In this content, I’m going to use the BAii plus calculator to resolve this amortization trouble. A $450,000 mortgage, financed at 2.Eight% compounded semi-annually, calls for month-end bills of $1850. How many payments are required? To resolve this, allow’s first set P/Y to 12 (since payments are made month-to-month), and then we set C/Y to two for semiannual compounding. Press 2nd Quit. Next input 2.8 I/Y for interest, 450,000 for Present Value, 1850 (negative) for payments (since payment is an outflow), and then 0 Future value since we plan to pay off the mortgage at the end. And when we compute N, we obtain 358.53. That is, 359 payments are required. The length of the 359th charge, however, might be smaller than the preceding payments of 1850. To find the principal repaid in the 10th payment, we press 2nd AMORT, set P1 to 10, ENTER, scroll down, set P2 to 10 as well, ENTER and when you scroll down, you see the outstanding balance after the 10th payment, when you scroll down again, you see the principal repaid in the 10th payment which is 823.06.
Mortgage Calculations using BA II Plus
Now we want to calculate the total interest paid in the 2nd year. Since payments are made monthly, the first year comprises of payments 1 to 12. The second-year will comprise the next 12 payments which will be 13th to the 24th. So we set P1 to 13 ENTER, and scroll down , and then set P2 to 24 ENTER. Scroll right down to interest, and we’ve got total interest of $12,126.51 for the second year.
Next, we want to calculate the first-rate predominant stability at the Quit of the primary 4 years. Since bills are made month-to-month. In 4 years, we’ve got 4 times 12, which equals 48 bills. So second AMORT, P1 can be any price less than forty-eight, but we need to make sure we set P2 to 48 ENTER. And while we scroll down, we see that the balance is 409122.15. To locate the full hobby paid inside the first five years, we set P1 to at least one, scroll down, and set P2 to 5 time 12, which equals 60.ENTER then scroll down to INTEREST, and we have 59,172.31. Next, we find the size of the final payment. Recall that N equals 358.53, So the final payment is the 359th payment.
So we set P1 and P2 to 359. And here that the amazing balance is 870.25 which actually represents an overpayment of account, if we make the regular payment of 1850. So to discover the desired remaining charge, we subtract the overpayment value of 870.25 from the everyday price of 1850, and that offers 979.75. So the last payment size is 979.75. So the size of the last payment is 979.75.