Question

4. The following table summarizes the financial data for a proposed new asset. The issue is how long the asset should be kept. Assume an interest rate of 12% and determine the economic service life of the asset before taxes End of year (n)Investment cost |O&M cost Salvage Value $14,000 $3,400 4600 5800 7200 8300 $8,000 6000 4000 2000 2 4

Please solve and show all working

Answer 1

Answer: 4

Compute Effective annual cost (EAC)
If sold in Year 1
Year Cahflows PV factor@12% PV cash flows
0 $ -14,000.00 1 $ -14,000.00
1 $ 4,600.00 0.892857143 $ 4,107.14
$ -9,892.86
EAC = -9892.86/0.89286 $ -11,080.00
If sold in Year 2
Year Cahflows PV factor@12% PV cash flows
0 -14000 1 -14000
1 -3400 0.892857143 -3035.714286
2 1400 0.797193878 1116.071429
1.69005102 -15919.64286
EAC = -15919.6429/1.6900 $ -9,419.62
If sold in Year 3
Year Cahflows PV factor@12% PV cash flows
0 -14000 1 -14000
1 -3400 0.892857143 -3035.714286
2 -4600 0.797193878 -3667.091837
3 -1800 0.711780248 -1281.204446
2.401831268 -21984.01057
EAC = -21984.0106/2.40183 $ -9,153.02
If sold in Year 4
Year Cahflows PV factor@12% PV cash flows
0 -14000 1 -14000
1 -3400 0.892857143 -3035.714286
2 -4600 0.797193878 -3667.091837
3 -5800 0.711780248 -4128.325437
4 -5200 0.635518078 -3304.694008
3.037349347 -28135.82557
EAC = -28135.8256/3.0373 $ -9,263.28
If sold in Year 5
Year Cahflows PV factor@12% PV cash flows
0 -14000 1 -14000
1 -3400 0.892857143 -3035.714286
2 -4600 0.797193878 -3667.091837
3 -5800 0.711780248 -4128.325437
4 -7200 0.635518078 -4575.730165
5 -8300 0.567426856 -4709.642902
3.604776202 -34116.50463
EAC = -34116.5046/3.60477 $ -9,464.25
If asset was sold at the end of year 3, it will leads to lower EAC.
Therefore, the asset shall be kept for 3 years
So, economic service life of the asset is 3 years