Engineering Economic Analysis

90 MORE INTERESTFORMULAS

p

r

O-1 2~3-4 5

t t t t t

A A A A A

I=

--

4

4

~-~~

P=$5000 n=5 i__8% A=unknown

=

Tb;eannualloan paymeQtis $1252.

In Example 4-3, with interest at 8%, a present sum of $5000 is equivalent to five equal.

end-of-period disbursements of $1252. This is another way of stating Plan 3 of Table 3-1.

The method for determining the annual payment that would repay $5000 in 5 years with

8% interest has now been explained. The calculation is simply

A=5000(Aj P,8%, 5) = 5000(0.2505) = $1252

If the capital recovery formula (Equation 4-7) is solved for the present sumP,we

obtain the uniform series present worth formula

[

(1 + i)n - 1

]

=A(PjA, i%, n)

P=A i(1 + i)n

(4-8)

and

[

(1 + i)n - 1

(PjA,i%,n)= i(1+i)n ]

which is the uniform series present worth factor.

An investor holds a time payment purchase contract on some machine tools. The contract calls
for.the payment of $140 at the end of eactlmonth for a5..yearperiod/I'hefirstpaYmetlt is due
in one month..fle offers to sell you the cOlltractfor $6800 cash today. lf you otherwise can make
1% per monThon yourmoQey, WqUId.you accept or reject tlJ,ewvestpr's.pffer?
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