We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > Math for Liberal Arts: Co-requisite Course

Annuities and Loans

  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. In this section, we will explore the math behind specific kinds of accounts that gain interest over time, like retirement accounts. We will also explore how mortgages and car loans, called installment loans, are calculated.  

Learning Objectives

The learning outcomes for this section include:
  • Calculate the balance on an annuity after a specific amount of time
  • Discern between compound interest, annuity, and payout annuity given a finance scenario
  • Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan
  • Determine which equation to use for a given scenario
  • Solve a financial application for time

Savings 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. Glass jar labeled An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship

[latex]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}[/latex]

For a savings annuity, we simply need to add a deposit, d, to the account with each compounding period:

[latex]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}+d[/latex]

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.

Example

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. Write an explicit formula that represents this scenario.

Answer: 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]{{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: [latex-display]P_0=0[/latex-display] [latex-display]P_1=(1.005)P_0+100=100[/latex-display] [latex-display]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[/latex-display] [latex-display]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[/latex-display] Continuing this pattern, after m deposits, we’d have saved: [latex-display]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[/latex-display] 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 -2 months. The last month's deposit (L) would have earned only one month's 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: [latex-display]1.005{{P}_{m}}=1.005\left(100{{\left(1.005\right)}^{m-1}}+100{{\left(1.005\right)}^{m-2}}+\cdots+100(1.005)+100\right)[/latex-display] Distributing on the right side of the equation gives [latex-display]1.005{{P}_{m}}=100{{\left(1.005\right)}^{m}}+100{{\left(1.005\right)}^{m-1}}+\cdots+100{{(1.005)}^{2}}+100(1.005)[/latex-display] Now we’ll line this up with like terms from our original equation, and subtract each side [latex-display]\begin{align}&\begin{matrix}1.005{{P}_{m}}&=&100{{\left(1.005\right)}^{m}}+&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&{}\\{{P}_{m}}&=&{}&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&+100\\\end{matrix}\\&\\\end{align}[/latex-display] Almost all the terms cancel on the right hand side when we subtract, leaving [latex-display]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex-display] Factor [latex]P_m[/latex] out of the terms on the left side. [latex-display]\begin{array}{c}P_m(1.005-1)=100{{\left(1.005\right)}^{m}}-100\\(0.005)P_m=100{{\left(1.005\right)}^{m}}-100\end{array}[/latex-display] Solve for Pm [latex-display]\begin{align}&0.005{{P}_{m}}=100\left({{\left(1.005\right)}^{m}}-1\right)\\&\\&{{P}_{m}}=\frac{100\left({{\left(1.005\right)}^{m}}-1\right)}{0.005}\\\end{align}[/latex-display] Replacing m months with 12N, where N is measured in years, gives [latex-display]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-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 savings annuity formula.

Annuity Formula

[latex-display]P_{N}=\frac{d\left(\left(1+\frac{r}{k}\right)^{Nk}-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.

Examples

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?

Answer: In this example,

d = $100 the monthly deposit
r = 0.06 6% annual rate
k = 12 since we’re doing monthly deposits, we’ll compound monthly
N = 20  we want the amount after 20 years
Putting this into the equation: [latex-display]\begin{align}&{{P}_{20}}=\frac{100\left({{\left(1+\frac{0.06}{12}\right)}^{20(12)}}-1\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{20}}=\frac{100\left({{\left(1.005\right)}^{240}}-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(3.310-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(2.310\right)}{\left(0.005\right)}=\$46200 \\\end{align}[/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.

This example is explained in detail here. https://youtu.be/quLg4bRpxPA  

Try It Now

A 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?

Answer:

d = $5              the daily deposit r = 0.03           3% annual rate k = 365            since we’re doing daily deposits, we’ll compound daily N = 10             we want the amount after 10 years [latex-display]P_{10}=\frac{5\left(\left(1+\frac{0.03}{365}\right)^{365*10}-1\right)}{\frac{0.03}{365}}=21,282.07[/latex-display]

Try It Now

 

Solving For Time

We can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.

