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An Introduction To Management Science: Quantitative Approaches To Decision Making, Revised, 13th Edition Solution Manual

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    Contents Preface Chapter 1. Introduction 2. An Introduction to Linear Programming 3. Linear Programming: Sensitivity Analysis and Interpretation of Solution 4. L inear Programming Applications in Marketing, Finance and Operations Management 5. Advanced Linear Programming Applications 6. Distribution and Network Models 7. Integer Linear Programming 8. Nonlinear Optimization Models 9. Project Scheduling: PERT/CPM 10. Inventory Models 11. Waiting Line Models 12. Simulation 13. Decision Analysis 14. Multicriteria Decision Problems 15. Forecasting 16. Markov Processes 17. Linear Programming: The Simplex Method 18. Simplex - Based Sensitivity Analysis and Duality 19. Solution Procedures for Transportation and Assignment Problems 20. Minimal Spanning Tree 21. Dynamic Programming Appendix A: Building Spreadsheet Models

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    Chapter 1 Introduction Learning Objectives 1. Develop a general understanding of the management science/operations research approach to decision making. 2. Realize that quantitative applications begin with a problem situation. 3. Obtain a brief introduction to quantitative techniques and their frequency of use in practice. 4. Understand that managerial problem situations have both quantitative and qualitative considerations that are important in the decision making process. 5. Learn about models in terms of what they are and why they are useful (the emphasis is on mathematical models). 6. Identify the step - by - step procedure that is used in most quantitative approaches to decision making. 7. Learn about basic models of cost, revenue, and profit and be able to compute the breakeven point. 8. Obtain an introduction to the use of computer software packages such as M icrosoft Excel in applying quantitative methods to decision making. 9. Understand the following terms: model infeasible solution objective function management science constraint operations research deterministic model fixed cost stochastic model variable cost feasible solution breakeven point Solutions:

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    1. Management science and operations research, terms used almost interchangeably, are broad disciplines that employ scientific methodology in managerial decision making or problem solving. Drawing upon a variety of disciplines (behavioral, mathematical, etc.), management science and operations research combine quantitative and qualitative considerations in order to establish policies and decisions that are in the best interest of the organization. 2. Define the problem Identify the alternatives Determine the criteria Evaluate the alternatives Choose an alternative For further discussion see section 1.3 3. See section 1.2. 4. A quantitative approach should be considered because the problem is large, complex, important, new and repetitive. 5. Models usually have time, cost, and risk advantages over experimenting with actual situations. 6. Model (a) may be quicker to formulate, easier to solve, and/or more easily understood. 7. Let d = distance m = miles per gallon c = cost per gallon,  Total Cost = 2 d c m       We must be willing to treat m and c as known and not subject to variation. 8. a. Maximize 10 x + 5 y s.t. 5 x + 2 y  40 x  0, y  0 b. Controllable inputs: x and y Uncontrollable inputs: profit (10,5), labor hours (5,2) and labor - hour availability (40) c.

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    P rofit : Labor Hours: 5/unit for x 2/ unit for y $10/unit for x $ 5/ unit for y 40 labor-hour capacit y Un con trol lable Inputs P roduct ion Quant ities x and y Con troll abl e Input P roject ed P rofit and check on product ion t ime const raint O u tpu t Max 10 x + 5 y s.t. 10 x y + 5 40  x y 0 0   Mathe matical Model d. x = 0, y = 20 Profit = $100 (Solution by trial - and - error) e. Deterministic - all un controllable inputs are fixed and known. 9. If a = 3, x = 13 1/3 and profit = 133 If a = 4, x = 10 and profit = 100 If a = 5, x = 8 and profit = 80 If a = 6, x = 6 2/3 and profit = 67 Since a is unknown, the actual values of x and profit are not known with certainty. 10. a. Total Units Received = x + y b. Total Cost = 0.20 x +0.25 y c. x + y = 5000 d. x  4000 Kansas City Constraint y  3000 Minneapolis Constraint e. Min 0.20 x + 0.25 y s.t. x + y = 5000 x  4000 y  3000 x , y  0 11. a. at $20 d = 800 - 10(20) = 600

