Solution Manual for Introduction to Management Science, 13th Edition

Chapter Two: Linear Programming: Model Formulation and Graphical Solution PROBLEM SUMMARY 1. Maximization (1–40 continuation), graphical solution 34. Maximization, graphical solution 35. Sensitivity analysis (2–34) 2. Minimization, graphical solution 36. Maximization, graphical solution 3. Sensitivity analysis (2–2) 37. Sensitivity analysis (2–36) 4. Minimization, graphical solution 38. Maximization, graphical solution 5. Maximization, graphical solution 39. Sensitivity analysis (2–38) 6. Slack analysis (2–5), sensitivity analysis 40. Minimization, graphical solution 7. Maximization, graphical solution 41. Sensitivity analysis (2–40) 8. Slack analysis (2–7) 42. Maximization, graphical solution 9. Maximization, graphical solution 43. Sensitivity analysis (2–42) 10. Minimization, graphical solution 44. Maximization, graphical solution 11. Maximization, graphical solution 45. Sensitivity analysis (2–44) 12. Sensitivity analysis (2–11) 46. Maximization, graphical solution 13. Sensitivity analysis (2–11) 47. Minimization, graphical solution 14. Maximization, graphical solution 48. Sensitivity analysis (2–47) 15. Sensitivity analysis (2–14) 49. Minimization, graphical solution 16. Maximization, graphical solution 50. Sensitivity analysis (2–49) 17. Sensitivity analysis (2–16) 51. Maximization, graphical solution 18. Maximization, graphical solution 52. Minimization, graphical solution 19. Standard form (2–18) 53. Sensitivity analysis (2–52) 20. Maximization, graphical solution 54. Maximization, graphical solution 21. Constraint analysis (2–20) 55. Sensitivity analysis (2–54) 22. Minimization, graphical solution 56. Maximization, graphical solution 23. Sensitivity analysis (2–22) 57. Sensitivity analysis (2–56) 24. Minimization, graphical solution 58. Multiple optimal solutions 25. Minimization, graphical solution 59. Infeasible problem 26. Sensitivity analysis (2–25) 60. Unbounded problem 27. Minimization, graphical solution 28. Maximization, graphical solution 29. Minimization, graphical solution 30. Maximization, graphical solution 31. Sensitivity analysis (2–30) 32. Minimization, graphical solution 33. Maximization, graphical solution 2-1 Copyright © 2019 Pearson Education, Inc. 6×1 + 6×2 ≥ 36 (phosphate, oz) x2 ≥ 2 (potassium, oz) x1,x2 ≥ 0 PROBLEM SOLUTIONS 1. a) x1 = # cakes x2 = # loaves of bread maximize Z = $10×1 + 6×2 subject to 3×1 + 8×2 ≤ 20 cups of flour 45×1 + 30×2 ≤ 180 minutes x1,x2 ≥ 0 b) b) 5. a) Maximize Z = 400×1 + 100×2 (profit, $) subject to 8×1 + 10×2 ≤ 80 (labor, hr) 2×1 + 6×2 ≤ 36 (wood) x1 ≤ 6 (demand, chairs) x1,x2 ≥ 0 2. a) Minimize Z = .05×1 + .03×2 (cost, $) subject to b) 8×1 + 6×2 ≥ 48 (vitamin A, mg) x1 + 2×2 ≥ 12 (vitamin B, mg) x1,x2 ≥ 0 b) 6. a) In order to solve this problem, you must substitute the optimal solution into the resource constraint for wood and the resource constraint for labor and determine how much of each resource is left over. Labor 3. The optimal solution point would change from point A to point B, thus resulting in the optimal solution x1 = 12/5 x2 = 24/5 Z = .408 4. a) Minimize Z = 3×1 + 5×2 (cost, $) subject to 8×1 + 10×2 ≤ 80 hr 8(6) + 10(3.2) ≤ 80 48 + 32 ≤ 80 80 ≤ 80 There is no labor left unused. 