Help

- Watch and Read
- - Let's Start Thinking
  - Introduction to Compound Interest
  - Different Compounding Periods
  - Calculating Present Value, Interest Rate, or Number of Compounding Periods
  - Wrap-Up
  - Alternative Format
- Review
- - Explore This 1
  - Check Your Understanding 1
  - Check Your Understanding 2
  - Check Your Understanding 3
  - Check Your Understanding 4
  - Check Your Understanding 5
  - Check Your Understanding 6
  - Check Your Understanding 7
  - Check Your Understanding 8
- Practise
- - Exercises
  - Answers
  - Solutions

Different Compounding Periods

The interest rates for different bank accounts, investments, and loans are almost always described as a yearly rate - that is, in $\%$ per year. However, the interest is often compounded more frequently than once per year. This is something that we need to keep in mind when calculating compound interest.

Recall

The compounding period is the length of time between interest payments.

For the formula $A=P(1+i)^n$, the compounding period affects:

$i$, since this represents the interest rate per compounding period.
$n$, since this represents the number of compounding periods.

Here are some common compounding periods and how they affect both $i$ and $n$:

Compounding Period	Description	Interest Rate ($i$)	Number of Compounding Periods ($n$)
Annually	Interest is compounded once each year	$i=$ yearly rate	$n=$ number of years
Semi-annually	Interest is compounded twice each year	$i=$ yearly rate $\div 2$	$n=$ number of years $\times 2$
Quarterly	Interest is compounded four times each year	$i=$ yearly rate $\div 4$	$n=$ number of years $\times 4$
Monthly	Interest is compounded each month ($12$ times per year)	$i=$ yearly rate $\div 12$	$n=$ number of years $\times 12$

In the next examples, we will calculate compound interest under different compounding periods.

Present Value and Future Value

$P$ can be thought of as the present value.
$A$ can be thought of as the future value.

Example 4 — Part A

Calculate the future value of the following:

$$5000$ in a bank account with an annual interest rate of $1.8\%$, compounded monthly for four years.

Solution — Part A

Monthly interest rate: $\dfrac{1.8\%}{12}=0.15\%$, so $ i= 0.0015 $

Example 4 — Part B

Calculate the future value of the following:

A loan of $$2800$, borrowed at an annual interest rate of $3.2\%$, compounded semi-annually for $10$ months.

Solution — Part B

Semi-annual interest rate: $\dfrac{3.2\%}{2}=1.6\%$, so $ i= 0.016 $

Paused Finished

Slide /

Check Your Understanding 3

Part 1: null Part 2: null

How Did I Do?Try Another

Check Your Understanding 4

Part 1: null

How Did I Do?Try Another

Example 5

Compare the amount of interest earned on a $$20~000$ investment at an annual interest rate of $4.5\%$ over $10$ years, if interest is compounded:

annually
semi-annually
quarterly
monthly

Solution

For all four of the scenarios, the principal invested is $P=20~000$.

Part A — Compounded Annually

One compounding period per year
Interest rate per compounding period:

$\dfrac{4.5\%}{1}=4.5\%$

$i=0.045$

Number of compounding periods:

$ n=10 \times 1\\ n=10$

Accumulated value of the investment:

$\begin{align*} A&=P(1+i)^n \\ &= 20~000(1.045)^{10} \\ &= 31~059.388 \end{align*}$

The investment has an accumulated value of $$31~059.39$.

Interest earned over $10$ years:

$31~059.39-20~000= 11~059.39$

The total amount of interest is $$11~059.39$.

Part B — Compounded Semi-Annually

Two compounding periods per year
Interest rate per compounding period:

$\dfrac{4.5\%}{2}=2.25\%$

$i=0.0225$

Number of compounding periods:

$n=10 \times 2\\ n=20$

Accumulated value of the investment:
$\begin{align*} A&=P(1+i)^n \\ &= 20~000(1.0225)^{20} \\ &= 31~210.184 \end{align*}$

The investment has an accumulated value of $$31~210.18$.

Interest earned over $10$ years:

$31~210.18-20~000= 11~210.18$

The total amount of interest is $$11~210.18$.

Part C — Compounded Quarterly

Four compounding periods per year
Interest rate per compounding period:

$\dfrac{4.5\%}{4}=1.125\%$

$i=0.01125$

Number of compounding periods:

$ n=10 \times 4\\ n=40$

Accumulated value of the investment:

$\begin{align*} A&=P(1+i)^n \\ &= 20~000(1.01125)^{40} \\ &= 31~287.537 \end{align*}$

The investment has an accumulated value of $$31~287.54$.

Interest earned over $10$ years:

$31~287.54-20~000= 11~287.54$

The total amount of interest is $$11~287.54$.

Part D — Compounded Monthly

Twelve compounding periods per year
Interest rate per compounding period:

$\dfrac{4.5\%}{12}=0.375\%$

$i=0.00375$

Number of compounding periods:

$n=10 \times 12\\ n=120$

Accumulated value of the investment:

$\begin{align*} A&=P(1+i)^n \\ &= 20~000(1.00375)^{120} \\ &= 31~339.8555 \end{align*}$

The investment has an accumulated value of $$31~339.86$.

Interest earned over $10$ years:

$31~339.86-20~000= 11~339.86$

The total amount of interest is $$11~339.86$.

