

Direct cost (Rs.) Normal Crash

A B C D E F G

_ _ _ A C A B, D, E

4 6 2 5 2 7 4

60 150 38 150 100 115 100

3 4 1 3 2 5 2

90 250 60 250 100 175 240

Indirect costs vary as follows: Days Cost (Rs.)

15 600

14 500

13 400

12 250

11 175

10 100

9 75

8 50

7 35

6 25

(i) Draw the network of the project. (ii) Determine the project duration which will result in minimum total project cost.

Chapter 15

Queueing Theory

15.1

Objective

The objective of this chapter is to discuss • The different situations which generate queueing problems • The various factors/components of a queueing system • A variety of performance measures (operating characteristics) of a queueing system • The probability distributions of arrivals and departures in a queueing system • Poisson queueing models with ﬁnite and inﬁnite capacity • The single server inﬁnite capacity queueing model with Poisson arrival and multiphase Erlang service • The single server inﬁnite capacity queueing model with Poisson arrival and general service • A mixed queueing model with random arrival rate which follows the Poisson distribution and deterministic service rate • Finding the optimum service level by minimizing the total cost of M/M/1 systems with ﬁnite and inﬁnite capacity.

15.2

Basics of Queueing Theory

15.2.1 Introduction A group of customers waiting to receive a service including those receiving the service is known as a waiting line or queue. Queueing theory involves the mathematical study of queues or waiting lines. It analyses a production and servicing process system exhibiting random variability in market demand (arrival times) and © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_15

403

404

15

Queueing Theory

service times. This subject had its beginning in the research work of Danish engineer A.K. Erlang. In the year 1909, Erlang experimented with fluctuating demand in telephone trafﬁc. After eight years he established a report addressing the delays in automatic dialing equipment. At the end of the Second World War, Erlang’s earlier work was extended and applied to solve general problems and the business applications of waiting lines. Queueing theory tries to answer questions involving the average number of customers in the queue, average number of customers in the system, mean waiting time in the queue, mean waiting time in the system, and so forth. The objective of a queueing model is to determine how to provide the service to the customers so as to maintain an economic balance between the cost (time) of the service and the cost (time) associated with the wait for the service by manipulating certain variables such as number of servers, rate of service and order of service.

15.2.2 Reason Behind the Queue The formation of waiting lines (or queues) is a common phenomenon which occurs whenever the current demand for a service exceeds the current capacity to provide that service; i.e. a customer does not get the service immediately but must wait, or the service facilities stand idle or the total number of service facilities exceeds the number of customers requiring service. Waiting lines or queues are a common occurrence both in our daily life and in various business and industrial situations. Most queueing problems involve the question of ﬁnding the level of services that a ﬁrm should provide. Table 15.1 shows examples of queueing services. For example: (i) Supermarkets must decide how many cash counters should be opened. (ii) Gasoline stations must decide how many pumps should be opened and how many attendants should be on duty. (iii) Manufacturing plants must determine the optimal number of machines to have running in each shift. (iv) Banks must decide how many teller windows to keep open to serve customers during certain hours of the day. The person waiting in a queue or receiving the service is called the customer, and the person by whom the customer is serviced is called a server. Queueing problems arise because the cost of service may not be cheap enough to ensure sufﬁcient service facilities so that no customer has to wait. Customers arrive at a counter according to a certain probabilistic law (Poisson’s input, Erlang input, etc.). On the other hand, the customers will be served by one or more servers following a certain principle (FCFS, i.e. ﬁrst come, ﬁrst served, random service, etc.). By providing additional service facilities, the customer’s waiting time or the queue can be reduced; conversely, by decreasing the number of service facilities, a long queue

15.2

Basics of Queueing Theory

405

Table 15.1 Examples of queueing services Situation

Customers/ arrivals

Service facility or servers

Service process

Flow of automobile trafﬁc through a road network Ships entering a port

Automobiles

Ships

Docks

Telephone calls Aeroplanes

Switchboard operators or electric switching operators Runways

Bank

Customer

Bank clerks, check drawn

Sale of railway tickets

Booking clerks

Maintenance and repair of machines Department store

Railway passengers Machine breakdown Customer

Landing or flying the planes Money deposited or withdrawn Selling tickets

Mechanics

Repairing Bill payment

Job interviewing

Applicant

Checking clerks and bag packers Interviewee

Telephone exchange Number of runways at airports

Selection of the applicant

will be formed. The service times are random variables governed by a given probabilistic law. After being served, a customer leaves the queue. The objective of the queueing model is to determine how to provide the service to the customers so as to minimize the total cost of service and the waiting time of customers by manipulating certain factors (variables) such as number of servers, rate of service and order of service.

15.2.3 Characteristics of Queueing Systems Any queueing system can be characterized by the following set of characteristics: • • • •

Input source Service mechanism Service discipline Capacity of the system.

406

15

Queueing Theory

Serviced customers

Queueing discipline

queue

Customers waiting in the

Customers outside the system and decide to join or not to join in the queue

The general framework of a queueing system is shown below. Input Service Service Queue source mechanism completed

Input source The input source is a device or group of devices that provides customers at the service facility for the service. An input source is characterized by the following factors: • Input size • Arrival pattern • Behaviour of the arrivals. Input size If the total number of customers requiring service is only a few, then the input size is said to be ﬁnite. But if the potential customers requiring service are sufﬁciently large in number, then the input source is considered to be inﬁnite. Also, if the customers arrive at the service facility in batches of ﬁxed size or variable size instead of one at a time, then the input is said to occur in bulk or batches. If the service is not available, they may be required to join the queue. Arrival pattern If the pattern by which customers arrive at the service system or time between successive arrivals (inter-arrival time) is uncertain, then the arrival pattern is measured by either mean arrival rate or inter-arrival time. These measures are characterized in the form of the probability distribution associated with these random processes. The most common stochastic queueing model assumes that the arrival rate follows a Poisson distribution or, equivalently, that the inter-arrival time follows an exponential distribution. Customer behaviour The customers generally behave in different ways as follows: (a) Balking: A customer may leave the queue because the queue is too long and he/ she has no time to wait or there is not sufﬁcient waiting space.

15.2

Basics of Queueing Theory

407

(b) Reneging: A customer may become impatient after waiting in the queue for some time and leave the queue. (c) Priorities: In certain applications some customers are served before others regardless of their order of arrival. These customers have priority over others. (d) Jockeying: Customers may change position from one queue to another, hoping to receive the service more quickly. Service mechanism The service mechanism is concerned with the manner in which customers are served. It can be characterized by observing different factors: (i) Availability of service facility: Service may be available only at certain times but not always. (ii) Capacity of service facility: The capacity is measured in terms of customers who can be served simultaneously. A queueing system may have one or more parallel service channels or a sequence of channels in series, as shown in Figs. 15.1, 15.2, 15.3 and 15.4. (iii) Service time or duration: Service time or duration may be constant or a random variable. In general, service time is not constant, since it is a common practice to use overtime or extra effort when an excessive queue condition is present.

Service facilities

Queue

Fig. 15.1 Single channel single phase queueing system

Queue

Service facility

Service facility

Fig. 15.2 Single channel multiple phase queueing system

Fig. 15.3 Multiple channel single phase queueing system

Service facility Queue Service facility

408

15

Queueing Theory

Service facility

Service facility

Service facility

Service facility

Queue

Fig. 15.4 Multiple channel multiple phase queueing system

A variety of probability distributions can be used to characterize the service pattern. The service pattern is assumed to be independent with respect to the arrival of customers. In the most common stochastic queueing model, it is assumed that service time follows an exponential distribution or, equivalently, that service rate follows a Poisson distribution. Service discipline The service discipline refers to the order or manner in which customers in the queue will be served. The most common disciplines are as follows: (i) First come, ﬁrst served (FCFS): According to this discipline the customers are served in the order of their arrival. It is also known as FIFO (ﬁrst input ﬁrst output). This service discipline may be seen at a cinema ticket window or a railway ticket window, etc. (ii) Last come, ﬁrst served (LCFS): This discipline may be seen in a big godown, where the units (items) which come last are taken out (served) ﬁrst. (iii) Serving in random order (SIRO): In this case, the arrivals are serviced randomly irrespective of their arrival in the system. (iv) Service on some priority procedure: Some customers are served before the others without considering their order of arrival; i.e. some customers are served on a priority basis. There are two types of priority services. (a) Preemptive priority In this case customers with the highest priority are allowed to enter the service immediately after entering the system, even if a customer with lower priority is already in service; i.e. lower priority customer service is interrupted to start service for a special customer. This interrupted service is resumed again after the highest priority customer is served. (b) Non-preemptive priority In this case the highest priority customer goes ahead of the queue, but service is started immediately on completion of the current customer service.

15.2

Basics of Queueing Theory

409

Capacity of the system In certain cases a queueing system is unable to accommodate more than the required number of customers at a time. No further customers are allowed to enter until the space becomes available to accommodate new customers. This type of situation is referred to as a ﬁnite source queue.

15.2.4 Important Deﬁnitions Queue length: Queue length is deﬁned by the number of persons (customers) waiting in the queue at any time. Average length of queue: Average length of queue is deﬁned by the number of customers in the queue per unit time. Waiting time: The time up to which a unit has to wait in the queue before it is taken into the service. Servicing time: The time taken for servicing of a unit. Busy period: The busy period of a server is the time during which the server remains busy in servicing. Thus, it is the time between the starting of service of the ﬁrst unit to the end of service of the last unit in the queue. Idle period: When all the customers in the queue are served, the idle period of the server begins, and it continues up to the time of arrival of a customer. Thus, the idle period of a server is the time during which he remains free because there is no customer present in the system. Mean arrival rate: The mean arrival rate in a queue is deﬁned as the expected number of arrivals occurring in a time interval of unit length. Mean servicing rate: The mean servicing rate for a particular servicing station is deﬁned as the expected number of services completed in a time interval of unit length, given that the servicing is going on throughout the entire time unit. Trafﬁc intensity: For a simple queue, the trafﬁc intensity is the ratio of the mean arrival and the mean servicing rate, i.e., Traffic intensity ¼

Mean arrival rate Mean servicing rate

15.2.5 The State of the System The state of a system involves the study of a system’s behaviour over time. The states of a system may be classiﬁed as follows:

410

15

Queueing Theory

(i) Transient state: A system is said to be in transient state when its operating characteristics are dependent on time. Thus, a queueing system is in transient state when the probability distributions of arrivals, waiting time and servicing time of the customers are dependent on time. This state occurs at the beginning of the operation of the system. (ii) Steady state: A system is said to be in steady state when its operating characteristics become independent of time. Thus, a queueing system is in steady state when the probability distributions of arrivals, waiting time and servicing time of the customers are independent of time. Let pn(t) denote the probability that there are n units in the system at time t. Then if the probability pn(t) remains the same as time passes, the system acquires steady state. Mathematically, in steady state, lim pn ðtÞ ¼ pn ðindependent of timeÞ;

t!1

which implies lim ddt pn ðtÞ ¼ ddptn ¼ 0: t!1 (iii) Explosive state: If the servicing rate is less than the arrival rate, the length of the queue will go on increasing with time and will tend to inﬁnity as t ! /. The system is then said to be in explosive state.

15.2.6 Probability Distribution in a Queueing System In the queueing system, it is assumed that the customer’s arrival is random and follows a Poisson distribution or, equivalently, that the inter-arrival times obey an exponential distribution. In most cases, service times are also assumed to be exponentially distributed. The basic assumptions (axioms) are as follows. Axiom 1 The probability that exactly one arrival will occur during the time interval (t, t + Dt) is equal to kDt + D(t), i.e. p1(Dt)= kDt + O(Dt), where k is a constant and independent of the total number of arrivals up to the time t, Dt is a small time interval and O(Dt) n denotes o a quantity which is of smaller order of magnitude than Dt such that lim

Dt!o

OðDtÞ Dt

¼ 0:

Axiom 2 The probability of more than one arrival during the time interval (t, t + Dt) is negligible and is denoted by O(Dt), i.e. p0 ðDtÞ þ p1 ðDtÞ þ OðDtÞ ¼ 1: Axiom 3 The number of arrivals in non-overlapping intervals is statistically independent.

15.2

Basics of Queueing Theory

411

Distribution of arrival (pure birth process) A process in which only arrivals are counted and no departure takes place is called a pure birth process. Stated in terms of queueing, a birth-death process usually arises by birth or arrival in the system and decreases by death or departure of serviced customers from the system. Property If the arrivals are completely random, then the probability distribution of the number of arrivals in a ﬁxed time interval follows a Poisson distribution. Proof Let there be n units in the system at time t. In order to derive the arrival distribution in queues, we will consider the preceding three axioms. Now we wish to determine the probability of n arrivals at time t, denoted by pn ðtÞ: Clearly, n will be an integer greater than or equal to zero. Case 1 When n = 0 If there is no unit in the system at time t þ Dt, there will be no unit at time t and no arrival during Dt: Therefore, the probability of no unit in the system at time t þ Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  kDtÞ þ OðDtÞ:

ð15:1Þ

Case 2 When n > 0 In this case, there may be the following mutually exclusive ways of having n units at time t þ Dt: (i) n arrivals in the system at time t, and no arrival during the time Dt (ii) (n – 1) arrivals in the system at time t and one arrival during the time Dt (iii) (n – 2) arrivals in the system at time t and two arrivals during the time Dt, and so on. Therefore, the probability of n units in the system at time t þ Dt is given by pn ðt þ DtÞ ¼ pn ðtÞð1  kDtÞ þ pn1 ðtÞkDt þ OðDtÞ: Equations (15.1) and (15.2) can be written as

ð15:2Þ

412

15

p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kp0 ðtÞ þ ; n¼0 Dt Dt pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn ðtÞ þ kpn1 ðtÞ þ ; Dt Dt

Queueing Theory

n [ 0:

Taking the limit as Dt ! 0 on both sides of these equations, we have the following system of differential difference equations: p00 ðtÞ ¼ kp0 ðtÞ;

n¼0

p0n ðtÞ ¼ kpn ðtÞ þ kpn1 ðtÞ;

n [ 0:

ð15:3Þ ð15:4Þ

Solution of differential difference equations From (15.3),

p00 ðtÞ p0 ðtÞ

¼ k:

Integrating, we have log p0 ðtÞ ¼ kt þ A, where A is an arbitrary constant. When t = 0, p0 ð0Þ ¼ 1, which implies A = 0. Hence, p0 ðtÞ ¼ ekt :

ð15:5Þ

Now substituting n = 1 in (15.4), we have p01 ðtÞ þ kp1 ðtÞ ¼ kekt

  * p0 ðtÞ ¼ ekt :

This is a ﬁrst-order linear differential equation. Solving and using p1 ðtÞ = 0 when t = 0, we have p1 ðtÞ ¼ ktekt : Now, substituting n = 2 in (15.4) and using the result of p1 ðtÞ, we have   p02 ðtÞ þ kp2 ðtÞ ¼ k ktekt : Solving this equation and using p2 ð0Þ ¼ 0 when t = 0, we have p2 ðtÞ ¼ ekt Similarly, for n = 3,

ðktÞ2 : 2!

ð15:6Þ

15.2

Basics of Queueing Theory

413

p3 ðtÞ ¼ ekt

ðktÞ3 : 3!

pm ðtÞ ¼ ekt

ðktÞm : jm

Let us assume that for n = m,

Now, we shall prove that this result is true for n = m + 1 also. For this purpose, substituting n = m + 1 in (15.4) and using the result m pm ðtÞ ¼ ekt ðktm!Þ , we have p0m þ 1 ðtÞ þ kpm þ 1 ðtÞ ¼ k

ðktÞm ekt : m!

On solving and using pm þ 1 ð0Þ ¼ 0 when t =0, we have pm þ 1 ð t Þ ¼

ðktÞm þ 1 kt e : ðm þ 1Þ!

Hence, in general, pn ð t Þ ¼

ðktÞn kt e ; n!

n ¼ 0; 1; 2; . . .

which is the probability mass function of the Poisson distribution. Therefore, the probability distribution of the number of arrivals in a ﬁxed time interval follows a Poisson distribution. Distribution of inter-arrival times The inter-arrival time is deﬁned as the time interval between two successive arrivals. Property of inter-arrival times If the arrival process in a queueing system follows a Poisson distribution, then the associated random variable deﬁned as inter-arrival time follows an exponential distribution and vice versa. Proof If n is the number of arrivals at time t, then the probability of n arrivals at time t is given by pn ð t Þ ¼

ðktÞn ekt : n!

ð15:7Þ

414

15

Queueing Theory

ΔT

T t0

t0 + T

First arrival

t0 + T + ΔT second arrival

Let t0 be the instant of an initial arrival. Let T be the random variable of inter-arrival time. If there is no arrival during the intervals ðt0 ; t0 þ T Þ and ðt0 þ T; t0 þ T þ DT Þ, then t0 þ T þ DT will be the instant of consecutive arrival. From (15.7), p0 ðTÞ ¼ probability of no arrival at time T ¼

ðkT Þ0 ekT ¼ ekT ; 0!

and p0 ðT þ DTÞ ¼ probability of no arrival at time T þ DT fkðT þ DT Þg0 ekðT þ DT Þ ¼ ekðT þ DT Þ ¼ ekT  ekDT 0! ¼ p0 ðT Þ½1  kDT þ OðDT Þ;

¼

where OðDT Þ contains the terms of second and higher powers of DT: ) p0 ðT þ DT Þ  p0 ðT Þ ¼ kp0 ðT ÞDt þ OðDT Þp0 ðT Þ or

p0 ðT þ DT Þ  p0 ðT Þ OðDT Þ ¼ kp0 ðT Þ þ p0 ðT Þ : D DT

Taking the limit as DT ! 0, we have d ½p0 ðT Þ ¼ kp0 ðT Þ: dT   d * ½p0 ðT Þ ¼ kekT *p0 ðT Þ ¼ ekT : dT

ð15:8Þ

According to probability theory, F 0 ð xÞ ¼ f ð xÞ, where F ð xÞ is the distribution function and f(x) is the probability density function. Hence, by a similar argument, we may write ddT ½p0 ðT Þ as f ðT Þ, where p0 ðT Þ is the probability distribution

15.2

Basics of Queueing Theory

415

function for no arrival at time T and f ðT Þ is the corresponding probability density function of T. Again, since the probability density function is always non-negative, we disregard the negative sign from the right-hand side of (15.8). Hence, f ðT Þ ¼ kekT , which is the probability density function of the exponential distribution. That is, the random variable T follows an exponential distribution. The converse of the property may be proved in a similar way. Distribution of departure (pure death process) Sometimes a situation may arise where no additional customer joins the system while service is continued for those who are in the queue. At time t let there be N (  1) customers in the system. It is clear that service will be provided at a rate l. The number of customers in the system at time t  0 is equal to N minus the total departure up to the time t. The distribution of the departure can be obtained with the help of the following basic axioms. Basic axioms (i) The probability of one departure during the time interval (t, t + Dt) is equal to lDt + O(Dt). (ii) The probability of more than one departure during the time interval (t, t + Dt) is negligible and is denoted by O(Dt). (iii) The numbers of departures in non-overlapping intervals are statistically independent.

15.2.7 Classiﬁcation of Queueing Models Generally, a queueing model may be speciﬁed in a symbolic form as (a/b/c):(d/e/f), where: a = Arrival distribution, i.e. type of arrival process b = Service time distribution, i.e. type of service process c = Number of servers d = Capacity of the system e = Service discipline f = Number of calling source capacity of input service. Generally, the last symbol f is not used. The standard notation for a and b is taken as M, which is called the Poisson (or Markovian) arrival or departure distribution or equivalently the exponential inter-arrival service time distribution. Ek = Erlangian or gamma inter-arrival for service time distribution with parameter k G = General service time distribution or departure distribution.

416

15

Queueing Theory

Thus, (M/Ek/1):(∞/FIFO/∞) deﬁnes a queueing system in which arrival follows the Poisson distribution; service time distributions are Erlangian; there is a single server; the capacity of the system is inﬁnite, i.e. the system can hold an inﬁnite number of customers; the service discipline is ﬁrst input ﬁrst output; and ﬁnally the source generating the arriving customers has an inﬁnite capacity. Generally queues with arrivals and departures start under a transient condition and gradually reach steady state after a sufﬁciently long time has elapsed, provided that the parameters of the system permit reaching steady state. Now we deﬁne some parameters for a steady state queueing system: pn = (steady state) Probability of n customers in the system Ls = Expected number of customers in the system Lq = Expected number of customers in the queue Ws = Expected waiting time in the system Wq = Expected waiting time in the queue

15.3

Poisson Queueing Models

15.3.1 Introduction In this section, we shall discuss single as well as multiserver queueing models with Poisson arrival and departure processes. According to the capacity of the queueing system, we can classify these models as follows: (i) Single server queueing model with inﬁnite capacity of the system, i.e., ðM=M=1Þ:ð1=FCFS=1Þ (ii) Single server queueing model with ﬁnite capacity of the system, i.e., ðM=M=1Þ:ðN=FCFS=1Þ (iii) Multiserver queueing model with inﬁnite capacity of the system, i.e., ðM=M=cÞ:ð1=FCFS=1Þ (iv) Multiserver queueing model with ﬁnite capacity of the system, i:e:ðM=M=cÞ:ðN=FCFS=1Þ: We shall also discuss the repair problem of machine breakdowns and the power supply problem.

15.3

Poisson Queueing Models

417

15.3.2 Model (M/M/1):(∞/FCFS/∞) Assumptions In this model, we shall discuss a single server queueing system under the following assumptions: (i) (ii) (iii) (iv) (v)

The The The The The

arrival rate follows a Poisson distribution. servicing rate follows a Poisson distribution. queueing system has only a single service channel. service discipline is ﬁrst come, ﬁrst served (FCFS). capacity of the system is inﬁnite.

Let k and l be the mean arrival rate and mean service rate of units respectively, and let n be the number of customers in the system. Hence, the probability of one arrival during time Dt is kDt þ OðDtÞ and the probability of one departure or service during Dt is lDt þ OðDtÞ: If there is no customer in the system at time t +Dt, there may be the following mutually exclusive cases: (i) No customer in the system at time t, no arrival in time Dt (ii) One customer in the system at time t, no arrival in time Dt, one service in time Dt (iii) No customer in the system at time t, one arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of no customers in the system at time t +Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞ½1  kDt þ OðDtÞ þ p1 ðtÞ½1  kDt þ OðDtÞ½lDt þ OðDtÞ þ OðDtÞ ¼ p0 ðtÞ  kp0 ðtÞDt þ lp1 ðtÞDt þ OðDtÞ or,

p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ þ : Dt Dt

Taking the limit as Dt ! 0 on both sides in the preceding expression, we have p00 ðtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ:

ð15:9Þ

To have n (>0) customers in the system at time t þ Dt, there may be the following mutually exclusive cases: (i) n customers in the system at time t, no arrival in time Dt, no service in time Dt (ii) (n – 1) customers in the system at time t, one arrival in time Dt, no service in time Dt (iii) (n + 1) customers in the system at time t, no arrival in time Dt, one service in time Dt, and so on.

418

15

Queueing Theory

Therefore, the probability of n customers in the system at time (t + Dt) is given by pn ðt þ DtÞ ¼ pn ðtÞf1  kDt þ OðDtÞgf1  lDt þ OðDtÞg þ pn1 ðtÞ ½kDt þ OðDtÞ½1  lDt þ OðDtÞ þ pn þ 1 ðtÞ½1  kDt þ OðDtÞ½lDt þ OðDtÞ þ OðDtÞ for n [ 0 ¼ pn ðtÞ þ ½ðk þ lÞpn ðtÞ þ kpn1 ðtÞ þ lpn þ 1 ðtÞDt þ OðDtÞ: ) pn ðt þ DtÞ  pn ðtÞ ¼ ½kpn1 ðtÞ  ðk þ lÞpn ðtÞ þ lpn þ 1 ðtÞDt þ OðDtÞ pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn1 ðtÞ  ðk þ lÞpn ðtÞ þ lpn þ 1 ðtÞ þ for n [ 0: or Dt Dt

Now taking the limit as Dt ! 0 of both sides in this equation, we have p0n ðtÞ ¼ kpn1 ðtÞ  ðk þ lÞpn ðtÞ þ lpn þ 1 ðtÞ;

n [ 0:

ð15:10Þ

Under the steady state condition of the system, i.e. lim pn ðtÞ ¼ pn and lim p0n t!1

ðtÞ ¼ 0, Eqs. (15.9) and (15.10) reduce to

t!1

kp0 þ lp1 ¼ 0

ð15:11Þ

kpn1  ðk þ lÞpn þ lpn þ 1 ¼ 0 for n [ 0;

ð15:12Þ

which are the steady state difference equations of the system. Solution of the Model From (15.11), kp0 þ lp1 ¼ 0; which implies k p1 ¼ p0 ¼ q p0 l

 *

 k ¼q : l

Now setting n = 1 in (15.12), we have kp0  ðk þ lÞp1 þ lp2 ¼ 0  k k þ 1 p1 or p2 ¼  p0 þ l l or p2 ¼ qp0 þ ðq þ 1Þqp0 ¼ q2 p0 : Again setting n = 2 in (15.12), we have

15.3

Poisson Queueing Models

419

kp1  ðk þ lÞp2 þ lp3 ¼ 0  k k þ 1 p2 or p3 ¼  p1 þ l l ¼ q2 p0 þ q3 p0 þ qp0 ¼ q3 p0 : Hence, p 3 ¼ q3 p 0 :

ð15:13Þ

Proceeding in this way, we will have p n ¼ qn p 0 : The value of p0 is obtained from the condition i.e. p0 ð1 þ q þ q2 þ . . .Þ ¼ 1 or

P1 n¼0

pn ¼ 1,

p0 ¼ ð1  qÞ; since q ¼ k=l\1:

ð15:14Þ

Therefore, (15.13) becomes pn ¼ ð1  qÞpn ;

n ¼ 0; 1; 2; . . .

ð15:15Þ

This gives the probability that there are n units in the system at any time. Equations (15.14) and (15.15) rather give the required probability distribution of the queue length. We note that the probability distribution in (15.15) depends only on the trafﬁc intensity ratio q ¼ k=l. Obviously in order for p0 to exist, it is required that ðq¼Þk=l\1, i.e. k\l; otherwise, the series will diverge. This is the stability condition for the M/M/1 model. Characteristics of the Model (i) Probability of queue size greater than or equal to N, i.e. prob (queue size  N) prob (queue size  NÞ ¼ pN þ pN þ 1 þ pN þ 2 þ      ¼ p 0 q N 1 þ q þ q2 þ     N qN k ¼ qN ¼ ½*p0 ¼ 1  q: ¼ p0 l 1q  N k N : ) prob (queue size  NÞ ¼ q ¼ l (ii) Expected line length Ls

420

15

Queueing Theory

Ls = expected number of customers in the system, i.e. expected line length or average number of customers in the system ¼

1 X

npn ¼

n¼0

1 X

nqn p0 ¼ p0

n¼0

1 X

nqn

n¼0

¼ p0 q½1 þ 2q þ 3q2 þ 4q3 þ    1 ¼ p0 qð1  qÞ2 ¼ ð1  qÞq ð1  qÞ2   q k k ¼ * q¼ ¼ 1q lk l

½* p0 ¼ 1  q:

(iii) Expected queue length Lq Lq=Pexpected number of customers in queue (i.e. expected queue length) ¼ 1 n¼1 ðn  1Þpn [Since there are (n − 1) customers in the queue excluding the one in service] ¼

1 X

npn 

n¼1

1 X

pn ¼ L s 

1 X

n¼1

! pn  p0

n¼0

q q ¼ Ls  ð1  p0 Þ ¼ Ls  q ¼ 1q   k2 k * q¼ ¼ l lðl  kÞ



Note: Ls ¼ Lq þ q ¼ Lq þ lk as Lq = Ls − q. (iv) Variance of queue length

From the deﬁnition of variance, VarðnÞ ¼ E n2  fE ðnÞg2 ¼

1 X

( n2 pn 

n¼1

1 X n¼1

)2 npn

2 q ¼ n q ð1  qÞ  1q n¼1 " 1 X q * Ls ¼ npn ¼ and pn ¼ qn ð1  qÞ 1  q n¼1 1 X

2 n





q *Ls ¼ 1q

15.3

Poisson Queueing Models

421

¼ qð1  qÞ þ 22 q2 ð1  qÞ þ 32 q3 ð1  qÞ þ 42 q4 ð1  qÞ þ       ¼ qð1  qÞ 1 þ 22 q þ 32 q2 þ 42 q3 þ     q2

¼ qð1  qÞS 

ð1  qÞ2

;

q2 ð1  qÞ2

q2 ð1  qÞ2

where S ¼ 1 þ 22 q þ 32 q2 þ 42 q3 þ   

Now, we have to calculate the simpliﬁed expression of S. S ¼ 1 þ 22 q þ 32 q2 þ 42 q3 þ    Integrating both sides with respect to q from 0 to q, we have Zq Zq   Sdq ¼ 1 þ 22 q þ 32 q2 þ 42 q3 þ    dq q¼0

q¼0

¼ q þ 2q2 þ 3q3 þ 4q4 þ 5q5 þ    ¼ qð1  qÞ2 Zq ) Sdq ¼ qð1  qÞ2 ¼ q¼0

q ð1  qÞ2

:

Differentiating both sides with respect to q, we have S¼ ¼

1 ð1  qÞ2 1þq ð1  qÞ3

þq

2 ð1  qÞ3

ð1Þ ¼

1 ð1  qÞ2

þ

2q ð1  qÞ3

¼

1  q þ 2q ð1  qÞ3

: )VarðnÞ ¼ qð1  qÞS 

q2

ð1  qÞ2 1þq q2 ¼ qð1  qÞ  : 3 ð1  qÞ2 ð1  qÞ qð1 þ qÞ q2 q ¼  ¼ 2 2 ð1  qÞ ð1  qÞ2 ð1  qÞ

(v) Probability density function of waiting time excluding the service time distribution

422

15

Queueing Theory

In a steady state, each customer has the same waiting time distribution. Let this distribution be a continuous distribution with probability density function w(w), and we denote by w(w) dw the probability that a customer begins to be served in the interval (w, w +dw), where w is measured from the time of his arrival. We suppose that a customer arrives at time w = 0 and his service begins in the interval (w, w +dw).

Customer A’s arriveal

A’s service start

w=0

w

w + dw

There may be two possibilities: (i) There is a ﬁnite probability p0 (the probability that the system is empty) that the waiting time is zero. (ii) Let there be n customers in the system, (n − 1) waiting, one in the service, when customer A arrives. Therefore, before the service of A begins, (n − 1) customers must leave in the time interval (0, w) and with the nth customer in (w, w +dw). As the server’s mean rate of service is l in unit time or lw in time w, and the service distribution is Poisson, we have prob [(n − 1) customers waiting are served in time w] ¼

ðlwÞn1 elw ðn  1Þ!

and prob [one customer is served in time dw] = l dw. ) wn ðwÞdw ¼ probability that a new arrival is taken into service after a time lying between w and w þ dw ¼ prob½ðn1Þcustomers waiting are served in time w  prob ½one customer is served in time dw ¼

ðlwÞn1 elw l dw: ðn  1Þ!

Let W be the waiting time of the customer who has to wait such that w  W  w þ dw: Since the queue length can vary between 1 and ∞, the probability density of the waiting time is given by

15.3

Poisson Queueing Models

423

wðwÞdw ¼ pðw  W  w þ dwÞ 1 X ½probability that there are n customers in the system  wn ðwÞdw ¼ n¼1

¼

1 X

qn ð1  qÞ

n¼1

ðlwÞn1 elw ldw ½* pn ¼ qn ð1  qÞ ðn  1Þ!

1 X ðlwÞn1 ðqlwÞn1 ¼ qð1  qÞelw ldw ðn  1Þ! ðn  1Þ! n¼1 n¼1

n1   k 1 1 X l lw k k lw k lw X ðkwÞn1 ¼k 1 1 e ldw e dw ¼ ðn  1Þ! ðn  1Þ! l l l n¼1 n¼1 " #   2 k lw kw ðkwÞ k lw kw ¼k 1 þ  ¼ k 1   e dw e dw 1 þ þ e 2! l 1! l  k ðlkÞw dw where w [ 0: e ¼k 1 l

¼ ð1  qÞelw ldw

1 X

qn

Hence, Z1 0

ZB  Z1  k ðlkÞw k ðlkÞw wðwÞdw ¼ k 1 dw ¼ lim k 1 dw e e B!1 l l 0

0

  ðlkÞw B k e ¼k 1 lim l B!1 ðl  kÞ 0    k 1 lk 1 k ¼ 6¼ 1: ¼k 1 0þ ¼k l ðl  kÞ l lk l

This happens because the case for which w = 0 is not included. Thus, Z1 prob½w [ 0 ¼

wðwÞdw ¼ q 0

and prob[w = 0] = prob[no customer in the system] = p0 = 1 − q. Since the sum of these probabilities of waiting time is 1, the complete distribution of the waiting time is: (a) Continuous for w  W  w + dw with probability density function w(w) given by

424

15

Queueing Theory

 k ðlkÞw wðwÞdw ¼ k 1  dw e l (b) Discrete for w = 0 with prob(w = 0) = 1  lk. Note that the probability that the waiting time exceeds w1 is  1 Z1 k k wðwÞdw ¼  eðlkÞw ¼ eðlkÞw1 ; l l w1

w1

which does not include the service time. (vi) Busy period distribution, i.e. the conditional density function for waiting time, given that a person has to wait Here we ﬁnd the probability density function for the distribution of total time (waiting plus service) that an arrival spends in the system. Let w(w/w >0) = probability density function for waiting time such that a person has to wait wðwÞ probðw [ 0Þ " * probðw [ 0Þ þ probðw ¼ 0Þ ¼ 1 wðwÞ ¼ q ) probðw [ 0Þ ¼ 1  probðw ¼ 0Þ ¼ 1  ð1  qÞ ¼ q ð15:16Þ

k 1  lk eðlkÞw ¼ ¼ ðl  kÞeðlkÞw k ¼

l

R1 R1 Now 0 wðw=w [ 0Þ ¼ 0 ðl  kÞeðlkÞw dw ¼ 1. Thus, (15.16) gives the required probability density function for the busy period. (vii) Mean or expected waiting time in the queue, i.e. Wq (average waiting time of an arrival in the queue) We have Z1 Wq ¼

wwðwÞdw 0

Z1 ¼ 0

 k ðlkÞw wk 1  dw e l

15.3

Poisson Queueing Models

425

1  ZB k eðlkÞw  l eðlkÞw dw ¼ wk 1  þ k Lt 1  B!1 l ðl  kÞ 0 k lk 0

½Integrating by parts taking was the first function #B  k eðlkÞw ¼ 0  0þk 1  Lt l B!1 ðl  kÞ2 #1 0  lk 1 ¼ k eðlkÞw l ðl  kÞ2 0

kðl  kÞ

k 1 ¼ Lq ; ¼ 0þ 1¼ 2 lðl  kÞ k lðl  kÞ

where Lq ¼

k2 : lðl  kÞ

(viii) Expected waiting time in the system Ws, i.e. average time that an arrival spends in the system Ws ¼ expected waiting time in the system ¼ expected waiting time in queue plus expected service time   1 1 Since the expected service time ¼ mean service time ¼ ¼ Wq þ l l   k 1 k þ * Wq ¼ ¼ lðl  kÞ l lðl  kÞ 1 : ¼ lk 1 k and Ls ¼ ; Ls ¼ kWs : As W s ¼ lk lk (ix) Expected length of non-empty queue, or average length of non-empty queue, i.e. L/L > 0. We have Lq Lq Lq ¼ ¼ P1 prob½an arrival has to wait, L [ 0 prob½ðn  1Þ [ 0 n¼2 pn Lq Lq Lq ¼ ¼ P1 ¼ p  ðp þ p Þ 1  p  qp 1  p ð1 þ qÞ n 0 1 0 0 0 n¼0 1 l ¼ : ¼ 1q lk

ðL=L [ 0Þ ¼

426

15

Queueing Theory

Note: From the preceding characteristics, we can write Ls ¼ Lq þ lk ; Ls ¼ kWs , Lq ¼ kWq and Ws ¼ Wq þ l1. These formulas are known as Little’s formula.

Example 1 Arrivals at a telephone booth are considered to have a Poisson distribution with an average time of 10 min between one arrival and the next. The length of a phone call is assumed to be distributed exponentially with mean 3 min. (a) What is the probability that a person arriving at the booth will have to wait? (b) What is the average length of queues that form from time to time? (c) The telephone company will install a second booth when convinced that an arrival would expect to have to wait at least 3 min for the phone. By how much time should the flow of arrivals be increased to justify a second booth? (d) Find the average number of units in the system. (e) What is the probability that an arrival has to wait more than 10 min before the phone is free? (f) Estimate the fraction of a day that the phone will be in use (or busy). Solution This is an (M/M/1):(∞/FCFS/∞) problem. 1 In this problem, the mean arrival rate is k ¼ 10 and the mean service time is 1 l ¼ 3. (a) Now the probability that a person arriving at the telephone booth will have to wait ¼ prob½w [ 0 ¼ 1  probðw ¼ 0Þ ¼ 1  p0 ¼ 1  ð1  qÞ ¼ q ¼

k ¼ 0:3: l

(b) The average length of the queues that form from time to time ¼ L=L [ 0 ¼

l ¼ 1:43 persons lk

(c) The installation of a second booth will be justiﬁed if the arrival rate is greater than the waiting time. Then the length of the queue will go on increasing. In that case, let k′ be the mean arrival rate. We know that the average waiting time of an interval in the queue is k0 Wq ¼ lðlk 0 : Þ Here, Wq ¼ 3;

1 k0  ) 3 ¼ 1 1 0 or; 3 3 3k

k0 ¼

1 1  k0 or k0 ¼ : 3 6

15.3

Poisson Queueing Models

427

1 1 Hence, the increase in mean arrival rate ¼ k0  k ¼ 16  10 ¼ 15 ¼ 0:067 arrival per minute. 1 Again, the increase in the flow of arrivals ¼ 15  60 ¼ 4 per hour. Therefore, the second booth is justiﬁed if the increase in arrival rate is 0.067 arrival per minute (or 4 persons per hour). k (d) The average number of units in the system is Ls ¼ lk ¼ 0:43 persons. (e) prob [waiting time of an arrival in the queue > 10]

Z1 ¼

Z1 wðwÞdw ¼

10

10

3B k k eðlkÞw 5 ðlkÞw ðl  kÞe dw ¼ lim ðl  kÞ B!1 l l ðl  kÞ

i kh ¼  0  eðlkÞ10 ¼ 0:03 ðapproxÞ l

10

(f) The fraction of a day that the phone will be busy = trafﬁc intensity ¼ q ¼ lk ¼ 0:3. Example 2 In a railway yard, goods trains arrive at a rate 30 trains per day. Assume that the inter-arrival time follows an exponential distribution, and the service time (the time taken to hump a train) distribution is also exponential with an average of 36 min. Calculate the following: (i) (ii) (iii) (iv) (v)

The average number of trains in the system The probability that the number of trains in the system exceeds 10 Expected waiting time in the queue Average number of trains in the queue If the input of trains increases to an average 33 per day, what will be the change in (i) and (ii)?

