Transcript Spreadsheet Modeling & Decision Analysis:

Spreadsheet Modeling
& Decision Analysis
A Practical Introduction to
Management Science
5th edition
Cliff T. Ragsdale
Chapter 15
Decision Analysis
Introduction to Decision Analysis
 Models help managers gain insight and
understanding, but they can’t make decisions.
 Decision making often remains a difficult task
due to:
– Uncertainty regarding the future
– Conflicting values or objectives
 Consider the following example...
Deciding Between Job Offers
 Company A
– In a new industry that could boom or bust.
– Low starting salary, but could increase rapidly.
– Located near friends, family and favorite sports
team.
 Company B
– Established firm with financial strength and
commitment to employees.
– Higher starting salary but slower advancement
opportunity.
– Distant location, offering few cultural or sporting
activities.
 Which job would you take?
Good Decisions vs. Good Outcomes
 A structured approach to decision making
can help us make good decisions, but
can’t guarantee good outcomes.
 Good decisions sometimes result in bad
outcomes.
Characteristics of Decision Problems
 Alternatives - different courses of action intended to
solve a problem.
– Work for company A
– Work for company B
– Reject both offers and keep looking
 Criteria - factors that are important to the decision
maker and influenced by the alternatives.
– Salary
– Career potential
– Location
 States of Nature - future events not under the decision
makers control.
– Company A grows
– Company A goes bust
– etc
An Example: Magnolia Inns
 Hartsfield International Airport in Atlanta, Georgia, is one of
the busiest airports in the world.
 It has expanded many times to handle increasing air traffic.
 Commercial development around the airport prevents it
from building more runways to handle future air traffic.
 Plans are being made to build another airport outside the
city limits.
 Two possible locations for the new airport have been
identified, but a final decision will not be made for a year.
 The Magnolia Inns hotel chain intends to build a new
facility near the new airport once its site is determined.
 Land values around both possible sites for the new airport
are increasing as investors speculate that property values
will increase greatly in the vicinity of the new airport.
 See data in file Fig15-1.xls
The Decision Alternatives
1) Buy the parcel of land at location A.
2) Buy the parcel of land at location B.
3) Buy both parcels.
4) Buy nothing.
The Possible States of Nature
1) The new airport is built at location A.
2) The new airport is built at location B.
Constructing a Payoff Matrix
See file Fig15-1.xls
Decision Rules
 If the future state of nature (airport location) were
known, it would be easy to make a decision.
 Failing this, a variety of nonprobabilistic decision
rules can be applied to this problem:
– Maximax
– Maximin
– Minimax regret
 No decision rule is always best and each has its
own weaknesses.
The Maximax Decision Rule
 Identify the maximum payoff for each alternative.
 Choose the alternative with the largest maximum
payoff.
 See file Fig15-1.xls
 Weakness
– Consider the following payoff matrix
Decision
A
B
State of Nature
1
2
30
-10000
29
29
MAX
30 <--maximum
29
The Maximin Decision Rule
 Identify the minimum payoff for each alternative.
 Choose the alternative with the largest minimum
payoff.
 See file Fig15-1.xls
 Weakness
– Consider the following payoff matrix
Decision
A
B
State of Nature
1
2
1000
28
29
29
MIN
28
29 <--maximum
The Minimax Regret Decision Rule
 Compute the possible regret for each alternative
under each state of nature.
 Identify the maximum possible regret for each
alternative.
 Choose the alternative with the smallest maximum
regret.
 See file Fig15-1.xls
Anomalies with the Minimax Regret Rule
 Consider the following payoff matrix
State of Nature
Decision
1
2
A
9
2
B
4
6
 The regret matrix is:
State of Nature
Decision
1
2
A
0
4
B
5
0
 Note that we prefer A to B.
 Now let’s add an alternative...
MAX
4 <--minimum
5
Adding an Alternative
 Consider the following payoff matrix
State of Nature
Decision
1
2
A
9
2
B
4
6
C
3
9
 The regret matrix is:
State of Nature
Decision
1
2
A
0
7
B
5
3
C
6
0
 Now we prefer B to A???
MAX
7
5 <--minimum
6
Probabilistic Methods
 At times, states of nature can be assigned probabilities
that represent their likelihood of occurrence.
 For decision problems that occur more than once, we can
often estimate these probabilities from historical data.
 Other decision problems (such as the Magnolia Inns
problem) represent one-time decisions where historical
data for estimating probabilities don’t exist.
 In these cases, subjective probabilities are often
assigned based on interviews with one or more domain
experts.
 Interviewing techniques exist for soliciting probability
estimates that are reasonably accurate and free of the
unconscious biases that may impact an expert’s
opinions.
 We will focus on techniques that can be used once
appropriate probability estimates have been obtained.
Expected Monetary Value
 Selects alternative with the largest expected
monetary value (EMV)
EMVi   rij p j
j
rij  payoff for alternative i under the jth state of nature
p j  the probability of the jth state of nature
 EMVi is the average payoff we’d receive if we faced the
same decision problem numerous times and always
selected alternative i.
 See file Fig15-1.xls
EMV Caution
 The EMV rule should be used with caution in onetime decision problems.
 Weakness
– Consider the following payoff matrix
Decision
A
B
Probability
State of Nature
1
2
15,000
-5,000
5,000
4,000
0.5
0.5
EMV
5,000 <--maximum
4,500
Expected Regret or Opportunity Loss
 Selects alternative with the smallest expected regret
or opportunity loss (EOL)
EOLi   gij p j
j
gij  regret for alternative i under the jth state of nature
p j  the probability of the jth state of nature
 The decision with the largest EMV will also have the
smallest EOL.
 See file Fig15-1.xls
The Expected Value
of Perfect Information
 Suppose we could hire a consultant who could predict
the future with 100% accuracy.
 With such perfect information, Magnolia Inns’ average
payoff would be:
EV with PI = 0.4*$13 + 0.6*$11 = $11.8 (in millions)
 Without perfect information, the EMV was $3.4 million.
 The expected value of perfect information is therefore,
EV of PI = $11.8 - $3.4 = $8.4 (in millions)
 In general,
EV of PI = EV with PI - maximum EMV
 It will always the the case that,
EV of PI = minimum EOL
A Decision Tree for Magnolia Inns
Land Purchase Decision
Buy A
-18
Buy B
-12
Airport Location
1
2
0
Buy A&B
-30
Buy nothing
0
Payoff
A 31
13
B 6
-12
A 4
-8
B 23
11
A 35
5
B 29
-1
A 0
0
B 0
0
3
4
Rolling Back A Decision Tree
Land Purchase Decision
Airport Location
0.4
Buy A
-18
EMV=-2
1
A 31
13
6
B 0.6
-12
0.4
Buy B
-12
EMV=3.4
2
0
EMV=3.4
A 4
-8
23
B 0.6
11
0.4
Buy A&B
-30
EMV=1.4
A 35
5
B 29
0.6
-1
3
0.4
Buy nothing
0
EMV= 0
Payoff
4
A 0
0
B 0
0
0.6
Alternate Decision Tree
Land Purchase Decision
Airport Location
0.4
Buy A
-18
EMV=-2
1
A 31
13
6
B 0.6
-12
0.4
Buy B
-12
EMV=3.4
2
0
EMV=3.4
A 4
-8
23
B 0.6
11
0.4
Buy A&B
-30
EMV=1.4
A 35
5
B 29
-1
3
0.6
Buy nothing
0
Payoff
0
Using PrecisionTree
 PrecisionTree is an Excel add-in for decision
trees.
 See file Fig15-14.xls
Completed Tree For Magnolia Inns
 See file Fig15-22.xls
Multi-stage Decision Problems
 Many problems involve a series of decisions
 Example
– Should you go out to dinner tonight?
– If so,
How much will you spend?
Where will you go?
How will you get there?
 Multistage decisions can be analyzed using
decision trees
Multi-Stage Decision Example:
COM-TECH
 Steve Hinton, owner of COM-TECH, is considering whether to
apply for a $85,000 OSHA research grant for using wireless
communications technology to enhance safety in the coal
industry.
 Steve would spend approximately $5,000 preparing the grant
proposal and estimates a 50-50 chance of receiving the grant.
 If awarded the grant, Steve would need to decide whether to
use microwave, cellular, or infrared communications
technology.
 Steve would need to acquire some new equipment depending
on which technology is used…
Technology
Equipment Cost
Microwave
$4,000
Cellular
$5,000
Infrared
$4,000
continued...
COM-TECH
(continued)
 Steve knows he will also spend money in R&D, but he doesn’t
know exactly what the R&D costs will be. Steve estimates the
following best case and worst case R&D costs and probabilities,
based on his expertise in each area.
Best Case
Worst Case
Cost
Prob.
Cost
Prob.
Microwave
$30,000 0.4
$60,000 0.6
Cellular
$40,000 0.8
$70,000 0.2
Infrared
$40,000 0.9
$80,000 0.1
 Steve needs to synthesize all the factors in this problem to
decide whether or not to submit a grant proposal to OSHA.
 See file Fig15-23.xls
Risk Profiles
 A risk profile summarizes the make-up of an EMV
 The $13,500 EMV for COM-TECH was created as follows:
Event
Probability
Receive grant, Low R&D costs
Payoff
0.5*0.9=0.45
$36,000
Receive grant, High R&D costs 0.5*0.1=0.05
-$4,000
Don’t receive grant 0.5
EMV
-$5,000
$13,500
 This can also be summarized in a decision tree.
 See file Fig15-24.xls
Analyzing Risk in a Decision Tree
 How sensitive is the decision in the COM-TECH
problem to changes in the probability estimates?
 We can use Solver to determine the smallest
probability of receiving the grant for which Steve
should still be willing to submit the proposal.
 Let’s go back to file Fig15-25.xls...
Other Risk Analysis Techniques
 Tornado Charts & Spider Charts
– See file Fig15-27.xls
 Strategy Tables
– See file Fig15-29.xls
Using Sample Information
in Decision Making
 We can often obtain information about the
possible outcomes of decisions before the
decisions are made.
 This sample information allows us to refine
probability estimates associated with various
outcomes.
Example: Colonial Motors
 Colonial Motors (CM) needs to determine whether to build
a large or small plant for a new car it is developing.
 The cost of constructing a large plant is $25 million and
the cost of constructing a small plant is $15 million.
 CM believes a 70% chance exists that demand for the
new car will be high and a 30% chance that it will be low.
 The payoffs (in millions of dollars) are summarized below.
Factory Size
Large
Small
Demand
High
Low
$175
$95
$125
$105
 See decision tree in file Fig15-31.xls
Including Sample Information
 Before making a decision, suppose CM conducts a consumer
attitude survey (with zero cost).
 The survey can indicate favorable or unfavorable attitudes
toward the new car. Assume:
P(favorable response) = 0.67
P(unfavorable response) = 0.33
 If the survey response is favorable, this should increase CM’s
belief that demand will be high. Assume:
P(high demand | favorable response)=0.9
P(low demand | favorable response)=0.1
 If the survey response is unfavorable, this should increase CM’s
belief that demand will be low. Assume:
P(low demand | unfavorable response)=0.7
P(high demand | unfavorable response)=0.3
 See decision tree in file Fig15-32.xls
The Expected Value
of Sample Information
 How much should CM be willing to pay to conduct the
consumer attitude survey?
Expected Value of
Sample Information
=
Expected Value with
Sample Information
-
Expected Value without
Sample Information
 In the CM example,
E.V. of Sample Info. = $126.82 - $126 = $0.82 million
Computing Conditional Probabilities
 Conditional probabilities (like those in the CM example) are
often computed from joint probability tables.
