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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 years, 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.

1.What is the expected annual savings and the expected PW?
2.Compute the PW for the pessimistic, most likely, and optimistic estimates of the annual savings. What is the expected PW?
3.Do the answers for the expected PW match? Why or why not?

1 Answer

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Answer:

See attached files

Step-by-step explanation:

answered
User Sanket Vetkoli
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