An Introduction To Management Science: Quantitative Approach, 15th Edition Solution Manual

Chapter 2 An Introduction to Linear Programming Learning Objectives 1. Obtain an overview of the kinds of problems linear programming has been used to solve. 2. Learn how to develop linear programming models for simple problems. 3. Be able to identify the special features of a model that make it a linear programming model. 4. Learn how to solve two variable linear programming models by the graphical solution procedure. 5. Understand the importance of extreme points in obtaining the optimal solution. 6. Know the use and interpretation of slack and surplus variables. 7. Be able to interpret the computer solution of a linear programming problem. 8. Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems. 9. Understand the following terms: problem formulation constraint function objective function solution optimal solution nonnegativity constraints mathematical model linear program linear functions feasible solution feasible region slack variable standard form redundant constraint extreme point surplus variable alternative optimal solutions infeasibility unbounded 2-1 Chapter 2 Solutions: 1. a, b, and e, are acceptable linear programming relationships. c is not acceptable because of 2B 2 d is not acceptable because of 3 A f is not acceptable because of 1AB c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 2. a. B 8 4 0 4 8 4 8 A b. B 8 4 0 A c. B Points on line are only feasible points 8 4 0 4 2-2 8 A An Introduction to Linear Programming 3. a. B (0,9) A 0 (6,0) b. B (0,60) A 0 (40,0) c. B Points on line are only feasible solutions (0,20) A (40,0) 0 4. a. B (20,0) (0,-15) 2-3 A Chapter 2 b. B (0,12) (-10,0) A c. B (10,25) Note: Point shown was used to locate position of the constraint line A 0 5. B a 300 c 200 100 b A 0 100 200 2-4 300 An Introduction to Linear Programming 6. 7A + 10B = 420 is labeled (a) 6A + 4B = 420 is labeled (b) -4A + 7B = 420 is labeled (c) B 100 80 60 (b) (c) 40 20 (a) A -100 -80 -60 -40 -20 0 20 40 60 80 100 7. B 100 50 A 0 50 100 150 2-5 200 250 Chapter 2 8. B 200 133 1/3 (100,200) A -200 0 -100 100 200 9. B (150,225) 200 100 0 (150,100) 100 -100 -200 2-6 200 300 A An Introduction to Linear Programming 10. B 5 4 Optimal Solution A = 12/7, B = 15/7 3 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 2 1 A 0 1 2 (1) × 5 (2) – (3) 4 5 2B 3B 10B 7B B = 6 = 15 = 30 = -15 = 15/7 3 A 5A 5A + + + – From (1), A = 6 – 2(15/7) = 6 – 30/7 = 12/7 2-7 6 (1) (2) (3) Chapter 2 11. B A = 100 Optimal Solution A = 100, B = 50 Value of Objective Function = 750 100 B = 80 A 0 100 200 12. a. B 6 5 4 Optimal Solution A = 3, B = 1.5 Value of Objective Function = 13.5 3 (3,1.5) 2 1 A (0,0) 1 2 3 2-8 4 (4,0) 5 6 An Introduction to Linear Programming b. B 3 Optimal Solution A = 0, B = 3 Value of Objective Function = 18 2 1 A (0,0) 1 c. 2 3 4 5 6 7 8 There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3). 13. a. B 8 6 Feasible Region consists of this line segment only 4 2 0 A 2 b. 4 The extreme points are (5, 1) and (2, 4). 2-9 6 8 9 10 Chapter 2 c. B 8 6 Optimal Solution A = 2, B = 4 4 2 0 A 2 14. a. 4 6 Let F = number of tons of fuel additive S = number of tons of solvent base Max s.t. 40F + 30S 2/5F + 1/ S 2 1/ S 5  200 Material 1  5 Material 2 3/ S 10  21 Material 3 3/ F 5 F, S  0 + 2 – 10 8 An Introduction to Linear Programming b. S F c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. 