John Deere expects the cost of a tractor part to increase by $15 per year beginning...
John Deere expects the cost of a tractor part to increase by $15 per year beginning 5 years from now. If the cost in years 1-4 is $80, determine the present worth in year 0 of the cost through year 10 at an interest rate of 10% per year.
Show the cash flow diagram, solve it in Excel,show formula
Homework Answers
NPV is calculated as shown in the below table:
| Year | CF | Discount Factor | Discounted CF | ||
| 0 | $ - | 1/(1+0.1)^0= | 1 | 1*0= | $ - |
| 1 | $ 80.00 | 1/(1+0.1)^1= | 0.909090909 | 0.909090909090909*80= | $ 72.73 |
| 2 | $ 80.00 | 1/(1+0.1)^2= | 0.826446281 | 0.826446280991735*80= | $ 66.12 |
| 3 | $ 80.00 | 1/(1+0.1)^3= | 0.751314801 | 0.751314800901578*80= | $ 60.11 |
| 4 | $ 80.00 | 1/(1+0.1)^4= | 0.683013455 | 0.683013455365071*80= | $ 54.64 |
| 5 | $ 80.00 | 1/(1+0.1)^5= | 0.620921323 | 0.620921323059155*80= | $ 49.67 |
| 6 | $ 95.00 | 1/(1+0.1)^6= | 0.56447393 | 0.564473930053777*95= | $ 53.63 |
| 7 | $ 110.00 | 1/(1+0.1)^7= | 0.513158118 | 0.513158118230706*110= | $ 56.45 |
| 8 | $ 125.00 | 1/(1+0.1)^8= | 0.46650738 | 0.466507380209733*125= | $ 58.31 |
| 9 | $ 140.00 | 1/(1+0.1)^9= | 0.424097618 | 0.424097618372485*140= | $ 59.37 |
| 10 | $ 155.00 | 1/(1+0.1)^10= | 0.385543289 | 0.385543289429531*155= | $ 59.76 |
| NPV = Sum of all Discounted CF | $ 590.78 | ||||
So the present worth will be 590.78
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