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Matlab FOR Loop help!

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Dylan Flores
Dylan Flores am 29 Jul. 2014
Bearbeitet: Ben11 am 30 Jul. 2014
_ I'm trying to get this function to work. But i keep getting an error. I am trying to determine the amount in the savings account for next 18 years, which is represented by x(k) in an array. Problem is that only the first value for the first month shows up in the array._ * _ * _
clc,clear
% Variable declaration
% x will be used to store the value of the new balance
% a will be used to store the value of the old balance
% i will be used to store the value of the interest rate
% c will be used to store the value of the user's contribution
% Variable initialization
a = input('Enter a value for the initial balance: ');
i = input('Enter a value for the interest rate: ');
c = input('Enter a value that you will contribute: ');
% Calculation
month = 1:12:216;
x=zeros(length(month));
for k=1:length(month);
x(k) = a + i(k) + c
end
  10 Kommentare
Sara
Sara am 29 Jul. 2014
Old balance is x(k-1) not a and I think interest means x(k-1)*i. The loop becomes:
x(1) = a;
for k = 2:numel(month)
x(k) = x(k-1)+ x(k-1)*i + c; %or x(k-1)*(i+1) + c;
end
If the interest is %, divide i by 100.
Dylan Flores
Dylan Flores am 29 Jul. 2014
Thanks! It works a lot better now!

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Antworten (1)

Ben11
Ben11 am 29 Jul. 2014
Bearbeitet: Ben11 am 30 Jul. 2014
Maybe try this:
month = 1:12:216;
x=zeros(1,length(month)); % otherwise you have a 18*18 array
x(1) = a;
for k=2:length(month);
x(k) = x(k-1) + (i/100)*x(k-1) + c; % add the amount present during the previous month. Oh and I divided your interest rate by 100.
end
  2 Kommentare
Dylan Flores
Dylan Flores am 29 Jul. 2014
Hey Thanks for helping out!
Ben11
Ben11 am 30 Jul. 2014
Bearbeitet: Ben11 am 30 Jul. 2014
You're welcome! It looks like Sara and I arrived at pretty much a similar solution; although I incorrectly used a instead of x(k-1) :)

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