Contents
Preface
Chapter
1. Introduction
2. An Introduction to Linear Programming
3. Linear Programming: Sensitivity Analysis and Interpretation of Solution
4. Linear 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

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:

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.

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P 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 + 5y
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

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at $70 d = 800 - 10(70) = 100
b. TR = dp = (800 - 10p)p = 800p - 10p2
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 + 30x
b. P = 40x - (1000 + 30x) = 10x - 1000
c. Breakeven point is the value of x when P = 0
Thus 10x - 1000 = 0
10x = 1000
x = 100
13. a. Total cost = 4800 + 60x
b. Total profit = total revenue - total cost
= 300x - (4800 + 60x)
= 240x - 4800
c. Total profit = 240(30) - 4800 = 2400
d. 240x - 4800 = 0
x = 4800/240 = 20
The breakeven point is 20 students.
14. a. Profit = Revenue - Cost
= 20x - (80,000 + 3x)
= 17x - 80,000
17x - 80,000 = 0
17x = 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 + 3x)
= 4000p - (80,000 + 3(4000)) = 0
4000p = 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 = 100,000x - (1,500,000 + 50,000x) = 0
50,000x = 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 6x + 4y
b. 50x + 30y  80,000
50x  50,000
30y  45,000
x, y  0
17. a. sj = sj - 1 + xj - dj
or sj - sj-1 - xj + dj = 0
b. xj  cj
c. sj  Ij

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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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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 3S + 9D.
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 5A + 2B + 0S1 + 0S2 + 0S3
s.t.
1A - 2B + 1S1 = 420
2A + 3B + 1S2 = 610
6A - 1B + 1S3 = 125
A, B, S1, S2, S3  0
18. a.
Max 4A + 1B + 0S1 + 0S2 + 0S3
s.t.
10A + 2B + 1S1 = 30
3A + 2B + 1S2 = 12
2A + 2B + 1S3 = 10
A, B, S1, S2, S3  0

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21. a. and b.
0
10
20
30
40
B
10 20 30 40
Optimal Solution
A
2A + 3B = 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
5A + 5(10 + A) = 400
5A + 50 + 5A = 400
10A = 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, these are binding
constraints.

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e. Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 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
5A + 4C = 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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23. a. Let E = number of units of the EZ-Rider produced
L = number of units of the Lady-Sport produced
Max 2400E + 1800L
s.t.
6E + 3L  2100 Engine time
L  280 Lady-Sport maximum
2E + 2.5L  1000 Assembly and testing
E, L  0
b.
c. The binding constraints are the manufacturing time and the assembly and testing time.
24. a. Let R = number of units of regular model.
C = number of units of catcher’s model.
Max 5R + 8C
s.t.
1R + 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 6000 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
25. 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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26. a. a. Let N = amount spent on newspaper advertising
R = amount spent on radio advertising
Max 50N + 80R
s.t.
N + R = 1000 Budget
N  250 Newspaper min.
R  250 Radio min.
N -2R  0 News  2 Radio
N, R  0
b.
500
5 00
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
27. Let I = Internet fund investment in thousands
B = Blue Chip fund investment in thousands
Max 0.12I + 0.09B
s.t.
1I + 1B  50 Available investment funds
1I  35 Maximum investment in the internet fund
6I + 4B  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
6I + 4B  320
This constraint is redundant; the available funds and the maximum Internet fund investment
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
6I + 4B  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
0.12I + 0.09B
Optimal Solution
10
20
30
40
50
60
6010 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.
28. a. Let W = number of jars of Western Foods Salsa produced
M = number of jars of Mexico City Salsa produced
Max 1W + 1.25M
s.t.
5W 7M  4480 Whole tomatoes
3W + 1M  2080 Tomato sauce
2W + 2M  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
29. 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: 2000B + 600D
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,
2000B + 600D = 1000(1 - B) + 1400(1 - D)
2000B + 600D = 1000 - 1000B + 1400 - 1400D
3000B + 2000D = 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 2000B + 600D
In addition, B  1 and D  1.

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The linear programming model is:
Max 2000B + 600D
s.t.
3000B + 2000D = 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
3000B + 2000D = 2400
2000B + 600D = 300
Optimal
Solution

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30. a. Let E = number of shares of Eastern Cable
C = number of shares of ComSwitch
Max 15E + 18C
s.t.
40E + 25C  50,000 Maximum Investment
40E  15,000 Eastern Cable Minimum
25C  10,000 ComSwitch Minimum
25C  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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An Introduction To Management Science: Quantitative Approaches To Decision Making, Revised, 13th Edition Solution Manual

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