[Math] The Newton-Raphson Method for finding a correct monthly interest rate

calculusderivativesnewton raphson

I am very new to this topic and just started to learn about this method. Trying to understand this method intuitively.

In this document I found and interesting real life 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
$$Ar=M\left(1-\frac1{(1+r)^n}\right).$$
A car loan of $10000$ dollars was repaid in $60$ monthly payments of $250$ dollars.
Use the Newton Method to find the monthly interest rate correct to $4$
significant figures.

Can somebody please explain intuitively why do we use this method in real life? If I understood correctly so far, we can make a guess and then find a very close number to the real answer, using this method.

Appreciate your time and other interesting examples, ideally with a code example in R/Python. Thanks!

Best Answer

For intuition, you might be interested in reading Why does Newton's method work? and Math Insight.

As for this specific example, we cannot find a nice closed-form solution, so we are stuck using numerical methods instead and will choose Newton's Method, but many other root finding methods will work.

Some of the goals of numerical methods are to be stable and have as fast as convergence as possible for each iteration of the algorithm. In this problem, we are given:

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 $$Ar=M\left(1-\dfrac1{(1+r)^n}\right).$$ A car loan of $10000$ dollars was repaid in $60$ monthly payments of $250$ dollars. Use the Newton Method to find the monthly interest rate correct to $4$ significant figures.

So, we know

$$Ar = M\left(1-\dfrac1{(1+r)^n}\right) \rightarrow 10000 r = 250 \left(1-\dfrac1{(1+r)^{60}}\right)$$

We want to solve this function for $r$, but there is no closed form solution, so Newton's Method it is. For the Newton iteration step below, we can write our function as

$$f(r) = 40 r + \dfrac1{(r+1)^{60}} - 1$$

Taking the derivative, we have

$$f'(r) = 40 - \dfrac{60}{(r+1)^{61}}$$

The Newton iteration is given by $x_{n+1} = x_n - \dfrac{f(x)}{f'(x)}$, so we have

$$r_{n+1} = r_n - \dfrac{40 r_n +\dfrac{1}{(r_n+1)^{60}}-1}{40-\dfrac{60}{(r_n+1)^{61}}} = $$

Now, we choose a starting value, say, $r_0 = 1$, and keep our fingers crossed and have the iteration

$r_0 = 1.0000000$
$r_1 = 0.0250000$
$r_2 = 0.0164861$
$r_3 = 0.0145644$
$r_4 = 0.0143962$
$r_5 = 0.0143948$
$r_6 = 0.0143948$

So, we find that the monthly interest rate is $1.439 \%$ to four significant figures.

Related Solutions

[Math] How to calculate APR using Newton Raphson

Let's collect the comments into an answer. I like single letter variables for formulas, so

T -- $amount;     or principal
n -- $numPay;     number of months
R -- $payment;    monthly rate
F -- $fee;
x -- $x;          monthly interest rate

The payment plan, as I understand it, has an initial debt or principal $T$, and additional a fee $F$. After a month $R+F$ are paid, then every month $R$, after the n-th payment, the debt is paid in full.

As stated, the interest on the fee for the one month is not raised, which means that $F$ instead of $F\cdot(1+x_{nominal})$ is paid after a month. In balance, this still increases the effective interest rate.

The balance after $n$ months, seen from the start of the debt, and according to the flow of payments, reads as

\begin{align}T+F&=(R+F)(1+x)^{-1}+R(1+x)^{-2}+\dots+R(1+x)^{-n}\\ &=F(1+x)^{-1}+R\frac{1-(1+x)^{-n}}{x}\\ Tx+F\frac{x^2}{1+x}&=R(1-(1+x)^{-n}) \end{align}

Define

$$f(x)=Tx(1+x)+Fx^2-R(1+x-(1+x)^{1-n})$$

with derivative

$$f'(x)=T(1+2x)+2Fx-R(1-(n-1)(1+x)^{-n})$$

for a Newton iteration with the initial guess

$$x_0=\frac{nR-T}{\frac{n(n+1)}2R+F}$$

Best Answer

Related Solutions

[Math] How to calculate APR using Newton Raphson

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