# 4.3: Annuities

- Page ID
- 130969

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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 previous section, we used simple and compound interest to calculate the value of money invested at a certain interest rate over a period of time. In these situations, the money was only transferred twice: once at the start of the loan from the lender to the borrower, and once at the end of the loan in the opposite direction. For some situations, this kind of borrowing or investing makes sense. However, for many people who are trying to save up for a large expense, such as a car, house, or retirement fund, it's not reasonable to invest a large lump sum in an account at its start, and simply wait for interest to accrue.

For example, let's say you wanted to save up for your retirement. You plan to retire in \(40\) years, and you'd like to save \(\$1,000,000\) by that point. Even with an excellent savings account rate of \(5\%\), and with interest compounded monthly, you'd need to initially deposit \(\$135,898.79\) in the account. This simply isn't an amount of money that most people have lying around to put into an account!

## Ordinary Annuities

Fortunately, there is another savings structure that allows someone to deposit smaller amounts of money regularly over time to save for a large savings goal. This structure is known as an *ordinary annuity*.

An **ordinary annuity** is an account into which a sequence of equal, regular payments are made, and that receives compound interest on those deposits. The deposits are made with the same regularity that the interest compounds. The formula that describes annuities is: \[B(t)= \left( \dfrac{n \cdot pymt}{r} \right)\cdot \left(\left(1+\frac{r}{n}\right)^{(nt)}- 1\right)\]

where \(B(t)\) is the balance at time \(t\), \(r\) is the annual interest rate, \(n\) is the number of times per year interest compounds, \(pymt\) is the periodic payment, and \(t\) is time in years.

Usually annuities have an agreed-upon time frame (meaning that a person agrees to pay the same amount regularly over some number of years). Once the time has elapsed, the annuity **matures**. The balance at that time is called the **future value** of the annuity.

Looking at the definition above, we see some familiar elements. The variables \(B(t), r, n\), and \(t\) are all familiar to us from our study of compound interest, and stand for exactly the same things. The only new variable here is \(pymt\), which stands for the periodic payment. In practice, this is usually a monthly payment, but can be paid at any schedule (quarterly, daily, etc.). The timing of the payment and the interest compounding are assumed to be the same in this text (though, in some real-world cases, they are not).

We also notice that part of the annuity equation is the compound interest equation! That is due to the fact that annuities are built on compound interest repeated many times. That is, each time a deposit is made, the compound interest is recalculated on the increased amount in the account. At the start of an annuity, the amount in the account is quite small. But over time, the account grows rapidly for two reasons. First, money is being deposited into the account at regular intervals. But also, interest is being compounded on those new deposits! Annuities are a great way to save money if you don't have a lot at the start, but can reliably make deposits into an account over a long period of time. Most Independent Retirement Accounts (IRAs) have the form of an ordinary annuity, potentially with some slight modifications.

Let's see an example that illustrate how ordinary annuities work, and why they are often more useful for people than savings accounts in saving for large expenditures.

A certain math professor would like to retire in \(10\) years. (Just kidding. There is no way that is going to happen.) She opens an ordinary annuity with \(3.6\%\) annual interest compounded monthly. She makes monthly payments of \(\$200\) into the annuity. Write the function describing the balance in the annuity at time t. How much will she have when the annuity matures?

###### Solution

This question uses an ordinary annuity, which means we need to identify the variables in the ordinary annuity formula before we can proceed. We see that the length of the annuity is \(10\) years, so we'll use \(t = 10\). The interest rate is \(3.6\%\), so we'll use \(r = .036\). Since the deposits are made monthly, that means that interest also compounds monthly, since the deposits and the compounding are made on the same schedule. Therefore, we'll use \(n = 12\). Finally, the monthly payment is \(\$200\), so we'll use \(pymt = 200\).

We can organize this information below, which is recommended whenever you are solving an annuity problem: \[\begin{align*} B &= \text{Balance} = \; ? \\ pymt &= \text{Payment} = 200 \\ n &= \text{monthly} = 12 \\ r &= \text{rate} = .036\\ t &= \text{time in years} = 10\\ \end{align*} \]

The first question asks us to find a function that describes the balance of the account after \(t\) years. To do this, we'll use the ordinary annuity formula, with all of the variables except for \(t\) plugged in. We have: \[ B(t) = \left( \dfrac{12 \cdot 200}{.036} \right)\cdot \left(\left(1+\frac{.036}{12}\right)^{(12t)}- 1\right)\]

Take a moment to note that the equation above actually gives us a lot of information: we could plug in any value of \(t\) and find the balance of the account after that number of years. However, we are interested in what happens when \(t = 10\) years. So, we'll plug in \(t = 10\). We have: \[B(10) = \left( \dfrac{12 \cdot 200}{.036} \right)\cdot \left(\left(1+\frac{.036}{12}\right)^{(12 \times 10)}- 1\right)\]

The value on the right side of the equation above is the answer. However, we need to calculate it to get a meaningful answer that corresponds to a dollar amount. There are numerous ways to do this. For the easiest way, see this video. We'll show a slightly more complicated way here that will lead to the same final answer, just for reference.

