Contemporary Business Mathematics with Canadian Applications [7 ed.] 013120128X

Citation preview

Contemporary Business Mathematics

Ft

with Canadian Applications

Seventh Edition

List of Formulas Formula 2.1

am X an = am + n

The rule for multiplying two powers having the same base

Formula 2.2

am

The rule for dividing two powers having the same base

Formula 2.3

(am)" = amn

The rule for raising a power to a power

Formula 2.4

(ab)m = ambm

The rule for taking the power of a product

Formula 2.5

(—\m - —

The rule for taking the power of a quotient

Formula 2.6

an = am ~ "

\b)

~ bm 1 a m = — am

The definition of a negative exponent

Formula 2.7

iff-iff

The rule for a fraction with a negative exponent

Formula 2.8

an = ^Ta

The definition of a fractional exponent with numerator 1

Formula 2.9

_i

1

The definition of a fractional exponent with numerator -1

Formula 2.10

a" = Vo"

The definition of a positive fractional exponent

Formula 2.11

a n

The definition of a negative fractional exponent

Formula 2.12

In (ab) — In a + In b

The relationship used to find the logarithm of a product

Formula 2.13

In \ -g) = In a — In b

The relationship used to find the logarithm of a quotient

Formula 2.14

In (ak) = k(ln a)

The relationship used to find the logarithm of a power

Formula 3.1A

PERCENTAGE = RATE

X

BASE

The basic percentage relationship

or NEW NUMBER = RATE

Formula 3.2

ORIGINAL NUMBER

X

ORIGINAL NUMBER

+ INCREASE _ NEW — DECREASE NUMBER

The relationship to use with problems of increase or decrease (problems of change)

Formula 3.3

AMOUNT OF CHANGE

RATE OF CHANGE

ORIGINAL NUMBER

Formula for finding the rate of change (rate of increase or decrease)

Formula 3.4

INCOME IN CURRENT DOLLARS REAL INCOME

CONSUMER PRICE INDEX

Formula for eliminating the effect of inflation on income

Formula 4.1

y = tnx

Formula 5.1

PROFIT = (VOLUME

+

b

Slope-y-intercept form of a linear equation

X

UNIT REVENUE) — (VOLUME

X

VARIABLE COST PER UNIT) — FIXED COST

Formula for finding profit when separating fixed and variable costs

Formula 5.2

CONTRIBUTION MARGIN PER UNIT = SELLING PRICE PER UNIT — VARIABLE COST PER UNIT

Formula for finding contribution margin per unit UNIT CONTRIBUTION MARGIN

Formula 5.3

CONTRIBUTION RATE =

UNIT SELLING PRICE

Formula 5.4

BREAK-EVEN VOLUME =

unit CONTRIBUTION MARGIN

Formula for finding contribution rate

FIXED COST

Formula for finding break-even volume based on unit contribution margin

ORIGINAL _ RESIDUAL

Formula 5.5

YEARLY

_

COST_VALUE

DEPRECIATION

~

n

Formula for finding the yearly depreciation when using the straight-line method

where n = the number of years in the life of the asset

DEPRECIATION PER UNIT OF PRODUCT

Formula 5.6

_ —

ORIGINAL _ RESIDUAL COST VALUE

Formula for finding the depreciation per unit of

n

product when using the units-of-product method

where n = the number of product units in the life of the asset

Formula 5.7

,

WEARING VALUE

Formula for finding the value of one part (constant of

k = -:-rn(n + 1)

proportion) when using the sum-of-the-years-digits

2

method

where Wearing Value

Formula 5.8

Original _ Residual Cost

d= 2 X — n where n = the number of years in

Value Formula for finding the rate of depreciation when using the simple declining-balance method

the life of the asset

Formula 6.1A

Formula 6.2

Formula 6.3A

AMOUNT

RATE OF

OF DISCOUNT

DISCOUNT

Finding the amount of discount when the list price X LIST PRICE

AMOUNT OF NET PRICE = LIST PRICE -

TRADE DISCOUNT

NET PRICE = 100% — % DISCOUNT FACTOR (NPF)

is known

Finding the net amount when the amount of discount is known Finding the net price factor (NPF)

Finding the net amount directly without computing the

Formula 6.4A

NET PRICE =

NET PRI,CE v X LIST PRICE FACTOR (NPF)

NET PRICE FACTOR

Formula 6.5A

(NPF) FOR THE DISCOUNT SERIES

NPF FOR THE

amount of discount

NPF FOR THE

FIRST DISCOUNT X SECOND DISCOUNT

x ‘ ' '

NPF FOR THE LAST DISCOUNT

Digitized by the Internet Archive in 2018 with funding from Kahle/Austin Foundation

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National Library of Canada Cataloguing in Publication Hummelbrunner, S. A. (Siegfried August) Contemporary business mathematics with Canadian Applications / S.A. Hummelbrunner, K. Suzanne Coombs. — 7th ed. ISBN 0-13-120128-X 1. Business mathematics. I. Coombs, K. Suzanne. II. Title. HF5691.H85 2005

650'.01'513

C2003-906883-8

Copyright © 2005, 2001, 1998, 1994, 1990, 1986, and 1982 Pearson Education Canada Inc., Toronto, Ontario

Pearson Prentice Hall. All rights reserved. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission, write to the Permissions Department. BA II PLUS and BA-35 Solar are trademarks of Texas Instruments Incorporated. HP-10B Business Calculator is a trademark of Hewlett-Packard Company. EL-733A is a trademark of Sharp Corporation. ISBN 0-13-120128-X Vice President, Editorial Director: Michael Young Senior Acquisitions Editor: Gary Bennett Director of Marketing: Bill Todd Senior Developmental Editor: Madhu Ranadive Production Editor: Joel Gladstone Copy Editor: Rodney Rawlings Production Coordinator: Janette Lush Page Layout: Pronk Permissions Research: Nicola Winstanley Art Director: Julia Hall Interior and Cover Design: Open House Incorporated/Sonya Thursby Cover Image: Corbis

6 789 10

10 09 08 07 06

Printed and bound in the United States.

Brief Contents Preface ... x Part

Mathematics Fundamentals and Business Applications ... 1 1 Review of Arithmetic ... 1 2 Review of Basic Algebra ... 44 Ratio, Proportion, and Percent ... 94 4 Linear Systems ... 147

Part

Mathematics of Business and Management... 183 L-"5^ Business Applications—Break-even Analysis and Depreciation ... 183 l.6 Trade Discount, Cash Discount, Markup, and Markdown ... 218 Simple Interest ... 268 8 Simple Interest Applications ... 310

Part

Mathematics of Finance and Investment... 345 uJT Compound Interest—Future Value and Present Value ... 345 10 Compound Interest—Further Topics ... 408 11 Ordinary Simple Annuities ... 439 12 Ordinary General Annuities ... 485 13 Annuities Due, Deferred Annuities, and Perpetuities ... 515 14 Amortization of Loans, Including Residential Mortgages ... 574 15 Bond Valuation and Sinking Funds ... 634 16 Investment Decision Applications ... 697 Appendix I: Review of Basic Algebra ... 735 Appendix II: Instructions and Tips for Four Preprogrammed Financial Calculator Models ... 755 Answers to Odd-Numbered Problems, Review Exercises, and Self-Tests ... 772 Index ... 787 License Agreement... 796 Spreadsheet Template Disk, including Appendix A: Linear Applications Appendix B: Finding the Rate of Interest (i) Without Preprogramming

Contents Preface ... x

Part

Mathematics Fundamentals and Business Applications ... 1 1 Review of Arithmetic ... 1 1.1 Basics of Arithmetic ... 2 1.2 Fractions ... 3 1.3 Percent ... 8 1.4 Applications—Averages ... 13 1.5 Applications—Payroll ... 20 1.6 Applications—Taxes ... 28 Business Math News Box ... 33 Review Exercise ... 34 Self-Test ... 36 Challenge Problems ... 38 Case Study 1.1: Business and the GST ... 38 Case Study 1.2: How Much Are You Worth? ... 41 Glossary ... 42 Useful Internet Sites ... 43

2 Review of Basic Algebra ... 44 2.1 Simplification of Algebraic Expressions ... 45 2.2 Integral Exponents ... 51 2.3 Fractional Exponents ... 58 2.4 Logarithms—Basic Aspects ... 62 Business Math News Box ... 68 2.5 Solving Basic Equations ... 69 2.6 Equation Solving Involving Algebraic Simplification ... 74 2.7 Solving Word Problems ... 80 Review Exercise ... 85 Self-Test ... 87 Challenge Problems ... 88 Case Study 2.1: Investing for a Rainy Day ... 89 Case Study 2.2: Expenses on the Road ... 90 Summary of Formulas ... 91 Glossary ... 92 Useful Internet Sites ... 93

3 Ratio, Proportion, and Percent... 94 3.1 Ratios ... 95 3.2 Proportions ... 100 3.3 The Basic Percentage Problem ... 106 3.4 Problems Involving Increase or Decrease ... 115 3.5 Problems Involving Percent ... 119 Business Math News Box ... 124 3.6 Applications—Currency Conversions ... 127

