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Topics in Finance

Introduction

Many college students seem not to expect that they will ever reach middle age, let alone retirement, and that is probably natural. However, it is incorrect, and financial decisions made quite early in adulthoood can do much in determining the quality of one's life thereafter. In this worksheet, we will introduce some topics which are critical to everyone: some basics about everyday finance. All of us will have loans to pay off, and most of us will at some time be dealing with a very large, long-term loan in the form of a mortgage on a home or other building. All of us need to plan for saving, especially for the long run. The basic principles of how we can do best for ourselves in these situations are not complicated, and they can be studied quite easily once the appropriate formulas are known. Moreover, the vast majority of college students are not exposed to these topics in their regular mathematics curriculum, even though a strong argument can be made that they are as important as anything else one learns in school.

There is little in this worksheet which absolutely requires a computer algebra system for analysis; our point is to demonstrate the ease of studying a new subject when the tools we have learned are at our disposal.

Terminology and Typical Scenarios

We begin by discussing the standard vocabulary and notation which is used in elementary finance. Most precalculus texts discuss the idea of interest and compounding, and you can refer to your text and/or instructor for those most basic ideas. What we will discuss in this worksheet are applications involving annuities . By an annuity we mean a series of equal payments, each of an amount called R, made at equal time intervals, with a total of n time intervals (so there are also n payments). We will study only what are called ordinary annuities , in which the each payment, including the first, is made at the end of the corresponding time period. For instance, if you win a lottery prize in the form of ten years of monthly payments of $1000 each, we will assume that the first payment is made one month after the prize is first awarded, and each subsequent payment is made a month later, with a total of 120 payments. We would say R = 1000 (units: dollars) and n = 120 (units: months). Or if you are repaying a loan in the form of 100 weekly payments of $75 each, we would assume that the first payment is made one week after the loan is awarded and say that R = 75 (dollars) and n = 100 (weeks). (Other forms of annuities, such as those where the first payment is made immediately, require slight modifications of the formulas which will follow, but do not change the fundamental principles we will uncover.)

One might be tempted to describe the lottery prize above as a "one hundred twenty thousand dollar prize", since it is in the form of 120 payments of $1000 each. However, that is somewhat deceptive. Suppose we try to see how much money you have after the last payment is made. Let's assume that you put each payment into an account earning interest at a certain monthly rate i. The first thousand dollars earns interest on 119 occasions, the second deposit earns interest 118 times, and so on, with the final payment not having time to earn any interest. The sum now in the account, called the future value of your annuity and denoted by S, will not be $120,000. It will be more than that, and possibly much more, depending on the interest rate.

Let's look at this another way as well. The so-called "$120,000 prize" is not paid to you all at once at the beginning of the ten years, and if you requested such a deal, you wouldn't get it. If you deposited $120,000 all at once and let interest compound on it for ten years, you'd have considerably more money than in the scenario described in the previous paragraph. The people paying you the prize do not need to have all $120,000 on hand. They can have a smaller lump sum, one on which they earn interest, and of just the right size such that as they collect their interest and also pay out monthly thousands to you, their account is depleted exactly as of the last payment to you at the end of the ten years. The single amount which could be used to pay your prize in this fashion is called the present value of your annuity and is denoted by P. It will be less than $120,000, and could be much less, depending on the interest rate.

To summarize what we have discussed so far, we can say the following. Each annuity, a series of n equal payments of amount R each, can be viewed as being equivalent to a certain future value S or to a certain present value P. S and P are related as follows: if the amount P were deposited today and earned interest for n time periods at the prevailing rate i (per time period), the amount in the account at the end of n interest payments would be S. So all three items -- P dollars today, the series of n payments of R dollars each, and S dollars n time periods from now -- can be thought of as having equivalent financial value. This forms the basis for what follows, the primary concept behind typical loan calculations and savings plans.

When one obtains a loan, one is given an amount of money P today and in exchange promises to pay back an annuity of n payments of R dollars each. That annuity should have a present value equal to P. When one embarks on a savings plan designed to accumulate an amount of money S at some future date, one may do so by paying an annuity of n payments of R dollars each into an account or other investment vehicle. That annuity should have a future value equal to S.