15. a. 2 – 11 Chapter 2 b. Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160. c. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem. 16. a. b. A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3S + 9D. Optimal Solution is S = 0 and D = 540. c. Department Cutting and Dyeing Sewing Finishing Inspection and Packaging Hours Used 1(540) = 540 5 /6(540) = 450 2 /3(540) = 360 1 /4(540) = 135 Max. Available 630 600 708 135 + 17. Max s.t. 5A + 2B + 0S1 1A 2A 6A – 2B + 3B – 1B + 1S1 + 0S2 0S3 + 1S2 1S3 + = = = 420 610 125 A, B, S1, S2, S3  0 18. a. Max s.t. 4A + 1B + 0S1 10A 3A 2A + 2B + 2B + 2B + 1S1 + 0S2 + 0S3 + 1S2 + 1S3 A, B, S1, S2, S3  0 2 – 12 = 30 = 12 = 10 Slack 90 150 348 0 An Introduction to Linear Programming b. B 14 12 10 8 6 Optimal Solution A = 18/7, B = 15/7, Value = 87/7 4 2 0 c. A 2 4 6 8 + 0S2 + 0S3 10 S1 = 0, S2 = 0, S3 = 4/7 19. a. Max s.t. 3A + 4B + 0S1 -1A 1A 2A + + + 2B 2B 1B + 1S1 + 1S2 + 1S3 A, B, S1, S2, S3  0 2 – 13 = 8 = 12 = 16 (1) (2) (3) Chapter 2 b. B 14 (3) 12 10 (1) 8 6 Optimal Solution A = 20/3, B = 8/3 Value = 30 2/3 4 2 (2) 0 A 2 c. 4 6 8 10 12 S1 = 8 + A – 2B = 8 + 20/3 – 16/3 = 28/3 S2 = 12 – A – 2B = 12 – 20/3 – 16/3 = 0 S3 = 16 – 2A – B = 16 – 40/3 – 8/3 = 0 20. a. Max s.t. 3A + 2B A 3A A A + B + 4B – S1 + S2 – S3 – – S4 B A, B, S1, S2, S3, S4  0 2 – 14 = = = = 4 24 2 0 An Introduction to Linear Programming b. c. S1 = (3.43 + 3.43) – 4 = 2.86 S2 = 24 – [3(3.43) + 4(3.43)] = 0 S3 = 3.43 – 2 = 1.43 S4 = 0 – (3.43 – 3.43) = 0 2 – 15 Chapter 2 21. a. and b. B 90 80 70 Constraint 2 60 50 40 Optimal Solution Constraint 1 Constraint 3 30 Feasible Region 20 10 2A + 3B = 60 A 0 10 c. 20 30 40 50 60 70 80 90 100 Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2, B = 10 + A Substituting for B in constraint 1 we obtain 5A + 5(10 + A) 5A + 50 + 5A 10A A = 400 = 400 = 350 = 35 B = 10 + A = 10 + 35 = 45 Optimal solution is A = 35, B = 45 d. Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints. 2 – 16 An Introduction to Linear Programming e. Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 1(35) + 3(45) = 170. Because 170 exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint. 22. a. C 3500 3000 2500 Inspection and Packaging 2000 Cutting and Dyeing 5 1500 Feasible Region 4 1000 Sewing 3 5A + 4C = 4000 500 0 2 1 500 1000 1500 2000 2500 Number of All-Pro Footballs A 3000 b. Extreme Point 1 2 3 4 5 Coordinates (0, 0) (1700, 0) (1400, 600) (800, 1200) (0, 1680) Profit 5(0) + 4(0) = 0 5(1700) + 4(0) = 8500 5(1400) + 4(600) = 9400 5(800) + 4(1200) = 8800 5(0) + 4(1680) = 6720 Extreme point 3 generates the highest profit. c. Optimal solution is A = 1400, C = 600 d. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints. e. New optimal solution is A = 800, C = 1200 Profit = 4(800) + 5(1200) = 9200 2 – 17 Chapter 2 23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced Max s.t. 2400E + 6E + 2E + 1800L 3L  2100 L  280 2.5L  1000 E, L  0 Engine time Lady-Sport maximum Assembly and testing b. L 700 Number of EZ-Rider Produced 600 Engine Manufacturing Time 500 400 Frames for Lady-Sport 300 Optimal Solution E = 250, L = 200 Profit = $960,000 200 100 Assembly and Testing 0 E 100 300 200 400 500 Number of Lady-Sport Produced c. 24. a. The binding constraints are the manufacturing time and the assembly and testing