4F Simple Interest Amortized Loans
We studied in Section B the socalled addon interest loan, in which the interest charged each payment period is based on the original amount borrowed. These loans are more commonly entered into for moderately priced consumer items. For much larger purchases – such as in buying homes or new cars, for example – the simple interest amortized loan is the standard. With this type of loan, interest is paid only on the outstanding balance. As the borrower makes payments on the loan, the outstanding principal on the loan decreases, and accordingly the amount of interest being paid likewise decreases. But the amount of each payment remains the same, so as time goes on more of each payment is used to pay down the principal and less is used to pay off the interest.
We use the following notation in our calculations with simple interest amortized loans:
p = payment amount  r = annual interest rate  
n = # of payments per year  R = r/n = periodic interest rate  
t = # of years  P = original principal  
N = n · t = total # of payments 
The formula for the amount of each payment on the loan is
In the vast majority of home mortgages, payments are made on a monthly basis. For such loans the number of payments per year is n = 12, while the periodic interest rate is the annual interest rate divided by 12, or R = r/12. The formula for the monthly payment then becomes
example 1
Lucky and Lucille borrowed $20,000 to buy a car, getting a simple interest amortized loan with a 12% annual interest rate, and with payments to extend over 3 years. We calculate their monthly payment, the sum of all their payments, and the total interest they will pay.
The principal is P = $20,000, the periodic interest rate is R = 12%/12 = 1% = .01, and the number of payments is N = 12 · t = 12 · 3 = 36. The monthly payment formula gives
As they will make 36 payments of this amount, the total sum of money they will pay to the lender over the life of the loan is
36 · $664.29 = $23,914.44 .
The total interest they will pay is the difference in what they pay back and the principal they borrowed,
$23,914.44 – $20,000.00 = $3,914.44 .
Now we prepare a socalled amortization schedule for the first three payments of the loan in the previous example. The periodic, or monthly, interest rate on the loan is R = 12%/12 = 1% = .01; thus Lucky and Lucille must each month pay 1% interest on the outstanding balance on the loan. Using the simple interest formula, we calculate the interest for the first month as
I = P · R · t = $20,000 · .01 · 1 = $200 .
This amount comes out of their first payment, and the remainder,
$664.29 – $200.00 = $464.29 ,
is applied to pay off the principal. Thus, after the first month the outstanding principal, or balance, is
$20,000.00 – $464.29 = $19,535.71 .
For the second month, the interest on the outstanding principal is
I = P · R · t = $19,535.71 · .01 · 1 = $195.36 .
This amount comes out of the second payment, and the remainder,
$664.29 – $195.36 = $468.93 ,
is applied to the principal; the new balance on the principal is the difference
$19,535.71 – $468.93 = $19,066.78 .
Finally, for the third month the interest is
I = P · R · t = $19,066.78 · .01 · 1 = $190.67 ,
the amount applied to the principal is
$664.29 – $190.67 = $473.62 ,
and the outstanding balance after the third month is
$19,066.78 – $473.62 = $18,593.16 .
We summarize the above calculations in an amortization table:
Payment Number 
Total Payment 
Amount to Interest 
Amount to Principal 
Balance $20,000.00 
1  $664.29  $200.00  $464.29  $19,535.71 
2  $664.29  $195.36  $468.93  $19,066.78 
3  $664.29  $190.67  $473.62  $18,593.16 
A complete amortization table would continue through all 36 payments, until after the last payment the balance reduces to zero.
example 2
The Nakashima family will borrow $240,000 to buy a house in Kailua on Enchanted Lake. From American Savings they can get a loan at a 7% annual interest rate, with monthly payments stretched over 30 years. We will calculate the
Finally, we will construct an amortization table for the first three months of the loan.
We substitute P = $240,000, r = 7% = .07, and t = 30 into the payment formula and find that the amount of the monthly payment is
In 30 years there will be 12 · 30 = 360 payments, and the total of all payments will be
360 · $1596.73 = $574,822.80 .
Finally, the interest the family will pay over the life of the loan is the difference
$574,822.80 – $240,000.00 = $334,822.80 .
Now, to make the table, we compute the first month's interest as
I = P · R · t = $240,000 · .07/12 · 1 = $1400 .
The amount of the first payment applied to principal is
$1596.73 – $1400.00 = $196.73 ,
and the balance on the principal after one month is
$240,000.00 – $196.73 = $239,803.27 .
For the second month the interest is
I = P · R · t = $239,803.27 · .07/12 · 1 = $1398.85 ,
the amount applied to principal is
$1596.73 – $1398.85 = $197.88 ,
and the new balance is
$239,803.27 – $197.88 = $239,605.39 .
We repeat this procedure for the third month. You can check that the calculations for the third month are as displayed in the table:
Payment Number 
Total Payment 
Amount to Interest 
Amount to Principal 
Balance $20,000.00 
1  $1596.73  $1400.00  $196.73  $239.803.27 
2  $1596.73  $1398.85  $197.88  $239,605.39 
3  $1596.73  $1397.70  $199.03  $239,406.36 
EXERCISES 4F
