Solutions to Case Problems Manual to Accompany An Introduction To management science Quantitative Approaches To Decision Making Twelfth Edition David R. Anderson University of Cincinnati Dennis J. Sweeney University of Cincinnati Thomas A. Williams Rochester Institute of Technology R. Kipp Martin University of Chicago South-Western Cincinnati, Ohio Contents Preface
Chapter 1: Introduction ¦ Scheduling a Golf League Chapter 2: An Introduction to Linear Programming ¦ Workload Balancing ¦ Production Strategy ¦ Hart Venture Capital Chapter 3: Linear Programming: Sensitivity Analysis and Interpretation of Solution ¦ Product Mix ¦ Investment Strategy ¦ Truck Leasing Strategy Chapter 4: Linear Programming Applications in Marketing, Finance and Operations Management ¦ Planning an Advertising Campaign ¦ Phoenix Computer ¦ Textile Mill Scheduling ¦ Workforce Scheduling ¦ Duke Energy Coal Allocation Chapter 6: Distribution and Network Models ¦ Solution Plus ¦ Distribution Systems Design Chapter 7: Integer Linear Programming ¦ Textbook Publishing ¦ Yeager National Bank ¦ Production Scheduling with Changeover Costs Chapter 8: Nonlinear Optimization Models ¦ Portfolio Optimization with Transaction Costs Chapter 9: Project Scheduling: PERT/CPM ¦ R. C. Coleman Chapter 10: Inventory Models ¦ Wagner Fabricating Company ¦ River City Fire Department Chapter 11: Waiting Line Models ¦ Regional Airlines ¦ Office Equipment, Inc. Chapter 12: Simulation ¦ Tri-State Corporation ¦ Harbor Dunes Golf Course ¦ County Beverage Drive-Thru Chapter 13: Decision Analysis ¦ Property Purchase Strategy ¦ Lawsuit Defense Strategy Chapter 14: Multicriteria Decision Problems ¦ EZ Trailers, Inc.
Chapter 15: Forecasting ¦ Forecasting Sales ¦ Forecasting Lost Sales Chapter 16: Markov Processes ¦ Dealer’s Absorbing State Probabilities in Black Jack Chapter 21: Dynamic Programming ¦ Process Design Preface The purpose of An Introduction to Management Science is to provide students with a sound conceptual understanding of the role management science pays in the decision-making process. The text emphasizes the application of management science by using problem situations to introduce each of the management science concepts and techniques. The book has been specifically designed to meet the needs of nonmathematicians who are studying business and economics. The Solutions to Case Problems Manual contains solutions to the case problems.
Note: The solutions to the end-of-chapter problems and learning objectives for each chapter are included in the Solutions Manual. Acknowledgements We would like to provide a special acknowledgement to Catherine J. Williams for her efforts in preparing the Instructor’s Manual. We are also indebted to our acquisitions editor Charles E. McCormick, Jr. and our developmental editor Alice C. Denny for their support during the preparation of this manual. David R. Anderson Dennis J. Sweeney Thomas A. Williams R. Kipp Martin Chapter 1 Introduction Case Problem: Scheduling a Golf League Note to Instructor: This case problem illustrates the value of the rational management science approach.
The problem is easy to understand and, at first glance, appears simple. But, most students will have trouble finding a solution. The solution procedure suggested involves decomposing a larger problem into a series of smaller problems that are easier to solve. The case provides students with a good first look at the kinds of problems where management science is applied in practice. The problem is a real one that one of the authors was asked by the Head Professional at Royal Oak Country Club for help with. Solution: Scheduling problems such as this occur frequently, and are often difficult to solve. The typical approach is to use trial and error.
An alternative approach involves breaking the larger problem into a series of smaller problems. We show how this can be done here using what we call the Red, White, and Blue algorithm. Suppose we break the 18 couples up into 3 divisions, referred to as the Red, White, and Blue divisions. The six couples in the Red division can then be identified as R1, R2, R3, R4, R5, R6; the six couples in the White division can be identified as W1, W2,…, W6; and the six couples in the Blue division can be identified as B1, B2,…, B6. We begin by developing a schedule for the first 5 weeks of the season so that each couple plays every other couple in its own division.
