Business Mathematics: Annuities and Their Applications - Amortization

A loan is amortized if both the principal and interest are paid off with a single periodic payment whose amount is fixed for the life of the loan.

The most common example of an amortized loan is a home mortgage which is typically paid off in monthly installments lasting from 15 to 30 years.
The amount of interest is computed monthly based on the principal balance and is taken from the monthly payment. The part of the monthly payment not used to pay interest is used to reduce the principal balance. Thus the following month, less interest is paid and the principal balance is reduced further.

To compute the monthly payment, we can think of an amortized loan as a simple annuity whose present value A is the amount of money borrowed.

The payment R can be computed from the formula discussed previously.
A = R × discount factor
by solving for R:



Example
Compute the monthly payment on a $10,000 loan at a 6% annual interest rate which is amortized over 15 years.

Thus the monthly payment should be $84.39.
Instead of calculating the monthly payments using the discount factor, as shown above, tables can be used to compute monthly payments. The table below is an abbreviated table of monthly loan payments.

Table: Monthly Payments per $1,000 Financed



Example
To compute the monthly payment on the loan in Example 17 using Table 6.3, we obtain 8.43857 corresponding to 15 years and 6.0%. Since this corresponds to the monthly payment per $1,000 borrowed, we must multiply by 10 for a $10,000 loan, 8.43857 × 10 = $84.39, to the nearest cent.


Example
Suppose we wish to compute the monthly payment on a $15,000 loan, at a rate of 8.5%, to be paid back monthly over a period of 5 years. First we look up (in Table 6.3) the monthly payment per $1,000 borrowed: $20.51653. Since our loan is $15,000, we multiply 15 × $20.51653 = $307.75.


Example
An amortization schedule shows the allocation of the monthly payment toward interest and principal. Consider a $100,000 15 year loan at 6% annual interest. The monthly payment is $843.86. Each month (1) interest is computed by multiplying the previous months balance by 0.5% (0.005), (2) interest is deducted from the monthly payment, and (3) the remainder of the monthly payment is used to reduce the principal balance.
Note how the interest, which is always 0.5% of the previous principal balance, decreases from month to month while the reduction to principal increases. The sum of these two numbers is always equal to the monthly payment of $843.86.

The monthly payment is mathematically computed so that the balance at the end of the loan period is $0.00. In actuality, the amount is never $0.00 due to rounding of the interest. To account for this, the last payment is usually slightly different from the others.


Example
Richard and Maddy want to buy a house that costs $240,000. They can afford to put a 20% down payment toward the cost of the house but must obtain a mortgage for the rest. We would like to determine how much they must pay each month if they elect a 30-year term and the prevailing interest rate is 7½%.
Since the down payment is 20% of the purchase price of the house, $48,000, they must finance $192,000. Table 6.3 gives a monthly payment of 6.99213 dollars per $1,000 borrowed. Their monthly payment is 192 × 6.99215 = $1,342.49 per month.


Example
We would like to compute how much Richard and Maddy (Example 21) pay in finance charges for the life of the loan.

First we compute how much they paid over the 30-year life of the loan:
30 × 12 = 360 payments
360 × $1,342.49 = $483,296.40

Now we subtract the amount borrowed:
$483,296.40 - $192,000.00 = $291,296.40
The total finance charge is $291,296.40.


Solved Problems

6.25   Find the monthly payment on an auto loan of $20,000 to be amortized over a 5-year period at a rate of 9%.
 

Solution
There are 60 monthly payments and the monthly interest rate i = 0.09/12 = 0.0075. The discount factor from Table 6.2 is 48.17337:


6.26   Calculate the monthly payment in Prob. 6.25 using Table 6.3.
Solution
From Table 6.3 the rate per $1,000 financed corresponding to 5 years and 9% is 20.7582. Since our auto loan is for $20,000 (20 × $1,000), we multiply 20 × 20.75836 = 415.1672. The monthly payment is $415,17.

6.27   Bill Murphy financed a $2,500 necklace for his wife Jane. If he will be making 36 monthly payments of $82.44, what rate of financing did he receive?
Solution
Since 2,500/1,000 = 2.5, there are 2.5 thousands in $2,500. His monthly payment per $1,000 is 82.44/2.5 = 32.976. Referring to Table 6.3 across the row labeled 3 years (36 months), we see that 32.976 corresponds closest to 11.5%. This is his annual rate of interest.

6.28   Anthony wants to buy a boat which costs $17,000. He has saved $5,000, which he will use as a deposit, and he will finance the rest of it by taking out a loan to be paid back in equal monthly installments, amortized over 5 years at an annual interest rate of 12%. What will his payment be?

Solution A
R = $17,000 = $5,000 = $12,000, i = 0.12/12 = 0.01, and there are 60 monthly payments. The discount factor from Table 6.2 is 44.95504:

His monthly payment will be $266.93.

Solution B
From Table 6.3, the monthly payment per $1,000 on a 5-year loan at 12% annual interest is $22.24445. Multiplying, 12 × $22.24445 = $266.93.

6.29   Construct an amortization schedule for a 3-year loan of $5,000 at 5% interest, which is to be repaid in quarterly installments over 3 years.
Solution
First we compute the amount to be paid each quarter. The discount factor (Table 6.2, n = 3 × 4 = 12, r = 0.05/4 = 0.0125) is 11.07931: