# Math Help - min repayment

1. ## min repayment

hello,

i am struggling a wee bit trying to find a formula that is behind this calculator

Minimum Repayments: DANGER! Don't get locked into debt....

it can be found by scrolling half way down.

the problem for me is the minimum for choosing say 3% or £10 in payments.

any help would be really appreciated

thanks

2. Don't understand what you're asking.

That calculator works fine.
I entered 1000 balance, interest of 1% per month,
payments of 2% of outstanding balance with 5 minimum, and got:
17 years 4 months, total cost of 848; which is exactly correct.

Opening balance owing: 1000.00
1st month: -20.00 +10.00 990.00
2nd month: -19.80 +9.90 980.10
and so on....

3. Hi Wilmer,

I am wondering if there is just one formula that could summarise this calculator without having to do the iterative method. I am creating a calculator similar to this for a website.

Thanks alot

4. Take this example:
a = amount owing ; 1000
i = interest monthly ; .01
p = percentage payment ; .03
m = mimimum payment ; 15

Can be treated as a saving account receiving negative interest (thus being depleted)
at rate of (p - i)%, or $1000 at rate -.02 per month in above example. 1000 - 20 = 980; 980 - 19.60 = 960.40; and so on... Required is calculation of the month where the balance owing is such that a switch to the minimum payment is required. In other words, ap(1 + i - p)^n = m ; solving for n, and converting to integer: n = INT{LOG[m / (ap)] / LOG(1 + i - p)} + 1 For above example, you'll get n = 35. Actually, at month 35, ~493.07 is owing. 493.07 * .03 = 14.79 ; hence the switch to minimum payment of 15. Let k = 1 - i + p (less typing!) The amount owing of 493.07 obtained from: ak^n Now required is the number of minimum payments of$15 that'll pay off this 493.07.
Obtained from formula m = [a i k^n] / [1 - 1 / (1 + i)^x]
The x is the number of months left; solving for x:
x = LOG[m / (m - a i k^n)] / LOG[1 + i]
Here you'll get x = 40.055....; means a partial 41st payment.
Actual balance owing after 40th minimum payment is .83 only.

So you'd round out the x to: INT(x) + 1

Now we have the number of months for the whole thing: n + x = 35 + 41 = 76

As far as the interest cost goes, I'm not sure that it can be calculated
by formula, or as a derivative of the above formulas.
I don't think it can...so iterative process required.
But I see nothing wrong with having such a process in a calculator:
most respectable calculators do!

Don't despair: perhaps someone else will come up with a direct way...

5. Originally Posted by Wilmer
a = amount owing ; 1000
i = interest monthly ; .01
p = percentage payment ; .03
m = mimimum payment ; 15

Required is calculation of the month where the balance owing is such
that a switch to the minimum payment is required.
In other words, ap(1 + i - p)^n = m ; solving for n, and converting to integer:
n = INT{LOG[m / (ap)] / LOG(1 + i - p)} + 1

For above example, you'll get n = 35.
Actually, at month 35, ~493.07 is owing.
493.07 * .03 = 14.79 ; hence the switch to minimum payment of 15.
Sorry. I have tried a few times to get 35 from that formula you gave but i am getting other values

Originally Posted by Wilmer
Now required is the number of minimum payments of \$15 that'll pay off this 493.07.
Obtained from formula m = [a i k^n] / [1 - 1 / (1 + i)^x]
The x is the number of months left; solving for x:
x = LOG[m / (m - a i k^n)] / LOG[1 + i]
Here you'll get x = 40.055....; means a partial 41st payment.
Actual balance owing after 40th minimum payment is .83 only.

So you'd round out the x to: INT(x) + 1
and again I am not getting 40.055.... for this one?

can you show me how you gain those values?

Thanks alot

6. Dunno what to tell you, dada.
Are you using a calculator?
How would I know why you're not getting those answers?
You're obviously coding something improperly.

n = INT{LOG[m / (ap)] / LOG(1 + i - p)} + 1

=INT{[LOG(15 / 30)] / LOG(.98)} + 1
= 35