Solution This is an (M/M/1):(1/FCFS/1) problem. 30 1 In this problem, the mean arrival rate k ¼ 30 trains/day ¼ 6024 trains/minute ¼ 48 1 trains/min, and the mean service time l ¼ 36 trains/min. 1

3 Hence, the trafﬁc intensity q ¼ lk ¼ 481 ¼ 36 48 ¼ 4 ¼ 0:75. 36

(i) The average number of trains in the system is given by q 0:75 Ls ¼ 1q ¼ 10:75 ¼ 0:75 0:25 ¼ 3 trains: (ii) The probability that the number of trains in the system exceeds 10 ¼ probðqueue size  10Þ ¼ q10 ¼ ð0:75Þ10 ¼ 0:056. (iii) The expected waiting time in the queue is given by 1=48 k Wq ¼ lðlk Þ ¼ 1 ð 1  1 Þ min ¼ 108 minor1 h 48 min: 36 36 48

428

15

Queueing Theory

(iv) The average number of trains in the queue is given by 2 ð481 Þ k2 Lq ¼ lðlk ¼ 94 ¼ 2:25or nearly 2 trains: ¼ 1 1 1 Þ 36ð3648Þ 33 (v) If the input of trains increases to 33 trains per day, then k ¼ 6024 trains/min 11

11 ¼ 480 trains/min. Hence, q ¼ lk ¼ 4801 ¼ 33 40. 36

33

q ¼ 14033 ¼ 33 In this case, Ls ¼ 1q 7 ¼ 5 trains (approx.) and prob(queue size 40   10 ¼ 0:2 (approx.).  10) ¼ q10 ¼ 33 40

15.3.3 Model (M/M/1):(N/FCFS/∞) Assumptions In this model, we shall discuss a single server ﬁnite queueing system under the following assumptions: (i) (ii) (iii) (iv) (v)

Customers arrive according to a Poisson fashion. The service rate follows a Poisson distribution. The queueing system has only a single service channel. The service discipline is FCFS (ﬁrst come, ﬁrst served). The maximum number of customers in the system is limited to N.

If a new customer arrives when N persons are already in the system, then the customer is restricted from entering, and is said to be lost from the system. This is often referred to as blocking. Model Formulation Let n be the customers in the system. Let k and l be the mean arrival and service rate of units respectively. In this model, ( kn ¼

k when n\N i.e.; ðn ¼ 0; 1; 2; :. . .N  1Þ 0 when n  N

and ln ¼ l ðindependent of nÞ: Therefore, the probability of one arrival during Dt is kn Dt þ OðDtÞ for n  N, and the probability of one departure or service during Dt is lDt þ OðDtÞ. To have 0 (zero) customer in the system at time t þ Dt, there may be the following mutually exclusive cases: (i) No customer in the system at time t, no arrival in time Dt (ii) One customer in the system at time t, no arrival in time Dt, one service in time Dt, and so on.

15.3

Poisson Queueing Models

429

Therefore, the probability of no customer in the system at time t + Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  kDtÞ þ p1 ðtÞð1  kDtÞlDt þ OðDtÞ ¼ p0 ðtÞ  kp0 ðtÞDt þ lp1 ðtÞDt þ OðDtÞ or

p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ þ Dt Dt

for n ¼ 0:

ð15:17Þ

To have n (>0) customers in the system at time t + Dt, there may be the following mutually exclusive cases: (i) n customers in the system at time t, no arrival in time Dt, no service in time Dt (ii) (n − 1) customers in the system at time t, one arrival in time Dt, no service in time Dt (iii) (n + 1) customers in the system at time t, no arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of n customers in the system at time (t + Dt) is given by pn ðt þ DtÞ ¼ pn1 ðtÞkn1 Dtð1  lDtÞ þ pn ðtÞð1  kn DtÞð1  lDtÞ þ pn þ 1 ðtÞð1  kn þ 1 DtÞlDt þ OðDtÞ or

pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kn1 pn1 ðtÞ  ðkn þ lÞpn ðtÞ þ lpn þ 1 ðtÞ þ : Dt Dt ð15:18Þ

When n < N, then kn1 ¼ k; kn ¼ k. Hence, from (15.18), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn1 ðtÞ  ðk þ lÞpn ðtÞ þ lpn þ 1 ðtÞ þ : Dt Dt

ð15:19Þ

When n = N, pn þ 1 ¼ 0; kn1 ¼ k; kn ¼ 0. From ∴ (15.18), we have pN ðt þ DtÞ  pN ðtÞ OðDtÞ ¼ kpN1 ðtÞ  lpN ðtÞ þ : Dt Dt Taking the limit as Dt ! 0 in Eqs. (15.17), (15.19) and (15.20), we have p00 ðtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ for n ¼ 0

p0n ðtÞ ¼ kpn1 ðtÞ  ðk þ lÞpn ðtÞ þ lpn þ 1 ðtÞ for 0\n\N p0n ðtÞ ¼ kpN1 ðtÞ  lpN ðtÞ for n ¼ N

ð15:20Þ

430

15

Queueing Theory

Under a steady state condition of the system, these equations reduce to kp0 þ lp1 ¼ 0 for n ¼ 0 kpn1  ðk þ lÞpn þ lpn þ 1 ¼ 0 kpN1  lpN ¼ 0

for 1  n\N

for n ¼ N:

ð15:21Þ ð15:22Þ ð15:23Þ

Solution of the Model From (15.21), we have p1 ¼ lk p0 ¼ qp0 where q ¼ lk . Setting n = 1 in (15.22), we have kp0  ðk þ lÞp1 þ lp2 ¼ 0  k k p1 ¼ qp1 ¼ q2 p0 : or p2 ¼  p0 þ 1 þ l l Setting n = 2 in (15.22), we have kp1  ðk þ lÞp2 þ lp3 ¼ 0  k k ) p3 ¼  p1 þ 1 þ p2 ¼ qp2 ¼ q3 p0 : l l Proceeding in this way, we have  n k p0 ¼ qn p0 for 1  n\N pn ¼ l  N1 k ) pN1 ¼ p0 ¼ qN1 p0 : l

ð15:24Þ

From (15.23), we have pN ¼ lk pN1 ¼ qN p0 ) pn ¼ qn p0 for 0  n  N: Since the capacity of the system is N,

ð15:25Þ

15.3

Poisson Queueing Models N X

431

pn ¼ 1

n¼0

or p0 or

1  qN þ 1 ¼ 1 for q 6¼ 1 and ðN þ 1Þp0 ¼ 1 for q ¼ 1 1q ( 1q 1qN þ 1 for q 6¼ 1 p0 ¼ : 1 for q ¼ 1 N þ1

From (15.9), we have ( pn ¼ q p0 ¼ n

ð1qÞqn 1qN þ 1 1 N þ1

for 0  n  N and q 6¼ 1 for 0  n  N and q ¼ 1

Note (i) This solution holds for any value of q, as the number of customers allowed in the system is controlled by the queue length (N – 1), not by the rates of arrival and departure k and l. So in this model we do not require the condition q\1, and hence the system is stable for q\1 as well as q  1. (ii) keff is the effective arrival rate in the system, i.e. the mean rate of customers entering the system. It equals the arrival rate k when all arriving customers can join the system; otherwise, keff \k. Since the probability that a customer does not join the system is pN, the probability that the customers join the system is ð1  pN Þ. Hence, keff ¼ kð1  pN Þ. (iii) In this model, customers arrive at the rate k, but not all arrivals can join the system. So the equations Ls ¼ kWs and Lq ¼ kWq are modiﬁed by redeﬁning k to include only those customers who actually join the system. Hence, for this model, Ls ¼ keff Ws and Lq ¼ keff Wq , where keff ¼ effective arrival rate for those who join the system.

Characteristics of the Model (i) Expected line length, i.e. average number of customers in the system The expected line length, i.e. the average number of customers in the system, is given by N 1q n 1q X q ¼ nqn 1  qN þ 1 1  qN þ 1 n¼0 n¼0 n¼0  1q  ¼ q þ 2q2 þ 3q3 þ    þ NqN : N þ 1 1q

Ls ¼ EðnÞ ¼

N X

npn ¼

N X

n

when q 6¼ 1

432

15

Queueing Theory

[Let S ¼ q þ 2q2 þ 3q3 þ    þ NqN . ) qS ¼ q2 þ 2q3 þ    þ NqN þ 1 . Subtracting the second equation from the ﬁrst, we have Sð1  qÞ ¼ q þ q2 þ q3 þ    n times  NqN þ 1 or or

q ð1  q N Þ  NqN þ 1 Sð1  qÞ ¼ 1q   qð1  qN Þ 1 N þ1  Nq S¼ 1q 1q   1q qð1  qN Þ N þ1  Nq ¼ =ð1  qÞ 1  qN þ 1 1q   1 q  qN þ 1  NqN þ 1 þ NqN þ 2 ¼ 1  qN þ 1 1q N N þ1 q½1  ðN þ 1Þq þ Nq  when q 6¼ 1: ¼ ð1  qÞð1  qN þ 1 Þ

For q ¼ 1; Ls ¼

N X

npn ¼

N X

n¼0

"

For q ¼ 1; pn ¼

1 N þ1

n 1 N ¼ ð1 þ 2 þ    þ NÞ ¼ N þ 1 N þ 1 2 n¼0 # N X and pn ¼ 1 : n¼0

(ii) Average number of customers in the queue The average number of customers in the queue ¼ Lq ¼

N X

ðn  1Þpn ¼

n¼1

¼

N X n¼0

npn 

N X n¼1

N X

npn 

N X n¼1

pn "

pn þ p0 ¼ Ls  1 þ p0 * Ls ¼

n¼0

N X n¼0

 q½1  ðN þ 1ÞqN þ NqN þ 1  1q  1  ð1  qÞð1  qN þ 1 Þ 1  qN þ 1     ¼ q2 1  NqN1 þ ðN  1ÞqN =ð1  qÞ 1  qN1 : ¼

(iii) Waiting time in the queue The waiting time in the queue is

# npn

15.3

Poisson Queueing Models

433

Lq Lq ¼ ½* keff ¼ kð1  pN Þ keff kð1  pN Þ Lq q2 ½1  NqN1 þ ðN  1ÞqN  h i¼ : N kð1  qN Þð1  qÞ k 1  ð1qÞq N þ1

Wq ¼

1q

(iv) Waiting time in the system The waiting time in the system is given by Ws ¼ Wq þ

 1 Ls or : l keff

Special case: If q\1 and N ! 1, then the steady state solution is pn ¼ ð1  qÞqn , and the resulting system reduces to the result ðM=M=1Þ:ð1=FCFS=1Þ. Example 3 In a car wash service facility, information gathering indicates that cars arrive for service according to a Poisson distribution with mean 5 per hour. The time for washing and cleaning for each car varies but is found to follow an exponential distribution with mean 10 min per car. The facility cannot handle more than one car at a time and has a total of 5 parking spaces. If the parking spot is full, newly arriving cars balk to seek services elsewhere. (a) How many customers is the manager of the facility losing due to the limited parking space? (b) What is the expected waiting time until a car is washed? Solution In this system, there may be 5 cars in the 5 parking spots and one can be serviced; i.e. the capacity of this system is N = 5 + 1 = 6. Hence, this problem is of the model (M/M/1):(6/FCFS/∞). 1 Here k = 5 per hour, l = 10  60 = 6 per hour ) q ¼ lk ¼ 56. 1q 1q 6 N Now pN ¼ p6 ¼ 1q 6 þ 1 q ¼ 0:0774 [Since pN ¼ 1qN þ 1 q ].

(a) Therefore, the rate at which the cars balk ¼ k  keff ¼ k  kð1  pN Þ ¼ kp6 ¼ 5  :0774 ¼ 0:387 car/h: Assuming 8 working hours per day, a manager will lose 0.387  8 = 3.096 = 3 cars a day.

434

15

Queueing Theory

(b) The expected waiting time until a car is washed is given by Ws ¼

Ls keff

Nþ1

þ 1Þq þ Nq  Here Ls ¼ q½1ðN = 2.29 h, ð1qÞð1qN þ 1 Þ and keff = k(1 − pN) = k(1 − p6) = 4.613. Hence, Ws ¼ kLeffs = 0.496 h. N

15.3.4 Model (M/M/C):(∞/FCFS/∞) Assumptions In this model, we shall discuss a multiserver queueing system under the following assumptions: (i) The arrival rate follows a Poisson distribution. (ii) This queueing system deals with queues which are being served by parallel service channels in which the service time of each server has an independently and identically distributed with exponential service time distribution. (iii) This queueing system has cð [ 1Þ service channels. (iv) The service discipline is ﬁrst come ﬁrst served. (v) The capacity of the system is inﬁnite. (vi) The length of the waiting line will depend on the number of occupied channels. Model Formulation Let n be the number of customers in the system. Therefore, according to assumption (i), the mean arrival rate is given by kn ¼ k for all n. For the mean service rate, if there are more than c customers in the system ðn [ cÞ, all the c servers will remain busy and each is providing service at a mean rate l, while ðn  cÞ customers will be waiting in the queue. Thus, the mean service rate is cl. If n = c, then all the service channels will be busy, each providing at a mean service rate l, and thus the mean rate of service is cl. If there are fewer than c customers in the system ðn\cÞ, only n of the c servers will be busy, and in this case ðc  nÞ servers will remain idle; there will be no customer waiting in the queue. Thus, the mean service rate is nl. Hence, for this model, kn ¼ k; n ¼ 0; 1; 2; . . .  nl; 0  n\c ; ln ¼ cl n  c

15.3

Poisson Queueing Models

435

i.e. the probability of one arrival during Dt is k. Dt + O(Dt), and the probability of one departure during Dt is nlDt þ OðDtÞ if 0\n\c and clDt þ OðDtÞ

if n  c:

To have zero customer in the system during time t + Dt, there may be the following mutually exclusive cases: (i) No customer in the system at time t, no arrival in time Dt (ii) One customer in the system at time t, no arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of no customer in the system at time (t + Dt) is given by or or

p0 ðt þ DtÞ ¼ p0 ðtÞð1  kDtÞ þ p1 ðtÞð1  kDtÞlDt þ OðDtÞ p0 ðt þ DtÞp0 ðtÞ ¼ kp0 ðtÞDt þ p1 ðtÞlDt þ OðDtÞ p0 ðt þ DtÞp0 ðtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ þ OðDtÞ Dt Dt :

ð15:26Þ

To have n customers in the system at time t + Dt, there may be the following mutually exclusive cases: (i) n customers in the system at time t, no arrival in time Dt, no service in time Dt (ii) (n − 1) customers in the system at time t, one arrival in time Dt, no service in time Dt (iii) (n + 1) customers in the system at time t, no arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of n customers in the system at time (t + Dt) is given by pn ðt þ DtÞ ¼ pn ðtÞð1  kDtÞð1  ln DtÞ þ pn1 ðtÞkDtð1  ln1 DtÞ þ pn þ 1 ðtÞð1  kDtÞln þ 1 DDt þ OðDtÞ pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn1 ðtÞ  ðk þ ln Þpn ðtÞ þ ln þ 1 pn þ 1 ðtÞ þ : or Dt Dt ð15:27Þ (a) When 0 < n < c, then ln ¼ nl; ln þ 1 ¼ ðn þ 1Þl. Hence, from (15.27), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn1 ðtÞ  ðk þ nlÞpn ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ þ : Dt Dt ð15:28Þ

436

15

Queueing Theory

(b) When n  c, then ln ¼ cl, ln þ 1 ¼ cl. Hence, from (15.27), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpn1 ðtÞ  ðk þ clÞpn ðtÞ þ clpn þ 1 ðtÞ þ : Dt Dt

ð15:29Þ

Now taking the limit as Dt ! 0, the differential difference equations for this system obtained from (15.26), (15.28) and (15.29) are as follows: p00 ðtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ

for n ¼ 0

p0n ðtÞ ¼ kpn1 ðtÞ  ðk þ nlÞpn ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ; for 0\n\c p0n ðtÞ ¼ kpn1 ðtÞ  ðk þ clÞpn ðtÞ þ clpn þ 1 ðtÞ; for n  c:

Under the steady state condition of the system, i.e. lim pn ðtÞ ¼ pn and lim p0n ðtÞ ¼ 0, t!1

t!1

the preceding three equations reduce to kp0 þ lp1 ¼ 0 for n ¼ 0 kpn1  ðk þ nlÞpn þ ðn þ 1Þlpn þ 1 ¼ 0 kpn1  ðk þ clÞpn þ clpn þ 1 ¼ 0

for for

ð15:30Þ 0\n\c

ð15:31Þ

n  c;

ð15:32Þ

which are the steady state differential difference equations of the system. Solution From (15.30), k kp0 þ lp1 ¼ 0 or, lp1 ¼ kp0 or; p1 ¼ p0 l Now setting n = 1 in (15.31), we have kp0  ðk þ lÞp1 þ 2lp2 ¼ 0 kþl k p1  p0 or p2 ¼ 2l 2l    k 1 k 2 k p1 ¼ ¼ p0 * p1 ¼ p 0 : 2l ð2Þ! l l  2 1 k p0 : ) p2 ¼ ð2Þ! l Again setting n = 2 in (15.31), we have

15.3

Poisson Queueing Models

437

kp1  ðk þ 2lÞp2 þ 3lp3 ¼ 0 k þ 2l k p2  p1 or, p3 ¼ 3l 3l  1 k 3 ¼ p0 : ð3Þ! l Hence, we have pn ¼ n!1

n k l

) pc1 ¼

p0 for 1  n < c.

 c1 1 k 1 k pc2 : p0 ¼ ðc  1Þ! l c  1l

Now setting n = c – 1 in (15.31), we have kpc2  ½k þ ðc  1Þlpc1 þ clpc ¼ 0 k þ ðc  1Þl k pc1  pc2 or pc ¼ cl cl k c1 c1 pc1  pc1 ½by ð15:8Þ ¼ pc1 þ cl c c  1 k c ðcqÞc k p0 where q ¼ : ¼ p0 ¼ ðcÞ! l cl ðcÞ! Setting n = c in (15.32), we have kpc1  ðk þ clÞpc þ clpc þ 1 ¼ 0 k þ cl k pc  pc1 or pc þ 1 ¼ cl cl   k k ¼ pc þ pc  pc * pc ¼ pc1 cl cl  c þ 1 1 k ¼ p0 : cðcÞ! l 1 Similarly, pc þ 2 ¼ c2 ðcÞ!

c þ 2 k l

p0 .

In general,  n 1 k p0 for n  c cnc ðcÞ! l  n cc k pn ¼ n p0 for n  c: c ðcÞ! l pn ¼

ð15:33Þ

438

15

Queueing Theory

Therefore,  1 k n p0 for 1  n\c pn ¼ ðnÞ! l  cc k n ¼ n p0 for n  c: c ðcÞ! l

Determination of p0 : 1 P

We have

pn ¼ 1

n¼0 1 P pn þ pn ¼ 1 n¼c n¼1 n n cP 1 1 P 1 k cc k p0 þ ðnÞ! l p0 þ cna ðcÞ! l p0 ¼ 1 n¼c n¼1 c1  1 n P 1 k n P cc k þ ðcÞ! p0 ¼1 ðnÞ! l cl n¼c n¼0  cP 1 1 P n 1 cc k qn ¼ 1 where q ¼ cl p0 ðnÞ! ðcqÞ þ ðcÞ!

p0 þ

or or or or

cP 1

n¼c

n¼0

p0 ¼ hPc1 1n ðcqÞ

or

n¼0 ðnÞ!

ð15:34Þ

þ

ðcqÞc 1 ðcÞ! 1q

i:

Hence, 8
c, a queue of n customers would consist of c customers being served together with a genuine queue of (n − c) waiting customers. Hence, Lq ¼

1 X

ðn  cÞpn

n¼c þ 1 1 X

cc qn p0 ¼ ðn  cÞ ðcÞ! n¼c þ 1 ¼

  cc n k q p0 where q ¼ * pn ¼ cl ðcÞ!

1 X cc qc p0 ðn  cÞqnc ðcÞ! n¼c þ 1

 ðcqÞc  p0 q þ 2q2 þ 3q3 þ    ðcÞ! q ¼ pc ð1  qÞ2 q ) Lq ¼ pc : ð1  qÞ2 ¼

Particular case: When c = 1; i.e. for a system with one channel only, Lq ¼

q ð1  qÞ

2

p1 ¼

q 2

ð1  qÞ

qð1  qÞ ¼

q2 : 1q

(ii) Average number of customers in the system We know that Ls ¼ Lq þ lk. ) Ls ¼

qpc ð1  qÞ

2

þ

k qpc ¼ þ cq l ð1  qÞ2

where q ¼

(iii) Average waiting time The average waiting time in the system is Average number of customers in the system rate of arrival Ls qpc =ð1  qÞ2 þ cq qpc cq ¼ ¼ ¼ þ : 2 k k k kð1  qÞ

Ws ¼

k cl

440

15

Queueing Theory

The average waiting time in the queue is Average number of customers in the queue rate of arrival Lq qpc ¼ ¼ : k kð1  qÞ2

Wq ¼

(iv) Expected length of non-empty queue, i.e. to ﬁnd L/L > 0

ðL=L [ 0Þ ¼

Lq prob[an arrival has to wait]

¼

Lq qpc =ð1  qÞ2 qpc =ð1  qÞ2 ¼ P1 ¼ P1 cc n probðn [ cÞ n¼c þ 1 pn n¼c þ 1 jc q p0

¼

qpc =ð1  qÞ2 qpc =ð1  qÞ2 q=ð1  qÞ2 ¼ ¼ P 2 3 2 ðcqÞ 1 p qnc pc ðq þ q þ q þ   Þ qð1 þ q þ q þ   Þ jc 0 n¼c þ 1

¼

2

q=ð1  qÞ2 q 1q

¼

1 : 1q

(v) Probability that a customer has to wait Probability that a customer has to wait ¼ probðn  cÞ ¼ c



1 X n¼c

pn ¼ 

1 c X c n¼c

jc

q n p0 ¼

1 cc X qn p0 jc n¼c

c p0 qc þ qc þ 1 þ    jc cc ðqcÞc 1 1 pc ¼ pc : ¼ p0 : ¼ p0 qc ð1 þ q þ q2 þ   Þ ¼ 1q 1q 1q jc jc ðqcÞc since pc ¼ p0 : jc

¼

(vi) Probability that on arrival a customer will not have to wait pc . ¼ 1  probðn  cÞ ¼ 1  1q Alternatively, we can ﬁnd this probability using the formula prob P (n < c) = c1 n¼0 pn .

15.3

Poisson Queueing Models

441

The probability that all channels will be occupied ¼probðn  cÞ ¼

pc : 1q

Example 4 A telephone exchange has two long-distance operators. The telephone company ﬁnds that, during the peak load, long-distance calls arrive in a Poisson fashion at an average rate of 15 per hour. The length of service on these calls is approximately exponentially distributed with mean length 5 min. (a) What is the probability that a subscriber will have to wait for this long-distance call during the peak hours of the day? (b) If the subscriber waits and is serviced in turn, what is the expected waiting time? Solution This problem is of the model (M/M/c):(∞/FCFS/∞). 1

1 1 k 4 Here c = 2 and k = 15 calls/h ¼ 15 60 ¼ 4 calls=min, l ¼ 5 and q ¼ cl ¼ 2:1 ¼ 5 8

5

and cq ¼ 2  58 ¼ 54. Now, p0 ¼ Pc1 ðcqÞ1n n¼0

jn

þ

ðcqÞc

jc ð1qÞ

¼P

1

1 n¼0

ð54Þ jn

n

ð 5Þ þ 4 5 j2 ð18Þ n

¼

1 1þ

5 4 j1

25

þ 163

3 ¼ 13 .

2: 8

(a) The probability that a subscriber will have to wait for his long-distance call ¼ probðn  cÞ ¼

pc 1q

¼

ðcqÞc p jc 0

1q

c ð 5Þ : 3 ¼ ðcqÞ p0 ¼  4 13 ¼ 25 52 ¼ 0:48. cð1qÞ 2ð58Þ 2

(b) The expected waiting time 52 5 Lq qpc q ðcqÞ2 3 8 4 ¼ Wq ¼ ¼ ¼ : p0 ¼  : 2 : 2 2 1 5 k jc j2 13 kð1  qÞ kð1  qÞ 4 18 ¼

5 25 125 8 16 3 ¼ 3:2 min: : ¼ :   1 3 2 2 13 39 4: 8

Example 5 A supermarket has two girls ringing up sales at the counters. If the service time for each customer is exponential with mean 4 min, and if the people arrive in a Poisson fashion at the counter at the rate 10 per hour, then: (a) What is the probability of having to wait for service? (b) What is the expected percentage of idle time for each girl? (c) If a customer has to wait, what is the expected length of his waiting time?

442

15

Queueing Theory

Solution This problem is of the model (M/M/c):(∞/FCFS/∞). 1 1 Here c = 2 k ¼ 10 60 ¼ 6 people per minute, l ¼ 4 people per minute. 1 1 k 2 1 ¼ 6 1 ¼ 61 ¼ ¼ cl 2: 4 2 6 3 1 2 ) cq ¼ 2  ¼ 3 3

) q¼

1 1 ¼ ¼ 12. 2 2 2 P1 ð23Þn ð23Þ2 ð 3Þ 3 1þ þ 2 þ n¼0 jn 2 j1 j2 ð113Þ 3 (a) The probability of having to wait for service

Now, p0 ¼ Pc1 ðcqÞ12 n¼0

jn

þ

ðcqÞ2

¼

jc ð1qÞ

ð23Þ :1 2

¼ probðn  2Þ ¼

P1 n¼2

pn ¼

pc 1q

¼

ðcqÞc p jc 0

1q

¼

j2 2 113

¼ 0:167.

(b) The expected number of girls who are idle h i 1 ¼ 2  p0 þ 1  p1 ¼ 2  12 þ 1  13 ¼ 1 þ 13 ¼ 43 p1 ¼ lk p0 ¼ 61  12 ¼ 13 . 4

Now the probability of any girl being idle expected number of idle girls 43 2 ¼ 2 ¼ 3 ¼ 0:67. total number of girls Hence, the expected percentage of idle time for each girl = 0.67  100% = 67%. ¼

(c) The expected length of the customer waiting time on the condition that the customer has to wait ¼ W=W [ 0 ¼ ¼

1 1 ¼ Mean service rate  mean arrival rate for n  c cl  k

1 ¼ 3 min: 2l  k

15.3.5 Model (M/M/C):(N/FCFS/∞) Assumptions In this model, we shall discuss a multiserver queueing system under the following assumptions:

15.3

Poisson Queueing Models

443

(i) (ii) (iii) (iv)

Customers arrive in a Poisson fashion. The service rate follows a Poisson distribution. This queueing system has cð [ 1Þ service channels. The maximum number of customers in the system is limited to N, where N [ c; the capacity of the system is ﬁnite. (v) The service discipline is ﬁrst come ﬁrst served.

Model Formulation Let n be the number of customers in the system and k, l be the mean arrival rate and mean service rate of units respectively. In this model,  kn ¼

k 0

when 0  n\N when n  N

and  ln ¼

nl when 0  n\c cl when c  n  N

Hence, the probability of one arrival during time Dt is kn Dt þ OðDtÞ, and the probability of one departure or service during time Dt is ln Dt þ OðDtÞ. To have zero customer in the system at time t + Dt, there may be the following mutually exclusive cases: (i) No customer in the system at time t, no arrival during time Dt (ii) One customer in the system at time t, no arrival during time Dt, one service during time Dt, and so on. Therefore, the probability of no customer in the system at time t + Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  k0 DtÞ þ p1 ðtÞð1  k1 DtÞl1 Dt þ OðDtÞ  Here k0 ¼ k or p0 ðt þ DtÞ  p0 ðtÞ ¼ kp0 ðtÞDt þ lp1 ðtÞDt þ OðDtÞ k1 ¼ k; l1 ¼ l p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ þ for n ¼ 0: or Dt Dt Taking the limit as Dt ! 0, we have p00 ðtÞ ¼ kp0 ðtÞ þ lp1 ðtÞ for n ¼ 0:

ð15:36Þ

To have n customers in the system at time t + Dt, there may be the following three cases:

444

15

Queueing Theory

(i) n customers in the system at time t, no arrival during time Dt, no service during time Dt (ii) (n – 1) customers in the system at time t, one arrival during time Dt, no service during time Dt (iii) (n + 1) customers in the system at time t, no arrival during time Dt, one service during time Dt, and so on. Therefore, the probability of n customers in the system at time (t + Dt) is given by pn ðt þ DtÞ ¼ pn ðtÞð1  kn DtÞð1  ln DtÞ þ pn1 ðtÞkn1 Dtð1  ln1 DtÞ þ pn þ 1 ðtÞð1  kn þ 1 DtÞln þ 1 Dt þ OðDtÞ or pn ðt þ DtÞ  pn ðtÞ ¼ ðkn þ ln Þpn ðtÞDt þ kn1 pn1 ðtÞDt þ ln1 pn þ 1 ðtÞDt þ OðDtÞ pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ ðkn þ ln Þpn ðtÞ þ kn1 pn1 ðtÞ þ ln þ 1 pn þ 1 ðtÞ þ : or Dt Dt

ð15:37Þ

(a) For 0 < n < c, kn1 ¼ kn ¼ k;

ln ¼ nl; ln þ 1 ¼ ðn þ 1Þl:

Hence, from (15.37), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ ðk þ nlÞpn ðtÞ þ kpn1 ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ þ : Dt Dt Taking the limit as Dt ! 0 of both sides, we have p0n ðtÞ ¼ ðk þ nlÞpn ðtÞ þ kpn1 ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ:

ð15:38Þ

(b) For c  n\N; kn1 ¼ k; kn ¼ k; ln ¼ cl; ln þ 1 ¼ cl: Hence, from (15.37), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ ðk þ clÞpn ðtÞ þ kpn1 ðtÞ þ clpn þ 1 ðtÞ þ : Dt Dt Taking the limit as Dt ! 0 of both sides, we have

15.3

Poisson Queueing Models

445

p0n ðtÞ ¼ ðk þ clÞpn ðtÞ þ kpn1 ðtÞ þ clpn þ 1 ðtÞ for c  n\N

ð15:39Þ

(c) For n = N, kn ¼ kN ¼ 0; kn1 ¼ kN1 ¼ k ln ¼ cN ¼ cl; pn þ 1 ðtÞ ¼ 0 . Hence, from (15.37), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ ðo þ clÞpn ðtÞ þ kpn1 ðtÞ þ : Dt Dt Taking the limit as Dt ! 0 of both sides, we have p0n ðtÞ ¼ clpn ðtÞ þ kpn1 ðtÞ

p0N ðtÞ ¼ clpN ðtÞ þ kpN1 ðtÞ

Under a steady state condition of the system, i.e. lim p0 ðtÞ t!1 n

ð15:40Þ

for n ¼ N

¼ 0, Eqs. (15.36), (15.38)–(15.40) reduce to:

lim pn ðtÞ ¼ pn and

t!1

kp0 þ lp1 ¼ 0 for n ¼ 0 kpn1  ðk þ nlÞpn þ ðn þ 1Þlpn þ 1 ¼ 0

ð15:41Þ ð15:42Þ

for 0\n\c

kpn1  ðk þ clÞpn þ clpn þ 1 ¼ 0 for c  n\N

ð15:43Þ

kpN1  clpN ¼ 0 for n ¼ N:

ð15:44Þ

Solution of the Model

pn ¼ where

1

8 cP 1 > ðcqÞn > > < n! þ n¼0 p0 ¼  cP 1 > > ðcqÞn > : n! þ n¼0

Characteristics of the Model (i) Average queue length The average queue length is

n n!c ðncqÞ p0 ; cq c! p0 ;

for 0  n\c ; for c  n  N

ðcqÞc 1qNc þ 1 c! 1q ðcqÞc c!

1

1 ðN  c þ 1 Þ

k for q ¼ cl 6¼ 1 k for q ¼ cl ¼1

:

446

15

Lq ¼

N X

ðn  cÞpn ¼

n¼c þ 1

¼ ¼

c

c p0 jc

N X

ðn  cÞpn ¼

n¼c

N X

ðn  cÞqn ¼

n¼c

N X

ðn  cÞ

n¼c c

c p0 qc jc

N X

Queueing Theory

cc qn p0 jc

ðn  cÞqnc

n¼c

N c ðcqÞc X p0 xqx putting n  c ¼ x jc x¼0

N c ðcqÞc X d p0 q ðqx Þ dq jc n¼0 ( ) c N c ðcqÞ d X x q ¼ p0 q dq x¼0 jc   ðcqÞc d 1  qNc þ 1 ¼ p0 q dq 1q jc 8 c ðcqÞ 1  qNc þ 1  ð1  qÞðN  c þ 1ÞqNc k > > 6¼ 1 for q ¼ > < jc p 0 q cl ð1  qÞ2 ¼ > > cc ðn  cÞðN  c þ 1Þ k > : p0 ¼1 for k ¼ Lj c cl

¼

(ii) Average number of customers in the system Ls = average number of customers in the system ¼ EðnÞ ¼

N X

npn ¼

n¼0

¼

N X

¼ Lq þ c

"

npn þ

N X

cpn þ

n¼c N X

pn 

n¼0

¼ Lq þ c  c

c1 X

#

c1 X

pn þ

n¼0

¼ Lq þ c  p0

pn þ

npn

c1 X

npn

n¼0

c1 X

npn ¼ Lq þ c þ

n¼0

c1 X ðc  nÞðqcÞn n¼0

npn

n¼0

n¼0

c1 X

N X n¼c

n¼0

ðn  cÞpn þ

n¼c

c1 X

jn

Calculation of keff: keff ¼ kð1  pN Þ Let c = expected number of idle servers

c1 X n¼0

ðn  cÞpn

15.3

Poisson Queueing Models

) c ¼

c X

447

ðc  nÞpn

n¼0

) c  c ¼ the expected number of busy servers. ) keff ¼ lðc  cÞ Lq ks and Wq ¼ Ws ¼ keff keff L

Now, Ws ¼ kkseff and Wq ¼ keffq . Special cases: (i) Let us take c ¼ 1; N ! 1 and consider q\1. Then we get p0 ¼ 1  q and pn ¼ qn p0 , n ¼ 0; 1; 2; . . . This result is exactly the same as that of model ðM=M=1Þ:ð1=FCFS=1Þ. (ii) Let us take c = 1. Then we get ( p0 ¼

1q 1qN þ 1 1 1þN ;

k ; q ¼ cl 6¼ 1 k q ¼ cl ¼ 1

and pn ¼ qn p0 ; 1  n  N. This result corresponds to that of the queueing model ðM=M=1Þ:ðN=FCFS=1Þ. k (iii) Let us take N ! 1 and consider q ¼ cl \1.

Then we get " p0 ¼

c1 X ðcqÞn n¼0

n!

ðcqÞc 1  qNc þ 1 þ c! 1q

#1

8 1 > < ðcqÞn p0 ; 0  n\c , and pn ¼ n! > : 1 cc qn p ; c  n  N 0 c! which is the steady state solution of the queueing model ðM=M=cÞ:ðN= FCFS=1Þ. Thus, we can consider this model as a general queueing model, as we can get the results of previous queueing models from this one.

448

15

Queueing Theory

Machine Repairing Problem Whenever a machine breaks down, it will result in a great loss to an organization. Thus, machine repair problems are very important problems in queueing theory. When a machine breaks down, at that point the machine starts to be repaired. During this time, if another machine breaks down, then it will be attended after completion of the repair of the ﬁrst machine. Thus, the broken-down machines form a queue and wait for their repair. There are various situations. There may be one or more than one mechanic. If there is one mechanic, we have a problem of single channel; if there is more than one mechanic, we have a multichannel problem. The machines may be repaired in a single phase or in k phases.

