Question
In: Economics
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 agoRelated Solutions
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 conventional benefit cost ratio method. MARR= %13
per year.
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 modified benefit cost ratio method. MARR= %13 per
year.
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Years 0 1 2 3 4 Investment Outlay Equipment cost ($350,000) Shipping and installation ($70,000)...
Years
0
1
2
3
4
Investment Outlay
Equipment
cost
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Shipping
and installation
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Increase
in inventory
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Increase
in accounts payable
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$
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$ 140,450
$ 152,210
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termination cash flow
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Project Cash Flows
Net cash flows
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$113,990
$96,350
$140,450
$205,460
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