Example 1

A traditional individual retirement account (IRA) is a special type of retirement account in which the money
you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA
earning 6% interest, how much will you have in the account after 20 years?

Solution

In this example,

$\begin{array}{lll}\mathrm{pymt}\hfill & =\$100\hfill & \text{the monthly deposit}\hfill \\ r\hfill & =0.06\hfill & \text{6\% annual rate}\hfill \\ n\hfill & =12\hfill & \text{since we’re doing monthly deposits, we’ll compound monthly}\hfill \\ t\hfill & =20\hfill & \text{we want the amount after 20 years}\hfill \end{array}$

Putting this into the annuity equation we can find the future value:

$$\begin{array}{l}\mathrm{FV}=\frac{100\left({\left(1+\frac{0.06}{12}\right)}^{20(12)}-1\right)}{\left(\frac{0.06}{12}\right)}\\ \mathrm{FV}=\frac{100\left({\left(1.005\right)}^{240}-1\right)}{\left(0.005\right)}\\ \mathrm{FV}=\frac{100\left(3.310-1\right)}{\left(0.005\right)}\\ \mathrm{FV}=\frac{100\left(2.310\right)}{\left(0.005\right)}=\$46200\end{array}$$

The account will grow to $46,200 after 20 years. 

Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months).  The
difference between what you end up with and how much you put in is the interest earned.  In this
case it is
$\$46,200-\$24,000=\$22,200$
.

A word on rounding. The work above rounded the different steps while doing the computation. It is always
better to do the entire computation at once. If you had done that the total FV would have been closer
to $46,204.09.

Example 2

You want to have $200,000 in your account when you retire in 30 years.  Your retirement account earns 8%
interest compounded monthly.  How much do you need to deposit each month to meet your retirement goal?

Solution

In this example, we’re looking for the monthly deposits into the account or pymt.

$\begin{array}{lll}r\hfill & =0.08\hfill & \text{8\% annual rate}\hfill \\ n\hfill & =12\hfill & \text{since we’re depositing monthly}\hfill \\ t\hfill & =30\hfill & \text{30 years}\hfill \\ \mathrm{FV}\hfill & =\$200,000\hfill & \text{the amount we want to have in 30 years}\hfill \end{array}$

In this case, we’re going to have to set up the equation, and solve for pymt.

$$\begin{array}{l}FV=\frac{pymt\left({\left(1+\frac{r}{n}\right)}^{\left(n\cdot t\right)}-1\right)}{\left(\frac{r}{n}\right)}\\ 200,000=\frac{pymt\left({\left(1+\frac{0.08}{12}\right)}^{\left(12\cdot 30\right)}-1\right)}{\left(\frac{0.08}{12}\right)}\\ 200,000\left(\frac{0.08}{12}\right)=pymt\left({\left(1+\frac{0.08}{12}\right)}^{\left(12\cdot 30\right)}-1\right)\\ pymt=\frac{200,000\left(\frac{0.08}{12}\right)}{\left({\left(1+\frac{0.08}{12}\right)}^{\left(12\cdot 30\right)}-1\right)}\\ pymt\approx \$134.20\end{array}$$

So you would need to deposit $134.20 each month to have $200,000 in 30 years if your account earns 8%
interest

Example 3

After retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement
account.  The account earns 6% interest.  How much will you need in your account when you retire?

Solution

In this example,

$\begin{array}{lll}\mathrm{pymt}\hfill & =\$1000\hfill & \text{the monthly withdrawal}\hfill \\ r\hfill & =0.06\hfill & \text{6\% annual rate}\hfill \\ n\hfill & =12\hfill & \text{since we’re doing monthly withdrawals, we’ll compound monthly}\hfill \\ t\hfill & =20\hfill & \text{since were taking withdrawals for 20 years}\hfill \end{array}$

We’re looking for P; how much money needs to be in the account at the beginning. Putting this
into the equation:

$\begin{array}{l}P=\frac{1000\left(1-{\left(1+\frac{0.06}{12}\right)}^{-20(12)}\right)}{\left(\frac{0.06}{12}\right)}\\ P=\frac{1000\times \left(1-{\left(1.005\right)}^{-240}\right)}{\left(0.005\right)}\\ P=\frac{1000\times \left(1-0.302\right)}{\left(0.005\right)}=\$139,600\end{array}$

You will need to have $139,600 in your account when you retire. 

Notice that you withdrew a total of $240,000 ($1000 a month for 240 months).  The difference between
what you pulled out and what you started with is the interest earned during that time from when you
started taking withdrawal till the last withdrawal. In this case it is
$\$240,000-\$139,600=\$100,400$
in interest.

A word on rounding. The work above rounded the different steps while doing the computation. It is always
better to do the entire computation at once. If you had done that the value for P would have been
closer to $139,580.77 (about a $1000 difference).

In this example, we’re looking for the amount you can withdraw each month, pymt.

$\begin{array}{lll}P\hfill & =\$500,000\hfill & \text{we are beginning with \$500,000}\hfill \\ r\hfill & =0.08\hfill & \text{8\% annual rate}\hfill \\ n\hfill & =12\hfill & \text{since we’re doing monthly withdrawals}\hfill \\ t\hfill & =30\hfill & \text{taking withdrawals for 30 years}\hfill \end{array}$

In this case, we’re going to have to set up the equation, and solve for d.

$$\begin{array}{l}P=\frac{pymt\left(1-{\left(1+\frac{r}{n}\right)}^{\left(-n\cdot t\right)}\right)}{\left(\frac{r}{n}\right)}\\ 500,000=\frac{pymt\left(1-{\left(1+\frac{0.08}{12}\right)}^{\left(-12\cdot 30\right)}\right)}{\left(\frac{0.08}{12}\right)}\\ 500,000\left(\frac{0.08}{12}\right)=pymt\left(1-{\left(1+\frac{0.08}{12}\right)}^{\left(-12\cdot 30\right)}\right)\\ pymt=\frac{500,000\left(\frac{0.08}{12}\right)}{\left(1-{\left(1+\frac{0.08}{12}\right)}^{\left(-12\cdot 30\right)}\right)}\\ pymt\approx \$3,368.82\end{array}$$

You would be able to withdraw $3,368.82 each month for 30 years. 

If we had ignored the interest being earned the $500,000 payments over 30 years would be
$\frac{\$500,000}{12\cdot 30}=\$1,388.89$
, which is less than half of the payments actually received. The ability for the savings to continue to earn
interest (grow) allows that $500,000 to be stretched into a much higher payment. The total of all the payments
would actually give you
$\$3,368.82\cdot 12\cdot 30=\$1,212,775.20$
over the 30 years.