6.3 Simple Interest

A sale paper titled 'Earn big with a locked-in rate'. The percentages for 2, 3, and 4 year C Ds are 1.00, 1.50, and 2.00 — **Figure 6.4** Interest is how savings earns money.Interest is how savings earns money. (credit: “Interest Rates” by Mike Mozart/Flickr, CC BY 2.0)

Learning Objectives

After completing this section, you should be able to:

Compute simple interest.
Understand and compute future value.
Compute simple interest loans with partial payments.
Understand and compute present value.

There is truth in the phrase “You need to have money to make money.” In essence, if you have money to lend, you can lend it at a cost to a borrower and make money on that transaction.

When money is borrowed, the person borrowing the money (borrower) typically has to pay the person or entity that lent the money (the lender) more than the amount of money that was borrowed. This extra money is the interest that is to be paid. Interest is sometimes referred to as the cost to borrow, the cost of the loan, or the finance cost.

This idea also applies when someone deposits money in a bank account or some other form of investment. That person is essentially lending the money to the bank or company. The money earned by the depositor is also called interest. The interest is typically based on the amount borrowed, or the principal.

The pairing of borrower and lender can take various forms. The borrower may be a consumer using a credit card or taking out a loan from a bank, the lender. Companies also borrow from lending banks. Someone who invests in a company’s stock is the lender in this case; the company is essentially the borrower.

In this section, we examine the basic building block of interest paid on loans and borrowed credit and also the returns on investments like bank accounts, simple interest.

Compute Simple Interest

Let’s get some terminology understood. Interest to be paid by a borrower is often expressed as an annual percentage rate, which is the percent of the principal that is paid as interest for each year the money is borrowed. This means that the more that is borrowed, the more that must be paid back. Sometimes, the interest to be paid back is simple interest, which means that the interest is calculated on the amount borrowed only.

The length of time until the loan must be paid off is the term of the loan. The date when the loan must be paid off is when the loan is due. The day that the loan is issued is the origination date. We’ll put this terminology to use in the following examples. Note that in this section we will use letters, called variables, to represent the different parts of the formulas we’ll be using. This will help keep our formulas and calculations manageable.

Simple Interest Loans with Integer Year Terms

Calculating simple interest is similar to the percent calculations we made in Understanding Percent and Discounts, Markups, and Sales Tax, but must be multiplied by the term of the loan (in years, if dealing with an annual percentage rate).

When the annual interest rate, the principal, and the number of years that the money is borrowed is known, the interest to be paid can be found and from there the total to be repaid can be calculated.

Example 1 Simple Interest on Loans with Integer Year Terms

Calculate the simple interest to be paid on a loan with the given principal, annual percentage rate, and number of years. Then, calculate the loan payoff amount.

Principal $P$ = $4,000, annual interest rate $r$ = 5.5%, and number of years $t$ = 4
Principal $P$ = $14,800, annual interest rate $r$ = 7.9%, and number of years $t$ = 7

Substitute the principal $P$ = $4,000, the decimal form of the annual interest rate $r$ = 0.055, and number of years $t$ = 4 into the formula for simple interest, and calculate.
$I = P \times r \times t = 4,000 \times 0.055 \times 4 = 880$ .
The simple interest, or cost of the loan, to be paid on the loan is $880.
The loan payoff amount, or the total to be repaid, is $T = P + I = 4,000 + 880 = 4,880$ , or $4,880.00.
Substitute the principal $P$ = $14,800, the decimal form of the annual interest rate $r$ = 0.079, and number of years $t$ = 7 into the formula for simple interest, and calculate.
$I = P \times r \times t = 14,800 \times 0.079 \times 7 = 8,184.4$ .
The simple interest, or cost to borrow, to be paid on the loan is $8,184.40.
The loan payoff amount, or the total to be repaid, is $T = P + I = 14,800 + 8,184.4 = 22,984.4$ , or $22,984.40.

Example 2 Simple Interest Equipment Loan

Riley runs an auto repair shop, and needs to purchase a new brake lathe, which costs $11,995. She takes out a two-year, simple interest loan at an annual interest rate of 14.9%. How much interest will she pay and how much total will she repay on the loan?

