The Time-Dependent Nature of Moneydbski/writings/Financial_Math.pdf · The Time-Dependent Nature of Money David Surowski Shanghai American School September 5, 2010

The Time-Dependent Nature of Money

David Surowski

Shanghai American School

September 5, 2010

Preface/Acknowledgement

The present set of notes were generated initially as ancillary materials for the course

Financial Mathematics, offered largely to seniors at Shanghai American School.

As the stated prerequisites for this course are very modest—Algebra I—I felt that the

text,1 while very elementary, might still be appropriate for this course. However, hav-

ing discovered that the least prepared student enrolled in this class already had taken

Algebra II, this book became—in my opinion—completely inappropriate in terms of

mathematical content (after all, this is a mathematics course!).

This mismatch between the students and the textbook left me with no other option

but to prepare my own materials for this course. Initially, I had planned for these

materials to be ancillary, that I could still use make occassional use of the text.2 How-

ever, it became increasingly clear that in order to provide any sort of mathematical

integrity to the course, the available textbook would have to be completely discarded

and that my “ancillary notes” would now become the primary instructional materials.

As a result, these notes evolved into its present form: a self-contained book.

This book is organized into five chapters: The Growth of Money, The Mathe-

matics of Accumulation, The Mathematics of Debt Repayment, Basic Eco-

nomic Theory, and Investment Instruments. One could, of course, argue that

the chapter on basic economic theory (a chapter on microeconomics) really doesn’t

belong. I won’t lobby strenuously at this time for it’s inclusion except to say that

at the time, the discussions seemed appropriate. What does set this chapter apart

from the others is that all other chapters are based on the fundamental and related

notions of exponential growth and geometric sums. Any treatment of financial math-

ematics must deal with these subjects; the above-mentioned textbook comes nowhere

near mentioning geometric sums. On such grounds alone one would be justified in

dismissing the text as inappropriate.

The common theme that binds all of this together is that defined by the present

and future values of money. In terms of these concepts, all of the important formulas

given in these notes follow. What I’ve tried to emphasize is that simple geometric

growth and decay model the future and present value of money, respectively, given

1W. H. Lange, T. G. Rousos, and R. D. Mason, Mathematics with Business Applications, Rourth Edition, Glencoe

McGraw-Hill, 1998, ISBN 0-020814730-8.2In the first chapter of this set of notes are occassional references to the text.

ii Preface/Acknowledgement

a single deposit or loan (to be paid in one installment). Geometric sums model the

growth of money when money is deposited on a regular basis or, likewise, when loans

are paid off in regular installments. Many of the topics taken up in elementary courses

on financial mathematics, such as savings accounts and annuities, home loans, and

bonds, are based on geometric sums.

At this time I feel that it’s appropriate to thank the six attentive students who

patiently endured my many ideosyncracies: Roohi Advani, Jason Hong, Christiana

Lilly, Sha-Way Tsai, Xiao-Lan Tsai, and Henry Tsao. I can only hope that not only

has their appreciation of mathematics increased, they will retain some of the more

practical aspects of the course for future applications.

David B. Surowski

Emeritus Professor of Mathematics

April 30, 2004

Contents

1 The Growth of Money 1

1.1 The Future Value of Money . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Credit Card Debt—a Comparitive Case Study . . . . . . . . . . . . . 6

1.3 Annual Yield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Doubling Time, Exponential and Logarithmic Functions . . . . . . . 11

1.5 Annual Yield—Again . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.6 Continuous Compounding . . . . . . . . . . . . . . . . . . . . . . . . 18

1.7 Present Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 The Mathematics of Accumulation 27

2.1 Geometric Sequences, Sums, and Series . . . . . . . . . . . . . . . . . 27

2.2 Simple Annuities and Sinking Funds . . . . . . . . . . . . . . . . . . 32

2.2.1 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2.2 Sinking Funds . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2.3 Annuities and Financial Planning . . . . . . . . . . . . . . . . 34

2.2.4 Using It All—Payout Annuities . . . . . . . . . . . . . . . . . 35

2.3 The Effect of Taxation—Fixed Deferred Annuities . . . . . . . . . . . 39

2.3.1 Payout Annuities and Taxation . . . . . . . . . . . . . . . . . 45

2.3.2 Payout Annuities with C.O.L.A. and Taxation . . . . . . . . . 48

2.4 Annuities—Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3 The Mathematics of Debt Repayment 59

3.1 Loan Repayment—Add-On Interest Loans . . . . . . . . . . . . . . . 59

3.2 Simple Interest Amortized Loans . . . . . . . . . . . . . . . . . . . . 60

3.3 Annual Percentage Rate—APR . . . . . . . . . . . . . . . . . . . . . 63

3.3.1 APR and the Graphing Calculator . . . . . . . . . . . . . . . 65

iii

iv CONTENTS

3.4 Simple-Interest Amortized Loans and Amortization Tables . . . . . . 69

3.4.1 Prepaid Finance Charges and APR . . . . . . . . . . . . . . . 80

4 Basic Economic Theory 95

4.1 Cost, Revenue, and Profit Functions . . . . . . . . . . . . . . . . . . 95

4.2 Marginal Cost, Revenue, and Profit . . . . . . . . . . . . . . . . . . . 99

4.2.1 Maximizing Profit—Cubic Cost Functions . . . . . . . . . . . 102

4.3 Supply and Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4.3.1 The Effect of Taxes on Equilibrium. . . . . . . . . . . . . . . . 108

4.3.2 Utility and Demand as Marginal Utility . . . . . . . . . . . . 109

4.3.3 Elasticity of Demand . . . . . . . . . . . . . . . . . . . . . . . 116

4.3.4 Shifts in Supply and Demand . . . . . . . . . . . . . . . . . . 121

4.4 Optimal Use of Resources—Linear Programming . . . . . . . . . . . . 122

4.4.1 Feasibility Regions and the Objective Function . . . . . . . . . 125

5 Investment Instruments 155

5.1 The Bond Market . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

5.1.1 Discount Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . 156

5.1.2 Coupon Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . 158

5.1.3 Reading a Bond Table . . . . . . . . . . . . . . . . . . . . . . 162

5.2 The Stock Market . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

5.2.1 What is a Mutual Fund? . . . . . . . . . . . . . . . . . . . . . 165

5.2.2 Analysis of Stocks . . . . . . . . . . . . . . . . . . . . . . . . . 166

5.2.3 A Little Bit of Mathematics—The P/E and PEG Ratios . . . 167

Index 173

Chapter 1

The Growth of Money

1.1 The Future Value of Money

The future value of money is simply the idea that because of interest (or other forms

of growth), the value of today’s unit of money (dollars, pounds, yuan, etc.) will grow

to a larger value in the future. When we speak of simple interest, we typically

refer to an investment as having interest only added to it once. The interest rate is

usually reported as an annual rate. For example, if we deposit $10,000 into a bank

that pays out simple interest of 5.5% at the end of the year, then after one year, this

deposit will grow to an amount

$10, 000(1 + .055) = $10, 550 .

The initial amount $10,000 is referred to as the principle and the added amount

of $550 is the interest. As another example, suppose that we purchase a 6-month

$2,000 certificate of deposit at an annual interst rate of 3.5%. If we cash in this CD

after it matures in six months, it will be worth

$2, 000

(1 +

.035

2

)= $2, 035 .

Note that the interest rate r = .035 has been divided by 2 to reflect the fact that the

interest rate for six months is just half the interest rate for one year.

As a slightly different example, suppose that I were to lend you a sum of money,

1

2 CHAPTER 1. THE GROWTH OF MONEY

say ¥10000. Let’s suppose that

(i) the interest rate on the loan is 12.6%;

(ii) the initiation date on the loan is September 5, and that

(iii) the due date on the loan is November 15.

At the maturity date of this loan, I’ll add on the required interest and you’ll pay the

principle plus the interest. The required interest would be carefully computed (by

me!) as follows. I’ll count the number of days from September 5 to November 15:

Sept. 6, Sept. 7, . . . , Nov. 14, Nov. 15.

This adds up to 71 days (you should check me on this). The simple interest would be

that fraction of a whole year occupied by 71 days. Therefore the amount you’ll owe

me on Nov. 15 is:

10000

(1 +

.126× 71

365

)≈ 10245.10 .

Of the above amount ¥10000 is the principle and ¥245.10 is the interest.

Compound interest is just the accumulated effect of applying simple interest

over and over. Let’s pick up the above example where $10,000 is invested in an

account bearing 5.5% annual interest. As already mentioned, this $10,000 will grow

to $10, 000(1 + .055) after one year. After two years, this will grow to

$10, 000(1 + .055)(1 + .055) = $10, 000(1 + .055)2 = $11, 130.25.

The trend is pretty clear, from which we can see that after n years, this initial amount

of $10,000 will grow to the amount $10, 000(1 + .055)n.

The vast majority of banking institutions compound interest more frequently than

just once per year. For example, if the annual interest rate is r, and the money is

compounded monthly, this means that the same interest rate of r is to be divided

into 12 equal parts. Thus, a deposit of P grow as follows:

P grows to P (1 + r12

) after one month,

1.1. THE FUTURE VALUE OF MONEY 3


)2 after two months,


)3 after three months,

and so on.

Thus, if we use monthly compounding, then after t months, the amount of P will

grow to the amount

A = P (1 + r12

)t .

The general idea should be clear, as we can compound as many times per year as

we like. In fact, most U.S. banks compound daily; thus, if r is the nominal interest

rate, a deposit of P will grow to

A = P (1 + r365

)t

after it has been in the bank for t days.1 Note that the above says that after n years,

our initial investment of P will grow to

A = P (1 + r365

)365n.

The general formula is this: If our investment grows at an annual rate of r and

is compounded k times per year, then after n years, this investment will grow to an

amount

A = P (1 + rk)kn .

Exercises for Section 1.1

(A) From your text:

page 156: 12–16

page 158: 7–14

page 161: 10–16

(B) Do the following problems:

1Actually, some banks will divide a year into 360 days instead of 365 days. The daily interest based on a 360-day

year is called ordinary interest; that based on a 365-day year is called exact interest. We shall typically only

consider exact interest.


1. If $12,000 is invested at 4.5% for 20 years, find the future value if the interest

is compounded:

(a) annually

(b) semiannually

(c) quarterly

(d) monthly

(e) daily

(f) every minute

(g) simply (i.e., simple interest)

2. If ¥132000 is invested at 6.9% for 30 years, find the future value if the

interest is compounded

(a) annually

(b) semiannually

(c) quarterly

(d) monthly

(e) daily

(f) every minute

(g) simply (i.e., simple interest)

For problems 3–5, find the interest on each debt and the total to be repaid:

3. $1,500 is borrowed at 21% simple interest, to be repaid in two years.

4. ¥15000 is borrowed at 19.6%, simple interest, to be repaid in 55 days.

5. $8,230 is borrowed at 22% simple interest, to be paid in 30 months.

6. Find the cost of each item in 5 years, assuming an inflation rate of 7%:

(a) cup of coffee, $.075

(b) Sunday newspaper, $1.25

(c) gallon of gas, $1.55

(d) TV set, $600

(e) small car, $14,000

(f) large car, $27,000

1.1. THE FUTURE VALUE OF MONEY 5

(g) yearly college tuition, $16,000

7. Suppose you deposit $8,000 for your child’s college education in an ac-

count that guarantees 7% annual interest, compounded monthly. How much

money will you have on hand?

8. A loan shark is someone who is willing to lend money to just about any-

one but who charges very high (usually unreasonable) interest. Only very

desperate people turn to loan sharks for loans. Anyway, here’s the prob-

lem, our loan shark has lent $10,000 at 15%. The interest on this loan will

compound every day, but with the following twist: for every day that

the money is not paid back, the interest doubles. Suppose that the poor

soul taking out this loan takes a week (7 days) to repay the loan. How much

interest did he end up paying?


1.2 Credit Card Debt—a Comparitive Case Study

A credit card is simply an instrument through which we can obtain short-term loans.

The convenience of credit cards is precisely what makes them so popular. Unlike more

traditional loans from a bank, there is no “loan application process” needed for each

credit card purchase.2 However, one pays for this convenience through higher interest

rates.

In this section we shall consider a very typical manner of computing credit card

debt and try to reconcile this with our understanding of compound interest. We start

by reviewing the average daily balance method of computing credit card debt as

described in your textbook.3 To aid us in this discussion, I’ve reproduced the text’s

credit card bill of the fictitious Scott McTique:

REFERENCE POSTING TRANSACTION DESCRIPTION PURCHASES & PAYMENTS &

DATE DATE ADVANCES CREDITS

1-32734 12/10 12/8 Housewares 25.85

2-44998 12/20 PAYMENT 70.00

BILLING PERIOD PREVIOUS PERIODIC AVERAGE DAILY FINANCE

BALANCE RATE BALANCE CHARGE

12/1–12/31 $125.80 2%

PAYMENTS & PURCHASES & NEW BALANCE MINIMUM PAYMENT

CREDITS ADVANCES PAYMENT DUE

$70.00 $25.85 $20.00 1/21

From the above bill, we see that the previous balance (that is, the amount owed

on 12/1) was $125.80. The periodic rate is, of course, just the monthly interest

rate; this means that the annual interest is 24%—a very high interest rate, indeed! If

Mr. McTique were not to make any further purchases during the month of December,

then on 1/1 he would owe

$125.80 (1 + .02) = $128.32 .

However, he has made some purchases and has made some payments against his

debt. The average daily balance method of computing the end-of-the-month bills

works like this.

Step 1: Compute the average daily balance. To to this you look at the individual bal-

2Of course, in order to get a credit card in the first place, one must apply. However, the requirements are minimal.3W. H. Lange, T. G. Rousos, and R. D. Mason, Mathematics with Business Applications, Glencoe McGraw-Hill,

New York, 1988.

1.2. CREDIT CARD DEBT—A COMPARITIVE CASE STUDY 7

ances and count the number of days for each balance and compute the “weighted

average” of these numbers. This is summarized pretty well in the text:

End-of-Day Number Sum of

Dates Payment Purchase Balance of Days Balances

12/1–12/9 $125.80 × 9 $1132.20

12/10 $25.85 $151.65 × 1 151.65

12/11–12/19 $151.65 × 9 $1364.85

12/20 $70.00 $81.65 × 1 81.65

12/21–12/31 $81.65 × 11 $898.15

Total 31 $3628.50

From the above we can now compute:

Average Daily Balance = Sum of Daily Balances÷ Number of Days

= $3628.50÷ 31

= $117.05

Step 2: Compute the finance charge from the average daily balance:

Finance Charge = Average Daily Balance×Monthly Interest Rate

= $117.05× .02

= $2.34

Step 3: Compute the end-of-the-month bill:

End-of-the-Month Bill = Unpaid Balance + Finance Charge + New Purchases

= $55.80 + $2.34 + $25.85

= $83.99


Now go back and write these number into the blank spaces on Mr. McTique’s

bill (page6).

Homework/Project: I want you to determine if, based on what you know about

compound interest, the above billing method is fair. Here’s how I want you to proceed.

Note that with an initial unpaid balance of $125.80, together with the fact that no

transactions were made (no payments or new purchases), then at the end of 1/9 (the

ninth day of the new billing cycle), the debt grows to

New Debt = $125.80

(1 +

.24

365

)9

= $126.56

Your homework project is to continue along these lines, and write a report on your

findings.

1.3. ANNUAL YIELD 9

1.3 Annual Yield

Consider again depositing $10,000 in an account that bears a nominal interest rate of

5.5%, but with daily compounding. Then, after one year, this initial deposit will

have grown to the new amount

A = $10, 000(1 + .055365

)365 = $10, 565.36 .

So, how much has the money really grown? It has grown more than the nominal rate

of 5.5%, for a 5.5% growth increases

from $10,000 to $10, 000(1 + .055) = $10, 550 .

This says that the “real growth,” or annual yield is (approximately) 5.65%. A good

way to see this is to argue this this:

1. Let r be the nominal interest rate (so r = .055) and let ra be the annual yield.

Then we have

$10, 000(1 + ra) = $10, 000(

1 +r

365

)365

= $10, 000

(1 +

.055

365

)365

.

2. Cancel off the factors of 10,000 and get

(1 + ra) =

(1 +

.055

365

)365

≈ 1.0565.

3. Subtract “1” from both sides and get

ra = 1.0565 = 1 = .0565.

This is where we got the annual yield of 5.65%.

If instead of daily compounding, as with the above example, the bank only com-

pounds monthly, this will have an effect on the annual yield. Thus, let’s continue

to assume that the nominal interest rate is 5.5%. Then after one year (with monthly

compounding) our $10,000 will grow to the amount

A = $10, 000(1 + .05512

)12 ≈ $10, 564.08 .

Not much difference from daily compounding, right? Anyway, we can see, therefore

(think about it!), that the annual yield in this case is about 5.65%.


The General Formula for Computing Annual Yield: If the bank advertises a

nominal interest rate of r, and compounding is done k times per year (so k = 365 for

daily compounding and k = 12 for monthly compounding), then the annual yield ra

is computed by

ra = (1 + rk)

k − 1 .


1. Recall the loan shark that lent an amount of money for seven days, whereas the

initial annual interest rate was 15%, but which doubled every day. Compute the

effective annual interest.

2. If the annual interest rate is 6.9%, find the effective annual interest rate of the

interest is compounded

(a) annually

(b) semiannually

(c) quarterly

(d) monthly

(e) daily

(f) every minute

1.4. DOUBLING TIME, EXPONENTIAL AND LOGARITHMIC FUNCTIONS 11

1.4 Doubling Time, Exponential and Logarithmic Functions

One of the most important questions asked by investors is this:

“Given the present rate of growth, how long will it take for my money to double in

value?”

To get a better grip on this question, let’s return to the above example where $10,000

was deposited in an account that bears a nominal interest rate of 5.5% with daily

compounding. Since “days” is the important unit here, our question becomes that

of finding the number, T , of days necessary for our investment to double. Therefore,

the problem becomes that of solving the equation

$10, 000(1 + .055365

)365T = $20, 000 .

The solution of the problem involves a few choice reminder about logarithms,

which are provided below. In Algebra II one considers exponential functions, which

are functions of the form

f(t) = at ,

where a > 0 is a real number. Indeed, we have already been considering exponential

functions in the context of the growth of money, as the function f(t) = (1 + rk)t is

such an exponential function (where we take a = 1 + rk).

The number “a” in the exponential function f(t) = at is called the base of the

exponential function. There are two important types of exponential functions:

1. a > 1: these are called exponen-

tial growth functions for pretty

much obvious reasons. As the vari-

able t gets larger and larger, so does

at; this is why we call it growth.

-10 -8 -6 -4 -2 2 4 6 8 10

2

4

6

8

10

y=aa>1

xxx


2. 0 < a < 1: The correspond-

ing exponential functions are called

these are called exponential de-

cay functions as they get closer and

closer to 0 as the variable x gets

large.

-10 -8 -6 -4 -2 2 4 6 8 10

2

4

6

8

10

y=a

0<a<1

xx

x

One quickly learns the following rules that govern the behavior of the exponential

functions:

Properties of Exponential Functions

sum−→ product au+v = auav

difference−→ quotient au−v = au/av

exponent of exponent−→ product of exponents (au)v = auv

Along with the exponential functions f(t) = at are the logarithmic functions,

written in the form g(t) = loga t. The definition is:

y = loga x means that x = ay .

Thus, we see that log3 9 = 2 because 32 = 9. Likewise log2(1/16) = −4 because 2−4 =

1/16. Note, finally, that for any positive base a, we have loga a = 1. There is a special

base that is often employed (see also the next section): this is the “magic number”

e, whose value is e = 2.71828 . . .. We refer to loge x as the natural logarithm of x

and usually write it as

loge x = lnx .

The logarithmic functions enjoy properties similar to those satisfied by the expo-

nential functions:


Properties of Exponential Functions

product−→ sum loga(uv) = loga u + loga v

quotient−→ difference loga(u/v) = loga u− loga v

exponent−→ coefficient (loga u)r = r loga u

The graphical relationship between the exponential and logarithmic functions is

depicted in the graph below:

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

y=a

y=log x

xx

x

a

Please note: We shall typically use the natural logarithm on our calculators for most

of our applications. As an application of the ideas considered above, let’s consider

the following

Example. Solve the equation

3x = 5.


Solution. To do this, the simplest approach is simple to take the natural logarithm

of both sides, resulting in

ln 3x = ln 5.

From the third listed property of the logarithm above, we get

x ln 3 = ln 5,

and so

x =ln 5

ln 3≈ 1.465

(where I used my TI-83 calculator to compute the fraction (ln 5)/(ln 3)).

We are now in a position to determine doubling time for investments. This is

probably best illustrated through the example at the beginning of this section.

Example. Assume that we invest P in an account that bears 5.5% compounded

daily. How long will it take for our investment to double?

Solution. We let T be the number of days required for our investment to double.

Thus, the equation needing to be solved is

P

(1 +

.055

365

)T

= 2P.

Clearly we can cancel the P s from both sides of the equation and arrive at the simpler

equation: (1 +

.055

365

)T

= 2.

Now take the natural log of both sides:

ln

(1 +

.055

365

)T

= ln 2.


We press on, using the third property listed above of the logarithm:

T ln

(1 +

.055

365

)= ln 2 and so

T =ln 2

ln(1 + .055

365

)≈ 4600 days

≈ 12.6 years

Just for fun, I’d like to give another somewhat nonstandard application of logarithms—

or at least of log10, the logarithm base 10. The idea is based on some simple calcula-

tions:

log10 1 = 0,

log10 10 = 1,

log10 100 = 2,

log10 1000 = 3,

and so on. Furthermore, by looking at the graph, we see that the logarithm function

is an increasing function. This means that if a < b, the log10 a < log10 b. This says,

for instance, that if x is a 2-digit whole number, then 10 ≤ x < 100, and so

1 = log10 10 ≤ log10 x < log10 100 = 2.

Likewise, if x is a 3-digit number, then we likewise conclude that

2 = log10 100 ≤ log10 x < log10 1000 = 3.

So what’s all of the above telling us? What could we conclude if, for instance, x

were a whole number whose logarithm satisfied 43 ≤ log10 x < 44. Well, this would

say precisely that 1043 ≤ x < 1044, which would say that x is a 44-digit number!

Anecdote. I asked my Algebra II students to try to determine how many digits are

in the number 3120. If you try to compute this on a calculator, then many calculators

will overflow without returning an answer.4 However, computing log10 3120 is easy for

most calculators supporting the common logarithm (i.e., log10). On computes that

log10 3120 ≈ 57.25, which says that 3120 contains exactly 58 digits!

4Actually, my TI-83 does return the answer 1.7970103E57, which says that the number has 58 digits, as already

predicted from the log.


As another example, we can consider how many digits are in the number 32231,

which will surely overflow on most hand-held calculators. But the logarithmic calcu-

lation will still yield a result, as one computes that

log10 32231 = 2231 log10 3 ≈ 1064.46,

which imples that the number 32231 has 1,065 digits.

Here’s a third very interesting application. You have probably heard of prime

numbers: they are the positive integers that cannot be factored into smaller integers.

The first few prime numbers are

2, 3, 5, 7, 11, 13, . . . , 2417, . . .

While it may seem that prime numbers are just a mathematical curiosity, they are

actually important in generating secure banking transactions5 as they involve creating

“trapdoor” functions that can be used to secure transfer of information (like your

personal information!). Perhaps we can return to this later. Anyway, the number

213466917 − 1 is the largest known prime number.6 The ideas discussed above can be

used to compute the number of digits in this number. (Try it! However, you’ll need

to convince yourselves first that 213466917 and 213466917 − 1 have the same number of

digits; now just compute 13466917 log10 2.)

5This is the so-called RSA public-key cryptosystem.6Prime numbers of the form 2p − 1, where p is itself a prime number, are called Mersenne primes. Do you see

why if a number of the form 2n − 1 is a prime, then n must be prime? Anyway, the number 13466917 is a prime,

and so 213466917 − 1 is a Mersenne prime. However, on Nov. 14, 2003, it was announced that there was even a larger

Mersenne prime discovered, namely 220996011 − 1.

1.5. ANNUAL YIELD—AGAIN 17

1.5 Annual Yield—Again

We recall from Section 1.3 that a nominal interest of r, compounded k times per year

provides an annual yield of

ra =(

1 +r

k

)k− 1.

Furthermore, in the suggested project of Section 1.2, you were to compare the

differences between the method of average daily balance versus a more “dy-

namic” method of computing the end-of-the-month bill, where the interest rate is

compounded daily.

However, we can see that if a nominal interest rate of 24% is used in both cases,

then the second method is likely to result in a higher monthly bill simply because the

effective annual interest rate is higher:

For monthly compounding, the effective annual interest rate is(1 +

.24

12

)12

− 1 ≈ .2682,

or 26.82%, whereas for daily compounding, the effective annual interest rate is(1 +

.24

365

)365

− 1 ≈ .2711,

which is 27.11%

From the above, we see that it would be unfair for the credit card company to

compound unpaid balances daily and still charge an annual interest rate of 24%.

Therefore, if we are to make a fair comparison, then the annual interest rate must

be reset to a new value so that the effective interest rates are the same. To do this,

we let x be the new (and more fair!) nominal annual interest rate. Since we want the

effective annual interest rate to be the same as for the 24% interest rate with monthly

compounding, we must have

(1 +

x

365

)365

− 1 = .2682,

which says that

(1 +

x

365

)365

= 1.2682.

Next, we take the 365th of both sides, which results in


1 +x

365=

365√

1.2682,

and so

x = 365(

365√

1.2682− 1)≈ .2377.

In other words, if the nominal interest rate of 23.77% is used with daily compounding,

this has the same effective annual interest rate as when 24% with monthly compound-

ing.

We can summarize the above, by turning around the formula given on page 10.

Let r be the nominal interest rate, and let ra be the effective annual yield based on

compounding k times per year. Then we can solve for r in terms of ra via:

r = k(

k√ra + 1− 1

).

(Don’t worry, I won’t ask you to memorize this!)

(A) Modified Homework/Project: You should return to the project given on

page 8 and once again make the comparisons, except that you should use the nominal

interest rate of 24% for the billing method based on the average daily balance, and

you should use the nominal interest rate of 23.77% for the method based on daily

compounding of the current debt.

1.6 Continuous Compounding

Here, we pick up our discussion in Section 1.1; recall that on page 3 we saw that an

amount P invested in an account that bears an annual interest rate of r, compounded

k times per year will grow to an amount of

A = P(

1 +r

k

)knafter n years. We analyzed this formula for simple interest (componding only once

per year), and also for monthly and daily compounding. Let’s compare and contrast

these results, together with the effective annual yield (see page 10) below:

1.6. CONTINUOUS COMPOUNDING 19

Comparative Annual Yields (AY)

Nominal Interest Rate Annual Yield Annual Yield

(monthly compounding) (daily compounding)

1.00% 1.005% 1.005%

1.25% 1.257% 1.258%

1.50% 1.510% 1.511%

1.75% 1.764% 1.765%

2.00% 2.018% 2.020%

2.25% 2.273% 2.275%...

......

6.00% 6.168 6.183

6.25% 6.432 6.449...

......

