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May 26th, 2013, 10:25 PM   #1
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Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 10,909
Thanks: 716

% 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
Above is a $1300 loan at interest of 1% per month,
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) = ?
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June 5th, 2013, 10:45 AM   #2
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 10,909
Thanks: 716

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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