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# 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 [latex-display]\displaystyle{P}_{{m}}={\left({1}+\frac{{r}}{{k}}\right)}{P}_{{{m}-{1}}}[/latex-display] For a savings annuity, we simply need to add a deposit, d, to the account with each compounding period: [latex-display]\displaystyle{P}_{{m}}={\left({1}+\frac{{r}}{{k}}\right)}{P}_{{{m}-{1}}}+{d}[/latex-display] 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 [latex-display]\displaystyle{P}_{{m}}={\left({1}+\frac{{0.06}}{{12}}\right)}{P}_{{{m}-{1}}}+{100}={\left({1.005}\right)}{P}_{{{m}-{1}}}+{100}[/latex-display] Assuming we start with an empty account, we can begin using this relationship: P0 = 0 P1 = (1.005)P0 + 100 = 100 P2 = (1.005)P1 + 100 = (1.005)(100) + 100 = 100(1.005) + 100 P3 = (1.005)P2 + 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: Pm = 100(1.005)m – 1 + 100(1.005)m – 2 + L + 100(1.005) + 100 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 months 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! To simplify things, multiply both sides of the equation by 1.005: 1.005 Pm = 1.005(100(1.005)m – 1 + 100(1.005)m – 1 + L + 100(1.005) +100) Distributing on the right side of the equation gives 1.005 Pm = 100(1.005)m + 100(1.005)m – 1 + L + 100(1.005)2 + 100(1.005) Now we'll line this up with like terms from our original equation, and subtract each side
 $\displaystyle{1.005}{P}_{{m}}$ = $\displaystyle{100}{\left({1.005}\right)}^{{m}}$ + $\displaystyle{100}{\left({1.005}\right)}^{{{m}-{1}}}$ + $\displaystyle{L}$ + $\displaystyle{100}{\left({1.005}\right)}$ $\displaystyle{P}_{{m}}$ = $\displaystyle{100}{\left({1.005}\right)}^{{{m}-{1}}}$ + $\displaystyle{L}$ + $\displaystyle{100}{\left({1.005}\right)}$ +
Almost all the terms cancel on the right hand side when we subtract, leaving 1.005PmPm = 100(1.005)m – 100 Solving for Pm [latex-display]\displaystyle{0.005}{P}_{{m}}={100}{\left({\left({1.005}\right)}^{{m}}-{1}\right)}[/latex-display] [latex-display]\displaystyle{P}_{{m}}=\frac{{{100}{\left({\left({1.005}\right)}^{{m}}-{1}\right)}}}{{0.005}}[/latex-display] Replacing m months with 12N, where N is measured in years, gives [latex-display]\displaystyle{P}_{{N}}=\frac{{{100}{\left({\left({1.005}\right)}^{{{12}{N}}}-{1}\right)}}}{{0.005}}[/latex-display] Recall 0.005 was r/k and 100 was the deposit d. 12 was k, the number of deposit each year. Generalizing this result, we get the saving annuity formula.

## Annuity Formula

[latex-display]\displaystyle{P}_{{N}}=\frac{{{d}{\left({\left({1}+\frac{{r}}{{k}}\right)}^{{{N}{k}}}-{1}\right)}}}{{{\left(\frac{{r}}{{k}}\right)}}}[/latex-display] PN 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. Compound interest assumes that you put money in the account once and let it sit there earning interest. Compound interest: One deposit Annuity: Many deposits.

### Example 7

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? In this example,  The monthly deposit $\displaystyle{d}=\{100}$ 6% annual rate $\displaystyle{r}={0.06}$ 12 months in 1 year $\displaystyle{k}={12}$ We want the amount after 20 years $\displaystyle{N}={20}$ Putting this into the equation: [latex-display]\displaystyle{P}_{{20}}=\frac{{{100}{\left({\left({1}+\frac{{0.06}}{{12}}\right)}^{{{20}{\left({12}\right)}}}-{1}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{20}}=\frac{{{100}{\left({\left({1.005}\right)}^{{240}}-{1}\right)}}}{{{\left({0.005}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{20}}=\frac{{{100}{\left({3.310}-{1}\right)}}}{{{\left({0.005}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{20}}=\frac{{{100}{\left({2.310}\right)}}}{{{\left({0.005}\right)}}}=\${46200}[/latex-display] 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 8

