Accumulating Interest

Date: 07/10/2003 at 18:25:25
From: Joey  
Subject: Accumulating interest

I'm taking a course on problem-solving, in which we were given the
following problem to solve:

Michael has $355.00 left over each month after paying all his monthly 
bills. Instead of spending his leftover money each month, Michael 
chooses to open a savings account offering 3% interest, accrued 
monthly. Calculate the amount of money Michael will have in his 
savings account at the end of each of the following periods:

a)  3 months
b)  6 months
c) 12 months

Here is what I tried so far:

 Mon  Amount
 ---  --------------------------------------------------
  1   (.03) * 355.00 = 365.65
  2   365.65 + 355.00 = 720.65 *.03 = 21.62 + 720.65 = $742.27
  3   742.27 + 355.00 = 1,097.27 * .03 = 32.92 + 1097.27 = 1130.19
  4   1130.19 + 355 = 1485.19 * .03 = 44.55 + 1485.19 = 1529.74

Am I on the right track?

- Joey

Date: 07/10/2003 at 19:33:08
From: Doctor Ian
Subject: Re: Accumulating interest

Hi Joey, 

One way to think about it is to imagine how you'd solve it using a
spreadsheet. Each row of the spreadsheet would represent what happens
at the end of a month. You have a current balance, which at first
would be a single deposit:

   +--------------+
   | Balance      | 
   | during month |
   +--------------+
   |            D |
   +--------------+

You compute some percentage of that, and add it to the balance to get
the new balance. 

   +--------------+---------------+
   | Balance      | After         |
   | during month | interest      |
   +--------------+---------------+
   |           D  |            iD |
   +--------------+---------------+

Note that I'm using i to mean (1 plus the interest rate), e.g., if the
interest rate is 3%, then i is 1.03.  Why?  Because that's how you
_add_ interest:

  x + 0.03*x = (1 + 0.03)*x

             = 1.03*x

After interest is paid, you add another deposit:

   +--------------+---------------+---------------+
   | Balance      | After         | After         |
   | during month | interest      | deposit       |
   +--------------+---------------+---------------+
   |            D |            iD |        iD + D |
   +--------------+---------------+---------------+

and what you have at the end becomes the new balance during the next
month:

   +--------------+---------------+---------------+
   | Balance      | After         | After         |
   | during month | interest      | deposit       |
   +--------------+---------------+---------------+
   |            D |            iD |        iD + D |
   +--------------+---------------+---------------+
   |       iD + D |     
   +--------------+

Now we need to pay interest on the balance during the month. Just as 
before, we do this by multiplying i by whatever is in the first 
column:

   +--------------+---------------+---------------+
   | Balance      | After         | After         |
   | during month | interest      | deposit       |
   +--------------+---------------+---------------+
   |            D |            iD |        iD + D |
   +--------------+---------------+---------------+
   |       iD + D |     i(iD + D) |     
   +--------------+---------------+

And then we add the new deposit:

   +--------------+---------------+---------------+
   | Balance      | After         | After         |
   | during month | interest      | deposit       |
   +--------------+---------------+---------------+
   |            D |            iD |        iD + D |
   +--------------+---------------+---------------+
   |       iD + D |     i(iD + D) | i(iD + D) + D |     
   +--------------+---------------+---------------+

Now we can simplify that last item to look like this:

   +--------------+---------------+---------------+
   | Balance      | After         | After         |
   | during month | interest      | deposit       |
   +--------------+---------------+---------------+
   |            D |            iD |        iD + D |
   +--------------+---------------+---------------+
   |       iD + D |     i(iD + D) | i^2D + iD + D |     
   +--------------+---------------+---------------+

Does this make sense so far?  If you actually _have_ a spreadsheet,
you can go ahead and set it up this way. 

Assuming it makes sense, can you see why the third column will look
like this?

                                 iD + D
  
                           i^2 + iD + D

                    i^3 + i^2D + iD + D

            i^4D + i^3D + i^2D + iD + D
                 .
                 .
  i^nD + i^(n-1)D + ... + i^2D + iD + D

At each step, you're multiplying everything you already had by i
(which is what you do to pay interest on it), and then adding D. We
can use the distributive property to factor out a D, leaving us with 

  (i^n + i^(n-1) + ... + i^2 + i + 1) * D

Now, on the one hand, this still looks kind of complicated, although
nowhere near as complicated as using actual numbers. But the good news 
is that a sequence like 

  a, ar, ar^2, ar^3, ar^4, ...

is a pretty well-known object, called a 'geometric sequence', and
there is a formula for finding the sum of one in a single step:

  a + ar + ar^2 + ... + ar^n = (a - ar^(n+1))/(1-r)

Since you're in a problem-solving class, there's a good chance that
the point of assigning the problem was to get you to see why solving
it in symbolic form is a better idea than just cranking out the
calculations. One of the reasons I didn't actually check your results
is that I couldn't really do that without going through all the steps
that you went through. And here's the thing: mathematics is a kind of 
applied laziness. A mathematician is someone who has realized that in 
many cases, it's harder to solve a single _particular_ problem than it 
is to solve _all_ the related problems at once, in one shot, which is 
what I did here.  

