Engineering Economic Analysis

Geometric Gradient 105

150

50

o t

0 -l-- --- -2-3-4 5 6

1!

p J

100

r

It is important that you closely examine the location ofJ.Based on the way the factor was

derived,therewillbe onezero valu~in thegradientserjesto therightofJ"(If this seems strange

or incorr~ct,review the beginning of this section on arithmetic gradients.)

J - G(PIG, i, n)

= 50(PIG, 10%,4) (Note:3 would be incorrect.)

==50(4.378) .·218.90

Then

P ..J(PIF,10%,2)

To o1;>tainthe present worth of theft.1turesumJ,use tbe(PIF,' i, n)factor. Combining, we

have

[I

P=50(P IG,10%,4)(P I}?,10%,2)

" .. 50(4.378)(0.8264)' $180.90

The value of Pis $1.80.90.

Geometric Gradient

Inthe preceding section, we saw that the arithmetic gradient is applicable where the period-

by-periodchange in a cash receipt or payment is a uniform amount.There are other situations

where the period-by-period change is auniform rate,g.For example, if the maintenance

costs for an automobile are $100 the first year and they increase at a uniform rate,g,of 10%

per year, the cash flow for the first 5 years would be as follows:

Year Cash Flow

1 100.00 - $100.00

(^2) 100.00 + 10%(100.00)=100(1 + 0.10)1 - 110.00. I
(^3) 110.00+ 10%(110.00)=100(1+ 0.10)2 - 121.00
(^4) 121.00 + 10%(121.00)=100(1 + 0.10)3 - 133.10
(^5) 133.10 + 10%(133.10)=100(1+ 0.10)4 - 146.41