Download Elements of a Decision Analysis

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Decision
Analysis
Decision Making
Under Uncertainty
Decision Analysis - 1
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Elements of a Decision Analysis
Bidding for a Government Contract at
SciTools
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Background Information
Bidding for a Government Contract at SciTools
• SciTools Incorporated specializes in scientific instruments
and has been invited to make a bid on a government
contract to provide these instruments this coming year
• SciTools estimates that it will cost $5000 to prepare a bid
and $95,000 to supply the instruments
• On the basis of past contracts, SciTools estimated the
probabilities of the low bid from competitors at a certain
dollar level
• In addition, they believe there is a 30% chance that there
will be no competing bids
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Decision Making Elements
• Although there is a wide variety of contexts in
decision making, all decision making problems have
three elements:
– the set of decisions (or strategies) available to the decision
maker
– the set of possible outcomes and the probabilities of
these outcome
– a value model that prescribes results, usually monetary
values, for the various combinations of decisions and
outcomes
• Once these elements are known, the decision maker
can find an “optimal” decision
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SciTools’ Problem
Bidding for a Government Contract at SciTools
• SciTools decision is whether to submit a bid and how
much they should bid (the bid must be greater than
$100,000 for SciTools to make a profit)
• Based on the estimated probabilities, SciTools should bid
either $115,000, $120,000, $125,000 (we’ll assume that
they will never bid less than $115,000 or more than
$125,000 due to the small profit margin or low chances of
winning the bid)
• The primary source of uncertainty is the behavior of the
competitors - will they bid and, if so, how much?
• The behavior of the competitors depends on how many
competitors are likely to bid and how the competitors
assess their costs of supplying the instruments
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SciTools’ Problem
Bidding for a Government Contract at SciTools
• The value model in this example is straightforward but in
other examples it is often complex
– If SciTools decides right now not to bid, then its monetary
values is $0 - no gain, no loss
– If they make a bid and are underbid by a competitor, then
they lose $5000, the cost of preparing the bid
– If they bid B dollars and win the contract, then they make a
profit of B - $100,000; that is, B dollars for winning the bid,
less $5000 for preparing the bid, less $95,000 for supplying
the instruments
– It is often convenient to list the monetary values in a payoff
table
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SciTools’ Payoff Tables
Bidding for a Government Contract at SciTools
Payoff Table for SciTools Bidding Example
Competitor’s Low Bid ($1000s)
Bid level
No Bid
<115
>115, <120
>120, <125
>125
No Bid
0
0
0
0
0
115
15
-5
15
15
15
120
20
-5
-5
20
20
125
25
-5
-5
-5
25
Probability
0.3
0.7(0.2)
0.7(0.4)
0.7(0.3)
0.7(0.1)
0.3
0.14
0.28
0.21
0.07
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SciTools’ Payoff Tables
Bidding for a Government Contract at SciTools
Alternative Payoff Table for SciTools Bidding Example
Monetary Value
SciTools
Wins
SciTools
Loses
Probability That
SciTools Wins
No Bid
NA
0
0.00
115
15
-5
0.86
120
20
-5
0.58
125
25
-5
0.37
SciTools Bid
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Risk Profiles for SciTools
Bidding for a Government Contract at SciTools
• A risk profile simply lists all possible monetary values
and their corresponding probabilities
• From the alternate payoff table we can obtain risk
profiles for SciTools
– For example, if SciTools bids $120,000 there are two possibly
monetary values, a profit of $20,000 or a loss of $5000, and their
probabilities are 0.58 and 0.42, respectively
• Risk profiles can be illustrated on a bar chart (the bars
above each possible monetary value measure the
probability of that value occurring)
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SciTools’ Expected Monetary Values (EMV)
Bidding for a Government Contract at SciTools
Alternative
EMV Calculation
EMV
No Bid
0(1)
Bid $115,000
15,000(0.86) +(-5000)(0.14)
$12,200
Bid $120,00
20,000 (0.58) + (-5000)(0.42)
$9500
Bid $125,000
25,000(0.37) + (-5000)(0.63)
$6100
$0
• EMV is a weighted average of the possible monetary
values, weighted by their probabilities
• What exactly does the EMV mean?
– It means that if SciTools were to enter many “gambles” like
this, where on each gamble the gains, losses and
probabilities were the same, then on average it would win
$12,200 per gamble
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Decision Tree Conventions
Bidding for a Government Contract at SciTools
•
•
•
A decision tree is a graphical tool that can represent a
decision problem with probabilities
Decision trees are composed of nodes (circles,
squares and triangles) and branches (lines)
The nodes represent points in time
-
A decision node (a square) is a time when the decision maker
makes a decision
A probability node (a circle) is a time when the result of an
uncertain event becomes known
An end node (a triangle) indicates that the problem is
completed - all decisions have been made, all uncertainty
have been resolved and all payoffs have been incurred
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Decision Tree Conventions
•
Bidding for a Government Contract at SciTools
Time proceeds from left to right
-
•
•
•
Branches leading out of a decision node represent the possible
decisions
Branches leading out of probability nodes represent the possible
outcomes of uncertain events
Probabilities are listed on probability branches (these
probabilities are conditional on the events that have
already been observed )
Individual monetary values are shown on the branches
where they occur, and cumulative monetary values are
shown to the right of the end nodes
Two values are often found to the right of each end node:
the top one is the probability of getting to that end node,
and the bottom one is the associated monetary value
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SciTools’ Decision Tree
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Decision Tree Folding Back Procedure
Bidding for a Government Contract at SciTools
• The solution for the decision tree is on the next slide
• The solution procedure used to develop this result is
called folding back on the tree
• Starting at the right on the tree and working back to the
left, the procedure consists of two types of calculations
– At each probability node we calculate EMV and write it below the
name of the node
– At each decision node we find the maximum of the EMVs and
write it below the node name
• After folding back is completed we have calculated
EMVs for all nodes
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Decision
Tree
Results
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The PrecisionTree Add-In
• This add-in enables us to build and label a decision tree, but it
performs the folding-back procedure automatically and then
allows us to perform sensitivity analysis on key input
parameters
• There are three options to run PrecisionTree:
– If Excel is not currently running , you can launch Excel and
PrecisionTree by clicking on the Windows Start button and
selecting the PrecisionTree item
– If Excel is currently running, the procedure in the previous bullet
will launch PrecisionTree on top of Excel
– If Excel is already running and the Desktop Tools toolbar is
showing, you can start
PrecisionTree
by clicking
on its icon
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Using PrecisionTree