CONTENTS

3.7 Applications—Index Numbers ... 130 3.8 Applications—Personal Income Taxes ... 134 3.9 Applications—Return on Mutual Funds ... 136 Review Exercise ... 139 Self-Test ... 141 Challenge Problems ... 143 Case Study 3.1: The Business of Taxes ... 143 Case Study 3.2: Trip of a Lifetime ... 144 Summary of Formulas ... 145 Glossary ... 145 Useful Internet Sites ... 146 4 Linear Systems ... 147 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables ... 148 4.2 Graphing Linear Equations ... 154 4.3 Graphing Linear Systems of Equations in Two Unknowns ... 169 Business Math News Box ... 172 4.4 Problem Solving ... 173 Review Exercise ... 178 Self-Test ... 179 Challenge Problems ... 179 Case Study 4.1: Finding the Right Combination ... 180 Case Study 4.2: What to Produce? ... 181 Summary of Formulas ... 182 Glossary ... 182 Useful Internet Sites ... 182 Mathematics of Business and Management ... 183 5 Business Applications—Break-even Analysis and Depreciation ... 183 5.1 Break-even Analysis ... 184 5.2 Finding the Effect of Changes to Cost-Volume-Profit ... 198 5.3 Depreciation ... 200 Business Math News Box ... 210 Review Exercise ... 211 Self-Test ... 212 Challenge Problems ... 213 Case Study 5.1: Planning for Production ... 214 Case Study 5.2: Calculating Car Costs ... 214 Summary of Formulas ... 215 Glossary ... 216 Useful Internet Sites ... 217 6 Trade Discount, Cash Discount, Markup, and Markdown ... 218 6.1 Merchandising ... 219 6.2 Trade Discount... 220 6.3 Multiple Discounts ... 224

CONTENTS

6.4 Cash Discount ... 231 6.5 Markup ... 239 6.6 Markdown ... 249 Review Exercise ... 259 Self-Test ... 261 Challenge Problems ... 263 Case Study 6.1: Focusing on Prices ... 263 Case Study 6.2: Putting a Price on the Table ... 264 Summary of Formulas ... 265 Glossary ... 266 Useful Internet Sites ... 267 7 Simple Interest... 268 7.1 Determining the Number of Days between Two Dates ... 269 7.2 Finding the Amount of Simple Interest ... 271 7.3 Finding the Principal, Rate, or Time ... 274 7.4 Computing Future Value (Maturity Value) ... 281 7.5 Finding the Principal (Present Value) ... 284 Business Math News Box ... 288 7.6 Computing Equivalent Values ... 289 Appendix: Determining the Number of Days Using Manual Techniques ... 300 Review Exercise ... 304 Self-Test ... 305 Challenge Problems ... 306 Case Study 7.1: Loans, Loans, Loans ... 306 Case Study 7.2: Pay Now or Pay Later? ... 307 Summary of Formulas ... 308 Glossary ... 308 Useful Internet Sites ... 309 8 Simple Interest Applications ... 310 8.1 Promissory Notes—Basic Concepts and Computations ... 311 8.2 Maturity Value of Interest-Bearing Promissory Notes ... 314 8.3 Present Value of Promissory Notes ... 316 8.4 Demand Loans ... 323 8.5 Lines of Credit and Credit Card Loans ... 328 Business Math News Box ... 333 8.6 Loan Repayment Schedules ... 333 Review exercise ... 338 Self-Test ... 339 Challenge Problems ... 340 Case Study 8.1: The Business of Borrowing ... 341 Case Study 8.2: Dealing with Debt ... 341 Summary of Formulas ... 342 Glossary ... 343 Useful Internet Sites ... 344

CONTENTS

Mathematics of Finance and Investment... 345 9 Compound Interest—Future Value and Present Value ... 345 9.1 Basic Concepts and Computations ... 346 9.2 Using the Formula for the Future Value of a Compound Amount FV = PV(1 + i)n ... 354 9.3 Present Value and Compound Discount ... 368 9.4 Application—Discounting Negotiable Financial Instruments at Compound Interest ... 374 Business Math News Box ... 381 9.5 Equivalent Values ... 382 Review Exercise ... 400 Self-Test ... 402 Challenge Problems ... 403 Case Study 9.1: What's in Your Best Interest? ... 404 Case Study 9.2: Planning Ahead ... 405 Summary of Formulas ... 406 Glossary ... 406 Useful Internet Sites ... 407 10 Compound Interest—Further Topics ... 408 10.1 Finding n and Related Problems ... 409 10.2 Finding / and Related Problems ... 417 10.3 Effective and Equivalent Interest Rates ... 422 Business Math News Box ... 433 Review Exercise ... 434 Self-Test ... 435 Challenge Problems ... 436 Case Study 10.1: Choosing a Credit Card ... 436 Case Study 10.2: Comparing Car Loans ... 437 Summary of Formulas ... 438 Glossary ... 438 Useful Internet Sites ... 438 11 Ordinary Simple Annuities ... 439 11.1 Introduction to Annuities ... 440 11.2 Ordinary Simple Annuity—Finding Future Value FV ... 443 11.3 Ordinary Simple Annuity—Finding Present Value PV ... 454 Business Math News Box ... 464 11.4 Ordinary Simple Annuities—Finding the Periodic Payment PMT ... 465 11.5 Finding the Term n of an Annuity ... 471 11.6 Finding the Periodic Rate of Interest / Using Programmed Financial Calculators ... 477 Review Exercise ... 479 Self-Test ... 480 Challenge Problems ... 481

CONTENTS

Case Study 11.1: Saving for Your Dream ... 481 Case Study 11.2: Getting the Picture ... 482 Summary of Formulas ... 483 Glossary ... 483 Useful Internet Sites ... 484 12 Ordinary General Annuities ... 485 12.1 Ordinary General Annuities—Finding the Future Value ... 486 12.2 Ordinary General Annuities—Finding the Present Value PV ... 494 12.3 Ordinary General Annuities—Finding the Periodic Payment PMT ... 497 12.4 Ordinary General Annuities—Finding the Term n ... 501 12.5 Ordinary General Annuities—Finding the Periodic Interest Rate /... 505 Business Math News Box ... 505 Review Exercise ... 508 Self-Test ... 509 Challenge Problems ... 510 Case Study 12.1: Cash-Back Options ... 510 Case Study 12.2: Fitness Finances ... 511 Summary of Formulas ... 513 Glossary ... 513 Useful Internet Sites ... 514 13 Annuities Due, Deferred Annuities, and Perpetuities ... 515 13.1 Simple Annuities Due ... 516 13.2 General Annuities Due ... 531 Business Math News Box ... 539 13.3 Deferred Annuities ... 540 13.4 Perpetuities ... 557 Review Exercise ... 565 Self-Test ... 569 Challenge Problems ... 571 Case Study 13.1: From Casino to College ... 571 Case Study 13.2: Setting Up Scholarships ... 572 Summary of Formulas ... 573 Glossary ... 573 Useful Internet Sites ... 573 Amortization of Loans, Including Residential Mortgages ... 574 14.1 Amortization Involving Simple Annuities ... 575 14.2 Amortization Involving General Annuities ... 593 14.3 Finding the Size of the Final Payment ... 605 14.4 Residential Mortgages in Canada ... 613 Business Math News Box ... 621 Review Exercise ... 626 Self-Test... 628 Challenge Problems ... 629

Case Study 14.1: Managing a Mortgage ... 630 Case Study 14.2: Steering the Business ... 631 Summary of Formulas ... 632 Glossary ... 632 Useful Internet Sites ... 633 15 Bond Valuation and Sinking Funds ... 634 15.1 Purchase Price of Bonds ... 635 15.2 Premium and Discount ... 648 Business Math News Box ... 660 15.3 Bond Schedules ... 661 15.4 Finding the Yield Rate ... 671 15.5 Sinking Funds ... 674 Review Exercise ... 689 Self-Test ... 692 Challenge Problems ... 693 Case Study 15.1: Investing in Bonds ... 693 Case Study 15.2: The Business of Bonds ... 694 Summary of Formulas ... 695 Glossary ... 695 Useful Internet Sites ... 696 16 Investment Decision Applications ... 697 16.1 Discounted Cash Flow ... 698 16.2 Net Present Value Method ... 706 Business Math News Box ... 715 16.3 Finding the Rate of Return on Investment... 716 Review Exercise ... 729 Self-Test ... 731 Challenge Problems ... 731 Case Study: 16.1: To Lease or Not to Lease? ... 732 Case Study: 16.2: Building a Business ... 733 Summary of Formulas ... 734 Glossary ... 734 Useful Internet Sites ... 734 Appendix I: Review of Basic Algebra ... 735 Appendix II: Instructions and Tips for Four Preprogrammed Financial Calculator Models ... 755 Answers to Odd-Numbered Problems, Review Exercises and Self-Tests ... 772 Index ... 787 License Agreement... 796 Spreadsheet Template Disk, including Appendix A: Linear Applications Appendix B: Finding the Rate of Interest (i) Without Preprogramming

Preface INTRODUCTION Contemporary Business Mathematics is intended for use in introductory mathe¬ matics of finance courses in business administration programs. In a more general application it also provides a comprehensive basis for those who wish to review and extend their understanding of business mathematics. The primary objective of the text is to increase the student’s knowledge and skill in the solution of practical financial and mathematical problems encountered in the business community. It also provides a supportive base for mathematical topics in finance, accounting, and marketing.

ORGANIZATION Contemporary Business Mathematics is a teaching text using the objectives approach. The systematic and sequential development of the material is supported by carefully selected and worked examples. These detailed step-by-step solutions presented in a clear and colourful layout are particularly helpful in allowing stu¬ dents, in either independent studies or in the traditional classroom setting, to care¬ fully monitor their own progress. Each topic in each chapter is followed by an Exercise containing numerous drill questions and application problems. The Review Exercise, Self-Test, and the Case Studies at the end of each chapter integrate the material studied. The first four chapters and Appendix I (Review of Basic Algebra) are intended for students with little or no background in algebra and provide an opportunity to review arithmetic and algebraic processes. The text is based on Canadian practice, and reflects current trends using avail¬ able technology—specifically the availability of reasonably priced electronic pocket calculators. Students using this book should have access to calculating equipment having a power function and a natural logarithm function. The use of such calculators eliminates the arithmetic constraints often associated with finan¬ cial problems and frees the student from reliance on financial tables. The power function and the natural logarithm function are often needed to determine values that will be used for further computation. Such values should not be rounded and all available digits should be retained. The student is encouraged to use the memory to retain such values. When using the memory the student needs to be aware that the number of digits retained in the registers of the calculator is greater than the number of digits displayed. Depending on whether the memory or the displayed digits are used, slight differences may occur. Such differences will undoubtedly be encountered when working through the examples presented in the text. However, they are insignificant and should not be of concern. In most cases the final answers will agree, whichever method is used. Students are encouraged to use preprogrammed financial calculators, though this is not essential. The use of preprogrammed calculators facilitates the solving of most financial problems and is demonstrated extensively in chapters 9 to 16.

