Question

The following payoff table shows the profit for a decision problem with two states of nature and two decision alternatives:

	State of Nature
Decision Alternative	s 1	s 2
d 1	6	3
d 2	9	1

1. Use graphical sensitivity analysis to determine the range of probabilities of state of nature s ₁ for which each of the decision alternatives has the largest expected value. If required, round your answers to two decimal places.
   

   d1 is optimal for probability of s 1	≥ ??
   d2 is optimal for probability of s 1	≤ ??

2. Suppose P(s₁) = 0.85 and P(s₂) = 0.15. What is the best decision using the expected value approach?
   

   Best decision:

   ??

3. Perform sensitivity analysis on the payoffs for decision alternative d₁. Assume the probabilities are as given in part (b), and find the range of payoffs under states of nature s₁ and s₂ that will keep the solution found in part (b) optimal. Is the solution more sensitive to the payoff under state of nature s₁ or s₂?
   
   The input in the box below will not be graded, but may be reviewed and considered by your instructor.

Answer 1

(a) Graphical representation:

Expected Value of d1 = p (6) +(1-p)3 = 6p+3-3p = 3p+3

Expected value of d2 = p (9) +(1-p)1 = 9p+1-p = 8p+1

Insert the graph using the following data in excel.

Probability	d1	d2
0.1	(3*0.1) +3 = 3.3	(8*0.1) +1 = 1.8
0.15	3.45	2.2
0.2	3.6	2.6
0.25	3.75	3
0.3	3.9	3.4
0.35	4.05	3.8
0.4	4.2	4.2
0.45	4.35	4.6
0.5	4.5	5
0.55	4.65	5.4
0.6	4.8	5.8
0.65	4.95	6.2
0.7	5.1	6.6
0.75	5.25	7
0.8	5.4	7.4
0.85	5.55	7.8
0.9	5.7	8.2
0.95	5.85	8.6
1	6	9

Here, below 0.4, d1 is optimal. Above 0.4, d2 is optimal.

3p+3 = 8p+1

8p-3p = 3-1

5p = 2

p = 2/5 = 0.4

d1 is optimal for probability s1≤0.4

d2 is optimal for probability of s1≥0.4

(b)

Expected value approach:

Best value will be for d2 as mentioned in the above table, for 0.85 the payoff from d1 = 5.55 and for d2, it is 8.2

d 1 = (0.85*6) +(0.15*3) = 5.1+ 0.45 = 5.55

d 2 = (0.85*9) +(0.15*1) = 7.65+0.15 = 7.80

(c)

The best decision was for d2.

S = payoff of d2 under s1.

EV(d2) = (0.85) *s + (0.15)1 ≥ 5.55

0.85s+0.15≥5.55

0.85s≥5.55-0.15

0.85s≥5.40

s = 5.40/0.85 = 6.3529 = 6.35

As long as the payoff for s1 is greater than 6.35, d2 will be optimal in this given payoff matrix.