Engineering Economic Analysis

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Relationships Between Compound Interest Factors 97

Combining, we have

P- [20(PI A,15%,3) +.10(P IF,. -... '~- -15%,2)](P I F,15%, 1)

= =$46.28

Relationships Between Compound Interest Factors

From the derivations,we see there are several simple relationships between the compound interest factors. They are summarized here.

Single Payment 1 Compound amount factor=Present worth factor 1 (FIP,i,n) = (PIF,i,n)

(4-9)

Uniform Series

. 1
Capital recovery factor=Present worth factor
1
(AIP,i,n) = (PIA,i,n)

1 Compound amount factor=Sinking fund factor 1 (FI A, i,n) = (AI F, i, n)

(4-10)

(4-11)

The uniform series present worth factor is simply the sum of thenterms of the single payment present worth factor n

(PIA,i,n) =L(PIF,i, J) 1=1

(4-12)

For example:

(PIA,5%, 4) = (PI F,5%, 1) + (PI F,5%, 2) + (PI F,5%,3) + (PI F,5%, 4)

3.546 = 0.9524 + 0.9070 + 0.8638 + 0.8227

= =$46.28