Use recurrence relations and the present value of an annuity to find loan repayments, outstanding balances and total interest paid

What this dot point is asking

NESA wants you to model a reducing-balance loan with a recurrence relation, derive the closed-form formula for the regular repayment using the present value of an annuity, compute outstanding balances and total interest, and split a single payment into interest and principal components.

The answer

The recurrence model

A loan of $P$ is repaid by equal payments $M$ at the end of each period. Interest of rate $r$ per period accrues on the outstanding balance. Let $B_n$ be the balance just after the $n$th payment, with $B_0 = P$.

Each period: add interest, then subtract the payment.

Iterating this recurrence and using a geometric series gives the closed form

The first term is what the loan would grow to without payments; the second term is the future value of the payments made so far. The difference is what is still owed.

The repayment formula (present value of an annuity)

The loan is fully repaid when $B_n = 0$. Setting the closed form to zero and solving for $M$:

Equivalently, the loan amount is the present value of the stream of $n$ payments:

This is the present-value-of-annuity formula. It is the discounted sum of the geometric series $M v + M v^2 + \cdots + M v^n$ with $v = (1 + r)^{-1}$.

Splitting a payment into interest and principal

The interest portion of the $k$th payment is the interest charged on the previous balance:

The principal portion is the rest:

Early in the loan, most of each payment goes to interest. Near the end, almost all goes to principal. The principal portion grows geometrically with ratio $(1 + r)$ across periods.

Total interest

Total amount paid over the loan is $M n$. Since the loan amount $P$ is returned, total interest is

Time to repay

If a borrower chooses $M$ instead of $n$, the loan term is

The expression inside the log must be positive, which requires $M > P r$, that is the payment must exceed the first period's interest. Otherwise the loan never finishes.

Worked examples

Standard repayment

\25000$car loan at$7.2%$per annum compounded monthly, repaid over$5$ years.

$r = \frac{0.072}{12} = 0.006$, $n = 60$.

$(1.006)^{-60} \approx 0.69892$, so $1 - 0.69892 = 0.30108$.

M = \frac{25000 \cdot 0.006}{0.30108} = \frac{150}{0.30108} \approx \498.21$.

Total paid: 60 \cdot 498.21 = \29892.60$. Total interest:$\$4892.60$.

Outstanding balance partway through

For the loan above, balance after $24$ payments:

$(1.006)^{24} \approx 1.15418$.

B_{24} = 25000 \cdot 1.15418 - 498.21 \cdot \frac{0.15418}{0.006} = 28854.50 - 498.21 \cdot 25.697 \approx 28854.50 - 12802.50 = \16052.00$.

After two years, roughly $36\%$ of principal has been repaid even though $40\%$ of payments have been made. This is the interest-front-loading effect.

Interest vs principal in one payment

For the same loan, the $25$th payment is split as

I_{25} = r \cdot B_{24} = 0.006 \cdot 16052.00 \approx \96.31$.

P_{25} = M - I_{25} = 498.21 - 96.31 \approx \401.90$.

So roughly \96$ of that month's payment is interest, and the rest goes to reducing the principal.

Solving for the term

You can afford \1500$a month on a$\$300000$ loan at $6\%$ per annum compounded monthly. How many months will it take?

$P r / M = 300000 \cdot 0.005 / 1500 = 1$, so $1 - P r / M = 0$, and the log is undefined. The payment exactly equals the interest, so the principal never reduces. The loan would never finish.

Bump the payment to \1700$:$P r / M = 0.88235$,$1 - P r / M = 0.11765$.

$n = \frac{-\ln 0.11765}{\ln 1.005} = \frac{2.13980}{0.004988} \approx 429.0$ months, or about $35.75$ years.

Common traps

Using the future-value formula for a loan. Loan repayments use the present-value-of-annuity formula. The future-value formula is for accumulating savings.

Forgetting to convert the rate. Monthly compounding requires the monthly rate $R / 12$. Using the annual rate gives a wildly wrong answer.

Mixing up the two forms. $M = \frac{P r (1 + r)^n}{(1 + r)^n - 1}$ and $M = \frac{P r}{1 - (1 + r)^{-n}}$ are the same. Pick one and use it consistently.

Treating principal repaid as $M \cdot$ fraction. Early payments are mostly interest; the principal-repaid pattern is geometric, not linear. Use $P - B_n$ for the amount of principal repaid after $n$ payments.

Ignoring the no-completion case. If $M \le P r$, the loan never finishes. The log formula will fail or give a negative result.

In one sentence

A reducing-balance loan satisfies $B_n = B_{n - 1}(1 + r) - M$ with closed form $B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}$, the repayment is set by the present-value-of-annuity formula $M = \frac{P r}{1 - (1 + r)^{-n}}$, and total interest is $M n - P$.