9.4 Annuities

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

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

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 $12N$ , where $N$ is measured in years, gives

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

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

Note

Annuity Formula

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

$P_N$ is the balance in the account after N 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.

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?

# Ch 9.4 - a savings annuity: you deposit d every period and it all earns interest.
def annuity_balance(d, r, k, N):
    i = r / k
    return d * ((1 + i) ** (N * k) - 1) / i

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

D, APR, K, YEARS = 100, 0.06, 12, 20     # the book's IRA
# TRY IT: D = 150. TRY IT: YEARS = 40 - the extra 20 years are worth far more
# than the extra $50 a month, and the table below shows why.

i = APR / K
end = annuity_balance(D, APR, K, YEARS)
paid = D * K * YEARS
print(f"${D}/month at {100 * APR:g}% compounded monthly for {YEARS} years")
print(f"   PN = {D}((1 + {i:.6g})^({YEARS}x{K}) - 1) / {i:.6g} = ${end:,.2f}")
print(f"   you deposited ${paid:,.2f} of your own money")
print(f"   interest earned ${end - paid:,.2f}, which is "
      f"{100 * (end - paid) / end:.0f}% of the final balance\n")

print("year   your deposits   balance      of which interest")
for y in range(0, YEARS + 1, 2):
    bal = annuity_balance(D, APR, K, y) if y else 0.0
    print(f"{y:>4}   {D * K * y:>13,.0f}   {bal:>9,.2f}   {bal - D * K * y:>16,.2f}")
print("The 'of which interest' column starts near zero and then takes over.\n")

# The formula run backwards: what deposit reaches a goal?
GOAL, G_APR, G_YEARS = 200000, 0.08, 30
d = deposit_needed(GOAL, G_APR, K, G_YEARS)
print(f"to reach ${GOAL:,} in {G_YEARS} years at {100 * G_APR:g}%, deposit "
      f"${d:,.2f} a month")
print(f"   over {G_YEARS} years that is only ${d * K * G_YEARS:,.2f} out of your pocket\n")
print("start saving   years of saving   monthly deposit needed for $200,000")
for start_age, retire in ((25, 65), (35, 65), (45, 65), (55, 65)):
    yrs = retire - start_age
    print(f"    age {start_age:>2}          {yrs:>2}                 "
          f"${deposit_needed(GOAL, G_APR, K, yrs):>9,.2f}")
print("Waiting ten years does not cost you a tenth more - look at the last row.")

Show solution

In this example,

$\begin{array}{ll} d = \$100 & \text{the monthly deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \\ N = 20 & \text{we want the amount after 20 years} \end{array}$

Putting this into the equation:

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

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

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

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

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?

Show solution

In this example,

We’re looking for $d$ .

$\begin{array}{ll} r = 0.08 & 8\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \\ N = 30 & \text{30 years} \\ P_{30}=\$ 200,000 & \text{The amount we want to have in 30 years} \end{array}$

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

$\begin{aligned} &200,000=\frac{d\left(\left(1+\frac{0.08}{12}\right)^{30(12)}-1\right)}{\left(\frac{0.08}{12}\right)} \\ &200,000=\frac{d\left((1.00667)^{360}-1\right)}{(0.00667)} \\ &200,000=d(1491.57) \\ &d=\frac{200,000}{1491.57}=\$ 134.09 \end{aligned}$

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

Your Turn

Try it Now 2

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

$\begin{array}{ll} d = \$5 & \text{the daily deposit} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 365 & \text{since we’re doing daily deposits, we’ll compound daily} \\ N = 10 & \text{we want the amount after 10 years} \end{array}$

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

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

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.