Quantitative analysis for management 10e by barry render , stair , hanna SM

Teaching Suggestion 2.7: Expected Value of a Probability Distribution.. The expected value is the average of the distribution and is computed by using the following formula: EX X · PX

Instructor’s Solutions Manual

Quantitative Analysis

for Management

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TEACHINGSUGGESTIONS

Teaching Suggestion 1.1: Importance of Qualitative Factors.

Section 1.2 gives students an overview of quantitative analysis In

this section, a number of qualitative factors, including federal

leg-islation and new technology, are discussed Students can be asked

to discuss other qualitative factors that could have an impact on

quantitative analysis Waiting lines and project planning can be

used as examples

Teaching Suggestion 1.2: Discussing Other Quantitative

Analysis Problems.

Section 1.2 covers an application of the quantitative analysis

ap-proach Students can be asked to describe other problems or areas

that could benefit from quantitative analysis

Teaching Suggestion 1.3: Discussing Conflicting Viewpoints.

Possible problems in the QA approach are presented in this

chap-ter A discussion of conflicting viewpoints within the organization

can help students understand this problem For example, how

many people should staff a registration desk at a university?

Stu-dents will want more staff to reduce waiting time, while university

administrators will want less staff to save money A discussion of

these types of conflicting viewpoints will help students understand

some of the problems of using quantitative analysis

Teaching Suggestion 1.4: Difficulty of Getting Input Data.

A major problem in quantitative analysis is getting proper input

data Students can be asked to explain how they would get the

in-formation they need to determine inventory ordering or carrying

costs Role-playing with students assuming the parts of the analyst

who needs inventory costs and the instructor playing the part of a

veteran inventory manager can be fun and interesting Students

quickly learn that getting good data can be the most difficult part

of using quantitative analysis

Teaching Suggestion 1.5: Dealing with Resistance to Change.

Resistance to change is discussed in this chapter Students can be

asked to explain how they would introduce a new system or

change within the organization People resisting new approaches

can be a major stumbling block to the successful implementation

of quantitative analysis Students can be asked why some people

may be afraid of a new inventory control or forecasting system

SOLUTIONS TODISCUSSIONQUESTIONS

1-1. Quantitative analysis involves the use of mathematical

equations or relationships in analyzing a particular problem In

most cases, the results of quantitative analysis will be one or morenumbers that can be used by managers and decision makers inmaking better decisions Calculating rates of return, financial ra-tios from a balance sheet and profit and loss statement, determin-ing the number of units that must be produced in order to breakeven, and many similar techniques are examples of quantitativeanalysis Qualitative analysis involves the investigation of factors

in a decision-making problem that cannot be quantified or stated

in mathematical terms The state of the economy, current or ing legislation, perceptions about a potential client, and similarsituations reveal the use of qualitative analysis In most decision-making problems, both quantitative and qualitative analysis areused In this book, however, we emphasize the techniques andapproaches of quantitative analysis

pend-1-2. Quantitative analysis is the scientific approach to managerialdecision making This type of analysis is a logical and rational ap-proach to making decisions Emotions, guesswork, and whim arenot part of the quantitative analysis approach A number oforganizations support the use of the scientific approach: the Institutefor Operation Research and Management Science (INFORMS), Decision Sciences Institute, and Academy of Management

1-3. Quantitative analysis is a step-by-step process that allows cision makers to investigate problems using quantitative techniques

de-The steps of the quantitative analysis process include defining theproblem, developing a model, acquiring input data, developing a so-lution, testing the solution, analyzing the results, and implementingthe results In every case, the analysis begins with defining the prob-lem The problem could be too many stockouts, too many bad debts,

or determining the products to produce that will result in the mum profit for the organization After the problems have been de-fined, the next step is to develop one or more models These modelscould be inventory control models, models that describe the debt sit-uation in the organization, and so on Once the models have beendeveloped, the next step is to acquire input data In the inventoryproblem, for example, such factors as the annual demand, the order-ing cost, and the carrying cost would be input data that are used bythe model developed in the preceding step In determining the prod-ucts to produce in order to maximize profits, the input data could besuch things as the profitability for all the different products, theamount of time that is available at the various production depart-ments that produce the products, and the amount of time it takes foreach product to be produced in each production department Thenext step is developing the solution This requires manipulation ofthe model in order to determine the best solution Next, the resultsare tested, analyzed, and implemented In the inventory control

maxi-1

C H A P T E R

Introduction to Quantitative Analysis

problem, this might result in determining and implementing a policy

to order a certain amount of inventory at specified intervals For the

problem of determining the best products to produce, this might

mean testing, analyzing, and implementing a decision to produce a

certain quantity of given products

1-4. Although the formal study of quantitative analysis and the

refinement of the tools and techniques of the scientific method

have occurred only in the recent past, quantitative approaches to

decision making have been in existence since the beginning of

time In the early 1900s, Frederick W Taylor developed the

prin-ciples of the scientific approach During World War II,

quantita-tive analysis was intensified and used by the military Because of

the success of these techniques during World War II, interest

con-tinued after the war

1-5. Model types include the scale model, physical model, and

schematic model (which is a picture or drawing of reality) In this

book, mathematical models are used to describe mathematical

re-lationships in solving quantitative problems

In this question, the student is asked to develop two

mathe-matical models The student might develop a number of models

that relate to finance, marketing, accounting, statistics, or other

fields The purpose of this part of the question is to have the

stu-dent develop a mathematical relationship between variables with

which the student is familiar

1-6. Input data can come from company reports and documents,

interviews with employees and other personnel, direct

measure-ment, and sampling procedures For many problems, a number of

different sources are required to obtain data, and in some cases it is

necessary to obtain the same data from different sources in order to

check the accuracy and consistency of the input data If the input

data are not accurate, the results can be misleading and very costly

to the organization This concept is called “garbage in, garbage out”

1-7. Implementation is the process of taking the solution and

in-corporating it into the company or organization This is the final

step in the quantitative analysis approach, and if a good job is not

done with implementation, all of the effort expended on the

previ-ous steps can be wasted

1-8. Sensitivity analysis and postoptimality analysis allow the

de-cision maker to determine how the final solution to the problem

will change when the input data or the model change This type of

analysis is very important when the input data or model has not

been specified properly A sensitive solution is one in which the

re-sults of the solution to the problem will change drastically or by a

large amount with small changes in the data or in the model When

the model is not sensitive, the results or solutions to the model will

not change significantly with changes in the input data or in the

model Models that are very sensitive require that the input data

and the model itself be thoroughly tested to make sure that both are

very accurate and consistent with the problem statement

1-9. There are a large number of quantitative terms that may not

be understood by managers Examples include PERT, CPM,

simu-lation, the Monte Carlo method, mathematical programming,

EOQ, and so on The student should explain each of the four terms

selected in his or her own words

1-10. Many quantitative analysts enjoy building mathematical

models and solving them to find the optimal solution to a problem

Others enjoy dealing with other technical aspects, for example, dataanalysis and collection, computer programming, or computations

The implementation process can involve political aspects, ing people to trust the new approach or solutions, or the frustrations

convinc-of getting a simple answer to work in a complex environment

Some people with strong analytical skills have weak interpersonalskills; since implementation challenges these “people” skills, it willnot appeal to everyone If analysts become involved with users andwith the implementation environment and can understand “wheremanagers are coming from,” they can better appreciate the difficul-ties of implementing what they have solved using QA

1-11. Users need not become involved in technical aspects of the

QA technique, but they should have an understanding of what the

limitations of the model are, how it works (in a general sense), the jargon involved, and the ability to question the validity andsensitivity of an answer handed to them by an analyst

1-12. Churchman meant that sophisticated mathematical tions and proofs can be dangerous because people may be afraid toquestion them Many people do not want to appear ignorant andquestion an elaborate mathematical model; yet the entire model,its assumptions and its approach, may be incorrect

solu-1-13. The breakeven point is the number of units that must besold to make zero profits To compute this, we must know the sell-ing price, the fixed cost, and the variable cost per unit

1-14. f⫽ 350 s ⫽ 15 v ⫽ 8

a) Total revenue⫽ 20(15) ⫽ $300Total variable cost⫽ 20(8) ⫽ $160b) BEP⫽ f/(s ⫺ v) ⫽ 350/(15 ⫺ 8) ⫽ 50 units

1-18. BEP⫽ f/(s ⫺ v)

500⫽ 1400/(s ⫺ 3) 500(s⫺ 3) ⫽ 1400

Total revenue⫽ 50(96) ⫽ 4800

1-21. f ⫽ 2400 s ⫽ ? v⫽ 25

BEP⫽ f/(s ⫺ v)

