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Paying off the Mortgage

Date: 10/08/98 at 00:53:23
From: Jody Holder 
Subject: College Algebra

This problem has to do with compound interest and mortgage rates.

You have a mortgage with required monthly payments of $370 per month 
for 30 years and an interest rate of 8% compounded monthly. If you 
instead pay $400 each month, when will you pay off the mortgage?

I really just need some type of formula then I can work out all of the 
problems I have, but I do need this one worked out also.

Please respond.

Date: 10/08/98 at 12:41:04
From: Doctor Rob
Subject: Re: College Algebra

Hi Jody,

I like to work these problems out with a table. First of all, the 
monthly interest rate is r = 8%/12 = 2/3% = .08/12 = .00666667 = 1/150. 
When interest is accrued on any amount, the new amount is the old 
amount times (1+r), 1 for the old amount and r for the interest added.

n    Principal             Interest         Payment   New principal
1       M                    r*M               P       M*(1+r) - P
2    M*(1+r)-P           r*[M*(1+r)-P]         P       M*(1+r)^2 -
                                                         P*(1+r) - P
3 M*(1+r)^2-P*(1+r)-P  r*(M*(1+r)^2-P*(1+r)-P) P       M*(1+r)^3 -             
                                                         P*(1+r)^2 - 
                                                         P*(1+r) - P

Perhaps you can already see the pattern: The principal due at the 
beginning of month n-1 is:

   M*(1+r)^n - P*[1 + (1+r) + (1+r)^2 + ... + (1+r)^(n-1)]

Now the part in brackets [] is a geometric series, and the sum can be
easily computed to be [(1+r)^n-1]/[(1+r)-1], so the principal due at 
the beginning of month n is:

   M*(1+r)^n - P*[(1+r)^n-1]/r

If at the beginning of month n the principal due is 0, then:

   M = P*[(1+r)^n-1]/[r*(1+r)^n]
     = P*[1-(1+r)^(-n)]/r

In your case, P = $370, r = 1/150, and n = 360. I computed M to be
$50,424.89. (Probably the correct value is $50,000.00, in which case 
the monthly payments P should be $366.88.) Using this figure for M, 
changing P to $400, and letting n be the variable, we get the 

             50424.89 = 400*[1-(1+1/150)^(-n)]/(1/150)
   50424.89/(400*150) = 1-(1+1/150)^(-n)
       (1+1/150)^(-n) = 1 - 50424.89/60000
                      = 0.1595851
  log[(1+1/150)^(-n)] = log(0.1595851)
       n*log(1+1/150) = -log(0.1595851)
                    n = -log(0.1595851)/[log(151)-log(150)]
                    n = 276.1932

so the loan will be paid off in 277 months, or 23 years, 1 month, and 
the last payment will be only $76.98. You save 84 monthly payments of 
$370 by making 276 extra payments of $30, and one payment of $76.98, 
for a net savings of $22,723.02.

These figures may be off by a few cents because of rounding. If the 
interest accrued is rounded to the nearest cent, or rounded down or up 
to a whole number of cents, the amounts will change slightly. My 
calculations were done using fractional cents with no rounding at all.

- Doctor Rob, The Math Forum
Associated Topics:
High School Interest

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