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Solution Manual For Introduction To Management Science, 12th Edition

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    Solution Manual For Introduction to Management Science Twelfth Edition Bernard W. Taylor III Virginia Polytechnic Institute and State University

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    Contents Chapter 1 Management Science ..................................................................................1-1 Chapter 2 Linear Programming: Model Formulation and Graphical Solution ...........2-1 Chapter 3 Linear Programming: Computer Solution and Sensitivity Analysis ..........3-1 Chapter 4 Linear Programming: Modeling Examples.................................................4-1 Chapter 5 Integer Programming ..................................................................................5-1 Chapter 6 Transportation, Transshipment, and Assignment Problems .......................6-1 Chapter 7 Network Flow Models ................................................................................7-1 Chapter 8 Project Management ...................................................................................8-1 Chapter 9 Multicriteria Decision Making....................................................................9-1 Chapter 10 Nonlinear Programming ...........................................................................10-1 Chapter 11 Probability and Statistics...........................................................................11-1 Chapter 12 Decision Analysis .....................................................................................12-1 Chapter 13 Queuing Analysis......................................................................................13-1 Chapter 14 Simulation .................................................................................................14-1 Chapter 15 Forecasting................................................................................................15-1 Chapter 16 Inventory Management .............................................................................16-1 Module A: The Simplex Solution Method ........................................................................A-1 Module B: Transportation and Assignment Solution Methods ....................................... B-1 Module C: Integer Programming: The Branch and Bound Method ............................... C-1 Module D: Nonlinear Programming Solution Techniques .............................................D-1 Module E: Game Theory ................................................................................................. E-1 Module F: Markov Analysis .............................................................................................F-1

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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 change (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. Break-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; price and volume analysis 27. Break-even analysis; profit analysis 28. Break-even analysis; profit analysis 29. Break-even analysis; profit analysis 30. Decision analysis 31. Expected value 32. Linear programming 33. Linear programming 34. Linear programming 35. Linear programming 36. Forecasting/statistics 37. Linear programming 38. Waiting lines 39. 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, $18, 000, $0.90, $3.20; TC 18, 000 (12, 000)(0.90) $28,800; TR (12, 000)($3.20) $38, 400; $38, 400 28,800 $9, 600 per year v c c p c vc vp Z = = = = = + = + = = = = = − = b) f v 18, 000 7,826 3.20 0.90 = = = − − 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 7,826 .652 65.2% 12, 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 18, 000 9, 729.7 cupcakes 2.75 0.90 It increases the break-even volume from 7,826 to 9, 729.7 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

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    1-3 Total break-even volume = 600 + 5,685.7 = 6,285.7 pizzas 5,685.7 Total time to break-even 30 20 314.3 days = + = 19. a) Cost of Regular plan = $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) − 350,000 − (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. c f = $26,000 c v = $0.67 p = $3.75 v = 26, 000 3.75 0.67 − = 8,442 slices Forecasted annual demand = (540)(52) = 28,080 Z = $105,300 – 44,813.6 = $60,486.40 26. Fixed cost ( c f ) = 100,000 Variable cost ( c v ) = $(.50)(.35) + (.35)(.50) + (.15)(2.30) = $0.695

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    1-4 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 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 27. a) Monthly fixed cost ( c f ) = cost of van/60 months + labor (driver)/month = (21,500/60) + (30.42 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) + (12.4 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 28. a) = = − − f v 500 30 14 c v p c v = 31.25 jobs b) (8 weeks)(6 days/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) 29. a) = = − − f v 700 35 3 c v p c v = 21.88 jobs 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 30. This is a decision analysis problem – the subject of Chapter 12. The payoff table is: Weather Conditions Decision Alternatives Good Bad $3.25 $12,800 $8,450 $4.00 $14,400 $5,275 The student’s decision depends on the degree of risk they are willing to assume. Chapter 12 includes decision criteria for this problem. 31. This problem uses expected value for the decision alternatives in problem 30. Expected value ($3.25) = ($12,800)(0.60) + ($8,450)(0.40) = $11,060 Expected value ($4.00) = ($14,400)(0.60) + ($5,275)(0.40) = $10,750 Although the decision to sell hotdogs for $3.25 results in the greatest expected value, the results are so close, Annie would likely be indifferent. 32. 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.

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    1-5 33. 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. 34. Labor usage Clay usage Profit Possible # bowls # mugs 12 x + 15 y < = 60 9 x +5 y < = 30 300 x + 250 y solution? 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 35. Maximize 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. 36. 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. 37. 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 it violates 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 $40. 38. This problem demonstrates the cost trade-off inherent in queuing analysis, the topic of Chapter 13. In this problem the cost of service, i.e., the cost of staffing registers, is added to the cost of customers waiting, i.e., the cost of lost sales and ill will, as shown in the following table. Registers staffed 1 2 3 4 5 6 7 8 Waiting time (mins) 20 14 9 4 1.7 1 0.5 0.1 Cost of service ($) 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 39. 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.

