Question

A loan of \(A\) dollars is repaid by making \(n\) equal monthly payments of \(M\) dollars, starting a month after the loan is made. It can be shown that if the monthly interest rate is \(r\), then
\[
A r=M\left(1-\frac{1}{(1+r)^{n}}\right) \text {. }
\]
A car loan of \(\$ 10000\) was repaid in 60 monthly payments of \(\$ 250\). Use the Newton Method to find the monthly interest rate correct to 4 significant figures.

Answer 1

Even quite commonplace money calculations involve equations that cannot be solved by 'exact' formulas. Let \(r\) be the interest rate. Then
\[
10000 r=250\left(1-\frac{1}{(1+r)^{60}}\right) .
\]
If we are going to work by hand, it is maybe worthwhile to simplify a bit to \(f(r)=0\) where
\[
f(r)=40 r+\frac{1}{(1+r)^{60}}-1
\]

and therefore
\[
f^{\prime}(r)=40-\frac{60}{(1+r)^{61}} .
\]
The Newton Method iteration is now easy to write down. In raw form it is
\[
r_{n+1}=r_{n}-\frac{40 r_{n}+1 /\left(1+r_{n}\right)^{60}-1}{40-60 /\left(1+r_{n}\right)^{61}} .
\]
Compute. Particularly if we do the work by hand, it is helpful to make a good choice of \(r_{0}\). If the interest rate were \(2.5 \%\) a month, the monthly interest on \(\$ 10,000\) would be \(\$ 250\), and so with monthly payments of \(\$ 250\) we would never pay off the loan. So the monthly interest rate must have been substantially under \(2.5 \%\). A bit of fooling around suggests taking \(r_{0}=0.015\). We then find that \(r_{1}=0.014411839\), \(r_{2}=0.014394797\) and \(r_{3}=0.01439477\). This suggests that to four significant figures the monthly interest rate is \(1.439 \%\).