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Engineering economy homework
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Eleven payments of $9,000 each are to be repaid monthly at the end of each month The monthly interest rate is 3%. a. What is the present equivalent (ie., Po) of these payments? b. Repeat Part (a) when the payments are made at the beginning of the month. Note that the present equivalent will be at the same time as the first monthly payment c. Explain why the present equivalent amounts in Parts (a) and (b) are different 昌Click the icon to view the interest and annuity table for discrete compounding when : 3% per month. a. The present equivalent of the payments is $ 89586(Round to the nearest dollar)
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| a) | |
| Monthly payment | $9000 |
| Number of payments | 11 |
| Monthly interest rate | 3% |
| Payment made at | end of the month |
| PVIFA @3% 11 months (table) | 9.2526 |
| Present equivalent (9.2526*9000) | 83273.4 |
| b) | |
| Monthly payment | $9000 |
| Number of payments | 11 |
| Monthly interest rate | 3% |
| Payment made at | first of the month |
| PVIFA @3% 10 months (table) + 1 (8.5302+1) | 9.5302 |
| Present equivalent (9.5302*9000) | 85771.8 |
| c) | |
| Difference between a & b | 2498.4 |
| Reason for the difference | |
| In case (a) the all payments were made at the end of the months, hence the Present worth factor to be calculated for 11 periods. But in case (b) the payments were made at the beginning of the month. Hence the present woth factor need not be calculated for the first month, rather 1 should be added to the present worth factor of the remaining 10 months. That is the reason for the difference. | |
