Engineering Economic Analysis

106 MORE INTERESTFORMULAS

0-1-2-3-4-5

:..

1.00 00 -----_

110.00---_

121.00-----....

133.10-.......

146.41'

From the table, we can see that the maintenance cost in any year is

$100(1 +gt-1

Stated in a more general form,

(4-22).

where g=uniformrateof cash flow increase/decrease from period

to period, that is, the geometric gradient

Al=value of cash flow at Year 1 ($100 in the example)

An=value of cash flow at any Yearn

Since the present worthPnof any cash flowAnat interest rateiis

(4-23)

we can ~ubstituteEquation 4-22 into Equation 4-23 to get

This may be rewritten as

P=A1(1+0-1 t (^1 +~)

X-l

x=1 1 +l

(4-24)

The present worth of the entire gradient series of cash flows may be obtained by expanding.

Equation 4-24:

P=A1(1+0-1 t (^1 +~)

X-l

x=1 1+l

(4-25)