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Solutions Manual for Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 8th Edition Ragsdale $15.49
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Solutions Manual for Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 8th Edition Ragsdale

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  • January 9, 2025
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  • 2024/2025
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SOLUTIONS MANUAL FOR
SPREADSHEET MODELING AND
DECISION ANALYSIS A
PRACTICAL INTRODUCTION TO
BUSINESS ANALYTICS 8TH
EDITION RAGSDALE

,Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics

Chapter 2 - Introduction to Optimization & Linear Programming : S-1
————————————————————————————————————————————


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 X1 and X2, 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 X1 - .3333 X2 = 75
d. linear, can be re-written as: 2.1 X1 + 1.1 X2 - 3.9 X3 ≤ 0
e. nonlinear

6.




Visit TestBankDeal.com to get complete for all chapters

, Chapter 2 - Introduction to Optimization & Linear Programming : S-2
————————————————————————————————————————————

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)




0 5 10 15 20 25 X1

, Chapter 2 - Introduction to Optimization & Linear Programming : S-3
————————————————————————————————————————————


9.




10.

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