time. Let R = number of units of regular model. C = number of units of catcher’s model. Max s.t. 5R + 8C 1R + 1/ R 2 1/ R 8 + 3/ C 2 1/ C 3 1/ C 4 +  900 Cutting and sewing  300 Finishing  100 Packing and Shipping R, C  0 2 – 18 . An Introduction to Linear Programming b. C 1000 F Catcher’s Model 800 600 C& 400 S P& Optimal Solution (500,150) S 200 R 0 200 400 600 800 1000 Regular Model c. 5(500) + 8(150) = $3,700 d. C&S 1(500) + 3/2(150) = 725 F 1/ (500) + 1/ (150) = 300 2 3 P&S 1/ (500) + 1/ (150) = 100 8 4 e. Department C&S F P&S 25. a. Usage 725 300 100 Slack 175 hours 0 hours 0 hours Let B = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund Max s.t. b. Capacity 900 300 100 0.06 B + 0.10 S B 0.06 B B + + 0.10 S S   = 0.3 0.075 1 Optimal solution: B = 0.3, S = 0.7 Value of optimal solution is 0.088 or 8.8% 2 – 19 Bond fund minimum Minimum return Percentage requirement Chapter 2 26. a. Let D = amount spent on digital advertising R = amount spent on radio advertising Max s.t. 50D + 80R D D + D R = 1000 Budget  250 Digital min. R  250 Radio min. -2R  0 Digital  2 Radio D, R  0 b. 27. Let I = Internet fund investment in thousands B = Blue Chip fund investment in thousands Max s.t. 0.12I + 0.09B 1I 1I 6I + 1B + 4B I, B  0    50 35 240 Available investment funds Maximum investment in the internet fund Maximum risk for a moderate investor 2 – 20 An Introduction to Linear Programming B Blue Chip Fund (000s) 60 Risk Constraint Optimal Solution I = 20, B = 30 $5,100 50 40 Maximum Internet Funds 30 20 10 Objective Function 0.12I + 0.09B Available Funds $50,000 0 I 10 30 20 40 50 60 Internet Fund (000s) Internet fund Blue Chip fund Annual return b. $20,000 $30,000 $ 5,100 The third constraint for the aggressive investor becomes 6I + 4B  320 This constraint is redundant; the available funds and the maximum Internet fund investment constraints define the feasible region. The optimal solution is: Internet fund Blue Chip fund Annual return $35,000 $15,000 $ 5,550 The aggressive investor places as much funds as possible in the high return but high risk Internet fund. c. The third constraint for the conservative investor becomes 6I + 4B  160 This constraint becomes a binding constraint. The optimal solution is Internet fund Blue Chip fund Annual return $0 $40,000 $ 3,600 2 – 21 Chapter 2 The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fund is still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. The remaining $10,000 could be invested in low-risk bonds or certificates of deposit. 28. a. Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced Max s.t. 1W + 1.25M 5W 3W + 2W + W, M  0 7M 1M 2M    4480 2080 1600 Whole tomatoes Tomato sauce Tomato paste Note: units for constraints are ounces b. Optimal solution: W = 560, M = 240 Value of optimal solution is 860 29. a. Let B = proportion of Buffalo’s time used to produce component 1 D = proportion of Dayton’s time used to produce component 1 Buffalo Dayton Maximum Daily Production Component 1 Component 2 2000 1000 600 1400 Number of units of component 1 produced: 2000B + 600D Number of units of component 2 produced: 1000(1 – B) + 600(1 – D) For assembly of the ignition systems, the number of units of component 1 produced must equal the number of units of component 2 produced. Therefore, 2000B + 600D = 1000(1 – B) + 1400(1 – D) 2000B + 600D = 1000 – 1000B + 1400 – 1400D 3000B + 2000D = 2400 Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can maximize the number of electronic ignition systems produced by maximizing the number of units of subassembly 1 produced. Max 2000B + 600D In addition, B  1 and D  1. 