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    at $70 d = 800 - 10(70) = 100 b. TR = dp = (800 - 10 p ) p = 800 p - 10 p 2 c. at $30 TR = 800(30) - 10(30) 2 = 15,000 at $40 TR = 800(40) - 10(40) 2 = 16,000 at $50 TR = 800(50) - 10(50) 2 = 15,000 Total Revenue is maximized at the $40 price. d. d = 800 - 10(40) = 400 units TR = $16,000 12. a. TC = 1000 + 30 x b. P = 40 x - (1000 + 30 x ) = 10 x - 1000 c. Breakeven point is the value of x when P = 0 Thus 10 x - 1000 = 0 10 x = 1000 x = 100 13. a. Total cost = 4800 + 6 0 x b. Total profit = total revenue - total cost = 300 x - (4800 + 6 0 x ) = 2 4 0 x - 4800 c. Total profit = 2 4 0(30) - 4800 = 2400 d. 24 0 x - 4800 = 0 x = 4800/2 4 0 = 20 The breakeven point is 20 students. 14. a. Profit = Revenue - Cost = 20 x - (80,000 + 3 x ) = 17 x - 80,000 17 x - 80,000 = 0 17 x = 80,000 x = 4706 Breakeven point = 4706 b. Profit = 17(4000) - 80,000 = - 12,000 Thus, a loss of $12,000 is anticipated. c. Profit = px - (80,000 + 3 x ) = 4000 p - (80,000 + 3(4000)) = 0 4000 p = 92,000 p = 23 d. Profit = $25.95 (4000) - (80,000 + 3 (4000)) = $11,800

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    Probably go ahead with the project although the $11,800 is only a 12.8% return on the total cost of $92,000. 15. a. Profit = 100,000 x - (1,500,000 + 50,000 x ) = 0 50,000 x = 1,500,000 x = 30 b. Build the luxury boxes. Profit = 100,000 (50) - (1,500,000 + 50,000 (50)) = $1,000,000 16. a. Max 6 x + 4 y b. 50 x + 30 y  80,000 50 x  50,000 30 y  45,000 x , y  0 17. a. s j = s j - 1 + x j - d j or s j - s j - 1 - x j + d j = 0 b. x j  c j c. s j  I j

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    Chapter 2 An Introduction to Linear Programming Learning Objectives 1. Obtain an overview of the kinds of problems linear programming has been used to solve. 2. Learn how to develop linear programming models for simple problems. 3. Be able to identify the special features of a model that make it a linear programming model. 4. Learn how to solve two variable linear programming models by the graphical solution procedure. 5. Understand the importance of extreme points in obtaining the optimal solution. 6. Know the use and interpretation of slack and surplus variables. 7. Be able to interpret the computer solution of a linear programming problem. 8. Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems. 9. Understand the following terms: problem formulation feasible region constraint function slack variable objective function standard form solution redundant constraint optimal solution extreme point nonnegativity constraints surplus variable mathematical model alternative optimal solutions linear program infeasibility linear functions unbounded feasible solution

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    Solutions: 1. a, b, and e, are acceptable linear programming relationships. c is not acceptable because of 2 2 B  d is not acceptable because of 3 A f is not acceptable because of 1 AB c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 2. a. 8 4 4 8 0 B A b. B A 8 4 4 8 0 c. B A 8 4 4 8 0 Points on line are only feasible points

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    3. a. B A 0 (0,9) (6,0) b. B A 0 (0,60) (40,0) c. B A 0 (0,20) (40,0) Points on line are only feasible solutions 4. a. B A (20,0) (0, - 15)

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    b. B A (0,12) ( - 10,0) c. B A 0 (10,25) Note: Point shown was used to locate position of the constraint line 5. B A 0 100 200 300 100 200 300 a b c

