How to write a simple but robust loop for the following example?

Suppose, I have a $100. Every time when I go to the coffee shop I spend 8.3 % of my money. So, at first when I spend 8.3% I will have (100- 0.083*100) = $91.7 remaining. The next day, I want to spend 8.3% of $91.7, after that I will have (91.7 - 0.083*91.7) = $84.0889 remaining, and so on.

So, for how many days I can use my money?

[Stephen23]

"So, for how many days I can use my money?"

You don't seem to have a lower limit to how much money you need to have/spend, so surely the answer is "forever" ?

[Att]

lower limit is zero. Any help?

[madhan ravi]

Did you see my answer?

[Jan]

@madhan: I do not see your answer.

@A T: This sounds like a homework and you can write a small code, which calculates the answer. You just need a while loop and for keeping teh standards fractinal cents are rounded, e.g. by round(x, 2). Please try to write this code. Post it here if there are problems and ask a specific question.

The lower limit is zero? Of cause you can do this an infinite number of times, if you do not consider the rounding to cents. I guess the minimal value is 1 cent.

[Att]

@Jan: madhan deleted his erroneous answer. Any help will be appreciated.

[OCDER]

Oh, the joy of doing homeworks that ask to solve unrealistic problems... Or is this a trick question?

When you have $1 left, you cannot buy anything in most coffee shops. You may get lucky with $2.00 drip coffee. Below that, you'll get a bunch of baristas with puzzled looks, something like this face: o_O?

You may be able to get sample coffee beans or literally, one drip of coffee. Also, prices of coffee do not scale continuously from $8.30 to 0.0001 cent. You're be better off doing the following:

Days of visit = floor($100 / (minimum price of an item in the store))

You should give the homework answer and the real-life answer.

[madhan ravi]

@Jan yes I deleted my answer since it was not the one OP was looking for.

[Stephen23]

"lower limit is zero. Any help?"

Then the answer is trivially easy: you can do this for infinite days, because you will never reach zero using the formula that you have given us.

Answer 1

[Jan]

What have you tried so far? The current information is that the lower limit is 0. Then the result is trivial: There is no limit in the number of iterations. So either change the lower limit to 1 cent or consider the question as solved. In other words: "$84.0889 remaining" is not meaningful.

At least this would be a start:

x = 100;
n = 0;
while x > 0.01  % One cent at least
    ...
end

Now implement subtracting 8.3%.

My suggestion of round was misleading, because rounding 0.06 after removing 8.3% results in 0.6 again, such that there is still an infinite number of iterations. I guess, the cent values are cropped by floor. To round the cent values, either do all calculations in cents, or use floor(x * 100) / 100.

The limit of 1 cent is reached on the 84th iteration.

Answer 2

[Image Analyst]

To answer, you need to know the price of the least expensive item in the coffee shop. I'm certain that the coffee shop does not sell anything for $0.01. In fact you might have trouble finding anything less than $0.50.

Anyway, like Jan said, 8.3% of 6 cents is less than half a cent, so that rounds to no money being spent, so you have to quit once you get to 6 cents because the requirement of spending 8.3% can no longer be met.

Here's a hint for a line of code you can put into the loop to print out your balance at each iteration:

fprintf('After %d purchases, you have %f left.\n', iteration, balance);

Unlike Jan, I get to 6 cents at 85 iterations, not 84. Maybe it's because he used floor() while I used round(). I think cash registers actually do use round() in the US. When I traveled in Europe it seemed everything is rounded to the nearest 5 euro cents, though I could be wrong.

[Jan]

You are right: With floor to cent values the limit of 1 cent is reached after 84 days.

Some European countries decided to avoid the 1 and 2 cent coins and round to the nearest 5c value. The idea is, that this saves time and energy for counting and transporting a large amount of almost non-value heavy coins.

[Image Analyst]

Honestly, I wish the US would get rid of the cent. They cost more to make than they're worth. No one even wants them - people won't even pick one up off the ground if they see one. And in the long run, the money paid will average out to the same amount anyway.

[Jan]

Feel free to send all your cents to me, or to a project, which needs the money more urgently than I do.

[OCDER]

PH = 0.75; % penny diameter
PT = 0.06; % penny thickness
Box = [23+11/16, 11.75, 3];
Grid = floor(Box1./[PH PH PT]);
PennyPerBox = prod(Grid);
PennyValue = PennyPerBox/100;
ShippingCost = 18.90; %FlatRate USPS 
MoneySent = PennyValue - Cost;
MassOfBox = PennyPerBox * 2.5 * 0.002204;
%You can send around 23250 pennies for $18.90 via USPS flat rate box, for a
%total of $213.60 monetary value left over. The mailman will not like
%carrying the 128-lb box though...

[Jan]

@OCDER: Thank you. I wil give the postman 2136 cents as a tip.