10.00% 10.471 10.516

10.25% 10.746 10.792

From the above table, we see that there is very little difference between the annual

yield obtained from monthly compounding as compared to daily compounding. At

this point, we ask the following question, seemingly only of an intellectual nature.

That is, what if we compound more and more frequently, rather than days, we use

hourly compounding, or even compound every minute or every second!

If we compound every hour, then the formula on page 3 can again be used. How-

ever, since every day has 24 hours and every year has 365 days, we see that the “k”

in the formula becomes k = 24 × 365 = 8760. This says that the initial amount P ,

invested at an annual interest rate of r will grow to the amount

A = P(

1 +r

8760

)8760n

after n years.

To understand the effect of this, we compute the annual yield (ra) for a couple of

values of r:

r = .02: ra =

(1 +

.02

8760

)8760

− 1 = .02020

r = .06: ra =

(1 +

.06

8760

)8760

− 1 = .06184


r = .10 ra =

(1 +

.10

8760

)8760

− 1 = .10517

Now compare these values with the corresponding values for monthly and daily com-

pounding, you’ll notice that there’s not much of a difference. This is not a bad thing,

as we’ll see, as it will give us a useful way to approximate the growth of money. The

point here is that for any value of r, there is a limiting value of the expressions

(1 +

r

k

)kas k gets larger and larger. This limiting value turns out to equal the number er,

where e is the number we already met back on page 12. We summarize:

er is the limiting value of(

1 +r

k

)k

as k gets increasing large. Since we make think of very large values of k as repre-

senting “continuous compounding,” we adopt this as a definition, and agree that if

we invest and amount P in an account that bears an annual interest rate

r compounded continuously, then after n years, this amount will grow to

the amount

A = Pern

Furthermore, given continuous compounding, the annual yield, ra now takes on a

very simple form, namely,

ra = er − 1 .

To drive home the point that this is a pretty good approximation to the “discrete”

compounding models, we retabulate the above, including a column for the annual

yield for the continuous compounding case.


Comparative Annual Yields (AY)

Nominal Annual Annual Annual

Interest Rate Yield Yield Yield

Rate

monthly daily continuous

compounding compounding compounding

1.00% 1.005% 1.005% 1.005%

1.25% 1.257% 1.258% 1.258%

1.50% 1.510% 1.511% 1.511%

1.75% 1.764% 1.765% 1.765%

2.00% 2.018% 2.020% 2.020%

2.25% 2.273% 2.275% 2.276%...

......

...

6.00% 6.168% 6.183% 6.184%

6.25% 6.432% 6.449% 6.449%...

......

...

10.00% 10.471% 10.516% 10.517%

10.25% 10.746% 10.792% 10.794%

From the above, one sees only a very slight difference between the continuous com-

pounding versus daily compounding, and then only at the higher percentage rates.

If we use the continuous model, then the calculation of doubling time for an in-

vestment becomes particularly easy. Again, the initial investment of P at an annual

interest rate of r will grow to twice its value, 2P after n years where

Pern = 2P.

Divide both sides by P and take “ln” of both sides:

rn = ln ern = ln 2,

and so the doubling time is given by

n =ln 2

r≈ .693

r.


Simple Example. Suppose that Alfred invests money in an account that bears an

annual interest rate of 6.25%, compounded continuously. How long will it take for

his money to double?

Solution. If n is the number of years needed for the investment to double, then by

the above formula we know that

n =ln 2

.0625≈ 11.09 years ;

rounded off, this is roughly 11 years.

I’d like to conclude this section with a discussion of the so-called “Rule of 72.” If

you get on the internet and initiate a search (say a “google” search), you’ll find literally

hundreds of references to this rule of thumb. Ask almost any financial advisor7 about

doubling time, and (s)he will very likely tell you that the way to calculate this is to

take the annual yield (i.e., the effective interest rate) r (this time, expressed as a

percent, and not as a decimal) and divide it into 72:

Rule of 72: Doubling time ≈ 72

r.

Exercises for Sections 1.4–1.6

1. If the annual interest rate is 6.9%, find the “doubling time” if the interest is

compounded

(a) annually

(b) semiannually

(c) quarterly

(d) monthly

(e) daily

(f) every minute

(g) continuously

7Mine is no different. When he told me about this, I asked why he didn’t just compute a logarithm. Perhaps as

expected, he just looked at me with a blank look on his face.


2. Do the same problem above, except that you are to find the “tripling time.”

3. What is the interest rate r needed for an investment to double in five years if

the interest compounds

(a) annually;

(b) monthly;

(c) daily;

(d) continuously?

4. Do the same problem above is the required doubling time is seven years.

5. Do the problem suggested in the notes, namely, find the number of digits needed

to write down the largest known prime number (213466917 − 1).

6. How many digits are needed to write down the number 33324253 − 2?

7. What would be the nominal interest rated needed to achieve an annual yield of

7.2% given that the interest is compounded

(a) annually;

(b) semiannually;

(c) monthly;

(d) daily;

(e) continuously?

8. Same question as above, but with the targeted annual interest rate being 10%.


9. Complete the table below

Rule of 72—Doubling Times

Effective Interest Rate Actual Rule of 72 % error

3%

3.5%

4.0%

5.0%

6.0%

6.5%

7.0%

8.0%

9.0%

10.0%

10.5%

11.0%

11.5%

12.0%

12.5%

18.0%

19.0%

20.0%

1.7. PRESENT VALUE OF MONEY 25

1.7 Present Value of Money

We have seen that a certain amount, P , of money, invested in an account that bears

an interest rate of r, compounded k times per year will grow to an amount

P(

1 +r

k

)ktafter t years. That is to say, P dollars today is worth P

(1 + r

k

)ktt years from now.

Of course, we can reverse our point of view and ask how much an amount A, given

to us t years in the future, is worth to us today? Again, if we base our answer on the

same investment assumption as above (interest of r, compounded k times per year),

then we see that A dollars t years in the future is worth

A(

1 +r

k

)−kttoday.

If instead of the above “discrete model,” (compounding k times per year) we adopt

the continuous model (so P dollars grows to Pert after t years), then in today’s dollars,

A dollars at a point t years in the future is worth

Ae−rt

today.

We will not belabor this concept any further here, except to make the following

observation, which shall be of crucial importance to our study of loan repayment.

Namely, suppose that we take out a loan (say to purchase an automobile), where the

interest rate is 6%, compounded monthly. Suppose further that our loan payments are

$150 and we pay this out over five years (= 60 payments). Therefore, over the course

of these five years, we will have paid out a total of 60× $150 = $9, 000. However, in

today’s dollars we will have paid out quite a bit less:

Total Paid out in Today’s Dollars =

150

(1 +

.06

12

)−1

︸︷︷︸first payment

+ 150

(1 +

.06

12

)−2

︸︷︷︸second payment

+ · · ·+ 150

(1 +

.06

12

)−60

︸︷︷︸last payment

.

To compute exactly how much less would require us to compute the sum of the


60 terms in the above, seemingly a very difficult task. Fortunately, the above sum is

just a geometric sum, which is the topic of our next chapter.


(A) Compute the present value of each of the quantities given below:

1. $10,000 five years from now with an interest rate of 5%, compounded monthly;

2. $10,000 five years from now with an interest rate of 5%, compounded daily;

3. $10,000 five years from now with an interest rate of 5%, compounded con-

tinuously.

(B) Do the same for the following quantities and interest rates.

1. ¥50000 six years from now with an interest rate of 4.6%, compounded daily;

2. ¥100000 four years from now with an interest rate of 3.4%, compounded

monthly;

3. ¥400000 ten years from now with an interest rate of 7.2%, compounded

daily;

4. ¥70000 two years from now with an interest of 6%, compounded continu-

ously.

Chapter 2

The Mathematics of Accumulation

Throughout Chapter 1 we studied how a single sum of money grows throughout time.

This sum of money could represent either a single deposit or a debt (which also grows

in time). In the present chapter, we shall consider the effect of either making regular

deposits to an account or making regular payments against a loan.

2.1 Geometric Sequences, Sums, and Series

A sequence of real numbers is called a geometric sequence if it can be represented

in the form

a, ar, ar2, ar3, . . . ,

where a and r are real numbers. The number r is called the ratio in the geometric

sequence.

Thus we see that the following are simple examples of geometric sequences:

1, 2, 4, 8, 16, . . . (Here, a = 1 and r = 2.)

4, 8, 16, 32, . . . (a = 4 and r = 2)

23, 2

9, 2

27, . . . (a = 2

3and r = 1

3)

1,−1, 1,−1, 1, . . . (a = 1 and r = −1)

12,−1

4, 1

8,− 1

16, . . . (a = 1

2and r = −1

2)

P, P(1 + r

k

), P(1 + r

k

)2, P(1 + r

k

)3, . . . (a = P and the ratio is

(1 + r

k

))

27

28 CHAPTER 2. THE MATHEMATICS OF ACCUMULATION

The last example should make it clear that we have been dealing with geometric

sequences all along.

What’s more interesting than a geometric sequence is a geomtric sum. These

are formed by adding the first n terms of a geometric sequence; thus, the expression

S = a + ar + ar2 + · · · arn−1

is a geometric sum. On page 25 we already encountered the geometric sum

150

(1 +

.06

12

)−1

+ 150

(1 +

.06

12

)−2

+ · · ·+ 150

(1 +

.06

12

)−60

which represents the present value of 60 loan payments of $150 each.

A similar example can be considered, as follows. Suppose that, starting today, I

deposit $80 into an account that pays 4.2% (compounded monthly). Thereafter, I

shall continue to make monthly deposits of $80. How much will I have in the bank two

years from now? We can analyze this problem along the following lines. Note that the

first deposit of $80 will remain in the bank for two full years, and will therefore grow

to an amount $80(1 + .042

12

)24. The next month’s deposit will also be $80, but it will

remain in the bank only for 23 months, and will therefore grow to $80(1 + .042

12

)23.

Likewise, the third deposit will grow to $80(1 + .042

12

)22. Assuming that a total of 24

deposits are made, then the last (the 24th) deposit will only have been in the bank for

one month at the end of the two-year period. Thus, this last deposit will only grow

to $80(1 + .042

12

).

Of course, after the 24 months, our total balance will be the sum of all of the above

amounts:

Amount in Bank =

$80(1 + .042

12

)24+ $80

(1 + .042

12

)23+ $80

(1 + .042

12

)22+ · · ·+ $80

(1 + .042

12

).

We see that the above is also a geometric sum with first term a = $80(1 + .042

12

)and

ratio r =(1 + .042

12

).

With the above having been said, we see that it would be very nice to be able to

2.1. GEOMETRIC SEQUENCES, SUMS, AND SERIES 29

actually compute geometric sums. In fact, this is quite simple, and requires a rather

standard trick. To this end, we start with the geometric sum

S = a + ar + ar2 + · · ·+ arn−1.

Therefore, we see that

rS = ar + ar2 + · · ·+ arn,

from which we easily see that

S − rS = a− arn.

From the above, we readily obtain

S =a− arn

1− r.

We shall conclude this section with a couple of calculations.

Example 1. Compute the geometric sum

1 +1

2+

1

4+ · · ·+ 1

210.

Solution. Simply note that in the geometric sum, we have first term a = 1 and

ratio r = 12. This sum involves n = 11 terms, and so the above formula gives us the

result

1 +1

2+

1

4+ · · ·+ 1

210=

1− 1211

1− 12

=211 − 1

210≈ 1.999.

Example 2. What about the above problem, where deposits of $80 are made every

month for two years?

Solution. In this case, we have a = $80(1 + .042

12

)and ratio r =

(1 + .042

12

). Since

n = 24 (there are 24 terms in this sum), we obtain


Amount =$80

(1 + .042

12

)− $80

(1 + .042

12

) (1 + .042

12

)24

1−(1 + .042

12

)=

($80

(1 + .042

12

)(12)

.042

)((1 +

.042

12

)24

− 1

)

≈ $2, 006.30.

Notice that $80× 24 = $1, 920, and so the total interest earned is

$2, 006.30− $1, 920 = $86.30.


1. The reader has probably encountered the so-called summation notation for

expressing sums:

n−1∑i=0

ari expresses the sum a + ar + ar2 + · · ·+ arn−1.

Now compute the following geometric sums

(a)5∑

i=0

23i,

(b)5∑

i=0

(23

)i,

(c)10∑i=1

(23

)i,

(d)10∑i=1

(−23

)i,

(e)10∑i=1

(−2)i,

(f)100∑i=1

(12

)i,

(g)100∑i=3

(13

)i,

2. Compute the following sums, and provide interpretations.

2.1. GEOMETRIC SEQUENCES, SUMS, AND SERIES 31

(a) $120(1 + .042

12

)24+ $120

(1 + .042

12

)23+ $120

(1 + .042

12

)22+ · · ·+ $120

(1 + .042

12

). (What

does this number represent?)

(b) $50(1 + .042

12

)60+$50

(1 + .042

12

)59+$50

(1 + .042

12

)58+ · · ·+$50

(1 + .042

12

). (What does

this number represent?)

(c) 150(1 + .06

12

)−1+ 150

(1 + .06

12

)−2+ · · ·+ 150

(1 + .06

12

)−60. (What does this number

represent?)


2.2 Simple Annuities and Sinking Funds

The present section discusses two closely related problems:

(i) If you make regular deposits into an interest-bearing account, how much will you

have after a certain period of time?

(ii) If your goal is to have a certain amount of money after a certain period of time,

and you can make periodic payments into an interest-bearing account, how much

should these periodic payments be?

The first problem determines what is called an annuity; the second problem de-

termines a sinking fund. We shall consider these separately.

2.2.1 Annuities

Let us begin with a simple problem. Suppose that you make regular—say monthly—

deposits into an interest-bearing account for a number of years. How much will there

be in the account? Such a sequence of payments is called an annuity. Furthermore,

the computation of the balance in this account is a classic application of geometric

sums.

Let us flesh this out with a specific example. Let’s put $80 per month for 20 years

into an account that bears a nominal interest rate of 3.5% (compounded monthly).

How much will we have after the 20 years?1

Assuming that the first deposit is made today, and that a total of 12 × 20 = 240

deposits are made, the total will be given by

Total = 80

(1 +

.035

12

)240

︸︷︷︸value of first deposit

+ 80

(1 +

.035

12

)239

︸︷︷︸value of second deposit

+ · · ·+ 80

(1 +

.035

12

)1

︸︷︷︸value of last deposit

The above is, of course, a geometric sum with first term 80(1 + .035

12

), common ratio(

1 + .03512

)and having 241 terms. We may use the formula given on page 29 and

compute:

1For both annuities and sinking funds we make the assumption that there is a deposit at the very beginning and

at the very end. Therefore, the total number of deposits after n years is 12n + 1.

2.2. SIMPLE ANNUITIES AND SINKING FUNDS 33

Total =80(1 + .035

12

)− 80

(1 + .035

12

)240

1−(1 + .035

12

) ≈ $27, 830.48.

Let’s develope a formula that is a bit more usable for annuities. That is to say,

assume that we make monthly deposits of an amount M into an account that bears

a nominal interest of r, compounded monthly. How much is present after n years?

The solution of this is as follows, and is based on geometric sums.

Total in Annuity = M(

1 +r

12

)1

+ M(

1 +r

12

)2

+ · · ·+ M(

1 +r

12

)12n

=M(1 + r

12

)−M

(1 + r

12

)12n

1−(1 + r

12

)=

(12M

r

)((1 +

r

12

)12n+1

− 1

).

So we summarize as follows.

The Basic Problem of Annuities . If regular (monthly) deposits of M are

made in an account that bears an interest rate of r (compounded monthly), how

much is present after n years? On the basis of the above, we see that this amount is

given by

Total in Annuity =

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

).

Note, finally, that over time annuities earn quite a bit of interest. That is, if we

make regular deposits of M into an annuity, then the interest earned over n years is

just the value of the annuity minus the total deposits (which is the total number of

deposits times M). In the above example, where $80/month is put into the annuity

bearing 3.5% for 20 years, then a total of 240 deposits have been made, for a total

of $19,200. Since the annuity is worth $27,830.48, this means that the total interest

earned over the 20 years is $27,830.48−$19,200=$8,630.48.


2.2.2 Sinking Funds

A sinking fund is really the same as an annuity except that the objective is to obtain

a pre-determined amount of money in the fund from regular deposits.

The Basic Problem of Sinking Funds . If the objective is to obtain an amount

P after n years in a fund that bears a nominal interest of r (compounded monthly),

what should the monthly deposits be?

The relevant equation is then

P = M(

1 +r

12

)+ M

(1 +

r

12

)2

+ · · ·+ M(

1 +r

12

)12n

=M(1 + r

12

)−M

(1 + r

12

)12n

1−(1 + r

12

)=

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

),

from which it follows immediately that

Required Monthly Deposits = M =

(rP

12(1 + r

12

))((1 +r

12

)12n

− 1

)−1

.

2.2.3 Annuities and Financial Planning

This subsection asks us to consider the “bigger” picture in financial planning, in

particular that which applies to retirement savings.

Typical Case Study . Mr. and Mrs. Smith project that 30 years from now (when

they retire) they will be able to live on approximately $2,500/month. Assume that at

that time they will be able to transfer their annuity into a saving account that pays

6.5% nominal interest, compounded monthly.


Question 1: How much will they need to have in this savings account to generate this

money?

Answer to Question 1. What we are looking for is an amount, P , of money that

each month will generate $2,500 in income. That is to say, we need for the monthly

interest on this amount to be $2,500:

Necessary Interest = $2, 500 = P

(1 +

.065

12

)− P,

which says that we must have

P =$2, 500(

.06512

) ≈ $461, 538.46.

Next, Mr. and Mrs. Smith have decided to try to earn the above amount by making

regular monthly deposits into a sinking fund that pays a nominal 8.2% (compounded

monthly).

Question 2: What will be the necessary monthly deposits into this sinking fund to

realize the objective?

Answer to Question 2. We simply use the formula for required monthly deposits

into a sinking fund. Thus,

Required Monthly Deposits =

((.082)($461,538.46)

12

)((

1 + .08212

)360 − 1) ≈ $295.31.

Thus, we see that Mr. and Mrs. Smith will have to start putting $295.35 each month

into their sinking fund to realize their retirement objectives.

2.2.4 Using It All—Payout Annuities

In the previous section, we calculated a two-stage process in providing for our retire-

ment:


Make monthly

deposits for many

years into an

annuity

retire;put proceeds

into a -savings account

monthly interest--- retirement expenses

Notice, however, that in the above model, only the interest from the proceeds of the

annuity is being used for our retirement living expenses. The principle generated

by the annuity is never touched. So what are we going to do with this money? A few

choices occur:

(a) Let our children inherit this money;

(b) will it to charity;

(c) us it ourselves!

We shall take (the admitedly selfish) option (c) as the basis of this subsection.

That is to say, we shall carefully plan the withdrawals of our interest + principal so

that we get it all! Of course, a further assumption needs to be made here, namely

that we have to estimate our own longevity. That is to say, if we are going to

arrange for a total payout of our retirement money over, say, 30 years, then we are

assuming that we will be living no longer than 30 years. If, in fact, we pass on before

30 years (and therefore before all the money is gone), then what’s left over become

part of our estate, to be inherited by our designated beneficiaries.

The idea, then is that we wish to have certain monthly payments over the course

of many years, with the idea that at the end of this time frame, all of the money

has been paid out. This defines what is called a payout annuity.

The Basic Problem of Payout Annuities . If we have an annuity at a given

nominal interest rate of r (compounded monthly), and if we wish to have equal

monthly payments of M from this annuity over n years, how much do we need to put

into this annuity to realize this objective?


The solution of this problem is deceptively simple, and is based on the idea of the

present value of money. So let’s say that our annuity starts next month with our

monthly payments of M , and that this continues for n years (or 12n months). Then

in today dollars, this payout is worth

P = M(

1 +r

12

)−1

+ M(

1 +r

12

)−3

+ · · ·+ M(

1 +r

12

)−12n

,

which, of course, is a geometric sum with 12n terms, first term M(1 + r

12

)−1and

ratio(1 + r

12

)−1. This tells us, therefore, that

P =M(1 + r

12

)−1(

1−(1 + r

12

)−12n)

1−(1 + r

12

)−1

=12M

r

(1−

(1 +

r

12

)−12n)

(after some simplification!).

Let’s see what this means in the context of some specific numbers. Suppose that we

can put money in a payout annuity with a nominal interest rate of 8.2% (compounded

monthly), and that we wish to payout $2,000 per month for 25 years. How much needs

to be put into this annuity? To solve this, we use M = $2, 000, r = .082 and n = 25.

We calculate P :

P =12($2, 000)

.082

(1−

(1 +

.082

12

)−300)

≈ $256, 481.51.

This means that in order to realize the $2,000/month payout for 25 years, we need

to start with the amount $254,740.79. Note incidently, that the total interest

earned from this payout annuity is the total amount paid out minus the initial

amount deposited. In this case, the total interest is 300× $2, 000− $254, 740.79 =

$345, 259.21.

To summarize: in order to realize a monthly payout of M for n years from a payout

annuity paying a nominal interest rate r, we need to start with an amout P , where

P =12M

r

(1−

(1 +

r

12

)−12n)

.


Before leaving this section, we should address the effects of cost of living on a

payout annuity. That is to say, if a payout annuity is to be made over the span of

decades, then it makes sense to build in annual increases in the monthly payments to

reflect increases in the cost of living.

The way this works is as follows. We shall, in the first year of our payout, take

a monthly amount M . However, in each of the subsequent years, this amount shall

increase by a cost of living adjustment (C.O.L.A.), which we shall denote by c. We

shall also let r be the nominal interest rate for the annuity; we shall also require

the annual yield ra for the annuity, which we recall from page 10 is related to r

by(1 + r

12

)12= 1 + ra. Therefore, while the first 12 monthly payments are for the

amount M , the next 12 monthly payments are for the amount M(1+c). We continue

increasing the payments in subsequent years, until the payout is complete. In order to

determine the amount that must be initially in this annuity to realize these payments,

we must, exactly as above, convert everything to present value. Thus, we let P be

this unknown amount and compute like mad:

P = M(

1 +r

12

)−1

+ M(

1 +r

12

)−2

+ · · ·+ M(

1 +r

12

)−12

+ M(1 + c)(

1 +r

12

)−13

+ M(1 + c)(

1 +r

12

)−14

+ · · ·+ M(1 + c)(

1 +r

12

)−24

+ M(1 + c)2(

1 +r

12

)−25

+ M(1 + c)2(

1 +r

12

)−26

+ · · ·+ M(1 + c)2(

1 +r

12

)−36

+ · · ·...

+ M(1 + c)n−1(

1 +r

12

)−12(n−1)−1

+ M(1 + c)n−1(

1 +r

12

)−12(n−1)−2

+ · · ·+ M(1 + c)n−1(

1 +r

12

)−12n

.

As is quite clear, the above formula is very complicated. However, it really is nothing

more than one geometric sum contained within another. The very persistent student

can sum the above; here I’ll just summarize the final result as follows.

The Basic Problem of Payout Annuities with C.O.L.A.

Suppose we have a payout annuity at a nominal interest rate r and an annual yield of

ra. Suppose that we wish to receive monthly payments for n years from this annuity,

2.3. THE EFFECT OF TAXATION—FIXED DEFERRED ANNUITIES 39

starting with an amount M , but increasing in each subsequent year by a cost of living

allowance c. Then we need to have an initial amount P in this annuity, where

P =12raM

r(ra − c)

[1−

(1 + c

1 + ra

)n].2

We shall conclude this subsection with a numerical example. Suppose that you

wish to set up a payout annuity through your bank for monthly payouts of $3,000/month

in the first year with an annual C.O.L.A. of 3% each year thereafter for 25 years. How

much should be deposited into the annuity? Assume that the nominal interest rate

on the payout annuity is 6%.

Note first that the annual yield is

ra =(

1 +r

12

)12

− 1 ≈ .0617.

Therefore, we can compute P by

P =(12)(.0617)(3000)

.06(.0617− .03)

[1−

(1.03

1.0617

)25]≈ $620, 476.86.

2.3 The Effect of Taxation—Fixed Deferred Annuities

Since in most cases, interest on money earned is taxable,3 we need to illustrate the

effect of taxation on the total yield of an annuity. We shall focus on two types of

annuities: before-tax and after-tax annuities. I will try to show that the first type

of annuity is typically the better choice.

Before-Tax Annuities, or Tax-Deferred Annuities. In such annuities the in-

terest on your annuity won’t be taxed until you withdraw the money from the

annuity.

After-Tax Annuities. In such annuities the interest on your annuity is taxed each

year and the taxes are withdrawn from your account. When you draw the money

out of the annuity, there will be no further taxes.

2Note that the formula on page 377 of D. B. Johnson and T. A. Mowry, Mathematics, A Practical Odyssey, Fifth

Edition, Thomson, Brooks/Cole, 2004 is incorrect, as it is missing the factor of ra/r.3An exception to this general rule is provided by the so-called Roth IRAs.


It doesn’t sound like there should be any difference, right? Surprisingly, there is.

In order to get a better understanding of this problem, let’s take the case of the

simplest annuity, namely one in which a single deposit of $50,000 is made and left

in the annuity for 30 years. Assume further that

(i) The interest on the annuity is 8%, compounded monthly, and that

(ii) you will be taxed at a rate of 31% (a relatively common tax rate).

In the after-tax annuity, the interest on your earnings will be taxed each year at

a rate of 31%. In order to simplify the math a bit, let’s first compute the effective

annual interest rate, ra:

ra =

(1 +

.08

12

)12

− 1 ≈ .083,

which is 8.3%. Next, since each year we lose 31% of our interest to taxes, this means

that we only get to keep 69% of our interest. That is to say, the real growth of our

money each year is only

69% of 8.3% = 69% × 8.3% ≈ 5.727%.

From the above, we see that after 30 years, and after deducting taxes each year, our

$50,000 will grow to the final amount:

$50, 000 (1 + .05727)30 ≈ $265, 795.29.

On the other hand, suppose that all of our earnings are tax-deferred until the

maturity of the account (in this case, 30 years). Then after 30 years, the initial

$50,000 will grow to an amount

$50, 000 (1 + .083)30 ≈ $546, 793.95,

of which $496,793.95 is interest. We now pay 31% tax on this interest, which is

$496, 793.95× .31 ≈ $154, 006.13.

Once we’ve paid this, this leaves a grand total of


$546, 793.95 − $154, 006.13 ≈ $392, 787.82,

a considerably larger amount than in the after-tax annuity!

We summarize the above into a useful formula. Assume that a one-time deposit

of P is made into an annuity that pays an effective interest rate of ra. Assume that

the annual income tax to be applied is t. Assume that this money is left in the annuity

for n years.

In the Fixed Deferred Annuity, this amount will grow to the amount

A = P + P (1− t) ((1 + ra)n − 1) .

In the After-Tax Annuity, this amount will grow to

A = P (1 + ra(1− t))n .

In the more typical case, namely that in which regular (usually monthly) deposits

are made to the annuity, then it is still true that the fixed deferred annuity provides

a lower over all tax liability.