120⫽ 2400/(s ⫺ 25) 120(s⫺ 25) ⫽ 2400

s⫽ 45

1-22. f ⫽ 11000 s ⫽ 250 v ⫽ 60

BEP⫽ f/(s ⫺ v) ⫽ 11000/(250 ⫺ 60) ⫽ 57.9

SOLUTION TOFOOD ANDBEVERAGES

ATSOUTHWESTERNUNIVERSITYFOOTBALLGAMES

The total fixed cost per games includes salaries, rental fees, and

cost of the workers in the six booths These are:

Salaries⫽ $20,000

Rental fees⫽ 2,400 ⫻ $2 ⫽ $4,800

Booth worker wages⫽ 6 ⫻ 6 ⫻ 5 ⫻ $7 ⫽ $1,260

Total fixed cost per game⫽ $20,000 ⫹ $4,800 ⫹ $1,260 ⫽ $26,060

The cost of this allocated to each food item is shown in the table:

The break-even points for each of these items are found by

com-puting the contribution to profit (profit margin) for each item and

dividing this into the allocated fixed cost These are shown in the

next table:

To determine the total sales for each item that is required to break

even, multiply the selling price by the break even volume The

results are shown:

Thus, to break even, the total sales must be $43,433.33 If the

at-tendance is 35,000 people, then each person would have to spend

$43,433.33/35,000⫽ $1.24 If the attendance is 60,000, then each

person would have to spend $43,433.33/60,000⫽ $0.72 Both of

these are very low values, so we should be confident that this food

and beverage operation will at least break even

Note: While this process provides information about break-even

points based on the current percent revenues for each product,

there is one difficulty The total revenue using the break-even

points will not result in the same percentages (dollar volume of

product/total revenue) as originally stated in the problem A more

complex model is available to do this (see p 284 Jay Heizer and

Barry Render, Operations Management, 7th ed., Upper Saddle

River, NJ: Prentice Hall, 2004)

TEACHINGSUGGESTIONS

Teaching Suggestion 2.1: Concept of Probabilities Ranging

From 0 to 1.

People often misuse probabilities by such statements as, “I’m

110% sure we’re going to win the big game.” The two basic rules

of probability should be stressed

Teaching Suggestion 2.2: Where Do Probabilities Come From?

Students need to understand where probabilities come from

Sometimes they are subjective and based on personal experiences

Other times they are objectively based on logical observations

such as the roll of a die Often, probabilities are derived from

his-torical data—if we can assume the future will be about the same as

the past

Teaching Suggestion 2.3: Confusion Over Mutually Exclusive

and Collectively Exhaustive Events.

This concept is often foggy to even the best of students—even if

they just completed a course in statistics Use practical examples

and drills to force the point home The table at the end of Example

3 is especially useful

Teaching Suggestion 2.4: Addition of Events That Are Not

Mutually Exclusive.

The formula for adding events that are not mutually exclusive is

P(A or B)  P(A)  P(B)  P(A and B) Students must understand

why we subtract P(A and B) Explain that the intersect has been

counted twice

Teaching Suggestion 2.5: Statistical Dependence with

Visual Examples.

Figure 2.3 indicates that an urn contains 10 balls This example

works well to explain conditional probability of dependent events

An even better idea is to bring 10 golf balls to class Six should be

white and 4 orange (yellow) Mark a big letter or number on each

to correspond to Figure 2.3 and draw the balls from a clear bowl to

make the point You can also use the props to stress how random

sampling expects previous draws to be replaced

Teaching Suggestion 2.6: Concept of Random Variables.

Students often have problems understanding the concept of

ran-dom variables Instructors need to take this abstract idea and

pro-vide several examples to drive home the point Table 2.2 has some

useful examples of both discrete and continuous random variables

Teaching Suggestion 2.7: Expected Value of a

Probability Distribution.

A probability distribution is often described by its mean and

variance These important terms should be discussed with such

practical examples as heights or weights of students But studentsneed to be reminded that even if most of the men in class (or theUnited States) have heights between 5 feet 6 inches and 6 feet 2inches, there is still some small probability of outliers

Teaching Suggestion 2.8: Bell-Shaped Curve.

Stress how important the normal distribution is to a large number

of processes in our lives (for example, filling boxes of cereal with

32 ounces of cornflakes) Each normal distribution depends on themean and standard deviation Discuss Figures 2.8 and 2.9 to showhow these relate to the shape and position of a normal distribution

Teaching Suggestion 2.9: Three Symmetrical Areas Under the

Normal Curve.

Figure 2.10 is very important, and students should be encouraged

to truly comprehend the meanings of 1, 2, and 3 standard tion symmetrical areas They should especially know that man-agers often speak of 95% and 99% confidence intervals, whichroughly refer to 2 and 3 standard deviation graphs Clarify that95% confidence is actually 1.96 standard deviations, while 3standard deviations is actually a 99.7% spread

devia-Teaching Suggestion 2.10: Using the Normal Table to Answer

Probability Questions.

The IQ example in Figure 2.11 is a particularly good way to treatthe subject since everyone can relate to it Students are typicallycurious about the chances of reaching certain scores Go through

at least a half-dozen examples until it’s clear that everyone can

use the table Students get especially confused answering

ques-tions such as P(X  85) since the standard normal table shows

only right-hand-side Z values The symmetry requires special care.

ALTERNATIVEEXAMPLES Alternative Example 2.1: In the past 30 days, Roger’s RuralRoundup has sold either 8, 9, 10, or 11 lottery tickets It never soldfewer than 8 nor more than 11 Assuming that the past is similar tothe future, here are the probabilities:

Alternative Example 2.2: Grades received for a course have aprobability based on the professor’s grading pattern Here are Pro-fessor Ernie Forman’s BA205 grades for the past five years

2

C H A P T E R

Probability Concepts and Applications

P(drawing a 3 from a deck of cards)  4/52  1/13

P(drawing a club on the same draw)  13/52  1/4These are neither mutually exclusive nor collectively exhaustive

Alternative Example 2.4: In Alternative Example 2.3 we

looked at 3s and clubs Here is the probability for 3 or club:

P(3 or club)  P(3)  P(club)  P(3 and club)

 4/52  13/52  1/52

 16/52  4/13

Alternative Example 2.5: A class contains 30 students Ten are

female (F) and U.S citizens (U); 12 are male (M) and U.S

citi-zens; 6 are female and U.S citizens (N); 2 are male and

Alternative Example 2.6: Your professor tells you that if you

score an 85 or better on your midterm exam, there is a 90% chance

you’ll get an A for the course You think you have only a 50%

chance of scoring 85 or better The probability that both your

score is 85 or better and you receive an A in the course is

P(A and 85)  P(A  85)  P(85)  (0.90)(0.50)  0.45

 45%

Alternative Example 2.7: An instructor is teaching two

sec-tions (classes) of calculus Each class has 24 students, and on the

surface, both classes appear identical One class, however, consists

of students who have all taken calculus in high school The

in-structor has no idea which class is which She knows that the

prob-ability of at least half the class getting As on the first exam is

only 25% in an average class, but 50% in a class with more math

background

A section is selected at random and quizzed More than halfthe class received As Now, what is the revised probability that the

class was the advanced one?