15.3.6 Model (M/M/R):(K/GD) K > R Assumptions In this model, we shall discuss a queueing system under the following assumptions: (i) This is a queueing system in which there are R service channels. (ii) The maximum limit imposed on incoming customers, say k, is ﬁxed. (iii) The arrivals (breakdowns) and services are assumed to follow a Poisson distribution with parameters k and l respectively. (iv) This system has a ﬁnite calling source. Model Formulation Let n be the number of machines in a Hence, the approximate probability of where 8 < nl ln ¼ Rl : 0

breakdown situation. a single service during the time Dt is lnDt, for n  R for R  n  k : for n [ k

If k is the rate of breakdown per machine, then the probability of a single arrival during the time Dt (when there are n machines in a breakdown situation) is approximately knDt for n  k, where ( kn ¼

ðk  nÞk for 0  n\k 0

for n  k

:

To have 0 customer in the system at time t + Dt, there may be the following three cases: (i) No customer in the system at time t, no arrival in time Dt

15.3

Poisson Queueing Models

449

(ii) One customer in the system at time t, no arrival in time Dt, one service in time Dt (iii) No customer in the system at time t, one arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of no customers in the system at time t + Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  k0 DtÞ þ p1 ðtÞð1  k1 DtÞl1 Dt þ OðDtÞ ¼ p0 ðtÞð1  kkDtÞ þ p1 ðtÞf1  ðk  1ÞkDtglDt þ OðDtÞ p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kkp0 ðtÞ þ lp1 ðtÞ þ : or Dt Dt

ð15:45Þ

To have n (>0) customers in the system at time t + Dt, there may be the following three cases: (i) n customers in the system at time t, no arrival in time Dt, no service in time Dt (ii) (n – 1) customers in the system at time t, one arrival in time Dt, no service in time Dt (iii) (n + 1) customers in the system at time t, no arrival in time Dt, one service in time Dt, and so on. Therefore, the probability of n customers in the system at time (t + Dt) is given by pn ðt þ DtÞ ¼ pn ðtÞð1  kn DtÞð1  ln DtÞ þ pn1 ðtÞkn1 Dtð1  ln1 Dt þ pn þ 1 ðtÞð1  kn þ 1 DtÞln þ 1 Dt þ OðDtÞ or

pn ðt þ DtÞ  pn ðtÞ ¼ ðkn þ ln Þpn ðtÞ Dt þ kn1 pn1 ðtÞ þ ln þ 1 pn þ 1 ðtÞ þ ½dividing both sides by Dt:

ð15:46Þ OðDtÞ Dt

When 0 < n < R, then kn1 ¼ fk  ðn  1Þgk kn ¼ ðk  nÞk; kn þ 1 ¼ fk  ðn þ 1Þgk ln ¼ nl; ln þ 1 ¼ ðn þ 1Þl Hence, from (15.46), we have pn ðt þ DtÞ  pn ðtÞ ¼ fðk  nÞk þ nlgpn ðtÞ þ fk  ðn  1Þgkpn1 ðtÞ Dt OðDtÞ þ ðn þ 1Þlpn þ 1 ðtÞ þ Dt

ð15:47Þ

450

15

Queueing Theory

For R  n < k, kn1 ¼ fk  ðn  1Þgk; kn ¼ ðk  nÞk; ln ¼ Rl; ln þ 1 ¼ Rl; Hence, from (15.46), we have pn ðt þ DtÞ  pn ðtÞ ¼ fðk  nÞk þ Rlgpn ðtÞ þ fk  ðn  1Þgkpn1 ðtÞ Dt OðDtÞ for R  n\k þ Rlpn þ 1 ðtÞ þ Dt

ð15:48Þ

For n = k, kn1 ¼ fk  ðn  1Þgk ¼ fk  ðk  1Þgk ¼ k½)n ¼ k kn ¼ ðk  nÞk ¼ 0 ln ¼ Rl Hence, from (15.46), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ ð0 þ RlÞpn ðtÞ þ kpn1 ðtÞ þ Dt Dt pk ðt þ DtÞ  pk ðtÞ OðDtÞ i:e: ¼ kpk1 ðtÞ þ Rlpk ðtÞ þ : Dt Dt

for n ¼ k;

ð15:49Þ

Now taking the limit as Dt ! 0 in (15.45), (15.47)–(15.49), we have p0n ðtÞ ¼ lkp0 ðtÞ þ lp1 ðtÞ for n ¼ 0 p0n ðtÞ ¼ fðk  nÞk þ nlgpn ðtÞ þ fk  ðn  1Þgkpn1 ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ for 0\n\R p0n ðtÞ ¼ fðk  nÞk þ Rlgpn ðtÞ þ fk  ðn  1Þgkpn1 ðtÞ þ Rlpn þ 1 ðtÞ for R  n\k p0n ðtÞ ¼ kpk1 ðtÞ þ Rlpk ðtÞ for n ¼ k:

Under a steady state condition of this system, i.e. lim pn ðtÞ ¼ pn and lim p0n ðtÞ ¼ 0, the preceding four equations reduce to

t!1

t!1

kkp0 þ lp1 ¼ 0 for n ¼ 0

ð15:50Þ

15.3

Poisson Queueing Models

451

fk  ðn  1Þgkpn1  fðk  nÞk þ nlgpn þ ðn þ 1Þlpn þ 1 ¼ 0 fk  ðn  1Þgkpn1  fðk  nÞk þ Rlgpn þ Rlpn þ 1 ¼ 0 kpk1  Rlpk ¼ 0

for 0\n\R ð15:51Þ

for R  n\k ð15:52Þ

for n ¼ k:

Solution From (15.50), lp1 ¼ kkp0 k or p1 ¼ k p0 or p1 ¼ kqp0 l

k where q ¼ : l

Now setting n = 1 in (15.51), we have kkp0  fðk  1Þk þ lgp1 þ 2lp2 ¼ 0 2lp2 ¼ fðk  1Þk þ lgp1  kkp0   k k 2p2 ¼ ðk  1Þ þ 1 p1  k p0 l l   k ¼ fðk  1Þq þ 1gp1  kqp0 * q ¼ l

or or

¼ ðk  1Þqp1 þ p1  p1 ½* p1 ¼ kqp0  ¼ ðk  1Þqp1 ¼ ðk  1Þqkqp0 ¼ kðk  1Þq2 p0 ! k kðk  1Þ 2 q p0 ¼ p2 ¼ q 2 p0 : 2 2 Again, setting n = 2 in (15.51), we have ðk  2Þðk  1Þkq3 p0 2  k ðk  2Þðk  1Þkq3 p0 p3 ¼ ¼ q3 p 0 : 3:2 3

3p3 ¼ or

 k By induction, pn ¼ qn p0 for 0  n  R. n Setting n = R in (15.52), we have

ð15:53Þ

452

15

Queueing Theory

fk  ðR  1ÞgkpR1  fðk  RÞk þ RlgpR þ RlpR þ 1 ¼ 0 or RpR þ 1 ¼ fðk  RÞq þ RgpR  ðk  R þ 1ÞqpR1   k k R ¼ fðk  RÞq þ Rg qR1 p0 q p0  ðk  R þ 1Þq R1 R   k k R ¼ fðk  RÞq þ Rg q p0  R qR p0 R R  k ¼ qR p0 ½ðk  RÞq þ R  R R  k ¼ ðk  RÞqR þ 1 p0 R  k ¼ ðR þ 1ÞqR þ 1 p0 Rþ1  k R þ 1 Rþ1 q ) pR þ 1 ¼ p0 R Rþ1 Again, setting n = R + 1 in (15.52), we have

ðk  RÞqpR  ðk  R þ 1Þq þ R pR þ 1 þ RpR þ 2 ¼ 0 or

RpR þ 2 ¼ fðk  R  1Þq þ RgpR þ 1  ðk  RÞqpR ¼ ðk  R  1ÞqpR þ 1 þ RpR þ 1  ðk  RÞqpR  ðk  R  1ÞðR þ 1Þ k ) pR þ 2 ¼ qR þ 2 p 0 R2 Rþ1 ðk  R  1ÞðR þ 1Þ jk qR þ 2 p 0 ¼ R2 j R þ 1j K  R  1 ðR þ 1ÞðR þ 2Þ jk qR þ 2 p 0 ¼ 2 R jR þ 2 jk  ðR þ 2Þ   k k ðR þ 1ÞðR þ 2Þ R þ 2 jR þ 2 R þ 2 ¼ q p ¼ q p0 0 R2 Rþ2 R þ 2 jR R2  k jR þ i R þ i ) pR þ i ¼ q p0 for R  R þ i\k: R þ i jR Ri

15.3

Poisson Queueing Models

453

From (15.53), we have Rlpk ¼ kpk1

 k k jk  1 k1 or Rpk ¼ pk1 ¼ q q p0 l k  1 jR RkR1    k k jk  1 k k k jk  1 k jk q p0 ¼ q p0 ¼ qk p0 : kR1 kR1 k jR R k jR RkR1 1 jR R  k jk ) pk ¼ qk p0 : k jR RkR

Thus, 8 k > n > for 0  n  R > < n q p0 : pn ¼  > k jn > n > q p for R  n  k : 0 n jR RnR Now, or or or

n P

pn ¼ 1

n¼0 R1 P

k P pn þ pn ¼ 1 n¼0  n¼R  R1 k P P k k jn qn p0 þ qn p0 ¼ 1 n n jR RnR n¼R n¼0 1   p0 ¼ : PR1 k n Pn k qn j n q þ n¼0 n¼R n n RnR jR

Characteristics of the Model (i) Average number of customers in the system The average number of customers in the system is given by Ls ¼ EðnÞ ¼

k X

npn ¼

n¼0

R1 X n¼0

k X n¼R

npn

! k k 1 X qn jn ¼ n q þ n nR jR n¼R n n jR R n¼0 ! ! " # n1 k n X X k k 1 q n j ¼ p0 n qn þ n nR jR n¼R n n R n¼0 R1 X

k

!

npn þ

n

454

15

Queueing Theory

(ii) Expected queue length The expected queue length is given by Lq ¼

k X

ðn  RÞpn ¼

n¼R þ 1

¼

k X

¼ Ls 

R X n¼0

R X

(

npn  R

¼ Ls  R 

npn  R 1 

pn  )

pn

n¼0 R X

npn þ R

n¼0

¼ Ls  R þ

k X

R X

n¼0

R X

R X

npn

n¼R þ 1

n¼0

(

k X

npn 

n¼R þ 1

npn 

n¼0

k X

R X

)

pn

n¼0

"

*

k X

# pn ¼ 1

n¼0

pn

n¼0

ðR  nÞpn

n¼0

¼ Ls  ðR  RÞ where R ¼

PR n¼0

ðR  nÞpn = expected number of idle machine repairmen.

(iii) Effective arrival rate (keff) In this queueing system, the arrivals occur with a rate k, but all arrivals do not join the system. This situation occurs when the maximum allowable queue length is reached. In that case, no new arrivals are allowed to join the queue. So, we shall deﬁne k considering those arrivals which join the system. This arrival rate is known as the effective arrival rate, and it is denoted by keff. Hence, the effective arrival rate is given by keff ¼ lðR  RÞ or keff ¼

k X

kn pn ¼ E½kn  ¼ E½kðk  nÞ½Since kn ¼ kðk  nÞ

n¼0

¼ kE ½ðk  nÞ ¼ k½k  E ðnÞ ¼ k½k  Ls : (iv) Expected waiting time The expected waiting time of a customer in the system is given by

15.3

Poisson Queueing Models

455

Ws ¼

Ls ; keff

whereas the expected waiting time of a customer in the queue is given by Wq ¼

Lq : keff

Example 6 There are 5 machines, each of which, when running, suffers breakdowns at an average rate of 2 per hour. There are two servicemen, and only one man can work on a machine at a time. If n machines are out of order when n > 2, then n – 2 of them wait until a serviceman is free. Once a serviceman starts work on a machine, the time to complete the repair has an exponential distribution with mean 5 min. Find the distribution of the number of machines out of action at a given time. Also, ﬁnd the average time an out-of-action machine has to spend waiting for the repairs to start. Solution Here k = total number of machines in the system = 5 R = number of servicemen = 2 k = 2 per hour, l = 1 per 5 min per hour. 2 Hence, q = k/ l = 12 ¼ 16. Let n be the number of machines out of order. Hence,

8 5 1n > > > < n 6 p0 pn ¼  > 5 jn 1n > > p0 : n 2n2 j2 6 8 5 1n > > > < n 6 p0 ¼   1 n > 5 > > p0 2jn 12 : n

for 0  n  2 for 2  n  5 for 0  n  2 ; for 2  n  5

where " p0 ¼

21   n X 1 5 n¼0

n

6

þ

5  X 5 n¼2



1 2j n n 12

n #1

The average number of machines out of action is given by

¼

648 : 1493

456

15

Lq ¼

5 X

ðn  2Þpn ¼

n¼2 þ 1

5 X

ðn  2Þpn ¼p3 þ 2p4 þ 3p5 ¼

n¼3

Queueing Theory

165 : 1493

The average time an out-of-action machine has to spend waiting for repairs to start is Wq = Lq/keff. But keff ¼

k X

kn pn ¼

n¼0

k X

kðk  nÞpn ¼ k

n¼0

5 X

ð5  nÞpn ¼

n¼0

6  2050 : 1493

Hence, Wq = 55/4100, h = 33/41 min.

15.3.7 Power Supply Model Model Formulation The power supply problem can be analysed as a queueing problem. The objective of this problem is to ensure the better utilization of a power supply to the customers. Let there be an electrical circuit supplying power to customers. Further, let the requirement and supply schedules follow a Poisson distribution with parameters k and l respectively. Let there be n customers in the queue at any time t and let C be the total number of customers. Then, kn ¼ ðc  nÞk and ln ¼ nl

) for 0  n  c:

Proceeding similarly as in other models, the probabilities that (i) there is no customer and (ii) there are n customers in the system at time t þ Dt are given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  k0 DtÞ þ p1 ðtÞð1  k1 DtÞl1 Dt þ OðDtÞ

ð15:54Þ

pn ðt þ DtÞ ¼ pn1 ðtÞkn1 Dtð1  ln1 DtÞ þ pn ðtÞð1  kn DtÞð1  ln DtÞ þ pn þ 1 ðtÞð1  kn þ 1 DtÞln þ 1 Dt þ OðDtÞ:

ð15:55Þ

and

From (15.1), we have

15.3

Poisson Queueing Models

457

p0 ðt þ DtÞ ¼ p0 ðtÞð1  ckDtÞ þ p1 ðtÞf1  ðc  1ÞkDtglDt þ OðDtÞ or

p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ ckp0 ðtÞ þ lp1 ðtÞ þ : Dt Dt

ð15:56Þ

From (15.55), we have pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kn1 pn1 ðtÞ  ðkn þ ln Þpn ðtÞ þ ln þ 1 pn þ 1 ðtÞ þ : Dt Dt ð15:57Þ When 0\n\c, kn1 ¼ fc  ðn  1Þgk;

kn ¼ ðc  nÞk; ln ¼ nl; ln þ 1 ¼ ðn þ 1Þl:

Hence, from (15.57), we have Pn ðt þ DtÞ  pn ðtÞ ¼ ða  n þ 1Þkpn1 ðtÞ  fðc  nÞk þ nlgpn ðtÞ Dt OðDtÞ ; 0\n\c: þ ðn þ 1Þlpn þ 1 ðtÞ þ Dt

ð15:58Þ

When n = c, kn1 ¼ fc  ðc  1Þgk ¼ k, kn ¼ ðc  cÞk ¼ 0 ln ¼ nl ¼ cl; pn þ 1 ðtÞ ¼ 0: Hence, from (15.57), we have pc ðt þ DtÞ  pc ðtÞ OðDtÞ ¼ kpc1 ðtÞ  clpc ðtÞ þ : Dt Dt

ð15:59Þ

Taking the limit as Dt ! 0, from (15.56), (15.58) and (15.59), we have p00 ðtÞ ¼ ckp0 ðtÞ þ lp1 ðtÞ for n ¼ 0

ð15:60Þ

p0n ðtÞ ¼ ðc  n þ 1Þkpn1 ðtÞ fðc  nÞk þ nlgpn ðtÞ þ ðn þ 1Þlpn þ 1 ðtÞ ð15:61Þ for 0\n\c p0c ðtÞ ¼ kpc1 ðtÞ  clpc ðtÞ for n ¼ c:

ð15:62Þ

Under a steady state condition of this system, i.e. lim pn ðtÞ ¼ pn and lim p0n ðtÞ ¼ 0, the preceding three equations reduce to

t!1

t!1

458

15

ckp0 þ lp1 ¼ 0

Queueing Theory

for n ¼ 0

ðc  n þ 1Þkpn1  fðc  nÞk þ nlgpn þ ðn þ 1Þlpn þ 1 ¼ 0;

ð15:63Þ 0\n\c ð15:64Þ

kpc1  clpc ¼ 0 for n ¼ c: Equations (15.63)–(15.65) are the steady state equations of the system. Solution of the Model From (15.63), we have p1 ¼ c

k l

p0 .

Now, setting n = 1 in (15.64), we have ckp0  fðc  1Þk þ lgp1 þ 2lp2 ¼ 0  k p0 or 2lp2 ¼ ckp0 þ fðc  1Þk þ lgc l  c ð c  1Þ k 2 or p2 ¼ p0 : l j2 Again, for n = 2, c ð c  1Þ ð c  2Þ p3 ¼ j3

 3 k p0 : l

In general, pn ¼

 c ð c  1Þ    ð c  n þ 1Þ k n p0 l jn

From (15.65), we have 1k pc1 cl

pc ¼

Þ2 or pc ¼ 1c lk : cðc1 jc 1 But

c1 k l

p0 ¼

c k l

c X n¼0

p0 :

pn ¼ 1

for 0\n\c:

ð15:65Þ

15.3

Poisson Queueing Models

or or or or

459

p0 þ p1 þ p2 þ . . . þ pc ¼ 1 "   c # ck cðc  1Þ k 2 k þ þ  þ p0 1 þ ¼1 l l l j2  k c ¼1 p0 1 þ l  c l p0 ¼ : kþl

Hence,   c c ð c  1Þ    ð c  n þ 1Þ k n l l kþl jn ! n  cn c k l ¼ ; kþl kþl n

pn ¼

which is a binomial distribution.

15.4

Non-poisson Queueing Models

15.4.1 Introduction When the arrivals and/or departures of a queueing system do not follow a Poisson distribution, they are called non-Poisson queueing systems. The development of these systems is more complicated, as the Poisson axioms do not hold. In studying non-Poisson queues, usually the following techniques are employed: (i) Phase technique (ii) Imbedded Markov chain technique (iii) Supplementary variable technique. In the phase technique, service is provided to a customer in several phases, say k in number. The technique by which non-Markovian queues are reduced to Markovian queues is termed the imbedded Markov chain technique. In the supplementary variable technique, one or more random variables are added to convert a non-Markovian process into a Markovian one. These techniques have been applied in studying queueing systems like M/Ek/1, M/G/1, G1/G/c, etc. In this chapter, we shall discuss M/Ek/1 and M/G/1 queueing systems only.

460

15

Queueing Theory

15.4.2 Model (M/Ek/1):(∞/FCFS/∞) Assumptions In this model, we shall discuss a single server non-Poisson queueing system under the following assumptions: (i) The arrivals follow a Poisson distribution. (ii) The service time has an Erlang type k distribution. Even though the service may not actually consist of k phases, it is convenient in analysing this system to consider the Erlang as being made up of k exponential phases, 1 each with mean kl , where l is the number of customers served per unit of time. (iii) The service discipline is ﬁrst come, ﬁrst served. (iv) The capacity of the system is inﬁnite. (v) This system consists of a single service channel in which there are k phases in the system (waiting or in service). (vi) A new arrival creates k phases of service, and departure of one customer reduces k phases of service. (vii) The distribution of the total service time of a customer in the system will be some combined distribution of time in all these phases. (viii) Each customer is served in k phases one by one, and a new service does not start until all k phases have been completed; i.e. each arrival increases the number of phases by k in the system.

Model Formulation If at any time there are m customers waiting in the queue with one customer in service which has to still pass through s phases, then the total number of phases n in the system (waiting and in service) will be n = mk + s. Figure 15.5 shows the multiphase service-based model. As l is the number of customers served per unit time, then kl will be the number 1 of phases served per unit time and kl will be the average time taken by the server at each phase. Therefore, kn ¼ k; constant arrival per unit time with ln ¼ kl phases served per unit time.

Queue

1

s k phases of services

Fig. 15.5 Multiphase service-based queueing system

depurture k

15.4

Non-poisson Queueing Models

461

Hence, the probability of one arrival during time Dt is kDt þ OðDtÞ and the probability of one departure (of k phases) during time Dt is klDt þ OðDtÞ. Let Pn ðtÞ be the probability that there are n phases in the system at time t. This model consists of a single service channel in which there are n phases in the system (waiting or in service). According to the assumption of the model, a new arrival creates k phases of service, and departure of one customer reduces k phases of service. To have zero phase in the system at time t þ Dt, there may be the following cases: (i) No phase in the system at time t, no arrival in time Dt (ii) One phase in the system at time t, no arrival in time Dt, service of one unit (of k phase) in time Dt, and so on. Therefore, the probability of no phase in the system at time t þ Dt is given by p0 ðt þ DtÞ ¼ p0 ðtÞð1  kDtÞ þ p1 ðtÞð1  kDtÞklDt þ OðDtÞ p0 ðt þ DtÞ  p0 ðtÞ OðDtÞ ¼ kp0 ðtÞ þ klp1 ðtÞ þ : or Dt Dt

ð15:66Þ

To have n (>0) phases in the system at time t þ Dt, there may be the following cases: (i) (n – k) phases in the system at time t, one arrival in time Dt, no service (of any phase) in time Dt (ii) n phases in the system at time t, no arrival in time Dt, no service (of any phase) in time Dt (iii) (n + 1) phases in the system at time t, no arrival in time Dt, service of one unit (of k phase) in time Dt, and so on. Therefore, the probability Pn ðt þ DtÞ that there are n (>0) phases in the system at time t þ Dt is given by pn ðt þ DtÞ ¼ pnk ðtÞkDtð1  klDtÞ þ pn ðtÞð1  kDtÞð1  klDtÞ þ pn þ 1 ðtÞð1  kDtÞklDt þ OðDtÞor pn ðt þ DtÞ  pn ðtÞ OðDtÞ ¼ kpnk ðtÞ  ðk þ klÞpn ðtÞ þ klpn þ 1 ðtÞ þ : Dt Dt

ð15:67Þ

Now taking the limit as Dt ! 0 in (15.1) and (15.67), we get the differential difference equations for the system as follows: p0n ðtÞ ¼ kpnk ðtÞ  ðk þ klÞpn ðtÞ þ klpn þ 1 ðtÞ and

p00 ðtÞ ¼ kp0 ðtÞ þ klp1 ðtÞ:

ð15:68Þ ð15:69Þ

462

15

Queueing Theory

Under the steady state condition of the system, i.e. lim pn ðtÞ ¼ pn and t!1

lim p0n ðtÞ ¼ 0, these two equations reduce to

t!1

kpnk  ðk þ klÞpn þ klpn þ 1 ¼ 0;

n1

ð15:70Þ

and kp0 þ klp1 ¼ 0;

n ¼ 0:

ð15:71Þ

k Let q ¼ kl ; then Eqs. (15.70) and (15.71) reduce to

p1 ¼ qp0 ;

n¼0

ð15:72Þ

and ð1 þ qÞpn ¼ qpnk þ pn þ 1 ; n  1:

ð15:73Þ

Now, to solve Eq. (15.73), we shall use the generating function technique. Solution of the Model Let the generating function be G(x), which is deﬁned by Gð xÞ ¼

1 X

pn x n ;

j xj  1:

ð15:74Þ

x¼0

Multiplying both sides of (15.73) by xn and then summing over from 1 to ∞ (since the equation holds for n  1), we have ð 1 þ qÞ

1 X

pn x n ¼ q

n¼1

or

ð1 þ qÞ

1 X

or or

or

ð 1 þ qÞ ð 1 þ qÞ

pnk xn þ

n¼1

1 X

pn þ 1 x n

1 X

pnk xn þ

n¼1

#

pn x þ p0  p0 ¼ q n

n¼1 1 X

1 X

n¼0

n¼k

pn xn  p0 ¼ q

1 X n¼1

pn xn þ qp0 ¼ p1 þ q

n¼1

"

1 X

1 X

1 X

pn þ 1 x n

½* qp0 ¼ p1 

n¼1

"

pnk x þ p1 þ n

n¼1

pnk xn þ

1 X n¼1

1 1X pn þ 1 x n þ 1 x n¼0

½since pnk ¼ 0; for n  k\0; i.e. for n\k 1 1 1 X X 1X pn x n  p0 ¼ q pj x j þ k þ pi x i ð 1 þ qÞ x n¼0 j¼0 i¼1

# pn þ 1 x

n

15.4

or or or

Non-poisson Queueing Models

463

½substituting n  k ¼ j; i:e. ¼ j þ k and n þ 1 ¼ i; i:e.n ¼ i  1 " # 1 1 1 X X 1 X n k j i pn x  p0 ¼ qx pj x þ pi x  p0 ð1 þ q Þ x i¼0 n¼0 j¼0 1 ð1 þ qÞGð xÞ  p0 ¼ qxk Gð xÞ þ ½Gð xÞ  p0  x p0 ð 1  x Þ G ð xÞ ¼ j xj  1 ð1  xÞ  qxð1  xk Þ   1 p0 1  xk  k  ¼ p0 1  qx ¼ 1x 1  qx 1x 1x  1 k m X 1x ¼ p0 ðqxÞm 1x m¼0

ð15:75Þ

½by binomial expansion: 1 X  m ðxqÞm 1 þ x þ x2 þ    þ xk1 ) G ð x Þ ¼ p0 m¼0

or

  1  xk ¼ 1 þ x þ x2 þ    þ xk1 Since 1x 1 X  m G ð x Þ ¼ p0 qm x þ x2 þ    þ xk : m¼0

Now we shall ﬁnd the value of p0. From (15.74) and (15.75), we have 1 X

pn x n ¼ p0

n¼0

1 X

 m qm x þ x2 þ    þ xk :

m¼0

This is an identity. Now setting x =1 on both sides, we have 1 X

pn ¼ p0

n¼0

1 X

qm ð1 þ 1 þ    þ k timesÞm

m¼0 1 X

or

1 ¼ p0

or

h i 1 ¼ p0 1 þ qk þ ðqk Þ2 þ ðqk Þ3 þ   

qm k m

m¼0

or or

1 k ; if qk\1 i:e.; if \1 i:e:; k\l 1  qk l p0 ¼ 1  kq: 1 ¼ p0

Substituting p0 ¼ 1  kq in (15.76), we have

ð15:76Þ

464

15 1 X

pn xn ¼ ð1  kqÞ

n¼0

or

1 X

1 X

ðqxÞm

m¼0

pn x ¼ ð1  kqÞ n

n¼0

1 X

ðqxÞ

 m 1  xk 1x

m

1x

 k m

ð 1  xÞ

Queueing Theory

ð15:77Þ m

m¼0

m rk Now 1  x ¼ ð1Þ x r r¼0 1 P mþs  1 s and ð1  xÞm ¼ x. s s¼0 Hence, from (15.77), we have 

 k m

1 X

or



m P

r

pn x ¼ ð1  kqÞ n

1 X

n¼0

n¼0

1 X

1 X

pn xn ¼ ð1  kqÞ

n¼0

" m m

q x " qm

m¼0

m X

ð1Þ

r¼0 m X 1 X

ð1Þr

r¼0 s¼0

!

# 1  X mþs  1 s x x s r s¼0 ! # m m þ s  1 m þ rk þ s x : s r

m

r

rk

Comparing the coefﬁcients of xn from both sides, we have pn ¼ ð1  kqÞ

X

 qm ð1Þr

m;r;s

m r



mþs  1 : s

[The summation is taken over all such values of m, r, s ð0  m\1; 0  r  m; 0  s\1Þ for which m þ rk þ s ¼ n.] Characteristics of the Model (i) Average (expected) number of phases in the system The average number of phases in the system is given by Lp ¼

1 X

npn

n¼0

From (15.7), we have ð1 þ qÞpn ¼ qpnk þ qn þ 1 ;

n  1:

Now multiplying both sides by n2 and then summing from 1 to ∞, we have

15.4

Non-poisson Queueing Models

ð1 þ qÞ ð1 þ q Þ

or

1 X

465

n2 pn ¼ q

1 X

1 X

n2 pnk þ

n¼1

n¼1

n¼1

1 X

1 X

1 X

n2 pn ¼ q

n¼1

n2 pnk þ

n2 pn þ 1 n2 pn þ 1

n¼0

n¼k

½since pnk ¼ 0 for n  k\0; i:e: for n\k: Now taking n + 1 = m and n − k = m in the ﬁrst and second summations on the right-hand side respectively, we have ð1 þ q Þ

1 X

n2 pn ¼ q

n¼0

¼q

1 X

1 X

1 X

ð m þ k Þ 2 pm þ

m¼0

ðn þ kÞ2 pn þ

n¼0

ðm  1Þ2 pm

m¼1 1 X

ð n  1Þ 2 pn  p0

n¼0

1   X   ¼ q n2 þ 2nk þ k 2 þ n2  2n þ 1 pn  p0 n¼0

¼ ð1 þ q Þ

1 X

n2 pn  2ð1  qk Þ

n¼0

or or

2ð1  qkÞ 2ð1  qkÞ

1 X n¼0 1 X

1 X

1  X npn þ 1 þ qk2 pn  p0

n¼0



npn ¼ 1 þ qk

 2

"

 p0 *

1 X

#

n¼0

pn ¼ 1

n¼0

  npn ¼ 1 þ qk 2  ð1  qkÞ

½* p0 ¼ 1  qk 

n¼0

or

1 X

npn ¼

n¼0

qkðk þ 1Þ 2ð1  qkÞ

2

qkðk þ 1Þ l ð k þ 1Þ

¼ 2ð1  qkÞ 2 1  k k

or

Lp ¼

l

k 6 * q ¼ lk 6 6 4 i:e.:; qk ¼

3 7 7 7 k5 l

kþ1 k ¼ : : 2 lk (ii) Average (expected) number of units in the queue We know that the average number of phases of one unit in service ¼ k þ2 1. Therefore, the time taken for their service ¼ k þ2 1 l1, as l1 is the mean service time per unit.

466

15

Hence, the expected number of phases in service = expected number of phases arrived during the time

Queueing Theory

kþ1 2l

þ1 ¼ k 2l k. Therefore, the expected number of units in the queue is given by

 1 Lp  expected number of phases in service k  1 kþ1 k kþ1 ¼  k k 2 lk 2l

Lq ¼

¼

kþ1 k2 : : 2k lðl  kÞ

(iii) Average (expected) number of units in the system From the relation between the expected number of units in the system and in the queue, we have Ls ¼ Lq þ or

Ls ¼

k l

kþ1 k2 k þ : 2k lðl  kÞ l

(iv) Expected waiting time in the system The expected waiting time in the system is given by 1 Ws ¼ L s k kþ1 k 1 þ : ¼ 2k lðl  kÞ l (v) Expected waiting time in the queue The expected waiting time in the queue is given by 1 Wq ¼ Lq k kþ1 k : ¼ 2k lðl  kÞ Example 7 In a factory cafeteria, the customers have to pass through three counters. The customers buy coupons at the ﬁrst counter, select and collect the snacks at the

15.4

Non-poisson Queueing Models

467

second counter and collect tea at the third. The server at each counter takes on average 1.5 min, although the distribution of service time is approximately exponential. If the arrival of customers to the cafeteria is approximately Poisson at an average rate of 6 per hour, calculate: (i) The average time that a customer spends waiting in the cafeteria (ii) The average time of getting the service (iii) The most probable time getting the service. Solution This is a queueing problem of the form ðM=Ek =1Þ:ð1=FCFS=1Þ. Here k = number of phases = 3 Service time per phase = 1.5 min ∴ Service time per customer = 1.5  3 = 4.5 min 1 ) l ¼ 4:5 customers/minute = 40 3 customers/h. Here k ¼ 6 customers/h. (i) The average waiting time spent by a customer in the cafeteria is given by kþ1 k 3þ1 6 40  h : ¼ 40 2k lðl  kÞ 2  3 3 3  6 9 h ¼ 2:45 min: ¼ 220

Wq ¼

(ii) The average time of getting the service ¼

1 3 ¼ h ¼ 4:5 min l 40

(iii) The most probable time spent in getting the service = the modal value of service time t = the modal value of the Erlang distribution. 31 31 1 Hence, the most probable time spent ¼ k1 lk ¼ 403 ¼ 403 ¼ 20 h = 3 min. 3

3

[The probability density function of the Erlang distribution with parameters l and k is given by f ðt; l; k Þ ¼

ðlkÞk k1 klt t e ; ðk  1Þ!

0  t\1; k ¼ 1; 2; . . .: 2

l 1 The mode of the distribution is at k1 kl and the mean = l, with variance ¼ k .]

468

15

Queueing Theory

15.4.3 Model (M/G/1):(1/GD) A precise description of this model is as follows: (i) Customers arrive according to a Poisson fashion at an average rate of k. (ii) The service time t is represented by any probability distribution with mean E ðtÞ ¼ l1 and variance vðtÞ ¼ r2 . (iii) This system consists of a single server. (iv) The service discipline is a general service discipline such as FCFS, LCFS, SIRO, etc. (v) An arriving customer who ﬁnds the servers idle begins to take the service immediately. (vi) All blocked customers wait until served. (vii) The server cannot be idle when there is a waiting customer. Here the probability that a customer leaves the system in ðt; t þ DtÞ is a quantity dependent on when the service of that particular customer begins. That is, it would involve not only t, but also another time variable, say t0 , measured from the beginning of the current service. Let q0 denote the queue length when the service of the nth customer terminates and let q1 be the queue length when the service of the ðn þ 1Þ th customer terminates. Further, let kðk ¼ 0; 1; 2; . . .Þ denote the number of customers who arrive during the service of the ðn þ 1Þth customer. Then the relation between q0 ; q1 and k is given by 8 > >
> : i:e. the queue has q0 units after serving n-th customer: This relation can be written as q1 ¼ q0  d þ k;

ð15:78Þ



0 if q0 ¼ 0 . 1 if q0 [ 0 Now, taking expectation on both sides of (15.78), we have

where d ¼

E ðq1 Þ ¼ E ðq0 Þ  EðdÞ þ EðkÞ: In a steady state system, E ðq1 Þ ¼ E ðq0 Þ. Hence, from (15.79), we have

ð15:79Þ

15.4

Non-poisson Queueing Models

469

E ðdÞ ¼ Eðk Þ:

ð15:80Þ

Now, squaring both sides of (15.78), we have q21 ¼ q20 þ k2 þ d2 þ 2kq0  2dq0  2dk or

q21 ¼ q20 þ k2 þ d þ 2kq0  2q0  2kd " # since d can take values 0 and 1 only, so d2 ¼ d and dq0 ¼ q0

ð15:81Þ :

Now taking expectation on both sides, we have       E q21 ¼ E q20 þ E k 2 þ E ðdÞ þ 2E ðkq0 Þ  2Eðq0 Þ  2E ðkdÞ:

ð15:82Þ

    Again, since E q21 ¼ E q20 is in a steady state system, from (15.82) we have   2Eðq0 Þ  2Eðkq0 Þ ¼ E k 2 þ E ðdÞ  2E ðkdÞ   or 2E ðq0 Þ  2Eðk ÞE ðq0 Þ ¼ E k2 þ E ðk Þ  2fE ðkÞg2 " # From (3), E ðdÞ ¼ E ðkÞ and q0 ; k and d are independent or

E ðq0 Þ ¼

ð15:83Þ

E ð k 2 Þ þ E ð k Þ  2fE ð k Þ g2 : 2f1  E ð k Þ g

Now, in order to determine E ðq0 Þ, the values of E ðkÞ and E ðk2 Þ must be computed. Since the arrivals in time t follow a Poisson distribution,     Eðk=tÞ ¼ kt; E k 2 =t ¼ ðktÞ2 ; V ðk=tÞ ¼ E k 2 =t  fE ðk=tÞg2 ; where k is the mean arrival rate. Hence, Z1 Z Eðk Þ ¼ Eðk=tÞf ðtÞdt ¼

Z

1

kt f ðtÞdt ¼ k

0

0

1

t f ðtÞdt ¼ kEðtÞ

0

Also,   E k2 ¼

Z1 Z1h i  2  E k =t f ðtÞdt ¼ ðktÞ2 þ kt f ðtÞdt 0

0

  ¼ k E t2 þ kE ðtÞ ¼ k2 V ðtÞ þ k2 fE ðtÞg2 þ kEðtÞ: 2

470

15

Queueing Theory

Substituting the values of E ðkÞ and E ðk2 Þ in (15.5), we have E ðq0 Þ ¼

k2 V ðtÞ þ k2 fE ðtÞg2 þ kE ðtÞ þ kE ðtÞ  2k2 fE ðtÞg2 2f1  kE ðtÞg

k2 V ðtÞ þ k2 fE ðtÞg2 þ 2kE ðtÞ  2k2 fE ðtÞg2 2f1  kE ðtÞg h i 2 k V ðtÞ þ fE ðtÞg2 þ kE ðtÞ ¼ 2f1  kE ðtÞg   2 2 1 k2 r2 þ l1 1 6 As E ðtÞ ¼ l where l is the mean o þk  4 ¼ n l 2 1  k  l1 service time and V ðtÞ ¼ r2   k2 r2 þ q2 k þ q As q ¼ ¼ l 2ð1  qÞ ¼

or

Ls ¼ q þ

ð15:84Þ

k2 r 2 þ q 2 : 2ð1  qÞ

This is known as the Pollaczek-Khintchine (P-K) formula. Alternative form of Pollaczek formula If the (n + 1)th customer has to wait for the time w units before taking the service, the length of the queue must be q1 during the time w + t. Hence, E ðq1 Þ ¼ kE ðw þ tÞ ¼ kE ðwÞ þ kE ðtÞ or kE ðwÞ ¼ E ðq1 Þ  kE ðtÞ, h i i.e. the average waiting time ¼ EðwÞ ¼ 1k E ðq1 Þ  lk where E ðtÞ ¼ l1 or

#   " since Eðq1 Þ ¼ E ðq0 Þ 1 k2 r2 þ q2 q E ðw Þ ¼ P þ k 2ð1  qÞ in the steady state

k2 r2 þ q2 : 2ð1  qÞk   E ð w Þ l k2 r 2 þ q 2 ¼ ) E ðtÞ 2ð1  qÞk   E ðwÞ q ¼ 1 þ l2 r2 : or E ðt Þ 2ð1  qÞ ¼

This formula was also suggested by Pollaczek. Hence, the average number of customers in the system is given by

ð15:85Þ

15.4

Non-poisson Queueing Models

471

Ls ¼ q þ

k2 r2 þ q2 : 2ð1  qÞ

The average number of customers in the queue is given by Lq ¼ Ls 

k k 2 r 2 þ q2 ¼ Ls  q ¼ : l 2ð1  qÞ

Example 8 In a heavy machine shop, the overhead crane is 75% utilized. Time study observations gave the average slinging time as 10.5 min with a standard deviation of 8.8 min. What is the average calling rate for the service of the crane, and what is the average delay in getting service? If the average service time is cut to 8.0 min with a standard deviation of 6.0 min, how much reduction will occur, on average, in the delay of getting service? Solution This is an (M/G/1):(1/FCFS/1) queueing system. Since the overhead crane is utilized 75%, q ¼ 0:75. Again, 60 ¼ 5:71 per h 10:5 k ¼ ql ¼ 0:75  5:71 ¼ 4:29 per h 8:8 r ¼ 8:8 min ¼ h: 60

Now. the average delay in getting the service is given by 2  2 þ ð0:75Þ2 k2 r2 þ q2 ð4:29Þ  8:8 60 ¼ ¼ 0:4468 h ¼ 26:8 min: Lq ¼ 2kð1  qÞ 2  4:29  ð1  0:75Þ

If the service time is cut to 8 min, then l¼

60 4:29 ¼ 7:5 per hour and q ¼ ¼ 0:571: 8 7:5

6 In this case, the standard deviation = 6.0 min ¼ 60 ¼ 0:1 h.

Then Lq ¼

ð4:29Þ2 ð0:1Þ2 þ ð0:571Þ2 h ¼ 0:1386 h ¼ 8:3 min: 2  4:29  ð1  0:571Þ

472

15

Queueing Theory

In the second case, a reduction of (26.8–8.3), i.e. 18.5 min, will occur on average in the delay of getting service. The average waiting time of a customer in queue is given by Lq k2 r2 þ q2 ¼ : k 2kð1  qÞ

Wq ¼

The average waiting time that a customer spends in the system is given by Ws ¼ Wq þ service time ¼

k2 r2 þ q2 1 þ : 2kð1  qÞ l

Efﬁciency of this queueing system The efﬁciency of this queueing system is measured by the ratio

Wq E ðt Þ

as follows:

Wq average waiting time in the queue : ¼ average service time E ðt Þ From this ratio, we conclude that the system will be better for a smaller ratio. Note: In particular, let G ¼ M. Then 1 1 ; E ðt Þ ¼ : 2 l l h i 2 1 1 k k l2 þ l2 q2 q

¼ qþ ; Ls ¼ þ ¼ k l 1  q 1  q 2 1l VarðtÞ ¼

which is the same result as obtained earlier for the queueing model (M/M/1):(1/ FCFS/1). Example 9 In a queueing system with a single server, arrivals are Poisson distributed with mean 5 per hour. Find the expected waiting time in the system if the service time distribution is normal with mean 3 min and standard deviation 2 min. Hints: Here k = 5 per hour, l ¼ 60 3 ¼ 20 per hour. r ¼ 2 min ¼ The expected waiting time in the system

2 h: 60

15.4

Non-poisson Queueing Models

473

¼ Ls ¼

15.5

k2 r2 þ q2 1 þ : 2kð1  qÞ l

Mixed Queueing Model and Cost

15.5.1 Introduction In this section, we shall discuss a queueing model with random arrival rate and deterministic service rate. This type of queueing model is known as a mixed queueing model. We shall also discuss the cost models of an M/M/1 system with ﬁnite and inﬁnite capacity to ﬁnd the optimum service level by minimizing the cost of the system.