This can be done fairly easily by trial and error. Shown below is the first 5-week schedule for the Red division. Week 1 R1 vs. R2 R3 vs. R4 R5 vs. R6 Week 2 R1 vs. R3 R2 vs. R5 R4 vs. R6 Week 3 R1 vs. R4 R2 vs. R6 R3 vs. R5 Week 4 R1 vs. R5 R2 vs. R4 R3 vs. R6 Week 5 R1 vs. R6 R2 vs. R3 R4 vs. R5 Similar 5-week schedules can be developed for the White and Blue divisions by replacing the R in the above table with a W or a B. To develop the schedule for the next 3 weeks, we create 3 new six-couple divisions by pairing 3 of the teams in each division with 3 of the teams in another division; for example, (R1, R2, R3, W1, W2, W3), (B1, B2, B3, R4, R5, R6), and (W4, W5, W6, B4, B5, B6).
Within each of these new divisions, matches can be scheduled for 3 weeks without any couples playing a couple they have played before. For instance, a 3-week schedule for the first of these divisions is shown below: Week 6 R1 vs. W1 R2 vs. W2 R3 vs. W3 Week 7 R1 vs. W2 R2 vs. W3 R3 vs. W1 Week 8 R1 vs. W3 R2 vs. W1 R3 vs. W2 A similar 3-week schedule can be easily set up for the other two new divisions. This will provide us with a schedule for the first 8 weeks of the season. For the final 9 weeks, we continue to create new divisions by pairing 3 teams from the original Red, White and Blue divisions with 3 teams from the other divisions that they have not yet been paired with. Then a 3week schedule is developed as above.
Shown below is a set of divisions for the next 9 weeks. CP – 1 Chapter 1 Weeks 9-11 (R1, R2, R3, W4, W5, W6) Weeks 12-14 (R1, R2, R3, B1, B2, B3) Weeks 15-17 (R1, R2, R3, B4, B5, B6) (W1, W2, W3, B1, B2, B3) (R4, R5, R6, B4, B5, B6) (W1, W2, W3, B4, B5, B6) (W4, W5, W6, R4, R5, R6) (W1, W2, W3, R4, R5, R6) (W4, W5, W6, B1, B2, B3) This Red, White and Blue scheduling procedure provides a schedule with every couple playing every other couple over the 17-week season. If one of the couples should cancel, the schedule can be modified easily. Designate the couple that cancels, say R4, as the Bye couple. Then whichever couple is scheduled to play couple R4 will receive a Bye in that week.
With only 17 couples a Bye must be scheduled for one team each week. This same scheduling procedure can obviously be used for scheduling sports teams and or any other kinds of matches involving 17 or 18 teams. Modifications of the Red, White and Blue algorithm can be employed for 15 or 16 team leagues and other numbers of teams. CP – 2 Chapter 2 An Introduction to Linear Programming Case Problem 1: Workload Balancing 1. Production Rate (minutes per printer) Line 1 Line 2 3 4 6 2 Model DI-910 DI-950 Profit Contribution ($) 42 87 Capacity: 8 hours ? 60 minutes/hour = 480 minutes per day Let D1 = number of units of the DI-910 produced D2 = number of units of the DI-950 produced 42D1 + 87D2 ? 480 ? 80 Line 1 Capacity Line 2 Capacity Max s. t. 3D1 + 6D2 4D1 + 2D2 D 1, D 2 ? 0 Using The Management Scientist, the optimal solution is D1 = 0, D2 = 80. The value of the optimal solution is $6960. Management would not implement this solution because no units of the DI-910 would be produced. 2. Adding the constraint D1 ? D2 and resolving the linear program results in the optimal solution D1 = 53. 333, D2 = 53. 333. The value of the optimal solution is $6880. Time spent on Line 1: 3(53. 333) + 6(53. 333) = 480 minutes Time spent on Line 2: 4(53. 333) + 2(53. 333) = 320 minutes Thus, the solution does not balance the total time spent on Line 1 and the total time spent on Line 2.
This might be a concern to management if no other work assignments were available for the employees on Line 2. 4. Let T1 = total time spent on Line 1 T2 = total time spent on Line 2 3. Whatever the value of T2 is, T1 ? T2 + 30 T1 ? T2 – 30 Thus, with T1 = 3D1 + 6D2 and T2 = 4D1 + 2D2 3D1 + 6D2 ? 4D1 + 2D2 + 30 3D1 + 6D2 ? 4D1 + 2D2 ? 30 CP – 3 Chapter 2 Hence, ? 1D1 + 4D2 ? 30 ? 1D1 + 4D2 ? ?30 Rewriting the second constraint by multiplying both sides by -1, we obtain ? 1D1 + 4D2 ? 30 1D1 ? 4D2 ? 30 Adding these two constraints to the linear program formulated in part (2) and resolving using The Management Scientist, we obtain the optimal solution D1 = 96. 667, D2 = 31. 667. The value of the optimal solution is $6815.
