Could you please explain in details to let me understand clearly... Thank you so
ID: 1129719 • Letter: C
Question
Could you please explain in details to let me understand clearly... Thank you so much.
2) In a replacement analysis for a vacuum seal on a spacecraft, the following data are known about the challenger the initial investment is S11,000; there is no annual maintenance cost for the first three years, however, it will be $2,000 in each of years four and five, and then S6,000 in the sixth year and increasing by S1,000 each year thereafter. The market value at the end of year k is 50% of year k-1 at all times, and MARR is 10% per year. (Hint: After a certain year the EUAC will keep increasing.) a) What is the economic life of this challenger? b) What is the EUAC corresponding to the economic life (20 pts) (30 pts) in part a?Explanation / Answer
a)We will calculate the EUAC for year 1,2,3 and so on
EUAC for year 1= 11000(A/P,10%,1)
=11000(1.1000)
= 12100$
EUAC for year 2= 11000(A/P,10%,2)
=11000(0.5762)
= 6338.2$
EUAC for year 3= 11000(A/P,10%,3)
=11000(0.4021)
= 4423.1$
EUAC for year 4= (11000+2000) (A/P,10%,4)
=13000* 0.3155
=4101.5$
EUAC for year 5= (11000+2000)(A/P.,10%,5)
=13000*0.2638
= 3429.4$
EUAC for year 6= (11000+6000)(A/P,10%,6)
=17000*0.2296
=3903.2$
Since the EUAC is increasing after year 5 so the economic life of this challanger is 5 years
b) The EUAC corresponding to economic life of 5 years is 3429.4$
Note the A/P value are taken from below table-
Interest Rate 10.00% n F/P P/F A/F A/P 1 1.100 0.9091 1.0000 1.1000 2 1.210 0.8264 0.4762 0.5762 3 1.331 0.7513 0.3021 0.4021 4 1.464 0.6830 0.2155 0.3155 5 1.611 0.6209 0.1638 0.2638 6 1.772 0.5645 0.1296 0.2296