Amortization

Study guide: Exam FM-2

Amortization is the process of repaying a loan through a series of installment payments, usually made at regular intervals: each payment first covers the interest due on the Outstanding Balance, and the rest of it repays principal, so the amount owed is progressively reduced to zero. The payments need not be level, but the common case is a loan $L$ repaid by $n$ level payments $P$ at the end of each period, at rate $i$ per period, where $P$ is determined by:

$P = \frac{L}{a_{\overline{n}|i}}$

Whatever the payment pattern, the loan equals the present value of its payments, and the interest in each payment is $i$ times the balance at the start of that period — the general relations are on Loan Amortization, and the row-by-row table is the Amortization Schedule.
With level payments, interest is charged on the declining balance, so the interest portion of each payment decreases over time while the principal portion increases — but the total payment remains constant.
At the end of the $n$ -th payment the loan is exactly paid off.
Level payments — interest in payment $k$ : $I_k = P \cdot (1 - v^{n-k+1})$ where $v = 1/(1+i)$
Principal in payment $k$ : $PR_k = P \cdot v^{n-k+1}$
Outstanding balance after payment $k$ : $OB_k = P \cdot a_{\overline{n-k}|i}$

Amortizing a Loan A $ $10{,}000$ loan is repaid with level annual payments over 4 years at $i = 5\%$ per year. Find the annual payment and the interest and principal portions of the first payment.

Answer $P = \frac{10000}{a_{\overline{4}|5\%}} = \frac{10000}{3.545951} = \$2{,}820.12$ Payment 1 interest: $I_1 = 10000 \times 0.05 = \$500.00$ Payment 1 principal: $PR_1 = 2820.12 - 500.00 = \$2{,}320.12$ Outstanding balance after payment 1: $OB_1 = 10000 - 2320.12 = \$7{,}679.88$ , which equals $P \cdot a_{\overline{3}|5\%} = 2820.118(2.723248) = \$7{,}679.88$ .