Line 1 is scheduled for 480 minutes and Line 2 for 450 minutes. The effect of workload balancing is to reduce the total contribution to profit by $6880 – $6815 = $65 per shift. 5. The optimal solution is D1 = 106. 667, D2 = 26. 667. The total profit contribution is 42(106. 667) + 87(26. 667) = $6800 Comparing the solutions to part (4) and part (5), maximizing the number of printers produced (106. 667 + 26. 667 = 133. 33) has increased the production by 133. 33 – (96. 667 + 31. 667) = 5 printers but has reduced profit contribution by $6815 – $6800 = $15. But, this solution results in perfect workload balancing because the total time spent on each line is 480 minutes. Case Problem 2: Production Strategy 1.
Let BP100 = the number of BodyPlus 100 machines produced BP200 = the number of BodyPlus 200 machines produced 371BP100 + 8BP100 5BP100 2BP100 -0. 25BP100 BP100, BP200 ? 0 + + + + 461BP200 12BP200 10BP200 2BP200 0. 75BP200 ? ? ? ? 600 450 140 0 Machining and Welding Painting and Finishing Assembly, Test, and Packaging BodyPlus 200 Requirement Max s. t. CP – 4 Solutions to Case Problems BP200 80 . Number of BodyPlus 200 70 Assembly, Test, and Packaging 60 50 40 30 20 10 0 10 20 Painting and Finishing BP100 30 40 50 60 70 80 90 100 Number of BodyPlus 100 Optimal Solution BodyPlus 200 Requirement Machining and Welding Optimal solution: BP100 = 50, BP200 = 50/3, profit = $26,233. 33. Note: If the optimal solution is rounded to BP100 = 50, BP200 = 16. 7, the value of the optimal solution will differ from the value shown. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as The Management Scientist. 2. In the short run the requirement reduces profits. For instance, if the requirement were reduced to at least 24% of total production, the new optimal solution is BP100 = 1425/28, BP200 = 225/14, with a total profit of $26,290. 18; thus, total profits would increase by $56. 85. Note: If the optimal solution is rounded to BP100 = 50. 89, BP200 = 16. 07, the value of the optimal solution will differ from the value shown.
The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as The Management Scientist. If management really believes that the BodyPlus 200 can help position BFI as one of the leader’s in high-end exercise equipment, the constraint requiring that the number of units of the BodyPlus 200 produced be at least 25% of total production should not be changed. Since the optimal solution uses all of the available machining and welding time, management should try to obtain additional hours of this resource. 3. CP – 5 Chapter 2 Case Problem 3: Hart Venture Capital 1.
Let S = fraction of the Security Systems project funded by HVC M = fraction of the Market Analysis project funded by HVC Max s. t. 1,800,000S 600,000S 600,000S 250,000S S S,M + + + + 1,600,000M 500,000M 350,000M 400,000M M 0 ? ? ? ? ? 800,000 700,000 500,000 1 1 Year 1 Year 2 Year 3 Maximum for S Maximum for M ? The solution obtained using The Management Scientist software package is shown below: OPTIMAL SOLUTION Objective Function Value = Variable ————-S M 2486956. 522 Reduced Costs —————–0. 000 0. 000 Value ————–0. 609 0. 870 Constraint ————-1 2 3 4 5 Slack/Surplus ————–0. 000 30434. 783 0. 000 0. 391 0. 130 Dual Prices —————–2. 83 0. 000 0. 522 0. 000 0. 000 OBJECTIVE COEFFICIENT RANGES Variable ———–S M Lower Limit ————–No Lower Limit No Lower Limit Current Value ————–1800000. 000 1600000. 000 Upper Limit ————–No Upper Limit No Upper Limit RIGHT HAND SIDE RANGES Constraint ———–1 2 3 4 5 Lower Limit ————–No Lower Limit 669565. 217 461111. 111 0. 609 0. 870 Current Value ————–800000. 000 700000. 000 500000. 000 1. 000 1. 000 Upper Limit ————–822950. 820 No Upper Limit No Upper Limit No Upper Limit No Upper Limit CP – 6 Solutions to Case Problems Thus, the optimal solution is S = 0. 609 and M = 0. 870.
