CHAPTER 9 DECISION ANALYSIS
9.1 Introduction to Decision Analysis
Elements:
(1) Decision maker (“we”),
(2) Choices (decisions, strategies),
(3) Payoffs
Selection problems. Define a binary variable yj for each
decision with yj = 1, if we make decision j, and 0
otherwise. A selection problem with n possible
decisions will then feature the constraint y1 + y2 + …+
yn = 1.
Multiple criteria (MCDM = multiple criteria decision
making vs multiple states of nature – decision analysis
= games against nature.
Example: Consider a problem with three decisions d1,
d2, and d3, and four states of nature s1, s2, s3, and s4.
The payoffs are shown below.
d1
d2
d3
s1
3
2
5
s2
2
0
2
s3
4
4
0
s4
6
1
3
This game is
(1) asymmetric: the decision maker is rational (looks at
the payoffs), while nature is a random player
(2) a simultaneous game (we do not know in advance
what state of nature will be chosen).
Consider the continuum between
certainty ― … ― risk ― …― uncertainty
Certainty: we know exactly which strategy nature will
play.
Risk: we know the probability distribution nature uses
to play her strategies (e.g., by way of past
observations)
Uncertainty: we do not know even the probability
distribution of nature’s strategies (e.g., new untried
product).
Certainty is trivial: Since we know what column
nature plays, all we have to do is choose the row that
leads to the highest payoff.
9.2 Visualizations of Decision Problems
Macro view: Influence diagrams.
Micro view: Decision trees.
Influence diagrams: Decision nodes, random nodes,
consequence nodes. Shows general interrelations
between decisions, chance events, & results/outcomes.
Example:
Decision
Random event
Add electronics General economic
department
conditions
Relocate
Local acceptance
department
of services
into a separate
building
Consequence
Profit
The broken arcs: possible influences (local acceptance
of an electronics department or store may be
influenced by the existence of an electronics
department in our department store and our
competitors’ reaction to our introduction of the
department).
Decision tree for the same problem:
9.3 Decision Rules Under Uncertainty and
Risk
Start with uncertainty. (Little input on our part, only
crude information will come out).
Example:
d1
d2
d3
d4
s1
2
0
2
2
s2
2
1
1
3
s3
5
7
1
4
Before commencing, check for dominances. One
decision (row) dominates another, if its payoffs are all
greater or equal than those of a single other row.
Here, d1 dominates d4. Dominated decisions can be
deleted. Column dominances do not exist.
Checking for dominances requires a total of ½m(m1)
pairs of comparisons. General procedure for all
decision rules: Determine anticipated outcomes for all
decisions.
Decision rules under uncertainty:
(1) Wald’s rule (pessimist’s rule, maximin rule): for
each decision, the anticipated outcome is the worstcase outcome. Among the anticipated outcomes,
choose the best. The decision that leads to this outcome
is chosen.
In our example, the anticipated outcomes are 2, 1, 1,
& 3 (in case we did not eliminate decision d4), the
maximum is 1, which belongs to d3. This is the chosen
decision.
(2) Optimist’s rule. For each decision, the anticipated
outcome is the state of nature with the highest payoff.
Among those, we choose the decision with the highest
anticipated payoff → a maximax rule.
In this example, the anticipated payoffs are 5, 7, 2, &
4. Choose d2 that―in our opinion―leads to a payoff of
7.
(3) Savage – Niehans rule (minimax regret criterion).
First generate regrets, by comparing each payoff with
the highest payoff given that state of nature (i.e., in
that column). Then apply Wald’s rule to the regret
matrix.
In this example, the regret matrix is
0
2
R= 
0
0

3 2
2 0

0 6
4 3
Applying Wald’s rule means choosing the worst
possible outcomes for each decision (i.e., the highest
regrets) & then choose the best (i.e., the lowest) among
them. This makes Wald’s rule applied to this regret (=
cost) matrix a minimax rule.
Decision making under risk.
Given probabilities for all states of nature, e.g., p = [.5,
.3, .2].
(4) Bayes’s rule: Choose the maximum of the expected
values (a weighted maxisum rule).
In the example, the expected payoffs (or expected
monetary values EMV):
EMV(d1) = 2(.5)  2(.3) + 5(.2) = 1.4,
EMV(d2) = 0(.5)  1(.3) + 7(.2) = 1.1,
EMV(d3) = 2(.5) + 1(.3) + 1(.2) = 1.5, &
EMV(d4) = 2(.5)  3(.3) + 4(.2) = .9.
The anticipated payoffs are 1.4, 1.1, 1.5, and 0.9 →
choose d3.
Newspaper Boy Problem. Decide how man papers to
purchase, given that the demand is uncertain.
Example: The demand is 10, 20, 30, or 40 newspaper
on any given day. Purchase a newspaper for 20¢ and
sell for 90¢. The salvage value of an unsold newspaper
is 5¢, while the opportunity cost for newspapers has
been estimated to be 15¢ for each newspaper that
could have been sold but was not due to the lack of
supply. Suppose that the purchasing strategies are 10,
20, 30, & 40.
Payoff matrix
d1
A=
d2
d3
d4
s1
s2
s3
s4
$7.00 5.50 4.00 2.50 
 5.50 14.00 12.50 11.00 


