Design and Management of Production Systems - Practice Book 9788881327775

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TABLE OF CONTENT CHAPTER 1. PERFORMANCE MEASUREMENT E E E

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CHAPTER 2. FABRICATION SYSTEMS E E

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CHAPTER 3. ASSEMBLY SYSTEMS E E

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CHAPTER 4. DEMAND FORECASTING E E

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CHAPTER 5. INVENTORY MANAGEMENT E E

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CHAPTER 6. AGGREGATE PRODUCTION PLANNING E

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CHAPTER 7. ON-DEMAND MANAGEMENT (MRP) E E E

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CHAPTER 8. SCHEDULING E

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CHAPTER 9. EXAMS’ EXAMPLES

2017) 2010)

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Chapter 1. Performance Measurement

Exercise 1 The turning department of Alpo Ltd operates 200 days/year with 2 shifts/day, from Monday to Friday, with 4 old manual lathes. Given the characteristics of the machines, an operator for each machine is strictly necessary for the whole duration of the shift (hours/shift). During the annual summer closure of the plant, the management decided to develop a performance monitoring of the turning department, so to identify possible production bonuses for the whole department. For the evaluation of the performance the department manager provided the following data regarding plant shutdowns: Annual CAUSE entity [h] Break of mandrel n°4 15 Lack of information 12 General micro-shutdowns 10 General failure 40 Servomechanism waiting time 12 Substitution/grinding tools 250 Strikes 15 Topped up of components feeders 20 Product change 95 Absenteeism 30 Plant cleaning 128 Physiological stops 300 Maintenance wait 10 Man-machine interference 40 Break of refrigerant fluid for machine 3 50 Union assembly 16 Machine setting 78 Lack of energy 24 Coffee break 60 Samples 14 Break of motor machine 2 100 Lack of material machine-side 280 Lack of material 120 Inappropriate chemical-physical of material 9 Shift change 4 Utensils and equipment change 350 Break of machine 1 mandrel 250 Components changeover 85 Spare parts wait 8 Tests/ trials 23 Maintenance 80 Lunch break 1,000 Stand-by for another machine wait 250 TOTAL 3,778

The department manager also provided the total standard set up times, equal to 900 h/year - highly reliable data, established based on machines’ performance, and defectiveness coefficient, about 2% of the total units produced. As Alpo Ltd produces a wide range of products, the department manager provided the management with an indication of a standard production rate of equivalent product (representative of the whole production) of 50 pcs/h, comprehensive of the processing time only – i.e. set up times are not included,

and of an indication the total annual production volume measured in equivalent units stored in the warehouse (only good units) of 430,024 units per year. Based on the available data evaluate the productivity of the workforce and the performance indexes for the turning department, and decide whether to give the production bonus to the whole department or not.

Solution Exercise 1 The productivity of a production system can be evaluated, in general term as:

And, focusing on the productivity of the workforce, it can be expressed as:

The productivity can be decomposed in:

Where the utilization and efficiency factors become:

For the evaluation of the productivity of the workforce for the turning department, we thus need to properly evaluate the different statuses of the systems, relating each cause of shutdowns to a specific typology of cause. The different typologies can be identified as follows: TSc = strikes; TMm = idle time due to lack of materials; TMo = idle time due to lack of orders; TO = organizational causes; TG = idle time due to breakdown; TM = idle time due to maintenance; TPr = time for tests/ trials; TS = time of set up. CAUSE Break of mandrel n°4 Lack of information General micro-shutdowns General failure Servomechanism waiting time Substitution/grinding tools Strikes Topped up of components feeders

Annual entity [h] 15 12 10 40 12 250 15 20

Cause’s type TG TO TS TG TO TS TSc TS

Product change Absenteeism Plant cleaning Physiological stops Maintenance wait Man-machine interference Break of refrigerant fluid for machine 3 Union assembly Machine setting Lack of energy Coffee break Samples Break of motor machine 2 Lack of material machine-side Lack of material Inappropriate chemical-physical of material Shift change Utensils and equipment change Break of machine 1 mandrel Components changeover Spare parts wait Tests/ trials Maintenance Lunch break Stand-by for another machine wait TOTAL

95 30 128 300 10 40

TS TO TS TO TM TO

50

TG

16 78 24 60 14 100 280 120

Tsc TS TMm TO TPr TG TMm TMm

9

TMm

4 350 250 85 8 23 80 1,000 250 3,778

TO TS TG TS TM TPr TM TO TO

Consequently:

It is now easy to evaluate times related to the different statuses of the turning department. Particularly, we need the Opening calendar Time, the Actual Production Time and the Actual Production Time in Std. hours. The starting point is the calculation of the Opening Calendar Time, based on the annual working calendar, the shift policy and the number of machines, that can be seen as a multiplicative coefficient for the available production hours.

For the evaluation of the Actual Production Time, since

We have to pass first from the definition of the Actual Utilization Time and then of the Theoretical Utilization Time since And

The Theoretical Utilization Time can be thus obtained subtracting from the Opening Calendar Time all the managerial and organizational causes (internal or external) that caused the shutdowns.

Breakdowns and maintenance are then considered for determining the Actual Utilization Time.

It is now possible to determine the Actual Production Time, which is the time actually dedicated to the production (both good units and scraps). It is obtained starting from the Actual Utilization Time subtracting times for tests/ trials.

If we then subtract set up times from the Actual Utilization Time, we can evaluate the Machining Time, in other words, the time used by the machines for producing the products:

As it can be observed, the effect of the managerial and organizational causes is more relevant than the one related to technological causes. We can now proceed with the evaluation of the productivity of the turning department. As said before, productivity can be break up in utilization and efficiency factors, which would allow having a better understanding of the contribution of each productive factor. The computation for utilization is quite immediate:

Regarding efficiency, the concept of standard hour results of fundamental importance. The standard time is the time requested by an operation according to a fixed reference point. Formally, workforce efficiency is evaluated as follow:

Where: = Standard processing time to produce good products; Standard processing time to produce scraps; = Standard set up time; = Actual production time to produce good products; = Actual production time to produce scraps; = Actual set up time. As the denominator has been previously evaluated, we now focus the attention on the numerator, for which it is necessary to compute the standard production time and the standard set up time for all the products. In the present case, the operation is simplified since the production mix has been already expressed into equivalent units. Hence, it is necessary only to evaluate the standard production time needed to produce both good units and scraps, considering the total volume produced and the standard production rate RS. As, in general terms:

We need to identify the total Q produced as the sum of QB (good units) and QS (scraps). Given

It is possible to compute the number of scraps as:

From that, it is possible to determine the standard production time both for good products and scraps.

Numerically:

Hence, the numerator of the workforce efficiency formula is:

Having used as the 900 h/year estimated by the department manager. The efficiency thus results:

Finally, the productivity can be computed as:

Or Other performance can be easily derived from the statuses of the system previously evaluated. These are the Availability and the Saturation:

Availability is high, meaning that failures’ effect is not particularly relevant. Machines’ saturation is also high, indicating in the specific that set up times do not take more than 10% of the production time. Deciding whether to give the production bonus to the whole department or not, we should evaluate how well the workforce is performing. The productivity value for the department is good, but not particularly high. The main cause can be identified in the U, which represents the impact of external causes on the production and it includes both organizational and managerial causes, as well as technical causes. The efficiency, which represents the performance of the productive factor (in this case workforce) compared to the standard, is slightly lower than 100%. The deviation from the standard is related both to the increase in the set-up times (900 h standard, 1,016 h actual) and to the increase in the actual machining times. As for the latter point, considering the produced volumes (both good units and scraps) and the standard times:

Compared with an actual machining time of 9,022 . Consequently, unless further indications (as new hires that would possibly reduce the actual processing time or increase the actual set up time due

to their inexperience), the management should not give any production bonus to the turning department.

Exercise 2 ACIB Ltd is a leader in the packaging sector. ACIB owns an old (but properly working) labelling machine. Historically, Code 5 has been the reference product for the company. Nevertheless, during the years, different products with different production rates have been included in the production, so that ACIB Ltd is struggling to have a proper view on the mix potentiality of the machine (mix production rate). Assuming the demand of the 20 codes currently produced by the labelling machine as constant (see table), please evaluate the mix potentiality and the mix potentiality expressed in equivalent units of Code 5. QB [pcs/y] Code 1 Code 2 Code 3 Code 4 Code 5 Code 6 Code 7 Code 8 Code 9 Code 10 Code 11 Code 12 Code 13 Code 14 Code 15 Code 16 Code 17 Code 18 Code 19 Code 20

Def. coeff. n° setup for code

RS [pcs/h]

T std setup [h/setup]

240,508

0.025

3

6,000

15

249,004

0.062

2

9,800

10

111,130

0.085

4

7,300

10

3,452,340

0.039

12

5,500

11

12,312

0.065

4

10,000

8

23,450

0.033

5

7,000

12

123,121

0.037

4

4,200

9

34,520

0.050

4

7,200

15

123,131

0.055

5

7,500

9

34,442

0.033

5

8,400

14

1,266,985

0.033

3

7,800

9

65,788

0.077

5

7,700

15

59,907

0.028

2

6,800

13

457,843

0.030

2

8,100

9

3,674,595

0.068

4

9,000

12

23,472

0.003

5

3,000

11

675,328

0.013

3

3,400

13

214,232

0.046

2

4,800

14

5,349,221

0.100

5

8,800

15

549,221

0.005

4

9,800

15

Solution Exercise 2 Standard mix potentiality

For the evaluation of the mix potentiality, the following formula is used:

Where: i = index of code i-th; n = Maximum number of codes considered; = Good units stored of code i-th; = Scraps stored of code i-th; = Total machining time necessary for the production (h/year); = Total standard set up time necessary for the production (h/year); The data related to and for each code are reported in the table. The for the different codes can be easily derived from the available data (related to the defectiveness coefficient). Indeed, can be easily evaluated as follows:

where is the defectiveness coefficient of every i-th code. The following numerical values are thus obtained[1]: Code 1 Code 2 Code 3 Code 4 Code 5 Code 6 Code 7 Code 8 Code 9 Code 10 Code 11 Code 12 Code 13

QB [pcs/y]

Def. coeff.

QS [pcs/y]

240,508

0.025

6,166.872

249,004

0.062

16,458.687

111,130

0.085

10,323.552

3,452,340

0.039

140,105.369

12,312

0.065

855.914

23,450

0.033

800.259

123,121

0.037

4,730.506

34,520

0.050

1,816.842

123,131

0.055

7,166.354

34,442

0.033

1,175.373

1,266,985

0.033

43,237.337

65,788

0.077

5,488.273

59,907

0.028

1,725.716

Code 14 Code 15 Code 16 Code 17 Code 18 Code 19 Code 20

457,843

0.030

14,160.093

3,674,595

0.068

268,103.498

23,472

0.003

70.628

675,328

0.013

8,894.898

214,232

0.046

10,329.845

5,349,221

0.100

594,357.889

549,221

0.005

2,759.905

As for the times, for the total standard set up time times related to the different codes, that is:

, it is necessary to sum all the standard set up

Where: = Number of set up related to the production of the code i-th [setup/year]; = Set up time for code i-th [h/setup]; k = Number of codes considered. Hence,

could be quantified as follows:

Starting from the pieces actually produced (both good ones and scraps), it is possible to calculate the standard processing time for the production of i-th code:

That, with numbers, is: Code 1 Code 2 Code 3 Code 4 Code 5 Code 6 Code 7 Code 8 Code

QB [pcs/y]

QS [pcs/y]

Qtot [pcs/y]

RS [pcs/h]

Ttr std [h/y]

240,508

6,166.872

246,674.872

6,000

41.11

249,004

16,458.687

265,462.687

9,800

27.09

111,130

10,323.552

121,453.552

7,300

16.64

3,452,340 140,105.369 3,592,445.369

5,500

653.17

12,312

855.914

13,167.914

10,000

1.32

23,450

800.259

24,250.259

7,000

3.46

123,121

4,730.506

127,851.506

4,200

30.44

34,520 123,131

1,816.842 7,166.354

36,336.842 130,297.354

7,200 7,500

5.05 17.37

9 Code 10 Code 11 Code 12 Code 13 Code 14 Code 15 Code 16 Code 17 Code 18 Code 19 Code 20

34,442 1,266,985

1,175.373

35,617.373

8,400

4.24

43,237.337 1,310,222.337

7,800

167.98

65,788

5,488.273

71,276.273

7,700

9.26

59,907

1,725.716

61,632.716

6,800

9.06

457,843

14,160.093

472,003.093

8,100

58.27

3,674,595 268,103.498 3,942,698.498

9,000

438.08

23,472

70.628

23,542.628

3,000

7.85

675,328

8,894.898

684,222.898

3,400

201.24

214,232

10,329.845

224,561.845

4,800

46.78

5,349,221 594,357.889 5,943,578.889

8,800

675.41

9,800

56.32

549,221

2,759.905

551,980.905

It is now possible to calculate the standard Pmix of the labelling machine.

Standard mix potentiality expressed in equivalent unit The procedure for the evaluation of the standard mix potentiality for the labelling machine expressed in equivalent units of Code 5 is not particularly different from the one previously implemented, but volumes of each code must be expressed in equivalent units of Code 5. Considering the standard production time, the production of a quantity of Code A at a standard production rate

is equivalent (from a productive resource point of view) to produce a quantity

Code B at a standard production rate

. Consequently,

of

in equivalent units of Code A, results:

Proceeding in this way for all the codes, it is possible to calculate the units of product i-th in equivalent units of Code 5 (which has a standard production rate of 10,000 pcs/h).

Code 1 Code 2 Code 3

QB [pcs/y] 240,508 249,004 111,130

RS [pcs/h] 6,000 9,800 7,300

QB eq. Prod. 5 [pcs eq./y] 400,846.667 254,085.714 152,232.877

Code Code Code Code Code Code Code 10 Code 11 Code 12 Code 13 Code 14 Code 15 Code 16 Code 17 Code 18 Code 19 Code 20

4 3,452,340 5,500 6,276,981.818 5 12,312 10,000 12,312.000 6 23,450 7,000 33,500.000 7 123,121 4,200 293,145.238 8 34,520 7,200 47,944.444 9 123,131 7,500 164,174.667 34,442 1,266,985

8,400

41,002.381

7,800 1,624,339.744

65,788

7,700

85,438.961

59,907

6,800

88,098.529

457,843

8,100

565,238.272

3,674,595 23,472

9,000 4,082,883.333 3,000

78,240.000

675,328

3,400 1,986,258.824

214,232

4,800

5,349,221 549,221

446,316.667

8,800 6,078,660.227 9,800

560,429.592

It is possible now to proceed as done previously and recalculate the different . The computation of the total standard processing times does not change, as well as the standard set up time. The following table reports the results: QB eq. [pcs/y]

QS eq. [pcs/y]

Code 1 400,846.667 10,278.120 Code 2 254,085.714 16,794.578 Code 3 152,232.877 14,141.852 Code 4 6,276,981.818 254,737.035 Code 5 12,312.000 855.914 Code 6 33,500.000 1,143.226 Code 7 293,145.238 11,263.109 Code 8 47,944.444 2,523.392 Code 9 164,174.667 9,555.139 Code 10 41,002.381 1,399.254 Code 11 1,624,339.744 55,432.483 Code 12 85,438.961 7,127.627

QTOT eq. [pcs/y] 411,124.786 270,880.292 166,374.729 6,531,718.853 13,167.914 34,643.226 304,408.347 50,467.836 173,729.806 42,401.635 1,679,772.227 92,566.588

Code 88,098.529 2,537.818 90,636.347 13 Code 14 565,238.272 17,481.596 582,719.868 Code 15 4,082,883.333 297,892.775 4,380,776.109 Code 16 78,240.000 235.426 78,475.426 Code 17 1,986,258.824 26,161.464 2,012,420.287 Code 18 446,316.667 21,520.510 467,837.177 Code 19 6,078,660.227 675,406.692 6,754,066.919 Code 20 560,429.592 2,816.229 563,245.821 TOTAL 24,701,434.195

The standard mix potentiality expressed in equivalent unit thus:

Exercise 3 The turning department of Alpo Ltd operates 200 days/year with 2 shifts/day, 8 hours/shift, from Monday to Friday, with 4 lathes. During the annual summer closure of the plant, the management decided to develop a performance monitoring of the turning department. For the evaluation of the performance the department manager provided the following data regarding plant shutdowns: Annual entity CAUSE [h] Break of mandrel n°4 15 General micro-shutdowns 10 General failure 40 Substitution/grinding tools 250 Strikes 15 Topped up of components feeders 20 Product change 95 Plant cleaning 128 Maintenance wait 10 Break of refrigerant fluid for machine 3 50 Union assembly 16 Machine setting 78 Lack of energy 24 Samples 14 Break of motor machine 2 100 Lack of material machine-side 280 Lack of material 120 Inappropriate chemical-physical of material 9 Utensils and equipment change 350 Break of machine 1 mandrel 250 Components changeover 85 Spare parts wait 8 Tests/ trials 23 Maintenance 80 TOTAL 2,070

The department manager also provided the total standard set up times, equal to 900 h/year - highly reliable data, established based on machines’ performance, and defectiveness coefficient, about 2% of the total units produced. As Alpo Ltd produces a wide range of products, the department manager, provided the management with an indication of a standard production rate of equivalent product (representative of the whole production) of 50 pcs/h, comprehensive of the only processing time only – i.e. set up times are not included, and an indication of the total annual production volume measured in equivalent units stored in the warehouse (only good units) of 430,024 units per year. Based on the available data, you are asked to evaluate the OEE.

Solution Exercise 3 OEE evaluates the Overall Effectiveness of an Equipment. OEE is computed by multiplying the three OEE factors: Availability, Quality and Performance:

A-Availability considers the impacts of down time loss:

Q-Quality considers quality losses (expressed with std times):

P-Performance considers speed losses:

We can identify four different times:

Where: T= opening calendar time TSc = strikes; TMm = idle time due to lack of materials; TMo = idle time due to lack of orders; TO = organizational causes; TG = idle time due to breakdown; TM = idle time due to maintenance; TPr = time for tests/ trials; TS = time of set up; = standard time for the production of a good piece; = standard time for the production of a scrap. It becomes easy to express the OEE factors as follows:

For properly measuring the OEE, it is necessary to assign each cause of shutdowns to a specific typology of cause. Annual entity Cause’s [h] type TG 15

CAUSE Break of mandrel n°4 General micro-shutdowns

10

TS

General failure

40

TG

250

TS

Strikes

15

TSC

Topped up of components feeders

20

TS

Product change

95

TS

128

TS

10

TM

50

TG

16

TSC

Machine setting

78

TS

Lack of energy

24

TMm

Samples

14

TPr

Substitution/grinding tools

Plant cleaning Maintenance wait Break of refrigerant machine 3 Union assembly

fluid

for

100

TG

Lack of material machine-side

280

TMm

Lack of material

120

TMm

Break of motor machine 2

Inappropriate chemical-physical of material Utensils and equipment change

9

TMm

350

TS

Break of machine 1 mandrel

250

TG

Components changeover Spare parts wait Tests/ trials Maintenance TOTAL

It is now easy to evaluate times:

85

TS

8

TM

23

TPr

80

TM

2,070

For the computation of the Net Operating Time and Fully Productive Time, it is necessary to evaluate the standard production time for all the products. In the present case, the operation is simplified since the production mix has been already expressed into equivalent units. Hence, it is necessary only to refer to the standard production rate of the equivalent product RS, the good units produced and the defectiveness coefficient.

It is now possible to evaluate the OEE:

That is equal to:

Exercise 4 (Exam 4th March 2010) Volumes Annual Popi is a small firm located in Bergamo producing Standard last quarter Demand products pressed in metal sheets with two main Press Code Rate of 2009 Forecast different product lines: silencers for the automotive [pcs/h] [pcs] 2010 [pcs] sector and panels for boilers. The productive 1 1 1,000 80,000 320,000 process is partly the same for the two products, at 1 2 500 200,000 800,000 least the first phase: the metal reel after being 1 3 1,000 250,000 1,000,000 mounted on a specific support, is moulded in the 2 4 3,000 200,000 1,500,000 printing press department. Later, the product is 2 5 4,000 600,000 2,500,000 transferred to a department for further operations 2 6 5,000 700,000 3,000,000 such as heat treatment (mufflers) or mechanical 2 7 2,000 120,000 1,000,000 ones (drilling for boilers panels). Regarding the printing press department, two presses are present, namely P1, recently acquired, and P2. The department operates three shifts a day, 8 hours per shift, 5 days a week, with a total of 12 operators coordinated by 3 leaders (one for each shift), who are in charge of only monitoring the operations on the machines and change the plate. The average standard set up time, evaluated on the previous year, is 3.5 h/setup. The average number of setups by quarter conducted on the two presses is equal to 20 setups for P1 and 10 for P2. Recently, a market study has been commissioned. The study has underlined several good scenarios for improvement in the year 2010 regarding a further development of the firm, since the possibility of growth in emerging markets and the well-known quality of the product. TABLE 1 Data of codes on the presses Causes P1 P2 [h] Considering this prospect, issues concerning the production [h] capacity emerged, and an analysis for the identification of the Lack of Personnel 2 65 assessment of the main production performance was requested. In Lack of Materials 400 Table 1 quarterly data are provided (70 working days, last quarter Break of the water 70 2009) with reference to two presses: P2 has been operating for circuit several bbbbbTABLE 2 Non-productive times years, while P1 Breakdown of the 90 has been recently installed and bbbbb mill standard production time just defined. Maintenance 97 20 Meeting for 20 Specific plates are installed on each press, so that a code produced Improvement on a press cannot be produced on the other one. In the same Tests/ trials 45 45 analysis, the main causes of unavailability for the two presses for Samples 75 63 the same period have been also identified (Table 2). Based on the available data it is requested to: 1) Determine the production capacity of the two presses in the considered quarter (last quarter of 2009), pointing out Pmix (calculated for P1 with reference to Code 3, for P2 with reference to Code 7), Utilization and Efficiency; 2) Considering the demand forecast for the 2010, verify that the installed production capacity is sufficient for fulfilling the demand forecasts, pointing out Pmix (calculated for P1 with reference to Code 3, for P2 with reference to Code 7), Utilization and Efficiency; 3) Comment on the performance of the two presses (and related causes) and propose possible actions for improving them.

Solution Exercise 4 Production capacity of the two presses Production capacity is evaluated as follows: To proceed, we have to determine the contribution of each term.

Pmix The Pmix is used in multi-products situations, evaluating the ratio between the quantities of the different codes to produce and the time that is needed to produce them including the setup time and defining as the average number of units that can be produced per time unit.