In other words, approximately 61% of the Security Systems project should be funded by HVC and 87% of the Market Analysis project should be funded by HVC. The net present value of the investment is approximately $2,486,957. 2. Security Systems Market Analysis Total Year 1 $365,400 $435,000 $800,400 Year 2 $365,400 $304,500 $669,900 Year 3 $152,250 $348,000 $500,250 Note: The totals for Year 1 and Year 3 are greater than the amounts available. The reason for this is that rounded values for the decision variables were used to compute the amount required in each year. To see why this situation occurs here, first note that each of the problem coefficients is an integer value.
Thus, by default, when The Management Scientist prints the optimal solution, values of the decision variables are rounded and printed with three decimal places. To increase the number of decimal places shown in the output, one or more of the problem coefficients can be entered with additional digits to the right of the decimal point. For instance, if we enter the coefficient of 1 for S in constraint 4 as 1. 000000 and resolve the problem, the new optimal values for S and D will be rounded and printed with six decimal places. If we use the new values in the computation of the amount required in each year, the differences observed for year 1 and year 3 will be much smaller than we obtained using the values of S = 0. 09 and M = 0. 870. 3. If up to $900,000 is available in year 1 we obtain a new optimal solution with S = 0. 689 and M = 0. 820. In other words, approximately 69% of the Security Systems project should be funded by HVC and 82% of the Market Analysis project should be funded by HVC. The net present value of the investment is approximately $2,550,820. The solution obtained using The Management Scientist software package follows: OPTIMAL SOLUTION Objective Function Value = Variable ————-S M Constraint ————-1 2 3 4 5 2550819. 672 Reduced Costs —————–0. 000 0. 000 Dual Prices —————–0. 000 2. 098 2. 164 0. 000 0. 00 Value ————–0. 689 0. 820 Slack/Surplus ————–77049. 180 0. 000 0. 000 0. 311 0. 180 OBJECTIVE COEFFICIENT RANGES Variable ———–S M Lower Limit ————–No Lower Limit No Lower Limit Current Value ————–1800000. 000 1600000. 000 Upper Limit ————–No Upper Limit No Upper Limit CP – 7 Chapter 2 RIGHT HAND SIDE RANGES Constraint ———–1 2 3 4 5 Lower Limit ————–822950. 820 No Lower Limit No Lower Limit 0. 689 0. 820 Current Value ————–900000. 000 700000. 000 500000. 000 1. 000 1. 000 Upper Limit ————–No Upper Limit 802173. 913 630555. 556 No Upper Limit No Upper Limit 4.
If an additional $100,000 is made available, the allocation plan would change as follows: Year 1 $413,400 $410,000 $823,400 Year 2 $413,400 $287,000 $700,400 Year 3 $172,250 $328,000 $500,250 Security Systems Market Analysis Total 5. Having additional funds available in year 1 will increase the total net present value. The value of the objective function increases from $2,486,957 to $2,550,820, a difference of $63,863. But, since the allocation plan shows that $823,400 is required in year 1, only $23,400 of the additional $100,00 is required. We can also determine this by looking at the slack variable for constraint 1 in the new solution. This value, 77049. 80, shows that at the optimal solution approximately $77,049 of the $900,000 available is not used. Thus, the amount of funds required in year 1 is $900,000 – $77,049 = $822,951. In other words, only $22,951 of the additional $100,000 is required. The differences between the two values, $23,400 and $22,951, is simply due to the fact that the value of $23,400 was computed using rounded values for the decision variables. The value of $22,951 is computed internally in The Management Scientist output and is not subject to this rounding. Thus, the most accurate value is $22,951. CP – 8 Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution Case Problem 1: Product Mix
