Question

An engineer borrows $10 000 to buy a personal computer. He must repay $218.94 a month for 5 years. What is the nominal annual interest rate, based upon continuous compounding?

Answer 1

Let us first find the monthly interest rate. Let it be i

Monthly payment=$218.94

Number of payments=n=5*12=60 months

Loan amount=$10000

We know that

Monthly payment=Loan amount*(A/P,i,60)

218.94=10000*(A/P,i,60)

(A/P,i,60)=218.94/10000=0.021894

First we estimate by trial and error method. Let us try at i=0.8%, 0.9% and 1%

(A/P,1,n) = 1 -

(A/P, 0.008, 60) = 1 0.008 - = 0.021051 (1+0.008)60

(A/P, 0.009,60) = 1 0.009 - = 0.021643 (1+0.009)60

0.01 (A/P, 0.01,60) = = 0.022244 (1+0.01)60

We can see that desired interest rate lies between 0.9% and 1%

Now let us try at i=0.95%

0.0095 (A/P,0.0095, 60) = 1 1- 4+0.0095700 -= 0.021942

We can see that desired interest rate lies between 0.9% and 0.95%

Let us interpolate now. We know that

y = y1 + 92–91 * (2 – 21) 12 - 11

Set y1=0.090, y2=0.0095, x1=0.021643, x2=0.021942, x=0.021894

y = 0.009+ 0.0095 - 0.009 -* (0.021894 – 0.021942) = 0.00942 0.021942 -0.021643*

or i=0.942%

Effective rate of interest=(1+0.942)^12-1=0.119084

Effective rate of interest in case of continuous compounding=e^r-1

So,

e^r-1=0.119084

r=ln(1+0.119084)=11.25% (Nominal rate of interest in case of continuous compounding)