First, we'll simplify the first set of parentheses, perform the division and addition in the second set of parentheses, and the multiplication in the exponent. Since we are midway through the problem, we'll write out the entire decimal expansion visible in the calculator window to get maximum accuracy in the final answer. We have: \[B(10) = 66666.66667 \cdot \left( \left(1.003\right)^{120} - 1 \right)\]

Next, we'll calculate the expression \((1.003)^{120} - 1\). We find: \[ B(10) = 66666.66667 \cdot .4235571693\]

Finally, we'll multiply these two values together. We get: \[B(10) = 28837.14462\]

That is, the balance after \(10\) years is \(\$28,837.14\).

There are several things to note about the example above. First of all: \(\$28,837.14\) is not enough money to have saved up to retire comfortable for the vast majority of people, so sadly, this math professor won't be able to retire in 10 years with this sort of savings plan. However, it does show how money can accumulate to a relatively large amount, even though the individual deposits are relatively small.

Moreover, the annuity described above is better than simply putting \(\$200\) in a shoebox each month for \(10\) years. Simply putting away \(\$200\) per month results in setting aside the same amount of money, but without the compound interest built in. If you simply set aside \(\$200\) each month without investing it, you'd be making \(120\) payments of \($200\) each, which would result in \(\$24,000\) after \(10\) years. This is less than the \(\$28,837.14\) in the annuity calculation. The difference is due to the compound interest that is calculated on each deposit.

The question above illustrates how ordinary annuities work in general. But what if you wanted to save up for a particular amount at a particular time, and you wanted to know how much to deposit each month to get there? That situation arises quite often -- any time that you are trying to save for a particular goal, you would ask yourself this question. Let's see an example to illustrate how this works.

Find the size of the monthly payments necessary to accumulate \(\$1\) million at \(5\%\) annual interest compounded monthly on an ordinary annuity over \(40\) years.

###### Solution

This problem refers to a situation earlier in this section -- you'd like to save \(\$1,000,000\) in \(40\) years. How do you get there? In this case, we'll use the ordinary annuity formula to find the value of \(pymt\), which will represent the monthly payment in this case.

The other variables in the ordinary annuity formula can also be identified in this question. We have that deposits are made monthly, so we'll use \(n = 12\). We are interested in the value after \(40\) years, so we'll use \(t = 40\). The interest rate, as before, is \(5\%\), so we'll use \(r = .05\). Finally, we would like the balance to be \(\$1,000,000\). This means that \(B(40) = 1000000\).

Now, instead of finding the function as we did in the previous problem, we'll go ahead and plug in our values for \(B(t), r, n\), and \(t\). And then we'll solve for \(pymt\). Using the ordinary annuity formula with our variables plugged in, we have:\[ 1000000 = \left( \dfrac{12 \cdot pymt}{.05} \right)\cdot \left(\left(1+\frac{.05}{12}\right)^{(12 \times 40)}- 1\right)\]

Let's start by simplifying on the right first. We can calculate the second set of parentheses and reduce the equation to: \[ 1000000 = \left( \dfrac{12 \cdot pymt}{.05} \right)\cdot (6.358417319)\]

Next, we'll divide both sides by the \(6.358417319\) value. We have: \[ \frac{1000000}{6.358417319} = \dfrac{12 \cdot pymt}{.05}\]

Next, we'll use Cross Multiplication to eliminate the fractions. We have: \[ 76.30100782 \times \; pymt = 50000\]

where the \(76.30100782\) value is obtained by multiplying \(12 \times 6.358417319\).

Finally, we'll use Division undoes Multiplication to find the value of \(pymt\): \[ pymt = \frac{50000}{76.30100782} = 655.2993391\]

That is, we should be making monthly payments of \(\$655.30\) to reach \(\$1,000,000\) after \(40\) years.

Spending \(\$655\) per month on a savings goal is a lot of money! However, for most people, that is still a more attainable goal to reach than an initial deposit of nearly \($136,000\) in a savings account to reach \($1,000,000\). Additionally, we can see the power of compound interest magnified here: if we simply deposited \(\$655.30\) into a shoebox every month for \(40\) years, we'd only end up with \(\$314,544\). That means that *almost \(\$700,000\) of additional money is generated through compound interest.* This is why many people use retirement accounts such as IRAs instead of simply setting aside money in a non-interest-bearing account.

We've seen how to save up for a particular goal. What happens we if we have a large amount of money, and would like to receive regular outflows from it, while the remaining amount in the account continues to accrue interest? We'll see how that works next.

## Payout Annuities

Let's say you have a large amount of money — say, the amount of money you've put in a retirement savings account — and you'd like to take out small amounts of money over time, while the remaining money in the account accrues compound interest. This is a structure known as a *payout annuity*, and it is the basic structure behind how most pension retirement accounts function. Here's the basic definition, which has a rather complicated formula.