15.5.2 Mixed Queueing Model For a queueing model, if either the arrival rate or the service rate is ﬁxed and the other one is random, then the model is known as a mixed queueing model. Model: (M/D/1):(∞/FCFS/∞) This queueing system involves a single server, Poisson arrivals and a deterministic distribution. The capacity of the system is inﬁnite. Here the service time is not a random variable but a constant. Without loss of generality, the service time can be taken as unity, l ¼ 1. It is also assumed that a steady state condition prevails, with q ¼ k\1 as l ¼ 1. In this scheme, the system is observed just after the completion of service of a customer. Hence, if n customers are in the system at the observation time, then at the time of the next observation, the number n0 in the system is given by n0 ¼



k if n ¼ 0 ; n  1 þ k; if n [ 0

where k ¼ 0; 1; 2; . . . is the number of arrivals during the service time which is taken as unity. Let qk be the probability that k persons arrive during the service time. Since the arrival rate follows a Poisson distribution, qk ¼

ek ðkÞk ; k!

k ¼ 0; 1; 2; . . .

474

15

Table 15.2 Transaction matrix

Queueing Theory

n0 ! n

0

1

2

3

4

0 1 2 3 4 .. . .. .

q0 q0 0 0 0 .. . .. .

q1 q1 q0 0 0 .. . .. .

q2 q2 q1 q0 0 .. . .. .

q3 q3 q2 q1 q0 .. . .. .

q4 q4 q3 q2 q1 .. . .. .

… … … … … .. . .. .

… … … … … .. . .. .

where k is the mean arrival rate. For different values of n and n0 , the transaction matrix between the end and the beginning of the service is given by Table 15.2. In the transaction matrix, q0 is the probability of no arrival during the service time. Let pn ðn ¼ 0; 1; 2; . . .Þ be the steady state probability that there are n customers in the system. Then the probabilities are given by p0 ¼ p0 q0 þ p1 q0 p1 ¼ p0 q1 þ p1 q1 þ p2 q0 p2 ¼ p0 q2 þ p1 q2 þ p2 q1 þ p3 q0  

 

 

The ðn þ 1Þth equation of the above is given by pn ¼ p0 qn þ p1 qn þ p2 qn1 þ p3 qn3 þ    þ pn þ 1 q0 ; or

pn ¼ p 0 q n þ

nX þ1

pj qn þ 1j ;

n ¼ 0; 1; 2; . . .

n ¼ 0; 1; 2; . . .:

j¼1

ð15:86Þ Applying a Z-transform to Eq. (15.86), we have P n [Z-transform of pn is deﬁned by Z ðpn Þ ¼ 1 n¼0 z pn ¼ PðzÞ ðsayÞ] 1 X n¼0

z n pn ¼

1 X n¼0

z n p0 qn þ

1 X n¼0

( zn

nX þ1 j¼1

) pj qn þ 1j

15.5

Mixed Queueing Model and Cost

¼

1 X

z pn qn þ n

n¼0

or

1 X

475

( z

n þ 11

n¼0

nX þ1

) pj qn þ 1j

j¼0

( ) 1 m X 1X m z pn ¼ p0 z qn þ z pj qmj  p0 qm ; where n ¼ m þ 1: z m¼1 n¼0 n¼0 j¼0

1 X

1 X

n

n

ð15:87Þ Let us consider rm ¼

m X

pj qmj

ð15:88Þ

j¼0

or

rm ¼ p0 qm þ p1 qm1 þ p2 qm2 þ    þ pm q0 ;

i.e. rm is the convolution of pm and qm . Hence, Z ðrm Þ ¼ Z ðqm ÞZ ðpm Þ or

RðzÞ ¼ QðzÞPðzÞ; where RðzÞ ¼ Z ðrm Þ and QðzÞ ¼ Z ðqm Þ:

ð15:89Þ

Then, from (15.87), we have PðzÞ ¼ p0 QðzÞ þ

1 1 1X p0 X zm rm  z m qm z m¼1 z m¼1

1 p0 fRðzÞ  r0 g  fQðzÞ  q0 g z z 1 p0 ¼ p0 QðzÞ þ RðzÞ  QðzÞ ½as r0 ¼ p0 q0  z z 1 p0 ¼ p0 QðzÞ þ PðzÞQðzÞ  QðzÞ ½as RðzÞ ¼ PðzÞQðzÞ z z   ðz  1ÞQðzÞ PðzÞ ¼ p0 : z  QðzÞ ¼ p0 QðzÞ þ

or Now,

 ðz  1ÞQðzÞ 0 Pð1Þ ¼ lim PðzÞ ¼ p0 lim ¼ form z!1 z!1 z  Q ðzÞ 0 ðz  1ÞQ0 ðzÞ þ QðzÞ ¼ p0 lim ½Using L'Hospital's rule z!1 1  Q0 ðzÞ Q ð 1Þ ¼ p0 1  Q 0 ð 1Þ p0 ½as Qð1Þ ¼ 1: ¼ 1  Q 0 ð 1Þ 

ð15:90Þ

476

15

Hence,

Queueing Theory

¼1

p0 1Q0 ð1Þ

or p0 ¼ 1  Q0 ð1Þ: Now we have to compute Q0 ð1Þ. )QðzÞ ¼

1 X

qn zn ; where qn is the probability of new arrivals:

n¼0

)QðzÞ ¼

1 k n X e k

n!

n¼0

¼ ek

zn ½Since the distribution of arrivals is Poisson; qn ¼

ek kn  n!

1 X

ðkzÞn n! n¼0

¼ ekðz1Þ : Hence, Q0 ð1Þ ¼ k i:e:; p0 ¼ 1  k. From (15.90), we have PðzÞ ¼

ð1  kÞðz  1ÞQðzÞ ð1  kÞðz  1Þekðz1Þ ð1  kÞð1  zÞ ¼ ¼ : z  Q ðzÞ 1  zekðz1Þ z  ekðz1Þ

Determination of pn In obtaining pn we observe that a general expression cannot be easily obtained in this case because of the complexity of PðzÞ. However, pn can be obtained by using the formula Pn ð0Þ ¼ n!pn ; where Pn ð0Þ ¼

h

i

@ n PðzÞ @zn z¼0 1 n n! P ð0Þ.



as PðzÞ ¼

P1 n¼0

 pn z n .

Therefore, pn ¼ Again, the expected number of customers in the system is given by 2

1 X

3

pn z 7 6 since PðzÞ ¼ 6 7 n¼0 7 n pn ¼ P ð 1Þ 6 Ls ¼ 6 7 1 X 4 0 n¼0 n1 5 )P ðzÞ ¼ npn z 1 X

or

Ls ¼

n

0

00

k þ Q ð1Þ kþk ¼ 2ð 1  kÞ 2ð 1  kÞ 2



n¼0

As PðzÞ ¼ p0



ðz  1ÞQðzÞ Z  QðzÞ

 :

15.6

15.6

Cost Models in Queueing System

477

Cost Models in Queueing System

In a queueing system, there are two types of conflicting costs: (i) Cost of offering the service to the customers (ii) Cost incurred due to delay in offering the service to the customers. These two types of costs conflict in nature. An increase in the existing service facility would reduce the customers’ waiting time; on the other hand, a decrease in the level of service would increase the customers’ waiting time. This means that a higher cost of offering the service would reduce the cost of waiting to the customers and vice versa. Therefore, the service level in a queueing system is a function of the service rate l, which balances the two conflicting costs. The objective is to ﬁnd the optimum service level such that the sum of these costs is minimized. Hence, the total cost TC of the queueing system is given by TC ¼ \cost for providing service facility[ þ \cost for waiting of customers[ TC ¼ C1 l þ C2 Ls where C1 = cost of service per customer per unit time C2 = cost for waiting of customers per unit time Ls = average number of customers in the queueing system. Now, for the queueing model (M/M/1):(1 /FCFS/1), Ls ¼

k lk

and TC ¼ C1 l þ C2

k : lk

Here TC is a function of the continuous variable l. Then, the necessary condition for TC to be optimum is given by dðTCÞ ¼ 0; dl C2 k which implies C1  ðlk ¼0 Þ2

ð15:91Þ

478

15

) l ¼ kþ

Queueing Theory

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ kC2 =C1 :

ð15:92Þ

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2kC2 ¼ ðlk kC2 =C1 . Then ddlTC 3 , which is positive for l ¼ k þ 2 Þ Hence, the optimum value of l is l ¼ k þ

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ kC2 =C1 :

ð15:93Þ

This result shows that the optimum value of l is dependent on C1 ; C2 and also on the arrival rate k. Again, for the queueing model (M/M/1):(N/FCFS/1), the capacity of the system is ﬁnite, and its value is N when the system is fulﬁlled; then newly arrived customers cannot join in the queue. As a result, the larger the value of N, the smaller will be the number of lost customers. In this case, the total cost TC of the system is given by TC ¼ hcost of service ratei þ hcost of waiting customer i or

þ hcost of servicing customersi þ hcost of lost customersi TC ¼ C1 l þ C2 Ls þ C3 N þ kpN C4 ; ð15:94Þ

where C3 = cost of servicing to each customer per unit time C4 = cost per lost customer and kpN = number of lost customers per unit time. Case 1 When q 6¼ 1 Ls ¼

q½1  ðN þ 1ÞqN þ NqN þ 1  ð1  qÞð1  qN þ 1 Þ

ð15:95Þ

ð1  qÞqN : 1  qN þ 1

ð15:96Þ

and pN ¼

In this case, TC ¼ C1 l þ

  C2 q½1  ðN þ 1ÞqN þ NqN þ 1  þ C3 N þ kC4 ð1  qÞqN = 1  qN þ 1 : N þ 1 ð1  qÞð1  q Þ ð15:97Þ

15.6

Cost Models in Queueing System

479

Case 2 When q = 1 Ls ¼

N 1 and pN ¼ : 2 N þ1

In this case, TC ¼ C1 l þ C2 N=2 þ C3 N þ kC4 =ðN þ 1Þ:

ð15:98Þ

As N is given, TC is a function of the continuous variable l only. The optimum value of l can be obtained easily by minimizing the total cost TC given by either (15.97) or (15.98).

15.7

Exercises

1. Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 min between one arrival and the next. The length of a phone call is assumed to be distributed exponentially with mean 3 min. (i) What is the probability that a person arriving at the booth will have to wait? (ii) Find the average number of units in the system. (iii) Estimate the fraction of a day that the phone will be in use. (iv) What is the probability that it will take a person more than 10 min altogether to wait for the phone and complete his call? 2. At a certain airport it takes exactly 5 min to land an aeroplane, once it is given the signal to land. Although incoming planes have scheduled arrival times, the wide variability in arrival times produces an effect which makes the incoming planes appear to arrive in a Poisson fashion at an average rate of 6 per hour. This produces occasional congestion in the airport, which can be dangerous and costly. Under these circumstances, how much time will a pilot expect to spend circling the ﬁeld waiting to land? 3. Patients arrive at a clinic according to a Poisson distribution at a rate of 30 patients per hour. The waiting room does not accommodate more than 14 patients. The examination time per patient is exponential with a mean rate of 20 per hour. (i) Find the effective arrival rate at the clinic. (ii) What is the probability that an arriving patient will not wait? (iii) What is the expected waiting time until a patient is discharged from the clinic?

480

15

Queueing Theory

4. A barber with a one-man operation takes exactly 25 min to complete one haircut. If customers arrive in a Poisson fashion at an average rate of one every 40 min, how long on average must a customer wait for service? Also ﬁnd the average time that a customer spends in the barbershop. 5. A road transport company has one reservation clerk on duty at a time. He handles information on bus schedules and makes reservations. Customers arrive at a rate of 8 per hour, and the clerk can service 12 customers on average per hour. (i) What is the average number of customers waiting for the service of the clerk? (ii) What is the average time a customer has to wait before getting service? (iii) The management is contemplating installing a computer system to handle the information and reservations. This is expected to reduce the service time from 5 to 3 min. The additional cost of having the new system works out to Rs. 400.00 per day. If the cost of goodwill of having to wait is estimated to be Re. 1.00 per minute spent waiting to be served, should the company install the computer system? Assume that the working time is 8 h per day. 6. At a one-man barbershop, customers arrive according to a Poisson distribution with a mean arrival rate of 5 per hour, and the hair-cutting time is exponentially distributed, with an average haircut taking 10 min. It is assumed that because of his excellent reputation, customers are always willing to wait for this barber. Calculate: (i) The average number of customers in the shop and the average number of customers waiting for a haircut (ii) The percentage of time an arrival can walk right in without having to wait (iii) The percentage of customers who have to wait prior to getting into the barber’s chair. 7. At a certain ﬁlling station, customers arrive in a Poisson process with an average time of 12 per hour. The time intervals between services follow an exponential distribution, and as such the mean time taken to service a unit is 2 min. Evaluate: (i) (ii) (iii) (iv) (v) (vi)

The The The The The The

probability that there is no customer at the counter probability that there are more than two customers at the counter probability that there is no customer to be served probability that a customer is being served, but nobody is waiting expected number of customers in the waiting line expected time a customer spends in the system.

8. A supermarket has a single cashier. During peak hours, customers arrive at a rate of 20 customers per hour. The average number of customers who can be processed by the cashier is 24 per hour. Calculate:

15.7

Exercises

(i) (ii) (iii) (iv) (v)

The The The The The

481

probability that the cashier is idle average number of customers in the queueing system average time a customer spends in the system average number of customers in the queue average time a customer spends in the queue waiting for service.

482

14.

15.

16.

17.

18.

19.

15

Queueing Theory

repair a machine and demands wages Rs. 50 per hour. The second mechanic B takes about 4 min in repairing a machine and demands wages at the rate of Rs. 75 per hour. Assuming that the rate of machine breakdown is Poisson distributed and the repair rate is exponentially distributed, which of the two mechanics should be engaged? In the production shop of a company, the breakdown of the machines is found to be Poisson distributed with an average rate of 3 machines per hour. The breakdown time at one machine costs Rs. 200 per hour to the company. The company is considering two repairmen for hire. One of the repairmen is slow but cheap, the other fast but expensive. The slow-cheap repairman demands Rs. 100 per hour and will repair the broken-down machines exponentially at the rate of 4 per hour. The fast-expensive repairman demands Rs. 150 per hour and will repair machines exponentially at an average rate of 6 per hour. Which repairman should be hired? At a railway station, only one train is handled at a time. The railway yard is sufﬁcient only for waiting of two trains to wait; then a newly arriving train is given the signal to leave the station. Trains arrive at the station at an average rate of 6 per hour, and the railway station can handle them on an average of 12 per hour. Assuming Poisson arrivals and an exponential service distribution, ﬁnd the steady state probabilities for the various numbers of trains in the system. Also ﬁnd the average waiting time of a new train coming into the yard. Assume that the goods trains are coming into a yard at the rate of 30 trains per day, and suppose that the inter-arrival times follow an exponential distribution. The service time for each train is assumed to be exponential with an average of 36 min. If the yard can admit 9 trains at a time (there being 10 lines, one of which is reserved for shunting purposes), calculate the probability that the yard is empty and ﬁnd the average queue length. If for a period of 2 h in the day (8 to 10 a.m.) trains arrive at the yard every 20 min but the service time continues to remain 36 min, then calculate for this period: (a) the probability that the yard is empty, (b) the average number of trains in the system, on the assumption that the line capacity of the yard is limited to 4 trains only. A petrol station has a single pump and space for not more than 3 cars (2 waiting, 1 being served). A car arriving when the space is ﬁlled to capacity goes elsewhere for petrol. Cars arrive according to a Poisson distribution at a mean rate of one every 8 min. Their service time has an exponential distribution with a mean of 4 min. The proprietor has the opportunity of renting an adjacent piece of land, which would provide space for an additional car to wait (he cannot build another pump). The rent would be Rs. 10 per week. The expected net proﬁt from each customer is Re. 0.50, and the station is open 10 h every day. Would it be proﬁtable to rent the additional space? A barbershop has two barbers and three chairs for waiting customers. Assume that customers arrive in a Poisson fashion at a rate of 5 per hour and that each

15.7

Exercises

483

barber services customers according to an exponential distribution with a mean of 15 min. Further, if a customer arrives and there are no empty chairs in the shop, he will leave. Find the steady state probabilities. What is the probability that the shop is empty? What is the expected number of customers in the shop? 20. A car servicing station has two bays where service can be offered simultaneously. Due to space limitations, only 4 cars are accepted for servicing. The arrival pattern is Poisson with 12 cars per day. The service time in both bays is exponentially distributed with l ¼ 8 cars per day per bay. Find the average number of cars in the service station, the average number of cars waiting to be serviced and the average time a car spends in the system. 21. A mechanic repairs four machines. The mean time between service requirements is 5 h for each machine and forms an exponential distribution. The mean repair time is 1 h and also follows the same distribution pattern. Machine downtime costs Rs. 25 per hour, and the mechanic costs Rs. 55 per day. Determine the following: (a) The probability that the service facility will be idle (b) The probability of various numbers of machines (0–4) to be repaired, and being repaired (c) The expected number of machines waiting to be repaired, and being repaired (d) The expected downtime cost per day.

22.

23.

24.

25.

Would it be economical to engage two mechanics, each repairing only two machines? A repairman is to be hired to repair machines which break down at an average rate of 6 per hour. The breakdown follows a Poisson distribution. The non-productive time of a machine is considered to cost Rs. 20 per hour. Two repairmen, Mr. X and Mr. Y, have been interviewed for this purpose. Mr. X charges Rs. 10 per hour, and he services breakdown machines at the rate of 8 per hour. Mr. Y demands Rs. 14 per hour, and he services at an average rate of 12 per hour. Which repairman should be hired? (Assume an 8-h shift per day.) A TV repairman ﬁnds that the time spent on his jobs has an exponential distribution with mean 30 min. If he repairs sets in the order in which they come in, and if the arrival of sets is approximately Poisson with an average of 10 per 8-h day, what is the repairman’s expected idle time each day? How many jobs are ahead of the average set just brought in? An insurance company has three claim adjusters in its branch ofﬁce. People with claims against the company are found to arrive in a Poisson fashion at an average rate of 20 per 8-h day. The amount of time that an adjuster spends with a claimant is found to have an exponential distribution with a mean service time of 40 min. Claimants are processed in the order of their appearance. How many hours a week can an adjuster except to spend with claimants? At a port, let there be six unloading berths and four unloading crews. When all the berths are full, arriving ships are diverted to an overflow capacity 20 km

484

15

Queueing Theory

down the river. Tankers arrive according to a Poisson process with a mean of one every 2 h. It takes an unloading crew, on average, 10 h to unload a tanker, the unloading time following an exponential distribution. (a) On average, how long does a tanker spend at the port? (b) What is the average arrival rate at the overflow capacity? 26. There is congestion on the platform of a railway station. The trains arrive at the rate of 30 trains per day. The waiting time for any train to hump is exponentially distributed with an average of 36 min. Calculate the following: (a) The mean queue size (b) The probability that the queue size exceeds 9. 27. A colliery working one shift per day uses a large number of locomotives which break down at random intervals; on average one fails per 8-h shift. The ﬁtter carries out a standard maintenance schedule on each faulty locomotive. Each of the ﬁve main parts of this schedule takes on average 12 h, but the time varies widely. How much time will the ﬁtter have for the other tasks, and what is the average time a locomotive is out of service? 28. If, for a period of 2 h in a day, trains arrive at the yard every 24 min and the service time is 36 min, then calculate for this period: (a) The probability that the yard is empty (b) The average queue length on the assumption that the line capacity of the yard is limited to 4 trains only. 29. At what average rate must a clerk at a supermarket work in order to ensure a probability of 0.09 that the customer will not have to wait longer than 10 min? It is assumed that there is only one counter, to which customers arrive in a Poisson fashion at an average of 15 per hour. The length of service by the clerk has an exponential distribution. 30. Customers arrive at a one-window drive-in bank according to a Poisson distribution with mean 10 per hour. The service time per customer is exponential with mean 5 min. The space in front of the window, including that for the serviced car, can accommodate 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? (d) How many spaces should be provided in front of the window so that all the arriving customers can wait in front of the window at least 90% of the time?

15.7

Exercises

485

31. In a self-service facility, arrivals occur according to a Poisson distribution with mean 50 per hour. The service time per customer is exponentially distributed with mean 5 min. (a) Find the expected number of customers in service. (b) What is the percentage of time the facility is idle? 32. An oil reﬁnery receives crude oil at an average rate of one tanker per day. The unloading facilities, which operate 24 h per day, can handle one tanker at a time, but can unload tankers at an average rate of 2 per day. Under the usual assumptions of Poisson arrival and exponential service times, determine the: (a) (b) (c) (d)

Average number of tankers in the system Average time spent by a tanker in the system Average waiting time of a tanker in the queue Percentage of time in which exactly two tankers are in the system.

33. A group of engineers has two terminals available to aid in their calculations. The average computing job requires 20 min of terminal time, and each engineer requires some computation about once every 0.5 h; that is, the mean time between a call for service is 0.5 h. Assume these jobs are distributed with an exponential distribution. If there are six engineers in the group, ﬁnd: (i) The expected number of engineers waiting to use one of the terminals (ii) The total lost time per day. 34. In a factory cafeteria, the customers (employees) have to pass through three counters. The customers buy coupons at the ﬁrst counter, select and collect the snacks at the second counter and collect tea at the third. The server at each counter takes on average 1.5 min, although the distribution of service time is approximately Poisson at an average rate of 6 per hour. Calculate: a. The average time of getting the service b. The average time a customer spends waiting in the cafeteria c. The most probable time in getting the service. 35. At a certain airport it takes exactly 5 min to land an aeroplane, once it is given the signal to land. Although incoming planes have scheduled arrival times, the wide variability in arrival times produces an effect which makes the incoming planes appear to arrive in a Poisson fashion at an average rate of 6 per hour. This produces occasional congestion at the airport, which can be dangerous and costly. Under these circumstances, how much time will a pilot expect to spend circling the ﬁeld waiting to land? 36. In a car manufacturing plant, a loading crane takes exactly 10 min to load a car into a wagon and again come back to position to load another car. If the arrival of cars is a Poisson stream at an average of one every 20 min, calculate the average waiting time of a car.

486

15

Queueing Theory

37. A barber runs his own saloon. It takes him exactly 25 min to complete one haircut. Customers arrive in a Poisson fashion at an average rate of one every 35 min. (a) For what percentage of time would the barber be idle? (b) What is the average time that a customer spends in the shop? 38. The repair of a lathe requires four steps to be completed one after another in a certain order. The time taken to perform each step follows an exponential distribution with a mean of 5 min and is independent of the other steps. The machine breakdown follows a Poisson process with a mean rate of 2 breakdowns per hour. Answer the following: (i) What is the expected idle time of the machine, assuming there is only one repairman available in the workshop? (ii) What is the average waiting time of a broken-down machine in the queue? (iii) What is the expected number of broken-down machines in the queue?

Chapter 16

Flow in Networks

16.1

Objectives

The objectives of this chapter are to: • Present ideas on graphs and their properties • Study flow and potential in networks • Discuss the solution procedure of a minimum path problem when all arc lengths are either non-negative or unrestricted in sign • Present ideas on network flow, maximum flow problems and generalized maximum flow problems • Discuss the source, the sink of a flow, the cut and its capacity in a network flow problem and also the proof of the max-flow min-cut theorem • Discuss the solution procedure of maximum flow problems • Derive the LPP formulation of a network.

16.2

Introduction

Network is an important topic in the areas of engineering as well as in applied mathematics. It arises in the different sectors of our daily life, such as transportation, electrical circuits and communication. Its representation is also widely used for problems in diverse areas like production, distribution, project planning, facilities location, resource management and ﬁnancial planning. In fact, a network representation provides such a powerful visual and conceptual aid for portraying the relationship between the components of the system that it is used in virtually every ﬁeld of scientiﬁc, social and economic endeavours. One of the most exciting developments in operations research in recent years has been the rapid advancement in both the methodology and applications of network optimization models. © Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_16

487

488

16 Flow in Networks

A number of algorithmic breakthroughs have had a major impact, employing ideas from computer science concerning data structure and efﬁcient data manipulation. Consequently, algorithms and software are now available and are being used to solve numerous problems on a routine basis that would have been completely intractable two or three decades ago. A network, in its more generalized and abstract sense, is called a graph. In recent years, graph theory has been a subject of much study and research by mathematicians and has found ever-increasing applications in diverse areas. In the ﬁeld of operations research, graph theory plays an important role in ﬁnding the optimal solution of the problem of choosing the best sequence of operations from a ﬁnite number of alternatives.

16.3

Graphs

A graph is a set of points (commonly called vertices or nodes) in space that are interconnected by a set of lines (called edges or arcs). For a graph G, the vertex set is denoted by V and the edge set by E. Common nomenclature denotes the number of vertices jV j by n and the number of edges jE j by m. In Fig. 16.1, a graph G with V ¼ fv1 ; v2 ; v3 ; v4 ; v5 g, E ¼ fe1 ; e2 ; . . .; e7 g; n ¼ 5 and m ¼ 7 is shown.   The elements of E are denoted either by ek ; k ¼ 1; 2; . . .; m or vi ; vj . If the values of both n and m are ﬁnite, then the graph G is said to be ﬁnite. The degree of a vertex v [denoted by d(v)] is the number of edges that have v as an end point. For example, in the graph of Fig. 16.1, there are two vertices v1 and v2 of odd degree (both have degree 3).     A self-loop is an edge vi ; vj where vi ¼ vj . Two edges vi ; vj and ðvr ; vs Þ are parallel edges if vi ¼ vr and vj ¼ vs . An arc with an arrow denoting the direction from vi to vj is called a directed arc. A graph with directed arcs is called a directed graph or digraph (Fig. 16.2). A subgraph of G(V, E) is deﬁned as a graph G1(V1, E1) with V1 V and E1 containing all those arcs of G which connect the vertices of G1. For example, in Fig. 16.1, if V1 ¼ ðv1 ; v2 ; v4 Þ and E1 ¼ ðe1 ; e4 ; e6 Þ, then G1(V1, E1) is a subgraph of G. Fig. 16.1 Graph

v1

e1

v2 e2

e7

v3

e4

e6

e3 v5

e5

v4

16.3

Graphs

Fig. 16.2 a Undirected graph, b directed graph

489

v1

(a)

v2

(b)

v3

A partial graph of G(V, E) is a graph G2(V2, E2) which contains all the vertices of G and some of its arcs ðE2 E Þ. An arc (directed or undirected) is said to be incident with a vertex in which it joins to some other vertex; thus, it connects two vertices. The directed arc ek ¼   vi ; vj is said to be incident from or going from vi and incident to or going to vj. We   call vi as the initial vertex and vj as the terminal vertex of the arc vi ; vj . Let V1 and  V2 be two subsets of V such that they have no common vertex, and let ek ¼ vi ; vj be an arc such that vi 2 V1 ; vj 2 V2 . Then ek is said to be incident from or going from V1 and incident to or going to V2. It is incident with both V1 and V2 and is said to connect them. In Fig. 16.3, if V1 ¼ fv2 ; v3 g and V2 ¼ fv6 ; v7 g, then e4 connects V1 and V2. It runs from V1 to V2 and is incident with both. Now we denote by XðVk Þ the set of arcs of G(V, E) incident with a subset Vk of V, by X þ ðVk Þ the set of arcs incident to Vk and by X ðVk Þ the set of arcs incident from Vk. Hence,

Fig. 16.3 Directed graph

490

16 Flow in Networks

X þ ðV1 Þ ¼ fe1 ; e5 g;

X ðV1 Þ ¼ fe2 ; e4 ; e6 g; XðV1 Þ ¼ fe1 ; e2 ; e4 ; e5 ; e6 g:

Similarly, X þ ðV2 Þ ¼ fe4 ; e8 g; X ðV2 Þ ¼ fe5 g; XðV2 Þ ¼ fe4 ; e5 ; e8 g. For a sequence of arcs ðe1 ; e2 ; . . .; ek ; ek þ 1 ; . . .; em Þ of a graph, if every intermediate arc ek has one vertex common with the arc ek−1 and another common with ek+1, then this sequence is called a chain. For example, the sequence ðe1 ; e6 ; e7 Þ in Fig. 16.3 is a chain. We may also denote a chain by the vertices which it connects; for example, this chain may also be written as ðv1 ; v2 ; v4 ; v5 Þ. A chain becomes a cycle if, in the sequence of arcs, no arc is used twice, and the ﬁrst arc has a common vertex with the last arc and this vertex is not common with any intermediate arc. For example, the chain ðe3 ; e4 ; e9 ; e5 Þ in Fig. 16.3 is a cycle. A path is a chain in which all the arcs are directed in the same sense such that the terminal vertex of the preceding arc is the initial vertex of the succeeding arc. In Fig. 16.3, the sequence of arcs ðe1 ; e6 ; e7 ; e8 Þ is a path. We may also denote the path in terms of the vertices as ðv1 ; v2 ; v4 ; v5 ; v7 Þ. Note that every path is a chain, but every chain is not a path. A circuit is a cycle in which all the arcs are directed in the same sense. The cycle ðe1 ; e3 ; e2 Þ is a circuit. A graph is said to be connected if for every pair of vertices there is a chain connecting the two. A graph is strongly connected if there is a path connecting every pair of vertices in it. A tree is deﬁned as a connected graph with at least two vertices and no cycles. It can be proved that a tree with n vertices has (n − 1) arcs and that every pair of vertices is joined by one and only one chain. If we delete an arc from a tree, the resulting graph is not connected, and if we add an arc, a cycle is formed. As the name indicates, a natural tree is the best example of a graphical tree, with the branches forming the arcs and the extremities of the branches forming the vertices (Fig. 16.4). A vertex which is connected to every other vertex of the graph by a path is called a centre of the graph. A graph may or may not have a centre, or it may have many centres. Every vertex of a strongly connected graph is a centre. A tree with a centre is called an arborescence. In Fig. 16.5, the centre is marked. In arborescence all the arcs incident with the centre are going away from it, and all other arcs are directed in the same sense.

Fig. 16.4 Tree

16.4

Minimum Path Problem

491

.

Centre Fig. 16.5 Graph with centre

16.4

Minimum Path Problem

  Let us consider a real number xij which is associated with each arc vi ; vj of a graph G(V, E), and also let va and vb be two vertices of the graph. There may be a number P of paths from va to vb. For each path, the length of the path is deﬁned as xij , where the summation is over the sequence of arcs forming the path. In this case, the objective of the problem is to ﬁnd the path of smallest length. The term ‘length’ is used here in a general sense of any real number associated with the arc and should not be regarded as a geometrical distance. It may be distance, time, cost, etc. There may be some abstract situations in which the length is not even non-negative. In general, xij is a real number, unrestricted in sign. Several techniques/algorithms have been suggested for solving the minimum path problem. Here, we shall describe two different techniques for solving the minimum path problem when (i) xij  0 and (ii) xij is unrestricted in sign. Case I When xij  0, i.e. when all arc lengths are non-negative. Let fi denote the minimum length of the path from the vertex va to vi. We have to ﬁnd fb. Obviously, fa = 0. Let Vp be a subset of V such that va 2 Vp and vb 62 Vp . First, we have to determine fi for every vi in Vp and then fi + xij for every vi in Vp and vj not in Vp such that vi ; vj is an arc incident from Vp. Let

  fr þ xrs ¼ Min fi þ xij ; i

where vr 2 Vp and vs 62 Vp : Then the length of the minimum path from va to vs is given by fs = fr + xrs. Now, from an enlarged subset Vp+1 of V denoted by Vp þ 1 ¼ Vp [ fvs g we repeat the process.

492

16 Flow in Networks

Suppose we start with p = 0 with V0 consisting of a single vertex va and fa = 0. Following the procedure, the sets V1 ; V2 ; . . .; Vp ; Vp þ 1 ; . . . are formed. As soon as we arrive at a set in this sequence which includes vb and fb has been obtained. If no such set can be found, there is no path connecting va to vb. We note that for solving these types of problems, drawing of the graph is not essential either to describe the problem or to ﬁnd its solution. The problem is completely enunciated if all the vertices, arcs and arc lengths are speciﬁed. In fact, in a large problem with vertices and arcs, drawing of a graph is neither practicable nor necessary. Case II: When xij is unrestricted in sign, i.e. when arc lengths are unrestricted in sign. Let va, vb be two vertices in the graph G(V, E) whose arc lengths are real numbers, positive, negative or zero. We have to ﬁnd the minimum path from va to vb. Let us assume that there is no circuit in the graph whose arc length is negative. If there is any such circuit, one can go around and around it and decrease the length of the path without limit, getting an unbounded solution. For solving the problem, we have to construct an arborescence A1 ðV1 ; E1 Þ; V1 V; E1 E, with centre va and V1 containing all the vertices of V which can be reached from va along a path and E1 containing from va along a path and some of the arcs of E which are necessary to construct the arborescence. If V1 contains vb, a path connects va to vb. In a particular arborescence, this path is unique. There may be several arborescences; therefore, several paths exist. If in any problem only one arborescence is possible, there is only one path from va to vb, and that is the solution. If V1 does not contain vb, there is no path from va to vb, and the problem has no solution. We now discuss the method of construction of an arborescence. Mark out the arcs going from vb. From the vertices so reached, mark out the arcs (not necessarily all of them) going out to other vertices. No vertex should be reached by more than one arc; that is, not more than one arc should be incident to any vertex. If there is a vertex to which no arc is incident, it cannot be reached from va and so is left out. No arc incident to va should be drawn (Fig. 16.6). Let fi denote the length of the path from va to any vertex vi in the arborescence. The arborescence determines fi uniquely for each vi in V, but fi is not necessarily   minimum. Let vj ; vi be an arc in G but not in arborescence A1. Then consider the length fj þ xji and compare it with fi. If fi  fj þ xji , make no change. If fi [ fj þ xji ,   delete the arc incident to vi in A1 and include the arc vj ; vi . This modiﬁes the arborescence from A1 to A2 and reduces fi to its new value fj þ xji , the reduction in the value of fi being fi  fj  xji . The lengths of the paths to the vertices going through vi are also reduced by the same amount. These adjustments are made, and thus the new values of fi for all vi in A2 are calculated. Now repeat the operations in A2; i.e. select a vertex and check whether any alternative arc gives a smaller path to it. If yes, modify A2 to A3 and adjust fi

16.4

Minimum Path Problem

Fig. 16.6 Graph with centre

493

(a) 6

7

3

5

4

2

8

0

1

6

7

3

5

4

2

0

1

(b)

(c) 6

7

3

5

4

2

0

1

accordingly. Finally, we reach an arborescence Ar which cannot further be changed by the described procedure. Theorem-1 The arborescence Ar marks out the minimum path to each vi from va, and fb in this arborescence is the minimum path to vb. Proof Let ðva ; v1 ; v2 ; . . .; vb Þ be any path in G from va to vb. Its length is xa1 þ x12 þ    þ xpb . The vertices in this path are in Ar because Ar contains all those vertices of G which can be reached  from va. By the property of Ar, for every vertex vi in Ar and for every arc vj ; vi in G,

494

16 Flow in Networks

fi  fj þ xji or

fi  fj  xji

because otherwise Ar could have been further modiﬁed. Writing these inequalities for all vertices of the above path, we have f1  fa  xa1 f2  f1  x12 ... fb  fp  xpb : Adding, we get fb  fa  xa1 þ a12 þ    þ xpb or

fb  xa1 þ a12 þ    þ xpb

½* fa ¼ 0:

Thus, it is proved that no path from va to vb in G can be smaller than fb. Since the path length fb is also in G, this path is the minimum. Note The path of maximum length can be found either by changing the signs of the lengths of all the arcs and then ﬁnding the minimum or by reversing the inequality fi [ fj þ xji to fi \fj þ xji .

