Engineering Economic Analysis

108 MORE INTERESTFORMULAS

The expression in the brackets of Equation 4-29 is the geometric series present worth

factor wherei=Jg.

(PIA, g, i,-it) ...

[

1 ~-(1+.g)n(1+i)-n

z-g ]

where i =J g (4-30)

In the special case ofi=g,Equation 4-29 becomes

(PIA,g,i,n) =[n(1+ i)-I] wherei=g (4-31)

The first-year maintenance cost for a new automobile is estimated to be $100, and it increases at

a uniform rate of 10% per year. Using an 8% interest rate, calculate the present worth of cost of

the first 5 years of maintenance.

STEP-BY-STEP, SOLUTION

PWof

Maintenance

$ 92.59

94.30

96.05

97.83

99.65

$480.42

IIIiiiI ::=;'

P=Al

[

1 - (1 +.g)n(1+i)-n

. Z- g ]

where i =J g

= 100.00

[

1 - (1.10)5(1.08)-5

]

=$480.42

-0.02

I

I

I

I

The present worth of cost of maintenance for the first 5 years is $480.42. I.

Maintenance

Yearn Cost (P/F,8%,n)

1 100.00 = 100.00 x 0.9259=

(^2) 100.00 + 10%(100.00) =110.00 x 0.8573=
(^3) 110.00 + 10%(110.00) =121.00 x 0.7938=
(^4) 121.00 + 10%(121.00) =133.10 x 0.7350=
(^5) 133.10 + 10%(133.10) =146.41 x 0.6806 =