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

A reducing-balance loan works by two actions every period, always in the same order: interest is charged on the current balance, then the fixed payment is subtracted. Because each payment chips a little off the principal, the balance the interest is charged on shrinks, so the interest portion of the next payment is smaller and a larger slice of the same fixed payment goes to principal. The balance therefore falls a little faster every period, tracing the gently steepening curve below all the way to zero.

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 summing the resulting geometric series gives the closed form

The first term $P(1 + r)^n$ is what the loan would grow to if no payments were ever made; the second term is the future value of the payments made so far (the same annuity sum used for savings). The difference is what is still owed. This closed form is the shortcut for "the balance after the $n$th payment" when $n$ is too large to iterate by hand.

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 (opening) 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, because $P_{k+1} = M - rB_k$ and $B_k = B_{k-1} - P_k$, so the part repaid each period multiplies by $(1+r)$.

Building the amortisation schedule, row by row

The safest way to see the split shift is to build a schedule one row at a time, carrying each closing balance down to be the next opening balance. Take the $25000 car loan at $0.6\%$ per month ($r = 0.006$) with the $497.39 monthly repayment found below, and build the first three months stage by stage. Each row follows the same three rules: interest is the opening balance times $r$; principal repaid is payment minus interest; closing balance is opening plus interest minus payment.

Stage 1, the first month. The opening balance is the whole loan, $25000. Interest is $0.006 \times 25000 = 150.00$, i.e. $150.00. The payment is $497.39, so principal repaid is $497.39 - 150.00 = 347.39$, i.e. $347.39, and the closing balance is $25000 + 150 - 497.39 = 24652.61$, i.e. $24652.61.

Stage 2, carry the closing balance down. Month $2$ opens at last month's close, $24652.61. Interest is $0.006 \times 24652.61 = 147.92$, i.e. $147.92, already lower because the balance is lower. Principal repaid is $497.39 - 147.92 = 349.47$, i.e. $349.47, slightly more than last month, and the closing balance is $24303.14.

Stage 3, repeat the pattern. Month $3$ opens at $24303.14. Interest is $0.006 \times 24303.14 = 145.82$, i.e. $145.82, principal repaid is $497.39 - 145.82 = 351.57$, i.e. $351.57, and the closing balance is $23951.57.

Stage 4, read the trend. Across the three rows the interest column falls ($150.00 \to 147.92 \to 145.82$) while principal repaid rises ($347.39 \to 349.47 \to 351.57$). The payment never changes, but its split shifts steadily from interest toward principal. Continue this to the end and the balance reaches zero at month $60$, with total interest $60 \times 497.39 - 25000 = 4843.40$, i.e. $4843.40.

Watching the interest-versus-principal split shift

The same fixed $497.39 payment is split very differently early and late in the loan. Each panel below shows a single payment as a bar: the accent block on the left is the interest portion $I_k = rB_{k-1}$, the muted block on the right is the principal portion $M - I_k$. The faded bars preview the payments still to come.

Stage 1, payment 1. With the full $25000 still owing, interest claims $150.00 of the payment and only $347.39 reduces the principal.

Stage 2, payment 24. Two years in, the balance has fallen to about $16061, so interest is only $98.76 and $398.63 now goes to principal.

Stage 3, payment 48. Four years in, interest is down to $37.22 and the bar is now overwhelmingly principal at $460.17.

Stage 4, the final payment. By payment $60$ barely a few dollars of interest remain, and almost the entire payment clears the last of the principal.

This front-loading is why making extra repayments early saves far more interest than the same dollar later: an extra dollar paid in month $1$ avoids interest for the whole remaining life of the loan, while the same dollar near the end avoids almost none.

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 fixes the payment $M$ rather than the term $n$, set $B_n = 0$ and solve for $n$:

The expression inside the log must be positive, which requires $M > P r$: the payment must exceed the first period's interest. If $M = Pr$ the balance never moves (an interest-only loan); if $M < Pr$ the balance actually grows and the loan never finishes. Because the last payment is usually a smaller partial payment, round the term up to the next whole period.

How exam questions ask about loan repayments

Each wording points to one of the tools above:

• "Find the monthly repayment / instalment." Use $M = \dfrac{Pr}{1 - (1 + r)^{-n}}$, the present-value-of-annuity formula rearranged for $M$.
• "Find the balance owing after the $n$th payment." Use the closed form $B_n = P(1 + r)^n - M\dfrac{(1 + r)^n - 1}{r}$, or run the recurrence if only a couple of rows are needed.
• "Complete the next row of the schedule" or "find the balance after the 3rd repayment." A recurrence / amortisation-table question: interest on the opening balance, principal equals payment minus interest, roll the closing balance forward.
• "How much of the $k$th payment is interest / pays off the loan?" A split question: $I_k = rB_{k-1}$, principal $= M - I_k$. Find $B_{k-1}$ first by table or formula.
• "Find the total interest paid over the life of the loan." Total interest $= nM - P$ (or add the actual payments if the last is partial).
• "After how many months is the loan repaid?" Set $B_n = 0$, solve for $n$ with logs, and round up.
• "Does the loan ever finish?" or "why does so little of an early payment reduce the principal?" Compare $M$ with $rP$ for completion; for the front-loading, note interest is charged on the large early balance, so most of $M$ is consumed by interest.

Edge case: interest-only versus reducing-balance

Take the same $25000 at $0.6\%$ per month. An interest-only payment would be $M = rP = 0.006 \times 25000 = 150$, i.e. $150 per month: it covers exactly the interest, so the balance stays at $25000 forever and no principal is repaid. The reducing-balance payment of $497.39 is $347.39 larger, and that extra is precisely what buys down the principal each month and clears the loan in $60$ payments. This is the general rule: a reducing-balance payment must exceed the interest-only payment $rP$, and the closer it sits to $rP$, the longer the loan runs.