Question

8.7 If an amount of money A is invested for k years at a nominal annual interest rate r (expressed as a decimal fraction), the value V of the in vestment after k years is given by V Ar/n where n is the number of compounding periods per year. Write a pro gram to compute V as n gets larger and larger, i.e. as the compounding periods become more and more frequent, like monthly, daily, hourly etc. Take A 1000, r 4% and k 10 years. You should observe that your output gradually approaches a limit. Hint: use a Tor loop which doubles n each time, starting withn= Also compute the value of the formula Aer for the same values of A,r and k (use the MATLAB function exp), and compare this value with the values of V computed above. What do you condude?

use matlab

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Answer 1

the maximum value for n is 50:

clc;clear all;close all;
A=1000; %investment
r=0.04; % interest rate
k=10; %years
for n=1:50;
V=A*(1+r/n)^(n*k);
end
display(V)
l=A*exp(r*k)

output:

V =

1.4916e+03

l =

1.4918e+03

for the max value for n=100:

clc;clear all;close all;
A=1000; %investment
r=0.04; % interest rate
k=10; %years
for n=1:100;
V=A*(1+r/n)^(n*k);
end
display(V)
l=A*exp(r*k)

output:

V =

1.4917e+03

l =

1.4918e+03

for n max value is 200:

clc;clear all;close all;
A=1000; %investment
r=0.04; % interest rate
k=10; %years
for n=1:200;
V=A*(1+r/n)^(n*k);
end
display(V)
l=A*exp(r*k)

output:

V =

1.4918e+03

l =

1.4918e+03

as increasing the value of n,at n=200 both the values are gets equal.