Engineering Economic Analysis

(Chris Devlin) #1
104 MORE INTERESTFORMULAS

gradient. Instead, we willsubtract an increasing gradient from an assumed uniform series of
payments.

6000
o t
0-1-2-3-4 - 0-1-2-3-4

24'

[


000 18,000

i

12,000

t 60;0
= 0-1-2-3-4

1 1 1 1
A' A' A' A'

A'= 24,000- 6000(AjG,10%,4)


=,24,000 -- 6000(1.38t)


= 15,714 rupees


The projected equivalent uniform maintenanceand repair cost is 15,714rupees per year.
oS

Computethe value ofPin the diagram. Use a 10% interest rate.


150

50

o t
0-1-2-3-4-5~6

100

r


p

~.,,_. "'.-; ",

SptlJ11()~' ,.

With the arithmetic gradient series present worth factor, we can compute a presentsumJ. I


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