Compound Interest

With compound interest, the interest you earn is added to your balance, and then it earns interest too. That “interest on interest” makes money grow exponentially rather than in a straight line. Almost every real savings account, investment, and loan uses compound interest.

Key ideas

The compound interest formula

$$A = P(1 + i)^n$$

• $A$ is the amount (also called the future value).
• $P$ is the principal (also called the present value).
• $i$ is the interest rate per compounding period, as a decimal.
• $n$ is the number of compounding periods.

Compounding periods

Interest can be added (compounded) more than once a year. Divide the annual rate by the number of periods per year to get $i$, and multiply the years by the periods per year to get $n$.

Compounded	Periods per year	$i$ for $6\%$ per year	$n$ for $5$ years
annually	$1$	$0.06$	$5$
semi-annually	$2$	$0.03$	$10$
quarterly	$4$	$0.015$	$20$
monthly	$12$	$0.005$	$60$

Simple vs. compound

The amounts at the end of each period form a geometric sequence with common ratio $1 + i$. Compared with simple interest’s arithmetic sequence, the difference grows bigger every year.

$1000 at $6\%$ per year for $30$ years: simple interest vs. interest compounded annually.

Present value

To find how much to invest now to reach a goal later, solve the formula for $P$:

$$P = \frac{A}{(1 + i)^n} = A(1 + i)^{-n}$$

Finding the rate or the time

• To find $i$, isolate $(1 + i)^n$, then take the $n$th root.
• To find $n$, use guess and check, a graph, or a TVM Solver (the finance app on a graphing calculator, or a spreadsheet’s financial functions). A TVM Solver has fields for $N$ (number of periods), $I\%$ (annual rate), $PV$, $PMT$ (regular payment, $0$ here), $FV$, and $P/Y$ and $C/Y$ (payment and compounding periods per year). It treats money you pay out as negative.

Worked examples

Example 1: Compounded annually

$2000 is invested at $5\%$ per year, compounded annually, for $10$ years. Find the amount and the interest earned.

Solution. $P = 2000$, $i = 0.05$, $n = 10$:

$$A = 2000(1.05)^{10} \approx 3257.79$$

The amount is $3257.79, so the interest is $3257.79 - 2000 = 1257.79$, or $1257.79.

Example 2: Compounded monthly

$5000 is invested at $6\%$ per year, compounded monthly, for $4$ years. Find the amount.

Solution. $i = \tfrac{0.06}{12} = 0.005$ and $n = 4 \times 12 = 48$:

$$A = 5000(1.005)^{48} \approx 6352.45$$

The amount is $6352.45.

Example 3: Present value

How much must be invested now at $4.8\%$ per year, compounded quarterly, to have $10 000 in $5$ years?

Solution. $i = \tfrac{0.048}{4} = 0.012$ and $n = 20$:

$$P = \frac{10\,000}{(1.012)^{20}} \approx 7877.52$$

$7877.52 must be invested now.

Example 4: Finding the rate and the time

(a) $3000 grows to $3900 in $6$ years, compounded annually. Find the annual rate.

(b) How long does it take $1000 to grow to $1500 at $6\%$ per year, compounded semi-annually?

Solution.

(a)

$$3000(1 + i)^6 = 3900 \quad\Rightarrow\quad (1 + i)^6 = 1.3 \quad\Rightarrow\quad 1 + i = 1.3^{\frac{1}{6}} \approx 1.0447$$

The rate is about $4.47\%$ per year.

(b) $i = 0.03$, and we need $1000(1.03)^n \ge 1500$, so $1.03^n \ge 1.5$. Guess and check:

• $1.03^{13} \approx 1.4685$ (not yet)
• $1.03^{14} \approx 1.5126$ (enough)

It takes $14$ half-year periods, which is $7$ years.

Common mistakes

Using the annual rate as $i$. For $6\%$ compounded monthly, $i = 0.005$, not $0.06$.

Using years as $n$. For $4$ years compounded monthly, $n = 48$, not $4$.

Mixing up present and future value. “How much will it grow to?” asks for $A$. “How much must I invest now?” asks for $P$.

Rounding $i$. For $5.4\%$ compounded monthly, $i = 0.0045$ exactly. Rounding to $0.005$ changes the answer noticeably.

Getting the TVM Solver signs wrong. Money you invest is entered as negative $PV$; the amount you get back shows as positive $FV$.

Practice

1. (Warm-up) Find $i$ and $n$ for each investment.

• (a) $6\%$ per year, compounded quarterly, for $5$ years
• (b) $3.6\%$ per year, compounded monthly, for $2$ years

Solution

(a) $i = 0.015$, $n = 20$.

(b) $i = 0.003$, $n = 24$.

2. (Warm-up) Find the amount when $1000 is invested at $4\%$ per year, compounded annually, for $3$ years.

Solution

$A = 1000(1.04)^3 \approx 1124.86$, so $1124.86.

3. (Warm-up) How much interest was earned in Question 2?

Solution

$1124.86 - 1000 = 124.86$, so $124.86.

4. (Core) Find the amount when $2500 is invested at $5.4\%$ per year, compounded monthly, for $6$ years.

Solution

$i = 0.0045$, $n = 72$:

$$A = 2500(1.0045)^{72} \approx 3454.11$$

The amount is $3454.11.

5. (Core) You need $8000 in $3$ years. How much should you invest now at $3\%$ per year, compounded semi-annually?

Solution

$i = 0.015$, $n = 6$:

$$P = \frac{8000}{(1.015)^6} \approx 7316.34$$

Invest $7316.34.

6. (Core) Compare $4000 invested for $5$ years at $5\%$ per year simple interest with the same amount at $5\%$ per year compounded annually.

Solution

Simple: $A = 4000(1 + 0.05 \times 5) = 5000$.

Compound: $A = 4000(1.05)^5 \approx 5105.13$.

Compounding earns $105.13 more.

7. (Core) $1500 grows to $1800 in $4$ years, compounded annually. Find the annual interest rate to two decimal places.

Solution

$$(1 + i)^4 = \frac{1800}{1500} = 1.2 \quad\Rightarrow\quad 1 + i = 1.2^{\frac{1}{4}} \approx 1.0466$$

The rate is about $4.66\%$ per year.

8. (Challenge) At $8\%$ per year compounded quarterly, how long does it take an investment to double? Compare with the “rule of 72”, which estimates the doubling time as $72 \div 8$ years.

Solution

$i = 0.02$, so we need $1.02^n \ge 2$:

• $1.02^{35} \approx 1.99989$ (just short)
• $1.02^{36} \approx 2.0399$ (doubled)

Interest is only added at the end of each quarter, so it doubles after $36$ quarters, which is $9$ years. The rule of 72 gives $72 \div 8 = 9$ years: a very good estimate.

9. (Challenge) Which is better for a saver: $6\%$ per year compounded annually, or $5.9\%$ per year compounded monthly? Compare what $1 grows to in one year.

Solution

$6\%$ annually: $1.06$.

$5.9\%$ monthly: $\left(1 + \tfrac{0.059}{12}\right)^{12} \approx 1.0606$.

The $5.9\%$ compounded monthly is slightly better: it works out to about $6.06\%$ per year, because the interest starts earning interest sooner.