16.5

Problem of Potential Difference

Potential plays an important role in electricity, fluid flow and other branches of mechanics. In graph theory, the potential in a network is deﬁned as follows. For each vertex in a graph G(V, E), there is associated a real number fi called its potential. For an arc of a graph, there is an associated potential difference. This potential difference is denoted by xij for an arc ek = (vi, vj) and is deﬁned by xij = fj – fi. Hence, the potential difference in a chain through the vertices v1, v2, …, vp is deﬁned as x12 þ x23 þ    þ xp1;p ¼ fp  f1 : It follows immediately that the potential difference in a cycle is zero. Again, it is obvious that xij = −xji. The problem of maximum potential difference in a network is stated as follows: The potential difference xab of two vertices va and vb of a graph G(V, E) is to be maximized subject to the conditions xij  cij for all the arcs (vi, vj), where cij are constants.

16.5

Problem of Potential Difference

495

In terms of potential, these constraints may be written as fj  fi  cij

for all the arcs ðvi ; vj Þ:

Hence, the problem reduces to the following: Maximize fb  fa subject to the constraints fj fi  cij

  for all arcs vi ; vj of the graph GðV; E Þ:

Now, if we consider the potential of the vertex va as zero, then this problem reduces to Maximize fb subject to the constraints   fj  fi  cij for all arcs vi ; vj of the graph GðV; E Þ; where fi is a real-valued function associated with the vertex vi in a graph with fa = 0. The problem is identical to the minimum path problem, as the inequalities of this problem are similar to the inequalities of the minimum path problem. The only difference between the two inequalities is that xij of the minimum path problem is replaced by cij here. Therefore, to solve the problem of potential difference, we have to solve the problem of minimum path from va to vb with cij as the length of the arc (vi, vj). A more general problem of maximum potential difference in a network is of the following type:   bij  fj  fi  cij for all arcs vi ; vj : The solution methodology remains the same, as each inequality of the above type can be written as fj  fi  cij fi  fj   bij :

16.6

Network Flow Problem

Like potential, flow in a network is also a familiar concept in electrical theory, flows of trafﬁc through a network of roads, flows of liquid through a network of pipelines, etc. The basic principle of flow in a network is that, for every vertex, the total flow in is equal to the total flow out. To extend this idea in a more abstract situation, it is required to give a precise deﬁnition of flow in a graph.

496

16 Flow in Networks

Let a real number xk be associated with every arc ek, k = 1, 2, …, m of a graph G (V, E) such that for every vertex vj, X X xk ¼ x ; k ¼ 1; 2; . . .; m; 1 k P where the left-hand side summation is taken for all arcs going to vj and the P right-hand side summation 1 is taken for all arcs going from vj. Then xk is said to be a flow in the arc ek and the set {xk}, k = 1, 2, …, m is said to be a flow in the graph G. Now we shall deﬁne a graph to state the problem of maximum flow in a network. Let G(V, E) be a graph with V as the set of (n + 2) vertices va, v1, v2, v3, …, vn, vb and E as the set of (m + 1) arcs e0, e1, e2, …, em. The vertices va and vb and the arc e0 play a special role in this graph. va is called the source, vb the sink, and the arc e0 connects vb to va. It is the only arc going from vb. For every arc ek, k = 1, 2, …, m (except e0), there is associated a real number ck (  0) called the capacity of the arc. Let {xk} be a flow in the graph G such that 0  xk  ck (k = 1, 2, …, m), where ck is the capacity of the flow of the arc ek. In this context, note that x0 as the flow in the arc e0 is deﬁned, but the capacity of the arc e0 is not deﬁned. So there is no constraint on x0. Since x0 is the only flow out at vb and flow in at va, then by the deﬁnition of flow, we have total flow in at vb ¼ total flow out at vb ¼ x0 ¼ total flow in at va ¼ total flow out at va Since all the flows out at va are equal to the flows in at vb, then va and vb are called the source and sink respectively. The arc e0 plays an important role in the flow network and is considered the return arc. Hence, our problem is to determine the maximum flow out at the source (=maximum flow in in the sink). More precisely, our problem is to determine the flow {xk} such that x0 is maximum subject to the constraints 0  xk  ck, k = 1, 2, …, m. Now we have an important question: How can we solve the maximum flow problem? The procedure for solving such a problem is given in the following algorithm. Algorithm Step 1 Set an initial feasible flow by guessing. If this is not possible, initially set xk = 0 for all k.

16.6

Network Flow Problem

497

Step 2 Divide the set V of vertices into two subsets W1 and W2 such that each vertex is either in W1 or in W2 but not in both. Initially, consider W1 = {va} and all other vertices in W2. Step 3 Transfer a vertex from W2 to W1 by the following procedure: For vi 2 W1, vj 2 W2, (a) Transfer vj to W1 if (vi, vj) is an arc ek with xk < ck; (b) Transfer vj to W1 if (vj, vi) is an arc ek with xk > 0; (c) Do not transfer vj to W1 otherwise. Repeat the process for transferring the vertices from W2 to W1. Finally, if vb is transferred to W1 by this procedure, the flow is not optimal. Go to Step 4. Step 4 Increase the value of xk in the arc of category (a) mentioned above in which xk < ck and decrease the value of xk in the arc of category (b) in which xk > 0 so that the flow remains feasible and at least one arc gets the capacity flow. Then go to Step 2. After repetitive operations of Steps 2, 3 and 4, a situation will appear when vb cannot be transferred to W1 by the operation of Step 3. In this situation, the flow is optimal. Deﬁnition If in the graph G(V, E) of the maximum flow problem, W2 is a subset of V such that vb 2 W2, va 62 W2, then the set of arcs X þ ðW2 Þ (arcs incident to W2) is said to be a cut. The capacity of the cut is the sum of the capacities of the arcs contained in the cut. Theorem 1 For any feasible flow {xk}, k = 1, 2, …, m, in the graph, the flow x0 in the return arc is not greater than the capacity of any cut in the graph. Proof Let X þ ðW2 Þ be any cut. Now, let us consider the flow in the arcs going to and going from W2. According to the deﬁnition of flow, the flow in is equal to the flow out. Therefore, X

xk ¼ x0 þ

X

x; 1 k

P P where and 1 denote the summations over the arcs going to and going from W2 (except e0). P Since xk  0 for all k, then xk  x0 . Also, xk  ck for all k. P P Therefore ck  x0 ; where ck is the capacity of the cut X þ ðW2 Þ. This proves the theorem. Theorem 2 The preceding algorithm solves the problem of the maximum flow. Proof Let us suppose that by successive applications of the algorithm, a situation will appear when no vertex of W2 can be transferred to W1 by the prescribed

498

16 Flow in Networks

procedure and vb 2 W2. Let the set of arcs X þ ðW2 Þ be a cut. Also let ek 2 X þ ðW2 Þ. This implies that ek is an arc (vr, vs), where vr 2 W1, vs 2 W2. In this case, the flow in this arc should be saturated, i.e. xk = ck; otherwise, it would have been possible to transfer vs from W2 to W1 by the criterion of Step 3(a), which is contrary to the hypothesis. Again, let ei 2 X þ ðW2 Þ, i 6¼ 0. This means that ui is an arc (vp, vq), where vp 2 W2, vq 2 W1. In this case, the flow in this arc should be zero; otherwise, it would have been possible to transfer vp from W2 to W1 by the criterion of Step 3(b), which is again contrary to the hypothesis. P Therefore, we conclude that the flow into W2 is ck , the summation being over all ek 2 X þ ðW2 Þ, and the flow out of W2 is only in the return arc e0 as it is the only arc going from W2 carrying a non-zero flow. Let the flow in e0 be y0. Then, from the deﬁnition of flow, X ck ¼ y0 ; P where ck is the total capacity of the cut obtained by the application of the algorithm. But from the theorem ‘For any feasible flow {xk}, k = 1, 2, …, m, in the graph, the flow x0 in the return arc is not greater than the capacity of any cut in the graph.’, for any flow x0 in e0, P ck ; x0  i:e: x0  y0 ; P where ck is the capacity of any cut. It follows that y0 = Max x0. This means that the maximal flow problem can be solved by the preceding algorithm.

Max-Flow Min-Cut Theorem Theorem 3 The maximum flow in a graph is equal to the minimum of the capacities of all possible cuts in it. Proof Let va and vb be the source and sink of the graph G(V, E) of a flow problem. Also, let W2 be a subset of V such that vb 2 W2, va 62 W2. Let X þ ðW2 Þ be any cut. Let us consider the flow in the arcs going to and going from W2. Now, from the deﬁnition of flow, the flow in should be equal to the flow out, P

i:e:

P

where and W2 (except e0).

1

X

xk ¼ x0 þ

X

x; 1 k

denote the summations over the arcs going to and going from

16.6

Network Flow Problem

499

P Since for all k, xk  0; xk  x0 . Again, xk  ck for all k. P P ck is the total capacity of the cut X þ ðW2 Þ, Therefore, ck  x0 where X or x0  ck : X Therefore; max x0  min ck : Again, we know that there is a cut corresponding to which the flow in e0 is equal to the cut capacity. Necessarily this flow should be maximum, and the corresponding cut capacity should be the least of all cut capacities. This proves the theorem.

16.7

Generalized Problem of Maximum Flow

The general form of the maximum flow problem is as follows: Let G(V, E) be a graph consisting of a source va, a sink vb, vertices vi, i = 1, 2, …, n, arcs ek, k = 1, 2, …, m and a return arc e0. Find a flow {xk} in G such that x0 is maximum subject to bk  xk  ck : Here xk can vary between an upper bound ck and a lower bound bk, both of which are real numbers and not necessarily non-negative. Thus, negative flow xk in an arc is admissible and is interpreted as a flow-xk in the reverse direction. The constraints on xk may be rewritten as 0  x k  bk  c k  bk or

0  yk  ck  bk ; where xk ¼ yk þ bk

or

y k ¼ x k  bk :

In terms of the new variables yk, the constraints are similar to those of the earlier maximum flow problem. However, if {xk} is a flow, {yk} is not necessarily a flow. From the deﬁnition of flow, {xk} is a flow if X

xk ¼

X

x 1 k

P P P P or if ykP þ bkP ¼ 1 yk þ 1 bk for every vertex vj, where and 1 are summations on arcs going to and going from vjP . From this equation, it is clear that for {y } to be a flow, the relation bk ¼ k P b must hold. But there is no reason that this relation should hold, in general. 1 k

500

16 Flow in Networks

To overcome this difﬁculty, a new vertex v0 in G, called a ﬁctitious vertex, is introduced, and the necessary number of arcs, called ﬁctitious arcs, are drawn according to the following rules. Rule 1: For an arc ek = (vi, vj) with bk > 0, form a cycle (v0, vj, vi, v0) by drawing the arcs (v0, vj) and (vi, v0). Then getting a flow in this cycle, assume a flow bk in each of the arcs (v0, vj), (vj, vi), (vi, v0). Rule 2: For an arc ek = (vi, vj) with bk < 0, form a cycle (v0, vi, vj, v0) by drawing the arcs (v0, vi) and (vj, v0). Then getting a flow in this cycle, assume a flow −bk in each of the arcs (v0, vi), (vi, vj), (vj, v0). Due to the inclusion of the ﬁctitious vertex and arcs in G(V, E), we get a new graph G1(V1, E1) with constant flows in the cycles so formed. These flows together form a flow in G1, as it is a linear combination of flows in the cycles. Now, we shall refer to this flow as the ﬁctitious flow in G1. Let {xk} be any flow in G. It is also a flow in G1. Let this flow be superimposed on the ﬁctitious flow. The result is a flow, as it is a linear combination of two flows. Denoting this flow by {yk}, we have  yk ¼

xk  bk ; for ek in G : the fictitious flow for all fictitious arcs

Let us now determine the flow {yk} in G1 such that y0 is maximum subject to 0  yk  ck  bk for ek 2 G and yk ¼ fictitious flow for fictitious arcs: This involves keeping the flow constant in some arcs of G1 and varying it in others until y0 is maximum. This can be done by our previous algorithm. Having determined the optimal flow {yk}, we have to calculate xk = yk + bk, and ﬁnally we get the optimal flow {xk} in G.

16.8

Formulation of an LPP of a Flow Network

Let us consider the flow network shown in Fig. 16.7. Also, let x0 be the flow of the return arc from vb to va. For different arcs, the flow variables are given in the following table: Arc Variable

(a, 1) x1

(1, 2) x2

(1, b) x3

(2, 1) x4

(2, 3) x5

(2, 4) x6

(2, b) x7

(3, a) x8

(3, 4) x9

(4, b) x10

(b, 2) x11

16.8

Formulation of an LPP of a Flow Network

501

x3

4

1

x1 5

x2

2

x4

x11

x5 5

2

x8

x6 x9

3

4

x7

2

a

b

3

6

5

x10 3

5

4

x0 Fig. 16.7 Representation of network flow

  Here vi ; vj is denoted by ði; jÞ: From the deﬁnition of flow, we know that the flow into the vertex is equal to the flow out from the vertex. Hence, for different vertices, the constraints are as follows: For vertex v1, x1 þ x4 ¼ x2 þ x3 ) x1 x2 x3 þ x4 ¼ 0: For vertex v2, x2 þ x11 ¼ x4 þ x5 þ x6 þ x7 ) x2 x4 x5 x6 þ x11 ¼ 0: For vertex v3, x5 ¼ x8 þ x9 ) x5 x8 x9 ¼ 0: For vertex v4, x6 þ x9 ¼ x10 ) x6 þ x9 x10 ¼ 0:

502

16 Flow in Networks

For vertex va, x0 þ x8 ¼ x1 ) x0 x1 þ x8 ¼ 0: For vertex vb, x3 þ x7 þ x10 ¼ x0 þ x11 ) x0 þ x11 x3 x7 x10 ¼ 0: For the network, the capacity constraints are as follows: x1  5, x2  6, x3  4, x4  2, x5  2, x6  5, x7  4, x8  5, x9  5, x10  3, x11  3 and xi  0; i = 1, 2, 3, …, 11. Hence, the LPP of the given network is as follows: Maximize f = x0 subject to x1 – x2 – x3 + x4

=0

– x4 – x5 – x6

x2

+ x11 – x8 – x9

x5

=0

+ x9 – x10

x6 x0 – x1

=0

+ x8

x0

– x3

=0

=0

– x7

- x10 +x11 = 0

x1  5, x2  6, x3  4, x4  2, x5  2, x6  5, x7  4, x8  5, x9  5, x10  3, x11  3, xi  0; i = 1, 2, 3, …, 11 and x0 is unrestricted in sign. Example 1 Find the minimum path from v0 to v8 in the graph (Fig. 16.8) in which the number along a directed arc denotes its length.

10

1

4

3

2

0

1

2

1

8

3 6

5

4

2

6

3

4

Fig. 16.8 Representation of directed flow network

10

7

6

1 8

2 5

1 7

8

16.8

Formulation of an LPP of a Flow Network

503

Solution Let V be the set of all of the given graph and Vp a subset of V. Let  vertices  xij be the arc length of an arc vi ; vj and fi be the minimum length of the path from the vertex v0 to a vertex vi of Vp. Also, let fs be the minimum length of the path from Vp to another vertex vs, not in Vp, through an arc incident from Vp,       i:e: fs ¼ min fi þ xij for all vi 2 Vp and vi ; vj 2 X Vp ; i

    where X Vp is the set of arcs incident from Vp. We denote vi ; vj as (i, j). Now to ﬁnd the minimum path, we construct Table 16.1. From Table 16.1, it is seen that the minimum path is either 0–1–2–3–6–8 or 0– 1–2–5–6–8 and the length of each path is 17. Example 2 Find the minimum path from v0 to v7 in the graph of Fig. 16.9. Solution From the graph, it is clear that there is no circuit whose length is negative. Now, we draw an arborescence A1 (see Fig. 16.10a) with centre v0 consisting of all those vertices of the graph which can be reached from v0 and the necessary number of arcs. From the graph, it is also clear that a number of arborescences can be drawn. A1 is one of them. Now the lengths fi of the paths from v0 to different vertices vi of A1 are as follows: f0 ¼ 0;

f1 ¼ 1; f2 ¼ 4; f3 ¼ 4  1 ¼ 5; f4 ¼ 3;

f7 ¼ 3 þ 2 ¼ 5; f5 ¼ 2; f6 ¼ 2 þ 2 ¼ 0: Let us consider the vertex v2. There is an arc ðv1 ; v2 Þ in G which is not in A1 such that f2 ¼ 4\f1 þ x12 ¼ 1 þ 2 ¼ 3: So, A1 will be unchanged. Now we consider the vertex v3. There is an arc ðv1 ; v3 Þ in G which is not in A1 such that f3 ¼ 5\f1 þ x13 ¼ 1 þ 2 ¼ 3: So, A1 will be unchanged. Now, for v4 in A2, arc ðv3 ; v4 Þ is in G, but not in A2 such that f4 ¼ 3 [ f3 þ x34 ¼ 5  4 ¼ 9: So we delete the arc ðv0 ; v4 Þ, include ðv3 ; v4 Þ, get another arborescence A2 (see Fig. 16.10b) with f4 = −9 and consequently f7 = −9 + 2 = −7. For v5 in A2, arc ðv4 ; v5 Þ is in G but not in A2 such that

504

16 Flow in Networks

Table 16.1 Calculation of minimum path   xij p Vp fi fi + xij X Vp 0

0

0

1

0

0

1

2

0 1

0 2

2

5

0 1 2 3 5

0 2 5 6 6

0 1 2 3 4 5 0 1 2 3 4 5 6

0 2 5 6 7 6 6 0 2 5 6 7 6 10

7

10

2

3

4

5

(0, (0, (0, (0, (0, (1, (1, (1, (0, (1, (1, (2, (2,

vs

Minimum path up to vs

2

1

0–1

5

2

0–1–2

6 6

3 5

0–1–2–3 0–1–2–5

7

4

0–1–2–5–4

10 10 10 11

10 10 10

6 7 6

0–1–2–3–6 0–1–2–5–4–7 0–1–2–5–6

7

17

17

8

0–1–2–3–6–8 0–1–2–5–6–8

10

20

1) 2) 3) 2) 3) 2) 4) 5) 3) 4) 5) 3) 5)

2 6 8 6 8 3 10 8 8 10 8 1 1

2 6 8 6 8 5 12 10 8 12 10 6 6

(1, 4)

10

12

6) 4) 6) 7)

4 1 4 5

10 7 10 11

(3, 6) (4, 7) (5,6) (5,7)

4 3 4 5

(6, 8) (7, 8)

(3, (5, (5, (5,

fs

16.8

Formulation of an LPP of a Flow Network

-5

-1

6

3

7

-2 -2

2

505

-4

2

-1 2

4

5

2

3

-2

2

-4 0

1

1

Fig. 16.9 Graph with known edge

(a)

(b)

6

3

7

2

2

-1 4

5

0

2

-4

1

0

1

1

1

(d)

6

5

-1

4

5

-2

(c)

2

-4

2

-4

-2

3

7

2 2

3

6

3

7

-4

2

6

-2 -2

-1

-2 4

2

3

7

-4

2

5

2

-4 0

-1

4

-4 1

0

1

1

Fig. 16.10 a Arborescence A1, b arborescence A2, c arborescence A3, d arborescence A4

f5 ¼ 2 [ f4 þ x45 ¼ 9  2 ¼ 11: So we delete ðv0 ; v5 Þ, include ðv4 ; v5 Þ, get another arborescence A3 (see Fig. 16.10c) with f5 = −11 and consequently f6 = −11 + 2 = −9. For v6 in A3, arc ðv4 ; v6 Þ is in G but not in A3 such that

506

16 Flow in Networks

Fig. 16.11 Arborescence A5

f6 ¼ 9 [ f4 þ x46 ¼ 9  2 ¼ 11: So we delete ðv5 ; v6 Þ, include ðv4 ; v6 Þ and get another arborescence A4 (see Fig. 16.10d) with f6 = −11. For v7 in A4, arc ðv6 ; v7 Þ is in G but not in A4 such that f7 ¼ 7 [ f6 þ x67 ¼ 11  1 ¼ 12: So we delete ðv4 ; v7 Þ, include ðv6 ; v7 Þ and get another arborescence A5 (see Fig. 16.11) with f7 = −12. Again, for v7 in A5, arc ðv3 ; v7 Þ is in G but not in A5 such that f7 ¼ 12\f3 þ x37 ¼ 5  5 ¼ 10: So, we leave A5 unchanged. From the above arborescence A5, it is observed that A5 cannot further be modiﬁed. Therefore, no alternative arc decreases the length of the path from v0 to any vertex. Hence, the minimum path from v0 to v7 contains the vertices v0 ; v2 ; v3 ; v4 ; v6 ; v7 with length −12. Example 3 Find the minimum path from v1 to v8 in the graph with arc lengths as follows [here (i, j) denotes the arc vi ; vj ]: Arc

(1, 2)

(1, 3)

(1, 4)

(2, 3)

(2, 6)

(2, 5)

(3, 5)

(3, 4)

−8

7

−3 (7, 3)

7 (7, 8)

Length Arc

−1 (4, 7)

4 (5, 6)

−11 2 (5, 8) (6, 3)

(6, 4)

(6, 7)

(6, 8)

Length

3

1

12

2

6

– 10

4

–2

2

Solution For the given problem, the corresponding graph G is given in Fig. 16.12. From the graph, it is clear that there is no circuit whose length is negative. Now, we draw an arborescence A1 (see Fig. 16.13) with centre v1 consisting of all those vertices of the graph which can be reached from v1 and the necessary arcs. Now the lengths fi of the paths from v1 to different vertices vi of A1 are as follows:

16.8

Formulation of an LPP of a Flow Network

507

-8 6

2 7

-1

-10

2 4

1

1 4

3

-3 7

12 5

8

2 -2

-11

2 7

4 Fig. 16.12 Graph

Fig. 16.13 Arborescence A1

8 -10

5

6 -8

7 3

-3 2

4

3 4

-1

-11 1

f1 ¼ 0;

f2 ¼ 1; f3 ¼ 4; f4 ¼ 11; f5 ¼ 1; f6 ¼ 9; f7 ¼ 8; f8 ¼ 19

Let us consider the vertex v3. This vertex is connected by the arcs ðv1 ; v3 Þ; ðv2 ; v3 Þ; ðv6 ; v3 Þ and ðv7 ; v3 Þ. Among these, arcs ðv2 ; v3 Þ; ðv6 ; v3 Þ and ðv7 ; v3 Þ are not in the arborescence A1 such that for the arcðv2 ; v3 Þ f3 ¼ 4 [ f2 þ x23 ¼ 1 þ 2 ¼ 1 f3 ¼ 4 [ f6 þ x63 ¼ 9 þ 4 ¼ 5 for the arcðv6 ; v3 Þ and f3 ¼ 4 [ f7 þ x73 ¼ 8  2 ¼ 10 for the arcðv7 ; v3 Þ:

508

16 Flow in Networks

Fig. 16.14 Arborescence A2

8 -10

5

6

7 -2

-8

3

-3 2

4

3

-1

-11 1

Hence, we delete the arc ðv1 ; v3 Þ, include ðv7 ; v3 Þ, get another arborescence A2 (see Fig. 16.14) with f3 ¼ 10 and consequently f5 ¼ 10  3 ¼ 13. Again for the vertex v4 in A2, arcs ðv3 ; v4 Þ and ðv6 ; v4 Þ are in G but not in A2 such that f4 ¼ 11 \ f3 þ x34 ¼ 10 þ 7 ¼ 3 for arc ðv3 ; v4 Þ and f4 ¼ 11 \ f6 þ x64 ¼ 9 þ 2 ¼ 7

for arc ðv6 ; v4 Þ:

So, A2 will not be changed. Now, for v5 in A2, arc ðv2 ; v5 Þ is in G but not in A2 such that f5 ¼ 13\f2 þ x25 ¼ 1 þ 7 ¼ 6: So A2 will not be changed. Again, for v6 in A2, arc ðv5 ; v6 Þ is in G but not in A2 such that f6 ¼ 9 [ f5 þ x56 ¼ 13 þ 1 ¼ 12: So we delete the arc ðv2 ; v6 Þ, include ðv5 ; v6 Þ, get another arborescence A3 (see Fig. 16.15) with f6 ¼ 12 and consequently f8 ¼ 12  10 ¼ 22. Now for vertex v7 in A3, arc ðv6 ; v7 Þ is in G but not in A3 such that f7 ¼ 8\f6 þ x67 ¼ 12 þ 6 ¼ 6:

16.8

Formulation of an LPP of a Flow Network

509

Fig. 16.15 Arborescence A3

8 -10 1 5

6

7 -2

3

-3 2

4

3

-1

-11 1

So we leave A3 unchanged. Again for vertex v8 in A3, arcs ðv5 ; v8 Þ and ðv7 ; v8 Þ are in G but not in A3 such that f8 ¼ 22\f5 þ x58 ¼ 13 þ 12 ¼ 1 and

f8 ¼ 22\f7 þ x78 ¼ 8 þ 2 ¼ 6:

So we leave A3 unchanged. From the arborescence A3, it is seen that the arborescence A3 cannot further be modiﬁed. Therefore, no alternative arc decreases the length of the path from v1 to any other vertex. Hence the minimum path from v1 to v8 is 1473568 with length −22. Example 4 Find the maximum potential difference x14 between v1 and v4 of the graph with the following data subject to the condition that for each arc xjk  cjk: V E cjk

1 2 ð1; 2Þ ð1; 3Þ 3 2

3 4 ð2; 3Þ ð2; 4Þ 2 1

ð3; 4Þ ð1; 4Þ 4 1

Solution Treating the values of cjk as the arc lengths of a graph, we construct the graph in Fig. 16.16. Now we draw an arborescence A1 with centre v1 consisting of all those vertices of the graph which can be reached from v1 and the necessary number of arcs (Fig. 16.17). Now the lengths fj of the paths from v1 to different vertices vj of A1 are as follows:

510

16 Flow in Networks

Fig. 16.16 Graph of the problem of Example 4

3

1

2 -2

1

2

-1 3

4 4

Fig. 16.17 Arborescence A1

3

1

-1

2 3

f1 ¼ 0;

2

4

f2 ¼ 3; f3 ¼ 2; f4 ¼ 1

Now, for the vertex v4 in A1, arcs (v2, v4) and (v3, v4) are in G, but not in A1 such that

and

f4 ¼ 1\f2 þ c24 ¼ 3 þ 1 ¼ 4 f4 ¼ 1\f3 þ c34 ¼ 2 þ 4 ¼ 6:

So we leave A1 unchanged. Hence, the maximum potential difference x14 = −1 with the optimal path (v1, v4). Example 5 Find the maximum potential difference between v1 and v4 in the graph G(V, E), where V E

1 ð1; 2Þ

subject to the constraints

2 ð1; 3Þ

3 ð2; 3Þ

4 ð3; 4Þ

ð4; 2Þ

ð1; 4Þ

16.8

Formulation of an LPP of a Flow Network

511

Fig. 16.18 Graph of the problem of Example 5

2 4

2 10 3 2

6 -2

-1

-6

3

1 3 2  f2 f1  3; 6  f3 f2  10; f4 f3   2 2  f2 f4 ; 1  f4 f1  6; f3 f1  7: Solution The given constraints can be written as follows: f2  f1  3;

f1  f2  2; f3  f2  10; f2  f3   6

f4  f3   2;

f4  f2  2; f4  f1  6; f1  f4   1; f3  f1  7

For the given problem, the graph is given in Fig. 16.18. The student may easily solve the problem. For this problem, the maximum potential difference between v1 and v4 is 3. Example 6 In the graph of Fig. 16.19, the numbers along the arcs are the values of ci. Find the maximum flow in the graph. For solving the given problem, let us assume that the initial flow is zero in each arc. Let W1 = {va} and W2 be the set of other vertices. Now, for an arc (va, v1), vav 2 W1, v1 2 W2, the flow is zero, which is less than its capacity. So we transfer v1 to W1. Again, for an arc (v1, v5), v1 2 W1, v5 2 W2, the flow is less than its capacity 4. So we transfer v5 to W1. For the arc (v5, vb), v5 2 W1, vb 2 W2, the flow is zero, which is less than its capacity. So vb is transferred to W1. As a result, the solution is not optimal. In this process, the corresponding chain is (va, v1, v5, vb). The least capacity in this chain is 3. So, in each arc of this chain and also in the return arc (vb, va), we increase the flow in such a way that the flow of at least one arc coincides with its capacity, keeping the flows as same in all other arcs. The modiﬁed flow is feasible, as in each arc it is either less than or equal to its capacity, and also at every vertex the total flow in is equal to the total flow out. The above process is repeated for every modiﬁed flow until it is not possible to transfer vb to W1. The flows of all iterations are shown in Table 16.2. In each feasible flow the number in parentheses indicates the corresponding arc forming the chain in which vb is transferred to W1. The asterisk indicates that the flow in the corresponding arc is equal to its capacity and will not further be increased.

512

16 Flow in Networks

1

1

3

4

4

2

0

2 2

a

1

2

5

5

b

1 1

2

1

3

2 6

1

Fig. 16.19 Flow in graph

Table 16.2 Solution of flow problem of Example 6 Arcs

Capacity (ci)

Feasible flows I II

III

IV

V

VI

(a, (a, (a, (1, (1, (1, (2, (2, (3, (3, (4, (4, (5, (5, (6, (b,

3 2 1 1 4 2 2 1 1 1 2 0 2 5 2 –

(0) 0 0 (0) 0 0 0 0 (0) 0 (0) 0* 0 (0) 0 0

3* (0) 0 1* 2 0 0 (0) 1* 0 1 0* 0 3 (0) 3

3* (1) 0 1* 2 0 (0) 1* 1* (0) (1) 0* 0 3 (1) 4

3* 2* (0) (1*) (2) 0 1 1* 1* 1* (2) 0* 0 (3) 2* 5

3* 2* 1* 0 3 0 1 1* 1* 1* 1 0* 0 4 2* 6

1) 2) 3) 4) 5) 6) 4) 6) 5) 6) 3) b) 2) b) b) a)

(1) 0 0 1* (0) 0 0 0 1* 0 1 0* 0 (1) 0 1

16.8

Formulation of an LPP of a Flow Network

513

From Table 16.2, it is seen that the maximum flow in the graph is 6. Note The preceding problem can be solved in different ways of vertex transfer from W2 to W1. Two different ways of solving the problem have been shown in Tables 16.3 and 16.4. Example 7 Find  the maximum flow in the graph with the following arcs and flow capacities. Arc vj ; vk is denoted as (j, k). va is the source and vb , the sink. Flow Flow

Arc Capacity Arc Capacity

(a, 1) 2 (2, 6) 1

(a, 2) 2

(a, 3) 2

(3, 4) 1

(1, 4) 1

(3, 5) 1

(1, 5) 1 (3, 6) 1

(1, 6) 1

(2, 4) 1

(4, b) 2

(5, b) 2

(2, 5) 1 (6, b) 2

For the given problem, ﬁrst of all we draw the graph, which is shown in Fig. 16.20. For solving the given problem, let us assume that the initial flow is zero in each arc. Let W1 = {va} and W2 be the set of other vertices. Now, for an arc (va, v1), va 2 W1, v1 2 W2, the flow is zero, which is less than its flow capacity. So we transfer v1 to W1. Again, for an arc (v1, v4), v1 2 W1, v4 2 W2, the flow is less than its flow capacity. So we transfer v4 to W1. For the arc (v4, vb), v4 2 W1, vb 2 W2, the flow is zero, which is less than its flow capacity. So vb is transferred to W1. As a

Table 16.3 Alternative solution of flow problem of Example 6 Arcs

Capacity (ci)

Feasible flows I II

III

IV

V

(a, (a, (a, (1, (1, (1, (2, (2, (3, (3, (4, (4, (5, (5, (6, (b,

3 2 1 1 4 2 2 1 1 1 2 0 2 5 2 –

(0) 0 0 0 (0) 0 0 0 0 0 0 0* 0 (0) 0 0

3* (1) 0 0 3 0 (0) 1* (0) 0 (0) 0* 0 (3) 1 4

3* 2* (0) 0 3 0 1 1* 1* (0) 1 0* 0 4 (1) 5

3* 2* 1* 0 3 0 1 1* 1* 1* 1 0* 0 4 2* 6

1) 2) 3) 4) 5) 6) 4) 6) 5) 6) 3) b) 2) b) b) a)

3* (0) 0 0 3 0 0 (0) 0 0 0 0* 0 3 (0) 3

514

16 Flow in Networks

Table 16.4 Alternative solution of flow problem of Example 6 Arcs

Capacity (ci)

Feasible flows I II

III

IV

V

(a, (a, (a, (1, (1, (1, (2, (2, (3, (3, (4, (4, (5, (5, (6, (b,

3 2 1 1 4 2 2 1 1 1 2 0 2 5 2 –

0 (0) 0 0 0 0 0 (0) 0 0 0 0* 0 0 (0) 0

0 2* (0) 0 0 0 1 1* (0) 1* 1 0* 0 (0) 2* 2

(0) 2* 1* 0 (0) 0 1 1* 1* 1* 1 0* 0 (1) 2* 3

3* 2* 1* 0 3 0 1 1* 1* 1* 1 0* 0 4 2* 6

1) 2) 3) 4) 5) 6) 4) 6) 5) 6) 3) b) 2) b) b) a)

1

1

2

0 (1) 0 0 0 0 (0) 1* 0 (0) (0) 0* 0 0 (1) 1

1

4

1

2

1 2

a

1

2 1

2

5

b

1 2

1 3

2

1

6

Fig. 16.20 Maximum flow and capacity

result, the solution is not optimal. In this process, the corresponding chain is (va, v1, v4, vb). The least flow capacity in this chain is 1. So, in each arc of this chain and also in the return arc (vb, va), we increase the flow in such a way that the flow of at

16.8

Formulation of an LPP of a Flow Network

515

Table 16.5 Solution of flow problem of Example 7 Arcs

Capacity

Feasible flows I II

III

IV

V

VI

VII

(a, (a, (a, (1, (1, (1, (2, (2, (2, (3, (3, (3, (4, (5, (6, (b,

2 2 2 1 1 1 1 1 1 1 1 1 2 2 2

(0) 0 0 (0) 0 0 0 0 0 0 0 0 (0) 0 0 0

2* (0) 0 1* 1* 0 0 (0) 0 0 0 0 1 (1) 0 2

2* (1) 0 1* 1* 0 0 1* (0) 0 0 0 1 2* (0) 3

2* 2* (0) 1* 1* 0 0 1* 1* 0 0 (0) 1 2* (1) 4

2* 2* (1) 1* 1* 0 0 1* 1* (0) 0 1* (1) 2* 2* 5

2* 2* 2* 1* 1* 0 0 1* 1* 1* 0 1* 2* 2* 2* 6

1) 2) 3) 4) 5) 6) 4) 5) 6) 4) 5) 6) b) b) b) a)

(1) 0 0 1* (0) 0 0 0 0 0 0 0 1 (0) 0 1

least one arc coincides with its flow capacity, keeping the flows as same in all other arcs. The modiﬁed flow is feasible, as in each arc it is either less than or equal to its flow capacity, and at every vertex the total flow in is equal to the total flow out. The above process is repeated for every modiﬁed flow until it is not possible to transfer vb to W1. The flows of all iterations are shown in Table 16.5. In each feasible flow the number in parentheses indicates the corresponding arc forming the chain in which vb is transferred to W1. The asterisk indicates that the flow in the corresponding arc is equal to its capacity and will not further be increased. From Table 16.5, it is seen that the maximum flow in the graph is 6. Example 8 Find the maximal flow in the graph of Fig. 16.21 with the constraints 2  x1  10;

4  x2  12; 2  x3  4; 0  x4  5; 0  x5  10:

Solution Since the lower bounds of the ﬁrst three variables x1, x2, x3 are non-zero real numbers, we introduce a ﬁctitious vertex v0. Now, for e1, b1 = 2. Therefore, we draw two ﬁctitious arcs (va, v1) and (va, v0) and assume a flow 2 in each of the arcs of the cycle (v0, v1, va, v0). See Fig. 16.22. Similarly, we draw the arcs (v0, v2) and (va, v0) and assume a flow 4 in each of the arcs of the cycle (v0, va, v2, v0). Since b3 = −2, we draw the arcs (v0, v2) and (v1, v0) and assume a flow 2 in each of the arcs of the cycle (v0, v2, v1, v0). In the other arcs e4 and e5, we assume the initial flow as zero.