Bidding for a Government Contract at SciTools
• Inputs: Enter the cost and probability inputs
• New tree: Click on the new tree
button (the far left button) on the
PrecisionTree toolbar, and then click
on any cell below the input section to
start a new tree. Click on the name
box of this new tree to open a dialog
box. Type in a descriptive name
for the tree.
• Decision nodes and branches:
To obtain decision nodes and
branches, click on the (only) triangle
end node to open the dialog box shown here
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The PrecisionTree Add-In
Bidding for a Government Contract at SciTools
• We’re calling this decision “Bid?” and specifying that there are two
possible decisions. The tree expands as shown here.
• The boxes that say “branch” show the default labels for these
branches. Click on either of them to open another dialog box where
you can provide a more descriptive name for the branch. Do this to
label the two branches “No” and “Yes.” Also, you can enter the
immediate payoff/cost for either branch right below it. Since there is
a $5000 cost of bidding, enter the formula =BidCost right below the
“Yes” branch in cell B19.
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The PrecisionTree Add-In
Bidding for a Government Contract at SciTools
• More decision branches: The top branch is completed; if SciTools
does not bid, there is nothing left to do. So click on the bottom end
node, following SciTools’ decision to bid, and proceed as in the
previous step to add and label the decision node and three
decision branches for the amount to bid. Note that there are no
monetary values below these decision branches because no
immediate payoffs or costs are associated with the bid amount
decision.
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The PrecisionTree Add-In
Bidding for a Government Contract at SciTools
• Probability nodes and branches: We now need a probability
node and branches from the rightmost end nodes to capture
the competition bids
– Click on the top one of these end nodes to bring up the dialog box
– Click on the red circle box to indicate that this is a probability
node. Label it “Any competing bid?”, specify two branches, and
click on OK.
– Label the two branches “No” and “Yes”.
– Repeat this procedure to form another probability node following
the “Yes” branch, call it “Win bid?”, and label its branches as
shown on the next slide.
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The PrecisionTree Add-In
Bidding for a Government Contract at SciTools
• Copying probability nodes and branches: You could
build the next node and branches or take advantage of
PrecisionTree’s copy and paste function. Decision trees
can be very “bushy”, but this copy and paste feature can
make them much less tedious to construct.
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•Labeling probability branches:
–First enter the probability of no
competing bid in cell D18 with the
formula =PrNoBid and enter its
complement in cell D24 with the
formula =1-D18
–Next, enter the probability that
SciTools wins the bid in cell E22
with the formula =SUM(B10:B12)
and enter its complement in cell
E26 with the formula =1-E22
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The PrecisionTree Add-In
Bidding for a Government Contract at SciTools
• For the monetary values, enter the formula
=115000-ProdCost in the two cells, D19 and E23, where
SciTools wins the contract
• Enter the other formulas on probability branches
– Using the previous step and the final decision tree as a guide,
enter formulas for the probabilities and monetary values on the
other probability branches, that is, those following the decision
to bid $120,000 or $125,000
• We’re finished! The completed tree shows the best strategy
and its associated EMV.
• Once the decision tree is completed, PrecisionTree has
several tools we can use to gain more information about the
decision analysis
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Risk Profile of Optimal Strategy
Bidding for a Government Contract at SciTools
• Click on the fourth button from the left on the PrecisionTree
toolbar (Decision Analysis) and fill in the resulting dialog box
(the Policy Suggestion option allows us to see only that part of
the tree that corresponds to the best decision)
• The Risk Profile option allows us to see a graphical risk of the
optimal decision (there are only two possible monetary
outcomes if
SciTools bids
$115,000)
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Sensitivity Analysis
Bidding for a Government Contract at SciTools
• We can enter any values not the input cells and watch how the tree
changes
• To obtain more systematic information, click on the PrecisionTree
sensitivity button
• The dialog box requires an EMV cell to analyze at the top and one or
more input cells in the middle
• The cell to analyze is usually the EMV cell at the far left of the
decision tree but it can be any EMV cell
• For any input cell (such as the production cost cell) we enter a
minimum value, a maximum value, a base value, and a step size
• When we click Run Analysis, PrecisionTree varies each of the
specified inputs and presents the results in several ways in a new
Excel file with Sensitivity, Tornado, and Spider Graph sheets
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Sensitivity Analysis: Sensitivity Chart
Bidding for a Government Contract at SciTools
• The Sensitivity sheet includes several charts, a typical one of
which appears here
• This shows how the EMV varies with the production cost for
both of the original decisions
• This type of graph is useful for seeing whether the optimal
decision changes over the range of input variable
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Sensitivity Analysis: Tornado Chart
Bidding for a Government Contract at SciTools
• The Tornado sheet shows how sensitive the EMV of the
optimal decision is to each of the selected inputs over the
ranges selected
• The production cost has the largest effect on EMV, and bid cost
has the smallest effect
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Sensitivity Analysis: Spider Chart
Bidding for a Government Contract at SciTools
• Finally, the Spider Chart shows how much the optimal EMV
varies in magnitude for various percentage changes in the
input variables
• Again, the production cost has a relatively large effect,
whereas the other two inputs have relatively small effects
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Sensitivity Analysis: Two-Way Analysis
Bidding for a Government Contract at SciTools
• Use a two-way analysis to see how the selected EMV varies as
each pair of inputs vary simultaneously
• For each of the possible values of production cost and
probability of no competitor bid, this chart indicates which bid
amount is optimal
• The optimal bid amount remains $115,000 unless the
production cost
and the probability
of no competing
bid are both large.
Then it becomes
optimal to
bid $125,000
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Multistage Decision Trees
Marketing a New Product at Acme
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Background Information
• Acme Company is trying to decide whether to
market a new product.
• As in many new-product situations, there is much
uncertainty about whether the product will “catchon”.
• Acme believes that it would be prudent to
introduce the product to a test market first.
• Thus the first decision is whether to conduct the
test market.
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Background Information -continued
• Acme has determined that the fixed cost of the test
market is $3 million.
• If they proceed with the test, they must then wait for
the results to decide if they will market the product
nationally at a fixed cost of $90 million.
• If the decision is not to conduct the test market, then
