6.4: Loans
- Page ID
- 67131
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\(P=\frac{PMT\left(1-\left(1+\frac{r}{k}\right)^{-tk}\right)}{\left(\frac{r}{k}\right)}\)
\(P\) is the balance in the account at the beginning (the principal, or amount of the loan).
\(PMT\) 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.
\(t\) 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 1
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,
\(\begin{array}{ll} PMT = \$200 & \text{the monthly loan payment} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 12 & \text{since we are doing monthly payments, we will compound monthly} \\ t= 5 & \text{since we are making monthly payments for 5 years} \end{array}\)
We’re looking for \(P\), the starting amount of the loan.
\[\begin{align*} P&=\frac{200\left(1-\left(1+\frac{0.03}{12}\right)^{-5(2)}\right)}{\left(\frac{0.03}{12}\right)} \\ P &=\frac{200\left(1-(1.0025)^{-60}\right)}{(0.0025)} \\P &=\frac{200(1-0.861)}{(0.0025)}=\$ 11,120\end{align*}\]
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 2
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 \(PMT\).
\(\begin{array}{ll} r = 0.06 & 6\% \text{ annual rate}\\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ t= 30 & \text{since we’re making monthly payments for 30 years} \\ P = \$140,000 & \text{the starting loan amount} \end{array}\)
In this case, we’re going to have to set up the equation, and solve for \(PMT\).
\[\begin{align*}140,000 &=\frac{PMT\left(1-\left(1+\frac{0.06}{12}\right)^{-30(12)}\right)}{\left(\frac{0.06}{12}\right)}\\ 140,000 &=\frac{PMT\left(1-(1.005)^{-360}\right)}{(0.005)} \\140,000 &=PMT(166.792) \\PMT &=\frac{140,000}{166.792}=\$ 839.37 \end{align*}\]
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.
To see what is happening to the loan balance over time, we could build an amortization schedule, a spreadsheet that shows the details of the payoff. To create one for the previous example, each month we’d calculated the interest charge as 3%/12 = 0.25% interest on the remaining loan balance. Payment beyond the interest charge goes towards principal and reduces the remaining loan balance. The first several months are shown here:
|
Month |
Payment |
Interest Charge |
Portion to Principal |
Remaining Loan Balance |
|---|---|---|---|---|
|
0 |
-- |
-- |
-- |
$140,000.00 |
|
1 |
$839.37 |
$350.00 |
$489.37 |
139,510.63 |
|
2 |
839.37 |
348.78 |
490.59 |
139,020.04 |
|
3 |
839.37 |
347.55 |
491.82 |
138,528.22 |
|
4 |
839.37 |
346.32 |
493.05 |
138,035.17 |
Notice that the interest charge decreases each month as the loan balance decreases.
Exercise 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?
- Answer
-
\(\begin{array}{ll} PMT = \text{ unknown} & \\ r = 0.16 & 16\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ t= 2 & \text{2 year to repay} \\ P = 3,000 & \text{the starting loan amount \$3,000 loan} \end{array}\)
\(3,000=\frac{PMT\left(1-\left(1+\frac{0.16}{12}\right)^{-2 \cdot 12}\right)}{\frac{0.16}{12}}\)
Solving for \(PMT\) 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.
Example \(\PageIndex{3}\)
Consider the $140,000 mortgage at 6% from the previous example. If the homeowner increased their payments to $1000 per month, how long will it take them to pay off the loan?
Solution
\(\begin{array}{ll} r= 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we are paying monthly} \\ P = \$140,000 & \text{the starting loan amount} \\ PMT = \$1000 & \text{the given payment amount} \end{array}\)
We are solving for \(t\) , the amount of time it will take to pay off.
