Three different alternatives shown in table below are being considered by Kal Te
ID: 1102019 • Letter: T
Question
Three different alternatives shown in table below are being considered by Kal Tech Engineering systems.
Assume that alternatives X and Z are replaced at the end of their lives.
Data
Alternative X
Alternative Y
Alternative Z
Initial Cost
$6,000
$1,000
$1,500
Uniform Annual Benefits
$810
$125
$ 230
Useful Life in Years
20
infinite
10
MARR
12% 12% 12%
Data
Alternative X
Alternative Y
Alternative Z
Initial Cost
$6,000
$1,000
$1,500
Uniform Annual Benefits
$810
$125
$ 230
Useful Life in Years
20
infinite
10
MARR
12% 12% 12%
Explanation / Answer
NPV x = -6000+810*PVIIFA(12,20)
where pvifa(r,n) = [1-(1+r)^-n]/r
NPV x = -6000+810*7.4694 = $50.214
NPV y = -1000+125/.12 = $41.67 { here (n is infinity so pvifa(r,n) = 1/r
NPV z = -1500+230*PVIFA(12,10) = -1500+230*5.6502 = -$200.454
NPV x is highest so X alternatiive is best
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