 4.00 12.50 21.00 19.50 
 2.50 11.00 19.50 28.00


Given probabilities of p = [.6, .2, .1, .1], the decisions
have expected payoffs of $5.95, $8.45, $8.95, and $8.45,
→buy 30 newspapers & expect a daily payoff of $8.95.
(5) Using target values T. Idea: Plot payoff values
against the probability that the value can be achieved.
The original payoff matrix was
d1
d2
d3
d4
s1
2
0
2
2
s2
2
1
1
3
s3
5
7
1
4
p = [.5, .3, .2].
d1: solid line
d2: broken line
d3: dotted line
Decision rules (by way of upper envelope):
If T < 2, any decision will achieve the target.
If T  [2, 1], d2 and d3 are best. Both will reach
the target with a probability of 1.
If T  [1, 1], d3 is best. It reaches the target with
a probability of 1.
If T  [1, 2], d1 is best. It reaches the target with
a probability of .7.
If T  [2, 5], d1 and d2 are best. Both achieve the
target with a probability of 0.2.
If T  [5, 7], d2 is best. It achieves the target with
a probability of .2.
If T > 7, none of the decisions will be able to reach
the target.
9.4 Sensitivity Analyses
“What if” individual payoffs aij change?
d1
d2
d3
d4
p
s1
2
0
2
2
.5
s2
2
1
1
3
.3
s3
5
7
1
4
.2
Suppose that we are uncertain about a23. Rewrite the
payoff as a23 = 7 +  with an unknown   [2, 3],
meaning that we expect the payoff to be between 5 &
10.
Expected payoffs:
 1 .4 
1.1  .2 
.
EMV = 
 1 .5 
 .9 


Clearly, d1 & d4 are dominated. The payoffs for the
remaining strategies are shown in the figure below.
This leads to
If  > 2 (i.e., a23 > 9), then decision d2 is best, &
if   2 (i.e., a23  9), decision d3 is best.
Another source of uncertainty
magnitude of the probabilities.
relates
to
the
Suppose that we are unsure about p1. Similar to the
above, we can use p1 +  with some unknown .
However: The sum of probabilities must equal 1, so if
p1 increases, the other probabilities must decrease by
. Assume that the other two probabilities decrease by
the same amounts, i.e.
p = [.5 + , .3  ½, .2  ½].
Given the same payoff matrix
d1
d2
d3
d4
p
s1
2
0
2
2
.5
s2
2
1
1
3
.3
s3
5
7
1
4
.2
we can compute the expected values as
1.4  .5 
 1.1  3 
.
EMV() = 
 1.5  1 
.9  1.5 


Suppose we estimate that p1 will be between .3 & .6.
(Alternatively, starting with p1 =.5, the change  
[.2, +.1].
Within this range, decision d3 dominates d1 & d4,
which can be deleted.
The expected monetary values for d2 & d3 are shown in
the figure below:
Decision rule:
If   .1 (i.e., p1  .4), then decision d2 is best, &
if  > .1 (i.e., p1 > .4), then decision d3 is best.
Different example: Same payoff matrix, but as p1  ,
p2  ⅔ & p3  ⅓. The expected payoffs are then
1.4  5  
3 

5
1.1  3  
 1 .5    .


