8.3: Annuities

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Learning Objectives

Understand what a financial annuity is.
Be able to calculate the future value or principle value of an annuity.

For most of us, we aren’t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a savings annuity. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.

An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship

$P_{m}=\left(1+\frac{r}{k}\right) P_{m-1}$

For a savings annuity, we simply need to add a deposit, $d$, to the account with each compounding period:

$P_{m}=\left(1+\frac{r}{k}\right) P_{m-1}+d$

Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general.

Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. In this example:

$r = 0.06$ (6%)

$k = 12$ (12 compounds/deposits per year)

$d = \$100$ (our deposit per month)

Writing out the recursive equation gives

$P_{m}=\left(1+\frac{0.06}{12}\right) P_{m-1}+100=(1.005) P_{m-1}+100$

Assuming we start with an empty account, we can begin using this relationship:

$P_{0}=0$

$P_{1}=(1.005) P_{0}+100=100$

$P_{2}=(1.005) P_{1}+100=(1.005)(100)+100=100(1.005)+100$

$P_{3}=(1.005) P_{2}+100=(1.005)(100(1.005)+100)+100=100(1.005)^{2}+100(1.005)+100$

Continuing this pattern, after $m$ deposits, we’d have saved:

$P_{m}=100(1.005)^{m-1}+100(1.005)^{m-2}+\cdots+100(1.005)+100$

In other words, after $m$ months, the first deposit will have earned compound interest for $m-1$ months. The second deposit will have earned interest for $m-2$ months. Last months deposit would have earned only one month worth of interest. The most recent deposit will have earned no interest yet.

This equation leaves a lot to be desired, though – it doesn’t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:

$1.005 P_{m}=1.005\left(100(1.005)^{m-1}+100(1.005)^{m-2}+\cdots+100(1.005)+100\right)$

Distributing on the right side of the equation gives

$1.005 P_{m}=100(1.005)^{m}+100(1.005)^{m-1}+\cdots+100(1.005)^{2}+100(1.005)$

Now we’ll line this up with like terms from our original equation, and subtract each side

$\begin{aligned} 1.005 P_{m} &=100(1.005)^{m}+100(1.005)^{m-1}+\cdots+\quad 100(1.005) \\ P_{m} &=\quad 100(1.005)^{m-1}+\cdots+\quad 100(1.005)+100 \end{aligned}$

Almost all the terms cancel on the right hand side when we subtract, leaving

$1.005 P_{m}-P_{m}=100(1.005)^{m}-100$

Solving for $P_m$

$0.005 P_{m}=100\left((1.005)^{m}-1\right)$

$P_{m}=\frac{100\left((1.005)^{m}-1\right)}{0.005}$

Replacing $m$ months with $12t$, where $t$ is measured in years, gives

$P_{t}=\frac{100\left((1.005)^{12 \mathrm{t}}-1\right)}{0.005}$

Recall 0.005 was $\frac{r}{k}$, while \$100 was the deposit $d$, and 12 was $k$, the number of deposit each year. Generalizing this result, we get the saving annuity formula.

Annuity Formula

$P_{t}=\frac{d\left(\left(1+\frac{r}{k}\right)^{k t}-1\right)}{\left(\frac{r}{k}\right)}$

$P_t$ is the balance in the account after t years.

$d$ is the regular deposit (the amount you deposit each year, each month, etc.)

$r$ is the annual interest rate in decimal form.

$k$ is the number of compounding periods in one year.

If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.

For example, if the compounding frequency isn’t stated:

If you make your deposits every month, use monthly compounding, $k=12$.

If you make your deposits every year, use yearly compounding, $k=1$.

If you make your deposits every quarter, use quarterly compounding, $k=4$.

Etc.

When do you use this

Annuities assume that you put money in the account on a regular schedule (every month, year, quarter, etc.) and let it sit there earning interest.

Compound interest assumes that you put money in the account once and let it sit there earning interest.

Compound interest: One deposit

Annuity: Many deposits.

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,

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

Putting this into the equation:

$P_{20}=\frac{100\left(\left(1+\frac{0.06}{12}\right)^{12(20)}-1\right)}{\left(\frac{0.06}{12}\right)}$

$P_{20}=\frac{100\left((1.005)^{240}-1\right)}{(0.005)}$

$P_{20}=\frac{100(3.310-1)}{(0.005)}$

$P_{20}=\frac{100(2.310)}{(0.005)}=\$ 46200$

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 $\$46,200 - \$24,000 = \$22,200$.

Example 2

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

Solution

In this example,

We’re looking for $d$.

$\begin{array}{ll} r = 0.08 & 8\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \\ t = 30 & \text{30 years} \\ P_{30}=\$ 200,000 & \text{The amount we want to have in 30 years} \end{array}$

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

$200,000=\frac{d\left(\left(1+\frac{0.08}{12}\right)^{12(30)}-1\right)}{\left(\frac{0.08}{12}\right)}$

$200,000=\frac{d\left((1.00667)^{360}-1\right)}{(0.00667)}$

$200,000=d(1491.57)$

$d=\frac{200,000}{1491.57}=\$ 134.09$

So you would need to deposit $\$134.09$ each month to have $\$200,000$ in 30 years if your account earns 8% interest

Try it Now 1

A more conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?

Answer

$\begin{array}{ll} d = \$5 & \text{the daily deposit} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 365 & \text{since we’re doing daily deposits, we’ll compound daily} \\ t = 10 & \text{we want the amount after 10 years} \end{array}$

$P_{10}=\frac{5\left(\left(1+\frac{0.03}{365}\right)^{365 \times 10}-1\right)}{\frac{0.03}{365}}=\$ 21,282.07$

We would have deposited a total of $\$ 5 \cdot 365 \cdot 10=\$ 18,250$, so $\$21,282.07 - \$18,250 = \$3,032.07$ is from interest

Solving for time

Often we are interested in how long it will take to accumulate money.

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in Chapter 5.

Example 3

If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?

Solution

This is a savings annuity problem since we are making regular deposits into the account.

$\begin{array}{ll} d = \$100 & \text{the monthly deposit} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \end{array}$

We don't know $t,$ but we want $P_{t}$ to be $\$ 10,000$

Putting this into the equation:

$\begin{array}{ll} 10,000=\frac{100\left(\left(1+\frac{0.03}{12}\right)^{12t}-1\right)}{\left(\frac{0.03}{12}\right)} & \text{ Simplifying the fractions a bit} \\ 10,000=\frac{100\left((1.0025)^{12 t}-1\right)}{0.0025} & \end{array}$

We want to isolate the exponential term, $1.0025^{12 t}$, so multiply both sides by 0.0025

$\begin{array}{ll} 25=100\left((1.0025)^{12 t}-1\right) & \text{Divide both sides by 100} \\ 0.25=(1.0025)^{12 t}-1 & \text{Add 1 to both sides} \\ 1.25=(1.0025)^{12 t} & \text{Now take the log of both sides} \\ \log (1.25)=\log \left((1.0025)^{12 t}\right) & \text{Use the exponent property of logs} \\ \log (1.25)=12 t \log (1.0025) & \text{Divide by \(12\log(1.0025)$ (to isolate $t$)} \\ \frac{\log (1.25)}{12 \log (1.0025)}=t & \text{Approximating to a decimal} \\ t=7.447\text{ years} & \end{array}\)

It will take about 7.447 years to grow the account to $\$ 10,000$.

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