6.1
Chapter 6: Overview of hard OR methodology Exercise solutions for Management Science, ISBN 1-4039-4174-2 © H.G. Daellenbach and D.C. McNickle, 2005. Palgrave Publishers Ltd. 1.
(a) High-level system diagram.
(b) Major boundary judgements: all output can be sold at given prices (fine if increase in output relatively small), operations not affected by weather (fine for climate controlled sheds, not necessarily true for compost production), additional raw materials for compost production available at same costs. (c) Decision maker: Gérard Mousse. Objective: achieve high net profits. Decision criterion: maximize profits. Performance measure: annual net profits. Alternative courses of action: number of flushes per cycle. Context: (see also (b) above) processes for making compost and growing mushrooms given, cost factors all given. (d) Influence diagram: see Figure 9-1. (e) Project proposal only given in outline form: Cover letter. Table of contents. (1) Introduction. (2) Summary of recommendations: yield data, picking time, for each individual flush to be collected, cost data on all phases of operation. Once data available, actual analysis no more than one or two days work. (3) Statement of problem. (4) Analysis: Data collection to determine average yield per flush, average picking rate per flush. Best done on a shed basis, with around 10 sheds covered for one full cycle each. Cost data collection using accounting records, covering all costs that are output or level of operation related. Set up a spreadsheet computing annual output of mushrooms, total annual operating costs, total annual revenue, individually for cycles covering n flushes, where n varies from 1 to 6. 5. Resources: This analysis can be completed in at most two days, with another day needed for writing up the report. Data collection may however take about two months of elapse time, requiring slightly more work for some pickers and Jennifer Bloom. 2.
(a) Decision maker: owner.. User: owner, other management staff Customers: other employees, firm’s customers and suppliers, possibly local community at large. Analyst: owner, possibly other management staff. (b) Decision maker: owner. Objective: achieve high return on capital invested by owners, (alternative: achieve high profits = surrogate for increase owner’s wealth). Decision criterion: maximize return on owner’s investment, (alternative: maximize profit). Performance measure: return on owner’s investment, (alternative: annual profit).
6.2
Alternative course of action: amount of equity and composition of liability funds, equipment choices, major operational decisions, such as final product composition, new products, raw material sources. (c) High-level systems diagram for sawmill as a profit making system.
(d) Owner mainly concerned with own benefits, concerns of low-level staff, community, etc. ignored, demand for products and availability of supplies predictable (otherwise profit maximization may not be a suitable criterion, survival of firm more appropriate). 3.
(a) The relationship between the defectives produced/hour and the time between adjustments is shown in the table below. Cumulative run time 6 min 12 min 24 min 36 min 48 min 60 min
Average number of defectives 0.2 0.8 2.0 4.2 7.0 11.0
Cumulative run time including adjustments 12 18 30 42 54 66
Number of cycles/hour
No of defectives/hour
5.00 3.33 2.00 1.43 1.11 0.91
1.00 2.67 4.00 6.00 7.78 10.00
