Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Extra Loan Payments, Amortization Tables


Date: 03/31/98 at 08:03:39
From: Hunter Haynes
Subject: accounting/finance

A 30-year home mortgage is taken out for $161,800 at an interest rate 
of 7.5%. The first payment on the loan is made in December, in the 
amount of $1,232.65. If one extra payment is made each year, for the 
same amount as a regular payment, how much money can be saved over the 
life of the 30-year mortgage and how much sooner can it be paid off?

This is a bonus question for our class. I don't know how to start 
since there are so many figures involved. If I could get the answers, 
maybe I could figure out how to work it--knowing what answers to work 
towards. Thanks!


Date: 03/31/98 at 15:19:26
From: Doctor Rob
Subject: Re: accounting/finance

Let P be the amount of the loan, or "principal." Let the monthly 
interest rate be i, and the number of months of the loan be n. In your 
case, P = $161,800, i = 7.5%/12 = .075/12 = .00625 (annual rate 
divided by 12 to get monthly rate), and n = 30*12 = 360. Let the 
balance due after m months be B(m). Then B(0) = P. The interest 
accrued during month m is B(m-1)*i, so the new balance due will be:

     B(m-1) + B(m-1)*i = B(m-1)*(1+i)

You want to make the same monthly payment R every month. Then at the 
end of the first month, you pay R, so that the new amount due at the 
end of month m is:

     B(m) = B(m-1)*(1+i) - R

You can use this formula over and over to compute B(1), B(2), B(3), 
....

There is a closed-form formula for B(m), however. Let a = 1 + i.  
Then the balance due after month m is given by the formula:

     B(m) = P*a^m - R*(a^m-1)/i

You can prove this formula by mathematical induction. Now to compute 
the correct amount for R, we want the balance after n months to have 
been reduced to zero. Thus B(n) = 0, so:

               0 = P*a^n - R*(a^n-1)/i
     R*(a^n-1)/i = P*a^n
       R*(a^n-1) = P*a^n*i
               R = P*a^n*i/(a^n-1)
               
Recall that in your case, i = .00625, a = 1.00625, P = $161,800, and
n = 360. According to this formula, R = $1,131.33 should be the 
monthly payment. (You'll probably need to use a calculator to figure 
out what a^n = 1.00625^360 is. I got 9.4215339.)  $1,232.65 would be 
the monthly payment on $176,291 at 7.5%, or on $161,800 at 8.4%.

There is a table, called an Amortization Table, that describes the 
payment of such a loan. In each line, we use the formula
B(m) = B(m-1)*(1+i) - R. At the beginning, you owe $161,800. 
After one month, you owe an additional .00625 of $161,800, or 
$1011.25, in interest accrued, so your debt is now $162,811.25. 
You pay $1,232.65, leaving a new balance of $161,578.60. This is 
the way the calculations go.

This formula does not quite apply here, because of the extra amount 
paid after the first month. Call the extra payment:

     S = $1232.65 - $1,131.33 = $101.32

Then:

     B(m) = P*a^m - S*a^(m-1) - R*[a^m-1]/i

Again you can prove this by mathematical induction.

Of course, we have to round interest and balances to the nearest cent, 
so some inaccuracies will creep into the calculations, amounting to a 
few cents at the very end, but this is a minor adjustment.

Here is the Amortization Table for the loan without the extra yearly
payments:

  m      B(m-1)      Interest     Payment         B(m)
  1   $161,800.00   $1,011.25    $1,232.65    $161,578.60
  2    161,578.60    1,009.87     1,131.33     161,457.14
  3    161,457.14    1,009.11     1,131.33     161,334.92
  4    161,334.92    1,008.34     1,131.33     161,211.93
  5    161,211.93    1,007.57     1,131.33     161,088.17
  6    161,088.17    1,006.80     1,131.33     160,963.64
  7    160,963.64    1,006.02     1,131.33     160,838.33
  8    160,838.33    1,005.24     1,131.33     160,712.24
  9    160,712.24    1,004.45     1,131.33     160,585.36
 10    160,585.36    1,003.66     1,131.33     160,457.69
 11    160,457.69    1,002.86     1,131.33     160,329.22
 12    160,329.22    1,002.06     1,131.33     160,199.95
 13    160,199.95    1,001.25     1,131.33     160,069.87
 ..        ...         ...          ...            ...
358      2,419.44       15.12     1,131.33       1,303.23
359      1,303.23        8.15     1,131.33         180.05
360        180.05        1.13       181.18            .00

Total payments are $101.32 + 359*$1,131.33 + $181.18 = $406,429.97

By paying $101.32 extra that first December, you save paying $950.15 
on the last payment.

If, in months 12, 24, 36, ..., (once a year), you pay an extra payment 
of T = $1,131.33, the formulas look a bit different. Now if we write
m = 12*s + t, with 0 <= t < 12, then:

   B = P*a^m - S*a^(m-1) - R*(a^m-1)/i - T*a^t*(a^[12*s]-1)/(a^12-1)

The Amortization Table for the loan with the extra yearly payments 
will look like this:

Month   Balance      Interest     Payment     New Balance
  1   $161,800.00   $1,011.25    $1,232.65    $161,578.60
  2    161,578.60    1,009.87     1,131.33     161,457.14
  3    161,457.14    1,009.11     1,131.33     161,334.92
  4    161,334.92    1,008.34     1,131.33     161,211.93
  5    161,211.93    1,007.57     1,131.33     161,088.17
  6    161,088.17    1,006.80     1,131.33     160,963.64
  7    160,963.64    1,006.02     1,131.33     160,838.33
  8    160,838.33    1,005.24     1,131.33     160,712.24
  9    160,712.24    1,004.45     1,131.33     160,585.36
 10    160,585.36    1,003.66     1,131.33     160,457.69
 11    160,457.69    1,002.86     1,131.33     160,329.22
 12    160,329.22    1,002.06     2,262.66     159,068.62
 13    159,068.62      994.18     1,131.33     158,931.47
 ..        ...         ...          ...            ...
275      9,695.51       60.60     1,131.33       8,624.78
276      8,624.78       53.90     2,262.66       6,416.02
277      6,416.02       40.10     1,131.33       5,324.79
278      5,324.79       33.28     1,131.33       4,226.74
279      4,226.74       26.42     1,131.33       3,121.83
280      3,121.83       19.51     1,131.33       2,010.01
281      2,010.01       12.56     1,131.33         891.24
282        891.24        5.57       896.81            .00

Total payments are:

   $101.32 + 258*$1,131.33 + 23*$2,262.66 + $896.81 = $344,922.45

Savings in interest:  $406,429.97 - $344,922.45 = $61,507.52. Loan 
paid off 360 - 282 = 78 months or 6.5 years early.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Interest

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/