61 Consumer Math

We have to work with money every day. While balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, when we start saving, planning for retirement, or need a loan, we need more mathematics.

Simple Interest

Discussing interest starts with the principal, or amount your account starts with. This could be a starting investment, or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.

One-time simple interest is only common for extremely short-term loans. For longer term loans, it is common for interest to be paid on a daily, monthly, quarterly, or annual basis. In that case, interest would be earned regularly. For example, bonds are essentially a loan made to the bond issuer (a company or government) by you, the bond holder. In return for the loan, the issuer agrees to pay interest, often annually. Bonds have a maturity date, at which time the issuer pays back the original bond value.

We can generalize this idea of simple interest over time.

Compound Interest

With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.

Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?

The 3% interest is an annual percentage rate (APR)—the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\displaystyle\frac{3\%}{12}\\[/latex] = 0.25% per month.

In the first month,

P₀ = $1000
r = 0.0025 (0.25%)
I = $1000 (0.0025) = $2.50
A = $1000 + $2.50 = $1002.50

In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.

In the second month,

P₀ = $1002.50
I = $1002.50 (0.0025) = $2.51 (rounded)
A = $1002.50 + $2.51 = $1005.01

Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding of interest gives us.

Calculating out a few more months:

To find an equation to represent this, if P_m represents the amount of money after m months, then we could write the recursive equation:

P₀ = $1000
P_m = (1 + 0.0025)P_m–1

You probably recognize this as the recursive form of exponential growth. If not, we could go through the steps to build an explicit equation for the growth:

P₀ = $1000
P₁ = 1.0025P₀ = 1.0025 (1000)
P₂ = 1.0025P₁ = 1.0025 (1.0025 (1000)) = 1.0025²(1000)
P₃ = 1.0025P₂ = 1.0025 (1.0025²(1000)) = 1.0025³(1000)
P₄ = 1.0025P₃ = 1.0025 (1.0025³(1000)) = 1.0025⁴(1000)

Observing a pattern, we could conclude

P_m = (1.0025)^m($1000)

Notice that the $1000 in the equation was P₀, the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year. Generalizing our result, we could write

[latex]\displaystyle{P_m}=P_0\left({1}+\frac{r}{k}\right)^m\\[/latex]

In this formula:

m is the number of compounding periods (months in our example)
r is the annual interest rate
k is the number of compounds per year.

While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If N is the number of years, then m = Nk. Making this change gives us the standard formula for compound interest.

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 most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.

Let us compare the amount of money earned from compounding against the amount you would earn from simple interest

As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.

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]\displaystyle{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]\displaystyle{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.

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]\displaystyle{P}_m=\left(1+\frac{0.06}{12}\right)P_{m-1}+100=(1.005)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}+\cdot\cdot\cdot{+}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. 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:

[latex]1.005{P}_m=1.005\left(100(1.005)^{m-1}+100(1.005)^{m-2}+100(1.005)+\cdot\cdot\cdot{+}100(1.005)+100\right)\\[/latex]

Distributing on the right side of the equation gives

[latex]1.005{P}_m=100(1.005)^m+100(1.005)^{m-1}+\cdot\cdot\cdot{+}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]1.005{P}_m=100(1.005)^m+100(1.005)^{m-1}+\cdot\cdot\cdot{+}100(1.005)\\[/latex]
[latex]P_m=100(1.005)^{m-1}+\cdot\cdot\cdot{+}100(1.005)+100\\[/latex]

Almost all the terms cancel on the right hand side when we subtract, leaving

[latex]1.005{P}_m-P_m=100(1.005)^m-100\\[/latex]

Solving for Pm

[latex]0.005P_m=100{\big(}(1.005)^m-1{\big)}\\[/latex]
[latex]\displaystyle{P_m}=\frac{100{\big(}(1.005)^m-1{\big)}}{0.005}\\[/latex]

Replacing m months with 12N, where N is measured in years, gives

[latex]\displaystyle{P_N}=\frac{100{\big(}(1.005)^{12N}-1{\big)}}{0.005}\\[/latex]

Recall 0.005 was [latex]\displaystyle\frac{r}{k}\\[/latex] and 100 was the deposit d. 12 was k, the number of deposit each year. Generalizing this result, we get the saving annuity formula.

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.

Payout 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.

Like with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.

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.

Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.

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?”

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

Which Equation to Use?

When presented with a finance problem (on an exam or in real life), you’re usually not told what type of problem it is or which equation to use.  Here are some hints on deciding which equation to use based on the wording of the problem.

The easiest types of problem to identify are loans.  Loan problems almost always include words like: “loan,” “amortize” (the fancy word for loans), “finance (a car),” or “mortgage” (a home loan).  Look for these words.  If they’re there, you’re probably looking at a loan problem.  To make sure, see if you’re given what your monthly (or annual) payment is, or if you’re trying to find a monthly payment.

If the problem is not a loan, the next question you want to ask is: “Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?”  If you’re letting the money sit in the account with nothing but interest changing the balance, then you’re looking at a compound interest problem.  The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest.

If you’re making regular payments or withdrawals, the next questions is: “Am I putting money into the account, or am I pulling money out?”  If you’re putting money into the account on a regular basis (monthly/annually/quarterly) then you’re looking at a basic Annuity problem.  Basic annuities are when you are saving money.  Usually in an annuity problem, your account starts empty, and has money in the future.

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.

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 to determine what approach will best allow you to solve the problem.

Solving for Time

Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level.

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.