Compound interest with Repayments

Date: 07/24/97 at 07:39:24
From: Worthington
Subject: Question about compound interest with repayments at regular 
intervals

Dear Sir,

I have a problem which is quite urgent and hard. Most people can work
out a compound interest problem, but I need a compound interest 
formula with repayments, such as in a bank situation.

Suppose a person came in asking to know

	1) How long the loan would be, (t)
	2) How much principle amount, (p)
	3) When the repayments are made, (r)
	4) Total amount of loan including interest, (a)
	5) Variable interest rate. (i)

It is however not as easy as it looks. The principle (p) will need to 
be multiplied by the daily interest rate (i), with monthly or
fortnightly or weekly repayments (r), and then interest calculated 
daily on the new amount, with a final total amount (a) to be able to 
input the interest rate.

Is this too much to ask?

Thanks.

Ian Worthington.

Date: 07/24/97 at 12:12:58
From: Doctor Rob
Subject: Re: Question about compound interest with repayments at 
regular intervals

Ian,

This is a commonly asked question. There are several variables here
which are not independent of each other. Another variable which you
have made a constant is the frequency with which interest is added
(daily in your case, but sometimes monthly, quarterly, or yearly -
rarely *continuously* [!]).  

The main information you need to work out the right formula is that 
with a given principal p, the interest which is to be added is i*p, 
so that the amount due after one day has increased from p to (1+i)*p.  
After d days, the amount due is p*(1+i)^d.

If a payment is then made of size r, the amount due is p*(1+i)^d-r.

If we repeat this process after 2*d days, the amount due then would be

   (1+i)^d*[p*(1+i)^d-r]-r = p*(1+i)^(2*d) - r*([1+i]^d + 1)

After 3*d days, the amount due would be

   p*(1+i)^(3*d) - r*([1+i]^[2*d] + [1+i]^d + 1)

and after 4*d days,

   p*(1+i)^(4*d) - r*([1+i]^[3*d] + [1+i]^[2*d] + [1+i]^d + 1)

      = p*(1+i)^(4*d) - r*([1+i]^[4*d] - 1)/([1+i]^d - 1)

From this you can see (and prove) that the pattern is that after k*d
days, the amount due is

   p*(1+i)^(k*d) - r*([1+i]^[k*d] - 1)/([1+i]^d - 1)

Now presumably t is a multiple of d days, so that after some payment,
the amount due is zero.  Set k = t/d.  Then

   0 = p*(1+i)^t - r*([1+i]^t - 1)/([1+i]^d - 1)

This gives an equation relating p, i, t, r, and d.  Then a is the 
total amount of money paid, which is r*k, or

   a = r*t/d.

The customer must give you four of the six quantities p, i, t, r, d, 
and a (but not the last four).  Then you can tell him the other two.  
Since t and d are integers, and t is an integer multiple of d, usually 
these two are among those given, not calculated.

For example, if the customer tells you he wants to pay r amount 
every d days for time t at interest rate i, you compute that
p = r*(1 - [1+i]^[-t])/([1+i]^d - 1) is the amount he can borrow, and
his total payments will be a = r*t/d.  On the other hand, if the 
customer wants to borrow p amount and pay every d days for time t at 
rate i, you compute r = p*([1+i]^d - 1)/(1 - [1+i]^[-t]) and 
a = r*t/d.  

It is somewhat more difficult if one of the variables you have to 
compute is i, but it still can be done.  Let (1+i)^d = x, and k = t/d, 
so

   0 = p*x^k - r*(x^k-1)/(x-1),
so
   0 = (p/r)*x^k - x^(k-1) - ... - x - 1

This is a polynomial equation in x whose real root bigger than 1 can
be found numerically.  Once we know x, i = x^(1/d) - 1.

I have ignored the consideration that amounts of money paid are always
in integer multiples of some unit of money: pence, pfennigs, centimes,
cents, yen, etc. This introduces rounding into these calculations, and
often the last payment made is not quite r in order to compensate for
this effect.

If this is not clear, or not what you wanted, write back and we can
try again.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/