The Pmix can be also expressed in terms of equivalent units of a reference product. We must define an equalizer coefficient in order to homologate the production of different products, i.e. to pass from quantities of different products to time. This can be done using the ration between the production rate of the two products. Indeed, since:

If we take the k-th product as the reference one:

Utilization, Efficiency, T Utilization and efficiency can be evaluated as follows:

While T represents the Opening Calendar Time. All of them can be evaluated starting from the statuses of the system. In particular, the Opening Calendar Time is the same for both the presses, while the Actual Production Time depends on the specific shutdowns for each press and the Actual Production Time in std. hours depends on the specific production on each press. For the evaluation of the Actual Production Time, since:

Where And It can be expressed as

Referring to the data reported in Table 2, we can specify for each shutdown the related type of cause: Causes Lack of Personnel Lack of Materials Break of the water circuit Breakdown of the mill Maintenance Meeting for Improvement Tests/ trials Samples

Type of causes T0 TMm TG

P1 [h]

P2 [h]

2

65 400 70

TG

90

TM TO

97

20 20

TPr TPr

45 75

45 63

For the evaluation of the Actual Production Time in std. hours, it is expressed as:

Where, in general terms:

We are given the standard production rate for each code, so that we can easily divide the volumes by the standard rate so to obtain the standard production time. Production capacity of P1 For P1 the reference product is Code 3, so that:

Given

The result is that for the Actual Production Time:

While for the Actual Production Time in std. hours:

Based on this:

Considering

=0 since no information is provided,

Production capacity of P2 For P2 the reference product is Code 7, so that:

Given

The result is that for the Actual Production Time:

While for the Actual Production Time in std. hours:

Based on this:

Considering

=0 since no information is provided,

Verification of the installed production capacity for the year 2010 We have to verify whether the installed capacity is able to meet the forecasted demand for the following year. If mix production rate (Pmix) and the plant opening time (T) are known, it is possible to estimate, for a future period, the production capacity (based on the hypothesis that in the future the impact of times is equal to the present one). Particularly: Pmix depends on the production mix: if enough data are provided, we can evaluate the future Pmix based on the forecasts of demand and, consequently, production; otherwise, we can use the current Pmix (approximation); T It depends on hexogen context and shift policy: if T is expected to remain the same or enough information about changes are provided, we can evaluate the future T; otherwise, we can use the current T (approximation); Utilization takes account of lost time for organizational issues, strikes, breakdown, maintenance, tests… etc. In the estimation of future Utilization, we usually consider the current scenario but TMo and TMm are considered equal to 0; Efficiency must be estimated on the basis of production policies, machine manufacturers’ data and expertise. For the estimation we can: i) consider the current efficiency; ii) evaluate a new efficiency where TMo and TMm are considered equal to 0; iii) evaluate a new efficiency based on improved standard times. The procedure is the same as the one followed in the previous section, but it is carried out for the entire year and not for only a quarter. We assume that all the data referred to the specific quarter will be replicated equally in the other quarters of the year. Production capacity of P1

As TMo and TMm for P1 in the year 2009 are already equal to 0, the Utilization and the Efficiency of the year 2010 will be the same if the year 2009.

We compared the CP obtained with the total number of equivalent units that should be processed on P1, which is:

The CP of P1 is thus enough to meet the production of the forecasted demand. Production capacity of P2

Given the total number of equivalent units that should be processed on P2, which is:

If

The CP of P2 is enough to meet the production of the forecasted demand. If

The CP of P2 is enough to meet the production of the forecasted demand. Comments Comments on P1 The utilization of P1 is not particularly high, and this is due mainly to the impact of breakdowns and maintenance. The external and organizational causes have a very low impact (only 2 h in total), so that

the Theoretical Utilization Time and Actual Utilization Time correspond. As for the estimation of efficiency, since the machine is new, the standard times provided for 2009 have not been developed on historical data but based on desk analysis. As historical data becomes available, Popi should better investigate the inefficiency causes. Comments on P2 The utilization of P2 is quite low: this time the impact of breakdowns and maintenance is lower, but we have highly impacting managerial issues. More in detail, the impact of the lack of materials is rather huge (400h out of 1680 h available). Hoping to be able to tackle this issue, the utilization in 2010 is expecting to increase considerably. As for the efficiency, the elimination of the managerial problems will help to increase the value. Nevertheless, the efficiency is very low both in the year 2009 and in the estimation for the year 2010. Popi should better investigate the inefficiency causes, also focusing on the setting standard time and why they are not met.

Exercise 5 (Exam 29th June 2010) MC Ltd produces high quality futsal shoes. Starting from the sales for year 2009, the sales manager, Mr Varo, proposed to the management to stop the production of Product 2. Mr Dona, newly hired in the logistic department, was appointed to evaluate this proposal, based on available data regarding sales and stock holding for every product. You are required to evaluate the proposal of Mr Varo (use 50% and 90% as thresholds) and make some remarks (correlated by hypothesis, if needed) regarding actions to be undertaken. Turnover (M€) Inventory (M€)

1

2

3

4

5

6

7

8

9

33

3

28

2

4

9

32

11

1

28

21

17

5

35

9

4

14

2

Solution Exercise 5 To evaluate the proposal of Mr Varo and understand the right decision to be taken we can apply the ABC-ABC analysis. This analysis, applied to the stock management, is useful to evaluate the rightness of the stock compared to the consumption, so to set appropriate management policies. In the ABC-ABC analysis, an ABC analysis is applied distinctly to the consumption value and stock value, and the classifications obtained are then merged. For the ABC-ABC analysis to be performed, both the consumption and the stock should be expressed in monetary value. Both for the stock value and the consumption value, we proceed according to the following steps: We order the items in decreasing order with reference to the percentage of their impact on the totality of the phenomenon; We evaluate the cumulative distribution; We assign the class; We carry out this operation both for the stock value and the consumption value; We cross the classifications, obtaining a matrix. Turnover ABC Analysis We follow the procedure described above. The thresholds provided by Mr Varo are 50% and 90% (placing each product in the class where it develops most of its value). Product 1 2 3 4 5 6 7 8 9

Turnover (M€) 33 3 28 2 4 9 32 11 1 123

Product 1 7 3 8 6 5 2 4 9

Turnover (M€) 33 32 28 11 9 4 3 2 1

% 26.829 % 26.016 % 22.764 % 8.943 % 7.317 % 3.252 % 2.439 % 1.626 % 0.813 %

Cumulative % 26.829 % 52.845 % 75.591 % 84.534 % 91.851 % 95.103 % 97.542 % 99.168 % 100 %

Class A A B B B C C C C

Inventory ABC Analysis We follow the procedure described above. The thresholds provided by Mr Varo are 50% and 90% (placing each product in the class where it develops most of its value). Product Inventory (M€) 1 28 2 21

Product Inventory (M€) % Cumulative % Class 5 35 25.956 % 25.956 % A 1 28 20.741 % 46.697 % A

3 4 5 6 7 8 9

17 5 35 9 4 14 2 135

2 3 8 6 4 7 9

21 17 14 9 5 4 2

15.556 % 12.593 % 10.370 % 6.667 % 3.704 % 2.963 % 1.481 %

62.253 % 74.846 % 85.216 % 91.883 % 95.587 % 98.550 % 100 %

B B B B C C C

ABC-ABC Analysis Crossing the classifications, we obtain the following matrix. Turnover B

A A

1

Inventory B C

5 3, 8, 6

7

C

2 4,9

Product 2 is positioned in the Inventory B and Turnover C cell. The products in this cell are characterized by irregular demand. We should evaluate the opportunity to move Product 2 in one of the adjacent cells (CC, BB, AC). Nevertheless, for this analysis we need to know in what lifecycle phase is Product 2. It’s rather likely, given the low turnover and the corresponding higher level of inventory, that Product 2 is an old product or is not selling as expected. It seems thus a reasonable suggestion to stop the production of Product 2.

Exercise 6 (Exam 06th February 2018) A firm wants to evaluate the performance of its warehouse. Data related to the annual consumption (thousands of pieces), to the annual stock level (thousands of pieces) and production cost (€/piece) for all the products (from 1 to 8) are provided. 1. Determine the distribution of products using ABC-ABC analysis (consumption and stock value). Use 50% and 75% as thresholds (place products where they develop the most part of their value); 2. Comment on the two most critical products.

Product 1 2 3 4 5 6 7 8

Average stock Unit Consumption level (k- cost (k- pieces) pieces) (€/pc) 35 35 60 50 25 40 30 30 50 40 20 60 70 90 25 35 25 65 40 55 35 40 80 30

Information Mature New Obsolete New Mature JIT Spare part Spare part

Solution Exercise 6 ABC-ABC Analysis For the ABC-ABC analysis to be performed, both the consumption and the stock should be expressed in monetary value. In the present case, consumption and average stock level are expressed in pieces. However, the monetary value for both consumption and stock level can be obtained multiplying the consumption and the stock level expressed in pieces by the unitary cost of each piece. Product 1 2 3 4 5 6 7 8

Consumption (k- pieces) 35 50 30 40 70 35 40 40

Average stock level (k- pieces) 35 25 30 20 90 25 55 80

Unit (€/pc) 60 40 50 60 25 65 35 30

cost

Consumption (k€) 2,100 2,000 1,500 2,400 1,750 2,275 1,400 1,200

Average stock (k€) 2,100 1,000 1,500 1,200 2,250 1,625 1,925 2,400

Both for the stock value and the consumption value, we proceed according to the following steps: We order the items in decreasing order with reference to the percentage of its impact on the totality of the phenomenon; We evaluate the cumulative distribution; We assign the class – in the specific case using 50% and 75% as thresholds (placing products where they develop the most part of their value); We carry out this operation both for the stock value and the consumption value; We cross the classifications, obtaining a matrix.

Consumption Product Consumption (k€) 1 2,100 2

2,000

3

1,500

4

2,400

5

1,750

6

2,275

7

1,400

8

1,200

Product Consumption % (k€) 4 2,400 16.410 % 6 2,275 15.556 % 1 2,100 14.359 % 2 2,000 13.67 5% 5 1,750 11.966 % 3 1,500 10.256 % 7 1,400 9.573 % 8 1,200 8.205 % 14,625

Cumulative % 16.410 %

Class

31.966 %

A

46.355 %

A

60.030 %

B

71.996 %

B

82.252 %

C

91.825 %

C

100%

C

A

Stock Product Stock (k€)

Product Stock (k€)

%

1

2,100

8

2,400

2

1,000

5

2,250

3

1,500

1

2,100

4

1,200

7

1,925

5

2,250

6

1,625

6

1,625

3

1,500

7 8

1,925 2,400

4 2

1,200 1,000 14,000

17.143 % 16.071 % 15,000 % 13.750 % 11.607 % 10.714 % 8.571 % 7.143 %

Class

33.214 %

A

48.214 %

A

61.964 %

B

73.571 %

B

84.285 %

C

92.856 % 100 %

C C

A

Consumption B C

A

Stock

Cumulative % 17.143 %

A

1

B

6

C

4

5

8 7

2

3

Comment on the two most critical products The most critical products are those products for which stock and consumption appear not right, also considering the information provided regarding the product in terms of lifecycle’s phases or the way production is management. The most critical products thus are:

Product 4: according to the information provided Product 4 is a new product; we should thus expect consumption to increase, but the stock level is already too low. There is quite a substantial risk of stock-out and the firm should increase the level of stock as soon as possible. Product 6: according to the information provided it should be managed with a JIT logic; nevertheless, instead of being placed in the cell CA (the appropriate one for JIT products), we find Product 6 in cell BA. This means that the current stock level of Product 6 is too high for the production logic in which the product is managed.

Chapter 2. Fabrication Systems

Exercise 1 Vitine Ltd operates exclusively in the manufacture of products for the furniture industry. Three main items are produced (codified as item-1, item-2 and item-3), according to the following production cycle: Item-1: cutting of the ribbon according to the desired shape, pressing, polishing; Item-2: pressing, cutting and tumbling; Item-3: pressing, polishing and tumbling. Vitine Ltd operates on a daily calendar of two shifts per day, 8 hours per shift. The production department is currently organized as a Job Shop. The production in the third shift, activated several years ago, has been definitely abandoned due to several strikes that had significantly slowed the production rate of the plant. Since the firm adopted the two shifts policy, the strike percentage has dropped to just 1%. Items’ sales forecasts and working calendar for 2009 are shown below: Demand for the Working period Item-1 Item-2 Item-3 days January 8,500,000 27,000,000 45,500,000 20 February 9,000,000 19,000,000 32,200,000 19 March 10,500,000 23,500,000 37,800,000 21 April 10,000,000 20,500,000 53,900,000 18 May 11,500,000 39,000,000 44,800,000 21 June 12,500,000 31,000,000 38,500,000 20 July 10,750,000 36,500,000 51,100,000 21 August 4,500,000 17,500,000 28,000,000 10 September 8,750,000 23,500,000 44,800,000 20 October 10,250,000 31,000,000 42,700,000 21 November 10,750,000 24,000,000 29,400,000 21 December 8,750,000 35,000,000 39,200,000 18 Global Demand 115,750,000 327,500,000 487,900,000 230

Time losses due to a non-optimal management of the production system (Lack of orders, Lack of material, Organizational time) affect globally the plant opening time for the 15%. As for machineries, the following data related to availability are accessible: Machineries availability Cutting Pressing Shearing Polishing Tumbling

0.80 0.95 0.90 0.85 0.90

The type of operations and the age of technologies (few automated machines) require the of workforce for the implementation of operations, as it can be inferred from the following table: Impact of human factor of performance Cutting 0.92 Pressing 0.95

Shearing Polishing Tumbling

0.91 0.76 0.94

The production of the items is quite standard, so that tests/ trials are not largely required. Nevertheless, for polishing activities the company is always trying to find new solutions, so to keep its market leadership. In the following table data about the tests/ trials related coefficient are reported: Impact of tests/ trials Cutting 0.95 Pressing 0.97 Shearing 0.99 Polishing 0.93 Tumbling 0.96

Cleaning and set-up operations are managed in the following way: Lot Size [pc/lot] Cutting Pressing Shearing Polishing Tumbling

Item-1 50,000 70,000

Item-2

Item-3

50,000 40,000

50,000

40,000 30,000

Set-up Time [s/lot] Item-1 Cutting 40 Pressing 25 Shearing Polishing 30 Tumbling

50,000 50,000

Item-2

Item-3

35 50

35 30 35

25

The processing time (excluded set-up time) for each piece is reported in the following table: Processing Time [sec/pc] Cutting Pressing Shearing Polishing Tumbling

Item1 0.10 0.05

Item2

Item3

0.07 0.05

0.01

0.05 0.05

0.04 0.06

The firm currently has 3 machineries for cutting, 6 for pressing, 3 for shearing, 5 for polishing and 7 tumbler machines for the removal of the remaining burrs; in the following table, the defectiveness rates for the different technologies are reported. Defectiveness Item- Itemrate 1 2

Item3

Cutting Pressing Shearing Polishing Tumbling

0.14 0.10

0.15 0.20

0.16 0.05

0.05 0.13 0.10

Based on forecasted sales volumes, evaluate the sizing of the production system and, if needed, indicate in which production unit it would be necessary to make changes and improvements.

Solution Exercise 1 The Job Shop sizing starts from the identification of the production cycle for each item, in terms of technologies. It can be summarized as follow: Operations Cutting Pressing Shearing Polishing Tumbling

Item- Item- Item1 2 3 x x x x x x x x x

Once identified the production cycle and the related technologies used for the production, it is necessary to evaluate the needed hours (for each technology) for fulfilling the forecasted demand for the following year, in terms of processing time and set-up time. Such estimation should be then revised considering factors that may increase the needed hours for production, such as yield, equipment availability, production time used for tests/ trials and type of operation (whether it is influenced by human factor or not). Hence, it is necessary to calculate:

Where: = Available hours in a year for production for each technology i-th (where index “i” represents the different type of operations: cutting, pressing, shearing, polishing, tumbling) [h/year]; = Processing time related to the technology i-th, for the production of one unit of product j-th (where index “j” represents items to be produced: item-1, item-2, item-3) [s/pcs]; = Product quantity to produce for item j-th [pcs/year]; = Defectiveness rate for product j-th on technology i-th; = Defectiveness rate for product j-th on technology k-th, where operations on technology kth are followers with respect to operations on technology i-th; = Setup time necessary for starting the production of item j-th on the technology i-th [s/lot]; = Annual number of lots of product j-th on technology i-th; = Availability of technology i-th; = Coefficient considering the impact of the human factor on performance Coefficient considering the impact of tests/ trials. As all the other necessary data are provided, it is necessary to evaluate the number of annual lots for the production of product j-th on technology i-th, as follows:

Where is the size of the production lot of the product j-th on the technology i-th. For example, considering the annual quantity of item-1 to be produced (115,750,000) and the size of the production lot of item-1 on cutting machines (50,000 pcs/lot), we have:

By proceeding in the same way for the other items and technologies: Number of set-ups [lots/year] Item-1 Item-2 Item-3 Cutting 2,315.00 Pressing 1,653.57 6,550.00 9,758.00 Shearing 8,187.50 Polishing 2,893.75 9,758.00 Tumbling 10,916.67 9,758.00

Since the request is to conduct a preliminary verification about the current state of the production system, and the company is interested in getting an understanding of the number of set-ups necessary for fulfill the production, values should not be rounded, considering also some decimal points. Is it possible now to proceed with the computation of the annual production hours needed (only production time, yield included) for each item on each technology. As an example, the calculation for item-1 on the cutting technology is reported:

Production hours needed (scraps included) [h/year] Item-1 Item-2 Item-3 Cutting 4,945.36 Pressing 2,126.51 9,857.67 1,821.98 Shearing 5,985.01 Polishing 1,913.86 6,923.51 Tumbling 4,788.01 9,035.19

Based on the number of annual lots and the unitary set-up time for the production of a single lot, it is possible to evaluate the annual hours to be used for the set up on the different technologies (in the following example, this procedure is implemented for cutting on item-1):

H set-ups needed [h/year] Cutting Pressing Shearing Polishing Tumbling

Item-1 25.72 11.48

Item-2

Item-3

63.68 113.72

94.87

24.11 75.81

81.32 94.87

Considering the following coefficients: Times for maintenance and for fixing failures (Availability of technologies) Times for test/ trial (Test/ trial coefficient) Dependence of operations from human related factors It is possible to evaluate the total needed hours for each technology: H needed for technology Item-1 Item-2 Item-3 Total [h/year] Cutting 7.109,68 - 7.109,68 Pressing 2.442,23 11.333,18 2.189,62 15.965,03 Shearing - 7.521,77 - 7.521,77 Polishing 3.225,76 - 11.659,56 14.885,32 Tumbling - 5.988,75 11.241,69 17.230,44

In order to evaluate the number of machines needed for each technology, it is essential to evaluate the available annual hours for production. Indeed, such hours, in comparison to the working calendar, would be reduced basing on shift policy, strikes and potential production hour lost for organizational causes. So: Where: i = Technology (i = 1,…,5); = Available hours in a year for production for each technology i-th (with ti shifts); = Opening hours in a year for production for each technology i-th (with ti shifts); = Coefficient considering time losses due to strikes; = Coefficient considering time losses due to non-optimal management of the production system. Since no further information is given, all the technologies are supposed to work on two shifts per day (8 hours/shift), according to the working calendar. Therefore, the available working hours are:

It is now very easy to compute the number of needed machines and the related saturation:

Cutting Pressing Shearing Polishing Tumbling

Theoretic Actual Actual number number saturation of of rate machines machines 2.30 3.00 67% 5.16 6.00 78% 2.43 3.00 72% 4.81 5.00 67% 5.56 6.00 83%

As the theoretical number of machines needed for the pressing is 5.15, in case of evaluation on an investment for modernization or for adding new machines, it would be necessary to figure out the convenience of having 5 or 6 machines. Anyhow, this is not the case, since 6 machines for the pressing are already available. Basing on the preliminary evaluation of the production system, and considering the annual demand, at first sight the production system does not seem to be too worrying:

Cutting Pressing Shearing Polishing Tumbling

Available Needed number number of of machines machines Δ 3 3 6 6 3 3 5 5 7 6 1

The result obtained are based on the analysis of the annual demand. Nevertheless, further considerations can be drawn looking at the monthly demand. The number of needed machines is expected to increase, since we are chasing the demand during the peak periods. Indeed, with the sizing looking at the annual demand, it is not certain that the production mix in a specific month would be perfectly fulfilled. Considering, for example, month 7 (July): H needed for technology in July Item- Item[h/month] 1 2 Cutting 660 Pressing 227 1,263

Item3

Total 660 229 1,719

Shearing Polishing Tumbling

838 300

838 1,221 1,521 667 1,177 1,845

From which it is possible to obtain the number of needed machines:

Comparing these results with the number of available machines in each department:

Cutting Pressing Shearing Polishing Tumbling

Available Needed machines machines Δ 3 3 6 7 -1 3 3 5 6 -1 7 7 -

As it can be easily inferred, the number of machines for pressing and polishing is not sufficient to fulfill the monthly demand of month 7. It is necessary for each technology to evaluate the most critical months. In general terms, the most critical months are those in which we have a peak of demand compared to the number of working days. In Table the critical months are highlighted.

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Total production for each technology/Working day Cutting Pressing Shearing Polishing Tumbling 425,000.00 4,050,000.00 1,350,000.00 2,700,000.00 3,625,000.00 473,684.21 3,168,421.05 1,000,000.00 2,168,421.05 2,694,736.84 500,000.00 3,419,047.62 1,119,047.62 2,300,000.00 2,919,047.62 555,55556 4,688,888.89 1,138,888.89 3,550,000.00 4,133,333.33 547,619.05 4,538,095.24 1,857,142.86 2,680,952.38 3,990,476.19 625,000.00 4,100,000.00 1,550,000.00 2,550,000.00 3,475,000.00 511,904.76 4,683,333.33 1,738,09524 2,945,238.10 4,171,428.57 450,000.00 5,000,000.00 1,750,000.00 3,250,000.00 4,550,000.00 437,500.00 3,852,500.00 1,175,000.00 2,677,500.00 3,415,000.00 488,095.24 3,997,619.05 1,476,190.48 2,521,428.57 3,509,523.81 511,904.76 3,054,761.90 1,142,857.14 1,911,904.76 2,542,857.14 486,111.11 4,608,333.33 1,944,444.44 2,663,888.89 4,122,222.22

The number of needed machines for each technology in the most critical months and the delta with the available number of machines for each technology of the firm is reported in the table below:

Needed Available number number of of Technology Month machines machines Δ Cutting Jun 3 3 Pressing Aug 7 6 1 Shearing Dec 4 3 1 Polishing Apr 7 5 2 8 7 Tumbling Aug 1

The saturation rate in the critical months can be calculated:

Technology Month Cutting Jun Pressing Aug Shearing Dec Polishing Apr Tumbling Aug

Needed number of Saturation machines rate 3 83% 7 80% 4 73% 7 65% 8 80%

It can also be computed which would be the saturation rate in the critical months on the available machines of the firm:

Technology Month Cutting Jun Pressing Aug Shearing May Polishing Apr Tumbling Aug

Available number of Saturation machines rate 3 83% 6 93% 3 98% 5 90% 7 91%

It can be observed that the saturation is higher than the threshold of 85% for every technology but the cutting. In the end, considering the sizing results obtained looking at the monthly demand, the following alternatives are possible: 1. Make an investment in additional machines in the department that need them (adding new machines or changing the current machines with newer and more performing ones); 2. Try to modify the monthly production planning and see whether the situation may be improved. However, this solution entails the involvement of the production management, and it would lead to an increasing in stock holding cost, or to turn to outsourcing activities; 3. Other more immediate solutions, as for example the increase of one shift or overtime shifts, do not seem to be viable, since the company has decided long time ago to abandon the adoption of the third shift (in order to avoid strikes).