Example

If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?

Answer: This is a savings annuity problem since we are making regular deposits into the account.

d = $100 the monthly deposit
r = 0.03 3% annual rate
k = 12 since we’re doing monthly deposits, we’ll compound monthly
We don’t know N, but we want PN to be $10,000. Putting this into the equation: [latex]10,000=\frac{100\left({{\left(1+\frac{0.03}{12}\right)}^{N(12)}}-1\right)}{\left(\frac{0.03}{12}\right)}[/latex]                        Simplifying the fractions a bit [latex-display]10,000=\frac{100\left({{\left(1.0025\right)}^{12N}}-1\right)}{0.0025}[/latex-display] We want to isolate the exponential term, 1.002512N, so multiply both sides by 0.0025 [latex]25=100\left({{\left(1.0025\right)}^{12N}}-1\right)[/latex]                             Divide both sides by 100 [latex]0.25={{\left(1.0025\right)}^{12N}}-1[/latex]                                     Add 1 to both sides [latex]1.25={{\left(1.0025\right)}^{12N}}[/latex]                                        Now take the log of both sides [latex]\log\left(1.25\right)=\log\left({{\left(1.0025\right)}^{12N}}\right)[/latex]                              Use the exponent property of logs [latex]\log\left(1.25\right)=12N\log\left(1.0025\right)[/latex]                          Divide by 12log(1.0025) [latex]\frac{\log\left(1.25\right)}{12\log\left(1.0025\right)}=N[/latex]                                               Approximating to a decimal N = 7.447 years It will take about 7.447 years to grow the account to $10,000.

This example is demonstrated here: https://youtu.be/F3QVyswCzRo

Payout Annuities

Removing Money from Annuities

In the last section you learned about annuities. In an annuity, you start with nothing, put money into an account on a regular basis, and end up with money in your account. In this section, we will learn about a variation called a Payout Annuity. With a payout annuity, you start with money in the account, and pull money out of the account on a regular basis. Any remaining money in the account earns interest. After a fixed amount of time, the account will end up empty. Black and white aerial shot of hands exchanging money Payout annuities are typically used after retirement. Perhaps you have saved $500,000 for retirement, and want to take money out of the account each month to live on. You want the money to last you 20 years. This is a payout annuity. The formula is derived in a similar way as we did for savings annuities. The details are omitted here.

Payout Annuity Formula

[latex-display]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\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

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?

Answer: In this example,

d = $1000 the monthly withdrawal
r = 0.06 6% annual rate
k = 12 since we’re doing monthly withdrawals, we’ll compound monthly
N = 20  since were taking withdrawals for 20 years
We’re looking for P0: how much money needs to be in the account at the beginning. Putting this into the equation: [latex-display]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-20(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-{{\left(1.005\right)}^{-240}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-0.302\right)}{\left(0.005\right)}=\$139,600 \\\end{align}[/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.

View more about this problem in this video. https://youtu.be/HK2eRFH6-0U

Try It Now

 

Try It Now

A donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year?

Answer:

d = unknown r = 0.04                       4% annual rate k = 1                           since we’re doing annual scholarships N = 20                         20 years P0 = 100,000               we’re starting with $100,000 [latex-display]100,000=\frac{d\left(1-\left(1+\frac{0.04}{1}\right)^{-20*1}\right)}{\frac{0.04}{1}}[/latex-display]
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.

Loans

Conventional Loans

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. Hand holding green pen, which has just written 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

[latex-display]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex-display]
  • P0 is the balance in the account at the beginning (the principal, or amount of the loan).
  • d 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.
  • 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.
  • Compound interest: One deposit
  • Annuity: Many deposits
  • Payout Annuity: Many withdrawals
  • Loans: Many payments

Example

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?

Answer: 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 P0, the starting amount of the loan. [latex-display]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\\end{align}[/latex-display] 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.