Note to Instructor: The difference between relevant and sunk costs is critical. The cost of the shipment of nuts is a sunk cost. Practice in applying sensitivity analysis to a business decision is obtained. You may want to suggest that sensitivity analyses other than the ones we have suggested be undertaken. 1. Cost per pound of ingredients Almonds $7500/6000 = $1. 25 Brazil $7125/7500 = $. 95 Filberts $6750/7500 = $. 90 Pecans $7200/6000 = $1. 20 Walnuts $7875/7500 = $1. 05 Cost of nuts in three mixes: Regular mix: . 15($1. 25) + . 25($. 95) + . 25($90) + . 10($1. 20) + . 25($1. 05) = $1. 0325 Deluxe mix . 20($1. 25) + . 20($. 95) + . 20($. 90) + . 20($1. 20) + . 20($1. 05) = $1. 07 Holiday mix: . 25($1. 5) + . 15($. 95) + . 15($. 90) + . 25($1. 20) + . 20($1. 05) = $1. 10 2. Let R = pounds of Regular Mix produced D = pounds of Deluxe Mix produced H = pounds of Holiday Mix produced Note that the cost of the five shipments of nuts is a sunk (not a relevant) cost and should not affect the decision. However, this information may be useful to management in future pricing and purchasing decisions. A linear programming model for the optimal product mix is given. The following linear programming model can be solved to maximize profit contribution for the nuts already purchased. Max s. t. 1. 65R 0. 15R 0. 25R 0. 25R 0. 10R 0. 25R R + + + + + + 2. 00D 0. 0D 0. 20D 0. 20D 0. 20D 0. 20D D H R, D, H ? 0 + + + + + + 2. 25H 0. 25H 0. 15H 0. 15H 0. 25H 0. 20H ? ? ? ? ? ? ? ? 6000 7500 7500 6000 7500 10000 3000 5000 Almonds Brazil Filberts Pecans Walnuts Regular Deluxe Holiday CP – 9 Chapter 3 The solution found using The Management Scientist is shown below. Objective Function Value = Variable ————-R D H 61375. 000 Reduced Costs —————–0. 000 0. 000 0. 000 Value ————–17500. 000 10624. 999 5000. 000 Constraint ————-1 2 3 4 5 6 7 8 Slack/Surplus ————–0. 000 250. 000 250. 000 875. 000 0. 000 7500. 000 7624. 999 0. 000 Dual Prices —————–8. 500 0. 000 0. 00 0. 000 1. 500 0. 000 0. 000 -0. 175 OBJECTIVE COEFFICIENT RANGES Variable ———–R D H Lower Limit ————–1. 500 1. 892 No Lower Limit Current Value ————–1. 650 2. 000 2. 250 Upper Limit ————–2. 000 2. 200 2. 425 RIGHT HAND SIDE RANGES Constraint ———–1 2 3 4 5 6 7 8 Lower Limit ————–5390. 000 7250. 000 7250. 000 5125. 000 6750. 000 No Lower Limit No Lower Limit -0. 000 Current Value ————–6000. 000 7500. 000 7500. 000 6000. 000 7500. 000 10000. 000 3000. 000 5000. 000 Upper Limit ————–6583. 333 No Upper Limit No Upper Limit No Upper Limit 7750. 000 17500. 000 10624. 999 9692. 307 3.
From the dual prices it can be seen that additional almonds are worth $8. 50 per pound to TJ. Additional walnuts are worth $1. 50 per pound. From the slack variables, we see that additional Brazil nut, Filberts, and Pecans are of no value since they are already in excess supply. Yes, purchase the almonds. The dual price shows that each pound is worth $8. 50; the dual price is applicable for increases up to 583. 33 pounds. 4. CP – 10 Solutions to Case Problems Resolving the problem by changing the right-hand side of constraint 1 from 6000 to 7000 yields the following optimal solution. The optimal solution has increased in value by $4958. 34. Note that only 583. 3 pounds of the additional almonds were used, but that the increase in profit contribution more than justifies the $1000 cost of the shipment. Objective Function Value = Variable ————-R D H Constraint ————-1 2 3 4 5 6 7 8 66333. 336 Reduced Costs —————–0. 000 0. 000 0. 000 Dual Prices —————–0. 000 0. 000 0. 000 5. 667 4. 333 0. 000 0. 000 -0. 033 Value ————–11666. 667 17916. 668 5000. 000 Slack/Surplus ————–416. 667 250. 000 250. 000 0. 000 0. 000 1666. 667 14916. 667 0. 000 OBJECTIVE COEFFICIENT RANGES Variable ———–R D H Lower Limit ————–1. 000 1. 976 No Lower Limit Current Value ————–1. 650 2. 000 2. 250 Upper Limit ————–1. 750 3. 300 2. 283
RIGHT HAND SIDE RANGES Constraint ———–1 2 3 4 5 6 7 8 Lower Limit ————–6583. 333 7250. 000 7250. 000 4210. 000 7250. 000 No Lower Limit No Lower Limit 0. 002 Current Value ————–7000. 000 7500. 000 7500. 000 6000. 000 7500. 000 10000. 000 3000. 000 5000. 000 Upper Limit ————–No Upper Limit No Upper Limit No Upper Limit 6250. 000 7750. 000 11666. 667 17916. 668 15529. 412 5. From the dual prices it is clear that there is no advantage to not satisfying the orders for the Regular and Deluxe mixes. However, it would be advantageous to negotiate a decrease in the Holiday mix requirement. CP – 11 Chapter 3 Case Problem 2: Investment Strategy 1.