PREFACE

NEW TO THIS EDITION In this seventh edition major revisions have been made to the text. To reflect cur¬ rent practices in Canada and to better suit the needs of users of this book, a num¬ ber of important changes have been made in content and organization. Specifically, in Chapter 3 (Ratio, Proportion, and Percent), the order for the sections has been changed. The section on proportions is followed by problems on proportions. The section on percent is followed by problems with percent. Currency conversion rates, index numbers, personal income tax rates, and mutual fund valuations have been updated. The major emphasis in Chapter 5 (Business Applications—Break-even Analysis and Depreciation) has been changed, with Break-even Analysis shown first, fol¬ lowed by Depreciation. The sections on break-even analysis have been reworked and expanded. A new section on cost-volume-profit relationships and sensitivity analysis has been added. New break-even problems have been added. The theory of break-even using formulas has been separated from the technique of graphing. The methods of depreciation have been updated and consolidated as well. In Chapter 6 (Trade Discount, Cash Discount, Markup, and Markdown), mer¬ chandising terminology has been updated, clarified, and standardized. Problems have been reordered and reworded to improve their logic. In Chapter 7 (Simple Interest), objectives have been simplified with a focus on outcomes. The manual calculation of days between dates has been placed in a chapter appendix. Financial calculator instructions have been increased and updated and the mathematical steps in solutions have been simplified. Consistent terminology has been used. In Chapter 8 (Simple Interest Applications), additional coverage of credit card loans has been included. Coverage of demand loans has been separated from cov¬ erage of lines of credit. The discussion of discounting promissory notes has been reduced and the computer application to develop a loan repayment schedule has been updated. In Chapter 9 (Compound Interest—Future Value and Present Value), compu¬ tations involving whole number and fractional exponents within future value and present value sections have been combined. The use of the terms FV and PV within the formulas has been introduced and used from the beginning of the chapter. In Chapter 10 (Compound Interest—Further Topics), calculator solutions have been added, along with many new problems. The sections on finding interest rates, and effective and equivalent interest rates, have been expanded as well. The topics for Chapters 11, 12, and 13 have been reorganized to emphasize the type of annuity; Chapter 11 focuses on Ordinary Simple Annuities while Chapter 12 focuses on Ordinary General Annuities. In each of these chapters, the sections have a similar order. The learner finds the future value, the present value, the peri¬ odic payment, the number of payments, and then the interest rate. Chapter 13 covers Simple Annuities Due, General Annuities Due, Deferred Annuities, and Perpetuities. In these chapters, an identification of the nominal interest rate per year and the number of compounding periods per year have been added. A new explanation of Down Payment has been added in Chapter 11. Many new exercise questions have been added in all three chapters, as well as additional calculator solutions.

PREFACE

In general, interest rates used reflect the current economic climate in Canada. Calculator tips and solutions have been updated or added. Spreadsheet instructions and Internet site references have been updated. More Pitfalls and Pointers have been added to assist in performing tasks and interpreting word problems, and sections have been rewritten to clarify the explanations. Many more word problems have been added and the problems have been ranked by difficulty, with an icon indicating the more difficult problems. New Business Math News Boxes and case studies have been added. Examples involving both business and personal situations are included. Where appropriate, the use of six decimal places has been standardized. The pedagogical elements of the previous edition have been retained. In response to requests and suggestions by users of the book, a number of new fea¬ tures for this edition have been included. They are described below.

FEATURES • A new colourful and student-friendly design has been created for the book, mak¬ ing it more accessible and less intimidating to learners of all levels. • Instructions for using preprogrammed financial calculators have been further expanded in this seventh edition. Although any preprogrammed financial calcu¬ lator may be used, this edition includes extensive instructions for using the Texas Instruments BA II Plus financial calculator. Equivalent instructions are given in Appendix II for the Texas Instruments BA-35 Solar, the Sharp EL-733A, and the Hewlett Packard 10B financial calculators. • To reduce the amount of “translation” required to go from the formulas in the text to the keystrokes on the preprogrammed financial calculator, the compounding and annuity formulas in Chapters 9 to 16 have been restated in this edition. From the beginning of Chapter 9, the P has been replaced with PV, S has been replaced with FV, and A has been replaced with PMT in these formulas. In addition, the compound¬ ing interval, C/Y, has been identified within the calculator solutions.

A new Spreadsheet Template Disk was developed by Bruce Coombs. It is a CD-ROM that accompanies the text and contains Word files and Excel spreadsheet files. It requires Windows 95 or newer versions of Windows to run. The disk con¬ tains 26 self-contained tutorials that show how to use one of Excel’s special spread¬ sheet functions to solve business problems. Each tutorial gives a brief description of the function, a general example showing how the function might be used, direc¬ tions for creating a spreadsheet template to use the function, and a list of questions

PREFACE

from the text that can be answered using the spreadsheet function. In addition, an Excel file containing ready-to-use examples of each function is included. The student can use these spreadsheets to enter values and obtain imme¬

Excel

diate results. The Spreadsheet Template Disk also contains spreadsheet templates for the spreadsheet applications described in the text: depreciation schedules (Chapter 5), loan repayment schedules (Chapter 8), accumula¬ tion of principal schedules (Chapter 9), amor¬ tization schedules (Chapter 14), and sinking fund schedules (Chapter 15). An Excel icon in the text highlights information on the use of Excel to solve problems and the Excel spread¬ sheet applications, directing students to the spreadsheet template disk. A spreadsheet icon in the text highlights both the questions in the text that can be solved using an Excel spreadsheet function. Appendix A and B are also included on the

9

Compound Interest—Future Value and Present Value

Spreadsheet Template Disk. OBJECTIVES

• A set of learning objectives is listed at the beginning of each chapter. • Each chapter opens with a description of a situation familiar to students to empha¬ size the practical applications of the material to follow.

• A Business Math News Box is presented in most chapters. This element consists of short excerpts based on material appear¬ ing in newspapers, magazines, or web¬ sites, followed by a set of questions. These boxes demonstrate how widespread busi¬ ness math applications are in the real world. Thirteen of the fifteen Business Math News Boxes are new or completely revised in this seventh edition.

PREFACE

• At least one new Pointers and Pitfalls box in each chapter emphasizes good prac¬ tices, highlights ways to avoid common errors, shows how to use a financial cal¬ culator efficiently, or gives hints for tack¬ ling business math situations to reduce math anxiety.

• A Did You Know? box in each chapter offers interesting practical and mathe¬ matical facts. Six of the sixteen Did You Know? boxes are new or revised in this seventh edition. • Numerous Examples with worked-out Solutions are provided throughout the book, offering easy-to-follow, step-bystep instructions. • Programmed solutions using the Texas Instruments BA II Plus calculator are offered for all examples in Chapters 9 to 16. Since this calculator display can be pre-set, it is suggested that the learner set the display to show six decimal places to match the mathematical calculations in the body of the text. Both mathematical and calculator solutions for all Exercises, Review Exercises, and Self-Tests are included in the Instructor’s Solutions Manual. An icon highlights information on the use of the BA II Plus Calculator. • Key Terms are introduced in the text in boldface type. A Glossary at the end of each chapter lists each term with its defi¬ nition and a page reference to where the term was first defined in the chapter.

PREFACE

• Main Equations are highlighted in the chapters and repeated in a Summary of Formulas at the ends of the chapters. Each main formula is presented in colour and labelled numerically (with the letter A suffix if equivalent forms of the for¬ mula are presented later). By contrast, equivalent formulae are presented in black and labelled with the number of the related main formula followed by the let¬ ter B or C. • A list of the Main Formulas begins on the inside of the front cover and continues at the end of the text for easy reference. • An Exercise set is provided at the end of each section in every chapter. If students choose, they can use the suggested Excel spreadsheet functions to answer those questions marked with the spreadsheet logo. In addition, each chapter contains a Review Exercise set and a Self-Test. If stu¬ dents choose, they can use an Excel spreadsheet function to answer those questions marked with a spreadsheet logo, but they must decide which function to use. Answers to all the odd-numbered Exercises, Review Exercises, and Self-Tests are given at the back of the book. Solutions for questions answered using Excel spreadsheet functions are given in the Instructor’s area of the Hummelbrunner Companion Website. • A set of Challenge Problems is provided in each chapter. These problems give users the opportunity to apply the skills learned in the chapter to questions that are pitched at a higher level than the Exercises. • Thirty-two Case Studies are included in the book, two near the end of each chap¬ ter. They present comprehensive realistic scenarios followed by a set of questions, and illustrate some of the important types of practical applications of the chapter material. At least one half of the

PREFACE

Case Studies represent personal-finance applications (indicated by a credit-card logo) while the rest are business applica¬ tions. Eight of the Case Studies are new or completely revised in this seventh edition. • A new set of Useful Internet Sites is pro¬ vided at the end of each chapter, with the URL and a brief description provided for each site. These sites are related to the chapter topic or to companies mentioned in the chapter, or they show how business math is important to the day-to-day operations of companies and industries.