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    6. 7 A + 10 B = 420 is labeled (a) 6 A + 4 B = 420 is labeled (b) - 4 A + 7 B = 420 is labeled (c) 80 40 40 80 B A 20 60 100 20 60 0 100 - 20 - 40 - 60 - 80 - 100 (c) (a) (b) 7. 0 B A 150 50 100 200 250 50 100

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    8. A 0 200 - 200 - 100 100 200 133 1 / 3 (100,200) B 9. B A 0 200 100 - 100 - 200 200 300 (150,225) (150,100) 100

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    10. 0 1 2 3 4 5 B 1 2 3 4 5 6 Optimal Solution A = 12/7, B = 15/7 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 A A + 2 B = 6 (1) 5 A + 3 B = 15 (2) (1) × 5 5 A + 10 B = 30 (3) (2) - (3) - 7 B = - 15 B = 15/7 From (1), A = 6 - 2(15/7) = 6 - 30/7 = 12/7

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    11. 0 100 B 100 200 Optimal Solution A = 100, B = 50 Value of Objective Function = 750 A A = 100 B = 80 12. a. (0,0) 1 2 3 4 5 B 1 2 3 4 5 6 Optimal Solution A = 3, B = 1.5 Value of Objective Function = 13.5 6 (4,0) (3,1.5) A

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    b. (0,0) 1 2 3 B 1 2 3 4 5 6 Optimal Solution A = 0, B = 3 Value of Objective Function = 18 A 7 8 9 10 c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3). 13. a. 0 2 4 6 8 B 2 4 6 8 Feasible Region consists of this line segment only A b. The extreme points are (5, 1) and (2, 4).

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    c. 0 2 4 6 8 B 2 4 6 8 Optimal Solution A A = 2, B = 4 14. a. Let F = number of tons of fuel additive S = number of tons of s olvent base Max 40 F + 30 S s.t. 2/5 F + 1 / 2 S  200 Material 1 1 / 5 S  5 Material 2 3 / 5 F + 3 / 10 S  21 Material 3 F , S  0

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    b. c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. 15. a. S F

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    b. Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160. c. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem. 16. a. A variety of objective functions with a slope greater than - 4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3 S + 9 D . b. Optimal Solution is S = 0 and D = 540. c. Department Hours Used Max. Available Slack Cutting and Dyeing 1(540) = 540 630 90 Sewing 5 / 6 (540) = 450 600 150 Finishing 2 / 3 (540) = 360 708 348 Inspection and Packaging 1 / 4 (540) = 135 135 0 17. Max 5 A + 2 B + 0 S 1 + 0 S 2 + 0 S 3 s.t. 1 A - 2 B + 1 S 1 = 420 2 A + 3 B + 1 S 2 = 610 6 A - 1 B + 1 S 3 = 125 A , B , S 1 , S 2 , S 3  0 18. a. Max 4 A + 1 B + 0 S 1 + 0 S 2 + 0 S 3 s.t. 10 A + 2 B + 1 S 1 = 30 3 A + 2 B + 1 S 2 = 12 2 A + 2 B + 1 S 3 = 10 A , B , S 1 , S 2 , S 3  0

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    b. 0 2 4 6 8 B 2 4 6 8 Optimal Solution A A = 18/7, B = 15/7, Value = 87/7 10 12 14 10 c. S 1 = 0, S 2 = 0, S 3 = 4/7 19. a. Max 3 A + 4 B + 0 S 1 + 0 S 2 + 0 S 3 s.t. - 1 A + 2 B + 1 S 1 = 8 (1) 1 A + 2 B + 1 S 2 = 12 (2) 2 A + 1 B + 1 S 3 = 16 (3) A , B , S 1 , S 2 , S 3  0

