May 26th, 2013, 09:25 PM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,434 Thanks: 570  % payment loan
Hope you have fun with this Code: MONTH PAYMENT INTEREST BALANCE 0 1,300.00 1 130.00 13.00 1,183.00 2 118.30 11.83 1,076.53 3 107.653 10.7653 979.6423 ======= 35.5953 and monthly payments of 10% of balance, with minimum payment = $100. Since 4th payment using 10% of balance = ~$97.96, then this payment will not be made, but instead the minimum payment will take effect. So, at end of month preceeding month of 1st minimum payment, interest of ~$35.60 has been paid. So in this example, we have: a = amount of loan (1300) m = minimum payment (100) j = percentage of balance (.10) i = interest per month (.01) k = interest paid up to month of 1st min. pay't (?) As a general case formula, k (in terms of a,m,j,i) = ? 
June 5th, 2013, 09:45 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,434 Thanks: 570  Re: % payment loan
Looks like this wasn't too popular!! u = m / (aj) v = 1  j + i n = CEILING[LOG(u) / LOG(v)] f = av^n k = i(a  f) / (j  i) ........................................... u = 100 / (1300*.1) = .76923.... v = 1  .10 + .01 = .91 n = CEILING[LOG(.76923...) / LOG(.91)] = 3 f = 1300 * .91^3 = 979.6423 k = .01(1300  979.6423) / (.10  .01) = 35.5953 

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