Present Values of Annuities

An annuity is a regular sequence of payments. We usually assume that the compounding period and the payment period coincide, so monthly payments go with monthly compounding. We will consider an annuity due: regular payments of size R are made at the end of each period for n periods. We assume an interest rate of i per period. We then ask for the value of these payments one period before the first payment is to be made. This is the typical situation in finding the relationship between the amount of money loaned and the size of the payments.

Here the picture looks like this:

So the sum of the present values of the payments is

$$\frac{R}{1+i}+\frac{R}{(1+i)^2}+\frac{R}{(1+i)^3}+ \cdots + \frac{R}{(1+i)^n}$$

a geometric series with n terms, common ratio $${\frac{1}{1+i}}$$, and first term $${\frac{R}{1+i}}$$. The formula for the sum of a geometric series gives

$$A = \left(\frac{R}{1+i}\right)\frac{1- \left(\frac{1}{1+i}\right)^n}{1-\frac{1}{1+i}}= R \frac{1-(1+i)^{-n}}{i}$$

Example:

A typical calculation asks what size loan you could get if you can afford to pay $1000 per month for 30 years at 8% annual interest: here $R=1000, i=\frac{.08}{12}, \mbox{ and }n=12\cdot 30=360$. Thus the amount you could borrow is

$$\$1000\frac{1-\left(1+\frac{.08}{12}\right)^{-360}}{\frac{.08}{12}}\approx 136283.49$$

Under these terms you would end up paying a total of $360,000, so the total interest paid would be $\$360,000-\$136,283.49=\$223,761.51.$ $\diamondsuit$

Often a payment is considered to consist of a portion which pays the interest on the outstanding balance of the loan for one period and a portion which reduces the balance due. Such calculations are needed for finding home interest deductions on taxes, for instance.

Example:

Payments of $1000 per month are being made on a 30 year mortgage at 8% annual interest (as in the previous example). What is the balance due right after the payment 40 has been made? Since there were 360 payments in the original loan and 40 of them have been made there are 320 left. So we should figure out what the loan would be which would result in 320 monthly payments of $1000 using 8% annual interest:

$$\$1000\frac{1-\left(1+\frac{.08}{12}\right)^{-320}}{\frac{.08}{12}}\approx \$132107.50$$

The interest portion of the payment 41 is then one month's interest on that outstanding balance, or

$$\$132,107.50 \cdot \frac{.08}{12}=\$880.72$$

The rest of the payment $\$1000-\$880.72=\$119.28$ goes to reduce the principle.

If we are interested in the amount of interest paid in the year from right after payment 40 to right after payment 52, we subtract the outstanding balance at the end of the year from the outstanding balance at the beginning:

$$\$132107.50-\$1000\frac{1-\left(1+\frac{.08}{12}\right)^{-308}}{\frac{.08}{12}}\approx \$1485.07$$

$\diamondsuit$

We can also figure out what the payments will be for a loan of a particular size:

Example:

If you decide to borrow $150000 under these same terms, then the payment will satisfy

$$R\frac{1-\left(1+\frac{.08}{12}\right)^{-360}}{\frac{.08}{12}}=150000$$

so that

$$R=\frac{150000}{\frac{1-\left(1+\frac{.08}{12}\right)^{-360}}{\frac{.08}{12}}}\approx 1100.65$$

$\diamondsuit$

Solving for either the interest rate or the number of payments is somewhat more difficult.

Example:

If you borrow $1000 at 8% compounded monthly and agree to pay it off with $10 per month payments, how long will it take to pay off the debt? Note that one month's interest on $1000 is $6.67, so this could easily take close to 200 months. To find an exact answer we solve

$$1000 = 10 \frac{1-\left(1+\frac{.08}{12}\right)^n}{\frac{.08}{12}}$$

for n:

$$\begin{eqnarray*}1000 & = & 10 \frac{1-\left(1+\frac{.08}{12}\right)^{-n}}{\frac... ...1-\frac{8}{12})}{\log\left(1+\frac{.08}{12}\right)}\\ &=& 165.34\end{eqnarray*}$$

Under these circumstances the final payment (the 166 $^{\mbox{st}}$) would be recalculated before it was made, since it will be less than the regular $10. $\diamondsuit$