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Solution Manual For Introduction To Management Science, 11th Edition

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    1 - 1 Chapter One: Management Science PROBLEM SUMMARY 1. Total cost, revenue, profit, and break - even 2. Total cost, revenue, profit, and break - even 3. Total cost, revenue, profit, and break - even 4. Break - even volume 5. Graphical analysis (1 − 2) 6. Graphical analysis (1 − 4) 7. Break - even sales volume 8. Break - even volume as a percentage of capacity (1 − 2) 9. Break - even volume as a percentage of capacity (1 − 3) 10. Break - even volume as a percentage of capacity (1 − 4) 11. Effect of price chang e (1 − 2) 12. Effect of price change (1 − 4) 13. Effect of variable cost change (1 − 12) 14. Effect of fixed cost change (1 − 13) 15. Break - even analysis 16. Effect of fixed cost change (1 − 7) 17. Effect of variable cost change (1 − 7) 18. Break - even analysis 19. Bre ak - even analysis 20. Break - even analysis 21. Break - even analysis; volume and price analysis 22. Break - even analysis; profit analysis 23. Break - even analysis 24. Break - even analysis; profit analysis 25. Break - even analysis; price and volume analysis 26. Break - even analysis; profit analysis 27. Break - even analysis; profit analysis 28. Break - even analysis; profit analysis 29. Linear programming 30. Linear programming 31. Linear programming 32. Linear programming 33. Forecasting/statistics 34. Linear programming 35. Waiting lines 36. Shortest route PROBLEM SOLUTIONS 1. a) = = = = = + = + = = = = = − = f v f v 300, $8,000, $65 per table, $180; TC $8,000 (300)(65) $27,500; TR (300)(180) $54,000; $54,000 27,500 $26,500 per month v c c p c vc vp Z b) f v 8,000 69.56 tables per month 180 65 c v p c = = = − − 2. a) = = = = = + = + = = = = = − = f v f v 12,000, $60,000, $9, $25; TC 60,000 (12,000)(9) $168,000; TR (12,000)($25) $300,000; $300,000 168,000 $132,000 per year v c c p c vc vp Z b) = = = − − f v 60,000 3,750 tires per year 25 9 c v p c 3. a) = = = = = + = + = = = = = − = − f v f v 18,000, $21,000, $.45, $1.30; TC $21,000 (18,000)(.45) $29,100; TR (18,000)(1.30) $23, 400; $23, 400 29,100 $5,700 (loss) v c c p c vc vp Z b ) 4. = = = = = = − − f v f v $25,000, $.40, $.15, 25,000 100,000 lb per month .40 .15 c p c c v p c f v 21,000 24,705.88 yd per month 1.30 .45 c v p c = = = − −

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    1 - 2 5. 6. 7. − − f v $25,000 = = = 1,250 dolls 30 10 c v p c 8. = = = = Break-even volume as percentage of capacity 3,750 .469 46.9% 8,000 v k 9. = = = = Break-even volume as percentage of capacity 24,750.88 .988 98.8% 25,000 v k 10. = = = = Break-even volume as percentage of 100,000 capacity .833 83.3% 120,000 v k 11. f v 60,000 2,727.3tires 31 9 per year; it reduces the break-even volume from 3,750 tires to 2,727.3 tires per year. c v p c = = = − − 12. = = = − − f v 25,000 55,555.55 lb .60 .15 per month; it reduces the break-even volume from 100,000 lb per month to 55,555.55 lb. c v p c 13. f v 25,000 65,789.47 lb .60 .22 per month; it increases the break-even volume from 55,555.55 lb per month to 65,789.47 lb per month. c v p c = = = − − 14. = = = − − f v 39,000 102,613.57 lb .60 .22 per month; it increases the break-even volume from 65,789.47 lb per month to 102,631.57 lb per month. c v p c 15. f v f v Initial profit: (9,000)(.75) 4,000 (9,000)(.21) 6,750 4,000 1,890 $860 per month; increase in price: (5,700)(.95) 4,000 (5,700)(.21) 5, 415 4,000 1,197 $218 per month; the dair Z vp c vc Z vp c vc = − − = − − = − − = = − − = − − = − − = y should not raise its price. 16. − f v 35,000 = = = 1,750 30–10 c v p c The increase in fixed cost from $25,000 to $35,000 will increase the break - even point from 1,250 to 1,750 or 500 dolls ; thus, he should not spend the extra $10,000 for advertising. 17. Original break - even point (from problem 7) = 1,250 New break - even point: = = = − − f v 17,000 1,062.5 30 14 c v p c Reduces BE point by 187.5 dolls. 18. a) = = = − − f v $27,000 5,192.30 pizzas 8.95 3.75 c v p c b) = 5,192.3 259.6 days 20 c) Revenue for the first 30 days = 30( pv − vc v ) = 30[(8.95)(20) − (20)(3.75)] = $3,120 $27,000 − 3,120 = $23,880, portion of fixed cost not recouped after 30 days. = = = − − f v $23,880 New 5,685.7 pizzas 7.95 3.75 c v p c Total break - even volume = 600 + 5,685.7 = 6,285.7 pizzas

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    1 - 3 5,685.7 Total time to break-even 30 20 314.3 days = + = 19. a) Cost of Regular pla n = $55 + (.33)(260 minutes) = $140.80 Cost of Executive plan = $100 + (.25)(60 minutes) = $115 Select Executive plan. b) 55 + ( x − 1,000)(.33) = 100 + ( x − 1,200)(.25) − 275 + .33 x = .25 x − 200 x = 937.50 minutes per month or 15.63 hrs. 20. a) = − 7,500 14,000 .35 p p = $0.89 to break even b) If the team did not perform as well as expected the crowds could be smaller; bad weather could reduce crowds and/or affect what fans eat at the game; the price she charges could affect demand. c) This will be a subjective answer, but $1.25 seems to be a reasonable price. Z = vp − c f − vc v Z = (14,000)(1.25) − 7,500 − (14,000)(0.35) = 17,500 − 12,400 = $5,100 21. a) c f = $1,700 c v = $12 per pupil p = $75 = − 1,700 75 12 v = 26.98 or 27 pupils b) Z = vp − c f − vc v $5,000 = v (75) − $1,700 − v (12) 63 v = 6,700 v = 106.3 pupils c) Z = vp − c f − vc v $5,000 = 60 p − $1,700 − 60(12) 60 p = 7,420 p = $123.67 22. a) c f = $350,000 c v = $12,000 p = $18,000 = − f v c v p c = − 350,000 18,000 12,000 = 58.33 or 59 students b) Z = (75)(18,000) − 350,0 00 − (75)(12,000) = $100,000 c ) Z = (35)(25,000) − 350,000 − (35)(12,000) = 105,000 This is approximately the same as the profit for 75 students and a lower tuition in part (b). 23. p = $400 c f = $8,000 c v = $75 Z = $60,000 + = − f v Z c v p c + = − 60,000 8,000 400 75 v v = 209.23 teams 24. Fixed cost ( c f ) = 875,000 Variable cost ( c v ) = $200 Price ( p ) = (225)(12) = $2 , 700 v = c f /( p – c v ) = 875 , 000/(2 , 700 – 200) = 350 With volume doubled to 700: Profit ( Z ) = (2 , 700)(700) – 875,000 – (700)(200) = $875,000 25. Fixed cost ( c f ) = 100,000 Variable cost ( c v ) = $(.50)(.35) + (.35)(.50) + (.15)(2.30) = $0.695 Price ( p ) = $6 Profit ( Z ) = (6)(45 , 000) – 100 , 000 – (45 , 000)(0.695) = $138,725 This is not the financial profit goal of $150,000. The price to achieve the goal of $150,000 is, p = ( Z + c f + vc v )/ v = (150 , 000 + 100 , 000 + (45 , 000)(.695))/45 , 000 = $6.25