The first step is to develop a linear programming model for maximizing return subject to constraints for funds available, diversity, and risk tolerance. Let G = Amount invested in growth fund I = Amount invested in income fund M = Amount invested in money market fund The LP formulation and optimal solution found using The Management Scientist are shown. MAX . 18G +. 125I +. 075M S. T. 1) 2) 3) 4) 5) 6) 7) G + I + M < 800000 . 8G -. 2I -. 2M > 0 . 6G -. 4I -. 4M < 0 -. 2G +. 8I -. 2M > 0 -. 5G +. 5I -. 5M < 0 -. 3G -. 3I +. 7M > 0 . 05G + . 02I -. 04M < 0 Funds Available Min growth fund Max growth fund Min income fund Max income fund Min money market fund Max risk OPTIMAL SOLUTION Objective Function Value = Variable ————-G I M 94133. 36 Reduced Costs —————–0. 000 0. 000 0. 000 Value ————–248888. 906 160000. 000 391111. 094 Constraint ————-1 2 3 4 5 6 7 Slack/Surplus ————–0. 000 88888. 898 71111. 109 0. 000 240000. 000 151111. 109 0. 000 Dual Prices —————–0. 118 0. 000 0. 000 -0. 020 0. 000 0. 000 1. 167 OBJECTIVE COEFFICIENT RANGES Variable ———–G I M Lower Limit ————–0. 150 -0. 463 0. 015 Current Value ————–0. 180 0. 125 0. 075 Upper Limit ————–No Upper Limit 0. 145 0. 180 CP – 12 Solutions to Case Problems RIGHT HAND SIDE RANGES Constraint ———–1 2 3 4 5 6 7 Lower Limit ————–0. 88 No Lower Limit -71111. 109 -106666. 641 -240000. 000 No Lower Limit -8000. 000 Current Value ————–800000. 000 0. 000 0. 000 0. 000 0. 000 0. 000 0. 000 Upper Limit ————–No Upper Limit 88888. 898 No Upper Limit 133333. 313 No Upper Limit 151111. 109 6399. 999 Rounding to the nearest dollar, the portfolio recommendation for Langford is as follows. Amount Invested $248,889 160,000 391,111 $800,000 Fund: Growth Income Money Market Total Yield = 94,133 / 800,000 = . 118 The portfolio yield is . 118 or 11. 8%. Note that the portfolio yield equals the dual price for the funds available constraint. 2. If Langford’s risk index is increased by . 05 that is the same as increasing the right-hand side of constraint 7 by . 005 (800,000) = 4000. Since this amount of increase is within the right-hand-side range, we would expect an increase in return of 1. 167 (4000) = 4668. The revised formulation and new optimal solution are shown below. Except for rounding, the value has increased as predicted; the new optimal allocation is Amount Invested $293,333 160,000 346,667 $800,000 Fund: Growth Income Money Market Total The portfolio yield becomes 98,800/800,000 = . 124 or 12. 4% MAX . 18G +. 125I +. 075M S. T. 1) 2) 3) 4) 5) 6) 7) G + I + M < 800000 . 8G -. 2I -. 2M > 0 . 6G -. 4I -. 4M < 0 -. 2G +. 8I -. 2M > 0 -. 5G +. 5I -. 5M < 0 -. 3G -. I +. 7M > 0 . 045G + . 015I-. 045M < 0 CP – 13 Chapter 3 OPTIMAL SOLUTION Objective Function Value = Variable ————-G I M 98800. 000 Reduced Costs —————–0. 000 0. 000 0. 000 Value ————–293333. 313 160000. 000 346666. 656 Constraint ————-1 2 3 4 5 6 7 3. Slack/Surplus ————–0. 000 133333. 328 26666. 666 0. 000 240000. 000 106666. 664 0. 000 Dual Prices —————–0. 124 0. 000 0. 000 -0. 020 0. 000 0. 000 1. 167 Since . 16 is in the objective coefficient range for the growth fund return, there would be no change in allocation. However, the return would decrease by (. 02) ($248,889) = $4978.
