Engineering Economic Analysis

I

78 INTERESTAND EQUIVALENCE

interest rate per interest period is111z%.For the total3-year duration, there are 12 interest periods.

Thus

P =$500 i=0.015'''' n''''::'(4x 3)= 12 F=unknown

F =P(1 +it= P(F /P, i, n)

= $500(1 + 0.015)12=$500(F/P, Jlh%, 12)

= $500(1.196) = $598

A $500 deposit now would yield $598 in 3 years.

Consider the following situation:

Year Cash Flow

o +P

1 0

2 0

3 -400

4 0

5 -600

Solve for Passuming a 12% interest rate and using the compound interest tables. Recall that

receipts have a plus sign and disbursements or payments have a negative sign. Thus, the diagram is:

p

R,oeip" (+) t

0-1-2-3-4-5

Disbursements ( ) ~

(^400)!
600
f = ,
I
P=400(P/F, 12%,3) +600(P/F, 12%,5)
= 400(0.7118) + 600(0.5674)

• $625.16
  J
  It i$ important to understandjust what the solution, $625.16, represents. We can say that $625.16
  is tbe amount of money that would need to be invested at 12% annual interest to allow for the
  withdraWalof $400 at the end of 3 years and $600 at the 'endof 5 years.
  ...