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### Compound Interest

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Date: 07/30/97 at 15:17:12
From: michael echols
Subject: Compound Interest

I would like to know how to calculate interest on a car. For example:
I want to buy a car that costs \$20,058.00.  The charges added on
are \$515.00 (destination charge) and \$1,193.23 (tax), and I place a
down payment of \$2000.00 on the car.  I am then financing \$19,766.23.
For 84 months, they are saying that my interest rate will be 7.6
percent. My total interest according to them is then \$5,782.90.

I would like to know and understand how they calculated that number.
Is there a formula?  (I=PRT)?

Thanks,
Corey Echols
```

```
Date: 08/01/97 at 15:04:29
From: Doctor Rob
Subject: Re: Compound Interest

There is a formula, but it is more complicated than that.

We will assume that the interest is compounded monthly.
Then if the annual percentage rate (APR) is 7.6 percent,
the monthly interest rate is 7.6/12 = 0.633333 percent,
if you figure it one way, and (1.076)^(1/12) - 1 = 0.612287
percent, if you figure it a second way.

Since we have two different rates, let the rate be r.
We will do the calculation with rate r, then substitute
these two rates at the end to see which way they computed.

The amount of interest accrued on a principal of P for one
month at interest rate r is P*r. Thus the total amount
owed at the end of one month is (1+r)*P. At that point,
you make a payment of R, which reduces the amount owed to
(1+r)*P - R.  This is the new amount that will accrue
interest in the second month.

The total amount owed after two months and two payments
is then

(1+r)^2*P - (1+r)*R - R

After three months, it is

(1+r)^3*P - (1+r)^2*R - (1+r)*R - R

and after four months,

(1+r)^4*P - (1+r)^3*R - (1+r)^2*R - (1+r)*R - R

I think you can figure out the formula after m months:

(1+r)^m*P - R*[(1+r)^(m-1) + ... + (1+r) + 1]

The part in [] can be rewritten, since it is the sum of
a geometric progression, and the result is

(1+r)^m*P - R*[(1+r)^m - 1]/[(1+r) - 1]
or
(1+r)^m*P - R*[(1+r)^m - 1]/r
or
(1+r)^m*(P - R*[1 - (1+r)^(-m)]/r)

Now at the end of 84 months, this balance should be zero,
so you get the equation

P = R*[1 - (1+r)^(-84)]/r

You have paid 84*R, of which P is principal and the rest
interest, so the total amount of interest you paid is I = 84*R - P.

Now we substitute in your numbers:  P = 19776.23, I = 5782.90.
This tells me (although you neglected to) that your monthly
payment was \$304.275357, or \$304.28.  Thus

19776.23 = 304.275357*[1 - (1+r)^(-84)]/r
so
64.9945174 = [1 - (1+r)^(-84)]/r

If r = .00633333, the value of the righthand side is 64.987069,
and if it is, then r = 65.517511, so it looks as if the first
method was the one they used to figure r from your APR. The actual
solution to this last equation is r = 0.00633036133, and
12*r = 0.075964336 or 7.5964336 percent, close enough to 7.6 percent.

Apparently they picked 7.6 percent, divided by 12, and rounded to
r = 0.00633. Then they used the above equation to find your monthly
payment, \$304.271115, rounded up to \$304.28.  Using this, they then
computed the total interest, \$5783.29. Since you were paying slightly
more than the required monthly payment, the last payment did not have
to be quite as large as the rest to pay off the car. In fact it should
be \$0.58 less, or \$303.70.  Thus the total interest was \$0.58 cents
less, or \$5782.71. This is off by \$0.19 from the figure you gave.

If they rounded down to \$304.27 for the monthly payment, the total
interest would be \$0.84 more before the correction, or \$5784.13, but
the last payment would have to be larger to compensate for the fact
that you were paying slightly less than the required monthly payment,
in fact \$0.07 larger. Then the total interest would come out to
\$5784.20. This is off by \$0.70 from the figure you gave.

These differences of a few cents are probably accounted for by
rounding or truncating the amounts of money in this calculation at
intermediate steps. At least you see roughly how they get the numbers
they do.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Interest

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