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QUESTION 1-LOAN AMORTIZATION (30 marks) Rohanna is planning on getting married and wants to purchase a...
QUESTION 1-LOAN AMORTIZATION Rohanna is planning on getting married and wants to purchase a house for $11 800 000. She made an agreement with the National Housing Trust (NHT) to provide 50% financing for the new home. It was agreed with the NHT that she will make monthly payments over fifteen (15) years at 6% per annum. The remainder of the funds required to cover the cost of the house will be borrowed from her commercial bank. The terms of the agreement for the commercial bank loan are, 10% down payment, and the balance is to be repaid at 8% interest over a ten (10) year period.
Required:
a. Calculate the annual end of year payments to be paid to the commercial bank.
b. Calculate the total amount of the payments and the amount of interest paid under each alternative. c.
C. prepare the amortization schedule for this loan.
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Expert Solution
a
| Yearly payment | = | [P × R × (1+R)^N ] / [(1+R)^N -1] | |
| Using the formula: | |||
| Loan amount | P | $ 5,310,000 | |
| Rate of interest per period: | |||
| Annual rate of interest | 8.000% | ||
| Frequency of payment | = | Once in 12 month period | |
| Numer of payments in a year | = | 12/12 = | 1 |
| Rate of interest per period | R | 0.08 /1 = | 8.0000% |
| Total number of payments: | |||
| Frequency of payment | = | Once in 12 month period | |
| Number of years of loan repayment | = | 10 | |
| Total number of payments | N | 10 × 1 = | 10 |
| Period payment using the formula | = | [ 5310000 × 0.08 × (1+0.08)^10] / [(1+0.08 ^10 -1] | |
| Yearly payment | = | $ 791,346.58 |
Payment is $791,346.58
2
Commercial bank:
| Total interest pay: | ||
| Total payments | = | 791346.58 × 10 |
| $ 7,913,465.85 | ||
| Less principle amount | $ 5,310,000.00 | |
| Interest payment- Finance charge | $ 2,603,465.85 |
Schedule for commercial bank loan:
| Period | Beginning liability | Uniform monthly payment | Interest owed | Principal payment | Total owed at end of month |
| N | A | C | B= A* 0.080000 | D=C-B | E=A-D |
| 1 | 5,310,000.00 | 791,346.58 | 424,800.00 | 366,546.58 | 4,943,453.42 |
| 2 | 4,943,453.42 | 791,346.58 | 395,476.27 | 395,870.31 | 4,547,583.10 |
| 3 | 4,547,583.10 | 791,346.58 | 363,806.65 | 427,539.94 | 4,120,043.17 |
| 4 | 4,120,043.17 | 791,346.58 | 329,603.45 | 461,743.13 | 3,658,300.03 |
| 5 | 3,658,300.03 | 791,346.58 | 292,664.00 | 498,682.58 | 3,159,617.45 |
| 6 | 3,159,617.45 | 791,346.58 | 252,769.40 | 538,577.19 | 2,621,040.26 |
| 7 | 2,621,040.26 | 791,346.58 | 209,683.22 | 581,663.36 | 2,039,376.90 |
| 8 | 2,039,376.90 | 791,346.58 | 163,150.15 | 628,196.43 | 1,411,180.47 |
| 9 | 1,411,180.47 | 791,346.58 | 112,894.44 | 678,452.15 | 732,728.32 |
| 10 | 732,728.32 | 791,346.58 | 58,618.27 | 732,728.32 | 0.00 |
NHT loan monthly payment and total payments:
| Monthly payment | = | [P × R × (1+R)^N ] / [(1+R)^N -1] | |
| Using the formula: | |||
| Loan amount | P | $ 5,900,000 | |
| Rate of interest per period: | |||
| Annual rate of interest | 6.000% | ||
| Frequency of payment | = | Once in 1 month period | |
| Numer of payments in a year | = | 12/1 = | 12 |
| Rate of interest per period | R | 0.06 /12 = | 0.5000% |
| Total number of payments: | |||
| Frequency of payment | = | Once in 1 month period | |
| Number of years of loan repayment | = | 15 | |
| Total number of payments | N | 15 × 12 = | 180 |
| Period payment using the formula | = | [ 5900000 × 0.005 × (1+0.005)^180] / [(1+0.005 ^180 -1] | |
| Monthly payment | = | $ 49,787.55 | |
| Total interest pay: | |||
| Total payments | = | 49787.55 × 180 | |
| $ 8,961,759.51 | |||
| Less principle amount | $ 5,900,000.00 | ||
| Interest payment- Finance charge | $ 3,061,759.51 |
Amortization schedule for first one year:
| Period | Beginning liability | Uniform monthly payment | Interest owed | Principal payment | Total owed at end of month |
| N | A | C | B= A* 0.005000 | D=C-B | E=A-D |
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