Exercise 2 Given the following part routing matrix: 1 A B C D E F G H I J

2

3 1

1

4

5

6 1

1

7

8

1

1

9

10 1

1 1 1

1

1

1

1

11

12

1 1

1

1 1

1

1

1

1

1 1 1

Design manufacturing cells, using to the following algorithms: 1. Single Linkage Clustering algorithm, threshold of restrictive allocation equal or higher than 0.5 and similarity coefficient:

2. Rank Order Clustering Algorithm. Comment the results.

Solution Exercise 2 SLC algorithm The threshold for allocation is particularly loose, allowing to group together two technologies if each of them manufactures at least the 50% of the total items manufactured on the two technologies. Iteration 1 It is necessary to calculate the similarity coefficients to be inserted in the following matrix: A B C D E F G H I J

A 1

B

C

D

E

F

G

H

I

J

1 1 1 1 1 1 1 1 1

Of course, given the symmetry of the matrix, it is enough to compute coefficients only for half of it. The similarity coefficient is computed as follows (with i,j, = A,…,J): = Number of parts visit both machines i-th and j-th; = Number of parts visit machine i-th but not j-th; = Number of parts visit machine j-th but not i-th. For example, for machines A and E

Or, for machines B and E

Proceeding in the same way for the other machineries, we obtain the following matrix: A B C D E F G H I J

A 1 0 0 0 0.14 0 1 0 0 0

B

C

D

E

F

G

H

I

J

1 0 0 0.8 0 0 0 0 0

1 1 0 0 0 0.25 0 0

1 0 0 0 0.25 0 0

1 0 0.14 0 0 0

1 0 0.67 0 0

1 0 0 0

1 0 0

1 1

1

The bold numbers indicate situations in which the technological affinity threshold is met, and the two corresponding technologies will be grouped together. Iteration 2 Aggregating the technologies basing on the results of Iteration 1, is it possible to obtain the new part routing matrix: A+G B+E C+D F+H I+J

1 0 0 0 1 0

2 0 1 0 0 0

3 1 1 0 0 0

4 0 1 0 0 0

5 0 0 0 1 0

6 1 0 0 0 0

7 0 1 0 0 0

8 0 1 0 0 0

9 0 0 1 0 0

10 1 0 0 0 0

11 0 0 1 1 0

12 0 0 0 0 1

This matrix will be the new baseline for the computation of the similarity coefficients. Proceeding in the same way as in Iteration 1, we obtain:

A+G B+E C+D F+H I+J

A+G 1 0.14 0 0 0

B+E C+D F+H I+J 1 0 0 0

1 0.25 1 0 0

1

As none of the coefficients satisfy the similarity threshold required, we stop. The obtained results are the followings: CELLS TECHNOLOGIES ITEMS AAA A+G 3, 6, 10 2, 3 ,4 BBB B+E ,7, 8

CCC DDD EEE

C+D 9, 11 F+H 1, 5, 11 I+J 12

As it can be noticed, items 3 and 11 are shared between two cells, particularly: cells AAA and BBB share the production of item 3; cells CCC and DDD share the production of item 11. In this case, SLC algorithm required a limited number of iterations for obtaining group technology. The number of items for each cell is quite limited, except for cell BBB, that manufactures a significant number of items (2,3,4,7,8) and could presents problems related to products flows. It is then worth to underline that cell EEE, that includes technologies I and J, is designed to produce item 12 only. ROC algorithm The ROC algorithm directly groups technologies in the different cells, specifying the items manufactured in each cell identified. The procedure, depending on the complexity of the items-technologies matrix, could require a significant number of iterations and does not guarantee that the obtained cells present a reduced number of technologies compared to the starting one. The identification of the cells is made up by 6 steps: STEP 1: associate a binary number to each row; STEP 2: order rows (top down) by decreasing value of the related binary number; STEP 3: associate a binary number to each column; STEP 4: order columns (left to right) by decreasing value of the related binary number; STEP 5: If in steps 2 and 4 no shifts are made, go to step 6, otherwise go back to step 1; STEP 6: stop. Iteration 1 STEP 1 A binary number is associated to each row, assigning to each column a value as follows:

Where: i = column i-th considered (i =1 … k); k = numbers of items contained in the items-technologies matrix (12 in this exercise). Hence, the items-technologies matrix is: Row 1 2 3 4 5 6 7 8 9 10

A B C D E F G H I J

2,048 1,024 512 1 2 3 1 1

256 4

128 5

64 6 1

1

32 7

16 8

1

1

8 9

4 10 1

1 1 1

1

1

1

1

1 12

1 1

1

1 1

1

2 11

1 1

1 1 1 1

A univocal score for each row, based on the technologies that manufacture each item:

+

+

+

+

+

+

+

+1

+

+ STEP 2 Starting from the obtained scores, rows are ordered according to decreasing values (top down). The order of reallocation of the rows is: h-f-e-b-a-g-c-d-i-j - there would have not been any difference in inverting the order of rows with the same score, as A-G, C-D, I-J. Thus, the obtained items-technologies matrix is: H F E B A G C D I J

1 1 1

2

3

4

1 1

1

1 1

1 1

5 1 1

6

7

8

1 1

1 1

9

1 1

10

11 1

12

1 1 1 1

1 1 1 1

STEP 3 A binary number is associated to each column, assigning to each row a value as follows:

Where: i = Row i-th considered (i =1 … k);

t = numbers of technologies contained in the items-technologies matrix (10 in this exercise). The items-technologies matrix is then 512 256 128 64 32 16 8 4 2 1

Column 1 H 1 F 1 E B A G C D I J

2

3

4

1 1

1

1 1

1 1

5 1 1

6

7

8

1 1

1 1

9

1 1

10

11 1

12

1 1 1 1

1 1 1 1

A univocal score for each column, based on the technologies that manufacture each item:

STEP 4 Starting from the obtained scores, columns are order according to decreasing values (left to right). Thus, the order of reallocation of the rows is: 1-5-11-2-4-7-8-3-6-10-9-12 (there is no difference between rows that have the same scores, like 1-5, 2-7-8). Hence, a new items-technologies matrix is obtained: H F E B A G C D I J

1 1 1

5 1 1

11 1

2

4

7

8

3

1 1

1 1

1 1

1 1

1 1 1

6

10

1 1

1 1

9

1 1

12

1 1 1 1

STEP 5 Since in step 2 and 4 changes have been applied on the rows/columns order, it is necessary to implement again the algorithm starting from step 1. In the following, only the results of each step are reported. Iteration 2 STEP 1 bis H F E B A G C D I J

1 1 1

5 1 1

11 1

2

4

1 1

7

1 1

8

1 1

3

1 1

6

10

9

12 3,584 3,072 496 480 28 28 514 514 1 1

1 1 1

1 1

1 1

1 1

1 1 1 1

STEP 2 bis H F C D E B A G I J

1 1 1

5 1 1

11 1

2

4

7

8

3

6

10

1 1

9

12

1 1 1 1

1 1

1 1

1 1

1 1 1

1 1

1 1 1 1

STEP 3 bis H F C D E B A G I J

1 1 1

5 1 1

11 1

2

4

7

8

3

6

10

1 1 1 1

1 1

1 1

1 1 1

768

12

1 1 1 1

768

9

1 1

1 1

704

48

48

48

48

44

12

12

192

1 1 3

11 1

9

2

4

7

8

3

6

10

12

1 1

1 1 1 1

1 1

1 1

1 1

1 1 1

1 1

STEP 4 bis H F C D E B A G I J

1 1 1

5 1 1

1 1

1 1

STEP 5 bis Since in step 2 and 4 changes have been applied on the rows/columns order, it is necessary to implement again the algorithm starting from Step 1. Iteration 3 STEP 1 ter H F C D E B A G I J

1 1 1

5 1 1

11 1

9

1 1

1 1

2

1 1

4

1 1

7

1 1

8

1 1

3

6

10

12

1 1 1

1 1

1 1 1 1

STEP 2 ter H F C

1 1 1

5 1 1

11 1

9

1

1

2

4

7

8

3

6

10

12

3,584 3,072 768 768 248 240 14 14 1 1

D E B A G I J

1

1 1 1

1 1

1 1

1 1

1 1 1

1 1

1 1 1 1

Rows have been ordered, it is now necessary to verify the order of the columns. STEP 3 ter H F C D E B A G I J

1 1 1

5 1 1

11 1

9

1 1

1 1

2

4

7

8

3

1 1

1 1

1 1

1 1

1 1 1

6

10

1 1

1 1

12

768

704

192

48

48

48

48

44

12

12

1 1 3

5 1 1

11 1

9

2

4

7

8

3

6

10

12

1 1

1 1 1 1

1 1

1 1

1 1

1 1 1

1 1

768

STEP 4 ter H F C D E B A G I J

1 1 1

1 1

1 1

STEP 5 ter Since in Step 2 and Step 4 no changes have been applied on the rows/columns order, the implementation of the algorithm could be stopped. STEP 6: STOP. Regarding the ROC algorithm, the final solution is the following: H F C D E B A G

1 1 1

5 1 1

11 1

9

1 1

1 1

2

4

7

8

3

1 1

1 1

1 1

1 1

1 1 1

6

10

1 1

1 1

12

I J

1 1

Cell AAA: Technologies H-F; Items 1-5 Cell BBB: Technologies C-D; Items 11-9 Cell CCC: Technologies E-B; Items 2-4-7-8 Cell DDD: Technologies A-G; Items 3-6-10 Cell EEE: Technologies I-J; Items 12 Some limited extra-cellular flows may be admitted (although should be avoided): in this case, we only have two cases, namely item 11 flows from cell BBB to cell AAA, item 3 flows from cell DDD to cell CCC. Further evaluation, also from an economic perspective, should be implemented so to avoid extracellular flows. In this case the ROC algorithm has found a solution corresponding to the one proposed by SLC algorithm.

Exercise 3 (Exam 1st February 2010) A company is planning to renovate the production equipment, currently consisting of NC lathes and NC polishers, organized in departments as a Job-Shop. The products produced by the firm are characterized as follow: Total Turning Time [min/pcs] Product 1 Product 2 Product 3 Product 4

Total Set up Set up Annual Lot Polishing Time Time Demand dimension Time (lathe) (polisher) [pcs/year] [pcs/lot] [min/pcs] [min/lots] [min/lots]

50

20

8,000

250

200

150

35

15

5,000

300

255

300

30

20

12,000

120

150

200

45

20

6,000

200

250

400

The availability of the machines is 85% for lathes and 90% for polishers. The firm operates 220 days per year, 8 hours per days, of which only 7 are for production. The complexity of the production makes strictly necessary the presence of one operator supervising each machine. Each turning and polishing machine has a purchase cost of 100,000 € and 200,000 € respectively (10 years; 10%; PVA = 6.145), and the cost of an operator is 25 €/h, both for the first and the second shifts (to be paid considering gross hours). Determine the best shifts policy (1 or 2 shifts) and the number of lathes and polishers to be purchased, under the hypothesis of apply the same shift policy to two departments.

Solution Exercise 3 One shift policy sizing To design the production system and evaluate the related costs, it is firstly necessary to evaluate the available time for production and the needed time for production, given the annual demand of products. For the evaluation of the available time for production on each technology (lathes and polishers), it is possible to proceed as following:

Where: = days available for production (in this case 220 days/year); = actual time for production (in this case 7 hours/day); = machine availability (in this case 85% for lathe and 90% for polisher) The availability hours are equal to 1,309 h/year for each lathe and 1,386 h/year for each polisher. The time needed for production, is the sum of processing time and set up time, as follows, given i=1-4 products; j=1 for lathe, j=2 for polisher: Computation of processing time on technology j-th

= Total processing time on technology j-th for product i-th (h/year); = Unitary processing time on technology j-th for product i-th (min/pc); = Annual demand for product i-th (pcs/year). Computation of set up time on technology j-th

=total set up time on technology j-th for the product i-th (h/year); = unitary set up time on technology j-th for product i-th (min/lot); = annual demand for product i-th (pcs/year); = average lot size (pcs/lot). In the following table, needed times for manufacturing the products are reported: [h/year] Product 1 Product 2 Product 3 Product 4 Total

Turning Time

Polishing Time

Tsetup Lathe

Tsetup Polisher

6,666.67

2,666.67

222.22

177.78

2,916.67

1,250.00

83.33

70.83

6,000.00

4,000.00

120.00

150.00

4,500.00

2,000.00

50.00

62.50

20,083.33

9,916.67

475.56

461.11

The time needed for production by each technology is the sum of processing time and set up time:

It is possible to evaluate the number of machines, the number of operators and the total cost of solution (one shift policy): Number of machines and operators

Checking the saturation of the obtained configuration, the saturation values are very critical and it is necessary to add three additional lathes and an additional polisher. Indeed:

Working with a one-shift policy, considering one operator for each machine, 29 operators are needed (20 for lathes, 9 for polishers). Economic evaluation It is now possible to evaluate the annual cost of the one shift policy solution:

j = technology index (1 for lathes, 2 for polishers); = equivalent annual cost for technology j-th ; = number of machine/operators for the j-th technology based on 1 shift policy; = annual cost of a single operator, calculated on gross hours (220 days/year for 8 h/day for unitary cost of operator, 25 €/h - 44,000 €/year)

Two shifts policy sizing Adopting the two shifts policy, and since no further information about the actual availability of technologies is given, it is reasonable to double the time available for production for each technology obtained in the evaluation of the one shift policy. Number of machines and operators

Nevertheless, looking at the saturation of these machines, it can be noted that the values are very critical and it is necessary to add three additional lathes and an additional polisher:

Working with a two shifts policy, considering one operator for each machine for each shift, 30 operators are needed (10 for lathes, 5 for polishers, both for two shifts). Economic Evaluation

In this case, the two shifts policy configuration seems to be less expensive, with a saving of 248,921.07 €/year (nearly the 13%). Interestingly, the total cost of the solutions is guided by the workforce cost, that is about 67% (one shift policy) and 80% (two shifts policy) of the total cost.

Chapter 3. Assembly Systems

Exercise 1 Mr. La Tazza, after working for many years in the precision manufacturing sector, decided to open a small firm for the assembly and packaging of small sized mechanical components, La Tazza Ltd. The firm operates as a full-service partner, as it is structured to produce assembled and finished products. Recently, new overseas competitors seemed to be interested in entering the market, threatening the firm’s leadership. Mr La Tazza thus deems as of fundamental importance to fulfill the production orders. The assembly/packaging process is exclusively manual, and, according to Mr La Tazza, processing times have a very high variability. Additionally, the saturation rate of the station should not be higher than 80%. On the basis of the information reported above, it is necessary to define the number of working stations and the operations implemented in each station. For line balancing, it is possible to adopt one of the following criteria, motivating the choice: 3. Ranked Positional Weighting 4. Maximum Incompletion Probability Allowed (PNC> Cu) = Set of k-th (direct or indirect) successors = Incompletion cost of r-th operation only In the following table

’s values evaluated for each operation are reported:

#Op. Tk Sk^2 Lk I'k Ik 1 14 1.3 0.25 0.35 1.55 2 6 0.7 0.15 0.2 0.7 3 4 0.8 0.2 0.3 0.5 4 3 1 0.45 0.5 0.7 5 5 0.5 0.1 0.2 0.2 6 12 0.4 0.15 0.2 0.2

As for the PNC, it is evaluated as:

Where:

With k = operation whose allocation to a given station is under evaluation; = sum of all the variances of the operations already allocated to a given station; = residual time of the given station, defined as the difference between the available cycle time and the sum of the average processing times of operations i-th already allocated to the station, until operation k-th included.

index allows, under the hypotheses of normal distribution and independent operations, to evaluate the completion probability for operation k-th. If an operation has

tending to 0 (usually Pk lower than 0.005), we defined it as a sure operation

and it can be allocated to the station without evaluating

.

Lastly, we defined a critical operation, an operation for which even in case the workstation is empty. It is now possible to implement the Kottas & Lau algorithm, whose procedure is reported in the following figure.

STEP 1 Station A; Available operations: 1-2 Operation 1 The residual time of the station, after the implementation of operation 1. is:

Where B is the set of operations previously allocated to Station A (in this case, none). Evaluating S as the sum of the variances of operations already allocated to the station under consideration, operation kth included:

it is possible to derive the following index:

In this case (referring to the table of quantiles) the operation is a sure one, having a completion probability of 99.5%. In the following table, the relevant information for operation 1 are reported: Tk

TRk S(si^2) Zk

Pk

14

4

>99.5%

1.30

3.51

PNCk

Ik

PNCk*Ik Lk

Allocation? Sure operation

1.55

Operation 2 Applying the previous procedure to operation 2, we obtain: Tk

TRk S(si^2) Zk

6

12

0.70

Pk

PNCk Ik

14.34 >99.5%

PNCk*Ik Lk

0.70

Allocation? Sure operation

Both the available operations result sure ones. According to the Kottas & Lau algorithm, we allocate operation 1 to station A, having the largest

index.

STEP 2 Station A; Available operations: 2-4-6 Operation 2 It is necessary to evaluate again indices for operation 2, since now operation 1 is allocated to station A. Tk TRk S(si^2) Zk 6

-2

2.00

Pk

PNCk

-1.41 7.93%

Ik

PNCk*Ik Lk

92.07% 0.70 0.644

Allocation? Non-desirable 0.15 operation

Operation 4 Tk TRk S(si^2) Zk 3

1

2.30

Pk

PNCk

Ik

PNCk*Ik Lk

0.66 74.54% 25.46% 0.70 0.178

Allocation? Desirable 0.45 operation

Operation 6 Tk TRk S(si^2) Zk 12 -8

1.70

Pk

PNCk

Ik

PNCk*Ik Lk

-6.14 99.5%

PNCk

Ik

PNCk*Ik Lk

0.70

Allocation? Sure operation

Operation 6 Tk 12

TRk S(si^2) Zk 6 0.40 9.49

Pk PNCk Ik PNCk*Ik Lk >99.5% 0.20

Allocation? Sure operation

Both the available operations result sure ones. According to the Kottas & Lau algorithm, we allocate operation 2 to station B, having the largest

index.

STEP 4 Station B; Available operations: 3-6 Operation 3 Tk 4

TRk S(si^2) Zk 8 1.50 6.53

Pk PNCk Ik PNCk*Ik Lk >99.5% 0.50

Allocation? Sure operation

Operation 6 Tk

TRk S(si^2) Zk

Pk

PNCk Ik

12

0

50.0%

50.0% 0.20 0.100

1.10

0.00

PNCk*Ik Lk

Allocation? Desirable 0.15 operation

According to the Kottas & Lau algorithm, between a desirable and sure operation, the sure should be allocated first. We thus allocate operation 3 to station B. STEP 5 Station B; Available operations: 5-6 Operation 5 Tk

TRk S(si^2) Zk

Pk

5

3

98.30% 1.70% 0.20 0.003

2.00

2.12

PNCk Ik

PNCk*Ik Lk

Allocation? Desirable 0.10 operation

Operation 6 Tk

TRk S(si^2) Zk

12

-4

1.90

Pk

-2.90 99.5% 0.20

Allocation? Sure operation

As 6 is the last operation, we close the station. In the following a synthesis table of the exercise, with all the logical steps, is reported: Stat. op.

ti Tk TRk

A A

0 14 0 6

1 2

4 12

(si^2) Zk 1.3 3.51 0.7 14.34

Pk >99.5% >99.5%

PNCk

Ik 1.55 0.7

PNCk*Ik Lk

Available operations Sure Check 2 Sure 1 largest Ik

Allocation?

A

1

0 14

4

1.3 3.508

A

2

14

6

-2

A

4

14

3

1

A

6

14 12

-8

1.7 -6.14 99.5% >99.5% >99.5% >99.5%

B

6

6 12

0

1.1

B B

3 5

6 10

4 5

8 3

1.5 6.532 2 2.12

B

6

10 12

-4

B C

5 6

10 5 0 12

3 6

2 -1.41 2.3

1.55

7.93% 92.07% 0.70

0.18 0.45

100%

0.2

0.2 0.15

0.2546

0.7 0.7 0.2 0.7 0.5

0.178 0.45

50.00% 50.00%

0.2

0.1 0.15

>99.5% 98.30%

1.70%

0.5 0.2

0.003

-2.9

99.5%

0.017

0.2 0.2

0

74.20% 25.80%

0.644 0.15

0.7

1.9

0.66

>99.5%

0.0034

0.1

0.1

Yes 2,4,6 NonCheck 4,6 desirable Desirable Check 6 NonAllocate 4 desirable Yes 2,6 Sure Check 6 Sure 2 largest Ik Yes 3,6 Sure Check 6 Better 3, Desirable because sure Yes 5,6 Desirable Check 6 Better 5, Nonbecause desirable desirable Yes 6 Sure, Yes END

Exercise 4 (Exam 29th June 2010) An assembly process is characterized by the following precedence diagram. The target cycle time (not modifiable) is equal to 5 minutes. The possible maximum Saturation Rate is 100%.

Evaluate the lower bound and design the line (number of stations) allocating operations to each workstation with a task-oriented approach[2], according to: 1) Ranked positional weight criterion; 2) Max operations following (MaxFol) criterion (sub-criterion MaxDur, then Alphabetical Order)

Solution Exercise 4 Lower Bound The Lower Bound can be easily evaluated:

Ranked Positional Weighting According to this criterion, operations are allocated according to a rank of decreasing priority index “positional weighting”. For each operation k-th, the following index is thus evaluated:

Where is the processing time of the operation k-th; is the set of successors of the operation k-th.

The k-th operation is assigned to the first feasible workstation, giving that the cycle time available should not be overcome. The first step is thus to evaluate the the set of successors. Operation A B C D E F G

for each operation of the line, passing from the identification of

Successors B; C; D; E; F; G. D; E; F; G. F; G. F; G. G. G. -

The PW result:

The rank according to which the operations will be allocated is thus: A- B- C- D- E- F- G. We can start know to assign the operations to the different workstations, remembering that

Station 1 1 2 2 3 4 4 5 5 5

Operation A B B C C D E E F G

Tk 3 4 4 5 5 4 2 2 1 1

TR 2 -2 1 -4 0 1 -1 3 2 1

Allocation? Yes No Yes No Yes Yes No Yes Yes Yes

Available B, C C, D, E D, E E, F F, G G -

Max Operations Following (sub-criterion MaxDur, then Alphabetical Order)

The MaxFol criterion (largest number of following operations) is similar to the Ranked Positional Weighting when weights are set to 1. The number of operations following the k-th operation is thus considered as priority index. Operations with larger number of following operations are prioritized and allocated as soon as possible The first step is thus to identify and evaluate the number of successors for each operation. Operation A B C D E F G

Successors B; C; D; E; F; G. D; E; F; G. F; G. F; G. G. G. -

Number 6 4 2 2 1 1 0

As it can be noticed from the table, there are some operations with the same number of successors, namely C and D, and E and F. It is necessary to apply a sub-criterion in order to prioritize operations with the same priority index. The sub- criterion to be use, according to the text, is the MaxDur, according to which the processing time of the k-th operation is considered as priority index and operations (those available) are ranked with descending order, setting as first the one with the largest processing time. As for the operation with the same number of successors: operation C has processing time of 5 minutes, while operation D of 4 minutes; operation E has processing time of 2 minutes, while operation F of 1 minute. The rank according to which the operations will be allocated is thus: A- B- C- D- E- F- G. We can start know to assign the operations to the different workstations, remembering that

Station 1 1 2 2 3 4 4 5 5 5

Operation A B B C C D E E F G

Tk 3 4 4 5 5 4 2 2 1 1

TR 2 -2 1 -4 0 1 -1 3 2 1

Allocation? Yes No Yes No Yes Yes No Yes Yes Yes

Available B, C C, D, E D, E E, F F, G G -

Chapter 4. Demand Forecasting

Exercise 1 KIDS firm manufactures toys for children and supplies them through its retail stores. With reference to the family product ROBOT 2000 monthly data regarding sales in the last two years (2007 e 2008) are reported in Table 1. 2007 2008

Jan 1,230 1,355

Feb 1,140 1,246

Mar Apr 890 1,260 1,030 1,354

May 1,320 1,452

Jun 1,450 1,478

Jul 1,520 1,642

Aug 670 805

Sep 1,345 1,402

Oct 1,548 1,753

Nov 1,768 1,837

Dec 3,576 3,768

TABLE 1 – Units of ROBOT 2000 sold in years 2007-2008 two-year period Assuming the applicability of Holt-Winters model and adopting the following values of smoothing coefficients: for average level; for trend; 0.5 for seasonality. It is requested to: 1. Calculate initial values for average, trend and seasonality basing on historical data of 2007-2008 two-years period; 2. Formulate the monthly demand forecasting for January and February 2009; 3. Assuming the actual demand for January 2009 equal to 1,453 units, formulate the monthly demand forecasting related to February, March, April, May and June 2009; 4. Having data related to the actual demand for February, March, April, May and June 2009 (Table 2), calculate, with reference to the above- mentioned months, the values of the following forecasting error measures ME, MAD, MAPE e MSE. 2009

Jan (1,453)

Feb 1,330

Mar 1,150

Apr 1,465

May 1,430

Jun 1,560

TABLE 2: units of ROBOT 2000 sold in the first semester of year 2009

Solution Exercise 1 Initialization of the model The initialization of the model is implemented through the calculation of average, trend and seasonality values, on the basis of historical data of years 2007-2008. Initialization of “Trend” The average sales values for years 2007 and 2008 are evaluated as follows:

2007

2008

M 1,476.42 1,593.50

Then, the monthly trend value is evaluated:

L is the seasonality period, which, in this case, is equal to 12. Initialization of “Average” The initial value of smoothing average is calculated starting from M2008 and adding L/2 times T0 value previously calculated:

Initialization of “Seasonality” The initial seasonality coefficients are calculated as the average of the seasonality coefficients in the considered month in years 2007 and 2008:

Where: and In table the values so determined are reported: Sjan

Sfeb

Smar

Sapr

Smay

Sjun

Sjul

Saug

Ssep

Soct

Snov

Sdec

2007 0.83310 0.77214 0.60281 0.85342 0.89406 0.98211 1.02952 0.45380 0.91099 1.04848 1.19749 2.42208 2008 0.85033 0.78193 0.64638 0.84970 0.91120 0.92752 1.03044 0.50518 0.87982 1.10009 1.15281 2.36461 S0jan

S0feb

S0mar

S0apr

S0may

S0jun

S0jul

S0aug

S0sep

S0oct

S0nov

S0dec

S0 0.84171 0.77703 0.62459 0.85156 0.90263 0.95481 1.02998 0.47949 0.89541 1.07429 1.17515 2.39334

Demand forecasting for January and February 2009 Holt-Winters model is implemented. With this model, at the end of period t, it is possible to calculate the forecast related to the general period t + m:

For January 2009, the forecast will be:

Period t = 0 is December 2008, thus the forecast horizon m is 1 month. For February 2009:

In this case, the forecast horizon m is equal to 2 months. After calculate the forecast value for the month, it is necessary to round the obtained result to the nearest integer.