All that remains is to know the exact relationship among the quantities S, P, R, n , and i .

Formulas for Annuities

We summarize the notation for annuities again here. An annuity is a series of n equal payments of R dollars each at n equal time intervals. Its future value is denoted by S (this would correspond to the total amount accumulated in a savings plan) and its present value is denoted by P (this would correspond to the amount borrowed in a loan, also called the original principal of the loan). In order to connect all these in a formula, we must also know the prevailing interest rate, which we call i . It is absolutely critical to note the following: the interest rate must be expressed for the time units involved in the annuity, and all time units must be made to match up in the formulas. If rate information is given as, say, "7.2% annually" and payments are made every month, then i equals (.072/12) per month, or i = .006. If the stated annual interest rate is 8% and payments are made quarterly -- every three months -- then i = .08/4 = .02. And should i turn out to be an "ugly" number with a nonterminating decimal expansion, then lots of accuracy should be kept. Answers in many problems need to be accurate to something like six significant figures, and exponentiation will occur in the formulas, so little or no rounding should be done. Of course, this is easy with a computer algebra system in use.

The formulas which relate the quantities discussed above are given just below. Their derivations are not particularly difficult but are not included here; if you are interested in learning about them, you can consult your instructor.

Future Value Formula For Annuities (FV)

> S=R*(((1+i)^n-1)/i);

>

Present Value Formula For Annuities (PV)

> P=R*((1-(1+i)^(-1*n))/i);

>

In a typical finance text, each of these formulas is now rearranged various ways to solve for each of the variables R and n . With Maple, we do not need to accumulate so many formulas; if a particular rearrangement is needed, we can have the system solve for R or n . For example, if we wish to solve for R in (FV) and n in (PV), we can do so quite simply, as illustrated here.

> solve(S=R*(((1+i)^n-1)/i),R);

>

> solve(P=R*((1-(1+i)^(-1*n))/i),n);

>

Notice in particular how nice a job the system has done on the second of those. Most precalculus students would find it a difficult task to solve accurately for the exponent n in either (FV) or (PV), while Maple is not fazed by such work. We'll use the various preceding formulas in a moment.

One might ask about solving for i in the preceding equations. Let's see what happens, using the future value formula (FV) as an example.

> solve(S=R*(((1+i)^n-1)/i),i);

>

The output from the system indicates that the desired value of i is a solution of a certain polynomial equation of degree n , and the system has not given us a formula for that solution. Indeed, if the positive integer n is anything other than a very small number, it is impossible to solve algebraically for i in the preceding equations. This is not a fault of this computer algebra system or any other; it is an unbreakable barrier of algebra that polynomial equations of degree greater than four do not have a general solution formula valid for all cases. However, if the values of S and R (or P and R) are known, we can obtain an (approximate) numerical solution by using the " fsolve " command.

With the preceding in hand, we are ready to proceed to some examples. The methods and formulas we use apply to a wide variety of savings plans and loans, from "Christmas club" savings at a bank to paying off a three-year car loan to much larger and longer-term commitments. We are primarily interested in the bigger fish, for they will have the most impact on your life. For savings plans, we are specifically interested in long-term projects such as retirement planning or saving for a child's college education. For loan repayment, we are most interested in home mortgages. We are fully aware that these may not sound particularly thrilling topics for most college students, and that they may not yet even seem relevant. However, they are in fact among the most important topics you can study in any subject, and knowing the basics of these as early as possible in your adult life can be of colossal importance in the quality of your life and your children's lives.

Examples

Example 1: An Automobile Loan You wish to borrow $10,000 to buy a car. The stated annual interest rate on your loan is to be 6.85%, and the term of the loan will be four years. If payments are made monthly, what will be the amount of each payment?