SUPPLEMENTS The following supplements have been carefully prepared to accompany this sev¬ enth edition. An Instructor’s Resource CD-ROM has been created for this text. It will include the Solutions Manual in PDF format, PowerPoint Lecture Slides, and Test Generator. Instructions to access each item is given on the CD-ROM. Each supple¬ ment is described in more detail below. • Instructor’s Solutions Manual provides complete mathematical and calculator solutions to all the Exercises, Review Exercises, Self-Tests, Business Math News Box questions, Challenge Problems, and Case Studies in the textbook • PowerPoint Lecture Slides are a brand new supplement to accompany the text. The slides present an outline of each chapter in the book, highlighting the major concepts taught. The presentation will include many of the figures and tables from the text and provides the instructor with a visually interesting summary of the entire book. • Test Generator, a special computerized version of the test bank, enables instruc¬ tors to edit existing questions, add new questions, and generate tests. The Test Generator is organized by chapter, with level of difficulty indicated for each question. • A Student’s Solutions Manual provides complete solutions to all the odd-num¬ bered Exercises, Review Exercises, and Self-Test questions in the textbook • A Companion Website provides students with self-test questions that include Case Studies, Multiple Choice, Fill-in-the-Blank, True/False, and Internet-based Exercises. Hints are included for most of the questions, and immediate feedback is available to assist students to study and test themselves. Links to other websites of interest and an Internet search of key business math terms have been updated for the Seventh Edition. New to this edition is information about Web-based cal¬ culators and the use of interactive charts and graphs that accompany them. Visit the Hummelbrunner Website at www.pearsoned.ca/hummelbrunner and then click on the Contemporary Business Mathematics with Canadian Applications, Seventh Edition cover.

PREFACE

• On-line Student Self-Tests is an exciting new Web-based student tutorial cre¬ ated for this Seventh Edition. Students will now be able to create multiple itera¬ tions of chapter quizzes online, using algorithmically generated exercises derived from their text. Created in Pearson’s Course Compass online course manage¬ ment shell, using TestGen software, the questions come with thorough tutorial feedback, linked back in to the textbook, so students can more readily master the concepts of business math.

PREFACE

ACKNOWLEDGEMENTS Special thanks must be given to Bruce M. Coombs of Kwantlen University College and Langara College for his help in creating and updating the Spreadsheet Template Disk and the Companion Website. He listened, suggested, proofed, and agonized over this edition throughout the entire process of updating. His contri¬ bution was invaluable. Also great thanks to Jim Roberts of Red River College, who checked and revised all material pertaining to the TI BA II Plus calculator. We would like to express our thanks to the many people who offered thought¬ ful suggestions and recommendations for updating and improving the book. We would particularly like to thank the following instructors for providing formal reviews for the Seventh Edition: Millie Atkinson (Mohawk College) Wayne Bemister (Red River College) Sam Boutilier (University College of Cape Breton) Ross Bryant (Conestoga College) Helen Catania (Centennial College) Michael R. Conte (Durham College) Laurel Donaldson (Douglas College) Tom Fraser (Niagara College) Brune Fullone (George Brown College) Frank Gruen (British Columbia Institute of Technology) Amoel Lisecki (Southern Alberta Institute of Technology) Judy Palm (Malaspina College) Harry Matsugu (Humber College) Colleen Quinn (Seneca College) David Roberts (Devry Institute, Calgary) Jim Roberts (Red River College) Aruna Sarkar (Algonquin College) Susan Vallery (Fleming College) Carol Ann Waite (Sheridan College) Donald J. Webber (College of the North Atlantic) Allen Zhu (Capilano College) We would also like to thank the many people at Pearson Education Canada who helped with the development and production of this book, especially to the production editor, Joel Gladstone; the production coordinator, Janette Lush; the copy editor, Rodney Rawlings; the proofreader, Lesley Mann; and to Ross Meacher, who completed a technical check of the entire manuscript. Special thanks to edi¬ tors Kelly Torrance and Madhu Ranadive for their supportive directions.

PEARSON

The Pearson Education Canada Canada

COMPANION

WEBSITE

A Great Way to Learn and Instruct Online The Pearson Education Canada Companion Website is easy to navigate and is organized to correspond to the chapters in this textbook. Whether you are a student in the classroom or a distance learner you will discover helpful resources for in-depth study and research that empower you in your quest for greater knowledge and maximize your potential for success in the course.

www.pearsoned.ca/hummelbrunner

]

,

W:

3 HR «3IST%I

Review of Arithmetic

OBJECTIVES Upon completing this chapter, you will be able to do the following: 1. Simplify arithmetic expressions using the basic order of operations. 2.

Determine equivalent fractions and convert fractions to decimals.

3.

Convert percents to common fractions and to decimals, and change decimals and fractions to percents.

4. Through problem solving, compute simple arithmetic and weighted averages. 5.

Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

6. Through problem solving, compute GST, sales taxes, and property taxes.

Operating a business means that you must collect the Goods and Services Tax (GST) on almost everything you sell, then remit the GST to the federal government. When you make purchases of goods and services, paying GST on those purchases, you are eligible for a GST refund. When you employ people in operating a business, you must determine the amounts to pay them in the form of salaries or wages, and you must deduct and pay payroll taxes such as Canada Pension Plan, Employment Insurance, and employee income taxes. You are responsible for paying your employees and submitting the tax amounts to the federal government. By using arithmetic and problem-solving approaches in this chapter, you should be able to determine the amounts owed.

CHAPTER 1: REVIEW OF ARITHMETIC

INTRODUCTION The basics of fraction, decimal, and percent conversions are vital skills for dealing with situations you face, not only as a small business owner, but as a consumer and investor. Although calculators and laptop computers have become common tools for solving business problems, it is still important to understand clearly the process behind the conversions between number forms, the rounding of answers, and the correct order of operations.

1.1 Basics of Arithmetic A. The basic order of operations To ensure that arithmetic calculations are performed consistently, we must follow the order of operations. If an arithmetic expression contains brackets as well as any or all of exponents, multiplication, division, addition, and subtraction, we use the following procedure: 1. Perform all operations inside a bracket first (the operations inside the bracket must be performed in proper order); 2. Perform exponents; 3. Perform multiplication and division in order as they appear from left to right; 4. Perform addition and subtraction. The following “BEDMAS” rule might help you to more easily remember the order of operations: BEDMAS Brackets Exponents Division Multiplication Addition Subtraction

(i) 9 — 4X2 = 9 — 8=1 (ii) 18^6 + 3X2 = 3 + 6 = 9

do multiplication before subtraction do multiplication and division before adding

(iii) (9 - 4) X 2 = 5 X 2 = 10

work inside the bracket first

(iv) (13 + 5) - 6 - 3 = 18 ^ 6 - 3 = 3-3 = 0

work inside the bracket first, then

(v) 18 = (6 + 3) X 2 = 18 - 9 X 2

work inside the bracket first, then

= 2X2

= 4

do division before subtraction

do division and multiplication in order

(vi) 18 = (3 X 2) + 3 = 18 = 6 + 3 -work inside the bracket first, then = 3 + 3 divide before adding = 6

PART 1:

MATHEMATICS FUNDAMENTALS AND BUSINESS APPLICATIONS

work inside the brackets

(vii) 8(9 - 4) - 4(12 - 5) = 8(5) - 4(7)

first, then multiply before

= 40-28 =

(viii)

12

subtracting

= (12 - 4) = (6 - 2)

the fraction line indicates

= 8 = 4

brackets as well as division

=

2

(ix) 128 = (2 X 4)2 -3 = 128 = 82 - 3

-work inside the bracket first,

= 128 = 64 - 3

do the exponent, then

= 2-3

divide before subtracting

=

-1

(x) 128 = (2 X 42) - 3 = 128 = (2 X 16) — 3 = 128 = 32 - 3 = 4-3 =

1

and do the exponent first, then multiply, then divide before subtracting

EXERCISE 1 .1 Simplify each of the following. 1. 12 + 6 = 3

2. (12 + 6) = 3

3. (3 X 8 — 6) = 2

4. 3X8 — 6 = 2

5. (7 + 4) X 5 - 2

6. 7 + 4X5 — 2

7. 5X3 + 2X4

8. 5(3 + 2) - 12 = 3

9. (3 X 9 - 3) = 6

10. 3 X (9 - 3) = 6

11. 6(7 - 2) - 3(5 - 3)

12. 8(9 - 6) + 4(6 + 5)

15.4(8 - 5)2 - 5(3 + 22)

16. (3 X 4 - 2)2 + (2 - 2 X 72)

1.2 Fractions A. Common fractions A common fraction is used to show a part of the whole. The fraction 2/j means two parts out of a whole of three. The number written below the dividing line is the whole and is called the denominator. The number written above the dividing line is the part and is called the numerator. The numbers 2 and 3 are called the terms of the fraction. A proper fraction has a numerator that is less than the denominator. An improper fraction has a numerator that is greater than the denominator.

f'CHAPTER

1:

REVIEW

OF

ARITHMETIC

3 8

denominator

than the denominator

6 5

numerator denominator

greater than the denominator

numerator

— a proper fraction, since the numerator is less

B. Equivalent fractions Equivalent fractions are obtained by changing the terms of a fraction without changing the value of the fraction. Equivalent fractions in higher terms can be obtained by multiplying both the numerator and the denominator of a fraction by the same number. For any frac¬ tion, we can obtain an unlimited number of equivalent fractions in higher terms. Equivalent fractions in lower terms can be obtained if both the numerator and denominator of a fraction are divisible by the same number or numbers. The process of obtaining such equivalent fractions is called reducing to lower terms. (i) Convert 3A into higher terms by multiplying successively by 2, 2, 3, 5, 5, and 11. 3 _ 3X2 _ 6 _ 6X2 _ 12 _ 12 X 3 _ 36 _ 36 X 5 _ 180 4_4X2~8~8X2_16~16X3_48_48X5_240 _ 180 X 5 _ 900 _ 900 X 11 _

9900

~ 240 X 5 ~ 1200 ~ 1200 X 11 ~ 13 200 3_6_22_36_180_ _900_ _

9900

huS 4 ~ 8 ~ 16 ~ 48 - 240 ~ 1200 ~ 13 200 (ii) Reduce 21%52 to lower terms. 210 _ 210 -t- 2 _ 105 252 ~~ 252

4-

2 “ 126

105 4 3 _ 35 ~ 126 4 3 “ 42 _ 35 -f- 7 _ _5_ - 42 + 7 ~ 6 The fractions 105/i26, 35/42, and % are lower-term equivalents of 210/252. The terms of the fraction % cannot be reduced any further. It represents the simplest form of the fraction 210/252. It is the fraction in lowest terms.