Step 1. Determine the variables, or parts of the formula. The principal $P$ is the cost of the brake lathe, so $P$ = $11,995. The interest rate Riley pays is 14.9%, or $r$ = 0.149 in decimal form. The length of the loan is two years, so $t$ = 2. We are first asked to find $I$ , the interest Riley will pay.

Step 2. Substitute the known variables into the formula for simple interest $I = P \times r \times t$ and solve for $I$ .

From Step 1 we have $I = P \times r \times t = 11995 \times 0.149 \times 2 = 3,574.51$ .

This tells us that the simple interest, or cost to borrow, to be paid on the loan is $3,574.51.

Step 3. Use the formula $T = P + I$ to determine the total amount Riley will repay, $T$ .

The total to be repaid is $T = P + I = 11,995 + 3,574.51 = 15,569.51$ , or $15,569.51.

Simple Interest Loans with Other Lengths of Terms

In the previous example and Your Turn exercise, the loans were paid back in one payment after an integer number of years. However, there are also loans lasting a length of time not equal to an integer number of years (like 1, 2, or 3 years or more), but in a number of months (like 4 months, 18 months, and so on). What model would apply to these situations?

When the loan is paid back after a term that is not an integer number of years but is instead a number of months, the term of the loan, or time, $t$ , is expressed as a fraction of the year. So for a 2-month loan, the time, in years, is 2/12 = 1/6. For a 5-month loan, the time in years is 5/12. For an 18-month term, the term in years is 18/12 = 1.5.

Those examples dealt in months. However, some loans are for days only (45 days, 60 days, 120 days). In such cases, we find the daily interest rate. The fraction we will use for the daily interest rate is the interest rate (as a decimal) divided by 365. This may be referred to as Actual/365. In order to find the term of the loan, divide the number of days in the term of the loan by 365.

Example 4 Loan for Moving Costs

David plans to move his family from Raleigh, North Carolina to Tempe, Arizona. His company will reimburse (pay after the move) David for the move. David does research and determines that movers will cost $5,600 to move his family’s belongings to Tempe. He takes out a simple interest, 45-day loan at 11.75% interest to pay this cost. How much interest will be paid on this 45-day loan, and what is David’s loan payoff amount?

This loan is in terms of days, so we will use the formula $I = P \times \frac{r}{365} \times t$ , where t is the number of days and $r$ is the annual interest rate.

The principal for the loan is the moving cost, or $P$ = 5,600. The annual interest rate that David will pay is 11.75%, which in decimal is 0.1175. The length of time for the loan is 45 days, so $t$ = 45.

Substituting these values into the formula and calculating, we find that the interest to be paid is $I = 5,600 \times \frac{0.1175}{365} \times 45 = 81.13$ , or $81.13 (remember, interest is almost always rounded up to the next cent).

The payoff for the loan is $5,681.13.

Understand and Compute Future Value

Money can be invested for a specific amount of time and earn simple interest while invested. The terminology and calculations are the same as we’ve already seen. However, instead of the total to be paid back, the investor is interested in the total value of the investment after the interest is added. This is called the future value of the investment.

Example 5 Simple Interest on a Deposit

In the following, determine how much interest was earned on the investment and the future value of the investment, if the investment yields simple interest.

Principal is $1,000, annual interest rate is 2.01%, and time is 5 years
Principal is $5,000, annual interest rate is 1.85%, and time is 30 years
Principal is $10,000, annual interest rate is 1.25%, and time is 18 months
Principal is $7,000, annual interest rate is 3.26%, and time is 100 days

