Annuities: Future Value

Most people don’t save by investing one big amount. They put in a little at a time: $100 a month, $500 a year. A series of equal, regular payments like this is an annuity. Its future value is what all those payments, plus their interest, add up to at the end.

Key ideas

Ordinary simple annuities

An annuity is a series of equal payments made at regular intervals. In this course you’ll work with ordinary simple annuities:

• ordinary: each payment is made at the end of its period
• simple: payments are made as often as interest is compounded (for example, monthly payments with monthly compounding)

Why it’s a geometric series

Each payment earns compound interest from the time it’s made until the end. The last payment is made right at the end, so it earns nothing; the first payment earns interest for $n - 1$ periods.

Four payments of $1000 at $5\%$ per year, compounded annually, each moved to the end of year $4$.

The future value is the sum $R + R(1 + i) + R(1 + i)^2 + \dots + R(1 + i)^{n - 1}$, a geometric series with first term $R$ and ratio $1 + i$.

The future value formula

Using the geometric series formula:

$$FV = \frac{R\big[(1 + i)^n - 1\big]}{i}$$

• $R$ is the regular payment.
• $i$ is the interest rate per period, and $n$ is the number of payments.

The interest earned is $FV - nR$: the future value minus the total of the deposits.

Finding the payment

To reach a savings goal, solve the formula for $R$:

$$R = \frac{FV \times i}{(1 + i)^n - 1}$$

A TVM Solver does all of these too: enter the payment as $PMT$ (negative, since you’re paying it in), with $PV = 0$.

Worked examples

Example 1: Monthly deposits

You deposit $200 at the end of every month for $3$ years into an account paying $6\%$ per year, compounded monthly. Find the future value and the interest earned.

Solution. $R = 200$, $i = \tfrac{0.06}{12} = 0.005$, $n = 36$:

$$FV = \frac{200\big[(1.005)^{36} - 1\big]}{0.005} \approx 7867.22$$

You deposit $36 \times 200 = 7200$ in total, so the interest is $7867.22 - 7200 = 667.22$.

The future value is $7867.22, including $667.22 of interest.

Example 2: Adding it up by hand

$1000 is deposited at the end of each year for $4$ years at $5\%$ per year, compounded annually. Find the future value by adding the payments, then check with the formula.

Solution. From the timeline above:

$$1000 + 1000(1.05) + 1000(1.05)^2 + 1000(1.05)^3 = 1000 + 1050 + 1102.50 + 1157.625 = 4310.125$$

With the formula:

$$FV = \frac{1000\big[(1.05)^4 - 1\big]}{0.05} \approx 4310.13$$

Both give $4310.13.

Example 3: Finding the payment

You want $15 000 in $5$ years. You’ll deposit equal amounts at the end of every quarter into an account paying $4\%$ per year, compounded quarterly. How much should each deposit be?

Solution. $i = 0.01$, $n = 20$:

$$R = \frac{15\,000(0.01)}{(1.01)^{20} - 1} \approx 681.23$$

Each deposit should be $681.23.

Example 4: Starting early

Two friends each save $100 at the end of every month at $6\%$ per year, compounded monthly. One saves for $40$ years; the other starts $10$ years later and saves for $30$ years. Compare their future values.

Solution. $i = 0.005$.

$40$ years ($n = 480$): $FV = \dfrac{100\big[(1.005)^{480} - 1\big]}{0.005} \approx 199\,149.07$

$30$ years ($n = 360$): $FV = \dfrac{100\big[(1.005)^{360} - 1\big]}{0.005} \approx 100\,451.50$

The early saver deposits only $12 000 more ($48 000 vs. $36 000) but ends up with almost twice as much, because those early deposits have decades longer to compound.

Common mistakes

Using the annual rate and years. As with compound interest, $i$ is the rate per period and $n$ is the number of payments.

Forgetting the $-1$. The formula is $\big[(1 + i)^n - 1\big]$, not $(1 + i)^n$.

Treating the future value as all interest. The interest is $FV - nR$. In Example 1, most of the $7867.22 is your own deposits.

Using the formula when payments are at the start of each period. The formula here is for payments at the end of each period. (Payments at the start earn one extra period of interest.)

Practice

1. (Warm-up) Find $i$ and $n$ for monthly deposits for $3$ years at $4.2\%$ per year, compounded monthly.

Solution

$i = \tfrac{0.042}{12} = 0.0035$ and $n = 36$.

2. (Warm-up) Find the future value of $500 deposited at the end of each year for $6$ years at $3\%$ per year, compounded annually.

Solution

$$FV = \frac{500\big[(1.03)^6 - 1\big]}{0.03} \approx 3234.20$$

The future value is $3234.20.

3. (Warm-up) In Question 2, how much was deposited in total, and how much interest was earned?

Solution

Deposits: $6 \times 500 = 3000$. Interest: $3234.20 - 3000 = 234.20$, so $234.20.

4. (Core) Find the future value of $75 deposited at the end of every month for $10$ years at $5.4\%$ per year, compounded monthly.

Solution

$i = 0.0045$, $n = 120$:

$$FV = \frac{75\big[(1.0045)^{120} - 1\big]}{0.0045} \approx 11\,898.82$$

The future value is $11 898.82.

5. (Core) You want $20 000 in $8$ years, making deposits at the end of every six months at $3\%$ per year, compounded semi-annually. Find the deposit.

Solution

$i = 0.015$, $n = 16$:

$$R = \frac{20\,000(0.015)}{(1.015)^{16} - 1} \approx 1115.30$$

Each deposit is $1115.30.

6. (Core) Compare saving $100 at the end of every month with saving $300 at the end of every quarter for $5$ years, both at $6\%$ per year (compounded monthly for the first, quarterly for the second). Why are the results different, even though both deposit $6000?

Solution

Monthly: $i = 0.005$, $n = 60$: $FV = \dfrac{100\big[(1.005)^{60} - 1\big]}{0.005} \approx 6977.00$.

Quarterly: $i = 0.015$, $n = 20$: $FV = \dfrac{300\big[(1.015)^{20} - 1\big]}{0.015} \approx 6937.10$.

The monthly plan earns about $40 more. Its money goes in sooner (some of each quarter’s $300 is deposited a month or two earlier) and interest is compounded more often.

7. (Core) How much interest is earned when $250 is deposited at the end of every month for $2$ years at $3.6\%$ per year, compounded monthly?

Solution

$i = 0.003$, $n = 24$:

$$FV = \frac{250\big[(1.003)^{24} - 1\big]}{0.003} \approx 6211.63$$

Deposits total $24 \times 250 = 6000$, so the interest is $211.63.

8. (Challenge) Use the geometric series formula $S_n = \dfrac{a(r^n - 1)}{r - 1}$ to derive the future value formula.

Solution

The future value is $R + R(1 + i) + \dots + R(1 + i)^{n - 1}$: a geometric series with $a = R$, $r = 1 + i$, and $n$ terms.

$$FV = \frac{R\big[(1 + i)^n - 1\big]}{(1 + i) - 1} = \frac{R\big[(1 + i)^n - 1\big]}{i}$$

9. (Challenge) You deposit $150 at the end of every month at $4.8\%$ per year, compounded monthly. Use guess and check to find how many deposits it takes to have at least $10 000.

Solution

$i = 0.004$. Try values of $n$ in $FV = \dfrac{150\big[(1.004)^n - 1\big]}{0.004}$:

• $n = 59$: $FV \approx 9959.19$ (not quite)
• $n = 60$: $FV \approx 10\,149.03$ (enough)

It takes $60$ deposits, which is $5$ years.