In a replacement analysis for a vacuum seal on a spacecraft the following data a
ID: 2806027 • Letter: I
Question
In a replacement analysis for a vacuum seal on a spacecraft the following data are known about the challenger. The initial investment is $11,000 , there is no annual maintenance cost for the first 3 years , however , it will be $3,000 in each of years four and five , and then $5,000 in the sixth year and increasing by $2,000 each year thereafter . The market value at the end of the year k is 50% of year k-1 at all times , the MARR is 15% per year.(Hint:After a certain year,the EUAC will keep increasing.)
a) What is the economic life of this challenger?
a) What is the economic life of this challenger?
b) What is the EUAC corresponding to the economic life in part a ?Explanation / Answer
Solution:
a and b.
Year 1
EUAC = $11,000 (A/P @ 15%, 1)
EUAC = $11,000 (1/0.8696)
EUAC = $12,650
Year 2
EUAC = $11,000 (A/P @ 15%, 2)
EUAC = $11,000 (1/1.6257)
EUAC = $6,766.32
Year 3
EUAC = $11,000 (A/P @ 15%, 3)
EUAC = $11,000 (1/2.2832)
EUAC = $4,817.80
Year 4
EUAC = [$11,000 + $3,000 (PVIF @ 15%, 4)] (A/P @ 15%, 4)
EUAC = [$11,000 + $3,000 (0.5718) ](0.3503)
EUAC = $4,454.20
Year 5
EUAC = [$11,000 + $3,000 (PVIF @ 15%, 4) + $3,000 (PVIF @ 15%, 5)] (A/P @ 15%, 5)
EUAC = [$11,000 + $3,000 (0.5718) + $3,000 (0.4972)] (0.2983)
EUAC = $4,238.11
Year 6
EUAC = [$11,000 + $3,000 (PVIF @ 15%, 4) + $3,000 (PVIF @ 15%, 5) + $5,000 (PVIF @ 15%, 6)] (A/P @ 15%, 6)
EUAC = [$11,000 + $3,000 (0.5718) + $3,000 (0.4972) + $5,000 (0.4323)] (0.2642)
EUAC = $4,324.56
After 5 years, EUAC starts increasing. Therefore, n = 5 is the economic life of the challenger.
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