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 1 year ago

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