Demand forecasting for February, March, April, May and June, basing on actual demand of January 2009 Considering an actual demand for January 2009 of 1,453 units, it is necessary to re-calculate average and trend related to period t = 1, that is January 2009. On the contrary, it is not necessary to re-calculate seasonality coefficients. Moreover, the data provided are not sufficient to re-calculate the seasonality coefficients for February- March- April- May, i.e. not enough historical data are provided.

It is now possible to compute demand forecasting for February, March, April, May and June 2009, employing the following formula: Where m = 1, 2, 3, 4, 5

Forecasting error measures The forecasting error for period t is defined as the difference between the actual demand and the forecasted one: Et = Dt - Pt Feb Actual demand 2009 1,330 Forecasted demand 1,305 2009 Error 25

Mar 1,150

Apr 1,465

May 1,430

Jun 1,560

1,056

1,449

1,546

1,646

94

16

-116

-86

Exercise 2 A food wholesaler would like to evaluate, at 31st December 2008, the demand forecasting for the following year, with reference to family product “cured meats”. Based on the available actual sales data of the last two years, the average is 1,450 and 1,210 kg/two-months period, respectively for years 2007 and 2008. The analysis of times series highlights a decreasing linear trend; the seasonality coefficients related to the different two-months periods are reported in Table 1. Starting from the available information, it is requested to compute the demand forecasting for the first and second two-months periods of the following year, using the Holt- Winters model. S Jan/Feb

S Mar/Apr

S May/Jun S Jul/Aug

S Sep/Oct

0.88

1.2

1.1

0.6

0.72

S Nov/Dec 1.5

TABLE 1 – Seasonality coefficients related to the different two-months periods

Solution Exercise 2 Initialization of the model The initialization of the model is implemented through the calculation of the initial values of average and trends, since seasonality is already provided. Initialization of “Trend” First, the two-months period trend is evaluated:

L is the seasonality period, which, in this case, is equal to 6. Initialization of “Average” The initial value of smoothing average is calculated starting from M2008 and adding L/2 times T0 value previously calculated:

Demand forecasting for the first and second two-months period of 2009 It is now possible to compute demand forecasting for the first and second two-months period of 2009, employing the following formula:

Chapter 5. Inventory Management

Exercise 1 Persian Ltd imports and sells carpets in Italy. Among the different carpets managed by the firm, the best-seller one is imported for 7 €/meter. Knowing that the annual demand of the best-seller carpet is 10,000 meters, the ownership rate is 15% and the ordering cost is 20 €/order line, you have to determine the Economic Order Quantity (EOQ) and the total cost related to the management of the carpet under consideration. It is also requested to determine the number of orders issuing in a year and the time between two orders’ issuing, considering that the store is open six days a week, and it is closed during holidays. Supposing an ordered quantity 20% smaller than the EOQ, you are asked to calculate all the annual costs related to the inventory management, analysing the differences with the previous case.

Solution Exercise 1 Determining the Economic Order Quantity (EOQ) For determining the EOQ in a fixed quantity ordered model, we should apply the following formula:

For which the following data are necessary: D, is the annual demand (D = 10,000 m); o, is the ordering cost (o = 20 €); cm, is the ownership rate (cm = 15%); P, is the price per unit (P = 7 € / m). It is thus possible to calculate EOQ as follows:

EOQ is an approximated value: it is based on an estimation of ordering cost and ownership cost, both of which, in turn, depends on an uncertain demand value. It is thus preferable to round the EOQ value to the nearest integer; if also type and size of packaging are taken into consideration, it turns out to be convenient to round the ordered quantity to a whole number of loading units (boxes or pallets). The annual total cost for inventory management is the sum of the cost of purchasing (CP), the cost of order issuing (CO) and the stock-holding costs (CSH). They can be easily evaluated as follows:

The annual total cost is:

Determining the annual number of orders and the time between two orders’ issuing The annual number of orders is evaluated as the ratio between annual demand and EOQ:

The time between two consecutive orders’ issuing (T) is evaluated as the ratio between the annual number of opening days of the store and the annual number of orders issued. Assuming a calendar of 305 working days per year, excluding Sundays as well as religious and national feast days[3]:

Order quantity 20% smaller than the EOQ Considering an order quantity 20% smaller than the EOQ previously identified:

The annual costs related to inventory management are:

As it can be easily inferred from the results, with the reduction of the quantity ordered, the ordering cost (inversely proportional to the entity of the order quantity) increases, while the stock holding cost (directly proportional to the entity of the order quantity) decreases; purchasing costs are kept constant since they do not depend on the quantity ordered.

Exercise 2 A regional warehouse purchases tools from different producers and distributes them to different retailers. The warehouse operates on a calendar of 5 days per week. Orders can be received only when the warehouse is open. Referring to the most sold drill model, it has been estimated a daily demand, distributed according to a normal distribution, with an average of 100 units and a standard deviation of 30 units. The procurement time for the regional warehouse is three working days. The warehouse purchases a drill for € 85. Suppose that the warehouse adopts a fixed-order quantity model: a) Knowing that the annual ownership rate is 11% and that ordering cost is equal to € 35, evaluate the Economic Order Quantity (EOQ) and the Re-Order Point (ROP) if the warehouse adopts a continuous control system and wants to maintain a service level (the probability of not suffering from stock out during lead time) equal to 92%; b) Assuming an initial availability of the warehouse of 40 drills, orders in progress of 440 units and no delayed orders, is it necessary to issue a new order? Suppose now that the warehouse adopts a fixed interval of order issuing control system: a) Evaluate the interval between reviews expressed in working days, considering a reordered quantity equal to the EOQ; b) Determine the objective level value; c) Assuming an initial availability of the warehouse of 40 drills, orders in progress of 440 units and no backlogs, how many units should be ordered?

Solution Exercise 2 Fixed-order quantity model a) Consider the case in which the warehouse adopts a fixed- order quantity model for managing the inventory, EOQ is equal to:

where the annual demand is computed under the hypothesis of 52 working weeks a year and 5 working days a week. The Re-Order Point is computed as:

Which is the sum of the average demand during the lead time and the safety stocks, subtracting the tardy orders. Safety stocks can be evaluated as:

Where

In this specific case, as lead time has no variability.

with µ=0.5 (hypothesizing a perfect statistical independent of the demand). The service level factor (k) is calculated by using the cumulative normal standard distribution for random variables that corresponds to the required service level (SL). In this case, as the desired service level is equal to 92%, k is equal to 1.41 Safety stocks thus result:

And the Re-Order Point can be then evaluated:

b) In the proposed situation, considering an order in progress of 440 units, initial availability of 40 units and not tardy orders, it is sufficient to compare the actual available quantity stocked and the ROP, in order to decide whether to issue a new order or not. It can be easily verified that the actual availability (Actual Availability = Order in Progress + Initial Availability = 440 + 40 = 480 units) is higher than the ROP so for now it is not necessary to issue any order. Fixed interval of order issuing model Suppose now that the warehouse adopts a fixed interval of the order issuing model. a) The interval of order issuing, considering an order quantity equal to EOQ, is:

b)

The objective level is the sum of safety stocks and the average demand during the period (LT+T). - the lead time plus the interval of order issuing.

The demand in (LT + T) is calculated as follows:

Safety stocks, in a fixed interval of order issuing model with

=0, are equal to:

LT

It is now possible to evaluate the objective level using the following formula:

c)

Lastly, the quantity that it would be necessary to order so to maintain the objective level is evaluated as the difference between the objective level and the availability:

Exercise 3 (Exam 7th September 2010) Spinotti manufactures components for audio products. The production is diversified in a limited range of models. In particular, data related to code SP-1 are available. SP-1 is the flagship product and it guarantees 35% of the whole Spinotti’s turnover. Assume the demand for SP-1 as known and distributed evenly during the firm’s working calendar. Annual demand Sale Price Raw material purchase price Stock holding expenses rate Opportunity cost rate Annual Spinotti’s working days

250,000 1 0.6 2.5% 10% 225

pcs/year €/pc €/pc yearly yearly days/year

Using the provided data, it is asked to: 1) Determine the EOQ for code SP-1, knowing that: a) Mr. Gallucci, production planning manager, and Mr. Masiello, the manager of the code SP-1 production line and Occupational Health and Safety manager, are hired by Spinotti: they work for 7.25 actual h/day and their costs for the firm are respectively around 45,000 € and 37,000 € per year; b) Every set up takes 2 hours, it is implemented by Mr. Masiello and has a cost of 15 €/setup, that is related to the solvents used for the cleaning of the line; c) The orders are issued by Mr. Gallucci within his normal tasks and the issuing takes about 15 minutes; d) The production line of SP-1 is not completely saturated and Mr. Masiello, during production stops, controls that the legislation related to Occupational Health and Safety are fully respected by every worker (indeed, he implements a sort of visual inspections through all the different production departments); e) The transfer of the products is entrusted to an external firm, that is paid by Spinotti 0.02 € for each unit produced. 2) Determine the Re-Order Point for code SP-1, considering that it takes 5 working days from when the production is launched is issued to the moment in which the goods are stocked.

Solution Exercise 3 1) In determining the EOQ it is of fundamental importance to consider which costs are relevant and different and which are not, keeping in mind that:

Where: D = Annual demand (pcs/year) o = Set up cost (€/setup); cm = Ownership rate (%/year); C = Variable unitary cost unit (€ / pc).

Regarding o, this cost is related only to the cost of the solvents used for the line cleaning by Mr. Masiello, and so it is equal to 15 €/setup. As Mr. Gallucci is in charge of manager of the code SP-1 production line, this activity is a part of his normal tasks and it is not differential for the setup cost. Regarding P, it is necessary to underline how the determining of EOQ descends from an optimization of the total annual cost in which it is considered the ownership cost of raw materials and the organization of the production: consequently, it is necessary to consider the raw material purchase price (0.6 €/pc), and not the sale price of the product (1 €/pc).

In determining ownership rate , it is necessary to consider both the stock holding expenses rate and the opportunity cost rate, which is 2.5+10 = 12.5 %/year. Lastly, regarding D, it is 250,000 pcs/year. Basing on these data:

2) The computation of the Re-Order Point then results to be very easy: Where: = Constant demand expressed in pcs/day = Lead time [days/order] It is important to note that and must be expressed in homogenous units: in this case, it has been decided to transform the annual demand in daily demand, and this was possible since the working days were known. In particular:

From which:

Exercise 4 A distributor manages a warehouse. The goods are purchased by producers and are then delivered to retail stores. Considering the product identified with the code Y, the weekly demand is distributed according to a normal distribution, with an average of 43 boxes and standard deviation equal to 5 boxes. The contract with the current supplier is about to expire and the distributor is seeking the best inventory management policy to be adopted. Indeed, the procurement from the producer can be managed according to two different policies: 1) Fixed-order quantity, with a lead time, distributed according to the normal distribution, of 17 days and standard deviation of 3 days; 2) Fixed- time period with an interval between reviews equal to 60 days, and lead time distributed according to a normal distribution of 6 days and a standard deviation of 2 days. Assuming the following hypotheses: The product is purchased at 10 €/box and it is sold for 20 €/box; The ordering cost is 20 € and 12 €, respectively for fixed order quantity and fixed-time period; The desired level service (i.e. the probability of avoiding an out of stock), is 95%; The transport from the warehouse to the retail store has a cost of 10 €/order; The annual amortization for the warehouse can be quantified in 50,000 €/year; The annual rate for stock holding expenses is 4%; you are asked to evaluate the total cost of the two available models, deciding which one the firm should adopt. From a financial perspective, the firm is currently into debt but still has debt capacity. To firms with a similar financial situation, a bank could issue a loan having a burden of 10%/year; whilst a firm investing its cash in bonds would have a yield of 6%/year. The distributor operates 52 weeks a year, 6 days a week.

Solution Exercise 4 Evaluation of the Fixed-Order Quantity Model The annual total cost is the sum of the cost of purchasing (CP), the cost of order issuing (CO) and the stock-holding costs (CSH). Considering a fixed-order quantity model, they can be easily evaluated as follows:

Where AIL is the Average Inventory Level. The CP can be easily computed since both P (Purchase Price) and D (Annual Demand) are provided:

For evaluating CO, it is necessary to determine the Economic Order Quantity (EOQ), for which the following data are necessary: Annual demand (D = 2,236 boxes/year); Ordering cost (o = 20 €); Ownership rate (m = 14%/year);

Price per unit (P = 10 € / box) – we consider, of course, the purchase cost. The ownership rate is evaluated as the sum of the holding stocks expenses rate (4%) and the opportunity cost rate. the rate applied by banks to firms in the same situation of the distributor under consideration (10%) (currently into debt, but with still debt capacity). It is, therefore, possible to calculate EOQ using the following formula:

And CO is then:

As for CSH, it is necessary to evaluate the Average Inventory Level

Where

for a level service equal to 95%. So that

And the costs related to stock holding are:

The annual total cost is:

Evaluation of the Fixed-time period model The annual total cost is the sum of the cost of purchasing (CP), the cost of order issuing (CO) and the stock-holding costs (CSH). Considering a fixed-time period model, they can be easily evaluated as follows:

Where AIL is the Average Inventory Level. The CP can be easily computed since both P (Purchase Price) and D (Annual Demand) are provided:

The CO can be calculated as:

As for CSH, it is necessary to evaluate the Average Inventory Level:

Where

for a service level equal to 95% So that:

The annual total cost is:

Comparison of the two policies To sum up, the cost breakdown for the two alternatives is as follows:

Costs [€/year]

breakdown EOQ

Purchasing costs (CP) Order issuing (CO) Stock-holding (CSH) Total costs

costs costs

Fixed period

22,360

22,360

176.76

62.4

230.44

351.64

22,767.2

22,774.04

–time

Note that the purchase cost has been calculated, however, it is not relevant for the choice of the model to be adopted, as the quantity purchased is the same regardless of the inventory management policy adopted. Please note that: The transport from the warehouse to the retail has been not considered because it is a cost sustained by the retailer; The annual amortization for the warehouse has been not consider because it is not related to choices related to inventory management.

Exercise 5 (Exam 1st September 2016) The “Charming Man” Ltd is a leather accessories manufacturer. The leather laptop bag, “Morrisey”, is the most sold product. The weekly demand for “Morrisey” is distributed according to a normal distribution, with an average of 140 bags and a standard deviation equal to 10 bags. The production is managed with a fixed quantity model with a lead time distributed according to a normal distribution, with an average of 10 days and standard deviation of 5 days. The unitary variable production cost is 300 €/bag and the sale price is 600 €/bag. The set-up cost is 1,200 € and the desired level service is 99% (k=2.33). The annual rate for stock holding expenses of 10% and the opportunity rate of 6%/year. Charming Man is open 52 weeks a year, 7 days a week. 1) Evaluate the total annual cost considering an infinite production rate. 2) Re-evaluate the total annual cost considering a finite production rate of 24 bags/day.

Solution Exercise 5 Evaluate the total cost The annual total cost is the sum of the cost of production (CP), the cost of orders (CO) and the stockholding costs (CSH). Considering a fixed-time period model, they can be easily evaluated as follows:

To proceed with the computation, we need to determine the Economic Order Quantity (EOQ), that is:

Where the Cm is evaluated as the sum of ownership rate and opportunity cost rate – 16 %/year. Regarding CSH, it is necessary to evaluate the Average Inventory Level (AIL)

As

for a level service equal to 99% Then

We can now evaluate AIL

And the costs:

Evaluate the total cost with a finite production rate A finite production rate means that replenishment is made incrementally. More in detail, since the consumption rate is D/H (with H= pre-defined working days [days/year]), during the replenishment period the inventory increases at a rate of (r-D/H), where r is the finite production rate. Regarding the costs, they are computed in the same way, but the formulas for the evaluation of EOQ and AIL change. Considering the EOQ, the formula becomes:

And so:

Considering AIL, the formula becomes:

Where

So, we obtain:

It is now possible to evaluate the costs:

Chapter 6. Aggregate Production Planning

Exercise 1 A firm has planned the production according to the following level plan: Demand Production Planning

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 200 100 200 200 200 300 300 200 200 200 100 200 200 200 200 200 200

200 200 200

200

200 200

200

The production capacity of the line is equal to 200 pieces/month and the production cost is equal to 10.00 €/piece. Stock holding cost and back-log (recoverable stock-out) cost are, respectively, 15.00 and 25.00 €/(piece*month). The production line works continuously so that every change in the production rate corresponds to high setup costs, equal to 1,800 €/setup. The firm can purchase the finished product from a sub-contractor for 30 €/piece, without any capacity constraint. It is required to: Evaluate the total cost of the given production plan; Evaluate a chase production plan, to verify if it would be possible to reduce the total cost, with a maximum production capacity of 200 pieces/month; Specify, if it exists: the set-up cost threshold that makes the two plans indifferent from a total cost perspective; the sub-contracting cost threshold that makes the two plans indifferent from a total cost perspective.

Solution Exercise 1 In general terms, the total cost of a production plan is given by the sum of the following costs: Stock Holding Cost; Stock-Out Cost (Back-log, Lost margin, Other penalties); Internal Production Cost or Direct Work Cost; Inefficiency Cost; Sub-Contracting Cost; Set-up Cost. The computation of the plan’s cost is developed with the support of a table, organized according to the followings: Plan: the quantity to be produced in each period (both through internal production and outsourcing) is reported; Demand: the demand for each period is reported; Cumulated Plan: for each period, the cumulative sum of the quantities produced until that period is reported; Cumulated Demand: for each period, the cumulative sum of the demand until that period is reported; Delta Production: the difference between production and demand in each period (not the cumulated ones); Inventory level – End of month (Eom): quantity in inventory at the end of each month. Average Inventory Level (AIL): quantity stocked, on average, in each period. For a given period t, it is evaluated as:

-

Inventory Cost: it is obtained multiplying the AIL by the unitary stock-holding cost; Stock-out: it the difference between the production and the demand if the difference is negative. In the present case, it is expressed in terms of back-log.

-

Stock-out Costs: it is obtained multiplying the back-log quantity by the unitary stock-out cost. In the present case, it is expressed in terms of back-log cost (€/piece*month) without penalties. Internal Production: the quantity that has been decided to produce internally; Internal Production Cost: it is obtained multiplying the internal production by the unitary production cost; Sub-contracting: the quantity that has been decided to purchase from the sub-contractor; Sub-contracting Cost: it is obtained multiplying the quantity purchased from the sub-contractor by the unitary sub-contracting cost; Set-up cost: it is the cost related to a change in the production rate (it is hypothesized the need of a set up at the beginning of the year).

Evaluation of the given production plan’s cost (LEVEL) The given plan, it makes leverage on the minimization of setup costs, but it may lead to higher stockholding costs. Indeed, as the provided plan is a level one, it considers a constant production in the different periods. In Table 1 the evaluation of all the costs is reported. Particularly, the total cost of the plan is 41,800 €. The back-log cost’s share of the total cost is 25%, due to 100 pieces that are not fulfilled for 4 periods. Evaluation of Alternative production plan (CHASE) The chase plan aims at matching the production rate to the demand rate along the planning horizon. Given the nature of the plan, it makes leverage on the minimization of stock-holding costs, but it may lead to higher setup costs and/or sub-contracting costs. In Table 2 the evaluation of the costs is reported. Particularly, the total cost of the chase plan is 37,000 €, 4,800 € less than the level plan previously evaluated. In this case, back-log costs have been eliminated (10,000 €), but a higher set-up cost should be paid (in periods 2, 3, 11 and 12), as well as a higher sub-contracting cost, as, so to fulfill the demand, the firm must purchase pieces for the sub-contractors in those periods in which the demand is higher than the internal production capacity.

Evaluation of threshold values for set up cost and sub-contracting cost As the set-up or sub-contracting costs may increase, keeping fixed all the other unitary costs, it would become less convenient to adopt a chase plan. In this case, the break-even point is obtained for: -

A set-up cost of 3,000 €/set up.

-

A sub-contracting cost of 54 €/piece.