the product can be marketed nationally with no delay.
• Acme’s unit margin, the difference between its selling
price and its unit variable cost, is $18 in both
markets.
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Background Information -continued
• Acme classifies the results in either market as great,
fair or awful.
• Each of these has a forecasted total units sold as (in
1000s of units) 200, 100 and 30 in the test market and
6000, 3000 and 900 for the national market.
• Based on previous test markets for similar products, it
estimates that probabilities of the three test market
outcomes are 0.3, 0.6 and 0.1.
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Background Information -continued
• Then based on historical data on products that were
tested then marketed nationally, it assesses the
probabilities of the national market outcomes given
each test market outcome.
– If the test market is great, the probabilities for the national
market are 0.8, 0.15, and 0.05.
– If the test market is fair. then the probabilities are 0.3, 0.5,
0.2.
– If the test market is awful, then the probabilities are 0.05,
0.25, and 0.7.
• Note how the probabilities of the national market
mirrors those of the test market.
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Elements of Decision Problem
• The three basic elements of this decision
problem are:
– the possible strategies
– the possible outcomes and their probabilities
– the value model
• The possible strategies are clear:
– Acme must first decide whether to conduct the test
market.
– Then it must decide whether to introduce the product
nationally.
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Contingency Plan
• If Acme decides to conduct a test market they
will base the decision to market nationally on the
test market results.
• In this case its final strategy will be a
contingency plan, where it conducts the test
market, then introduces the product nationally if
it receives sufficiently positive test market results
and abandons the product if it receives negative
test market results.
• The optimal strategies from many multistage
decision problems involve similar contingency
plans.
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Conditional Probabilities continued
• The probabilities of the test market outcomes and
conditional probabilities of national market
outcomes given the test market outcomes are
exactly the ones we need in the decision tree.
• However, suppose Acme decides not to run a test
market and then decides to market nationally.
Then what are the probabilities of the national
market outcomes? We cannot simply assess
three new probabilities.
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Conditional Probabilities continued
• These probabilities are implied by the given
probabilities. This follows from the rule of conditional
probability.
• If we let T1, T2, and T3 be the test market outcomes
and N be any national market outcomes, then by the
addition rule of probability and the conditional
probability formula
P(N) = P(N and T1) + P(N and T2) and P(N and T3)
=P(N|T1)P(T1) + P(N|T2)P(T2) + P(N|T3)P(T3)
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Conditional Probabilities continued
• This is sometimes called the law of probabilities.
• We determine the probabilities as 0.425 for a great
market, 0.37 for a fair market and 0.205 for an awful
market.
• The monetary values are the fixed costs of test
marketing or marketing nationally and these are
incurred as soon as the “go ahead” decisions are
made.
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ACME.XLS
• This file contains the inputs for the decision tree.
• The only calculated values in this spreadsheet
are in row 28, which follow from the equation.
The formula in cell B28 is
=SUMPRODUCT(B22:B24,$B$16:$B$18)
then it is copied across row 28.
• The creation of the tree is then straightforward to
build and label.
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Inputs
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Interpreting the Tree
• The interpretation of the tree is fairly
straightforward if we realize that each value just
below each node name is an EMV.
• Each of these EMVs have been calculated with
the folding back procedure.
• We can also see Acme’s optimal strategy by
following the “TRUE” branches from left to right.
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Optimal Strategy
• Acme should first run a test market and if the results
are great then they should market it nationally.
• If the test results are fair or awful they should
abandon the product.
• The risk profile for the optimal strategy can be seen
on the next slide.
• The risk profile (created by clicking on
PreceisionTree’s “staircase” button and selecting
Statistics and Risk Profile options) that there is a
small chance of two possible large losses, there is a
70% chance of a moderate loss and there is a 24%
chance of a nice profit.
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Optimal Strategy -- continued
• One might argue that the large potential (70%)
chance of some loss should persuade Acme to
abandon the product right away - without a test
market.
• This is what “playing the averages” with EMV is
all about.
• Since the EMV of this optimal strategy is greater
than 0, Acme should go ahead with this strategy
if it is an EMV maximizer.
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Expected Value of Sample
Information
• The role of the test market in this example is to
provide Acme with information.
• Information usually costs, in this case its fixed cost is
$3 million.
• From the decision tree we can see that the EMV from
test marketing is $807,000 better than the decision
not to test market. The most Acme would be willing to
pay for the test marketing is $3.807 million.
• This value is called the expected value of sample
information or EVSI.
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Expected Value of Sample
Information -- continued
• In general we can write the following expression:
EVSI=EMV with free information - EMV without
information
• For the Acme example the EVSI is $3.807 million.
• The reason for the term sample is that the
information does not remove all uncertainty about
the future.
• That is, even after the test market results are in,
there is still uncertainty about the national
market.
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Expected Value of Perfect
Information
• We can go one step further and ask how much
perfect information is worth.
• Perfect could be imagined as an envelope containing
the final outcome.
• No such envelope exists but if it did how much would
Acme be willing to pay for it?
• This question can be answered with the decision tree
shown on the next slide.
• Folding back produces an EMV of $7.65 million.
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Expected Value of Perfect
Information -- continued
• This value is called the expected value of
perfect information, or EVPI.
• The EVPI may appear to be irrelevant because
perfect information is almost never available - at
any price.
• However, it is often used as an upper bound on
EVSI for any potential sample information.
• That is, no sample information can ever be worth
more than the EVPI.
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Bayes’ Rule
Drug Testing College Athletes
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Background Information
• If an athlete is tested for a certain type of drug use,
the test will come out either positive or negative.
• However, these tests are never perfect. Some
athletes who are drug-free test positive (false
positives) and some who are drug users test negative
(false negatives). We will assume that
– 5% of all athletes use drugs
– 3% of all tests on drug-free athletes yield false positives
– 7% of all tests on drug users yield false negatives.
