Calculate the number of years the money will last?

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my program is supposed to calculate the # of years the money will last for each retiree.. each year 2.5% interest is added onto the balance and the individuals withdraws 4.5%. So i created a While loop but the loop just keeps repeating, it doesn't stop. I think my main problem is that i dont no what to write for the conditional statement of While loop. Here's what i have so far
No_of_elements = 7;% the number of retirees
min_years = 40;
max_years = 50;
*% # of years of employment
r = round(min_years + (max_years-min_years).*rand(No_of_elements,1))
% deposit '
x = (400000-300000 * r-min(r)) ./ max(r) - min(r) + 300000
W = 0; % the initial withdrawal amount
years = 0;
while x >= W
format bank
% the interest
i = x*0.025
% the balance after the interest is added
B= x+i
% the amount of withdrawal
W = B*0.045
% the balance after the withdrawal
x = B-W
years = years+1;
if x<=0
fprintf ('the number of years: %d \n',years)
end
end
  3 Kommentare
nada Elmoghany
nada Elmoghany am 8 Apr. 2015
Bearbeitet: nada Elmoghany am 8 Apr. 2015
oh sorry about that, that was a typo.. its supposed to W not Z, and the B1 is supposed to B. Sorry
dpb
dpb am 8 Apr. 2015
Suggest you then edit your code to reflect the changes...

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Jonathan Campelli
Jonathan Campelli am 8 Apr. 2015
Nada, your While loop was not breaking because your x values were never reaching zero. I've crafted a nested while loop that stops looping when your "bank account" dips below 1 Cent, or $0.01:
No_of_elements = 7;% the number of retirees
min_years = 40;
max_years = 50;
r = round(min_years + (max_years-min_years).*rand(No_of_elements,1)) % deposit '
x = (400000-300000 * r-min(r)) ./ max(r) - min(r) + 300000
W = 0.01; % the initial withdrawal amount
years = zeros(1,numel(x))';
for count=1:numel(x)
while x(count)>W
x(count)=x(count)+(x(count)*0.025)-(x(count)*0.045);
years(count)=years(count)+1;
end
end
years
  2 Kommentare
nada Elmoghany
nada Elmoghany am 8 Apr. 2015
Thank you very much for your help... That question was honestly frustrating me. Thanks again
Jonathan Campelli
Jonathan Campelli am 9 Apr. 2015
You're welcome. I'm glad I could be of help.

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