(See (b) below for meaning of a, b, m, n, t) [Time lost (min/h)] = [Number of adjustments/h][Time per adjustment (min/adj)] = na [Productive time (min/hr)] = [1 - (Time lost (min/h))] = 60 - na [Output (units/h)] = [Productive time (min/hr)][Machine rate (units/min)] = (60 - na)p, where p = 1/[Machine time (min/unit)] [Defectives produced/h] = function (n, machine characteristics) = 10t - 1 = 10/n -1 = b/n - 1 [Good output/h] = [Total output/h] - [Defectives/h] = (60 - na)p - (b/n - 1) [Parts reworked/h] = [Fraction reworked][Defectives/h] = w(b/n - 1) [Saleable parts/h] = [Good output] + [Parts reworked] = (60 - na)p - (b/n - 1) + w(b/n - 1) [Total profit] = [Unit selling price][Saleable parts/h] - [Unit material cost][Output/h] [Unit rework cost][Parts reworked/h] - [Labour cost] = [Revenue] - [Material cost] - [Rework cost] - [Labour cost] = R[(60 - na)p + (w - 1)(b/n - 1)] - Cm[(60 - na)p] - Cw[w(b/n - 1)] - CL
6.3
(b) Excel spreadsheet CH6EX3 (reproduced below) THE MACHINE ADJUSTMENT PROBLEM Machine time (m) 2 Machine rate (1/m) 0.5 Rate of defectives (b) 10 Time per adjustment (a) 6 Fraction defectives reworked (w) 0.5 (t) Adjustments/h (n) Output/min Defectives/h Salable output/h Revenue/h £ Cost/h £ Profit/h £
12 5.00 15 1 14.5 304.5 260 44.50
18 3.33 20 2 19 399 342 57.00
min/unit units/min units/hour min/adj
Selling price/unit Material cost/unit Reworking cost/unit Labour cost/hour
Time between adjustments (adj/min) 24 36 48 2.50 1.67 1.25 22.5 25 26.25 3 5 7 21 22.5 22.75 441 472.5 477.75 384 428 452 57.00 44.50 25.75
£21 £16 £4 £18
60 1.00 27 9 22.5 472.5 468 4.50
Net profit contribution as a function of time between adjustments $60.00 $50.00
Range where optimal solution lies
$40.00 $30.00 $20.00 $10.00 $0.00 10
20
30
40
50
60
70
Time between adjustments
(c) The optimal time between adjustments is 20.8. The loss in profit by rounding this number to 20 is £57.897 - £57.833 = £0.064. (d) (1) (2) (3) (4)
All times and rates remain constant or their averages remain constant. The rate of defectives is a liner function of t = 1/n, and remains stable over time. All output produced can be sold at a constant price. Stopping the machine at the end of the day does not effect its state, i.e., next day is just a continuation of previous day. (1) and (2) are reasonable. (3) may change with economic conditions or increased capacity. (4) may not be valid.
6.4
4. (a) Influence diagram for ELMO problem.
(b) Operation of that machine largely independent of rest of firm’s operation, machine speed only aspect that affects reject rate. (c) The relationship between average rejects produced/hour and machine speed is shown in the table below. Machine speed 30 40 50 60 70 80 90
(Machine speed)2 900 1600 2500 3600 4900 6400 8100
Actual average rejects/hour 1.17 2.00 3.00 4.67 6.00 8.00 10.17 2
Theoretical average rejects = speed2/800 1.13 2.00 3.13 4.50 6.13 8.00 10.13
2
[Average rejects/hour] = [Machine speed] /800 = s /800 2
[Fraction of rejects] = [Average rejects/hour]/[Machine speed] = (s /800)/s = s/800 [Size of batch] = [Coils required]/[1 - fraction of rejects] = 10,000/(1 - s/800) [Hours to produce batch] = [Size of batch]/[Machine speed] = [10,000/(1 - s/800)]/s = 2 10,000/(s - s /800) [Labour and Machine cost]= [Labour cost/hour + Machine cost/hour][Hours to produce batch] 2 = (L + M)[(10,000/(s - s /800)] [Raw material cost] = [Unit raw material cost][Size of batch] = R[10,000/(1 - s/800)] [Total cost] = [Labour and Machine cost] + [Raw material cost] 2 = (L + M)[(10,000/(s - s /800)] + R[10,000/(1 - s/800)] Additional boundary judgements: (1) Average rejects/h is a quadratic function of machine speed (or is a linear function of the square of machine speed). (Reasonable from data) (2) If machine is not used, workers engaged productively elsewhere, otherwise labour cost may not be relevant. (Reasonable for any sizeable firm)