• The question then is what we can conclude from a
positive or negative test result.
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Solution
• Let D and ND denote that a randomly chosen
athlete is or is not a drug user, and let T+ and Tindicate a positive or negative test result.
• We know the following probabilities
– First, since 5% of all athletes are drug users, we know
that P(D) = 0.05 and P(ND) = 0.95. These are called
prior probabilities because they represent the chance
that an athlete is or is not a drug user prior to the results
of a drug test.
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Solution -- continued
– Second, from the information on drug test accuracy, we
know the conditional probabilities P(T+|ND) = 0.08 and P(T|D)= 0.03.
– But a drug-free athlete either tests positive or negative, and
the same is true for a drug user. Therefore, P(T-|ND) = 0.92
and P(T+|D) = 0.97.
– These four conditional probabilities of test results given drug
user status are often called the likelihoods of the test
results.
• Given these priors and likelihoods we want posterior
probabilities such as P(D|T+) or P(ND|T-).
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Solution -- continued
• These are called posterior probabilities
because they are assessed after the drug test
results. This is where Baye’s rule enters.
• Bayes’ Rule says that a typical posterior
probability is a ratio. The numerator is a
likelihood times a prior, and the denominator is
the sum of likelihoods times priors.
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DRUGBAYES.XLS
• This file shows how easy it
is to implement Bayes’ rule
in a spreadsheet.
• The given priors and
likelihoods are listed in the
ranges B5:C5 and B9:C10.
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Calculations
• We calculate the products of likelihoods and priors in
the range B15:C16. The formula in cell B15 is
=B$5*B9 and it is copied to the rest of B15:C16
range.
• Their row sums are calculated in the range D15:D16.
These represent the unconditional probabilities of the
two possible results. They are also the denominator
of Bayes’ rule.
• Finally we calculate the posterior probabilities in the
range B21:C22. The formula in B21 is =B15/$D15
and it is copied to the rest of the range B21:C22.
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Resulting Probabilities
• A negative test result leaves little doubt that the
athlete is drug-free; this probability is 0.996.
• A positive test result leaves a lot of doubt of
whether the athlete is drug-free. The probability
that the athlete uses drugs is 0.620.
• Since only 5% of athletes use drugs it takes a lot
of evidence to convince us otherwise. This plus
the fact that the test produces false positives
means the athletes that test positive still have a
decent chance of being drug-free.
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Background Information
• The administrators at State University are trying
to decide whether to institute mandatory drug
testing for the athletes.
• They have all the same information of priors and
likelihoods as the previous example.
• They want to use a decision tree approach to see
whether the benefits outweigh the costs
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Decision Alternatives
• We will assume that there are only two
alternatives:
– perform drug testing on all athletes
– don’t perform any drug testing.
• In the former case we assume that if an athlete
tests positive, this athlete is barred from sports.
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Monetary Values
• The “monetary” values are more difficult to assess.
They include:
– the benefit B from correctly identifying a drug user and
barring him or her from sports
– the cost C1 of the test itself for a single athlete (materials and
labor)
– the cost C2 of falsely accusing a nonuser (and barring him or
her from sports)
– the cost C3 of not identifying a drug user (either by not
testing at all or by obtaining a false negative)
– the cost C4 of violating a nonuser’s privacy by performing the
test
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Monetary Values -- continued
• Only C1 is a direct monetary cost that is easy to
measure.
• The other costs and the benefit are real, and
they must be compared on some scale to enable
administrators to make a rational decision.
• We will do this by comparing everything to C1 to
which we will assign a value 1.
• There is a lot of subjectivity so sensitivity
analysis on the final decision is a must.
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Benefit-Cost Table
• Before constructing the decision tree it is useful
to form a benefit-cost table for both alternatives
and all possible outcomes.
• All benefits in this table have a positive sign and
all costs have a negative sign.
Net Benefit for Drug-Testing Example
Don’t Test
Perform Test
D
ND
D and T+
ND and T+
D and T-
ND and T-
-C3
0
B – C1
-(C1 + C2 + C4)
- (C1 + C3)
- (C1 + C4)
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DRUG.XLS
• This file provides the data to create the decision
tree with PrecisionTree.
• First, we enter the input data and then along with
Bayes’ rule calculations from before we can
create the tree and enter the links to the values
and probabilities.
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Timing
• Before interpreting the tree we need to discuss the
timing (from left to right).
• If drug testing is performed, the result of the drug test
is observed first (a probability node).
• Each result leads to an action (bar from sports or
don’t), and then the eventual benefit or cost depends
on whether the athlete uses drugs (again a
probability node).
• If no drug testing is performed, then there is no
intermediate test result node or branches.
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Interpretation
• First, we discuss the benefits and costs.
– The largest cost is falsely accusing (and barring) a nonuser.
– This is 50 times as large as the cost of the test.
– The benefit of identifying a drug user is only half this large,
and the cost of not identifying a user is 40% as large as
barring a nonuser.
– The violation of privacy of a nonuser is twice as large as the
cost of the test.
• Based on these values, the decision tree implies that
drug testing should not be performed. The EMVs are
both negative; thus costs outweigh benefits.
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Interpretation -- continued
• What would it take to change this decision?
• Most people in society would agree that the costs of
falsely accusing a nonuser should be the largest cost.
In fact, with legal costs we might argue that it should
be more than 50 times the cost of the test.
• On the other hand, if the benefit of identifying a user
and/or the cost C3 for not identifying a user increase,
the testing might be the preferred alternative.
• We can test this by varying the benefits and the
costs.
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Interpretation -- continued
• Other than benefits and costs, the only thing that
we might vary is the accuracy of the test,
measured by the error probabilities.
• Even when each error probability was decreased
to 0.01, the no-testing alternative was still
optimal.
• In summary, based on a number of reasonable
assumptions and parameter settings, this
example has shown that it is difficult to make a
case for mandatory drug testing.
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Decision
Analysis
Attitudes Toward Risk and
Multi-attribute Decision Making
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Utility Functions
• A utility function is a mathematical function that
transforms monetary values – payoffs and costs
– into utility values.
• Most individuals are risk averse, which means
intuitively that they are willing to sacrifice some
EMV to avoid risky gambles.
