1. Decision Analysis
MBA 8040
Dr. Satish Nargundkar

2. 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)

3. 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

4. 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.

5. 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. S1
(Poor) S2
(Avg) S3
(Good)
A1 (10 units)
300
350
400
A2 (20 units)
-100
600
700
A3 (40 units)
-1000
-200
1200
Probabilities
0.30
0.60
0.10

6. 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 S2 S3 EV A1
300
350
400
340 A2
-100
600
700 A3
-1000
-200
1200
-300
Probabilities
0.30
0.60
0.10
The best alternative under this criterion is A2, with a maximum EV of 400, which is better than the other two EVs.

7. Expected Opportunity Loss (EOL)
Compute the weighted average of the opportunity losses for each alternative to yield the EOL.
Opportunity Loss (Regret) Table S1 S2 S3
EOL A1
250
800
230 A2
400
500
170 A3
1300
870
Probabilities
0.30
0.60
0.10
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?

8. 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.
300*0.3 = 90
600*0.6 = 360
1200*0.1 = 120
_____________
Sum =
Thus EVUPI = 570 S1
(Poor) S2
(Avg) S3
(Good)
A1 (10 units)
300
350
400
A2 (20 units)
-100
600
700
A3 (40 units)
-1000
-200
1200
Probabilities
0.30
0.60
0.10

9. 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 = 170. Thus, EVPI = EVUPI – Evmax = EOLmin

10. Decision Tree
| 300
0.3
340
0.6
| 350
0.1
| 400 A1
| -100
0.3 A2
0.6
| 600 A2
400
400
0.1
| 700 A3
0.3
| -1000
0.6
| -200
-300
0.1
| 1200

11. 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.

12. Consultant’s Track Record
Instances of actual
occurrence of
states of nature S1 S2 S3
Favorable 20 60 70
Unfavorable 80 40 30
100
Previous
Economic
Forecasts

13. Probabilities P(F/S1) = 0.2 P(U/S1) = 0.8 P(F/S2) = 0.6 P(U/S2) = 0.4
F= Favorable U=Unfavorable

14. Joint Probabilities S1 S2 S3 Total Favorable 0.06 0.36 0.07 0.49S1 S2 S3
Total
Favorable
0.06
0.36
0.07
0.49
Unfavorable
0.24
0.03
0.51
Prior
Probabilities
0.30
0.60
0.10
1.00

15. 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

16. Solution
Solve the decision tree using the posterior probabilities just computed.