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

C. Converting common fractions into decimal form Common fractions are converted into decimal form by performing the indicated division to the desired number of decimal places or until the decimal terminates or repeats. We place a dot above a decimal number to show that it repeats. For example, 0.5 stands for 0.555....

(i) — = 9 -5- 8 = 1.125 8 (ii) — = 1 h- 3 = 0.3333... = 0.3 3 (hi) — = 7

5

- -

6

6

= 1.1666... = 1.16

D. Converting mixe d numbers to decimal form Mixed numbers are numbers consisting of a whole number and a fraction, such as 5%. Such numbers represent the sum of a whole number and a common fraction and can be converted into decimal form by changing the common fraction into decimal form.

5— = 5 + — = 5 + 0.75 = 5.75 4 4

6— = 6 + — = 6 + 0.666... = 6.6666... = 6.6 3

3

7— = 7 + — = 7 + 0.083333... - 7.083333... = 7.083 12 12

E. Rounding Answers to problems, particularly when obtained with the help of a calculator, often need to be rounded to a desired number of decimal places. In most business problems involving money values, the rounding needs to be done to the nearest cent, that is, to two decimal places. While different methods of rounding are used, for most business purposes the following procedure is suitable. 1. If the first digit in the group of decimal digits that is to be dropped is the digit 5 or 6 or 7 or 8 or 9, the last digit retained is increased by 1. 2. If the first digit in the group of decimal digits that is to be dropped is the digit 0 or 1 or 2 or 3 or 4, the last digit is left unchanged.

CHAPTER

1:

REVIEW

OF

ARITHMETIC

Round each of the following to two decimal places. 7.38

-- drop the digit 4

(ii) 7.385 -

7.39

-round the digit 8 up to 9

(iii) 12.9448

12.94

(iv) 9.32838

9.33

(v) 24.8975

24.90

(i) 7.384

-discard 48 -— round the digit 2 up to 3 -round the digit 9 up to 0; this requires rounding 89 to 90

(vi) 1.996

2.00

- round the second digit 9 up to 0; this requires rounding 1.99 to 2.00

(vii) 3199.99833 -

—- 3200.00 this requires rounding 3199.99 to 3200.00

F. Complex fractions Complex fractions are mathematical expressions containing one or more fractions in the numerator or denominator or both. Certain formulas used in simple inter¬ est and simple discount calculations result in complex fractions. When you encounter such fractions, take care to use the order of operations properly.

420

_

420

420 1400

1600 Xg| 1600 X °-875 (ii) 500 1 + 0.16 X

225\ 365 j

= 500(1 + 0.098630) = 500(1.098630) = 549.32 (hi) 1000 1

= 0.3

0.18 X

multiply 0.16 by 225 and divide by 365 add inside bracket

288\

365/

= 1000(1 - 0.142027) = 1000(0.857973) = 857.97

824 (iv) 1 + 0.15 X

(v)

1755

73

824

824

1 + 0.03

1.03

800

365

=

1755

l - 0.21 X ^ ~ 1 ~ 0.120822

1755 0.879178

1996.18

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

k When using a calculator to compute business math formulas involving complicated denomina¬ tors, consider using the reciprocal key (j. T ]

te simplify calculations. For example,

or

to calculate part (v) of Example 1.2F above, the following calculator sequence would apply: 1755 1 - 0.21 X

.

—

210

365

0.21

±

JO D

-

—

+

1

V

... *

]

[

x

] 1755

The result is 1996.18.

EXERCISE 1.2 JO*

©

©

Reduce each of the following fractions to lowest terms. 24

28

210

L 36

2* 56

3‘ 360

360

360

144

5‘ 225

6‘ 315

7’ 360

360 8l 288

25 9 365

115

365 n. ?3

12 365

219

dot above a decimal number to show that it repeats.

L 8 11 5-t

©

10‘ 365

Convert each of the following fractions into decimal form. If appropriate

11

e

330 4 360

7 2'4 7 6. 9

5

5

3-3

4’6

13

19

7-T2

8> 15

Convert each of the following mixed numbers into decimal form. 1.3f

2. 3f

3. 8j

4. 16f

5. 33}

6.83}

7/. 7— 9

8-7li

Round each of the following to two decimal places. 1. 5.633

2. 17.449

3. 18.0046

4. 253.4856

5. 57.69875

6. 3.09475

7. 12.995

8. 39.999

CHAPTER

1:

REVIEW

OF

ARITHMETIC

Simplify each of the following. 54

1.

0.12 X 3. 620

4400 X

1 + 0.14 X

250 250

365

2358

8.

330 365

1 + 0.12 X

3460 1 - 0.18 X

292

6. 8500 ( 1 - 0.17 X

365

1 + 0.15 X

146 365

4. 375 ( 1 + 0.16 X

5. 2100 ( 1 - 0.135 X ^

7.

264

2.

225 365

2901

10.

270 365

146 365

1 - 0.165 X

73 365

1.3 Percent A. The meaning of percent Fractions are used to compare the quantity represented by the numerator with the quantity represented by the denominator. The easiest method of comparing the two quantities is to use fractions with denominator 100. The preferred form of writing such fractions is the percent form. Percent means “per hundred,” and the symbol % is used to show “parts of one hundred.” percent means hundredths % means j()0 Accordingly, any fraction involving “hundredths” may be written as follows: (i) as a common fraction (ii) as a decimal (iii) in percent form

13

100 0.13 13%

B. Changing percents to common fractions When speaking or writing, we often use percents in the percent form. However, when computing with percents, we use the corresponding common fraction or decimal fraction. To convert a percent into a common fraction, replace the symbol % by the symbol

Then reduce the resulting fraction to lowest terms.

PART

1:

MATHEMATICS

FUNDAMENTALS

24

(i) 24%

BUSINESS

APPLICATIONS

replace % by —

100 6_

reduce to lowest terms

25 (ii) 175%

AND

175 =

7 X 25

7_

100

4 X 25

4

~

6.25

(iii) 6.25%

100 625

multiply by 100 to change the

10 000 125

numerator to a whole number

= _25_ = __5_

reduce gradually or in one step

2000 ~~ 400 ~ 80 J_ 16 (iv) 0.8% =

0.8

8

1

100

1000

125

0.025

25

1

100

100 000

4000

(v) 0.025% J_ 4

(vi) \% -

replace % by —-— ^

100

J

100

1

1

invert and multiply

= 4X 100 400

(vii) ~^% -

1 3 = 8X 100

100

800

1 , 33t (viii) 33y% = —

- replace % by —

100

-

3

____ convert the mixed number 33^

100 1

=M 3

into a common fraction

i 100

“ 3

2 (ix) 216-% -

216f

650

13
0

29. 137y%

30. |%

31. 0.875%

32. 2j%

33. 33}%

34. 166j%

35. 16y%

36. 116~%

37. 183~%

38. 83y%

39. 133}%

40. 66y%

Change each ot the following percents into a common fraction in lowest terms. 3. 175%

4. 5%

5. 37y%

6. 75%

7. 4%

8. 225%

10. 125%

11. 40%

12. 87y%

13. 250%

14. 2%

15. 12y%

16. 60%

17. 2.25%

18. 0.5%

19. jo/o

20. 33y%

21.1%

22. 66y%

23. 6.25%

24. 0.25%

25. 16y%

26. 7.5%

27. 0.75%

28. j%

29. 0.1%

30. |%

33. 133y%

34. 183y%

9. 8%

-p 0s

2. 62y%

00 OJ

1. 25%

35. 1663%

32. 2.5% 36. 116y%

Express each of the following as a percent. 1. 3.5

2. 0.075

3. 0.005

4. 0.375

5. 0.025

6. 2

7. 0.125

8. 0.001

9. 0.225

10. 0.008

11. 1.45

12. 0.0225

13. 0.0025

14. 0.995

15. 0.09

16. 3

17-I

184

19. f

20 -L-

21-200

22-1

23-io

25

26. | Q 30 ~ 400

27.f

800 4

29. j

7

U‘

3>-f

200

24-l

28-U 32. j

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

1.4 Applications—Averages A. Basic problems When calculators are used, the number of decimal places used for intermediate values often determines the accuracy of the final answer. To avoid introducing rounding errors, keep intermediate values unrounded. The local cooperative received 36 3A tonnes of feed at $240 per tonne. Sales for the following five days were 3% tonnes, 4% tonnes, 7% tonnes, 516 tonnes, and 63/s tonnes What was the value of inventory at the end of Day 5? Total sales (in tonnes) = 3-| + 4~ + 7y + 5y + 6-| = 3.625 + 4.75 + 7.666667 + 5.5 + 6.375 = 27.916667 Inventory (in tonnes) = 36.75 — 27.916667 = 8.833333 Value of inventory = 8.833333 X 240 = 2119.999992 = $2120.00 Complete the following excerpt from an invoice. Amount

Unit Price

Quantity 72

$0,875

45

66j(

54

83j(

42

$1.33

32

$1,375

$

$

Total

$ 63.00

72 X $0,875 45 X 66j4 = 45 X $0,666

= 45 X $0.666667 z= $ 30.00 54 X 83y P

D. Zero exponent A zero exponent results when using the law of division of powers on powers with equal exponents. 35 4. 35 =

35-5

- 3°

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

The result may be interpreted as follows. 3^ _ 3X3X3X3X3 35 “ 3 X 3 X 3 X 3 X 3

Similarly, ,

a6 + a6 = a6

6 = a0

... a6 (a)(a)(a)(a)(a)(a) a0 — a0 = —r = , ,, w u w ,, , a6 (a){a)(a)(a)(a)(a)

.

and since

- 1

a0 = 1 In general, any number raised to the exponent zero is 1, except zero itself. The expression 0° has no meaning and is said to be undefined.