6.5
(d) Excel spreadsheet CH6EX4 reproduced below. COIL WINDING MACHINE SPEED SETTING PROBLEM Raw material cost Labour cost Machine running cost Total to produce Speed coils/hour (s) Average rejects/hour Fraction of rejects Good output Size of batch Hrs to produce batch Material cost £ Labour/machine cost £ Total cost £
£2.60 £22
/coil /hour
£6
/hour
10,000
coils
30
40
50
60
70
83
90
1.13
2.00
3.13
4.50
6.13
8.61
10.13
0.04
0.05
0.06
0.08
0.09
0.10
0.11
28.88
38.00
46.88
55.50
63.88
74.39
79.88
10,389.61
10,526.32
10,666.67
10,810.81
10,958.90
11,157.60
11,267.61
346.32
263.16
213.33
180.18
156.56
134.43
125.20
27,012.99
27,368.42
27,733.33
28,108.11
28,493.15
29,009.76
29,295.77
9,696.97
7,368.42
5,973.33
5,045.05
4,383.56
3,764.01
3,505.48
36,709.96
34,736.84
33,706.67
33,153.15
32,876.71
32,773.77
32,801.25
The optimal machine speed is 83 coils/hour with a total cost of £32,773.77. 5. (a) Excel spreadsheet for LOD model 1 CH6EX5a (b) Excel spreadsheet for LOD model 1 CH6EX5b 6. Excel spreadsheet CH6EX6 answers both (a) and (b). Note that since all orders in multiples of four drums, EOQ should also be in multiples of 4 drums, requiring additional calculations. (a) The optimal inventory replenishment size for a cutoff point of 8 drums is 10.8 drums. Rounded up to 12 drums gives a total relevant cost of $2,015.20 and rounded down to 8 drums gives a total relevant cost of $2,046.20. The best stock replenishment size is therefore 12 drums. This has a total annual EOQ cost of $782. The same answer could have been obtained by using the square root formula, expression 6-2 to find the EOQ and then check again whether rounding to 8 or 12 is better: EOQ = [2(280)15/0.18(400)] ½ = 10.80 (b) The best cutoff point for special production runs is 12. The optimal stock replenishment is 14.94. Rounded up to 16 gives a total relevant cost of $1,985.30 and rounded down to 12 gives a total relevant cost of $2,008.80. The best stock replenishment size is therefore 16 drums. 7. Excel spreadsheet CH6EX7 answers both (a) and (b). (a) For a cutoff point L = 12, the EOQ is 16.84. Since demand is in multiples of 2, comparison of Q=16 and Q=18 shows that Q = 16 is cheaper, i.e. $759 versus $760. Total cost = $2100. (b) The optimal cutoff point is 6 drums and the optimal stock replenishment size is 11.73, rounded to 12, with a total annual relevant cost of $2020.80. 8. (a) The cost of preparing a batch of 3600 cans is £54 consisting of £36 for the 2 hours of technicians time and £18 for the two machine operators time. (b) s = £54, v = cost of ingredients + labour cost + cost of can and label = £1.60 + (1/3600 h/l)(2)(£18.00) + 0.15 = £1.76/l, r = 0.18, D = 180,000 EOQ = = 7833.4 Total relevant cost = = £2481.65 (c) The best batch size that is a multiple of 1152 and closest to 7833.49 is 8064 cans. This has a total relevant cost of 0.5(8064)(1.76)(0.18) + (54)(180000)/(8064) = £2482.69 9. (a) Total annual cost of proposed policy = (0.5)(500)(60)(0.2) + (200)(750)/500 = £3300 (b) Optimal policy: expected demand = 750 + (1/3)(750) = 1000. EOQ = = 182.57 TRC(150) = (0.5)(150)(60)(0.2) + (200)(1000)/150 = £2233.33, TRC(200) = (0.5)(200)(60)(0.2) + (200)(1000)/200 = £2200 He can do better by ordering 200 sets five times a year. This is a saving of £3300 - 2200 = £1100. The proposed policy of ordering 500 sets twice a year has a much higher holding cost than setup cost and therefore has a higher cost.
