Loans

Section 5.4 Loans

Objectives

Solve applications involving amortized loans
Create amortization schedules
Compute the remaining loan balance

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.

Subsection 5.4.1 Loans

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.

Definition 5.4.1. Loans Formula.

The loans formula is

\begin{equation*} P_0=\frac{d\left(1 - \left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)} \end{equation*}

$P_0$ 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.

Note 5.4.2. 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 5.4.3.

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,

$d = \$200$ 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 $P_0\text{,}$ the starting amount of the loan.

\begin{equation*} P_0=\frac{200\left(1 - \left(1+\frac{0.03}{12}\right)^{-5\cdot 12}\right)}{\left( \frac{0.03}{12}\right)} \end{equation*}

\begin{equation*} P_0 = \frac{200\left(1-\left(1.0025\right)^{-60}\right)}{0.0025} \end{equation*}

\begin{equation*} P_0 = 11,120 \end{equation*}

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 5.4.4.

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.

$r = 0.06$ 6% annual rate

$k = 12$ since we’re paying monthly

$N = 30$ 30 years

$P_0 = \$140,000$ the starting loan amount

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

\begin{equation*} 140000 = \frac{d\left(1 - \left(1 + \frac{0.06}{12}\right)^{-30\cdot 12}\right)}{\left(\frac{0.06}{12}\right)} \end{equation*}

\begin{equation*} 140000 = \frac{d\left(1 - \left(1.005\right)^{-360}\right)}{0.005} \end{equation*}

\begin{equation*} 140000 = d \left(166.792 \right) \end{equation*}

\begin{equation*} d = \frac{140000}{166.792}=\$839.37 \end{equation*}

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.

Problem 5.4.5. Try It Now.

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.

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.

Subsection 5.4.2 Amortization Schedules

Consider a 5-year loan for $10,000 with yearly payments. If the interest rate is 12%, we can compute the yearly payments. But how much interest are we paying each year? How much of our payments are going toward the principal of the loan?

We can create a table, also called an amortization schedule to show how the interest is spread out over the payments of the loan.

First, using the loans formula, we can solve for $d\text{,}$ or the amount of each payment.

\begin{equation*} 10000 = \frac{d\left(1 - \left(1 + \frac{.12}{1}\right)^{-5 \cdot 1}\right)}{\left(\frac{.12}{1}\right)} \end{equation*}

\begin{equation*} 10000 = \frac{d \left(1 - \left(1.12\right)^{-5}\right)}{.12} \end{equation*}

\begin{equation*} 10000 = d \left(3.6048 \right) \end{equation*}

\begin{equation*} d = 10000 \div 3.6048 = 2774.08 \end{equation*}

We will make 5 annual payments of $2,774.08. That means, we are paying a total of $2774.08 \times 5 = 13870.40$ to pay off this loan. Since we borrowed $10,000, the difference is interest. $13,870.40 - $10,000 = $3,870.40.

Since we owe more money at the beginning of the loan, we should expect to pay more interest in beginning. We can see how this works in Table 5.4.6 below.

Table 5.4.6.

Payment	Amount of Payment	Interest	Paid to Principal	New Balance
1	$2,774.08	$1,200	$1,574.08	$8,425.92
2	$2,774.08	$1,011.11	$1,762.97	$6,662.95
3	$2,774.08	$799.55	$1,974.53	$4,688.42
4	$2,774.08	$562.61	$2,211.47	$2,476.95
5	$2,774.08	$297.23	$2,476.85	$0.10

Before we make our first payment, we owe $10,000. We will pay 12% annual interest on that amount:

\begin{equation*} I = \$10000\cdot .12 \cdot 1 = \$1200 \end{equation*}

which is $1,200. That means $1,200 of our first payment is interest. The rest of the $2,774.08 will go toward the principal of the loan.

Since $2,774.08 - $1,200 = $1,574.08, the balance (or principal) of the loan will be reduced by $1,574.08

After our first payment, the balance of the loan is $10,000 - $1,574.08 = $8,425.92

This information is summarized in the first row of the amortization schedule in Table 5.4.6.