Details of this example are examined in this video. https://youtu.be/5NiNcdYytvY  

Try It Now

Try It Now

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:

d =                               unknown r = 0.16                       16% annual rate k = 12                         since we’re making monthly payments N = 2                           2 years to repay P0 = 3,000                   we’re starting with a $3,000 loan [latex-display]\begin{array}{c}3000=\frac{{d}\left(1-\left(1+\frac{0.06}{12}\right)^{-2*12}\right)}{\frac{0.16}{12}}\\\\3000=20.42d\end{array}[/latex-display] Solve for d to get monthly payments of $146.89 Two years to repay means $146.89(24) = $3525.36 in total payments.  This means Janine will pay $3525.36 - $3000 = $525.36 in interest.

Calculating the 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. Pair of glasses resting on a Mortgage Loan Statement 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

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?

Answer: 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

d = $1,000 the monthly loan payment
r = 0.06 6% annual rate
k = 12 since we’re doing monthly payments, we’ll compound monthly
N = 10  since we’re making monthly payments for 10 more years
[latex-display]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-10(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\left(1-{{\left(1.005\right)}^{-120}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\left(1-0.5496\right)}{\left(0.005\right)}=\$90,073.45 \\\end{align}[/latex-display] The loan balance with 10 years remaining on the loan will be $90,073.45.

This example is explained in the following video: https://youtu.be/fXLzeyCfAwE
  Oftentimes 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

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?

Answer: 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
P0 = $180,000 the starting loan amount
We set up the equation and solve for d. [latex-display]\begin{align}&180,000=\frac{d\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&180,000=\frac{d\left(1-{{\left(1.00333\right)}^{-360}}\right)}{\left(0.00333\right)}\\&180,000=d(209.562)\\&d=\frac{180,000}{209.562}=\$858.93 \\\end{align}[/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.
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 for 25 more years
[latex-display]\begin{align}&{{P}_{0}}=\frac{858.93\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-25(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&{{P}_{0}}=\frac{858.93\left(1-{{\left(1.00333\right)}^{-300}}\right)}{\left(0.00333\right)}\\&{{P}_{0}}=\frac{858.93\left(1-0.369\right)}{\left(0.00333\right)}=\$155,793.91 \\\end{align}[/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.

More explanation of this example is available here: https://youtu.be/-J1Ak2LLyRo

Solving for Time

Recall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.

Try It Now

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:

d = $30                        The monthly payments r = 0.12                       12% annual rate k = 12                         since we’re making monthly payments P0 = 1,000                   we’re starting with a $1,000 loan We are solving for N, the time to pay off the loan [latex-display]1000=\frac{30\left(1-\left(1+\frac{0.12}{12}\right)^{-N*12}\right)}{\frac{0.12}{12}}[/latex-display]
Solving for N gives 3.396. It will take about 3.4 years to pay off the purchase.

FYI

Home loans are typically paid off through an amortization process, amortization refers to paying off a debt (often from a loan or mortgage) over time through regular payments. An amortization schedule is a table detailing each periodic payment on an amortizing loan as generated by an amortization calculator. If you want to know more, click on the link below to view the website “How is an Amortization Schedule Calculated?” by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.

Which Formula to Use?

  Now that we have surveyed the basic kinds of finance calculations that are used, it may not always be obvious which one to use when you are given a problem to solve. Here are some hints on deciding which equation to use, based on the wording of the problem.

Loans

The easiest types of problems to identify are loans.  Loan problems almost always include words like loan, amortize (the fancy word for loans), finance (i.e. a car), or mortgage (a home loan). Look for words like monthly or annual payment. 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.

Loans Formula

[latex-display]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex-display]
  • P0 is the balance in the account at the beginning (the principal, or amount of the loan).
  • d 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.
  • N is the length of the loan, in years.
 