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? In this example, We're looking for d.  8% annual rate $\displaystyle{r}={0.08}$ 12 months in a year $\displaystyle{k}={12}$ 30 years $\displaystyle{N}={30}$ The amount we want to have in 30 years $\displaystyle{P}_{{30}}=\{200},{000}$ In this case, we're going to have to set up the equation, and solve for d. [latex-display]\displaystyle{200},{000}=\frac{{{d}{\left({\left({1}+\frac{{0.08}}{{12}}\right)}^{{{30}{\left({12}\right)}}}-{1}\right)}}}{{{\left(\frac{{0.08}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{200},{000}=\frac{{{d}{\left({\left({1.00667}\right)}^{{360}}-{1}\right)}}}{{{0.00667}}}[/latex-display] [latex-display]\displaystyle{200},{000}={d}{\left({1491.57}\right)}[/latex-display] [latex-display]\displaystyle{d}=\frac{{{200},{000}}}{{{1491.57}}}=\${134.09}[/latex-display] So you would need to deposit $134.09 each month to have$200,000 in 30 years if your account earns 8% interest

## Payout Annuity Formula

[latex-display]\displaystyle{P}_{{0}}=\frac{{{d}{\left({1}-{\left({1}+\frac{{r}}{{k}}\right)}^{{-{N}{k}}}\right)}}}{{{\left(\frac{{r}}{{k}}\right)}}}[/latex-display] P0 is the balance in the account at the beginning (starting amount, or principal). d is the regular withdrawal (the amount you take out each year, each month, etc.) r is the annual interest rate (in decimal form. Example: 5% = 0.05) k is the number of compounding periods in one year. N is the number of years we plan to take withdrawals Like with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.

## When Do You Use This?

Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest. Compound interest: One deposit Annuity: Many deposits. Payout Annuity: Many withdrawals

### Example 9

After retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement account. The account earns 6% interest. How much will you need in your account when you retire? In this example,  The monthly withdrawal $\displaystyle{d}=\{1000}$ 6% annual rate $\displaystyle{r}={0.06}$ 12 months in a year $\displaystyle{k}={12}$ We're taking withdrawals for 20 years $\displaystyle{N}={20}$ We're looking for P0; how much money needs to be in the account at the beginning. Putting this into the equation: [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{20}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1.005}\right)}^{{-{{240}}}}\right)}}}{{{\left({0.005}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{0.302}\right)}}}{{{\left({0.005}\right)}}}=\${139},{600}[/latex-display] You will need to have $139,600 in your account when you retire. Notice that you withdrew a total of$240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is$240,000 – $139,600 =$100,400 in interest.

## Evaluating Negative Exponents on Your Calculator

With these problems, you need to raise numbers to negative powers. Most calculators have a separate button for negating a number that is different than the subtraction button. Some calculators label this (–) , some with +/–. The button is often near the = key or the decimal point. If your calculator displays operations on it (typically a calculator with multiline display), to calculate 1.005–240 you'd type something like: 1.005 ^ (–) 240 If your calculator only shows one value at a time, then usually you hit the (–) key after a number to negate it, so you'd hit: 1.005 yx 240 (–) = Give it a try — you should get 1.005–240 = 0.302096

### Example 10

You know you will have $500,000 in your account when you retire. You want to be able to take monthly withdrawals from the account for a total of 30 years. Your retirement account earns 8% interest. How much will you be able to withdraw each month? In this example, We're looking for d.  8% annual rate $\displaystyle{r}={0.08}$ 12 months in a year $\displaystyle{k}={12}$ 30 years $\displaystyle{N}={30}$ We are beginning with$500,000 $\displaystyle{P}_{{0}}=\{500},{000}$
In this case, we're going to have to set up the equation, and solve for d. [latex-display]\displaystyle{500},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.08}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.08}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{500},{000}=\frac{{{d}{\left({1}-{\left({1.00667}\right)}^{{-{360}}}\right)}}}{{{\left({0.00667}\right)}}}[/latex-display] [latex-display]\displaystyle{500},{000}={d}{\left({136.232}\right)}[/latex-display] [latex-display]\displaystyle{d}=\frac{{{500},{000}}}{{{136.232}}}=\${3670.21}[/latex-display] You would be able to withdraw$3,670.21 each month for 30 years.

### Example 12

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? In this example, We're looking for d.  6% annual rate $\displaystyle{r}={0.06}$ 12 months in a year $\displaystyle{k}={12}$ 30 years $\displaystyle{N}={30}$ Starting loan amount $\displaystyle{P}_{{0}}=\{140},{000}$ In this case, we're going to have to set up the equation, and solve for d. [latex-display]\displaystyle{140},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{140},{000}=\frac{{{d}{\left({1}-{\left({1.005}\right)}^{{-{360}}}\right)}}}{{{\left({0.005}\right)}}}[/latex-display] [latex-display]\displaystyle{140},{000}={d}{\left({166.792}\right)}[/latex-display] [latex-display]\displaystyle{d}=\frac{{{140},{000}}}{{{166.792}}}=\${839.37}[/latex-display] 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.

