5.1 Mathematics of Finance

Chapter Overview

In this chapter, you will learn to:

1. Solve financial problems that involve simple interest.
2. Solve problems involving compound interest.
3. Find the future value of an annuity, and the amount of payments to a sinking fund.
4. Find the present value of an annuity, and an installment payment on a loan.

Simple Interest and Discount

SIMPLE INTEREST It costs to borrow money. The rent one pays for the use of money is called the interest. The amount of money that is being borrowed or loaned is called the principal or present value. Simple interest is paid only on the original amount borrowed. When the money is loaned out, the person who borrows the money generally pays a fixed rate of interest on the principal for the time period he keeps the money. Although the interest rate is often specified for a year, it may be specified for a week, a month, or a quarter, etc. The credit card companies often list their charges as monthly rates, sometimes it is as high as 1.5% a month.

DISCOUNTS AND PROCEEDS Banks often deduct the simple interest from the loan amount at the time that the loan is made. When this happens, we say the loan has been discounted. The interest that is deducted is called the discount, and the actual amount that is given to the borrower is called the proceeds. The amount the borrower is obligated to repay is called the maturity value.

Compound Interest

In the, we did problems involving simple interest. Simple interest is charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, the interest is paid (or charged) not only on the principal, but also on the past interest, and we say the interest is compounded.

Suppose we deposit $200 in an account that pays 8% interest. At the end of one year, we will have $$\$\text{200}+\$\text{200}\left(\text{.}\text{08}\right)=\$\text{200}\left(1+\text{.}\text{08}\right)=\$\text{216}$$.

Now suppose we put this amount, $216, in the same account. After another year, we will have $$\$\text{216}+\$\text{216}\left(\text{.}\text{08}\right)=\$\text{216}\left(1+\text{.}\text{08}\right)=\$\text{233}\text{.}\text{28}$$.

So an initial deposit of $200 has accumulated to $233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of $\$\text{16}\left(\text{.}\text{08}\right)=\$1\text{.}\text{28}$, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest.

Now suppose we leave this amount, $233.28, in the bank for another year, the final amount will be $$\$\text{233}\text{.}\text{28}+\$\text{233}\text{.}\text{28}\left(\text{.}\text{08}\right)=\$\text{233}\text{.}\text{28}\left(1+\text{.}\text{08}\right)=\$\text{251}\text{.}\text{94}$$.

Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.

After one year, we had

$$\$\text{200}\left(1+\text{.}\text{08}\right)=\$\text{216}$$

After two years, we had

$$\$\text{216}\left(1+\text{.}\text{08}\right)$$

But $\$\text{216}=\$\text{200}\left(1+\text{.}\text{08}\right)$, therefore, the above expression becomes

$$\begin{array}{|c|}\hline \$\text{200}\left(1+\text{.}\text{08}\right)\\ \hline\end{array}\left(1+\text{.}\text{08}\right)=\$\text{233}\text{.}\text{28}$$

After three years, we get

$$\begin{array}{|c|}\hline \$\text{200}\left(1+\text{.}\text{08}\right)\left(1+\text{.}\text{08}\right)\\ \hline\end{array}\left(1+\text{.}\text{08}\right)$$

Which can be written as

$$\$\text{200}{\left(1+\text{.}\text{08}\right)}^3=\$\text{251}\text{.}\text{94}$$

Suppose we are asked to find the total amount at the end of 5 years, we will get

$$\$\text{200}{\left(1+\text{.}\text{08}\right)}^5=\$\text{293}\text{.}\text{87}$$

We summarize as follows:

Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to one-fourth of 8%, or 2%.

Now if we deposit $200 in the bank, after one quarter we will have $\$\text{200}\left(1+\frac{\text{.}\text{08}}{4}\right)$ or $204.

After two quarters, we will have $\$\text{200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^2$ or $208.08.

After one year, we will have $\$\text{200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^4$ or $216.49.

