Question

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.)

Answer #1

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.

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?

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?

A project is being considered that has the first cost of
$12,500, creates $5,000 in annual cost savings, requires $3,000 in
annual operating costs and has a salvage value of $2,000 after a
project life of 3 years with 10% annual interest. Calculate the
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A project activity has an optimistic time estimate of three
days, a most likely time estimate of eight days, and a pessimistic
time estimate of 10 days. The expected time (in days) of this
activity is:
7
7.5
8
8.5
10

1/ A retrofit project is estimated to cost $150,000. The
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upgrade of $10,000 at year 15. The retrofit is expected to save 100
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b/ What is the equivalent cost of the saved energy?

A project has the following tasks and duration estimates given.
These three tasks are done in succession on a project.
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Optimistic Estimate
Most Likely Estimate
Pessimistic Estimate
Alpha
30 days
40 days
50 days
Beta
20 days
25 days
28 days
Charlie
4 days
10 days
12 days
Delta
25 days
30 days
50 days
What is the expected duration for the project? (3 points)
What is the variance for the project? (3 points)
What is the standard deviation...

A project with a life of 10 has an initial fixed asset
investment of $41,580, an initial NWC investment of $3,960, and an
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the life of the project and has no salvage value.
If the required return is 15 percent, what is the project's
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A project with a life of 10 has an initial fixed asset
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the life of the project and has no salvage value.
If the required return is 20 percent, what is the project's
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Multiple Choice
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If the required return is 11 percent, what is the project's
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Multiple Choice
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$-41,332.63
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