Lesson idea for geometric series: Loan Amortization!

Here’s a non-trivial application of the concept of geometric series with some actual context: calculating monthly payments on a mortgage.

The question:

You take out a loan for $50000 at an annual interest rate of 12%. If you want to pay off the loan in exactly 30 years, how much would you have to pay each month?

The details:

Interest is compounded monthly, so at the end of each month, the TOTAL amount that is owed is the remaining principal plus 1% interest. The monthly payment is then subtracted from this and the result becomes the starting amount for the next month, and the process is repeated each month. (The only reason I picked 12% as the interest rate is to make the monthly rate a nice round number. Real-life interest rates are–hopefully–much lower. At the end, students can compare results with different interest rates!)

An interesting concept:

Loan amortization leads to the surprising result that even if you pay the same amount each month, the principal decreases very slowly at first, and much more quickly towards the end of the loan period.

Here’s why: the monthly payment will pay off the interest from that month plus some amount of the principal. Early on in the loan period, while there is still a large amount of principal remaining to be paid off, the dollar value of the interest is higher than it will be later on (since the interest is being calculated from a higher principal). So, early on, each monthly payment will cover that month’s interest with only a small amount left to start paying down the principal. Over time, the principal gets smaller, so the interest each month on that principal gets smaller. Thus, if the payment is the same each month,  less and less of that payment will simply be covering that month’s interest and more and more will be going towards reducing the principal. The further into the repayment period, the FASTER the principal will be reduced. This is pretty surprising.

The problem:

At the beginning of the 30 years, how can I determine a fixed monthly payment that will actually result in exactly zero remaining balance at the end of 30 years (360 monthly payments in the future)?

A slightly easier question:

If I pay a fixed amount (say $700) each month, with the same principal ($50000) and same interest rate (12%), will I ever pay off the entire loan and, if so, how many months will this take?

One way to solve the “slightly easier question:”

Calculate month-by-month:

total owed at the beginning of next month = total owed at the beginning of this month + this month’s interestthis month’s payment.

Total owed at the end of month 1:

= $50000 + $50000*(1%)$700

= $50000 + $500$700

= $49800    (the payment of $700 went to cover $500 of interest and $200 of principal).

Total owed at the end of month 2:

= $49800 + $49800*(1%)$700

= $49800 + $498$700

= $49598 (the payment of $700 went to cover $498 of interest and $202 of principal)

This $49598 is the starting amount for the third month, and so on. Each month, as the principal slowly declines, less and less interest is charged each month, so more and more of the monthly payment goes towards paying off the principal.

Paying $700 a month, you would have to repeat this process 126 times to eventually reach a balance of zero. Thus, it will take 126 months (which is 10.5 years) to pay off this loan.

Doing this process repeatedly until getting a balance of zero (which would take 126 steps in this example) is a good exercise for any budding computer programmers in your class, or a good opportunity to have the whole class practice using Excel!

Now a solution to the real problem:

Recall that we wish to calculate the monthly payment that will allow the loan to be paid off is exactly 30 years. One way to do this would be to do exactly what we did above, pick a payment, and figure out how many months it would take to pay off the loan. If it will take less than 360 months, try a lower payment next time; if it will take more than 360 months, try a larger payment next time. Repeat this until you narrow in on a monthly payment that takes exactly 360 months to reach a balance of zero. This method will actually get you pretty close very quickly (the magic of a binary search)!

However, there is a more elegant mathematical solution involving geometric series:

A geometric series is a list of terms added together where each term is some constant multiple of the previous term. For example:   1 + 3  + 9 + 27 + 81 is a geometric series with each term being exactly 3 times the previous term. The series 4 + 2 + 1 + 1/2 + 1/4 + 1/8 is a geometric series with each term being exactly 1/2 of the previous term.

So how does this relate to the loan payments? First, some setup: let’s call the amount owed at the end of the first month, T1, the total at the end of the second month T2, and so on. Also, let p be the monthly payment (which we are trying to calculate).

