Engineering Economic Analysis

80 INTERESTAND EQUIVALENCE

Since we have already derived the relationship

p~-- - --=(1 +~. --".-.i) -n F

we write

p :- 400(1 + 0.12)-3 + 600(1 +0.12r5

$625.17

This is virtually the same amount computed from the first statement of this example. [The slight

diffe;renceisdue to the rounding in the compound interest tables. For example, (1 + 0.12)-5 =

0.56'742I!bu~!hecompound in,teresJtable for J2%~hows 0.5674.]!!: ,. ~ ~

-

Both problems in Example 3-8 have been solved by computing the value ofPthat is

equivalent to $400 at the end of Year 3 and $600 at the end of Year 5. In the first problem,

we received +Pat Year0 and were obligated to payout the $400 and $600 in later years.

In the second (alternate formation) problem, the reverse was true. We paid-P at Year 0

and would receive the $400 and $600 sums in later years. In fact, the two problems could

represent the buyer and seller of the same piece of paper. The seller would receive +Pat

Year 0 while the buyer would pay - P.Thus, while the problems looked different, they

could have been one situation examined first from the viewpoint of the seller and then from

that of the buyer. Either way, the solution is based on an equivalencecomputation.

The second set of cash flows in Example 3-8 was:

Year

o 1 2 3 4 5

Cash Flow

-P

o

o

+400

o

+600

Ata 12% interest rate,P was computed to be $625.17. Suppose the interest rate is increased

to 15%. Will the value ofPbe larger or smaller?

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On~ can considerPas a sum of money invested at 15% from which one is to obtain $400 at the

endlof 3 years and $600 at the end of 5 years. At 12%, the requiredPis $625.17. AtJ5%, Pwill

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