To determine the interest in that second payment, we have to take 12% of the new balance ($8.425.92):

\begin{equation*} I = \$8,425.92 \cdot .12 \cdot 1 = \$1,011.11 \end{equation*}

In the second payment, $2,774.08 - $1,011.11 = $1,762.97 will reduce the balance of the loan. That means, the balance of the loan after the second payment is: $8.425.92 - $1,792.97 = $6,662.95

To find the interest in the third payment we can take 12% of this new balance--$6,662.95.

We can continue this pattern to complete the rest of the table.

The remaining balance of 10 cents at the end of the loan is a result of rounding that was used to compute the loan payments. A lender wouldn't actually do any rounding, and the entire balance of the loan would be paid off at the end of the 5 years.

Example 5.4.7.

Create the first two lines of an amortization schedule for a 30-year mortgage for $150,000 at 6% interest with monthly payments.

Solution.

First we need to find the amount of the monthly payment (or d) using our loans formula:

\begin{equation*} 150000 = \frac{d\left(1 - \left(1 + \frac{0.06}{12}\right)^{-30\cdot 12}\right)}{\left(\frac{0.06}{12}\right)} \end{equation*}

\begin{equation*} 150000 = \frac{d\left(1 - \left(1.005\right)^{-360}\right)}{0.005} \end{equation*}

\begin{equation*} 150000 = d \left(166.792 \right) \end{equation*}

\begin{equation*} d = \frac{150000}{166.792}=\$899.32 \end{equation*}

Now we can compute the amount of interest owed during that first month of the loan. Using our simple interest formula,

\begin{equation*} I = P_0 r t \end{equation*}

\begin{equation*} I = 150000 \left( 0.06 \right) \left(\frac{1}{12}\right) \end{equation*}

\begin{equation*} I = 750 \end{equation*}

Recall, that $t$ is measured in years. One month = $\frac{1}{12}$ year.

This means $750 of our first payment is interest. The rest, $899.32 - $750 = $149.32, will go toward the balance of the loan.

The new balance of the loan can be obtained by subtracting $149.32 from $150,000 to get $149,850.68.

The first line of the amortization schedule is:

Table 5.4.8.

Payment	Amount of Payment	Interest	Paid to Principal	New Balance
1	$899.32	$750	$149.32	$149,850.68

Now we repeat the process to fill out the second line of table. To compute the amount of interest owed during the second month of the loan, we can use the simple interest formula,

\begin{equation*} I = P_0 r t \end{equation*}

\begin{equation*} I = 149,850.68\left( 0.06 \right) \left(\frac{1}{12}\right) \end{equation*}

\begin{equation*} I = 749.25 \end{equation*}

Notice that we still $\frac{1}{12}$ year for $t$ because we are interested in the interest that we owe over one month (specifically the second month) of the loan.

This means $749.25 of our second payment is interest. The rest, $899.32 - $749.25 = $150.07, will go toward the balance of the loan.

The new balance of the loan can be obtained by subtracting $150.07 from $149,850.68 to get $149,700.61.

Now we have the first two lines of the amortization schedule in the table.

Table 5.4.9.

Payment	Amount of Payment	Interest	Paid to Principal	New Balance
1	$899.32	$750	$149.32	$149,850.68
2	$899.32	$749.25	$150.07	$149,700.61

Subsection 5.4.3 Remaining Loan Balance

With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.

To determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don’t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will not have paid off $12,000 of the loan balance.

To determine the remaining loan balance, we can think “how much loan will these loan payments be able to pay off in the remaining time on the loan?”

Example 5.4.10.

If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?

Solution.

