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Formula for Mortgage

Date: 7/4/96 at 18:59:48
From: Anonymous
Subject: Mortgage Loan Formula

I am unable to find a formula for loans. The only formulas I have are 
for interest. What I need is a formula (like A=P(1+r/m)^mt).
Any help would be appeciated.

Date: 7/5/96 at 11:34:52
From: Doctor Anthony
Subject: Re: Mortgage Loan Formula

We will let L = loan, n = number of months for repayment, starting at 
end of first month, r = percentage interest rate per year, (take r/12 
as monthly rate). P = amount of repayment per month (starting at end 
of first month).

If we consider the loan first, this would increase by a factor 
(1 + r/1200) per month, so after n months the value of the loan would 
have increased to L(1 + r/1200)^n

Now consider the repayments.  These are $P per month, but the value of 
the earlier repayments also increases at a compound rate (1 + r/1200).  
Thus after the second repayment, the value of the repayments is:

P + P(1 + r/1200)  and after three months it would be:
P + P(1 + r/1200) + P(1 + r/1200)^2   and after n months it would be:
P{1 + (1+r/1200) + (1+r/1200)^2 + ..... + (1+r/1200)^(n-1)}

This is a geometric series with n terms, first term = 1 and common 
ratio (1+r/1200), so the sum of n terms is given by 

   P{(1+r/1200)^n - 1}/{(1+r/1200)-1)}
 = P{(1+r/1200)^n - 1}/(r/1200)
 = 1200P{(1+r/1200)^n - 1}/r

We must now equate the total repayments to total value of the loan, 
and this gives:

  1200P{(1+r/1200)^n - 1}/r  =  L(1+r/1200)^n

                     P = Lr(1+r/1200)^n/[1200{(1+r/1200)^n - 1}]

Example.  Find the monthly repayments on a loan of $20,000 over 15 
years at 12 percent per year compound interest.

Here we have n = 12*15 = 180 months, r = 12, and L = 20000.
We want to find P.

1+r/1200 = 1 + 12/1200 = 1.01   and the above formula becomes

       P = {20000*12*1.01^180}/{1200*(1.01^180 - 1)}

         = {20000*12*5.99}/{1200*(5.99 - 1)}

         = 1437600/5988
         = $240.08

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

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