2 – 22 An Introduction to Linear Programming The linear programming model is: Max s.t. 2000B + 600D 3000B B + 2000D = 2400 1 1 0 D B, D The graphical solution is shown below. D 1.2 1.0 30 .8 00 B+ 20 .6 00 D =2 40 .4 0 Optimal Solution 2000B + 600D = 300 .2 B 0 .2 .4 .6 .8 1.0 Optimal Solution: B = .8, D = 0 Optimal Production Plan Buffalo – Component 1 Buffalo – Component 2 Dayton – Component 1 Dayton – Component 2 .8(2000) = 1600 .2(1000) = 200 0(600) = 0 1(1400) = 1400 Total units of electronic ignition system = 1600 per day. 2 – 23 1.2 Chapter 2 30. a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch Max s.t. 15E + 18C 40E 40E + 25C 25C 25C E, C  0     50,000 15,000 10,000 25,000 Maximum Investment Eastern Cable Minimum ComSwitch Minimum ComSwitch Maximum b. C Number of Shares of ComSwitch 2000 Minimum Eastern Cable 1500 Maximum Comswitch 1000 Maximum Investment 500 Minimum Conswitch 0 500 1000 1500 Number of Shares of Eastern Cable E c. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) d. Optimal solution is E = 625, C = 1000 Total return = $27,375 2 – 24 An Introduction to Linear Programming 31. B 6 Feasible Region 4 2 A 0 2 4 6 3A + 4B = 13 Optimal Solution A = 3, B = 1 Objective Function Value = 13 32. A B A B A 2 – 25 Chapter 2 Extreme Points (A = 250, B = 100) (A = 125, B = 225) (A = 125, B = 350) Objective Function Value 800 925 1300 Surplus Demand 125 — — Surplus Total Production — — 125 33. a. xB2 6 4 2 0 xA1 2 4 6 Optimal Solution: A = 3, B = 1, value = 5 b. (1) (2) (3) (4) 3 + 4(1) = 7 2(3) + 1 = 7 3(3) + 1.5 = 10.5 -2(3) +6(1) = 0 Slack = 21 – 7 = 14 Surplus = 7 – 7 = 0 Slack = 21 – 10.5 = 10.5 Surplus = 0 – 0 = 0 2 – 26 Slack Processing Time — 125 — An Introduction to Linear Programming c. B A Optimal Solution: A = 6, B = 2, value = 34 34. a. B x2 4 3 Fe asible Re gion 2 1 (21/4, 9/4) (4,1) x1A 0 1 2 3 4 5 b. There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4) c. The optimal solution is A = 4, B = 1 2 – 27 6 Chapter 2 35. a. Min s.t. 6A + 4B + 0S1 2A 1A + + 1B 1B 1B – S1 + – 0S2 + 0S3 S2 + S3 = = = A, B, S1, S2, S3  0 b. The optimal solution is A = 6, B = 4. c. S1 = 4, S2 = 0, S3 = 0. 36. a. Let Max s.t. T = P = number of training programs on teaming number of training programs on problem solving 10,000T + 8,000P + + P P 2P T T 3T     8 10 25 84 T, P  0 2 – 28 Minimum Teaming Minimum Problem Solving Minimum Total Days Available 12 10 4 An Introduction to Linear Programming b. P Minimum Teaming Number of Problem-Solving Programs 40 30 Minimum Total 20 Days Available Minimum Problem Solving 10 0 10 20 Number of Teaming Programs c. There are four extreme points: (15,10); (21.33,10); (8,30); (8,17) d. The minimum cost solution is T = 8, P = 17 Total cost = $216,000 30 37. Mild Extra Sharp Regular 80% 20% Zesty 60% 40% 8100 3000 Let R = number of containers of Regular Z = number of containers of Zesty Each container holds 12/16 or 0.75 pounds of cheese Pounds of mild cheese used = = 0.80 (0.75) R + 0.60 (0.75) Z 0.60 R + 0.45 Z Pounds of extra sharp cheese used = = 0.20 (0.75) R + 0.40 (0.75) Z 0.15 R + 0.30 Z 2 – 29 T Chapter 2 Cost of Cheese = = = = Cost of