E, Negative exponents A negative exponent results when the exponent of the divisor is greater than the exponent of the dividend. 43 h- 45 = 43-5

= 4”2 The result may be interpreted as follows.

4X4X4

_1_

4X4X4X4X4

4X4

43

43 — 45 = — —-

45

42

Similarly, q5_(a) (a) (a) (a) (a)

and since

a8

(a)(a){a)(a){a)(a){a)(a) _

1 (a)(a)(a)

a

=

_ J_ a3

Formula 2.6

In general, a base raised to a negative exponent is equivalent to '1' divided by the same base raised to the corresponding positive exponent.

PART

(0 2 3

1:

MATHEMATICS

1

1

23

8

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

__ 1 j. (ii) (-3)~2 =

(-3)2 ~~ 9

n~4

1

(iii)

1 j

1_\4

1

256

= y X -j— = 256

256

(iv)

H)J

-125 27

125

•125

Note: Since

1 -27

(—5)3

3_\-3

AX3

33

27

in - (tr

- Formula 2.7

(v) (-4)° = 1 (vi) (1.05)-2 =

1

1

1.052

1.1025

(vii) (1 + i)->° =

= 0.9070295

1

(1 + i)10

(viii) (1 + iT1 = YT7 (ix) (1 + i)o - 1

EXERCISE 2.2 ^

O

Evaluate each of the following. 1. 34

2. I5

3.

‘■H)’

7. (0.5)2

* (f)‘ 9. (-4)°

>nr o

(-2)4

10. m°

11. 3~2

■nr

15.

l.or1

1. 25 X 23

2. (—4)3 X (-4)

3. 47 -s- 44

4. ( —3)9 -i- ( — 3)

5. (23)5

6. [ (

7. a4 X a'0

8. mn -i- m7

67 X 63 11.

4)3]6

10. (-1)3(-1)7(12.

8. (-0.1)3 12. ( —5)~3

Simplify.

9. 34 X 36 X 3

4. (-1)12

(xW) X7

16. (1.05)°

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

IH1)5 ih-c

14. i — i 16.

4-

|-

17. (1.02580)(1.02570)

18. 1.005240 4- 1.005150

19. [1.0420]4

20.

21. (1 + i)I00(l + z')100

22. (1 - r)2(l - r)2 (1 - r)2

23. [(1 + z')80]2

24. [(1 - r)4T

25. (ab)5

26. (2 xyY

27. (m3n)s

28.

29. 23 X 25 X 2“4

30. 52 4- 5-3 32.

_|X5H3

a3b2 N 4

x

1 + i

2.3 Fractional Exponents A. Radicals When the product of two or more equal factors is expressed in exponential form, one of the equal factors is called the root of the product. The exponent indicates the number of equal factors, that is, the power of the root. For example, 25 = 52

5 is the second power root (square root) of 25

8 = 23

2 is the third power root (cube root) of 8

81 = 34

3 is the fourth (power) root of 81

a5

-► a is the fifth root of a5

7"

~ 7 is the nth root of 7"

xn

► x is the nth root of x"

The operational symbol for finding the root of an expression is V

. This

symbol represents the positive root only. If the negative root is desired, a minus sign is placed in front of the symbol; that is, the negative root is represented by — V The power of a root is written at the upper left of the symbol, as in

V

or

. .

The indicated root is called a radical, the power indicated is called the index, and the number under the symbol is called the radicand. In V32, the index is 5 the radicand is 32 the radical is V32 When the square root is to be found, it is customary to omit the index 2. The symbol V

Js understood to mean the positive square root of the radicand. V49 means ^49 or 7

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

In special cases, like those shown in Example 2.3A, the radicand is an integral power of the root. The root can readily be found by expressing the radicand in exponential form; the index of the root and the exponent are the same. the radicand 64 is expressed in exponential form as a square

(i) V64 = Vs2 =

(ii) S/32 = =

8

one of the two equal factors 8 is the root

V25

express the radicand 32 as the fifth power

2

one of the five equal factors 2 is the root

of 2

(iii) D).125 = S/Off = 0.5 In most cases, however, the radicand cannot be easily rewritten in exponential form. The arithmetic determination of the numerical value of these roots is a labo¬ rious process. But computating the root is easily accomplished using electronic calculators equipped with a power function. To use the power function described in the Pointers and Pitfalls box on page

53, first rewrite the radical so it appears in exponential form. Then, as you will see in Formula 2.8 in the next section, av"= S/a. Use the opposite of this formula to rewrite a radical into exponential form: S/a = a'/n. For example, V32 can be written as 321/5. Now use the power function. The key sequence is

32

CD CD 1 m 5 CD CD The s°lution is 2-

(Instead of using brackets, you could first calculate 1 -s- 5, save the result in the calculator’s memory, then recall the result from memory after pressing (j^J.) The problems in Example 2.3B are intended to ensure that students are able to use the power function. They should be done using an electronic calculator.

(i) Vl425 = 37.7492

--Check 37.74 922 = 1425

(ii) Vl2 960 = 6.6454 --Check 6.6454s = 12 960 (iii) Do 000 = 2.0268 --Check 2.026815 = 40 010 (due to rounding) (iv) D 048 576 = 2 --Check 220 = 1 048 576 --Check 0.0722 = 0.005184

(v) VO.005184 = 0.072

(vi) V0.038468 = 0.6279 --Check 0.62797 = 0.03848 (due to rounding) --Check 1.01545 = 1.954213

(vii) D.954213 = 1.015 (viii) S/0.022528= 0.9 (ix)

VYb =

--Check 0.936 = 0.022528

V64 = 8

(x) D"6 = N/l5 625 = 25

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

Note: Many of the solutions for problems such as those above involve repeatiP ing and continuous numbers. In order to standardize, the number of decimal places shown in solutions will be set at six or less. The financial calculator can be set to show a maximum of six decimal places by following the steps: DEC

Format

6 (Enter

( 2nd ] iQuitJ

It is important to note that, even though the calculator display shows six or less decimal places, the complete, non-rounded quantity is used in continuous calcu¬ lations on the calculator.

B. Fractional exponents Radicals may be written in exponential form and fractional exponents may be represented in radical form according to the following definitions. (a) The exponent is a positive fraction with numerator 1. -Formula 2.8

42

Vi

=

=

2

27^ = V27 = W = 3 625^ = V625 = W = 5

(b) The exponent is a negative fraction with numerator a n

8

_i

= —- = 1

Formula 2.9 1

1

3= —

1.

83 ~ V8 ~ V23 ~~ 2 1

1

1

1

2437

V243

V^

3

243's = -

(c) The exponent is a positive or negative fraction with numerator other than 1.

=

a"

-Formula 2.10

=

1

1 m

16i = 27i

VW

Formula 2.11

= (^Tb)3 = (^)3 = (2)3 = 8

= V27^ = (V27)4

=

(^)4 =

36 2 = (V36)3

(V65)3

63

(3)4

=

81

216

For calculators, convert fractional exponents into decimals and compute the answer using the power function.

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

(i) 364 = 36L5 = 216 (ii) 3^ = 3'-25 = 3.948222 (iii) ^12 = 124 = 120-2 = 1.6437518 (iv) V3255 = 3258 = 3250-625 = 37.147287 (v) ^ 1.075 = 1.0754 = \ _075o.i666667 = 1.0121264

EXERCISE 2.3

^

O

Use an electronic calculator equipped with a power function to compute each of the following, correct to four decimals.

1. V5184

2. Y/205.9225

3. V2187

4. V I. 1046221

5. V4.3184

6. V'0.00001526

7. ^1.0825

8. VI. 15

12/-

Compute each of the following. 1. 30254

2. 240U

3. 525.218754

4. 21.6^

5. V1.1257

6. ^1.095

7. 4~3

8. 1.06 12

-p

9. 11.

1.0360 - 1 0.03

(1

ft 10.

12. (1 + 0.045)-12

+ 0.08)10

13. 26.50 (1 + 0.043)

14. 350.00 (1 + 0.05)

15. 133.00

16. 270.00

1 - 1.05 36 0.05

(1 + 0.043)30 - 1 0.043 (1 + 0.05) 20 0.05

_

1

1 - (1 + 0.056)-'2 0.056

r i1 —- (i(1 + 0.035)-8 0.035

17. 5000.00 (1 + 0.0275)-20 + 137.50

18. 1000.00 (1 + 0.03)-16 + 300.00

1 - (1 + 0.0275)"20 0.0275 1 - (1 + 0.03) -16 0.03

27 CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

2.4 Logarithms—Basic Aspects A. The concept of logarithm In Section 2.2 and Section 2.3, the exponential form of writing numbers was discussed. 64 = 26

the number 64 is represented as a power of 2

243 = 35 10 0 00 = 104 l

the number 243 is represented as a power of 3 - the number 10 000 is represented as a power of 10

5 = 1253 —- the number 5 is represented as a power of 125 0.001 = 10“3 —- the number 0.001 is represented as a power of 10 In general, when a number is represented as a base raised to an exponent, the exponent is called a logarithm. A logarithm is defined as the exponent to which a base must be raised to produce a given number. Accordingly, 64 = 26

6 is the logarithm of 64 to the base 2, written 6 = log264

243 = 35 10 000 = 104 , 5 = 1253

5 is the logarithm of 243 to the base 3, written 5 = log,243 - 4 is the logarithm of 10 000 to the base 10, written -

0.001 = 10“3

4= logi010 000 Vi is the logarithm of 5 to the base 125, written

Vi

= log]255

—3 is the logarithm of 0.001 to the base 10, written -3 = log100.001

In general, if N = by (exponential form)

(

then y = log(j N logarithmic form).

