Chapter 3: Math of Finance
Learning Outcomes
- 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
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.
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.
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.
Solution:
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]{{P}_{m}}=\left(1+\frac{0.06}{12}\right){{P}_{m-1}}+100=\left(1.005\right){{P}_{m-1}}+100[/latex]
Assuming we start with an empty account, we can begin using this relationship:
[latex]P_0=0[/latex]
[latex]P_1=(1.005)P_0+100=100[/latex]
[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[/latex]
[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[/latex]
Continuing this pattern, after m deposits, we’d have saved:
[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[/latex]
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. 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]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]
Distributing on the right side of the equation gives
[latex]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]
Now we’ll line this up with like terms from our original equation, and subtract each side
[latex]\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]
Almost all the terms cancel on the right hand side when we subtract, leaving
[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex]
Factor [latex]P_m[/latex] out of the terms on the left side.
[latex]\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]
Solve for Pm
[latex]\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]
Replacing m months with 12N, where N is measured in years, gives
[latex]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-1\right)}{0.005}[/latex]
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]P_{N}=\frac{d\left(\left(1+\frac{r}{k}\right)^{Nk}-1\right)}{\left(\frac{r}{k}\right)}[/latex]
- 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?
Solution:
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]P_{N}=\frac{d\left(\left(1+\frac{r}{k}\right)^{Nk}-1\right)}{\left(\frac{r}{k}\right)}[/latex]
[latex]P_{20}=\frac{100\left(\left(1+\frac{0.06}{12}\right)^{240}-1\right)}{\left(\frac{0.05}{12}\right)}[/latex]
(Notice we multiplied N times k before putting it into the exponent. It is a simple computation and will make it easier to enter into Desmos:
The account will grow to $46,204.09 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 $46,204.09 – $24,000 = $22,204.09.
This example is explained in detail here. Notice that each part was worked out separately and rounded. The answer above where we used Desmos is more accurate as the rounding was left until the end. You can work the problem either way, but be sure if you do follow the video below that you round out far enough for an accurate answer.
Try It
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?
Solution:
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]P_{10}=\frac{5\left(\left(1+\frac{0.03}{365}\right)^{365*10}-1\right)}{\frac{0.03}{365}}=21,282.07[/latex]
Financial planners typically recommend that you have a certain amount of savings upon retirement. If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.
Example
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.
| r = 0.08 | 8% annual rate |
| k = 12 | since we’re depositing monthly |
| N = 30 | 30 years |
| P30 = $200,000 | The amount we want to have in 30 years |
In this case, we’re going to have to set up the equation, and solve for d.
[latex]\begin{align}&200,000=\frac{d\left({{\left(1+\frac{0.08}{12}\right)}^{30(12)}}-1\right)}{\left(\frac{0.08}{12}\right)}\\&200,000=\frac{d\left({{\left(1.00667\right)}^{360}}-1\right)}{\left(0.00667\right)}\\&200,000=d(1491.57)\\&d=\frac{200,000}{1491.57}=\$134.09 \\\end{align}[/latex]
So you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.
View the solving of this problem in the following video.
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.
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]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex]
- 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?
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]\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]
You will need to have $139,600 in your account when you retire.
The problem above was worked in sections, but remember you can entire the entire problem all at once in your Desmos calculator and avoid rounding.
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.
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
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.
| r = 0.08 | 8% annual rate |
| k = 12 | since we’re withdrawing monthly |
| N = 30 | 30 years |
| P0 = $500,000 | we are beginning with $500,000 |
In this case, we’re going to have to set up the equation, and solve for d.
[latex]\begin{align}&500,000=\frac{d\left(1-{{\left(1+\frac{0.08}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.08}{12}\right)}\\&500,000=\frac{d\left(1-{{\left(1.00667\right)}^{-360}}\right)}{\left(0.00667\right)}\\&500,000=d(136.232)\\&d=\frac{500,000}{136.232}=\$3670.21 \\\end{align}[/latex]
You would be able to withdraw $3,670.21 each month for 30 years.
A detailed walkthrough of this example can be viewed here.
Try It
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?
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]100,000=\frac{d\left(1-\left(1+\frac{0.04}{1}\right)^{-20*1}\right)}{\frac{0.04}{1}}[/latex]
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.
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]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex]
- 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?
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]\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]
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.
Click here to view this video.
Example
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 |
| P0 = $140,000 | the starting loan amount |
In this case, we’re going to have to set up the equation, and solve for d.
[latex]\begin{align}&140,000=\frac{d\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&140,000=\frac{d\left(1-{{\left(1.005\right)}^{-360}}\right)}{\left(0.005\right)}\\&140,000=d(166.792)\\&d=\frac{140,000}{166.792}=\$839.37 \\\end{align}[/latex]
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.
View more about this example here.
Try It
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?
Solution:
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]\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]
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.
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?
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 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]\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]
The loan balance with 10 years remaining on the loan will be $90,073.45.
This example is explained in this video:
Oftentimes 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
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?
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 |
| P0 = $180,000 | the starting loan amount |
We set up the equation and solve for d.
[latex]\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]
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]\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]
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:
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.
Attributions
This chapter contains material taken from Math in Society (on OpenTextBookStore) by David Lippman, and is used under a CC Attribution-Share Alike 3.0 United States (CC BY-SA 3.0 US) license.
This chapter contains material taken from of Math for the Liberal Arts (on Lumen Learning) by Lumen Learning, and is used under a CC BY: Attribution license.