P(regular class chosen) 0.5

P(advanced class chosen) 0.5

A probability distribution of the results would be:

Alternative Example 2.9: Here is how the expected outcome

can be computed for the question in Alternative Example 2.8

 x3P(x3) x4P(x4)  x5P(x5)

 5(0.4)  4(0.3)  3(0.2)  2(0.1)  1(0)

 4.0

Alternative Example 2.10: Here is how variance is computed

for the question in Alternative Example 2.8:

1

Alternative Example 2.11: The length of the rods coming out

of our new cutting machine can be said to approximate a normal

distribution with a mean of 10 inches and a standard deviation of

0.2 inch Find the probability that a rod selected randomly will

have a length

a of less than 10.0 inches

b between 10.0 and 10.4 inches

c between 10.0 and 10.1 inches

d between 10.1 and 10.4 inches

e between 9.9 and 9.6 inches

f between 9.9 and 10.4 inches

g between 9.886 and 10.406 inches

First compute the standard normal distribution, the Z-value:

Next, find the area under the curve for the given Z-value by using

a standard normal distribution table

2-1. There are two basic laws of probability First, the probability

of any event or state of nature occurring must be greater than or equal

to zero and less than or equal to 1 Second, the sum of the simple

probabilities for all possible outcomes of the activity must equal 1

5, or 6 These six outcomes are collectively exhaustive becausethey include all possible outcomes Again, it is assumed that thedie will not land and stay on one of its edges

2-3. Probability values can be determined both objectively andsubjectively When determining probability values objectively,some type of numerical or quantitative analysis is used When de-termining probability values subjectively, a manager’s or decisionmaker’s judgment and experience are used in assessing one ormore probability values

2-4. The probability of the intersection of two events is tracted in summing the probability of the two events to avoiddouble counting For example, if the same event is in both of theprobabilities that are to be added, the probability of this event will

be included twice unless the intersection of the two events is tracted from the sum of the probability of the two events

sub-2-5. When events are dependent, the occurrence of one eventdoes have an effect on the probability of the occurrence of theother event When the events are independent, on the other hand,the occurrence of one of them has no effect on the probability ofthe occurrence of the other event It is important to know whether

or not events are dependent or independent because the probabilityrelationships are slightly different in each case In general, theprobability relationships for any kind of independent events aresimpler than the more generalized probability relationships fordependent events

2-6. Bayes’ theorem is a probability relationship that allowsnew information to be incorporated with prior probability values

to obtain updated or posterior probability values Bayes’ theoremcan be used whenever there is an existing set of probability valuesand new information is obtained that can be used to revise theseprobability values

2-7. A Bernoulli process has two possible outcomes, and the probability of occurrence is constant from one trial to the next

If n independent Bernoulli trials are repeated and the number

of outcomes (successes) are recorded, the result is a binomial distribution

2-8. A random variable is a function defined over a sample space

There are two types of random variables: discrete and continuous

The distributions for the price of a product, the number ofsales for a salesperson, and the number of ounces in a food con-tainer are examples of a probability distribution

2-9. A probability distribution is a statement of a probabilityfunction that assigns all the probabilities associated with a randomvariable A discrete probability distribution is a distribution of dis-crete random variables (that is, random variables with a limited set

of values) A continuous probability distribution is concerned with

a random variable having an infinite set of values

a The probability of getting a 4-inch nail is

2-10. The expected value is the average of the distribution and is

computed by using the following formula: E(X)  X · P(X) (this

is for a discrete probability distribution)

2-11. The variance is a measure of the dispersion of the

distribu-tion The variance of a discrete probability distribution is

com-puted by the formula

V  [X  E(X)]2· P(X)

2-12. The purpose of this question is to have students name three

business processes they know that can be described by a normal

distribution Answers could include sales of a product, project

completion time, average weight of a product, and product

de-mand during lead or order time

2-13. This is an example of a discrete probability distribution It

was most likely computed using historical data It is important to

note that it follows the laws of a probability distribution The total

sums to 1, and the individual values are less than or equal to 1

2-14.

2-16. The distribution of chips is as follows:

Green 10White 2

Total  20

a The probability of drawing a white chip on the first draw is

b The probability of drawing a white chip on the first drawand a red one on the second is

=1020

1020



 2 20

820

P W( ) 2   20

=

⎛

⎝⎜ ⎞⎠⎟

0.10 30300

=

⎛

⎝⎜ ⎞⎠⎟

0.30 90300

=

⎛

⎝⎜ ⎞⎠⎟

0.25 75300

=

⎛

⎝⎜ ⎞⎠⎟

0.27 80300

e We first calculate P(TH)  0.25, then calculate P(HT) 

(0.5)(0.5)  0.25 To find the probability of either one

occur-ring, we simply add the two probabilities The solution is 0.50

f At least one head means that we have either HT, TH, or

HH Since each of these have a probability of 0.25, their total

probability of occurring is 0.75 On the other hand, the plement of the outcome “at least one head” is “two tails.”

com-Thus, we could have also computed the probability from 1 

P(TT)  1  0.25  0.75

P(C) 90300

P(of receiving a C) =no students receiving aa C

total no students

c The probability that an employee who is not involved in

an exercise program will get a cold is

 31

d No If they were independent, then

P(C  E)  P(C), but

 0.2Therefore, these events are dependent

2-19. The probability of winning tonight’s game is

 0.6The probability that the team wins tonight is 0.60 The probability

that the team wins tonight and draws a large crowd at tomorrow’s

game is a joint probability of dependent events Let the probability

of winning be P(W) and the probability of drawing a large crowd

being a large crowd at tomorrow’s game is 0.54

2-20. The second draw is not independent of the first because the

probabilities of each outcome depend on the rank (sophomore or

junior) of the first student’s name drawn Let

610

number of winsnumber of games12

20

P C( ),

P C( )Number of people who had colds

Total nummber of employees

2-21. Without any additional information, we assume that there

is an equally likely probability that the soldier wandered into

either oasis, so P(Abu Ilan)  0.50 and P(El Kamin)  0.50.

Since the oasis of Abu Ilan has 20 Bedouins and 20 Farimas (atotal population of 40 tribesmen), the probability of finding aBedouin, given that you are in Abu Ilan, is 20/40  0.50 Simi-larly, the probability of finding a Bedouin, given that you are in ElKamin, is 32/40  0.80 Thus, P(Bedouin  Abu Ilan)  0.50,

P(Bedouin El Kamin)  0.80

We now calculate joint probabilities:

P(Abu Ilan and Bedouin)

 P(Bedouin  Abu Ilan)  P(Abu Ilan)

P(El Kamin  Bedouin)

The probability the oasis discovered was Abu Ilan is now only0.385 The probability the oasis is El Kamin is 0.615

2-22. P(Abu Ilan) is 0.50; P(El Kamin) is 0.50.

P(2 Bedouins  Abu Ilan)  (0.50)(0.50)  0.25

P(2 Bedouins  El Kamin)  (0.80)(0.80)  0.64

P(Abu Ilan and 2 Bedouins)

 P(2 Bedouins  Abu Ilan) P(Abu Ilan)

 (0.25)(0.50)

 0.125

P(El Kamin and 2 Bedouins)

 P(2 Bedouins  El Kamin) P(El Kamin)

Total probability of finding 2 Bedouins is 0.125  0.32  0.445.

P(Abu Ilan  2 Bedouins)

P(El Kamin  2 Bedouins)

These second revisions indicate that the probability that the oasis

was Abu Ilan is 0.281 The probability that the oasis found was El

Kamin is now 0.719

2-23. P(adjusted)  0.8, P(not adjusted)  0.2.

P(pass adjusted)  0.9,

P(pass  not adjusted)  0.2

P(adjusted and pass)

 P(pass  adjusted)  P(adjusted)

 (0.9)(0.8)  0.72

P(not adjusted and pass)

 P(pass  not adjusted)  P(not adjusted)

 (0.2)(0.2)  0.04Total probability that part passes inspection

a The probability that K will win every game is

P  P(K over MB) and P(K over M)

 (0.4)(0.2 )  0.08

b The probability that M will win at least one game is

P(M over K)  P(M over MB)  P(M over K)

 0.12  0.32

 0.44

d Probability 1  winning every game

 1  answer to part (a)

f No They do not appear to be a very good team

2-26. The probability of Dick hitting the bull’s-eye:

P(D)  0.90The probability of Sally hitting the bull’s-eye:

as-2-27. In the sample of 1,000 people, 650 people were fromLaketown and 350 from River City Thirteen of those with cancerwere from Laketown Six of those with cancer were from RiverCity

a The probability of a person from Laketown having cancer:

 0.020

 150

P(cancer|Laketown) 13

650

(MB Mama’s

Boys, K the Killers, and

The probability of a person having cancer:

 0.019Not independent

b I would rather live in River City

P(L) 1  0.22  0.78

2-29. Parts (a) and (c) are probability distributions because the

probability values for each event are between 0 and 1, and the sum

of the probability values for the events is 1

4,300 4,700

4,500 4,700

4,700 4,900

The area to the left of 475 is 0.8413 from Table 2.5, where ␴  1.

The area to the right of 475 is 1 0.8413  0.1587 Thus, the

probability of the oven getting hotter than 475 is 0.1587

To determine the probability of the oven temperature beingbetween 460 and 470, we need to compute two areas

The area between X1and X2is 0.7881  0.6554  0.1327 Thus,

the probability of being between 460 and 470 degrees is  0.1327

2-38. ␮  4,700; ␴  500

470 45025

2025

460 45025

1025

a The sale of 5,500 oranges (X 5,500) is the equivalent of

some Z value which may be obtained from

Z



 1.6The area under the curve lying to the left of 1.6␴  0.94520.

Therefore, the area to the right of 1.6␴  1  0.94520, or 0.0548.