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    1 - 4 The volume to achieve the goal of $150,000 is, v = ( Z + c f )/( p − c v ) = (150 , 000 + 100 , 000)/(6 − .695) = 47,125 26. a) Monthly fixed cost ( c f ) = cost of van/60 months + labor (driver)/month = (21,500/60) + (3 0.42 days/month)($8/hr) (5 hr/day) = 358.33 + 1 , 216.80 = $1,575.13 Variable cost ( c v ) = $1.35 + 15.00 = $16.35 Price ( p ) = $34 v = c f /( p − v c ) = (1,575.13)/(34 − 16.35) v = 89.24 orders/month b) 89.24/30.42 = 2.93 orders/day − Monday through Thursday Double for weekend = 5.86 orders/day − Friday through Sunday Orders per month = approximately (18 days) (2.93 orders) + (1 2.4 days)(5.86 orders) = 125.4 delivery orders per month Profit = total revenue − total cost = vp – ( c f + vc v ) = (125.4)(34) − 1,575.13 – (125.4)(16.35) = 638.18 27. a ) = = − − f v 500 30 14 c v p c v = 31.25 jobs b ) (8 weeks)(6 d ays/week)(3 lawns/day) = 144 lawns Z = (144)(30) − 500 − (144)(14) Z = $1,804 c ) (8 weeks)(6 days/week)(4 lawns/day) = 192 lawns Z = (192)(25) − 500 − (192)(14) Z = $1,612 No, she would make less money than (b) 28. a ) = = − − f v 700 35 3 c v p c v = 21.88 job s b ) (6 snows)(2 days/snow)(10 jobs/day) = 120 jobs Z = (120)(35) − 700 − (120)(3) Z = $3,140 c ) (6 snows)(2 days/snow)(4 jobs/day) = 48 jobs Z = (48)(150) − 1800 − (48)(28) Z = $4,056 Yes, better than (b) d ) Z = (120)(35) − 700 − (120)(18) Z = $1,340 Yes, still a profit with one more person 29. There are two possible answers, or solution points: x = 25, y = 0 or x = 0, y = 50 Substituting these values in the objective function: Z = 15(25) + 10(0) = 375 Z = 15(0) + 10(50) = 500 Thus, the solution is x = 0 and y = 50 This is a simple linear programming model, the subject of the next several chapters. The student should recognize that there are only two possible solutions, which are the corner points of the feasible solution space, only one of which is optimal. 30. The solution is computed by solving simultaneous equations, x = 30, y = 10, Z = $1,400 It is the only, i.e., “optimal” solution because there is only one set of values for x and y that satisfy both constraints simultaneously.

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    1 - 5 31. Labor usage Clay usage Profit Possible # bowls # mugs 12 x + 15 y < = 60 9 x +5 y < = 30 300 x + 250 y s olution? 0 1 15 5 250 yes 1 0 12 9 300 yes 1 1 27 14 550 yes 0 2 30 10 500 yes 2 0 24 18 600 yes 1 2 42 19 800 yes 2 1 39 23 850 yes 2 2 54 28 1100 yes, best solution 0 3 45 15 750 yes 3 0 36 27 900 yes 1 3 57 24 1050 yes 3 1 51 32 1150 no 2 3 69 33 1350 no 3 2 66 37 1400 no 3 3 81 42 1650 no 4 0 48 36 1200 no 0 4 60 20 1000 yes 1 4 72 29 1300 no 4 1 63 41 1450 no 2 4 84 38 1600 no 4 2 78 46 1700 no

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    1 - 6 32. M aximize Z = $30 x AN + 70 x AJ + 40 x BN + 60 x BJ subject to x AN + x AJ = 400 x BN + x BJ = 400 x AN + x BN = 500 x AJ + x BJ = 300 The solution is x AN = 400, x BN = 100, x BJ = 300, and Z = 34,000 This problem can be solved by allocating as much as possible to the lowest cost variable, x AN = 400, then repeating this step until all the demand has been met. This is a similar logic to the minimum cell cost method . 33. This is virtually a straight linear relationship between time and site visits ; thus, a simple linear graph would result in a forecast of approximately 34,500 site visits. 34. Determine logical solutions: Cakes Bread Total Sales 1 . 0 2 $12 2 . 1 2 $22 3 . 3 1 $36 4 . 4 0 $40 Each solution must be checked to see if i t violate s the constraints for baking time and flour. Some possible solutions can be logically discarded because they are obviously inferior. For example, 0 cakes and 1 loaf of bread is clearly inferior to 0 cakes and 2 loaves of bread. 0 cakes and 3 loaves of bread is not possible because there is not enough flour for 3 loaves of bread. Using this logic, there are four possible solutions as shown. The best one, 4 cakes and 0 loaves of bread, results in the highest total sales of $40. 36. The shortest route problem is one of the topics of Chapter 7. At this point, the most logical “trial and error” way that most students will probably approach this problem is to identify all the feasible routes and compute the total distance for each, as follows: 1 - 2 - 6 - 9 = 228 1 - 2 - 5 - 9 = 213 1 - 3 - 5 - 9 = 211 1 - 3 - 8 - 9 = 276 1 - 4 - 7 - 8 - 9 = 275 Obviously inferior routes like 1 - 3 - 4 - 7 - 8 - 9 and 1 - 2 - 5 - 8 - 9 that include additional segments to the routes listed above can be logically eliminated from consideration. As a result, the route 1 - 3 - 5 - 9 is the shortest. An additional aspect to this problem could be to have the students look at these routes on a real map and indicate which they think might “practically” be the best route. In this case, 1 - 2 - 5 - 9 would likely be a better route, because even though it’s two miles farther it is Interstate highway the whole way, whereas 1 - 3 - 5 - 9 encompasses U.S. 4 - lane highways and state roads. 3 5 . This problem demonstrates the cost trade - off inherent in queuing analysis, the topic of Chapter 13. In this problem the cost of service, i.e., the cost of staffing registers, is added to the cost of customers waiting, i.e., the cost of lost sales and ill will, as shown in the following table . Registers s taffed 1 2 3 4 5 6 7 8 Waiting t ime (mins) 20 14 9 4 1.7 1 0.5 0.1 Cost of service ($) 60 120 180 240 300 360 420 480 Cost of waiting ($) 850 550 300 50 0 0 0 0 Total cost ($) 910 670 480 290 300 360 420 480 The total minimum cost of $290 occurs with 4 registers staffed

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    1 - 7 CASE SOLUTION: CLEAN CLOTHES CORNER LAUNDRY a) = = = − − f v 1,700 2,000 items per month 1.10 .25 c v p c b) Solution depends on number of months; 36 used here. $16,200 ÷ 36 = $450 per month, thus monthly fixed cost is $2,150 = = = − − f v 2,150 2,529.4 items per month 1.10 .25 c v p c 529.4 additional items per month c) Z = vp − c f − vc v = 4,300(1.10) − 2,150 − 4,300(.25) = $1,505 per month After 3 years, Z = $1,955 per month d) = = = − − = − − = − − = f v f v 1,700 2,297.3 .99 .25 3,800(.99) 1,700 3,800(.25) $1,112 per month c v p c Z vp c vc e) With both options: Z = vp − c f − vc v = 4,700(.99) − 2,150 − 4,700(.25) = $1,328 She should purchase the new equipment but not decrease prices. CASE SOLUTION: OCOBEE RIVER RAFTING COMPANY = f Alternative 1: $3,000 c = $20 p = v $12 c = = = − − f 1 v 3,000 375 rafts 20 12 c v p c = f Alternative 2: $10,000 c = $20 p = v $8 c = = = − − f 2 v 10,000 833.37 20 8 c v p c If demand is less than 375 rafts, the students should not start the business. If demand is less than 833 rafts, alternative 2 should not be selected, and alternative 1 should be used if demand is expected to be between 375 and 833.33 rafts. If demand is greater than 833.33 rafts, which alternative is best? To determine the answer, equate the two cost functions. 3,000 + 12 v = 10,000 + 8 v 4 v = 7,000 v = 1,750 This is referred to as the point of indifference between the two alternatives. In general, for demand lower than this point (1,750) the alternative should be selected with the lowest fixed cost; for demand greater than this point the alternative with the lowest variable cost should be selected. (This general relationship can be observed by graphing the two cost equations and seeing where they intersect.) Thus, for the Ocobee River Rafting Company, the following guidelines should be used: demand < 375, do not start business; 375 < demand < 1,750, select alternative 1; demand > 1,750, select alternative 2 Since Penny estimates demand will be approximately 1,000 rafts, alternativ e 1 should be selected. Z = v p − c f − vc v = (1,000)(20) − 3,000 − (1,000)(12) Z = $5,000 CASE SOLUTION: CONSTRUCTING A DOWNTOWN PARKING LOT IN DRAPER a) The annual capital recovery payment for a capital expenditure of $4.5 million over 30 years at 8% is, (4,500,000)[0.08(1 + .08) 30 ] / (1 + .08) 30 − 1 = $399,723.45