A decrease to . 14 is outside the objective function coefficient range forcing us to resolve the problem. The new formulation and optimal solution is as follows. MAX . 14G +. 125I +. 075M S. T. 1) 2) 3) 4) 5) 6) 7) G + I + M < 800000 . 8G -. 2I -. 2M > 0 . 6G -. 4I -. 4M < 0 -. 2G +. 8I -. 2M > 0 -. 5G +. 5I -. 5M < 0 -. 3G -. 3I +. 7M > 0 . 05G + . 02I-. 04M < 0 OPTIMAL SOLUTION Objective Function Value = Variable ————-G I M 85066. 664 Reduced Costs —————–0. 000 0. 000 0. 000 Value ————–160000. 016 293333. 313 346666. 688 CP – 14 Solutions to Case Problems Constraint ————-1 2 3 4 5 6 7 4.
Slack/Surplus ————–0. 000 0. 000 160000. 000 133333. 313 106666. 688 106666. 688 0. 000 Dual Prices —————–0. 106 -0. 010 0. 000 0. 000 0. 000 0. 000 0. 833 Since the current optimal solution has more invested in the growth fund than the income fund, adding this requirement will force us to resolve the problem with a new constraint. We should expect a decrease in return as is shown in the following optimal solution. MAX . 18G +. 125I +. 075M S. T. 1) 2) 3) 4) 5) 6) 7) 8) G + I + M < 800000 . 8G -. 2I -. 2M > 0 . 6G -. 4I -. 4M < 0 -. 2G +. 8I -. 2M > 0 -. 5G +. 5I -. 5M < 0 -. 3G -. 3I +. 7M > 0 . 05G + . 02I-. 04M < 0 G – I < 0
OPTIMAL SOLUTION Objective Function Value = Variable ————-G I M Constraint ————-1 2 3 4 5 6 7 8 93066. 656 Reduced Costs —————–0. 000 0. 000 0. 000 Dual Prices —————–0. 116 0. 000 0. 000 0. 000 0. 000 0. 000 1. 033 0. 012 Value ————–213333. 313 213333. 313 373333. 313 Slack/Surplus ————–0. 000 53333. 324 106666. 664 53333. 324 186666. 656 133333. 328 0. 000 0. 000 Note that the value of the solution has decreased from $94,133 to $93,067. This is only a decrease of 0. 2% inyield. Since the yield decrease is so small, Williams may prefer this portfolio for Langford. 5. It is possible a model such as this could be developed for each client.
The changed yield estimates would require a change in the objective function coefficients and resolving the problem if the change was outside the objective coefficient range. CP – 15 Chapter 3 Case Problem 3: Truck Leasing Strategy 1. Let xij = number of trucks obtained from a short term lease signed in month i for a period of j months yi = number of trucks obtained from the long-term lease that are used in month i Monthly fuel costs are 20 ($100) = $2000. Monthly Costs for Short-Term Leased Trucks Note: the costs shown here include monthly fuel costs of $2000. Decision Variables x11, x21, x31, x41 x12, x22, x32 x13, x23 x14 Cost $4000 + $2000 = $6000 2 ($3700) + $2000 = $9400 3 ($3225) + $2000 = $11,675 4 ($3040) + $2000 = $14,160
Monthly Costs for Long-Term Leased Trucks Since Reep Construction is committed to the long-term lease and since employees cannot be laid off, the only relevant cost for the long-term leased trucks is the monthly fuel cost of $2000. MIN 6000X11 + 9400X12 + 11675X13 + 14160X14 + 6000X21 + 9400X22 + 11675X23 + 6000X31 + 9400X32 + 6000X41 + 2000Y1 + 2000Y2 + 2000Y3 + 2000Y4 S. T. 1) 2) 3) 4) 5) 6) 7) 8) X11 + X12 X21 + X22 X31 + X32 X41 + X32 Y1 < 1 Y2 < 2 Y3 < 3 Y4 < 1 + + + + X13 X23 X23 X23 + + + + X14 X14 X22 X14 + + + + Y1 X13 X14 Y4 = + + = 10 X12 + Y2 = 12 X13 + Y3 = 14 8 CP – 16 Solutions to Case Problems Objective Function Value = Variable ————-X11 X12 X13 X14 X21 X22 X23 X31 X32 X41 Y1 Y2 Y3 Y4 Constraint ————-1 2 3 4 5 6 7 8 151660. 