For the Fixed Deferred Annuity the tax is simply applied to the overall interest

of the annuity. Thus, if monthly deposits are made for n years into a deferred annuity

with a nominal interest rate of r, then the tax is just applied to the overall interest

earned. In turn, this is just the final value of the annuity, less the 12× n deposits of

M . From the formula on page 33, we see that

total interest earned =

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

)− 12nM.

This means that if the tax rate is t, then

total tax liability =

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

)t− 12nMt.

Finally, the total in the annuity after taxes have been taken out is therefore


Total in Annuity

After Taxes=

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

)(1− t) + 12nMt.

Since the basic problem in the context of sinking funds is to determine the needed

monthly payments to generate required amount, we shall denote by P the after-tax

total in the above formula and solve for M :

M =P(

12(1+ r12)

r

)((1 + r

12

)12n − 1)

(1− t) + 12nt

.

The tax analysis for the After-Tax Annuities requires a rather more careful

analysis. (The uninterested reader can just skip to the main results below.) We shall

continue to assume that

(a) The nominal interest rate on the annuity is r;

(b) monthly deposits of M are made

(c) for n years (a total of 12n deposits);

(d) the tax rate is t and that the tax is paid at the end of each year from the account.

The main complicating feature of this analysis is the fact that there are monthly

events (the deposits) and yearly events (the paying of taxes out of this account).

Therefore, we shall start by computing the amount Y such that a single yearly deposit

of Y (at the beginning of each year) will grow to the same amount as will 12 deposits

of M . In other words, we need to find Y such that

Y (1 + ra) = M(

1 +r

12

)12

+ M(

1 +r

12

)11

+ · · ·+ M(

1 +r

12

),

where, as usual, ra is the effective interest rate: ra =(1 + r

12

)12− 1. After computing

the above geometric sum, and using the above defintion of ra, we get:


Y (1 + ra) =12M

(1 + r

12

) ((1 + r

12

)12 − 1)

r

=12M

(1 + r

12

)ra

r,

Which implies that

Y =12M

(1 + r

12

)ra

r(1 + ra).

Numerical Example. Just to see how this works, suppose that instead of making

monthly deposits of $100 into an account bearing a nominal interest rate of 6%, let’s

compute the equivalent yearly deposit. In this case, we have M = $100 and r = .06,

and so ra = (1 + .0612

)12 − 1 ≈ .06168. Therefore, we have

Y =1200

(1 + .06

12

).06168

.06(1 + .06168)≈ $1, 167.73.

Again, what this says is that $1,167.73 deposited at the beginning of the year will

earn—at the effective interest rate ra ≈ .06168—the same amount as will depositing

$100 each month for 12 months at the interest rate of 6% (compounded monthly).

The next step is to determine the effect of removing the taxes from this account

on a yearly basis. We argue as follows:

1. Start with an initial deposit of Y .

2. This will grow to an amount of Y (1 + ra) by the end of the year.

3. The interest on this growth is Y ra.

4. The tax on this interest is Y rat; this gets paid out of the account, leaving

5. Y (1 + ra)− Y rat = Y (1 + ra(1− t)) at the end of year 1.

6. At the beginning of year 2 we make a new deposit of Y , bringing the balance to

Y (1 + ra(1− t)) + Y .

7. We repeat this cycle of events and infer that at the beginning of year 3 this has

in it the amount

(Y (1 + ra(1− t)) + Y ) (1 + ra(1− t)) + Y =

Y [(1 + ra(1− t))2 + (1 + ra(1− t)) + 1] .


This pattern will continue to repeat, leading to the fact that at the beginning of year

n there will be a total of

A = Y[(1 + ra(1− t))n−1 + · · ·+ (1 + ra(1− t)) + 1

].

Since the geometric sum can be easily computed, we obtain, finally that

A =Y [(1 + ra(1− t))n − 1]

ra(1− t).

We have already seen above that in terms of the monthly deposits M ,

Y =12M

(1 + r

12

)ra

r(1 + ra).

From this, we conclude that after n years, the amount, P , in our after-tax

annuity with nominal interest rate r, effective interest rate ra, and monthly

deposits M is given by

P =12M

(1 + r

12

)((1 + ra(1− t))n − 1)

r(1− t)(1 + ra).

Since we are really after the required monthly payments M , we solve everything for

M :

M =Pr(1− t)(1 + ra)

12(1 + r

12

)((1 + ra(1− t))n − 1)

.

Another Numerical Example. Assume that we wish to set up an annuity which

carries a 6.5% nominal interest rate. Assume that this annuity is, unfortunately, not

tax deferred, but rather, is an after-tax annuity, and that the tax rate on our returns

will be 31%. If we expect to pay into this annuity for 30 years, and that we wish

to have $500,000 at the end of these 30 years, how much must we contribute each

month?

Solution. We use the above and solve for M :

M =$500, 000(.065)(.69)(1 + ra)

12(1 + .065

12

) [(1 + ra(.69))30 − 1

]≈ $688.65.


From the above computation, we see that realizing the goal of $500,000 after 30 years

in an after-tax account that pays 6.5% is probably a bit unrealistic, as it will require

monthly deposits of roughly $689.15.

Numerical Example, Continued. Let’s do the same work-up, but this time

assume that the 6.5% annuity is tax deferred. This time we use the formula given on

page 42 in order to compute M :

$500, 000 =

(12M

(1 + r

12

)r

)((1 +

r

12

)12n

− 1

)(1− t) + 12nMt

=

(12M

(1 + .065

12

).065

)((1 +

.065

12

)360

− 1

)(.69) + 360M(.31)

If we solve for M in the above the result is M ≈ $568.83. In summary, we see again

that it is beneficial to try to put out money in a tax-deferred annuity rather than in

an after-tax annuity.

2.3.1 Payout Annuities and Taxation

As expected, payout annuities will also carry a tax liability. Even though deposits

are not being made to a payout annuity (the opposite is true, right?), there is still

interest being earned, and this interest is what’s being taxed. As payout annuities

are commonly used for generating retirement funds, one doesn’t typically see them on

a before-tax basis (i.e., having the tax applied at the end of the life of the annuity).

Thus, it is typical for the tax to be applied to the yearly interest generated by the

payout annuity, with the tax funds subtracted from the principle each year.

We shall conduct the analysis in a fashion parallel with that on page 43. Thus,

we assume that an initial principle, P , is deposited in the payout annuity, that r is

the nominal interest rate, that we draw monthly payments of M until the annuity

is completely paid out after n years. Assume that the tax rate is t. Again, we shall

find it necessary to convert monthly payouts of M into equivalent yearly payouts of Y

(where the interest rate is now the effective interest rate ra). To make this conversion,

recall the formula on page 37:


P =12M

r

(1−

(1 +

r

12

)−12n).

If instead of montly payouts of M we took annual payouts of Y , then the correspond-

ing formula for P would be

P =Y

ra

(1− (1 + ra)

−n) .Since these two formulas will give the same result for P , one has

Y

ra

(1− (1 + ra)

−n) =12M

r

(1−

(1 +

r

12

)−12n),

which implies that

Y =12Mra

r.

We now analyze as follows:

1. Start with an initial deposit of P .

2. This will grow to an amount of P (1 + ra) by the end of the year.

3. The interest on this growth is Pra.

4. The tax on this interest is Prat; this gets paid out of the account; since there is

also the payout of Y , we have

5. A = P (1 + ra)− Prat− Y = P (1 + ra(1− t))− Y in the payout annuity at the

end of year 1.

6. We repeat this analysis and obtain the amount

A = (P (1 + ra(1− t))− Y )(1 + ra(1− t))− Y =

P (1 + ra(1− t))2 − Y (1 + (1 + ra(1− t)))

at the end of year 2.


7. We repeat this cycle of events. For brevity, set x = 1 + ra(1− t) and infer that

at the end of year m the amount remaining in the payout annuity is

A = Pxm − Y(1 + x + x2 + · · ·+ xm−1

)= Pxm − Y

(xm − 1

x− 1

).

That is to say, after m months, the amount in the payout annuity is

A = P (1 + ra(1− t))m − Y

((1 + ra(1− t))m − 1

ra(1− t)

).

Since the annuity completely pays out after n years, we conclude that

0 = P (1 + ra(1− t))n − Y

((1 + ra(1− t))n − 1

ra(1− t)

),

from which we can determine the needed initial principle P in terms of everything

else. By way of summary, if we are to draw monthly deposits of M for n years from

a payout annuity at a nominal interest rate of r, and if the tax liability is t, then

the amount P needed in this account is given by

P =Y ((1 + ra(1− t))n − 1)

ra(1− t)((1 + ra(1− t))n.

Since

Y =12Mra

r,

we may rewrite this as

P =12M ((1 + ra(1− t))n − 1)

r(1− t) (1 + ra(1− t))n.

A slightly more convenient form is given below:

P =

(12M

r(1− t)

)[1−

(1

1 + ra(1− t)

)n].

Numerical Example. Suppose that we plan to need roughly $3,000/month for

25 years for our retirement years. Assume further that our vehicle will be a payout

annuity that pays a nominal interest rate of 5% and that our tax rate will be 15%.

How much money will we need to put in this annuity?


Solution. We apply the above expression for P , using the approximation ra ≈.0511:4

P =

($36, 000

(.05)(.85)

)[1−

(1

1 + (.0511)(.85)

)25]≈ $554, 455.96.5

2.3.2 Payout Annuities with C.O.L.A. and Taxation

In this final subsection, we shall consider the effect of taxation on a payout annuity

with a built-in C.O.L.A. Thus, assume that we have a payout annuity with nominal

interest rate r, effective interest rate ra, and that a cost of living of c is built into

payouts in subsequent years. Assume that we are to receive monthly payments for

n years, with the first year’s payments being M each month, and that the tax rate

is t. The problem is to find the necessary initial principle P that will sustain these

payments.

As above, we work with the equivalent first-year annual payout Y , where as above,

Y and M are related by

Y =12Mra

r.

The analysis now follows pretty predictable lines:

1. Exactly as in the case of the payout annuity with no cost of living adjustment,

the amount in the annuity after the first year is A = P (1 + ra) − Prat − Y =

P (1 + ra(1− t))− Y .

2. Since at the end of the second year, we will receive an amount (1 + c)Y , rather

than just Y , the amount in the annuity after the second year is then

A = (P (1 + ra(1− t))− Y )(1 + ra(1− t))− (1 + c)Y =

P (1 + ra(1 − t))(1 + ra(1 − t)) − Y (a + ra(1 − t) + (1 + c)). As

above, we set x = 1 + ra(1− t), so that the above expression can be written as

4It sometimes happens that an approximate value can generate significant roundoff errors, especially is relatively

high powers of expressions containing this approximation are applied. Note that in the above expression, the 25th

power of (1 + ra(1− t))25 is being computed.5If one uses the exact value of ra (namely, using (1 + r

12)12 − 1), then the computed value for P is $554,824.57,

a difference of less than $400.


A = Px2 − Y (x + (1 + c)).

This represents the balance in the annuity at the end of the second year.

3. At the end of the third year, the amount decreases to the amount

A = (Px2 − Y (x + (1 + c)))x− (1 + c)2Y

= Px3 − Y (x2 + (1 + c)x + (1 + c)2).

4. At this stage the pattern should be clear, so that after m years, there will be

A = Pxm − Y (xm−1 + (1 + c)xm−2 + · · ·+ (1 + c)m−2x + (1 + c)m−1

= Pxm − Y

[xm − (1 + c)m

x− (1 + c)

].

5. Since the annuity pays out after n years, we conclude that

0 = P (1 + ra(1− t))m − Y

[(1 + ra(1− t))m − (1 + c)m

ra(1− t)− c

].

6. As usual, since the annuity completely pays out after n years, we may solve for

P . So we summarize: if we we are to draw monthly amounts for n years, with the

first year’s monthly payments being M , from a payout annuity at a nominal

interest rate of r, C.O.L.A. of c, and if the tax liability is t, then the amount P

needed in this account to sustain these payments is

P =Y ((1 + ra(1− t))n − (1 + c)n)

(ra(1− t)− c)((1 + ra(1− t))n

=Y

ra(1− t)− c

[1−

(1 + c

1 + ra(1− t)

)n]

Since Y =12Mra

r, the above becomes

P =12Mra

r(ra(1− t)− c)

[1−

(1 + c

1 + ra(1− t)

)n].


Numerical Example. We shall continue the example on page 47, retaining all

of the hypotheses except that we shall add a C.O.L.A. of 3%. Applying the above

formula, we have

P ≈ $36, 000(.0511)

(.05)((.0511)− .03)

[1−

(1.03

1 + (.0511)(.85)

)25]≈ $757, 852.37.6

2.4 Annuities—Summary

In this chapter we have discussed the general concept of an annuity, especially how

it relates to retirement planning. We have seen how, in this context, these annuities

divide very naturally into two types: a sinking fund into which we contribute money

during our working careers, followed by a payout annuity from which we draw

money during our retirements. We indicate this below in the following schematic

fashion:

Sinking Fund

retire;put proceeds P

into a -Payout Annuity

make deposits

666

take monthly payments???

We have seen that it’s best to find a tax-deferred annuity into which to make

contributions during our working careers. Once we retire, the proceeds (P ) of this

annuity (less the taxes) are then put into a payout annuity, from which we shall

draw monthly payments for our retirement expenses.

We have seen how to allow for both taxation and C.O.L.A. on the payout annuity;

this is important as what one needs to compute is the amount P that will go into

the payout annuity. In turn, once we know P , then we can proceed to determine the

monthly payments needed to accumulate P in our sinking fund.

Exercises for Chapter 2 (In the exercises below, unless otherwise stated assume6If one carries through the exact value of ra, then the resulting calculation for P ≈ $758, 298.94.

2.4. ANNUITIES—SUMMARY 51

that all compounding is done on a monthly basis.)

1. Compute the value of the annuity, given the listed information.

(a) Monthly deposit = $1,000, r = 3%, n = 10 years.

(b) Monthly deposit = ¥2000, r = 4.2%, n = 15 years.

(c) Monthly deposit = ¥500, r = 2.2%, n = 10 years.

(d) Monthly deposit = $200, r = 5%, n = 20 years.

(e) Monthly deposit = $1,200, r = 3.6%, n = 20 years.

2. For each of the annuities above, compute the total earned interest.

3. For each of the annuities in Problem 1, compute the present value of the

annuity. (Use the technique of Section 1.7; such calculations are often used for

financial planning.)

4. For the annuities in Problems 1 and 2 above, how would you calculate the value

of the annuity if the compounding is done daily instead of monthly? (I haven’t

given you any formulas for this; you’ll need to try to figure this out on your

own.)

5. For each of the annuities below, assume that a 31% tax is imposed on the interest

on the account. Compute the total amount of taxes paid over the time interval

(n years) given:

(a) Monthly deposit $200, r = 5%, n = 10 years, given that this account is

i. Tax deferred;

ii. After-Tax.

(b) Monthly deposit $500, r = 7.2%, n = 20 years, given that this account is

i. Tax deferred;

ii. After-Tax.

(c) Monthly deposit $300, r = 6%, n = 25 years, given that this account is

i. Tax deferred;

ii. After-Tax.


(For the tax liability on Tax-deferred accounts, see page 41, for after-tax

accounts, just compute the difference between the balance in the annuity

without taxes (page 33) and the total in the annuity with taxes (page 44).)

6. For each of the sinking funds below, calculate the needed monthly deposits to

obtain the given final balance.

(a) Final Balance = $12,000, r = 3%, n = 10 years.

(b) Final Balance = ¥50000, r = 4.2%, n = 15 years.

(c) Final Balance =$80,000, r = 2.2%, n = 18 years.

(d) Final Balance = ¥200000, r = 3.2%, n = 25 years.

(e) Final Balance = $120,000, r = 5.6% n = 20 years.

(f) Final Balance = $400,000, r = 7.2%, n = 30 years.

(g) Final Balance = $600,000, r = 6.4%, n = 30 years.

7. For each of the sinking funds listed in Problem 6 above, calculate the needed

monthly deposits to obtain the given after-tax balance, given that the sinking

fund is tax-deferred and that the tax rate is 15%. (See page 42.)



fund is tax-deferred and that the tax rate is 31%. (See page 42.)



fund is after-tax and that the tax rate is 15%. (See page 44.)



fund is after-tax and that the tax rate is 31%. (See page 44.)

11. Mr. and Mrs. Drysdale have decided to set up a college fund for their new-born

baby. Assuming that they will need $100,000 for their child’s college education

when the child becomes 18 years of age, and assuming that they can set up a

sinking fund at a nominal interest rate of 7.2%, what will the necessary monthly

deposits be?


12. Answer Question 11 given that the sinking fund is tax deferred with a tax rate

of 20%.

13. Answer Question 11 given that the sinking fund is an after-tax fund with a tax

rate of 20%.

14. For each of the payout annuities below, calculate the needed initial principle in

order to realize the given monthly payments:

(a) Monthly Payment = $3,000, r = 3%, n = 10 years.

(b) Monthly Payment = ¥50000, r = 4.2%, n = 15 years.

(c) Monthly Payment =$2,500, r = 2.2%, n = 18 years.

(d) Monthly Payment = ¥20000, r = 3.2%, n = 25 years.

(e) Monthly Payment = $5,000, r = 5.6%, n = 20 years.

(f) Monthly Payment = $4,000, r = 7.2%, n = 30 years.

(g) Monthly Payment = $3,500, r = 6.4%, n = 30 years.

15. For each of the payout annuities in Problem 14, calculate the needed initial

principle in order that the first year’s monthly payments are as above, but in

subsequent years there is a C.O.L.A. of 3%. (See page 39.)


principle in order that the first year’s monthly payments are as above, but that

the funds are being taxed at 15%. (See page 47.)



subsequent years there is a C.O.L.A. of 3% and that the funds are being taxed

at 15%. (See page 47.)



subsequent years there is a C.O.L.A. of 3% and that the funds are being taxed

at 31%. (See page 47.)


19. When Billy Carter turned 15, he got a job at his parents fast-food restaurant. At

this time, his parents told him that if he put some of his savings into an IRA,7

then they would “match” his contributions. As a result of this, Billy decided

to contribute $800, starting immediately. However, when Billy turned 21 years

old, his parents stopped matching his contributions, so Billy increased his own

contributions to $1,500/year, and continued to make these contributions up to

(and including) his 65th birthday. His IRA paid a nominal interest of 7.75%

(simple interest-no compounding). Determine the following:

(i) the amount in Billy’s IRA at age 65

(ii) the present value of Billy’s IRA at age 65

(iii) Billy’s total contributions

(iv) Billy’s parents’ total contributions

(v) the future value of Billy’s IRA had he waited until age 18 before starting

his contributions

(vi) the future value of Billy’s IRA had he waited until age 25 before starting

his contributions.

(vii) Answer all of the above questions if the IRA compounds accounts monthly.

20. Assume exactly the same situation as in Problem 19, except that the annuity

compounds monthly. Assume that this is an tax-deferred annuity, and that

Billy will pay the tax on all of the interest when he withdraws the money at age

65. Assuming a 15% tax rate, how much will Billy have after taxes?

21. Assume that all hypotheses of Problem 20 hold, except that the account is a

taxable account, with tax computed on an after-tax basis. How much will Billy

have after he withdraws his money?

22. Tom McCarter, now aged 18, wants to retire when he reaches the age of 55, but

figures that he will need $3,500/month for his living expenses. He can start an

annuity now which pays an attractive 8.8%, but at his retirement he will need

to move his money into a traditional account that will pay 6%, after which time

he will live solely on the interest earned on this account.

7An “IRA,” or Individual Retirement Account is like an annuity except that the contributions can be at any

time, as opposed to on a regular basis.


(i) How much will he need to start contributing to his annuity?

(ii) If Tom decides to postpone his retirement until age 60, and if all other as-

sumptions remain valid, how much will he need to contribute to his annuity?

(iii) Same question, except that Tom has decided to wait until age 65 before

retiring.

23. Answer questions (i) and (ii) of Problem 22 above, where you now assume that

the 8.8% sinking fund is tax-deferred, with a 31% tax levied on the interest of

this account.


the 8.8% sinking fund is an after-tax account, with a 31% tax levied on the

interest of this account.


(a) the 8.8% sinking fund is tax-deferred, with a 31% tax levied on the interest

of this account, and

(b) the interest generated by the traditional 6% account is taxed at 15%,

26. We continue to follow Tom McCarter, the retirement objectives as in Problem 22

(to retire at age 55 and draw $3,500/month). Assume that his 8.8% sinking fund

is tax-deferred (tax rate = 31%) and that at age 55, he will transfer his money

into a payout account that has interest taxed at 15%, and where the payout is

complete after 25 years. What does he need to do?

27. Answer Question 26 where, in addition to the hypotheses given, Tom wishes to

build in a 3% cost of living adjustment on his retirement proceeds. What should

he do?

28. Mr. S has just agreed to buy a piece of property in Timbuktu. Here are the

details.

(i) He will pay a “property use fee” of $200/month, with the first payment being

at the start of next month;

(ii) At the end of six years, his monthly payments will cease, but at that time

he will pay $100,000 for the final purchase of the property.


In order to make preparations for the final $100,000 payout, Mr. S will initiate

a sinking fund at a nominal interest rate of 6.5%, compounded monthly. Now

answer the following:

(a) What is the present value of all money paid out by Mr. S?

(b) What will his monthly contributions to the sinking fund be in order to

realize his final payout amount?

29. Our client, Mr. Math, has the following attributes.

(a) He is now 20 years old.

(b) He wants to retire at age 60.

(c) He expects to need $4,500/month (for 30 years) for his retirement expenses.

(d) He wants to set up a sinking fund to prepare for retirement, and the best

one he can find is a tax-deferred account (31% tax bracket).

(e) The interest on the proceeds of his payout annuity will be 15%.

What should he do?

30. Suppose that Bill McTravel is 20 years old with great travel ambitions. He is

going to start paying into a 9.0% annuity and has the following plans:

(i) He will retire at age 60, after which point he will make no further contribu-

tions to his retirement annuity.

(ii) Upon his retirement he will travel the world for five years.

(iii) He estimates his total travel needs to run $150,000.

(iv) At age 65 he expects to have to put what’s left from his annuity into another

payout annuity that only pays 7.4%.

(v) He expects to need $3,500/month for his retirement expenses.

(vi) He expects to be dead by age 75 (very fast life style!).

Assume that his tax rate will be 15% on all proceeds. How should he proceed?


Project: Write a Letter to Your Client

You are to be in the role of a financial advisor, helping a recently-employed 23-year

old professional (who will be earning a pretty hefty salary) plan for her retirement.

Here are the basics of your recommendation.

1. That your client will put a certain amount, M , of money each month, starting

immediately, into a Fixed Deferred Annuity.

2. That the nominal interest rate on the above fixed deferred annuity will be set at

a fairly generous 7.3%.

3. That your client will want to retire at age 60, after which point there will

be no additional income (other than what is generated by the proceeds of

this annuity).

4. That at age 60, the proceeds—less the taxes on the interest—of the fixed deferred

annuity will be transferred to a payout annuity, which expects to bear at a

nominal interest rate of 6%.

5. That a 31% tax shall apply to the proceeds of the above 7.3% annuity.

6. That your client will be travelling the world from retirement at age 60 until age

65, and expects to require monthly proceeds in the amount $4,500 to cover all

expenses. Assume that all retirement interest shall be taxed at 15%.

7. That once your client fully retires at age 65, $3,500 per month will be sufficient

to cover expenses; however:

8. Assume that your client will live for another 20 years after full retirement.

9. That a Cost-of-Living Adjustment of 3% per year shall be applied to the

monthly payments after retirement.

10. That at the end of the retirement payout, there will remain $150,000 for inheri-

tance or charitable contributions.

Present your recommendations on a mock-up letterhead, bearing the (fictitious)

name of the financial company you represent. This should be a typed letter to your

client (give your client a fictitious name), explaning what will be necessary to realize


the stated objectives, which means that the monthly contributions M need to be

explicitly calculated. Be prepared to handle follow-up letters from your client where

alternative plans might be suggested.

Chapter 3

The Mathematics of Debt

Repayment

3.1 Loan Repayment—Add-On Interest Loans

The basic problem is simple. We borrow—with interest—a certain sum of money, P ,

to be repaid in time in regular installments (usually monthly). Such a loan is called

an amortized loan. In this chapter we shall consider two types of amortized loans,

the add-on interest loan and the simple-interest amortized loan.

The simplest type of amortized loan is the add-on interest loan, as follows. An

amount A is borrowed, to be repaid over the course of n years, with the annual interest

rate at r. The interest rate on this loan is simply computed as nr (interest of r over

n years), and the total interest is just A × nr. We simply add on this interest, so

that the total amount to be repaid is just

Total to Be Repaid = A + Arn .

It is typical for add-on interest loans to be repaid monthly, so that a loan to be repaid

over n years will involve 12n equal monthly payments:

Monthly payments =A + Arn

12n.

Add-on interest loans are often found in the context of automobile financing (es-

pecially through auto dealerships as opposed to through financial institutions).

59

60 CHAPTER 3. THE MATHEMATICS OF DEBT REPAYMENT

Example. Suppose that are to purchase an automobile for $10,000 by making a

down payment of $3,000 with the remaining amount to be financed through an add-

on interest loan through the car dealership. Assume further that the loan is to be

paid back in four years at an annual interest rate of 3.8%. Then the total to be repaid

will be

Total to Be Repaid = $7, 000 + $7, 000× 4× .038 = $8, 064.

The monthly payments are computed by dividing this amount into 48 equal monthly

payments:

Monthly Payments =$8, 064

48= $168.

Note that in this example a total of $1,064 was paid out in interest.

3.2 Simple Interest Amortized Loans

The vast majority of financial institutions use a method of loan repayment that,

although more complication that the notion of add-on loans, is more closely related

to the notion of a sinking fund, and is therefore a bit more honest (see Section 3.3).

In order to make this discussion as explicit as possible, let us recall the discussion on

page 25 where payments of $150 were made each month for 60 months. Furthermore,

at an interest rate of 6%, we saw that the present value of this activity was given

by the geometric sum

Present Value =

150

(1 +

.06

12

)−1

︸︷︷︸first payment

+ 150

(1 +

.06

12

)−2

︸︷︷︸second payment

+ · · ·+ 150

(1 +

.06

12

)−60

︸︷︷︸last payment

.

Next, while it was Exercise 2c of Section 2.1 to compute this sum, I’ll give the result

here: one computes that:

3.2. SIMPLE INTEREST AMORTIZED LOANS 61

150

(1 +

.06

12

)−1

+ 150

(1 +

.06

12

)−2

+ · · ·+ 150

(1 +

.06

12

)−60

≈ $7, 758.83.

So let’s try to understand what has just been computed. All of the money paid

out has been shown to equal—in today’s dollars—the amount $7,758.83. What does

this mean? This means that if we borrowed $7,758.83 today and then paid out $150

each month, over the next 60 months, then these payments would—again, in today’s

dollars—exactly match the borrowed amount. In other words, the 60 payments of

$150 each will exactly pay back the debt!

Of course, the above approach to the problem of loan payments is artificial and a

bit backwards. The basic problem of loan repayment can be stated as follows:

The Basic Problem of Simple Interest Amortized Loans . Suppose that

a simple interest amortized loan of A is taken out with a nominal interest rate of r

and which is to be repaid in k monthly repayments (starting with next month). What

will the monthly payments (M) be?