\[\begin{align*} 140,000 &= \frac{1000\left(1 - \left(1 + \frac{0.06}{12} \right)^{ - 12t} \right)}{\left(\frac{0.06}{12} \right)} \\ 140,000 &= \frac{1000\left(1 - \left(1.005 \right)^{ - 12t} \right)}{\left(0.005 \right)}\end{align*}\]
Multiply both sides by 0.005
\[700 = 1000\left(1 - \left(1.005\right)^{ -12t} \right)\nonumber\]
Divide both sides by 1000
\[0.7 = 1 - \left(1.005 \right)^{ - 12t}\nonumber\]
Subtract 1 from both sides
\[ - 0.3 = - \left(1.005 \right)^{ - 12t}\nonumber\]
Multiply by -1
\[0.3 = \left(1.005 \right)^{ - 12t}\nonumber\]
Take the log of both sides
\[\log \left(0.3 \right) = \log \left(\left(1.005\right)^{ - 12t} \right)\nonumber\]
Use the exponent property of logs
\[\log \left(0.3 \right) = - 12t\log \left(1.005 \right)\nonumber\]
Divide by \(-12\log(1.005)\)
\[t = \frac{\log \left(0.3 \right)}{-12\log \left(1.005 \right)} \approx 20.12\nonumber\]
The loan will be paid off in about 20 years.
By increasing the payment by about 20%, they were able to decrease the length of the loan by about 33%, and reduce interest paid by more than $60,000.
Exercise \(\PageIndex{2}\)
Joel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?
- Answer
-
\(\begin{array}{ll} PMT = \$30 & \text{The monthly payments} \\ r= 0.12 &12\% \text{ annual rate} \\ k = 12 & \text{ since we are making monthly payments} \\ P = 1,000 & \text{ we are starting with a \$1,000 loan}\end{array}\)
We are solving for \(t\), the time to pay off the loan
\[1,000 = \frac{30\left(1 - \left(1 + \frac{0.12}{12} \right)^{-12t} \right)}{\frac{0.12}{12}} \nonumber\]
Solving for \(t\) gives 3.396. It will take about 3.4 years to pay off the purchase.
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 4
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 \(P\) when
\(\begin{array}{ll} PMT = \$1,000 & \text{the monthly loan payment} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ t= 10 & \text{since we’re making monthly payments for 10 more years} \end{array} \)
\[\begin{align*} P=\frac{1000\left(1-\left(1+\frac{0.06}{12}\right)^{-10(2)}\right)}{\left(\frac{0.06}{12}\right)}\\ P=\frac{1000\left(1-(1.005)^{-120}\right)}{(0.005)} \\ P=\frac{1000\left(1-(1.005)^{-120}\right)}{(0.005)} \\ P=\frac{1000(1-0.5496)}{(0.005)}=\$ 90,073.45 \end{align*}\]
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
A couple purchases a home with a \(\$ 180,000\) mortgage at \(4 \%\) for 30 years with monthly payments. their mortgage be after 5 years?
Solution
First we will calculate their monthly payments.
We are looking for \(PMT\).
\(\begin{array}{ll} r = 0.04 & 4\% \text{ annual rate} \\ k = 12 & \text{since they’re payingmonthly} \\ t= 30 & \text{30 years} \\ P = \$180,000 & \text{the starting loan amount} \end{array} \)
We set up the equation and solve for \(PMT\).
\[\begin{align*} 180,000 &=\frac{PMT\left(1-\left(1+\frac{0.04}{12}\right)^{-30(12)}\right)}{\left(\frac{0.04}{12}\right)} \\ 180,000 &=\frac{PMT\left(1-(1.00333)^{-360}\right)}{(0.00333)} \\ 180,000 &=PMT(209.562) \\ PMT &=\frac{180,000}{209.562}=\$ 858.93\end{align*}\]
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.
\(\begin{array}{ll} PMT = \$858.93 & \text{the monthly loan payment we calculated above} \\ r = 0.04 & 4\% \text{ annual rate} \\ k = 12 & \text{since they’re payingmonthly} \\ t= 25 & \text{since they’d be making monthly payments for 25 more years} \end{array} \)
\[ \begin{align*} P &=\frac{858.93\left(1-\left(1+\frac{0.04}{12}\right)^{-25(12)}\right)}{\left(\frac{0.04}{12}\right)} \\ P &=\frac{858.93\left(1-(1.00333)^{-300}\right)}{(0.00333)} \\ P &=\frac{858.93(1-0.369)}{(0.00333)} \approx \$ 162,758\end{align*}\]
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.
Important Topics of this Section
Find the amount of a loan given the payments
Find the payments required given the loan amount
Find the remaining balance of a loan