The principal is $P$ = $1,000, the annual interest rate, in decimal form, is 0.0201, and the term is 5 years, or $t$ = 5. Since the term is an integer number of years, the interest earned on the investment is $I = P \times r \times t = 1,000 \times 0.0201 \times 5 = 100.5$ , or the interest earned was $100.50.
To find the future value, we use the formula $F V = P + I$ . Substituting the values and calculating, we find the future value of the investment to be $F V = P + I = 1,000 + 100.5 = 1100.5$ . The future value of the investment at the end of 5 years is $1,100.50.
Notice that the future value could have been calculated directly with $F V = P + P \times r \times t$
The principal is $P$ = $5,000, the annual interest rate, in decimal form, is 0.0185, and the term is 30 years, or $t$ = 30. Since the term is an integer number of years, the interest earned on the investment is $I = P \times r \times t = 5,000 \times 0.0185 \times 30 = 2,775$ , or the interest earned was $2,775.00. To find the future value, we use the formula $F V = P + I$ . Substituting the values and calculating, we find the future value of the investment to be $F V = P + I = 5,000 + 2,775 = 7,775$ . The future value of the investment at the end of 30 years is $7,775.00.
The principal is $P$ = $10,000, the annual interest rate, in decimal form, is 0.0125, and the term is 18 months. Since the term is in months, we have to write the months in terms of years. For 18 months, we use 18/12 as $t$ . The interest earned on the investment is $I = P \times r \times t = 10,000 \times 0.0125 \times \frac{18}{12} = 187.5$ , or the interest earned was $187.50. To find the future value, we use the formula $F V = P + I$ . Substituting the values and calculating, we find the future value of the investment to be $F V = P + I = 10,000 + 187.5 = 10,187.5$ . The future value of the investment at the end of 18 months is $10,187.50.
Principal is $7,000, annual interest rate is 3.26%, and time is 100 days. The principal is $P$ = $7,000, the annual interest rate, in decimal form, is 0.0326, and the term is 100 days. Since the term is in days, we have to write the time using actual/365, or $t$ = 100/365. The interest earned on the investment is $I = P \times r \times t = 7,000 \times 0.0326 \times \frac{100}{365} = 62.52$ , or the interest earned was $62.52. To find the future value, we use the formula $F V = P + I$ . Substituting the values and calculating, we find the future value of the investment to be $F V = P + I = 7,000 + 62.52 = 7,062.52$ . The future value of the investment at the end of 100 days is $7,062.52.

You may have noticed that for these problems, the future value was rounded down. When the future value is paid, the amount is typically rounded down.

A certificate of deposit (CD) is a savings account that holds a single deposit (the principal) for a fixed term at a fixed interest rate. Once the term of the CD is over, the CD may be redeemed (cashed in or withdrawn) and the owner of the CD receives the original principal plus the interest earned. The deposit often cannot be withdrawn until the term is up; if it can be withdrawn early, there is often a penalty imposed to do so.

Work It Out

The reason CD (certificate of deposit) rates look so small is because they are extremely safe investments. Though overall interest rates for CDs change over time and individual returns vary with the terms of the CD, investors are offered predictable interest income for their investments.

To investigate this yourself, search online to determine the strengths and weaknesses of CDs (investopedia.com offers good, basic information on investing). Then, online, identify five national banks and two local banks who offer CDs.

Track the interest rates for the CDs at various terms (1 year, 3 years, 5 years) for each of the banks you found that offer CDs.
Calculate the amount of interest earned for a $10,000 deposit for each CD at each of the terms.
Compare the results from the various banks, CDs, and terms and decide which is the best investment. You may want to consider both the length of time that the money is locked up, and the return.

Paying Simple Interest Loans with Partial Payments

In every example above, there was one payment for the loan, or one withdrawal for the investment. However, for many loans (house, car, in-ground swimming pool), the loan will be paid back in two or more payments. Such a payment is called a partial payment, because they only pay off part of the loan.

When a partial payment is made, some of the payment pays for the principal, but the rest of the payment pays for interest on the principal. When making the first partial payment, the interest is calculated on the principal for the time between the origination date of the loan and the date of the payment. If another partial payment is made, the interest is calculated based on the remaining principal and the time between the previous partial payment and the current partial payment date.

Example 7 Interest Paid in a Partial Payment on a Loan

A simple interest loan for $6,500 is taken out at 12.6% annual percentage rate. A partial payment is made 45 days into the loan period. How much of the partial payment will be for interest?
A simple interest loan for $13,700 is taken out at 6.55% annual interest rate. A partial payment is to be made after 60 days. How much of the partial payment will be for interest?