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    2 - 1 Chapter Two: Linear Programming: Model Formulation and Graphical Solution PROBLEM SUMMARY 1 . Maximization ( 1 – 2 8 continuation), graphical solution 2 . Maximization, graphical solution 3. Minimization, graphical solution 4. Sensitivity analysis ( 2 – 3) 5. Minimization, graphical solution 6. Maximization, graphical solution 7. Slack analysis ( 2 – 6) 8. Sensitivity analysis ( 2 – 6) 9. Maximization, graphical solution 1 0. Slack analysis ( 2 – 9) 11 . Maximization, graphical solution 12 . Minimization, graphical solution 1 3. Maximization, graphical solution 1 4. Sensitivity analysis ( 2 – 1 3) 1 5. Sensitivity analysis ( 2 – 1 3) 1 6. Maximization, graphical solution 1 7. Sensitivity analysis ( 2 – 1 6) 1 8. Maximization, graphical solution 1 9. Sensitivity analysis ( 2 – 1 8) 2 0. Maximization, graphical solution 21 . Standard form ( 2 – 2 0) 22 . Maximization, graphical solution 2 3. Standard form ( 2 – 22 ) 2 4. Maximization, graphical solution 2 5. Constraint analysis ( 2 – 2 4) 2 6. Minimization, graphical solution 2 7. Sensitivity analysis ( 2 – 2 6) 2 8. Sensitivity analysis ( 2 – 2 6) 2 9. Sensitivity analysis ( 2 – 22 ) 30. Minimization, graphical solution 3 1 . Minimization, graphical solution 3 2 . Sensitivity analysis ( 2 – 3 1 ) 33. Minimization, graphical solution 34. Maximization, graphical solution 35. Minimization, graphical solution 36. Maximization, graphical solution 37. Sensitivity analysis ( 2 – 34) 38. Minimization, graphical solution 39. Maximization, graphical solution 40. Maximization, graphical solution 4 1 . Sensitivity analysis ( 2 – 38) 4 2 . Maximization, graphical solution 43. Sensitivity analysis ( 2 – 40) 44. Maximization, graphical solution 45. Sensitivity analysis ( 2 – 4 2 ) 46. Minimization, graphical solution 47. Sensitivity analysis ( 2 – 44) 48. Maximization, graphical solution 49. Sensitivity analysis ( 2 – 46) 50. Maximization, graphical solution 5 1 . Sensitivity analysis ( 2 – 48) 5 2 . Maximization, graphical solution 53. Minimization, graphical solution 54. Sensitivity analysis ( 2 – 53) 55. Minimization, graphical solution 56. Sensitivity analysis ( 2 – 55) 57. Maximization, graphical solution 58. Minimization, graphical solution 59. Sensitivity analysis ( 2 – 5 2 ) 60. Maximization, graphical solution 6 1 . Sensitivity analysis ( 2 – 54) 6 2 . Multiple optimal solutions 63. Infeasible problem 64. Unbounded problem PROBLEM SOLUTIONS 1 . a) x 1 = # cakes x 2 = # loaves of bread maximize Z = $ 1 0 x 1 + 6 x 2 subject to 3 x 1 + 8 x 2 ≤ 2 0 cups of flour 45 x 1 + 30 x 2 ≤ 1 80 minutes x 1 , x 2 ≥ 0

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    2 - 2 b) 2 . a) M aximize Z = 6 x 1 + 4 x 2 (profit, $) subject to 1 0 x 1 + 1 0 x 2 ≤ 1 00 (line 1 , hr) 7 x 1 + 3 x 2 ≤ 4 2 (line 2 , hr) x 1 , x 2 ≥ 0 b) 3. a) M inimize Z = .05 x 1 + .03 x 2 (cost, $) subject to 8 x 1 + 6 x 2 ≥ 48 (vitamin A, mg) x 1 + 2 x 2 ≥ 1 2 (vitamin B, mg) x 1 , x 2 ≥ 0 b) 4. The optimal solution point would change from point A to point B, thus resulting in the optimal solution x 1 = 12 /5 x 2 = 2 4/5 Z = .408 5. a) M inimize Z = 3 x 1 + 5 x 2 (cost, $) subject to 1 0 x 1 + 2 x 2 ≥ 2 0 (nitrogen, oz) 6 x 1 + 6 x 2 ≥ 36 (phosphate, oz) x 2 ≥ 2 (potassium, oz) x 1 , x 2 ≥ 0 b) 6. a) M aximize Z = 400 x 1 + 1 00 x 2 (profit, $) subject to 8 x 1 + 1 0 x 2 ≤ 80 (labor, hr) 2 x 1 + 6 x 2 ≤ 36 (wood ) x 1 ≤ 6 (demand, chairs) x 1 , x 2 ≥ 0

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    2 - 3 b) 7. 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 8 x 1 + 1 0 x 2 ≤ 80 hr 8(6) + 1 0(3. 2 ) ≤ 80 48 + 3 2 ≤ 80 80 ≤ 80 There is no labor left unused. Wood 2 x 1 + 6 x 2 ≤ 36 2 (6) + 6(3. 2 ) ≤ 36 12 + 1 9. 2 ≤ 36 3 1 . 2 ≤ 36 36 − 3 1 . 2 = 4.8 There is 4.8 lb of wood left unused. 8. The new objective function, Z = 400 x 1 + 500 x 2 , is para llel to the constraint for labor, which results in multiple optimal solutions. Points B ( x 1 = 30/7, x 2 = 3 2 /7) and C ( x 1 = 6, x 2 = 3. 2 ) are the alternate optimal solutions, each with a profit of $4,000. 9. a) M aximize Z = x 1 + 5 x 2 (profit, $) subject to 5 x 1 + 5 x 2 ≤ 2 5 (flour, l b) 2 x 1 + 4 x 2 ≤ 1 6 (sugar, l b) x 1 ≤ 5 (demand for cakes) x 1 , x 2 ≥ 0 b) 1 0. 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 resou rce is left over. Flour 5 x 1 + 5 x 2 ≤ 2 5 lb 5(0) + 5(4) ≤ 2 5 2 0 ≤ 2 5 2 5 − 2 0 = 5 There are 5 lb of flour left unused. Sugar 2 x 1 + 4 x 2 ≤ 1 6 2 (0) + 4(4) ≤ 1 6 1 6 ≤ 1 6 There is no sugar left unused. 11 . 12 . a) M inimize Z = 80 x 1 + 50 x 2 (cost, $) subject to 3 x 1 + x 2 ≥ 6 (antibiotic 1 , units) x 1 + x 2 ≥ 4 (antibiotic 2 , units) 2 x 1 + 6 x 2 ≥ 12 (antibiotic 3, units) x 1 , x 2 ≥ 0 b)

