MS11 SM02

# MS11 SM02 - Chapter 2 An Introduction to Linear Programming...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 2 An Introduction to Linear Programming Learning Objectives 1. 2. 3. 4. 5. 6. 7. 8. Obtain an overview of the kinds of problems linear programming has been used to solve. Learn how to develop linear programming models for simple problems. Be able to identify the special features of a model that make it a linear programming model. Learn how to solve two variable linear programming models by the graphical solution procedure. Understand the importance of extreme points in obtaining the optimal solution. Know the use and interpretation of slack and surplus variables. Be able to interpret the computer solution of a linear programming problem. Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems. 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 9. 2-1 Chapter 2 Solutions: 1. a, b, and e, are acceptable linear programming relationships. 2 c is not acceptable because of −2 x2 d is not acceptable because of 3 x1 f is not acceptable because of 1x1x2 c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 2. a. x2 8 4 0 b. x2 8 4 8 x1 4 0 c. x2 8 4 8 x1 Points on line are only feasible points 4 0 4 8 x1 2-2 An Introduction to Linear Programming 3. a. x2 (0,9) 0 b. x2 (0,60) x1 (6,0) 0 c. x2 (40,0) x1 Points on line are only feasible points (0,20) x1 (40,0) 0 4. a. 2-3 Chapter 2 x2 (20,0) x1 (0,-15) b. x2 (0,12) (-10,0) x1 c. x2 (10,25) Note: Point shown was used to locate position of the constraint line 0 5. x1 2-4 An Introduction to Linear Programming x2 300 a c 200 100 b 0 100 200 300 x1 6. For 7x1 + 10x2, slope = -7/10 For 6x1 + 4x2, slope = -6/4 = -3/2 For z = -4x1 + 7x2, slope = 4/7 x2 100 80 60 (b) (c) 40 20 (a) x1 0 20 40 60 80 100 -100 7. -80 -60 -40 -20 2-5 Chapter 2 x2 100 50 0 50 100 150 200 250 x1 8. x2 200 133 1 /3 (100,200) x1 -200 9. -100 0 100 200 2-6 An Introduction to Linear Programming x2 (150,225) 200 100 (150,100) 0 100 200 300 x1 -100 -200 10. 2-7 Chapter 2 x2 5 4 Optimal Solution x1 = 12/7, x2 = 15/7 3 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 2 x1 +2 x2 1 =6 0 1 2 3 5x 1 + 3x 2 = 15 4 5 6 x1 5x1 (1) x 5 (2) - (3) 5x1 + + + - 2x2 3x2 10x2 7x2 x2 = = = = = 6 15 30 -15 15/7 (1) (2) (3) From (1), x1 = 6 - 2(15/7) = 6 - 30/7 = 12/7 11. 2-8 .. An Introduction to Linear Programming x2 x1= 100 Optimal Solution x1 = 100, x2 = 50 Value of Objecive Function = 750 100 x2 = 80 2x 1 +4 x 1 =4 00 x1 0 12. a. x2 6 100 200 5 4 Optimal Solution x1 = 3, x2 = 1.5 3 Value of Objective Function = 13.5 2 (3,1.5) 1 x1 1 2 3 4 (4,0) 5 6 (0,0) b. 2-9 Chapter 2 x2 3 Optimal Solution x1 = 0, x2 = 3 Value of Objective Function = 18 2 1 (0,0) 1 c. 13. a. x2 7 6 (2) 2 3 4 5 6 7 8 9 10 x1 There are four extreme points: (0,0), (4,0), (3,1.5), and (0,3). Redundant Constraint 5 4 3 2 (1) (3) Optimal Solution x1 = 2, x 2 = 2 Value of objective function = 10 1 (0,0) b. Yes, constraint 2. The solution remains x1 = 2, x2 = 2 if constraint 2 is removed. x1 1 2 3 4 5 14. a. 2 - 10 An Introduction to Linear Programming x2 8 6 4 F easible Region consists of this line segment only 2 0 b. c. 2 The extreme points are (5, 1) and (2, 4). 