To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 45 days.
The principal is $6,500. The annual interest rate, in decimal form, is 0.126.
The interest paid for 45 days is found by substituting the values for principal $P$ , rate $r$ , and time $t$ into the formula $I = P \times \frac{r}{365} \times t$ .
Calculating, we have $I = P \times \frac{r}{365} \times t = 6,500 \times \frac{0.126}{365} \times 45 = 100.9726$ . Rounding up, the portion of the partial payment that will be paid for interest is $100.98.
To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 60 days.
The principal is $13,700. The annual interest rate, in decimal form, is 0.0655.
The interest paid for those 60 days is found by substituting those values into the formula $I = P \times \frac{r}{365} \times t$ . Calculating, we have $I = P \times \frac{r}{365} \times t = 13,700 \times \frac{0.0655}{365} \times 60 = 147.5096$ . Rounding up, the portion of the partial payment that will be paid for interest is $147.51.

Remaining Balance

The previous examples demonstrated how to determine the interest paid in a partial payment. Using this, we can determine the remaining balance after a partial payment.

Step 1: determine the amount of the payment, $P$ , that is applied to interest, $I$ .

Step 2: subtract the amount paid in interest from the payment, $(P - I)$ . This is the amount applied to the balance.

Step 3: subtract the amount applied to the balance (the value obtained in Step 2) from the balance of the loan, $B - (P - I)$ . This is the remaining balance after the partial payment.

Example 8 Determining the Remaining Balance on a Loan After a Partial Payment

A simple interest loan for $45,500 is taken out at 11.8% annual percentage rate. A partial payment of $20,000 is made 50 days into the loan period. After this payment, what will the remaining balance of the loan be?
A simple interest loan for $150,000 is taken out at 5.85% annual percentage rate. A partial payment of $50,000 is made 70 days into the loan period. After this payment, what will the remaining balance of the loan be?

The principal is $45,500, which will be treated as the balance, $B$ , of the loan. The annual simple interest rate, in decimal form, is 0.118. The time is $t$ = 50 days.
Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formula $I = P \times \frac{r}{365} \times t$ and calculate. Calculating, the amount of the payment that is applied to interest is $I = 45,500 \times \frac{0.118}{365} \times 50 = 735.4795$ . Rounding up, we have $I$ = $735.48.
Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $2,000. The amount that is applied to the balance is $P - I = $ 20,000 - $ 735.48 = $ 19,264.52$ .
Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or $B - (P - I) = $ 45,500 - $ 19,264.52 = $ 26,235.48$ .
The remining balance after the partial payment is $26,235.48.
The principal is $150,000, which will be treated as the balance, $B$ , of the loan. The annual simple interest rate, in decimal form, is 0.0585. The time is $t$ = 70 days.
Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formula $I = P \times \frac{r}{365} \times t$ and calculate. Calculating, the amount of the payment that is applied to interest is $I = 150,000 \times \frac{0.0585}{365} \times 70 = 1,682.8767$ . Rounding up, we have $I$ = $1,682.88.
Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $50,000. The amount that is applied to the balance is $P - I = $ 50,000 - $ 1,682.88 = $ 48,317.12$ .
Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or $B - (P - I) = $ 150,000 - $ 48,317.12 = $ 101,682.88$ .
The remining balance after the partial payment is $101,682.88.

Loan Payoff

Finally, we will determine the amount to be paid at the end of the loan. To do so, we apply the formula for the loan payoff to the remaining balance. However, the length of time for that remaining balance is the time between the partial payment and the day the loan is paid off.

Step 1: Determine the remaining balance after the partial payment.

Step 2: Calculate the number of days between the partial payment and the date the loan is paid off. This will be the time $t$ in the payment formula.

Step 3: Calculate the amount to be paid at the end of the loan, or the payoff amount, using $Payoff = P + P \times \frac{r}{365} \times t$ , where $P$ is the remaining balance and $t$ is the time found in Step 2.

Example 9 Finding Loan Pay Off After a Partial Payment

Laura takes out an $18,400 loan for 120 days at 17.9% simple interest. She makes a partial payment of $7,500 after 45 days. What is her payoff amount at the end of the loan?

The initial balance, or principal, of her loan is $18,400. The interest rate in decimal form is 0.179. Her partial payment of $7,500 is made after 45 days. Using these values, we can determine how much of the partial payment is applied to the balance. From there, we can determine her final loan payoff after 120 days.

