With the estimates shown below, Sarah needs to determine the trade-in (replacement) value of machine X that will render its AW equal to that of machine Y at an interest rate of 9% per year. Determine the replacement value.
Machine X |
Machine Y |
|
Market Value, $ |
? |
92,000 |
Annual Cost, $ per Year |
−55,500 |
−40,000 for year 1,increasing by 2000 per year thereafter. |
Salvage Value |
15,500 |
17,000 |
Life, Years |
3 |
5 |
The replacement value is $ ________.
ANSWER:
We will equate the aw of machine x and machine y equal.
i = 9%
aw of machine x = market value(a/p,i,n) + annual cost + salvage value(a/f,i,n)
aw of machine x = market value(a/p,9%,3) - 55,500 + 15,500(a/f,9%,3)
aw of machine x = market value * 0.3951 - 55,500 + 15,500 * 0.3051
aw of machine x = 0.3951 market value - 55,500 + 4,729.05
aw of machine x = 0.3951 market value - 50,771
pw of machine y = market value + annual cost(p/a,i,n) + increase in cost(p/g,i,n) + salvage value(p/f,i,n)
pw of machine y = -92,000 - 40,000(p/a,9%,5) - 2,000(p/g,9%,5) + 17,000(p/f,9%,5)
pw of machine y = -92,000 - 40,000 * 3.89 - 2,000 * 7.111 + 17,000 * 0.6499
pw of machine y = -92,000 - 155,600 - 14,222 + 9,098.6
pw of machine y = -252,723
aw of machine y = pw of machine y(a/p,i,n)
aw of machine y = -252,723(a/p,9%,5)
aw of machine y = -252,723 * 0.2571
aw of machine y = -64,975.2
aw of machine x = aw of machine y
0.3951 market value - 50,771 = -64,975.2
0.3951 market value = -64,975.2 + 50,771
0.3951 market value = -14,204.2
market value = -14,204.2 / 0.3951
market value = -35,950.9
so the market value is $35,950.9 that will make the aw of both the machines equal.
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