mild + Cost of extra sharp 1.20 (0.60 R + 0.45 Z) + 1.40 (0.15 R + 0.30 Z) 0.72 R + 0.54 Z + 0.21 R + 0.42 Z 0.93 R + 0.96 Z Packaging Cost = 0.20 R + 0.20 Z Total Cost = (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z) = 1.13 R + 1.16 Z Revenue = 1.95 R + 2.20 Z Profit Contribution = Revenue – Total Cost = (1.95 R + 2.20 Z) – (1.13 R + 1.16 Z) = 0.82 R + 1.04 Z Max s.t. 0.82 R + 1.04 Z 0.60 R + 0.15 R + R, Z  0 0.45 Z 0.30 Z   8100 3000 Mild Extra Sharp Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280 38. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame Min s.t. 7.50S + 9.00P 0.10S + 0.30P 0.06S + 0.12P S + P S, P  0  6  3 = 30 carbon fiber (at least 20% of 30 yards) kevlar (no more than 10% of 30 yards) total (30 yards) 2 – 30 An Introduction to Linear Programming b. P Professional Grade (yards) 50 40 total Extreme Point S = 10 P = 20 30 Feasible region is the line segment 20 kevlar carbon fiber 10 Extreme Point S = 15 P = 15 S 0 10 20 30 40 50 60 Standard Grade (yards) c. Extreme Point (15, 15) (10, 20) Cost 7.50(15) + 9.00(15) = 247.50 7.50(10) + 9.00(20) = 255.00 The optimal solution is S = 15, P = 15 d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50. e. At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%). 39. a. Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund Min s.t. 8S + 50S 5S + + 3M 100M 4M M S, M,  0    1,200,000 Funds available 60,000 Annual income 3,000 Minimum units in money market 2 – 31 Chapter 2 Units of Mo ney Mark et Fu nd x2 M 20 000 + 3M = 62,000 8x8S 1 + 3×2 = 62,000 15 000 Optim al Solution . 10 000 50 00 0 50 00 10 000 15 000 20 000 x1S Units of Stock Fun d Optimal Solution: S = 4000, M = 10000, value = 62000 40. b. Annual income = 5(4000) + 4(10000) = 60,000 c. Invest everything in the stock fund. Let P1 = gallons of product 1 P2 = gallons of product 2 Min s.t. 1P1 + 1P1 + 1P1 + 1P2 1P2 2P2    P1 , P2  0 2 – 32 30 20 80 Product 1 minimum Product 2 minimum Raw material An Introduction to Linear Programming P2 Feasible Region +1 1 60 1P = P2 55 Number of Gallons of Product 2 80 40 20 Use 8 (30,25) 0 0g als. 40 20 60 80 Number of Gallons of Product 1 P1 Optimal Solution: P1 = 30, P2 = 25 Cost = $55 41. a. Let R = number of gallons of regular gasoline produced P = number of gallons of premium gasoline produced Max s.t. 0.30R + 0.50P 0.30R 1R + + 0.60P 1P 1P    18,000 50,000 20,000 R, P  0 2 – 33 Grade A crude oil available Production capacity Demand for premium Chapter 2 b. P Gallons of Premium Gasoline 60,000 50,000 Production Capacity 40,000 30,000 Maximum Premium 20,000 Optimal Solution R = 40,000, P = 10,000 $17,000 10,000 Grade A Crude Oil 0 R 10,000 20,000 30,000 40,000 50,000 60,000 Gallons of Regular Gasoline Optimal Solution: 40,000 gallons of regular gasoline 10,000 gallons of premium gasoline Total profit contribution = $17,000 c. Constraint 1 2 3 d. Value of Slack Variable 0 0 10,000 Interpretation All available grade A crude oil is used Total production capacity is used Premium gasoline production is 10,000 gallons less than the maximum demand Grade A crude oil and production capacity are the binding constraints. 2 – 34 An Introduction to Linear Programming 42. B x2 14 Sa tisfies Constraint #2 12 10 8 Infe asibility 6 4 Sa tisfies Constraint #1 2 0 4 2 6 8 10 12 x1A 43. Bx 2 4 Unbounded 3 2 1 0 1 2 3 x1A 44. a. xB2 Objective Function Optimal Solution (30/16 , 30/16 ) Value = 60/16 4 2 0 b. 2 New optimal solution is A = 0, B = 3, value = 6. 