Step 1: Determine the remaining balance after the partial payment. Using the partial payment process outlined in the previous example, we first find that the amount of the partial payment that is applied to the balance. Their interest paid in the partial payment is $I = P \times \frac{r}{365} \times t = 18,400 \times \frac{0.179}{365} \times 45 = 406.0603$ , or $406.07 (remember to round up!). Using this and that the loan amount was for $18,400, the remaining balance on the loan after the partial payment is $B - (P - I) = $ 18,400 - ($ 7,500 - $ 406.07) = $ 11,306.07$ .

Step 2: The number of days between the partial payment and the date that the loan is to be paid off is 120 – 45 = 75. This means that the time between the partial payment and the final payment is 75 days.

Step 3: To calculate the payoff amount, use $payoff = P + P \times \frac{r}{365} \times t$ , with $P$ = $11,306.07 (the remaining balance), $t$ = 75 (from Step 2) and $r$ = 0.179. The payoff amount, then, is $payoff = 11,306.07 + 11,306.07 \times \frac{0.179}{365} \times 75 = 11,721.9165$ . Rounding up, the payoff amount is $11,721.92.

Repeated Partial Payments

Car loans and mortgages (loans for homes) are paid off through repeated partial payments, most often monthly payments. Since car loans are often 3 to 6 years, and mortgages 15 to 30 years, calculating each individual monthly payment one at a time is time consuming and tedious. Even a 3-year loan would involve applying the above steps 36 times! Fortunately, there is a formula for determining the amount of each partial payment for monthly payments on a simple interest loan.

Example 10 Calculating Car Payments

Desiree buys a new car, by taking a loan out from her credit union. The balance of her loan is $27,845.00. The annual interest rate that Desiree will pay is 7.3%. She plans to pay this off over 4 years. How much will Desiree’s monthly payment be?

To use the formula for monthly payments, we need the principal, the interest rate, and the number of years. The principal is $27,845. The annual rate, in decimal form, is 0.073. Dividing 0.073 by 12 gives the monthly interest rate $0.073 / 12 equals 0.00608 3 ¯$ . She takes the loan out for 4 years, which is $t = 12 \times 4 = 48$ months. Substituting these values into the formula, $A = P \times \frac{r \times {(1 + r)}^{t}}{{(1 + r)}^{t} - 1}$ , we calculate:

$\begin{matrix} A & = & 27,845 \times \frac{0.00608 \bar{3} \times {(1 + 0.00608 \bar{3})}^{48}}{{(1 + 0.00608 \bar{3})}^{48} - 1} \\ = & 27,845 \times \frac{0.00608 \bar{3} \times {(1.00608 \bar{3})}^{48}}{{(1.00608 \bar{3})}^{48} - 1} \\ = & 27,845 \times \frac{0.00608 \bar{3} \times 1.337918996}{1.337918996 - 1} \\ = & 27,845 \times \frac{0.008139007}{0.337918996} \\ = & 27,845 \times 0.024085675 \\ = & 670.6656 \end{matrix}$

Using the formula and rounding up to the next cent, we see that Desiree’s monthly payment will be $670.67.

Tech Check

The calculation of payments is long, and involves many steps. However, most spreadsheet programs, including Google Sheets, have a payment function. In Google Sheets, that function is PMT. To find the payment for an installment loan (like for a car), you need to enter the interest rate per period, the number of payments, and the loan amount. From Your Turn 6.36, the rate was 0.0476/12, the number of payments was 60, and the loan amount was $23,660. In Google Sheets, select any cell and enter the following:

=PMT(0.0476/12,60,23660)

And click the enter key. Immediately, in the cell you selected, the payment of $443.90 appears, though with a negative sign. The negative sign indicates it is a payment out of an account. Since we want to know the payment amount, we ignore the negative sign. The result, with the formula in the formula bar, is shown in Figure 6.5.

Figure 6.5 Google Sheets payment function

In general, to use the PMT function in Google Sheets, enter

=PMT( $r$ /12, $t$ *12, $P$ )

where $r$ is the annual interest rate, $t$ is the number of years, and $P$ is the principal of the loan.