Write each of the following numbers in exponential form and in logarithmic form using the base indicated. (i) 32 base 2 (iii) 256 base 4

(ii) 81 base 3 (iv) 100 000 base 10 (vi) 3 base 27

(v) 6 base 36 (vii) 0.0001 base 10

(viii) tt base 2

Exponential Form

Logarithmic Form

(i) Since 32 = 2X2X2X2X2 32 = 25

5 = log,32

(ii) Since 81 = 3X3X3X3 81 = 34

4 = log381

(iii) Since 256 = 4X4X4X4 256 = 44

4 = log4256

(iv) Since 100 000 = 10 X 10 X 10 X 10 X 10 100 000 = 105

5 = log1Q100 000

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS TjJ

(v) Since 6 = \/36

6

=

36^

I = l0§366

(vi) Since 3 = V27 3 = 273 (vii) Since 0.0001 =

0.0001

1 10 000

1 104 -4 = log1Q0.0001

= 1(T4

(viii) Since — =

=

2

3 = log2-g

~

B. Common logarithms While the base b may be any positive number other than 1, only the numbers 10 and e are used in practice. Logarithms with base 10 are called common logarithms. Obtained from the exponential function x = 10’, the notation used to represent common logarithms is y — log x. (The base 10 is understood and so is not written.) By definition then, the common logarithm of a number is the exponent to which the base 10 must be raised to give that number. log 1000 = 3

since 1000 = 103

log 1 000 000 = 6

since 1 000 000 = 106

log 0.01 = —2

since 0.01 = 10 3

log 0.0001 = —4

since 0.0001 = 10“4

log 1 = 0

since 1 - 10°

Historically, common logarithms were used for numerical calculations that were required in problems involving compound interest. However, with the avail¬ ability of electronic calculators equipped with a power function, the need for common logarithms as a computational tool has disappeared. Accordingly, this text gives no further consideration to common logarithms.

C. Natural logarithms The most common exponential function is y = ex where e = limit (l + ^)" = 2.718282 approximately. n—'

The logarithmic form of this function is x = logfy but is always written as x = In y and called the natural logarithm.

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

Electronic calculators equipped with the universal power function are generally equipped as well with the ex function and the In: x function (natural logarithm function). This latter function eliminates any need for common logarithms and can be used when solving equations for which the unknown quantity is an expo¬ nent. As you will see in Chapters 9 through 16, the natural logarithm function would be used to solve for n, the number of compounding periods, in mathemat¬ ics of finance applications using the algebraic method.

In some calculator models, such as the Texas Instruments BA II PLUS, the natural logarithm function is found directly by entering the number and pressing the ; LN j key. In other calculators, the natural logarithm function is found indirectly by entering the number and pressing combinations of keys. For example: For the Hewlett Packard 10B calculator, enter the number, then press the keys I * ) For the Sharp EL-733A, enter the number, then press 2nd F j

i/x

I 2

.

j .

For the Texas Instruments BA-35 Solar, enter the number, then press

[2nd

Use an electronic calculator equipped with the natural logarithm key [

LN

to

determine the value of each of the following. (i) In 2 (iv) In 1

(ii) In 3000

(iii) In 0.5

(v) In 0.0125 •

(vi) In 2.718282

(i) To evaluate In 2 (using the Texas Instruments BA II PLUS), 1. Key in 2._ 2. Press LN

.

3. Read the answer in the display. In 2 = 0.693147 (ii) To evaluate In 3000 (using the Sharp EL-733A), key in 3000, press 2ndF press [ 1 lx j , and read the answer in the display. In 3000 = 8.006368 (iii) In 0.5 = -0.693147 (iv) In 1 = 0 (v) In 0.0125 = -4.382027 (vi) In 2.718282 = 1 Note:

1. The natural logarithm of 1 is zero. 2. The natural logarithm of a number greater than 1 is positive, e.g., In 2 = 0.693147. 3. The natural logarithm of a number less than 1 is negative, e.g., In 0.5 = -0.693147.

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS^

D. Useful relationships The following relationships are helpful when using natural logarithms: 1. The logarithm of a product of two or more positive numbers is the sum of the logarithms of the factors. In (ab) = In a + In b

Formula 2.12A

In (abc) = In a + In b + In c

Formula 2.12B

Notice that Formula 2.12B is an extension of Formula 2.12A. 2. The logarithm of the quotient of two positive numbers is equal to the loga¬ rithm of the dividend (numerator) minus the logarithm of the divisor (denominator). In f

) = In a — In b

Formula 2.13

3. The logarithm of a power of a positive number is the exponent of the power multiplied by the logarithm of the number. In (ak) = k(\n a)

-Formula 2.14

4. (i) In e = 1

since e = e1

(ii) In 1 = 0

since 1 = e°

Use an electronic calculator equipped with the natural logarithm function to evaluate each of the following. (i) ln[3(15)(36)] 5000 \ (ii) In

1.045 /

(iii) ln[1500(1.056)] (iv) In [ 5000( 1.045“1) ]

, r / 4000 (y) ^ [\X07TT (vi) ln[10 000(1.0125'17)] (vii) In [1.00 e7) (viii) ln[2.00 £T0'6] (ix) In 600

1 1.046 0.04

1 - 1,0625~12 (x) In

0.0625

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

(i) In[3( 15)(36)] = In 3 + In 15 + In 36 = 1.098612 + 2.708050 + 3.583519 =>■ 7.390181 Note: You can verify the answer by first simplifying. In 3(15)(36) = In 1620 = 7.390181 / 5000 \ (ii) In

= In 5000 - In 1.045

V 1.045 )

= 8.517193 - 0.044017 = 8.473176 (iii) ln[5000(1.045-1)] = In 5000 + In 1.045-1 = In 5000 - 1 (In 1.045) = 8.517193 - 1(0.044017) = 8.473176 (iv) ln[1500(1.056)] = In 1500 + In 1.056 = In 1500 + 6(ln 1.05) = 7.313220 + 6(0.048790) = 7.313220 + 0.292741 = 7.605961 (v) In

( 4000 \

= In 4000 - In 1.0712

Vl-0712/

= 8.294050 - 12(0.067659) = 8.294050 - 0.811904 = 7.482146

(vi) In[10 000(1.0125“17)] = In 10 000 - 17(ln 1.0125) = 9.210340 - 17(0.0124225) = 9.210340 - 0.211183 = 8.999157 (vii) ln[1.00e7] = In 1.00 + In e7 = In 1.00 + 7(ln e) = 0 + 7(1) = 7 (viii) ln[2.00e'0'6] = In 2.00 + In e~0-6 = In 2.00 - 0.6(ln e) = 0.693147 - 0.6 = 0.093147 (ix) In 600

, , / 1.046 - 1 /1.04s - IV = In 600 + In —^r:— V 0.04 j V 0.04 = In 600 + ln(1.046 - 1) - In 0.04 = In 600 + In (1.265319 - 1) - In 0.04 = In 600 + In 0.265319 - In 0.04 = 6.396930 - 1.326822 - (-3.218876) = 6.396930 - 1.326822 + 3.218876 = 8.288984

PART

(x) In

1:

MATHEMATICS

7 1 - 1.0625"12 \

A

0.0625

)

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS 'l

In (1 - 1.0625-12) - In 0.0625 In (1 - 0.483117) - In 0.0625 In 0.516883 - In 0.0625 -0.659939 - (-2.772589) -0.659939 + 2.772589 2.112650

EXERCISE 2.4 Express each of the following in logarithmic form. 1. 29 = 512 2. 37 = 2187

4. 10“5 = 0.00001 5. e2> = 18 6. e~ix = 12 Write each of the following in exponential form. 1. log232 = 5

2- 'os.'il =

“4

3. Ioglp10 = 1 4. In e2 = 2 Use an electronic calculator equipped with a natural logarithm function to evaluate each of the following. 1. In 2 2. In 200 3. In 0.105 4. In [300( 1.1015) ] 5. In 6.

/ 2000 \

V

1.09 V

In 850

/i.or120y V o.oi /.

S’ CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

What Is the Real Cost of Purchasing a House? When you purchase a new house, have you considered the cost of taxes and levies on that new house? What percent of the total cost to you is the cost of these ancillary fees? In a survey done in 2001 by the Canada Mortgage and Housing Corporation, the portion paid in these charges depends on where you are purchasing the house.

