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VCE 12 General 2023

5.05 Compound interest investments and loans

Lesson

Compound interest

Whereas simple interest is calculated on the principal (i.e. the initial) amount, meaning the amount of interest remains constant or fixed, compound interest is calculated on the current value of the investment or loan per period of time. Because of this, compound interest growth is another example of a geometric sequence.

For example, if you invest \$500 for 3 years at a rate of 6\% per annum, the investment would grow as follows:

\text{Beginning of year } 1\$500
\text{End of year } 1\$500 \times 1.06=\$500 \times 1.06^1
\text{End of year } 2(\$500 \times 1.06) \times 1.06=\$500 \times 1.06^2
\text{End of year } 3((\$500 \times 1.06) \times 1.06) \times 1.06=\$500 \times 1.06^3
\text{End of year } n\$500 \times 1.06^n

After just a couple of calculations, a pattern can be seen. The size of the investment after n years can be written in the form \$500 \times 1.06^n. This pattern holds true for all forms of compound interest and is expressed with the following formula.

The compound interest formula is: V_n=V_0 \times \left(1+\dfrac{r}{100} \right)^{n} where

  • n is the number of periods (can be years, months, weeks),

  • V_n is the future value (final amount of our investment after n periods),

  • V_0 is the present value (the initial principal amount), and

  • r is the percentage interest rate per compounding period.

Note: This formula, is often written in the form A=P \left(1+r\right)^{n}. In this form, A refers to the future value of the investment after n periods, while P is the principal amount invested, r refers to the rate per period and n refers to the total number of periods. This formula also gives us the total amount (ie. the principal and interest together). If we just want to know the value of the interest, we can work it out by subtracting the principal from the total amount of the investment, I=A-P.

Examples

Example 1

Tina has \$900 in a savings account which earns compound interest at a rate of 2.4\% p.a. If interest is compounded monthly, how much interest does Tina earn in 17 months?

Worked Solution
Create a strategy

Use the formula A=P(1+r)^n to find the final amount and I=A-P to find the interest.

Apply the idea

We are given that P=\$900,\, n=17 \text{ months} and r=\dfrac{2.4}{100\times 12}=0.002.

\displaystyle A\displaystyle =\displaystyle 900(1+0.002)^{17}Substitute the values
\displaystyle =\displaystyle \$931.09Evaluate
\displaystyle I\displaystyle =\displaystyle 931.09 - 900Subtract the principal from the final amount
\displaystyle =\displaystyle \$31.09Evaluate
Idea summary

The compound interest formula:

\displaystyle V_n=V_0 \times \left(1+\dfrac{r}{100} \right)^{n}
\bm{n}
is the number of periods (can be years, months, weeks)
\bm{V_n}
is the future value (final amount of our investment after n periods),
\bm{V_0}
is the present value (the initial principal amount)
\bm{r}
is the percentage interest rate per compounding period.

Alternative formula:

\displaystyle A=P \left(1+r \right)^{n}
\bm{A}
is the final amount of money
\bm{P}
is the principal
\bm{r}
is the interest rate per period
\bm{n}
is the number of compounding periods

Model compound interest using a recurrence relation

The recurrence relation used for modelling compound interest is the same as that of a reducing balance depreciation except that now, whatever is owed or invested, is increasing rather than decreasing over time.

Consider for example investing \$1000 at 10\% interest compounded annually. What will the value of the investment be after three years?

One approach is to repeatedly increasing each year by 10\%. In other words, the balance at the end of each year is 110\% of the previous year:

\text{End of year } 1\$1000 \times 1.10 = \$1100
\text{End of year } 2\$1100 \times 1.10 = \$1210
\text{End of year } 3\$1210 \times 1.10 = \$1331

Another approach is to apply the compound interest formula:\begin{array}{c} &V_0 &= &\$1000 \\ &n &= &3 \\ &r &= &10 \\ \end{array}

We can now substitute these values into the formula.

\displaystyle V_n\displaystyle =\displaystyle V_0 \times \left(r + \dfrac{1}{100}\right)^{n}
\displaystyle =\displaystyle 1000 \times \left(1 + \dfrac{10}{100}\right)^{3}
\displaystyle =\displaystyle 1000 \times (1.1)^{3}
\displaystyle =\displaystyle \$1331

Both approaches give us the same result. The investment is worth \$1331 at the end of the three year period.

The following recurrence relation can be used: V_0=k, \, V_{n+1}=R \times V_n where

  • V_{n+1} is the value of the loan or investment after n periods,

  • R equals 1 + \dfrac{r}{100}, expressed as a decimal, where r\% is the ineterest rate

  • k is the initial amount, or principal amount

Unlike simple interest, compound interest is an example of geometric growth.

