Engineering Economic Analysis

Applying Present Worth Techniques. 155

the $200 would remain in the account. At the end of the second year, the $200 balance

would again earn 4%($200) =$8. This $8 could also be withdrawn and the account would

still have $200. This procedure could be continued indefinitelyand the bank account would

always contain $200.

The year-by-year situation would be depicted like this:

Year1: $200 initialP~ 200 + 8= 208

Withdrawali P= - 8

Year2: $200 ~ 200 + 8 = 208

Withdrawali P=- 8

$200

and so on

Thus, for any initial present sumP, there can be an end-of-period withdrawal ofA

equal toi Peach period, and these withdrawals may continue forever without diminishing

the initial sumP.This gives us the basic relationship:

For n= 00, A=Pi

This relationship is the key to capitalized cost calculations. Earlier we defined capitalized

cost as the present sum of money that would need to be set aside at some interest rate to

yield the funds to provide the desired task or service forever. Capitalized cost is therefore

thePin the equationA=i P.It follows that:

Capitalized cost

A

P=-

i

(5-2)

If we can resolve the desired task or service into an equivalentA,the capitalized cost may

be computed. The following examples illustrate such computations.

How much should one set aside to pay $50 per year for maintenance on a gravesite if interest is

assumed to be 4%? For perpetual maintenance, the principal sum must remain undiminished after

making the annual disbursement.

'SOLUTION. ,

Capitalized costP=Annual disbursementInterest ratei A.

~ P= 50'.

'. 0.04= $1250

==One ~hould$et aside. $1250.= .~ !!IC:'''=== =====- = IiiII:I='==;;:::::1:::11<= 111I ~