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    2 - 4 1 3. a) M aximize Z = 300 x 1 + 400 x 2 (profit, $) subject to 3 x 1 + 2 x 2 ≤ 1 8 (gold, oz) 2 x 1 + 4 x 2 ≤ 2 0 (platinum, oz) x 2 ≤ 4 (demand, bracele ts) x 1 , x 2 ≥ 0 b) 1 4. The new objective function, Z = 300 x 1 + 600 x 2 , is parallel to the constraint line for platinum, which results in multiple optimal solutions. Points B ( x 1 = 2 , x 2 = 4) and C ( x 1 = 4, x 2 = 3) are the alternate optimal solutions, each w ith a profit of $3,000. The feasible solution space will change. The new constraint line, 3 x 1 + 4 x 2 = 2 0, is parallel to the existing objective function. Thus, multiple optimal solutions will also be present in this scenario. The alternate optimal solutions are at x 1 = 1 .33, x 2 = 4 and x 1 = 2 .4, x 2 = 3. 2 , each with a profit of $ 2 ,000. 1 5. a) Optimal solution: x 1 = 4 necklaces, x 2 = 3 bracelets. The maximum demand is not achieved by the amount of one bracelet. b) The solution point on the graph which corresponds to no bracelets being produced must be on the x 1 axis where x 2 = 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 x 1 + 2 x 2 = 1 8. Transforming this constraint into the form y = a + b x enables us to compute the slope: 2 x 2 = 1 8 − 3 x 1 x 2 = 9 − 3/ 2 x 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 x 2 = Z − 300 x 1 x 2 = Z /400 − 3/4 x 1 The profit for a necklace would hav e to increase to $ 600 to result in a slope of − 3/ 2 : 400 x 2 = Z − 600 x 1 x 2 = Z /400 − 3/ 2 x 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 . 1 6. a) M aximize Z = 50 x 1 + 40 x 2 (profit, $) subject to 3 x 1 + 5 x 2 ≤ 1 50 (wool, yd 2 ) 1 0 x 1 + 4 x 2 ≤ 2 00 (labor, hr) x 1 , x 2 ≥ 0 b) 1 7. The feasible solution space changes from the area 0ABC to 0AB'C' , as shown on the following graph.

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    2 - 5 The extreme points to e valuate are now A, B' , and C' . A: x 1 = 0 x 2 = 30 Z = 1 , 2 00 * B': x 1 = 1 5.8 x 2 = 2 0.5 Z = 1 ,6 1 0 C': x 1 = 2 4 x 2 = 0 Z = 1 , 2 00 Point B' is optimal 1 8. a) M aximize Z = 2 3 x 1 + 73 x 2 subject to x 1 ≤ 40 x 2 ≤ 2 5 x 1 + 4 x 2 ≤ 12 0 x 1 , x 2 ≥ 0 b ) 1 9. a) No, not this winter, but they might after they recover equipment costs, which should be after the 2 nd winter. b ) x 1 = 55 x 2 = 1 6. 2 5 Z = 1 ,85 1 No, profit will go down c ) x 1 = 40 x 2 = 2 5 Z = 2 ,435 Profit will increase slightly d ) x 1 = 55 x 2 = 2 7.7 2 Z = $ 2 ,073 Profit will go down from (c) 2 0. 21 . M aximize Z = 1 .5 x 1 + x 2 + 0 s 1 + 0 s 2 + 0 s 3 subject to x 1 + s 1 = 4 x 2 + s 2 = 6 x 1 + x 2 + s 3 = 5 x 1 , x 2 ≥ 0 A: s 1 = 4, s 2 = 1 , s 3 = 0 B: s 1 = 0, s 2 = 5, s 3 = 0 C: s 1 = 0, s 2 = 6, s 3 = 1 22 . 2 3. M aximize Z = 5 x 1 + 8 x 2 + 0 s 1 + 0 s 3 + 0 s 4 subject to 3 x 1 + 5 x 2 + s 1 = 50

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    2 - 6 2 x 1 + 4 x 2 + s 2 = 40 x 1 + s 3 = 8 x 2 + s 4 = 1 0 x 1 , x 2 ≥ 0 A: s 1 = 0, s 2 = 0, s 3 = 8, s 4 = 0 B: s 1 = 0, s 2 = 3. 2 , s 3 = 0, s 4 = 4.8 C: s 1 = 2 6, s 2 = 2 4, s 3 = 0, s 4 = 1 0 2 4. 2 5. It changes the optimal solution to point A ( x 1 = 8, x 2 = 6, Z = 112 ), and the constraint, x 1 + x 2 ≤ 1 5, is no longer part of the sol ution space boundary. 2 6. a) Minimize Z = 64 x 1 + 4 2 x 2 (labor cost, $) subject to 1 6 x 1 + 12 x 2 ≥ 450 (claims) x 1 + x 2 ≤ 40 (workstations) 0.5 x 1 + 1 .4 x 2 ≤ 2 5 (defective claims) x 1 , x 2 ≥ 0 b) 2 7. Changing the pay for a full - time claims processor from $64 to $54 will change the solution to point A in the graphical solution where x 1 = 2 8. 12 5 and x 2 = 0, i.e., there will be no part - time operators. Changing the pay for a part - time operator from $4 2 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 ,67 1 .95 . 2 8. Eliminating the constraint for defective claims would result in a new solution, x 1 = 0 and x 2 = 37.5, where only part - time operators would be hired. 2 9. The solution becomes infeasible; there are not enough workstations to handle the increase in the volume of claims. 30. 3 1 . 3 2 . The problem becomes infeasible.

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    2 - 7 33. 34. 35. 36. a) Maximize Z = $4. 1 5 x 1 + 3.60 x 2 (profit, $) subject to 1 2 1 2 1 1 2 2 1 2 115 (freezer space, gals.) 0.93 0.75 90 (budget, $) 2 or 2 0 (demand) 1 , 0 x x x x x x x x x x +  +   −   b) 37. No additional profit, freezer space is not a binding constraint. 38. a) M inimize Z = 2 00 x 1 + 1 60 x 2 (cost, $) subject to 6 x 1 + 2 x 2 ≥ 12 (high - grade ore, tons) 2 x 1 + 2 x 2 ≥ 8 (medium - grade ore, tons) 4 x 1 + 12 x 2 ≥ 2 4 (low - grade ore, tons) x 1 , x 2 ≥ 0 b)

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    2 - 8 39. a) M aximize Z = 800 x 1 + 900 x 2 (profit, $) subject to 2 x 1 + 4 x 2 ≤ 30 (stamping, days) 4 x 1 + 2 x 2 ≤ 30 (coating, days) x 1 + x 2 ≥ 9 (lots) x 1 , x 2 ≥ 0 b) 40. a) M aximize Z = 30 x 1 + 70 x 2 (profit, $) subject to 4 x 1 + 1 0 x 2 ≤ 80 (assembly, hr) 1 4 x 1 + 8 x 2 ≤ 1 1 2 (finishing, hr) x 1 + x 2 ≤ 1 0 (inventory, units) x 1 , x 2 ≥ 0 b) 4 1 . The slope of the original objective function is computed as follows: Z = 30 x 1 + 70 x 2 70 x 2 = Z − 30 x 1 x 2 = Z /70 − 3/7 x 1 slope = − 3/7 The slope of the new objective function is computed as follows: Z = 90 x 1 + 70 x 2 70 x 2 = Z − 90 x 1 x 2 = Z /70 − 9/7 x 1 slope = − 9/7 Th e 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 : x 1 = 0 C : x 1 = 5.3 x 2 = 8 x 2 = 4.7 Z = 560 Z = 806 B : x 1 = 3.3 D : x 1 = 8 x 2 = 6.7 x 2 = 0 Z = 766 Z = 7 2 0 4 2 . a) Maximize Z = 9 x 1 + 12 x 2 (profit, $ 1 ,000s) subject to 4 x 1 + 8 x 2 ≤ 64 (grapes, tons) 5 x 1 + 5 x 2 ≤ 50 (storage space, yd 3 ) 1 5 x 1 + 8 x 2 ≤ 12 0 (processing time, hr) x 1 ≤ 7 (demand, Nectar) x 2 ≤ 7 (demand, Red) x 1 , x 2 ≥ 0 b)