Compound interest rates are usually given as per annum, meaning the interest rate per year. This rate is called the nominal interest rate.

Although the annual interest rate is quoted, the compounding period may be different. For example, interest may be calculated monthly meaning that there are 12 interest periods within one year. The nominal interest rate therefore has to be converted to a compounding interest rate.

\text{Term}\text{Compounding periods per year} \left(n\right)\text{Note}
\text{quarterly}4\text{there are } 3 \text{ months in each quarter}
\text{monthly}12\text{even though not all months} \\ \text{have the same amount of days}
\text{fortnightly}26\text{even though there are not exactly} \\ 26 \text{ fortnightly periods}
\text{weekly}52\text{even though there are not exactly } 52 \text{ weeks}
\text{daily}365\text{even though there are leap years}

Exploration

Experiment with changing the values below to observe what happens when you change investments, interest rates, and periods.

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The longer the duration of an investment, the higher will be its future value. The higher the interest rate of an investment, the higher will be its future value. The longer the compounding period, the more will be the effect on the future value.

Examples

Example 2

The balance of an investment, in dollars, at the end of each month where interest is compounded monthly is given by A_{n+1}=1.025A_{n},\, A_{0}=4000.

a

State the monthly interest rate.

Worked Solution
Create a strategy

Use the recurrence relation: V_{n+1}=R \times V_{n}, where V_{0}=a

Apply the idea

Comparing the given expresion for A_{n+1}=1.025A_{n} to the recursive form V_{n+1}=R \times V_{n}, 1.025 represents the compounding factor and is equal to 1 + \dfrac{r}{100}, where r is the interest rate per month.

\displaystyle 1 + \dfrac{r}{100}\displaystyle =\displaystyle 1.025Write the equation
\displaystyle \dfrac{r}{100}\displaystyle =\displaystyle 0.025Subtract 1 from both sides
\displaystyle r\displaystyle =\displaystyle 2.5\%Multiply both sides by 100
b

Use the sequences facility on your calculator to determine the balance at the end of the first year.

Round your answer to the nearest cent.

Worked Solution
Create a strategy

Use the sequences facility on your calculator and enter the recursive rule and value of n.

Apply the idea

We need to determine the balance after 1 year so n=12.

You should get: A_{13}= 5379.56.

The balance at the end of the first year 5 years will be \$5379.56.

c

Use the compound interest formula to determine the balance at the end of the first year and confirm the answer from the previous part.

Worked Solution
Create a strategy

Substitute the value of n into the compound interest formula, V_n=V_0 \times \left(1+\dfrac{r}{100} \right)^{n}.

Apply the idea

From the given, we know that V_0=4000 and 1 + \dfrac{r}{100}=1.025. We need to determine the balance after 1 year so n=12.

\displaystyle V_{n}\displaystyle =\displaystyle V_0 \times \left(1+\dfrac{r}{100} \right)^{n}Write the formula
\displaystyle V_{12}\displaystyle =\displaystyle 4000 \times (1.025)^{12}Substitute the values
\displaystyle =\displaystyle \$5379.56Evaluate
d

Use the sequences facility on your calculator to determine at the end of which month and year the investment is first worth double the initial amount invested.

A
January 2012
B
June 2012
C
March 2011
D
May 2012
E
April 2013
Worked Solution
Create a strategy

Input the recurrence relation given in the question into your calculator. Find the first value of n that is greater than or equal to 8000 which the doubled amount of the initial amount invested.

Apply the idea

By looking through the list generated by your calculator, you should find that the first value greater than or equal to \$8000 is \$8185.64 which occurs when n=29.

So the investment will double after 29 months.

The n value represents the number of months that have passed since the beginning of January 2010. So the date 29 months after January 2010 is on May 2012.

Option D is the correct answer.

Idea summary

The following recurrence relation can be used:

\displaystyle V_0=k, \, V_{n+1}=R \times V_n
\bm{V_{n+1}}
is the value of the loan or investment after n periods,
\bm{R}
equals 1 + \dfrac{r}{100}, expressed as a decimal, where r\% is the ineterest rate
\bm{k}
is the initial amount, or principal amount

Unlike simple interest, compound interest is an example of geometric growth.

Compound interest rates are usually given as per annum, meaning the interest rate per year. This rate is called the nominal interest rate.

Simple and compound interest graphs

This image shows the pattern of a compound and simple interest. Ask your teacher for more information.

Simple interest is calculated only on the principal (that is, the initial amount) so the amount of interest being added to a loan or investment remains constant or fixed. Compound interest is interest earned on the principal amount plus interest on the interest already earned. In general, we observed the following pattern.