Given Level Plan Month Jan Feb Mar Apr May Jun Demand [pcs/month] 200 100 200 200 200 300 Cumulated [pcs] 200 300 500 700 900 1,200 Demand Plan [pcs/ 200 200 200 200 200 200 month] Cumulated [pcs] 200 400 600 800 1,000 1,200 Plan Delta PD [pcs/ 0 100 0 0 0 -100 month] Inv. Eom [pcs/ 0 100 100 100 100 0 month] AIL [pcs/ 0 50 100 100 100 50 month] Inventory [€/ month] 0 750 1500 1500 1500 750 Cost Back-Log [pcs/ 0 0 0 0 0 0 month] Back-Log [€/ month] 0 0 0 0 0 0 Cost Internal [pcs] 200 200 200 200 200 200 Production Internal [€/ month] 2,000 2,000 2,000 2,000 2,000 2,000 Production Cost Sub[pcs/ 0 0 0 0 0 0 contracting month] Sub[€/ month] 0 0 0 0 0 0 contracting Cost Set-up [€/ month] 1,800 0 0 0 0 0 Cost Total Cost

[€]

Unitary Cost

[€/pc]

Tot Jul Aug Sep Oct Nov Dec 300 200 200 200 100 200 1,500 1,700 1,900 2,100 2,200 2,400 200

200

200

200

200

200

2,4

2,4

1,400 1,600 1,800 2,000 2,200 2,400 -100

0

0

0

100

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

100

100

100

100

0

0

2,500 2,500 2,500 2,500

0

0 10,0

6,0

200

200

200

200

200

200

2,4 p 2,000 2,000 2,000 2,000 2,000 2,000 24,0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1,8

41,8 17.4 €/p TABLE 1 Evaluation of the Total cost of the given level plan

Month Demand Cumulated Demand Plan Cumulated Plan Delta PD Inv. Eom AIL Inventory Cost Back-Log Back-Log Cost Internal Production Internal Production Cost Subcontracting Subcontracting Cost Set-up Cost Total Cost Unitary Cost

Jan

Feb

Mar

Chase Plan Apr May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

To 300

300

200

200

200

100

200

[pcs/month] [pcs]

200

100

200

200

200

200

300

500

700

900 1,200 1,500 1,700 1,900 2,100 2,200 2,400

[pcs/month] [pcs]

200 200

100 300

200 500

200 700

200 300 300 200 200 200 100 200 900 1,200 1,500 1,700 1,900 2,100 2,200 2,400

[pcs/month] [pcs/month] [pcs/month] [€/month]

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

[pcs/month] [€/month]

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

200

100

200

200

200

200

200

200

200

200

100

200

[€/month]

2

2

2,000 1,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 1,000 2,000

[pcs/month]

0

0

0

0

0

100

0

0

0

0

0

[€/month]

0

0

0

0

0 3,000 3,000

0

0

0

0

0

1,800 1,800 1,800

0

0

0

0

0 1,800 1,800

[€/month]

2

100

0

0

[€] [€/pc]

15 TABLE 2 Evaluation of the Total cost of the chase plan

Exercise 2 (Exam 29th June 2010) Evaluate the best production strategy for the production plan given in the following table: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand [pcs/month] 200 200 400 400 400 500 500 400 400 400 200 200 Forecast Production [pcs/month] 200 200 200 800 800 800 200 200 200 200 200 200 Plan

The capacity of the production line, considering a one-shift policy, is 200 units/month (only on-off mode). There is however the possibility of adopting a second shift, whose production capacity, without extra cost, is up to 200 units/month. The production cost is 20.00 €/piece. Stock-holding and back-log costs are, respectively, 10.00 e 40.00 €/(piece*month). The production line works continuously so every change in the production rate corresponds to a high set-up cost of 3,000 €/set-up. Historically, the firm has always tried to minimize the total set-up cost (a set-up is however needed at the beginning of the year). The firm can purchase the final product from a sub-contractor for 50 €/piece, without any capacity constraint. It is then requested to formulate the optimal production plan.

Solution Exercise 2 Evaluation of the best production strategy for the given plan The best production strategy for the given plan is the one that minimizes the total cost of the plan. As the plan is, in this case, given, the costs related to stock-holding and back-log would not vary regardless of how we fulfill the demand. Indeed, the only leverage we have to minimize the total cost, if a plan is given, is the organization of the production within internal production and sub-contracting. Notably, the unitary cost for internal production (20 €/pc) is lower than the corresponding one for sub-contracting (50 €/pc), and, consequently, it is more convenient to produce internally than to turn to sub-contract. As for the internal production, we have the possibility to adopt a one-shift or a two-shifts policy. In the former case, we would produce 200 pcs/month, in the latter up to 400 pcs/month. Shifting from one case to the other a set-up is needed (changing in the production capacity). It is then interesting to evaluate the production quantity that makes it more convenient to turn to sub-contracting rather than produce internally using a two-shifts policy. Therefore, assuming the internal production capacity of one shift is not sufficient for the implementation of the plan, we compare the costs born using the production capacity in the second shift with those born turning to sub-contract, that is to evaluate: Where: = Number of units to be produced; = Unitary cost for sub-contracting (50 €/pcs); = Unitary cost in case of production in overtime hours (in this case, equal to the unitary production cost in ordinary time, 20 €/pc); = The set-up cost to be born in case of turning to overtime hours for the production, which leads to two variations of the internal production rate (3,000 €/set-up). Consequently, it results preferable to turn to sub-contract in case it should be necessary to turn to overtime hours for production lower than 200 pieces. The evaluation of the given plan is reported in Table 1.

Month Demand Cumulated Demand Plan Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Back-log Back-log Cost Internal Production Internal Production cost Subcontracting Subcontracting cost Set up cost

Current Plan Apr May 400 400 1,200 1,600

[pcs/month] [pcs]

Jan 200 200

Feb 200 400

Mar 400 800

[pcs/month]

200

200

200

800

800

[pcs]

200

400

600

1,400

2,200

[pcs/month] [pcs/month]

0 0

0 0

-200 0

400 200

400 600

[pcs/month] [€/month]

0 0

0 0

0 0

100 1,000

400 4,000

[pcs/month] [€/month]

0 0

0 -200 0 8,000

0 0

0 0

0 0

0 0

0 0

0 0

[pcs/month]

200

200

400

400

400

200

200

200

[€/month] 4,000 4,000 4,000

8,000

8,000

400

400

200

Jun Jul Aug Sep Oct Nov Dec 500 500 400 400 400 200 200 2,100 2,600 3,000 3,400 3,800 4,000 4,200 800

200

200

200

200

200

200

3,000 3,200 3,400 3,600 3,800 4,000 4,200 300 900

-300 600

-200 400

-200 200

-200 0

0 0

0 0

750 750 500 300 100 7,500 7,500 5,000 3,000 1,000

0 0

0 0

0 0

0 0

0 0

200

200

200

8,000 4,000 4,000 4,000 4,000 4,000 4,000

[pcs/month]

0

0

0

400

0

0

0

0

0

0

[€/month]

0

0

0 20,000 20,000 20,000

0

0

0

0

0

0

[€/month] 3,000

0

0

0 3,000

0

0

0

0

0

Total Cost

[€]

Unitary Cost

[€/pc]

3,000

0

Table 1. Evaluation of the given plan

Evaluation of the optimal production plan It is now required to formulate the optimal production plan for the fulfillment of the demand. To do so, it is necessary to make some remarks (some of them have been already done previously): internal production in a one-shift policy is more advisable, in general terms than the other two options, as it entails cost considerably lower than in the other two cases. Moreover, there is not a period in which demand is lower than the production capacity in a one-shift policy (200 pieces); unitary stock-holding cost is equal to a quarter of the back-log cost (10 €/pc versus 40 €/pc): stock-out is convenient only if it is not possible to produce a unit and stock it for more than four periods; sub-contracting is, in this specific case, more expensive than back-log: if stock-out is recoverable the next period and, in case it is not possible/convenient to use the internal production, it would be more convenient not to fulfill the demand; Formulating the optimal production plan, it is appropriate to proceed as follows: 1. Aim at fulfilling all the production capacity available the first shift; 2. Use all the production capacity of the second shift in those periods where the demand is equal to or higher than 400 pcs/month (from March to October included). Therefore, the production plan to be defined is:

Demand Production Plan

[pcs/month] [pcs/month]

Jan 200 200

Feb 200 200

Plan to be defined Mar Apr May 400 400 400 400 400 400

T Jun 500 400

Jul 500 400

Aug 400 400

Sep 400 400

Oct 400 400

Nov 200 200

Dec 200 200

It is possible to note that, compared to a demand of 4,200 pcs/year, there is a difference of 200 pieces. Thus, keeping in mind the aforementioned remarks, among several possibilities, two main alternatives should be considered: 1. Alternative 1: produce 200 units in period 2 and stock them until periods 6 and 7, when they are needed. The plan related to Alternative 1 is reported in Table 2. 2. Alternative 2: purchase from the sub-contractor 100 units in period 6 and 100 units in period 7 (completely level production plan), as the back-log is not recoverable in the next periods. The plan related to Alternative 2 is reported in Table 3. The optimal production plan is thus obtained with Alternative 2.

Month

Jan

Feb

Mar

Plan- Alternative 1 Apr May Jun

Oct

Nov

Dec

500

400

400

400

200

200

200

200

400

Cumulated Demand Plan

[pcs]

200

400

800 1,200 1,600 2,100 2,600 3,000 3,400 3,800 4,000 4,200

[pcs/month]

200

400

400

Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Back-log Back-log Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs]

200

600 1,000 1,400 1,800 2,200 2,600 3,000 3,400 3,800 4,000 4,200

Total Cost

[€]

Unitary Cost

[€/pc]

0 200

0 200

400

400

400

400

400

200

200

4

4

[pcs/month] [pcs/month]

0 0

-100 100

-100 0

0 0

0 0

0 0

0 0

0 0

[pcs/month] [€/month]

0 100 200 200 200 150 0 1,000 2,000 2,000 2,000 1,500

50 500

0 0

0 0

0 0

0 0

0 0

[pcs/month] [€/month]

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

[pcs/month]

200

400

400

400

400

400

400

400

400

400

200

200

4

4,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 4,000 4,000

84

[€/month]

0 200

400

500

Sep

[pcs/month]

400

400

Aug

Demand

200 200

400

T Jul

[pcs/month]

0

0

0

0

0

0

0

0

0

0

0

0

[€/month]

0

0

0

0

0

0

0

0

0

0

0

0

3,000 3,000

0

0

0

0

0

0

0

0 3,000

0

[€/month]

9,0

0

9,0 102 24

Table 2. Alternative 1

Month Demand

Plan- Alternative 2 Apr May Jun 400 400 500

To

[pcs/month]

Jan 200

Feb 200

Mar 400

Cumulated Demand Plan

[pcs]

200

400

800 1,200 1,600 2,100 2,600 3,000 3,400 3,800 4,000 4,200

[pcs/month]

200

200

400

Cumulated Plan Delta PD Inventory Eom

[pcs]

200

400

800 1,200 1,600 2,100 2,600 3,000 3,400 3,800 4,000 4,200

0 0

0 0

[pcs/month] [pcs/month]

0 0

400

0 0

400

0 0

500

0 0

Jul 500

500

0 0

Aug 400

400

0 0

Sep 400

400

0 0

Oct 400

400

0 0

Nov 200

200

0 0

Dec 200

200

0 0

4,2

4,2

AIL Inventory cost Back-log Back-log Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs/month] [€/month]

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

[pcs/month] [€/month]

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

[pcs/month]

200

200

400

400

400

400

400

400

400

400

200

200

Total Cost

[€]

99,0

Unitary Cost

[€/pc]

23.5 €

[€/month]

4,0

4,000 4,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 4,000 4,000 80,0

[pcs/month]

0

0

0

0

0

100

100

0

0

0

0

0

[€/month]

0

0

0

0

0 5,000 5,000

0

0

0

0

0 10,0

[€/month]

3,000

0 3,000

0

0

0

0

0 3,000

0

Table 3. Alternative 2

0

0

2

9,0

Exercise 3 The current production plan for the firm A is reported in the following table: Demand Production Planning

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 240 120 240 240 240 360 360 240 240 240 120 240 240 240 240 240 240

360 360 0

240

240 240

240

The production line works continuously so every change in the production rate leads to high setup costs, equal to 2,000 €/set-up. The production capacity of the line is equal to 200 pieces/month, and the cost of internal production is 25 €/piece. The firm has also the possibility to purchase the final product from a sub-contractor for 14 €/piece, without any capacity constraint. The current stock holding cost (under a hypothesis of infinity warehouse capacity) and back-log cost are, respectively, equal to 10.00 and 60.00 €/(piece*month). The inventory level at the beginning of the year should be considered as null. In the warehouse, a quality control activity for the units arriving from the sub-contractor is established, quantifiable in 10 €/piece. It is required to evaluate the minimum total cost and unitary cost for the implementation of the given production plan.

Solution Exercise 3 To evaluate the minimum total cost and unitary cost, we have to understand the different costs related to the plan on which we can act, i.e. we have to identify the production strategy that minimizes the cost of the given production plan. As the plan is, in this case, given, the costs related to stock-holding and back-log would not vary regardless of how we fulfill the demand. The only leverage we have to minimize the total cost, if a plan is given, is to organize the production within internal production and sub-contracting. Analyzing the cost, we can observe that: The internal production cost is 25 €/piece. The sub-contractor price is 14 €/piece, to which we have to sum 10€/piece for the quality control, for a total sub-contractor price is 24 €/piece. As can be observed the unitary cost of sub-contracting is lower than producing internally. If we produce internally, then, we should have to consider a set-up in August and a set-up in September for the change in the production rate, assuming the set-up in January as mandatory and not differential. It is then clear that it is more convenient to fulfill the demand buying all the quantity required from the sub-contractor and not make use of internal production. The evaluation of the minimum total cost and the unitary cost is reported in the Table.

To Month Demand Cumulated Demand Plan Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Back-log Back-log Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs/month] [pcs]

Jan 240 240

Feb 120 360

Mar 240 600

Apr May Jun Jul Aug Sep Oct Nov Dec 240 240 360 360 240 240 240 120 240 840 1,080 1,440 1,800 2,040 2,280 2,560 2,680 2,880

[pcs/month] [pcs]

240 240

240 480

240 720

240 240 360 360 0 240 240 240 240 960 1,200 1,560 1,920 1,920 2,160 2,400 2,640 2,880

[pcs/month] [pcs/month]

0 0

120 120

0 120

0 120

0 120

-240 0

0 0

0 0

120 0

0 0

[pcs/month] [€/month]

0 0

60 120 120 120 120 120 600 1,200 1,200 1,200 1,200 1,200

60 600

0 0

0 0

0 0

0 0

[pcs/month] [€/month]

0 0

0 0

0 0

0 0

0 0

0 0

0 120 120 120 0 7,200 7,200 7,200

0 0

0 0 21,6

[pcs/month]

0

0

0

0

0

0

0

0

0

0

0

0

[€/month]

0

0

0

0

0

0

0

0

0

0

0

0

240

240

240

240

240

360

360

0

240

240

240

240

[pcs/month]

0 120

0 120

2,8

2,8

7,2

0

2,8

[€/month]

5,760 5,760 5,760 5,760 5,760 8,640 8,640

0 5,760 5,760 5,760 5,760 69,

[€/month]

2,000

0

Total Cost

[€]

Unitary Cost

[€/pc]

0

0

0

0

0

0

0

0

0

0 2,00 99,9 34 €

Exercise 4 You are given the plan in the following table: Demand Forecast [pcs/month] Production Plan [pcs/month]

Jan 400 400

Feb Mar Apr May Jun Jul 400 1200 800 800 1,000 1,000 400 800 1,600 1,600 1,600 400

Aug 800 400

Sep Oct 800 800 400 400

Nov 400 400

Dec 400 400

The capacity of the production line is 800 pcs/month and the internal production cost is 10.00 €/piece. Stock-holding and stock-out (meant as lost margin) costs are, respectively, 5.00 €/(piece*month) and 30.00 €/piece. The production line works continuously so every change in the production rate corresponds a high set-up cost of 3,000 €/set-up and a set-up is needed at the beginning of the year. There is the possibility to purchase the final product from a sub-contractor for 15 €/piece, without any capacity constraint and the inventory level at the beginning of the year should be considered as null. Evaluate the best production strategy for the given production plan and comment on the obtained results, suggesting improvements.

Solution Exercise 4 Evaluation of the best production strategy The best production strategy for the given plan is the one that minimizes the total cost of the plan. As the plan is, in this case, given, the costs related to stock-holding and stock-out would not vary regardless of how we fulfill the demand. The only leverage we have to minimize the total cost, if a plan is given, is to organize the production within internal production and sub-contracting. Notably, the unitary cost for internal production (10 €/pc) is lower than sub-contracting (15 €/pc), and, consequently, it is more convenient to produce internally than to turn to sub-contract. We will thus produce internally according to the maximum capacity of the line, make use of sub-contracting for the remaining quantity. As pinpointed in the text, the stock-out is meant as lost margin and it is thus no recoverable and no further penalties have to be considered. The evaluation of the minimum total cost and unitary cost for the given plan is reported in Table 1.

Month Demand

Given Plan Apr May

Jan

Feb

Mar

400

400

1,200

800

800

400

800

2,000

2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

400

800

1,600

1,600

1,600

400

800

1,600

3,200

4,800

6,400 6,800 7,200 7,600 8,000 8,400 8,800

[pcs/month] [pcs/month]

0

0

-400

800

800

0

0

0

800

[pcs/month] [€/month]

0

0

0

0

0

[pcs/month] [€/month]

0

0

0

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs]

Total Cost

[€]

Unitary Cost

[€/pc]

[pcs/month]

[pcs/month]

400

Jun

Jul

Aug

Sep

Oct

Nov

Dec

1,000 1,000

800

800

800

400

400

400

400

400

400

-400

-400

0

0

1,600

2,200 1,600 1,200

800

400

400

400

400

1,200

1,900 1,900 1,400 1,000

600

400

400

0

2,000

6,000

9,500 9,500 7,000 5,000 3,000 2,000 2,000

400

0

0

0

0

0

0

0

0

0

0 12,000

0

0

0

0

0

0

0

0

0

800

400

400

400

400

400

400

800

800

800

4,000 4,000

8,000

8,000

8,000

800

800

-600

400

-400

400

600

400

[€/month]

[pcs/month]

8,000 4,000 4,000 4,000 4,000 4,000 4,000

0

0

0

800

0

0

0

0

0

0

0

0

0 12,000 12,000 12,000

0

0

0

0

0

0

3,000

0

0 3,000

0

0

0

0

0

[€/month]

[€/month]

3,000

0

0

Table 1. Evaluation of the minimum total cost and unitary cost for the given plan.

Comments and suggested improvements Although the total production equals the total demand, we can notice an unbalance, as in March we face a stock-out of 400 pieces, while, at the end of the considered planning horizon, we have an average of 400 pieces in inventory, as November and December), that are not needed. Indeed, we cannot use the stock to replenish the inventory for the following year, as the inventory at the beginning of the year should be considered as null. We should thus focus our attention on: -

Avoid the stock-out, i.e. produce enough quantity in January, February and March to cover the stock-out in March;

-

Avoid the stock-holding at the end of the year, reducing the production from April to December.

Stock-out Alternative 1 To tackle the stock-out issue we can produce 1,200 pieces in March instead of the current 800 pieces. The stock-out indeed has a cost of 12,000 €, while the production of 400 more pieces in March making use to sub-contracting (the capacity of the line is saturated) has a cost of 400 [pcs]*15 [€/pcs] = 6,000 €. The evaluation of Alternative 1 for the stock-out related issue is reported in Table 2. Alternative 2 Another possible solution is to produce internally exploiting the residual internal capacity available in February, thus producing 800 pieces against the actual 400 pieces. The set-up is not differential as it would have just been shifted from March to February. The internal production cost becomes 400 [pcs]*10 [€/pcs] = 4,000 €, to which, however, the stock-holding should be added. The AIL in February and March is 200 pcs equals to a stock-holding cost of 2,000 €. The evaluation of Alternative 2 for the stock-out related issue is reported in Table 3. Alternative 1 and Alternative 2 lead both to a total cost of 161,000 € and a unitary cost of 17.5 [€/pc].

Month Demand

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs]

Total Cost

[€]

Unitary Cost

[€/pc]

[pcs/month]

Stock-out Alternative 1 Mar Apr May Jun

Jan

Feb

400

400 1,200

800

800

400

800 2,000

2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

400 1,200

1,600

1,600

1,600

400

800 2,000

3,600

5,200

6,800 7,200 7,600 8,000 8,400 8,800 9,200

[pcs/month] [pcs/month]

0

0

0

800

800

0

0

0

800

[pcs/month] [€/month]

0

0

0

0

0

[pcs/month] [€/month]

0

[pcs/month]

Jul

Aug

Sep

Oct

Nov

Dec

1,000 1,000

800

800

800

400

400

600

400

-600

400

400

400

400

400

-400

-400

-400

0

0

1,600

2,200 1,600 1,200

800

400

400

400

400

1,200

1,900 1,900 1,400 1,000

600

400

400

0

2,000

6,000

9,500 9,500 7,000 5,000 3,000 2,000 2,000

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

400

400

800

800

800

800

400

400

400

400

400

400

4,000 4,000 8,000

8,000

8,000

800

800

[€/month]

[pcs/month]

400

8,000 4,000 4,000 4,000 4,000 4,000 4,000

0

0

800

0

0

0

0

0

0

0

0 6,000 12,000 12,000 12,000

0

0

0

0

0

0

0 3,000

0

0

0

0

0

[€/month]

[€/month]

3,000

0 3,000

0

0

Table 2. The evaluation of Alternative 1 for the stock-out related issue.

Month Demand

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom

[pcs]

[pcs/month]

[pcs/month] [pcs/month]

Stock-out Alternative 2 Mar Apr May Jun

Jan

Feb

400

400 1,200

800

800

400

800 2,000

2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

800

800

1,600

1,600

1,600

400 1,200 2,000

3,600

5,200

6,800 7,200 7,600 8,000 8,400 8,800 9,200

0

400

-400

800

800

0

400

0

800

1,600

Jul

Aug

Sep

Oct

Nov

Dec

1,000 1,000

800

800

800

400

400

600

400

-600

400

400

400

400

400

-400

-400

-400

0

0

2,200 1,600 1,200

800

400

400

400

AIL Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs/month] [€/month]

0

[pcs/month] [€/month]

0

0

0 400

Total Cost

[€]

Unitary Cost

[€/pc]

[pcs/month]

200

200

400

1,200

1,900 1,900 1,400 1,000

600

400

400

0 1,000 1,000

2,000

6,000

9,500 9,500 7,000 5,000 3,000 2,000 2,000

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

800

800

800

800

800

400

400

400

400

400

400

4,000 8,000 8,000

8,000

8,000

800

800

[€/month]

[pcs/month]

8,000 4,000 4,000 4,000 4,000 4,000 4,000

0

0

0

800

0

0

0

0

0

0

0

0

0 12,000 12,000 12,000

0

0

0

0

0

0

0 3,000

0

0

0

0

0

[€/month]

[€/month]

3,000 3,000

0

0

0

Table 3. The evaluation of Alternative 2 for the stock-out related issue.

Stock-holding Alternative 1 As for the stock-holding issue, we should try to avoid the production of the extra 400 pieces as soon as possible, to lower the monthly AIL from April to December. We can thus reduce the planned production in April from the actual 1,600 pieces to 1,200 pieces. This action would reduce the annual stock holding cost from 46,000 € to 37,000 € and the sub-contracting cost by 400 [pcs]*15 [€/pcs] = 6,000 €. The evaluation of Alternative 1 for the stock-holding related issue is reported in Table 4. Alternative 2 We could keep the stocks from April to November, not produce in December, and use the available stocks for fulfilling the demand of December, minimizing the number of set-ups needed (if we stopped the production in November and then produced again in December we would have had an extra set-up). This action would reduce the annual stock holding cost from 46,000 € to 45,000 € and the internal production cost by 400 [pcs]*10 [€/pcs] = 4,000 €. However, we need an extra set-up in December, which leads to an additional 3,000 €. The evaluation of Alternative 2 for the stock-holding related issue is reported in Table 5. Alternative 1 thus results better than Alternative 2. The former indeed leads to a total cost of 144,000 € and a unitary cost of 17.143 €/pc, while the latter to a total cost of 165,000 € and a unitary cost of 19.643 €/pc. Stock-out and Stock-holding To tackle the two issues together, we can combine Alternative 1 for the stock-holding related issue, as its unitary cost is lower, and one of the two alternatives for the stock-out related issue, as they have the same unitary cost. By way of example, the evaluation of the combination of the two Alternative 1 is reported in Table 6.