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    2 - 9 43. a) 1 5(4) + 8(6) ≤ 12 0 hr 60 + 48 ≤ 12 0 1 08 ≤ 12 0 12 0 − 1 08 = 12 hr left unused b) Points C and D would be eliminated and a new optimal solution point at x 1 = 5.09, x 2 = 5.45, and Z = 111 . 2 7 would result. 44. a) M aximize Z = . 2 8 x 1 + . 1 9 x 2 1 2 2 1 1 2 96 cans 2 , 0 x x x x x x +    b) 45. The model formulation would become, maximize Z = $0. 2 3 x 1 + 0. 1 9 x 2 subject to x 1 + x 2 ≤ 96 – 1 .5 x 1 + x 2 ≥ 0 x 1 , x 2 ≥ 0 The solution is x 1 = 38.4, x 2 = 57.6, and Z = $ 1 9.78 The discount would reduce profit. 46. a) M inimize Z = $0.46 x 1 + 0.35 x 2 subject to .9 1 x 1 + .8 2 x 2 = 3,500 x 1 ≥ 1 ,000 x 2 ≥ 1 ,000 .03 x 1 − .06 x 2 ≥ 0 x 1 , x 2 ≥ 0 b) 47 . a) M inimize Z = .09 x 1 + . 1 8 x 2 subject to .46 x 1 + .35 x 2 ≤ 2 ,000 x 1 ≥ 1 ,000 x 2 ≥ 1 ,000 .9 1 x 1 − .8 2 x 2 = 3,500 x 1 , x 2 ≥ 0 b) 477 − 445 = 3 2 fewer defective items 48. a) M aximize Z = $ 2 . 2 5 x 1 + 1 .95 x 2 subject to 8 x 1 + 6 x 2 ≤ 1 ,9 2 0 3 x 1 + 6 x 2 ≤ 1 ,440 3 x 1 + 2 x 2 ≤ 7 2 0 x 1 + x 2 ≤ 2 88 x 1 , x 2 ≥ 0

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    2 - 10 b) 49. A new constraint is added to the model in 1 2 1.5 x x  The solution is x 1 = 1 60, x 2 = 1 06.67, Z = $568 50. a) M aximize Z = 400 x 1 + 300 x 2 (profit, $) subject to x 1 + x 2 ≤ 50 (available land, acre s) 1 0 x 1 + 3 x 2 ≤ 300 (labor, hr) 8 x 1 + 2 0 x 2 ≤ 800 (fertilizer, tons) x 1 ≤ 2 6 (shipping space, acres) x 2 ≤ 37 (shipping space, acres) x 1 , x 2 ≥ 0 b) 5 1 . The feasible solution space changes if the fertilizer constraint changes to 2 0 x 1 + 2 0 x 2 ≤ 800 tons. The new solution space is A'B'C'D' . Two of the constraints now have no effect. The new optimal solution is point C ' : A': x 1 = 0 * C': x 1 = 2 5.7 1 x 2 = 37 x 2 = 1 4. 2 9 Z = 11 , 1 00 Z = 1 4,57 1 B': x 1 = 3 D': x 1 = 2 6 x 2 = 37 x 2 = 0 Z = 12 ,300 Z = 1 0,400 5 2 . a) Maximize Z = $7,600 x 1 + 22 ,500 x 2 subject to x 1 + x 2 ≤ 3,500 x 2 /( x 1 + x 2 ) ≤ .40 . 12 x 1 + . 2 4 x 2 ≤ 600 x 1 , x 2 ≥ 0

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    2 - 11 b) 53. a) M inimize Z = $(.05)(8) x 1 + (. 1 0)(.75) x 2 s ubject to 5 x 1 + x 2 ≥ 800 1 2 5 1.5 x x = 8 x 1 + .75 x 2 ≤ 1 , 2 00 x 1 , x 2 ≥ 0 x 1 = 96 x 2 = 3 2 0 Z = $6 2 .40 b) 54. The new solution is x 1 = 1 06.67 x 2 = 2 66.67 Z = $ 6 2 .67 If twice as many guests prefer wine to beer, then the Robinsons would be approximately 1 0 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 Robinson s 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 1 0 % of the beer. However, if twice as many guests prefer wine , then there would logically be no waste for wine but only for beer. This amount “logically” would be the waste from 2 66.67 bottles, or $ 2 0, and the amount from the additional 53 bottles, $3.98, for a total of $ 2 3.98. 55. a ) Minimize Z = 3700 x 1 + 5 1 00 x 2 subject to x 1 + x 2 = 45 (3 2 x 1 + 1 4 x 2 ) / ( x 1 + x 2 ) ≤ 21 . 1 0 x 1 + .04 x 2 ≤ 6 1 1 2 .25 ( ) x x x  + 2 1 2 .25 ( ) x x x  + x 1 , x 2 ≥ 0 b)

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    2 - 12 56. a) No, the solution would not change b) No, the solution would not change c) Yes, the solution would change to China ( x 1 ) = 22 .5, Brazil ( x 2 ) = 22 .5, and Z = $ 1 98,000. 57. a) x 1 = $ invested in stocks x 2 = $ invested in bonds maximize Z = $0. 1 8 x 1 + 0.06 x 2 (average annual return) subject to x 1 + x 2 ≤ $7 2 0,000 (available funds) x 1 /( x 1 + x 2 ) ≤ .65 (% of stocks) . 22 x 1 + .05 x 2 ≤ 1 00,000 (total possible loss) x 1 , x 2 ≥ 0 b) 58. x 1 = exams assigned to Brad x 2 = exams assigned to Sarah minimize Z = . 1 0 x 1 + .06 x 2 subject to x 1 + x 2 = 12 0 x 1 ≤ (7 2 0/7. 2 ) or 1 00 x 2 ≤ 50(600/ 12 ) x 1 , x 2 ≥ 0 59. If the constraint for Sarah’s time became x 2 ≤ 55 with an additional hour then the solution point at A would move to x 1 = 65, x 2 = 55 and Z = 9.8. If the constraint for Brad’s time became x 1 ≤ 1 08.33 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. One more hour of Sarah’s time would reduce the number of regraded exams from 1 0 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. 2 0 fewer regraded exams, whereas the marginal value of Brad’s is zero. 60. a) x 1 = # cups of Pomona x 2 = # cups of Coastal M aximize Z = $ 2 .05 x 1 + 1 .85 x 2 subject to 1 6 x 1 + 1 6 x 2 ≤ 3,840 oz or (30 gal. × 12 8 oz) (. 2 0)(.06 2 5) x 1 + (.60)(.06 2 5) x 2 ≤ 6 lbs. Colombian (.35)(.06 2 5) x 1 + (. 1 0)(.06 2 5) x 2 ≤ 6 lbs. Kenyan (.45)(.06 2 5) x 1 + (.30)(.06 2 5) x 2 ≤ 6 lbs. Indonesian x 2 / x 1 = 3/ 2 x 1 , x 2 ≥ 0

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    2 - 13 b) Solution: x 1 = 87.3 cups x 2 = 1 30.9 cups Z = $4 21 .09 6 1 . 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. One more pound of Colombian would increase sales from $4 21 .09 to $463. 2 0. 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. 6 2 . Multiple optimal solutions; A and B alternate optimal 63. 64.

