Question

Bob Sparrow purchases steak from a local meatpacking house. The meat is purchased on Monday at $2.00 per pound, and the shop sells the steak for $3.00 per pound. Any steak left over at the end of the week is sold to a local zoo for $0.50 per pound. The possible demands for steak and the probability of each are shown in the following table:

Demand (lbs.) Probability

20 0.2

21 0.3

22 0.5

Bob must decide how much steak to order in a week. Bob wants to maximize expected value. What is his expected value when purchasing optimally? [Hint: construct a payoff table for each of his decisions and each state of nature.] A) 20 B) 20.5 C) 20.25 D) 21 9.

What is Bob Sparrow’s Expected Value of Perfect Information? A) 20.5 B) 1.3 C) 0.8 D) 1.05

Answer 1

The possible outcomes for profit for each initial stock values of 20, 21, and 22 are computed here as:
E(stock of 20) = 20*(3 - 2) = 20 as all will be sold definitely.

P(stock of 21) = 0.2*[ 20*(3 - 2) + 1*(0.5 - 2) ] + 0.8*(21*(3-2)) = 0.2*(20 - 1.5) + 0.8*(21) = 20.5

P(stock of 22) = 0.2*(20*(3 -2) + 2*(0.5 - 2)] + 0.3*[21*(3 - 2) + 1*(0.5 - 2)] + 0.5*22*(3 -2)
= 0.2*(17) + 0.3*(19.5) + 11 = 20.25

Therefore as the maximum profit here is with stock of 21, the maximum profit is given as 20.5

Therefore b) 20.5 is the correct profit when purchased optimally.

The expected profit under certainty is computed here as:
= 0.2*(20)*(3-2) + 0.3*21*(3-2) + 0.5*22*(3-2)
= 4 + 6.3 + 11
= 21.3

Therefore the expected value of perfect information is computed here as:
= Expected profit under certainty - Expected profit when optimal decision is taken
= 21.3 - 20.5

Therefore c) 0.8 is the required value here.