Decision Analysis
EMBA 8150
Dr. Satish Nargundkar
Basic Terms
 Decision Alternatives (eg. Production quantities)
 States of Nature (eg. Condition of economy)
 Payoffs ($ outcome of a choice assuming a state of nature)
 Criteria (eg. Expected Value)
What kinds of problems?
 Mutually Exclusive vs. Mix of alternatives
 Single vs. Multiple criteria
Assumptions
 Alternatives known
 States of Nature and their probabilities are known.
 Payoffs computable under different possible scenarios
Decision Environments
Ignorance – Probabilities of the states of nature are unknown, hence
assumed equal
Risk/Uncertainty – Probabilities of states of nature are known
Certainty – It is known with certainty which state of nature will occur.
Trivial problem.
Example – Decisions under Risk
Assume the following payoffs in $ thousand for 3 alternatives –
building 10, 20, or 40 condos. The payoffs depend on how many
are sold, which depends on the economy. Three states of nature
are considered - a Poor, Average, or Good economy at the time
the condos are completed. Probabilities of the states of nature are
known, as shown below.
A1 (10 units)
A2 (20 units)
A3 (40 units)
Probabilities
S1
(Poor)
S2
(Avg)
S3
(Good)
300
-100
350
600
400
700
-1000
0.30
-200
0.60
1200
0.10
Expected Values
When probabilities are known, compute a weighed average of
payoffs, called the Expected Value, for each alternative and
choose the maximum value.
Payoff Table
S1
A1
300
-100
A2
-1000
A3
Probabilities 0.30
S2
S3
EV
350
600
-200
0.60
400
700
1200
0.10
340
400
-300
The best alternative under this criterion is A2, with a maximum EV
of 400, which is better than the other two EVs.
Expected Opportunity Loss (EOL)
Compute the weighted average of the opportunity losses for each
alternative to yield the EOL.
Opportunity Loss (Regret) Table
S1
A1
A2
A3
Probabilities
S2
S3
0
250 800
400
0
500
1300 800
0
0.30 0.60 0.10
EOL
230
170
870
The best alternative under this criterion is A2, with a minimum
EOL of 170, which is better than the other two EOLs.
Note that EV + EOL is constant for each alternative! Why?
EVUPI: EV with Perfect Information
If you knew everytime with certainty which state of nature was
going to occur, you would choose the best alternative for each
state of nature every time. Thus the EV would be the weighted
average of the best value for each state. Take the best times the
probability, and add them all.
A1 (10 units)
A2 (20 units)
A3 (40 units)
Probabilities
S1
(Poor)
S2
(Avg)
S3
(Good)
300
-100
350
600
400
700
-1000
0.30
-200
0.60
1200
0.10
300*0.3 = 90
600*0.6 = 360
1200*0.1 = 120
_____________
Sum =
570
Thus EVUPI = 570
EVPI: Value of Perfect Information
If someone offered you perfect information about which state of
nature was going to occur, how much is that information worth to
you in this decision context?
Since EVUPI is 570, and you could have made 400 in the long
run (best EV without perfect information), the value of this
additional information is 570 - 400 = 170.
Thus,
EVPI = EVUPI – Evmax
= EOLmin
Decision Tree
0.3
340
0.6
0.1
A1
0.3
0.6
A2
A2
400
400
A3
0.1
| 300
| 350
| 400
| -100
| 600
| 700
0.3
0.6
| -1000
0.1
| 1200
-300
| -200
Sequential Decisions
 Would you hire a consultant (or a psychic) to get more info about states
of nature?
 How would additional information cause you to revise your
probabilities of states of nature occurring?
 Draw a new tree depicting the complete problem.
Consultant’s Track Record
Instances of actual
occurrence of
states of nature
Previous
Economic
Forecasts
Favorable
Unfavorable
S1
20
80
100
S2
60
40
100
S3
70
30
100
Probabilities
 P(F/S1) = 0.2
 P(U/S1) = 0.8
 P(F/S2) = 0.6
 P(U/S2) = 0.4
 P(F/S3) = 0.7
 P(U/S3) = 0.3
 F= Favorable
U=Unfavorable
Joint Probabilities
S1
S2
S3
Total
Favorable
0.06
0.36
0.07
0.49
Unfavorable
0.24
0.24
0.03
0.51
Prior
Probabilities
0.30
0.60
0.10
1.00
Posterior Probabilities
 P(S1/F) = 0.06/0.49 = 0.122
 P(S2/F) = 0.36/0.49 = 0.735
 P(S3/F) = 0.07/0.49 = 0.143
 P(S1/U) = 0.24/0.51 = 0.47
 P(S2/U) = 0.24/0.51 = 0.47
 P(S3/U) = 0.03/0.51 = 0.06
Solution
 Solve the decision tree using the posterior probabilities just computed.