Question

3. When interest is compounded continuously, the following equation represents the growth of your savings:
P = Poert

Solve the following question in Matlab:

When interest is compounded continuously, the following equation represents the growth of your savings:

Where
P is the current balance
Po is the initial investment
r is the growth rate, expressed as a decimal fraction
t is the time invested in years.
Use the meshgrid function to determine all the possible amounts that could be in your account at the end of 10 years given the following initial investments ($1000, $2000, $3500, $1500, $500) and growth or interest rates (0.15, 0.20, 0.1).

Answer 1

ANSWER :

MATLAB CODE :

%% Finding Final Amount Present in ACCOUNT whose interest rate is
%% Compounded
clear
clc
P_oi = [1000 2000 3500 1500 500];% Initial amount
r = [0.15 0.2 0.1];% Interest rate
t = 10;% time period
m=1;
for i=1
for j=1:5
P_o = P_oi(i,j);
disp('Initial Investment And Differt Interest rates')
fprintf('\n Initial Investment P_o = %d\n\n',P_o);
for k = 1:3
P = P_o*exp(r(i,k)*t);
P(i,k)= P;
fprintf(' interest rate r = %f\nP after 10 years = %d\n\n\n',r(i,k),P(i,k));
end
end
end
[P_o1,r1] = meshgrid(500:500:3500,0.05:0.05:0.35)
Q1 = P_o1*exp(r1*t);
Q = [Q1]
surf(P_o1,r1,Q)

Output :

Initial Investment And Differt Interest rates

Initial Investment P_o = 1000

interest rate r = 0.150000
P after 10 years = 4.481689e+003

interest rate r = 0.200000
P after 10 years = 7.389056e+003

interest rate r = 0.100000
P after 10 years = 2.718282e+003

Initial Investment And Differt Interest rates

Initial Investment P_o = 2000

interest rate r = 0.150000
P after 10 years = 8.963378e+003

interest rate r = 0.200000
P after 10 years = 1.477811e+004

interest rate r = 0.100000
P after 10 years = 5.436564e+003

Initial Investment And Differt Interest rates

Initial Investment P_o = 3500

interest rate r = 0.150000
P after 10 years = 1.568591e+004

interest rate r = 0.200000
P after 10 years = 2.586170e+004

interest rate r = 0.100000
P after 10 years = 9.513986e+003

Initial Investment And Differt Interest rates

Initial Investment P_o = 1500

interest rate r = 0.150000
P after 10 years = 6.722534e+003

interest rate r = 0.200000
P after 10 years = 1.108358e+004

interest rate r = 0.100000
P after 10 years = 4.077423e+003

Initial Investment And Differt Interest rates

Initial Investment P_o = 500

interest rate r = 0.150000
P after 10 years = 2.240845e+003

interest rate r = 0.200000
P after 10 years = 3.694528e+003

interest rate r = 0.100000
P after 10 years = 1.359141e+003

P_o1 =

500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500
500 1000 1500 2000 2500 3000 3500

r1 =

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500
0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000
0.1500 0.1500 0.1500 0.1500 0.1500 0.1500 0.1500
0.2000 0.2000 0.2000 0.2000 0.2000 0.2000 0.2000
0.2500 0.2500 0.2500 0.2500 0.2500 0.2500 0.2500
0.3000 0.3000 0.3000 0.3000 0.3000 0.3000 0.3000
0.3500 0.3500 0.3500 0.3500 0.3500 0.3500 0.3500

Q =

1.0e+005 *

2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166
2.3166 2.3166 2.3166 2.3166 2.3166 2.3166 2.3166

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