Calculating Interest and Excel Functions:

Apply the Annual Compound Interest Formula

Simple interest only earns a fixed amount of interest based on the original principal amount. On the other hand, compound interest is calculated by taking the interest earned and adding it to the principal amount for the next interest earning period of time. Compound interest grows with each compounding period.

Here is a visual representation of the difference between simple interest and compound interest. Notice that in the simple interest images, the blocks representing interest always stay the same size. However, the blocks representing interest in the compound interest column get bigger with each period. This also makes the principal of the next period grow bigger as well.

Two columns. The first column illustrates simple interest and that the principal at the beginning of each period is the same size. At the end of each period, the amount of interest earned is the same as in all the previous periods. The second column illustrates compound interest and shows the principal amount at the beginning of each period gets a little bigger each time because it includes the interest amount from all the previous periods. This makes the interest amount for each period bigger than the previous interest amount earned as well.

W12 Simple and Compound Interest: Accessible document

We will start by just compounding one time per year, assuming that our rate is an annual rate.

Annual Compound Interest Formula:

${\text{A}}={\text{P}}(1+{\text{r}})^{\text{t}}$

A = Amount (ending amount)

P = Principal (beginning amount)

r = Interest rate when compounding one time per year as a decimal

t = Time in years

Video Source (03:56 mins) | Transcript

The next video uses simple interest to show an example of how compound interest works.

Video Source (06;41 mins) | Transcript

The following video will show another example of using simple interest to calculate compound interest, this time using excel to do those calculations. Then it will introduce the compound interest formula, which makes these calculations much faster.

Video Source (08:13 mins) | Transcript

Compound interest adds the interest earned in the previous period to the principal amount, so the interest from previous periods also earns interest. This is an important tool to understand as it is used daily in loans, credit cards, and investments (including savings and some checking accounts). Spend the time you need to understand this principle because it will help you in your life.

Practice Problems

Use the information given in Question 1 to solve Questions 2 and 3:

If you invest $100 in an account that pays 10% simple interest annually for one year, what is the total amount of money that you would have at the end of one year?

Suppose that you take the total amount of money that you had after the first year in the previous problem and invest it in the same account for one more year. How much money would you have at the end of the second year?

Suppose that you take the total amount of money that you had after the second year in the previous problem and invest it in the same account for yet another year. How much money would you have at the end of the third year?

Use the annual compound interest formula for the following questions:

How much will you have at the end of three years if you invest $100 in an account that pays 10% annual interest compounded once a year? (You should get the same answer as in question 3.)

If you invest $1200 in an account that pays 8% interest compounded annually, what is the total amount of money that you would have at the end of three years?

If you invest $1200 in an account that pays 9% interest compounded annually, what is the total amount of money that you would have at the end of 25 years?

Solutions

$100 + $100 × 10% = $100 + $100 × 0.10 = $100 + $10 = $110

$110 + $110 × 10% = $110 + $110 × 0.10 = $110 + $11 = $121

$121 + $121 × 10% = $121 + $121 × 0.10 = $121 + $12.10 = $133.10

$100$\text{(1 + 10%)}^{3}$ = $100$\text{(1+0.1)}^{3}$ = $100$\text{(1.1)}^{3}$ = $100(1.331) = $133.10

$1200$\text{(1+8%)}^{3}$ = $1200$\text{(1+0.08)}^{3}$ = $1200$\text{(1.08)}^{3}$ = $1200(1.259712) = $1511.65 (Written Solution)

We are investing $1200 that pays 8% compounded annually for 3 years, so we will use the formula:

${\text{A=P(1+r)}}^{\text{t}}$

The first thing that we need to do is determine each of the following:

Principle = P = $1200

Rate = r = 0.08 (we must write it as a decimal for the formula)

Time = t = 3

Then we plug in each value into the formula:

${\text{A=}{\color{MediumVioletRed}{\text{P}}}(1 + {\color{DodgerBlue}{\text{r}}})}^{\color{MediumSeaGreen}{\text{t}}}$

${\text{A=}{\color{MediumVioletRed}{\text{1200}}}(1 + {\color{DodgerBlue}{\text{0.08}}})}^{\color{MediumSeaGreen}{\text{3}}}$

Add 1 to 0.08:

${\text{A=}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.08}}})}^{\color{MediumSeaGreen}{\text{3}}}$

Raise 1.08 to the third power:

${\text{A=}}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.259712}}})$

${\text{A=}}{\color{MediumVioletRed}{1511.6544}}$

So we will have $1511.65 at the end of 3 years.

$1200$\text{(1+9%)}^{25}$ = $1200$\text{(1+0.09)}^{25}$= $1200$\text{(1.09)}^{25}$ = $1200(8.623081) = $10347.70 (Written Solution)

We are investing $1200 that pays 9% compounded annually for 25 years, so we will use the formula:

${\text{A=P(1+r)}}^{\text{t}}$

The first thing that we need to do is determine each of the following:

Principle = P = $1200

Rate = r = 0.09 (we must write it as a decimal for the formula)

Time = t = 25

Then we plug in each value into the formula:

${\text{A=}{\color{MediumVioletRed}{\text{P}}}(1 + {\color{DodgerBlue}{\text{r}}})}^{\color{MediumSeaGreen}{\text{t}}}$

${\text{A=}{\color{MediumVioletRed}{\text{1200}}}(1 + {\color{DodgerBlue}{\text{0.09}}})}^{\color{MediumSeaGreen}{\text{25}}}$

Add 1 to 0.09:

${\text{A=}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.09}}})}^{\color{MediumSeaGreen}{\text{25}}}$

Raise 1.09 to the twenty-fifth power:

${\text{A=}}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{8.62308}}})$

Then multiply 1200 by 8.62308

${\text{A=}}{\color{MediumVioletRed}{10347.696792}}$

So we will have $10,347.70 at the end of 25 years.