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 13 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? 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 P0 when  The monthly loan payment $\displaystyle{d}=\{1},{000}$ 6% annual rate $\displaystyle{r}={0.06}$ 12 months in a year $\displaystyle{k}={12}$ 10 years $\displaystyle{N}={10}$ [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{10}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1.005}\right)}^{{-{{120}}}}\right)}}}{{{\left({0.005}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{0.5496}\right)}}}{{{\left({0.005}\right)}}}=\${90},{073.45}[/latex-display] The loan balance with 10 years remaining on the loan will be $90,073.45 Often times answering remaining balance questions requires two steps: 1. Calculating the monthly payments on the loan 2. Calculating the remaining loan balance based on the remaining time on the loan ### Example 14 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? First we will calculate their monthly payments. We're looking for d.
 4% annual rate $\displaystyle{r}={0.04}$ 12 months in a year $\displaystyle{k}={12}$ 30 years $\displaystyle{N}={30}$ Starting loan amount $\displaystyle{P}_{{0}}=\{180},{000}$
We set up the equation and solve for d. [latex-display]\displaystyle{180},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.04}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.04}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{180},{000}=\frac{{{d}{\left({1}-{\left({1.00333}\right)}^{{-{360}}}\right)}}}{{{\left({0.00333}\right)}}}[/latex-display] [latex-display]\displaystyle{180},{000}={d}{\left({209.562}\right)}[/latex-display] [latex-display]\displaystyle{d}=\frac{{{180},{000}}}{{{209.562}}}=\${858.93}[/latex-display] 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.  Monthly loan payment $\displaystyle{d}=\{858.93}$ 4% annual rate $\displaystyle{r}={0.04}$ 12 months in a year $\displaystyle{k}={12}$ 25 years $\displaystyle{N}={25}$ [latex-display]\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{\left({1}+\frac{{0.04}}{{12}}\right)}^{{-{25}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.04}}{{12}}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{\left({1.00333}\right)}^{{-{300}}}\right)}}}{{{\left({0.00333}\right)}}}[/latex-display] [latex-display]\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{0.369}\right)}}}{{{\left({0.00333}\right)}}}=\${155},{793.91}[/latex-display] The loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91 Over that 5 years, the couple has paid off$180,000 – $155,793.91 =$24,206.09 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 –$24,206.09 = $27,329.71 of what they have paid so far has been interest. ### Try it Now Answers 2.  Daily deposit $\displaystyle{d}=\{5}$ 3% annual rate $\displaystyle{r}={0.03}$ 365 days in a year $\displaystyle{k}={365}$ 10 years $\displaystyle{N}={10}$ $\displaystyle{P}_{{10}}=\frac{{{5}{\left({\left({1}+\frac{{0.03}}{{365}}\right)}^{{{365}\times{10}}}-{1}\right)}}}{{\frac{{0.03}}{{365}}}}=\{21},{282.07}$ We would have deposited a total of$5 × 365 × 10 = $18,250, so$3,032.07 is from interest. 3.
 $\displaystyle{d}={u}{n}{k}{n}{o}{w}{n}$ 4% annual rate $\displaystyle{r}={0.04}$ 1 per year $\displaystyle{k}={1}$ 20 years $\displaystyle{N}={20}$ Starting with $100,000 $\displaystyle{P}_{{0}}={100},{000}$ $\displaystyle{100},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.04}}{{1}}\right)}^{{-{20}\times{1}}}\right)}}}{{\frac{{0.04}}{{1}}}}$ Solving for d gives$7,358.18 each year that they can give in scholarships. It is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely. If this donor had specified that, $100,000(0.04) =$4,000 a year would have been available. 4.
 $\displaystyle{d}={u}{n}{k}{n}{o}{w}{n}$ 16% annual rate $\displaystyle{r}={0.16}$ 12 months in a year $\displaystyle{k}={12}$ 2 years $\displaystyle{N}={2}$ Starting with $3,000 $\displaystyle{P}_{{0}}={3000}$ $\displaystyle{3000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.16}}{{12}}\right)}^{{-{2}\times{12}}}\right)}}}{{\frac{{0.16}}{{12}}}}$ 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.
David Lippman, Math in Society, "Finance," licensed under a CC BY-SA 3.0 license.