After three years, we will have $\$\text{200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^{\text{12}}$ or $253.65, etc.

$$\begin{array}{ll}\text{The original amount}& \text{\$200}=\text{\$200}\\ \text{The amount after one quarter}& \text{\$200}\left(1+\frac{\text{.}\text{08}}{4}\right)=\$\text{204}\\ \text{The amount after two quarters}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^2=\$\text{208}\text{.}\text{08}\\ \text{The amount after one year}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^4=\text{216}\text{.}\text{49}\\ \text{The amount after two years}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^8=\$\text{234}\text{.}\text{31}\\ \text{The amount after three years}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^{\text{12}}=\$\text{253}\text{.}\text{65}\\ \text{The amount after five years}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^{\text{20}}=\$\text{297}\text{.}\text{19}\\ \text{The amount after}t\text{years}& \text{\$200}{\left(1+\frac{\text{.}\text{08}}{4}\right)}^{\mathrm{4t}}\end{array}$$

Therefore, if we invest a lump-sum amount of $P$ dollars at an interest rate $r$, compounded $n$ times a year, then after $t$ years the final amount is given by

$$A=P{\left(1+\frac{r}{n}\right)}^{\text{nt}}$$

Interest can be compounded yearly, semiannually, quarterly, monthly, daily, hourly, minutely, and even every second. But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts. When interest is compounded "infinitely many times", we say that the interest is compounded continuously. Our next objective is to derive a formula to solve such problems, and at the same time put things in proper perspective.

Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be $\$1\left(1+1\right)=\$2$.

If the interest is compounded semiannually, in one year we will have $\$1{\left(1+1/2\right)}^2=\$2\text{.}\text{25}$

If the interest is compounded quarterly, in one year we will have $\$1{\left(1+1/4\right)}^4=\$2\text{.}\text{44}$, etc.

We show the results as follows:

We have noticed that the $1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name "$e$" after the great mathematician Euler.

In mathematics, we say that as $n$ becomes infinitely large the expression ${\left(1+\frac{1}{n}\right)}^n$ equals $e$.

Therefore, it is natural that the number $e$ play a part in continuous compounding. It can be shown that as $n$ becomes infinitely large the expression ${\left(1+\frac{r}{n}\right)}^{\text{nt}}=e^{\text{rt}}$.

Therefore, it follows that if we invest $\$P$ at an interest rate $r$ per year, compounded continuously, after $t$ years the final amount will be given by $A=P\cdot e^{\text{rt}}$.

Next we learn a common-sense rule to be able to readily estimate answers to some finance as well as real-life problems. We consider the following problem.

By doing a few similar calculations we can construct a table like the one below.

The pattern in the table introduces us to the law of 70.

It is a good idea to familiarize yourself with the law of 70, as it can help you to estimate many problems mentally.

We summarize the concepts learned in this chapter in the following table:

Annuities and Sinking Funds

In and, we did problems where an amount of money was deposited lump sum in an account and was left there for the entire time period. Now we will do problems where timely payments are made in an account. When a sequence of payments of some fixed amount are made in an account at equal intervals of time, we call that an annuity. And this is the subject of this section.

To develop a formula to find the value of an annuity, we will need to recall the formula for the sum of a geometric series.

A geometric series is of the form: $a+\text{ar}+{\text{ar}}^2+{\text{ar}}^3+\text{.}\text{.}\text{.}+{\text{ar}}^n$.

The following are some examples of geometric series.

$$3+6+\text{12}+\text{24}+\text{48}$$

$$2+6+\text{18}+\text{54}+\text{162}$$

$$\text{37}+3\text{.}7+\text{.}\text{37}+\text{.}\text{037}+\text{.}\text{0037}$$

In a geometric series, each subsequent term is obtained by multiplying the preceding term by a number, called the common ratio. And a geometric series is completely determined by knowing its first term, the common ratio, and the number of terms.

In the example, $$a+\text{ar}+{\text{ar}}^2+{\text{ar}}^3+\text{.}\text{.}\text{.}+{\text{ar}}^{n-1}$$ the first term of the series is $a$, the common ratio is $r$, and the number of terms are $n$.

In your algebra class, you developed a formula for finding the sum of a geometric series. The formula states that the sum of a geometric series is

$$\frac{a\left[r^n-1\right]}{r-1}$$

We will use this formula to find the value of an annuity.

Consider the following example.

When the payments are made at the end of each period rather than at the beginning, we call it an ordinary annuity.

If the payment is made at the beginning of each period, rather than at the end, we call it an annuity due. The formula for the annuity due can be derived in a similar manner. Reconsider Example 18, with the change that the deposits be made at the beginning of each month.