Favorable Response
Unfavorable Response
Total
High
Demand
0.600
0.100
0.700
Low
Demand
0.067
0.233
0.300
 The joint probabilities indicate:
P(F  H)  0.6, P(F  L) = 0.067
P(U  H) = 0.1, P(U  L)  0.233
 The marginal probabilities indicate:
P(F)  0.667, P(U) = 0.333
P(H) = 0.700, P(L)  0.300
Total
0.667
0.333
1.000
Computing Conditional Probabilities
(cont’d)
Favorable Response
Unfavorable Response
Total
High
Demand
0.600
0.100
0.700
 In general,
P(A|B) =
Low
Demand
0.067
0.233
0.300
Total
0.667
0.333
1.000
P(A  B)
P(B)
 So we have,
P(H  F)
0.60
P(H|F) =

 0.90
P(F)
0.667
P(L|F) =
P(L  F) 0.067

 0.10
P(F)
0.667
P(H  U)
0.10
P(H|U) =

 0.30
P(U)
0.333
P(L|U) =
P(L  U) 0.233

 0.70
P(U)
0.333
Bayes’s Theorem
 Bayes’s Theorem provides another definition of conditional
probability that is sometimes helpful.
P(B|A)P(A)
P(A|B) =
P(B|A)P(A) + P(B|A)P(A)
 For example,
P(H|F) =
P(F|H)P(H)
(0857
. )(070
. )

 090
.
P(F|H)P(H) + P(F|L)P(L) (0857
. )(070
. )  (0223
. )(030
. )
Utility Theory
 Sometimes the decision with the highest EMV is not the
most desired or most preferred alternative.
 Consider the following payoff table,
Decision
A
B
Probability
State of Nature
1
2
150,000
-30,000
70,000
40,000
0.5
0.5
EMV
60,000 <--maximum
55,000
 Decision makers have different attitudes toward risk:
Some might prefer decision alternative A,
Others would prefer decision alternative B.
 Utility Theory incorporates risk preferences in the decision
making process.
Common Utility Functions
Utility
risk averse
1.00
risk neutral
0.75
risk seeking
0.50
0.25
0.00
Payoff
Constructing Utility Functions
 Assign utility values of 0 to the worst payoff and 1 to the best.
 For the previous example,
U(-$30,000)=0 and U($150,000)=1
 To find the utility associated with a $70,000 payoff identify the
value p at which the decision maker is indifferent between:
Alternative 1: Receive $70,000 with certainty.
Alternative 2: Receive $150,000 with probability p and lose
$30,000 with probability (1-p).
 If decision maker is indifferent when p=0.8:
U($70,000)=U($150,000)*0.8+U(-30,000)*0.2=1*0.8+0*0.2=0.8
 When p=0.8, the expected value of Alternative 2 is:
$150,000*0.8 + $30,000*0.2 = $114,000
 The decision maker is risk averse. (Willing to accept $70,000
with certainty versus a risky situation with an expected value of
$114,000.)
Constructing Utility Functions (cont’d)
 If we repeat this process with different values in
Alternative 1, the decision maker’s utility function
emerges (e.g., if U($40,000)=0.65):
Utility
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
-30
-20
-10
0
10
20
30
40
50
60
70
Payoff (in $1,000s)
80
90
100
110
120
130
140
150
Comments
 Certainty Equivalent - the amount that is equivalent
in the decision maker’s mind to a situation involving
risk.
(e.g., $70,000 was equivalent to Alternative 2 with p = 0.8)
 Risk Premium - the EMV the decision maker is
willing to give up to avoid a risky decision.
(e.g., Risk premium = $114,000-$70,000 = $44,000)
Using Utilities to Make Decisions
 Replace monetary values in payoff tables with utilities.
 Consider the utility table from the earlier example,
Decision
A
B
Probability
State of Nature
1
2
1
0
0.8
0.65
0.5
0.5
Expected
Utility
0.500
0.725 <--maximum
 Decision B provides the greatest utility even though it the
payoff table indicated it had a smaller EMV.
