Section 4D

Loan Payments, Credit Cards, and Mortgages

...

Generalities

So far, we have discussed investments and savings plans

It is good when one has extra money to save and invest. However ...

many people have to borrow instead. We now consider loans and ways to pay them off.

Basic ideas

While you owe something to a bank, the bank charges interest.

Assume that you borrow an amount and promise to pay back after a certain period of time.

How much do you have to pay then?

That is the same investment problem: the bank invests money into the borrower. Same formulas apply: APR, compounding periods, etc. as we discussed

Typically both banks and people are reluctant to do that for good reasons.

And that situation is not frequent in real life.

Basic ideas

People want to pay their loans off in portions with regular payments, and banks want the loans to be paid this way.

Example. Assume that you borrow \( \$ 1,200 \) at APR=\( 12 \% \) compounded monthly, which is 1% per month.

After 1 month, you owe to the bank

\( 1200 + 0.01 \times 1200 = 1200 +12= \$ 1,212 \)

If you pay \( \$ 12 \) at the end of the first month, then you again owe \( \$ 1,200 \). After another month, you again owe same \(\$ 1,212\), pay \( \$ 12 \), and you are back on square one with your debt of \( \$ 1,200 \).

In this way, you may pay \(\$12\) a month forever.

Basic ideas

Assume now that you want to pay your debt of \(\$ 1,200 \) off in 6 months.

You pay at the end of 1-st month: \(200 + 0.01 \times 1200 = \$ 212 \)

You now owe only \(\$ 1,000 \).

You pay at the end of 2-nd month: \(200 + 0.01 \times 1000 = \$ 210 \)

You now owe only \(\$ 800\).

Every month you pay \(\$200\) plus the interest charged. This scheme is good, but not quite convenient.

Basic terminology

The amount owed at any moment of time is called the loan principal.

Interest is charged on the principal

The loan term is the time you have to pay back the loan in full

Installment loan is a loan paid off with equal regular payments

Loan Payment formula (Installment Loans)

\( PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } } \)

\(PMT =\) regular payment amount

\(P=\) staring loan principal

\(APR =\) annual percentage rate (in decimal)

\(n=\) number of payment periods per year

\(Y=\) loan term in years

Exercise 14 p 250

You borrowed \(\$ 15,000 \) at an APR of \( 7 \%\), which you are paying off with monthly payments of \( \$ 190 \) for 10 years.

(a) Identify the amount borrowed, the annual interest rate, the number of payments per year, the loan term, and the payment amount.

Solution. The amount borrowed: \(P=15000 \), the annual interest rate: \(APR = 0.07 \), the number of payments per year: \(n=12\), the loan term: \(Y=10\), the payment amount \(PMT=190\).

Exercise 14 p 250 ... continued

(b) How many total payments does the loan require? What is the total amount paid over the term of the loan?

Solution. The loan requires \(n \times Y = 10 \times 12 = 120 \) payments. With 120 payments of \(\$190 \), the amount will be \(120 \times 190 = \$22,800 \)

Exercise 14 p 250 ... continued

(c) Of the total amount paid, what percentage is paid toward the principal, and what percentage is paid toward the interest?

Solution. The total amount paid is \(\$22,800 \), while the loan principal was \(\$ 15,000 \). Thus the percentage paid towards the principal is \( \frac{15000}{22800} \approx 0.658 = 65.8 \% \). The rest \(100\% - 65.8\%=34.2\%\) is paid toward the interest.

Exercise 16 p 250

Consider a student loan of \(\$ 12,000\) at a fixed APR of 7% for 10 years.

(a) Calculate the monthly payment

Solution. We use the Loan Payment formula \( PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } } \)
with \(P=12000\), \(APR =0.07\), \(n=12\), \(Y=10\):
\( PMT= \frac{ 12000 \times \left( \frac{0.07}{12} \right) } {1 - \left( 1 + \frac{0.07}{12} \right)^{ \left( -12 \times 10 \right) } } \approx \$ 139.33\)

Exercise 16 p 250 ... continued

Consider a student loan of \(\$ 12,000\) at a fixed APR of 7% for 10 years.

(b) Determine the total amount paid over the term of the loan.

Solution. \(n \times Y \times PMT = 12 \times 10 \times 139.33 = \$ 16,719.6 \)

Exercise 16 p 250 ... continued

Consider a student loan of \(\$ 12,000\) at a fixed APR of 7% for 10 years.

(c) Of the total amount paid, what percentage is paid toward the principal, and what percentage is paid toward the interest?

Solution.

The total amount paid is \(\$ 16,719.6 \), while the loan principal was \(\$ 12,000 \). Thus the percentage paid towards the principal is \( \frac{12000}{16719.6} \approx 0.718 = 71.8 \% \). The rest \(100\% - 71.8\%=28.2\%\) is paid toward the interest.

Exercise 26 p 251

For a student loan of \(\$ 24,000 \) at a fixed APR of 8% for 15 years, make a table showing the amount of each monthly payment which goes toward principal and interest for the first three months.

Solution. We start with calculating the monthly payments using the Loan Payment formula
\( PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } } \)
with \(P=24000\), \(APR =0.08\), \(n=12\), \(Y=15\):
\( PMT= \frac{ 24000 \times \left( \frac{0.08}{12} \right) } {1 - \left( 1 + \frac{0.08}{12} \right)^{ \left( -12 \times 15 \right) } } \approx \$ 229.36 \)

Exercise 26 p 251 ... continued

End of first month:

The interest: \(24000 \times \frac{0.08}{12} = \$ 160 \)

Payment: \(\$ 229.36 =\$ 160 + \$ 69.36\)

New principal: \(\$ 24.000 - \$ 69.36 = \$ 23,930.64 \)

End of second month:

The interest: \(23930.64 \times \frac{0.08}{12} = \$ 159.54 \)

Payment: \(\$ 229.36 =\$ 159.54 + \$ 69.82\)

New principal: \(\$ 23,930.64 - \$ 69.82 = \$ 23,860.82 \)

... same pattern continues ...

Exercise 38 p 251

Compare the monthly payments and total cost for the two options of a \(\$ 75,000 \) loan.

Option (1): a 30-year loan at an APR of 8%

Option (2): a 15-year loan at an APR of 7%

Solution. We start with calculating the payments
\( PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } } \)
for both options, \(P=75000\), \(n=12\).
for option (1), \(APR =0.08\), \(Y=30\).
for option (2), \(APR =0.07\), \(Y=15\).
Results: (1):\( PMT = \$ 550.32 \) and (2): \( PMT = \$674.12 \)

Exercise 38 p 251 ... continued

Total payments = \(PMT \times n \times Y \)

Option (1): \( 550.32 \times 12 \times 30 = \$ 198,115.20 \)

Option (2): \( 674.12 \times 12 \times 15 = \$ 121,341.60 \)

While option (1) offers lower monthly payment (\( \$ 550.32 \) vs \( \$ 674.12 \)), the total amount paid with option (2) is smaller.