4 6 8 x1 x 2 8 6 O p t i m a l S o l u t i o n x = 2 , x = 4 1 2 4 x 1 + 2 x 2 = 1 0 2 x 0 1 2 4 6 8 2 - 11 Chapter 2 15. a. D 600 Optima l Solu tion S = 3 00, D = 4 20 500 400 Value of Objective Function = 1 0,560 (540,252) 300 200 100 0 100 200 300 400 500 600 700 S 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 20(708) + 9(0) = 14,160. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem. A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make extreme point 5 the optimal solution. For example, one possibility is 3S + 9D. Optimal Solution is S = 0 and D = 540. Dept. C&D S F I&P Hours Used 1(540) = 540 5/ (540) = 450 6 2/ (540) = 360 3 1/ (540) = 135 4 Max. Available 630 600 708 135 Slack 90 150 348 — c. 16. a. b. c. 17. Max s.t. 1x1 2x1 6x1 - 2x2 + 3x2 - 1x2 + + 1/ x 23 1x3 3x3 x1, x2, x3, s1, s2, s3 ≥ 0 + 1s1 + 1s2 + 1s3 = 420 = 610 = 125 5x1 + 2x2 + 8x3 + 0s1 + 0s2 + 0s3 2 - 12 An Introduction to Linear Programming 18. a. Max s.t. 10x1 3x1 2x1 b. x2 14 12 + 2x2 + 2x2 + 2x2 + 1s1 + 1s2 + 1s3 = 30 = 12 = 10 4x1 + 1x2 + 0s1 + 0s2 + 0s3 x1, x2, s1, s2, s3 ≥ 0 10 8 6 4 2 x1 0 c. 19. a. Max s.t. -1x1 1x1 2x1 + 2x2 + 2x2 + 1x2 + 1s1 + 1s2 + 1s3 x1, x2, s1, s2, s3 ≥ 0 = 8 (1) (2) (3) 3x1 + 4x2 + 0s1 + 0s2 + 0s3 s1 = 0, s2 = 0, s3 = 4/7 2 4 6 8 10 Optimal Solution x1 = 18/7, x2 = 15/7, Value = 87/7 = 12 = 16 2 - 13 Chapter 2 b. x2 14 (3) (1) 12 10 8 6 Optima l Solu tion x 1 = 2 0/3 x 2 = 8 /3 Value = 30 2/3 4 2 (2) 0 2 4 6 8 10 12 x1 c. s1 = 8 + x1 - 2x2 = 8 + 20/3 - 16/3 = 28/3 s2 = 12 - x1 - 2x2 = 12 - 20/3 - 16/3 = 0 s3 = 16 - 2x1 - x2 = 16 - 40/3 - 8/3 = 0 20. 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 2 - 14 An Introduction to Linear Programming b. L 700 Engine M anufacturing Time 600 Number of EZ-Rider Produced 500 400 300 Frames for Lady-Sport Optima l Solu tion E = 2 50, L = 2 00 Profit = \$ 960,000 200 100 Assembly and Testing 0 100 200 300 400 50 0 E Number of Lad y-Sport Produced c. 21. a. The binding constraints are the manufacturing time and the assembly and testing time. Let F = number of tons of fuel additive S = number of tons of solvent base Max s.t. 2/5 F 3/ + 1/ S 2 1/ S 5 5 F F, S ≥ 0 + 3/ 10 S ≤ 200 ≤ ≤ 5 21 Material 1 Material 2 Material 3 40F + 30S 2 - 15 Chapter 2 b. S x2 70 60 teri Ma al 3 Tons of Solvent Base 50 40 30 Material 2 Optimal Solution (25,20) 1 20 M 10 Feasible R egion ate ria l F1 x 0 . 10 20 30 40 50 Tons of Fuel Additive c. d. 22. a. Material 2: 4 tons are used, 1 ton is unused. No redundant constraints. Let R = number of units of regular model. C = number of units of catcher’s model. Max s.t. 1R 1/ R 2 1/ R 8 R, C ≥ 0 + + + 3/ C 2 1/ C 3 1/ C 4 ≤ ≤ ≤ 900 300 100 Cutting and sewing Finishing Packing and Shipping 5R + 8C 2 - 16 An Introduction to Linear Programming . b. C 1000 800 Catcher's Model F 600 C& S 400 Optimal Solution (500,150) S P& 200 R 0 c. d. 5(500) + 8(150) = \$3,700 C & S 1(500) + 3/2(150) = 725 F 1/2(500) + 1/3(150) = 300 P & S 1/8(500) + 1/4(150) = 100 e. Department C&S F P&S 23. a. Capacity 900 300 100 Usage 725 300 100 Slack 175 hours 0 hours 0 hours 200 400 600 800 1000 Regular Model Let B = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund Max s.t. B 0.06 B + 0.10 S ≥ ≥ 0.3 0.075 Bond fund minimum Minimum return 0.06 B + 0.10 S 2 - 17 Chapter 2 B + S = 1 Percentage requirement b. Optimal solution: B = 0.3, S = 0.7 Value of optimal solution is 0.088 or 8.8% 24. a. a. Let N = amount spent on newspaper advertising R = amount spent on radio advertising Max s.t. N N + R = 1000 ≥ 250 R ≥ 250 N N, R ≥ 0 ≥ 2R Budget Newspaper min. Radio min. News ≥ 2 Radio 50N + 80R b. R 1000 Newspaper Min Budget Optimal Solution N = 666.67, R = 333.33 Value = 60,000 N = 2R 500 Radio Min Feasible region is this line segment N 0 25. 5 00 1000 Let I = Internet fund investment in thousands B = Blue Chip fund investment in thousands 2 - 18 An Introduction to Linear Programming Max s.t. 0.12 I 1I 1I 6I + 0.09 B 1B ≤ ≤ 50 35 240 Available investment funds Maximum investment in the internet fund Maximum risk for a moderate investor + + I, B ≥ 0 4B B 60 Risk Constraint Optima l Solu tion I = 2 0, B = 3 0 \$5,100 Blue Chip F und (000s) 50 40 Max im um Internet Funds 30 20 Available F unds \$50,000 I 10 20 30 40 50 60 10 0 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 \$35,000 \$15,000 2 - 19 Chapter 2 Annual return \$ 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 \$0 Blue Chip fund \$40,000 Annual return \$ 3,600 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. H 3000 U.S . O il M ax Ri sk 2000 Optima l Solu tion U = 8 00, H = 1 200 Total Annual Return = 8400 1000 Fu nd sA va ila b le 0 1000 2000 3000 U 26. a. Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced Max s.t. 5W 3W 2W + + 7M 1M 2M ≤ ≤ ≤ 4480 2080 1600 Whole tomatoes Tomato sauce Tomato paste 1W + 1.25M W, M ≥ 0 Note: units for constraints are ounces 2 - 20 An Introduction to Linear Programming b. Optimal solution: W = 560, M = 240 Value of optimal solution is 860 27. a. Let B = proportion of Buffalo's time used to produce component 1 D = proportion of Dayton's time used to produce component 1 Maximum Daily Production Component 1 Component 2 2000 1000 600 1400 Buffalo Dayton 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. The linear programming model is: Max s.t. 2000B 3000B B + 600D + 2000D D B, D = 2400 ≤1 ≤1 ≥0 The graphical solution is shown below. 2 - 21 Chapter 2 D 1.2 1.0 .8 30 00 B + .6 20 00 D = 24 .4 Opti mal Solution .2 2000 B + 6 00 D = 3 00 00 B 0 .2 .4 .6 .8 1.0 1.2 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. 28. a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch 15E 40E 40E + 18C + 25C ≤ ≥ ≥ ≤ 50,000 15,000 10,000 25,000 Maximum Investment Eastern Cable Minimum ComSwitch Minimum ComSwitch Maximum Max s.t. 25C 25C E, C ≥ 0 2 - 22 An Introduction to Linear Programming b. C 2000 Minimum E astern Ca ble Number of S hares o f ComSw itch 1500 1000 Maxi mum C om swit ch Maximu m Inve stment 500 Mini mum C om swit ch 0 500 1000 1500 Number of Shares of Eastern Cable E c. d. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) Optimal solution is E = 625, C = 1000 Total return = \$27,375 29. 2 - 23 Chapter 2 x2 6 Feasible Region 4 2 0 2 Optimal Solution x1 = 3, x 2 = 1 Objective Function Value = 13 30. B A 60 Proces 4 x 61 3x1 + 4 2 = 13 x x2 MinimumA x1 = 125 Min x1 50 (125, Produ 40 30 20 (125, 10 A Bx 1 0 10 20 30 40 (250, 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 Slack Processing Time — 125 — 2 - 24 An Introduction to Linear Programming 31. a. x2 6 4 2 x1 0 2 Optimal Solution: x1 = 3, x2 = 1, value = 5 b. (1) (2) (3) (4) c. x2 6 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 4 6 4 2 x1 0 2 4 6 Optimal Solution: x1 = 6, x2 = 2, value = 34 