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    b. 0 2 4 6 8 B 2 4 6 8 Optimal Solution A A = 20/3, B = 8/3 Value = 30 2/3 10 12 14 10 12 (2) (1) (3) c. S 1 = 8 + A – 2 B = 8 + 20/3 - 16/3 = 28/3 S 2 = 12 - A – 2 B = 12 - 20/3 - 16/ 3 = 0 S 3 = 16 – 2 A - B = 16 - 40/3 - 8/3 = 0 20. a. Max 3 A + 2 B s.t. A + B - S 1 = 4 3 A + 4 B + S 2 = 24 A - S 3 = 2 A - B - S 4 = 0 A , B , S 1 , S 2 , S 3 , S 4  0

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    b. c. S 1 = (3.43 + 3.43) - 4 = 2.86 S 2 = 24 - [3(3.43) + 4(3.43)] = 0 S 3 = 3.43 - 2 = 1.43 S 4 = 0 - (3.43 - 3.43) = 0

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    21 . a. and b. 0 10 20 30 40 B 10 20 30 40 Optimal Solution A 2 A + 3 B = 60 50 60 70 50 60 80 70 80 90 90 100 Constraint 3 Constraint 2 Constraint 1 Feasible Region c. Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2, B = 10 + A Substituting for B in constraint 1 we obtain 5 A + 5(10 + A ) = 400 5 A + 50 + 5 A = 400 10 A = 350 A = 35 B = 10 + A = 10 + 35 = 45 Optimal solution is A = 35, B = 45 d. Because the optimal solution occurs at the intersection of constraints 1 and 2 , t hese are binding constraints.

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    e. Constraint 3 is the nonbinding constraint. At the optimal solution 1 A + 3 B = 1(35) + 3(45) = 170. Because 170 exceeds the right - hand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint. 22. a. 0 500 1000 1500 2000 C 500 1000 1500 2000 Feasible Region A 5 A + 4 C = 4000 2500 3000 3500 2500 3000 Inspection and Packaging Cutting and Dyeing Sewing Number of All-Pro Footballs 5 4 3 2 1 b. Extreme Point Coordinates Profit 1 (0, 0) 5(0) + 4(0) = 0 2 (1700, 0) 5(1700) + 4(0) = 8500 3 (1400, 600) 5(1400) + 4(600) = 9400 4 (800, 1200) 5(800) + 4(1200) = 8800 5 (0, 1680) 5(0) + 4(1680) = 6720 Extreme point 3 generates the highest profit. c. Optimal solution is A = 1400, C = 600 d. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints. e. New optimal solution is A = 800, C = 1200 Profit = 4(800) + 5(1200) = 9200

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    2 3 . a. Let E = number of units of the EZ - Rider produced L = number of units of the Lady - Sport produced Max 2400 E + 1800 L s.t. 6 E + 3 L  2100 Engine time L  280 Lady - Sport maximum 2 E + 2.5 L  1000 Assembly and testing E, L  0 b. c. The binding constraints are the manufacturing time and the assembly and testing time. 2 4 . a. Let R = number of units of regular model. C = number of units of catcher’s model. Max 5 R + 8 C s.t. 1 R + 3 / 2 C  900 Cutting and sewing 1 / 2 R + 1 / 3 C  300 Finishing 1 / 8 R + 1 / 4 C  100 Packing and Shipping R , C  0 0 L Profit = $960,000 Optimal Solution 100 200 300 400 500 600 700 100 200 300 400 500 E Engine Manufacturing Time Frames for Lady-Sport Assembly and Testing E = 250, L = 200 Number of Lady-Sport Produced Number of EZ-Rider Produced

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    b. . 800 400 400 800 C R 200 600 1000 200 600 0 1000 Optimal Solution (500,150) F C & S P & S Regular Model Catcher's Model c. 5(500) + 8(150) = $3,700 d. 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 Capacity Usage Slack C & S 900 725 175 hours F 300 300 0 hours P & S 100 100 0 hours 2 5 . a. Let B = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund Max 0.06 B + 0.10 S s.t. B  0.3 Bond fund minimum 0.06 B + 0.10 S  0.075 Minimum return B + S = 1 Percentage requirement b. Optimal solution: B = 0.3, S = 0.7 Value of optimal solution is 0.088 or 8.8%

