Question

An energy efficiency project has a first cost of $400,000, a life of 10 years, and no salvage value. Assume that the interest rate is 10%, The most likely value for annual savings is $50,000. The optimistic value for annual savings is $80,000 with a probability of 0.2. The pessimistic value is $40,000 with a probability of 0.25.

PW for Optimistic ($80,000)= $91,600
PW for Most Likely ($50,000)=$92,750
PW for Pessimistic ($40,000) =$154,200
PW values (above) come from the calculation:
PW-(Annual Savings)(P/A, 1096, 10)-$400,000
a) What is the expected annual savings and the expected PW?
b) Compute the PW for the pessimistic, most likely, and optimistic estimates of the annual savings. What is the expected PW?
c) Do the answers for the expected PW match? Why or why not?

Answer 1

a) What is the expected annual savings and the expected PW?

Expected annual savings = ($80,000 x 20%) + ($50,000 x 55%) + ($40,000 x 25%) = $53,500

b) Compute the PW for the pessimistic, most likely, and optimistic estimates of the annual savings. What is the expected PW?

PW of expected annual savings = ($53,500 x 6.145) - $400,000 = $328,757.50 - $400,000 = $71,242.50

c) Do the answers for the expected PW match? Why or why not?

Expected PW = ($91,600 x 20%) + (-$92,750 x 55%) + (-$154,200 x 25%) = -$71,242.50

Yes, they match.