Solution This problem, like all involving loans, concerns an annuity and its present value. You are being given $10,000 today and are agreeing to reciprocate by paying back a four-year annuity having the same present value, $10,000. Since the payments are to be monthly, months are the time unit to use throughout the problem. (In all of the problems we will do, we will assume that interest is calculated at each time period. Annuities where the interest time periods differ from the payment periods are quite rare and require some modifications to the formulas we've presented.) For this loan, we know that P = 10,000 (dollars), n = 48 (months), and the monthly interest rate is found to be i = .0685/12. The monthly payment R (dollars) is the unknown. So we need to solve (PV) for R with numerical values given for the other variables. This can be done a number of ways: one can first solve algebraically for R and then substitute in the other values, one can go directly back to (PV), and so forth. We'll go directly back to (PV) to begin, solve it for R, and substitute in the given information.

> P=R*((1-(1+i)^(-1*n))/i);

> solve(%,R);

> subs(n=48,P=10000,i=.0685/12,%);

>

The monthly payment will be $238.77. Examples and calculations like this one will allow you to compare various loan plans, determine how much you can borrow, and so forth.

Example 2: Planning For College A couple wishes to have a sum of $100,000 available when their only child begins college. We will assume that they make quarterly payments into an account (that is, a payment every three months) and that the account earns interest at a stated annual rate of 7%. What is the amount of each payment if

(a) They begin as soon as they learn they are expecting a child and thus make 19 years of payments;

(b) They begin after the child passes the fifth grade and thus make 7 years of payments?

Solution This problem involves the future value S = 100,000 (dollars) of an annuity, so we will use (FV), and we again need to find the amount R of each payment. Note that the time periods here are quarter-years, so that the interest rate per time period is i = .07/4, and that in part (a) there are n = 76 time periods and that in part (b) there are n = 28 time periods. We begin with part (a).

> solve(S=R*(((1+i)^n-1)/i),R);

>

> subs(S=100000.,n=76.,i=.07/4,%);

>

In the first situation, when savings is done for 19 years, the quarterly payment is just $639.20. This is quite a reasonable sum; many couples can, with some budgeting, put aside some $200 per month. And the stunning sum of $100,000 is then available for their child's college education. To complete part (b), we choose to copy and re-execute the line in which (FV) is solved for R; this helps us in that we can use the " % " symbol, which refers to the most recently executed command, without fear of confusion. (What would happen if we skipped this step? You might wish to experiment and find out.)

> solve(S=R*(((1+i)^n-1)/i),R);

> subs(S=100000.,n=28.,i=.07/4,%);

>

Ouch! The quarterly payment has more than quadrupled, to a huge $2798.15, an amount that few families can afford. The change from 19 years of accumulation to 7 years of accumulation has caused a devastating effect in the savings plan. The moral, of course, is to begin saving as early as possible. The earliest payments have time to earn the most interest, and the final amounts resulting from each payment grow exponentially with time. The following example will also illustrate this, perhaps in more dramatic fashion.

Example 3: Planning For Retirement Two employees of XYZ Corporation are in the process of saving for their retirements. The corporation, like many, has a so-called "matching" plan wherein your contributions to your own retirement account are matched by equal employer contributions (provided you remain with the company for a considerable length of time). Employee Dave, a dedicated saver, has been contributing $250 a month beginning at age 23 and plans to retire at age 63. Employee Charlie lived the carefree single life for a while and did not begin contributing until he was 42, planning to retire at age 64. He contributes the same $250 per month. All of their contributions, and the company's matching ones, are being invested into an account which has a guaranteed annual rate of return of 8.08%. How much does each have at the time of his retirement?

Solution Again we have a situation involving future values. In this problem it is S which is unknown, so we can use (FV) in its original form. The time periods are months, and i = .0808/12. For each employee, each payment is R = 500 (dollars), counting both his contribution and his employer's. We begin with Charlie, whose 22 years of contributions give n = 264 (months).

> S=R*(((1+i)^n-1)/i);

> subs(R=500,n=264,i=.0808/12,%);

>

Charlie has saved $362,413.48. While this may sound like a great deal of money, we might consider the situation for a moment. If he is fully retired, his income from other sources may be very small indeed. If he intends to live in a style similar to that of his earlier years, that sum of money may not last long. Should he live well into his 80s, as many people now do, there may be some lean years ahead, and that is a sad and intolerable prospect.

