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

An Introduction To Management Science: Quantitative Approach, 15th Edition Solution Manual is a structured study guide that simplifies complex textbook topics.

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    1 - 1 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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    Chapter 1 1 - 2 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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    Introduction 1 - 3 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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    Chapter 1 1 - 4 at $70 d = 800 - 10(70) = 100 b. at $26 d = 800 - 10(26) = 540 at $27 d = 800 - 10(27) = 530 I f the firm increases the per unit price from $26 to $27 , the number of units the firm can sell falls by 10. at $42 d = 800 - 10(42) = 380 at $43 d = 800 - 10(43) = 370 I f the firm increases the per unit price from $ 42 6 to $ 43, the number of units the firm can sell falls by 10. This suggest s that the relationship between the per - unit price and annual demand for the product in units is linear between $20 and $70 and that annual demand for the product decreases by 10 units when the price is increased by $1. c . TR = dp = (800 - 10 p ) p = 800 p - 10 p 2 d . 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. e . d = 800 - 10(40) = 400 units TR = $16,000 12. a. TC = 2 000 + 6 0 x b. P = 8 0 x - ( 2 000 + 6 0 x ) = 2 0 x - 2 000 c. Breakeven point is the value of x when P = 0 Thus 2 0 x - 2 000 = 0 2 0 x = 2 000 x = 100 13. a. Total cost = 96 00 + (2*60) x = 9600 + 120 x b. Total profit = total revenue - total cost = 6 00 x - ( 96 00 + 120 x ) = 48 0 x - 96 00 c. Total profit = 48 0(30) - 96 00 = 48 00 d. 480 x - 9600 = 0 x = 96 00/ 48 0 = 20 The breakeven point is 20 students. 14. a. Profit = Revenue - Cost = 46 x - ( 16 0,000 + 6 x ) = 40 x - 16 0,000 40 x - 16 0,000 = 0 40 x = 16 0,000 x = 4 000 Breakeven point = 4 000

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    Introduction 1 - 5 b. Profit = 40 ( 38 00) - 16 ,000 = - 8 000 Thus, a loss of $ 8 000 is anticipated. c. Profit = px - (160,000 + 6 x ) = 3800 p - (160,000 + 6(3800)) = 0 3800 p = 182,800 p = 48.105 or $48.11 d. Profit = $ 50 .95 ( 38 00) - ( 16 0,000 + 6 ( 38 00)) = $1 0 ,8 1 0 Probably go ahead with the project although the $1 0 ,8 1 0 is only a 5.9 8% return on the total cost of $ 182,800 . 15. a. Profit = 3 00,000 x - ( 4 ,500,000 + 1 50,000 x ) = 0 1 50,000 x = 4 ,500,000 x = 30 b. Build the luxury boxes. Profit = 300 ,000 (50) - (4,500,000 + 150,000 (50)) = $3,000,000 16. a. Max 6 x + 4 y b. 50 x + 30 y  80 0 ,000 50 x  5 0 0,000 30 y  45 0 ,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 18. a. maximize (3.10 – 0.30) x 1 + (3.10 – 0.40) x 2 + (3.10 – 0.48) x 3 + (3.20 – 0.30) y 1 + (3.20 – 0.40) y 2 +(3.20 - 0.48) y 3 = maximize 2.80 x 1 + 2,70 x 2 + 2.62 x 3 +2.90 y 1 + 2.80 y 2 +2.72 y 3 b. (1) x 1 + y 1 ≤ 12,000 (2) x 2 + y 2 ≤ 20,000 (3) x 3 + y 3 ≤ 24,000 c. x 1 ≥ .35( x 1 + x 2 + x 3 ) or .65 x 1 - .35 x 2 - .35 x 3 ≥ 0 ( Blend A must be composed of at least 35% of oil from the Texas well ) x 2 ≤ .50( x 1 + x 2 + x 3 ) or - .50 x 1 + .50 x 2 - .50 x 3 ≤ 0 ( Blend A must be composed of no more than 50% of oil from the Oklahoma well ) x 3 ≥ .15( x 1 + x 2 + x 3 ) or - .15 x 1 - .15 x 2 + .85 x 3 ≥ 0 ( Blend A must be composed of at least 15 % of oil from the California well ) y 1 ≥ .20( y 1 + y 2 + y 3 ) or .80 y 1 - .20 y 2 - .20 y 3 ≥ 0 ( Blend B must be composed of at least 2 0% of oil from the Texas well )

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    Chapter 1 1 - 6 y 2 ≥ .3 0( y 1 + y 2 + y 3 ) or - .30 y 1 + .70 y 2 - .30 y 3 ≥ 0 ( Blend B must be composed of at least 30% of oil from the Oklahoma well ) y 3 ≤ .40( y 1 + y 2 + y 3 ) or - .40 y 1 - .40 y 2 + .60 y 3 ≤ 0 ( Blend B must be composed of no more than 40% of oil from the California well ) x 1 + x 2 + x 3 ≥ 20,000 ( Long - term contracts require at least 20,000 gallons of Blend A to be produced ) y 1 + y 2 + y 3 ≥ 20,000 ( Long - term contracts require at least 20,000 gallons of Blend B to be produced ) 19 . a. Profit per contemporary cabinet is $ 90 – [ 2.0 ($15) + 1.5 ($12) + 1.3 ($18)] = $18.60 Profit per farmhouse cabinet is $85 – [ 2.5 ($15) + 1.0 ($12) + 1.2 ($18)] = $13.90 So the objective function is maximize 18.6 x + 13.9 y b. 2.0 x 1 + 2.5 y 1 ≤ 3000 hours available in carpentry 1.5 x 2 + 1.0 y 2 ≤ 1500 hours available in carpentry 1.3 x 3 + 1.2 y 3 ≤ 1500 hours available in carpentry c. x ≥ 500 y ≥ 650 20. a. maximize 7000 x + 4000 y b. 500 x + 250 y ≤ 100,000 c. x ≤ 20 d. y ≥ 50 e. , or 2 x – y ≥ 0 f. If the constraint s for the maximum number of television ads to be used from part(c) and the minimum number of internet ads to be used from part (d) are both satisfied, then television ads can be at most . That is, the constraint for the stipulated ratio of television ads to Internet ads cannot be satisfied. Therefore, the problem as stated is infeasible.

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    2 - 1 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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    Chapter 2 2 - 2 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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    An Introduction to Linear Programming 2 - 3 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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    Chapter 2 2 - 4 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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    An Introduction to Linear Programming 2 - 5 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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    Chapter 2 2 - 6 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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    An Introduction to Linear Programming 2 - 7 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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    Chapter 2 2 - 8 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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    An Introduction to Linear Programming 2 - 9 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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