Therefore, the probability of sales being greater than 5,500

oranges is 0.0548

5 500 4 700500,  ,

Xμσ

b

450 X1 X2

4,700 5,500

Z area  0.6554probability  0.6554c

4 500 4 700500

d

4 900 4 700500

Area to the right of 81,000  0.9332, from Table 2.5, where Z 

1.5 Thus, the area to the left of 81,000  1  0.9332  0.0668 

the probability that sales will be fewer than 81,000 packages

2-42. The time to complete the project (X) is normally

distrib-uted with ␮  40 and ␴  5 A penalty must be paid if the project

takes longer than the due date (or if X due date)

a) P(X  40)  1  (X 40)  1  P(Z (40  40)/5)

 1  P(Z 0)  1  0.5  0.5 b) P(X  43)  1  P(X 43)  1  P(Z (43  40)/5)

 1 13

81,000 87,000

454,000 457,000 460,000

␮  457,000

Ninety percent of the time, sales have been between 460,000 and

454,000 pencils This means that 10% of the time sales have

exceeded 460,000 or fallen below 454,000 Since the curve is

symmetrical, we assume that 5% of the area lies to the right of

460,000 and 5% lies to the left of 454,000 Thus, 95% of the area

under the curve lies to the left of 460,000 From Table 2.5, we

note that the number nearest 0.9500 is 0.94950 This corresponds

to a Z value of 1.64 Therefore, we may conclude that the Z value

corresponding to a sale of 460,000 pencils is 1.64

Using Equation 2-12, we get

2-46 Let S steroids present

N steroids not present

TP test is positive for steroids

TN test is negative for steroids

2-47 Let G market is good

P market is poor

PG test predicts good market

PP test predicts poor market

P(G )  0.70 P(P )  0.30

P(PG | G )  0.85 P(PP | G )  0.15

P(PG | P )  0.20 P(PP | P )  0.90

2-48 Let W candidate wins the election

L candidate loses the election

PW poll predicts win

PL poll predicts loss

2-49 Let S successful restaurant

U unsuccessful restaurant

PS model predicts successful restaurant

PU model predicts unsuccessful restaurant

2-50. Let D  Default on loan; D'  No default; R  Loanrejected; R  Loan approved Given:

P(D)  0.2P(D')  0.8P(R | D)  0.9P(R' | D')  0.7(a) P(R | D')  1  0.7  0.3(b)

2-51 (a) F0.05, 5, 10 3.33(b) F0.05, 8.7 3.73(c) F0.05, 3, 5 5.41(d) F0.05, 10 4 5.96

2-52 (a) F0.01, 15, 6 7.56(b) F0.01, 12, 8 5.67(c) F0.01, 3, 5 12.06(d) F0.01, 9, 7 6.72

2-53 (a) From the appendix, P(F3,4  6.59)  0.05, so

P(F  6.8) must be less than 0.05

P D R P R D P D

P R D P D P R D P D

( ' | ) ( | ') ( ')

( | ') ( ') ( | ) ( ) ( )

15.5214

2-54 (a) From the appendix, P(F5,4  15.52)  0.01, so

P(F  14) must be greater than 0.05

This area is 1  0.7257, or 0.2743 Therefore, the probability ofselling more than 265 boats  0.2743.

For a sale of fewer than 250 boats:

X 250

␮  250

␴  25

However, a sale of 250 boats corresponds to ␮  250 At this

point, Z 0 The area under the curve that concerns us is that half

of the area lying to the left of ␮  250 This area  0.5000 Thus,

the probability of selling fewer than 250 boats  0.5

2-57. ␮  0.55 inch (average shaft size)

earlier, the area to the left of ␮ is  0.5000.

We are concerned with the area between ␮ and ␮  1␴ This

is given by the difference between 0.8413 and 0.5000, and it is0.3413 Thus, the probability of a shaft size between 0.55 inch and0.65 inch  0.3413

0 10

0 10

 = 0.55 X = 0.65

0.55 0.65

2-58. Greater than 0.65 inch:

area to the left of 1␴  0.8413

area to the right of 1␴  1  0.8413

 0.1587Thus, the probability of a shaft size being greater than 0.65 inch is0.1587

(b) From the appendix, P(F6,3  27.91)  0.01, so P(F  30) must be less than 0.01

(c) From the appendix, P(F10,12  4.30)  0.01, so P(F  4.2) must be greater than 0.01

(d) From the appendix, P(F2,3  30.82)  0.01, so P(F  35) must be less than 0.01

(e) From the appendix, P(F2,3  30.82)  0.01, so P(F  35) must be greater than 0.01

280 25025



Xμσ

From Table 2.5, the area under the curve corresponding to a Z of

1.20  0.8849 Therefore, the probability that the sales will be less

Z26525025

From Table 2.5, we find that the area under the curve to the left of

Z 0.60 is 0.7257 Since we want to find the probability of selling

more than 265 boats, we need the area to the right of Z 0.60

The shaft size between 0.53 and 0.59 inch:

Since Table 2.5 handles only positive Z values, we need to

calcu-late the probability of the shaft size being greater than 0.55 

0.02  0.57 inch This is determined by finding the area to the left

of 0.57, that is, to the left of 0.2 From Table 2.5, this is 0.5793

The area to the right of 0.2␴ is 1  0.5793  0.4207 The area

to the left of 0.53 is also 0.4207 (the curve is symmetrical) The

area to the left of 0.4␴ is 0.6554 The area between X1and X2is

0.6554  0.4207  0.2347 The probability that the shaft will be

between 0.53 inch and 0.59 inch is 0.2347

0 02

0 10

0 04

0 10

X1μσ

0.53  = 0.55 0.59

0.45 0.55

Thus, we need to find the area to the left of 1␴ Again, since Table

2.5 handles only positive values of Z, we need to determine the

area to the right of 1␴ This is obtained by 1  0.8413  0.1587

(0.8413 is the area to the left of 1␴) Therefore, the area to the left

of 1␴  0.1587 (the curve is symmetrical) Thus, the probability

that the shaft size will be under 0.45 inch is 0.1587

[probability that Marie willwin all four games againstJan]

Probability that Marie will be number one is 04694  003906 .05086

2-60. Probability one will be fined 

= 1−(05)( ) ( )50 55−(15)( ) ( )51 54

4

4 25 4 75 0 003906( )( ) ( )  0464

SOLUTION TOWTVX CASE

1 The chances of getting 15 days of rain during the next 30 dayscan be computed by using the binomial theorem The problem iswell suited for solution by the theorem because there are two andonly two possible outcomes (rain or sun) with given probabilities(70% and 30%, respectively) The formula used is:

Probability of r successes

where

n the number of trials (in this case, the number of days  30),

r the number of successes (number of rainy days  15),

p probability of success (probability of rain  70%), and

q probability of failure (probability of sun  30%)

The probability of getting exactly 15 days of rain in the next 30days is 1.06%

2 Joe’s assumptions concerning the weather for the next 30 daysstate that what happens on one day is not in any way dependent onwhat happened the day before; what this says, for example, is that

if a cold front passed through yesterday, it will not affect whathappens today

But there are perhaps certain conditional probabilities ated with the weather (for example, given that it rained yesterday,the probability of rain today is 80% as opposed to 70%) Not beingfamiliar with the field of meteorology, we cannot say preciselywhat these are However, our contention is that these probabilities

associ-do exist and that Joe’s assumptions are fallacious

TEACHING SUGGESTIONS

Teaching Suggestion 3.1: Using the Steps

of the Decision-Making Process.

The six steps used in decision theory are discussed in this chapter

Students can be asked to describe a decision they made in the last

semester, such as buying a car or selecting an apartment, and

de-scribe the steps that they took This will help in getting

stu-dents involved in decision theory It will also help them realize

how this material can be useful to them in making important

per-sonal decisions

Teaching Suggestion 3.2: Importance of Defining the Problem

and Listing All Possible Alternatives.

Clearly defining the problem and listing the possible alternatives can

be difficult Students can be asked to do this for a typical

decision-making problem, such as constructing a new manufacturing plant

Role-playing can be used to make this exercise more interesting

Many students get too involved in the mathematical proaches and do not pay enough attention to the importance of

ap-carefully defining the problem and considering all possible

alter-natives These initial steps are important Students need to realize

that if they do not carefully define the problem and list all

alterna-tives, most likely their analyses will be wrong

Teaching Suggestion 3.3: Categorizing Decision-Making Types.