• If a person is indifferent, then the expected
utilities from the two options must be equal. We
will call the resulting value the indifference
value.
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Example
• John Jacobs owns his own business.
• Because he is about to make an important decision
where large losses or large gains are at stake, he
wants to use the expected utility criterion to make his
decision.
• He knows that he must first assess his own utility
function, so he hires a decision analysis expert,
Susan Schilling, to help him out.
• How might the session between John and Susan
proceed?
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Solution
• Susan first asks John for the largest loss and largest
gain he can imagine.
• He answers with the values $200,000 and $300,000,
so she assigns utility values U(-200,000) = 0 and
U(300,000) = 1 as anchors for the utility function.
• Now she presents John with the choice between two
options
– Option 1: Obtain a payoff of z (really a loss if z is negative).
– Option 2: Obtain a loss of $200,000 or a payoff of $300,000,
depending on the flip of a fair coin.
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Solution -- continued
• Susan reminds John that the EMV of option 2 is
$50,000.
• He realizes this, but because he is quite risk
averse, he would far rather have $50,000 for
certain than take the gamble for option 2.
• Therefore the indifference value of z must be less
than $50,000.
• Susan then poses several values of z to John.
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Solution -- continued
• Would he rather have $10,000 for sure or take option
2? He says he would rather take this $10,000.
• Would he rather pay $5000 for sure or take option 2.
He says he would rather take option 2.
• By this time, we know the indifference value of z must
be less than $10,000 and greater than -$5000.
• With a few more questions of this type, John finally
decides on z=$5000 as his indifference value. He is
indifferent between obtaining $5000 for sure and
taking the gamble in option 2.
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Solution -- continued
• We can substitute these values into the equation:
U(5000) = 0.5U(-200,000) + 0.5U(300,000) = 0.5(0) + 0.5(1) = 0.5
• Note that John is giving up $45,000 in EMV because
of his risk aversion.
• The EMV of the gamble in option 2 is $50,000, and he
is willing to accept a sure $5000 in its place.
• The process would then continue. For example, since
she now knows U(5000) and U(300,000), Susan
could ask John to choose between these options:
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Solution -- continued
– Option 1: Obtain a payoff of z.
– Option 2: Obtain a payoff of $5000 or a payoff of $300,000,
depending on the flip of a fair coin.
• If John decides that his indifference value is now z =
$130,000, then with the equation we know that
U(130,000) = 0.5U(5000) + 0.5U(300,000) = 0.5(0.5) + 0.5(1) = 0.75
• Note that John is now giving up $22,500 in EMV
because the EMV of the gamble in option 2 is
$152,500. By continuing in this manner, Susan can
help John assess enough utility values to
approximate a continuous utility curve.
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Incorporating Attitudes
Toward Risk
Deciding Whether to Enter Risky
Ventures at Venture Limited
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Background Information
• Venture Limited is a company with net sales of $30
million. The company currently must decide whether
to enter one of two risky ventures or do nothing.
• The possible outcomes of the less risky venture are
$0.5 million loss, a $0.1 million gain, and a $1 million
gain.
• The probabilities of these outcomes are 0.25, 0.50,
and 0.25.
• The possible outcomes of the most risky venture is
$1 million loss, a $1 million gain, and a $3 million
gain.
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Background Information -continued
• The probabilities of these outcomes are 0.35,
0.60, and 0.05.
• If Venture Limited can enter at most one of the
two risky ventures, what should it do?
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Solution
• We will assume that Venture Limited has an
exponential utility function.
• An exponential utility function has only one
adjustable numerical parameter, and there are
straightforward ways to discover the most
appropriate value of this parameter for a
particular individual or company.
• Also, based on Howard’s guidelines, we will
assume that the company’s risk tolerance is
6.4% of its net sales, or 1.92 million.
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Solution -- continued
• We can substitute into the equation to find the utility of
any monetary outcome.
• For example, the gain from doing nothing is $0, and
its utility is U(0) = 1 e-0/1.92 = 1-1 = 0. As another
example, the utility of a $1 million loss is U(-1) = 1 – e(-1)/1.92 = 1 – 1.683 = - 0.683.
• These are the values we use (instead of monetary
values) in the decision tree.
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Using PrecisionTree
• Fortunately, PrecisionTree takes care of all the
details.
• After we build a decision tree and label it in the usual
way, we click on the name of the tree to open the
dialog box shown here.
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Using PrecisionTree
• We then fill in the utility function information as
shown in the upper right section of the dialog
box.
• This says to use an exponential function with risk
tolerance 1.92.
• It also indicates that we want the expected
utilities (as opposed to EMVs) to appear in the
decision tree.
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VENTURE.XLS
• We build our tree exactly the same way and link
probabilities and monetary values to its branches in the
usual way.
• For example, there is a link in cell C22 to the monetary
value in cell A10. However, the expected values shown
in the tree are expected utilities, and the optimal
decision is the one with the largest expected utility.
• In this case the expected utilities for doing nothing,
investing in the less risky venture, and investing in the
more risky venture are 0,0.0525, and 0.0439.
Therefore, the optimal decision is to invest in the less
risky venture.
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Solution -- continued
• Note that the EMVs of the three decisions are $0,
$0.175 million, and $0.4 million. The later of these
are calculated in row 14 as the usual “sumproduct” of
monetary values and probabilities.
• How sensitive is the optimal decision to the key
parameter, the risk tolerance?
• We can answer this by changing the risk tolerance
and watching how the tree changes.
• So the optimal decision depends heavily on the
attitudes toward risk of Venture Limited’s top
management.
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Certainty Equivalents
• Now suppose that Venture Limited has only two options.
• It can enter the less risky venture or receive a certain
dollar amount x and avoid the gamble altogether.
• We want to find the dollar amount x such that the company
is indifferent between these two options.
• If it enters the risky venture, its expected utility is 0.0525,
calculated above. If he receives x dollars for certain, its
(expected) utility is U(x) = 1 – e-x/1.92.
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Certainty Equivalents -continued
• To find the value x where it is indifferent between
the two options, we set 1 – e-x/1.92 equal to 0.0525
or e-x/1.92 = 0.9475, and solve for x.
• Taking natural logarithms of both sides and
multiplying by –1.92, we obtain x = 1.92ln(0.9475) ~ $0.104 million.
• This value is called the certainty equivalent of
the risky venture.
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Certainty Equivalents -continued
• The company is indifferent between entering the less