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    2 - 14 CASE SOLUTION: METROPOLITAN POLICE PATROL The linear programming model for this case problem is M inimize Z = x /60 + y /45 subject to 2 x + 2 y ≥ 5 2 x + 2 y ≤ 12 y ≥ 1 .5 x x, y ≥ 0 The objective function coefficients are determined by dividing the distance traveled, i.e., x /3, by the travel speed, i.e., 2 0 mph. Thus, the x coefficient is x /3 ÷ 2 0, or x /60. In the first two constraints, 2 x + 2 y represents the formula for the perimeter of a rectangle. The graphical solution is displayed as follows. 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. CASE SOLUTION: “THE POSSIBILITY” RESTAURANT The linear programming model formulation is Maximize = Z = $ 12 x 1 + 1 6 x 2 subject to x 1 + x 2 ≤ 60 . 2 5 x 1 + .50 x 2 ≤ 2 0 x 1 / x 2 ≥ 3/ 2 or 2 x 1 − 3 x 2 ≥ 0 x 2 /( x 1 + x 2 ) ≥ . 1 0 or .90 x 2 − . 1 0 x 1 ≥ 0 x 1 x 2 ≥ 0 The graphical solution is shown as follows. Changing the objective function to Z = $ 1 6 x 1 + 1 6 x 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 x 2 − . 1 0 x 1 ≥ 0 to .80 x 2 − . 2 0 x 1 ≥ 0 has no effect on the solution. CASE SOLUTION: ANNABELLE INVESTS IN THE MARKET x 1 = no. of shares of inde x fund x 2 = no. of shares of internet stock fund Maximize Z = (. 1 7)( 1 75) x 1 + (. 2 8)( 2 08) x 2 = 2 9.75 x 1 + 58. 2 4 x 2 subject to 1 2 1 2 2 1 1 2 175 208 $120, 000 .33 2 , 0 + =    x x x x x x x x x 1 = 2 03 x 2 = 406 Z = $ 2 9,69 1 .37 2 1 Eliminating the constraint .33 x x  will have no effect on the solution. 1 2 Eliminating the constraint 2 x x  will change the solution to x 1 = 1 49, x 2 = 45 1 .55, Z = $30,73 1 .5 2 .

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    2 - 15 Increasing the amount available to invest (i.e., $ 12 0,000 to $ 12 0,00 1 ) will increase profit from Z = $ 2 9,69 1 .37 to Z = $ 2 9,69 1 .6 2 or approximately $0. 2 5. Increasing by another dollar will increase profit by another $0. 2 5, and increasing the amount available by one more dollar will again increase profit by $0. 2 5. This indicates that for each extra dollar invested a return of $0. 2 5 might be expected with this investment strategy. Thus, the marginal value of an extra dollar to invest is $0. 2 5, which is also referred to as the “shadow” or “dual” price as described in Chapter 3.

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    3 - 1 Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis PROBLEM SUMMARY 1. QM for Windows 2. QM for Windows and Excel 3. Excel 4. Excel solution 5. Excel 6. Graphical solution; sensitivity analysis 7. Model formulation 8. Graphical solution; sensitivity analysis (3 – 7) 9. Sensitivity analysis (3 – 7) 10. Model formulation 11. Graphical solution; sensitivity analysis (3 – 10) 12. Sensitivity analysis (3 – 10) 13. Model formulation 14. Graphical solution; sensitivity analysis (3 – 13) 15. Computer solution; sensitivity analysis (3 – 13) 16. Model formulation 17. Graphical solution; sensitivity analysis (3 – 16) 18. Computer solution; sensitivity analysis (3 – 16) 19. Model formulation 20. Graphical solution; sensitivity analysis (3 – 19) 21. Computer solution; sensitivity analysis (3 – 19) 22. Model formulation 23. Graphical solution; sensitivity analysis (3 – 22) 24. Computer solution; sensitivity analysis (3 – 22) 25. Model formulation 26. Graphical solution; sensitivity analysis (3 – 25) 27. Computer solution; sensitivity analysis (3 – 25) 28. Model formulation 29. Graphical solution; sensitivity analysis (3 – 28) 30. Computer solution; sensitivity analysis (3 – 28) 31. Model formulation; graphical solution 32. Computer solution; sensitivity analysis (3 – 31) 33. Model formulation 34. Graphical solution; computer solution; sensitivity analysis (3 – 33). 35. Model formulation 36. Graphical solution; sensitivity analysis (3 – 3 5 ) 37. Computer solution; sensitivity analysis (3 – 3 5 ) 38. Model formulation; computer solution 39. Model formulation; computer solution 40. Computer solution; sensitivity analysis 41. Model formulation 42. Graphical solution; sensitivity analysis (3 – 41) 43. Computer solution; sensitivity analysis (3 – 41) 44. Model formulation; computer solution 45. Sensitivity analysis (3 – 44) 46. Model formulation; computer solution 47. Sensitivity analysis (3 – 46) 48. Model formulation 49. Computer solution; sensitivity analysis (3 – 48) 50. Model formulation 51. Computer solution; sensitivity analysis (3 – 50) 52. Model formulation 53. Computer solution; sensitivity analysis (3 – 52) 54. Model formulation 55. Computer solution; sensitivity analysis (3 – 54) 56. Computer solution 57. Model formulation 58. Sensitivity analysis (3 – 57) 59. Model formulation 60. Computer solution; sensitivity analysis (3 – 59) 61. Model formulation; graphical solution 62. Computer solution; sensitivity analysis (3 – 61) PROBLEM SOLUTIONS 1. 2. QM for Windows establishes a “template” for the linear programming model based on the user’s specification of the type of objective function, the number of constraints, and number of variables. Then the model parameters are input and the problem is solved. In Excel the model “template” must be developed by the user.

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    3 - 2 3. Set Target cell: B13 Changing cells: B10:B12 Profit: = B10 * C4 + B11 * D4 + B12 * E4 Constraints: B10:B12 ≥ 0 G6 ≤ F6 G7 ≤ F7 4. x 1 = 0 x 2 = 9 Z = $54 5. F6: = C6 * B12 + D6 * B13 F7: = C7 * B12 + D7 * B13 F8: = C8 * B12 + D8 * B13 F9: = C9 * B12 + D9 * B13 G6: = E6 − F6 G7: = E7 − F7 G8: = E8 − F8 G9: = E9 − F9 B14: = C4 * B12 + D4 * B13 x 1 = 0 x 2 = 5.2 Z = 81.6 6. The slope of the constraint line is − 70/60. The optimal solution is at point A where x 1 = 0 and x 2 = 70. To change the solution to B, c 1 must increase such that the slope of the objective function is at least as great as the slope of the constraint line, − c 1 /50 = − 70/60 c 1 = 58.33 Alternatively, c 2 must decrease such that the slope of the objective function is at least as great as the slope of the constraint line, – 30/ c 2 = – 70/60 c 2 = 25.71 Thus, if c 1 increases to greater than 58.33 or c 2 decreases to less than 25.71, B will become optimal. 7. a) x 1 = no. of basketballs x 2 = no. of footballs maximize Z = 12 x 1 + 16 x 2 subject to 3 x 1 + 2 x 2 ≤ 500 4 x 1 + 5 x 2 ≤ 800 x 1 , x 2 ≥ 0 b) maximize Z = 12 x 1 + 16 x 2 + 0 s 1 + 0 s 2 subj ect to 3 x 1 + 2 x 2 + s 1 = 500 4 x 1 + 5 x 2 + s 2 = 800 x 1 , x 2 , s 1 , s 2 ≥ 0 8. a) A: 3(0) + 2(160) + s 1 = 500 s 1 = 180 4(0) + 5(160) + s 2 = 800 s 2 = 0 B: 3(128.5) + 2(57.2) + s 1 = 500 s 1 = 0 4(128.5) + 2(57.2) + s 2 = 800 s 2 = 0 C: 2(167) + 2(0) + s 1 = 500 s 1 = 0 4(167) + 5(0) + s 2 = 800 s 2 = 132 b) Z = 12 x 1 + 16 x 2 and, x 2 = Z /16 − 12 x 1 /16 The slope of the objective function, − 12/16, would have to become steeper (i.e., greater) than the slope of the constraint line 4 x 1 + 5 x 2 = 800, for the solution to change. The profit, c 1 , for a basketball that would change the solution point is, − 4/5 = − c 1 /16 5 c 1 = 64 c 1 = 12.8 Since $13 > 12.8 the solution point would change to B where x 1 = 128.5, x 2 = 57.2. The new Z value is $2,585.70. For a football, − 4/5 = − 12/ c 2 4 c 2 = 60 c 2 = 15 Thus, if the profit for a football decreased t o $15 or less, point B will also be optimal (i.e., multiple optimal solutions). The solution at B is x 1 = 128.5, x 2 = 57.2, and Z = $2,400.