Decision-making types are discussed in this chapter; decision

making under certainty, risk, and uncertainty are included

Stu-dents can be asked to describe an important decision they had to

make in the past year and categorize the decision type A good

ex-ample can be a financial investment of $1,000 In-class discussion

can help students realize the importance of decision theory and its

potential use

Teaching Suggestion 3.4: Starting the EVPI Concept.

The material on the expected value of perfect information (EVPI)

can be started with a discussion of how to place a value on

infor-mation and whether or not new inforinfor-mation should be acquired

The use of EVPI to place an upper limit on what you should pay

for information is a good way to start the section on this topic

Teaching Suggestion 3.5: Starting the Decision-Making

Under Uncertainty Material.

The section on decision-making under uncertainty can be started

with a discussion of optimistic versus pessimistic decision makers

Students can be shown how maximax is an optimistic approach,

while maximin is a pessimistic decision technique While few

peo-ple use these techniques to solve real problems, the concepts and

general approaches are useful

Teaching Suggestion 3.6: Decision Theory

and Life-Time Decisions.

This chapter investigates large and complex decisions Duringone’s life, there are a few very important decisions that have amajor impact Some call these “life-time decisions.” Students can

be asked to carefully consider these life-time decisions and howdecision theory can be used to assist them Life-time decisions in-clude decisions about what school to attend, marriage, and the first job

Teaching Suggestion 3.7: Popularity of Decision Trees

Among Business Executives.

Stress that decision trees are not just an academic subject; they are

a technique widely used by top-level managers Everyone ates a graphical display of a tough problem It clarifies issues andmakes a great discussion base Harvard business students regularlyuse decision trees in case analysis

appreci-Teaching Suggestion 3.8: Importance of Accurate

Tree Diagrams.

Developing accurate decision trees is an important part of thischapter Students can be asked to diagram several decision situa-tions The decisions can come from the end-of-chapter problems,the instructor, or from student experiences

Teaching Suggestion 3.9: Diagramming a Large Decision

Problem Using Branches.

Some students are intimidated by large and complex decisiontrees To avoid this situation, students can be shown that a largedecision tree is like having a number of smaller trees or decisionsthat can be solved separately, starting at the end branches of thetree This can help students use decision-making techniques onlarger and more complex problems

Teaching Suggestion 3.10: Using Tables to Perform

Bayesian Analysis.

Bayesian analysis can be difficult; the formulas can be hard to remember and use For many, using tables is the most effectiveway to learn how to revise probability values Once students un-derstand how the tables are used, they can be shown that the for-mulas are making exactly the same calculations

ALTERNATIVEEXAMPLES Alternative Example 3.1: Goleb TransportGeorge Goleb is considering the purchase of two types of industrialrobots The Rob1 (alternative 1) is a large robot capable of perform-ing a variety of tasks, including welding and painting The Rob2 (al-ternative 2) is a smaller and slower robot, but it has all the capabilities

3

C H A P T E R

Decision Analysis

of Rob1 The robots will be used to perform a variety of repair

opera-tions on large industrial equipment Of course, George can always do

nothing and not buy any robots (alternative 3) The market for the

re-pair could be either favorable (event 1) or unfavorable (event 2)

George has constructed a payoff matrix showing the expected returns

of each alternative and the probability of a favorable or unfavorable

market The data are presented below:

This problem can be solved using expected monetary value The

equations are presented below:

EMV (alternative 1)  ($50,000)(0.6)  ($40,000)(0.4)

 $14,000EMV (alternative 2)  ($30,000)(0.6)  ($20,000)(0.4)

 $10,000EMV (alternative 3)  0

The best solution is to purchase Rob1, the large robot

Alternative Example 3.2: George Goleb is not confident about

the probability of a favorable or unfavorable market (See

Alterna-tive Example 3.1.) He would like to determine the equally likely

(Laplace), maximax, maximin, coefficient of realism (Hurwicz), and

minimax regret decisions The Hurwicz coefficient should be 0.7

The problem data are summarized below:

The Laplace (equally likely) solution is computed averaging the

payoffs for each alternative and choosing the best The results are

shown below Alternatives 1 and 2 both give the highest average

return of $5,000

Average (alternative 1)  [$50,000  ($40,000)]/2

 $5,000Average (alternative 2)  [$30,000  ($20,000)]/2

 $5,000Average (alternative 3)  0

The maximin decision (pessimistic) maximizes the minimum

pay-off outcome for every alternative: these are 40,000; 20,000;

and 0 Therefore, the decision is to do nothing

The maximax decision (optimistic) maximizes the maximum

payoff for any alternative: these maximums are 50,000; 30,000;

and 0 Therefore, the decision is to purchase the large robot

The Hurwicz approach uses a coefficient of realism value of0.7, and a weighted average of the best and the worst payoffs foreach alternative is computed The results are as follows:

Weighted average (alternative 1)  ($50,000)(0.7)

 ($40,000)(0.3)

 $23,000Weighted average (alternative 2)  ($30,000)(0.7)

 ($20,000)(0.3)

 $15,000Weighted average (alternative 3)  0The decision would be alternative 1

The minimax regret decision minimizes the maximum tunity loss The opportunity loss table for Goleb is as follows:

Alternative Example 3.3: George Goleb is considering the sibility of conducting a survey on the market potential for indus-trial equipment repair using robots The cost of the survey is

pos-$5,000 George has developed a decision tree that shows the all decision, as in the figure on the next page

over-This problem can be solved using EMV calculations Westart with the end of the tree and work toward the beginning com-puting EMV values The results of the calculations are shown inthe tree The conditional payoff of the solution is $18,802

Alternative Example 3.4: George (in Alternative Example 3.3)would like to determine the expected value of sample information(EVSI) EVSI is equal to the expected value of the best decisionwith sample information, assuming no cost to gather it, minus theexpected value of the best decision without sample information

Because the cost of the survey is $5,000, the expected value of thebest decision with sample information, assuming no cost to gather

it, is $23,802 The expected value of the best decision withoutsample information is found on the lower branch of the decisiontree to be $14,000 Thus, EVSI is $9,802

Alternative Example 3.5: This example reveals how the

condi-tional probability values for the George Goleb examples (above)have been determined The probability values about the survey aresummarized in the following table:

survey result The calculations are presented in the following

two tables

Probability revision given a positive survey result

Is Mark still a risk seeker?

U($10,000)  0.8U($0)  0.9U($10,000)  1Using the data above, we can determine the expected utility ofeach alternative as follows:

U(Mark plays the game)  0.45(1)  0.55(0.8)  0.89U(Mark doesn’t play the game)  0.9

Thus, the best decision for Mark is not to play the game with anexpected utility of 0.9 Given these data, Mark is a risk avoider

FirstDecisionPoint

SecondDecisionPoint

ResultsFavorable

ResultsNegative

Conduct Market Survey

Do Not Conduct Survey

Rob1Rob2

Favorable Market (0.871)Unfavorable Market (0.129)Favorable Market (0.871)Unfavorable Market (0.129)

$45,000–$45,000

–$5,000

$25,000–$25,000

Rob1Rob2

Favorable Market (0.158)Unfavorable Market (0.842)Favorable Market (0.158)Unfavorable Market (0.842)

$45,000–$45,000

–$5,000

$25,000–$25,000

Rob1Rob2

Favorable Market (0.60)Unfavorable Market (0.40)Favorable Market (0.60)Unfavorable Market (0.40)

$50,000–$40,000

$30,000–$20,000

$0

$ –5,000

$18,802

$14,000

Figure for Alternative Example 3.3

Probability given a negative survey result

3-10. The purpose of Bayesian analysis is to determine rior probabilities based on prior probabilities and new information.

poste-Bayesian analysis can be used in the decision-making processwhenever additional information is gathered This information canthen be combined with prior probabilities in arriving at posteriorprobabilities Once these posterior probabilities are computed,they can be used in the decision-making process as any other prob-ability value

3-11. The expected value of sample information (EVSI) is theincrease in expected value that results from having sample infor-mation It is computed as follows:

EVSI  (expected value with sample information)

 (cost of information)  (expected value withoutsample information)

3-12. The overall purpose of utility theory is to incorporate a cision maker’s preference for risk in the decision-making process

de-3-13. A utility function can be assessed in a number of differentways A common way is to use a standard gamble With a standardgamble, the best outcome is assigned a utility of 1, and the worstoutcome is assigned a utility of 0 Then, intermediate outcomes areselected and the decision maker is given a choice between havingthe intermediate outcome for sure and a gamble involving the bestand worst outcomes The probability that makes the decision makerindifferent between having the intermediate outcome for sure and agamble involving the best and worst outcomes is determined Thisprobability then becomes the utility of the intermediate value Thisprocess is continued until utility values for all economic conse-quences are determined These utility values are then placed on autility curve