risky venture and receiving $0.104 million to avoid it.
• Although the EMV of the less risky venture is $0.175
million, the company acts as if it is equivalent to a
sure $0.104 million.
• In this sense, the company is willing to give up the
difference in EMV, $71,000, to avoid a gamble.
• By a similar calculation, the certainty equivalent of
the more risky venture is approximately $0.086
million, when in fact its EMV is a hefty $0.4 million!
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Certainty Equivalents -continued
• So in this case it is willing to give up the difference in
EMV, $314,000, to avoid this particular gamble.
• Again, the reason is that the company dislikes risk.
• We can see these certainty equivalents in
PrecisionTree by adjusting the Display box to show
Certainty Equivalent.
• The tree then looks like the one on the next slide.
• The certainty equivalents we just discussed appear in
cells C24 and C32.
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Multi-attribute Decision Making
Using AHP to Select a Job
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Background Information
• Jane is about to graduate from college and is trying to
determine which job to accept from among three
options.
• She plans to choose among the offers by determining
how well each job offer meets the following four
objectives:
– Objective 1: High starting salary
– Objective 2: Quality of life in city where job is located
– Objective 3: Interest of work
– Objective 4: Nearness of job to family
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Value structure properties
How do we determine proper objectives?
• Completeness (covers all concerns)
• Non-redundant (no overlap)
• Decomposable (can consider each objective
without considering others)
• Operability (must be understandable)
• Number of objectives (smaller generally better)
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Objectives Assessment
Determine attribute measures for each objective
• Salary (measure annual salary in $K & assume that
other benefits and cost of living in different locations are
similar)
Job 1: $40K
Job 2: $36K
Job 3: $28K
• Quality of life (based on subjective 1-10 rating done by
Jane using reference material)
Job 1: 5
Job 2: 6
Job 3: 8
• Work interest (based on subjective 1-10 rating using job
description and interview)
Job 1: 3
Job 2: 9
Job 3: 6
• Nearness to family (measured in travel time)
Job 1: 3 hours Job 2: 1 hour Job 3: 20 minutes
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Additive Weighting Approach
• Develop a value function
– Transform attribute measures from quantitative/qualitative
measures into a linear scaled measure
– Assess and normalize weights for each attribute
– Each alternative receives a value score (sum of attribute
measures weighted by objective weights)
– Preferred alternative has the largest value score
• Simple Additive Weighting Template
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Transforming Attribute Measures
• We want scales with the following properties:
– Interval scale (difference between values has meaning)
– Consistent units of measurement
– Scales in the same direction
• We will use a linear proportional scale:
Max Attribute: Rescaled Score =
Min Attribute:
Attribute Value
Max Attribute Value
Min Attribute Value
Rescaled Score =
Attribute Value
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Assess & Normalize Weights
• Weights should be based on importance relative to the
range of values for each attribute
• How much more/less important are the following
attributes relative to each other?
– Salary (ranging from $28K to $40K)
– Quality of life (ranging from 5 to 8 on a 10 point scale)
– Work interest (ranging from 3 to 9 on a 10 point scale)
– Nearness to family (ranging from 20 min to 3 hr)
• Rank order attributes, give most important attribute a top
score, other scores relative to best
• Normalize: divide each weight by the sum of weighted
scores (sum of normalized weights equal 1)
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Determine Value Score
• Use weights of salary=10, work interest=5, near to
family=3 and quality of life=2
• Final value scores: Job 1=.66, Job 2=.82, Job 3=.77
• Preferred option is Job 2 (although Job 3 is close
enough to examine these two jobs against each other)
• Sensitivity analysis is important due to subjectivity
• Always present and rationalize preferred option based
on objectives (why was Job 2 preferred? –best for
work interest, overall strong on all objectives)
• Job 3 does well due to dominance in nearness to family
(may consider rescaling attribute measure or
importance if this does not seem reasonable)
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Analytic Hierarchy Process
(AHP)
• Transform objective measures from quantitative or
qualitative measures using pairwise comparisons and
an intensity scale of relative importance
• Assess weights using the same pairwise comparison
method
• Each alternative receives a weighted score (sum of
objective measures weighted by objective weights)
• Preferred alternative has the largest weighted score
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Solution
• To illustrate how AHP works, suppose that Jane is
facing three job offers and must determine which offer
to accept.
• In this example there are four objectives, as listed
earlier.
• For each objective, AHP generates a weight (by a
method to be described shortly).
• By convention, the weights are always chosen so that
they sum to 1.
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Pairwise Comparison Matrices
• To obtain the weights for the various objectives,
we begin by forming a matrix A, known as the
pairwise comparison matrix.
• The entry in row i and column j of A, labeled aij,
indicates how much more (or less) important
objective i is than objective j.
• “Importance” is measured on an integer-valued
1-9 scale with each number having the
interpretation shown in the table.
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Pairwise Comparison Matrices - continued
Interpretation of Values in Pairwise Comparison Matrix
Value of aij
Interpretation
1
Objective i and j are equally important.
3
Objective i are slightly more important than j.
5
Objective i are strongly more important than j.
7
Objective i are very strongly more important
than j.
9
Objective i are absolutely more important than j.
• The phrases in this table, such as “strongly more
important than,” are suggestive only.
• They simply indicate discrete points on a continuous
scale that can be used to compare the relative
importance of any two objectives.
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Pairwise Comparison Matrices - continued
• For example, if a13 = 3, then objective 1 is slightly
more important than objective 3. If aij = 4, a value not
in the table, then objective i is somewhere between
slightly and strongly more important than objective j.
• If objective i is less important then objective j, we use
the reciprocal of the appropriate index.
• For example, if objective i is slightly less important
than objective j, the aij = 1/3.
• Finally, for all objectives i, we use the convention that
aij = 1.
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Pairwise Comparison Matrices - continued
• For consistency, it is necessary to set aij = 1/aij.
• This simply states that if objective 1 is slightly
more important than objective 3, then objective 3
is slightly less important than job 1.
• It is usually easier to determine all aij’s that are
greater than 1 and then use the relationship aij =
1/aij to determine the remaining entries in the
pairwise comparison matrix.
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Pairwise Comparison Matrices - continued
• To illustrate, suppose that Jane has identified the
following pairwise comparison matrix for her four
objectives:
1 5 2
4