Month Demand

Stock-holding Alternative 1 Mar Apr May Jun

Jan

Feb

400

400

1,200

800

800

400

800

2,000 2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

400

800 1,200

1,600

1,600

400

800

1,600 2,800

4,400

6,000 6,400 6,800 7,200 7,600 8,000 8,400

[pcs/month] [pcs/month]

0

0

-400

400

800

0

0

0

400

1,200

[pcs/month] [€/month]

0

0

0

200

800

0

0

0 1,000

4,000

[pcs/month] [€/month]

0

0

400

0

0

0

0

0

0

0

0 12,000

0

0

0

0

0

800

800

800

400

400

8,000 8,000

8,000

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[pcs]

Total Cost

[€]

Unitary Cost

[€/pc]

[pcs/month]

[pcs/month]

400

400

800

Jul

Aug

Sep

Oct

Nov

Dec

1,000 1,000

800

800

800

400

400

600

400

400

400

400

400

400

-600

-400

-400

-400

0

0

1,800 1,200

800

400

0

0

0

1,500 1,500 1,000

600

200

0

0

7,500 7,500 5,000 3,000 1,000

0

0

0

0

0

0

0

0

0

400

400

400

400

[€/month] 4,000 4,000 [pcs/month]

400

800

8,000 4,000 4,000 4,000 4,000 4,000 4,000

0

0

0

800

0

0

0

0

0

0

0

0

0 6,000 12,000 12,000

0

0

0

0

0

0

3,000

0

0 3,000

0

0

0

0

0

[€/month]

[€/month]

3,000

0

0

Table 4. The evaluation of Alternative 1 for the stock-holding related issue.

Month Demand

Stock-holding Alternative 2 Mar Apr May Jun

Jan

Feb

400

400

1,200

800

800

400

800

2,000

2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

400

800

1,600

1,600

1,600

400

800

1,600

3,200

4,800

6,400 6,800 7,200 7,600 8,000 8,400 8,400

[pcs/month] [pcs/month]

0

0

-400

800

800

0

0

0

800

[pcs/month]

0

0

0

400

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom AIL

[pcs]

[pcs/month]

Jul

Aug

Sep

Oct

Nov

Dec

1,000 1,000

800

800

800

400

400

600

400

-600

400

400

400

400

0

-400

-400

-400

0

-400

1,600

2,200 1,600 1,200

800

400

400

0

1,200

1,900 1,900 1,400 1,000

600

400

200

Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting Subcontracting cost Set-up cost

[€/month]

0

0

0

2,000

6,000

[pcs/month] [€/month]

0

0

400

0

0

0

0

0

0

0

0

0

0

0 12,000

0

0

0

0

0

0

0

0

0

800

400

400

400

400

400

0

8,000 4,000 4,000 4,000 4,000 4,000

0

Total Cost

[€]

Unitary Cost

[€/pc]

[pcs/month]

400

400

800

800

800

4,000 4,000

8,000

8,000

8,000

800

800

9,500 9,500 7,000 5,000 3,000 2,000 1,000

[€/month]

[pcs/month]

0

0

0

800

0

0

0

0

0

0

0

0

0 12,000 12,000 12,000

0

0

0

0

0

0

3,000

0

0 3,000

0

0

0

0 3,000

[€/month]

[€/month]

3,000

0

0

Table 5. The evaluation of Alternative 2 for the stock-holding related issue.

Month Demand

[pcs/month]

Cumulated Demand Plan

[pcs]

Cumulated Plan Delta PD Inventory Eom AIL Inventory cost Stock-out Stock-out Cost Internal Production Internal Production Cost Subcontracting

[pcs]

[pcs/month]

Stock-out Alternative 1 and Stock-holding Alternative 1 Jan Feb Mar Apr May Jun Jul Aug

Sep

Oct

Nov

Dec

400

400 1,200

800

800

800

800

400

400

400

800 2,000 2,800

3,600

4,600 5,600 6,400 7,200 8,000 8,400 8,800

400

400 1,200 1,200

1,600

1,600

400

800 2,000 3,200

4,800 6,4700 6,800 7,200 7,600 8,000 8,400 8,800

1,000 1,000

400

400

[pcs/month] [pcs/month]

0

0

0

400

800

0

0

0

400

1,200

[pcs/month] [€/month]

0

0

0

200

800

0

0

0 1,000

4,000

[pcs/month] [€/month]

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

400

400

800

800

800

800

400

400

4,000 4,000 8,000 8,000

8,000

[pcs/month]

600

400

800

T

400

400

400

-600

-400

-400

-400

0

0

1,800 1,200

800

400

0

0

0

1,500 1,500 1,000

600

200

0

0

7,500 7,500 5,000 3,000 1,000

0

0

0

0

0

0

0

0

0

400

400

400

400

[€/month]

[pcs/month]

0

0

400

400

800

8,000 4,000 4,000 4,000 4,000 4,000 4,000 800

0

0

0

0

0

0

Sub[€/month] contracting cost Set-up cost [€/month] Total Cost

[€]

Unitary Cost

[€/pc]

0

3,000

0 6,000 6,000 12,000 12,000

0 3,000

0

0

0

0

0

0

0

0

0 3,000

0

0

0

0

0

9 1

Table 6. The evaluation of the combination of the two Alternatives 1.

Exercise 5 The Elettro firm operates in the electronics sector and uses the Wagner Within algorithm as a base for the aggregate production plan. Referring to code ELE103 and under the following hypotheses: planning horizon of 6 periods, each period consists of two weeks; orders received for the first quarter as reported in the following table; every production launch is related to a cost of staring of the production of 300 €; variable cost of the product 200 €/pc; annual stock holding rate equal to 15% of the variable cost. Based on previous data, it is requested to: evaluate the optimal production plan for the first quarter of the year 2009; evaluate, separately, set up costs and stock holding costs. Period 1 2 3 4 5 6 Orders 180 260 200 220 300 200 [units] TABLE 1 – Received orders for product ELE103 for the first quarter of the year 2009

Solution Exercise 5 Evaluation of the optimal production plan 1 1 2 … … … T

2

3

4

5

6

F(t)

For the evaluation of the optimal plan, it necessary, firstly, to define … the previous table, in which: ,,, … T is the planning horizon; … F(t,j) … t is the period until which the … … … … considered horizon extents to (t … F(T,j) … … F(T,T) included between 1 and T); j is the last period in which the last set up is realized (j included between 1 and T, and can not, of course, be higher than t; F(t,j) is the cost of the plan, until period t, with the last setup realized in j F(t) = minj F(t,j) F(1,1) F(2,1) … F(t,1) … F(T,1)

F(t,j) is equal to:

Where: a(j) is the set-up cost in the period j; i(h) is the stock holding cost in period h; D(k) is the demand in period k. For the specific problem, is possible to proceed as following: F (1,1) = a(1) = 300€ and, of course, F(1) = 300€ F (2,1) = a(1) + i(1) * D(2) = 300 + 1.25*260 = 625 €

F(1) F(2) … F(t) … F(T)

The stock holding cost for one period is

F (2,2) = F(1) + a(2) = 300 + 300 =600€ And thus F (2) = 600 For what concerns period 3: F (3,2) = F(1) + a(2) + i(2)*D(3) = 300 + 300 + 1,25 * 200 = 850€ F (3,3) = F(2) + a(3) = 600 + 300 = 900€ And thus F (3) = 850 € According to Planning Horizon Theorem, is it now possible to exclude F (3,1), since F(2) is equal to F(2,2). Therefore, for the production planning of the following periods (i.e. from period 4 on), it will be possible to avoid the evaluation of alternatives planning to have the production of those periods (i.e. from period 4 on) in period 1. For this reason: F (4,2) = F (1) + a(2) + i(2)*D(3) + i(2) *D(4) + i(3)*D(4) = 300 + 300 + 1,25*200 + 1,25*220 + 1,25*220 = 1,400€ F (4,3) = F (2) + a (3) + i(3)*D(4) = 600 + 300 + 1,25*220 =1,175€ F (4,4) = F (3) + a (4) =850 + 300 = 1,150€ And, thus F (4) = 1,150€ F(5,4) = F(3) + a(4) + i(4)*D(5) = 850 + 300 + 1,25*300 = 1,525€ F(5,5) = F(4) + a(5) = 1,150 +300 =1,450 € And thus F(5) = 1,450€ F(6,5) = F(4) + a(5) + i(5)*D(6) = 1,150 + 300 + 1,25*200 = 1,700€ F(6,6) = F(5) + a(6) = 1,450 + 300 = 1,750€ And so F(6) = 1,700€, F(6) is the minimum cost of the plan. 1 2 3 4 5 6

1 300 625 1125 1950 3450 3820

2

3

4

5

6

600 850 1,400 2525 3525

900 1,175 1925 2675

1,150 1,525 2025

1,450 1,700

1,750

F(t) 300 600 850 1,150 1,450 1,700

It is now possible to detail the production plan, proceeding backward, starting from the last period to the first one: F(6) = 1,700 = F(6,5). The optimal plan is obtained with a production launch in periods 5, in which the production realized covers both period 5 and 6. At this point, knowing the policy for period 5, period 4 is considered. F(4) = 1,150 = F(4,4). The optimal plan is obtained producing in period 4 only the requirements for this period. At this point period 3 is considered. F(3) = 850 = F(3,2). The optimal plan is obtained producing in period 2 both for periods 3 and 2. Then period 1 is considered.

F(1) = 300. The optimal plan is obtained producing in period 1 the requirements for this period. Therefore, the obtained plan is reported in the following table: Optimal Plan

1

2

3

4

5

6

180

460

0

220

500

0

Evaluation of the plan related costs The relevant costs for the Wagner-Whitin plan are setup and stock-holding costs: 1 Optimal plan 180 Set up 1 Units stocked

2 460 1 200

3 0

4 220 1

5 500 1

6 0

200

Total setup cost is equal to the number of set up for the unitary cost of set up: CSU = NSU * UCS = 4 *300 = 1,200 € Total stock holding costs are equal to the number of units stocked in a period for the unitary stockholding cost: CMS = NUS * UCSH = 400 *1,25 = 500€ Therefore, the total cost of the plan is: 1,200 + 500 = 1,700 € (as already calculated).

Chapter 7. On-demand Management (MRP)

Exercise 1 Adaptor Ltd produces laptops transformers. The production process of the different codes is mainly based on the assembly phase of components that are either produced by Adaptor Ltd or ordered from external suppliers. One of the most popular products is the universal adaptor “All in”, whose bill of material, coefficients of use and all data related to production policy are reported in the following figure and table. The quality control implemented within the firm guarantees the compliance of all the materials stocked in the warehouse.

It is requested to: 1) Determine the plan of orders for the second semester of 2009 for each code within the bill of material; 2) Discuss the solution and qualitatively point out possible corrective actions. Month Requirements all ok

July 200

Code

Type

All ok Cable Transformer Block Metal Plastic

Production Purchase Production Production Purchase

August 50

September 300

October 150

November 200

Initial Coefficient of Lead time Defectiveness Safety inventory use [months] Coefficient stocks [units] 1 100 10% 3 2 700 4% 1 1 250 2% 1 1 50 2% 2 3 1,290 1%

Decembe 300 Re-order Policy

0 0 0 0 0

L4 EOQ=60 L4 L4 EOQ=50

Solution Exercise 1 Plan of orders The definition of the plan of orders is developed through the use of an on-demand management approach, thanks to the implementation of the MRP procedure (Material Requirement Planning). MRP is implemented for each of the components included in the bill of material, starting from the values of: Gross Total Requirements: it is determined starting from the requirements of the higher-level item (parent), then multiplied by the coefficient of use; Initial Inventory: is the quantity available in the warehouse in period 0. Then, for each period, it is necessary to evaluate: Initial availability: in period 1 is the difference between Initial Inventory and Safety Stocks; in the other periods, it is evaluated as the sum of Initial Availability and Lot- size Requirements minus the Gross Total Requirements of the previous period; Net Requirements: it is equal to the difference between Gross Total Requirements and Initial Availability; Scraps-adjusted Net Requirements: it is evaluated starting from Net Requirements multiplied by (1 + coefficient of defectiveness); Orders in Progress: orders issued previously, whose arrival is expected in the specific period;

Order in Progress adjusted Net Requirements: it is evaluated from the Scraps-adjusted Net Requirements considering the Orders in Progress. Lot size Requirements: it is the requirement expressed according to the re-order policy. Indeed, according to the specific policy, it could be equal to the Order in Progress adjusted Net Requirements (L4L policy) or multiple of a fixed quantity (lot policy); Orders to Issue: Order to be issued in a specific period. For each period i-th it is equal to the Lot size Requirements of period i-th + n, where n indicates the period of lead time. All ok Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

Cable Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 200 0

2 50 0

3 300 0

4 150 0

5 200 0

6 300 0

100 100 110 0 110 110 55

0 50 55 0 55 55 330

0 300 330 0 330 330 165

0 150 165 0 165 165 220

0 200 220 0 220 220 330

0 300 330 0 330 330 0

1 165 0

2 990 0

3 495 0

4 660 0

5 990 0

6 0 0

370 0 0 0 0 0 600

205 785 817 0 817 1,200 600

368 127 133 0 133 600 1200

449 211 220 0 220 600 0

365 625 60 0 650 1200 0

526 0 0 0 0 0 0

100

110

0 330 700

1,200

At the end of the second period the Initial Availability for period 3 is evaluated as follows:

So, Net Requirements for period 3 will be 495-368=127, the Scrap and OC Adjusted Requirements will be 133, and an order of 600 units will be unit (EOQ=600). Similarly, at the end of the third period the Initial Availability for period 4 is evaluated as follows:

Transformer block Gross Total Requirements SS Initial Inventory Initial Availability

0 110

1 55 0

2 330 0

3 165 0

4 220 0

5 330 0

6

140

85

0

0

0

0

0

250

Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

Metal Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

Plastic Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0 0

0 0 0 0 0 250

245 250 0 250 250 169

165 169 0 169 169 225

220 225 0 225 225 337

330 337 0 337 337 0

0 0 0 0 0 0

1 250 0

2 169 0

3 225 0

4 337 0

5 0 0

6

50 200 204 0 204 204 173

0 169 173 0 173 173 230

0 225 230 0 230 230 344

0 337 344 0 344 344 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

1 500 0

2 338 0

3 450 0

4 674 0

5 0 0

6 0 0

1,290 0 0 0 0 0 1,000

790 0 0 0 0 0 0

452 0 0 0 0 0 0

2 672 679 0 679 1,000 0

318 0 0 0 0 0 0

318 0 0 0 0 0 0

0

50

204

0 0 1,290

Discussion on the solution The plan presents several infeasibilities, as for Code All ok, Cable and Metal it would be necessary to issue an order in Period 0. In general terms, in case of an infeasibility, possible solutions are the following: Use the safety stocks, if sufficient, and then replenish them as soon as possible - in this specific case, it would have not been possible as there are not safety stocks; Issue a compelling order, thus reducing the lead time (if the order is issued to a supplier, it might be necessary to pay an extra cost to the suppliers); Ask external suppliers if the code is usually produced within the firm, or ask to other suppliers if the code is usually purchased; Use safety stocks of parent item - in this specific case, it would have not been possible as there are not safety stocks; Postpone requirements, i.e. to verify with the customers if they are willing to accept a delay (maybe it would be necessary to offer customers a further discount).

Exercise 2 GAMMA Ltd produces units for the precision manufacturing sector. Focusing on the item Premium 800F, for its production several components are necessary, among the others, the components B, W and X (Figure 1). The management of components and units is on-demand. Assuming the following data: MPS of Premium 800F for a planning horizon of six weeks, as reported in Table 1; Coefficient of use for components/units as reported in Table 2; Managements parameters (lead time, safety stocks, re-order policy) of components/ units as reported in Table 2; Initial inventory of components/units at the period of evaluation of material requirements, as reported in Table 2; and disregarding coefficient of defectiveness both for products and process, it is requested to determine for codes B, W, and X within the bill of material, the plan of orders for the following next weeks. Week MPS [pieces/week]

1 0

2 0

3 4 5 6 12 7 10 15 TABLE 1: MPS of Premium 800F

FIGURE 1: Bill of material of Premium 800F

Code Type

B W X

Coefficient of use

Lead time [weeks] 1 2

Initial Safety Re-order inventory stocks policy [units] 100 30 EOQ=100 150 10 EOQ=200

Production 10 Purchase 2 for Robot 800F; 1 for B Purchase 2 1 400 0 L4L TABLE 2: Data of components/units B, W e X

Solution Exercise 2 Components W and X are used at different levels of Serie 800f’s Bill Of Materials (BOM), we have to apply the “Low-level code” rule. It is necessary to identify the lowest level of the BOM at which a code is found (e.g. code W can be found at level 1 of BOM and level 2 of BOM), and then re-classify it at the lowest level (e.g. code W is then re-classified at level 2 of the BOM). This is needed to correctly evaluate the Gross Total Requirements of the code - only once and at the lowest level possible - as the sum of all the requirements related to its parent codes. LEVEL Level 0 Level 1

CODE Premium 800F B

Level 2 Level 3

W X

The details for the different codes according to the MRP approach will follow the order reported in the previous table: B, W, X. Starting from code B:

B Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 0 30

2 0 30

3 120 30

4 70 30

5 100 30

6 150 30

70 0 0 0 0 0 0

70 0 0 0 0 0 100

70 50 50 0 50 100 100

50 20 20 0 20 100 100

80 20 20 0 20 100 100

80 70 70 0 70 100 0

100

The gross total requirements for code W are calculated as the sum of the gross requirements deriving from the MRP of both B and 800 RF.

W Gross Requirements from Premium 800F Gross Requirements from B Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 0 0 0 10

2 0 100 100 10

3 24 100 124 10

4 14 100 114 10

5 20 100 120 10

6 30 0 30 10

140 0 0 0 0 0 200

140 0 0 0 0 0 0

40 84 84 0 84 200 200

116 0 0 0 0 0 0

2 118 118 0 118 200 0

82 0 0 0 0 0 0

150

0

Lastly, it is possible to proceed with the evaluation of the plan of orders for X, that depends only on W’s ones. X Gross Total Requirements SS Initial Inventory Initial Availability

0

1 400 0

2 0 0

3 400 0

4 0 0

5 0 0

6 0 0

400

0

0

0

0

0

400

Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

0 0 0 0 0 0

0 0 0 0 0 400

400 400 0 400 400 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

Without following an approach structured according to the levels, an incorrect plan of orders would have been defined, leading to an excess of stocks both for X and W.

Exercise 3 Consider the following Bill of Material:

Determine the plan of orders in the considered time horizon (period 1-8) for each code within the bill of material, having the following information regarding gross requirements and about the single items: Week Gross internal requirements (X) Gross external requirements (Y)

1 70 200

2 120 80

3 50 50

4 40 100

5 30 90

X Lead time [weeks] Re-order policy Initial inventory Orders in progress (good units) Safety stock Scrap rate

6 170 80

7 150 100

8 100 100

Y

4 EOQ=100 400 200 (week 3) 10 0%

Z

3 EOQ=800 1000 700 (week 2) 250 20%

2 L4L 250 50 (week 1) 150 10%

Please consider that only good units can be stocked in the warehouse.

Solution Exercise 3 First of all, the “Low-level code” rule is applied. It is necessary to identify the lowest level of the BOM at which a code is needed to calculate the Gross Total Requirements of the code - only once and at the lowest level possible - as the sum of all the requirements related to its parent codes. LEVEL Level 0 Level 1 Level 2

CODE X Z Y

We start evaluating the MRP for Ian- Code X (Level 0):

X Gross Total Requirements SS Initial Inventory Initial Availability

0

1 70 10

2 120 10

3 50 10

4 40 10

5 30 10

6 170 10

7 150 10

8 100 10

390

320

200

350

310

280

110

60

400

Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0 0 0 0 0 0

0 0 0 0 0 0

0 0 200 0 0 100

0 0 0 0 0 100

0 0 0 0 0 0

0 0 0 0 0 0

40 40 0 40 100 0

40 40 0 40 100 0

We continue evaluating the MRP fo2r Code z (Level 1): Z Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 0 150

2 0 150

3 220 150

4 220 150

5 0 150

6 0 150

7 0 150

8 0 150

100 0 0 50 0 0 77

150 0 0 0 0 0 242

150 70 77 0 77 77 0

0 220 242 0 242 242 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

250

We move to the last level (Level 2), evaluating the MRP for Code Y. The gross total requirement for Code Y is calculated as the sum of the gross requirements deriving from the MRP of Code X, Code Z, and considering the gross external requirement provided in the data for Code Y. Y Gross Requirements – From X Gross Requirements – From Z Gross External Requirements Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 0 231 200 431 250

2 0 726 80 806 250

3 100 0 50 150 250

4 100 0 100 200 250

5 0 0 90 90 250

6 0 0 80 80 250

7 0 0 100 100 250

8 0 0 100 100 250

750 0 0 0 0 0 800

319 487 585 700 0 0

213 0 0 0 0 0

63 137 165 0 165 800

529 0 0 0 0 0

439 0 0 0 0 0

359 0 0 0 0 0

259 0 0 0 0 0

1000

Exercise 4 (Exam 18th July 2016) Month 1 Requirement Code A 5

2 8

3 9

4 5 6 10 12 10

Blue Monday Ltd produces pens. The production process of the different products is mainly based on the assembly phase of components that are either produced by Blue Monday Ltd or ordered from external suppliers. One of the most popular products is the pen “Ian”, identified with the code A, whose bill of material and data are reported in the following figure and table. It is requested to: 1) Determine for each code within the bill of material the plan of orders for the six months considered. 2) Discuss the solution and point out possible corrective actions. Please keep in mind that another firm asked Blue Monday Ltd 200 units of Code C in Month 1, and Blue Monday is going to deliver this quantity. Code Type A B C D

Production Purchase Production Purchase

Lead Coefficient Initial Defectiveness Safety Time Policy of use Inventory coefficient Stocks (months) 0 1 5 10% 0 L4L 5 2 1,900 0% 100 EOQ=400 20 1 700 10% 200 L4L 10 3 1,500 5% 0 EOQ=500

Solution Exercise 4 Plans of orders First of all, the “Low-level code” rule is applied. It is necessary to identify the lowest level of the BOM at which a code is needed to calculate the Gross Total Requirements of the code - only once and at the lowest level possible - as the sum of all the requirements related to its parent codes. LEVEL Level 0 Level 1 Level 2

CODE Ian- Code A C D, B

We start evaluating the MRP for Ian- Code A (Level 0):

A Gross Total Requirements SS Initial Inventory

0

5

1 5 0 0

2 8 0 0

3 9 0 0

4 10 0 0

5 12 0 0

6 10 0 0

Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

5 0 0 0 0

0 8 9 0 9 9 10

9

0 9 10 0 10 10 11

0 10 11 0 11 11 14

0 12 14 0 14 14 11

0 10 11 0 11 11 0

4 280 200 0 0 280 308 0 308 308 242

5 220 200 0 0 220 242 0 242 242 0

6 0 200 0 0 0 0 0 0 0 0

We continue evaluating the MRP for Code C (Level 1): C Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1 380 200 0 500 0 0 0 0 0 88

700

2 200 200 0 120 80 88 0 88 88 242

3 220 200 0 0 220 242 0 242 242 308

We move to the last level (Level 2), evaluating the MRP for Code D and Code B. The gross total requirements for Code B are calculated as the sum of the gross requirements deriving from the MRP of Ian- Code A and Code C. D Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

B (From A and C) Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements

-1

0

1,500

2,000

0

1,900

3,500

1 485 100 0 1,800 0 0 0 0

1 880 0 0 1,500 0 0 0 0 0 2,500

2 1,260 100 0 1,315 0 0 0 0

2 2,420 0 0 620 1,800 1,890 0 1,890 2,000 0

3 1,595 100 0 55 1,540 1,540 0 1,540

3 3,080 0 0 104 2,976 3,125 0 3,125 3,500 0

4 1,280 100 0 60 1,220 1,220 0 1,220

4 2,420 0 0 357 2,063 2,167 0 2,167 2,500 0

5 0 0 0 317 0 0 0 0 0 0

5 55 100 0 380 0 0 0 0

6 0 0 0 317 0 0 0 0 0 0

6 0 100 0 325 0 0 0 0

Lot-size Requirements Orders to issue

0

1,600

0 1,600

1,600 0

1,600 0

0 0

0 0

Discussion of the results and possible corrective actions The plan of orders for Code D has two infeasibilities, as, according to the plan, it would be necessary to issue an order in period 0 and another in period -1. To eliminate the infeasibilities, it is possible to take different corrective actions, among them: Use the safety stocks of the parent item (as, Code D unfortunately, has no safety stocks). Parent items are Ian- Code A and Code C, but only Code C has safety stocks. Using the safety stocks of Code C, however, would eliminate only the infeasibility in period -1: C Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue D Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

700

0

1,500

3,000

1 380 200

2 200 0

3 220 0

4 280 0

0 500 0 0 0 0 0 0

0 320 0 0 0 0 0 110

0 120 100 110 0 110 110 308

0 0 280 308 0 308 308 462

1 0 0 0 1,500 0 0 0 0 0 5,000

2 1,100 0 0 1,500 0 0 0 0 0 0

3 3,080 0 0 400 2,680 2,814 0 2,814 3,000 0

4 4,620 0 0 177 4,443 4,666 0 4,666 5,000 0

5 220 200[4] 0 0 220+200 462 0 462 462 0 5 0 0 0 318 0 0 0 0 0 0

6 0 200 0 0 0 0 0 0 0 0 6 0 0 0 318 0 0 0 0 0 0

Decide to not fulfill the order of 200 parts of Code C in month 1. This action, however, would eliminate only the infeasibility in period -1, and might have negative impact of firm’s image: C Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

700

1 180 200 0 500 0 0 0 0 0 0

2 200 200 0 320 0 0 0 0 0 110

3 220 200 0 120 100 110 0 110 110 308

4 280 200 0 0 280 308 0 308 308 242

5 220 200 0 0 220 242 0 242 242 0

6 0 200 0 0 0 0 0 0 0 0

D Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

1,500

3,000

1 0 0 0 1,500 0 0 0 0 0 2,500

2 1,100 0 0 1,500 0 0 0 0 0 0

3 3,080 0 0 400 2,680 2,814 0 2,814 3,000 0

4 2,420 0 0 177 2,243 2,356 0 2,356 2,500 0

5 0 0 0 137 0 0 0 0 0 0

6 0 0 0 137 0 0 0 0 0 0

Decide not to fulfill the order of 200 parts of Code C in period 1 and use safety stocks of Code C. This action, however, would eliminate only the infeasibility in period -1: C Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

D Gross Total Requirements SS Initial Inventory Initial Availability Net Requirements Scrap-adjusted Requirements Orders in Progress (OC) OC-adjusted Requirements Lot-size Requirements Orders to issue

0

700

0

1,500

1,000

1 180 0 0 700 0 0 0 0 0 0

2 200 0 0 520 0 0 0 0 0 0

3 220 0 0 320 0 0 0 0 0 198

4 280 0 0 100 180 198 0 198 198 462

1 0 0 0 1,500 0 0 0 0 0 4,500

2 0 0 0 1,500 0 0 0 0 0 0

3 1,980 0 0 1,500 480 504 0 504 1,000 0

4 4,620 0 0 472 4,148 4,355 0 4,355 4,500 0

5 220 200 0 0 220+200 462 0 462 462 0

6 0 200

5 0 0 0 137 0 0 0 0 0 0

6 0 0 0 137 0 0 0 0 0 0

0 0 0 0 0 0 0

Issue a compelling order, thus reducing the lead time (it might be necessary to pay an extra-cost to the suppliers); Ask to other suppliers; Postpone requirements, i.e. to verify with the customers if they are willing to accept a delay (maybe it would be necessary to offer customers a discount).