10×1 + 2×2 ≥ 20 (nitrogen, oz) 2-2 Copyright © 2019 Pearson Education, Inc. Sugar Wood 2×1 + 6×2 ≤ 36 2(6) + 6(3.2) ≤ 36 12 + 19.2 ≤ 36 31.2 ≤ 36 36 − 31.2 = 4.8 2×1 + 4×2 ≤ 16 2(0) + 4(4) ≤ 16 16 ≤ 16 There is no sugar left unused. There is 4.8 lb of wood left unused. 9. b) The new objective function, Z = 400×1 + 500×2, is parallel to the constraint for labor, which results in multiple optimal solutions. Points B (x1 = 30/7, x2 = 32/7) and C (x1 = 6, x2 = 3.2) are the alternate optimal solutions, each with a profit of $4,000. 7. a) Maximize Z = x1 + 5×2 (profit, $) subject to 5×1 + 5×2 ≤ 25 (flour, lb) 2×1 + 4×2 ≤ 16 (sugar, lb) x1 ≤ 5 (demand for cakes) x1,x2 ≥ 0 b) 10. a) Minimize Z = 80×1 + 50×2 (cost, $) subject to 3×1 + x2 ≥ 6 (antibiotic 1, units) x1 + x2 ≥ 4 (antibiotic 2, units) 2×1 + 6×2 ≥ 12 (antibiotic 3, units) x1,x2 ≥ 0 b) 8. In order to solve this problem, you must substitute the optimal solution into the resource constraints for flour and sugar and determine how much of each resource is left over. 11. a) Maximize Z = 300×1 + 400×2 (profit, $) subject to 3×1 + 2×2 ≤ 18 (gold, oz) 2×1 + 4×2 ≤ 20 (platinum, oz) x2 ≤ 4 (demand, bracelets) x1,x2 ≥ 0 Flour 5×1 + 5×2 ≤ 25 lb 5(0) + 5(4) ≤ 25 20 ≤ 25 25 − 20 = 5 There are 5 lb of flour left unused. 2-3 Copyright © 2019 Pearson Education, Inc. b) The profit for a necklace would have to increase to $600 to result in a slope of −3/2: 400×2 = Z − 600×1 x2 = Z/400 − 3/2×1 However, this creates a situation where both points C and D are optimal, ie., multiple optimal solutions, as are all points on the line segment between C and D. 14. a) Maximize Z = 50×1 + 40×2 (profit, $) subject to 12. 3×1 + 5×2 ≤ 150 (wool, yd2) 10×1 + 4×2 ≤ 200 (labor, hr) x1,x2 ≥ 0 The new objective function, Z = 300×1 + 600×2, is parallel to the constraint line for platinum, which results in multiple optimal solutions. Points B (x1 = 2, x2 = 4) and C (x1 = 4, x2 = 3) are the alternate optimal solutions, each with a profit of $3,000. b) The feasible solution space will change. The new constraint line, 3×1 + 4×2 = 20, is parallel to the existing objective function. Thus, multiple optimal solutions will also be present in this scenario. The alternate optimal solutions are at x1 = 1.33, x2 = 4 and x1 = 2.4, x2 = 3.2, each with a profit of $2,000. 13. a) Optimal solution: x1 = 4 necklaces, x2 = 3 bracelets. The maximum demand is not achieved by the amount of one bracelet. 15. b) The solution point on the graph which corresponds to no bracelets being produced must be on the x1 axis where x2 = 0. This is point D on the graph. In order for point D to be optimal, the objective function “slope” must change such that it is equal to or greater than the slope of the constraint line, 3×1 + 2×2 = 18. Transforming this constraint into the form y = a + bx enables us to compute the slope: The feasible solution space changes from the area 0ABC to 0AB’C’, as shown on the following graph. 