## Loan Repayments

#### Problem

If £2000 is borrowed, interest is charged at an annual rate of 12%, and the loan is to be fully repaid after thirty-six fixed monthly payments, how much should each monthly repayment be?

#### Solution

We shall begin by finding a general formula for fixed payments also receiving compound interest.

If the rate of growth for each period is $r$ and the fixed payment is $p$ then the balance after $n$ periods is given by the recurrence relation: $u_n = r u_{n-1} + p$.

If the initial balance is $u_0$, then we can recursively apply this relation.

\begin{align}u_1 &= r u_0 + p\\\\u_2 &= r u_1 + p\\&= r(r u_0 + p) + p\\&= r^2 u_0 + rp + p\\\\u_3 &= r u_2 + p\\&= r(r^2 u_0 + rp + p) + p\\&= r^3 u_0 + r^2 p + rp + p\end{align}

This leads to, $u_n = r^n u_0 + p(r^{n-1} + r^{n-2} + ... + r^2 + r + 1)$.

Writing $S_n = r^{n-1} + r^{n-2} + ... + r^2 + r + 1$, $rS_n = r^n + r^{n-1} + ... + r^2 + r$.

Therefore $S_n(r - 1) = r^n - 1$, giving $S_n = \dfrac{r^n - 1}{r - 1}$.

Hence $u_n = r^n u_0 + \dfrac{p(r^n - 1)}{r - 1}$.

For example, if we start with a balance of £100, receive interest at a rate of 5% per annum, and make regular yearly payments of £50 for ten years, the balance after ten years, $u_{10} = 1.05^10 \times 100 + \dfrac{50(1.05^10 - 1)}{1.05 - 1} \approx$ £791.78. As £$100 + 10 \times 50 =$ £600 has been invested, this would represent a growth of about 32% in our original investment.

Now we return to the original problem...

As the annual interest is 12%, the monthly rate, $r = 1.12^{1/12} \implies r^{36} = 1.12^3$, $u_0 = 2000$, and we are aiming for $u_{36} = 0$.

Solving $0 = 1.12^3 \times 2000 + \dfrac{p(1.12^3 - 1)}{1.12^{1/12} - 1}$, we get $p \approx$ £65.84.

Thirty-six fixed payments of £65.84 is £2370.24, which means that the interest charged on the full loan is about 18.5%.

Problem ID: 260 (09 Jan 2006)     Difficulty: 4 Star

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