5 Loans
In the last section, you learned about payout annuities.
In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.
One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you’re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.
Loans Formula
[latex]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex]
P0 is the balance in the account at the beginning (the principal, or amount of the loan).
d is your loan payment (your monthly payment, annual payment, etc)
r is the annual interest rate in decimal form.
k is the number of compounding periods in one year.
N is the length of the loan, in years
Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.
When do you use this
The loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.
Compound interest: One deposit
Annuity: Many deposits.
Payout Annuity: Many withdrawals
Loans: Many payments
Example 11
You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?
In this example,
d = $200 the monthly loan payment
r = 0.03 3% annual rate
k = 12 since we’re doing monthly payments, we’ll compound monthly
N = 5 since we’re making monthly payments for 5 years
We’re looking for P0, the starting amount of the loan.
[latex]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\ \end{align}[/latex]
You can afford a $11,120 loan.
You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $12,000-$11,120 = $880 interest total.
Example 12
You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?
In this example,
We’re looking for d.
r = 0.06 6% annual rate
k = 12 since we’re paying monthly
N = 30 30 years
P0 = $140,000 the starting loan amount
In this case, we’re going to have to set up the equation, and solve for d.
[latex]\begin{align}&140,000=\frac{d\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&140,000=\frac{d\left(1-{{\left(1.005\right)}^{-360}}\right)}{\left(0.005\right)}\\&140,000=d(166.792)\\&d=\frac{140,000}{166.792}=\$839.37 \\ \end{align}[/latex]
You will make payments of $839.37 per month for 30 years.
You’re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 – $140,000 = $162,173.20 in interest over the life of the loan.
Try it Now 4
Janine bought $3,000 of new furniture on credit. Because her credit score isn’t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?