8
 .9  3  
Decision rule:
If  <.15 (i.e., p1 < .35), then decision d2 is optimal,
if   .15 (i.e., p1  .35), then decision d3 is optimal.
9.5 Decision Trees and the Value of
Information
Value of information. Extreme case first:
Expected value of perfect information (EVPI):
Difference between the expected payoff with perfect
information minus the expected payoff without
information (beyond the prior probabilities p).
Previous example:
d1
d2
d3
d4
p
s1
2*
0
2*
2*
.5
s2
2
1
1*
3
.3
s3
5
7*
1
4
.2
p = [.5, .3, .2].
The best responses to nature’s strategies (in this
sequential game, we assume that while we cannot
influence nature’s choice, we know about it before we
make our move) are shown with asterisks. The
expected payoff with perfect information EPPI is then
EPPI = 2(.5) + 1(.3) + 7(.2) = 2.7.
As the expected payoff without perfect information
(i.e., the expected monetary value of the best strategy
EMV*) was 1.5 (achieved by using d3), the expected
value of perfect information is then
EVPI = EPPI  EMV* = 2.7  1.5 = 1.2.
Now imperfect information.
Image a forecasting institute that uses indicators I1, I2,
… to forecast the states of nature s1, s2, … . Clearly,
the indicators & the states of nature should be related.
(typical examples are demand & wholesaler’s orders,
the changes of the supply of money & the rate of
inflation, etc.).
The relations between indicators & states of nature
may be given in the form of conditional probabilities.
P(I|s):
s1
I1 .6
I2 .4
s2
.9
.1
s3
.2
.8
Notice: The state s1 is linked only very weakly to the
indicators.
Decision tree:
From left to right, the tree depicts the sequence of
events. Decision modes (we decide) are squares, event
nodes (nature makes a random choice) are circles, &
terminal nodes (there are no further muves) as
triangles.
Note: the lower part of the tree equals what we have
already done in the matrix when we determined the
EMV* strategy.
Numbers that are needed:
(1) Payoffs (at the terminal nodes)
(2) Indicator probabilities P(I) at nature’s choice of
indicator
(3) Posterior probabilities P(s|I) at nature’s choice of
the state of nature.
While the payoffs are known, the probabilities in (2) &
(3) are not.
However, it is possible to use the known probabilities
P(I|s) to compute the indicator probabilities P(I) & the
posterior probabilities P(s|I). For that purpose, we use
Bayes’s theorem: P(si|I1) = P(I1|si)P(si)/P(I1) & similar
for all other indicators.
Here:
For I1, we compute P(I1) and P(s|I1)
s P(s)
s1
.5
s2
.3
s3
.2
P(I1|s) P(I1|s)P(s)
.6
.30
.9
.27
.2
.04
P(I1) = .61
P(s|I1)
.4918
.4426
.0656
For I2, we compute P(I2) and P(s|I2)
s
s1
s2
s3
P(s)
.5
.3
.2
P(I2|s) P(I2|s)P(s)
.4
.20
.1
.03
.8
.16
P(I2) =.39
P(s|I2)
.5128
.0769
.4103
The complete decision tree is then as follows:
The numbers next to the nodes are computed by
backward recursion. The recursion starts at the
triangular terminal nodes & works backwards to the
root of the tree.
Two rules apply:
(1) Back into a decision node: choose the best strategy,
i.e., the one with the highest (expected) payoff.
(2) Back into an event node: compute the expected
value of all successor nodes.
Here, the result is the payoff with imperfect information
EPII = 2.05.
Now compare: without information (beyond prior
probabilities), we can get EMV* = 1.5. With additional
information, we can get EPII = 2.05.
Hence the expected value of perfect information
EVSI = 2.05  1.5 = 0.55.
A standardized measure is efficiency E, which is
E = .55/1.2 = .4583.
The same example with different numbers:
s1
P(I|s): I1 .9
I2 .1
s2
.6
.4
s3
.2
.8
Note: This matrix is the same as before with some
columns exchanges. However, now s1 is strongly linked
to the indicators, & P(s1) = .5, so the result should be
(at least somewhat) better.
Computation of P(I1) and P(s|I1):
s
s1
s2
s3
P(s)
.5
.3
.2
P(I1|s) P(I1|s)P(s)
.9
.45
.6
.18
.2
.04
P(I1) = .67
P(s|I1)
.6716
.2687
.0597
Computation of P(I2) and P(s|I2):
s
s1
s2
s3
P(s)
.5
.3
.2
P(I2|s) P(I2|s)P(s)
.1
.05
.4
.12
.8
.16
P(I2) =.33
P(s|I2)
.1515
.3636
.4848
We then obtain EPII = 2.12, so that
EVSI = 2.12  1.5 = .62 & E = .62/1.2 = .5167.
Extreme examples: Random indicators result in the
prior probabilities equaling the posterior probabilities
& EVSI = 0.
On the other hand, while a forecast that is always
correct will have EVSI = EVPI & E = 1, a forecast that
is always wrong has the same features → it is the
consistency of the indicators that is important, not
their actual meaning.
9.6. Utility Theory
Expected values are meaningful, only if decision
makers are risk neutral. This means, they should be
indifferent to either
(1) receiving $100,000 in cash, no questions asked, or
(2) playing the lottery with a 50% of winning $200,000
& a 50% chance of winning nothing.
Most people would prefer (1) i.e., they are not risk
neutral.
Once the utilities have been determined, they can be
used instead of payoffs. All procedures remain
unchanged.