3-14. When a utility curve is to be used in the decision-makingprocess, utility values from the utility curve replace all monetaryvalues at the terminal branches in a decision tree or in the body of

a decision table Then, expected utilities are determined in thesame way as expected monetary values The alternative with thehighest expected utility is selected as the best decision

3-15. A risk seeker is a decision maker who enjoys and seeksout risk A risk avoider is a decision maker who avoids risk even ifthe potential economic payoff is higher The utility curve for a riskseeker increases at an increasing rate The utility curve for a riskavoider increases at a decreasing rate

3-16. a Decision making under uncertainty

3-17. Using the maximin criterion, the best alternative is theTexan (see table above) because the worst payoff for this ($18,000) is better than the worst payoffs for the other decisions

3-18. a Decision making under risk—maximize expectedmonetary value

SOLUTIONS TODISCUSSIONQUESTIONS

3-1. The purpose of this question is to make students use a

per-sonal experience to distinguish between good and bad decisions

A good decision is based on logic and all of the available

informa-tion A bad decision is one that is not based on logic and the

avail-able information It is possible for an unfortunate or undesiravail-able

outcome to occur after a good decision has been made It is also

possible to have a favorable or desirable outcome occur after a bad

decision

3-2. The decision-making process includes the following steps:

(1) define the problem, (2) list the alternatives, (3) identify the

pos-sible outcomes, (4) evaluate the consequences, (5) select an

evalua-tion criterion, and (6) make the appropriate decision The first four

steps or procedures are common for all decision-making problems

Steps 5 and 6, however, depend on the decision-making model

3-3. An alternative is a course of action over which we have

complete control A state of nature is an event or occurrence in

which we have no control An example of an alternative is

decid-ing whether or not to take an umbrella to school or work on a

par-ticular day An example of a state of nature is whether or not it

will rain on a particular day

3-4. The basic differences between decision-making models

under certainty, risk, and uncertainty depend on the amount of

chance or risk that is involved in the decision A decision-making

model under certainty assumes that we know with complete

confi-dence the future outcomes Decision-making-under-risk models

assume that we do not know the outcomes for a particular decision

but that we do know the probability of occurrence of those

out-comes With decision making under uncertainty, it is assumed that

we do not know the outcomes that will occur, and furthermore, we

do not know the probabilities that these outcomes will occur

3-5. The techniques discussed in this chapter used to solve

deci-sion problems under uncertainty include maximax, maximin, equally

likely, coefficient of realism, and minimax regret The maximax

decision-making criterion is an optimistic decision-making criterion,

while the maximin is a pessimistic decision-making criterion

3-6. For a given state of nature, opportunity loss is the difference

between the payoff for a decision and the best possible payoff for

that state of nature It indicates how much better the payoff could

have been for that state of nature The minimax regret and the

mini-mum expected opportunity loss are the criteria used with this

3-7. Alternatives, states of nature, probabilities for all states of

nature and all monetary outcomes (payoffs) are placed on the

deci-sion tree In addition, intermediate results, such as EMVs for

mid-dle branches, can be placed on the decision tree

3-8. Using the EMV criterion with a decision tree involves

starting at the terminal branches of the tree and working toward

the origin, computing expected monetary values and selecting the

best alternatives The EMVs are found by multiplying the

proba-bilities of the states of nature times the economic consequences

and summing the results for each alternative At each decision

point, the best alternative is selected

3-9. A prior probability is one that exists before additional

in-formation is gathered A posterior probability is one that can be

computed using Bayes Theorem based on prior probabilities and

additional information

b EMV (Sub 100)  0.7(300,000)  0.3(–200,000)

 150,000EMV (Oiler J)  0.7(250,000)  0.3(–100,000)

 145,000EMV (Texan)  0.7(75,000)  0.3(–18,000)

 47,100Optimal decision: Sub 100

c Ken would change decision if EMV(Sub 100) is lessthan the next best EMV, which is $145,000 Let X payoff for Sub 100 in favorable market

(0.7)(X)  (0.3)(200,000)  145,0000.7X  145,000  60,000  205,000

X  (205,000)/0.7  292,857.14The decision would change if this payoff were less than 292,857.14,

so it would have to decrease by about $7,143

3-19. a The expected value (EV) is computed for each

alternative

EV(stock market)  0.5(80,000)  0.5(20,000)  30,000EV(Bonds)  0.5(30,000)  0.5(20,000)  25,000EV(CDs)  0.5(23,000)  0.5(23,000)  23,000Therefore, he should invest in the stock market

b EVPI  EV(with perfect information)

 (Maximum EV without P, I)

 [0.5(80,000)  0.5(23,000)]  30,000

 51,500  30,000  21,500Thus, the most that should be paid is $21,500

3-20. The opportunity loss table is

b Best decision: deposit $10,000 in bank

3-22. a Expected value with perfect information is1,400(0.4)  900(0.4)  900(0.2)  1,100; the maxi-mum EMV without the information is 900 Therefore,Allen should pay at most EVPI  1,100 – 900  $200

b Yes, Allen should pay [1,100(0.4)  900(0.4) 900(0.2)] 900  $80

3-23. a Opportunity loss table

3-26. Cost of produced case  $5

Cost of purchased case  $16

Selling price  $15

b Produce 300 cases each day.

3-27. a The table presented is a decision table The basis for

the decisions in the following questions is shown in thetable below

b Maximax decision: Very large station

c Maximin decision: Small station

d Equally likely decision: Very large station

e Criterion of realism decision: Very large station

f Opportunity loss table:

$0

Payoff1

3-28. EMV for node 1  0.5(100,000)  0.5(40,000) 

$30,000 Choose the highest EMV, therefore construct the clinic

Money recovered from each unsold case  $3

E QUALLY C RIT OF

3-29. a.

Favorable Market (0.82)Unfavorable Market (0.18)

$95,000–$45,000

Survey Negative (0.

45)

Conduct Market Surv ey

$69,800

Favorable Market (0.11)Unfavorable Market (0.89)

$95,000–$45,000

–$5,000CONSTRUCT

$100,000–$40,000

$0CONSTRUCT CLINIC

 10,450  $40,050  –$29,600EMV(node 4)  $30,000

EMV(node 1)  (0.55)($69,800)  (0.45)(–$5,000)

 38,390  2,250  $36,140The EMV for using the survey  $36,140

EMV(no survey)  (0.5)($100,000)  (0.5)(–$40,000)

 $30,000The survey should be used

c EVSI  ($36,140  $5,000)  $30,000  $11,140

Thus, the physicians would pay up to $11,140 for the survey

Favorable MarketUnfavorable Market2

Favorable MarketUnfavorable Market3

Large Shop

No Shop

Small Shop

Favorable MarketUnfavorable Market4

Favorable MarketUnfavorable Market5

Large Shop

No Shop

Small Shop

Favorable MarketUnfavorable Market6

Favorable MarketUnfavorable Market7

Large Shop

No Shop

Small Shop

Favorable Survey

UnfavorableSurvey

No Survey

Market Survey

1

3-31.

a EMV(node 2)  (0.9)(55,000)  (0.1)(–$45,000)

 49,500  4,500  $45,000EMV(node 3)  (0.9)(25,000)  (0.1)(–15,000)

 22,500  1,500  $21,000EMV(node 4)  (0.12)(55,000)  (0.88)(–45,000)

 6,600  39,600  –$33,000EMV(node 5)  (0.12)(25,000)  (0.88)(–15,000)

 3,000  13,200  –$10,200EMV(node 6)  (0.5)(60,000)  (0.5)(–40,000)

 30,000  20,000  $10,000EMV(node 7)  (0.5)(30,000)  (0.5)(–10,000)

 15,000  5,000  $10,000EMV(node 1)  (0.6)(45,000)  (0.4)(–5,000)

 27,000  2,000  $25,000Since EMV(market survey) > EMV(no survey), Jerry should con-

duct the survey Since EMV(large shop | favorable survey) is

larger than both EMV(small shop | favorable survey) and EMV(no

shop | favorable survey), Jerry should build a large shop if the

sur-vey is favorable If the sursur-vey is unfavorable, Jerry should build

nothing since EMV(no shop | unfavorable survey) is larger than

both EMV(large shop | unfavorable survey) and EMV(small shop |

unfavorable survey)

Favorable Market (0.9)

Unfavorable Market (0.1) 2

Favorable Market (0.9)

Unfavorable Market (0.1) 3

Favorable Market (0.12)