1 / 5 1 1 / 2 1 / 2

A
1 / 2 2 1
2 


1
/
4
2
1
/
2
1


• The rows and columns of A each correspond to
Jane’s four objectives: salary, quality of life,
interest of work, and nearness to family.
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Pairwise Comparison Matrices - continued
• The entries in this matrix have built-in pairwise
consistency because we require aij = 1/aij for each i
and j.
• However, they may not be consistent when three
alternatives are considered simultaneously.
• For example, Jane claims that salary is strongly more
important than quality of life (a12 = 5) and that salary
is very slightly more important than interesting work
(a13 = 2). But she also says that interesting work is
very slightly more important than quality of life (a32 =
2).
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Pairwise Comparison Matrices - continued
• The question is whether these ratings are all
consistent with one another.
• They are not, at least not exactly.
• It can be shown that some of Jane’s pairwise
comparisons, slight inconsistencies are common
and fortunately do not cause serious difficulties.
• An index that can be used to measure the
consistency of Jane’s preferences will be
discussed later in this section.
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Determining the Weights
• Although the ideas behind AHP are fairly intuitive,
the mathematical reasoning required to derive
the weights for the objectives is quite advanced.
Therefore, we simply describe how it is done.
• Starting with the pairwise comparison matrix A,
we find the weights for Jane’s four objectives
using the following two steps.
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Determining the Weights -continued
1. For each of the columns of A, divide each entry in
the column by the sum of the entries in the column.
This yields a new matrix (call it Anorm, for
“normalized”) in which the sum of the entries in each
column is 1.
For Jane’s pairwise comparison matrix, this step
0.5128 0.5000 0.5000 0.5333
yields
0.1026 0.1000 0.1250 0.0667

Anorm  
0.2564 0.2000 0.2500 0.2667


0
.
1282
0
.
2000
0
.
1250
0
.
1333


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Determining the Weights -continued
2. Estimate wi, the weight for objective i, as the
average of the entries in row i of Anorm. For
Jane’s matrix this yields
0.5128  0.5000  0.5000  0.5333
 0.5115
4
0.1026  0.1000  0.1250  0.0667
w2 
 0.0986
4
w1 
0.2564  0.2000  0.2500  0.2667
 0.2433
4
0.1282  0.2000  0.1250  0.1333
w4 
 0.1466
4
w3 
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Determining the Weights -continued
• Intuitively, why does wi approximate the weight for
objective 1 (salary)? Here is the reasoning.
• The proportion of weight that salary is given in pairwise
comparisons of each objective to salary is 0.5128.
Similarly, 0.50 represents the proportion of total weight
that salary is given in pairwise comparisons of each
objective to quality of life.
• Therefore, we see that each of the four numbers
averaged to obtain wi represents a measure of the total
weight attached to salary. Averaging these numbers
should give a good estimate of the proportion of the
total weight given to salary.
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Determining the Score of Each
Decision Alternative on Each
Objective
• Now that we have determined the weights, we
need to determine how well each job scores on
each objective.
• To determine these scores, we use the same
scale described in the table to construct a
pairwise comparison matrix for each objective.
• Consider the salary objective, for example.
Suppose that Jane assesses the following
pairwise comparison matrix.
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• We denote this matrix as A1 because it reflects her
comparisons of the three jobs with respect to the first
objective of salary.
2 4
 1
A1  1 / 2 1 2
1 / 4 1 / 2 1
• The rows and columns of this matrix correspond to
the three jobs. For example, the first row means that
Jane believes job 1 is superior to job 2 (and even
more superior than job 3) in terms of salary.
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• To find the relative scores of the three jobs on salary,
we now apply the same two-step procedure we did
earlier to the salary pairwise comparison matrix A1.
That is we first divide each column entry by the
column sum to obtain
0.5714 0.5714 0.5714
A1,norm  0.2857 0.2857 0.2857
0.1429 0.1429 0.1429
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• Then we average the numbers in each row to obtain
the vector of scores for the three jobs on salary,
denoted by S1:
.05714 
S1  0.2857
0.1429
• That is, the scores for jobs 1, 2, and 3 on salary are
0.5714, 0.2857, and 0.1429. In terms of salary, job 1
is clearly the favorite.
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• Next we repeat these calculations for Jane's
other objectives.
• Each of these objective requires a pairwise
comparison matrix, which we will denote as A2,
A3, A4.
• Suppose that Jane’s pairwise comparison matrix
for quality of life is
1 1 / 2 1 / 3
A2  2 1 1 / 3
3 3
1 
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• Then the corresponding normalized matrix is
0.1667 0.1111 0.2000
A2,norm  0.3333 0.2222 0.2000
0.5000 0.6667 0.6000
and by averaging, we obtain
0.1593
S 2  0.2519
0.5889
• Here, Job 3 is the clear favorite. However this might
not have much impact because Jane puts relatively
little weight on quality of life.
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• For interest of work, suppose the pairwise
comparison matrix is 1 1 / 7 1 / 3
A3  7 1
3 1 / 3
3
1



