Long First Month Mortgage Payment Calculation
Date: 06/22/2001 at 00:15:28 From: Genevieve Chan Subject: Long 1st month mortgage payment calculation Dr. Math, All the mortgage payment formulae seem to deal with level pay whole monthly payments, given interest, loan principal, and length of the loan. But in the real world, if one applies for a loan, the first payment generally is not exactly one month from the time the loan is taken. For example, imagine a loan is taken out on Jan. 15, 2001. The first payment is due Jan. 25, 2001. Thereafter payment is due on the 25th of each month. The principal is 120,000, interest is 8%. The length of the loan is 15 years (180 payments). Can this problem be solved with one formula? If not how would you calculate the monthly payment amount, given that the payments have to be a uniform amount for the entire loan life? That is, the monthly payment amount is the same from payment 1 to 180. Thank you in advance. Regards.
Date: 06/22/2001 at 06:55:07 From: Doctor Mitteldorf Subject: Re: Long 1st month mortgage payment calculation Dear Genevieve, In the real world, most banks and lenders duck this mathematical problem by treating the first month differently: the borrower gets his full loan, but then immediately pays back the partial month's interest in advance. This is called "prepaid interest." The remaining loan is an even number of payment periods, and is calculated in the standard way. The bank actually reaps a small benefit from this practice, because for the partial month the payment is in advance, whereas any other way of treating the partial month would follow the standard practice of charging interest in arrears. If you want to do the calculation with level payments including the partial months, it is not any more difficult than the calculation for an even number of months. The trick is to think in terms of present values, and to use the principle that => the present value of all payments as of the loan date is equal to the amount of the loan. Let's take your example. The nominal interest rate is 8%, so the monthly interest is 8/12 of 1% or 0.006667. So each month, the present value of your payment is discounted by another factor of 1/(1+0.006667) = 0.993377. For the partial month, there are varying, slightly different ways to compute the discounting factor; for now, I'll use the simplest and calculate it as 10/30 of a full month's interest, so 8/36 of 1% is 0.002222, and 1/(1+0.002222) = 0.997782. Now let's call the unknown payment amount p. The first payment is made after 10/30 of a month, so its present value is discounted by the factor 0.99777. The second payment is discounted by 0.99777*0.993377, and the third by 0.99778*0.993377*0.993377, the fourth by 0.99777*0.993377*0.993377*0.993377, etc. So the sum of the present values of the payments is p*0.99778*(1 + 0.993377 + 0.993377^2 + 0.993377^3 + ... + 0.993377^179) We can find the sum of the series in parentheses using the same standard technique we always use. Call the sum S. Multiply S times 0.99333, and notice that the product is almost the same, just missing the first term and with an extra term tacked on at the end: S = 1 + 0.993377 + 0.993377^2 + 0.993377^3 + ... + 0.993377^179 0.993378*S = 0.993377 + 0.993377^2 + 0.993377^3 + ... + 0.993377^180 Therefore 0.993378*S = S - 1 + 0.993378^180. Solve this for S to find S = (1-0.993378^180) / 0.0066225 = 105.338 Going back to our original equation with p, we have p * 0.997782 * 105.338 = $120,000 and this gives p = $1141.72 The upshot is that we do the same calculation as for a standard 180-payment loan, but that the payment amount is reduced by a factor of 0.993378/0.997782. The numerator of this factor is the discounting factor you would have used for a full month; the denominator is the discounting factor you actually use for the partial month. Read more about loan calculations at our FAQ page, Loans and Interest http://mathforum.org/dr.math/faq/faq.interest.html and in the High School Interest area of our Dr. Math archives: http://mathforum.org/dr.math/tocs/interest.high.html - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum