Question 18.17: Uncertain N Requires Probability Distribution for PW for Exa...
Uncertain N Requires Probability Distribution for PW for Exact Answer
Aproject has a first cost of $22,000, no salvage value, and a net annual benefit of $4000. The firm uses an interest rate of 12%. The benefits may last 5, 10, or 15 years, with respective probabilities of 20%, 60%, and 20%. What is the expected value of the PW?
An approximate answer can be obtained by using the expected value for the number of years, 10 years. This is the best single estimate of the project’s life. Since 5 and 15 years are equidistant from 10 years and they have the same probability, the probability distribution is symmetric, and the mean (average) is the point of symmetry. Mathematically, the average life of 10 years equals .2 · 5 + .6 · 10 + .2 · 15. The PW for this life is:
PW = −22,000 + 4000(P/A,12%,10) = $601
The exact answer is obtained by finding the PW for each possible life, then finding the expected value, as shown in Exhibit 18.11.
EXHIBIT 18.11 Exact expected value computations with uncertain N
| P.PW^{2} | P.PW | PW | P | N |
| $11,494,312 | −$1516 | −$7581 | .2 | 5 |
| 216,721 | 361 | 601 | .6 | 10 |
| 5,497,810 | 1049 | 5243 | .2 | 15 |
| 17,208,843 | Expected values −$106 | |||
σ_{PW} = \sqrt{17,208,843 − (−106^{2})} = \$4147
The exact value of the expected PW is negative, and it indicates a different recommendation than the approximate value does. That is, that the project should not be undertaken. Notice also that the standard deviation of the PW is relatively large.
It should be noted, however, that the approximation is not that far off. The $707 difference in calculated PWs is small relative to the uncertainty in the original values.