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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, alternative 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 This is part of the annual fixed cost. The other part of the fixed cost is the employee annual salaries of $140,000. Thus, total fixed costs are, $399,723.45 + 140,000 = $539,723.45 = − = − = f v 539,723.45 3.20 0.60 207,585.94 parked cars per year c v p c

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    1-8 b) If 365 days per year are used, then the daily usage is, 207,585.94 568.7 or approximately 569 cars 365 per day = This seems like a reachable goal given the size of the town and the student population. CASE SOLUTION: A BUS SERVICE FOR DRAPER Fixed cost (3 buses) = 1,200,000 Total Variable Cost = 591,300 Annual Revenue = 648,240 Passengers/bus/trip = 37 Passenger fare = 4 Trips/bus/day = 4 Number of buses = 3 Days/year = 365 Total annual revenue = 648,240 = (37)(4)(4)(3)(365) Bus operating hrs/day = 18 Operating cost/hr = 90 Days/year = 365 Total annual variable cost = 591,300 = (18)(90)(365) (a) First year loss = (1,143,060.00) (b) Years to break even: Loss in year 1 = –1,143,060.0 Not possible to break even (c) 45 passengers per trip: Annual Revenue = 788,400 First year loss = (1,002,900) Not possible to break even 50 passengers per trip: Annual revenue = 876,000 First year loss = (915,300) Break even year: (3.215) years (d) Decrease in trips: Annual revenue = 657,000 Total variable cost = 443,475 First year loss = (986,475) Break even year: (5.62) years Bus operating hrs/day = 13.5 Operating cost/hr. = 90 Days/year = 365 Total annual variable cost = 443,475 (e) $1,200,000 Grant: Fixed Cost = 0 First Year Revenue = 56,940

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    2-1 Chapter Two: Linear Programming: Model Formulation and Graphical Solution PROBLEM SUMMARY 1. Maximization (1–28 continuation), graphical solution 2. Minimization, graphical solution 3. Sensitivity analysis (2–2) 4. Minimization, graphical solution 5. Maximization, graphical solution 6. Slack analysis (2–5), sensitivity analysis 7. Maximization, graphical solution 8. Slack analysis (2–7) 9. Maximization, graphical solution 10. Minimization, graphical solution 11. Maximization, graphical solution 12. Sensitivity analysis (2–11) 13. Sensitivity analysis (2–11) 14. Maximization, graphical solution 15. Sensitivity analysis (2–14) 16. Maximization, graphical solution 17. Sensitivity analysis (2–16) 18. Maximization, graphical solution 19. Standard form (2–18) 20. Maximization, graphical solution 21. Constraint analysis (2–20) 22. Minimization, graphical solution 23. Sensitivity analysis (2–22) 24. Sensitivity analysis (2–22) 25. Sensitivity analysis (2–22) 26. Minimization, graphical solution 27. Minimization, graphical solution 28. Sensitivity analysis (2–27) 29. Minimization, graphical solution 30. Maximization, graphical solution 31. Minimization, graphical solution 32. Maximization, graphical solution 33. Sensitivity analysis (2–32) 34. Minimization, graphical solution 35. Maximization, graphical solution 36. Maximization, graphical solution 37. Sensitivity analysis (2–36) 38. Maximization, graphical solution 39. Sensitivity analysis (2–38) 40. Maximization, graphical solution 41. Sensitivity analysis (2–40) 42. Minimization, graphical solution 43. Sensitivity analysis (2–42) 44. Maximization, graphical solution 45. Sensitivity analysis (2–44) 46. Maximization, graphical solution 47. Sensitivity analysis (2–46) 48. Maximization, graphical solution 49. Minimization, graphical solution 50. Sensitivity analysis (2–49) 51. Minimization, graphical solution 52. Sensitivity analysis (2–51) 53. Maximization, graphical solution 54. Minimization, graphical solution 55. Sensitivity analysis (2–54) 56. Maximization, graphical solution 57. Sensitivity analysis (2–56) 58. Maximization, graphical solution 59. Sensitivity analysis (2–58) 60. Multiple optimal solutions 61. Infeasible problem 62. Unbounded problem

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    2-2 PROBLEM SOLUTIONS 1. a) x 1 = # cakes x 2 = # loaves of bread maximize Z = $10 x 1 + 6 x 2 subject to 3 x 1 + 8 x 2 ≤ 20 cups of flour 45 x 1 + 30 x 2 ≤ 180 minutes x 1 , x 2 ≥ 0 b) 2. a) Minimize 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 ≥ 12 (vitamin B, mg) x 1 , x 2 ≥ 0 b) 3. The optimal solution point would change from point A to point B, thus resulting in the optimal solution x 1 = 12/5 x 2 = 24/5 Z = .408 4. a) Minimize Z = 3 x 1 + 5 x 2 (cost, $) subject to 10 x 1 + 2 x 2 ≥ 20 (nitrogen, oz) 6 x 1 + 6 x 2 ≥ 36 (phosphate, oz) x 2 ≥ 2 (potassium, oz) x 1 , x 2 ≥ 0 b) 5. a) Maximize Z = 400 x 1 + 100 x 2 (profit, $) subject to 8 x 1 + 10 x 2 ≤ 80 (labor, hr) 2 x 1 + 6 x 2 ≤ 36 (wood) x 1 ≤ 6 (demand, chairs) x 1 , x 2 ≥ 0 b) 6. a) In order to solve this problem, you must substitute the optimal solution into the resource constraint for wood and the resource constraint for labor and determine how much of each resource is left over. Labor 8 x 1 + 10 x 2 ≤ 80 hr 8(6) + 10(3.2) ≤ 80 48 + 32 ≤ 80 80 ≤ 80 There is no labor left unused.

