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	s1	s2
d1	10	1
d2	7	3

(a)	Suppose P(s1)=0.2 and P(s2)=0.8. What is the best decision using the expected value approach? Round your answer in one decimal place.
	The best decision is decision alternative (- Select your answer -d1 /d2) with an expected value of.
(b)	Perform sensitivity analysis on the payoffs for decision alternative d1. Assume the probabilities are as given in part (a), and find the range of payoffs under states of nature s1 and s2 that will keep the solution found in part (a) optimal. Is the solution more sensitive to the payoff under state of nature s1 or s2? Round your answer in two decimal places.
	The solution is more sensitive to (Select your answer -s1 / s2), as an increment of .......... for d1(s2) causes a break even between the decision alternatives, whereas it takes an increment of .......... , to provide the same effect for d1(s1).

Answer 1

a) First we have to calculate the expected value to find the best decision alternative

P(s1) = 0.2

P(s2) = 0.8

For d1

Expected value = 10*0.2+1*0.8

= 2+0.8

= 2.8

For d2

E value = 7*0.2+3*0.8

= 1.4+2.4

= 3.8

Hence we can say that best decision alternative is d2 with an expected value of 3.8

b) let P be the probability of s1 and the probability of the state of nature s2 is 1-P

By the given payoff table expected values are

EV(d1) = P*10 + 1*(1-P)

= 10P -P+1

= 9P+1

Also for d2

EV(d2) = 7P + 3(1-P)

= 7P +3 -3P

= 4P + 3

The value of P can be calculated by equating both expected value

Then the value of

9P+1 = 4P+3

5P = 2

P = 2/5

P = 0.4

Here by taking a graph we can say that

0.2s + 0.8*3 >= 2.8

0.2s >= 2.8- 2.4

0.2s >= 0.4

S>= 0.4/0.2

S >= 2

Hence the solution is more sensitive to S2 as an increment of 0.8 for

d1 (s2) causes a break even between the decision alternative whereas it takes an increment of 1.8 to provide the same effect for d1 (s1).