Question

# The energy efficiency project described in Problem 10-23 has a first cost of \$150,000, a life...

The energy efficiency project described in Problem 10-23 has a first cost of \$150,000, a life of 10 years, and no salvage value. Assume that the interest rate is 8%. a) What is the equivalent uniform annual worth for the expected annual savings? b) Compute the equivalent uniform annual worth for the pessimistic, most likely, and optimistic estimates of the annual savings. What is the expected value of the equivalent uniform annual worth? c) Do the answers to (a) and (b) match? Why or why not? (Problem #23: Annual savings due to an energy project have a most likely value of \$30,000. The high estimate of \$40,000 has a probability of .25 and the low estimate of \$20,000 has a probability of .35.)

a. EUAB for the expected savings= 40000*0.25 + 20000*0.35 + 30000*(1-0.25-0.35)

= \$29000

EAC of the capital investment = Capital cost *(A/P,8%,10 years)

= 150000* 0.149

= \$22350

Annual worth= EUAB-EUAC

29000- 22350 = \$6650

b.1. Annual worth for the pessimist= EUAB for pessimist- Capital cost

= 20,000- 22350 = \$ - 2350

2. Annual worth for optimist= EUAB of optimist- Capital cost

= 40000- 22350 = \$17650

3. Annual worth of most likely case= 30000- 22350 = \$7650

Expected value= -2350*0.35+ 17650*0.25+ 7650*0.4

= \$6650

c) Answers are same due to the presence and consideration of probabilities. In the first case we first calculated the expected value after considering probabilities. In the second we first computed the present worth and then multiplied with probabilities to arrive at the same value.

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