Using the same interest rate as above (12% yearly), recall that total owed at the beginning of next month = total owed at the beginning of this month + this month’s interest this month’s payment, so the total at the end of the first month could be written like this: T1 = 50000 + 50000*(1%)p.

For simplicity, we can combine the first two terms and write 50000*1.01. This just represents the principal and that month’s interest combined (this is the total amount you owe). Thus, T1=50000*1.01p.

The total at end of the second month is T1 plus interest minus the monthly payment: T2= T1*1.01p.

Recalling that we already have an expression for T1, we can plug that in the expression above to calculate T2 = T1*1.01p = (50000*1.01 – p)*1.01p. Simplifying a bit gives: T2 = 50000*(1.01)2-(1.01)pp.

Now, T3=T2*1.01p = (50000*(1.01)2-(1.01)p – p) *1.01p = 50000*(1.01)3 -(1.01)2p-(1.01)pp.

To calculate T4, again take T3 and multiply by 1.01 and subtract. After simplifying, T4=50000*(1.01)4 -(1.01)3p – (1.01)2p -(1.01)pp.

Recall that 1=1.011, so the very last term becomes (1.01)0p instead of p (you’ll see why this matters in a moment), and we have T4=50000*(1.01)4 – (1.01)3p – (1.01)2p – (1.01)1p – (1.01)0p.

Let’s look a little more closely and try to break down what this tells us. The first term, 50000*(1.01)4 can be thought of as the total amount we would owe on the loan after 4 months is we didn’t pay at all—this is the just the principal (50000) plus the interest (1%) added each month for 4 months.

The other terms just tell you how much you have saved by making a payment of p each month since the loan began. The last one, (1.01)0p, which is really just p since (1.01)0=1, is just the payment made this month (which is subtracted from the total you will owe next month). The next to last term (1.01)1p is LAST month’s payment (plus an extra month’s interest that would have been paid on that amount, if you hadn’t already paid it last month).

The next term, (1.01)2p, is the payment you made TWO months ago (plus the TWO extra months of interest that would have been paid on this amount had you not paid it two months ago). Likewise, the (1.01)3p term is the payment you made THREE months ago (plus the extra THREE extra months of interests that would have been paid on this amount had you not paid it three months ago).

Ignoring, for a moment, the leftmost term, we have the following series:  -(1.01)3p – (1.01)2p – (1.01)1p – (1.01)0p. Each term is exactly 1.01 times the term to its right, so this is a geometric series! There is a term in this series for every month you make a payment. We’re assuming this is a 30 year mortgage, so there will be 360 months and 360 payments, so by the time you have paid off the loan, this series looks like: – (1.01)359p – (1.01)358p – (1.01)357p – … – (1.01)2p – (1.01)1p – (1.01)0p, where the ellipsis indicates lots more terms in the middle.

For simplicity, I’ll multiply through by -1 and call the resulting series S. Thus, S =  (1.01)359p + (1.01)358p + (1.01)357p + … + (1.01)2p + (1.01)1p + (1.01)0p.

Now to calculate T360 (the total amount of money still owed after 360 months) , we have to put the “principal” term (that we had temporarily been ignoring) back in, so T360 = 50000*1.01360 – (1.01)359p – (1.01)358p – (1.01)357p – … – (1.01)2p – (1.01)1p – (1.01)0p = 50000*1.01360 – S.

However, we already know what T360 should be! After 360 months (30 years), our loan should be paid off, so the remaining balance should be zero. Thus, T360 = $0.

Since T360 = 50000*1.01360 – S and T360=0, we now know that 0 = 50000*1.01360 – S. Adding S to both sides gives S = 50000*1.01360, evaluating the expression the right gives S = 1797482.

Now, solve for p, the monthly payment:  S= 1797482 and S = (1.01)359p + (1.01)358p + (1.01)357p + … + (1.01)2p + (1.01)1p + (1.01)0p, so we have 1797482 = (1.01)359p + (1.01)358p + (1.01)357p + … + (1.01)2p + (1.01)1p + (1.01)0p. Call that equation 1.