2×2 = 18 − 3×1 x2 = 9 − 3/2×1 From this equation the slope is −3/2. Thus, the slope of the objective function must be at least −3/2. Presently, the slope of the objective function is −3/4: 400×2 = Z − 300×1 x2 = Z/400 − 3/4×1 The extreme points to evaluate are now A, B’, and C’. A: *B’: x1 = 0 x2 = 30 Z = 1,200 x1 = 15.8 x2 = 20.5 Z = 1,610 2-4 Copyright © 2019 Pearson Education, Inc. C’: 18. x1 = 24 x2 = 0 Z = 1,200 Point B’ is optimal 16. a) Maximize Z = 23×1 + 73×2 subject to x1 ≤ 40 x2 ≤ 25 x1 + 4×2 ≤ 120 x1,x2 ≥ 0 19. b) Maximize Z = 5×1 + 8×2 + 0s1 + 0s3 + 0s4 subject to 3×1 + 5×2 + s1 = 50 2×1 + 4×2 + s2 = 40 x1 + s3 = 8 x2 + s4 = 10 x1,x2 ≥ 0 A: s1 = 0, s2 = 0, s3 = 8, s4 = 0 B: s1 = 0, s2 = 3.2, s3 = 0, s4 = 4.8 C: s1 = 26, s2 = 24, s3 = 0, s4 = 10 20. 17. a) No, not this winter, but they might after they recover equipment costs, which should be after the 2nd winter. b) x1 = 55 x2 = 16.25 Z = 1,851 21. No, profit will go down c) x1 = 40 x2 = 25 Z = 2,435 Profit will increase slightly d) x1 = 55 x2 = 27.72 Z = $2,073 It changes the optimal solution to point A (x1 = 8, x2 = 6, Z = 112), and the constraint, x1 + x2 ≤ 15, is no longer part of the solution space boundary. 22. a) Minimize Z = 64×1 + 42×2 (labor cost, $) subject to 16×1 + 12×2 ≥ 450 (claims) x1 + x2 ≤ 40 (workstations) 0.5×1 + 1.4×2 ≤ 25 (defective claims) x1,x2 ≥ 0 Profit will go down from (c) 2-5 Copyright © 2019 Pearson Education, Inc. 25. b) 23. a) Changing the pay for a full-time claims solution to point A in the graphical solution where x1 = 28.125 and x2 = 0, i.e., there will be no part-time operators. b) Changing the pay for a part-time operator from $42 to $36 has no effect on the number of full-time and part-time operators hired, although the total cost will be reduced to $1,671.95. c) 26. The problem becomes infeasible. 27. Eliminating the constraint for defective claims would result in a new solution, x1 = 0 and x2 = 37.5, where only part-time operators would be hired. d) The solution becomes infeasible; there are not enough workstations to handle the increase in the volume of claims. 28. 24. 2-6 Copyright © 2019 Pearson Education, Inc. b) 29. 30. a) Maximize Z = $4.15×1 + 3.60×2 (profit, $) subject to 33. a) Maximize Z = 800×1 + 900×2 (profit, $) subject to 2×1 + 4×2 ≤ 30 (stamping, days) 4×1 + 2×2 ≤ 30 (coating, days) x1 + x2 ≥ 9 (lots) x1,x2 ≥ 0 x1 + x2 ≤ 115 (freezer space, gals.) 0.93 x1 + 0.75 x2 ≤ 90 (budget, $) x1 2 ≥ or x1 − 2 x2 ≥ 0 (demand) x2 1 b) x1 ,x2 ≥ 0 34. a) Maximize Z = 30×1 + 70×2 (profit, $) subject to 4×1 + 10×2 ≤ 80 (assembly, hr) 14×1 + 8×2 ≤ 112 (finishing, hr) x1 + x2 ≤ 10 (inventory, units) x1,x2 ≥ 0 b) 31. No additional profit, freezer space is not a binding constraint. 