Engineering Economic Analysis

314 UNCERTAINTY IN FUTURE EVENTS

The first cost of the project in Example 10-5 is $25,000. Use the expected values for annual

benefits and life to estimate the present worth. Use an interest rate of 10%.

EVbenefit=5000(0.3) + 8000(0.6) + 10,000(0.1)=$7300

EVlife=6(2/3) + 9(1/3) =7 years

The PW using these values is

PW(EV) =-25,000 +7300(P /A, 10%,7) = -25,000 +6500(4.868) =$10,536

[Note:This is the present worth of the expected values, PW(EV), not the expected value of the

present worth, EV(PW). It is an easy value to calculate that approximates the EV(PW), which

will be computed using the joint probability distributioI).found in Example 10.6.]

Example 10-7 is a simple way to approximate the project's expected PW. But the true

expected value of the PW is somewhatdifferent.To findit, we must use thejoint probability

distribution for benefit and life, and the resulting probability distribution function for PW

that was derived in Example 10-6. Example 10-8 shows the expected value of the PW or

the EV(PW).

Use the probability distributionfunction of the PW that was derived in Example 10-6 to calculate

the EV(PW). Does this indicate an attractive project?

The table from Example 10-6 can be reused with one additional column for the weighted values

of the PW(=PW x probability). Then, the expected value of the PW is calculated by summing

the column of present worth values that have been weighted by their probabilities.

Annual Life Joint

.';" Benefit Probability (years) Probability Probability

$ 5,000 30% 6 66.7% 20.0%

8,000 60 6 66.7 40.0

10,000 10 6 66.7 6.7

5,000 30 9 33.3 10.0

8,00000 9 33.3 20.0

10,000 10 9 33.3 -3.3

,. 100.0%

PW x Joint

PW Probability

-$ 3,224 -$ 645

9,842 3,937

18,553 1,237

3,795 380

21,072 4,214

32,590 1,086

EV(PW) =$10,209