8.5: Amortized Loans

In Section 8.2, we learned how to find the future value of a lump sum, and in Section 8.4, we learned how to find the future value of an annuity. With these two concepts in hand, we will now learn how to amortize a loan, and how to find the present value of an annuity.

Amortize a Loan

If a person or business needs to buy or pay for something now (a car, a home, college tuition, equipment for a business) but does not have the money, they can borrow the money as a loan.

They receive the loan amount now, called the principal, P, (or present value), and are obligated to pay back the principal in the future over a stated amount of time (term of the loan), as regular periodic payments, PMT, plus interest.

Example \(\PageIndex{1}\)

Find the monthly payment for a car costing $15,000 if the loan is amortized over five years at an interest rate of 9%.

Solution

Consider the following scenario:

Two people, Mr. Cash and Mr. Credit, go to buy the same car that costs $15,000. Mr. Cash pays cash and drives away, but Mr. Credit wants to make monthly payments for five years.

Our job is to determine the amount that Mr. Credit needs to pay each month for 5 years. We reason as follows:

If Mr. Credit pays PMT dollars per month, then the PMT dollar payment deposited each month at 9% for 5 years should yield the same amount as the $15,000 lump sum deposited in an annuity for 5 years.

Again, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like them to be the same.

Since Mr. Cash is paying a lump sum of $15,000, the future value \(F\) is given by the lump sum formula (8.2), and it is

\[F =\ 15,000\left(1+\dfrac{.09}{12}\right)^{60} \nonumber\]

Mr. Credit wishes to make a sequence of payments of \(PMT\) dollars per month, and the future value is given by the annuity formula (8.4), and this value is

\[F = PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \nonumber\]

We set the two future amounts equal to eachother and solve for the unknown value, PMT.

\[\begin{array}{l}
\ 15,000\left(1+\dfrac{.09}{12}\right)^{60}=PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \\ \ 15,000(1.5657)=PMT(75.4241) \\
\ 311.3792=PMT
\end{array} \nonumber\]

Therefore, the monthly payment needed to repay the loan is $311.38 for five years.

The formula used above (and restated here), for finding payments on an amortized loan, can appear cumbersome.

\(\begin{array}{l}
\ P\left(1+\dfrac{r}{m}\right)^{m t}=PMT\frac{\left[\left(1+\dfrac{r}{m}\right)^{m t}-1\right]}{\dfrac{r}{m}}
\end{array} \nonumber\)

If we do the necessary algebra to solve this equation for PMT, we can use the new formula to find the payments. The algebra has been omitted and the new formula is stated in the box below.

Amortization Formula

\(PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\)

\(P\) is the balance in the account at the beginning (the principal, or amount of the loan)

\(r\) is the annual interest rate in decimal form

\(t\) is the length of the loan, in years

\(m\) is the number of compounding periods in one year

\(PMT\) is the loan payment (monthly payment, annual payment, etc.)

*Notes:
The compounding frequency is not always explicitly given, but is determined by how often you make payments.
We will round payments on a loan up to the next cent.

Example \(\PageIndex{2}\)

You want to take out a $340,000 mortgage (home loan). The interest rate on the loan is 3.5%, and the loan is for 30 years. How much will your monthly payments be? How much interest will you pay over the life of the loan?

Solution

We’re looking for \(PMT\).

\(\begin{array}{ll} P = \$340,000 & \text{the starting loan amount} \\ r = 0.035 & 3.5\% \text{ annual rate} \\ t = 30 & \text{since we’re making monthly payments for 30 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=340000\cdot \dfrac{\left(\dfrac{0.035}{12}\right)}{\left[1-\left( 1+\dfrac{0.035}{12}\right)^{-12(30)}\right]} \\
\ PMT=1526.7519
\end{array} \nonumber\]

You will make payments of $1526.76 per month for 30 years. (Remember to round up for payments.)

You will pay a total of \(\$ 1526.76\) per month for 360 months which equals \(\$ 549,633.60\) to the loan company.
The total paid over the life of the loan is \(\$ 549,633.60 - \$ 340,000=\$ 209,633.60\).