32. a. 2 - 25 Chapter 2 x2 4 3 Feasible R egion 2 (21/4, 9/4) 1 (4,1) x1 0 1 2 3 4 5 6 b. c. 33. a. There are two extreme points: (x1 = 4, x2 = 1) and (x1 = 21/4, x2 = 9/4) The optimal solution is x1 = 4, x2 = 1 Min s.t. 6x1 2x1 1x1 + 4x2 1x2 1x2 1x2 + 0s1 s1 + 0s2 + 0s3 = 12 10 4 + + - - s2 + s3 = = x1, x2, s1, s2, s3 ≥ 0 b. c. 34. a. The optimal solution is x1 = 6, x2 = 4. s1 = 4, s2 = 0, s3 = 0. Let T = number of training programs on teaming P = number of training programs on problem solving Max s.t. 10,000T T P T 3T + + P 2P + 8,000P ≥8 ≥ 10 ≥ 25 ≤ 84 Minimum Teaming Minimum Problem Solving Minimum Total Days Available T, P ≥ 0 2 - 26 An Introduction to Linear Programming b. P 40 Minim um Te aming Number o f Problem-Solv ing Pro grams 30 Min im um Total 20 Days Av ailable 10 Mini mum P ro blem S o lvin g 0 10 20 Number of Teaming Pro grams 30 T c. d. There are four extreme points: (15,10); (21.33,10); (8,30); (8,17) The minimum cost solution is T = 8, P = 17 Total cost = \$216,000 35. Regular Mild Extra Sharp 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 0.20 (0.75) R + 0.40 (0.75) Z 0.15 R + 0.30 Z Pounds of extra sharp cheese used = = 2 - 27 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 = 1.95 R + 2.20 Z Revenue 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. 0.60 R 0.15 R + + 0.45 Z 0.30 Z ≤ ≤ 8100 3000 Mild Extra Sharp 0.82 R + 1.04 Z R, Z ≥ 0 Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = \$13,280 36. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame 7.50S 0.10S 0.06S S S, P + 9.00P + 0.30P + 0.12P + P ≥0 ≥6 ≤3 = 30 carbon fiber (at least 20% of 30 yards) kevlar (no more than 10% of 30 yards) total (30 yards) Min s.t. 2 - 28 An Introduction to Linear Programming b. P 50 Professional Grad e (yards) 40 total Extreme Point S = 10 P = 20 Feasible r egion is the line segment kevlar 30 20 carbon fiber Extreme Point S = 15 P = 15 10 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. 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%). Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund Min s.t. 50S 5S + + 100M 4M M S, M, ≥ 0 ≤ 1,200,000 ≥ ≥ Funds available 8S + 3M e. 37. a. 60,000 Annual income 3,000 Minimum units in money market 2 - 29 Chapter 2 M x2 Units of Money Market Fund 20000 8S + 3M = 62,000 8x1 + 3x2 = 62,000 15000 Optimal Solution . 10000 5000 S x1 0 5000 10000 15000 20000 Units of Stock Fund Optimal Solution: S = 4000, M = 10000, value = 62000 b. c. 38. Annual income = 5(4000) + 4(10000) = 60,000 Invest everything in the stock fund. Let P1 = gallons of product 1 P2 = gallons of product 2 Min s.t. 1P1 1P1 + 1P2 + 2P2 P1, P2 ≥ 0 ≥ ≥ ≥ 30 20 80 Product 1 minimum Product 2 minimum Raw material 1P1 + 1P2 2 - 30 An Introduction to Linear Programming P2 80 Number of Gallons of Product 2 60 Feasible Region 55 1P 1 +1 P2 = 40 20 (30,25) 0 Us e8 0g a ls . 40 20 60 80 Number of Gallons of Product 1 P1 Optimal Solution: P1 = 30, P2 = 25 Cost = \$55 39. a. Let R = number of gallons of regular gasoline produced P = number of gallons of premium gasoline produced Max s.t. 0.30R 1R + + 0.60P 1P 1P R, P ≥ 0 ≤ ≤ ≤ 18,000 50,000 20,000 Grade A crude oil available Production capacity Demand for premium 0.30R + 0.50P 2 - 31 Chapter 2 b. P 60,000 Gallons o f Premiu m Gasoli ne 50,000 Production Capacity 40,000 30,000 Max imu m P rem ium Optima l Solu tion R = 4 0,000, P = 1 0, 000 \$17,000 Grade A Crud e Oil 0 10,000 20,000 30,000 40,000 50,000 60,000 R 20,000 10,000 Gallons of Regular G asoline Optimal Solution: 40,000 gallons of regular gasoline 10,000 gallons of premium gasoline Total profit contribution = \$17,000 c. Constraint 1 2 3 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 d. Grade A crude oil and production capacity are the binding constraints. 