Let us now move to employee Dave and make the corresponding calculation, in which R = 350 and n = 480.

> S=R*(((1+i)^n-1)/i);

> subs(R=500,n=480,i=.0808/12,%);

>

Start picking out a new set of golf clubs, Dave, because you are retiring on a nest egg of about $1,786,437. That is approximately five times as big as Charlie's, and you can probably live a comfortable life indefinitely by just living off the income obtained by investing this sum. For instance, if Dave puts that sum into secure government bonds paying just 5% annually, his annual income will be over $89,000 -- forever.

As you can see from the details of that example, the enormous difference between the two employees' plans is due entirely to the length of time each is saving. The forty years of saving dwarfs the shorter plan. To accumulate the same amount in the 22-year plan, Charlie would have to contribute well over $1000 per month.

The moral of the story in savings plans, as illustrated in Examples 2 and 3, is simply this: In any situation involving saving, the most important thing to do is to begin saving as early as possible . The earliest payments earn interest for the longest time and thus contribute the most to the final sum. This will be further illustrated in the following example.

Example 4: Accumulated Savings Graphed Over Time Let's return to Example 3 and look at exactly how an employee's accumulated savings grows over time. The actual value of R here is not really important. If we graph S for, say, R = 1, the same shape will occur for any other value of R, with its vertical coordinates changed by that factor. So let us create a function of n which describes how much has accumulated in a savings plan in which one dollar a month is contributed and earns interest at the same monthly rate i = .0808/12 as used in Example 3. Recall that n represents months, and note that a reasonable domain might be from zero (no savings plan at all) to, say, 600 (representing 50 years). We'll use the future value formula to define a function and then look at its graph.

Solution Letting SAV be the name of the function which describes the total amount accumulated, we have the following.

> SAV:=n->1*(((1+.0808/12)^n-1)/(.0808/12));

> plot(SAV,0..600,color=green,thickness=4);

>

The graph is a dramatic illustration of the exponential growth of the accumulated savings. For each dollar per month that is put into the savings plan, you have approximately $2000 saved after 400 months, over $4000 saved over 500 months, and so on.

Example 5: A Home Mortgage And Accelerated Payoff Plan A family wishes to borrow $180,000 to buy a home. To keep their payments as low as possible, they select a 30-year term (generally the longest term available), and the prevailing annual interest rate is 6.9%.

(a) What is the amount of each monthly payment? On the first payment, how much is interest,
and how much goes to decreasing the principal still owed?

(b) On almost all mortgages, the option exists to pay extra in any given month, with that extra
payment going entirely toward decreasing the principal. If this family pays $200 extra every
month, how long does it take them to pay off the entire loan?

Solution This problem involves an annuity with a present value of $180,000. In part (a), we need to solve (PV) for the monthly annuity payment R. The annuity is 360 months long, and the monthly interest rate is .069/12.

> P=R*((1-(1+i)^(-1*n))/i);

> solve(%,R);

> subs(P=180000,n=360,i=.069/12,%);

>

The monthly payment is $1185.48. Since $180,000 was borrowed, at the end of one month the interest owed was P i = ($180,000)(.069/12) = $1035, and the remainder of the payment, $150.48, goes to decreasing principal. After one payment, the family has paid back $150.48 of the loan (and therefore still owes $179,849.52). As time goes by and the principal begins to decrease, less and less interest is owed each month, and more and more principal is paid each month. This leads us to the second part of the problem; perhaps paying off extra principal early will accelerate this process and decrease the term of the loan.

We still have an annuity with a present value of $180,000 and a monthly interest rate of .069/12, but each payment R now is $1385.48. We need to solve (PV) for n and then use all that information.

> P=R*((1-(1+i)^(-1*n))/i);

> solve(%,n);

> subs(P=180000,R=1385.48,i=.069/12,%);

> evalf(%);

>

The mortgage will be completely paid off in 240 months, or 20 years. The term of the loan has been shrunk from 30 years to 20, and the total paid in interest has been slashed dramatically, by the extra payment each month.