Example \(\PageIndex{3}\)

Jack goes to a car dealer to buy a new car for $18,000 at 2% APR with a five-year loan. The dealer quotes him a monthly payment of $425. Verify that this is the correct monthly payment.

P = $18,000, r = 0.02, t = 5, m = 12

\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=18000\cdot \dfrac{\left(\dfrac{0.02}{12}\right)}{\left[1-\left( 1+\dfrac{0.02}{12}\right)^{-12(5)}\right]} \\
\ PMT=315.4996
\end{array} \nonumber\]

Figure \(\PageIndex{3}\): Calculation for PMT

Jack should have a monthly payment of $315.50, not $425.

What should the total principal and interest be with the $315.50 monthly payment? \(\$ 315.50(12)(5) =\$ 18,930\)
The $315.50 payment per month has a total of $930 in interest paid over the 2-year loan period.

However, the dealer is trying to get Jack to pay $425 per month. This equates to \(\$ 425(12)(5) =\$ 25,500\) which is significantly more than the calculation above. And notice the difference in the the total interest charges; $25,500 - $18,000 = $7,500. This means that the quoted rate of 2% APR is not accurate, or the quoted price of $18,000 is not accurate, or both.

Try it Now 1

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? What is the total interest charged?

Answer

\(\begin{array}{ll} P = 3,000 & \text{the starting loan amount \$3,000 loan} & \\ r = 0.16 & 16\% \text{ annual rate} \\ t = 2 & \text{2 year to repay} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=3000\cdot \dfrac{\left(\dfrac{0.16}{12}\right)}{\left[1-\left( 1+\dfrac{0.16}{12}\right)^{-12(2)}\right]} \\
\ PMT=146.8893
\end{array} \nonumber\]

Janine will need to make monthly payments of \(\$ 146.89\).

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

Example \(\PageIndex{4}\)

With a fixed rate mortgage, you are guaranteed that the interest rate will not change over the life of the loan. Suppose you need $250,000 to buy a new home. The mortgage company offers you two choices: a 30-year loan with an APR of 6% or a 15-year loan with an APR of 5.5%. Compare your monthly payments and total loan cost to decide which loan you should take. Assume no difference in closing costs.

Option 1: First calculate the monthly payment:

Figure \(\PageIndex{5}\): Calculation for PMT for Example \(\PageIndex{5}\), Option 1

The monthly payment for a 30-year loan at 6% interest is $1498.88.

Now calculate the total cost of the loan over the 30 years:

The monthly payments are $1498.88 and the total cost of the loan is $539,596.80.

Option 2: First calculate the monthly payment:

Figure \(\PageIndex{6}\): Calculate PMT for Example \(\PageIndex{5}\), Option 2

The monthly payment for a 15-year loan at 5.5% interest is $2042.71.

Now calculate the total cost of the loan over the 15 years:

The monthly payments are $2042.71 and the total cost of the loan is $367,687.80.

Therefore, the monthly payments are higher with the 15-year loan, but you spend a lot less money overall.

Present Value of an Annuity

The present value of an annuity is the amount of money we would need now in order to be able to make the annuity payments in the future. Often, we know how much we can afford to pay for each regular payment, so we need to find how much money we can borrow.

Example \(\PageIndex{5}\)

Jordan can afford $400 per month as a car payment. The car dealership offers an auto loan at 12% interest for 4 years. What is the present value of the car? In other words, what loan amount can Jordan afford at $400 per month?

Solution

In this example,

\(\begin{array}{ll} PMT = \$400 & \text{the monthly loan payment} \\ r = 0.12 & 12\% \text{ annual rate} \\ t = 4 & \text{since we’re making monthly payments for 4 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ 400=P\cdot \dfrac{\left(\dfrac{0.12}{12}\right)}{\left[1-\left( 1+\dfrac{0.12}{12}\right)^{-12(4)}\right]} \\
\ P=15189.5838
\end{array} \nonumber\]

Jordan will pay a total of $15,189.58 ($400 per month for 48 months) to the car dealership. The difference between the amount paid and the amount of the loan is the interest paid. In this case, Jordan is paying \(\$ 19,200 -\$ 15,189.58=\$ 4,010.42\) interest total.

