9.5 Payout Annuities

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.

Note

Payout Annuity Formula

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

$P sub 0$ 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: $5 % equals 0.05$ )

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

$N$ 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.

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?

# Ch 9.5 - a payout annuity: you START with the money and draw it down.
# Same formula as a loan, seen from the other side of the desk.
def principal_needed(d, r, k, N):
    i = r / k
    return d * (1 - (1 + i) ** (-N * k)) / i

def withdrawal_from(P0, r, k, N):
    i = r / k
    return P0 * i / (1 - (1 + i) ** (-N * k))

D, APR, K, YEARS = 1000, 0.06, 12, 20     # the book's retirement example
# TRY IT: YEARS = 30. TRY IT: APR = 0.02 - a bad market costs you a LOT of
# starting capital for the same monthly cheque.

i = APR / K
need = principal_needed(D, APR, K, YEARS)
taken = D * K * YEARS
print(f"to withdraw ${D:,}/month for {YEARS} years from an account paying {100*APR:g}%")
print(f"   P0 = {D}(1 - (1 + {i:.6g})^-({YEARS}x{K})) / {i:.6g} = ${need:,.2f}")
print(f"   you will withdraw ${taken:,.2f} in total")
print(f"   ${taken - need:,.2f} of that is interest the account earned "
      "while you were spending it\n")

print("year   withdrawn so far   interest earned that year   balance left")
balance = need
for year in range(1, YEARS + 1):
    earned = 0.0
    for _ in range(K):
        earned += balance * i
        balance = balance * (1 + i) - D
    if year % 2 == 0 or year == 1:
        print(f"{year:>4}   {D * K * year:>16,.0f}   {earned:>25,.2f}   {balance:>12,.2f}")
print("The account runs to (almost exactly) zero in the final year. That is the")
print("whole design of a payout annuity. Notice the balance with 10 years still to")
print("go: that is the same $90,073.45 the book computes in section 9.7.\n")

P0, W_APR, W_YEARS = 500000, 0.08, 30
print(f"the other direction: ${P0:,} at {100 * W_APR:g}% has to last {W_YEARS} years")
print(f"   monthly withdrawal = ${withdrawal_from(P0, W_APR, K, W_YEARS):,.2f}\n")
print("how long the money lasts, at various monthly withdrawals:")
for w in (2500, 3000, 3335, 3670, 4000, 5000):
    monthly_interest = P0 * W_APR / K
    if w <= monthly_interest:
        print(f"   ${w:>6,}/month  ->  forever: it is less than the "
              f"${monthly_interest:,.2f} of interest the account earns each month")
    else:
        bal, months = P0, 0
        while bal > 0 and months < 12000:
            bal = bal * (1 + W_APR / K) - w
            months += 1
        print(f"   ${w:>6,}/month  ->  lasts {months / 12:.1f} years")

Show solution

In this example,

$\begin{matrix} d = $ 1000 & the monthly withdrawal \\ r = 0.06 & 6 % annual rate \\ k = 12 & since we’re doing monthly withdrawals, we’ll compound monthly \\ N = 20 & since were taking withdrawals for 20 years \end{matrix}$

We’re looking for $P sub 0$ ; how much money needs to be in the account at the beginning.

Putting this into the equation:

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

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

$P_{0} = \frac{1000 \times (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.

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?

Show solution

In this example,

We’re looking for d.

$\begin{matrix} r = 0.08 & 8 % annual rate \\ k = 12 & since we’re doing monthly withdrawals \\ N = 30 & since were taking withdrawals for 30 years \\ P_{0} = $ 500, 000 & we are beginning with $ 500, 000 \end{matrix}$

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

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

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

$500, 000 equals d open parenthesis 136.232 close parenthesis$

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

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

Your Turn

Try it Now 3

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{matrix} d = unknown \\ r = 0.04 & 4 % annual rate \\ k = 1 & since we’re doing annual scholarships \\ N = 20 & since were taking withdrawals for 20 years \\ P_{0} = $ 100, 000 & we are starting with $ 100, 000 \end{matrix}$

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

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 open parenthesis 0.04 close parenthesis equals $ 4, 000$ a year would have been available.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.