Example 6: Term Of Loan Graphed As A Function Of Payment Size Returning to Example 5, let's see just how the time to pay off the mortgage is affected by the size of the monthly payment R, from its original value of $1185.48 for a 30-year term to some higher value (say, $1800). We'll keep P and i unchanged, create a function we call TERM, and display it as R varies.

Solution

> TERM:=R->-ln((-180000*(.069/12)+R)/R)/ln(1+(.069/12));

>

> plot(TERM,1185.48..1800,color=navy,thickness=3);

>

Note how quickly the term of the loan decreases, particularly for the first extra dollars contributed each month. Extra monthly payment of just $100 (so that R is approximately $1285) cuts the loan term under 24 years, and extra payment of $300 (so that R is approximately $1485) gets the loan paid off in under 17.5 years.

Since many families do not keep their first homes through the length of the entire mortgage, we should note here that the time to reach partial payoff goals is accelerated in equally dramatic fashion by extra payments toward principal. If one wishes to study, say, the time required to pay off half the loan, or how much of the loan is paid off two-thirds of the way through the term, and so forth, one will find the same stunning effects of early payments toward principal. Space does not permit us a detailed discussion here, but the formulas and procedures are no more difficult than those we have presented. Consult your instructor if you wish to pursue these topics.

Those examples lead us to the second key principle of personal financial planning: When debts are owed and gathering interest, pay off as much as you can as soon as you can . The higher the interest rate, the more important this is.

Homework Problems

As always, complete these homework problems in a Maple worksheet with full explanation and commentary included.

Problem 1. A college student needs to buy a car and can afford monthly payments of $180 for the next 3.5 years. If the prevailing annual interest rate is 8.7%, how much can she borrow?

Problem 2. A woman wins a so-called "ten million dollar" prize in the state lottery. It is actually 20 annual payments of $50,000 each. What lump sum payment today would have the same present value? Assume that any investments earn interest at an annual rate of 7.72%.

Problem 3. Kristen is 25, has just begun her new job, and is planning for retirement. She can make monthly contributions to a savings plan which earns interest at an annual rate of 7.47%, and for each dollar she contributes, the company contributes fifty cents. She wishes to have 1.5 million dollars available to her when she retires. How much does she have to contribute each month if

(a) She contributes from age 25 to age 65?

(b) She contributes from age 34 to age 65?

(c) She contributes from age 43 to age 65?

What additional comments might you offer?

Problem 4. A professor borrows $90,000 to buy a modest home. He gets a 25-year mortgage, and each monthly payment is $884. What monthly interest rate, and what annual interest rate, is in effect?

Problem 5. In all the loan problems which have been done above, the quantity -P i + R has always been positive. Specifically, in the problems in which we have found the number of payments needed to complete the payoff of a loan, we have had to take a logarithm of ((-P i + R)/R), and this is defined only when -P i + R is positive. In the context of loans, what exactly does this requirement mean? What would it mean for the payoff process if this quantity were not positive?

Problem 6. A couple secures a 30-year mortgage in which they borrow $165,000 at a stated annual rate of 7.00%.

(a) What is the size of each monthly payment?

(b) If they pay an extra $150 each month, how long would it take to pay off the entire loan?

(c) If they want to pay off the entire loan in 15 years instead of 30, what is the size of each payment?

Problem 7. Around 1980, interest rates on loans and mortgages were sometimes as high as 18%. Repeat Problem 5 with the stated annual interest rate changed to 18%. Make any appropriate comments about the effect of high interest rates on consumers. Note: Many credit cards charge interest at annual rates of 18% or even higher. If you have significant unpaid balances on such cards, the interest payments are draining you, and you should immediately seek advice on how to pay off the debts.

Problem 8. A couple is saving for their daughter's college education. They have determined that they can set aside $175 per month now and for the forseeable future, and they have access to an investment account which will pay interest at a stated annual rate of 8.6%.

(a) How much will they accumulate if they use this plan for 18 years?

(b) How much will they accumulate if they use this plan for nine years?

(c) Suppose that they use this plan for 18 years and also discover later that they can set aside an
additional $125 per month into the same account. If they make these additional contributions
for ten years, how much is accumulated in all their savings?