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Thickness Installation Cost annual savings Maintenance once in each 2 years 2cm 217000 68000 4600 5cm...

Thickness Installation Cost annual savings Maintenance once in each 2 years
2cm 217000 68000 4600
5cm 446000 125000 0

Useful lives of both alternatives are 10 years. Find the best alternative using discounted payback period when MARR= %13 per year

Solutions

Expert Solution

Take the case when thickness is 2 cm

Following schedule can be generated

Year , n     Cash Flow, CF Discounted cash flow, DCF= CF/(1+13%)^n Cumulative discounted cash flow
0 -217000 -217000.00 -217000.00
1 68000 60176.99 -156823.01
2 63400 49651.50 -107171.51
3 68000 47127.41 -60044.10
4 63400 38884.41 -21159.69
5 68000 36907.68 15747.98
6 63400 30452.19 46200.18
7 68000 28904.12 75104.30
8 63400 23848.54 98952.84
9 68000 22636.17 121589.01
10 63400 18676.90 140265.91


Discounted payback period in case when thickness is 2 cm=4+21159.69/36907.68=4.57 years

Now consider

Take the case when thickness is 5 cm

Following schedule can be generated

Year , n     Cash Flow, CF Discounted cash flow, DCF= CF/(1+13%)^n Cumulative discounted cash flow
0 -446000 -446000.00 -446000.00
1 125000 110619.47 -335380.53
2 125000 97893.34 -237487.20
3 125000 86631.27 -150855.93
4 125000 76664.84 -74191.08
5 125000 67844.99 -6346.09
6 125000 60039.82 53693.72
7 125000 53132.58 106826.30
8 125000 47019.98 153846.29
9 125000 41610.60 195456.89
10 125000 36823.54 232280.43

Discounted payback period in case when thickness is 5 cm=5+6346.09/60039.82=5.11 years

Discounted payback period is lower in case thickness is 2 cm. It should be selected.

Rahul Sunny answered 2 years ago
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