The Exponential Utility Function
 The exponential utility function is often used to model classic
risk averse behavior:
-x/R
U( x ) = 1- e
U(x)
1.00
0.80
R=200
R=100
0.60
0.40
R=300
0.20
0.00
-0.20
-0.40
-0.60
-0.80
-50
-25
0
25
50
75
100
125
150
175
200
225
250
275
300
325
350
x
Incorporating Utilities in PrecisionTree
 PrecisionTree will automatically convert monetary values to
utilities using the exponential utility function.
 We must first determine a value for the risk tolerance
parameter R.
 R is equivalent to the maximum value of Y for which the
decision maker is willing to accept the following gamble:
Win $Y with probability 0.5,
Lose $Y/2 with probability 0.5.
 Note that R must be expressed in the same units as the
payoffs!
 On PrecisionTree’s ‘Tree Settings’ dialog box specify
Function 'Exponential'
 See file Fig15-37.xls
Multicriteria Decision Making
 Decision problem often involve two or more
conflicting criterion or objectives:
– Investing:
risk vs. return
– Choosing Among Job Offers:
salary, location, career potential, etc.
– Selecting a Camcorder:
price, warranty, zoom, weight, lighting, etc.
– Choosing Among Job Applicants:
education, experience, personality, etc.
 We’ll consider two techniques for these types of
problems:
– The Multicriteria Scoring Model
– The Analytic Hierarchy Process (AHP)
The Multicriteria Scoring Model
 Score (or rate) each alternative on each criterion.
 Assign weights the criterion reflecting their
relative importance.
 For each alternative j, compute a weighted
average score as:
ws
i ij
i
wi = weight for criterion i
sij = score for alternative i on criterion j
 See file Fig15-38.xls
The Analytic Hierarchy Process (AHP)
 Provides a structured approach for determining the
scores and weights in a multicriteria scoring model.
 We’ll illustrate AHP using the following example:
– A company wants to purchase a new payroll and
personnel records information system.
– Three systems are being considered (X, Y and Z).
– Three criteria are relevant:
Price
User support
Ease of use
Pairwise Comparisons
 The first step in AHP is to create a pairwise comparison matrix
for each alternative on each criterion using the following values:
Value
Preference
1
Equally Preferred
2
Equally to Moderately Preferred
3
Moderately Preferred
4
Moderately to Strongly Preferred
5
Strongly Preferred
6
Strongly to Very Strongly Preferred
7
Very Strongly Preferred
8
Very Strongly to Extremely Preferred
9
Extremely Preferred
 Pij = extent to which we prefer alternative i to j on a given criterion.
 We assume Pji = 1/Pij
 See price comparisons in file Fig15-42.xls
Normalization & Scoring
 To normalize a pairwise comparison matrix,
1) Compute the sum of each column,
2) Divide each entry in the matrix by its column sum.
 The score (sj) for each alternative is given by the
average of each row in the normalized comparison
matrix.
 See file Fig15-42.xls
Consistency
 We can check to make sure the decision maker
was consistent in making the comparisons.
 The consistency measure for alternative i is:
P s
ij j
Ci 
where
j
si
Pij = pairwise comparison of alternative i to j
sj = score for alternative j
 If the decision maker was perfectly consistent each
Ci should equal to the number of alternatives in
the problem.
Consistency (cont’d)
 Typically, some inconsistency exists.
 The inconsistency is not deemed a problem provided
the Consistency Ratio (CR) is no more than 10%
CI
CR 
 0.10
RI
where,
 Ci

CI =   n  n / (n  1)
 i

n  the number of alternatives
RI =
for n =
0.00
0.58
0.90
1.12
1.24
1.32
1.41
2
3
4
5
6
7
8
Obtaining Remaining Scores & Weights
 This process is repeated to obtain scores for
the other criterion as well as the criterion
weights.
 The scores and weights are then used as
inputs to a multicriteria scoring model in the
usual way.
 See file Fig15-42.xls
End of Chapter 15