516

16 Flow in Networks

1

e4

e1 e3

b

a

e5

e2

2

Fig. 16.21 Graph with constraints

1

0

b

a 2

Fig. 16.22 Maximum flow in modiﬁed graph

Then the initial flow is as follows: Arcs

(va, v0)

(v0, v2)

(v0, v1)

(va, v2)

(va, v1)

(v2, v1)

(v1, vb)

(v2, vb)

Flow

6

6

0

-4

-2

2

0

0

Now expressing the constraints in terms of yi (where yi ¼ xi  bi ), we have 0  y1  8; 0  y2  8; 0  y3  6; 0  y4  5; 0  y5  10.

16.8

Formulation of an LPP of a Flow Network

Table 16.6 Solution of flow problem of Example 8

517

Arcs

Capacity (ci)

Feasible flows I II

III

(a, (a, (a, (0, (0, (1, (2, (2, (b,

– 8 8 – – 5 6 10 –

6 (−2) −4 0 6 (0) 2 0 0

6 3 6 0 6 5* 2 10* 15*

0) 1) 2) 1) 2) b) 1) b) a)

6 3 (−4) 0 6 5* 2 (0) 5

Keeping the flow in the ﬁctitious arcs as same, we shall apply the algorithm for ﬁnding the maximum flow in the modiﬁed graph. The iterations for determining the modiﬁed flow are shown in Table 16.6. From Table 16.6, we get the optimal solution for the modiﬁed graph as y1 ¼ 3;

y2 ¼ 6; y3 ¼ 2; y4 ¼ 5; y5 ¼ 10:

Hence, x1 = y1 + 2 = 5, x2 = y2 + 4 = 10, x3 = −2 + y3 = 0, x4 = y4 = 5, x5 = y5 = 10 and the maximum flow = 15, satisfying the constraints of the original graph. Note For this problem, there is also an alternative solution. This solution is y1 = 5, y2 = 4, y3 = 0, y4 = 5, y5 = 10; i.e. x1 = 7, x2 = 8, x3 = −2, x4 = 5, x5 = 10. The negative flow −2 in the arc (v2, v1) can be interpreted as the positive flow 2 in the arc (v1, v2).

16.9

Exercises

1. Formulate the minimal cost and flows in a network. Hence,  ﬁnd the minimal cost of the network shown in Fig. 16.23, where, for cij ; dij ; cij represents the flow capacity from vertex i to vertex j, and dij represents the cost per unit flow from vertex i to vertex j. 2. Formulate the LPP of the network G shown in Fig. 16.24. 3. Find the maximum flow in the graph with the following   arcs and arc capacities, where the flow in each arc is non-negative. Arc vj ; vk is denoted as ðj; kÞ. va is the source and vb, the sink.

518

16 Flow in Networks

1

(15, 4)

(14, 1)

b a

(13, 6)

(11, 2)

(8, 2)

3

(16, 1)

2

(17, 3)

Fig. 16.23 Flow network

4 1

5

6 3

6

a

2

b

2 4

3

5

2

4 5

3

5

Fig. 16.24 Flow network

Arc

(a, 1)

(a, 2)

(a, 3)

(1, 4)

(1, 5)

(1, 6)

(2, 4)

(2, 6)

Capacity Arc Capacity

3 (3, 5) 1

2 (3, 6) 1

1 (4, 3) 2

1 (4, b) 0

4 (5, 2) 2

2 (5, b) 5

2 (6, b) 2

1

16.9

Exercises

519

4. Find the maximum flow in the graph with the following arcs and arc capacities, where the flow in each arc is non-negative. Arc (vj, vk) is denoted as (j, k). va is the source and vb is the sink. Arc

(a, 1)

(a, 2)

(a, 3)

(1, 4)

(1, 5)

(1, 6)

(2, 4)

(2, 5)

(2, 6)

Capacity Arc Capacity

2 (3, 4) 1

2 (3, 5) 1

2 (3, 6) 1

1 (4, b) 2

1 (5, b) 2

1 (6, b) 2

1

1

1

5. Find the maximum non-negative flow in the network described below. Arc (vj, vk) is denoted as (j, k). va is the source and vb is the sink. Arc

(a, 1)

(a, 2)

(1, 2)

(1, 3)

(1, 4)

(2, 4)

(3, 2)

(3, 4)

(4, 3)

Capacity Arc Capacity

8 (3, b) 10

10 (4, b) 9

3

4

2

8

3

4

2

6. Find the maximum flow in the network with the following data, where the flow in arcs is not necessarily non-negative. The arc (vj, vk) is denoted as (j, k), and the flow limit (b, c) means that the constraint on the flow xi is bi  xi  ci . va is the source and vb is the sink. Arc

(a, 1)

(a, 2)

(1, 2)

(1, 3)

(2, 4)

(3, 4)

(3, b)

(4, b)

ðbi ; ci Þ

(0, 10)

(0, 5)

(−2, 3)

(7, 10)

(−3, 5)

(−1, 1)

(0, 8)

(0, 4)

Chapter 17

Inventory Control Theory

17.1

Objectives

The objectives of this chapter are to: • Discuss the concepts of inventory as well as the various forms of inventory and reasons for maintaining inventories • Deﬁne the different terminologies of inventory like demand, replenishment, constraints, shortages, lead time, planning/time horizon and various types of inventory costs • Derive the single item purchasing and manufacturing inventory models with and without shortages • Determine the optimal order quantity for single item price break inventory models and multi-item purchasing model with different constraints like investment, average inventory and space constraints • Introduce the probabilistic models for discrete (the well-known newspaper boy problem) and continuous cases • Discuss the fuzzy inventory model.

17.2

Introduction

In a broad sense, inventory is deﬁned as an idle resource of an enterprise/company/ manufacturing ﬁrm. It can be deﬁned as a stock of physical goods, commodities or other economic resources which are used to meet customer demand or production requirements. This means that the inventory acts as a buffer stock between a supplier and a customer.

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_17

521

522

17

Inventory Control Theory

The inventory or stock of goods may be kept in any one of the following forms: (i) (ii) (iii) (iv)

Raw materials Semi-ﬁnished goods (work-in-process inventory) Finished (or produced) goods Maintenance, repair and operating (MRO) supply items.

In any sector of an economy, the control and maintenance of inventory is a common problem to all organizations. Inventories of physical goods are maintained in government and non-government establishments, e.g. agriculture, industry, military, business, etc. Some reasons for maintaining inventories are as follows: (i) To conduct the smooth and efﬁcient running of business (ii) To provide customers service by meeting their demands from stock without delay (iii) To avail price discounts for bulk purchasing (iv) To maintain more stable operations and/or workforce (v) To take ﬁnancial advantage of transporting/shipping economics (vi) To plan an overall operating strategy through decoupling of successive stages in the chain of acquiring goods, preparing products, shipping to branch warehouses and ﬁnally serving customers (vii) To motivate customers to purchase more by displaying a large number of goods in the showroom/shop (viii) To take advantage in purchasing of some raw materials and some commonly used physical goods (such as paddy, wheat, etc.) whose prices seasonally fluctuate. In this context, it is more proﬁtable to procure a sufﬁcient quantity of these raw materials/commonly used physical goods when their prices are low to be used later during the high price season or when need arises. Production/inventory planning and control is essentially concerned with the design, operation and control of an inventory system in any sector of a given economy. The problem of inventory control is primarily concerned with the following fundamental questions: (i) (ii) (iii) (iv)

Which items should be carried in stock? Which items should be produced? How many of each of these items should be ordered/produced? When should an order be placed? Or when should an item be produced? What type of inventory control system should be used?

In practice, it is a formidable task to determine a suitable inventory policy. Regarding these questions, an inventory problem is a problem of making optimal decisions. In other words, an inventory problem deals with decisions that optimize either the cost function (total or average cost) or the proﬁt function (total or average proﬁt) of the inventory system. However, there are certain types of problems, such as those relating to the storage of water in a dam, in which one has no control over

17.2

Introduction

523

the replenishment of inventory. The supply of inventory of water in a dam depends on rainfall; the organization operating the dam has no control over it. This chapter presents mathematical models of different inventory systems. However, this type of model is based on different assumptions and approximations. It is difﬁcult both to devise and operate with an exact/accurate model as nobody knows what the real situation is. Therefore, it is almost impossible to construct a realistic model with accuracy. For this reason, some approximations and simpliﬁcations must be considered during the model formulation. The solution of the inventory problem is a set of speciﬁc values of variables that minimizes the total (or average) cost of the system or maximizes the total (or average) proﬁt of the system.

17.2.1 Types of Inventories There are different types of inventories, namely: (i) transportation inventories, (ii) fluctuation inventories, (iii) anticipation inventories, (iv) decoupling inventories and (v) lot-size inventories. Transportation inventories These arise due to transportation of inventory items to various distribution centres and customers from the various production centres. When the transportation time is long, the items under transport cannot be served to customers. These inventories exist solely because of transportation time. Fluctuation inventories These have to be carried because sales and production times cannot be predicted accurately. In real-life problems, there are fluctuations in the demand and lead times that affect the production of items. Anticipation inventories These build up in advance by anticipating or foreseeing the future demand for the season of large sales due to a promotion programme or a plant shutdown period. Decoupling inventories These inventories are used to reduce the interdependence of various stages of the production system. Lot-size inventories Generally, the rate of consumption is different from the rate of production or purchasing. Therefore, items are produced in larger quantities which result in lot-size inventories, also called cycle inventories.

524

17.3

17

Inventory Control Theory

Basic Terminologies in Inventory

The inventory system depends on several system factors and parameters such as demand, replenishment rate, shortages, constraints, various types of costs, etc. Demand Demand is deﬁned as the number of units of an item required by the customer in a unit time and has the dimension of a quantity. It may be known exactly or in terms of probabilities, or it may be completely unknown. The demand pattern of items may be either deterministic or probabilistic. Problems in which demand is known and ﬁxed are called deterministic problems. Problems in which the demand is assumed to be a random variable are called stochastic or probabilistic problems. In the case of deterministic demand, it is assumed that the quantities needed over subsequent periods of time are known exactly. Further, the known demand may be ﬁxed or variable with time, stock level or selling price of an item, etc. Probabilistic demand occurs when requirements over a certain period of time are not known with certainty, but their pattern can be described by a known probability distribution. In some cases, demand may also be represented by uncertain data in a non-stochastic sense, i.e. by vague/imprecise data. This type of demand is termed fuzzy demand, and the system is called a fuzzy system. Replenishment Replenishment refers to the amount of quantities that are scheduled to be put into inventories, at the time when decisions are made about ordering these quantities or at the time when they are actually added to stock. Replenishment can be categorized according to size, pattern and lead time. Replenishment size may be constant or variable, depending upon the type of inventory system. It may depend on time, demand and/or on-hand inventory level. The replenishment patterns are usually instantaneous, uniform or in batch. The replenishment quantity again may be probabilistic or fuzzy in nature. Constraints Constraints are the limitations imposed on the inventory system. They may be imposed on the amount of investment, available space, the amount of inventory held, average instantaneous expenditure, number of orders, etc. Fully backlogged/partially backlogged shortages During a stock-out period, sales or goodwill may be affected either by a delay or by complete refusal in meeting the demand. If the unfulﬁlled demand for the goods is satisﬁed completely at a later date, then it is a case of a fully backlogged shortage. That is, it is assumed that no customer balks during this period, and the demand of all the waiting customers is met at the beginning of the next period gradually after the commencement of the next production.

17.3

Basic Terminologies in Inventory

525

Generally, it is observed that during the stock-out period some of the customers wait for the product and others balk away. When this happens, the phenomenon is called a partially backlogged shortage. Lead time The time gap between the time of placing an order or production start and the time of arrival of goods in stock is called the lead time. It may be a constant or a variable. Again, variable lead time may be probabilistic or imprecise. Planning/time horizon The time period over which the inventory level will be controlled is called the time/ planning horizon. It may be ﬁnite or inﬁnite depending upon the nature of the inventory system for the commodity. Deterioration/damageability/perishability Deterioration is deﬁned as decay, evaporation, obsolescence or loss of utility or marginal value of a commodity that results in a decreasing usefulness from the original condition. Vegetables, foodgrains and semiconductor chips are examples of products that can deteriorate. Damageability is deﬁned by the damage when items are broken or lose their utility due to accumulated stress, bad handling, a hostile environment, etc. The amount of damage by stress varies with the size of the stock and the duration for which the stress is applied. Items made of glass, china, clay, ceramic, mud, etc. are examples of damageable products. Perishable items are those which have a ﬁnite lifetime (ﬁxed or random). Fixed lifetime products (e.g. human blood) have a deterministic shelf life, while the random lifetime scenario is closely related to the case of an inventory which experiences continuous physical depletion due to deterioration or decay. Various types of inventory costs Inventory costs are the costs associated with the operation of an inventory system and result from action or lack of action on the part of management in establishing the system. They are basic economic parameters to any inventory decision model. Purchase or unit cost The purchase or unit cost of an item is the unit purchase price of obtaining the item from an external source or the unit production cost for the internal production. It may also depend upon the demand. When production is done in large quantities, it results in reduction of production costs per unit. Also, when quantity discounts are allowed for bulk orders, the unit price is reduced and dependent on the quantity purchased or ordered.

526

17

Inventory Control Theory

Ordering/setup cost The ordering or setup cost originates from the experience of issuing a purchase order to an outside supplier or from internal production setup costs. The ordering cost includes clerical and administrative costs, telephone charges, telegrams, transportation costs, loading and unloading costs, etc. Generally, this cost is assumed to be independent of the quantity ordered or produced. In some cases, it may depend on the quantity of goods purchased because of price breaks quantity discounts or transportation costs. Holding or carrying cost The holding or carrying cost is the cost associated with the storage of the inventory until its use or sale. It is directly proportional to the amount/quantity in the inventory and the time for which the stocks are held. This cost generally includes costs related to insurance, taxes, obsolescence, deterioration, renting of warehouse, light, heat, maintenance and interest on the money locked up. Shortage cost or stock-out cost The shortage cost or stock-out cost is the penalty incurred for being unable to meet a demand when it occurs. This cost arises due to shortage of goods, lost sales for delay in meeting the demand or total inability to meet the demand. In the case where the unfulﬁlled demand for the goods can be satisﬁed at a later date (backlogging case), this cost depends on both the shortage quantity and the delaying time. On the other hand, if the unfulﬁlled demand is lost (no backlogging case), the shortage cost becomes proportional to the shortage quantity only. In both cases, there is a loss of goodwill, which cannot be quantiﬁed for the development of the mathematical model. Disposal cost When an amount of some units of an item remains in excess at the end of the inventory cycle, and if this amount is sold at a lower price in the next cycle to derive some advantages like clearing the stock or winding up the business, the revenue earned through such a process is called the disposal cost. Salvage values During storage, some units are partially spoiled or damaged; i.e. some units lose their utility partially. In a developing country, it is normally observed that some of these units are sold at a reduced price (less than the purchase price) to a section of customers, and this gives some revenue to management. This revenue is called the salvage value.

17.4

Classiﬁcation of Inventory Models

Inventory problems (models) may be classiﬁed into two categories.

17.4

Classiﬁcation of Inventory Models

527

(i) Deterministic inventory models These are the inventory models in which demand is assumed to be known constant or variable (dependent on time, stock level, selling price of the item, etc.). Here, we shall consider deterministic inventory models for known constant demand. Such models are usually referred to as economic lot-size models or economic order quantity (EOQ) models. There are different types of models under this category, namely (a) (b) (c) (d) (e) (f)

Purchasing inventory model with no shortage Manufacturing inventory model with no shortage Purchasing inventory model with shortages Manufacturing model with shortages Multi-item inventory model Price break inventory model.

(ii) Probabilistic inventory models These are the inventory models in which the demand is a random variable with a known probability distribution. Here, the future demand is determined by collecting data from past experience.

17.5

Purchasing Inventory Model with No Shortage (Model 1)

In this model, we want to derive the formula for the optimum order quantity per cycle of a single product so as to minimize the total average cost under the following assumptions and notation: (i) Demand is deterministic and uniform at a rate D units of quantity per unit time. (ii) Production is instantaneous (i.e. production rate is inﬁnite). (iii) Shortages are not allowed. (iv) The lead time is zero. (v) The inventory planning horizon is inﬁnite, and the inventory system involves only one item and one stocking point. (vi) Only a single order is placed at the beginning of each cycle, and the entire lot is delivered in one batch. (vii) The inventory carrying cost C1 per unit quantity per unit time and the ordering cost C3 per order are known and constant. (viii) T is the cycle length and Q is the ordering quantity per cycle. Let us assume that an enterprise purchases an amount of Q units of an item at time t = 0. This amount will be depleted to meet customer demand. Ultimately, the stock level reaches zero at time t = T. The inventory situation is shown in Fig. 17.1.

528

17

Inventory Control Theory

B Q

t=0

Q

t = 2T

A t=T

O

Time

Fig. 17.1 Pictorial representation of purchasing inventory model with no shortage

Clearly;

Q ¼ DT

ð17:1Þ

Now, the inventory carrying cost for the entire cycle T is C1  (area of DAOB) = 12 C1 QT, and the ordering cost for the said cycle T is C3. Hence, the total cost for time T is given by X ¼ C3 þ

1 C1 QT: 2

Therefore, the total average cost is given by CðQÞ ¼ or or

C3 1 þ QC1 2 T C3 D 1 þ C1 Q CðQÞ ¼ Q 2 CðQÞ ¼



X T

 Q * Q ¼ DT ) T ¼ : D

ð17:2Þ

The optimum value of Q which minimizes C(Q) is obtained by equating the ﬁrst derivative of C(Q) with respect to Q to zero, dC 1 C3 D ¼ 0 or, C1  2 ¼ 0 dQ 2 Q rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D : or Q ¼ C1 i:e:;

2 Now, ddCðQÞ ¼ 2CQ33D, which is positive for Q ¼ Q2

qﬃﬃﬃﬃﬃﬃﬃﬃ

2C3 D C1 .

17.5

Purchasing Inventory Model with No Shortage (Model 1)

529

Hence, C(Q) is minimum for which the optimum value of Q is rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D : Q ¼ C1 

ð17:3Þ

This is known as either the economic lot-size formula or the EOQ formula. The qﬃﬃﬃﬃﬃﬃﬃ  corresponding optimum time interval is T  ¼ QD ¼ C2C1 D3 , and the minimum cost pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ per unit time is given by Cmin ¼ CQ3D þ 12 C1 Q ¼ 2C1 C3 D. This model was ﬁrst developed by Ford Harris (1915) of the Westinghouse Corporation, USA, in the year 1915. He derived the well-known classical lot-size Formula (17.3). However, according to Erlenkotter (1989), the earliest model was formulated by Harris in the year 1913. This formula was also developed independently by R. H. Wilson after a few years; thus, it has been called the Harris-Wilson formula. Remark (i) The total inventory time units for the entire cycle T is 12 QT, so the average inventory at any time is 12 QT=T ¼ 12 Q. (ii) Since C1 [ 0, from f ðQÞ ¼ 12 C1 Q it is obvious that the inventory carrying cost is a linear function of Q with a positive slope; i.e. for a smaller average inventory, the inventory carrying costs are lower. In contrast, gðQÞ ¼ CQ3 D; i.e. the ordering cost increases as Q decreases. (iii) In the preceding model, if we always maintain an inventory B on hand as buffer stock, then the average inventory at any time is 12 Q þ B. Therefore, the   total cost per unit time is CðQÞ ¼ 12 Q þ B C1 þ C3 D Q. As before, we obtain the optimal values of Q and T as follows: rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D 2C3   Q¼Q ¼ and T ¼ T ¼ : C1 DC1 (iv) In the preceding model, if the ordering cost is taken as C3 þ bQ (where b is the purchase cost per unit quantity) instead of a ﬁxed ordering cost, then there is no change in the optimum order quantity. Proof In this case, the average cost is given by 1 D CðQÞ ¼ C1 Q þ ðC3 þ bQÞ: 2 Q As the necessary condition for the optimum of C(Q) in (17.4), we have

ð17:4Þ

530

17

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D C ðQÞ ¼ 0 implies Q ¼ C1 0

Hence, Q ¼

Inventory Control Theory

and C00 ðQÞ [ 0:

qﬃﬃﬃﬃﬃﬃﬃﬃ

2C3 D C1 .

This shows that there is no change in Q* in spite of the change in the ordering cost. Example 1 An engineering factory consumes 5000 units of a component per year. The ordering, receiving and handling costs are Rs. 300 per order, the trucking cost is Rs. 1200 per order, the interest cost is Rs. 0.06 per unit per year, the deterioration and obsolescence costs are Rs. 0.004 per unit per year and the storage cost is Rs. 1000 per year for 5000 units. Calculate the economic order quantity and minimum average cost. Solution In the given problem, we have demand (D) = 5000 units. Ordering cost=replenishment cost ¼ ordering; receiving; handling costs and trucking costs ¼ Rs: ð300 þ 1200Þ ¼ Rs:1500 per order: Inventory carrying cost ¼ interest costs þ deterioration and obsolescence costs þ storage costs   1000 ¼ 0:06 þ 0:004 þ rupees per unit per year 5000 ¼ Rs: 0:264 per unit per year:

Hence, the economic order quantity is given by rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D 2  1500  5000 Q ¼ ¼ ¼ 753:8 ðapprox:Þ: C1 0:264 

Also, the minimum average cost is pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C1 C3 D ¼ Rs: 2  0:264  1500  5000 ¼ Rs: 1989:97 ðapprox:Þ:

17.6

Manufacturing Model with No Shortage (Model 2)

(Economic lot-size model with ﬁnite rate of replenishment and without shortage) In this model, we shall derive the formula for the optimum production quantity per cycle of a single product so as to minimize the total average cost under the following assumptions and notation:

17.6

Manufacturing Model with No Shortage (Model 2)

531

(i) (ii) (iii) (iv)

Demand is deterministic and uniform at a rate D units of quantity per unit time. Shortages are not allowed. The lead time is zero. The production rate or replenishment rate is ﬁnite, say, K units per unit time (K > D). (v) The production-inventory planning horizon is inﬁnite, and the production system involves only one item and one stocking point. (vi) The inventory carrying cost C1 per unit quantity per unit time and the setup cost C3 per production cycle are known and constant. (vii) T is the cycle length and Q is the economic lot size. In this model, each production cycle time T consists of two parts t1 and t2 , where: (i) t1 is the period during which the stock is increasing at a constant rate K – D units per unit time. (ii) t2 is the period during which there is no replenishment (or production), but inventory is decreasing at the rate of D units per unit time. Further, it is assumed that S is the stock available at the end of time t1 which is expected to be consumed during the remaining period t2 at the consumption rate D. The pictorial representation of the model is shown in Fig. 17.2. Therefore, ðK  DÞt1 ¼ S S : KD Since the total quantity produced during the production period t1 is Q, or

t1 ¼

ð17:5Þ

) Q ¼ Kt1 or Q ¼ K

S KD ; which implies S ¼ Q: KD K

ð17:6Þ

Again, Q = DT, i.e. T ¼ Q D.

A K–D

D S

O

t1

M

t2

B

Time

Fig. 17.2 Pictorial representation of manufacturing inventory model with no shortage

532

17

Inventory Control Theory

Now the inventory carrying cost for the entire cycle T is ðDOABÞC1 ¼ 12 TSC1 , and the setup cost for time period T is C3. Therefore, the total cost for the entire cycle T is given by X ¼ C3 þ 12 C1 ST. Therefore, the total average cost is given by CðQÞ ¼ XT C3 1 þ C1 S 2 T C3 D 1 K  D þ C1 Q or CðQÞ ¼ Q 2 K

or CðQÞ ¼



 KD * Q ¼ DT and S ¼ Q : K

ð17:7Þ

The optimum of Q which minimizes C(Q) is obtained by equating the ﬁrst derivative of C(Q) with respect to Q to zero, dC ¼0 dQ C3 D 1 K  D ¼0 or  2 þ C1 Q 2 K rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 DK : or Q ¼ : C1 K  D i:e:

2

Again; ddQC2 ¼ 2CQ33D ¼ a positive quantity for Q ¼

ð17:8Þ

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 DK C1 : KD:

Hence, C(Q) is minimum for which the optimum value of Q is rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 DK : Q ¼ : C1 K  D 

ð17:9Þ

The corresponding time interval is Q T ¼ ¼ D

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 K ; C1 DðK  DÞ

ð17:10Þ

and the minimum average cost is given by Cmin

1K  D C3 D C1 Q þ  ¼ ¼ 2 K Q

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ KD 2C1 C3 D : K

ð17:11Þ

Remark (i) For this model, Q*, T* and Cmin can be written in the following form:

17.6

Manufacturing Model with No Shortage (Model 2)

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D 1 Q ¼ ; C1 1  D=K

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 1 T ¼ : DC1 1  D=K

533

ð17:12Þ

If K ! ∞, i.e. the production rate is inﬁnite, this model reduces to Model 1. Therefore, when K ! ∞, then Q*, T* and Cmin reduce to the expressions for Q*, T* and Cmin of Model 1.

17.7

Purchasing Inventory Model with Shortage (Model 3)

In this model, we shall derive the optimal order level and the minimum average cost under the following assumptions and notation: (i) Demand is deterministic and uniform at a rate D units of quantity per unit time. (ii) Production is instantaneous (i.e. production rate is inﬁnite). (iii) Shortages are allowed and fully backlogged. (iv) The lead time is zero. (v) The inventory planning horizon is inﬁnite, and the inventory system involves only one item and one stocking point. (vi) Only a single order will be placed at the beginning of each cycle, and the entire lot is delivered in one batch. (vii) The inventory carrying cost C1 per unit quantity per unit time, the shortage cost C2 per unit quantity per unit time and the ordering cost C3 per order are known and constant. (viii) Q is the lot size per cycle, S1 is the initial inventory level after fulﬁlling the backlogged quantity of the previous cycle and Q − S1 is the maximum shortage level. (ix) T be the cycle length or scheduling period whereas t1 is the no shortage period. The inventory situation is shown in Fig. 17.3. According to the assumptions of (viii) and (ix), we have Q = DT. Regarding the cycle length or scheduling period of the inventory system, two cases may arise: Case 1: The cycle length or scheduling period T is constant. Case 2: The cycle length or scheduling period T is a variable. For Case 1, T is constant; i.e. the inventory is to be replenished after every time period T. As t1 is the no shortage period, S1 ¼ Dt1 or t1 ¼ S1 =D.

534

17

Inventory Control Theory

Now, the inventory carrying cost during the period 0 to t1 is 1 1 C1 ðarea of DOABÞ ¼ C1 S1 t1 ¼ C1 S21 =D: 2 2 Again the shortage cost during the interval (t1 , T) is 1 C2 ðarea of DACDÞ ¼ C2 ðQ  S1 ÞðT  t1 Þ 2   1 Q  S1 ¼ C2 ðQ  S1 Þ2 =D * T  t1 ¼ : 2 D Hence the total average cost of the system is given by (see Fig. 17.3) " # 1 S21 1 ðQ  S1 Þ2 C1 þ C2 =T: C¼ 2 D 2 D

ð17:13Þ

Since the setup cost C3 and time period T are constant, the average setup cost C3 =T, also being constant, will not be considered in the cost expression. Since T is constant, Q = DT is also constant. Hence, the preceding expression for the average cost is a function of a single variable S1 . Thus, we can easily minimize expression (13) with respect to S1 , as in Model 1. In this case; S1 ¼

C2 Q C2 DT C1 C2 Q C1 C2 DT ¼ and Cmin ¼ ¼ : C1 þ C2 C1 þ C2 C1 þ C2 C1 þ C2 ð17:14Þ

B S1 A t1 T

D Q - S1 C

Fig. 17.3 Pictorial representation of purchasing inventory model with shortage

Time

17.7

Purchasing Inventory Model with Shortage (Model 3)

535

For Case 2, the cycle length or scheduling period T is a variable. As in Case 1, the average cost of the inventory system will be " # 1 S21 1 ðQ  S1 Þ2 C ¼ C3 þ C1 þ C2 =T; ð17:15Þ 2 D 2 D where Q = DT. Here, the average cost C is a function of two independent variables T and S1. Now, for the optimal value of C, we have @C ¼0 @S1 Thus,

and

@C ¼ 0: @T

@C DT : ¼ 0 gives S1 ¼ C2 @S1 ðC1 þ C2 Þ

ð17:16Þ

Again, @C C1 S21 DT  S1 C2 ðDT  S1 Þ2 C3 ¼ 0 gives   þ C2  2 ¼ 0: 2 @T 2D T T 2D T2 T Setting S1 ¼ C2 ðC1DT þ C2 Þ in this expression and simplifying, we have sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ  : T ¼T ¼ C1 C2 D Then, S1 ¼

S1

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C2 C3 D ¼ : C1 ðC1 þ C2 Þ

ð17:17Þ

ð17:18aÞ

ð17:18bÞ

Obviously, for the values of T and S1 given by (17.18a) and (17.18b), @2C @2S [ 0; [0 2 @T 2 @S1

 2 2 @2C @2C @ C  and [ 0: @S1 @T @S1 dT

Hence, C is minimum for the values of T and S1 given by (17.18a) and (17.18b).

536

17

Inventory Control Theory

Therefore, the optimum order quantity for minimum cost is given by sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ 2C3 ðC1 þ C2 ÞD ¼ Q ¼ DT  ¼ D C1 C2 D C1 C2 and Cmin

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C1 C2 C3 D ¼C ¼ : ðC1 þ C2 Þ 

ð17:19Þ

ð17:20Þ

Remark (i) If C1 ! ∞ and C2 > 0, inventories are prohibited. In this case S1 ¼ 0 and qﬃﬃﬃﬃﬃﬃﬃﬃ each lot size Q ¼ 2CC32D is used to ﬁll the backorders. (ii) If C2 ! ∞ and C1 > 0, then shortages are prohibited. In this case, S1 ¼ qﬃﬃﬃﬃﬃﬃﬃﬃ Q ¼ 2CC31D and each batch Q* is used entirely for inventory. (iii) If shortage costs are negligible, then C1 > 0 and C2 ! 0. In this case, S1 ! 0 and Q* ! ∞. (iv) If the inventory carrying costs are negligible, then C1 ! 0 and C2 > 0. In this case, Q* ! ∞ and S1 ! ∞; i.e. S1 ! Q*. Thus, due to very small inventory carrying costs, a large lot size should be ordered and used to meet the future demand. (v) When the inventory carrying costs and shortage costs are equal, i.e. when C1 = C2, C1 Cþ1 C2 ¼ 12. pﬃﬃﬃqﬃﬃﬃﬃﬃﬃﬃﬃ In this case, Q ¼ 2 2CC31D, which shows that the lot size is √2 times the lot size of Model 1. Example 2 The demand for an item is 18,000 units per year. The inventory carrying cost is Rs. 1.20 per unit time, and the cost of shortages is Rs. 5.00. The ordering cost is Rs. 400.00. Assuming that the replenishment rate is instantaneous, determine the optimum order quantity, shortage quantity and cycle length. Solution For the problem, it is given that demand (D) = 1800 units per year, the carrying cost (C1) = Rs. 1.20 per unit, the shortage cost (C2) = Rs. 5.00 and the ordering cost (C3) = Rs. 400 per order. The optimum order quantity Q* is given by sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 ÞD 2  400  ð1:2 þ 5Þ  18000 ¼ 3857 units: ¼ Q ¼ C1 C2 1:2  5

17.7

Purchasing Inventory Model with Shortage (Model 3)

537

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C2 C3 D The optimum shortage quantity Q  S1 ¼ 3857  C1 ðC1 þ C2 Þ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  5  400  18; 000 ¼ 3857  ¼ 746 units ðapprox:Þ 1:2  ð1:2 þ 5Þ 

3857 ¼ 0:214 year (approx.) The optimal cycle length T  ¼ QD ¼ 18000

Example 3 The demand for an item is deterministic and constant over time and is equal to 600 units per year. The unit cost of the item is Rs. 50.00, while the cost of placing an order is Rs. 100.00. The inventory carrying cost is 20% of the unit cost of the item, and the shortage cost per month is Re. 1. Find the optimal ordering quantity. If shortages are not allowed, what would be the loss of the company? Solution It is given that D = 600 units/year. C1 = 20% of Rs. 50.00 = Rs. 10.00 C2 = Re 1.00 per month, i.e. Rs. 12.00 per year C3 = Rs. 100.00 per order. When shortages are allowed, the optimal ordering quantity Q* is given by sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 ÞD ¼ 148 units Q ¼ C1 C2 

and the minimum cost per year is CðQ Þ ¼

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C1 C2 C3 D=ðC1 þ C2 Þ ¼ Rs: 809:04.

If shortages are not allowed, then the optimal order quantity is Q ¼

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D ¼ 109:5 units C1

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ and the relevant average cost is given by CðQ Þ ¼ Rs: 2C1 C3 D ¼ Rs: 1095:44. Therefore, if shortages are not allowed, the loss of the company will be Rs. (1095.44–809.04), i.e. Rs. 286.40.

538

17.8

17

Inventory Control Theory

Manufacturing Model with Shortage

(Economic lot-size model with ﬁnite rate of replenishment and shortages) In this model, we shall derive the formula for the optimum production quantity, shortage quantity and cycle length of a single product by minimizing the average cost of the production system under the following assumptions and notation: (i) The production rate or replenishment rate is ﬁnite, say K units per unit time (K > D). (ii) The production-inventory planning horizon is inﬁnite, and the production system involves only one item and one stocking point. (iii) Demand of the item is deterministic and uniform at a rate D units of quantity per unit time. (iv) Shortages are allowed. (v) The lead time is zero. (vi) The inventory carrying cost C1 per unit quantity per unit time, the shortage cost C2 per unit quantity per unit time and the setup cost C3 per setup are known and constant. (vii) T is the cycle length of the system; i.e. T is the interval between production cycles. (viii) Q is the economic lot size. Let us assume that each production cycle of length T consists of two parts t12 and t34 which are further subdivided into t1 and t2 , t3 and t4 , where (i) inventory is building up at a constant rate K − D units per unit time during the interval ½0; t1 ; (ii) at time t ¼ t1 ; the production is stopped and the stock level decreases due to meeting customer demand only up to the time t12 ¼ t1 þ t2 ; (iii) shortages are accumulated at a constant rate of D units per unit time during the time t3 , i.e. during the interval ½t12 ; t12 þ t3 : (iv) Shortages are being ﬁlled immediately at a constant rate K – D units per unit time during the time t4 , i.e., during the interval ½t12 þ t3 ; T: (v) The production cycle then repeats itself after the time T ¼ t1 þ t2 þ t3 þ t4 : Again, let the inventory level be S1 at t ¼ t1 and at the end of time t ¼ t1 þ t2 , the stock level reaches zero. Now shortages start, and we assume that shortages are built up of quantity S2 at time t ¼ t1 þ t2 þ t3 , and then these shortages are ﬁlled up to the time t = t1 + t2 + t3 + t4. The pictorial representation of the inventory situation is given in Fig. 17.4. Now our objectives are to ﬁnd the optimal value of Q; S1 ; S2 ; t1 ; t2 ; t3 ; t4 and T with the minimum average total cost. The inventory carrying cost over the time period T is given by 1 1 Ch ¼ C1  DOAC ¼ C1  OC  AB ¼ C1 ðt1 þ t2 ÞS1 ; 2 2

17.8

Manufacturing Model with Shortage

539

A

K–D D C O

t1

t2

B

F

H t3

t4

Time

K–D

D E

Fig. 17.4 Pictorial representation of manufacturing inventory model with shortage

and the shortage cost over time T is given by 1 1 Cs ¼ C2  DCEF ¼ C2  CF  EH ¼ C2 ðt3 þ t4 ÞS2 : 2 2 Hence, the total average cost of the production system is given by C ¼ ½C3 þ Ch þ Cs =T:

ð17:21Þ

From Fig. 17.4, it is clear that S1 ¼ ðK  DÞt1

or; t1 ¼

S1 : K D

ð17:22Þ

Again, S1 ¼ Dt2

or; t2 ¼

S1 : D

or, t3 ¼

S2 D

Now, in a stock-out situation, S2 ¼ Dt3

ð17:23Þ

540

17

Inventory Control Theory

S2 : KD

ð17:24Þ

and S2 ¼ ðK  DÞt4

or, t4 ¼

Since the total quantity produced over the time period T is Q, Q ¼ DT; where D is the demand rate or Dðt1 þ t2 þ t3 þ t4 Þ ¼ Q   S1 S1 S2 S2 þ þ þ or D ¼ Q: KD D D KD

ð17:25Þ

After simpliﬁcation, we have S1 þ S 2 ¼

KD Q: K

ð17:26Þ

Again t1 þ t2 ¼

K S1 DðK  DÞ

and t3 þ t4 ¼

K S2 : DðK  DÞ

ð17:27Þ

Now substituting the values of t1 + t2, t3 + t4 and T = Q/D in (17.21), we have CðQ; S1 ; S2 Þ ¼

 DC3 1 K  C1 S1 þ C2 S22 þ : 2Q K  D Q

ð17:28Þ

Using (17.26), this equation reduces to "  # 2 1 K KD DC3 CðQ; S2 Þ ¼ C1 Q  S2 þ C2 S22 þ : 2Q K  D K Q

ð17:29Þ

Now, for the extreme values of C(Q, S2), we have @C @C ¼ 0; ¼ 0: @Q @S2 @C KD Q ¼ 0 implies S2 ¼ C1 : @Q K C1 þ C2

ð17:30Þ

17.8

Manufacturing Model with Shortage

Again, @C ¼ 0 gives Q ¼ @S2

541

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ KD  : C1 C2 KD

ð17:31Þ

For the values of Q and S2 given in (17.31) and (17.30), it can easily be veriﬁed that  2 2 @2C @2C @2C @2C @ C [ 0; [ 0 and  [ 0: 2 2 2 2 @Q @Q @S2 @Q@S2 @S2 Hence, C(Q, S2) is minimum and the optimal values of Q and S2 are given by sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ KD Q ¼ C1 C2 KD 

ð17:32Þ

and S2

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C1 C3 DðK  DÞ ¼ K C2 ðC1 þ C2 Þ

Q T ¼ ¼ D

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃsﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ K C1 C2 DðK  DÞ

KD  S1 ¼ Q  S2 ¼ K

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C2 C3 DðK  DÞ : K C1 ðC1 þ C2 Þ

Now Cmin ¼ CðQ



; S2 Þ

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C1 C2 C3 DðK  DÞ : ¼ K C1 þ C2

ð17:33Þ

ð17:34Þ

ð17:35Þ

ð17:36Þ

Remarks (i) In this model, if we assume that the production rate is inﬁnite, i.e. K ! ∞, then the optimal quantities obtained by taking K ! ∞ in (17.32), (17.34) and (17.36) are sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ Q ¼ D; C1 C2 sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ 2C1 C2 C3 D  and Cmin ¼ T ¼ : C1 C2 D C1 þ C2

542

17

Inventory Control Theory

This means that Model 4 reduces to Model 3 if K ! ∞. (ii) If shortages are not allowed in Model 4, then it reduces to Model 3. In this case, taking C2 ! ∞ in (17.32), (17.34) and (17.36), we obtain the required expressions of Model 3, which are as follows: sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 KD Q ¼ ; C1 ðK  DÞ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 K 2C1 C3 DðK  DÞ  : T ¼ and Cmin ¼ C1 DðK  DÞ K 

Example 4 The demand for an item in a company is 18,000 units per year. The company can produce the item at a rate of 3000 per month. The cost of one setup is Rs. 500, and the holding cost of one unit per month is Rs. 0.15. The shortage cost of one unit is Rs. 20 per month. Determine the optimum manufacturing quantity and the shortage quantity. Also determine the manufacturing time and the time between setups. Solution For this problem, it is given that C1 = Rs. 0.15 per month, C2 = Rs. 20 per month, C3 = Rs. 500.00 per setup, K = 3000 per month and D = 18,000 units per year, i.e. 1500 units per month. The optimum manufacturing quantity Q* is given by sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃrﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 ðC1 þ C2 Þ KD 2  500  ð0:15 þ 20Þ 3000  1500 ¼ Q ¼ C1 C2 KD 0:15  20 3000  1500 ¼ 4489 units ðapprox:Þ The optimum shortage quantity is given by S2 ¼ C1

Q D

KD Q ¼ 17 units ðapprox:Þ: K ðC1 þ C2 Þ

The manufacturing time = ¼ 4489 1500 ¼ 3 months.