Chapter 8. Scheduling

Exercise 1 (Exam 4th March 2010) A manufacturing firm has to schedule, at the 4th of March, the production plan for some jobs (J1 – J4) that must be delivered during the month of March. Due delivery dates and processing times on machines M1, M2 and M3 are reported in the following table. In developing the scheduling, it must be considered that it is not possible to process more than one job at the same time and it is not possible to interrupt the processing of a job once it has begun. The firm has always scheduled the production on the first machine following the S/OPN dispatching rule. Eng. Corri, newly hired in the production department, decided to evaluate the jobs’ scheduling using an operative planning based-approach, seeking to obtain a better solution M1

J1 J2 J3 J4

1 5 6 2

M2

M3

8 4 9 6

10 2 7 9

Requested delivery date 24/03/2010 27/03/2010 31/03/2010 1) 30/03/2010

Compute MAK and average lateness for the two solutions (S/OPN and planning-based approach); 2) In what production conditions the dispatching rules can provide better results than operative planning (scheduling)?

Solution Exercise 1 MAK and Average Lateness S/OPN At present, the production scheduling of the firm is implemented according the Slack on Operation (S/OPN) rule. It is thus necessary to determine the S/OPN index for each job that has to be processed on machine M1, remembering that:

where: i= i-th job considered; j = j-th machine whose index is under computation; = Planned delivery date for job i-th; = Time in which the index is evaluated (in this case march the 4th, 2010); = Residual processing time for job i-th job on machine k-th (S is the set of machines that have not processed job i-th yet, machine j-th under consideration included); = Number of residual operations (included operation on the jth machine) necessary to end the processing of job i-th.

For example, considering J1 and M1:

Consequently, the difference between due date and time now of J1 with reference to M1 is 20 days. The slack is 1 day and the S/OPN index is 0.33. Similarly, for M1, indices are the one reported in the table below:

M1 J1 J2 J3 J4

Slack 1 12 5 9

S/OPN 0.33 4.00 1.67 3.00

The sequence on machine M1 is thus: 1-3-4-2. Similarly, indices S/OPN for M2 and M3 can be calculated, but it is of fundamental importance to remember that pre-emption (stopping of the processing of one job on a machine in order to process another one that is more urgent) and passing (overtaking between jobs waiting to be processed, for urgency reasons) are not allowed. These two hypotheses are very strong and the no passing fixes the sequence found for M1 on the other machines (M2 and M3) - same sequence on the machines. The processing sequences for the other two machines will be the same as M1. The make span (MAK) is 37 days and the Gantt diagram, necessary for determining the MAK for the considered scheduling, is the following (note that day 1 is the 4th of March, 2010):

It is now possible to calculate the average lateness. The average lateness is the average delay (expressed in days) of the actual delivery date of a job with respect to the requested due date. In particular, observing the Gantt diagram, it is possible to compute the following actual delivery dates:

J1 J2 J3 J4

Actual delivery date 23/03/2010 10/04/2010 30/03/2010 08/04/2010

Requested due date 24/03/2010 27/03/2010 31/03/2010 30/03/2010

Lateness (days) -1 14 -1 9

The average lateness is (-1+14-1+9)/4= 5.25 days/job. Optimal Scheduling Aiming at optimizing the production scheduling using an operative planning-based approach, and as the production needs three machines, we can try to implement the Johnson’s algorithm. According to this algorithm, having the target of minimizing MAK and with a flow shop with 3 machines, the optimal solution still belongs to the set of “permutation schedule”. Johnson, indeed, demonstrated that, if the

same optimal sequence is obtained in the two sub-problems, that is M1-M2 and M2-M3, then this sequence is a priori the optimal solution also for the complete problem involving all 3 machines. Under the same hypotheses of flow-shop, considering the first two machines (M1-M2) and implementing the Johnson’s algorithm, the following sequence is obtained: 1-4-3-2. Considering then the second pair of machines (M2-M3), always implementing the Johnson’s algorithm, the following sequence is obtained: 4-1-3-2. As the two sequences do not coincide, it is not possible to identify affirm a priori the optimal solution. In what production conditions the dispatching rules can provide better results than operative planning Dispatching rules are based on the choice of processing first the job with the highest priority (assigned according to the chosen rule), without pre-ordering the other jobs; operative planning algorithm, on the contrary, are based on sequencing, that means to schedule a priori all the jobs that have to be processed by each machine. It is then clear that dispatching rules can guarantee better results than the operative planning approach (scheduling) where a high planning flexibility is requested, since they allow to obtain sub-optimal solutions; nevertheless, a first critical factor for the evaluation of the obtained results is the identification of the parameter that should be optimized by the rule.

Exercise 2 (Exam 15th February 2010) A manufacturing firm has to schedule the production of some jobs. The forecasted delivery dates and the processing times are reported in the following table. In developing the scheduling, it must be considered that it is not possible to process more than one job at the same time and it is not possible to interrupt the processing of a job once it has begun. The firm has always aimed at delivering as many jobs as possible within the forecasted due. Which is, according to this policy, the optimal scheduling? Eng. Telli, newly hired in the production department, estimated, on the basis of historical data, the cost of each day of delay compared to the due date for each job. Which could be an optimal scheduling that, respecting firm’s delivery policy, would improve firm’s performance? Job Duration [n] [days] 1 4 2 5 3 9 4 3 5 5 6 5 7 7

Forecasted Due Date [day] 6 20 8 10 4 25 15

Penalty for delay [€/day] 100 90 70 80 100 60 5

Solution Exercise 2 Optimal Scheduling according to the actual policy The schedule of production is based on an operative planning-based approach, in particular on the Hodgson’s algorithm, whose objective is to minimize the number of tardy jobs. It is possible to proceed with the implementation of the algorithm. Iteration 1 STEP 1: Two sets are defined: L, the set of the tardy jobs; and E, the set of non-tardy jobs. At this point L is empty, while E contains all the jobs - from 1 to 7. STEP 2: Jobs in set E are ranked according to increasing delivery date. Consequently, the first job in the set will be 5, followed by job 1, then 3, and so on. Job [n] 1 2 3 4 5 6 7

Due [days]

date Ranking

6 20 8 10 4 25 15

E 5 1 3 4

2 6 3 4 1 7 5

L

7 2 6

STEP 3: Identify the first tardy job in the set E. The first tardy job is searched starting from the first job in E, which is, in this case, 5. Job [n] 5

Duration [days] 5

Forecasted Due date [day] 4

Actual delivery date [day] 5

Delay? Yes

STEP 4: Job 5 is moved to set L. E 1 3 4 7 2 6

L 5

STEP 5: Go back to step 3. Iteration 2 STEP 3: Identify the second tardy job within set E (that now contains all the jobs from 1 to 7, except from 5), starting from the first job (now it is job 1). Job 1 needs a processing time of 4 days and the due date is in 6 days, thus it is not tardy. Proceeding with the analysis of the jobs in set E, ranked according to increasing due date, the following job is job 3, which needs a processing time of 9 days and the due date is in 8 days. Additionally, since job 1 will be processed before job 3, the actual delivery date will be on day 13. Consequently, this job is tardy. Job [n] 1 3

Duration [days] 4 9

Forecasted Due date [day] 6 8

Actual delivery date [day] 4 13

Delay? No Yes

STEP 4: Job 3 is moved to set L.

E 1 4 7 2 6

L 5 3

STEP 5: Go back to step 3. Iteration 3 STEP 3: Identify the third tardy jobs within set E (that now contains all the jobs from 1 to 7, but 5 and 3), starting from the first job (still job 1).

Job 1 needs a processing time of 4 days and the due date is in 6 days, so it is not tardy. Proceeding with the analysis of the jobs in set E, ranked according to increasing due date, the following job is job 4, which needs a processing time of 3 days and the due date is in 10 days.As already stated for job 3, actual delivery date of job 4 will be on day 7, as it is necessary to process job 1 first. Nevertheless, job 4 would not be tardy. We proceed with the following job that, according to increasing due date, is job 7. This job must be delivered within day 15: considering the previously processed jobs (job 1 and job 4) it would be actually delivered on day 14: 4 days required for job 1, 3 days for job 4, 7 days for job 7, thus job 7 will not be tardy. Proceeding in the same way for job 2 and job 6, it is possible to note that none of them would be tardy. Job [n] 1 4 7 2 6

Duration [days] 4 3 7 5 5

Forecasted Due date [day] 6 10 15 20 25

Actual delivery date [day] 4 7 14 19 24

Delay? No No No No No

The jobs processing sequence is the following: the jobs contained in set E (1-4-7-2-6), with a MAK of 24 days, followed by the jobs contained in set L (3 and 5) in any order: indeed, Hodgson’s algorithm does not provide any indication related to tardy jobs’ sequence. Optimal sequence for improving performance Eng. Telli estimated penalties related to every day of delay in the delivery of each job: consequently, considering the constraints related to the operative planning-based approach (Hodgson’s algorithm), it would be possible to properly schedule job 5 and job 3 so to minimize penalties. In both cases, the first of the two jobs to be processed will be processed after day 24, i.e. after the delivery of jobs of set E (14-7-2-6). Alternative 1: sequence 1-4-7-2-6 + 5-3 Starting to process job 5 on day 25, it will be delivered within day 29, with a delay of 25 days. Considering that the penalty for every day of delay has been estimated by Eng. Telli in 100 €/days, the total penalty for job 5’s delay is 25 x 100 = 2,500 €. After job 5, job 3 can be processed on day 30 and delivered on day 38, with 30 days of delay. Proceeding similarly, the total penalty for job 3’s delay is 30 x 70 = 2,100 €. Consequently, choosing this sequence total penalties for delivery delays are 2,500 + 2,100 = 4,600 € Job Duration [n] [days] 5 3

5 9

Due date [day] 4 8

Actual delivery date [day] 29 38

Delay? Penalty Penalty [days] [€/day] [€] 25 30

100 70

2,500 2,100

Alternative 2: sequence 1-4-7-2-6 + 3-5 The procedure to follow is the same as in Alternative 1. The table below shows the results. Job Duration [n] [days] 3 5

9 5

Due date [day] 8 4

Actual Delay? Penalty Penalty delivery date [days] [€/day] [€] [day] 33 25 70 1,750 38 34 100 3,400

It is clear that Alternative 2 is more expensive than Alternative 1, as the total penalty amounts to 1,750 + 2,100 = 5,150 €. This consideration was, however, easily deducible considering the daily penalty for the two jobs: being equal the days in delay (in both cases the jobs will be finished on day 38), it is preferable to minimize the delay of job 5, as it has a higher daily penalty than job 3.

Exercise 3 (Exam 18th July 2016) A manufacturing firm has to schedule, at the 10th of May, the production plan for some jobs (J1 – J4) that must be delivered according to the due dates provided. Due dates and processing times on machines M1, M2 and M3 are reported in the table. In developing the scheduling, it must be considered that it is not possible to process more than one job at the same time and it is not possible to interrupt the processing of a job once it has begun. 1) Schedule the production according to the OPNDD rule, dividing in equal parts and compute MAK and average lateness; 2) Schedule the production according to the OPNDD rule, dividing operations proportional to the processing time and compute MAK and average lateness. Processing (days) JOB J1 J2 J3 J4

Due Date

Cycle

M1-M225 May M3 01 June M2-M3 M1-M326 May M2 28 May M3-M1

Time

M1

M2

M3

6

4

3

7

5

4

3

4 5

2

Solution Exercise 3 In applying the Operation Due Date, we schedule first the operation with the earliest OPNDD, i.e. the one with the expected operation’s end date closest to time now. The flow time can be divided equally among operations or proportionally to the processing time. OPNDD rule dividing operations in equal parts In the case of OPNDD dividing the interval between the moment when the job enters the system and the due date among all the operations in equal parts, we assign to each operation of the job-th the same OPNDD value. In particular, once evaluated the number of operations needed by the job- th, the OPNDD would be:

For Job1, for example, the OPNDD is equal to:

The results for all the jobs on each machine are the followings:

JOB J1 J2 J3

M1 5 5.33

OPNDD M2 5 11 5.33

M3 5 11 5.33

J4

9

-

9

On every machine, the precedence sequence is thus the same: J1, J3, J4, J2. The related Gantt diagram is the following:

The MAK is equal to 22 days. The lateness for each job is: JOB Lateness J1 0 J2 -10 J3 6 J4 -3

So, the average lateness is:

OPNDD rule dividing operations proportionally to the processing time In the case of OPNDD dividing the interval between the moment when the job enters the system and the due date among all the operations proportionally to the processing time, we assign to each operation of the job-th a different OPNDD value. In particular, considering the processing time required by every operation needed by the job- th, , the OPNDD would be:

For Job1 - Operation on M1, for example, the OPNDD is equal to:

The results for all the jobs on each machine are the followings: JOB J1 J2 J3 J4

M1 6.9 5.8 12.9

OPNDD M2 4.6 12.8 5.8

M3 3.5 9.2 4.4 5.1

The precedence sequences. in this case, are different on each machine: M1: J3, J1, J4 M2: J1, J3, J2 M3: J1, J3, J4, J2 The related Gantt diagram is the following:

The MAK is equal to 18 days. The lateness for each job is: JOB Lateness J1 3 J2 -10 J3 -5 J4 -3

So, the average lateness is:

Exercise 4 (Exam 31st January 2017) A manufacturing firm has to schedule the production of 5 Jobs (J1, J2, J3, J4 and J5). Due delivery dates, processing times (comprehensive of set up times), and set up times on machine M1, M2, M3, M4 and M5, all expressed in hours, are reported in tables. The rule for loading the production system is SPT (Shortest Processing Time), but the scheduling on each machine is developed with a different dispatching rule: LPT (Longest Processing Time) on M1; EDD (Earliest Due Date) on M2; MSUT (Minimum Set Up Time) on M3; FIFO (First In First Out) on M4; SPT (Shortest Processing Time) on M5 (for all the dispatching rules, use the processing time reported in table). At the time 0, starting point for the development of the scheduling, the firm is still processing some WIPs (J6, J7, J8). All these WIPs have already started their last operation, percentage of completeness at time 0 of these last operations is 50% for J6 and J7, 75% for J8. In developing the scheduling, it must be considered that it is not possible to process more than one job at the same time and it is not possible to interrupt the processing of a job once it has begun. It is requested to: schedule the production and represent it on the Gantt; compute MAK, average lateness and average tardiness. -

J1 J2 J3 J4 J5 J6 J7 J8

J1 J2 J3 J4 J5 J6 J7 J8

Processing Time (h) M1 M2 M3 M4 2 4 7 5 5 3 3 1 5 2 3 1 2 4 10 8 7 5 12 2 4 9

Sequence First Second Operation Operation M1 M4 M4 M2 M5 M4 M3 M4 M2 M5 M1 M3 M2 M4 M5 M2

M5 5 1 2 4 3 4 12

J1 J2 J3 J4 J5 J6 J7 J8

Set up Time (h) M1 M2 M3 M4 1 2 3 1 1 2 2 0 2 0 1 0 1 2 1 5 2 5 0 2 1 3

Third Operation M5 M5 M1 M5 M3 M4 M3 M1

M5 2 0 0 1 1 1

Due Date (h) (from time 0) J1 25 J2 16 J3 8 J4 20 J5 9

4

Fourth Operation M3

Fifth Operation M2

M1

M2

M5

Solution Exercise 4 First of all, we have to determine the precedence sequence for each machine, according to the different dispatching rules, Job 6, Job 7 and Job 8 have already started their last operation, this means that they are, at the time of the scheduling of the production, processed by the system. Since it is not possible to interrupt the processing of a job once it has begun, we will not consider these jobs in the definition of the precedence sequence for the different machines. Scheduling of the production Precedence Sequence

All the dispatching rules are static ones. This means that the priority index of each job can be computed once for all when job enters the queue at the machine. M1- Longest Processing Time First the job with the longest processing time on the considered machine. The sequence is: J4 (5h), J3 (3h), J1 (2h) M2- Earliest Due Date First the job with the earliest due date, The sequence is: J5 (9h), J2 (16h), J4 (20h), J1 (25h) M3- Minimum Set Up Time First the job with the minimum set-up time on the machine. The sequence is: J4 (1h), J5 (2h), J1 (3h) M4- First In First Out First the job that arrive first at the machine. The sequence is determined during the scheduling. M5- Shortest Processing Time First the job with the minimum residual processing time. The sequence is: J2 (1h), J3 (5h), J5 (6h), J4 (11h), J1 (15h) Gantt In representing the scheduling on the Gantt diagram, we have first of all to remember that J6, J7 and J8 have, at the time of the scheduling development, already started their last operation, and so are processed by the system. We also know the percentage of completeness of these operations, and we can evaluate the time needed by each job to be completed. Last on J6 J7 J8

M5 M3 M1

operation

Processing Time Percentage of of the last completeness operation (h) 4 50% 12 50% 4 75%

Remaining time for the last operation at t=0 (h) 2 6 1

We can already insert these jobs in the system and then proceed with the scheduling on the available machines:

We can now start with the scheduling of J1, J2, J3, J4 and J5.

t=0 Machine Job M1 Busy J5 is the first it can be M2 processed M3 Busy The first job available in the M4 system is J2

M5

Busy

Machine Job The first is J4, but it cannot be processed as other operations are required before; the second is J3, but it cannot be processed as M1 other operations are required before; the third is J1 it can be processed t=1 M2 Busy M3 Busy M4 Busy M5 Busy

Machine Job M1 Busy The first is J2 but it cannot be processed as it is processed on M4; the second is J4 but it cannot be processed because other M2 operations are required before; the third is J1 but it cannot be it is t=2 processed on M1 M2 waits M3 Busy M4 Busy The first is J2, but it cannot be processed because other operations M5 are required before; the second is J3 it can be processed

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3, but it cannot be processed because other operations are required before M1 waits t=3 M2 The first job is J2 it can be processed M3 Busy M4 The first job available in the system is J1 M5 Busy

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3, but it cannot be processed because other operations are required before M1 waits Busy t=4 M2 M3 Busy M4 Busy The first is J2, but it cannot be processed as it is processed on M2; M5 the second is J5 it can be processed

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3, but it cannot be processed because other operations are required before M1 waits t=6 M2 Busy M3 The first job is J4 it can be processed M4 Busy M5 Busy

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3, but it cannot be processed because other operations are required before M1 waits M2 Busy t=7 M3 Busy M4 Busy The first is J2, but it cannot be as it is processed on M2; the second M5 is J4, but it cannot be processed because other operations are required before; the third is J1, as it is processed on M4 M5 waits

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3, but it cannot be processed because other operations are required before M1 waits The first is J4, but it cannot be processed as it is processed on M4; t=8 M2 the second is J1, but it cannot be processed because other operations are required before M2 waits M3 Busy M4 The first job available in the system is J3 M5 The first is J2 it can be processed

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before; the second is J3 it can be processed The first is J4, but it cannot be processed because other operations M2 are required before; the second is J1, but it cannot be processed t=9 because other operations are required before M2 waits M3 The first is J5 it can be processed M4 The first job available in the system is J4 The first is J4, but it cannot be processed because other operations M5 are required before; the third is J1 it can be processed

Machine Job M1 Busy The first is J4, but it cannot be processed because other operations M2 are required before; the second is J1, but it cannot be processed t=10 because other operations are required before M2 waits M3 Busy M4 END M5 Busy

t=12 Machine Job

M1

M2 M3 M4 M5

The first is J4, but it cannot be processed because other operations are required before M1 waits The first is J4, but it cannot be processed because other operations are required before; the second is J1, but it cannot be processed because other operations are required before M2 waits Busy END Busy

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before M1 waits The first is J4, but it cannot be processed because other operations M2 are required before; the second is J1, but it cannot be processed t=13 because other operations are required before M2 waits The first is J1, but it cannot be processed as it is processed on M3 M5 M3 waits M4 END M5 Busy

Machine Job The first is J4, but it cannot be processed because other operations M1 are required before M1 waits The first is J4, but it cannot be processed because other operations t=14 M2 are required before; the second is J1, but it cannot be processed because other operations are required before M2 waits M3 The first is J1 it can be processed M4 END M5 The first is J4 it can be processed

t=18 Machine Job M1 The first is J4 it can be processed The first is J4, but it cannot be processed as it is processed on M1; M2 the second is J1, but it cannot be processed because other operations are required before M2 waits M3 Busy M4 END

M5

END

Machine Job M1 Busy The first is J4, but it cannot be processed as it is processed on M1; t=21 M2 the second is J1 it can be processed M3 END M4 END M5 END

Machine M1 t=23 M2 M3 M4 M5

Job END Busy END END END

Machine Job M1 END The first is J4 t=25 M2 processed M3 END M4 END M5 END

t=27 Machine Job M1 END M2 END

it can be

M3 M4 M5

END END END

Compute MAK, average lateness, average tardiness The MAK, as it can be seen from the Gantt, is equal to 27 h.

