How many years do i need to get the amount?

Imagine that you went to the bank and deposited $20,000 in an account that earns 6% interest every year, with each year’s interest being deposited back into the account. Write a MATLAB program that computes the number of years it would take to accumulate $500,000.

The only thing I know is that 6% = 0.06 and I need to use a for loop, with an if statement, to solve this problem. I couldn't even start the program

So I have come this far but I don't know if it correct or not:

y = 1;
r = 0.06;
p = 20000;
A = 500000;

function t = final result
t = log (A/p)/y*log (r/y);

Answer 1

[Image Analyst]

Wasn't the simple interest formula discussed in the class? Like if P is your principal (initial) amount, then after 1 year you have P + 0.06*P, and after 2 years you have (P + 0.06*P) + 0.06 * (P + 0.06*P) and so on. There is a formula you can use to get the final amount without a for loop, but if you're required to do a for loop you can do

finalAmount = 20000  % Amount you're starting with.
for y = 1 : totalNumberOfYears
  finalAmount = finalAmount + ........
end

I think that should be plenty of hint to get you started. Use inputdlg() to ask the user for the number of years, principle amount, interest rate, etc.

[Hamzah]

is it like for example:

for i = 1:n 
    sum = sum + i
end ???

sorry but im really bad at this so bare with me

[Image Analyst]

No. Why would you add the number of years to the dollar amount???? that doesn't make any sense. You have to add the amount of money it gained during that year. I really can't do that for you or else I've done the whole thing for you. Just think. If I have $1000 for a year and it gained 6% interested, what is the dollar amount of the interest? What do I multiply by what to get the answer?

You also have to have a way to bail out once you've reached 500000. You can do that with break.

if finalAmount > 500000
   break;
end

Also, don't use "sum" as the name of a variable since it's a built-in function.

[Image Analyst]

Still can't figure it out? Here's a little more:

totalNumberOfYears = 100;
finalAmount = 20000  % Amount you're starting with.
for y = 1 : totalNumberOfYears
  finalAmount = finalAmount + finalA.......FINISH THIS LINE
  fprintf('After %d years, the balance is $%.2f\n', y, finalAmount);
  if finalAmount > 500000
    break;
  end
end

You should be able to get it now since only a few characters are missing.

Answer 2

[Roger Stafford]

This is a problem in compound interest, the compounding occurring once each year. Think it through like this. After one year you have 20000*1.06, after two years it amounts to 20000*1.06*1.06, and so forth. Does that give you a hint as to how to solve the problem?

[Hamzah]

i know that compound interest formula is A = p(1+r/n)^(n*t)

and the other thing i think its: for i = 1:n %for example sum = sum + i end

sorry but im really bad at this so bare with me

Answer 3

[Star Strider]

You can always just ‘brute-force’ your solution and see what fzero, our reliable friend in intractable situations, (although this is hardly intractable) comes up with:

A = 500000;                     % Workspace Variables & Constants
p =  20000;                     % Workspace Variables & Constants
r = 0.06;                       % Workspace Variables & Constants
n = 1;                          % Workspace Variables & Constants
C = @(t) p*(1+r/n).^(n*t);      % Compound Interest Formula (As Anonymous Function)
t = fzero(@(t) (C(t)-A), 5);    % Time until Retirement (Years)

[Roger Stafford]

There is no need for 'fzero'. By taking the logarithm of the above equation, the variable 't' can be solved directly and then a 'ceil' done. Actually there's no need for matlab. A scientific calculator can do it easily.

[Star Strider]

I completely agree.

I was suggesting a way of getting the correct answer as guidance, since OP’s analytic solution (the last paragraph in OP’s edited Question) is not the same result I got when I solved it. (Mine was a one-line derivation that yielded the correct result.)

[Image Analyst]

It may not be the way we'd solve it but we've all seen homework problems where the instructor specifies a certain method be used. He says he needs to use a for loop and if statement for some reason (presumably the instructor wants that so as to teach them for loops or something): "The only thing I know is that 6% = 0.06 and I need to use a for loop, with an if statement, to solve this problem."

Answer 4

[kobi]

i=0.06; 
a=20000;
n=a;

for y=1:a n=n*i+n;
    if n>500000 
        display(n);
        display(y);
        break;
    end
end

there you go buddy, answer is 56 :)