SOLUTION MANUAL SPREADSHEET
MODELING AND DECISION ANALYSIS
A PRACTICAL INTRODUCTION TO
BUSINESS ANALYTICS 8TH EDITION
s-analytics-8th-edition-ragsdale-solutions-manual/
Chapter 2 - Introduction to Optimization & Linear Programming : S-1
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Chapter 2
Introduction to Optimization & Linear Programming
1. If an LP model has more than one optimal solution it has an infinite number of alternate optimal solutions.
In Figure 2.8, the two extreme points at (122, 78) and (174, 0) are alternate optimal solutions, but there are
an infinite number of alternate optimal solutions along the edge connecting these extreme points. This is
true of all LP models with alternate optimal solutions.
2. There is no guarantee that the optimal solution to an LP problem will occur at an integer-valued extreme
point of the feasible region. (An exception to this general rule is discussed in Chapter 5 on networks).
3. We can graph an inequality as if they were an equality because the condition imposed by the equality
corresponds to the boundary line (or most extreme case) of the inequality.
4. The objectives are equivalent. For any values of X
and X
1
2
, the absolute value of the objectives are the
same. Thus, maximizing the value of the first objective is equivalent to minimizing the value of the second
objective.
5. a. linear
b. nonlinear
c. linear, can be re-written as: 4 X
- .3333 X
1
2
= 75
d. linear, can be re-written as: 2.1 X
+ 1.1 X
- 3.9 X
1
2
3
≤ 0
e. nonlinear
6.
Chapter 2 - Introduction to Optimization & Linear Programming : S-2
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7.
8.
X2
20
(0, 15) obj = 300
15
(0, 12) obj = 240
10
(6.67, 5.33) obj =140
5
(11.67, 3.33) obj = 125
(optimal solution)
X1
0
5
10
15
20
25
Chapter 2 - Introduction to Optimization & Linear Programming : S-3
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9.
10.
Chapter 2 - Introduction to Optimization & Linear Programming : S-4
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11.
12.
Chapter 2 - Introduction to Optimization & Linear Programming : S-5
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13. X
= number of softballs to produce, X
= number of baseballs to produce
1
2
MAX 6 X
+ 4.5 X
1
2
ST 5X
+ 4 X
1
2
≤ 6000
6 X
+ 3 X
1
2
≤ 5400
4 X
+ 2 X
1
2
≤ 4000
2.5 X
+ 2 X
1
2
≤ 3500
1 X
+ 1 X
1
2
≤ 1500
X
, X
1
2
≥ 0
14. X
= number of His chairs to produce, X
= number of Hers chairs to produce
1
2
MAX 10 X
+ 12 X
1
2
ST 4 X
+ 8 X
1
2
≤ 1200
8 X
+ 4 X
1
2
≤ 1056
2 X
+ 2 X
1
2
≤ 400
4 X
+ 4 X
1
2
≤ 900
1 X
- 0.5 X
1
2
≥ 0
X
, X
1
2
≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-6
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15. X
= number of propane grills to produce, X
= number of electric grills to produce
1
2
MAX 100 X
+ 80 X
1
2
ST 2 X
+ 1 X
1
2
≤ 2400
4 X
+ 5 X
1
2
≤ 6000
2 X
+ 3 X
1
2
≤ 3300
1 X
+ 1 X
1
2
≤ 1500
X
, X
1
2
≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-7
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16. X
= number of generators, X
= number of alternators
1
2
MAX 250 X
+ 150 X
1
2
ST 2 X
+ 3 X
1
2
≤ 260
1 X
+ 2 X
1
2
≤ 140
X
, X
1
2
≥ 0
17. X
= number of generators, X
= number of alternators
1
2
MAX 250 X
+ 150 X
1
2
ST 2 X
+ 3 X
1
2
≤ 260
1 X
+ 2 X
1
2
≤ 140
X
1
≥ 20
X
2
≥ 20
d. No, the feasible region would not increase so the solution would not change -- you'd just have extra
(unused) wiring capacity.