Solve the Amortization Formula for P, Present Value

In the example above, notice we were given the monthly payment and asked to find the the present value, P. It would be helpful to solve the amortization formula for the present value first.

\(PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\)

---Since P is multiplied by a fraction, solve for P by multiplying both sides of the equation by the reciprocal of that fraction.

\(PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\cdot {\dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}}\)

---Simply the right side of the equation, then rewrite as,

\(P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}\)

Present Value of an Annuity Formula

Use this formula when you know the payment and you want to find the present value, P.

\(P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}\)

\(P\) is the balance in the account at the beginning (the principal, or amount of the loan)

\(r\) is the annual interest rate in decimal form

\(t\) is the length of the loan, in years

\(m\) is the number of compounding periods in one year

\(PMT\) is the loan payment (monthly payment, annual payment, etc.)

*Note: The compounding frequency is not always explicitly given, but is determined by how often you make payments.

Example \(\PageIndex{6}\)

Grace buys an iPad from a rent-to-own business on credit with payments of $30 a month for four years at 14.5% APR compounded monthly. If Grace had bought the iPad from Best Buy or Amazon it would have cost $500. What is the price that Grace paid for the iPad at the rent-to-own business? How much interest was paid over the life of the loan? What is the better option?

PMT = $30, r = 0.145, t = 4, m = 12

\[\begin{array}{l}
\ P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)} \\
\ P=30\cdot \dfrac{\left[1-\left(1+\dfrac{0.145}{12}\right)^{-12(4)}\right]}{\left(\dfrac{0.145}{12}\right)} \\
\ P=1087.8254
\end{array} \nonumber\]

Figure \(\PageIndex{5}\): Calculation of Present Value

The price Grace paid for the iPad was $1,087.83.

That’s a lot more that $500!

Also, the total amount paid over the course of the loan was . Therefore, the total amount of interest paid was $1440 - $1087.83 = $352.17.

Example \(\PageIndex{7}\)

Suppose you have won a lottery that pays $1,000 per month for the next 20 years. But, you prefer to have the entire amount now. If the interest rate is 8%, how much will you accept?

Solution

This classic present value problem needs our complete attention because the rationalization we use to solve this problem will be used again in the problems to follow.

Consider, for argument purposes, that two people Mr. Cash, and Mr. Credit have won the same lottery of $1,000 per month for the next 20 years. Mr. Credit is happy with his $1,000 monthly payment, but Mr. Cash wants to have the entire amount now.

Our job is to determine how much Mr. Cash should get. We reason as follows:

If Mr. Cash accepts P dollars, then the P dollars deposited at 8% for 20 years should yield the same amount as the $1,000 monthly payments for 20 years. In other words, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like the future values to equal.

Since Mr. Cash is receiving a lump sum of \(x\) dollars, its future value is given by the lump sum formula we studied in Section 6.2, and it is

\[\mathrm{A}=\mathrm{P}(1+.08 / 12)^{240} \nonumber\]

Since Mr. Credit is receiving a sequence of payments, or an annuity, of $1,000 per month, its future value is given by the annuity formula we learned in Section 6.3. This value is

\[\mathrm{A}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \nonumber\]

The only way Mr. Cash will agree to the amount he receives is if these two future values are equal. So we set them equal and solve for the unknown.

\[\begin{array}{l}
\mathrm{P}(1+.08 / 12)^{240}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \\
\mathrm{P}(4.9268)=\$ 1000(589.02041) \\
\mathrm{P}(4.9268)=\$ 589020.41 \\
\mathrm{P}=\$ 119,554.36
\end{array} \nonumber\]

The present value of an ordinary annuity of $1,000 each month for 20 years at 8% is $119,554.36

The reader should also note that if Mr. Cash takes his lump sum of \(\mathrm{P}\) = $119,554.36 and invests it at 8% compounded monthly, he will have an accumulated value of \(\mathrm{A}\)=$589,020.41 in 20 years.

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.

Try it Now 5

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.

1. 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?
2. 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?
3. 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?
4. 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?
5. 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

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