Unfavorable Market (0.88) 5

Favorable Market (0.5) Unfavorable Market (0.5) 7

Large Shop

No Shop

Small Shop

FavorableSurvey (0.6)

Unfavorable Survey (0.4)

No Survey

MarketSurvey

–$15,000

$0

$30,000

$60,000 –$40,000

This becomes:

P(45,000)  (1  P)(–5,000) ≥ $10,000Solving gives

45,000P  5,000  5,000P ≥ 10,00050,000P≥ 15,000

P≥ 0.3Thus, the probability of a favorable survey could be as low as0.3 Since the marketing professor estimated the probability

at 0.6, the value can decrease by 0.3 without causing Jerry tochange his decision Jerry’s decision is not very sensitive tothis probability value

InformationUnfavorable

(0.5)

A1

Gather MoreInformation

A 2

Do Not Ga

ther More Info

(0.1) (0.9)

Payoff

$12,000 –$23,000

$2,000 –$13,000 –$3,000 3

(0.6) (0.4)

$12,000 –$23,000

$2,000 –$13,000

–$3,000 5

(0.3) (0.7)

$15,000 –$20,000

$5,000 –$10,000

$0 7

A1: gather more information

A2: do not gather more information

Decisions: do not gather information; build quadplex

3-33. I1: favorable research or information

I2: unfavorable research

S1: store successful

S2: store unsuccessful

P(S1)  0.5; P(S2)  0.5P(I1| S1)  0.8; P(I2| S1)  0.2P(I1| S2)  0.3; P(I2| S2)  0.7

a P(successful store | favorable research)  P(S1| I1)

b P(successful store | unfavorable research)  P(S1| I2)

3-34. I1: favorable survey or information

I2: unfavorable survey

S1: facility successful

S2: facility unsuccessfulP(S1)  0.3; P(S2)  0.7P(I1| S1)  0.8; P(I2| S1)  0.2P(I1| S2)  0.3; P(I2| S2)  0.7P(successful facility | favorable survey)  P(S1| I1)

P(successful facility | unfavorable survey)  P(S1| I2)

Good economy 0.2Fair economy 0.3Poor economy 0.5

10,0002,000–5,000

6,0004,0000

b EMV(A)  10,000(0.2)  2,000(0.3)

 (5,000)(0.5)  100EMV(B)  6,000(0.2)  4,000(0.3)  0(0.5)

 2,400Fund B should be selected

c Let X  payout for Fund A in a good economy

EMV(A)  EMV(B)X(0.2)  2,000(0.3)  (–5,000)(0.5)  2,4000.2X  4,300

X  4,300/0.2  21,500Therefore, the return would have to be $21,500 for Fund A in agood economy for the two alternatives to be equally desirablebased on the expected values

3-35. a

3-36. a.

3

1

Survey Favorable

Survey Unfavorable

Produce Razor

Do Not Produce Razor

Favorable Market

Unfavorable Market

4 Produce Razor

Do Not Produce Razor

Study Unfavorable

Produce Razor

Do Not Produce Razor

Favorable Market

Unfavorable Market

6 Produce Razor

Do Not Produce Razor

Favorable Market

Unfavorable Market

7 Produce Razor

Do Not Produce Razor

Payoff

$95,000 –$65,000 –$5,000

$95,000 –$65,000 –$5,000

$80,000 –$80,000 –$20,000

$80,000 –$80,000 –$20,000

$100,000 –$60,000

$0 Study

b S1: survey favorable S2: survey unfavorable S3: study favorable S4: study unfavorable S5: market favorable S6: market unfavorable

 51,200EMV(node 7)  100,000(0.5)  (60,000)(0.5)  20,000EMV(conduct survey)  59,800(0.45)  (–5,000)(0.55)

 24,160EMV(conduct pilot study)  62,400(0.45)  (20,000)(0.55)

 17,080EMV(neither)  20,000

Therefore, the best decision is to conduct the survey If it is able, produce the razor If it is unfavorable, do not produce the razor

favor-3-37. The following computations are for the decision tree thatfollows

EU(node 3)  0.95(0.78)  0.5(0.22)  0.85EU(node 4)  0.95(0.27)  0.5(0.73)  0.62EU(node 5)  0.9(0.89)  0(0.11)  0.80EU(node 6)  0.9(0.18)  0(0.82)  0.16EU(node 7)  1(0.5)  0.55(0.5)  0.78EU(conduct survey)  0.85(0.45)  0.8(0.55)  0.823EU(conduct pilot study)  0.80(0.45)  0.7(0.55)  0.745EU(neither test)  0.81

Therefore, the best decision is to conduct the survey Jim is a riskavoider

1

Survey

Produce Razor

Do Not Produce Razor

Market Favorable (0.78)

Market Unfavorable (0.22)

4 Produce Razor

Do Not Produce Razor

Do Not Produce Razor

Market Favorable (0.89)

Market Unfavorable (0.11)

6 Produce Razor

Do Not Produce Razor

Market Favorable (0.18)

Market Unfavorable (0.82)

7 Produce Razor

Do Not Produce Razor

Utility 0.95 0.5 0.8 0.95 0.5 0.8 0.9 0 0.7

0.9 0 0.7

1 0.55 0.81

Favorable (0.45)

Survey Unfavorable (0.55)

Study Favorable (0.45)

Study Unfavorable (0.55)

3-38. a P(good economy | prediction of

good economy) P(poor economy | prediction of good economy) 

P(good economy | prediction ofpoor economy) 

P(poor economy | prediction ofpoor economy) 

b P(good economy | prediction of good economy) 

P(poor economy | prediction ofgood economy) 

P(good economy | prediction ofpoor economy)  0 2 0 7

0 2 0 7 0 9 0 3 0 341

( ) ( )+ ( )= .

0 1 0 3

0 8 0 7 0 1 0 3 0 051

( ) ( )+ ( )= .

0 8 0 7

0 8 0 7 0 1 0 3 0 949

( ) ( )+ ( )= .

0 9 0 6

0 2 0 6 0 9 0 4 0 75

( ) ( )+ ( )= .

0 2 0 6

0 2 0 6 0 9 0 4 0 25

( ) ( )+ ( )= .

0 1 0 4

0 8 0 6 0 1 0 4 0 077

( ) ( )+ ( )= .

0 8 0 6

0 8 0 6 0 1 0 4 0 923

( ) ( )+ ( )= .

P(poor economy | prediction ofpoor economy) 

3-39. The expected value of the payout by the insurance pany is

com-EV  0(0.999)  100,000(0.001)  100The expected payout by the insurance company is $100, but thepolicy costs $200, so the net gain for the individual buying thispolicy is negative (–$100) Thus, buying the policy does not maxi-mize EMV since not buying this policy would have an EMV of 0,which is better than –$100 However, a person who buys this pol-icy would be maximizing the expected utility The peace of mindthat goes along with the insurance policy has a relatively high util-ity A person who buys insurance would be a risk avoider

0 9 0 3

0 2 0 7 0 9 0 3 0 659

( ) ( )+ ( )= .

Do Not Construct Clinic

Favorable Market (0.82)

Unfavorable Market (0.18)

3 Construct Clinic

Do Not Construct Clinic

0.99 0 0.7 0.99 0 0.7

Favorable (0.55)

Survey Unfavorable (0.45)

U = 0.76

U = 0.8118

U = 0.1089

4 Construct Clinic

Do Not Construct Clinic

Favorable Market (0.5)

Unfavorable Market (0.5)

1.0 0.1 0.9

U = 0.55

Utility

$95,000 –$45,000 –$5,000

$95,000 –$45,000 –$5,000

$100,000 –$40,000

$0 Payoff

The expected utility with no survey (0.9) is higher than the

ex-pected utility with a survey (0.7615), so the survey should be not

used The medical professionals are risk avoiders

3-41. EU(large plant | survey favorable)  0.78(0.95)

 0.22(0)  0.741EU(small plant | survey favorable)  0.78(0.5)  0.22(0.1)

 0.412EU(no plant | survey favorable)  0.2

EU(large plant | survey negative)  0.27(0.95)  0.73(0)

 0.2565EU(small plant | survey negative)  0.27(0.5)  0.73(0.10)

 0.208EU(no plant | survey negative)  0.2

EU(large plant | no survey)  0.5(1)  0.5(0.05)  0.525

EU(small plant | no survey)  0.5(0.6)  0.5(0.15)  0.375

EU(no plant | no survey)  0.3

EU(conduct survey)  0.45(0.741)  0.55(0.2565)  0.4745

EU(no survey)  0.525

John’s decision would change He would not conduct the survey

and build the large plant

3-42. a Expected travel time on Broad Street  40(0.5) 

