Chapter 4.7, Problem 33E

To determine

To calculate: The monthly interest rate.

Answer to Problem 33E

The interest rate is

  $x\approx 0.0076286$

Explanation of Solution

Given information:

The annuity formula $A=\frac{R}{i}\left[1-{(1+i)}^{-n}\right]$ where A represents the present value of annuity, R represent the payment amount, i is the interest rate per time period and n is the number of payments that will be made.

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Formula: $s=r\theta$ where s is arc length, r is the radius of the circle and $\theta$ is the central angle.

Law of cosine $c^2=a^2+b^2-2ab\cos \theta$where a , b , c are sides of triangle .

Calculation:

Consider the,

A =$18000

R =$375

Payments are to be made every month for 5 years.

This implies

  $\begin{array}{l}1year=12months\\ 5year=12\times 5months\end{array}$

  $=60months$

Therefore, n =60

  $=18000=\frac{375}{i}\left[1-{(1+i)}^{-60}\right]$ (i)

Re-placing i by x we get,

  $\begin{array}{l}=18000=\frac{375}{x}\left[1-{(1+x)}^{-60}\right]\\ =18000x=375\left[1-{(1+x)}^{-60}\right]\\ $ =\frac{18000x}{375}=1-{(1+x)}^{-60}$=48x=1-\frac{1}{{(1+x)}^{60}}\\ =48x=\frac{{(1+x)}^{60}-1}{{(1+x)}^{60}}\end{array}$ (ii)

  $\begin{array}{}$ =48x{(1+x)}^{60}={(1+x)}^{60}-1$$ =48x{(1+x)}^{60}-{(1+x)}^{60}+1=0$\end{array}$ . (iii)

Hence, we get

  $\begin{array}{}$ g(x)=48{(1+x)}^{60}-{(1+x)}^{60}+1$$ g(x)=48{(1+x)}^{60}-{(1+x)}^{60}+1$\end{array}$

Now, let initial approximation be $x_1=0.1$

For $n=1$

  $x_2=x_1-\frac{g(x_1)}{g\text{'}(x_1)}$

  $x_2=x_1-\frac{48{(1+x_1)}^{60}-{(1+x_1)}^{60}+1}{2880{(1+x_1)}^{59}-60{(1+x_1)}^{59}}$

  $\begin{array}{l}{x}_2=(0.1)-\frac{48{(1+0.1)}^{60}-{(1+0.1)}^{60}+1}{28801{(1+0.1)}^{59}-60{(1+0.1)}^{59}}\\ x_2=(0.1)-\frac{48{(1.1)}^{60}-{(1.1)}^{60}+1}{28801{(1.1)}^{59}-60{(1.1)}^{59}}\end{array}$

The second approximation is $x_2=0.85101094$ .

Following with the iterations after 14 iterations will get the approximate value of x .

Hence , the monthly interest to be pay is :-

  $x\approx 0.0076286$