Q K

¼ 4489 3000 ¼ 1:5 months, and the time between setups

17.9

17.9

Multi-item Inventory Model

543

Multi-item Inventory Model

In earlier, we have discussed inventory models for single items, or each item separately, but if there exists a relationship among the items under some limitations, then it is not possible to consider them separately. Thus, after constructing the average cost expression in these models, we shall use the method of Lagrange multipliers to minimize the average cost. In all such problems, ﬁrst of all we shall solve the problem ignoring the limitations and then consider the effect of limitations. Now, we shall develop a multi-item inventory model under the following assumptions and notations: (i) There are n items with instantaneous production; i.e. the production rate of each item is inﬁnite. (ii) Shortages are not allowed. For the ith (i = 1, 2, …, n) item: (iii) Di is the uniform demand rate. (iv) The inventory carrying cost Cli per unit quantity per unit time and the ordering cost C3i per order are known and constant. (v) Ti is the cycle length. Let Qi be the ordering quantity of the ith item. Then;

Qi ¼ Di Ti or; Ti ¼ Qi =Di :

Now, the total inventory time units for the ith item is 12 Qi Ti . Hence, the inventory carrying cost for the ith item over the inventory cycle is 1 C 2 1i Qi Ti . Therefore, the average cost for the ith item is 

 1 Ci ¼ C3i þ C1i Qi Ti =Ti 2 1 or Ci ¼ C3i Di =Qi þ C1i Qi : 2 Hence, the total average cost for n items is given by C¼

n X

Ci

i¼1

 n  X 1 or C ¼ C3i Di =Qi þ C1i Qi : 2 i¼1 Here C is a function of Q1 ; Q2 ; . . .; Qn .

544

17

Inventory Control Theory

For optimum values of Qi (i = 1, 2, …, n), we must have @C=@Qi ¼ 0; 1 i:e: C1i  C3i Di =Q2i ¼ 0 2 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di : or Qi ¼ C1i Hence, the optimum value of Qi is Qi

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di ¼ C1i

ð17:37Þ

17.9.1 Limitation on Inventories If there is a limitation on inventories that requires that the average number of all units in inventory should not exceed k units of all types, then the problem is to minimize the cost C subject to the condition that: 1 2

Pn

i¼1

Qi  k [Since the average number of inventory at any time for an item is

Qi .]

1 2

or

n X

Qi  2k  0:

i¼1

Here two cases may arise: Case I When

1 2

n P i¼1

Qi  k

In this case, the optimal values Q*i (i = 1, 2,…,n) given by Qi

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di ¼ C1i

satisfy the constraint directly. Case II When

1 2

n P i¼1

Qi [ k

In this case, we have to solve the following problem:

ð17:38Þ

17.9

Multi-item Inventory Model

Minimize

545

n  X 1

2

i¼1

C1i Qi þ C3i Di =Qi



subject to the constraint (17.38). To solve this problem, we shall use the Lagrange multiplier method. The corresponding Lagrangian function is L¼

n  X 1 i¼1

2

"



C1i Qi þ C3i Di =Qi þ k

n X

# Qi  2k ;

i¼1

where k ð [ 0Þ is the Lagrange multiplier. The necessary conditions for L to be minimum are @L @L ¼ 0: ¼ 0; i ¼ 1; 2; . . .; n and @Qi @k Now, from

@L @Qi

¼ 0 we have 1 C3i Di C1i  2 þ k ¼ 0; 2 Qi

i ¼ 1; 2; . . .; n



or

Again, from

@L @k

1 2C3i Di 2 : Qi ¼ C1i þ 2k

¼ 0 we have n X

Qi  2k ¼ 0 or

i¼1

n X

Qi ¼ 2k:

i¼1

Hence, the optimum value of Qi is Qi ¼

and



2C3i Di C1i þ 2k

n X

12

Qi ¼ 2k:

ð17:39Þ ð17:40Þ

i¼1

To obtain the values of Qi from (17.39), we ﬁnd the optimal value of k* of k by the successive trial and error method or the linear interpolation method, subject to the condition given by (17.40). Equation (17.40) implies that Qi must satisfy the inventory constraint in an equality sense.

546

17

Inventory Control Theory

17.9.2 Limitation of Floor Space (or Warehouse Capacity) Here we shall discuss the multi-item inventory model with the limitation of warehouse floor space. Let A be the maximum storage area available for n different items, ai be the storage area required per unit of the ith item and Qi be the amount ordered for the ith item. Thus, the storage requirement constraint becomes n X

ai Qi  A;

Qi [ 0

i¼1

or

n X

ai Qi  A  0:

ð17:41Þ

i¼1

Now two possibilities may arise: Case 1 When

n P i¼1

ai Qi  A

In this case, the optimal values Qi (i = 1, 2, …, n) given by Qi ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di C1i

satisfy

the constraint (17.41) directly. Hence, these optimal values Qi are the required values. Case 2 When

n P i¼1

ai Qi [ A

In this case, we have to solve the problem as follows: n

P 1 Minimize C ¼ 2 C1i Qi þ C3i Di =Qi i¼1

subject to the constraint (17.41). To solve it, we shall use the Lagrange multiplier method, and the corresponding Lagrangian function is L¼

n  X 1 i¼1

2



C1i Qi þ C3i Di =Qi þ k

"

n X

# ai Qi  A ;

i¼1

where k (>0) is the Lagrange multiplier. The necessary conditions for L to be minimum are @L ¼ 0; @Qi @L ¼ 0: and @k

i ¼ 1; 2; . . .; n

17.9

Multi-item Inventory Model

Now from

@L @Qi

547

¼ 0; we have 1 C3i Di C1i  2 þ k ai ¼ 0; 2 Qi

Again, from

@L @Qi

i ¼ 1; 2; . . .; n:

ð17:42aÞ

¼ 0; we have n X

ai Qi  A ¼ 0

ð17:42bÞ

i¼1

Solving (17.42a) and (17.42b), we have the optimal values of Qi as Qi



12 2C3i Di ¼ ; C1i þ 2 k ai

i ¼ 1; 2; . . .; n

ð17:43Þ

and n X

ai Qi ¼ A:

ð17:44Þ

i¼1

To obtain the values of Qi from (17.43), we ﬁnd the optimal value k* from k by the successive trial and error method or the linear interpolation method subject to the condition given by (17.44). Equation (17.44) implies that Qi must satisfy the inventory constraint in the equality sense.

17.9.3 Limitation on Investment In this case, there is an upper limit M on the amount to be invested on inventory. Let C4i be the unit price of the ith item. Then n X

C4i Qi  M:

ð17:45Þ

i¼1

Now two possibilities may arise: Case I When

n P i¼1

C4i Qi  M

In this case, the constraint is satisﬁed by Q*i automatically. Hence, the optimal values of Q*i (i = 1, 2, …, n) are given by Qi ¼

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di : C1i

548

17

Case II When

n P i¼1

Inventory Control Theory

C4i Qi [ M

In this case, our problem is as follows: C¼

Minimize

n  X 1

2

i¼1

C1i Qi þ C3i Di =Qi



subject to the constraint (17.45). To solve it, we shall use the Lagrange multiplier method, and the corresponding Lagrangian function is L¼

n  X 1 i¼1

2



"

C1i Qi þ C3i Di =Qi þ k

n X

# C4i Qi  M ;

i¼1

where k (>0) is the Lagrange multiplier. The necessary conditions for L to be minimum are @L ¼ 0; i ¼ 1; 2; . . .; n @Qi @L ¼ 0: and @k Now from

@L @Qi

¼ 0, we have Qi ¼

Again, from

@L @k



12 2C3i Di ; i ¼ 1; 2; . . .; n: C1i þ 2kC4i

¼ 0, we have n X

C4i Qi  M ¼ 0 or

i¼1

n X

C4i Qi ¼ M:

i¼1

Hence, the optimum value of Qi is given by Qi

and



2C3i Di ¼ C1i þ 2k C4i n X i¼1

12

C4i Qi ¼ M:

ð17:46Þ ð17:47Þ

17.9

Multi-item Inventory Model

549

Thus, the values of Qi are obtained from (17.46) subject to the condition given by (17.47) where the optimal value k* of k can be found by either the successive trial and error method or the linear interpolation method. Example 5 A workshop produces three machine parts A, B, C, and the total storage space available is 640 m2. Obtain the optimal lot size for each item from the following data:

Cost per unit (Rs.) Storage space required (m2/unit) Ordering cost (Rs.) (C3) No. of units required/year

Items A

B

C

10 0.60 100 5000

15 0.80 200 2000

5 0.45 75 10,000

The carrying charge on each item is 20% of unit cost. Solution Considering one year as one unit of time, we have: carrying charge of A (C11) = Rs. (20% of 10) = Rs. 2.00 carrying charge of B (C12) = Rs. (20% of 15) = Rs. 3.00 carrying charge of C (C13) = Rs. (20% of 5) = Re. 1.00. Now, without considering the effect of restriction on storage space availability, the qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ optimal value Qi of the ith item is given by Qi ¼ 2CC3i1iDi ; i ¼ 1; 2; 3. rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2Di C3i 2  5000  100 ) ¼ ¼ ¼ 707 2 C11 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  2:000  200 ¼ 516 Q2 ¼ 3 p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Q3 ¼ 2  10; 000  75 ¼ 1225: Q1

Then the total storage space required for these values of Q*i (i = 1, 2, 3) is 3 X

ai Qi ¼ a1 Q1 þ a2 Q2 þ a3 Q3

i¼1

¼ ð0:60707 þ 0:80516 þ 0:451225Þ m2 ¼ 1388:25 m2 :

550

17

Inventory Control Theory

This storage space is greater then the available storage space of 640 m2. Therefore, we shall try to ﬁnd a suitable value of k* by the trial and error method for computing Q*i by using Qi ¼

and

3 P i¼1



2C3i Di C1i þ 2k ai

12

ai Qi ¼ 640.

If we take k* = 5, rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i Di 2  5000  100 ¼ 354 ¼ ¼ 2 þ 2  5  0:60 C1i þ 2  5  a1 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  2000  200 Q2 ¼ ¼ 270 3 þ 2  5  0:80 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  10000  75 ¼ 522: Q3 ¼ 1 þ 2  5  0:45 Q1

and

Hence, the corresponding storage space is 0.60354 + 0.80270 + 0.45552 = 663.30 m2. This storage space is greater than the available storage space of 640 m2. Again, if we take k* = 6, then rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  5000  100 ¼ 330 ¼ 2 þ 2  6  0:60 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  2000  200 Q2 ¼ ¼ 252 3 þ 2  6  0:80 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  10000  75 ¼ 484: Q3 ¼ 1 þ 2  6  0:45 Q1

and

Hence, the corresponding storage space is 0.60330 + 0.80352 + 0.45484 = 617.40 m2, which is less then the available storage space of 640 m2. Hence, it is clear that the most suitable value of k* lies between 5 and 6. Let us assume that the required storage space will be 640 m2 for k* = x. Now considering the linear relationship between the value of k* and the required storage space, we have x6 640617:4

or

56 ¼ 663:3617:4

x  6 ¼ 22:6 45:9 ) k ¼ 5:5:

or x ¼ 5:5 ðapprox:Þ

17.9

Multi-item Inventory Model

551

For this value of k*, rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  5000  100 ¼ 341 ¼ 2 þ 2  5:5  0:60 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  2000  200 Q2 ¼ ¼ 272 2 þ 2  5:5  0:80 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  10000  75 ¼ 502: Q3 ¼ 1 þ 2  5:5  0:45 Q1

And

Hence, the optimal lot sizes of the three machine parts A, B, C are Q1 ¼ 341 units, Q2 ¼ 272 units and Q3 ¼ 502 units. Example 6 A company producing three items has a limited inventory of averagely 750 items of all types. Determine the optimal production quantities for each item separately, when the following information is given: Product

1

2

3

Holding cost (Rs.) Ordering cost (Rs.) Demand

0.05 50 100

0.02 40 120

0.04 60 75

Solution Neglecting the restriction of the total value of the inventory level, we get the qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ optimal values Q*i for the ith item, which is given by Qi ¼ 2CC3i1iDi ; i ¼ 1; 2; 3. Hence;

And

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  50  100 ¼ ¼ 447 0:05 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  40  120 Q2 ¼ ¼ 693 0:02 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  60  75 ¼ 474: Q3 ¼ 0:04 Q1

Therefore, the total average inventory is (447 + 693 + 474)/2 units = 807 units. But the average inventory is 750 units. Therefore, we have to determine the value of parameter k* by the trial and error method for computing Qi by using Qi

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3i D 1X  Qi ¼ 750: ¼  and C1i þ 2 k 2

552

17

Inventory Control Theory

Now, for k* = 0.005, Q1

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  50  100 ¼ 408; Q2 ¼ 566 and Q3 ¼ 424: ¼ 0:05 þ 2  0:005

Therefore, the total average inventory is (408 + 566 + 424)/2 = 699 units, which is less than the given average inventory of items. Again, for k* = 0.003, Q1 ¼ 423; Q2 ¼ 608; Q3 ¼ 442; and the average inventory = (423 + 608 + 442) = 737, which is less than 750 units. Again, for k* = 0.002, we have Q1 ¼ 430, Q2 ¼ 632, Q3 ¼ 452, and the average inventory = (430 + 632 + 452) = 757, which is greater than 750 units. Therefore, the most suitable value of k* lies between 0.002 and 0.003. Let us assume that for k* = x the average inventory will be 750. Now, considering the linear relationship between k* and the average inventory, we have x  0:003 0:003  0:002 13  0:001 ¼ or x  0:003 ¼ 750  737 737  757 20 or x ¼ 0:00235 or k ¼ 0:00235: For k* = 0.00235, Q1 ¼ 428; Q2 ¼ 623; Q3 ¼ 449. Hence, the optimal production quantities for products 1, 2 and 3 are 428, 623 and 449 units respectively.

17.10

Inventory Models with Price Breaks

In the earlier discussion, we assumed that the unit production cost or unit purchase cost is constant, so, we did not need to consider this cost in the analysis. However, in reality, it is not always true that the unit cost of an item is independent of the quantity procured or produced. Again, discounts are offered by the supplier or wholesaler or manufacturer for the purchase of large quantities. Such discounts are referred to as quantity discounts or price breaks. In this section, we shall consider a class of inventory in which cost is a variable. When items are purchased in bulk, some discount price is usually offered by the supplier. Let us assume that the unit purchase cost of an item is pj when the purchased quantity lies between bj − 1 and bj (j = 1, 2, …, m). Explicitly, we have:

17.10

Inventory Models with Price Breaks

Range of quantity to be purchased b0 \Q\b1 b1  Q\b2 b2  Q\b3 ... bj1  Q\bj ... bm1  Q\bm

553

unit purchase cost p1 p2 p3 ... pj ... pm

In general, b0 = 0 and bm = ∞ and p1 > p2 >  > pj >  > pm. The values b1, b2, b3, …, bm − 1 are termed as price breaks, as the unit price lies in the intervals between these values. The problem is to determine an economic order quantity Q which minimizes the total cost. In this model, the assumptions are as follows: (i) (ii) (iii) (iv)

The demand rate is known and uniform. Shortages are not permitted. Production for the supply of commodities is instantaneous. The lead time is zero.

17.10.1

Purchasing Inventory Model with Single Price Break

Let D be the demand rate, C1 be the holding cost per unit quantity per unit time and C3 the ﬁxed ordering cost per order. Also, let p1 be the purchasing cost per unit quantity if the ordered quantity is less than b and p2 (p2 < p1) be the purchasing cost per unit quantity if the ordered quantity is greater than or equal to b quantities, i.e. Range of quantity to be purchased

Unit purchase cost

0 b, then the optimum lot size is Q*. (ii) If Q* < b, then go to Step 2.

for the case Q > b and then

17.10

Inventory Models with Price Breaks

Step 2. Evaluate Q* by the formula Q ¼ 0



00

555

qﬃﬃﬃﬃﬃﬃﬃﬃ

2C3 D C1

for the case Q < b and evaluate

C ðQ Þ and C ðbÞ. (i) If C 0 ðQ Þ\C00 ðbÞ, then Q* is the optimum lot size. (ii) Otherwise, b is the optimal lot size.

17.10.2

Purchase Inventory Model with Two Price Breaks

Range of quantity to be purchased

Unit purchase cost

0 < Q < b1 b1  Q < b2 b2  Q

p1 p2 p3

The procedure used involving one price break is extended to the case with two price breaks. Working rule: Step 1. Compute Q* for Q > b2 (say Q*3). If Q*3 > b2, then the optimal lot size is Q*3; otherwise, go to Step 2. Step 2. Compute Q* for b1  Q*< b2 (say Q*2). Since Q*3 < b2, then Q*2 is also less than b2. In this case there are two possibilities: either Q*2  b1 or Q*2 < b1. If Q*2  b1, then compare the cost C(Q*2) and C(b2) to obtain the optimum lot size. The quantity with lower cost will naturally be the optimum one. If Q*2 < b1, then go to Step 3. Step 3. If Q*2 < b1, then compute Q*1 for the case 0 < Q < b1 and compare the cost C(Q*1), C(b1) and C(b2) to determine the optimal lot size. The quantity with lower cost will naturally be the optimum one. Example 7 Find the optimum order quantity for a product for which the price breaks are as follows: Range of quantity to be purchased 0\Q\100 100  Q\200 200  Q

Purchase cost per unit Rs: 20:00 Rs: 18:00 Rs: 16:00

The monthly demand for the product is 400 units. The storage cost is 20% of the unit cost of the product and the cost of ordering is Rs. 25.00 per order.

556

17

Inventory Control Theory

Solution Here D = 400 units/month, C3 = Rs. 25.00 per order C1 = 20% of purchase cost per unit = 0.2 time of purchase cost per unit Let

Q3

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D for Q [ 200 ¼ C1 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 2  25  400 ¼ 79: ¼ 0:2  16

Since Q*3 < 200, Q*3 is not the optimum order quantity. Therefore, we have to proceed to calculate Q*2. Now

Q2

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D for 100  Q\200 ¼ C1 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  25  400 ¼ 75: ¼ 0:2  18

Again, since Q*2 < 100, Q*2 is not the optimum order quantity. Now we have to proceed to calculate Q*1. Q1

rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2C3 D for 0\Q\100 ¼ C1 ﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  25  400 ¼ ¼ 71: 0:2  20

Since 0 < Q*1 < 100, then we have to compute C(Q*1), C(100), C(200). Now 1 C3 D CðQ1 Þ ¼ C0 ðQ1 Þ ¼ p1 D þ C1 Q1 þ  2 Q1 1 25  400 ¼ 20  400 þ  0:2  20  71 þ 2 71 ¼ Rs: 8282:85 1 C3 D C1  100 þ 2 100 1 25  400 ¼ 18  400 þ  0:2  18  100 þ 2 100 ¼ Rs: 7480:00

Cð100Þ ¼ C 00 ð100Þ ¼ p2 D þ

17.10

Inventory Models with Price Breaks

557

and 1 C3 D C1  200 þ 2 200 1 25  400 ¼ 16  400 þ  0:2  16  200 þ 2 200 ¼ Rs: 6770:00:

Cð200Þ ¼ C 000 ð200Þ ¼ p3 D þ

Since C(200) < C(100) < C(Q*1), the optimal order quantity is 200 units, i.e. Q* = 200.

17.11

Probabilistic Inventory Model

Here we consider the situations when the demand is not known exactly, but the probability distribution of the demand is somehow known. The control variable in such cases is assumed to be either the scheduling period or the order level or both. The optimum order levels will thus be derived by minimizing the total expected cost rather than the actual cost involved.

17.11.1

Single Period Inventory Model with Continuous Demand (No Replenishment Cost Model)

In this model, we have to ﬁnd the optimum order quantity so as to minimize the total expected cost with the following assumptions: (i) The scheduling period T is ﬁxed and known. Hence, we do not include the setup cost in our derivation, as it is a prescribed constant. (ii) Production is instantaneous. (iii) The lead time is zero. (iv) The demand is uniformly distributed over the period. (v) Shortages are allowed and fully backlogged. (vi) The holding cost C1 per unit quantity per unit time and the shortage cost C2 per unit quantity per unit time are known and constant. Model Formulation: Let x be the amount on hand before an order is placed. Also, let Q be the level of inventory at the beginning of each period, and r units is the demand per time period. Depending on the amount r, two cases may arise: Case I: r  Q Case II: r > Q

558

17

Fig. 17.5 Pictorial representation of single period model with continuous demand (Case I)

Inventory Control Theory

B

Q

M Q-r

O Fig. 17.6 Pictorial representation of single period model with continuous demand (Case II)

T

A

B

Q A O

C

r-Q M

The inventory situation for the case of r  Q is shown in Fig. 17.5 and the situation for the case of r > Q is shown in Fig. 17.6. In the ﬁrst case, r  Q as shown in Fig. 17.5, no shortage occurs. In the second case, r > Q as shown in Fig. 17.6, both the costs are involved. Discrete case: when r is a discrete random variable Let the demand for D = r units be estimated at a discontinuous rate with probability p(r), r = 1, 2, …, n, … That is, we may expect the demand for one unit with probability p(1), 2 units with probability p(2) and so on. Since all the possibilities / P are to be considered, we must have pðrÞ ¼ 1 and p(r)  0. r¼1

We also assume that r will only be non-negative integers. Case I: In this case, r  Q. Thus, there is no shortage, and the total inventory is represented by the total area OAMB ¼ 12 ðQ þ QrÞT ¼ ðQr=2ÞT (see Fig. 17.5).

17.11

Probabilistic Inventory Model

559

Hence, the holding cost for the time period T is C1 (Q − r/2) T. This is the holding cost when r (  Q) units is the demand rate in one period. But, the probability of the demand of r units is p(r). Hence, the expected value of the cost is C1 (Q − r/2) T p(r). Now, r can have only values less than Q. Hence, the total expected cost when r < Q is equal to Q

X r C1 Q  TpðrÞ: 2 r¼0 Case II: In this case, r > Q, both the holding and shortage costs are involved. Here, the holding cost is 12 C1 Qt1 and the shortage cost is 12 C2 ðr  QÞt2 where t1 and t2 represent the no-shortage and shortage cases and t1 + t2 = T. Now, from the similar triangles OBC and ACM in Fig. 17.6, we have t1 t2 ¼ Q rQ t1 t2 t1 þ t2 t1 t2 T or ¼ ¼ or; ¼ ¼ Q rQ r Q rQ r QT ðr  QÞT and t2 ¼ : or t1 ¼ r r Hence the expected cost is   / X 1 1 C1 Qt1 pðrÞ þ ðr  QÞC2 t2 pðrÞ 2 2 r¼Q þ 1 " # / X Q2 ðr  QÞ2 ¼ TpðrÞ þ C2 TpðrÞ : C1 2r 2r r¼Q þ 1 Therefore, the average expected cost is given by TECðQÞ ¼ C1

Q 1 1 X X X r Q2 ðr  QÞ2 pðrÞ þ C2 pðrÞ: Q  pðrÞ þ C1 2 2r 2r r¼Q þ 1 r¼Q þ 1 r¼0

ð17:49Þ The problem is now to ﬁnd Q, so as to minimize TEC(Q). Let an amount Q + 1 instead of Q be produced. Then the average total expected cost is TEC(Q þ 1Þ ¼ C1

Q þ1 X r¼0

Qþ1 

/ / X X r ðQ þ 1Þ2 ðr  Q  1Þ2 pðrÞ þ C1 pðrÞ þ C2 pðrÞ: 2 2r 2r r¼Q þ 2 r¼Q þ 1

560

17

Inventory Control Theory

But C1

Q þ1 X r¼o

Qþ1 

Q X r r pðrÞ ¼ C1 Q þ 1  pðrÞ 2 2 r¼0   Qþ1 þ C1 Q þ 1  pðQ þ 1Þ 2 Q Q X X r Qþ1 ¼ C1 pðQ þ 1Þ: Q  pðrÞ þ C1 pðrÞ þ C1 2 2 r¼0 r¼0

Again, C1

1 1 X X ðQ þ 1Þ2 ðQ þ 1Þ2 ðQ þ 1Þ2 pðrÞ ¼ C1 pðrÞ  C1 pðQ þ 1Þ 2r 2r 2ðQ þ 1Þ r¼Q þ 2 r¼Q þ 1

¼ C1

1 X Q2 þ 2Q þ 1 Qþ1 pðrÞ  C1 pðQ þ 1Þ 2r 2 r¼Q þ 1

¼ C1

1 1 1 X X Q2 Q C1 X pðrÞ pðrÞ þ pðrÞ þ C1 r r 2r 2 r¼Q þ 1 r¼Q þ 1 r¼Q þ 1



C1 ðQ þ 1ÞpðQ þ 1Þ: 2

Similarly, C2

1 X ðr  Q  1Þ2 pðrÞ 2r r¼Q þ 2

¼ C2

1 X ðr  Q  1Þ2 pðrÞ  0 2r r¼Q þ 1

¼ C2

1 1 1 1 X X X ðr  QÞ2 Q C2 X pðrÞ pðrÞ þ : pðrÞ  C2 pðrÞ þ C2 r 2r 2 r¼Q þ 1 r r¼Q þ 1 r¼Q þ 1 r¼Q þ 1

Substituting these values in TEC(Q + 1) and then simplifying, we get "

#   1 1 X pðrÞ TECðQ þ 1Þ ¼ TECðQÞ þ ðC1 þ C2 Þ pðrÞ þ Q þ  C2 : 2 r¼Q þ 1 r r¼0 Q X



Q X

1 If we set pðrÞ þ Q þ 2 r¼0

 X 1 pðrÞ ¼ LðQÞ; r r¼Q þ 1

ð17:50Þ

17.11

Probabilistic Inventory Model

561

then TECðQ þ 1Þ ¼ TECðQÞ þ ðC1 þ C2 ÞLðQÞC2

ð17:51Þ

Similarly, substituting Q – 1 in place of Q in (17.49), we have TECðQ  1Þ ¼ TECðQÞ  ðC1 þ C2 ÞLðQ  1Þ þ C2 :

ð17:52Þ

For optimal Q, we must have TECðQ þ 1Þ  TECðQÞ [ 0 i.e. ðC1 þ C2 ÞLðQ Þ  C2 [ 0 [From (17.51)] or

LðQÞ [

C2 : C1 þ C2

ð17:53Þ

Again, for optimal Q, we have TECðQ  1Þ  TECðQÞ [ 0 or  ðC1 þ C2 ÞLðQ  1Þ þ C2 [ 0 C2 or LðQ  1Þ\ : C1 þ C2

ð17:54Þ

Combining (17.53) and (17.54) for the optimal value of Q*, we have LðQ  1Þ\

C2 \LðQ Þ; C1 þ C2

ð17:55Þ

  P1 P pðrÞ 1 where LðQÞ ¼ Q r¼0 pðrÞ þ Q þ 2 r¼Q þ 1 r . Using the relation (17.55), we ﬁnd the range of optimum values of Q. In these cases, Q* need not be unique. If C1 Cþ2 C2 ¼ LðQ Þ, then both Q* and Q* + 1 are the optimal values. Similarly, if optimal values.

C2 C1 þ C2

¼ LðQ  1Þ, then both Q* and Q* − 1 are the

Continuous case: when r is a continuous random variable When the uncertain demand is estimated as a continuous random variable, the cost expressions of inventory, holding and shortage costs involve integrals instead of summation signs.

562

17

Inventory Control Theory

Let f(r) be the known probability density function for demand r. The discrete point probabilities by the probability differential f(r)dr for a small

p(r) are replaced  d r d r interval, say, r  ; r þ . In this case, we have 2

2

Z1

f ðrÞdr ¼ 1 and f ðrÞ  0:

0

Case 1 When r  Q

  Proceeding as before for r  Q, the holding cost is C1 Q  2r t and there is no shortage cost. Case 2 When r > Q Proceeding as before for r > Q, the holding cost is

C1 Q 2 t 2r

and the shortage cost is

2 t C2 ðrQÞ 2r .

TECðQÞ ¼

ZQ

r C1 Q  f ðrÞdr þ 2

dTECðQÞ ¼ C1 dQ

# Q2 ðr  QÞ2 þ C2 f ðrÞdr C1 2r 2r

Q

0

)

Z/ "

ZQ 0

þ

r dr Q f ðrÞdr þ C1 ½ Q  f ðrÞ 2 dQ 0

Z/  Q

( )  C1 C2 c1 Q2 C2 ðr  QÞ2 dr 1 2Q  2ðr  QÞ f ðrÞdr þ þ f ðrÞ dQ Q 2r 2r 2r 2r

After simpliﬁcation, we have dTECðQÞ ¼ ðC1 þ C2 Þ dQ

ZQ

f ðrÞdr þ ðC1 þ C2 Þ

Z1

Qf ðrÞ dr  C2 : r

Q

0

ðQÞ The necessary condition for TEC(Q) to be optimum is dTEC dQ ¼ 0

i.e. ðC1 þ C2 Þ

R Q 0

f ðrÞdr þ ðC1 þ C2 Þ ZQ or 0

ð17:56Þ

R/

f ðrÞdr þ

Q

Q f ðrÞ r

Z1 Q

for Q ¼ Q ,

 C2 ¼ 0

Q f ðrÞ C2 ¼ : r C1 þ C2

ð17:57Þ

17.11

Probabilistic Inventory Model

563

Again from (17.56), d2 TECðQÞ ¼ ðC1 þ C2 Þf ðQÞ þ ðC1 þ C2 Þ dQ2

Z1

f ðrÞ dr  ðC1 þ C2 Þf ðQÞ: r

Q

After simpliﬁcation, we have d2 TECðQÞ ¼ ðC1 þ C2 Þ dQ2

Z1

f ðrÞ dr ¼ a positive quantity: r

Q

Hence, Eq. (17.57) gives the optimum value of Q for the minimum expected average cost. Example 8 A contractor of second-hand motor trucks maintains a stock of trucks every month. The demand of the trucks occurs at a relatively constant rate but not with a constant amount. The demand follows the following probability distributions: Demand (r)

0

1

2

3

4

5

6 or more

Probability [p(r)]

0.40

0.24

0.20

0.10

0.05

0.01

0.0

The holding cost of an old truck in stock for one month is Rs. 100.00, and the penalty for a truck not being supplied on demand is Rs. 1000.00. Determine the optimal size of the stock for the contractor. Solution

PQ

P1

Q +

1

0.3878

0.5

0.1939

0.40

0.5937

0.2400

0.1478

1.5

0.2217

0.64

0.8617

0.20

0.1000

0.0478

2.5

0.1195

0.84

0.9595

3

0.10

0.0333

0.0145

3.5

0.0575

0.94

0.9907

4

4

0.05

0.0125

0.0020

4.5

0.0090

0.99

0.9990

5

5

0.01

0.0020

0.0000

5.5

0.0000

1.0

1.0000

6 or more

6 or more

0.00

0.0000

0.0000

6.5

0.0000

1.0

1.0000

Q

r

p(r)

pðrÞ r

0

0

0.40

1

1

0.24

2

2

3

pðrÞ r¼Q þ 1 r

1 2

ðQ þ 12Þ

PQ þ

1 pðrÞ 2 r¼Q þ 1 r

r¼0

pðrÞ

Here C1 = Rs. 100.00 and C2 = Rs. 1000.00. )

C2 1000 10 ¼ ¼ 0:9090: ¼ C1 þ C2 100 þ 1000 11

LðQÞ ¼ ðQ þ 12Þ P1 PQ pðrÞ r¼Q þ 1 r þ r¼0 pðrÞ

564

17

Inventory Control Theory

Now for the optimal value of Q (say, Q*), we must have LðQ  1Þ\

where LðQÞ ¼

PQ

r¼0

C2 \LðQ Þ; C1 þ C2

  P/ pðrÞ pðrÞ þ Q þ 12 r¼Q þ 1 r .

From the table, it is clear that for Q = 2, the preceding inequality is satisﬁed; i.e. L(1) < 0.9090 < L(2) or L(2 − 1) < 0.9090 < L(2). Hence, Q* = 2; i.e. the optimum stock level of trucks is 2.

17.11.2

Single Period Inventory Model with Instantaneous Demand (no Replenishment Cost Model)

In this model, we have to ﬁnd the optimum order quantity which minimizes the total expected cost under the following assumptions: (i) T is the constant interval between orders (T may also be considered as unity, e.g. daily, weekly, monthly, etc.). (ii) Q is the stock level at the beginning of each period T. (iii) The lead time is zero. (iv) The holding cost C1 per unit quantity per unit time and the shortage cost C2 per unit quantity per unit time are known and constant. (v) r is the demand at each interval T. Model Formulation: In this model it is assumed that the total demand is ﬁlled at the beginning of the period. Thus, depending on the amount r demanded, the inventory position just after the demand occurs may be either positive (surplus) or negative (shortage); i.e. there are two cases: Case 1: r  Q Case 2: r > Q The corresponding inventory situations are shown in Figs. 17.7 and 17.8. Discrete case: when r is discrete Let r be the estimated demand at an instantaneous rate with probabilities p(r). Then there is only a holding cost and no shortage cost. Here the holding cost is C1(Q − r).

17.11

Probabilistic Inventory Model

565

r r≤ Q (No shortage)

Q

Inventory

Q-r Time

T Fig. 17.7 Pictorial representation of single period model with no shortage

Q r

r > Q (No inventory) Time Shortage

r-Q

T Fig. 17.8 Pictorial representation of single period model with no inventory

In the second case, the demand r is ﬁlled at the beginning of the period. There is only a shortage cost, no holding cost. Therefore, the shortage cost in this case is C2 (r − Q). Therefore, the total expected cost for this model is TECðQÞ ¼ C1

Q X r¼0

ðQ  rÞpðrÞ þ C2

/ X r¼Q þ 1

ðr  QÞpðrÞ:

ð17:58Þ

566

17

Inventory Control Theory

Our problem is now to ﬁnd Q so that TEC(Q) is minimum. Let an amount Q + 1 instead of Q be ordered. Than the total expected cost given in (17.58) reduces to TECðQ þ 1Þ ¼ C1

Q X

/ X

ðQ þ 1  rÞpðrÞ þ C2

ðr  Q  1ÞpðrÞ:

r¼Q þ 2

r¼0

On simpliﬁcation, we have TECðQ þ 1Þ ¼ C1

Q X

/ X

ðQ  rÞpðrÞ þ C2

ðr  QÞpðrÞ þ C1

r¼Q þ 1

r¼0

¼ TECðQÞ þ ðC1 þ C2 Þ

Q X

Q X

pðrÞ  C2

r¼0

"

/ X

pðrÞ  C2 Since

pðrÞ ¼ 1 

r¼Q þ 1

r¼0

) TECðQ þ 1Þ  TECðQÞ ¼ ðC1 þ C2 Þ

Q X

/ X r¼Q þ 1

Q X

pðrÞ #

pðrÞ :

r¼0

pðrÞ  C2 :

ð17:59Þ

r¼0

Similarly, when an amount Q−1 instead of Q is ordered, then we have TECðQ  1Þ ¼ TECðQÞ  ðC1 þ C2 Þ

Q1 X

pðrÞ þ C2

r¼0

or TECðQ  1Þ  TECðQÞ ¼ ðC1 þ C2 Þ

Q1 X

ð17:60Þ pðrÞ þ C2 :

r¼0

For optimal Q*, we must have TECðQ þ 1Þ  TECðQ Þ [ 0: From (17.59), we have ðC1 þ C2 Þ

Q X

pðrÞ  C2 [ 0

r¼0 Q X

C2 or pðrÞ [ : C 1 þ C2 r¼0 Similarly, for optimal Q*, TEC(Q* − 1) − TEC(Q*) > 0.