6.6
10. Excel spreadsheet CH6EX10 answers both (a) and (b) and is reproduced below. Q-ELECTRONICS GUARANTEE PROBLEM INPUTS: Selling price
$35
/chip
Production cost Labour cost
$25 $10
/chip /replacement
Chips sold
2000
/year
Time Replacements Revenue Production cost Labour cost Profit
3 1 $70,000 $50,025 $10 $19,965
Increase guarantee from 6 months to: Increase Costs
9 months 12 months 15 months 18 months $105 $245 $455 $700
Extra
4 1 $70,000 $50,025 $10 $19,965
5 2 $70,000 $50,050 $20 $19,930
6 3 $70,000 $50,075 $30 $19,895
11
26
46
72
7 3 $70,000 $50,075 $30 $19,895
8 4 $70,000 $50,100 $40 $19,860
9 6 $70,000 $50,150 $60 $19,790
10 7 $70,000 $50,175 $70 $19,755
* Extra sales based on a gross profit = selling price – production cost = $10. Note that these extra sales will also produce guarantee claims. As a simple approximation, for example, for a guarantee period of 18 months, there are an additional 27 - 2 =25 replacements for 1000 chips. For resulting in 25(70/1000) = 1.75 further replacements, rounded to 2. 11. (a) Influence diagram for pump assembly system. Planned repl'ment size Production rate
Time to assemble pumps
Actual stock increase
Demand while assembling
Daily demand
Annual number of stock repl'ments Production setup cost/batch
Average stock level
Average stock investment
Annual setup cost
Annual stock investment
Unit product value
Investment holding cost
Total relevant cost
(b) [Number of replenishments/year] = [Yearly demand]/[Planned replenishment size] = D/Q [Days to assemble pumps] = [Planned replenishment size]/[Production rate] = Q/b [Demand while assembling] = [Daily demand] [Days to assemble pumps] = aQ/b [Actual stock increase] = [Planned replenishment size] - [Demand while assembling] = Q - aQ/b [Average stock level] = [Actual stock increase]/2 = (Q - aQ/b)/2 = 0.5Q(1 - a/b) [Annual holding cost] = [Average stock level] [Product value] [Holding cost] = 0.5Q(1 - a/b)vr [Annual setup cost] = [Number of replenishments/year] [Setup cost/replenishment] = Ds/Q [Total annual cost] = [Annual holding cost] + [Annual setup cost] = 0.5Q(1 - a/b)vr + Ds/Q (c) (1) Demand occurs at a constant rate. (reasonable approximation on average) (2) Setup times and production rates are constant or vary little, so that averages are a good approximation. (reasonable)
6.7
(d) Excel Spreadsheet CH6EX11 is reproduced below. Pump Assembly Problem Demand
1250
pumps/year
Daily demand
3.42
pumps/day
EOQ
110.72
TRC
£5,125.64
Annual setup cost
Annual holding cost
Product value
£216.00
/pump
Production setup cost Investm't holding cost
£227.00 £0.25
/production run /dollar invested/year
Production rate
24
pumps/day
Planned size of repl'ments
Number of repl'ments per year
1250
1.00
52.08
178
1,072
536
£227
£28,934
1100
1.14
45.83
157
943
472
£258
£25,462
£25,712
1000
1.25
41.67
143
857
429
£284
£23,147
£23,431
900
1.39
37.50
128
772
386
£315
£20,833
£21,148
800
1.56
33.33
114
686
343
£355
£18,518
£18,873
700
1.79
29.17
100
600
300
£405
£16,203
£16,608
600
2.08
25.00
86
514
257
£473
£13,888
£14,361
500
2.50
20.83
71
429
214
£568
£11,574
£12,141
400
3.13
16.67
57
343
171
£709
£9,259
£9,968
300
4.17
12.50
43
257
129
£946
£6,944
£7,890
Days to assemble pumps
Demand while assembling
Actual stock increase
Average stock level
Total relevant cost £29,161
200
6.25
8.33
29
171
86
£1,419
£4,629
£6,048
100
12.50
4.17
14
86
43
£2,838
£2,315
£5,152
50
25.00
2.08
7
43
21
£5,675
£1,157
£6,832
Optimal replenishment size is 110.72 with total relevant cost of £5125.64. 12. (a) Use Excel spreadsheet FIG6-5 Setup cost % Difference EOQ Total relevant cost % difference in cost