32. a) Minimize Z = 200×1 + 160×2 (cost, $) subject to 6×1 + 2×2 ≥ 12 (high-grade ore, tons) 2×1 + 2×2 ≥ 8 (medium-grade ore, tons) 4×1 + 12×2 ≥ 24 (low-grade ore, tons) x1,x2 ≥ 0 2-7 Copyright © 2019 Pearson Education, Inc. b) b) 35. The slope of the original objective function is computed as follows: Z = 30×1 + 70×2 70×2 = Z − 30×1 x2 = Z/70 − 3/7×1 slope = −3/7 The slope of the new objective function is computed as follows: Z = 90×1 + 70×2 70×2 = Z − 90×1 x2 = Z/70 − 9/7×1 slope = −9/7 37. a) 15(4) + 8(6) ≤ 120 hr 60 + 48 ≤ 120 108 ≤ 120 120 − 108 = 12 hr left unused b) Points C and D would be eliminated and a new optimal solution point at x1 = 5.09, x2 = 5.45, and Z = 111.27 would result. 38. a) Maximize Z = .28×1 + .19×2 x1 + x2 ≤ 96 cans x2 ≥2 x1 The change in the objective function not only changes the Z values but also results in a new solution point, C. The slope of the new objective function is steeper and thus changes the solution point. A: x1 = 0 x2 = 8 Z = 560 C: x1 = 5.3 x2 = 4.7 Z = 806 B: D: x1 = 8 x2 = 0 Z = 720 x1 = 3.3 x2 = 6.7 Z = 766 x1 ,x2 ≥ 0 b) 36. a) Maximize Z = 9×1 + 12×2 (profit, $1,000s) subject to 4×1 + 8×2 ≤ 64 (grapes, tons) 5×1 + 5×2 ≤ 50 (storage space, yd3) 15×1 + 8×2 ≤ 120 (processing time, hr) x1 ≤ 7 (demand, Nectar) x2 ≤ 7 (demand, Red) x1,x2 ≥ 0 2-8 Copyright © 2019 Pearson Education, Inc. 39. The model formulation would become, maximize Z = $0.23×1 + 0.19×2 subject to x1 + x2 ≤ 96 –1.5×1 + x2 ≥ 0 x1,x2 ≥ 0 The solution is x1 = 38.4, x2 = 57.6, and Z = $19.78 The discount would reduce profit. 40. a) Minimize Z = $0.46×1 + 0.35×2 subject to .91×1 + .82×2 = 3,500 x1 ≥ 1,000 x2 ≥ 1,000 .03×1 − .06×2 ≥ 0 x1,x2 ≥ 0 b) 477 − 445 = 32 fewer defective items b) 42. a) Maximize Z = $2.25×1 + 1.95×2 subject to 8×1 + 6×2 ≤ 1,920 3×1 + 6×2 ≤ 1,440 3×1 + 2×2 ≤ 720 x1 + x2 ≤ 288 x1,x2 ≥ 0 b) 41. a) Minimize Z = .09×1 + .18×2 subject to .46×1 + .35×2 ≤ 2,000 x1 ≥ 1,000 x2 ≥ 1,000 .91×1 − .82×2 = 3,500 x1,x2 ≥ 0 2-9 Copyright © 2019 Pearson Education, Inc. 43. A new constraint is added to the model in 45. The feasible solution space changes if the fertilizer constraint changes to 20×1 + 20×2 ≤ 800 tons. The new solution space is A’B’C’D’. Two of the constraints now have no effect. x1 ≥ 1.5 x2 The solution is x1 = 160, x2 = 106.67, Z = $568 The new optimal solution is point C’: A’: x1 = 0 x2 = 37 Z = 11,100 B’: x1 = 3 x2 = 37 Z = 12,300 44. a) Maximize Z = 400×1 + 300×2 (profit, $) subject to x1 + x2 ≤ 50 (available land, acres) 10×1 + 3×2 ≤ 300 (labor, hr) 8×1 + 20×2 ≤ 800 (fertilizer, tons) *C’: x1 = 25.71 x2 = 14.29 Z = 14,571 D’: x1 = 26 x2 = 0 Z = 10,400 46. a) Maximize Z = $7,600×1 + 22,500×2 subject to x1 + x2 ≤ 3,500 x2/(x1 + x2) ≤ .40 .12×1 + .24×2 ≤ 600 x1,x2 ≥ 0 x1 ≤ 26 (shipping space, acres) x2 ≤ 37 (shipping space, acres) x1,x2 ≥ 0 b) b) 2-10 Copyright © 2019 Pearson Education, Inc. wine, then there would logically be no waste for wine but only for beer. This amount “logically” would be the waste from 266.67 bottles, or $20, and the amount from the additional 53 bottles, $3.98, for a total of $23.98. 