The solution is formally identical with that of the sinking funds problem, except

that

(i) one typically does not make a payment at the very beginning (i.e., at the initiation

of the loan)—the first payment is (typically) deferred until one month later.

(ii) the geometric sum contains negative exponents, reflective of the fact that present

values are computed, rather than future values.

The main idea that we use to compute M is to understand that the net sum of all of

the future payments of M each should equal the present value of the loan, namely A.

That is to say we require that

A = M(

1 +r

12

)−1

+ M(

1 +r

12

)−2

+ · · ·+ M(

1 +r

12

)−k.

The next step is to recognize that the right-hand side of the above equality is a

geometric sum with first term M(1 + r

12

)−1, ratio

(1 + r

12

)−1and k terms in all. We

now apply the formula for computing a geometric sum on page 29. The result is


A =

(M(1 + r

12

)−1 −M(1 + r

12

)−(k+1)

1−(1 + r

12

)−1

)

=M −M

(1 + r

12

)−k(r12

)=

12M

r

(1−

(1 +

r

12

)−k).

This allows us to solve for M in terms of A and everything else:

M =rA

12(

1−(1 + r

12

)−k) .

Example 1. Suppose that we wish to borrow ¥100000, with a nominal interest rate

of 4%, to be paid in monthly installments and to be paid off after 10 years. What

will the monthly payments be?

Solution. We have A = 100000, r = 0.4 and k = 10× 12 = 120 months. Using the

above formula, we get

M =(r = .04)(100000)

12(

1−(1 + .04

12

)−120) ≈ ¥1012.45

That is to say, a 10-year loan of ¥100000 to be repaid in monthly payments at an

interest rate of 4% will incur monthly payments of ¥1012.45. Note that the total of

all of these payments gives a total pay out of

Total Paid = ¥1012.45× 120 = ¥121494,

which means that the total interest paid on this loan is the difference between this

amount and the amount borrowed:

Total Interest Paid = ¥121494− ¥100000 = ¥21494.

But wait a minute! Let’s compare this result with what the monthly payments

would have been had an add-on loan been taken out for the same amount and at

the same interest rate. Here, the total amount to be paid off would be

3.3. ANNUAL PERCENTAGE RATE—APR 63

Total to pay back = ¥100000 + ¥100000× .04× 10 years = ¥140000,

which is considerably higher than the total payback of ¥121494 with the simple

interest amortized loan. That is to say, even though the two loans both report a 4%

annual interest rate, the total interest paid out for the add-on loan is considerably

higher!

3.3 Annual Percentage Rate—APR

As we saw in the last subsection, even when the nominal interest rates on an add-on

interest loan and a simple interest amortized loan are the same, the total interest paid

on these loans can be quite different. This can put the unwary consumer at a great

disadvantage when making comparisons. Therefore, in response to this, in 1968 the

United States Congress passed the Truth-in-Lending Act, which made it necessary

for all lending institutions to report what is now called Annual Percentage Rate,

or APR for short. The APR of a loan is the nominal interest rate that would be used

for a simple interest amortized loan to pay of the same debt. In other words, the APR

of a loan is calculated as follows:

1. Compute the Principle + Interest (A+I) that is to be paid off on the existing

loan.

2. Divide by the number of months to find the monthly payment M .

3. Find the interest rate rAPR for a simple interest amortized loan that would result

in the same monthly payment.

We can clarify this with a simple example; however, the computations will

turn out to be so complicated that graphing calculators will be required!

Thus, assume that we take out a four-year add on interest loan of A = $1, 000 at a

percentage rate of 5%. Therefore, the add-on interest is 4× 5% = 20%, which means

that the total interest is I = $1, 000 × .20 = $200. The principle plus interest is

then A + I = $1, 200 which results in monthly payments of M = $1, 200/48 = $25.


Now ask this question: what would be the required interest rate, rAPR, on a four-year

simple interest amortized loan that would also result in $25 monthly payments. Thus,

we use the formula for M given on page 62 (where we remember that the quantity k

is the total number of monthly payments and A is the amount borrowed; thus k = 48

and A = $1, 000 in this example):

M = $25 =$1, 000rAPR

12(

1−(1 + rAPR

12

)−48) .

This simplifies slightly to

25× 12

1000= .3 =

rAPR(1−

(1 + rAPR

12

)−48) .

The above equation can be solved for rAPR; the result is r ≈ 9.24%. That is to say,

the APR on the above add-on loan with advertised interest rate of 5% actually has

an annual percentage rate of 9.24%! This rate must—by the Truth in Lending

Act—be conspicuously reported in order not to deceive the consuming public.

The Basic Problem of APR . Suppose that we have an n-year loan of A

advertised as having an interest rate of r and whose monthly payments are M . Then

the APR (rAPR) of this loan is calculated by solving the equation

M =ArAPR(

1−(1 + rAPR

12

)−12n) .

I’m not quite ready to box this expression in, as I’d like to simplify it just a bit more.

Indeed, if we define a new unknown by setting

x = 1 +rAPR

12,

then the above expression can be written as

M =A(x− 1)

(1− x−12n).

If we multiply the numerator and denominator of the fraction on the right by x12n,

then this simplifies just a bit more:


M =A(x12n+1 − x12n)

(x12n − 1).

Finally, set a = M/A and multiply both sides through by x12n − 1:

(x12n − 1)a = (x12n+1 − x12n),

which results in the hard polynomial equation:

x12n+1 − (a + 1)x12n + a = 0 ,

which must be solved for x in order to find the APR! Of course, once we find x, then

rAPR = 12(x− 1).

We summarize . In finding the APR, rAPR, for an n-year loan of A with monthly

payments M , we do this:

1. We set x =rAPR

12+ 1;

2. we set a = M/A;

3. we solve the hard polynomial equation:

x12n+1 − (a + 1)x12n + a = 0;

4. we recover rAPR by rAPR = 12(x− 1).

3.3.1 APR and the Graphing Calculator

In the previous subsection we saw that the problem of finding APR reduced to solving

a very difficult polynomial equation; so difficult in fact that we shall have to resort to

using technology (e.g., the TI-83 graphing calculator) for finding the solution. Before

giving the graphing calculator techniques, we shall go into a little of the mathematics;

this is necessary to find the approximate window settings. However, this discussion

uses a little bit of calculus; students wishing to skip this discussion are free to do so.


We saw from the above that the problem of finding APR demanded that we be

able to solve a polynomial equation of the form

p(x) = xn+1 − (a + 1)xn + a = 0,

where n is a rather large exponent, and where a is a (typically small) positive number.

Note first that there is an almost obvious zero of the above polynomial, namely x = 1.

However, this value of x says that rAPR = 12(x−1) = 0, which clearly is not relevant

to our discussion. Indeed, in order to get a positive annual percentage, we must

look for a zero, x, of the above polynomial with x > 1.

To determine this sought-after value of x, we begin by calculating the derivative

of the polynomial p(x) and setting equal to zero in the interest of finding the relative

extrema of this polynomial. We have

p′(x) = (n + 1)xn − n(a + 1)xn−1 = xn−1((n + 1)x− n(a + 1)).

We set this equal to 0:

xn−1((n + 1)x− (a + 1)) = 0,

which gives x = 0 (uninteresting to us) and

x =n(a + 1)

n + 1.

To determine the nature of the above critical point, it is convenient to use the

second derivative test. Thus, we compute

p′′(x) = n(n + 1)xn−1 − n(n− 1)(a + 1)xn−2

= nxn−2((n + 1)x− (n− 1)(a + 1)),

and note, therefore, that


p′′(n(a + 1)

n + 1

)= n

(n(a + 1)

n + 1

)n−2((n + 1)

(n(a + 1)

n + 1

)− (n− 1)(a + 1)

)= n

(n(a + 1)

n + 1

)n−2

(n(a + 1)− (n− 1)(a + 1))

= n

(n(a + 1)

n + 1

)n−2

(a + 1) > 0.

The above result says, by the second derivative test, that the polynomial function

p(x) assumes a relative minimum at the critical value x = n(a + 1)/(n + 1). Since

we have already seen that p(x) = 0, it must happen that

p

(n(a + 1)

n + 1

)< 0.

(Of course, this could be seen directly.) Since p(a + 1) = a > 0, we see that the zero,

x0, of interest to us must satisfy

n(a + 1)

n + 1< x0 < a + 1.

Note, finally that

p

(n(a + 1)

n + 1

)= a− 1

n + 1

(n

n + 1

)n

(a + 1)n+1 .

This is all depicted in the graph below:


-

6

•

•

•

•

•

x

y

(0, a)

��

(1, 0)

��

(n(a+1)n+1

, a−(

1n+1

)n(a + 1)n+1

)6

(x0, 0)

HHY

(a + 1, a) -

y = xn+1 − (a + 1)xn + a

Before proceeding any furthermore, one should observe that the above graph is rather

out of scale. In fact, what happens in practice is that a is quite a bit larger than

the absolute value∣∣a− 1

n+1

(n

n+1

)n(a + 1)n+1

∣∣; this comment applies to determining

adequate window settings in the discussion below:

TI-83 Technology. The above gives us an idea as to how to proceed to find the

relevant zero x0 of the polynomial p(x). First of all, graph the polynomial p(x) using

the following window settings

Xmin = n(a+1)n+1

(needs to be computed first)

Xmax = a + 1

Ymin = a− 1n+1

(n

n+1

)n(a + 1)n+1 (needs to be computed first)

Ymax =∣∣a− 1

n+1

(n

n+1

)n(a + 1)n+1

∣∣If we return to the example at the beginning of this section, we have n = 48, a =

3.4. SIMPLE-INTEREST AMORTIZED LOANS AND AMORTIZATION TABLES 69

M/A = .025, and so the window settings are thus:

Xmin ≈ 1.004

Xmax = 1.025

Ymin ≈ −.0004

Ymax = .0004

It’s reasonable to expect that certain small adjustments might be necessary (you’ll

just need to experiment). For example in graphing the above, it’s probably advisable

to lower Xmax a bit—say to about 1.01.

Once the graph is complete, use the “calc” button on your TI-83 calculator (menu

option “zero”) to complete the necessary calculation. The calculator may take several

seconds before returning the answer of x = 1.0077015. This says that

rAPR = 12(x− 1) = .092418,

which amounts to saying that the APR is approximately 9.24%.

(Whew!)

3.4 Simple-Interest Amortized Loans and Amortization Ta-

bles

We begin this discussion with the following question. Suppose that we take out a

simple-interest amortized loan for an amount A for n years at the nominal interest rate

r. Then each of the monthly payments against this loan consists of two components:

(i) the amount that is applied against the loan itself (or as is more commonly stated,

is applied against the principal of the loan), and

(ii) the amount that pays the interest on the loan.

Let us take a simple example of a $150,000 loan for 30 years at 6.4%. Using the

basic formula for the monthly payments given on page 62, we see that the monthly


payments will be $938.26. Next, notice that after one month, the amount owed on

the loan has grown to

New debt, after first month = $150, 000(1 +.064

12) = $150, 800.

Therefore, the first payment on the loan covers the $800 interest with the remaining

amount—$138.26—being applied against the principal on the loan. This means that

after the first monthly payment, the new principal on the loan has shrunk to

New principal = $150, 000− $138.26 = $149, 871.74.

Let’s now consider the second monthly payment. Thus, after the second month,

the new principal will grow to a new amount

New debt, after second month = $149.871.74(1 +.064

12) = $150, 671.06.

From this we see that the second month’s interest is the difference $150, 671.06 −$149, 871.74 = $799.32, which is slightly less than the first month’s interest. There-

fore, of the second monthly payment of $938.26, $799.32 covers the interest and

$138.94 applies toward the principal.

If we continue the above calculations over the life of the loan, we obtain what is

called an amortization table. We can tablulate the first 12 and the last 6 entries

of the amortization table for the above loan:


Loan Amortization1

Number Principal Balance Payments Interest Paid Principal Applied New Balance

1 $150,000.00 $938.26 $800.00 $138.26 $149,861.74

2 $149,861.74 $938.26 $799.26 $139.00 $149,722.74

3 $149,722.74 $938.26 $798.52 $140.48 $149,583.01

4 $149,583.01 $938.26 $797.78 $141.23 $149,442.52

5 $149,442.52 $938.26 $797.03 $141.99 $149,301.29

6 $149,301.29 $938.26 $796.27 $142.74 $149,159.31

7 $149,159.31 $938.26 $795.52 $143.50 $149,016.56

8 $149,016.56 $938.26 $794.76 $144.27 $149,873.06

9 $149,873.06 $938.26 $793.99 $145.04 $149,728.79

10 $149,728.79 $938.26 $793.22 $145.81 $149,583.75

11 $149,583.75 $938.26 $792.45 $146.59 $149,437.94

12 $149,437.94 $938.26 $791.67 $147.37 $149,291.35

......

......

......

354 $6,4291 $938.26 $34.29 $903.97 $5,525.94

355 $5,525.94 $938.26 $29.47 $908.79 $4,617.16

356 $4,617.16 $938.26 $24.62 $913.63 $3,703.52

357 $3,703.52 $938.26 $19.75 $918.51 $2,785.02

358 $2,785.02 $938.26 $14.85 $923.41 $1,861.61

359 $1,861.61 $938.26 $9.93 $928.33 $933.28

360 $933.28 $938.26 $4.98 $933.28 $0.00

The most important—and most obvious—feature of this table is that toward the

beginning of the loan, the interest payments are large and the payments against the

principal are rather low. This gives the impression that the loan is going to be paid

off very slowly. However, as time wears on, the interest payments continue to decrease

the the payments against the principal continue to increase.

The amortization schedule can be helpful in many respects; let us consider a few

application in the example below.

Example. Suppose that a house is purchased by John Deare for $150,000. Mr.

Deare had on hand $30,000 for a down payment, meaning that he had to finance

$120,000 for the purchase of the home. Assume that his bank was able to offer him a

loan at 6.3% with a choice of taking out a 15-year mortgage or a 30-year mortgage.

1This table was taken from a java script written by PFS Mortage Loans of Santa Cruz; see

http://fdc.mortgage101.com/partner-scripts/1170.asp?p=pfsml


Assume also that after five years, Mr. Deare expects to sell the house for 25% more

than he paid for it. After the sale of his house, he will pay off his loan and keep the

remaining profit. Analyze his profit.

30-year mortgage: First of all, one computes his monthly payments to be $742.77.

Next, if we appeal to the relevant amortization schedule, we find that after five

years (60 payments), Mr. Deare will still owe $112,071.26 on his $120,000 loan.

Therefore, when he sells his house, the sale price will be $150,000(1+.25)=$187,500.

When he pays off his loan ($112,071.26), he will still have $75,428.74. Of course,

in order to interpret this money appropriately, we should compute the present

value of this money, computed at the time of his home purchase:

Present value = $75, 426.74(1 +.063

12)−60 = $55, 092.43.

Of course, we still need to subtract his $30,000 down payment in order to compute

the profit:

Present Value of After-sales profit = $55, 092.43− $30, 000 = $25, 092.43.

Of course, we can take a different view of this situation. Let’s not forget that Mr.

Deare paid $742.77/month for 60 months. On the surface, we would think that

this means a total of 60×$742.77 = $44, 566.20. However, since we are computing

all values at the beginning of the loan, the present value of this money spent is2

Present value of Monthly Payments =

$742.77(1 +.063

12)−1 + $742.77(1 +

.063

12)−2 + · · ·+ $742.77(1 +

.063

12)−60 = $38, 144.35.

We’re ready to summarize the above findings. First of all, due to the sale of the

house, Mr. Deare made a profit—in present dollars—of $25,092.43. However,

for the monthly payments he spent—again in present dollars—$38,144.35.

2Note that the calculation below is exactly the same as calculating the money needed in a five-year 6.3% payout

annuity to generate $742.77/month.


Thus, we can regard the difference between these two sums as being his real

expenditure in present dollars:

Real Expenditure for the Purchase of the Home =

$38, 144.35− $25, 092.43 = $13, 051.92 .

15-year mortgage: As expected, we start by computing the monthly payments: they

are $1,032.18. Also, since Mr. Deare is paying off a rather sizable debt in half

the time, we would expect the monthly payments for the 15-year mortgage to

be quite a bit higher than for the 30-year mortgage. Next, we find that after

five years (60 payments), Mr. Deare will still owe $91,722.45 on his $120,000

loan. Therefore, when he sells his house at $187,500 and pays off the loan of

$91,722.45, he will still have $95,777.55. As above, we compute the present value

of this profit:

Present value = $95, 777.55(1 +.063

12)−60 = $69, 955.01.

Of course, we still need to subtract his $30,000 down payment in order to compute

the profit:

Present Value of After-sales profit = $69, 955.01− $30, 000 = $39, 955.01.

Next we calculate the present value of all of his monthly payments:

Present value of Monthly Payments =

$1, 032.18(1 +.063

12)−1 + $1, 032.18(1 +

.063

12)−2 + · · ·+ $1, 032.18(1 +

.063

12)−60 = $53, 006.78.

Finally, we compute the difference between the present value of his stream of

monthly payments and the present value of his after-sales profit on the sale of

his home:


Real Expenditure for the Purchase of the Home =

$53, 006.78− $39, 955.01 = $13, 051.77 .

Note that the real expenditures are the same! Perhaps this may have come as a

surprise, but the fact that the interest rate used was the same in each case meant

that end costs turned out to be the same (regardless of the rate of payments).

However, in practice it turns out that the mortgage rates for 15-year mortgages

tend to be a bit less than for the 30-year mortgages3 and so the end costs have

the potential be being a bit less in this case.

Unpaid balance without using an Amotization Schedule

Notice that in the above discussion, our analyses were largely based on the computa-

tions of interest and unpaid balances. In turn, these were extracted from amortization

schedules (available from the web). This is a bit of overkill, as we can easily work out

the formula for the unpaid balance on a simple-interest amotized loan, as follows.

The Basic Problem of Unpaid Balance . Assume that a simple-interest amor-

tized loan in the amount A is taken out at an interest rate r and to be repaid in n

years. We would like to compute the amount that is still owed on the loan—the

unpaid balance—after k years. Assume that (using the formula on page 62) the

monthly payments needed to pay off this loan after n years is M . Note first that—in

today’s dollars—the amount of the loan that will be repaid after k years is given

by the difference between the amount borrowed, A, and the present value of all of the

monthly payments:

Present Value of Amount Owed =

A−M(

1 +r

12

)−1

−M(

1 +r

12

)−2

− · · · −M(

1 +r

12

)−12k

.

3As of the time of the writing of these notes, interest rates for 30-year loans were in the vicinity of 5.5%, and

interest rates on 15-year mortgages were approximately 4.8%


Therefore, the amount owed after k years is this amount, measured k years from now:

Unpaid Balance after k years

=

(A−M

(1 +

r

12

)−1

−M(

1 +r

12

)−2

− · · · −M(

1 +r

12

)−12k)(

1 +r

12

)12k= A

(1 +

r

12

)12k−M

(1 +

r

12

)12k−1

−M(

1 +r

12

)12k−2

− · · · −M

= A(

1 +r

12

)12k−M

((1 + r

12

)12k − 1r12

).

Example. Suppose that Mr. and Mrs. Cooper took out a 30-year $150,000 loan at

5.8%. Find the unpaid balance on the loan

after the first year;

after the second year; and

after the third year.

Next, take this information to compute the amount of interest paid

during the first year;

during the second year; and

during the third year.

Note first that, using the formula for monthly payments on page 62, we see that the

monthly payments on this loan are $880.13. Next, using the above formula for unpaid

balance, we find that

unpaid balance after the first year = $148, 088.15;

unpaid balance after the second year = $146, 062.42; and

unpaid balance after the third year = $143, 916.03.

Before computing the total interest paid in each of the first three years, we show

first how to do this in general. The idea is that if we know that if we are paying off

a loan with monthly payments of M , and if after k years the unpaid balance is Uk,


then a total of A− Uk has been paid off against the loan. The difference of the total

payments, 12kM − (A− Uk) is the total interest paid during the first k years:

Total interest paid in k years = 12kM − (A− Uk).

Therefore, we see that the total interest paid during the k-th year is the difference

between the total interest paid during the first k years and the total difference paid

during the first k − 1 years. This simplifies to

Interest paid during the k-th year = 12M + Uk − Uk−1 .

Therefore, we find that

interest paid in the first year = 12× $880.13− $148, 099.15 = $8, 649.71

interest paid in the second year = 12 × $880.13 + $146, 062.42 − $148, 099.15 =

$8, 524.83

interest paid in the third year = 12×$880.13+$143, 916.03−$146, 062.42 = $8, 415.17

Before closing this section, we mention the notion of equity in a home. This is

defined roughly as the amount of the home that belongs to the owner. Thus, if on

a $150,000 home the unpaid balance at a given time is still $100,000, we would say

that the owner has $50,000 equity in the home.

Example. Suppose that Mr. Corbett bought a $180,000 home, putting a down

payment of $30,000 and borrowing the balance of $150,000 using a 30-year 6.5%

simple-interest amortized loan. Find his equity after 5 years.

Using the formula for unpaid balance on a $150,000 6.5% 30-year loan, we find

that after five years, Mr. Corbett still owes $140,416.47. Thus says that his equity in

the home (don’t forget the down payment!) is

equity on home = $30, 000 + $150, 000− $140, 416.47 = $39, 583.53.


Suppose that Mr. Corbett sells his home after five years for $200,000. What is his

equity? Here, one simply computes the difference of what the house will sell for,

namely $200,000, and the unpaid balance:

equity on home = $200, 000− $140, 416.47 = $59, 583.53.

Correspondingly, if Mr. Corbett were to sell his home at a loss after five years, say

for $160,000, then his equity would be less:

equity on home = $160, 000− $140, 416.47 = $19, 583.53.

Tax implications

As we saw throughout Chapter 2 interest earned on money will typically carry a tax

liability. However, in the present case— i.e., that of loans—we are actually paying

interest (to the lender) rather than earning it. As a result, we have every reason

to expect that the lender will be paying tax on this money. In order to facilitate

the balance of this discussion, let’s take the specific situation in which one borrows

$150,000 for 30 years at an interest rate of 6.4%. First, we compute that the monthly

payments are $938.26; from page 76 we see that the interest paid during the first year

is $9,550.46.

Thus, it is reasonable to assume that the lender will be paying tax on this $9,550.47.

In an ideal world4 we (the borrower) should not also have to pay money on this money,

for then the government would be “double taxing” the same amount of money. There-

fore, when figuring our own income tax, this sum of money would be subtracted

from our total earnings before computing our tax liability.5 This says, for example,

that if the borrower were in the 31% tax bracket, then his total tax savings would be

Tax Savings = 31% × $9, 550.47 = $2, 960.65.

Let’s return to John Deare and the total cost of his home ownership (see page 71).

4The extent to which the tax policies of various countries reflect an “ideal world” varies greatly!5In the United States, we would only be eligible to subtract this amount if we were to “itemize our deductions.”

In case we don’t qualify to itemize our deductions, then the government would, in fact, double tax the interest paid.


For the purpose of this analysis, we assume that the tax savings shall accumulate at

the end of each of the five years. Also, we need to be sure to compute the present

value of these savings as the total cost of the home ownership is computed at the

time of the initiation of the loan.

30-year mortgage: We recall that the monthly payments on this loan are $742.77.

Next, we can compute the total interest paid in each of the first five years:

interest paid during year 1: $7,520.23;




interest paid during year 5: $7,122.19.

From this, we can compute the yearly tax savings (assuming that Mr. Deare is

in the 31% tax bracket):

tax savings during year 1: $7, 520.23× 31% = $2, 331.27;




tax savings during year 5: $7, 122.19× 31% = $2, 207.88.

Therefore, in present dollars, his savings due to his tax break is the sum of the

present values of each of the above tax savings:

$2, 331.27

(1 +

.063

12

)−1

+ $2, 303.27

(1 +

.063

12

)−2

+ $2, 273.45

(1 +

.063

12

)−3

+

$2, 241.69

(1 +

.063

12

)−4

+ $2, 207.88

(1 +

.063

12

)−5

= $11, 182.43

We now subtract this from the expenditure computed on page 73 and determine

that Mr. Deare’s real expenditure for owning his home for five years

is only $1,869.49.


15-year mortgage: In this case the monthly payments are $1,032.18. The total

interest paid in each of the first five years is as follows:





interest paid during year 5: $5,998.58.

From this, we can compute the yearly tax savings (again assuming that Mr.

Deare is in the 31% tax bracket):





tax savings during year 5: $5, 998.58× 31% = $1, 859.56.

Therefore, in present dollars, the savings due to his tax break is the sum of the

present values of each of the above tax savings:

$2, 299.64

(1 +

.063

12

)−1

+ $2, 199.76

(1 +

.063

12

)−2

+ $2, 093.41

(1 +

.063

12

)−3

+

$1, 980.16

(1 +

.063

12

)−4

+ $1, 859.56

(1 +

.063

12

)−5

= $10, 275.88

From the above work, we infer that that Mr. Deare’s real expenditure for

owning his home for five years is $13,051.77-$10,275.88=$2,775.89 (which

is slightly greater than for the 30-year mortgage6).

6Note that this runs counter to many people’s intuition. Many people simply assert that if one can afford the higher

payments, the shorter mortgage is better as the principle is paid off more quickly. However, the effect of taxation

is often ignored in this “quasi-analysis.” Again, however, we remind the reader that the shorter mortgages typically

incur a lower interest rate, which will change the above comparison; we’ll investigate this further in the exercises.


3.4.1 Prepaid Finance Charges and APR

Assume that simple-interest amortized loan in the amount A is assumed for a period

of n years. If the monthly payments are M , then the total amount that will be paid

out in n will be

Total Payments = 12n×M.

The amount that this amount exceeds the borrowed amount A is what is called the

finance charge:

Finance Charge = 12nM − A.

Example. If a simple-interest amortized loan of $120,000 is assumed at 7.2% for 30

years, then the monthly payments will be $814.55. Over the period of 30 years, a

total of 360× $814.55 = $293, 238.00 will be paid out, resulting in a finance charge of

$293, 238.00 − $120, 000.00 = $173, 238.00.

However, when purchasing a home, there are many other charges that typically

apply to the loan. These fall into two categories: those that fall into the Prepaid

Finance Charges and those that do not.7 To get an idea of what this might look

like, consider the following tables of charges8:

7The determination of which charges fall into which category is dictated by the Truth in Lending Act, referred to

earlier.8See, for example, David B. Johnson and Thomas A. Mowry, MATHEMATICS, A Practical Odyssey, Thomson,

Brooks/Cole, 2004, ISBN 0-534-40059-0, page 367.


