8.4: Payout Annuities

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Learning Objectives

Understand the difference between a savings annuity and a payout annuity.
Be able to calculate the future value and present value of a payout annuity.
Be able to correctly identify when a problem will use the savings annuity formula vs the payout annuity formula.

In the last section you learned about annuities. In an annuity, you start with nothing, put money into an account on a regular basis, and end up with money in your account.

In this section, we will learn about a variation called a payout annuity. With a payout annuity, you start with money in the account, and pull money out of the account on a regular basis. Any remaining money in the account earns interest. After a fixed amount of time, the account will end up empty.

Payout annuities are typically used after retirement. Perhaps you have saved $500,000 for retirement, and want to take money out of the account each month to live on. You want the money to last you 20 years. This is a payout annuity. The formula is derived in a similar way as we did for savings annuities. The details are omitted here.

Payout Annuity Formula

$P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-kt}\right)}{\left(\frac{r}{k}\right)}$

$P_0$ is the balance in the account at the beginning (starting amount, or principal).

$d$ is the regular withdrawal (the amount you take out each year, each month, etc.)

$r$ is the annual interest rate (in decimal form. Example: $5\% = 0.05$)

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

$t$ is the number of years we plan to take withdrawals

Like with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.

When do you use this

Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.

Compound interest: One deposit

Annuity: Many deposits.

Payout Annuity: Many withdrawals

Example 1

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,

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

We’re looking for $P_0$; how much money needs to be in the account at the beginning.

Putting this into the equation:

$P_{0}=\frac{1000\left(1-\left(1+\frac{0.06}{12}\right)^{-12(20)}\right)}{\left(\frac{0.06}{12}\right)}$

$P_{0}=\frac{1000 \left(1-(1.005)^{-240}\right)}{(0.005)}$

$P_{0}=\frac{1000 (1-0.302)}{(0.005)}=\$ 139,600$

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. In this case it is $\$ 240,000-\$ 139,600=\$ 100,400$ in interest.

Evaluating negative exponents on your calculator

With these problems, you need to raise numbers to negative powers. Most calculators have a separate button for negating a number that is different than the subtraction button. Some calculators label this [(-)], some with [+/-] . The button is often near the = key or the decimal point.

If your calculator displays operations on it (typically a calculator with multiline display), to calculate $1.005^{-240}$ you'd type something like: 1.005 [^] [(-)] 240 [enter].

If your calculator only shows one value at a time, then usually you hit the (-) key after a number to negate it, so you'd hit: 1.005 [$y^{x}$] 240 [(-)] [enter].

Give it a try - you should get $1.005^{-240}=0.302096$

Example 2

You know you will have $500,000 in your account when you retire. You want to be able to take monthly withdrawals from the account for a total of 30 years. Your retirement account earns 8% interest. How much will you be able to withdraw each month?

Solution

In this example,

We’re looking for d.

$\begin{array} {ll} r = 0.08 & 8\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly withdrawals} \\ t = 30 & \text{ since were taking withdrawals for 30 years} \\ P_0 = \$500,000 & \text{we are beginning with }\$500,000 \end{array}$

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

$500,000=\frac{d\left(1-\left(1+\frac{0.08}{12}\right)^{-12(30)}\right)}{\left(\frac{0.08}{12}\right)}$

$500,000=\frac{d\left(1-(1.00667)^{-360}\right)}{(0.00667)}$

$500,000=d(136.232)$

$d=\frac{500,000}{136.232}=\$ 3670.21$

You would be able to withdraw $3,670.21 each month for 30 years.

Try it Now 1

A donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year?

Answer

$\begin{array} {ll} d = \text{?} & \\ r = 0.04 & 4\% \text{ annual rate} \\ k = 1 & \text{since we’re doing annual scholarships} \\ t = 20 & \text{ since were taking withdrawals for 20 years} \\ P_0 = \$100,000 & \text{we are starting with } \$100,000 \end{array}$

$100,000=\frac{d\left(1-\left(1+\frac{0.04}{1}\right)^{-1(20)}\right)}{\frac{0.04}{1}}$

Solving for $d$ gives $7,358.18 each year that they can give in scholarships.

It is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely. If this donor had specified that, $\$100,000(0.04) = \$4,000$ a year would have been available.

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

Alicia wants to withdraw $4000 a month in a retirement. If she has $700,000 saved for retirement, in an account that earns 6% interest, how long will her money last?

Solution

$\begin{array} {ll} d= \$4000 & \text{The monthly withdrawals} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly withdrawals} \\ P_0 = \$700,000 & \text{we are beginning with }\$700,000 \end{array}$

We are solving for $t$, the time before her account reaches a zero balance.

$700,000=\frac{4000\left(1-\left(1+\frac{0.06}{12}\right)^{-12t}\right)}{\frac{0.06}{12}}$

$700,000=\frac{4000\left(1-\left(1.005\right)^{-12t}\right)}{.005}$

$3500=4000\left(1-\left(1.005\right)^{-12t}\right)$

$0.875=1-\left(1.005\right)^{-12t}$

$0.875-1=-\left(1.005\right)^{-12t}$

$-0.125=-\left(1.005\right)^{-12t}$

$0.125=\left(1.005\right)^{-12t}$

$\ln(0.125)=\ln(\left(1.005\right)^{-12t})$

$\ln(0.125)=-12t\ln\left(1.005\right)$

$\frac{\ln(0.125)}{-12\ln\left(1.005\right)}=t$

Thus, solving for $t$ gives about $34.743$. Her money should last 34.7 years

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