Paying off the MortgageDate: 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 equation: 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 http://mathforum.org/dr.math/ |
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