47. a) Minimize Z = $(.05)(8)x1 + (.10)(.75)x2 subject to 5×1 + x2 ≥ 800 5 x1 = 1.5 x2 8×1 + .75×2 ≤ 1,200 x1, x2 ≥ 0 x1 = 96 x2 = 320 Z = $62.40 49. a) Minimize Z = 3700×1 + 5100×2 subject to x1 + x2 = 45 (32×1 + 14×2) / (x1 + x2) ≤ 21 .10×1 + .04×2 ≤ 6 b) x1 ≥ .25 ( x1 + x2 ) x2 ≥ .25 ( x1 + x2 ) x1, x2 ≥ 0 b) 48. The new solution is 50. a) No, the solution would not change x1 = 106.67 x2 = 266.67 Z = $62.67 b) No, the solution would not change c) If twice as many guests prefer wine to beer, then the Robinsons would be approximately 10 bottles of wine short and they would have approximately 53 more bottles of beer than they need. The waste is more difficult to compute. The model in problem 53 assumes that the Robinsons are ordering more wine and beer than they need, i.e., a buffer, and thus there logically would be some waste, i.e., 5% of the wine and 10% of the beer. However, if twice as many guests prefer Yes, the solution would change to China (x1) = 22.5, Brazil (x2) = 22.5, and Z = $198,000. 51. a) x1 = $ invested in stocks x2 = $ invested in bonds maximize Z = $0.18×1 + 0.06×2 (average annual return) subject to x1 + x2 ≤ $720,000 (available funds) x1/(x1 + x2) ≤ .65 (% of stocks) .22×1 + .05×2 ≤ 100,000 (total possible loss) x1,x2 ≥ 0 2-11 Copyright © 2019 Pearson Education, Inc. One more hour of Sarah’s time would reduce the number of regraded exams from 10 to 9.8, whereas increasing Brad by one hour would have no effect on the solution. This is actually the marginal (or dual) value of one additional hour of labor, for Sarah, which is 0.20 fewer regraded exams, whereas the marginal value of Brad’s is zero. b) 54. a) x1 = # cups of Pomona x2 = # cups of Coastal Maximize Z = $2.05×1 + 1.85×2 subject to 16×1 + 16×2 ≤ 3,840 oz or (30 gal. × 128 oz) (.20)(.0625)x1 + (.60)(.0625)x2 ≤ 6 lbs. Colombian (.35)(.0625)x1 + (.10)(.0625)x2 ≤ 6 lbs. Kenyan (.45)(.0625)x1 + (.30)(.0625)x2 ≤ 6 lbs. Indonesian x2/x1 = 3/2 x1,x2 ≥ 0 b) Solution: x1 = 87.3 cups x2 = 130.9 cups Z = $421.09 52. x1 = exams assigned to Brad x2 = exams assigned to Sarah minimize Z = .10×1 + .06×2 subject to x1 + x2 = 120 x1 ≤ (720/7.2) or 100 x2 ≤ 50(600/12) x1,x2 ≥ 0 53. If the constraint for Sarah’s time became x2 ≤ 55 with an additional hour then the solution point at A would move to x1 = 65, x2 = 55 and Z = 9.8. If the constraint for Brad’s time became x1 ≤ 108.33 with an additional hour then the solution point (A) would not change. All of Brad’s time is not being used anyway so assigning him more time would not have an effect. 55. a) The only binding constraint is for Colombian; the constraints for Kenyan and Indonesian are nonbinding and there are already extra, or slack, pounds of these coffees available. Thus, only getting more Colombian would affect the solution. 2-12 Copyright © 2019 Pearson Education, Inc. One more pound of Colombian would increase sales from $421.09 to $463.20. 58. Increasing the brewing capacity to 40 gallons would have no effect since there is already unused brewing capacity with the optimal solution. b) If the shop increased the demand ratio of Pomona to Coastal from 1.5 to 1 to 2 to 1 it would increase daily sales to $460.00, so the shop should spend extra on advertising to achieve this result. 