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

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

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 =2d c
m
 
 
 
We must be willing to treat m and c as known and not subject to variation.
8. a. Maximize 10x + 5y
s.t.
5x + 2y  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.

Introduction
1 - 3P rofit :
Labor Hours: 5/unit forx
2/ unit fory
$10/unit forx
$ 5/ unit fory
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 10x + 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 uncontrollable 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 = 0.20x +0.25y
c. x + y = 5000
d. x  4000 Kansas City Constraint
y  3000 Minneapolis Constraint
e. Min 0.20x + 0.25y
s.t.
x + y = 5000
x  4000
y  3000
x, y  0
11. a. at $20 d = 800 - 10(20) = 600

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
If 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
If the firm increases the per unit price from $426 to $43, the number of units the firm can sell falls by 10.
This suggests 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 - 10p)p = 800p - 10p2
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 = 2000 + 60x
b. P = 80x - (2000 + 60x) = 20x - 2000
c. Breakeven point is the value of x when P = 0
Thus 20x - 2000 = 0
20x = 2000
x = 100
13. a. Total cost = 9600 + (2*60)x = 9600 + 120x
b. Total profit = total revenue - total cost
= 600x - (9600 + 120x)
= 480x - 9600
c. Total profit = 480(30) - 9600 = 4800
d. 480x - 9600 = 0
x = 9600/480 = 20
The breakeven point is 20 students.
14. a. Profit = Revenue - Cost
= 46x - (160,000 + 6x)
= 40x - 160,000
40x - 160,000 = 0
40x = 160,000
x = 4000
Breakeven point = 4000

Introduction
1 - 5
b. Profit = 40(3800) - 16,000 = -8000
Thus, a loss of $8000 is anticipated.
c. Profit = px - (160,000 + 6x)
= 3800p - (160,000 + 6(3800)) = 0
3800p = 182,800
p = 48.105 or $48.11
d. Profit = $50.95 (3800) - (160,000 + 6 (3800))
= $10,810
Probably go ahead with the project although the $10,810 is only a 5.98% return on the total cost of
$182,800.
15. a. Profit = 300,000x - (4,500,000 + 150,000x) = 0
150,000x = 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 6x + 4y
b. 50x + 30y  800,000
50x  500,000
30y  450,000
x, y  0
17. a. sj = sj - 1 + xj - dj
or sj - sj-1 - xj + dj = 0
b. xj  cj
c. sj  Ij
18. a. maximize (3.10 – 0.30)x1 + (3.10 – 0.40) x2 + (3.10 – 0.48)x3 + (3.20 – 0.30)y1 + (3.20 – 0.40)y2
+(3.20-0.48)y3
= maximize 2.80x1 + 2,70 x2 + 2.62x3 +2.90y1 + 2.80y2 +2.72y3
b. (1) x1 + y1 ≤ 12,000
(2) x2 + y2 ≤ 20,000
(3) x3 + y3 ≤ 24,000
c. x1 ≥ .35(x1 + x2 + x3) or .65x1 - .35x2 - .35x3 ≥ 0 (Blend A must be composed of at least 35% of oil
from the Texas well)
x2 ≤ .50(x1 + x2 + x3) or -.50x1 + .50x2 - .50x3 ≤ 0 (Blend A must be composed of no more than 50%
of oil from the Oklahoma well)
x3 ≥ .15(x1 + x2 + x3) or -.15x1 - .15x2 + .85x3 ≥ 0 (Blend A must be composed of at least 15% of oil
from the California well)
y1 ≥ .20(y1 + y2 + y3) or .80y1 - .20y2 - .20y3 ≥ 0 (Blend B must be composed of at least 20% of oil
from the Texas well)

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Chapter 1
1 - 6
y2 ≥ .30(y1 + y2 + y3) or -.30y1 + .70y2 - .30y3 ≥ 0 (Blend B must be composed of at least 30% of oil
from the Oklahoma well)
y3 ≤ .40(y1 + y2 + y3) or -.40y1 - .40y2 + .60y3 ≤ 0 (Blend B must be composed of no more than 40%
of oil from the California well)
x1 + x2 + x3 ≥ 20,000 (Long-term contracts require at least 20,000
gallons of Blend A to be produced)
y1 + y2 + y3 ≥ 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 18.6x + 13.9y
b. 2.0x1 + 2.5y1 ≤ 3000 hours available in carpentry
1.5x2 + 1.0y2 ≤ 1500 hours available in carpentry
1.3x3 + 1.2y3 ≤ 1500 hours available in carpentry
c. x ≥ 500
y ≥ 650
20. a. maximize 7000x + 4000y
b. 500x + 250y ≤ 100,000
c. x ≤ 20
d. y ≥ 50
e. , or 2x – y ≥ 0
f. If the constraints 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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