Question

# An energy efficiency project with a first cost of \$150,000, life of 10 years, and with...

An energy efficiency project with a first cost of \$150,000, life of 10 years, and with no salvage value has a most likely value of \$30,000. The high estimate of 40,000 has a probability of 0.2, and the low estimate of \$20,000 has a probability of 0.3.

Assume that interest rates are 8%. What is the present worth based on the probabilities given?

First Cost of the project = 150,000

Life = 10 years

Interest = 8%

Annual Benefits – most likely = 30,000

High estimates = 40,000 with probability of 0.2

Low estimates = 20,000 with probability of 0.3

Step 1

Calculate the expected annual benefits

High Estimates = 40,000 * 0.2 = 8,000

Low Estimates = 20,000 * 0.3 = 6,000

Most likely estimates = 30,000 * (1 – 0.3 – 0.2) = 15,000

Expected Annual Benefits = 8,000 + 6,000 + 15,000 = 29,000

Step 2

Calculate the Present Worth

PW = -150,000 + 29,000 (P/A, 8%, 10)

PW = -150,000 + 29,000 (6.71008) = 44,592.32

The present worth of the system is 44,592.32

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