• Then the same type calculations show that the
scores for jobs 1, 2, and 3 on interest of work are
0.0882
S3  0.6687
 0.2431
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Determining the Score of Each
Decision Alternative on Each
Objective -- continued
• Finally, suppose the pairwise comparison matrix
for nearness to family is
1 1 / 4 1 / 7 
A4  4 1 1 / 2
7 2
1 
• In this case the scores for jobs 1, 2, and 3 on
nearness to family are
0.0824
S 4   0.3151
0.6025
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Determining the Best
Alternative
• Let’s summarize what we determined so far.
– Jane first assesses a pairwise comparison matrix A that
measures the relative importance of each of her objectives to
one another.
– From this matrix we obtain a vector of weights that
summarizes the relative importance of the objectives.
– Next, Jane assesses a pairwise comparison matrix Ai for
each objective i. This matrix measures how well each job
compares to other jobs with regard to this objective. For each
matrix Ai we obtain a vector of scores Si that summarizes
how the jobs compare in terms of achieving objective i.
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Determining the Best
Alternative -- continued
• The final step is to combine the scores in the Si
vectors with the weights in the w vector.
• Actually we have already done this. Note that the
columns of the Table are the Si vectors we just
obtained.
• If we form a matrix S of these score vectors and
multiply this matrix by w, we obtain a vector of
overall scores for each job, as shown on the next
slide.
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Determining the Best
Alternative -- continued
0.5115
0
.
5714
0
.
1593
0
.
0882
0
.
0824


0.3415
0.0986
  0.3799
S w  0.2857 0.2519 0.6687 0.3151  


0.2433
0.1429 0.5889 0.2431 0.6025
0.2786


0.1466
• These are the same overall scores that we obtained
earlier. As before, the largest of these overall scores is
for job 2, so AHP suggests Jane should accept this job.
Job 1 follows closely behind, with job 3 somewhat
farther behind.
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Checking the Consistency
• As mentioned earlier, any pairwise comparison
matrix can suffer from inconsistencies.
• We now describe a procedure to check for
inconsistencies.
• We illustrate this on the A matrix and its
associated vector of weights w. The same
procedure can be used on any of the Ai matrices
and their associated “weights” vector Si.
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Checking the Consistency -continued
1. Compare Aw. For the example, we obtain
 1
1 / 5
Aw  
1 / 2

1 / 4
5 2
4  0.5115 2.0774
1 1 / 2 1 / 2 0.0986 0.3958


2 1
2  0.2433 0.9894
 
 

2 1 / 2 1  0.1466 0.5933
2. Find the ratio of each element of Aw to the
corresponding weight in w and average these
ratios. For the example, this calculation is
2.0774 0.3958 0.9894 0.5933



0.5115 0.0986 0.2433 0.1466  4.0477
4
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Checking the Consistency -continued
• Compute the consistency index (labeled CI) as
CI 
( Step 2 result )  n
n 1
where n is the number of objective. For the
example this is
4.4077  4
CI 
 0.0159
4 1
• Compare CI to the random index (labeled RI) in
the table on the next slide for the appropriate
value of n.
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Checking the Consistency -continued
Random Indices for Consistency Check for AHP Example
N
2
3
4
5
6
7
8
9
10
RI
0
0.58
0.90
1.12
1.24
1.32
1.41
1.45
1.51
• To be perfectly consistent decision maker, each
ratio in Step 2 should equal n. This implies that a
perfectly consistent decision maker has CI = 0.
• The values of RI in the table give the average
values of CI if the entries in A were chosen at
random.
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Checking the Consistency -continued
• If the ratio of CI to RI is sufficiently small, then the
decision maker’s comparisons are probably
consistent enough to be useful.
• Saaty suggests that if CI/RI < 0.10, then the degree
of consistency is satisfactory, whereas if CI/RI > 0.10,
serious inconsistencies exist and AHP may not yield
meaningful results.
• In Jane’s example, CI/RI = 0.159/0.90 = 0.0177,
which is much less than 0.10. Therefore Jane’s
pairwise comparison matrix A does not exhibit any
serious inconsistencies.
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AHP Template
• The Template file will perform all of the calculations,
including consistency indices and ratios.
• The next slide shows the pairwise judgments used by
Jane
• Using Jane’s information, Job 2 has the highest
preference score, followed by Job 1 and then Job 3
• The results match the additive weighting approach for
the highest preference, but the order of Job 1 and Job 3
are changed (due to subjectivity in weighting & scaling)
• Jane’s pairwise comparisons are reasonably consistent
for all matrices
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