15(0.5)  27.5 minutes Broad Street has a lower pected travel time

ex-Congestion (0.5)

NoCongestion (0.5)1

Expressway

BroadStreet

c Lynn is a risk avoider

1.0 0.8 0.6 0.4 0.2 0

Time (minutes)

3-43. Selling price  $20 per gallon; manufacturing cost 

$12 per gallon; salvage value  $13; handling costs  $1 pergallon; and advertising costs  $3 per gallon From this informa-tion, we get:

marginal profit  selling price minus the manufacturing, handling,and advertising costs

marginal profit  $20  $12  $1  $3  $4 per gallon

If more is produced than is needed, a marginal loss is incurred

marginal loss  $13  $12  $1  $3  $3 per gallon

In addition, there is also a shortage cost Coren has agreed to fulfillany demand that cannot be met internally This requires that Corenpurchase chemicals from an outside company Because the cost ofobtaining the chemical from the outside company is $25 and theprice charged by Coren is $20, this results in

shortage cost  $5 per gallon

In other words, Coren will lose $5 for every gallon that is sold thathas to be purchased from an outside company due to a shortage

a A decision tree is shown below:

b The computations are shown in the following table These

numbers are entered into the tree above The best decision is to

Probabilities 0.2 0.3 0.4 0.1

c EVwPI  (0.2)(2,000)  (0.3)(4,000)  (0.4)(6,000)

 (0.1)(8,000)  $4,800EVPI  EVwPI  EMV  $4,800  $3,300  $1,500

3-44. If no survey is to be conducted, the decision tree is fairly

straightforward There are three main decisions, which are

build-ing a small, medium, or large facility Extendbuild-ing from these

decision branches are three possible demands, representing the

possible states of nature The demand for this type of facility could

be either low (L), medium (M), or high (H) It was given in the

problem that the probability for a low demand is 0.15 The

proba-bilities for a medium and a high demand are 0.40 and 0.45,

respec-tively The problem also gave monetary consequences for building

a small, medium, or large facility when the demand could be low,

medium, or high for the facility These data are reflected in the

fol-lowing decision tree

Stock 500

Stock 1,000

Stock 2,000

Stock 1,500

(0.2) (0.3) (0.4) (0.1) (0.2) (0.3) (0.4) (0.1)

500 1,000 1,500 2,000

500 1,000 1,500 2,000

500 1,000 1,500 2,000 500 1,000 1,500 2,000

$2,000 = (500)(4) –$500 = (500)(4) – (500)(5)

$500 = (500)(4) – (500)(3)

$4,000 = (1,000)(4)

$1,500 = (1,000)(4) – (5)(500) –$1,000 = (1,000)(4) – (5)(1,000)

Decision Tree–No Survey

Small $500,000

Large $580,000

Medium $670,000

(0.15) (0.40) (0.45) (0.15) (0.40) (0.45) (0.15) (0.40) (0.45)

With no survey, we have: EMV(Small)  500,000;

EMV(Medium)  670,000; and EMV(Large)  580,000

The medium facility, with an expected monetary value of

$670,000, is selected because it represents the highest pected monetary value

ex-If the survey is used, we must compute the revised ties using Bayes’ theorem For each alternative facility, threerevised probabilities must be computed, representing low,medium, and high demand for a facility These probabilitiescan be computed using tables One table is used to compute theprobabilities for low survey results, another table is used for

Medium

Large

LMHLMHLMH

Small

Medium

Large

LMHLMHLMH

Small

Medium

Large

LMHLMHLMH

450,000450,000450,000150,000650,000750,000–250,000350,000950,000

450,000450,000

650,000750,000150,000

350,000950,000–250,000

450,000450,000450,000

650,000750,000150,000

350,000950,000–250,000

450,000Decision Tree–Survey

$495,000 Low (0.310)

$821,000High (0.3

25)

$646,000Medium(0.365)

medium survey results, and a final table is used for high survey

re-sults These tables are shown below These probabilities will be used

in the decision tree that follows

For low survey results—A1:

P(A1)  0.310For medium survey results—A2:

P(A2)  0.365For high survey results—A3:

P(A3)  0.325When survey results are low, the probabilities are P(L) 

0.339; P(M)  0.516; and P(H)  0.145 This results in

EMV(Small)  450,000; EMV(Medium)  495,000; and

EMV(Large)  233,600

When survey results are medium, the probabilities are P(L) 

0.082; P(M)  0.548; and P(H)  0.378 This results in EMV

(Small)  450,000; EMV(Medium)  646,000; and EMV(Large) 

522,800

When survey results are high, the probabilities are P(L) 

0.046; P(M)  0.123; and P(H)  0.831 This results in

EMV(Small)  450,000; EMV(Medium)  710,100; and

EMV(Large)  821,000

If the survey results are low, the best decision is to build the

medium facility with an expected return of $495,000 If the survey

results are medium, the best decision is also to build the medium

plant with an expected return of $646,000 On the other hand, if

the survey results are high, the best decision is to build the large

facility with an expected monetary value of $821,000 The

ex-pected value of using the survey is computed as follows:

EMV(with Survey)  0.310(495,000)  0.365(646,000)

 0.325(821,000)  656,065Because the expected monetary value for not conducting the sur-

vey is greater (670,000), the decision is not to conduct the survey

and to build the medium-sized facility

3-45. a.

Mary should select the traffic circle location (EMV  $250,000)

b Use Bayes’ Theorem to compute posterior probabilities

P(SD | SRP) = 0.78; P(SD ¯¯¯¯ | SRP) = 0.22 P(SM | SRP) = 0.84; P(SM ¯¯¯¯¯ | SRP) = 0.16 P(SC | SRP) = 0.91; P(SC ¯¯¯¯ | SRP) = 0.09 P(SD | SRN) = 0.27; P(SD ¯¯¯¯ | SRN) = 0.73 P(SM | SRN) = 0.36; P(SM ¯¯¯¯¯ | SRN) = 0.64 P(SC | SRN) = 0.53; P(SC ¯¯¯¯ | SRN) = 0.47

Example computations:

These calculations are for the tree that follows:

EMV(2)  $171,600  $28,600  $143,000EMV(3)  $226,800  $20,800  $206,000EMV(4)  $336,700  $20,700  $316,000EMV(no grocery A)  –$30,000

EMV(5)  $59,400  $94,900  –$35,500EMV(6)  $97,200  $83,200  $14,000EMV(7)  $196,100  $108,100  $88,000EMV(no grocery B)  –$30,000

 $224,800EMV(D)  (best of two alternatives)

Decision

Point

Second Decision Point

Purchase MarketSurvey

Survey ResultsPositive (0.6)Survey ResultsNegative (0.4)

Do Not Purchase Market Survey

$220,000 –$130,000

–$230,000

$270,000 –$130,000

$370,000

–$30,000

$220,000 –$130,000

–$230,000

$270,000 –$130,000

$370,000

–$30,000

$250,000 –$100,000

–$200,000

$300,000 –$100,000

3-46. a Sue can use decision tree analysis to find the best

solu-tion The results are presented below In this case, the best decision

is to get information If the information is favorable, she should

build the retail store If the information is not favorable, she should

not build the retail store The EMV for this decision is $29,200

In the following results (using QM for Windows), Branch 1

(1–2) is to get information, Branch 2 (1–3) is the decision to not get

information, Branch 3 (2–4) is favorable information, Branch 4

(2–5) is unfavorable information, Branch 5 (3–8) is the decision to

build the retail store and get no information, Branch 6 (3–17) is the

decision to not build the retail store and to get no information,

Branch 7 (4–6) is the decision to build the retail store given favorable

information, Branch 8 (4–11) is the decision to not build given

favor-able information, Branch 9 (6–9) is a good market given favorfavor-able

information, Branch 10 (6–10) is a bad market given favorable formation, Branch 11 (5–7) is the decision to build the retail storegiven unfavorable information, Branch 12 (5–14) is the decisionnot to build the retail store given unfavorable information, Branch

in-13 (7–12) is a successful retail store given unfavorable information,Branch 14 (7–13) is an unsuccessful retail store given unfavorableinformation, Branch 15 (8–15) is a successful retail store given that

no information is obtained, and Branch 16 (8–16) is an ful retail store given no information is obtained

b The suggested changes would be reflected in Branches 4 and 5 The decision stays the same, but the EMV

increases to $46,000 The results are provided in the tables that follow:

Results for 3-46 a