2 – 35 4 xA1 Chapter 2 45. a. B A A A 46. B A b. Feasible region is unbounded. c. Optimal Solution: A = 3, B = 0, z = 3. d. An unbounded feasible region does not imply the problem is unbounded. This will only be the case when it is unbounded in the direction of improvement for the objective function. Let N = number of sq. ft. for national brands G = number of sq. ft. for generic brands Problem Constraints: N N + G G    2 – 36 200 120 20 Space available National brands Generic An Introduction to Linear Programming Extreme Point 1 2 3 N 120 180 120 G 20 20 80 a. Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the generic brand. b. Alternative optimal solutions. Any point on the line segment joining extreme point 2 and extreme point 3 is optimal. c. Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic brand. 2 – 37 Chapter 2 47. Bx2 P ro 600 s ce s ing 500 e Tim 400 300 Alte rnate optim a (125,225) 200 100 (250,100) 0 100 200 300 400 x1A Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100). Cost = 3(125) + 3(225) = 1050 Cost = 3(250) + 3(100) = 1050 or The solution (A = 250, B = 100) uses all available processing time. However, the solution (A = 125, B = 225) uses only 2(125) + 1(225) = 475 hours. . Thus, (A = 125, B = 225) provides 600 – 475 = 125 hours of slack processing time which may be used for other products. 2 – 38 An Introduction to Linear Programming 48. Possible Actions: i. Reduce total production to A = 125, B = 350 on 475 gallons. ii. Make solution A = 125, B = 375 which would require 2(125) + 1(375) = 625 hours of processing time. This would involve 25 hours of overtime or extra processing time. iii. Reduce minimum A production to 100, making A = 100, B = 400 the desired solution. 49. a. Let P = number of full-time equivalent pharmacists T = number of full-time equivalent physicians The model and the optimal solution are shown below: MIN 40P+10T S.T. 1) 2) 3) P+T >=250 2P-T>=0 P>=90 Optimal Objective Value 5200.00000 Variable P T Value 90.00000 160.00000 2 – 39 Reduced Cost 0.00000 0.00000 Chapter 2 Constraint 1 2 3 Slack/Surplus 0.00000 20.00000 0.00000 Dual Value 10.00000 0.00000 30.00000 The optimal solution requires 90 full-time equivalent pharmacists and 160 full-time equivalent technicians. The total cost is $5200 per hour. b. Pharmacists Technicians Current Levels 85 175 Attrition 10 30 Optimal Values 90 160 New Hires Required 15 15 The payroll cost using the current levels of 85 pharmacists and 175 technicians is 40(85) + 10(175) = $5150 per hour. The payroll cost using the optimal solution in part (a) is $5200 per hour. Thus, the payroll cost will go up by $50 50. Let M = number of Mount Everest Parkas R = number of Rocky Mountain Parkas Max s.t. 100M + 150R 30M 45M 0.8M + + – 20R 15R 0.2R    7200 Cutting time 7200 Sewing time 0 % requirement Note: Students often have difficulty formulating constraints such as the % requirement constraint. We encourage our students to proceed in a systematic step-by-step fashion when formulating these types of constraints. For example: M must be at least 20% of total production M  0.2 (total production) M  0.2 (M + R) M  0.2M + 0.2R 0.8M – 0.2R  0 2 – 40 An Introduction to Linear Programming The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 100(65.45) + 150(261.82) = $45,818. If we think of this situation as an on-going continuous production process, the fractional values simply represent partially completed products. If this is not the case, we can approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with a corresponding profit of $45,650. 