00 Reduced Costs —————–3515. 000 3725. 000 0. 000 0. 000 2810. 000 210. 000 0. 000 0. 000 915. 000 3515. 000 0. 000 0. 000 0. 000 0. 000 Dual Prices ——————2485. 000 -3190. 000 -6000. 000 -2485. 000 485. 000 1190. 000 4000. 000 485. 000 Value ————–0. 000 0. 000 3. 000 6. 000 0. 000 0. 000 1. 000 1. 000 0. 000 0. 000 1. 000 2. 000 3. 000 1. 000 Slack/Surplus ————–0. 000 0. 000 0. 000 0. 000 0. 000 0. 000 0. 000 0. 000 OBJECTIVE COEFFICIENT RANGES Variable ———–X11 X12 X13 X14 X21 X22 X23 X31 X32 X41 Y1 Y2 Y3 Y4 Lower Limit ————–2485. 000 5675. 000 10760. 000 13950. 000 3190. 000 9190. 000 10485. 000 3190. 000 8485. 000 2485. 00 No Lower Limit No Lower Limit No Lower Limit No Lower Limit Current Value ————–6000. 000 9400. 000 11675. 000 14160. 000 6000. 000 9400. 000 11675. 000 6000. 000 9400. 000 6000. 000 2000. 000 2000. 000 2000. 000 2000. 000 Upper Limit ————–No Upper Limit No Upper Limit 11885. 000 15075. 000 No Upper Limit No Upper Limit 11885. 000 6915. 000 No Upper Limit No Upper Limit 2485. 000 3190. 000 6000. 000 2485. 000 CP – 17 Chapter 3 RIGHT HAND SIDE RANGES Constraint Lower Limit ————————-1 4. 000 2 11. 000 3 13. 000 4 2. 000 5 0. 000 6 1. 000 7 0. 000 8 0. 000 2. 3. Current Value ————–10. 000 12. 000 14. 000 8. 000 1. 000 2. 000 3. 000 1. 000
Upper Limit ————–11. 000 13. 000 No Upper Limit 11. 000 7. 000 3. 000 4. 000 7. 000 The total cost associated with the leasing plan is $151,660. If Reep Construction is willing to consider the possibility of layoffs, we need to include driver costs of $3200 per month. Replacing the coefficients for y1, y2, y3, and y4 in our previous linear program with $5200 and resolving resulted in the following leasing plan: Month Leased 1 2 3 4 Length of Lease (Months) 1 0 0 0 0 2 0 0 0 _ 3 4 2 _ _ 4 6 _ _ _ In addition, in month 3, two of the trucks from the long-term leases were used. The total cost of this leasing plan is $165,410. To see what effect a o layoff policy has, we can set y1 = 1, y2 = 2, y3 = 3, y4 = 1 and resolve the linear program using objective coefficients of $5200 for y1, y2, y3, and y4. The new optimal solution forces us to use all the available trucks from the long-term lease; the optimal leasing plan is shown below. Month Leased 1 2 3 4 Length of Lease (Months) 1 0 0 1 0 2 0 0 0 _ 3 3 1 _ _ 4 6 _ _ _ The total cost associated with this solution is $174,060. Thus, if Reep maintains their current policy of no layoffs they will incur an additional cost of $174,060 – $165,410 = $8,650. CP – 18 Chapter 4 Linear Programming Applications in Marketing, Finance and Operations Management Case Problem 1: Planning an Advertising Campaign
The decision variables are as follows: T1 = number of television advertisements with rating of 90 and 4000 new customers T2 = number of television advertisements with rating of 40 and 1500 new customers R1 = number of radio advertisements with rating of 25 and 2000 new customers R2 = number of radio advertisements with rating of 15 and 1200 new customers N1 = number of newspaper advertisements with rating of 10 and 1000 new customers N2 = number of newspaper advertisements with rating of 5 and 800 new customers The Linear Programming Model and solution using The Management Scientist follow: MAX 90T1+55T2+25R1+20R2+10N1+5N2 S. T. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 1T1
