next up previous
Next: About this document ... Up: Basic Mathematics of Finance Previous: Present Values of Annuities

Future Values of Annuities

Another common situation is a sequence of payments made to accumulate a sum in the future. Reflecting the human tendancy to good intentions, we assume that the payments of size R start at the end of a payment period and then continue until right after the $n^{\mbox{th}}$ payment is made at which time we cash in the account and ask for the balance. A typical example might be the good intentions to save regularly for a car or to have a lump sum to use for retirement.

Here the picture looks like this:


\begin{picture}(380,180)(-20,-138)\usebox{\timeline}\usebox{\timelabels}\put(0,-......}\put(0,-10){\usebox{\nmforward} }\put(0,-10){\usebox{\nforward} }\end{picture}

So the sum of the future values of the payments is

\begin{displaymath}R+R(1+i)+R(1+i)^2+R(1+i)^3+ \cdots + R(1+i)^{n-1}\end{displaymath}



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

\begin{displaymath}S = R\frac{1- (1+i)^n}{1-(1+i)}= R \frac{1-(1+i)^{n}}{-i}= R \frac{(1+i)^n-1}{i}\end{displaymath}



Example:

Let us consider the future value of $1000 paid at the end of each month into an account paying 8% annual interest for 30 years. How much will accumulate? This is a future value calculation with R=1000, n=360, and $i=\frac{.08}{12}$. This account will accumulate
\begin{displaymath}\$1000 \frac{(1+\frac{.08}{12})^{360}-1}{\frac{.08}{12}} \approx \$1,490,359.45\end{displaymath}


Note that this is much larger than the sum of the payments, since many of those payments are earning interest for many years. $\diamondsuit$

We can also figure out what payment would need to be made each month to achieve a financial goal.

Example:

How much do you have to resolve to save at the end of each month in order to accumulate $24000 in 4 years if you can get 6% annual interest compounded monthly? Here we know S=24000, n=48 and $i=\frac{.06}{12}$, and we solve for R:
\begin{eqnarray*}\$24000 &=& R \frac{(1.005)^{48}-1}{.005}\\R &=& \$24000 \frac{.005}{(1.005)^{48}-1} \\&=& \$443.64 \end{eqnarray*}


Here the total of the payments is $\$443.64 \cdot 48 = \$21294.72$ the rest comes from the interest earned. $\diamondsuit$

next up previous
Next: About this document ... Up: Basic Mathematics of Finance Previous: Present Values of Annuities

Larry Stout
2000-10-09