12 –33.33% 41.53 2,392.31 –18.35%
18 0% 50.87 2,929.97 0%
24 33.33% 58.74 3,383.23 15.47%
36 100% 71.94 4,143.60 41.42%
12% –33.33% 62.30 2,392.31 –18.25%
18% 0% 50.87 2,929.97 0%
24% 33.33% 44.05 3,383.23 15.47%
30% 66.67% 39.40 3,782.57 29.10%
£16.8 –20% 9.11 –£15.20 –126.25%
£18.9 –10% 16.54 £17.33 –70.7%
£21 0% 20.78 £57.90 0%
£26.25 25% 27.05 £168.23 190.55%
0.4 –20% 19.50 £53.90 –6.91%
0.5 0% 20.78 £57.90 0%
0.6 20% 22.36 £62.30 7.60%
0.75 0.5% 25.58 £69.89 20.71%
(b) Use Excel spreadsheet FIG6-5 Investment holding % difference EOQ Total relevant cost % difference in cost
13. (a) Use Excel spreadsheet CH6EX13 Net selling price % difference Time between adjustments Profit % difference in cost
(b) Use Excel spreadsheet CH6EX13 Defectives reworked % difference Time between adjustments Profit % difference in cost
6.8
(c) Use Excel spreadsheet CH6EX13 Slope of defectives function % difference Time between adjustments Profit % difference in cost
6 –40% 26.83 £58.93 1.77%
8 –20% 23.24 £58.91 1.74%
10 0% 20.78 £57.90 0%
5,000 –50% 83 i16,386.89 –50%
10,000 0% 83 i32,773.77 0%
20,000 100% 83 i65,547.55 100%
12 20% 18.97 £56.27 –2.82%
14. (a) Use Excel spreadsheet CH6EX14 Order size % difference Optimal speed Total cost % difference in cost
(b) Use Excel spreadsheet CH6EX14 Rate of Defectives % difference Optimal speed Total cost % cost increase
0.001 –20% 94 i31,985.35 –2.41%
0.00125 0% 83 i32,773.77 0%
0.0015 20% 75 i33,502.35 2.22%
15. (a) Use Excel spreadsheet FIG6-5 Used wrong value of $18 for setup cost Setup cost - true value 12 Error in setup cost 50% Should have ordered 41.53 Cost of optimal order size $2,392.31 Actual cost incurred $2442 Difference in cost $49.69 % cost increase 2.08%
24 –25% 58.74 $3,383.23 $3418 $34.77 1.03%
(b) Use Excel spreadsheet FIG6-5 Used wrong value of 18% for holding cost Holding cost - true value 12% % error in holding cost 50% Should have ordered 62.30 Cost of optimal order size $2,392.31 Actual cost incurred $2442 Difference in cost $49.69 % cost increase 2.08%
24% –25% 44.05 $3,383.23 $3418 $34.77 1.03%
(c) Use Excel spreadsheet FIG6-5 Error in setup cost -80% -20% 50% 100%
% Increase in costs 34% 0.6% 2.1% 6.1%
Setup cost &Holding cost [$12 & 12%] [$15 & 25%] [$24 & 15%]
Error in ratio of setup cost and holding costs 0% 66.67% –35.5%
% Increase in costs 0% about 3% about 3%
40,000 300% 83 i131,095.09 300%
6.9
16. (a) Use Excel spreadsheet CH6EX13 Used wrong value of £21 for net selling price Selling price - true value % error in selling price Time between adjustments should have been Profit if true value used Actual profit incurred Difference in profit % profit decrease
£26.25 –20% 27.05 £168.23 £163.45 £0.22 2.84%
(b) Use Excel spreadsheet CH6EX13 Used wrong value of 0.5 for the rate of defectives reworked Rework rate - true value 0.4 % error in rework rate 25% Time between adjustments should have been 19.5 Optimal profit £53.90 Actual profit incurred £53.71 Difference in profit £0.19 % profit decrease 0.35%
0.75 –33.33% 25.58 £69.89 £68.37 £1.52 2.17%
(c) Use Excel spreadsheet CH6EX13 Used wrong value of 10 for the slope of defectives function Slope of defectives function - true value 8 % error 25% Time between adjustments should have been 23.24 Optimal profit £67.04 Actual profit incurred £66.56 Difference in profit £0.48 % profit decrease 0.72%
17. (a) Use Excel spreadsheet CH6EX14 Used wrong value of 10,000 for order size Order size - true value 20,000 % error in order size –50% Speed should have been 83 Optimal total cost i65,547.55 Actual cost incurred i65,547.55 Difference in cost 0 % cost increase 0%
