Engineering Economic Analysis

92 MORE INTERESTFORMULAS

P= A(P/A,i,n)

$6800= $140(P/A, i, 60)

. .~.;... .. .... ._~ .-
6800
(P/A, i,60) = - 140 = 48.571

We know the value of the uniform series present worth factor, but we do not know the interest

ratei.As a result, we need to look through several compound interest tables and then compute

the rate of returniby interpolation. Entering values from the tables in the appendix, we find

(PIA, i, 60)

5L726:lJ

48.571.-.J d

48.174

The rate of return, which is between 1/2%and %%, may indeed be computedby a linear interpola-

tiOh.The interestformulas are not linear, so a linear interpolation will not give an exact solution.

To minimize the error, the interpolation should be computed usinginterest rates as close to the

correct answer as possible. [Sincea/b =e/d, a=bee/d)], we write

Rate of return i=0.5%+a

=0.5% +bee/d)

(

51.726 - 48.571

=0.50%+ 0.25%.. 51.72 - 48.174 (^6) )
(
3.155
)
=0.50% + 0.25% ~ =0.50%+0.22%
3.552

=0.72% per month

The)monthly rate of return on our investment would be 0.72% per month.

Using a 15% interest rate, compute the value ofFin the following cash flow:

Year Cash Flow

1 + 100

2 + 100

3 +100

4 0

5 -F

-.. -.... - '