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    3 - 3 c) If the constraint line for rubber changes to 3 x 1 + 2 x 2 = 1,000, it moves outward, eliminating points B and C . However, since A is the optimal point, it will not change and the optimal solution remains the same, x 1 = 0, x 2 = 160, and Z = 2,560. There will be an increase in slack, s 1 , to 680 lbs. If the constraint line for leather changes to 4 x 1 + 5 x 2 = 1,300, point A will move to a new location, x 1 = 0, x 2 = 250, Z = $4,000. 9. a) For c 1 the upper limit is computed as − 4/5 = − c 1 /16 5 c 1 = 64 c 1 = 12.8 and the lower limit is unlimited. For c 2 the lower limit is, − 4/5 = − 12/ c 2 4 c 2 = 60 c 2 = 15 and the upper limit is unlimited. Summarizing, ∞ ≤ c 1 ≤ 12.8 15 ≤ c 2 ≤ ∞ For q 1 the upper limit is ∞ since no matter how much q 1 increases the optimal solution point A will not change. The lower limit for q 1 is at the point where the constraint line ( 3 x 1 + 2 x 2 = q 1 ) intersects with point A where x 1 = 0, x 2 = 160, 3 x 1 + 2 x 2 = q 1 3(0) + 2(160) = q 1 q 1 = 320 For q 2 the upper limit is at the point where the rubber constraint line (3 x 1 + 2 x 2 = 500) intersects with the leather constraint line (4 x 1 + 5 x 2 = 800) along the x 2 axis, i.e., x 1 = 0, x 2 = 250, 4 x 1 + 5 x 2 = q 2 4(0) + 5(250) = q 2 q 2 = 1,250 The lower limit is 0 since that is the lowest point on the x 2 axis the constraint line can decrease to. Summarizing, 320 ≤ q 1 ≤ ∞ 0 ≤ q 2 ≤ 1,250 c) Z = 2,560.000 Variable Value Reduced Cost x 1 0.00 0.80 0 x 2 160.000 0.000 Constraint Slack/Surplus Shadow Price c 1 180.00 0.00 c 2 0.00 3.20 The shadow price for rubber is $0. Since there is slack rubber left over at the optimal point, extra rubber would have no marginal value. The shadow price for leather is $3.20. For each additional ft. 2 of leather that the company can obtain profit would increase by $3.20, up to the upper limit of the sensitivity range for leather (i.e., 1,250 ft. 2 ). 10. a) x 1 = no. of units of A x 2 = no. of units of B maximi ze Z = 9 x 1 + 7 x 2 subject to 12 x 1 + 4 x 2 ≤ 60 4 x 1 + 8 x 2 ≤ 40 x 1 , x 2 ≥ 0 b) maximize Z = 9 x 1 + 7 x 2 + 0 s 1 + 0 s 2 subject to 12 x 1 + 4 x 2 + s 1 = 60 4 x 1 + 8 x 2 + s 2 = 40 x 1 , x 2 , s 1 , s 2 ≥ 0 b) Objective Coefficient Ranges Variables Lower Limit Current Va lues Upper Limit Allowable Increase Allowable Decrease x 1 No limit 12.000 12.800 0.800 No limit x 2 15.000 16.000 No limit No limit 1.000 Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease c 1 320.000 500.000 No limit No limit 180.000 c 2 0.000 800.000 1,250.000 450.000 800.000

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    3 - 4 11. a) A: 12(0) + 4(5) + s 1 = 60 s 1 = 40 4(0) + 8(5) + s 2 = 40 s 2 = 0 B: 12(4) + 4(3) = 60 s 1 = 0 4(4) + 8(3) + s 2 = 40 s 2 = 0 C: 12(5) + 4(0) + s 1 = 60 s 1 = 0 4(5) + 8(0) + s 2 = 40 s 2 = 20 b ) The constraint line 12 x 1 + 4 x 2 = 60 would move inward resulting in a new location for point B at x 1 = 2, x 2 = 4, which would still be optimal. c) In order for the optimal solution point to change from B to A the slope of the objective function must be at least as flat as the slope of the constraint line, 4 x 1 + 8 x 2 = 40, which is − 1/2. Thus, the profit for product B would have to be, − 9/ c 2 = − 1/2 c 2 = 18 If the profit for product B is increased to $15 the optimal solution point will not change, although Z would change from $57 to $81. If the profit for product B is increased to $20 the solution point will change from B to A , x 1 = 0, x 2 = 5, Z = $100. 12. a) For c 1 the upper limit is computed as, − c 1 /7 = − 3 c 1 = 21 and the lower limit is, − c 1 /7 = − 1/2 c 1 = 3.50 For c 2 the upper limit is, − 9/ c 2 = − 1/2 c 2 = 18 and the lower limit is, − 9/ c 2 = − 3 c 2 = 3 Summarizing, 3.50 ≤ c 1 ≤ 21 3 ≤ c 2 ≤ 18 b) Z = 57.000 Variable Value Reduced Cost x 1 4.000 0.000 x 2 3.000 0.000 Constraint Slack/Surplus Shadow Price c 1 0.000 0.550 c 2 0.000 0.600 Objectiv e Coefficient Ranges Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease x 1 3.500 9.000 21.000 12.000 5.500 x 2 3.000 7.000 18.000 11.000 4.000 Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease c 1 20.000 60.000 120.000 60.000 40.000 c 2 20.000 40.000 120.000 80.000 20.000 c) The shadow price for line 1 time is $0.55 per

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    3 - 5 hour, while the shadow price for line 2 time is $0.60 per hour. The company wou ld prefer to obtain more line 2 time since it would result in the greater increase in profit. 13. a) x 1 = no. of yards of denim x 2 = no. of yards of corduroy maximize Z = $2.25 x 1 + 3.10 x 2 subject to 5.0 x 1 + 7.5 x 2 ≤ 6,500 3.0 x 1 + 3.2 x 2 ≤ 3,000 x 2 ≤ 510 x 1 , x 2 ≥ 0 b) maximize Z = $2.25 x 1 + 3.10 x 2 + 0 s 1 + 0 s 2 + 0 s 3 subject to 5.0 x 1 + 7.5 x 2 + s 1 = 6,500 3.0 x 1 + 3.2 x 2 + s 2 = 3,000 x 2 + s 3 = 510 x 1 , x 2 , s 1 , s 2 , s 3 ≥ 0 14. a) 5.0(456) + 7.5(510) + s 1 = 6,500 s 1 = 6,500 − 6,105 s 1 = 395 lbs. 3.0(456) + 3.2(510) + s 2 = 3,000 s 2 = 0 hrs. 510 + s 3 = 510 s 3 = 0 Therefore demand for corduroy is met. b) In order for the o ptimal solution point to change from B to C the slope of the objective function must be at least as great as the slope of the constraint line, 3.0 x 1 + 3.2 x 2 = 3,000, which is − 3/3.2. Thus, the profit for denim would have to be, − c 1 /3.0 = − 3/3.2 c 1 = 2.91 If the profit for denim is increased from $2.25 to $3.00 the optimal solution would change to point C , where x 1 = 1,000, x 2 = 0, Z = 3,000. Profit for corduroy has no upper l imit that would change the optimal solution point. c) The constraint line for cotton would move inward as shown in the following graph where point C is optimal. 15. Z = 2,607.000 Variable Value Reduced Cost x 1 456.000 0.000 x 2 510.000 0.000 Constraint Slack/Surplus Shadow Price c 1 395.000 0.000 c 2 0.000 0.750 c 3 0.000 0.700

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    3 - 6 Objective Coefficient Ranges Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease x 1 0.000 2.250 2.906 0.656 2.250 x 2 2.400 3.100 No limit No limit 0.700 Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease c 1 6,015.000 6,500.000 No limit No limit 395.000 c 2 1,632.000 3,000.000 3,237.000 237.000 1,368.000 c 3 0.000 510.000 692.308 182.308 510.000 a) The company should select 237 additional hours of processing time, with a shadow price of $0.75 per hour. Cotton has a shadow price of $0 because there is already extra (slack) cotton availa ble and not being used, so any more would have no marginal value. b) 0 ≤ c 1 ≤ 2.906 6,105 ≤ q 1 ≤ ∞ 2.4 ≤ c 2 ≤ ∞ 1,632 ≤ q 2 ≤ 3,237 0 ≤ q 3 ≤ 692.308 The demand for corduroy can decrease to zero or increase to 692.308 yds. without changi ng the current solution mix of denim and corduroy. If the demand increases beyond 692.308 yds., then denim would no longer be produced and only corduroy would be produced. 16. x 1 = no. of days to operate Mill 1 x 2 = no. of days to operate Mill 2 minimize Z = 6,000 x 1 + 7,000 x 2 subject to 6 x 1 + 2 x 2 ≥ 12 2 x 1 + 2 x 2 ≥ 8 4 x 1 + 10 x 2 ≥ 5 x 1 , x 2 ≥ 0 17. a) 6(4) + 2(0) − s 1 = 12 s 1 = 12 2(4) + 2(0) − s 2 = 8 s 2 = 0 4(4) + 10(0) − s 3 = 5 s 3 = 11 b) The slope of the objective function, − 6,000/7,000 , must become flatter (i.e., less) than th e slope of the constraint line, 2 x 1 + 2 x 2 = 8, for the solution to change. The cost of operating Mill 1, c 1 , that would change the solution point is, − c 1 /7,000 = − 1 c 1 = 7,000 Since $7,500 > $7,000, the solution point will change to B , where x 1 = 1, x 2 = 3, Z = $28,500. c ) If the constraint line for high - grade aluminum changes to 6 x 1 + 2 x 2 = 10, it moves inward but does not change the optimal variable mix. B remains optimal but moves to a new location, x 1 = 0.5, x 2 = 3.5, Z = $27,500. 18. Z = 24,000 Variable Value x 1 4.000 x 2 0.000 Constraint Slack/Surplus Shadow Price c 1 12.000 0.000 c 2 0.000 − 3,000.000 c 3 11.000 0.000

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    3 - 7 Objective Coefficient Ranges Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease x 1 0.000 6,000.000 7,000.000 1,000.000 6,000.000 x 2 6,000.000 7,000.000 No limit No limit 1 ,000.000 Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease c 1 No limit 12.000 24.000 12.000 No limit c 2 4.000 8.000 No limit No limit 4.000 c 3 No limit 5.000 16.000 11.000 No limit a) There is surplus high - grade and low - grade aluminum so the shadow price is $0 for both. The shadow price for medium - grade aluminum is $3,000, indicating that for every ton that this constraint could be reduced, cost will decrease by $ 3,000. b) 0 ≤ c 1 ≤ 7,000 ∞ ≤ q 1 ≤ 24 6,000 ≤ c 2 ≤ ∞ 4 ≤ q 2 ≤ ∞ ∞ ≤ q 3 ≤ 16 c) There will be no change. 1 9 . x 1 = no. of acres of corn x 2 = no. of acres of tobacco maximize Z = 3 00 x 1 + 520 x 2 subject to x 1 + x 2 ≤ 410 105 x 1 + 210 x 2 ≤ 52,500 x 2 ≤ 100 x 1 , x 2 ≥ 0 20. a) x 1 = 320, x 2 = 90 320 + 90 + s 1 = 410 s 1 = 0 acres uncultivat ed 90 + s 3 = 100 s 3 = 10 acres of tobacco allotment unused b) At point D only corn is planted. In order for point D to be optimal the slope of the objective function will have to be at least as great (i.e., steep) as the slope of the constraint line, x 1 + x 2 = 410, which is − 1. Thus, the profit for corn is computed as, − c 1 /520 = − 1 c 1 = 520 The profit for corn must be greater than $520 for the Bradleys to plant only corn. c) If the constraint line changes from x 1 + x 2 = 410 to x 1 + x 2 = 510, it will move outward to a l ocation which changes the solution to the point where 105 x 1 + 210 x 2 = 52,500 intersects with the axis. This new point is x 1 = 500, x 2 = 0, Z = $150,000. d) If the constraint line changes from x 1 + x 2 = 410 to x 1 + x 2 = 360, it moves inward to a location which changes the solution point to the intersection of x 1 + x 2 = 360 and 105 x 1 + 210 x 2 = 52,500. At this point x 1 = 260, x 2 = 100, and Z = $130,000. 21. Z = 142,800.000 Variable Value x 1 320.000 x 2 90.000 Constraint Slack/Surplus Shadow Price c 1 0.000 80.000 c 2 0.000 2.095 c 3 10.000 0.000

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    3 - 8 Objective Coefficient Ranges Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease x 1 260.000 300.000 520.000 220.000 40.000 x 2 300.000 520.000 600.000 80.000 220.000 Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease c 1 400.000 410.000 500.000 90.000 10.000 c 2 43,050.000 52,500.000 53,550.000 1,050.000 9,450.000 c 3 90.000 100.000 No limit No limit 10.000 a ) No, the shadow price for land is $80 per acre indicating that profit will increase by no more than $80 for each additional acre obtained. The maximum price the Bradleys should pay is $80 and the most they should obtain is at the upper limit of the sensitivity range for land. This limit is 500 acres, or 90 additional acres. Beyond 90 acres the shadow price would change. b) The shadow price for the budget is $2.095. Thus, for every $1 borrowed they could expect a profit increase of $2.095. If they borrowed $1,000 it would change the amount of corn and tobacco they plant; x 1 = 310.5 acres of corn and x 2 = 99.5 acres of tobacco. 22. x 1 = no. of sausage biscuits x 2 = no. of ham biscuits maximize Z = .60 x 1 + .50 x 2 subject to .10 x 1 ≤ 30 .15 x 2 ≤ 30 . 04 x 1 + .04 x 2 ≤ 16 .01 x 1 + .024 x 2 ≤ 6 x 1 , x 2 ≥ 0 23. a) x 1 = 300, x 2 = 100, Z = $230 .10(300) + s 1 = 30 s 1 = 0 leftover s ausage .15(100) + s 2 = 30 s 2 = 15 lbs. leftover ham . 01(300) + .024(100) + s 4 = 6 s 4 = 0.6 hr. b) The slope of the objective function, − 6/5, must become flatter (i.e., less) than the slope of the constraint line, .04 x 1 + .04 x 2 = 16, for the solution to change. The profit for ham, c 2 , that would change the solution point is, − 0.6/ c 2 = − 1 c 2 = .60 Thus, an increase in profit for ham of 0.60 will create a second optimal so lution point at C where x 1 = 257, x 2 = 143, and Z = $240. (Point D would also continue to be optimal, i.e., multiple optimal solutions.) c) A change in the constraint line from .04 x 1 + .04 x 2 = 16 to .04 x 1 + .04 x 2 = 18 would move the line outward, eliminating both points C and D . The new solution point occurs at the intersection of 0.01 x 1 + .024 x 2 = 6 and .10 x = 30. This point is x 1 = 300, x 2 = 125, and Z = $242.50. 24. Z = 230.000 Variable Value x 1 300.000 x 2 100.000 Constraint Slack/Surplus Shadow Price c 1 0.000 1.000 c 2 15.000 0.000 c 3 0.000 12.500 c 4 0.600 0.000

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