Question

The following payoff table provides profits based on various possible decision alternatives and various levels of demand at Amber Gardner's software firm:

		Demand Level
		0.3	0.7
		Low	High
Alternative	A	$10,000	$30,000
	B	$5,000	$40,000
	C	($2,000)	$50,000
*Profits in $ thousands

a. Plot the expected-value lines on a graph. (Answered below)

Alternative		Demand Level
		0	1
	A	$ 10,000.00	$ 30,000.00
	B	$   5,000.00	$ 40,000.00
	C	$ (2,000.00)	$ 50,000.00

560,000.00 550,000.00 $40,000.00 530,000.00 520,000.00 $10,000.00 0 S(10,000.00)

b. Is there any alternative that would never be appropriate in terms of maximizing expected profit? Explain on the basis of your graph. (see graph in part a)

c. For what range of P(High Demand) would alternative A be the best choice if the goal is to maximize expected profit?

d. For what range of P(High Demand) would alternative B be the best choice if the goal is to maximize expected profit?

e. Compute the expected values for each alternative if the probability of low demand level is 0.30. Which of the the options is best under this probability?

f. Using the probability of low demand as 0.3 (therefore probability of high demand = 0.7), compute the EVPI. Explain the significance of this number (i.e. what does it mean?).

Answer 1

a)

1) Expected profit Probability tligg 2 Low 0 4 0.05 5 0.1 -10000 A4+B4 30000 5000*A4+40000 B4 -2000 A4+50000 B4 =1-A4 1-A5 = 1-A6 -10000 A6+B6 30000 5000 A6+40000 B6 -2000 A6+50000 B6 10000 A7+B7 30000 10000 A8+B8 30000 5000 A8+40000 B8 E-2000 A8+50000 B8 8 0.25 9 0.3 10 0.35 F1-A8 -1-A10 1-A11 1-A12 -1-A13 1-A14 1-A15 =1-A16 12 0.45 13 0.5 14 0.55 15 0.6 16 0.65 17 0.7 18 0.75 19 0.8 20 0.85 21 0.9 22 0.95 23 24 10000 A10+B10 30000 5000 A10+40000 B1E-2000 A10+50000 B10 10000 A11+B11 30000 5000 A11+40000 B1E-2000 A11+50000 B11 10000 A12+B12 30000 5000 A12+40000 B1E-2000 A12+50000 B12 10000 A13+B13 30000 5000 A13+40000 B1-2000 A13+50000 B13 10000 A14+B14 30000 5000 A14+40000 B1 2000 A14+50000 B14 10000 A15+B15 30000 5000 A15+40000 B1-2000 A15+50000 B15 -10000 A16+B16 30000 5000 A16+40000 81-2000 A16+50000 816 10000 A17+B17 30000 5000 A17+40000 81 2000 A17+50000 B17 10000 A18+B18 30000 5000 A18+40000 B1-2000 A18+50000 B18 10000 A19+B19 30000 5000 A19+40000 B1-2000 A19+50000 B19 10000 A20+B20 30000 5000 A20+40000 B2-2000 A20+50000 B20 10000 A21+B21 30000 5000 A21+40000 B2-2000 A21+50000 B21 10000 A22+B22 30000 5000 A22+40000 B2 2000 A22+50000 B22 10000 A23+B23 300005000 A23+40000 B2-2000 A23+50000 B23 -1-A18 1-A19 -1-A20 1-A21 1-A22 1-A23

G23 0 Probability 2 LowHigh A Expected profit 30000 0 2900038250 47400 0.9 0 2600033000 0.75 0.7 0 0.6 0.55 0.4 0.45 13 05 5002500 24000 14 0550.45 1900020750 21400 0.4 1800019000 16 0.65 0.3517000 1725016200 0.6 0.8 0 15000 13750 8400 HIGH DEMAND PROBABILITY A B -C 0 0 5800 0.1 0 10000 5000 3

b)

There is no such alternative. Every alternative has the highest payoff at some probability range

c)

Alternative A is best for H=0 and above

Let the upper limit be p

When P(high) =p, A and B would be same and after that B would give better payoff

To find p, Expected payoff of A = expected payoff of B

10000*(1-p)+30000*p = 5000*(1-p)+40000*p

p = 0.333

d)

The lower limit is found is part c.

At the upper limit, ::

At this probability payoff of B and C would be same

Let probability of high demand be p

5000*(1-p)+40000*p = -2000*(1-p)+50000*p

p = 0.412

Therefore, B is best for 0.333<= P(high) <= 0.412

e)

1) Expected profit 2 Low High A 0.3 0.7 24000 29500 34400

Can be seen from the table in the first part

f)

Expected value with out perfect information = max(24000,29500,34400) = 34400

When there is perfect information, under each demand condition, best possible alternative is chosen.

Expected value with perfect information = 0.3*10000+0.7* 50000 = 38000

EVPI = 38000-34400= 3600