J1 J2 J3 J4 J5 Average

Due Date (h) 25 16 8 20 9

Delivery date (h) 25 9 12 27 13

Lateness (h) 0 -7 4 7 4

Tardiness (h) 0 0 4 7 4

1.6

3

Exercise 5 (Exam 1st February 2010) A manufacturing firm has to schedule the production of some jobs that must be delivered next week. Jobs, to be produced, must be manufactured following a fixed sequence on the three machineries, with the processing times reported in table. It must be considered that: It is not possible to process more than one job at the same time; It is not possible to interrupt the processing of a job once it has begun. Which is the scheduling that could be implemented, being sure, a priori, it is the optimal one for the considered jobs? J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

M3 3 2 7 8 6 5

Solution Exercise 5 To be sure, a priori, to identify the optimal scheduling for the considered jobs it is necessary to adopt an algorithmic optimization. In a flow shop composed by 3 machines, as the one proposed here, we can apply the Johnson's algorithm extended to 3 machines, checking then the optimality of the sequence. Indeed, in the case of 3 machines, Johnson’s algorithm is applied to M1 and M2 and then to M2 and M3, finding then 2 sequences (one for M1-M2; one for M2-M3): if the two sequences are the same, then the sequence found is optimal, otherwise nothing can be said about optimality. Johnson’s algorithm on M1-M2 We start implementing the Johnson’s algorithm on M1-M2. Step 1 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

The shortest processing time is the one of J6 on M1: J6 _ _ _ _ _

Step 2 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

Followed by J5 on M1: J6 J5 _ _ _ _

Step 3 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

Followed by J3 on M1: J6 J5 J3 _ _ _ Step 4 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

Followed by J4 on M1: J6 J5 J3 J4 _ _ Step 5 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

Followed by J1 on M1: J6 J5 J3 J4 J1 _

Step 6 J1 J2 J3 J4 J5 J6

M1 5 7 3 4 2 1

M2 6 8 5 7 4 3

Followed by J2 on M1: J6 J5 J3 J4 J1 J2 Johnson’s algorithm on M2-M3 We start implementing the Johnson’s algorithm on M2-M3. Step 1 J1 J2 J3

M2 6 8 5

M3 3 2 7

J4 J5 J6

7 4 3

8 6 5

The shortest processing time is the one of J2 on M3: _ _ _ _ _ J2 Step 2 J1 J2 J3 J4 J5 J6

M2 6 8 5 7 4 3

M3 3 2 7 8 6 5

Followed by J1 on M3: _ _ _ _ J1 J2 Step 3 J1 J2 J3 J4 J5 J6

M2 6 8 5 7 4 3

M3 3 2 7 8 6 5

Followed by J6 on M2: J6 _ _ _ J1 J2 Step 4 J1 J2 J3 J4 J5 J6

M2 6 8 5 7 4 3

M3 3 2 7 8 6 5

Followed by J5 on M2: J6 J5 _ _ J1 J2 Step 5 J1 J2 J3 J4 J5 J6

M2 6 8 5 7 4 3

M3 3 2 7 8 6 5

Followed by J3 on M2: J6 J5 J3 _ J1 J2 Step 5 J1 J2

M2 6 8

M3 3 2

J3 J4 J5 J6

5 7 4 3

7 8 6 5

Followed by J4 on M2: J6 J5 J3 J4 J1 J2 Optimal scheduling sequence As the two sequences are the same, J6 J5 J3 J4 J1 J2 is the optimal a priori sequence for the considered jobs.

Chapter 9. Exams’ examples

Exam 26th of June, 2017 Exercise 1[5] (7 points) STAR is a small manufacturing company, located in the county of Bergamo, 40 km from Milano. It is the typical Italian SME, active since almost 50 years, in the production of specialized machines. After the crisis of 2008-2010, the company – thank to some relevant decisions taken by the new manager – is registering a good period, with an increasing demand and a stable portfolio of new orders. For this reason, the board of the company has decided to invest in a new plant, to be installed out of the town in which currently the old plant is (based on the first production department created in the ‘60ies). Part of the new plant should be dedicated to some mechanical products and treatments and the main idea is to organise such part as the conventional job shop layout. In this area the company should produce 6 product types (P1, P2, P3, P4, P5, P6), using 5 main production shops (M1, M2, M3, M4, M5). Processing times are hereafter defined [data in minutes]: P1: (M1, 5), (M2, 3) P2: (M3, 4), (M4, 7), (M5, 10) P3: (M1, 3), (M2, 1), (M5, 2) P4: (M1, 5), (M2, 2), (M3, 5), (M5, 1) P5: (M3, 9), (M4, 2), (M5, 8) P6: (M4, 4), (M5, 11) Setup times are based on the different machines. M1 takes normally 15 minutes / setup, M2 20 minutes/setup, M3 30 minutes/setup, M4 10 minutes/setup, M5 8 minutes/setup. The company has decided to work with a pre-defined dimension of lot: P1 should be produced in batches of 250 pieces, P2 500 pieces, P3 1,000 pieces, P4 250 pieces, P5 500 pieces, P6 500 pieces. The yearly demand of the different products should be the following: P1 5,000 pieces/year, P2 15,000 pieces/year, P3 20,000 pieces/year, P4 15,000 pieces/year, P5 2,500 pieces/year, P6 30,000 pieces/year. At minimum, the production system will work 2 shifts of 8 hours per day, 300 days/year. The following coefficients should be used in the design of the job shop: TR = 0.9 (Trial Rate) R= 0 (Scrap Rate); A = 0.95 (Availability); HC = 0.9 (Human Coefficient); SE = 0.8 (Scheduling Efficiency). The planned machines cost as follow: M1 = 50,000 € M2 = 45,000 € M3 = 55,0000 € M4 = 90,000 € M5 = 300,000 € For each machine, a direct cost of 10,000 €/year should be considered (for energy, maintenance, etc.). Each machine needs at least one operator to work. The cost of one operator is 45,000 €/year Let’s define the best solution for the production system, knowing that: M1 and M2 shops could work only 2 shifts a day; If needed, M3, M4, M5 shops could work up to 24 hours/day, with 3 shifts (1 shift more); Rate of return is equal to 0; Life time T of 2 years; Saturation can be considered up to 100%; All the costs related to the warehouses can be neglected (stock holding cost equal to 0) Exercise 1 – Solution The first step is the identification of the production cycle for each item, in terms of technologies: M1 M2 M3

P1 x x

P2

x

P3 x x

P4 x x x

P5

x

P6

M4 M5

x x

x

x x

x

x x

We proceed then with the evaluation of the needed hours (for each technology) for fulfilling the forecasted demand (for the detailed steps please refer to Chapter 2, Exercise 1. Starting with the evaluation of the number of lots, considering for example the annual quantity of P1 to be produced (5,000 pcs) and the size of the production lot of P1 on M1 (250 pcs/lot), we have:

By proceeding in the same way for the other products and machines:

Computation of the number of P1 set up M1 20 M2 20 M3 M4 M5

P2

30 30 30

P3

P4

20 20

60 60 60

20

60

P5

P6

5 5 5

60 60

Proceeding with the evaluation of the production hours needed, considering for example the production of P1 on the M1, we have:

By proceeding in the same way for the other products and machines:

Production hours needed M1 M2 M3 M4 M5

P1 416.67 250

P2

1,000 1,750 2,500

P3 1,000 333.33

666.67

P4 1,250 500 1,250 250

P5

P6

375 83.33 333.33

2,000 5,500

As for the set-up, considering for example the production of P1 on the M1, we have:

H set ups needed M1 M2

P1 5 6.667

P2

P3 5 6.667

P4 15 20

P5

P6

M3 M4 M5

15 5 4

30 2.667

2.5 0.83 0.667

8

10 8

It is possible thus to evaluate the total needed hours for each technology, considering also the times for maintenance and for fixing failures, times for test/ trial and the dependence of operations from human related factors. H needed technology

for M1 M2 M3 M4 M5

P1

P2

547.97 333.55 1319.04 2280.70 3254.06

P3

P4

1306.04 441.85

1643.92 675.76 1663.42

869.83

335.28

P5

P6

490.58 1009.38 434.05

2612.09 7157.89

We can proceed with the evaluation of the number of machines needed for each technology, passing from the evaluation of the available annual hours for production.

Two-shifts policy Given that the entire production system will work 2 shifts of 8 hours per day according to the working calendar of 300 days/year, all the technologies are supposed to work on two eight-hour shifts per day. Therefore, the working hours actually available are:

The number of needed machines is thus:

M1 M2 M3 M4 M5

Theoretic number machines 0.91 0.38 0.90 1.30 3.14

Actual Actual of number of saturation rate machines 1.00 73% 1.00 30% 1.00 72% 2.00 53% 4.00 63%

We can proceed with the economic assessment of the identified solution (based on 2 eight-hour shifts per day). Cost of the solution with 2 shifts

M1

M2

M3

M4

M5

Total

Ns = Number of shifts Nm = Number of machines

2

2

2

2

2

-

1

1

1

2

4

9

Machines

No = Ns * Nm = Number of operators C1= Cost of machines’ acquisition

2

2

2

4

8

18

Operators

50,000

45,000

55,000

180,000

1,200,000

1,530,000

€

10,000

10,000

10,000

20,000

40,000

90,000

€/y

C2= Cost of machines (other direct costs) C3= Cost of operators

Shifts/day

90,000

90,000

90,000

180,000

360,000

810,000

€/y

CI=C1 + C2 + C3= Total cost first year CII= C2 + C3 = Total cost second year

150,000

145,000

155,000

380,000

1,600,000

2,430,000

€/y

100,000

100,000

100,000

200,000

400,000

900,000

€/y

CI + CII =Total cost of 2 years

250,000

245,000

255,000

580,000

2,000,000

3,330,000

€

Three-shifts policy Although M1 and M2 shops can work only 2 shifts a day (as in the previous configuration), M3, M4, M5 can work up to 24 hours/day (so with a three-shifts policy). It is thus needed to re-evaluate the available annual hours for production for M3, M4 and M5, that become:

As the number of shifts and machines remains the same for M1 and M2, it is possible to proceed with the computation of the new number of needed machines only for M3, M4 and M5 departments and the related saturation:

M3 M4 M5

Theoretic number machines 0.60 0.87 2.09

Actual of number machines 1.00 1.00 3.00

Actual of saturation rate 48% 70% 56%

We can proceed with the economic assessment of the identified solution, based on 2 shifts per day for M1 and M2 and 3 shifts per day for M3, M4 and M5. Cost of the solution with 3 shifts (where possible) Ns = Number of shifts

M1

Nm = Number of machines No = Ns * Nm = Number of operators C1= Cost of machines’ acquisition

M2

M3

M4

M5

Total

2

2

3

3

3

-

1

1

1

1

3

7

Shifts/day Machines

2

2

3

3

9

19

Operators

50,000

45,000

55,000

90,000

900,000

1,140,000

€

C2= Cost of machines (other direct costs) C3= Cost of operators

10,000

10,000

10,000

10,000

30,000

70,000

€/y

90,000

90,000

135,000

135,000

405,000

855,000

€/y

CI=C1 + C2 + C3= Total cost first year CII= C2 + C3 = Total cost second

150,000

145,000

200,000

235,000

1,335,000

2,065,000

€/y

100,000

100,000

145,000

145,000

435,000

925,000

€/y

250,000

245,000

345,000

380,000

1,770,000

2,990,000

€

year CI,II= CI + CII =Total cost of 2 years

Best solution The solution with 3 shifts (where possible) appears more convenient than the 2shifts configuration, as its total cost is 2,990,000 € against 3,330,000 €. However, this doesn’t mean that it is also the best possible solution. Indeed, comparing the costs of the solution with 3 shifts (where possible) with those of the solution with 2 shifts, it is possible to select the best economic solution for each specific technology and thus the best configuration of the entire plant from an economic viewpoint. Cost of the best solution

M1

M2

M3

M4

M5

Total

Ns = Number of shifts Nm = Number of machines

2

2

2

3

3

-

1

1

1

1

3

7

Machines

No = Ns * Nm = Number of operators C1= Cost of machines’ acquisition

2

2

2

3

9

18

Operators

50,000

45,000

55,000

90,000

900,000

1,140,000

€

10,000

10,000

10,000

10,000

30,000

70,000

€/y

C2= Cost of machines (other direct costs) C3= Cost of operators CI=C1 + C2 + C3= Total cost first year

Shifts/day

90,000

90,000

90,000

135,000

405,000

810,000

€/y

150,000

145,000

155,000

235,000

1,335,000

2,020,000

€/y

CII= C2 + C3 = Total cost second year CI,II= CI + CII =Total cost of 2 years

100,000

100,000

100,000

145,000

435,000

880,000

€/y

250,000

245,000

255,000

380,000

1,770,000

2,900,000

€

Exercise 2[6] (7 points) Green Tea Muffins (GTM) is a start-up established by three young and talented Engineers recently graduated at Politecnico di Milano. The mission of the company is to let the whole world discover their huge gamut of new and mouthwatering flavours of muffins, created merging creativity and tradition. The best-selling item is – of course – the Green Tea flavored muffin. Demand for such item has been quite constant in the past, with an average of 200 pieces/day and a weekly standard deviation of 50 pieces. In order to optimize the production batches, the muffins are produced to stock, and shipped to the customer whenever an order is received. An airtight wrap guarantees that the product would remain fresh and fragrant for a reasonably long shelf-life. GTM has debts, yet it could borrow further money. The financial burden of loans granted by banks is 15% / year. At the moment, there are no other investments which could be deemed interesting by the company management. Knowing that: • Muffins are produced in a small laboratory, operated by Mario, the Muffin Specialist. Mario is an employee of the company, hired with a permanent position, with a gross salary of 20$ per hour. • In order to be compliant with the strict “Good Manufacturing Practices” for food processing, in the laboratory, only one flavour of muffin is produced at a time (this is not a problem, though, as the production volumes are still quite low for the company, and the laboratory is not in use most of the time; when the laboratory is not in use, Mario doesn’t perform any other activity). • To start a production batch of any flavour, a thorough cleaning of all the tools is required. The cleaning takes one working day, and is performed by Mario, with the help of Beppe, the handyman. Beppe is an hourly worker and he is paid only when there are activities to do. The daily wage of Beppe is 50$. • During cleaning, solvents and other consumable materials are used. Their cost is estimated in around 20$ per cleaning. • Furthermore, the day after the cleaning, there is a warm up of the oven and the other instrument. Production during warm-up would be not-conforming; as a consequence, the laboratory is not producing any muffin in the whole day. During warm-up, only Mario operates the laboratory (the aid of Beppe is required only during cleaning). • Starting from day 3 (day 1 = cleaning; day 2 = warm up), the plant is able to produce good products at full rate. The production rate is 500 muffins a day. • Rental of the production laboratory costs GTM 4,000$ a month, while machinery depreciation is 2,000 $ a month. • The following raw materials are employed for one single muffin: food ingredients 1.5 $ / piece, and wrapping and packaging materials 1 $ / piece. • Energy consumption of the laboratory: 10 $ / day (incurred both in the days of actual production and in the warm-up day). Producing during only one fraction of a day will result in a proportional saving in energy (i.e.: if in a day only 250 pieces are to be produced, after half-working day the plant is stopped and – consequently – no more energy is used). • The finished products are stocked in a nearby warehouse operated by a logistics provider (as GTM does not have any suitable storage area internally). Every month, GTM pays for the stock holding service 0.5% of the average value of the stock in that very month. • The service level required is 95% (k = 1.65). Considering the working calendar of the company, the following relations hold: 1 year = 10 months = 40 weeks = 200 days. With the data available, calculate the EOQ, safety stock, the average stock and reorder point for the Green Tea muffins. O D P Cm

80 40,000 2.52 0.21

r H EOQ EOQ rounded EOQ (r) EOQ rounded

500 200 3,477.6 3,478.0 4,489.6 (r)

k SS SS rounded Average stock ROP

4,490 1.65 52.2 53

1,400 453

Exercise 3[7] (7 points) Two plans (refer table below), available for a company to fulfil the demand, are to be assessed and the best one to be chosen. Month Demand Plan A Plan B

1 200 175 280

2 210 175 250

3 205 175 265

4 220 175 250

5 145 175 275

6 105 175 155

7 115 175 155

8 0 175 0

9 225 175 135

10 225 175 125

11 215 175 105

12 235 175 105

The line maximum production capacity is 250 pcs/month, and the internal variable production cost is 10.00 €/pc. The production rate can be changed each period, if needed. However, production rate can be modified only according to lots equal to or multiple of 50 pieces. Productivities strongly vary when production rate varies (i.e. inefficiency cost). It was estimated that the variable cost per piece increases when production rate increases with respect to the rate of 50 pieces per period. The unit variable cost increases for the overall production of 0.5 €/pc for a production between 51 and 100 pieces/period included, 1 €/pc for a production between 101 and 150 pieces/period included, 1.5 €/pc for a production between 151 and 200 pieces/period included, 2 €/pc for a production between 201 and 250 pieces/period included. The production line runs continuously. At every change in the production rate the company incurs in a relevant setup cost of 3,000.00 €. At the beginning of the year one set up is to be considered, regardless any production planning decision, due to maintenance. The company can buy the finished product directly from a subcontractor with no quantity limits at the price of 11.00 €/pc. A quality control has to be made on purchased products and it costs 5 €/pc. Due to the high level of defectiveness, only 70% of the purchased products is compliant with minimum quality standards. The stock holding cost is 2 €/pc*month, and there is also a (not recoverable) stock-out fee of 9 €/pc. You are requested to: • Assess the two plans (filling BOTH the tables in all their parts) and chose the best one; • Which is the set- up cost [€/set up] for PLAN B, under hypothesis of divesting production and of avoiding also the set up in January, that makes subcontracting more convenient than internal production (i.e. company adopts only subcontracting to have availability of products to fulfil the demand)? • Which is stock-out fee [€/pc] for PLAN A, under hypothesis of divesting production, that makes subcontracting more convenient than internal production? 1. € Plan Ctot 33,687.14 A Cu € 16.04 € Plan Ctot 43,708.57 B Cu € 20.81 2. 5,517.86 € 3. None

Exam 17th of July, 2017 Exercise 1[8] (7 points) EIASME is a medium-sized manufacturing company, with several production plants spread around Europe, resulting from several acquisitions done in the last 5 years. It produces plastic components for the automotive industry. EIASME is currently registering an increase of demand and it is planning to add an extra stamping department to one of its plants, in Poland. The stamping department should be composed by 5 main types of machines (M1, M2, M3, M4, M5) and it should be able to deliver 6 main product families (P1, P2, P3, P4, P5, P6). Processing times of the mentioned products are hereafter defined [data in minutes]: • P1: (M1, 5), (M2, 3) • P2: (M3, 4), (M4, 7), (M5, 10) • P3: (M1, 3), (M2, 1), (M5, 2) • P4: (M1, 5), (M2, 2), (M3, 5), (M5, 1) • P5: (M3, 9), (M4, 2), (M5, 8) • P6: (M4, 4), (M5, 11) The board of the company is interested to evaluate the alternative of a production layout organised in manufacturing cells. Starting from the provided data, please define the reference product/machine matrix and determine a feasible cell-based layout. For the definition of the cells, please use the ROC (Rank Order Clustering) technique.

1 cell 2 cell

Machines Products type 4,3,1 1,2,5,3 2,5,6 5,3,4

Exercise 2[9] (7 points) Task Duration U-spin-me-round is an Italian SME producing small engines Standard for food mixers, hair-driers and other similar appliances for Task Precedence Mean Deviation households and communities. The company wants to [seconds] [seconds] integrate vertically, expanding downstream into the final A 100 10 assembly of the finished products. For this reason, you are B A 90 6 appointed to design a new final assembly line for the line of C A 80 22 wall-hair-driers dedicated to swimming-pools, gyms and D A 90 14 hotels. E B 70 8 x F(x) x F(x) The first assembly line to x F(x) F B,C 35 6 1.0 0.84 2.0 0.98 be introduced will be 0.0 0.50 G D 105 15 1.1 0.86 2.1 0.98 dedicated to best-selling 0.1 0.54 H C,E 55 8 0.2 0.58 1.2 0.88 2.2 0.99 product, called I E,F 50 6 1.3 0.90 2.3 0.99 “SuperPhon-y”. The 0.3 0.62 L F,G 50 8 1.4 0.92 2.4 0.99 assembly line would 0.4 0.66 M H,I,L 65 10 1.5 0.93 2.5 0.99 work 5 days a week, on 0.5 0.69 0.6 0.73 1.6 0.95 2.6 1.00 two shifts of 7.5 hours 0.7 0.76 1.7 0.96 2.7 1.00 each. The production 0.8 0.79 1.8 0.96 2.8 1.00 capacity should be 0.9 0.82 1.9 0.97 2.9 1.00 enough to face the current SuperPhon-y demand, steady at 1,200 pieces/week. The assembly process of SuperPhon-y has been decomposed into the following elementary tasks: 1. Start drawing the assembly graph of SuperPhon-y. 2. Calculate the cycle time. 3. Design a manual assembly line adopting the Maximum Incompletion Probability Allowed criteria (with max probability = 5%), adequately commenting all steps followed and showing all calculations. Please use (MANDATORILY) the following Cumulative Normal Distribution table (noteworthy values, rounded figures).

Workstation 1 2 3

4 5

Task A B D C G E H F I L M

Hp: Sub- criterion MaxDur; Station Oriented.

Exercise 3[10] (7 points) A firm wants to evaluate the performance of its warehouse. Data related to the monthly consumption (thousands of pieces), to the monthly stock level (thousands of pieces) and production cost (€ per piece) for all the products (from 1 to 8) are provided. 1. Determine the distribution of products using ABC-ABC analysis (consumption and stock value). Use 50% and 75% as thresholds for both of them (place products where they develop the most part of their value);

2. Comment products 5, 6, 7, 8; 3. Calculate the monthly STI (Stock Turnover Index) and daily CI (Coverage Index) for product 3 and 4 and comment the results obtained). Consumption Product (thousands pieces) 1 35 2 50 3 30 4 40 5 70 6 35 7 40 8 40

Average stock of level (thousands of pieces) 35 25 30 20 90 25 55 80

Unit (€/pc)

cost

60 40 50 60 25 65 35 30

Information Mature New Old JIT Mature New Spare part Old

Consumption A B C A 1 5 8 Stock B 6 7 C 4 2 3 Product STI 3 (month) 4 CI 3 (days) 4

[1]

Value 1 2 30 15

In the implementation of the procedure, QS values have been rounded to the next integer.

[2] Task-oriented approach: when the remaining idle time is not sufficient to assign the k-th operation, a new workstation is set.

[3] The 305 days have been calculated starting from the 365 annual days and subtracting 52 days of holidays plus 8 days of national and religious feast (1st of January; Easter; Easter Monday; 25th of April; 1st of May; 2nd of June; 1st of November; 25th of December) [4] The safety stocks are replanish at the beginning of the period, as soon as the issued order arrives. [5] Exercise developed by Sergio Terzi and Claudio Sassanelli. [6] Exercise developed by Alessandro Brun. [7] Exercise developed by Enrico Cagno and Alessandra Neri. [8] Exercise developed by Sergio Terzi and Claudio Sassanelli. [9] Exercise developed by Alessandro Brun. [10] Exercise developed by Enrico Cagno and Alessandra Neri.