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Apologies if this is a too trivial question but I'm teaching myself and can't get my answer to match the one in my text book.

The task is to calculate the monthly repayments of a £500 loan to be repaid in two years. Interest on the remaining debt is calculated monthly and charged at 11% p.a. First repayment a month after loan given.

Here's my attempt: First I figured the monthly interest charge, M, as $$M = 1.11^\frac {1}{12}$$ After the first month, if a repayment of $\chi$ is made the remaining debt would be $$ 500M - \chi $$ After two months $$ (500M - \chi)M - \chi = $$

$$500M^2 - \chi M - \chi $$ After n months $$ 500M^n - \chi M^{n-1} - \chi M^{n-2} ... \chi M^1 - \chi$$ Or $$ 500M^n - \frac{\chi (M^n - 1)}{M - 1} $$

I reckon this should equal zero after 24 repayments so, rearranging $$ \chi = \frac{500M^{24} (M - 1)}{M^{24} - 1} $$ which comes to £23.18 but the answer given is £23.31. I've tried different numbers of charges/payments and the nearest I got was $$ \chi = \frac{500M^{25} (M - 1)}{M^{24} - 1} $$ equalling £23.38 Can anyone see where I'm going wrong? I guess it could be a typographical error but it'd be the only one I've spotted (so far.) Here's the question exactly as stated in case I'm missing something there

A bank loan of £500 is arranged to be repaid in two years by equal monthly instalments. Interest, calculated monthly, is charged at 11% p.a. on the remaining debt. Calculate the monthly repayment if the first repayment is to be made one month after the loan is granted.

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If the interest is payed m times a year then you usually use the period interest rate $i_m=\frac{i}{m}$. With $m=12$ and $i=0.11$ we get $i_{12}=\frac{0.11}{12}$ Therefore the equation is

$$500\cdot \left(1+\frac{0.11}{12} \right)^{24}=x\cdot \frac{\left(1+\frac{0.11}{12} \right)^{24}-1}{\frac{0.11}{12} }$$

$$x=500\cdot \left(1+\frac{0.11}{12} \right)^{24}\cdot \frac{0.11}{12\cdot \left(\left(1+\frac{0.11}{12} \right)^{24}-1\right)}=23.30391\approx 23.30$$

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  • $\begingroup$ Ah, got it! Thanks for explaining that, I was going spare trying to see the mistake. I thought "12th root of 1.11 multiplied 12 times works out to 11% p.a". but I get how things are calculated now. More expensively. Well, that's usury. $\endgroup$ – Andy Offgrid Dec 13 '18 at 22:02
  • $\begingroup$ @AndyOffgrid You´re welcome. $\endgroup$ – callculus Jul 8 at 18:28
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Let's look at this in general. We need to pay someone $x$ money units over $n$ payments with an interest rate of $r>1$ after each payment. We want to pay an equal amount, say $y$ money units, every payment. Let $x_i$ denote the amount of money we need to pay after $i$ payments. Then we find

$$ x_0=x \\x_1=r(x_0-y)=rx-ry \\x_2=r(x_1-y)=r^2x-(r^2+r)y \\... \\x_i=r^ix-(r^i+r^{i-1}+...+r)y. $$

This gets quite messy. However, there is a formula for $r^i+r^{i-1}+...+r$. To derive this, let's say that $S=r^i+r^{i-1}+...+r$. Then $rS=r^{i+1}+r^i+r^{i-1}+...+r^2$, so $rS+r=r^{i+1}+S$. Hence $r-r^{i+1}=S-rS=(1-r)S$, so we find $S=\frac{r-r^{i+1}}{1-r}$. Plugging this into $x_i$ we find

$$x_i=r^ix-\frac{r-r^{i+1}}{1-r}y.$$

We want to be done with paying after $n$ payments, so we need to find $y$ such that $x_n=0$. So we finally just need to do some algebra.

$$ 0=x_n=r^nx-\frac{r-r^{n+1}}{1-r}y \\r^nx=\frac{r-r^{n+1}}{1-r}y \\y=\frac{1-r}{r-r^{n+1}}r^nx. $$

In your case $x=500$ and $n=24$ and $r=1.11^{\frac1{12}}$, so you can just plug it into the formula.

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  • $\begingroup$ Thanks for your very complete and prompt answer. Too prompt in fact - I should have made it clearer that I'd only posted to see how the MathJax was being formatted as I can't seem to get a preview on my phone. Anyway, as you can see (now that I've finished grappling the MathJax and finished my post) I had made a mistake but even using your formula the result doesn't agree with the text book. Perhaps I'm reading the question wrong? It's copied it in full at the bottom of my post. Despite Occam's razor I'd like too exhaust all avenues before concluding it's a typo. $\endgroup$ – Andy Offgrid Dec 12 '18 at 20:01

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