2 - 32 An Introduction to Linear Programming 40. x2 14 12 10 8 6 4 Satisfies Constraint #1 2 x1 Infeasibility Satisfies Constraint #2 0 41. 2 4 6 8 10 12 x2 4 3 2 1 x1 Unbounded 0 1 2 3 2 - 33 Chapter 2 42. a. x2 Objective Function 4 Optimal Solution (30/16, 30/16) Value = 60/16 2 0 2 b. c. 4 x1 New optimal solution is x1 = 0, x2 = 3, value = 6. Slope of constraint is -3/5 Slope of objective function when c1 = 1 is -1/c2 Set slopes equal: -1/c2 = -3/5 -5 = -3c2 c2 = 5/3 Objective function needed: max x1 + 5/3x2 43. a. x2 1 Fsible a R go 8 6 4 2 OtiveFunco3 jc= Cstrain#1 n Cstrain#2 n x1 6 8 1 2 4 OptimaSoun x 1 =3, V=3 le x 2 =0 2 - 34 An Introduction to Linear Programming b. c. d. Feasible region is unbounded. Optimal Solution: x1 = 3, x2 = 0, z = 3. 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 44. Problem Constraints: N N G G 200 + G ≤ ≥ ≥ 200 120 20 Space available National brands Generic Minimum National Shelf Space 100 Minimum Generic 0 Extreme Point 1 2 3 a. 100 N 120 180 120 G 20 20 80 200 N Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the generic brand. Alternative optimal solutions. Any point on the line segment joining extreme point 2 and extreme point 3 is optimal. Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic brand. b. c. 2 - 35 Chapter 2 45. B x2 g sin ces Pro e Tim 600 500 400 300 Alternate optima 200 (125,225) 100 (250,100) A x1 0 100 200 300 400 Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100). Cost or Cost = 3(250) + 3(100) = 1050 = 3(125) + 3(225) = 1050 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 - 36 An Introduction to Linear Programming 46. B 600 500 400 300 Original Feasible Solution 200 Feasible solutions for constraint requiring 500 gallons of production 100 A 0 Possible Actions: i. Reduce total production to A = 125, B = 350 on 475 gallons. 100 200 300 400 600 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. 2 - 37 Chapter 2 47. a. Sx 2 70 60 Tons of Solvent Base 50 40 30 5 20 10 1 0 10 20 30 40 50 Tons of Fuel Additive b. Yes. New optimal solution is F = 18.75, S = 25. Value of the new optimal solution is 40(18.75) + 60(25) = 2250. An optimal solution occurs at extreme point 3, extreme point 4, and any point on the line segment joining these two points. This is the special case of alternative optimal solutions. For the manager attempting to implement the solution this means that the manager can select the specific solution that is most appropriate. 4 3 Feasible R egion 2 F x1 c. 48. a. S 40 Min imu m S Points satisfying minimum produc tion requirements Tons of Solvent Base 30 20 Points satisfying material requirements Mi ni mu m F 10 0 10 20 30 40 50 F Tons of Fuel Additive 2 - 38 An Introduction to Linear Programming There are no points satisfying both sets of constraints; thus there will be no feasible solution. b. Materials Material 1 Material 2 Material 3 Minimum Tons Required for F = 30, S = 15 2/5(30) + 1/2(15) = 19.5 0(30) + 1/5(15) = 3 3/5(30) + 3/10(15) = 22.5 Tons Available 20 5 21 Additional Tons Required 1.5 Thus RMC will need 1.5 additional tons of material 3. 