Engineering Economic Analysis

r

Uniform Series Compound Interest Formulas 93

100 100 100

t t t

0-1-2-3-4-5

. .. .... 1
F

We see that the cash flow diagram is not the same as the diagram shown in Example 4-1, which

is a sinking fund factor diagram.

A A A

t t t

0-1-2---'---3

1

F

Since the diagrams do not agree, the problem is more difficult than those We've discussed so

far. The general approach to use in this situation is to convert the cash flow fronl its present

form into standard forms, for which we have compound interest factors and compound interest

tables.

On~ way to solve this problem is to consider the cash flow as a series of single paYJ:l1entsP

and thet}to compute their sumIJ,. F.In other words, the cash flow is bJ;okeninto three parts, each,

one of Whichwe can solve.

"

100 100 100

t t t

0 1)-2---,3 4 5=

_w

100

t

+ 0 1"--'2.=3~4--.,5

{

. - ,.. ....

F I

I

1"

FW FI + F'z+ F3 ..JOO(FI P,15%, 4) +lQO(f I P,15%, 3) + 100(FI P,15%, 2) I

= 100(1.749) -I- 100(1.521) + 100(1.322)

$459.20

"

II

"

The val~~ofFin theilfustrated cash flow is $459.20.

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