3.2: Annuities and Loans
Annuities
For most of us, we aren’t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a savings annuity. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.
An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship
\[P_{m}=\left(1+\dfrac{r}{k}\right) P_{m-1} \nonumber \]
For a savings annuity, we simply need to add a deposit, \(d\), to the account with each compounding period:
\[P_{m}=\left(1+\dfrac{r}{k}\right) P_{m-1}+d \nonumber \]
Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general.
Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. In this example:
\(r\)=0.06(6%)
\(k\)=12 (12 compounds/deposits per year)
\(d\)=$100 (our deposit per month)
Writing out the recursive equation gives
\[P_{m}=\left(1+\dfrac{0.06}{12}\right) P_{m-1}+100=(1.005) P_{m-1}+100 \nonumber \]
Assuming we start with an empty account, we can begin using this relationship:
\(P_0 = 0\)
\(P_1 = (1.005) P_0+100 = 100\)
\(P_2 = (1.005) P_1+100 = (1.005)(100)+100 = 100(1.005)+100 \)
\(P_3 = (1.005) P_2+100 = (1.005)(100(1.005)+100)+100 = 100(1.005)2 + 100(1.005)+100 \)
Continuing this pattern, after m deposits, we’d have saved:
\[P_{m}=100(1.005)^{m-1}+100(1.005)^{m-2}+\ldots+100(1.005)+100 \nonumber \]
In other words, after m months, the first deposit will have earned compound interest for \(m-1\) months. The second deposit will have earned interest for \(m-2\) months. Last month’s deposit would have earned only one month worth of interest. The most recent deposit will have earned no interest yet.
This equation leaves a lot to be desired, though – it doesn’t make calculating the ending balance any easier! Fortunately, however, there is a simpler way to add up all of these amounts in a formula, which comes from the idea called “geometric series.” Here is the formula:
Annuity Formula
\[P_{N}=\dfrac{d\left(\left(1+\dfrac{r}{k}\right)^{N k}-1\right)}{\left(\dfrac{r}{k}\right)} \nonumber \]
\(P_{N}\) is the balance in the account after \(N\) years.
\(d\) is the regular deposit (the amount you deposit each year, each month, etc.)
\(r\) is the annual interest rate in decimal form.
\(k\) is the number of compounding periods in one year.
If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.
For example, if the compounding frequency isn’t stated:
If you make your deposits every month, use monthly compounding, \(k\)=12.
If you make your deposits every year, use yearly compounding, \(k\)=1.
If you make your deposits every quarter, use quarterly compounding, \(k\)=4.
Etc.
When do you use this?
Annuities assume that you put money in the account on a regular schedule (every month, year, quarter, etc.) and let it sit there earning interest.
In the previous section, the compound interest formula assumed that you put money in the account once and let it sit there earning interest.
Remember, an annuity usually has many deposits (of the same amount).
Example \(\PageIndex{1}\)
A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?
Solution
In this example,
\(d\)=$100
\(r\)=0.06
\(k\)=12
\(N\)=20
Putting this into the equation:
\(P_{20}=\dfrac{100\left(\left(1+\dfrac{0.06}{12}\right)^{20(12)}-1\right)}{\left(\dfrac{0.06}{12}\right)}\)
\(P_{20}=\dfrac{100\left((1.005)^{240}-1\right)}{(0.005)}\) (Here, 0.005=0.5% is the interest rate per period, and 240 is the total number of periods)
\(P_{20}=\dfrac{100(3.310-1)}{(0.005)} \)
\(P_{20}=\dfrac{100(2.310)}{(0.005)}=\$ 46200\)
The account will grow to $46,200 after 20 years.
Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned. In this case it is $46,200 - $24,000 = $22,200.
Example \(\PageIndex{2}\)
You want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?
Solution
In this example,
We’re looking for \(d\).
\(r\)=0.08 8% annual rate
\(k\)=12 since we’re depositing monthly
\(N\)=30 30 years
\(P_{30}\)= $200,000 The amount we want 30 years
In this case, we’re going to have to set up the equation, and solve for \(d\).