49. a. Let P = number of full-time equivalent pharmacists T = number of full-time equivalent physicians The model and the optimal solution obtained using The Management Scientist is shown below: MIN 40P+10T S.T. 1) 2) 3) 1P+1T>250 2P-1T>0 1P>90 OPTIMAL SOLUTION Objective Function Value = Variable -------------P T Constraint -------------1 2 3 Value --------------90.000 160.000 Slack/Surplus --------------0.000 20.000 0.000 5200.000 Reduced Costs -----------------0.000 0.000 Dual Prices ------------------10.000 0.000 -30.000 The optimal solution requires 90 full-time equivalent pharmacists and 160 full-time equivalent technicians. The total cost is \$5200 per hour. b. Current Levels Pharmacists Technicians 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 2 - 39 Chapter 2 50. Let M = number of Mount Everest Parkas R = number of Rocky Mountain Parkas Max s.t. 30M 45M 0.8M + + 20R 15R 0.2R ≤ ≤ ≥ 7200 7200 0 Cutting time Sewing time % requirement 100M + 150R 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 R 500 Sewing 400 % Requirement 300 Optimal Solution (65.45,261.82) 200 100 Cutting Profit = \$30,000 M 0 100 200 300 400 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 2 - 40 An Introduction to Linear Programming 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. .25 C .20 N .25 C 4C + .40 N 6N ≥ ≥ ≥ 30,000 Current Min 10,000 New Min 0 Current vs. New .03C + .04N ≤ 1,200,000 Budget C, N, ≥ 0 Current Min. Current ≥ 2 N ew N 200,000 Budget .0 3 C + .0 4 N = 60 100,000 00 Optima l Solu tion C = 2 25,000, N = 5 0,000 Value = 8 ,750 New Min. 0 100,000 200,000 300,000 C 2 - 41 Chapter 2 52. Let S = number of standard size rackets O = number of oversize size rackets Max s.t. 0.8S 10S 0.125S + + S, O, ≥ 0 O 500 400 300 Time 200 100 Alloy S Optimal Solution (384,80) % Requirement 0.2O 12O 0.4O ≥ 0 % standard Time Alloy ≤ 4800 ≤ 80 10S + 15O - 0 53. a. Let 100 200 300 400 500 R = time allocated to regular customer service N = time allocated to new customer service Max s.t. R 25R -0.6R + + + N 8N N ≤ ≥ ≥ 80 800 0 1.2R + N R, N ≥ 0 b. OPTIMAL SOLUTION Objective Function Value = 90.000 2 - 42 An Introduction to Linear Programming Variable -------------R N Constraint -------------1 2 3 Value --------------50.000 30.000 Slack/Surplus --------------0.000 690.000 0.000 Reduced Costs -----------------0.000 0.000 Dual Prices -----------------1.125 0.000 -0.125 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 Max s.t. M1 M2 M1 M2 40 M1 + 50 M2 ≤ ≤ ≥ ≥ ≤ 15 10 5 5 1000 M-100 maximum M-200 maximum M-100 minimum M-200 minimum Raw material available 160 M1 + 345M2 M1, M2 ≥ 0 2 - 43 Chapter 2 b. OPTIMAL SOLUTION Objective Function Value = Variable -------------M1 M2 Constraint -------------1 2 3 4 5 5450.000 Reduced Costs -----------------0.000 0.000 Dual Prices -----------------0.000 145.000 0.000 0.000 4.000 Value --------------12.500 10.000 Slack/Surplus --------------2.500 0.000 7.500 5.000 0.000 The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200. 2 - 44 ...
View Full Document

Ask a homework question - tutors are online