51. Let C = number sent to current customers N = number sent to new customers Note: Number of current customers that test drive = .25 C Number of new customers that test drive = .20 N Number sold = .12 ( .25 C ) + .20 (.20 N ) = .03 C + .04 N Max s.t. .03C + .04N .25 C .20 N .25 C – .40 N 4C + 6N C, N,  0 30,000  10,000  0   1,200,000 2 – 41 Current Min New Min Current vs. New Budget Chapter 2 Current Min. N 200,000 Current  2 New Budget .03 C +. 04 N =6 100,000 00 0 Optimal Solution C = 225,000, N = 50,000 Value = 8,750 New Min. 0 52. Let 100,000 200,000 S = number of standard size rackets O = number of oversize size rackets Max s.t. 10S + 15O 0.8S 10S 0.125S + + S, O,  0 0.2O 12O 0.4O 2 – 42    0 4800 80 % standard Time Alloy 300,000 C An Introduction to Linear Programming 53. a. Let R = time allocated to regular customer service N = time allocated to new customer service Max s.t. 1.2R + N R 25R -0.6R + + + N 8N N    80 800 0 R, N,  0 b. Optimal Objective Value 90.00000 Variable R N Value 50.00000 30.00000 Reduced Cost 0.00000 0.00000 Constraint 1 2 3 Slack/Surplus 0.00000 690.00000 0.00000 Dual Value 1.12500 0.00000 -0.12500 Optimal solution: R = 50, N = 30, value = 90 HTS should allocate 50 hours to service for regular customers and 30 hours to calling on new customers. 54. a. Let M1 = number of hours spent on the M-100 machine M2 = number of hours spent on the M-200 machine Total Cost 6(40)M1 + 6(50)M2 + 50M1 + 75M2 = 290M1 + 375M2 Total Revenue 25(18)M1 + 40(18)M2 = 450M1 + 720M2 Profit Contribution (450 – 290)M1 + (720 – 375)M2 = 160M1 + 345M2 2 – 43 Chapter 2 Max s.t. 160 M1 + 345M2 M1 M2 M1 40 M1 + M2 50 M2      15 10 5 5 1000 M-100 maximum M-200 maximum M-100 minimum M-200 minimum Raw material available M1, M2  0 b. Optimal Objective Value 5450.00000 Variable M1 M2 Value 12.50000 10.00000 Reduced Cost 0.00000 145.00000 Constraint 1 2 3 4 5 Slack/Surplus 2.50000 0.00000 7.50000 5.00000 0.00000 Dual Value 0.00000 145.00000 0.00000 0.00000 4.00000 The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200. 55. Mr. Krtick’s solution cannot be optimal. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made. 56. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible (it had an optimal solution). Every solution that was feasible is still feasible when we change the constraint to lessthan-or-equal-to, since the new constraint is satisfied at equality (as well as inequality). In summary, we have relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as strict inequality). 57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-thanor-equal to constraint as shown below. Constraints 2,3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible. 2 – 44 An Introduction to Linear Programming Original Problem: Modified Problem: 58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of products needed so that everyone’s top three choices are included. There are only two possible outcomes relative to the boss’ new constraint. First, suppose the minimum number of products is 15. Then the new constraint makes the problem infeasible. 2 – 45