ð17:61Þ

17.11

Probabilistic Inventory Model

567

From (17.60), we have  ðC1 þ C2 Þ

 Q 1 X

pðrÞ þ C2 [ 0

r¼0

or

 Q 1 X

r¼0

ð17:62Þ

C2 pðrÞ\ : C1 þ C2

Thus, combining (17.61) and (17.62), we have  Q 1 X

pðrÞ\

r¼0

Q X C2 \ pðrÞ C1 þ C2 r¼0

ð17:63Þ

C2 or pðr\Q  1Þ\ \pðr  Q Þ; C1 þ C2 

where pðr  Q Þ represents the probability for r\Q . Continuous case: where r is a continuous variable Let the demand r be a continuous variable with the probability density function f(r). Then proceeding as before, the total expected cost for this model is ZQ Z/ TECðQÞ ¼ C1 ðQ  rÞf ðrÞdr þ C2 ðr  QÞf ðrÞdr: Q

0

dTECðQÞ ¼ C1 ) dQ

ZQ 0

f ðrÞdr  C2

Z/

f ðrÞdr

Q

½Using Leibniz's rule for differentiation under the integral sign 2 3 ZQ Z / ZQ ¼ C1 f ðrÞdr  C2 4 f ðrÞdr  f ðrÞdr 5 0

0

ZQ ¼ C2 þ ðC1 þ C2 Þ f ðrÞdr: 0

For the optimum value of Q, we must have dTECðQÞ ¼ 0; dQ

0

568

17

i:e: ðC1 þ C2 Þ

RQ 0

f ðrÞdr ¼ C2 Z or

Q

f ðrÞdr ¼

0

Inventory Control Theory

C2 : C1 þ C2

ð17:64Þ

Moreover, it can be proved that d2 TECðQÞ ¼ ðC1 þ C2 Þf ðQÞ [ 0: dQ2 Therefore, the optimum value of Q, i.e. Q*, is given by (17.64). Example 9 A newspaper boy buys papers for Rs. 1.40 each and sells them for 2.00. He cannot return the unsold newspapers. Daily demand has the following distribution. No. of customers

23

24

25

26

27

28

29

30

31

32

Probability

0.01

0.03

0.06

0.10

0.20

0.25

0.15

0.10

0.05

0.05

If each day’s demand is independent of the previous day’s demand, how many papers should he ordered each day? Solution Let Q be the number of newspapers ordered per day and r be the demand for them, i.e. the number that are actually sold per day. In this problem, the holding cost per paper per day is C1 = Rs. 1.40, and the shortage cost per paper per day is C2 = Rs. (2.00–1.40) = Rs. 0.60. Now to obtain the optimal solution, we shall ﬁnd the cumulative probability of daily demand of newspapers. Calculations are given in the following table. r

23

24

25

26

27

28

29

30

31

32

p(r) PQ

0.01 0.01

0.03 0.04

0.06 0.10

0.10 0.20

0.20 0.40

0.25 0.65

0.15 0.80

0.10 0.90

0.05 0.95

0.05 1.00

r¼23

pðrÞ

The required optimum value for Q* is determined by  Q 1 X

pðrÞ\

r¼23

or

 Q 1 X

r¼23

Q X C2 \ pðrÞ C1 þ C2 r¼23

pðrÞ\0:3\

Q X r¼23

pðrÞ

 *

 C2 0:60 3 ¼ ¼ 0:3 : ¼ C1 þ C2 1:40 þ 0:60 10

17.11

Probabilistic Inventory Model

569

From the table, we have 27 X

pðrÞ ¼ 0:40 [ 0:3 and

r¼23

)

26 X

pðrÞ ¼ 0:20\0:3:

r¼23

26 X

pðrÞ\0:3\

r¼23

27 X

pðrÞ:

r¼23

Hence, Q* = 27; i.e. the optimum number of newspapers to be ordered is 27. Example 10 The demand for a certain product has a rectangular distribution between 4000 and 5000. Find the optimal order quantity if the storage cost is Re. 1.00 per unit and the shortage cost is Rs. 7.00 per unit. Solution Here C1 = Re. 1.00, C2 = Rs. 7.00, C3 = 0. The required optimum value for Q* is determined by ZQ f ðrÞdr ¼

C2 : C1 þ C2

ð17:65Þ

0

Since the distribution is rectangular, the probability density function is given by 1 , a  r  b. Here a = 4000 and b = 5000. f ðrÞ ¼ ba Hence, from (17.65), we have ZQ 4000

ZQ

1 7 dr ¼ or 5000  4000 1þ7

1 7 dr ¼ 1000 8

4000

Q  4000 7 ¼ or Q ¼ 4875: or 1000 8 Hence, the optimal order quantity is Q ¼ 4875 units. Example 11 An ice cream company sells one type of its ice cream by weight. If the product is not sold on the day it is prepared, it can be sold for a loss of Rs. 0.50 per pound. But there is an unlimited market for one-day-old ice cream. On the other hand, the company makes a proﬁt of Rs. 3.20 on every pound of ice cream sold on the day it is prepared. If daily orders form a distribution with f(x) = 0.02 − 0.0002 x, 0  x  100, how many pounds of ice cream should the company prepare every day?

570

17

Inventory Control Theory

Solution It is given that C1 = Rs. 0.50 and C2 = Rs. 3.20. Let Q be the amount of ice cream prepared every day. The required optimum for Q* is determined by Q R 0

or

f ðxÞdx ¼ C1 Cþ2 C2 Q R 0

or

ð0:02  0:0002xÞdx ¼ 0:5 32 þ 3:2

0:0002Q2 þ 0:04Q þ 1:730 ¼ 0:

On solving, Q* = 136.7 and 63.5. But Q* = 136.7 is not possible, as 0  x  100. Hence, Q* = 63.5 lb is the optimum quantity; i.e. the company will prepare 63.5 lb of ice cream per day.

17.12

Fuzzy Inventory Model

In this section, we shall discuss the solution procedure of a fuzzy non-linear programming problem. A deterministic or crisp non-linear programming problem may be deﬁned as follows: Minimize g0 ðx; C0 Þ subject to gi ðx; Ci Þ  bi ; and x  0;

i ¼ 1; 2; . . .; m

ð17:66Þ

where x ¼ ðx1 ; x2 ; . . .; xn ÞT is a variable vector, f, gi’s are algebraic expressions in x with coefﬁcients C0 ¼ ðC01 ; C02 ; . . .; C0l0 Þ and Ci ¼ ðCi1 ; Ci2 ; . . .; Cili Þ respectively. Introducing fuzziness in crisp parameters, the Problem (17.66) reduces to

 f0 Minimize g0 x; C subject to

 Ci  bei ; gi x; f and x  0;

i ¼ 1; 2; . . .; m

ð17:67Þ

17.12

Fuzzy Inventory Model

571

where the wavy bar (*) represents the fuzziness of the parameters. According to fuzzy set theory, the fuzzy objective, coefﬁcients and constraints are deﬁned by their membership functions, which may be either linear or non-linear. According to Bellman and Zadeh (1970) and following the techniques of Carlsson and Korhonen (1986) and Trappy et al. (1988), the problem (17.67) is transformed into a new optimization problem as follows: Maximize a subject to

 1 gi x; l1 Ci ðaÞ  lbi ðaÞ; and

i ¼ 0; 1; 2; . . .; m

ð17:68Þ

x  0;

n o where lCi ðxÞ ¼ lCi1 ðxÞ; lCi2 ðxÞ; . . .; lCili ðxÞ are the membership functions of the fuzzy coefﬁcients, and lbi ðxÞ; i ¼ 1; 2; . . .; m are the membership functions of the fuzzy objective and fuzzy constraints. Here the additional variable a is known as the aspiration level. Therefore, the Lagrangian function Lða; x; kÞ is given by Lða; x; kÞ ¼ a 

m n  o X 1 ki gi x; l1 Ci ðaÞ  lbi ðaÞ ; i¼0

where k ¼ ðk0 ; k1 ; k2 ; . . .; km ÞT is the Lagrange multiplier vector. Now the Kuhn-Tucker necessary conditions are @L ¼ 0; @xj @L ¼0 @a n

j ¼ 1; 2; . . .; n

 o 1 ki gi x; l1 Ci ðaÞ  lbi ðaÞ ¼ 0

 1 i ¼ 0; 1; 2; . . .; m gi x; l1 Ci ðaÞ  lbi ðaÞ  0;

ð17:69Þ

ki  0: Solving the problem (17.69), the corresponding optimal solution can be obtained. In a deterministic EOQ model, the problem is to determine the order level Q (>0) which minimizes the average total cost C(Q), i.e.

572

17

Inventory Control Theory

Min CðQÞ ¼ C1 Q=2 þ C3 D=Q ð17:70Þ

subject to AQ  B; Q [ 0; where C3 = ordering cost per order, C1 = holding cost per unit quantity per unit time, D = demand per unit time, B = maximum available warehouse space (in square feet), A = the space required by each unit (in square units).

When the preceding objective goal, costs and available storage area become fuzzy, the problem (17.70) is transformed to: g ¼C f1 Q=2 þ C f3 D=Q CðQÞ

Min

ð17:71Þ

subject to e AQ  B;

Q [ 0:

So the corresponding fuzzy non-linear programming problem is Max a subject to 1 1 l1 C1 ðaÞQ=2 þ lC3 ðaÞD=Q  lC0 ðaÞ

AQ  l1 B ðaÞ;

Q [ 0;

a 2 ½0; 1;

where lC3 ðxÞ; lC1 ðxÞ; lC0 ðxÞ and lB ðxÞ are the membership functions for the fuzzy ordering cost, holding cost, objective goal and storage area respectively. Here the Lagrangian function is n o 1 1 Lða; Q; k1 ; k2 Þ ¼ a  k1 l1 C1 ðaÞQ=2 þ lC3 ðaÞD=Q  lC0 ðaÞ    k2 AQ  l1 B ðaÞ ;

ð17:72Þ

where k1 and k2 are Lagrange multipliers. Now, we consider three combinations of different types of membership functions to represent the fuzzy goal, costs and warehouse space. Fuzzy goal, costs and storage area represented by linear membership functions In this case, lCi ði ¼ 1; 3Þ is given by

17.12

Fuzzy Inventory Model

lCi ðuÞ ¼ 1

573

for u [ Ci

¼ 1  ðCi  uÞ=Pi ¼0

for Ci Pi  u  Ci ði ¼ 1; 3Þ for u\Ci Pi :

Graphically, this function is represented in Fig. 17.9. Again; lC0 ðCðQÞÞ ¼ 1

for CðQÞ\C0

¼ 1  ðCðQÞ  C0 Þ=P0 ¼0

for C0  CðQÞ  C0 þ P0 for CðQÞ [ C0 þ P0 :

Its pictorial representation is given in Fig. 17.10. The linear membership function for the storage area is given by lB ðAQÞ ¼ 1

for AQ\B

¼ 1  ðAQBÞ=P ¼0

for B  AQ  B þ P for AQ [ B þ P:

μCi ( u ) 1

u

Ci − Pi Ci Fig. 17.9 Linear membership function of Ci

μC ( C ( Q ) ) 0

O

C0

e ðQÞ Fig. 17.10 Linear membership function of C

C0 + P0

C(Q)

574

17

Inventory Control Theory

μ B ( AQ ) 1

B+P

B

O

AQ

Fig. 17.11 Linear membership function of AQ

This membership function is depicted in Fig. 17.11. Here P0, P and the Pi’s (i =1, 3) are the maximally acceptable violations of the aspiration levels C0, B and the Ci’s (i = 1, 3). Considering the nature of these parameters, we assume the membership function to be non-decreasing for fuzzy inventory costs and non-increasing for the fuzzy goal and storage area. Hence, l1 Ci ðaÞ ¼ Ci  ð1  aÞPi ; l1 C0 ðaÞ l1 B ðaÞ

ði ¼ 1; 3Þ

¼ C0 þ ð1  aÞP0 ¼ B þ ð1  aÞP:

From (17.72), we have Lða; Q; k1 ; k2 Þ ¼ a  k1 ½fC1  ð1  aÞP1 gQ=2 þ fC3  ð1  aÞP3 gD=Q  C0  ð1  aÞP0   k2 fAQ  B  ð1  aÞPg:

Now, the Kuhn-Tucker necessary conditions are @L ¼0 @a @L ¼0 @Q fC1  ð1  aÞP1 gQ=2 þ fC3  ð1  aÞP3 gD=Q  C0  ð1  aÞP0 ¼ 0 AQ  B  ð1  aÞP ¼ 0:

17.12

Fuzzy Inventory Model

575

Solving these equations, the expression for the optimal order quantity is Q ¼ fB þ ð1  a ÞPg=A; where a is a root of   P1 P2 ð1  aÞ3  C1 P2  2BPP1  2AP0 P ð1  aÞ2    2BC1 P  P1 B2  2DA2 P3  2APC0  2AP0 B ð1  aÞ    C1 B2 þ 2DA2 C3  2AC0 B ¼ 0:

Fuzzy goal and storage area represented by parabolic concave membership functions and costs by linear membership functions In this case, the membership functions lCi ðuÞ; i ¼ 1; 3 for the fuzzy inventory cost will be the same as deﬁned in the preceding section. The membership function lC0 ðCðQÞÞ for the fuzzy goal is deﬁned as follows: lC0 ðCðQÞÞ ¼ 1

for CðQÞ\C0

¼ 1  ½fCðQÞ  C0 g=P0  ¼0

2

for C0  CðQÞ  C0 þ P0 for CðQÞ [ C0 þ P0 :

It is graphically represented in Fig. 17.12. Again, the membership function for the storage area is deﬁned as lB ðAQÞ ¼ 1

for AQ\B 2

¼ 1  ½ðAQ  BÞ=P ¼0

for B  AQ  B þ P for AQ [ B þ P:

Fig. 17.12 Parabolic e ðQÞ membership function of C

C0

C0 + P0

576

17

Inventory Control Theory

Proceeding exactly as before, the optimal order quantity is given by n pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ o Q ¼ B þ ð1  a ÞP =A; where a is a root of   3 P1 P2 ð1  aÞ2 þ 2BPP1 ð1  aÞ2  C1 P2  P1 B2  2A2 DP3  2AP2 ð1  aÞ   1  ð2BPC1  2AC0 P  2APBÞð1  aÞ2  C1 B2 þ 2A2 DC3  2AC0 B ¼ 0:

17.13

Exercises

1. The demand rate of a particular item is 12,000 units per year. The ordering cost per order is Rs. 350.00, and the holding cost is Rs. 0.20 per unit per month. If no shortage is allowed and the replenishment is instantaneous, determine (i) the optimal lot size, (ii) the optimum scheduling period, (iii) the minimum total average cost. 2. The annual requirement for a product is 3000 units. The ordering cost is Rs. 100.00 per order. The cost per unit is Rs. 10.00. The carrying cost per unit per year is 30% of the unit cost. (a) Find the EOQ. (b) If a new EOQ is found by using the ordering cost as Rs. 80.00, what would be the further savings in cost? 3. The demand rate for an item of a company is 18,000 units per year. The company can produce at the rate of 3000 units per month. The setup cost is Rs. 500.00 per order, and the holding cost is 0.15 per units per month. Calculate: (i) Optimum manufacturing quantity, (ii) the time of manufacture, (iii) cycle length, (iv) the optimum annual cost if the cost of an item is Rs. 2.00 per unit. 4. A contractor has to supply 10,000 bearings per day to an automobile manufacturer. He ﬁnds that when he starts a production run he can produce 25,000 bearings per day. The holding cost of a bearing in stock is Rs. 0.02 per year. The setup cost of a production is Rs. 18.00. How frequently should production runs be made? 5. The demand for an item is 18,000 units per year. The holding cost is Rs. 1.20 per unit per unit item, and the shortage cost is Rs.5.00. The ordering cost is Rs. 400.00. Assuming the replenishment rate as instantaneous, determine the optimum order quantity along with the total minimum cost of the system. 6. Find the optimum order level which minimizes the total expected cost under the following assumptions: (i) t is the constant interval between orders.

17.13

Exercises

(ii) (iii) (iv) (v)

577

Q is the stock (in discrete units) at the beginning. d is the estimated random instantaneous demand at a discontinuous rate. C1 and C2 are the holding and shortage costs per item per t time unit. The level time is zero.

7. Let F(x) be the probability density function of the demand x in a prescribed scheduling period of T weeks. The demand is assumed to occur with a uniform pattern, and the probability distribution is continuous. The unit carrying cost and the shortage cost are respectively C1 money units and C2 money units, both per unit per week. There is no setup cost for the system. Obtain the expression for optimal control. 8. Let the probability density of demand of a certain item during a week be f ðxÞ ¼

0:1; 0  x  10 : 0; Otherwise

This demand is assumed to occur with a uniform pattern over the week. Let the unit carrying cost of the item in inventory be Rs. 2.00 per week and the unit shortage cost be Rs. 8.00 per week. Determine the optimal order level of the inventory and the total minimum cost of the inventory system. 9. The annual demand for a product is 500 units. The cost of storage per unit per year is 10% of the unit cost. The ordering cost is Rs. 180 for each order. The unit cost depends upon the amount ordered. The range of amount ordered and the unit price are as follows: Range of amount ordered

Unit cost (Rs.)

1  q < 500 500  q < 1500 1500  q < 3000 3000  q

25.00 24.80 24.60 24.40

Find the optimal order quantity. 10. Show that when determining the optimal inventory level which minimizes the total expected cost in case of continuous (non-discrete) quantities, the condition to be satisﬁed is FðS0 Þ ¼

C2 ; C1 þ C2

where FðS0 Þ ¼

RS0

f ðrÞdr,

0

f (r) = the probability density function of the requirement of r quantity C2 = shortage cost per unit per unit time

578

17

Inventory Control Theory

C1 = holding cost per unit per unit time. 11. An ice cream company sells one of its types of ice cream by weight. If the product is not sold on the day it is prepared, it can be sold at a loss of 50 paise per pound. There is, however, an unlimited market for one-day-old ice cream. On the other hand, the company makes a proﬁt of Rs. 3.20 on every pound of ice cream sold on the day it is prepared. Past daily orders form a distribution with f ðxÞ ¼ 0:020:0002x;

0  x  100:

How many pounds of ice cream should the company prepare every day? 12. The following data describe three inventory items. Determine the economic order quantity for each of the three items to be accommodated within a total available storage area of 25 square feet. Item

Setup cost (Rs.)

Demand (units per day)

Unit holding cost per unit time (Rs.)

Storage area required per unit (ft2)

1 2 3

10 5 15

2 4 4

0.3 0.1 0.2

1 1 1

Find the economic lot size for the inventory model with ﬁnite replenishment rate, shortages allowed but fully backlogged, uniform ﬁnite demand and zero lead time so that the total average cost is minimum. 13. A shop produces three items in lots. The demand rate for each item is constant and can be assumed to be deterministic. No backorders are allowed. The pertinent data for the items are given in the following table: Item

I

II

III

Carrying cost (Rs.) Setup cost (Rs.) Cost per unit (Rs.) Yearly demand rate (units)

20 50 6 10,000

20 40 7 12,000

20 60 5 7500

Determine approximately the economic order quantity for the three items subject to the condition that the total value of average inventory levels of these items does not exceed Rs. 1000.00. 14. Find the optimum order quantity for a product for which the price breaks are as follows: Quantity

0  Q1 < 100 100  Q2 < 200 200  Q3

Rs. 20 Rs. 18 Rs. 16

17.13

Exercises

579

The monthly demand for the product is 400 units. The storage cost is 20% of the unit cost of the product, and the cost of ordering is Rs. 25.00 per month. 15. A newspaper boy buys papers for Rs. 1.70 each and sells them for Rs. 2.00 each. He cannot return unsold newspapers. The daily demand has the following distribution: No. of customers

23

24

25

26

27

28

29

30

31

32

Probability

0.01

0.03

0.06

0.10

0.20

0.25

0.15

0.10

0.05

0.05

If each day’s demand is independent of the previous day’s demand, how many papers should he order each day? 16. A small shop produces three machine parts 1, 2, 3 in lots. The shop has only 700 m2 of storage space. The appropriate data for the three items are presented in the following table: Item

1

2

3

Demand (unit per year) Setup cost (Rs.) Cost per unit (Rs.) Floor space required (sq.mt/unit)

2000 100 10 0.50

5000 200 20 0.60

10000 75 5 0.30

The shop uses an inventory carrying charge of 20% of the average inventory valuation per annum. If no stock-outs are allowed, determine the optimal lot size for each item.

Chapter 18

Mathematical Preliminaries

18.1

Objective

The objective of this chapter is to discuss some mathematical background of several topics, like matrices, determinants, vectors, probability and Markov processes, which will be very helpful for studying the different subject matters of this book.

18.2

Matrices

A rectangular array of mn elements (m, n being positive integers), arranged in m rows and n columns as shown 2

a11 6 a21 6 A¼6 6  4  am1

a12 a22   am2

3    a1n    a2n 7 7   7 7   5    amn

0

a11 B a21 B or A ¼ B B  @  am1

a12 a22   am2

    

1 a1n a2n C C  C C  A amn

is called matrix of order (m  n) or an m by n matrix. The numbers aij are called the elements of the matrix. A useful, easy way of writing the matrix shown below:     A ¼ aij mn or aij mn ;

i ¼ 1; 2; . . .; m;

j ¼ 1; 2; . . .; n:

In general, matrices are denoted by boldface uppercase letters A, B, C, . . ..

© Springer Nature Singapore Pte Ltd. 2019 A. K. Bhunia et al., Advanced Optimization and Operations Research, Springer Optimization and Its Applications 153, https://doi.org/10.1007/978-981-32-9967-2_18

581

582

18

Mathematical Preliminaries

Square matrix: If m = n, the matrix A is called a square matrix of order n (or of order m). Row matrix: If m = 1, i.e. the matrix A has only one row, then A is called a row matrix (or sometimes called an n-dimensional row vector). Column matrix: If n = 1, i.e. the matrix A has only one column, then A is called a column matrix (or sometimes called an m-dimensional column vector). Equality of matrices: Two matrices A and B are said to be equal if A and B are of the same order and  each element ofA is the same as the corresponding element of B; i.e. if A ¼ aij mn and B ¼ bij mn , then A ¼ B only when aij ¼ bij for each i ¼ 1; 2; . . .; m and each j ¼ 1; 2; . . .; n. It is to be noted that the question of equality between two matrices of different orders does not arise.   Null/zero or void matrix: A matrix A ¼ aij mn is called a null matrix or a void matrix of order m  n, if every element of A is 0 (zero), i.e. aij ¼ 0 for each i ¼ 1; 2; . . .; m and j ¼ 1; 2; . . .; n. The null matrix is denoted by Omn or simply O. Diagonal element of a matrix A   Let A ¼ aij mn be a square matrix of order n. The diagonal through the element at the left-hand top corner of A is called the principal (or leading) diagonal or simply the diagonal of A, and the elements of this diagonal are called the diagonal elements of A. 2 3 a11 a12    a1n 6 a21 a22    a2n 7 6 7 Thus, if A ¼ 6 .. .. 7, .. .. 4 . . 5 . . an1

an2

   ann

then the diagonal elements of A are a11 ; a22 ; . . .; ann . Diagonal matrix: If all the elements of A other than the diagonal elements are zero, i.e. aij ¼ 0 for all i, j with i 6¼ j, then the matrix A is called a diagonal matrix and is denoted by ða11 ; a22 ; a33 ; . . .; ann Þ. Scalar matrix: If A is a diagonal matrix with all the diagonal elements are same, i.e. a11 ¼ a22 ¼    ¼ ann and aij ¼ 0 for all i, j with i 6¼ j, then A is said to be a scalar matrix. Identity matrix or unit matrix: If A be the scalar matrix with each diagonal element as unity, i.e. a11 ¼ a22 ¼    ¼ ann ¼ 1 and aij ¼ 0 for all i, j with i 6¼ j, then A is said to be a unit matrix and is denoted by In. Upper triangular matrix: If all the elements below the diagonal elements are zero, i.e. aij ¼ 0 for i > j, then the matrix is called an upper triangular matrix.

18.2

Matrices

583

Lower triangular matrix: If all the elements above the diagonal elements are 0, i.e. aij ¼ 0 for (i < j), then the matrix is called a lower triangular matrix. Triangular matrix: A matrix A is called a triangular matrix if it is either an upper triangular or a lower triangular matrix. Here it is mentioned that a matrix is a diagonal matrix if and only if it is both upper triangular and lower triangular. Addition of matrices Two matrices A and B are said to be conformable for addition of A and B with same order (i.e. addition of a matrix  A with a matrix Bis deﬁned only when they are of the same order). If A ¼ aij mn and B ¼ bij mn are two matrices, then C = A + B is a matrix of order m  n, whose elements cij are given by cij ¼ aij þ bij for all i and j: 

   2 1 2 3 4 7 and B ¼ , 5 3 4 1 2 4      2 1 2 3 4 7 5 then C ¼ A þ B ¼ þ ¼ 5 3 4 1 2 4 4 2 3   1 2 3 1 2 3 4 5 If A ¼ 2 3 4 and B ¼ , 1 4 1 5 3 2 Thus, if A ¼

 5 5 . 5 8

then these two matrices are not conformable for addition.   Negative of a matrix: Let A ¼ aij mn be a matrix. Then the matrix B of order m  n is called the negative of A if A þ B ¼ Omn . Clearly,  B is the matrix aij mn and we denote the negative of A by −A, i.e. A ¼ aij mn .     If A and B are two matrices of the same order, say A ¼ aij mn , B ¼ bij mn , then we deﬁne the difference between A and B to be a matrix of order m  n denoted by A  B and given by A  B ¼ A þ ðBÞ.       ) A  B ¼ aij mn þ bij mn ¼ aij  bij mn : Multiplication   by a scalar: Let A ¼ aij mn be a matrix of order (m  n) and c be a scalar. Then the scalar multiple c A is deﬁned to be a matrix of order (m  n), each element of which is c times the corresponding element of A,

584

18

Mathematical Preliminaries

  i:e: c A ¼ caij mn : 2

3 If c = 5, A ¼ 4 9 2

3 2 3 6 15 30 1 5, then c A ¼ 4 45 5 5. 0 10 0

Some algebric properties of matrices (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)

A + B = B + A (matrix addition is commutative) A + (B + C) = (A + B) + C (matrix addition is associative) ðc1 þ c2 ÞA ¼ c1 A þ c2 A, where c1 and c2 are scalars cðA þ BÞ ¼ cA þ cB, where c is a scalar ð1ÞA ¼ A A  B ¼ A þ ðBÞ cðAÞ ¼ ðcÞA, where c is a scalar ðAÞ ¼ A ¼ 1  A cðA  BÞ ¼ c A þ cðBÞ ¼ c A þ ðcÞB ¼ c A  c B ðc1  c2 ÞA ¼ c1 A  c2 A:

Multiplication of matrices Let A and B be two matrices. Then the product of A and B denoted by A B is deﬁned if and only if the number of columns of A is equal to the number of rows of B.     Thus, if A ¼ aij mp and B ¼ bij pn matrix, then the product   C ¼ A B ¼ cij will exist and will be an m  n matrix where the elements cij of C are given by p X cij ¼ aik bkj ; i ¼ 1; 2; . . .; m; j ¼ 1; 2; . . .; n: k¼1

 Let A ¼

10 21

1 3

0

and 23

1 0 B ¼ @ 1 2 2 1

1 0 3 A . 2 33

1 0 1 0 0 1 B C ) AB ¼ @ 1 2 3 A 2 1 3 2 1 2  1  1 þ 0  ð1Þ þ ð1Þ  2 1  0 þ 0  2 þ ð1Þ  1 1  0 þ 0  3 þ ð1Þ  ð2Þ ¼ 2  1 þ 1  ð1Þ þ 3  2 2  0þ1  2þ3  1 2  0 þ 1  3 þ 3  ð2Þ  1 1 2 ¼ : 7 5 3 

1

0

18.2

Matrices

585

Note (i) Matrix multiplication satisﬁes the associative and the distributive laws. For the matrices A, B, C, ðABÞC ¼ AðB CÞ; ðA þ BÞC ¼ AC þ BC

and

AðB þ C Þ ¼ AB þ AC

provided that multiplication and/or addition are deﬁned. (ii) For any two matrices A, B: • Neither AB nor BA may be deﬁned. • AB may be deﬁned but BA is not deﬁned and vice versa. • AB and BA are both deﬁned if the number of rows of A = number of columns of B and number of columns of A = number of rows of B. (iii) Matrix multiplication is not, in general, commutative; i.e. for two matrices A, B, the equality be true (even if AB and BA are deﬁned).  AB = BA need  not 1 2 5 6 Let A ¼ B¼ 3 4 1 2   5þ2 6þ4 7 10 AB ¼ ¼ and 15 þ 4 18 þ 8 19 26   5 þ 18 10 þ 24 23 34 BA ¼ ¼ : 1þ6 2þ8 7 10 Thus, AB 6¼ BA. (iv) If A is a square matrix of order (n  n) or n (say), then A commutes with In , the identity matrix. If On be the null matrix and A be any scalar matrix then • A In ¼ In A ¼ A • On A ¼ AOn ¼ On Transpose of a matrix Let A be a matrix of order m  n (say). Then the matrix of order n  m obtained by interchanging the rows and columns of A is called the transpose of A, to be denoted by AT or A0 .     So for A ¼ aij mn , AT ¼ aji nm . In this process, the ith row of A becomes 2   1 1 2 3 Let A ¼ , ) AT ¼ 4 2 4 5 6 23 3

the ith column of AT . 3 4 55 . 6 32

586

18

Mathematical Preliminaries

Some properties: (i) (ii) (iii) (iv)

T

ðAT Þ ¼ A ðcAÞT ¼ c AT , where c is a scalar ðA þ BÞT ¼ AT þ BT ðABÞT ¼ BT AT

  Symmetric matrix: A square matrix A ¼ aij nn of order n is said to be symmetric if A ¼ AT , i.e. aij ¼ aji ; 8 i; j.   Skew-symmetric matrix: A square matrix A ¼ aij nn of order n is said to be a skew-symmetric matrix if A ¼ AT , i.e. aij ¼ aji 8 i; j. Some properties: (i) Every diagonal matrix is symmetric. (ii) For any matrix A, A AT and AT A are symmetric matrices. (iii) For a square matrix A, A þ AT is symmetric, while A  AT is a skew-symmetric matrix. (iv) The diagonal elements of a skew-symmetric matrix are all zero. Orthogonal matrix: A square matrix A is said to be orthogonal if AT A ¼ A AT ¼ I (identity matrix). Submatrix: A matrix formed by omitting some rows and columns of a matrix is known as a submatrix of the original matrix. Partitioned matrices: Let A be a matrix given by 2 6 a11 6 6 a21 6 A ¼ 6  6 6 4 a31

a12

a41

a42

a22  a32

.. . .. .

 .. . .. .

3 a13

a14

a23 

a24 

a33

a34

a15 7 7 a25 7 7  7 7 7 a35 5

a43

a44

a45

Now, if we write 

A11 A21

a ¼ 11  a21 a ¼ 31 a41

then A can be written as

  a12 a A12 ¼ 13 a22   a23 a a32 A22 ¼ 33 a42 a43

a14 a24 a34 a44

 a15 a25  a35 a45

18.2

Matrices

587



A11 A¼ A21

 A12 : A22

Therefore, the matrix A has been partitioned into four submatrices, Aij ; i ¼ 1; 2; j ¼ 1; 2.

18.3

Determinant

There is a number associated with every square matrix A which is called the determinant of A and is usually denoted by j Aj or det A. The determinant of A of order n is deﬁned as X ð1Þa1j a2j    anp ; j Aj ¼ the sum being taken over all the permutations of the second subscripts. Note If A, B are square matrices of the same order (say, n), then (i) j Aj ¼ jAT j (ii) jA Bj ¼ j Aj jBj (iii) jk Aj ¼ kn j Aj, for any scalar k. Minor: The minor Mij of the element aij of a matrix A is the determinant obtained by striking out the ith row and jth column of matrix A. 2 3 a11 a12 a13 Let A ¼ 4 a21 a22 a23 5. a31 a32 a33

a22 a23

Then the minor of a11 is M11 ¼

a32 a33

a11 a12

. and the minor of a23 is M23 ¼

a31 a32

Cofactor: The cofactor of the element aij of the matrix A, denoted by Aij , is the coefﬁcient of aij in the expansion of A, i:e:

Aij ¼ ð1Þi þ j Mij :

Note The determinant of a square matrix A of order n can be expressed as n P aij Aij ; i ¼ 1; 2; . . .; n; j Aj ¼ j¼1

588

18

Mathematical Preliminaries

where Aij is the cofactor of aij . n P Obviously, j Aj ¼ aij Aij ; j ¼ 1; 2; . . .; n. i¼1

Properties of determinants: (i) The value of a determinant remains unaltered if its rows are changed into columns and its columns into rows. (ii) If any two rows (or columns) of a determinant are interchanged, the determinant retains its absolute value but changes its sign. (iii) If two rows or columns of a determinant are identical, then the determinant vanishes. (iv) If all the elements of any row (or of any column) are multiplied by the same quantity, the determinant is multiplied by the same quantity. Adjoint or adjugate of a square matrix:   Let A ¼ aij nn be a square matrix, and let Aij denote the cofactor of aij in j Aj (for   i; j ¼ 1; 2; . . .; nÞ. Let B ¼ Aij nn . Then the transpose BT of B is called the adjoint or adjugate of A and denoted by Adjð AÞ or Adj A. Properties: (i) (ii) (iii) (iv)

A  (Adj AÞ ¼ (Adj AÞ  A ¼ j Aj  In AdjðAT Þ ¼ ðAdj AÞT AdjðkAÞ ¼ kn1 Adjð AÞ, where k is a scalar jAdj Aj ¼ j Ajn1 ; if j Aj 6¼ 0

Inverse of a matrix: Let A be a square matrix of order n. A square matrix B (if it exists) is said to be an inverse of A if AB = BA = In. If such an inverse B of A exists, then A is said to be invertible. Singular matrices: A square matrix A is said to be singular if j Aj ¼ 0. Non-singular matrix: A square matrix A is said to be non-singular if j Aj 6¼ 0. Note (i) A square matrix is invertible if it is non-singular. (ii) An invertible matrix has a unique inverse. Adj A (iii) A1 ¼ j Aj ; if j Aj 6¼ 0: Some properties of the inverse of a matrix If A, B are two non-singular matrices of the same order, then 1

(i) ðA1 Þ ¼ A (ii) ðABÞ1 ¼ B1 A1

18.3

Determinant

589

(iii) ðATÞ1 ¼ ðA1 Þ (iv) ðk AÞ1 ¼ k1 A1 , for any scalar k 6¼ 0. T

Note If A1 ; A2 ; . . .; Am are m invertible matrices, then each A1 ; A2 ; . . .; Am is 1 1 1 also invertible, and ðA1 A2 . . .; Am Þ1 ¼ A1 m Am1 . . .; A2 A1 . Integral powers of a square matrix Let A be a square matrix of order n. Then the integral powers of A are deﬁned as follows: (i) (ii) (iii) (iv)

If A is non-singular, then A0 ¼ In A1 ¼ A Am ¼ Am1 A m If A is non-singular, Am ¼ ðA1 Þ ¼ ðAm Þ1 , where m is a positive integer.

18.4

Rank of a Matrix

Let A be an (m  n) non-null matrix. The rank of matrix A will be k ( min (m, n)), if the determinant of every square submatrix of A of order (k + 1) vanishes while there exists at least one square submatrix of order k whose determinant does not vanish. That is, the rank of A is the order of the largest square submatrix in A whose determinant does not vanish. The rank of A is denoted by r(A). 0

1 If A ¼ @ 2 3

1

1 2 3 2

4 6 A; then r ð AÞ ¼ 1; since j Aj ¼

2 4

3 6 6 9

3

6

¼ 0 9

and all square submatrices of order 2 also vanish. Hence, r(A) = 1.  If A ¼

3 1 ; then rðAÞ ¼ 2; since j Aj 6¼ 0: 2 4

Properties: (i) (ii) (iii) (iv)

The The The The

rank rank rank rank

of of of of

matrix A = the rank of AT . a non-singular matrix of order n is n. a singular square matrix of order n is less than n. a null matrix is deﬁned to be zero or inﬁnite.

590

18.5

18

Mathematical Preliminaries

Solution of a System of Linear Equations

The system of linear equations a11 x1 þ a12 x2 þ a21 x1 þ a22 x2 þ     an1 x1 þ an2 x2 þ

   þ a1n xn    þ a2n xn        þ ann xn

9 ¼ b1 > > > ¼ b2 > =  >  > > > ; ¼ bn

ð18:1Þ

can be written in matrix form as Ax ¼ b   where A ¼ aij nn is the coefﬁcient matrix, x ¼ ðx1 ; x2 ; . . .; xn ÞT T

and

1

b ¼ ðb1 ; b2 ; . . .; bn Þ . The system has a unique solution x ¼ A b, provided j Aj 6¼ 0, i.e. the coefﬁcient matrix A is non-singular. 2 3 a11 a12    a1n b1 6 a21 a22    a2n b2 7 6 7 7 Note Let us consider a matrix A ¼ ½A b ¼ 6 6 . . . . . . . . . . . . . . . 7: 4 ... ... ... ... ...5 an1 an2    ann bn The matrix A is called the augmented matrix of the system of Eq. (18.1).   It is obvious that r A ¼ n, since j Aj 6¼ 0 and there does not exist a square submatrix of A of order (n + 1).   Thus, it follows that the system (18.1) has a unique solution if r ð AÞ ¼ r A ¼ n. If r ð AÞ\n, i.e. if A is singular, then two cases arise:   (i) If r ð AÞ ¼ r A \n, then the system (18.1) has an inﬁnite number of solutions.   (ii) If r ð AÞ\r A  n, then the system (18.1) does not have any solutions. In this case, the system (18.1) is inconsistent.   The system (18.1) is consistent if and only if r ð AÞ ¼ r A , in which the necessary and sufﬁcient condition for the unique solution of the system (18.1) is   r ð AÞ ¼ r A