Compound Interest with monthly Contributions, Unable to vectorize.

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Billy
Billy 2012-11-13
Hello, I would like help with vectorizing this code so that I can plot the answer. I have been able to get individual answers for each month, however, I cannot write this function in terms of months, so thats why I can not get my answer into a vector. Here's what I have so far.
%
New_balance=1000
for month=[0:18*12]
New_balance=New_balance*1.005+100
end
It is monthly interest of .5% for 18 years with 100 every month and an initial $1000. I want to be able to plot(months,New_balance), including the 0th month of $1000. Thank You.

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-11-13
n = 18*12+1;
New_balance([1,n],1)=[1000;0];
for j1=2:n
New_balance(j1)=New_balance(j1-1)*1.005+100;
end
or
n = 18*12;
New_balance = filter(1,[1 -1.005],[1000;ones(n,1)*100]);
  2 个评论
Billy
Billy 2012-11-13
编辑：Billy 2012-11-13
You have the part of the answer that I want, but my directions tell me after the loop, I should be able to show the final amount at the end of 18 years. Your answer fits what I described, but I just noticed this detail. Do you know how I would be able to extract the final value by placing something in or after the for loop? Is there a way to define the final value as a variable? Thanks.
Oleg Komarov
Oleg Komarov 2012-11-13
New_balance(end)
@Billy: invest some time into the getting started guide (the chapter on matrix manipulations).

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更多回答（3 个）

C.J. Harris
C.J. Harris 2012-11-13
You don't even need to vectorise, just calculate the answer directly. The answer below is slightly lower than your answer. But in your calculation you appear to accrue interest in month zero (time of deposit).
P = 100;
i = 0.005;
A = 1000;
n = 18*12;
FV1 = ( (1 + i)^ n ) * A;
FV2 = P * ( ( (1 + i)^n - 1) / i );
FV = FV1 + FV2;
  1 个评论
Billy
Billy 2012-11-13
I should have mentioned, that I need to use a for loop for this problem and so I cannot just find the answer at the end of 18 years. I have to have a matrix of 216 values so that I will be able to plot.

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Alex
Alex 2012-11-13
You could try a 1-pole IIR filter.
Oleg Komarov
Oleg Komarov 2012-11-13
n = 18*12; % number of periods
i = 0.005; % per period interest
I = 1000; % initial investment
CF = 100; % recurrent cash flow
s = (1 + i).^(0:n); % compounding factor
capital = I*s + [0 cumsum(CF*s(1:end-1))];
plot(0:n,capital)

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