(b) Use Excel spreadsheet CH6EX14 Used wrong value of 0.00125 for rate of defectives Rate of defectives - true value 0.001 % error 25% Speed should have been 94 Optimal total cost i31,985.35 Actual cost incurred i32,032.16 Difference in cost i46.81 % cost increase 0.15%
12 –33.33% 18.97 £49.51 £49.24 £0.27 0.55%
6.10
18. The contradiction is the consequence of the evaluation to proceed being made on a different cost/benefit basis, i.e., total costs and benefits at the beginning of the project, but only incremental costs and benefits, once the analysis has been completed, but not yet implemented. The implication of this is that whether or not a project should be abandoned or continued should be made as early as possible, so as to avoid incurring costs that may never be recovered by the benefits of implementing the recommendations. An early assessment, even based on fairly inaccurate and vague data, may nevertheless indicate that even optimistic benefits may incur a high risk of not recovering further costs to be incurred and it may be prudent/wise to abandon the project at that point (or at least advise the problem owner of this tentative assessment to make an informed decision). 19. Outline of a project report for the ELMO case: Table of contents Introductory statement: Report reports the result of an analysis as to the best speed setting for the new coil winding machine for a production run of 10,000 motors, based on preliminary findings of the number of failures produced on six trial runs at speeds of 30 to 90 coils/hour. Executive Summary: Based on an approximate functional relation of the failure rate as a function of the speed setting derived from the trial runs for the motor in question, the best speed setting was found to be 83 at a total cost of i32,774. The optimal speed of 83 is independent of the lot size. This result is relatively robust in the sense that any setting between 70 and 90 deviates from this optimum by less than i100. Similarly, an error in the failure function of up to 25% causes a negligible increase in the actual cost incurred of only around 0.15%. It is recommended that the speed be set at the optimal level found for this type motor, and that future runs for other motors use the same model, modified for data. Statement of the Problem: (Summary of some of the wording in exercise 4) Analysis: Brief description of how failure rate function was derived and of the model used. Major findings: Reproduction of the spreadsheet shown in the answer to exercise 4 (c) and exercise 17 (b). Recommendations: For this problem, this section can be dropped if the ‘Major Findings’ clearly identify the optimal solution, given that the ‘Executive Summary’ already summarizes the recommendations. 20. Student's own words! 21. Establishing internal and external validity is important for the decision maker/user of the solution since it will strongly contribute towards establishing credibility of the model and confidence in the solution in the minds of these people. Without confidence and credibility implementation and proper continued use is unlikely to occur. It is important/crucial for the problem analyst for a number of reasons: (a) It is the only means to establish the correctness of the model and its solution and its validity in terms of the issue addressed (to the extent that this can be done!), (b) this steps allows the analyst to learn from any errors or misinterpretations made, and is thus important to increase her/his expertise, (c) it increases the confidence of the analyst in her/his abilities, and (d) it is part of the ethics of good mathematical modelling.