Typical Fees Associated with a Home Loan

Fees Included in Prepaid Fees Not Included in Prepaid

Finance Charges Finance Charges

2 points $ 2,500.00 appraisal fee $ 100.00

prorated interest $ 1,100.10 credit report $ 65.00

prepaid mortgage closing fee $ 720.00

insurance $ 450.00 title insurance $ 254.10

loan fee $ 1,500.00 recording fee $ 25.00

document notary fee

preparation fee $ 100.00 tax and insurance $ 723.14

tax service fee $ 25.00 escrow $ 684.80

processing fee $ 75.00

subtotal $ 5,750.10 subtotal $ 2,572.04

We won’t go into many of the specifics, except to say that the “2 points” entry

refers to a fee where one point is equal to 1% of the loan. Therefore, we see that 1

point = $1,250 and so the loan amount was for $125,000.9 All we shall focus is the

total amount contained in the prepaid finance charges. The Truth-in-Lending Act

defines the Legal Loan Amount by the formula:

Legal Loan Amount = Amount Borrowed− Prepaid Finance Charges .

A good way of thinking of the above is as follows. We borrow a certain amount

from the bank, but then they essentially ask us to pay for some of the financing up

front. What’s left over is the legal loan amount, or the amount that we’re actually

borrowing. Therefore, we see that the above table defines a legal loan amount of

$125, 000.00− $5, 750.10 = $119, 249.90.

Next, suppose that the above loan of $125,000 was a 30-year loan at 6.7%. Then we

will have computed monthly payments to be $806.60. The legal definition of annual

percentage rate—APR—in this setting is the percentage rate that we would be

paying in order that 360 monthly payments of $806.60 would exactly pay off the legal

loan amount of $119,249.90. (See pages 64–65.)

Example. When Mr. S purchased his home in 1998, the loan amount was for

9Actually, this type of point is more properly called an “origination point,” as opposed to a “discount point.” We

won’t go into the details of discount points here.


$106,000 at 7.125% to be paid back in 30 years. However, the prepaid finance charges

came to approximately $1,800.00 Find the APR.

To solve this, we first find the monthly payments on the above loan; they are

$714.14. Next, note that the legal loan amount is $106, 000.00−$1, 800.00 = $104, 200.00.

Finally, we recall from Section 3.3 that we need to solve the “hard polynomial prob-

lem”

x12n+1 − (a + 1)x12n + a = 0,

where a = M/A = $714.14/$104, 200.00 ≈ .00685. On the TI-83 calculator, we graph

the above polynomial (with n = 30) and using the window settings as described on

page 68:

Xmin = 360×1.00685361

≈ 1.00406

Xmax = 1.00685

Ymin = a− 1361

(360361

)360(1.00685)361 ≈ −.005147

Ymax = .005147.

Once we have the polynomial x361 − 1.00685x360 + 0.00685 graphed with the above

window settings, we can use the “2nd Calc” button (choose the Zero option) and

arrive at the approximate value x ≈ 1.00608. This gives the final result:

rAPR = 12(x− 1) ≈ .07296;

that is, the approximate APR on Mr. S’s home purchase was 7.3%.

We close this section with a simple observation. Namely, if we know the nominal

interest rate, the APR, and the borrowed amount for a simple-interest amortized loan,

we can estimate the prepaid finance charges. That this is quite simple is illustrated

in the example below.

Example. Suppose that Mr. Tafel took out a 15-year $135,000 loan at 6.2% nominal

interest and that the APR was 6.9%. Approximate his prepaid finance charges.

Note that the monthly payments on the above loan are $1,153.85. Next, we recall

the formula on page 62 relating monthly interest and borrowed amount, where we

use .069 for r and $1,153.85 for M . This gives a value of A, where


A =12M

(1−

(1 + r

12

)−180)

r≈ $129, 174.83.

This says that monthly payments of $1,153.85 at 6.9% for 15 years pay off a loan of

$129,174.83. The difference between this and the actual borrowed amount represents

the prepaid finance charges; they are

$135, 000− $129, 174.83 = $5, 825.17.


1. Suppose that Xiao-Lan Tsai wishes to buy a late-model automobile from Cal

Worthington’s Dodge (in Southern California). The negotiated price of the car

is $12,000 and Xiao Lan has $3,000 available for a down payment. Since she has

very little credit history in the United States, she is unable to find a bank for

financing, therefore, she must settle for financing through the auto dealership.

She is offered an add-on loan for four years at an annual rate of 5.2%, to be

repaid in monthly installments.

(a) What will Xiao-Lan’s monthly payments be?

(b) What will be the total interest she pays?

2. Suppose that an add-on loan of ¥50000 is taken out at an annual percentage

rate of 4.3%.

(a) What will the monthly payments be on a 4-year loan?

(b) What will be the total interest paid on a 4-year loan?

(c) What will the monthly payments be on a 5-year loan?

(d) What will be the total interest paid on a 5-year loan?


1. Sha-Way Tsai wants to buy a used (but very nearly new) car in the United States

that will cost $15,829.32. He has two possibilities for his financing:


(a) A four-year add-on interest loan at 734% and which requires a down payment

of $1,000.

(b) A simple-interest amortized loan through his uncle’s bank at 878% and which

requires a 10% down payment.

(a) Find the monthly payment for each loan.

(b) Find the total interest to be paid for each loan.

(c) Which loan should Sha-Way choose? Why?

2. Christiana Lilly has her eyes on an older used car with a negotiated price of

$4,000. There are two possibilities for her financing:

(a) A three-year add-on interest loan at 6% with a required $300 down payment.

(b) A loan through her father’s credit union; this would be a simple interest

amortized loan at 9.5% which requires a 10% down payment.

(a) Find the monthly payment for each loan.

(b) Find the total interest to be paid for each loan.

(c) What should Christiana do?

3. Suppose that Mr. and Mrs. Smith purchase a home for $165,000. They have

on hand $40,000 for the down payment and the remaining is borrowed from

their bank. They have two choices, take out a 30-year mortgage with an interest

rate of 6.2% or a 15-year mortgage with an interest rate of 5.9%. Compute the

monthly payments in each case.

4. Mr. “S” and his wife purchased a home in 1998 for $132,000, of which $106,000

was financed through the bank. Given that the interest rate was 718% and that

it was a 30-year mortgage, what were the monthly payments? What would the

monthly payments have been had Mr. “S” and his wife taken out a 15-year

mortgage with a nominal interest rate of 678%?



1. Compute the APR for the add-on interest loan offered Sha-Way Tsai in Ex-

ercise 3.2.1, above. Does knowing this provide another way of working Exer-

cise 3.2.1?

2. Compute the APR for the add-on interest loan offered Christiana Lilly in Ex-

ercise 3.2.2, above. Does knowing this provide another way of working Exer-

cise 3.2.2?

3. Compute the APR for the following add-on interest loans:

(a) A = $12, 000, n = 5 (years ), r = 6.1%.

(b) A = 8, 000, n = 4 (years ), r = 4.8%.

(c) A = 15, 000, n = 5 (years ), r = 3.8%.


1. Use either an amortization schedule or the formula we derived for unpaid balance

to find

(a) the first year’s total interest payments, and

(b) the principal balance at the end of the first year

for loan amounts and interests as follows:

(i) 30-year loan of $200,000 at 6.5%

(ii) 30-year loan of $134,000 at 7.0%

(iii) 30-year loan of $125,000 at 4.6%

(iv) 15-year loan of $100,000 at 6.0%

(v) 15-year loan of $180,000 at 5.9%

2. Compute the owner’s equity after five years if the home were purchased according

to the following:

Home Purchase Price: $200,000

Down Payment: $40,000

Term of Loan: 30 years


Mortgage Interest Rate: 5.8%


to the following:






to the following:






to the following:


Estimated Value of Home: $240,000





to the following:


itemEstimated Value of Home: $165,000





7. Recall that Mr. and Mrs. Smith purchased a home for $165,000. They had on

hand $40,000 for the down payment and the remaining was borrowed from their

bank. Suppose that they opted for the 30-year mortgage at 6.2%. Using either

a loan amortization javascript or the formula for unbalance balance, answer the

following:

(a) How much money on the loan will still be owed after the first year (12

payments)?

(b) How much interest will be paid on the loan during the first year (12 pay-

ments)?

(c) How much money on the loan will still be owed after the first two years?

(d) How much money on the loan will still be owed after the first 15 years?

(e) How much total interest will be paid on the loan over the first 15 years?

(f) How much interest will be paid during the 15-th year?

(g) How much equity do they have in the home after 15 years?

(h) Suppose that Mr. and Mrs. Smith plan to stay in their house for only five

years, after which point they will try to sell the house. Assume also that the

value of their house will appreciate at an average of 5% per year. Based on

this, what will be the total cost of owning their house for five years? (In this

problem, ignore any tax breaks that they might enjoy as a result of owning

this home.)

(i) Assume that the interest rate for the 15-year mortgage is 5.9%. If the same

conditions as in (h) above hold, what will be the total cost of owning their

house for five years?

(j) What are your conclusions?

8. Assume that Mr. and Mrs. Smith are in the 22% tax bracket. Recompute

your answers in (h) and (i) above to reflect their tax breaks. What are your

conclusions?

9. Recall above Mr. S’s home purchase in 1998. He sold the home for $147,500 in

2002 after about four years’ worth of payments (48 payments). Complete the

following table:


Cost of Owning a Home

Purchase Price of Home:

Amount Financed:

Monthly Payments:

Sale Price of Home:

Profit from Sale Total Monthly Payments Total Expenses

in Present Dollars in Present Dollars in Present Dollars

10. Mr. S’s taxes were approximately 16% during the ownership of his home. Com-

plete the following table:



Amount Financed:

Monthly Payments:

Sale Price of Home:

Profit from Sale Total Monthly Payments Tax Break Total Expenses

in Present Dollars in Present Dollars in Present Dollars in Present Dollars

11. The house that Mr. S purchased in 1998 was approximately 20 years old. How


would you expect this to influence the total cost of owning the home for four

years? More specifically, let’s estimate the home repairs and maintainence costs

during each of the four years:

Cost for year 1: $1,200

Cost for year 2: $900



Determine a revised estimate for the present (that is, 1998) value of the cost of

Mr. S’s home ownership.

12. Suppose that Max Alero is about to purchase a home for $175,000. He has

$40,000 on hand for a down payment and can get a 30-year loan at 6.4%. Max

expects to live in his home for approximately six years before selling the house

and moving elsewhere. Assuming that houses in Max’s area are appreciating at

an annual rate of 8%, compute the following:



Amount Financed:

Monthly Payments:

Sale Price of Home:

Profit from Sale Total Monthly Payments Total Expenses

in Present Dollars in Present Dollars in Present Dollars

13. Mr. Alero’s taxes will be approximately 18% during the ownership of his home.

Complete the following table:




Amount Financed:

Monthly Payments:

Sale Price of Home:

Profit from Sale Total Monthly Payments Tax Break Total Expenses

in Present Dollars in Present Dollars in Present Dollars in Present Dollars

14. Assume that Mr. Alero’s home is new. Assume that the repairs/maintainence

on his house over the first six years are estimated as follows:







Based on the above estimate, determine a revised estimate of the total cost to

Mr. Alero for owning his home for six years.

15. Max Alero also finds that he can rent a very good apartment in this general area

for $600/month. Calculate, in present dollars, the total expenses associated with

his apartment rental. Do you draw any conclusions?

16. Assume that Max Alero (see the above problem) can find apartment housing in

his area for $600/month, but that the rental rate is likely to go up by 10%/year.

Assuming that he is to live in the area for six years, compute the total rental

cost (in present dollars; use 6.4% as the nominal interest rate).


17. Suppose that Mr. Thompson is considering borrowing $145,000 for the purchase

of a new home. He can take out a 30-year mortgage at 6.4%, and has earnings

that put him in the 25% tax bracket. He plans to live in his home for five years

and hopes to sell it for $200,000.

(a) What is the total cost of Mr. Thompson’s home ownership for the five years?

(b) What would necessary interest rate be for a 15-year mortgage in order that

the total cost of his home ownership would be the same as in part (a)? (This

may take some time.10)

Exercises for Section 3.4.1

1. Mr. and Mrs. Lee take out a 30-year 6.8% loan of $145,000. The prepaid finance

charges associated with this loan amount to $4,789.24. Estimate the APR for

this loan.

2. Ms. McKay purchased a $200,000 home. She put down $50,000 and borrowed the

remainder through her bank, which offered a 30-year 6.4% mortgage. The pre-

paid finance charges added up to $6,721.10. Estimate the APR on Ms. McKay’s

loan.

3. Suppose you have the following choice:

(i) take out a 15-year 6.7% $100,000 loan with an APR of 7.1%, or

(ii) take out a 15-year 6.7% $100,000 loan with an APR of 7.15%.

(a) What can you say about the monthly payments for the two loans?

(b) What can you say about the actual costs of the two loans?

(c) Which is a better loan from the point of view of the consumer?

4. If a 30-year 6.5% loan of $150,000 carries an APR of 7.4%, how much are the

prepaid finance charges?

5. If a 15-year 5.9% loan of $136,000 carried an APR of 6.2$, how much are the

prepaid finance charges?

10I am indebted to Mr. Taullie Thompson at SAS for suggesting this problem.


Project: Variations on a home loan.11 Mr. Williams, in a somewhat vain

attempt to impress friends, decided to purchase a home in the exclusive subarbs for

$245,000. Unfortunately, he had no money on hand for a down payment, so here’s

what he did. First of all, he was able to borrow 80% of the loan from his bank, using

a 30-year 11.5% simple-interest amortized loan. For the remaining 20% of the loan,

he was able to convince the seller to take a second mortgage for the remaining 20%.

What this means is that the seller lends him the additional 20%; the terms of this

second mortgage are that

1. Mr. Williams will pay the seller monthly interest-only payments at 12%

nominal interest rate. (Thus, each month, Mr. Williams will pay the seller the

amount .01× 20% of $245, 000).

2. At the end of five years, Mr. Williams will pay to the seller the full amount that

he borrowed (the 20% of $245, 000); this is often called a balloon payment.

Now answer the following questions:

(a) How much did Mr. Williams borrow from the bank and how much did he borrow

from the seller?

(b) Find Mr. Williams’ monthly payments to the bank.

(c) Find Mr. Williams’ monthly payments to the seller.

(d) Assume that Mr. Williams is to save for his ballon payment with a sinking fund

that earns 7%. How much money each month must be put into this sinking fund

each month to have the necessary money on hand for the balloon payment (five

years from now).

(e) What are Mr. Williams’ total monthly payments for the first five years?

(f) Assume that at the conclusion of the five years, he sells his home for $420,000.

Compute the present value of the total cost of owning his home. (Use 7% interest

as the basis by which to compute present values. Also, ignore the effect of any

tax breaks.)

11The specifics of this project were fairly commonplace during the inflationary period of the late 1970s and early

1980s.


(g) Compute the present value of Mr. Williams’ home ownership if he is in the 31%

tax bracket. (Remember to combine the interest paid to both sources.)


Chapter 4

Basic Economic Theory

4.1 Cost, Revenue, and Profit Functions

The main problem of economics is the description and analysis of the production,

distribution, and the consumption of goods and services. The production and dis-

tribution of the goods/services is the province of what we shall call a “company,”

(or “producer”) and the consumption is by what we (obviously) call “consumers.”

There is a delicate balance—or tension—here inasmuch as the company will only

produce when it’s sufficiently profitable to do so. At the same time, the rest of us—

the consumers—will buy the output of the company only when (a) we need or want

the company’s product, and (b) when the product comes at a fair price. As we’ll see

below, these considerations lend themselves quite well to mathematical analysis.

We’ll begin from the point of view of the producer and give a simple mathematical

description of the company’s profit obtained by selling a product. A necessary ingre-

dient for the company to turn a profit is the revenue obtained by selling a product.

This is, of course, just the money taken in as the result of selling x items.1 So if the

company is to charge a flat rate of p dollars per item, the the result of selling x items

provides a revenue of

R(x) = xp (dollars) .

The graph of a hypothetical revenue function is depicted below:

1Of course, an “item” must be taken in the generalized sense. If we’re talking about services, then it would be

reasonable to think of an “item” as being a unit of time (say one hour of service).

95

96 CHAPTER 4. BASIC ECONOMIC THEORY

-

6

��

$1000

$2000

50 100

R = R(x)

x (units)

R (Revenue)

From the above graph, we infer that when x = 50, the total revenue is R ≈ $700.

This tells us—since the revenue function is linear—that the units are being sold at

approximately $14 per unit.

Next, we need to take into the account the cost of producing x units of output.

Note, however, that this cost can be divided into two principal sources: the fixed

cost and the variable cost. The fixed cost includes the cost of setting up the

business (or company) in the first place, but can also account for regularly-occurring

costs such as utilities, payroll, and related costs. If the company is a large-scale

factory, then all of the equipment that will be used in producing the output is part of

the fixed cost; in this example, the fixed cost is likely to be quite high. On the other

hand, if the company basically sells “human services,” (say a consulting company),

then the fixed costs are likely to involve the physical building (office) and utilities, the

needed office supplies (chairs, desks, computers, etc.), and some of the payroll. Thus,

there is a fixed cost even before the first unit of output is produced. Next, there is

the variable cost, which is the cost of actually delivering x units of the good being

consumed.

4.1. COST, REVENUE, AND PROFIT FUNCTIONS 97

Economy of Scale.

In practice, it is frequently more efficient to produce a larger number of goods than a

smaller number of goods.2 This means that the cost of producing items has a graph

that is more steeply sloped near the level of zero production and then tapers off a

bit. However, it’s possible to “overproduce,” at which point the cost can rise sharply

again (perhaps generated by the overtime pay for the employees or the extra wear

and tear on the equipment). These considerations result in a cost function that often

has the following characteristic “S” shape:

-

6

x (units)

C (Cost)

economy of scale

� -

C = C(x)

2What we mean by “efficient” is contained in the notion of marginal cost; we’ll encounter this in sectionmarginal.


Finally, the profit function is given, simply, by the difference between the revenue

and the cost functions:

Profit P (x) = R(x) − C(x) .

The Break-Even Point

Whenever there is a fixed cost in the production of goods or services (which is almost

always), it is clear that a certain minimal number of the items much be sold in order

cover the fixed costs. Thus, if there is a million dollar startup cost in setting up

a consulting office, and if the consulting fees are $150/hour, then many hours of

consulting must be sold in order even to cover the fixed costs.

The minimum number of units that need to be sold in order to match the cost

of production is called the break-even point; this is indicated on the graph below.

Where the revenue graph is above the cost graph, a profit is realized.

-

6

��

x (units)

y

y = R(x)

y = C(x)

break-even point

4.2. MARGINAL COST, REVENUE, AND PROFIT 99

4.2 Marginal Cost, Revenue, and Profit

Marginal Revenue. If a company charges a price of p dollars per unit of output,

then the amount p is sometimes referred to as the marginal revenue. Note that p

can be interpreted mathematically as the slope of the graph of the revenue function

(see again the graph on page 96).

The concepts of marginal cost and marginal profit aren’t quite as simple to define

as the quantities defined can depend on the level, x, of the output.

Marginal Cost. If a company produces x units of output, then the corresponding

marginal cost of so doing is defined as the cost of producing the x+1-st item. What

this means is that the marginal cost is computed via the difference:

MC(x) = C(x + 1)− C(x).

Geometrically, this quantity can be approximated by the slope of the tangent

line to the graph of the cost function.3

-

6

��

x (units)

C (Cost)

slope of line = marginal cost

C = C(x)

x units of output

3Thus, the student who has taken some calculus can see why methods of calculus are often applied to economics

and finance.


From the above geometrical interpretation of the marginal cost, we see that from the

point of view of “economy of scale,” the most cost-efficient level of output occurs

where the slope of the line tangent to the cost curve is a minimum:

-

6

((((((((

((((((((

((((((((

((((

x (units)

C (Cost)

slope of line =

minimal marginal cost

C = C(x)

x units of output

(optimal output)

Marginal Profit. Since profit is just defined as the difference P (x) = R(x)−C(x),

we see likewise that the marginal profit is the difference

MP (x) = p−MC(x),

where we recall that the marginal revenue is simply the unit cost p of each item.

As with the marginal cost, we may also interpret MP (x) as the profit obtained

by selling the x + 1st item, given a production level of x. From the point of view of

the seller, the objective is to set the production level to that for which the marginal

profit is zero. We can see this as follows. First of all, when we study the revenue and

cost curves together, we see that those production levels for which the cost exceeds

the revenue will result in a loss (i.e., a negative profit) for the seller. Similarly, where


the revenue exceeds the cost, the seller realizes a net profit. This is indicated below:

-

6

��

x (units)

y

y = R(x)

y = C(x)

loss region

(negative profit)

R(x) < C(x)

profit region

(positive profit)

R(x) > C(x)

We now graph the profit function P (x) = R(x)− C(x) below:

-

6

optimal output level

@@@@@@@R

x units of output

y

y = P (x) = R(x) − C(x)


In the above graph we have identified the highest point on the graph (i.e., the

maximum profit); at the same time, note that the slope of the tangent line is zero.

In other words, the maximum profit will occur when the marginal profit is zero.

Intuitively, this makes sense. Indeed, if the seller is operating at a level where the

marginal profit is positive, then raising the output level one unit will raise the total

profit—which is what the seller wants. Thus, this level of output won’t be optimal.

Similarly, if the marginal profit is negative, then reducing the level of output by one

unit will also raise the overall profit. The best is, then, to operate as closely as

possible to the level x which realizes zero marginal profit: MP (x) = 0.

4.2.1 Maximizing Profit—Cubic Cost Functions

In the above, we have argued that typical cost functions can be modeled by cubic

functions. In this subsection we shall provide a trick for maximizing profit when

the (cubic) cost function C(x) is known as a cubic polynomial4 and when a revenue

function R(x) = xp is given. For students already having had calculus, this is all

very trivial stuff; those not having had calculus should simply regard this as a handy

trick.

The trick is founded on the derivative, which will associate a quadratic function

with a cubic function. This is given by the formula

d

dx

(ax3 + bx2 + cx + d

)= 3ax2 + 2bx + c.

Based on the above, we have the following useful facts.

Useful Fact 1. Give the cubic cost function C(x), the derivative ddxC(x) gives the

marginal cost at any point x, i.e., MC(x) = ddxC(x).

Useful Fact 2. Assume that the profit function is given by P (x) = R(x)−C(x),

where C(x) is a cubic cost function and R(x) = xp. Then the maximum of P (x)

occurs at the largest zero of ddxP (x).

Discussion. What’s going on is this. Inspection of the cost/revenue graphs given

above (and in the exercises) shows that the maximum profit occurs when the separa-

tion of the revenue curve above the cost curve is the widest. Closer inspection reveals4On page 133, we consider an example where we show how to use a TI-83 calculator to approximate cost data by

a cubic polynomial.


that this separation occurs when the marginal cost is equal to the marginal (i.e., per

unit) revenue. Indeed, this makes sense as if the marginal profit at a given level, x,

of output is zero, then there is no benefit in increasing (or decreasing) production.

That is, we are already operating at an optimal level.

The fact that we must take the largest zero of ddxC(x) is a consequence of the

fact that this quadratic will tend to have two zeros. The smaller zero (which might

be negative) will occur where the cost has a maximum separation above the revenue

function. This would be, of course, where the loss is at a maximum.

Simple Example. Suppose that we are given the production of items with daily

revenue and cost functions as below:

R(x) = $784x and C(x) = x3 − 60x2 + 1400x + 1000,

where x represents 100s of units produced (per day), and C(x) is measured in dollars.

Determine the production level that maximizes profit.

Solution. We have that the profit function is given by

P (x) = −x3 + 60x2 − 616x− 1000.

The derivative is therefore given by

d

dxP (x) = −3x2 + 120x− 616;

we set this equal to zero and compute the larger of the two solutions:

−3x2 + 120x− 616 = 0 =⇒ x =−120−

√1202 − 4 · (−3) · (−616)

2 · (−3)≈ 33.95.

This says that the maximum profit is realized by setting the daily production level

at about 3,395 items/day. Note that at this level of production, the profit realized is

P (33.95) = −33.953 + 60 · 33.952 − 616 · 33.95− 1000 ≈ $8, 112.13

per day.


What might have occurred to the reader in studying the above example is that if we

were simply to increase the per unit price of each item, then since P (x) = xp−C(x),

the profit will obviously increase. Indeed, in this—and all examples up to this point—

we have assumed that at a given price, the company can sell everything at the given

price. Experience tells us that this is obviously not true; rather the propensity of

consumers to purchase at a given price is the subject of supply and demand, taken

up in the next section.

Before finishing this subsection, we shall return to the initial interpretation of the

derivative, namely that

Derivative of the Cost Function = Marginal Cost .

From the above simple observation, we can determine the output that is optimal in

sense of giving minimal marginal cost, and therefore the “best” economy of scale.

Since the derivative of a cubic function is a quadratic function, we are faced with

minimizing a quadratic function. We remind the readers how this is carried out.5 If

a quadratic function

Q(x) = ax2 + bx + c

is given, and if we know that a > 0, then the graph is a parabola whose vertex

designates the minimum of the function (the y-coordinate of the vertex) as well as

where this minimum value occurs (the x-coordinate of the vertex). The idea is based

on completing the square:

Q(x) = ax2 + bx + c

= a

(x2 +

b

ax

)+ c

= a

(x2 +

b

ax +

b2

4a2− b2

4a2

)+ c

= a

(x +

b

2a

)2

+

(c− b2

4a

)From the above, one sees immediately that

5This is standard material in any algebra course!

4.3. SUPPLY AND DEMAND 105

The minimum of Q(x) occurs at x = − b

2awith minimum value

(c− b2

4a

).

Simple Example, Continued. We consider the example taken up on page 103,

where the cost function

C(x) = x3 − 60x2 + 1400x + 1000

was given. Compute the output level where the marginal cost of production is a

minimum.

Solution Here, we begin as above and compute the derivative of the cost function;

after doing this, however, we complete the square:

d

dxC(x) = 3x2 − 120x + 1400

= 3 (x− 20)2 + 200.

This implies that the optimal level of production (in terms of ecomony of scale)

is 2,000 units/day. Furthermore, at this level of production, the marginal cost is

$200/item.

4.3 Supply and Demand

In the previous discussions, we have studied a company’s profit under the simplifying

assumption that all items produced will be bought at the given price. Of course, were

this really the case, then all the producer has to do is to continue to raise the price,

thereby increasing the marginal revenue and thereby increasing the marginal (and

therefore total) profit! However—and as experience tells us—the price of an item is

not altogether determined by the seller; rather it is typically the market itself that

determines the price. Let us explain.

We begin with a few general remarks. If a company produces a certain good—

widgets—to be sold to the public, then our collective experience tells us that we are

generally more inclined to buy these widgets if the price is relatively low and are


relatively less inclined to buy these widgets if the price is high. This information

might be summarized in what is called a demand curve; a typical such curve would

follow the general shape below:

-

6

Demand

x (units)

p (price/unit)

Note that the curve says that if the price per unit is high, then the demand for

this quantity is low. Correspondingly, if the price per unit is low, then more of this

quantity is demanded. This makes sense!