Here is a summary of the total interest earned on an investment of $$20~000$ at an annual interest rate of $4.5\%$ over $10$ years, at different compounding periods:

Compounding Frequency	Annually	Semi-Annually	Quarterly	Monthly
Interest Rate per Compounding Period	$4.5\%$	$2.25\%$	$1.125\%$	$0.375\%$
Number of Compounding Periods	$10$	$20$	$40$	$120$
Total Interest Earned	$$11~059.39$	$$11~210.18$	$$11~287.54$	$$11~339.86$

The investment will earn the least interest if interest is compounded annually, and it will earn the most interest if interest is compounded monthly.

When interest is compounded more frequently at the same annual rate over the same length of time, more interest is earned in total.

Example 6

What annual interest rate, compounded annually, is equivalent to an annual interest rate of $6\%$, compounded monthly? Check your answers by graphing.

Solution

If the two interest rates are equivalent, the same amount of principal, $P$, invested for the same length of time, should accumulate to the same value under both compounding periods.

Consider a one-year investment.

For an investment at $6\%$, compounded monthly:

$\dfrac{6\%}{12}=0.5\%$, so $i=0.005$.
There are $12$ compounding periods, so $n=12$.
$A=P(1.005)^{12}$

For an investment at the unknown interest rate, $i$, compounded annually:

There is only one compounding period, so $n=1$ and the interest rate $i$ is not divided.
$A=P(1+i)^1$, or $A=P(1+i)$

Since both investments should accumulate to the same value:

\[P(1.005)^{12}=P(1+i)\]

This equation can be solved for $i$. Divide both sides of the equation by $P$:

$(1.005)^{12}=1+i$, or $1+i=(1.005)^{12}$

So $i=(1.005)^{12}-1$, or $i \approx 0.0617$.

Therefore, an annual interest rate of approximately $6.17\%$, compounded annually, is equivalent to an annual interest rate of $6\%$, compounded monthly.

Check by Graphing

Let's compare these two interest rates by comparing the graphs of these two investments.

Consider a principal investment of $$100$ at both of these interest rates: $6\%$ compounded monthly and $6.17\%$ compounded annually.

Here is the graph of the exponential function $f(x)=100(1.005)^x$.

$f(x)=100(1.005)^x$ represents an investment of $$100$, at an annual interest rate of $6\%$ compounded monthly, where $x$ represents the number of months.

Here is the graph of the exponential function $g(x)=100(1.0617)^x$.

$g(x)=100(1.0617)^x$ represents an investment of $$100$, at an annual interest rate of $6.17\%$ compounded annually, where $x$ represents the number of years.

Although these monthly and annual interest rates are equivalent, these graphs are very different. However, remember that for $f(x)$, $x$ represents the number of months, while for $g(x)$, $x$ represents the number of years. $x=100$ on the first graph does not represent the same time period as $x=100$ on the second graph. Let's adjust the scales and compare the graphs again, keeping in mind that $12$ months is equal to $1$ year.

Here is the graph of the exponential function $f(x)=100(1.005)^x$.

$f(x)=100(1.005)^x$ represents an investment of $$100$, at an annual interest rate of $6\%$ compounded monthly, where $x$ represents the number of months.

Here is the graph of the exponential function $g(x)=100(1.0617)^x$.

$g(x)=100(1.0617)^x$ represents an investment of $$100$, at an annual interest rate of $6.17\%$ compounded annually, where $x$ represents the number of years.

For $f(x)$, the value of the investment after one year is given by $f(12)$ (since $x$ is in months). From the graph, we can see that $f(12)=106.17$, meaning the investment has a value of $$106.17$ after $1$ year.

For $g(x)$, the value of the investment after one year is given by $g(1)$ (since $x$ is in years). From the graph, we can see that $g(1)=106.17$, meaning that the investment also has a value of $$106.17$ after $1$ year with this interest rate.

Similarly, $f(24)=112.72$ and $g(2)=112.72$. Both represent the value of the investment after $2$ years (or $24$ months). The same type of comparison can be made for $f(36)$ and $g(3)$, and for $f(48)$ and $g(4)$. The values are almost identical, although they are not exactly the same because the interest rate $6.17\%$ has been rounded.

However, we can see from the graphs that an annual interest rate of $6\%$, compounded monthly, is approximately equivalent to an annual interest rate of $6.17\%$, compounded annually.

Compounding Period	Description	Interest Rate (\(i\))	Number of Compounding Periods (\(n\))
Annually	Interest is compounded once each year	\(i=\) yearly rate	\(n=\) number of years
Semi-annually	Interest is compounded twice each year	\(i=\) yearly rate \(\div 2\)	\(n=\) number of years \(\times 2\)
Quarterly	Interest is compounded four times each year	\(i=\) yearly rate \(\div 4\)	\(n=\) number of years \(\times 4\)
Monthly\(\)	Interest is compounded each month (\(12\) times per year)	\(i=\) yearly rate \(\div 12\)	\(n=\) number of years \(\times 12\)

Compounding Frequency	Annually	Semi-Annually	Quarterly	Monthly
Interest Rate per Compounding Period	\(4.5\%\)	\(2.25\%\)	\(1.125\%\)	\(0.375\%\)
Number of Compounding Periods	\(10\)	\(20\)	\(40\)	\(120\)
Total Interest Earned	\($11~059.39\)	\($11~210.18\)	\($11~287.54\)	\($11~339.86\)

Different Compounding Periods

Slide Notes

Glossary

All Slides

Navigation

Present Value and Future Value

Example 4 — Part A

Solution — Part A

Example 4 — Part B

Solution — Part B

Check Your Understanding 3

Check Your Understanding 4

Example 5

Solution

Example 6

Solution