A **payout annuity** is an account that pays out equal payments each period over a specified amount of time, while the remaining balance in the annuity continues to earn compound interest. The formula that describes payout annuities is: \[ P= \left( \dfrac{n \cdot pymt}{r} \right)\cdot \left( \dfrac{ \left(\left(1+\frac{r}{n}\right)^{(nt)}- 1\right)}{\left(1 + \frac{r}{n} \right)^{(nt)}} \right)\] where \(P\) is the principal amount invested in the account, \(r\) is the annual interest rate, \(n\) is the number of payouts each year, \(pymt\) is the amount paid out each period, and \(t\) is time in years.

Now, the right side of the equation above is quite intimidating. However, you can see some familiar parts cropping up -- the compound interest equation is there, as well as parts of the ordinary annuity formula. Basically, this formula compares the compound interest formula to the ordinary annuity formula, and calculates the difference. That is, we can use it to find the principal amount invested given a desired periodic payout.

Let's see a final example that shows how to use this formula.

Marissa has a payout annuity that pays out \(\$2,000\) each quarter. If the money is invested at \(7.2\%\) annual interest, compounded quarterly, and is intended to continue for \(15\) years, how much was originally deposited?

###### Solution

This is a payout annuity question, so we'll start by identifying the variables in the payout annuity formula. We have that \(pymt = 2000\) since there is a \)(\$2000\) payout each quarter. Since the payouts are quarterly, we have \(n = 4\). We have that the interest rate is \(7.2\%\), and thus we'll use \(r = .072\). Finally, we have that \(t = 15\), since this annuity lasts \(15\) years. Therefore, using the payout annuity formula, we have: \[ P= \left( \dfrac{4 \cdot 2000}{.072} \right)\cdot \left( \dfrac{ \left(\left(1+\frac{.072}{4}\right)^{(4 \times 15)}- 1\right)}{\left(1 + \frac{.072}{4} \right)^{(4 \times 15)}} \right)\]

As before, you can enter this into Wolfram Alpha as a complicated expression, but you may also want to know how to calculate this in another way. Indeed, it's not so bad. First, we'll simplify the expressions that look like \((1 + \frac{.072}{4})^{(4 \times 15)}\). We have \[ P= \left( \dfrac{4 \cdot 2000}{.072} \right)\cdot \left( \dfrac{ \left(\left(1.018\right)^{60}- 1\right)}{\left(1.018 \right)^{60}} \right)\]

Next we'll calculate the expressions of the form \((1.018)^{60}\) and the first set of parentheses. We get that \[ P = (111111.1111) \cdot \dfrac{\left(2.916531558 - 1 \right)}{2.916531558}\]

Next, we can subtract on the top of the fraction:\[ P = (111111.1111) \cdot \dfrac{1.916531558}{2.916531558}\]

Next, we'll simplify the fraction by dividing: \[ P = (111111.1111) \cdot (.6571269743)\]

Finally, multiplying, we get that: \[P = 73014.10825 \]

That is, Marissa will need to initially deposit \(\$73,014.11\) in the account to have the desired payout.

While this is a large amount of money, the result of having such an investment is a payout of \(\$2000\) each quarter for \(15\) years. That would be a total of \(\$120,000\) that is paid out. So, on an initial investment of \(\$73,014.11\), that is a very good deal!

## Exercises

Almost all of the answers on this homework are dollar values. In the middle of a problem, if you calculate a decimal, please do not round. However, at the end of the problem, please do round to two decimal places (the cents place).

- Linda is \(25\) when she opens a retirement savings annuity. She deposits \(\$100\) each month into the account, which has \(2.4\%\) annual interest compounded monthly.
- Write a function \(B(t)\) that describes the balance in Linda's account \(t\) years after she opens it.
- Using the function above, determine the amount of money in the account when Linda is \(65\).

- Kai needs to have \(\$800,000\) saved by the time that they retire. They plan to retire in \(45\) years, and they find an ordinary annuity with \(1.2\%\) annual interest compounded monthly. What size monthly deposits will they need to make to reach their goal?
- When Sarah retires, she wishes to make a retirement income of \(\$2500\) per month. She has found a payout annuity that has \(4.8\%\) annual interest compounded monthly. She'd like to have this income for \(30\) years after she retires. What amount will she need to initially deposit in the payout annuity to make this guaranteed retirement income?
- Jethro intends to retire in \(15\) years. He can afford to put \(\$250\) per month into an account that accrues \(1.8\%\) annual interest compounded monthly to save for his retirement.
- Is this situation better described by an ordinary annuity or a payout annuity? Explain in one sentence.
- Find out how much Jethro will have if he saves \(\$250\) per month using this account. Show your work.

- Jaime has just retired and has managed to save \(\$1,200,000\) for retirement. He plans to live on this money for \(40\) years. He has the money in an account out of which he plans to make regular monthly withdrawals while the remaining balance accrues \(3.6\%\) annual interest compounded monthly.
- Is this situation better described by an ordinary annuity or a payout annuity? Explain in one sentence.
- Find out how much Jaime will be able to withdraw monthly using this account. Show your work.