Formula for MortgageDate: 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. Thanks, Patrick 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 Check out our web site! http://mathforum.org/dr.math/ |
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