Spreadsheet Modeling And Decision Analysis: A Practical Introduction To Business Analytics, 8th Edition Solution Manual

Chapter 2 – Introduction to Optimization & Linear Programming : S-1 ———————————————————————————————————————————— Chapter 2 Introduction to Optimization & Linear Programming 1. If an LP model has more than one optimal solution it has an infinite number of alternate optimal solutions. In Figure 2.8, the two extreme points at (122, 78) and (174, 0) are alternate optimal solutions, but there are an infinite number of alternate optimal solutions along the edge connecting these extreme points. This is true of all LP models with alternate optimal solutions. 2. There is no guarantee that the optimal solution to an LP problem will occur at an integer-valued extreme point of the feasible region. (An exception to this general rule is discussed in Chapter 5 on networks). 3. We can graph an inequality as if they were an equality because the condition imposed by the equality corresponds to the boundary line (or most extreme case) of the inequality. 4. The objectives are equivalent. For any values of X1 and X2, the absolute value of the objectives are the same. Thus, maximizing the value of the first objective is equivalent to minimizing the value of the second objective. 5. a. b. c. d. e. 6. linear nonlinear linear, can be re-written as: 4 X1 – .3333 X2 = 75 linear, can be re-written as: 2.1 X1 + 1.1 X2 – 3.9 X3 0 nonlinear Chapter 2 – Introduction to Optimization & Linear Programming : S-2 ———————————————————————————————————————————— 7. 8. X2 20 (0, 15) obj = 300 15 (0, 12) obj = 240 10 (6.67, 5.33) obj =140 (11.67, 3.33) obj = 125 (optimal solution) 5 0 5 10 15 20 25 X1 Chapter 2 – Introduction to Optimization & Linear Programming : S-3 ———————————————————————————————————————————— 9. 10. Chapter 2 – Introduction to Optimization & Linear Programming : S-4 ———————————————————————————————————————————— 11. 12. Chapter 2 – Introduction to Optimization & Linear Programming : S-5 ———————————————————————————————————————————— 13. X1 = number of softballs to produce, X2 = number of baseballs to produce MAX ST 14. 6 X1 + 4.5 X2 5X1 + 4 X2 6000 6 X1 + 3 X2 5400 4 X1 + 2 X2 4000 2.5 X1 + 2 X2 3500 1 X1 + 1 X2 1500 X1, X2 0 X1 = number of His chairs to produce, X2 = number of Hers chairs to produce MAX ST 10 X1 + 12 X2 4 X1 + 8 X2 1200 8 X1 + 4 X2 1056 2 X1 + 2 X2 400 4 X1 + 4 X2 900 1 X1- 0.5 X2 ≥ 0 X1 , X2 ≥ 0 Chapter 2 – Introduction to Optimization & Linear Programming : S-6 ———————————————————————————————————————————— 15. X1 = number of propane grills to produce, X2 = number of electric grills to produce MAX ST 100 X1 + 80 X2 2 X1 + 1 X2 2400 4 X1 + 5 X2 6000 2 X1 + 3 X2 3300 1 X1 + 1 X2 1500 X1 , X2 ≥ 0 Chapter 2 – Introduction to Optimization & Linear Programming : S-7 ———————————————————————————————————————————— 16. X1 = number of generators, X2 = number of alternators MAX ST 17. 250 X1 + 150 X2 2 X1 + 3 X2 260 1 X1 + 2 X2 140 X1, X2 0 X1 = number of generators, X2 = number of alternators MAX ST 250 X1 + 150 X2 2 X1 + 3 X2 260 1 X1 + 2 X2 140 X1 20 X2 20 d. No, the feasible region would not increase so the solution would not change — you’d just have extra (unused) wiring capacity. Chapter 2 – Introduction to Optimization & Linear Programming : S-8 ———————————————————————————————————————————— 18. X1 = proportion of beef in the mix, X2 = proportion of pork in the mix MIN ST 19. .85 X1 + .65 X2 1X1 + 1 X2 = 1 0.2 X1 + 0.3 X2 0.25 X1, X2 0 T= number of TV ads to run, M = number of magazine ads to run MIN ST 500 T + 750 P 3T + 1P  14 -1T + 4P  4 0T + 2P  3 T, P  0 Chapter 2 – Introduction to Optimization & Linear Programming : S-9 ———————————————————————————————————————————— 20. X1 = # of TV spots, X2 = # of magazine ads MAX ST 15 X1 + 25 X2 5 X1 + 2 X2 8 0.2 X1 + 0.25 X2> 6 0.15 X1 + 0.1 X2> 5 X1, X2 0 (cost) (copper) (zinc) (magnesium) Chapter 2 – Introduction to Optimization & Linear Programming : S-10 ———————————————————————————————————————————— 22. R = number of Razors produced, Z = number of Zoomers produced MAX ST 23. 70 R + 40 Z R + Z  700 R – Z  300 2 R + 1 Z  900 3 R + 4 Z  2400 R, Z  0 P = number of Presidential desks produced, S = number of Senator desks produced MAX 103.75 P + 97.85 S ST 30 P + 24 S  15,000 1 P + 1 S  600 5 P + 3 S  3000 P, S  0 Chapter 2 – Introduction to Optimization & Linear Programming : S-11 ———————————————————————————————————————————— 24. X1 = acres planted in watermelons, X2 = acres planted in cantaloupes MAX ST 256 X1 + 284.5 X2 50 X1 + 75 X2 6000 X1 + X2 100 X1, X2 0 X2 (0, 80) obj = 100 75 (60, 40) obj =26740 (optimal 50 25 (100, 0) obj = 0 0 25. 25 50 75 100 125 X1 D = number of doors produced, W = number of windows produced MAX ST 500 D + 400 W 1 D + 0.5 W  40 0.5 D + 0.75 W  40 0.5 D + 1 W  60 D, W  0 Chapter 2 – Introduction to Optimization & Linear Programming : S-12 ———————————————————————————————————————————— 26. X1 = number of desktop computers, X2 = number of laptop computers MAX ST 600 X1 + 900 X2 2 X1 + 3 X2 300 X1 80 X2 75 X1, X2 0 Case 2-1: For The Lines They Are A-Changin’ 1. 200 pumps, 1566 labor hours, 2712 feet of tubing. 2. Pumps are a binding constraint and should be increased to 207, if possible. This would increase profits by $1,400 to $67,500. 3. Labor is a binding constraint and should be increased to 1800, if possible. This would increase profits by $3,900 to $70,000. 4. Tubing is a non-binding constraint. They’ve already got more than they can use and don’t need any more. 5. 9 to 8: profit increases by $3,050 8 to 7: profit increases by $850 7 to 6: profit increases by $0 6. 6 to 5: profit increases by $975 5 to 4: profit increases by $585 4 to 3: profit increases by $390 7. 12 to 13: profit changes by $0 13 to 14: profit decreases by $760 14 to 15: profit decreases by $1,440 Chapter 2 – Introduction to Optimization & Linear Programming : S-13 ———————————————————————————————————————————— 8. 16 to 17: profit changes by $0 17 to 18: profit changes by $0 18 to 19: profit decreases by $400 9. The profit on Aqua-Spas can vary between $300 and $450 without changing the optimal solution. 10. The profit on Hydro-Luxes can vary between $233.33 and $350 without changing the optimal solution.