Engineering Economic Analysis

308 UNCERTAINTY IN FUTURE EVENTS

than on the other two. This equation was developed as an approximation with the beta

distribution.

Optimistic value + 4(Most likely value) + Pessimistic value

Mean value= 6 (10-1)

This approach is illustrated in Example 10-4.

Solve Example 10-3 by using Equation 10-1. Compute the resulting mean rate of return.

.SOLUTION

Compute the mean for each parameter:

Mean cost =[950+ 4 x 1000 + 1150]/6=1016.7

Mean net annual benefit- [210 + 4 x 200 + 170]/6=196.7

Mean useful life=[12 + 4 x 10 + 8]/6=10.0

Mean salvage life=100/6=16.7

Compute the mean rate of return:

PW of cost =PW of benefit

$1016.7 =196.7(j>/A,IRRbeta,10) + 16.7(j>IF,IRRbeta, 10)

IRRbeta= 14.2%

Iioo

Example 10-3 gave a most likely rate of return (15.1%) that differed from the mean

rate of return (14.2%) computed in Example 10-4. These values are different because the

former is based exclusively on the most likely values and the latter takes into account the

variability of the parameters.

In examining the data, we see that the pessimistic values are further away from the most

likely values than are the optimistic values. This is a common occurrence. For example, a

savings of 10 to 20% may be the maximum possible, but a cost overrun can be 50%, 100%,

or even more. This causes the resulting weighted mean values to be less favorable than the

most likely values. As a result, the mean rate of return, in this example, is less than the rate

of return based on the most likely values. '

PROBABILITY

We all have used probabilities. For example, what is the probability of getting a "head"

when flipping a coin? Using a model that assumes that the coin is fair, both the head and

tail outcomes occur with a probability of 50%, orY2'This probability is the likelihood {'