Extra Loan Payments, Amortization TablesDate: 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/ |
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