5.3: Loans

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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

$P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-N k}\right)}{\left(\frac{r}{k}\right)}$

$P_0$ 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?

Solution

In this example,

$\begin{array}{ll} d = \$200 & \text{the monthly loan payment} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ N = 5 & \text{since we’re making monthly payments for 5 years} \end{array}$

We’re looking for $P_0$, the starting amount of the loan.

$P_{0}=\frac{200\left(1-\left(1+\frac{0.03}{12}\right)^{-5(12)}\right)}{\left(\frac{0.03}{12}\right)}$

$P_{0}=\frac{200\left(1-(1.0025)^{-60}\right)}{(0.0025)}$

$P_{0}=\frac{200(1-0.861)}{(0.0025)}=\$ 11,120$

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?

Solution

In this example,

We’re looking for $d$.

$\begin{array}{ll} r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ N = 30 & \text{since we’re making monthly payments for 30 years} \\ P_0 = \$140,000 & \text{the starting loan amount} \end{array}$

In this case, we’re going to have to set up the equation, and solve for $d$.

$140,000=\frac{d\left(1-\left(1+\frac{0.06}{12}\right)^{-30(12)}\right)}{\left(\frac{0.06}{12}\right)}$

$140,000=\frac{d\left(1-(1.005)^{-360}\right)}{(0.005)}$

$140,000=d(166.792)$

$d=\frac{140,000}{166.792}=\$ 839.37$

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?

Answer

$\begin{array}{ll} d = \text{ unknown} & \\ r = 0.16 & 16\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ N = 2 & \text{2 year to repay} \\ P_0 = 3,000 & \text{the starting loan amount \$3,000 loan} \end{array}$

$3,000=\frac{d\left(1-\left(1+\frac{0.16}{12}\right)^{-2 \times 12}\right)}{\frac{0.16}{12}}$

Solving for $d$ gives $\$ 146.89$ as monthly payments.

In total, she will pay $\$ 3,525.36$ to the store, meaning she will pay $\$ 525.36$ in interest over the two years.

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Loans Formula

When do you use this

Example 11

Solution

Example 12

Solution

Try it Now 4