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    2-3 Wood 2 x 1 + 6 x 2 ≤ 36 2(6) + 6(3.2) ≤ 36 12 + 19.2 ≤ 36 31.2 ≤ 36 36 − 31.2 = 4.8 There is 4.8 lb of wood left unused. b) The new objective function, Z = 400 x 1 + 500 x 2 , is parallel to the constraint for labor, which results in multiple optimal solutions. Points B ( x 1 = 30/7, x 2 = 32/7) and C ( x 1 = 6, x 2 = 3.2) are the alternate optimal solutions, each with a profit of $4,000. 7. a) Maximize Z = x 1 + 5 x 2 (profit, $) subject to 5 x 1 + 5 x 2 ≤ 25 (flour, lb) 2 x 1 + 4 x 2 ≤ 16 (sugar, lb) x 1 ≤ 5 (demand for cakes) x 1 , x 2 ≥ 0 b) 8. In order to solve this problem, you must substitute the optimal solution into the resource constraints for flour and sugar and determine how much of each resource is left over. Flour 5 x 1 + 5 x 2 ≤ 25 lb 5(0) + 5(4) ≤ 25 20 ≤ 25 25 − 20 = 5 There are 5 lb of flour left unused. Sugar 2 x 1 + 4 x 2 ≤ 16 2(0) + 4(4) ≤ 16 16 ≤ 16 There is no sugar left unused. 9. 10. a) Minimize 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) 11. a) Maximize Z = 300 x 1 + 400 x 2 (profit, $) subject to 3 x 1 + 2 x 2 ≤ 18 (gold, oz) 2 x 1 + 4 x 2 ≤ 20 (platinum, oz) x 2 ≤ 4 (demand, bracelets) x 1 , x 2 ≥ 0

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    2-4 b) 12. 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 with a profit of $3,000. The feasible solution space will change. The new constraint line, 3 x 1 + 4 x 2 = 20, is parallel to the existing objective function. Thus, multiple optimal solutions will also be present in this scenario. The alternate optimal solutions are at x 1 = 1.33, x 2 = 4 and x 1 = 2.4, x 2 = 3.2, each with a profit of $2,000. 13. 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 = 18. Transforming this constraint into the form y = a + bx enables us to compute the slope: 2 x 2 = 18 − 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 have 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 . 14. a) Maximize Z = 50 x 1 + 40 x 2 (profit, $) subject to 3 x 1 + 5 x 2 ≤ 150 (wool, yd 2 ) 10 x 1 + 4 x 2 ≤ 200 (labor, hr) x 1 , x 2 ≥ 0 b) 15. The feasible solution space changes from the area 0ABC to 0AB'C' , as shown on the following graph. The extreme points to evaluate are now A, B' , and C' . A: x 1 = 0 x 2 = 30 Z = 1,200 * B': x 1 = 15.8 x 2 = 20.5 Z = 1,610

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    2-5 C': x 1 = 24 x 2 = 0 Z = 1,200 Point B' is optimal 16. a) Maximize Z = 23 x 1 + 73 x 2 subject to x 1 ≤ 40 x 2 ≤ 25 x 1 + 4 x 2 ≤ 120 x 1 , x 2 ≥ 0 b) 17. 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 = 16.25 Z = 1,851 No, profit will go down c) x 1 = 40 x 2 = 25 Z = 2,435 Profit will increase slightly d) x 1 = 55 x 2 = 27.72 Z = $2,073 Profit will go down from (c) 18. 19. Maximize 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 2 x 1 + 4 x 2 + s 2 = 40 x 1 + s 3 = 8 x 2 + s 4 = 10 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 = 26, s 2 = 24, s 3 = 0, s 4 = 10 20. 21. It changes the optimal solution to point A ( x 1 = 8, x 2 = 6, Z = 112), and the constraint, x 1 + x 2 ≤ 15, is no longer part of the solution space boundary. 22. a) Minimize Z = 64 x 1 + 42 x 2 (labor cost, $) subject to 16 x 1 + 12 x 2 ≥ 450 (claims) x 1 + x 2 ≤ 40 (workstations) 0.5 x 1 + 1.4 x 2 ≤ 25 (defective claims) x 1 , x 2 ≥ 0

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