Spreadsheet Modeling And Decision Analysis Homework Answers
Chapter 2 - Introduction to Optimization & Linear Programming : S-1
Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 7th Edition Cliff Ragsdale Solutions ManualDownload:http://testbanklive.com/download/spreadsheet-modeling-and-decision-analysis-a-practical-introduction-to-business-analytics-7th-edition-cliff-ragsdale-solutions-manual/Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 7th Edition Cliff Ragsdale Test Bank Download:http://testbanklive.com/download/spreadsheet-modeling-and-decision-analysis-a-practical-introduction-to-business-analytics-7th-edition-cliff-ragsdale-test-bank/
Introduction to Optimization & Linear
1. If an LP model has more than one optimal solution it has an
number of alternate optimal
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
, 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
= 75d. linear, can be re-written as: 2.1 X
+ 1.1 X
- 3.9 X
0e. nonlinear 6.
UPDATE (1/15/2013):Cliff Ragsdale was kind enough to include the modification I describe below in the 7th edition of his book (it’s now problem 32 in Chapter 6). He even named a character after me! Thanks, Cliff!
When I teach the OR class to MBA students, I adopt Cliff Ragsdale’s textbook entitled “Spreadsheet Modeling and Decision Analysis“, which is now in its sixth edition. I like this book and I’m used to teaching with it. In addition, it has a large and diverse collection of interesting exercises/problems that I use both as homework problems and as inspiration for exam questions.
One of my favorite problems to assign as homework is problem number 30 in the Integer Linear Programming chapter (Chapter 6). (This number refers to the 6th edition of the book; in the 5th edition it’s problem number 29, and in the 4th edition it’s problem number 26.) Here’s the statement:
The emergency services coordinator of Clarke County is interested in locating the county’s two ambulances to maximize the number of residents that can be reached within four minutes in emergency situations. The county is divided into five regions, and the average times required to travel from one region to the next are summarized in the following table:
The population in regions 1, 2, 3, 4, and 5 are estimated as 45,000, 65,000, 28,000, 52,000, and 43,000, respectively. In which two regions should the ambulances be placed?
I love this problem. It exercises important concepts and unearths many misconceptions. It’s challenging, but not impossible, and it forces students to think about connecting distinct—albeit related—sets of variables; a common omission in models created by novice modelers. BUT, in its present form, in my humble opinion, it falls short of the masterpiece it can be. There are two main issues with the current version of this problem (think about it for a while and you’ll see what I mean):
- It’s easy for students to eyeball an optimal solution. So they come back to my office and say: “I don’t know what the point of this problem is; the answer is obviously equal to …” Many of them don’t even try to create a math model.
- Even if you model it incorrectly, that is, by choosing the wrong variables which will end up double-counting the number of people covered by the ambulances, the solution that you get is still equal to the correct solution. So when I take points off for the incorrect model, the students come back and say “But I got the right answer!”
After a few years of facing these issues, I decided I had had enough. So I changed the problem data to achieve the following (“evil”) goals:
- It’s not as easy to eyeball an optimal solution as it was before.
- If you write a model assuming every region has to be covered (which is not a requirement to begin with), you’ll get an infeasible model. In the original case, this doesn’t happen. I didn’t like that because this isn’t an explicit assumption and many students would add it in.
- If you pick the wrong set of variables and double-count the number of people covered, you’ll end up with an incorrect (sub-optimal) solution.
These improvements are obtained by adding a sixth region, changing the table of distances, and changing the population numbers as follows:
The new population numbers (in 1000’s) for regions 1 through 6 are, respectively, 21, 35, 15, 60, 20, and 37.
I am now much happier with this problem and my students are getting a lot more out of it (I think). At least I can tell you one thing: they’re spending a lot more time thinking about it and asking me intelligent questions. Isn’t that the whole purpose of homework? Maybe they hate me a bit more now, but I don’t mind practicing some tough love.
Feel free to use my modification if you wish. I’d love to see it included in the 7th edition of Cliff’s book.
Note to instructors: if you want to have the solution to the new version of the problem, including the Excel model, just drop me a line: tallys at miami dot edu.
Note to students: to preserve the usefulness of this problem, I cannot provide you with the solution, but if you become an MBA student at the University of Miami, I’ll give you some hints.