From the point of view of the producer, there is another curve—the supply

curve—that will dictate his tendency to supply the product at a given price. Obvi-

ously, if the consumers are not willing to pay very much for a widget, then there’s

not much benefit for the producer to put very many widgets on the market. How-

ever, if the consumer is willing to pay more for these widgets, then the producer will

correspondingly supply more. Again, we can say this with a graph:


-

6

Supply

x (units)

p (price/unit)

Equilibrium. It is natural to assume—and we shall do so—that the market forces

dictated by the supply and demand will cause the production level and price to agree

with both of these constraints. This agreement gives rise to and equilibrium price

and an equilibrium production, as indicated on the graph below:

-

6

(x∗, p∗)

•

SupplyDemand

x (units)

p (price/unit)

In the above, x∗ represents the equilibrium level of production and p∗ represents the


equilibrium price.

4.3.1 The Effect of Taxes on Equilibrium.

Two types of taxes that one often encounters are sales taxes and specific taxes.

Sales taxes typically are for a percentage of the sales price (as the name indicates)

and are charged directly to the consumers. Specific taxes are typically per-unit taxes

and are collected from the producers. Typical examples of specific taxes are those

in the U.S. on gasoline, alcohol, and cigarettes. We shall only consider the effect on

equilibrium prices by specific taxes.

In order to fix our ideas, we consider a particularly simple and idealized pair of

supply and demand functions—linear ones, given by the equations

S(p) = 5p− 200 and D(p) = 500− 2p,

where p is unit price in dollars and the units are measured in 1000s of items/day.

-

6

��

HHHHH

HHHHHH

HHHHHH

HHHHHH

HHHH

Supply

Demand

v(x∗, p∗)

x (units)

p (price/unit)

The equilibrium point (x∗, p∗) is determined by setting supply = demand: 5p−200 =

500 − 2p, from which one concludes immediately that the equilibrium price is p∗ =


$100/item, at which price the the equilibrium production will be 300,000 units/day.

Suppose now that the government decides to place a specific tax of $8/item; we

would like to compute the effect this has on our over-all picture. Remember, this is

a tax that is supposed to be paid by the producer.

The first step in our analysis is to observe that instead of receiving p dollars per

unit, the producer will only get p−8 dollars per unit. At this new price, the producer

is only willing to supply S(p− 8) = 5(p− 8)− 200 = 5p− 240 (thousand) units/day.

That is to say, 5p − 240 is the new supply; setting supply = demand now produces

the equation

(supply) S(p− 8) = 5p− 240 = 500− 2p = D(p) (demand),

which now yields the new equilibrium price p ≈ $105.71 per unit. In other words,

the new equilibrium price is now $105.71, which is almost 6 dollars higher than the

original equilibrium price. Furthermore, since the new price—$105.71—is the price

that consumers are willing to pay at the given supply, we see that the consumers

end up paying $5.71 of the specific tax, and the producer pays only $2.39 of

the specific tax. In other words, the market forces—and not the unethical behavior

of the producers—have passed most of the tax onto the consumers!

4.3.2 Utility and Demand as Marginal Utility

One means often used by economists to understand the principle of demand is to

consider a single consumer (or homogeneous group of consumers—say the members

of a family) and consider the utility that a particular commodity has. First of all,

utility can be thought of as being a measure of “happiness” or satisfaction that a

consumer derives from owning a certain quantity of a product. One might consider,

for example, the (total) utility for a typical teenage consumer to own a number

of music CDs. We would expect that the more CDs that this teenager owns, the

happier (s)he would be. Thus, if we were to graph this situation, the result would

be an increasing graph. However, after a while, the teenager’s level of happiness will

saturate after the number of CDs owned becomes very large, producing what would

be called diminishing marginal utility. This is all indicated below:


-

6

UtilityHappiness

(Number of CDs Owned)

Note that the diminishing marginal utility is indicated by the fact that the curve

levels off as the number of CDs owned increases.

In the present situation it makes sense to replace “happiness” by “price,” as it

is reasonable to assume that the above happiness comes at a price—the greater the

happiness, the greater the price. Thus, we can quantify the above two axes of the

above graph in order to draw more quantifiable conclusions. One such quantification

might result in the following:


-

6

100

200

10 20

Total UtilityPrice

x (Number of CDs Owned)

We interpret the above graph as follows. The typical teenager is willing to pay

roughly $75 for the first 10 CDs and is willing to pay roughly $120 for first 20 CDs.

We can get a better understanding by looking more closely at the marginal utilities

at the levels x = 10 and at x = 20:6

6Recall that marginal quantities are approximated by slopes of tangent lines.


-

6

100

200

,,,,,,,,,,

��

slope ≈ 5

slope ≈ 313

10 20

Total Utility

Price


From the above, one can approximate the slopes of the above two tangent lines: the

line tangent to the utility curve at x = 10 has slope ≈ 5, and the line tangent to the

utility curve at x = 20 has slope ≈ 313. That is,

• the marginal utility at x = 10 is roughly $5, which means that the teenager

in question would pay roughly $5 for the 11-th CD; similarly,

• the marginal utility at x = 20 is roughly $3.33, which means that the teenager

in question would pay roughly $3.33 for the 21-st CD.

In order to relate the above discussion to the notion of demand, we consider the

following general question.

Suppose that a consumer has a given total utility function p = U(x) for a com-

modity and that the given commodity is priced at a price p0. How many items of this

commodity should the consumer purchase?

Naturally, in order to answer this question, we need to know how many items of

the commodity are already owned by the consumer.


We shall address the above question by first continuing the example of the teenager

and the CDs. Before doing this, however, we shall find it helpful to tabulate some of

the marginal utilities (again, determined by approximating the slopes of the tangent

lines):

Marginal Utility of CD Purchases

Number of CDs Owned Marginal Utility

0 $10.00

5 $6.67

10 $5.00

15 $4.00

20 $3.33

25 $2.86

30 $2.50

35 $2.22

40 $2.00

Consider now the following situations:

(a) The teenager presently owns no CDs and the CDs go on sale for $7 each.

(b) The teenager presently owns 10 CDs and the CDs go on sale for $7 each.

(c) The teenager presently owns no CDs and the CDs go on sale for $5 each.

(d) The teenager presently owns five CDs and the CDs go on sale for $5 each.

(e) The teenager presently owns no CDs and the CDs go on sale for $2 each.

(f) The teenager presently owns 20 CDs and the CDs go on sale for $2 each.

The general principle to be applied here is that of the so-called optimal purchase

rule, which says

Optimal Purchase Rule: If the sale price of an item is less than the marginal

utility at a given level, purchase one more of this item.


We now take up each of the above situations in turn, which will reinforce the

optimal purchase rule.

(a) The teenager presently owns no CDs and the CDs go on sale for $7 each. Dis-

cussion: Here, the marginal utility is $10, which is greater than purchase price.

Therefore, the teenager should keep purchasing CDs until the marginal utility

falls to below $7. Thus, we can assume that by following this rule, the teenager

will buy 4–5 CDs (let’s say 5), but no more, as the marginal utility once the

teenager already owns five CDs has dropped to $6.67.

(b) The teenager presently owns 10 CDs and the CDs go on sale for $7 each. Dis-

cussion: If the teenager already owns 10 CDs, then since the marginal utility

at that level is only $5.00, there will be no motivation to purchase any additional

CDs at $7 each.

(c) The teenager presently owns no CDs and the CDs go on sale for $5 each. Dis-

cussion: Here we see that the optimal purchase rule will prompt the teenage

to buy 10 CDs (maybe even the 11-th).

(d) The teenager presently owns five CDs and the CDs go on sale for $5 each.

Discussion: The teenager will buy five (or six) more CDs.

(e) The teenager presently owns no CDs and the CDs go on sale for $2 each. Dis-

cussion: Since the marginal utility at the level x = 40 CDs is $2.00, we see

that the teenager (if (s)he abides by the optimal purchase rule) will run out and

purchase 40 (or 41) CDs.

(f) The teenager presently owns 20 CDs and the CDs go on sale for $2 each. Dis-

cussion: Here, the teenager will buy 20–21 new CDs.

Next, we shall consider the optimal purchase rule from a different perspective.

Namely, if the items (in the present case CDs) are sold at a given price p, and if the

consumer buys x of these, then we call the difference between the consumer’s total

utility and the price of purchasing x items the consumer surplus. Of course, this

amount might be negative. This terminology makes sense, as the consumer’s total

utility in buying x items is the amount that the consumer would theoretically pay for


these x items. The amount C = xp is the actual amount that the consumer would pay

for these x items; thus the consumer surplus is the money “saved” by the consumer

in purchasing x items.

We illustrate this below for the teenager who is to purchase CDs at a fixed price of

$5 per CD. The graph below roughly indicates the consumer surplus at the approxi-

mate levels of x = 4, x = 10, and x = 16 CDs. If the teenager is to determine the

number of CDs to buy on the basis of maximizing consumer surplus, then we can see

that, graphically, x = 5 will yield the optimal purchase, which will yield roughly a

$30 consumer surplus.

-

6

%%%%%%%%%%%

C = 5x (What the consumer actually pays)

100

200

10 20

Total Utility (What the consumer iswilling to pay)

Price


A close look at the above graph will convince the reader that at this maximum

consumer surplus, the slope of the tangent to the total utility curve is the same as

the slope of the cost curve. That is to say, at the optimal buying level,

Marginal Utility = Unit Sales Price ,

which is precisely the optimal purchase rule referred to above. In other words, the

optimal purchase rule is the same as maximizing consumer surplus!

Finally, we note that at a given price p the consumer who uses the above optimal

rule will buy—that is, demand—x items, where

x = MU(p) (Marginal Utility).

Put differently, this says that


(Demand) D(p) = MU(p) (Marginal Utility);

since marginal utility decreases for increasing prices, we have another way of seeing

why demand curves are downward-sloping curves!

4.3.3 Elasticity of Demand

We begin this discussion by comparing two items frequently purchased by consumers:

flour and home furnature. While both will have downward-sloping demand curves,

there is one very important difference. Since we consider flour a necessity, small

changes in the price are not likely to have very significant changes in the demand for

this staple. Furnature, on the other hand, is something that we won’t buy if the price

goes up very much. Correspondingly, if furnature goes on sale, we might buy some

pieces of furnature even if we don’t need it. Thus, we might say that the demand for

flour is inelastic whereas the demand for furnature is elastic.

A way of quantifying the above is to measure the percentage change in demand

produced by a certain percentage change in price. For example, suppose that the price

for a commodity rises from $5 to $6 and that the corresponding demand decreases

from 15,000 (units/day) to 10,000 (units/day). Thus, the price has increased by an

percentage amount:

Percentage Price Increase =$6− $5

$5= 20%.

The corresponding change in demand is

Percentage Demand Decrease =15, 000− 10, 000

15, 000= 33

1

3%.

Economists typically combine the above results into a single number, called the elas-

ticity of demand:

Elasticity of Demand =percentage decrease in demand

percentage increase in price.

The above is usually expressed more mathematically, by expressing it as

Elasticity of Demand E =

∣∣∣∣∆x/x

∆p/p

∣∣∣∣ ,


where p represents price, x represents demand and the notation ∆ represents “change.”

The absolute value symbols are put in just to guarantee that the elasticity E is always

a positive number. It is customary to call the demand elastic if E > 1 and inelastic

if E < 1.

For example, we can consider the various elasticities for a given commodity:7

Price p 10 11 12 13 14 15

Demand x 800 758 712 662 608 550

Revenue R = xp 8000 8338 8544 8606 8512 8250

Elasticity E 0.5 0.64 0.81 1.02 1.29 1.64

Thus, we see from the above table that the elasticity of demand for a commodity will,

in general, depend on the price. We’ll return to this observation shortly.

We shall note in passing (this is for the benefit for those students having had some

calculus!) that the elasticity of demand can be related to the marginal demand by

noting that the ratio ∆x∆p

is the marginal demand. Since it can be approximated by

the slope of the line tangent to the demand curve at the price p, we may approximate

the elasticity by the price derivative of demand:

Elasticity of Demand E =

∣∣∣∣px · dxdp∣∣∣∣ .

In summary, we may conclude that if the elasticity of a commodity is less than

1, then small changes in price do not significantly change the demand. On the other

hand, if the elasticity is greater than 1, then small price changes can cause large

changes in demand.

Example. Suppose that one determines that if the owner of a small clothing store

were to raise the price of his $100 suits to $115, then the change in demand will

decrease by 10%. What is the elasticity of demand at the price level of $100? Should

the owner raise the price?

Solution. Note first of all that the relative price increase is

7This example is taken from D. Hughes-Hallet, et. al, Applied Calculus, Second Edition, John Wiley & Sons,

2002, ISBN 0-471-27-18-4, page 195.


Relative Price Increase =$115− $100

$100= 15%.

Since the elasticity of demand is percentage decrease in demand divided by the per-

centage increase in price, we see that the elasticity is 10/15 = 0.67. We may intuitively

interpret this result as follows. Namely, since the elasticity is less than one, the con-

sumers want these suits badly enough that a even a price increase won’t significantly

diminish their demand for these suits. More quantitatively, the new revenue for the

seller will become

R = (x− .10x)(p + .15p) = (0.90)(1.15)xp = 1.035xp,

that is to say, despite the fact that the demand will have decreased, the revenue will

increase. Thus, the seller would benefit by raising the price of his suits. Note that

in terms of percentages, the seller’s revenue actually increases by 3.5%.

The above example suggests a selling strategy based on elasticity. Namely, if the

elasticity is less than one, then raising the price will not adversely affect the demand

and the total revenue will increase. Correspondingly, if the elasticity is greater than

one, then any further price increases will result in large decreases in demand and

corresponding decreases in revenue. In other words, we have the following

Optimal Pricing Rule. The seller’s optimal selling strategy is to fix the price as

closely as possible to a level where the elasticity is one.

Example. Note that the table on page 117 clearly illustrates a maximum revenue

near where the elasticity is one.

Example. Suppose that the demand for a certain commodity roughly followes a

linear function of the form x = D(p) = 1200− 3.1p, where x is in 1000s of items per

week, and p is the price per 100 items. We may visualize the graph below:


-

6

QQQQQQQQQQQQQQQQQQQQ

Demand

D(p) = 1200− 3.1p

p (price per 100 units)

x (1000s)

For this function it is fairly simple to compute the elasticity of demand. First of

all, for any linear function L = L(p) = mp + b, one can compute the relative change

very easily. If p changes from an amount p1 to an amount p2, then the change ∆p

is given by ∆p = p2 − p1. Corresponding to this change in p is the change ∆L in L:

∆L = (mp2 + b)− (mp1 + b) = m(p2 − p1). Therefore, the relative change has a very

simple expression:

∆L

∆p=

m(p2 − p1)

p2 − p1

= m.

From the above, we may now compute the elasticity of demand for the above linear

demand function:

E(p) =

∣∣∣∣px · ∆x

∆p

∣∣∣∣ =

∣∣∣∣ 3.1p

1200− 3.1p

∣∣∣∣ . (4.1)

Notice that the above demand makes sense only when it is positive, namely when

p ≤ 12003.1≈ 387. We shall consider prices only in this region and solve for the regions

of inelasticity and elasticity:

Region of inelasticity:3.1p

1200− 3.1p< 1,


Region of elasticity:3.1p

1200− 3.1p> 1.

Since all quantities are positive, we may be “naive” in our approach to the above

inequalities:

3.1p

1200− 3.1p< 1 ⇐⇒ 3.1p < 1200− 3.1p

⇐⇒ 6.2p < 1200

⇐⇒ p <1200

6.2≈ 194.

In an entirely similar fashion, the region of elasticity would occur when p > 194.

We summarize in a picture:

-

6


Demand

D(p) = 1200− 3.1p

p (price per 100 units)

x (1000s)

(region of inelasticity) (region of elasticity)

p = 194(E = 1)

From the above discussion, we see that the optimal pricing (from the point of view

of maximizing revenue) is p = 194, which translates to $1.94/unit.

We may easily generalize the above derivation to any linear demand function,

resulting in the following optimal pricing rule.

Optimal Pricing Rule—linear demand functions. Assume that we are given

a linear demand function D(p) = b−mp. Then,

(a) if p < b/2m, then E(p) < 1, i.e., the demand is inelastic;


(b) if p > b/2m, then E(p) < 1, i.e., the demand is elastic; and

(c) if p = b/2m, then E(p) = 1, which give the optimal price.

4.3.4 Shifts in Supply and Demand

We close our brief discussion on supply and demand with a few simple concluding

remarks. Namely that for a given commodity, there is no single supply (or demand)

curve dictating the behavior of this commodity. Many factors can affect the sup-

ply/demand curve, sometimes causing an abrupt shift in this curve. For example, if

popcorn is the commodity and if there is a sudden shortage of popping corn (say, due

to a lower than expected corn harvest), then the supplier of pop corn will be paying

more for his corn and will only be able to supply the same amount of popcorn at a

higher price. This would result in a shift of the supply curve to the right:

-

6

Popcorn supply curve

after shortage-

Original popcorn

supply curve�

x (units)

p (price/unit)

Demand can be likewise affected. For example, the demand for popcorn at a given

price in the supermarket would be far less than the demand for popcorn at the same

price in a movie theater (where prices for snacks and beverages is typically much

higher). This would be visualized as follows:


-

6

Demand for

popcorn in a

supermarket

-

Demand for

popcorn in a

movie theater

�

x (units)

p (price/unit)

Thus, we see that supply and demand are very “dynamic;” they respond—sometimes

instantly8—to world events.

4.4 Optimal Use of Resources—Linear Programming

In order to motivate the central idea, we begin with a simple problem. Suppose that

a manufacturer is to produce two products, say sailboats and motorboats. Suppose

next that the sale of each sailboat will result in a profit of $1200 and that the sale of

each motorboat will result in a profit of $1400. Of course, everything else being equal,

there would appear not to be much point in manufacturing sailboats, as it seems that

making and selling motorboats is more profitable. However, there are constraints

to this problem. Namely, we assume that the manufacture of each of these boats

requires two stages: Stage A and Stage B. More specifically, let’s assume that

The manufacture of each sailboat requires 20 hours of stage A work and 30 hours of

stage B work, and that

the manufacture of each motorboat requires 30 hours of stage A work and 10 hours

of stage B work.8A very good example is the supply and demand for gasoline when there is turmoil in the Middle East.

4.4. OPTIMAL USE OF RESOURCES—LINEAR PROGRAMMING 123

Next, we assume that this small company can only allocate a maximum of 1000

hours/week to the stage A manufacturing process and a maximum of 870 hours/week

to the stage B process.

Question: What is the optimal weekly production level for these two types of

boats? That is to say, how many sailboats should be manufactured each week and

how many motorboats should be manufactured each week in order to maximize profit

subject to the above constraints?

Discussion: We can, without the introduction of any new mathematics, set up the

above problem. We start by letting

x = number of sailboats produced each week, and

y = number of motorboats produced each week.

It is convenient to tabulate the information as follows:

Stage A Stage B

Sailboats (x) 20 30

Motor Boats (y) 30 10

Max Available

Hours (Weekly) 1000 870

Maximize: P (x, y) = 1200x + 1400y

Thus, we want to maximize the profit function

P (x, y) = 1200x + 1400y,

subject to the constraints

20x + 30y ≤ 1000

30x + 10y ≤ 870

x ≥ 0

y ≥ 0.

We won’t attempt to solve the above problem until somewhat later in this section.


We consider a similar problem below. Suppose that we have three juices available,

apple juice, orange juice, and grapefruit juice. From various mixtures of these juices

we are going to make some “new” juices—each of which come in 4 litre bottles—as

follows:

Classic Mix, made by mixing one litre of apple juice, one litre of orange juice, and

two litres of grapefruit juice.

Party Mix, made by mixing two litres of apple juice, one litre of orange juice and

one litre of grapefruit juice.

Orange Zap, made by mixing one litre of apple juice, two litres of orange juice,

and one litre of grapefruit juice.

The suppliers of the individual flavors—apple, orange, and grapefruit have limited

the daily amounts that they can provide:

the apple juice supplier can supply at most 150 litres of apple juice per day;

The orange juice supplier can supply at most 200 litres of orange juice per day; and

the grapefruit supplier can supply at most 170 litres of grapefruit juice per day.

Finally, assume that we can make a profit of $0.75 per (four-litre) bottle of Classic

Mix, a profit of $0.60 per bottle of Party Mix, and a profit of $0.85 per bottle of

Orange Zap.

Question: What is the optimal daily production level of the above fruit juice

mixes?

Discussion: As above, we start by setting

x = number of bottles of Classic Mix produced daily

y = number of bottles of Party Mix produced daily, and

z = number of bottles of Orange Zap produced daily

We shall find it convenience to tabulate the needs below:


Apple Juice Orange Juice Grapefruit Juice

Classic Mix (x) 1 1 2

Party Mix (y) 2 1 1

Orange Zap (z) 1 2 1

Max Daily Supply 150 200 170

Maximize: P (x, y, z) = 0.75x + 0.60y + 0.85z

Therefore, we see that the problem the above problem is to maximize the function

P (x, y, z) = 0.75x + 0.60y + 0.85z

subject to the constraints

x + 2y + z ≤ 150

x + y + 2z ≤ 200

2x + y + z ≤ 170

x ≥ 0

y ≥ 0

z ≥ 0.

4.4.1 Feasibility Regions and the Objective Function

Both of the problems discussed above share a common feature. Both are concerned

with the optimization (in this case, maximization) of an objective function, sub-

ject to (linear) constraints. The set of all variables that satisfy the constraints is

called the feasibility region. Linear programming is the mathematics involved

in solving such problems.

In the case in which only two variables are involved (as in the first example consid-

ered above), the feasibility region can easily be graphed as a region in the plane. In

the case of the first example considered above, we get the following feasibility region:


20x + 30y ≤ 1000

30x + 10y ≤ 870

x ≥ 0

y ≥ 0.

-

6

BBBBBBBBBBBBBB

QQQQQQQQ

x

y

Feasibility

Region

-

25 50 75

25

50

75

In general, the two-dimensional linear programming problem problem is simply

the problem of optimizing (i.e., finding the maximum or the minimum of) a linear

objective function of either of the forms:

F (x, y) = ax + by

subject to a number of constraints of the form

rx + sy ≥ t, or rx + sy ≤ t.

Finding the Feasibility Region.

First of all, we need to recall from our algebra classes how a typical linear constraint

is to be graphed. Consider, for example the linear inequality 3x + 5y ≤ 150. We

recall that the graph of the equality 3x + 5y = 150 is a straight line, indicated as

below:

-

6

bbbbbbbbbbb

3x + 5y = 150

x

y

25 50 75

25

50

75


Next, in order to graph the full inequality, all we need to do is to determine which

“side” of the above straight line a typical point (x, y) satisfying 3x+ 5y ≤ 150 would

fall. Furthermore, it is easy to see that if any given point (x, y) falls on a particular

side of the line 3x + 5y = 150, then all of the points satisfying the inequality

would fall on the same side! In many cases, we can just choose the origin (0, 0)

as our “test point;” since

3 · 0 + 5 · 0 ≤ 150,

we may conclude that the graph of 3x + 5y ≤ 150 consists of all points that lie on

the same side of the line 3x + 5y = 150 as does the origin. We depict this graph as

follows:

-

6

��

��

��

��

��

��

��

��

��

bbbbbbbbbbb

3x + 5y ≤ 150

x

y

25 50 75

25

50

75

Next, suppose that we wish to graph two simultaneous inequalities, say

3x + 5y ≤ 150

3x− y ≤ 75

In this case, we would superimpose the graphs of the two inequalities, depicted

below:

-

6

��

��

��

��

��

��

��

��

��

bbbbbbbbbbb

��

PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP PPPP

PP

3x + 5y ≤ 150

3x− y ≤ 75�

��

��x

y

25 50 75

25

50

75


The graph of the two simultaneous inequalities is then the “intersection” of the two

“half-planes.” If we were to further restrict the above region to include only to non-

negative values of the variables, then the resulting region would then appear as a

(convex9) polygon. If we add the inequalities x ≥ 0 and y ≥ 0 to the above two

inequalities, the resulting picture is as follows:

-

6

bbbbb

��

3x + 5y ≤ 150

3x− y ≤ 75

x

y

25 50 75

25

50

75

The “Corner Point Principle”—Two Variables.

In this section we shall describe the principle which is basic to the entire theory.

Again, we are considering a linear objective function of the form P (x, y) = ax + by,

subject to linear constraints. As indicated above, the feasibility region will be a convex

region in the plane with some “corner points;” these will occur at the intersection of

the lines defining the linear inequalities. Note now that for any particular value of

the objective function, say where P (x, y) = c, we have a straight line. We continue

the example considered above and assume that we have an objective function of the

form

P (x, y) = 3x + 2y.

We now graph a few values of this objective function on the same graph as the

feasibility region.

9A region is called convex if for any two points in the region, the line joining these two points in also wholly

contained in the region. Our feasibility regions can easily be shown to be convex as they are the intersection of convex

regions (half-planes).


-

6JJJJJJJJJJJJJJJ

JJJJJJJJJJJJJ

JJJJJJJJJJ

bbbbb

��

3x + 2y = 78

3x + 2y = 112.5

3x + 2y = 150��

��

��

x

y

25 50 75

25

50

75

As we can see from the above graph, the values of the linear objective function increase

as the individual lines “move toward the northeast.” The values of this objective func-

tion increase to a maximum value of P (x, y) = 112.5 after which the corresponding

lines no longer intersect the feasibility region. Thus, we see that the maximum value

of the objective function P (x, y) = 3x + 2y occurs at one of the “corners” of the

feasibility region.10

Not too much thought is required to be convinced that this always happens:

The Corner Point Principle. Given the linear objective function P (x, y) =

ax + by, the maximum (or minimum) value of this function must occur at one of the

“corners” of the feasibility region.

10The actual coordinates of this corner point can be obtained by solving the simultaneous linear equations

3x− y = 75

3x + 5y = 150.

The solution is x = 175/6 and y = 25/2. Note that

P (175/6, 25/2) = 3(175/6) + 2(25/2) = 225/2 = 112.5.


Example. We return to the example on page 122 concerning sailboats and motor-

boats. The objective function to be optimized is P (x, y) = 1200x + 1400y, and the

linear constraints define the feasibility region

-

6

(P (0, 33 13 ) = 46, 666 2

3 )

(P (23, 18) = 52, 800)

(P (29, 0) = 40, 600)

(P (0, 0) = 0)

BBBBBB

QQQ

QQ

QQQQ

r

r

r

(29, 0)

(23, 18)

(0, 33 13 )

r(0, 0)

x

y

Feasibility

Region

-

From the above, we conclude that the production level that will maximize profit is

to produce 23 sailboats each week and 18 motorboats each week. Furthermore, not

that at this production level, the weekly profit realized is

P (23, 18) = $52, 800.

The Corner Point Principle—More than Two Variables.

When there are more than two variables—as is the case with the mixed fruit juice

problem of page 124, the problem becomes considerably more complicated, both

conceptually and computationally. First of all, in the case where there are three

variables, say x, y, and z, then a linear inequality of the form rx + sy + tz ≤q has as a graph in 3-dimensional space a “half-plane” (points lying on one side

of a plane). Thus, the feasibility region is then viewed as a 3-dimensional convex

polyhedron. Furthermoremore, the Corner Point Principle continues to be valid:

the objective function has a maximum (or minimum) at one of the vertices of the

convex polyhedron.


