Handbook of Civil Engineering Calculations

ent costs. Where extraordinary payments occur, simply add the present worth of these
payments, as was done in the present case with respect to repairs at the end of the fourth
year. In the special case where/= f, the present worth of costs for the first life is P - L

• Nc.

COSTCOMPARISON WITH

ANTICIPATED INFLATION

Two alternative machines have the following cost data as based on present costs:

Machine A Machine B

First cost, $ 45,000 80,000

Salvage value, $ 3,000 2,000

Life, years 4 6

Annual maintenance, $ 8,000 6,000

Determine which machine is more economical, using an interest rate of 10 percent and
annual inflation rate of 7 percent.

Calculation Procedure:

1. Establish the method of cost comparison
   The present-worth method is suitable here. Select an analysis period of 12 years, which
   encompasses three lives of machine A and two lives of machine B.


2. Compute the present worth of costs of machine A
   for the first life
   Refer to the equation given at the conclusion of the preceding calculation procedure. Set h
   = (1.07/1.1O)^4 = 0.89529. By reversing the sequence in the last two terms of the equation,
   present worth = $45,000 - $3000(0.89529) + $8000(1.07)0 - 0.89529)7(0.10 - 0.07) =
   $72,190.


3. Compute the present worth of costs of machine A
   for the first three lives
   Refer to step 4 of the preceding calculation procedure. Thus, PW = $72,190[1 +
   (1.07/1.1O)^4 + (1.07/1.1O)^8 ] = $72,190(2.69684) = $194,680.


4. Compute the present worth of costs of machine B
   for the first Ufe
   Set h = (1.07/1.1O)^6 = 0.84712. The present worth of costs for the first life = $80,000 -
   $2000(0.84712) + $60000.07)0 ~ 0.84712)7(0.10 - 0.07) = $111,020.


5. Compute the present worth of costs of machine B
   for the first two lives
   PW = $111,020(1 + 0.84712) = $205,070.


6. Determine which machine is preferable
   Machine A has the lower cost and so is preferable.