Interest-Bearing Accounts

Accounts that gain interest fall into two main categories.  The first is on where you put money in an account once and let it sit, the other is where you make regular payments or withdrawals from the account as in a retirement account. Interest
  • 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. Look for words like compounded, or APY. Compound interest assumes that you put money in the account once and let it sit there earning interest.

COMPOUND INTEREST

[latex-display]P_{N}=P_{0}\left(1+\frac{r}{k}\right)^{Nk}[/latex-display]
  • PN is the balance in the account after N years.
  • P0 is the starting balance of the account (also called initial deposit, or principal)
  • r is the annual interest rate in decimal form
  • k is the number of compounding periods in one year
    • If the compounding is done annually (once a year), k = 1.
    • If the compounding is done quarterly, k = 4.
    • If the compounding is done monthly, k = 12.
    • If the compounding is done daily, k = 365.
  • The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest.

SIMPLE INTEREST OVER TIME

[latex-display]\begin{align}&I={{P}_{0}}rt\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\end{align}[/latex-display]
  • I is the interest
  • A is the end amount: principal plus interest
  • [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
  • r is the interest rate in decimal form
  • t is time
The units of measurement (years, months, etc.) for the time should match the time period for the interest rate.
  Annuities
  • 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. 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.

ANNUITY FORMULA

[latex-display]P_{N}=\frac{d\left(\left(1+\frac{r}{k}\right)^{Nk}-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.
 
  • 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. 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.

PAYOUT ANNUITY FORMULA

[latex-display]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\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
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 determine what approach will best allow you to solve the problem.

Try It Now

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?

    Answer: This is a payout annuity problem. She can pull out $1833.60 a quarter.

  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?

    Answer: This is a savings annuity problem. He will have saved up $7,524.11

  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? 

    Answer: This is compound interest problem. She would need to deposit $22,386.46.

  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? 

    Answer: This is a loans problem. She can buy $4,609.33 of new equipment

  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: This is a savings annuity problem. You would need to save $200.46 each month

In the following video, we present more examples of how to use the language in the question to determine which type of equation to use to solve a finance problem. https://youtu.be/V5oG7lLTECs In the next video example, we show how to solve a finance problem that has two stages, the first stage is a savings problem, and the second stage is a withdrawal problem. https://youtu.be/CNkvwMuLuis

Try It Now

Try It Now

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • Bad Credit? We can help!. Authored by: BookMama. Located at: https://www.flickr.com/photos/myloonyland/430367107. License: CC BY: Attribution. License terms: 2.0.
  • Math in Society. Authored by: David Lippman. Located at: http://www.opentextbookstore.com/mathinsociety/. License: CC BY-SA: Attribution-ShareAlike.
  • Retirement. Authored by: Tax Credits. Located at: https://www.flickr.com/photos/76657755@N04/7027606047/. License: CC BY: Attribution.
  • Savings Annuities. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Savings annuities - solving for the deposit. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Question ID 6691, 6688. Authored by: Lippman, David. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Determining The Value of an Annuity. Authored by: Sousa, James (Mathispower4u.com). License: CC BY: Attribution.
  • Determining The Value of an Annuity on the TI84. Authored by: Sousa, James (Mathispower4u.com). License: CC BY: Attribution.
  • Payment. Authored by: Sergio Marchi. Located at: https://www.flickr.com/photos/magic74/2411468004/. License: CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives.
  • Payout Annuities. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Payout annuity - solve for withdrawal. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Question ID 6681, 6687. Authored by: David Lippman. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Payout Annuity Formula - Part 1, Part 2. Authored by: Sousa, James (Mathispower4u.com). License: CC BY: Attribution.
  • approved-finance-business-loan-1049259. Authored by: InspiredImages. License: CC0: No Rights Reserved.
  • Car loan. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Calculating payment on a home loan. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Question ID 6684, 6685. Authored by: Lippman, David. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Identifying type of finance problem. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Multistage finance problem. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Question ID 67287, 67306, 67133. Authored by: Abert, Rex. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.