This image shows the graph of a compound and simple interest between money and time. Ask your teacher for more information.

With simple interest the balance will increase with a constant step up. Graphing the balance at the start of each period will form a straight line pattern. For compound interest instead of the value of your investment increasing in a straight line the steps become larger and larger, a graph of the balance will grow exponentially and look something like the the image.

Over time, simple interest will continue to grow by the same amount each year, while compound interest will grow faster and faster.

Notice:

  • The simple interest graph is a straight line and the compound interest graph is a smooth curve.

  • Both graphs are increasing.

  • The simple interest line is increasing at a constant rate and the compound interest curve is increasing at an increasing rate.

  • Both graphs have the same y-intercept (present value or principal).

Knowing the basic shape of the curve that each type of investment makes will help us think about key points in the life cycle of an investment, and compare investment strategies.

The value of an investment earning simple interest is calculated using the formula A=P+Prn which is a linear equation in terms of n, the number of periods. Meanwhile, compound interest uses the formula A=P(1+r)^{n} which is non-linear equation in terms of n.

Examples

Example 3

When \$2250 is deposited a bank offers two types of savings accounts.

4
8
12
16
20
24
28
\text{n (years)}
1
2
3
4
5
\text{FV } \left(\$1000\right)
  • After depositing \$2250 the account accrues 2.9\% simple interest per annum

  • After depositing \$2100 the account accrues 2.8\% compound interest per annum

a

Complete the equation which describes the future value FV of the simple interest account after n years.

FV=⬚ + ⬚n

Worked Solution
Create a strategy

Simple interest is calculated only on the principal so the amount of interest being added to a loan or investment remains constant or fixed.

Apply the idea

We are given that V_0=2250 and r = 2.9\% = 0.029. So we have:

\displaystyle FV\displaystyle =\displaystyle V_0 + V_0 \times rnWrite the equation
\displaystyle FV\displaystyle =\displaystyle 2250 + 2250 \times 1.029nSubstitute the values
\displaystyle =\displaystyle 2250 + 65.25nEvaluate
b

Complete the equation which describes the future value FV of the compound interest account after n years.

FV=⬚ (⬚)^n

Worked Solution
Create a strategy

Compound interest is interest earned on the principal amount plus interest on the interest already earned.

Apply the idea

We are given that V_0=2100 and r = 2.9\% = 0.028. So we have:

\displaystyle FV\displaystyle =\displaystyle V_0 \times (1+ 0.029)^nWrite the equation
\displaystyle =\displaystyle 2250 \times (1+ 0.029)^nSubstitute the values
\displaystyle =\displaystyle 2250(1.029)^nEvaluate
c

Using the graphs provided, which account would have a greater balance in the 6th year?

Worked Solution
Create a strategy

Find the point on each graph that lines up with x=6.

Apply the idea
4
8
12
16
20
24
28
\text{n (years)}
1
2
3
4
5
\text{FV } \left(\$1000\right)

From the graphs we can see that when x=6 the value on Simple Interest is higher than the value on Compounded Interest.

So Simple interest account has a greater balance in the 6th year.

d

Using the graphs provided, which account would have a greater balance in the 29th year?

Worked Solution
Create a strategy

Find the point on each graph that lines up with x=29.

Apply the idea
4
8
12
16
20
24
28
\text{n (years)}
1
2
3
4
5
\text{FV } \left(\$1000\right)

From the graphs we can see that when x=29 the value on Simple Interest is higher than the value on Compounded Interest.

So Compound interest account has a greater balance in the 29th year.

Idea summary

Simple interest is calculated only on the principal (that is, the initial amount) so the amount of interest being added to a loan or investment remains constant or fixed. Compound interest is interest earned on the principal amount plus interest on the interest already earned. In general, we observed the following pattern.

Outcomes

U3.AoS2.6

demonstrate the use of a recurrence relation to determine the depreciating value of an asset or the future value of an investment or a loan after 𝑛 time periods for the initial sequence

U3.AoS2.7

use a rule for the future value of a compound interest investment or loan, or a depreciating asset, to solve practical problems

U3.AoS2.3

the concepts of financial mathematics including simple and compound interest, nominal and effective interest rates, the present and future value of an investment, loan or asset, amortisation of a reducing balance loan or annuity and amortisation tables

U3.AoS2.4

the use of first-order linear recurrence relations to model compound interest investments and loans, and the flat rate, unit cost and reducing balance methods for depreciating assets, reducing balance loans, annuities, perpetuities and annuity investments

U3.AoS2.9

use technology with financial mathematics capabilities, to solve practical problems associated with compound interest investments and loans, reducing balance loans, annuities and perpetuities, and annuity investments

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