So, in the case of an annuity due, to find the future value, we increase the number of periods $n$ by 1, and subtract one payment.

$$\text{The Future Value of an Annuity due}=\frac{m\left[{\left(1+r/n\right)}^{\text{nt}+1}-1\right]}{r/n}-m$$

Most of the problems we are going to do in this chapter involve ordinary annuity, therefore, we will down play the significance of the last formula. We mentioned the last formula only for completeness.

Finally, it is the author's wish that the student learn the concepts in a way that he or she will not have to memorize every formula. It is for this reason formulas are kept at a minimum. But before we conclude this section we will once again mention one single equation that will help us find the future value, as well as the sinking fund payment.

Present Value of an Annuity and Installment Payment

In, we learned to find the future value of a lump sum, and in, we learned to find the future value of an annuity. With these two concepts in hand, we will now learn to amortize a loan, and to find the present value of an annuity.

Let us consider the following problem.

We now consider another problem that involves the same logic.

Miscellaneous Application Problems

We have already developed the tools to solve most finance problems. Now we use these tools to solve some application problems.

One of the most common problems deals with finding the balance owed at a given time during the life of a loan. Suppose a person buys a house and amortizes the loan over 30 years, but decides to sell the house a few years later. At the time of the sale, he is obligated to pay off his lender, therefore, he needs to know the balance he owes. Since most long term loans are paid off prematurely, we are often confronted with this problem. Let us consider an example.

Most of the other applications in this section's problem set are reasonably straight forward, and can be solved by taking a little extra care in interpreting them. And remember, there is often more than one way to solve a problem.

Classification of Finance Problems

We'd like to remind the reader that the hardest part of solving a finance problem is determining the category it falls into. So in this section, we will emphasize the classification of problems rather than finding the actual solution.

We suggest that the student read each problem carefully and look for the word or words that may give clues to the kind of problem that is presented. For instance, students often fail to distinguish a lump-sum problem from an annuity. Since the payments are made each period, an annuity problem contains words such as each, every, per etc.. One should also be aware that in the case of a lump-sum, only a single deposit is made, while in an annuity numerous deposits are made at equal spaced time intervals.

Students often confuse the present value with the future value. For example, if a car costs $15,000, then this is its present value. Surely, you cannot convince the dealer to accept $15,000 in some future time, say, in five years. Recall how we found the installment payment for that car. We assumed that two people, Mr. Cash and Mr. Credit, were buying two identical cars both costing $15, 000 each. To settle the argument that both people should pay exactly the same amount, we put Mr. Cash's cash of $15,000 in the bank as a lump-sum and Mr. Credit's monthly payments of $x$ dollars each as an annuity. Then we make sure that the future values of these two accounts are equal. As you remember, at an interest rate of 9%

the future value of Mr. Cash's lump-sum was $\$\text{15},\text{000}{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}$, and

the future value of Mr. Credit's annuity was $\frac{x\left[{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}-1\right]}{\text{.}\text{09}/\text{12}}$.

To solve the problem, we set the two expressions equal and solve for $x$.

The present value of an annuity is found in exactly the same way. For example, suppose Mr. Credit is told that he can buy a particular car for $311.38 a month for five years, and Mr. Cash wants to know how much he needs to pay. We are finding the present value of the annuity of $311.38 per month, which is the same as finding the price of the car. This time our unknown quantity is the price of the car. Now suppose the price of the car is $y$, then

the future value of Mr. Cash's lump-sum is $y{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}$, and

the future value of Mr. Credit's annuity is $\frac{\$\text{311}\text{.}\text{38}\left[{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}-1\right]}{\text{.}\text{09}/\text{12}}$.

Setting them equal we get,

$$\begin{array}{l}y{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}=\frac{\$\text{311}\text{.}\text{38}\left[{\left(1+\text{.}\text{09}/\text{12}\right)}^{\text{60}}-1\right]}{\text{.}\text{09}/\text{12}}\\ y\left(1\text{.}\text{5657}\right)=\left(\$\text{311}\text{.}\text{38}\right)\left(\text{75}\text{.}\text{4241}\right)\\ y\left(1\text{.}\text{5657}\right)=\$\text{23},\text{485}\text{.}\text{57}\\ y=\$\text{15},\text{000}\text{.}\text{04}\end{array}$$

We now list six problems that form a basis for all finance problems. Further, we classify these problems and give an equation for the solution.