Luckily, this is a geometric series, so we can use a cool algebra trick to solve for p. First, create equation 2 by multiplying both sides of equation 1 by an extra factor of 1.01. So, equation 2: 1797482*1.01 = (1.01)360p + (1.01)359p + (1.01)358p + … + (1.01)3p + (1.01)2p + (1.01)1p. Note that this just has the effect of increasing the exponents on each term on the right by 1.

Now, subtract equation 1 from equation 2:

On the left side, we get 1797482*1.011797482 = 1797482*.01=17974.82.

On the right side something interesting happens. For most of the terms on the right side of equation 2, the exact same term also appears in equation 1, so when we subtract, most of the terms fall away. We are left with (1.01)360p(1.01)0p. Evaluating the exponents and simplifying gives us 34.95p. (This step exactly models the steps to derive the “formula” for summing a geometric series. In fact, that’s what we just did!)

Setting the left side equal to the right side, we have 17974.82 = 34.95p. Divide both sides by 34.95, and we get that p = $514.31. If we pay $514.31 each month, we’ll pay off the loan in exactly 30 years! We just solved our equation to calculate the monthly payment by simplifying a geometric series!

It is interesting to think about what that geometric series actually represents. It does NOT represent the payment each month, the interest each month, the balance at the end of each month, etc. It DOES represent the amount of money you have paid, plus the future interest that was never charged on that amount since it had already been paid.

In other words, S,  the sum of our geometric series, above, represents the total amount you would have paid by waiting until the very end of the loan and paying it all off as one lump sum (although your house would certainly be foreclosed long before that!). In this case, S=$1797482, which is almost 1.8 million dollars! Paying our calculated monthly payment of $514.31,  you will pay 514.31*360=$185151.60 over the life of the loan–a much better deal!

Conclusion:

Call a bank! Find out their rates and loan details and have the students calculate a monthly payment. Ask the bank to see if their numbers match!

There are also lots of free amortization calculators available online. Have you students do the calculations and then plug their principal, interest rate, and number of months into one of these websites to see if their results match the software’s results!

Possible extensions:

Students can explore the implications of changing the principal, down-payment, interest rate, number of months, etc.

What happens if the amount someone pays in a particular month is less than the interest charged that month?

What happens if the amount someone pays in a particular month is exactly equal to the amount of interest charged that month?

Students could also do some research on loans such as adjustable rate mortgages.

HT: Gautam Kaul’s Introduction to Finance course on Coursera.
P.S. I wish wordpress had a LaTex editor!
Advertisements
This entry was posted in Lesson ideas and tagged , , , , . Bookmark the permalink.

5 Responses to Lesson idea for geometric series: Loan Amortization!

  1. zacksg1 says:

    The expression you are referring to is just the expression for T4 described above (look for the lines that say “T4=”) with the very first term–the principle–removed. This was derived from each month multiplying the remaining cost by 1.01 (to add the 1% monthly interest) and then subtracting that month’s payment. This is repeated each month (for four months in the example you are talking about).

    Intuitively, you can think of this expression as first calculating the entire amount that WOULD be owed if NO payment had been made in the first four months–that is the first term that we were temporarily ignoring in that section–and then subtracting out all of the payment (plus interest). There are negative signs since this is money that is no longer owed….

    Like

  2. Stephen Denney says:

    In the paragraph that say, ” Ignoring for a moment the left most term we have the following series, -(1.01)^3 – (1.01^2…” I don’t understand where the Negative sign comes from in front of (1.01)^3.
    Please explain.

    Thank You

    Like

  3. zacksg1 says:

    Ahhh! Two days after posting this, I figured out how to do LaTex typesetting on the blog. That would have greatly simplified the typing of the math in this post! Oh, well…..there is always next time!

    Like

  4. zacksg1 says:

    http://www.teachforever.com/2012/08/math-in-real-world-should-i-take-this.html
    Here is an example of an activity where students could take an actual loan advertisement and first check it to make sure the payment actually makes sense based on the principal, APR, and time. Then, as suggested by the author of that post, the students could discuss whether or not this loan is actually a good idea!

    Like

What do you think? Anything to add? Anything you disagree with? Please add your comments below....

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s