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    2 6 . a. a. Let N = amount spent on newspaper advertising R = amount spent on radio advertising Max 50 N + 80 R s.t. N + R = 1000 Budget N  250 Newspaper min. R  250 Radio min. N - 2 R  0 News  2 Radio N , R  0 b. 500 5 0 0 R N 0 1000 1000 Optimal Solution N = 666.67, R = 333.33 Value = 60,000 Newspaper Min Budget Radio Min Feasible region is this line segment N = 2 R 2 7 . L et I = Internet fund investment in thousands B = Blue Chip fund investment in thousands Max 0.12 I + 0.09 B s.t. 1 I + 1 B  50 Available investment funds 1 I  35 Maximum investment in the internet fund 6 I + 4 B  240 Maximum risk for a moderate investor I, B  0

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    Internet fund $20,000 Blue Chip fund $30,000 Annual return $ 5,100 b. The third constraint for the aggressive investor becomes 6 I + 4 B  320 This constraint is redundant; the available funds and the maximum Internet fund investmen t constraints define the feasible region. The optimal solution is: Internet fund $35,000 Blue Chip fund $15,000 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 6 I + 4 B  160 This constraint becomes a binding constraint. The optimal solution is Internet fund $0 Blue Chip fund $40,000 Annual return $ 3,600 0 B Objective Function 0.12 I + 0.09 B Optimal Solution 10 20 30 40 50 60 60 10 20 30 40 50 I Risk Constraint Maximum Internet Funds Available Funds $50,000 I = 20, B = 30 $5,100 Internet Fund (000s) Blue Chip Fund (000s)

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    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. 2 8 . a. Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced Max 1 W + 1.25 M s.t. 5 W 7 M  4480 Whole tomatoes 3 W + 1 M  2080 Tomato sauce 2 W + 2 M  1600 Tomato paste W , M  0 Note: units for constraints are ounces b. Optimal solution: W = 560, M = 240 Value of optimal solution is 860 2 9 . 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 Buffalo 2000 1000 Dayton 600 1400 Number of units of component 1 produced: 2000 B + 600 D 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, 2000 B + 600 D = 1000(1 - B ) + 1400(1 - D ) 2000 B + 600 D = 1000 - 1000 B + 1400 - 1400 D 3 000 B + 2000 D = 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 2000 B + 600 D In addition, B  1 and D  1.

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    The linear programming model is: Max 2000 B + 600 D s.t. 3000 B + 2000 D = 2400 B  1 D  1 B , D  0 The graphical solution is shown below. Optimal Solution: B = .8, D = 0 Optimal Production Plan Buffalo - Component 1 .8(2000) = 1600 Buffalo - Component 2 .2(1000) = 200 Dayton - Component 1 0(600) = 0 Dayton - Component 2 1(1400) = 1400 Total units of electronic ignition system = 1600 per day. 0 D B .2 .4 .6 .8 1.0 1.2 .2 .4 .6 .8 1.0 1.2 3000 B + 2000 D = 2400 2000 B + 600 D = 300 Optimal Solution

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    30 . a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch Max 15 E + 18 C s.t. 40 E + 25 C  50,000 Maximum Investment 40 E  15,000 Eastern Cable Minimum 25 C  10,000 ComSwitch Minimum 25 C  25,000 ComSwitch Maximum E, C  0 b. c. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) d. Optimal solution is E = 625, C = 1000 Total return = $27,375 0 500 1000 500 1000 C E 1500 1500 2000 Minimum Eastern Cable Maximum Comswitch Minimum Conswitch Maximum Investment Number of Shares of Eastern Cable Number of Shares of ComSw itch

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