Engineering Economic Analysis

Arithmetic Gradient 101

A man has purchased a new automobile. He wishes to set aside enough money in a bank account

to pay the maintenance on the car for the firstS years-.-Ithas been estimated that the maintenance

cost of an automobile is as follows:

Year

1

2

3

4

5

Maintenance Cost

$120

150

180

210

240

Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest.

How much should the car owner deposit in the bank now?

SOLU)"ION

210

240

0-1-2-3-4-5

t

p

The cash flow may be broken into its two components:

240

210

180

60

30

o! t

0-1-2-3-4-5 =0-1-2~3 4-5 +0-1-2 ,;:;-3~4-5

1 t A~120 1

150

If"1

120 120 120 120 120

r r r r r

G::; 30

p

Both components represent cash flows for which compound interest factors have been derived. The
first is a uniform series present worth, and the second is an arithmetic gradient series present worth:

P=A(P j A,5%, 5) +G(I' jG, 5%,5)

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