While the above is not too difficult to visualize in three dimensions, it becomes

vastly more difficult—if not impossible—to visualize in higher dimensions. Even

more difficult are the computations needed to determine the corner points and then

to evaluate the objective function at these corner points. As already mentioned above,

solving such problems is the subject of linear programming, within which are a

number of efficient methods for finding optimal solutions.11 We won’t go into this;

rather, the student is directed to the internet, where a number of useful Javascript

codes to solve such problems can be found.12

Exercises for Sections 4.1 and 4.2

1. Suppose that it has been determined that the cost for a company to produce x

items of output is C(x) = 10, 000 + 2x and that the company sells the manufac-

tured items for $5/item.

(a) Write down the revenue function as a function of output, x.

(b) What are the fixed costs of production?

(c) Does the cost function indicate any region of economy of scale?

(d) On the same coordinate axes, graph the cost and revenue functions, indicat-

ing the break-even point.

(e) Note that the assumptions of this problem suggest that the more units are

sold, the greater is the profit. Would you agree that this is a reasonable

model?

2. Below are depicted the graphs of the cost and revenue functions of the company

Corbett Consulting; the product sold is consulting. Study the graph below and

answer the given questions:

11The so-called “simplex method” is the most commonly studied method and has been in use for nearly 60 years.

In 1984 a much more efficient method for solving linear programming problems was invented by Bell Laboratories

mathematician Natendra Karmarkar.12One particularly easy-to-use Javascript can be found at

http://people.hofstra.edu/faculty/Stefan Waner/RealWorld/simplex.html.


-

6

,,,,,,,,,,,,,,,,,,,,,,,,

2 4

C = C(x)

R = R(x)

x (hundreds of man-hours/week)

y

one unit =

$10,000/week

(a) Estimate the cost of one hour of consulting from Corbett Consulting.

(b) Estimate the fixed cost incurred in setting up Corbett Consulting.

(c) Estimate the (first) break-even point.

(d) Note that there is a second break-even point. Estimate this weekly level of

output and give an interpretation.

(e) Estimate the total man-hour output (per week) that will maximize the profit.

3. We continue our analysis of Corbett Consulting, focusing on some of the marginals.

(a) Estimate the marginal cost of production at the first break-even point.

(b) Estimate the marginal cost of producation at the second break-even point.

(c) Estimate the marginal cost of production at the optimal level of production

(i.e., that level of production that maximizes profit).

4. Assume that a company’s team of analyists has determined that its daily cost

of producing gadgets can be modeled by the cubic equation

C(x) = 0.4 + 3x− 1.2x2 + 0.2x3,

where x is measured in millions of units of output per day, and where the cost

is measured in terms of $10,000/day.


(a) Using your graphing calculator, plot this cost function. Use the range 0 ≤x ≤ 4 for the independent variable (x) and use the range 0 ≤ C ≤ 5 for the

dependent variable. Lable the axes appropriately.

(b) Using the graph, estimate the total cost of producing one million gadgets

per day.

(c) Estimate the cost of producing the 1,000,000-th item on a given day.

(d) Estimage the average cost of producing 1,000,000 items per day.13

(e) Give a rough indication of the region of “economy of scale” on your graph.

(f) Would you say that at one million gadgets/day the company is under-

producing, over-producing, or neither.

(g) How about a daily production level of two million gadgets?

(h) Give a rough estimate of the optimal daily production level, based on the

notion of economy of scale.

(i) Estimate the marginal cost of production, given that the company maintains

the optimal daily production level.

5. Mr. Phillip McTernan is confronted with a decision: whether to stay in academe

and continue to be a man of letters or to abandon all of his beliefs and enter

into business for himself. His business venture would be to sell anthologies of

Shakespeare’s works. Not completely without business savvy, Mr. McTernan has

visited other mid-sized companies also in the business of selling such anthologies

and determined that these other companies have incurred the following average

weekly costs of production.

13I stuck this question in to help the students understand the difference between average cost and marginal cost.


Average Weekly Average Weekly

Production (1000s) Cost ($10,000)

Company A 16.47 48.325

Company B 6.12 37.447

Company C 19.06 49.914

Company D 12.47 45.388

Company E 14.88 47.085

Company F 36.71 112.880

Company G 21.86 52.993

Mr. McTernan, holding fast in his belief that cost curves should generally be

cubic in nature, decided to use his TI-83 calculator to find a cubic function

that “fits” these data as closely as possible. We outline below how to do this

(presumably this is also what Mr. McTernan did):

Step 1: Enter the production levels as a set:

{16.47, 6.12, 19.06, 12.47, 14.88, 36.71, 21.86} → X

Step 2: Enter the costs as a set:

{48.325, 37.447, 49.914, 45.388, 47.085, 112.880, 52.993} → Y

Step 3: On the calculator, push the stat button.

Choose the CALC menu option.

Toggle down to option 6: CubicReg.

After CubicReg appears on the main screen of your calculator,

push 2nd , LIST , and choose the list variable “X” that you de-

fined above. After the “X”, type a “comma” (,) from your calcula-

tor.

Push 2nd , LIST , and choose the list variable “Y” that you defined

above. This will result in the display

CubicReg LX , LY


Now just hit ENTER ; after some time the regression coefficients

will be displayed. For the data entered above, the display looks like

CubicReg

y = ax3 + bx2 + cx + d

a = .0053063455

b = −.2437944339

c = 4.366325356

d = 18.62182183

Now help Mr. McTernan complete his analysis by performing the following tasks

(or answering the following questions):

(a) Graph the above-determined cubic cost function on the interval 0 ≤ x ≤40 (remembering that x measures 1000s of anthologies sold, and that the

dependent variable is measured in units of $10,000/week).

(b) Estimate the cost of producing the 1,000-th anthology in a given week.

(c) Give a rough indication of the region of “economy of scale” on the graph.

(d) Would you say that at one thousand anthologies/week, Mr. McTernan is

under-producing, over-producing, or neither.

(e) How about a weekly production level of two thousand anthologies?

(f) Give a rough estimate of the optimal weekly production level of anthologies,

based on the notion of economy of scale.

(g) Suppose that Mr. McTernan’s market analysts say that there is a pretty

good market for these anthologies if the unit price is held to about $25

each. Sketch the resulting revenue function on the same axes on which you

graphed the cost function.

(h) What are the break-even points? What weekly level of output would you

say is optimal for Mr. McTernan’s company?

(i) If Mr. McTernan maintains the level of output you suggested in (f) above,

what would be the resulting weekly profit?

(j) Estimate the weekly production level of anthologies that will maximize

profit.


6. The daily cost and revenue functions for a small shoe factory are given below.

��

25 50 75 100 125 150

2,000

4,000

6,000

8,000

10,000

$12,000

daily number of pairs of shoes

C = C(x)

R = R(x)

(a) Estimate the first break-even point.

(b) Estimate the marginal cost at this break-even point.

(c) Estimate the second break-even point.

(d) Estimate the marginal cost at this break-even point.

(e) Estimate the sales price of each pair of shoes.

(f) Estimate the total daily profit realized by selling 50 pairs of shoes per day.

(g) Estimate the cost of making the 50-th pair of shoes on a given day. (Hint:

You can approximate this by estimating the marginal cost at an output level

of 50 pairs of shoes per day.)

(h) Estimate the profit on making (and selling) the 50-th pair of shoes on a

given day.

(i) Estimate the cost of making the 100-th pair of shoes on a given day.

(j) Estimate the profit on making (and selling) the 100-th pair of shoes on a

given day.

(k) Estimate the level of output that will maximize the company’s profit.

(l) Estimate the marginal cost at this level of output.

(m) Estimate the marginal profit at this level of output.


7. Consider the “Simple Example” taken up on page 103. Now determine the

optimal levels of production given the revenue functions

(a) R(x) = 600x.

(b) R(x) = 850x.

8. Assume a cost function C(x) = 0.01x3 − 0.6x2 + 13x, where x is in units and

C(x) is in dollars.

(a) What is the fixed cost?

(b) Find the maximum profit if the per-unit price is set at $7.25.

(c) Suppose that we fix the production at x = 34 and that if the price is $7.25

per unit, then all 34 items will sell. Assume, however, that for each $1.00

increase in the per-unit price, 2 fewer items will be sold. Should the price

be raised, and if so, by how much?

9. For the cost function in Exercise 8, compute the level of output that minimizes

marginal cost.

10. A weekly cost function for delivering boxed lunches has been determined to be

C(x) = 0.04x3 − 3.1x2 + 74.2x + 95.8,

where x measures the number of boxed lunches (in 1000s), and C(x) is measured

in units of $100.

(a) Graph the above cost function, using the ranges 0 ≤ x ≤ 60 and 0 ≤ C(x) ≤2000.

(b) Compute the output that minimizes marginal cost.

(c) Compute the minimal marginal cost (note that it is negative).

(d) Find those values of x for which the marginal cost is ≤ 0.



1. For each pair of supply and demand functions,

(i) Graph the functions on the same coordinate axes, and

(ii) Compute the equilibrium price and demand.

(a) S(p) = 2p− 100 and D(p) = 1000− 3p, where p is in dollars and the supply

and demand and measured in 1000 items/day.

(b) S(p) = 3.4p−300.5 and D(p) = 504−5.4p, where p is in units of 100 dollars

and supply and demand are in units of 100/week.

(c) S(p) = 1.1x2 − 10.2 and D(p) = 22.4− x2, where p is in dollars and supply

and demand are measured in terms of 100/day.

(d) S(p) = 1.2√x and S(p) = 5.4/p, where p is measured in 1000 dollars and

supply and demand are in units of 1,000,000/week.

2. Study the supply and demand graph below. (Assume that the price is in dollars

and the supply and demand are in 100s of units per day.)

-

6

1

2

��

10 30

Demand

Supply

x (units)

p (price/unit)

(a) Estimate the equilibruim price.

(b) Estimate the equilibrium supply (demand).


(c) Describe what would happen should the price per unit of the items raises to

$30.

(d) Describe what would happen should the price per unit of the items falls to

$20.

3. Suppose that we have demand and supply functions given by

D(p) = 2432− 20.1p and S(p) = 10.3p− 507,

where p is measured in dollars and supply and demand are measured in hundreds

of items/day.

(a) Find the equilibruim price and daily quantity supplied (= demanded).

(b) Suppose that a specific tax of $5 is levied by the local government. on each

100 items produced. Find the new equilibrium price and quantity supplied.

Also, graph the new supply function S(p− 5) on your graph in part (a).

(c) How much of the $5 tax is paid by the consumers and how much is paid by

the suppliers?

(d) What is the total daily tax revenue received by the local government?

4. Recall the supply and demand functions given on page 108

S(p) = 5p− 200 and D(p) = 500− 2p,

where p is the unit price and supply and demand are measure in 1000s of units per

day. This time, however, rather than a specific tax per unit, the city government

decides to levy a 5.2% sales tax. Thus, the consumer now pays p + 0.052p for

each item, making the new demand function D(p+ 0.052p) instead of just D(p).

(a) By solving S(p) = D(p + 0.052p), determine the new equilibrium price and

output.

(b) How much of the sales tax is paid by the consumer and how much is paid

by the producer?

(c) Determine the city’s daily tax revenue due to sales.


5. 14 Suppose that we assume that the total utility of a typical woman who pur-

chases dresses is roughly as follows:

Dresses Total Utility Marginal Utility

0 $0

1 $110

2 $210

3 $290

4 $360

5 $410

6 $440

7 $460

(a) Sketch a graph of the consumer’s total utility function.

(b) In the third column of the above table, determine the marginal utilities at

each level.

(c) If the consumer already owns three dresses and dresses go on sale for $50

each, approximately how many additional dresses should the consumer pur-

chase?

(d) If the consumer owns only one dress to begin with, and if dresses go on sale

for $50 each, how many dresses should the consumer purchase?

(e) For the typical woman, if the cost function for dresses is C = 50x (dollars

for x dresses), how many dresses should this persons own, and what will be

the total consumer surplus?

(f) Indicate the cost function also on your graph in part (a) as well as the

consumer surplus at the optimal purchase level.

6. Consider the following hypothetical total utility curve:

14This problem is adapted from John S. Morton, Advanced Placement Economics, Microeconomics: Student Ac-

tivities, Fourth Printing, National Council on Economic Education, 2001.


-

6

500

1000

5 10

Total Utility

Price

x (Number of Items Owned)

(a) Fill in the following table with appropriate approximations.

Items Total Utility Marginal Utility

0 $0

5 $110

10 $210

15 $290

20 $360

25 $410

30 $440

35 $460

(b) According the the Optimal Purchase Rule, how many additional items should

each person below purchase:

(i) A person owning none and the items on sale for $50.

(ii) A person owning none and the items on sale for $100.

(iii) A person owning none and the items on sale for $30.

(iv) A person owning five items and the items on sale for $60.

(v) A person owning five items and the items on sale for $20.

(vi) A person owning 10 items and the items on sale for $10.

(c) Suppose now that a person (who initially owns none of these items) purchases


a number of items at the given price. Compute the consumer surplus (or

deficit) in each case.

Price Number Purchased Consumer Surplus

$10 20

$10 40

$10 100

$20 20

$20 40

$20 100

$40 20

$40 40

$60 20

$60 40

(d) If the cost function for these items is C = 40x (dollars for x items), how

many items should a typical consumer own, and what would be the total

consumer surplus?

(e) Indicate the cost function also on the graph above, as well as the consumer

surplus at the optimal purchase level.

7. Suppose that the elasticity of a certain commodity is E = 2.1. Find the per-

centage increase (or decrease) on the demand of this commodity given

(a) a 3% price increase,

(b) a 2% price decrease.

8. Suppose that the elasticity of a certain commodity is E = 0.75. Compute the

the percentage increase (or decrease) on the demand of this commodity given

(a) a 5% price increase,

(b) a 3.5% price decrease.

9. Suppose that the elasticity of a quantity satisfies E(2.1) = 0.75 and that E(4.25) =

1.5. Compute the the percentage increase (or decrease) on the demand of this

commodity given that the price changes from


(a) $2.10 to $2.50;

(b) $4.25 to $4.75;

(c) $4.25 to $4.00

10. If we go to any supermarket, we’re likely to find many different brands of facial

soap. Assuming that these brands are all similarly popular, what would you say

about the elasticity of demand for any one of these brands of soap (> 1, < 1, or

= 1)?

11. For the next couple of examples, use the fact that the derivative of a quadratic

function Q(x) = ax2 + bx + c is given by

d

dxQ(x) = 2x + b.

Now assume that the demand for a given commodity is given by x = D(p) =

1000 − 10p2, where p is measured in dollars and D(p) is measured in units of

10,000 items per day.

(a) Graph this demand function.

(b) Determine the function E = E(p). (Use the fact that E(p) = | p1000−10p2

dxdp|, 0 ≤

p ≤ 10.)

(c) Find that value of p for which E(p) = 1.

(d) On you graph above, indicate the region of inelasticity and the region of

elasticity. (See the graph on page 120.)

12. Suppose that the demand for Chinese cabbage in a certain city is D(p) = 4500−3.4p2, where p is measured in yuan (¥) and D is measured in 1,000 kg/day.

(a) Find the price which makes D(p) = 0.

(b) Graph the demand function.

(c) Compute E = E(p).

(d) Determine the price range where the demand is inelastic.

(e) Determine the price range where the demand is elastic.

(f) Determine the price which maximizes revenue.


13. For the products in the table below, state whether you think that the demand

is elastic or inelastic:

Air travel Hotel Lodgings

Sugar Chocolate Candy

Automobiles Beef

Laundry Soap CD players

Rice Tennis Racquets

Color TVs Jewelry

Furnature Watermelons

14. (For those having studied calculus.) Suppose that x = D(p) is a demand function

such that E(p) = 1. Show that D(p) = k/p, for some constant k. (Hint: the

equation ∣∣∣∣px dx

dp

∣∣∣∣ = 1,

can be written without absolute values as

dx

dp= −x

p.

Now solve the differential equation by separating the variables.)

15. (For those having studied calculus.) Suppose that x = D(p) is a demand function

such that E(p) = r, for some constant r. Show that D(p) = k/pr, for some

constant k. (Follow the hint to the above problem.)

16. (How economists compute elasticity for linear functions.) Consider the linear

demand function depicted below. Show that E(p) (see equation 4.1, page 119)

can be computed by the geometrical formula

E(p) =d1

d2

,

where d1 and d2 are as indicated below:


-

6


s� d1

- � d2-

(p, x)

Demand

x = D(p)

p

x

17. In each case an event and a commodity are given. Check the appropriate re-

sponses.

(a) The price of sugar increases. We would expect that for chocolate candy,

shift to the left shift to the right won’t change

Supply curve will

Demand curve will

(b) Sugar is shown to have unexpected health benifits. We would expect that

for chocolate candy,


Supply curve will

Demand curve will

(c) The wheat harvest in the U.S. is at record levels. We would expect that for

bread,


Supply curve will

Demand curve will

(d) Complex carbohydrates are shown to be linked to stomach cancer. We would

expect that for bread,



Supply curve will

Demand curve will

(e) The Saudi government is overthrown. We would expect that for gasoline,


Supply curve will

Demand curve will

(f) A vast oil reserve is descovered in Alaska. We would expect that for gaso-

line,


Supply curve will

Demand curve will

Exercises for Sections 4.4

1. Using appropriately-scaled coordinate axes, graph the following (systems of)

linear inequalities.

(a) 5x + 7y ≤ 350

(b) 5x + 7y ≥ 350

(c) 3x− 4y ≥ 1000

(d) 3x− 4y ≥ 1000, x ≥ 0, y ≤ 0

(e) 5x + 7y ≤ 350, x ≥ 0, y ≥ 0

(f) 5x + 7y ≤ 134, x + 6y ≤ 82, x ≥ 0, y ≥ 0

(g) x + 2y ≤ 40, 2x− y ≤ 55, x ≥ 0, y ≥ 0

(h) x + 2y ≤ 80, 3x + y ≤ 90, −x + 2y ≤ 5, x ≥ 0, y ≥ 0.

2. Express each constraint as a linear inequality.

(a) Each cake (x) requires 3 cups of flour and each bisquit (y) requires one cup

of flours. There is a total of 150 cups of flour available

(b) Sailboats (x) require 150 hours of labor and motorboats (y) require 200

hours of labor. There is a limitation of 1,500 hours of labor.


(c) Bottles of Drink A (x) require 6 oz of grapefruit juice, bottles of Drink B

(y) require 10 oz of grapefruit juice, and bottles of Drink C (z) require 5 oz

of grapefruit juice. There is a limit of 2,000 oz of grapefruit juice.

(d) Coffee blend A (x) requires 10 oz of Columbian coffee beans and coffee blend

B (y) requires 13 oz of Columbian coffee beans. The maximum availability

of Columbian coffee beans is 150 pounds (1 lb = 16 oz).

3. We are to produce two types of candy bars: the chocolate delight and the peanut

cluster. The chocolate delight candy bar is made of a mixture of 5 oz of chocolate

and 1 oz of peanuts, and the peanut cluster is made from a mixture of 3 oz of

peanuts and 3 oz of chocolate. The total daily availability of chocolate is 50 lbs

and the total daily availability of peanuts is 34 lbs. Complete the table below

and graph the feasibility region.

Chocolate Peanuts

Chocolate Delight (x)

Peanut Cluster (y)

Max Available (daily)

4. A tailor in Suzhou makes long and short cashmier overcoats. The long cashmier

coat requires 3 hours to cut the pattern and 4 hours for the sewing. The short

cashmier coat requires 4 hours for the cutting and 2 hours for the sewing. The

tailor’s employees can only dedicate 90 hours per week to the cutting and 100

hours per week to the sewing. Complete the table and graph the feasibility

region.

Cutting Sewing

Long Coats (x)

Short Coats (y)

Max Available

5. My auto repair shop offers two types of services: oil changes (x) and tune-ups (y).

Each oil change requires 10 minutes of a Grade B technician and 2 minutes of a

Grade A technician; each tune-up requires 45 minutes of a Grade B technician


and 30 minutes of a Grade A technician. My service shop has available 120 hours

of Grade B technician work each day and 66 hours of Grade A technician work

each day. My shop makes a profit of $12 on each oil change and a profit of $60

on each tune-up. Complete the table below, and graph the feasibility region. On

the same graph, indicate the graphs of

P (x, y) = 130

P (x, y) = 160

P (x, y) = 190

Grade A Grade B

Oil Changes (x)

Tune-Ups (y)

Max Available

Hours (daily)

Maximize: P (x, y) =

6. My clockworks shop makes three types of grandfather clocks: the Great Grand-

father (x), the American Classic (y), and the European Traditional (z). Each

of these clock require time in the wood shop, the metal shop, and in the clock-

works shop. Each Great Grandfather requires 1,000 minutes in the wood shop,

1500 minutes in the metal shop, and 750 minutes in the clockworks shop. Each

American Classic requires 1,250 minutes in the wood shop, 750 minutes in the

metal shop, and 600 mintes in the clockworks shop. Each European Traditional

requires 1,500 minutes in the wood shop, 500 minutes in the metal shop and 550

minutes in the clockworks shop. The total amount of labor (weekly) in these

shops is 420 hours in the wood shop, 500 hours in the metal shop, and 380 hours

in the clockworks shop. The profit for the sales of these clocks is $240 for the

Great Grandfather, $170 for the American Classic, and $300 for the European

Traditional. Complete the table below.


Wood Shop Metal Shop Clockworks Shop

Great Grandfather (x)

American Classic (y)

European Traditional (z)

Max Available

Hours (Weekly)

Maximize: P (x, y, z) =

7. Solve each of the two-dimensional linear programming problems using the Corner

Point Principle.

(a) Objective function:

maximize P (x, y) = 5x + 7y, subject to

2x + 3y ≤ 82

5x + 2y ≤ 95

x ≥ 0

y ≥ 0

(b) Objective function: maximize P (x, y) = 6x + 7y, subject to

3x + 5y ≤ 45

4x + 3y ≤ 38

x ≥ 0

y ≥ 0

(c) maximize P (x, y) = 3x + 2y, subject to

3x + 5y ≤ 45

4x + 3y ≤ 38

x ≥ 0

y ≥ 0


(d) maximize P (x, y) = 6x + 5y, subject to

3x + 4y ≤ 75

5x + 3y ≤ 81

y ≤ 15

x ≥ 0

y ≥ 0

8. Assume, in Problem 3 above, that the profit made on each Chocolate Delight

bar is $.30 and that the profit on each Peanut Cluster bar is $.45. Compute the

daily output and the resulting daily profit resulting from the sales of these candy

bars.

9. Again consider Problem 3, assume that the profit from selling a Peanut Cluster

bar is 25% greater than the profit from selling a Chocolate Delight bar. Compute

the optimal daily output.

10. In Problem 3, if the profit from selling a Chocolate Delight bars is twice the

profit from selling a Peanut Cluster bar, compute the optimal daily output.

11. In Problem 4, assume that the profit from selling a long cashmere coat is 200

and the profit from selling a short cashmere coat is 150. Compute the optimal

output and the profit realized from this output.

12. In Problem 4, assume that the profit from selling a short cashmere coat is 50%

more than the profit realized from selling a long cashmere coat. What would

you conclude?

13. Solve the linear programming problem stated in Problem 5.

14. Using a convenient Javascript (for example, the one given in the footnote on

page 131) solve the linear programming program given in Problem 6.15

15You’ll notice that the optimal values for x, y, and z aren’t all whole numbers. Since we’re certainly not going to

make a fraction of a clock, you’ll need to decide how to apply this answer.


(a) Objective function:

maximize P (x, y, z, w) = 25x + 45y + 39z + 32w, subject to

3x + 2y + 5z + 6w ≤ 128

4x + 5y + z + 2w ≤ 110

x + 7y + 9z + 3w ≤ 125

x ≥ 0

y ≥ 0

z ≥ 0

w ≥ 0

(b) In the above problem suppose that we leave everything the same, except that

we change the coefficient of z from 39 to 332. Will this change anything?

More generally, explain why the answer to this linear programming problem

is completely independent of the coefficient of z.

15. Objective function:

maximize P (x, y, z, w) = 37x + 26y + 25z + 33w + 43u subject to

23x + 4y + 12w + 10u ≤ 185

84y + 43z ≤ 237

10y + 55z ≤ 150

10x + 33w + 9u ≤ 213

14x + 16y + 34u ≤ 176

x ≥ 0

y ≥ 0

z ≥ 0

w ≥ 0


16. Objective function:

maximize P (x, y, z, w) = 334x + 216y + 319z + 178w subject to

10x + 5y + z ≤ 60

2x + 8y + 5z + 4w ≤ 65

3x + 8z + 2w ≤ 49

5x + 2y + 11w ≤ 54

x ≥ 0

y ≥ 0

z ≥ 0

w ≥ 0

17. 16 A bakery is to produce six different types of bakery items, listed in the chart

below. Each item requires a certain amount of labor, has a certain cost associated

with its ingredients, and each has a corresponding per-item profit. The bakery

has a weekly budget of $2,000 and a work force providing a total of 360 hours

per week. What is the weekly output that will maximize profit?

Per-Item Cost of Per-Item

Labor Ingredients Profit

Nine-Grain Bread (x) 50 min $1.05 $0.60

Sourdough Bread (y) 50 min $0.95 $0.70

Chocolate Cake (z) 35 min $2.00 $2.50

Poppyseed Cake (u) 30 min $1.55 $2.00

Blueberry Muffins (doz) (v) 15 min $1.60 $1.00

Apple-Cinnamon

Muffins (doz) (w) 15 min $1.30 $1.40

Weekly Maxima


16This, and the following problem, are adapted from D. B. Johnson and T. A. Mowry, Mathematics, A Practical

Odyssey, Fifth Edition, Thomson, Brooks/Cole, 2004, ISBN 0-534-40059-0.


18. A television distributor handles two lines of televisions, the Packard and the

Bell. the Packard some in three sizes, and the Bell comes in two sizes (tabulated

below). The monthly purchasing budget for this distributor is $57,000. These

televisions are stored in a 9,000 cubic foot storage area; the sizes of the individual

televisions are tabluated below. If the profits are as given, what is the optimal

ordering strategy for this distributor?

Size Cost Profit

Packard 18” (x) 30 cubic ft $150 $175

Packard 24” (y) 36 cubic ft $200 $200

Packard Rear Projection (z) 63 cubic ft $450 $720

Bell 18” (u) 28 cubic ft $150 $170

Bell 30” (v) 38 cubic ft $225 $230

Monthly Maxima



Chapter 5

Investment Instruments

In this chapter we shall take up speculative investments, i.e., those whose rates

of return are not guaranteed. We’ll focus primarily on the two most common such

investments instruments: stocks and bonds. Since bonds are easier to understand

and are a bit more predictable, we’ll consider these first.

5.1 The Bond Market

Suppose that a company wishes to raise a large amount of money for the purpose of

capital improvements. If the company doesn’t have enough money on hand it will,

obviously, need to borrow this money. Such borrowing typically takes the form of

bonds. Thus, a company might “sell” $1,000 (or greater denominations) bonds with

the promise that whoever buys such bonds will be paid back (with interest). How

the interest is paid varies with the type of bond; we consider two common varieties

of bonds below.

