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Replacement Decisions: equivalent annual cost

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A certain machine costs $25,000 to purchase and install. It has salvage values and operating costs as shown in the table in the attached file. The salvage value of $20,000 listed at time 0 reflects the loss of the installation costs at the time of installation. The MARR is 12%.

Life in Years Salvage value Operating cost
0 $20,000.00
1 16000.00 $3,000.00
2 12800.00 3225.00
3 10240.00 3466.88
4 8192.00 3726.89
5 6553.60 4006.41
6 5242.88 4306.89
7 4194.30 4629.90
8 3355.44 4977.15
9 2684.35 5350.43
10 2147.48 5751.72
11 1717.99 6183.09
12 1374.39 6646.83
13 1099.51 7145.34
14 879.61 7681.24
15 703.69 8257.33
16 562.95 8876.63
17 450.36 9542.38
18 360.29 10258.06

a) What is the economic life of the machine?
b) What is the equivalent annual cost over that life?

Now assume that the MARR is 5%.
c) What is the economic life of the machine?
d) What is the equivalent annual cost over that life?
e) Explain the effect of decreasing the MARR.

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The economic life of the machine, the equivalent annual cost are calculated and the effect of decreasing the MARR is examined.

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