To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we’re looking for $P_0$ when

$d = \$1,000$ the monthly loan payment

$r = 0.06$ 6% annual rate

$k = 12$ since we’re doing monthly payments, we’ll compound monthly

$N = 10$ since we’re making monthly payments for 10 more years

\begin{equation*} P_0=\frac{1000\left(1 - \left(1+\frac{0.06}{12}\right)^{-10\cdot 12}\right)}{\left( \frac{0.06}{12}\right)} \end{equation*}

\begin{equation*} P_0 = \frac{1000\left(1-\left(1.005\right)^{-120}\right)}{0.005} \end{equation*}

\begin{equation*} P_0 = 90,073.45 \end{equation*}

The loan balance with 10 years remaining on the loan will be $90,073.45

Often times answering remaining balance questions requires two steps:

Calculating the monthly payments on the loan
Calculating the remaining loan balance based on the remaining time on the loan

Example 5.4.11.

A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?

Solution.

First we will calculate their monthly payments. We’re looking for d.

$r = 0.04$ 4% annual rate

$k = 12$ since they’re paying monthly

$N = 30$ years

$P_0 = \$180,000$ the starting loan amount

We set up the equation and solve for d.

\begin{equation*} 180000 = \frac{d\left(1 - \left(1 + \frac{0.04}{12}\right)^{-30\cdot 12}\right)}{\left(\frac{0.04}{12}\right)} \end{equation*}

\begin{equation*} 180000 = \frac{d\left(1 - \left(1.00333\right)^{-360}\right)}{0.00333} \end{equation*}

\begin{equation*} 180000 = d \left(209.562 \right) \end{equation*}

\begin{equation*} d = \frac{180000}{209.562}=\$858.93 \end{equation*}

Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.

$d = \$858.93$ the monthly loan payment we calculated above

$r = 0.04$ 4% annual rate

$k = 12$ since they’re doing monthly payments

$N = 25$ since they'd be making monthly payments for 25 more years

\begin{equation*} P_0=\frac{858.93\left(1 - \left(1+\frac{0.04}{12}\right)^{-25\cdot 12}\right)}{\left(\frac{0.04}{12}\right)} \end{equation*}

\begin{equation*} P_0 = \frac{858.93\left(1-\left(1.00333\right)^{-300}\right)}{0.00333} \end{equation*}

\begin{equation*} P_0 = 162,758 \end{equation*}

The loan balance after 5 years, with 25 years remaining on the loan, will be $162,758

Over that 5 years, the couple has paid off $180,000 - $162,758 = $17,242 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 - $17,242 = $34,292.80 of what they have paid so far has been interest.

Subsection 5.4.4 Which Equation to Use?

When presented with a finance problem (on an exam or in real life), you're usually not told what type of problem it is or which equation to use. Here are some hints on deciding which equation to use based on the wording of the problem.

The easiest types of problem to identify are loans. Loan problems almost always include words like: "loan", "amortize" (the fancy word for loans), "finance (a car)", or "mortgage" (a home loan). Look for these words. If they're there, you're probably looking at a loan problem. To make sure, see if you're given what your monthly (or annual) payment is, or if you're trying to find a monthly payment.

If the problem is not a loan, the next question you want to ask is: "Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?" If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a compound interest problem. The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest.

If you're making regular payments or withdrawals, the next questions is: "Am I putting money into the account, or am I pulling money out?" If you're putting money into the account on a regular basis (monthly/annually/quarterly) then you're looking at a basic Annuity problem. Basic annuities are when you are saving money. Usually in an annuity problem, your account starts empty, and has money in the future.

If you're pulling money out of the account on a regular basis, then you're looking at a Payout Annuity problem. Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty in the future.

Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and to determine what approach will best allow you to solve the problem.

Problem 5.4.12. Try It Now.

For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.

Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?
Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?
Keisha is managing investments for a non-profit company. They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?
Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?
How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?

Answer.

This is a payout annuity problem. She can pull out $1833.60 a quarter
This is a savings annuity problem. He will have saved up $7,524.11
This is compound interest problem. She would need to deposit $22,386.46
This is a loans problem. She can buy $4,609.33 of new equipment.
This is a savings annuity problem. You would need to save $200.46 each month