Understand and Compute Present Value for Simple Interest Investments

When finding the future value of an investment, we know how much is deposited, but we have no idea how much that money will be worth in the future. If we set a goal for the future, it would be useful to know how much to deposit now so an account reaches the goal. The amount that needs to be deposited now to hit a goal in the future is called the present value.

Understanding what this tells you is important. When you find the present value, that is how much you need to invest now to reach the goal $F V$ , under the conditions (time and rate) at which the money will be invested.

Example 11

Compute the present value of the investment described. Interpret the result.

$F V$ = $10,000, $t$ = 15 years, annual simple interest rate of 5.5%
$F V$ = $150,000, $t$ = 20 years, annual simple interest rate of 6.25%
$F V$ = $250,000, $t$ = 486 months, annual simple interest rate of 4.75%

The future value is $F V$ = $10,000. The time of the investment is in years, so $t$ = 15. The annual, simple interest rate is 5.5%, which in decimal form is 0.055. We substitute those values into the formula and calculate. $P V = \frac{F V}{(1 + r t)} = \frac{10,000}{(1 + (0.055) \times (15))} = \frac{10,000}{(1 + 0.825)} = \frac{10,000}{(1.825)} = 5,479.452$ . Rounding up, we see that the present value of $10,000 invested at a simple annual interest rate of 5.5% for 15 years is $5,479.46. This means that $5,479.46 needs to be invested so that, after 15 years at 5.5% interest, the investment will be worth $10,000.
The future value is $F V$ = $150,000. The time of the investment is in years, so $t$ = 20. The annual, simple interest rate is 6.25%, which in decimal form is 0.0625. We substitute those values into the formula and calculate. $P V = \frac{F V}{(1 + r t)} = \frac{150,000}{(1 + (0.0625) \times (20))} = \frac{150,000}{(1 + 1.25)} = \frac{150,000}{(2.25)} = 66,666. \bar{6}$ . Rounding up, we see that the present value of $150,000 invested at a simple annual interest rate of 6.25% for 20 years is $66,666.67. This means that $66,666.67 needs to be invested so that, after 20 years at 6.25% interest, the investment will be worth $150,000.
The future value is $FV$ = $250,000. The time of the investment is 486 months. This needs to be converted to years. To do so, divide the number of months by 12, giving $y e a r s = \frac{486}{12} = 40.5$ , so $t$ = 40.5 years. The annual, simple interest rate is 4.75%, which in decimal form is 0.0475. We substitute those values into the formula and calculate. $P V = \frac{F V}{(1 + r t)} = \frac{250,000}{(1 + (0.0475) \times (40.5))} = \frac{150,000}{(1 + 1.92375)} = \frac{150,000}{(2.92375)} = 129,954.5159$ . Rounding up, we see that the present value of $250,000 invested at a simple annual interest rate of 4.75% for 486 months is $129,954.52. This means that $129,954.52 needs to be invested so that, after 486 months at 4.75% interest, the investment will be worth $150,000.

Key Terms

Interest
Principal
Annual percentage rate
Simple interest
Term
Due
Origination date
Payoff amount
Future value
Partial payment
Present value

Key Concepts

Interest is money that is paid by a borrower for the privilege of borrowing the money.
Simple interest is computed by substituting the principal, interest rate, and number of years into the formula $I = P \times r \times t$
The payoff for a loan is the amount of principal remaining on a loan plus the interest that accumulated on the loan since the last payment.
The future value of an investment yielding simple interest is the original principal plus the interest earned on the investment.
When making a partial payment, some of the payment pays off all the accumulated interest, while the remainder of the payment is applied to the principal of the loan.
Finding the present value of an investment is used to determine how much should be invested now in order to achieve a specific goal.

Formulas

$I = P \times r \times t$

$T = P + I$

$T = P + P \times r \times t$

$I = P \times \frac{r}{365} \times t$

$F V = P + I = P + P \times r \times t$

$A = P \times \frac{r \times {(1 + r)}^{t}}{{(1 + r)}^{t} - 1}$

$P V = \frac{F V}{(1 + r t)}$

Adapted from Contemporary Mathematics by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.