What if I want to change the interest rate?  No problem - I just 
change the value of the associated variable in my formula. What if I 
want to change the amount of each deposit? The same. And it's just a 
little bit more work to solve the problem in a way that lets me adjust 
for how often interest is compounded, e.g., weekly or daily. 

(And as a bonus, I didn't even have to write as much, because I was
using single letters instead of strings of digits to represent my
numbers.)

But if you just solve the specific problem, using numbers, then to
solve a slightly different problem is just as much work as the first
time around.  

By the way, 3% accrued monthly works out to an annual interest rate of
nearly 43 percent. Where is this bank?  I'd like to open an account
there!

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 07/11/2003 at 07:45:57
From: Joey 
Subject: Accumulating interest

The way I see doing this is: you start out with the amount which he 
has left over monthly which in my initial problems case is $355 and 
multiply it by the interest rate of .03.

Then for the second month, take your balance from the previous month
and add the deposit for the current month giving whatever and then
multiply that by 1.03 and you have your amount for that month.

Then just keep following these steps for the rest of the months. How
exactly is that wrong is what I do not understand. I am calculating
interest on his account balance after every month. Do you see what I
am saying?

Date: 07/11/2003 at 09:48:45
From: Doctor Ian
Subject: Re: Accumulating interest

Hi Joey, 

The problem with your method is that if you make the deposit _before_
computing the interest, you're paying interest immediately on the
deposit.  You don't get interest on money until after the bank has had
a chance to use it. That's what they're paying interest for.  

So on January 1, you put $355 into the bank. Then on February 1, you
put another $355 in, and the bank pays you interest... but only on the
first $355. You don't get interest on the second $355 until March, at 
which time you put in another deposit.

What makes these calculations a pain (and the reason people have
developed formulas to avoid having to do them) is that you can't just
compute the interest based on the current day's balance. You only get
interest on the money that's been there for a while. 

(Otherwise I could put $10,000 into an account the day before interest
gets paid, collect the interest, and withdraw all the money the next
day. No bank would be able to stay in business if this were the case.) 

Here is another way to look at it. Suppose you deposit $100 on the
first day of each month during the course of a year, but each time you 
start a new account instead of combining it all in one account. The 
bank pays you 5% interest per month on each account. 

At the end of the year, the account you opened on December 1 will have
accrued interest for one month:

  Dec. acct:  $100 * 1.05

What about the November account?  At the end of November, you would
get 5% interest; and you'd get it again at the end of December:

  Dec. acct:     $100 * 1.05
  Nov. acct:    ($100 * 1.05) * 1.05

And the October account?  You'll collect interest on that three times:

  Dec. acct:     $100 * 1.05
  Nov. acct:    ($100 * 1.05) * 1.05
  Oct. acct:   (($100 * 1.05) * 1.05) * 1.05

And so on.  Using exponents to tidy things up, 

  Dec acct:  $100 * 1.05^1
  Nov acct:  $100 * 1.05^2
  Oct acct:  $100 * 1.05^3
  Sep acct:  $100 * 1.05^4
  Aug acct:  $100 * 1.05^5
  Jul acct:  $100 * 1.05^6
  Jun acct:  $100 * 1.05^7
  May acct:  $100 * 1.05^8
  Apr acct:  $100 * 1.05^9
  Mar acct:  $100 * 1.05^10
  Feb acct:  $100 * 1.05^11
  Jan acct:  $100 * 1.05^12

Factoring out the original $100, that all adds up to 

  $100 * (1.05^1 + 1.05^2 + ... + 1.05^12)

which is our old friend, the geometric series. 

Now, this has to be the same amount of money as we'd get the other
way.  All we've done is change the situation around a little so that
we're not mixing deposits and interests. The cost of that is having
more accounts. But that's not _really_ a cost, because the accounts
are imaginary. They just exist long enough to find a general solution
to the class of problems, and then they disappear.

However, even though there is no cost, there is a big benefit, which
is that we get to solve 12 extremely simple problems rather than one
complicated one.  

Does it make more sense now? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/