\(200,000=\dfrac{d\left(1+\dfrac{0.08}{12}\right)^{30(12)}-1}{\dfrac{0.08}{12}}\)
\(200,000=\dfrac{d\left((1.00667)^{360}-1\right)}{(0.00667)}\) The right-hand side turns out to be \(1491.57d\).
\(d= 200,0001491.57=\$134.09\)
So you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.
Try it Now 2
A more conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?
Loans
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.
Loans Formula
\[P_{0}=\dfrac{d\left(1-\left(1+\dfrac{r}{k}\right)^{-N k}\right)}{\left(\dfrac{r}{k}\right)} \nonumber \]
\(P_{0}\) is the balance in the account at the beginning (the principal, or amount of the loan).
\(d\) is your loan payment (the amount of each payment, made monthly, annually, 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.
The compound interest formula in Section 3.1 assumes just one deposit. An annuity assumes regular deposits, and a loan assumes regular payments.
Example \(\PageIndex{3}\)
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 \(P_0\), the starting amount of the loan.
\(P_{0}=\dfrac{200\left(1-\left(1+\dfrac{0.03}{12}\right)^{-5(12)}\right)}{\left(\dfrac{0.03}{12}\right)} \)
\(P_{0}=\dfrac{200\left(1-(1.0025)^{-60}\right)}{(0.0025)} \)
\(P_{0}=\dfrac{200(1-0.861)}{(0.0025)}=\$ 11,120 \)
You can afford a $11,120 loan. (You may have to give up buying your dream car for now.)
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 \(\PageIndex{4}\)
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 \(d\).
\[\begin{array}{l}
140,000=\dfrac{d\left(1-\left(1+\dfrac{0.06}{12}\right)^{-30(12)}\right)}{\left(\dfrac{0.06}{12}\right)} \\
140,000=\dfrac{d\left(1-(1.005)^{-360}\right)}{(0.005)} \\
140,000=d(166.792) \\
d=\dfrac{140,000}{166.792}=\$ 839.37
\end{array} \nonumber \]
You will make payments of $839.37
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. Yes, over the next 30 years, you will be spending more money on the interest than on the house you are about to buy. This is the reality!
Try it Now 3
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?
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 \(\PageIndex{5}\)
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,
\(N\)=10 since we’re making monthly payments for 10 more years
\[\begin{array}{l}
P_{0}=\dfrac{1000\left(1-\left(1+\dfrac{0.06}{12}\right)^{-10(12)}\right)}{\dfrac{0.06}{12}} \\
P_{0}=\dfrac{1000\left(1-(1.005)^{-120}\right)}{(0.005)} \\
P_{0}=\dfrac{1000(1-0.5496)}{(0.005)}=\$ 90,073.45
\end{array} \nonumber \]
The loan balance with 10 years remaining on the loan will be $90,073.45
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 \(\PageIndex{6}\)
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? (The answer may be depressing.)
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 30 years
\(P_0\)=$180,000 the starting loan amount
We set up the equation and solve for \(d\).
\[\begin{array}{l}
180,000=\dfrac{d\left(1-\left(1+\dfrac{0.04}{12}\right)^{-30(12)}\right)}{\left(\dfrac{0.04}{12}\right)} \\
180,000=\dfrac{d\left(1-(1.00333)^{-360}\right)}{(0.00333)} \\
180,000=d(209,562) \\
d=\dfrac{180,000}{209.562}=\$ 858.93
\end{array} \nonumber \]
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 over those 25
\[\begin{array}{l}
P_{0}=\dfrac{858.93\left(1-\left(1+\dfrac{0.04}{12}\right)^{-25(12)}\right)}{\left(\dfrac{0.04}{12}\right)} \\
P_{0}=\dfrac{858.93\left(1-(1.00333)^{-300}\right)}{(0.00333)} \\
P_{0}=\dfrac{858.93(1-0.369)}{(0.00333)} \approx \$ 162,758
\end{array} \nonumber \]
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 $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.
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. In basic annuities, you are saving money. Usually in an annuity problem, your account starts empty and has money 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 then to determine what approach will best allow you to solve the problem.
Try it Now 4
For each of the following scenarios, determine if it is a compound interest problem with one deposit, a savings annuity problem, or a loans problem. Then solve each problem.
- 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?
Reference
- References (6)
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0