56. a) x1 = 16 in. pizzas x2 = hot dogs Maximize Z = $22×1 + 2.35×2 Subject to $10×1 + 0.65×2 ≤ $1,000 324 in2 x1 + 16 in2 x2 ≤ 27,648 in2 x2 ≤ 1,000 x1, x2 ≥ 0 b) Multiple optimal solutions; A and B alternate optimal 59. 60. 57. a) x1 = 35, x2 = 1,000, Z = $3,120 Profit would remain the same ($3,120) so the increase in the oven cost would decrease the season’s profit from $10,120 to $8,120. b) x1 = 35.95, x2 = 1,000, Z = $3,140 Profit would increase slightly from $10,120 to $10, 245.46. c) x1 = 55.7, x2 = 600, Z = $3,235.48 Profit per game would increase slightly. 2-13 Copyright © 2019 Pearson Education, Inc. The graphical solution is shown as follows. CASE SOLUTION: METROPOLITAN POLICE PATROL The linear programming model for this case problem is Minimize Z = x/60 + y/45 subject to 2x + 2y ≥ 5 2x + 2y ≤ 12 y ≥ 1.5x x, y ≥ 0 The objective function coefficients are determined by dividing the distance traveled, i.e., x/3, by the travel speed, i.e., 20 mph. Thus, the x coefficient is x/3 ÷ 20, or x/60. In the first two constraints, 2x + 2y represents the formula for the perimeter of a rectangle. Changing the objective function to Z = $16×1 + 16×2 would result in multiple optimal solutions, the end points being B and C. The profit in each case would be $960. Changing the constraint from .90×2 − .10×1 ≥ 0 to .80×2 −.20×1 ≥ 0 has no effect on the solution. The graphical solution is displayed as follows. CASE SOLUTION: ANNABELLE INVESTS IN THE MARKET x1 = no. of shares of index fund x2 = no. of shares of internet stock fund Maximize Z = (.17)(175)x1 + (.28)(208)x2 = 29.75×1 + 58.24×2 subject to 175×1 + 208×2 = $120, 000 x1 ≥ .33 x2 The optimal solution is x = 1, y = 1.5, and Z = 0.05. This means that a patrol sector is 1.5 miles by 1 mile and the response time is 0.05 hr, or 3 min. x2 ≤2 x1 x1, x2 > 0 CASE SOLUTION: “THE POSSIBILITY” RESTAURANT The linear programming model formulation is Maximize = Z = $12×1 + 16×2 subject to x1 + x2 ≤ 60 .25×1 + .50×2 ≤ 20 x1/x2 ≥ 3/2 or 2×1 − 3×2 ≥ 0 x2/(x1 + x2) ≥ .10 or .90×2 − .10×1 ≥ 0 x1x2 ≥ 0 x1 = 203 x2 = 406 Z = $29,691.37 x2 ≥ .33 x1 will have no effect on the solution. Eliminating the constraint x1 ≤2 x2 will change the solution to x1 = 149, x2 = 451.55, Z = $30,731.52. Eliminating the constraint 2-14 Copyright © 2019 Pearson Education, Inc. Increasing the amount available to invest (i.e., $120,000 to $120,001) will increase profit from Z = $29,691.37 to Z = $29,691.62 or approximately $0.25. Increasing by another dollar will increase profit by another $0.25, and increasing the amount available by one more dollar will again increase profit by $0.25. This indicates that for each extra dollar invested a return of $0.25 might be expected with this investment strategy. Thus, the marginal value of an extra dollar to invest is $0.25, which is also referred to as the “shadow” or “dual” price as described in Chapter 3. 2-15 Copyright © 2019 Pearson Education, Inc.