Not only do companies often finance their projects through the issuing of bonds,

governmental units: cities, states, and even the federal government often issues bonds

to raise needed capital. The U.S. Government issues bonds for major capital projects

(including that of waging war!). Similarly states and cities issue bonds. In the case of

cities and states, the repayment of the value (and interest) on the bonds is typically

financed through paying increased taxes. Thus, prior to this happening, the city or

state will have a bond election to be decided by the voters. That is, if the voter

feel that the project to be financed (often education) is worthwhile, then they will

approve the bond, allowing the government to collect increased taxes for the purpose

155

156 CHAPTER 5. INVESTMENT INSTRUMENTS

of financing the payback-plus-interest on the bonds.

We turn now to two commonly-occurring types of bonds.

5.1.1 Discount Bonds

A discount or zero-coupon bond is a bond that carries a certain face-value, but

is sold at a discount to the buyer, with the discount generally tied to the rate of

inflation. The discount provides for a certain rate of return to the invester and is

called the yield to maturity (YTM).

Example. Suppose that a $1,000 10-year discount bond is purchased with a YTM of

7%. To compute the discount price, one simply computes the present value of $1,000

ten years hence: assuming simple interest, we have

Present Value = $1, 000(1.07)−10 = $508.35.

That is to say, this $1,000 bond would be bought today for $508.35; after 10 years it

would be redeemed for the face value of $1,000.

Bonds are typically sold on the open market as investment instruments. To see

why this might happen, consider the result of a change in the interest rate.

Example. Suppose that we bought a $1,000 10-year discount bond with a YTM of

7%. Suddenly, the rate of inflation falls to 5%. What is the new present value of this

bond? The answer is simple:

Present Value = $1, 000(1.05)−10 = $613.91.

That is, the net effect of the falling interest rate resulted in our bond’s present value

increasing over $100.

From the above discussion, we can see why people might be buying bonds on the

open market: they are speculating on the fluctuations in interest rates. If they think

that interest rates might start falling, they would be inclined to purchase bonds; if

they think that interest rates might start rising, they might try to sell their bonds.

5.1. THE BOND MARKET 157

In any case, we can see why people tend to buy bonds when the interest rates are

high.

Yield to Maturity for Discount Bonds.

The YTM for bonds tends to vary over time, and is determined by (and determines!)

the selling price of the bond. Assume that a $1,000 10-year discount bond is purchased

for $400. The YTM is then the percentage rate that makes the present value of

this bond equal to $400. However, we must note that in determining the YTM

of a discount bond, we must also know the time to maturity for the bond. Thus,

assume that the above $1,000 discount bond was purchased with eight years left

before maturity. What percentage rate would yield a present value of $400 for $1,000

eight years from now? We just solve an equation:

$400 = $1, 000(1 + r)−8,

and so

(1 + r)8 =$1, 000

$400.

Finally, take the 8-th roots of both sides and get

1 + r =8

√$1, 000

$400≈ 1.1214,

forcing r ≈ 0.1214, which is an approximate interest rate of 12.14% .

Note that we can reverse the above analysis. Suppose that a $1,000 discount bond

with six years left to maturity has a current yield of 9%. What is its present value?

This is fairly simple, as we need only compute the present value of $1,000 at 9% taken

six years from now:

Present Value = $1, 000(1.09)−6 = $596.27.


5.1.2 Coupon Bonds

A coupon bond, like the discount bond, also has a face value. However, there

are regular interest payments (usually semiannually) at an interest rate called the

coupon rate.

Example. Suppose that a 10-year $1,000 coupon bond is purchased for its face

value and that the coupon rate is 8%. Then the holder of this bond is guaranteed

semiannual payments of 4% of $1,000 (or $40). When the bond matures, the holder

of the bond will be paid $1,000 by whoever issued the bond.

It is instructive to note that if the inflation rate is equal to the coupon rate, then

the present value of the above bond is worth exactly the $1,000 it was sold for. This

is a routine calculation of the present value of a stream of payments:

Present Value = $40

(1 +

.08

2

)−1

+ $40

(1 +

.08

2

)−2

+ · · · $40

(1 +

.08

2

)−20

︸︷︷︸(present value of coupon payments)

+ $1, 000

(1 +

.08

2

)−20

︸︷︷︸(present value of bond)

= $543.61 + $456.39

= $1, 000,

entirely as expected! Note more generally that if the interest rate remains unchanged,

the present value of a coupon bond will remain unchanged. Thus, the only

way for a bond to change in value is for the interest rate to change.

As in the case of the discount bond, the value of the bond increases when the going

interest rate (or inflation rate) decreases.

Example. Suppose that after buying the 10-year $1,000 (8%) coupon bond the

interest suddenly fell to 6%. The holder of this bond will continue to enjoy semiannual

interest payments of $40, and the face value of the bond is still $1,000. This time, we

get a present value of the bond as follows:


Present Value = $40

(1 +

.06

2

)−1

+ $40

(1 +

06

2

)−2

+ · · · $40

(1 +

.06

2

)−20

+ $1, 000

(1 +

.06

2

)−20

= $1, 148.77.

So, again, we see that the lowering of the interest rate has increased the (present)

value of the bond.

YTM for Coupon Bonds.

For coupon bonds, we compute the YTM in a manner similar to that of discount

bonds. However, as we’ll see, the mathemtics is quite a bit more complicated. As

in the case of discount bonds, the YTM and the current price of the bond serve to

determine each other.

Example. Suppose that a $1,000 coupon bond with a coupon rate of 612% has eight

years remaining to maturity and sells for $914. Compute the current yield of this

coupon bond. Again, this problem is to compute the interest rate needed for the

present value of an income stream to equal $914 (note that the coupon payments are

semiannual payments of 314% of $1,000, or $32.50):

$914 = $32.50(

1 +r

2

)−1

+ $32.50(

1 +r

2

)−2

+ · · · $32.50(

1 +r

2

)−16

︸︷︷︸(present value of coupon payments)

+ $1, 000(

1 +r

2

)−16

︸︷︷︸(present value of bond)

=

($65

r

)(1−

(1 +

r

2

)−16)

+ $1, 000(

1 +r

2

)−16

.

Here’s where it gets pretty hairy! For convenience, we set

x = 1 + r2,

P = $914 (price of the bond),


C = $32.50 (semiannual coupon payments),

M = $1, 000 (maturity value of the bond).

Then the above equation can be written as

P =C

x− 1

(x16 − 1

x16

)+

M(x− 1)

x16(x− 1),

and so

Px16(x− 1)− Cx16 −M(x− 1) + C = 0.

This gives another hard polynomial problem (compare with page 65):

Px17 − (P + C)x16 −Mx + M + C = 0.

Note that the above equation has the obvious, but uninteresting, solution x = 1.

(This solution is uninteresting since it gives r = 0.) The nontrivial solution of this

equation can be verified (by calculator) to be x ≈ 1.039874. Since x = 1 + r2, this

implies that r ≈ .07975, and so the YTM for this bond is 7.98%. Note that this

makes sense as by now our intuition is starting to tell us that when the price goes

down, the yield goes up.

We conclude this section by summarizing a couple of useful formulas that relate

the discount (or premium)1 prices of a bond with the YTM.

Basic Problem of Coupon Bonds. Assume that a coupon bond has a par, or

face value of M . Assume that there are n years to maturity, that C denotes the

semiannual coupon payments, and that the YTM is r. Then the present value, P , of

the bond is given by

P =C

x− 1

(x2n − 1

x2n

)+

M

x2n, x = 1 +

r

2.

If the present price P of the bond is known, then the YTM r can be solved by setting

x = 1 + r2

and solving the polynomial equation

Px2n+1 − (P + C)x2n −Mx + M + C = 0 .

1When a bond is sold at an amount less than the face value of the bond, it is said to be sold at a discount. When

a bond is sold at an amount more than the face value of the bond, it is said to be sold at a premium.


TI-83 Technology. Given that x = 1 is an uninteresting solution to the above

hard polynomial problem (since it gives r = 0), we know that the solution in question

is somewhere greater than one (and that the graph of the polynomial equation is

similar in nature to that on page 68.) Since we know the coupon rate, we can safely

assume that the YTM is likely not to be more than twice the coupon rate. Thus, the

window settings can take the form

Xmin = 1

Xmax = 1 + twice the coupon rate

The Ymin and Ymax settings will vary with the problem, but a rule of thumb that

one might use is to try Ymin = −M/100 and Ymax = M/100.

Example. Suppose that a $2,500 bond with a 4.2% coupon rate (with 12 years left

until maturity) is purchased for $1,800. What is the YTM?

Solution. Here we have the parameters

P = $2, 100 (the price of the bond),

M = $2, 500 (the face value2 of the bond),

C = 0.021× $2, 500 = $52.50 (the semiannual coupon payments).

Therefore the polynomial equation to solve is

Px2n+1 − (P + C)x2n −Mx + M + C = 0,

(with n = 12). Specifically, the equation is

2100x25 − 2152.5x24 − 2500x + 2552.5 = 0.

The solution of this equation yields x = 1.0304 which is a YTM of 6.04%. Apparantly,

this bond is very attractively discounted!

2Sometimes “face value” is referred to as the “par” or “par value” of the bond.


5.1.3 Reading a Bond Table

The table3

Coupon Mat. date bid $ Yld %

Corporate

AGT Lt 8.800 Sep 22/25 100.46 8.75

Air Ca 6.750 Feb 02/04 94.00 9.09

AssCap 5.400 Sep 04/01 100.01 5.38

Avco 5.750 Fun 02/03 100.25 5.63

Bell 6.250 Dec 01/03 101.59 5.63

Bell 6.500 May 09/05 102.01 5.95

BMO 7.000 Jan 28/03 106.55 6.04

BNS 5.400 Apr 01/03 100.31 5.24

BNS 6.250 Jul 16/07 100.56 5.95

CardTr 5.510 Jun 21/03 100.52 5.27

Cdn Pa 5.850 Mar 30/09 93.93 6.83

Clearn 0.000 May 15/08 88.50 8.61

CnCrTr 5.625 Mar 24/05 99.78 5.68

Coke 5.650 Mar 17/04 99.59 5.80

We proceed to explain the above table. First of all, notice that this is a table of

corporate bonds, as opposed to other types of bonds (e.g., municipal, school, or

treasury bonds).

(i) The first column lists the company that issued the bond.

(ii) The second column lists the coupon rate for the given bond. Therefore, the

coupon rate for the first bond is 8.8%.

(iii) The third column lists the maturity date of the bond. The first bond matures

on Sept. 22, 2025.

(iv) The fourth column represents the percentage of par value (sometimes called the

bid price at which the bond is trading. For example, the first bond is trading at

100.46% of its par value.3This is taken from www.investopedia.com/university/bonds .

5.2. THE STOCK MARKET 163

Note that if we are given the coupon rate c, the time to maturity and the bid

price b, we can calculate the bond’s YTM. Indeed, we recall that the calculation of

a bond’s YTM (call this r) results from setting x = 1 + r/2, and solving the hard

polynomial equation:

Px2n+1 − (P + C)x2n −Mx + M + C = 0,

where M is the par value, C is the value of the semiannual coupon payments, and n

is the number of years to maturity. Note that C = cM/2 and P = bM . Therefore,

the above equation becomes

bMx2n+1 − (bM + cM/2)x2n −Mx + M + cM/2 = 0;

we can divide through by M and get (after multiplying through by 2)

2bx2n+1 − (2b + c)x2n − 2x + 2 + c = 0 .

In solving this polynomial, use the same window settings as suggested on page 161

except that the Ymin and Ymax should be set initially to roughly ±0.02. The output

returns the value x ≈ 1.0256, from which we conclude that the YTM is 5.12%.

Example. Assume that a bond with a coupon rate of 4.8% has a bid price of 98.03

and has 10 years to maturity. Find its YTM.

Solution. Here, we have b = .9803, n = 10, and c = 0.048. The above equation

becomes

1.9606x21 − 2.0088x20 − 2x + 2.048 = 0;

5.2 The Stock Market

In the most general terms, investing consists of buying and selling two types of secu-

rities: debt securities and equity securities. Bonds are the primary examples of

debt securities: the purchaser of a bond is essentially making a loan and then realizing


a profit (or loss) by means of interest rate variations or selling the bond. Stocks, on

the other hand, represent equity securities inasmuch as the purchaser of a stock is

essentially buying a share (or shares) of the company and therefore becoming a part

owner in the company.

In the case of stocks, the mathematical analysis is vastly more complicated than

with bonds as there are far more variables to consider (including psychological issues).

Basically, however, here’s how the process works. Suppose I were to start up a small

coffee shop—say here in Shanghai. Perhaps I can get enough initial capital through

a bank or other types of loans to start up my business. However, it might happen

that my business is successful to the point that large-scale expansion is likely to be

profitable. However, as this expansion will require even larger sums of capital, I have

to make a choice: either to borrow more money, or to “go public” and sell shares

(i.e., stocks) in my company.4 When I first offer such stocks in my company—called

an initial public offering, or IPO for short—I’m doing so in what is called the

primary market. Anyone who buys shares of my stock in this market automatically

becomes a part owner in my company and will therefore share in my profits and suffer

my losses.

Of course, this is O.K. with me as I needed the money to grow in the first place. I

still think that it’s better to share huge profits with others rather than to keep small

profits all to myself.

The shareholders in my company (that is, the people that bought stocks in my

company) might after a while decide that they no longer wish to keep these stocks:

they will therefore sell these stocks in the secondary market.5

From this point on, people who buy and sell stocks in my coffee corporation are

trading ownership in the hopes of making profits. Naturally, if my coffee corporation

is well run, and consistently turns profits, then the price of shares in my company

can expect to increase. The converse is obviously also true. However there are other

factors—too numerous to itemize—that can effect the price of shares.6

4The actually mechanics of a company “going public” is a rather complicated legal process and won’t be discussed

in these notes.5Transactions in the secondary market are typically carried out in what are called stock exchanges, like the New

York Stock Exchange.6Consider the effect of a sudden coffee crop failure in Columbia!


5.2.1 What is a Mutual Fund?

Buying the right stock or stocks is arguably more an art than a science and requires

research and more than just a little luck. While we’ll make a few comments about

some of the quantitative aspects of stocks below, one should be advised that the most

successful stock traders are those who take the time to stay well informed—and read

the financial news regularly!

For those who don’t have the time or inclination to devote the time to buy and sell

individual stocks, a mutual fund might be the most sensible answer. A mutual fund

is simply a collection of stocks, “bundled together” to form a single product. This

allows for a bit of diversity in the stocks you own, but typically the fund itself has an

overarching philosophy (growth, income, blend, index, and so forth). For example, the

White Oaks Fund (Ticker: WOGSX) is a growth fund and is composed primarily

of relatively aggressive stocks with promise of growth in their values. An index fund

has stocks that mirror a certain benchmark, such as the Standard and Poor’s 500

Index (the “S and P 500”). An example of an index fund is USAA S & P 500 fund

(Ticker: USSPX).7

In order to get a bit more detailed information on the performance of a mutual

fund, there are hundreds of web sites available. Two of my personal favorites are

Quicken.com,

http://www.quicken.com ,

and Morningstar.com, at

http://www.morningstar.com .

Once you link to either of the above web pages, you’ll find an area to enter the “ticker

symbol” of the mutual fund. Having done this, you will then be linked to a summary

of the performance of this fund. The most basic items in these summaries are the

net asset value, or NAV, which is the price of one share of the entire mutual fund,

and a general indication of whether the NAV has been increasing or decreasing over

7I own shares in both of the above-mentioned mutual funds. The index fund has done much better than the growth

fund!


time.

As should be obvious, choosing a good mutual fund requires a bit of initial research.

For additional reading, there’s a nice web site (Stocks Digest) available, at

http://investsmart.coe.uga.edu/C001759/ .

(Scroll down to the information on mutual funds.)

5.2.2 Analysis of Stocks

Given that only about 25% of mutual funds will, in a given year, perform better

than the over-all stock market, one can see why so many people avoid mutual funds

altogether. The alternative, of course, is to buy stock directly, creating a “portfolio”

of stock which you—the investor—will manage yourself.

There are essentially two types of analysis typically used in buying stocks. One is

called fundamental analysis, which is based on an in-depth analysis of the company

whose stocks you’re considering buying (or selling). This can be very labor intensive,

but certain shortcuts are available on the internet. The above-mentioned website,

Quicken.com, does a decent job of summarizing a company’s fundamentals.

Example. We’ll take a stock that I own, Starbucks Coffee (Ticker: SBUX) as an

example. Link to Quicken.com and enter SBUX where you see “Quotes & Research”

(Enter symbol). Having done this, you’ll be taken to the general information about

the performance of this stock, including in particular, its most recent price and various

information on its trends. Within the left-hand banner of this page you’ll find a link

(under Company Data) called Fundamentals; click on this and you’ll find the usual

sort of statistics and information that go into the fundamental analysis of a company.

We won’t go into all of the items contained on this page; however, if you want to

become a good investor, this is something that you’ll eventually want to come to

grips with.

A second mode of analysis is called technical analysis, which is based on charts

depicting the growth of this stock. People more inclined to engage in fundamental

analysis often scoff at those who do technical analysis, often calling it based more on


superstition rather than on good solid information. Nonetheless, a good website for

information on technical analysis is ClearStation, at

http://clearstation.etrade.com/ .

Example. We can continue the example started above, namely the analysis of stock

in Starbucks (SBUX). Enter this ticker in the space provided (next to Symbol Search)

and click on “Go.” Here you’ll see a myriad of graphs on the basis of which technical

analysis is done. There is also information related to the “Fundamentals” of the

company, but this is not as extensive or as well organized as given by Quicken.com.

5.2.3 A Little Bit of Mathematics—The P/E and PEG Ratios

The P/E ratio, or price-to-earnings ratio plays an important role in virtually every

investor’s fundamental analysis of a stock. This is defined by the formula

P/E =price of the stock

company’s earnings per share over the past 12 months.

The numerator is pretty self explanatory: it’s the current price of one share of stock.

The denominator is computed as follows. We first let EPS denote the earnings per

share over the last 12 months. Then

EPS =company’s net profit over the past 12 months

total number of shares issued by the company.

As a simple example, suppose that a company made a net profit of $2 million over the

past 12 months and that it has 2 million outstanding shares of stock, then its EPS is

$1 per share. Next, if one share of this stock is selling at $35, the the stock’s P/E is

$35/$1 = 35. Investors often refer to this number by saying that “the stock is trading

at 35 times earnings.” Wall Street analysts generally agree that a stock is “fairly

valued” if its P/E is approximately the company’s projected growth in earnings.8

Thus, in the present case, if it appears that the company’s annual growth in earnings

8Here’s where a little gray area slips in. Projected growth is typically estimated by financial “experts,” but even

they can made substantial mistakes in their predictions.


(i.e., profit) will be in the 40% range, then many analysts would conclude that since

the price-to-earnings ratio (35) is less than the projected growth (40%), then the stock

is undervalued, and therefore is being sold at an attractive price. Conversely, if it

appeared that the company’s growth would not keep up with the stock’s P/E—say

it were in the 20% range—then analysts would call the stock overvalued, in which

case one wouldn’t consider buying the stock (or might even consider trying to sell

of some of the stock if he/she owned any). The ratio that divides projected growth

into the P/E is called the PEG ratio, or the price-to-earnings-to-growth ratio.

Thus, in the cases considered above, a P/E of 35 and a 40% growth rate would result

in a PEG of 35/40 = .875, and a P/E of 35 and a growth rate of 20% results in a

PEG of 35/20 = 1.75.

Example. Let’s look again at my Starbucks stock from the point of view of the

Quicken.com web page. Once we’ve linked to the fundamentals of this stock, we find

already that the P/E is 48.80, and that the PEG has already been calculated for us;

it’s given to be PEG = 2.1. On the basis of what’s been said above, it would appear

that the stock is considerably overvalued. Yet, in order to appreciate how many

other factors are likely to be in play here, try clicking on the Analyst Ratings under

Analysis in the left-hand banner. You’ll see that there is not a strong consensus

either way (to buy to to sell). The moral of this story is that any given mathematical

indicator is only one factor out of many in the analysis of a stock!



1. Compute a fair purchase price of a $1,000 discount bond in each case given below.

(a) The YTM is 6% and the time to maturity is 10 years.

(b) The YTM is 5.6% and the time to maturity is 8 years.

(c) The YTM is 4.3% and the time to maturity is 5 years.

2. Suppose that a $2,000 discount bond with 8 years remaining to maturity was

purchased at the (fair) price of $1,350. Compute the YTM.

3. Suppose that a $25,000 10-year discount bond with a YTM of 5.6% was pur-

chased.

(a) If a fair price was paid, what was the purchase price?

(b) Suppose that after holding this bond for one year, the interest rate suddenly

falls to 5.0%.

(i) What is the value of this bond now?

(ii) If you sell the bond at this new fair price, what would be your profit?

4. Suppose that you are holding a $3,000 discount bond with seven years to matu-

rity. Suppose that three years ago, you paid $1,775 for this bond.

(a) What is the YTM for this bond?

(b) Suppose that your financial consultant told you that the interest rate was

likely to move to roughly 6.2%, and that he had someone interested in buying

your bond for for $2,000. Should you sell it? Why?

5. We saw from the reading above (page 158) that the present value of a $1,000

10-year coupon bond with a coupon rate of 8% is $1,000. What will be the

present value of this same bond two years from now (assuming no changes in the

interest rate)? What about three years from now? Does this surprise you?

6. Suppose that you purchase a $2,500 10-year coupon bond with a coupon rate

of 5.3%. The day after you purchase this bond the interest rate falls to 4.3%.

What is the present value of this bond now?

7. Assume the same conditions as in the above problem, except that the interest

rate rises to 5.8%. What is the new present value of the bond?


8. Compute the YTD for bonds given the following information:

(a) Coupon rate = 3.4%, par value = $1,000, purchase price = $946.00, years

to maturity = 10 years.

(b) Coupon rate = 5.6%, par value = $5,000, purchase price = $4,340.34 years


(c) Coupon rate = 3.4%, par value = $1,000, purchase price $1,045.34, years to

maturity = 10 years.

(d) Coupon rate = 3.4%, par value = $10,000, purchase price $10,934.09, years


9. Compute the YTM for the bonds given below, assuming that the table was

created April 27, 2004:

Coupon Mat. date bid $ Yld %

Corporate

Scorp 8.210 Apr 27/25 100.46

SASinc 6.750 Oct 27/14 94.00

RuCh 5.800 Oct 27/09 100.01

Willinc 5.610 Apr 27/11 100.25


1. About three years ago, I purchased about 100 shares of the White Oaks Mutual

Fund (WOGSX). Using Quicken.com, try to answer the following questions.

(a) What is the present price of one share of this mutual fund (the “NAV”)?

(b) Has this mutual fund seen growth in its NAV over the past year? The past

two years?

(c) Use the graphing capabilities (see left-hand banner) to try to estimate the

NAV of this mutual fund three years ago.

(d) Roughly how much money did I spend for the 100 shares of this mutual

fund, and roughly how much money is 100 shares of this mutual fund worth

today? What are my profits (losses)?


(e) Do you find anything in this report that might discourage you from buying

any shares of this mutual fund?

2. About three years ago I purchased about 1,000 shares of the mutual fund USAA S

& P 500 fund (Ticker: USSPX). Use Quicken.com and answer the same questions

as in Question 1.

3. Would you consider buying shares in the PBHG Growth Fund (PBHGX)? Use

Quicken.com to guide your analysis.

4. You’re probably aware that the value of technology-based stocks dropped very

suddenly during the early part of 2001. This happened to one of my stocks,

ADC Telecommunications, Inc. (Ticker: ADCT, which specializes in broadband

technology). Estimate the percentage drop in the price of one share of this stock

from June, 2001 to the present.

5. What are the present P/E and PEG of ADC Telecommunications, Inc.?

6. Try to find the ticker for Coca Cola and then find the present P/E and PEG for

Coca Cola. Does this look like an attractive stock?


Index

P/E ratio, 167

after-tax annuity, 39, 42

amortization

table, 70

annual percentage rate, 63

basic problem of, 64

annual rate, 1

annual yield, 9, 17

annuity, 32

after-tax, 39, 42


before-tax, 39

fixed deferred, 41

payout, 36

APR, 63


average daily balance, 6

balloon payment, 92

before-tax annuity, 39

bid price, 162

bond

discount, 156

face-value, 156

yield to maturity, 156

zero-interest, 156

bond election, 155

bond table, 162

bonds, 155

par value, 160

common logarithm, 15

compound interest, 2

compounding

monthly, 2

constraints, 122, 125

consumer surplus, 114

continuous compounding, 19, 20

convex, 128

Corner Point Principle, 130

cost of living, 38

adjustment, 38

cost of living adjustment


cost of living adjustment (C.O.L.A.), 38

coupon bond, 158

coupon rate, 158

credit card, 6

debt security, 163

demand, 112

elasticity of, 116

derivative, 143

discount, 160

discount bond

YTM, 157

diversity, 165

173

174 INDEX

doubling time, 21

rule of 72, 22

earnings per share, 167

elastic, 116, 117

elasticity of demand, 116

EPS, 167

equity, 76

equity security, 163

exact interest, 3

feasibility region, 125

finance charge, 7, 80

fixed deferred annuity, 41

fundamental analysis, 166

geometric sequence, 27

ratio, 27

geometric sum, 28

hard polynomial equation, 65

hard polynomial problem, 160

inelastic, 116, 117

initial public offering, 164

interest

exact, 3

ordinary, 3

simple, 1

interest rate

nominal, 3

interest-only payments, 92

investments

speculative, 155

IPO, 164

legal loan amount, 81

linear inequality, 126

linear programming, 125, 131

problem of, 126

loan, 59

add-on interest, 59

amortized, 59

simple-interest amortized, 59, 60


loan shark, 5

logarithm

common, 15

logarithmic functions, 12

logarithms, 11

marginal utility, 112

diminishing, 109

Mersenne primes, 16

monthly compounding, 2

mutual fund, 165

NAV, 165

net asset value, 165

nominal interest rate, 3

objective function, 125, 126

optimal pricing rule, 120

optimal purchase rule, 113

ordinary interest, 3

payout annuity, 36


PEG ratio, 168

periodic rate, 6

points, 81

premium, 160

INDEX 175

prepaid finance charges, 80

present value, 25, 60

present value of money, 37

price-to-earnings ratio, 167

price-to-earnings-to-growth ratio, 168

primary market, 164

prime numbers, 16

principle, 1

public key cryptosystem, 16

ratio, 27

rule of 72, 22

sales tax, 108

secondary market, 164

security, 163

debt, 163

equity, 163

selling strategy, 118

sequence

geometric, 27

simple interest, 1

simplex method, 131

sinking fund, 32, 34


specific tax, 108

speculative investments, 155

stock

overvalued, 168

undervalued, 168

stock exchange, 164

stocks, 155, 164

sum

geometric, 28

summation notation, 30

tax savings, 78

technical analysis, 166

ticker, 165

Truth-in-Lending Act, 63, 81

unpaid balance, 74


utility, 109

yield to maturity, 156

Documents

The Time-Dependent Nature of Moneydbski/writings/Financial_Math.pdf · The Time-Dependent Nature of Money David Surowski Shanghai American School September 5, 2010