Chapter 2 - Introduction to Optimization & Linear Programming : S-8
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18. X
= proportion of beef in the mix, X
= proportion of pork in the mix
1
2
MIN
.85 X
+ .65 X
1
2
ST 1X
+ 1 X
= 1
1
2
0.2 X
+ 0.3 X
1
2
≤ 0.25
X
, X
1
2
≥ 0
19. T= number of TV ads to run, M = number of magazine ads to run
MIN 500 T + 750 P
ST 3T + 1P ≥ 14
-1T + 4P ≥ 4
0T + 2P ≥ 3
T, P ≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-9
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20. X
= # of TV spots, X
= # of magazine ads
1
2
MAX 15 X
+ 25 X
1
2
(profit)
ST 5 X
+ 2 X
< 100
(ad budget)
1
2
5 X
+ 0 X
1
2
≤ 70
(TV limit)
0 X
+ 2 X
1
2
≤ 50
(magazine limit)
X
, X
1
2
≥ 0
X2
40
(0,25)
(10,25)
30
15X1+25X2=775
20
(14,15)
10
15X1+25X2=400
(14,0)
20
10
X1
21. X
= tons of ore purchased from mine 1, X
= tons of ore purchased from mine 2
1
2
MIN 90 X
+ 120 X
(cost)
1
2
ST 0.2 X
+ 0.3 X
> 8
(copper)
1
2
0.2 X
+ 0.25 X
> 6 (zinc)
1
2
0.15 X
+ 0.1 X
> 5 (magnesium)
1
2
X
, X
1
2
≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-10
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22. R = number of Razors produced, Z = number of Zoomers produced
MAX 70 R + 40 Z
ST R + Z ≤ 700
R – Z ≤ 300
2 R + 1 Z ≤ 900
3 R + 4 Z ≤ 2400
R, Z ≥ 0
23. P = number of Presidential desks produced, S = number of Senator desks produced
MAX 103.75 P + 97.85 S
ST 30 P + 24 S ≤ 15,000
1 P + 1 S ≤ 600
5 P + 3 S ≤ 3000
P, S ≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-11
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24. X
= acres planted in watermelons, X
= acres planted in cantaloupes
1
2
MAX 256 X
+ 284.5 X
1
2
ST 50 X
+ 75 X
1
2
≤ 6000
X
+ X
1
2
≤ 100
X
, X
1
2
≥ 0
X2
(0, 80) obj = 22760
100
75
(60, 40) obj =26740
(optimal solution)
50
25
(100, 0) obj = 25600
0
X1
0
25
50
75
100
125
25. D = number of doors produced, W = number of windows produced
MAX 500 D + 400 W
ST 1 D + 0.5 W ≤ 40
0.5 D + 0.75 W ≤ 40
0.5 D + 1 W ≤ 60
D, W ≥ 0
Chapter 2 - Introduction to Optimization & Linear Programming : S-12
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26. X
= number of desktop computers, X
= number of laptop computers
1
2
MAX 600 X
+ 900 X
1
2
ST 2 X
+ 3 X
1
2
≤ 300
X
1
≤ 80
X
2
≤ 75
X
, X
1
2
≥ 0
Case 2-1: For The Lines They Are A-Changin’
1. 200 pumps, 1566 labor hours, 2712 feet of tubing.
2. Pumps are a binding constraint and should be increased to 207, if possible. This would increase profits
by $1,400 to $67,500.
3. Labor is a binding constraint and should be increased to 1800, if possible. This would increase profits
by $3,900 to $70,000.
4. Tubing is a non-binding constraint. They’ve already got more than they can use and don’t need any
more.
5. 9 to 8: profit increases by $3,050
8 to 7: profit increases by $850
7 to 6: profit increases by $0
6. 6 to 5: profit increases by $975
5 to 4: profit increases by $585
4 to 3: profit increases by $390
7. 12 to 13: profit changes by $0
13 to 14: profit decreases by $760
14 to 15: profit decreases by $1,440
Chapter 2 - Introduction to Optimization & Linear Programming : S-13
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8. 16 to 17: profit changes by $0
17 to 18: profit changes by $0
18 to 19: profit decreases by $400
9. The profit on Aqua-Spas can vary between $300 and $450 without changing the optimal solution.
10. The profit on Hydro-Luxes can vary between $233.33 and $350 without changing the optimal solution.
Spreadsheet Modeling
& Decision Analysis
A Practical Introduction to
Business Analytics
th
8
edition
Cliff T. Ragsdale
Chapter 2
Introduction to Optimization
and Linear Programming
Introduction
We all face decision about how to use
limited resources such as:
– Oil in the earth
– Land for dumps
– Time
– Money
– Workers
Mathematical Programming...
MP is a field of management science that
finds the optimal, or most efficient, way of
using limited resources to achieve the
objectives of an individual of a business.
a.k.a. Optimization
Applications of Optimization
Determining Product Mix
Manufacturing
Routing and Logistics
Financial Planning
Characteristics of
Optimization Problems
Decisions
Constraints
Objectives
General Form of an Optimization Problem
MAX (or MIN): f0
(X1
, X2
, …, Xn
)
Subject to:
f1
(X1
, X2
, …, Xn
)=bk
:
fm
(X1
, X2
, …, Xn
)=bm
Note: If all the functions in an optimization are linear,
the problem is a Linear Programming (LP) problem
Linear Programming (LP) Problems
MAX (or MIN): c1
X1
+ c2
X2
+ … + cn
Xn
Subject to: a11
X1
+ a12
X2
+ … + a1n
Xn
=bk
:
am1
X1
+ am2
X2
+ … + amn
Xn
= bm
An Example LP Problem
Blue Ridge Hot Tubs produces two types of hot
tubs: Aqua-Spas & Hydro-Luxes.
Aqua-Spa
Hydro-Lux
Pumps
1
1
Labor
9 hours
6 hours
Tubing
12 feet
16 feet
Unit Profit
$350
$300
There are 200 pumps, 1566 hours of labor,
and 2880 feet of tubing available.
5 Steps In Formulating LP Models:
1. Understand the problem.
2. Identify the decision variables.
X1
=number of Aqua-Spas to produce
X2
=number of Hydro-Luxes to produce
3. State the objective function as a linear
combination of the decision variables.
MAX: 350X1
+ 300X2
5 Steps In Formulating LP Models
(continued)
4. State the constraints as linear combinations
of the decision variables.
1X1
+ 1X2

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