Municipality

Price

Levies, Fees, Charges, and Taxes

Transaction Costs

Total($)

Total as Percent

Charlottetown

96 800

9 540

3 928

13 468

13.9

Halifax

90 000

9 447

5 110

14 557

16.2

123 200

12 921

4 555

17 476

14.2

5 170

28 588

24.6 20.9

Moncton Saint-Hubert

116 400

23 418

Mississauga

217 900

36 117

9 369

45 486

London

129 900

18 999

6 276

25 275

19.5

Winnipeg

117 000

12 123

4 618

16 741

14.3

Regina

119 000

16 463

4 551

21 014

17.7

Calgary

121 000

10 616

4 387

15 003

12.4

Edmonton

129 500

13 341

4 595

17 936

13.8

Surrey

269 900

37 348

11 321

48 669

18.0

Burnaby

479 200

43 193

21 906

65 099

13.6

Kelowna

154 600

22 121

6 615

28 736

18.6

Whitehorse

145 000

10 931

5 256

16 187

11.2

Yellowknife

165 000

8 733

5 604

14 337

8.7

QUESTIONS 1. What proportion of transaction costs would you be paying in Burnaby, compared to Moncton? 2. What proportion of levies, fees, and taxes would you be paying in Mississauga, compared to Kelowna? 3. Comparing the prices, levies, and transaction costs for Regina and London, in which city do you pay the most in fees? 4. What is the price of a new home in your city? How do the levies and transaction costs compare to those across the country? Source: Adapted from Canada Mortgage and Housing Corporation, "Levies, Fees, Charges, Taxes and Transaction Costs on New Housing," Research Highlights, November 2001, Canada Mortgage and Housing Corporation site, www.cmhcschl.gc.ca/publications/en/rh-pr/socio/socio94-e.pdf, accessed August 11, 2003. All rights reserved. Reproduced with the consent of CMHC. All other uses and reproductions of this material are expressly prohibited.

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS'*;

2.5 Solving Basic Equations A. Basic terms and concepts 1. An equation is a statement of equality between two algebraic expressions. 7x — 35 3fl — 4 = 11 — 2a 5(2k - 4) = -3(k + 2) 2. If an equation contains only one variable and the variable occurs with power 1 only, the equation is said to be a linear or first-degree equation in one unknown. The three equations listed above are linear equations in one unknown. 3. The two expressions that are equated are called the sides or members of an equation. Every equation has a left side (left member) and a right side (right member). In the equation 3a — 4 = 11 — 2a, 3a — 4 is the left side (left member) and 11 — 2a is the right side (right member). 4. The process of finding a replacement value (number) for the variable, which when substituted into the equation makes the two members of the equation equal, is called solving the equation. The replacement value that makes the two members equal is called a solution or root of an equation. A linear or firstdegree equation has only one root and the root, when substituted into the equation, is said to satisfy the equation. The root (solution) of the equation 3a — 4 = 11 — 2a is 3 because when 3 is substituted for a the left side 3a — 4 = 3(3) — 4 = 9 — 4 = 5 and the right side 11 — 2a =11 — 2(3) = 11—6 = 5. Thus, for a = 3, Left Side = Right Side and 3 satisfies the equation. 5. Equations that have the same root are called equivalent equations. I hus, 6x + 5 = 4x + 17, 6x = 4x + 12, 2x = 12, and x = 6 are equivalent equations because the root of all four equations is 6; that is, when 6 is substituted for x, each of the equations is satisfied. Equivalent equations are useful in solving equations. They may be obtained (a) by multiplying or dividing both sides of the equation by a number other than zero; and (b) by adding or subtracting the same number on both sides of the equation. 6. When solving an equation, the basic aims in choosing the operations that will generate useful equivalent equations are to (a) isolate the terms containing the variable on one side of the equation (this is achieved by addition or subtraction); and (b) make the numerical coefficient of the single term containing the variable equal to +1 (this is achieved by multiplication or division).

CHAPTER

B.

2:

REVIEW

OF

BASIC

ALGEBRA

Solving equations using addition If the same number is added to each side of an equation, the resulting equation is equivalent to the original equation. x

5 — 4

add 3

x~5+3=4+3

or add 5

x~5+5=4+5

original equation equivalent equations

Addition is used to isolate the term or terms containing the variable when terms that have a negative coefficient appear in the equation. (i)

x - 6 = 4

add 6 to each side of the equation

x — 6 + 6 = 4 + 6

to eliminate the term -6 on the left

x — 10 (ii)

(iii)

—2x = —3 — 3x —2x + 3x = —3 — 3x + 3x x = —3

side of the equation

— add 3xto each side to eliminate the term -3x on the right side

—x — 5 = 8 — 2x —x — 5 + 5 = 8 — 2x+ 5--on the left side ~x = 13 — 2x —x + 2x = 13 —2x + 2x x = 13

— add 2x to eliminate the term -2x

C. Solving equations using subtraction If the same number is subtracted from each side of an equation, the resulting equa¬ tion is equivalent to the original equation. x + 8 = 9 subtract 4

x+8 — 4 = 9 — 4

or subtract 8

x+8 — 8 = 9 — 8

original equation equivalent equations

Subtraction is used to isolate the term or terms containing the variable when terms having a positive numerical coefficient appear in the equation. (i)

x + 10 = 6 x + 10 - 10 = 6 - 10 x = —4

(ii)

subtract 10 from each side of the equation

7x = 9 + 6x 7x — 6x = 9 + 6x — 6x x = 9

subtract 6x from each side to eliminate the term 6x on the right side

PART

(iii)

1:

MATHEMATICS

FUNDAMENTALS

6x + 4 = 5x — 3

AND

BUSINESS

APPLICATIONS

subtract 4 from each side to eliminate

6x + 4 — 4 = 5x—3 — 4

the term 4 on the left side

6x = 5x — 7 6x — 5x = 5x — 7 — 5x x = —7 the right side

D. Solving equations using multiplication If each side of an equation is multiplied by the same non-zero number, the resulting equation is equivalent to the original equation. — 3x = 6 multiply by 2 or multiply by — 1

-original equation

6x — 12 3x = — 6

-equivalent equations

Multiplication is used in solving equations containing common fractions to eliminate the denominator or denominators.

(i)

original equation multiply each side by 2 to eliminate the denominator solution

(ii)

~\x = 2

4(~ix) =4(2) —

original equation multiply each side by 4 to eliminate the denominator

-lx = 8 (-l)(-x) = (—1)(8) x = —8 (iii)

multiply by (-1) to make the coefficient of the term in x positive

-jx = -2 (-7)(—yx) = (-7)(-2) x — 14

multiply by (-7) to eliminate denominator and to make the coefficient of x equal to +1

CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

E. Solving equations using division If each side of an equation is divided by the same non-zero number, the resulting equation is equivalent to the original equation. 15x = 45 5x = 15

divide by 3 or divide by 5

3x = 9

or divide by 15

original equation equivalent equations

x — 3

Division is used in solving equations when the numerical coefficient of the single term containing the variable is an integer or a decimal fraction. (i)

12x = 36 12x 12

36 12

x = 3 (h)

--original equation divide each side by the numerical coefficient 12 -solution

— 7x = 42 — 7x

42

divide each side by the numerical

-7

-7

coefficient -7

x = —6 (hi)

0.2x = 3 0.2x

3

0.2 ~ 0.2 x — 15 (iv)

x — 0.3x = 14 0.7x = 14 0.7x 14 0.7 ~ 0.7 x = 20

F. Using two or more operations to solve equations When more than one operation is needed to solve an equation, the operations are usually applied as follows. (a) First, use addition and subtraction to isolate the terms containing the vari¬ able on one side of the equation (usually the left side).

(b)

Second, after combining like terms, use multiplication and division to make the coefficient of the term containing the variable equal to +1.

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

(i) multiply by 5 to eliminate the denominator -3x = 60 — 3x

60

-3

-3

divide by -3

x = —20 (ii)

7x — 5 = 15 + 3x 7x - 5 + 5 = 15 + 3x + 5 7x = 20 + 3x 7x — 3x = 20 + 3x — 3x

add 5 combine like terms subtract 3x

4x = 20 x = 5 (iii)

divide by 4

3x + 9 — 7x = 24 — x — 3 9 — 4x = 21 —x

combine like terms

9 — 4x — 9 = 21— x — 9 — 4x = 12 — x —4x + x = 12 — x + x —3x = 12 x = —4

G. Checking equations To check the solution to an equation, substitute the solution into each side of the equation and determine the value of each side. 3 (i) For —yx = 12, the solution shown is x = —20. It

Right Side = 12 Since the Left Side = Right Side, -20 is the solution to the equation. (ii) For

lx -

5 = 15 + 3x, the solution shown is x = 5.

Check LS = 7x — 5 = 7(5) - 5 = 35 - 5 = 30 RS = 15 +3x— 15 + 3(5) = 15 + 15 = 30 Since the LS = RS, 5 is the solution. (iii) For

3x

+ 9 - 7x = 24 - x - 3, the solution shown is

Check LS = 3(—4) + 9 - 7( — 4) = -12 + 9 + 28 = 25 RS = 24 - (-4) - 3 = 24 + 4- 3 = 25 Since LS = RS, —4 is the solution.

x

= -4.

E' CHAPTER

2:

REVIEW

OF

BASIC

ALGEBRA

EXERCISE 2.5 Solve each of the following equations. 1. 15x = 45

2. -7x = 35

3. 0.9x = 72

1

4. 0.02x = 13 3 7-^ =-21

6. -jx = 7 8. — ~x = — 32

9.x - 3 = -7

10. — 2x = 7 — 3x

11. x + 6 = -2

12. 3x = 9 + 2x

13. 4 — x = 9 — 2x

14. 2x + 7 = x — 5

15. x + 0.6x = 32

16. x — 0.3x = 210

17. x - 0.04x = 192

18. x + 0.07x = 64.20

Solve each of the following equations and check your solution. 1. 3x + 5 = lx — 11

2. 5 — 4x = —4 — x

3. 2 — 3x — 9 = 2x — 7 + 3x

4. 4x — 8 — 9x = 10 + 2x — 4

2.6 Equation Solving Involving Algebraic Simplification A. Solving linear equations involving the product of integral constants and binomials To solve this type of equation, multiply first, then simplify. (i) 3(2x - 5) = -5(7 - 2x) 6x — 15 = —35 + lOx 6x — lOx = — 35 + 15 —4x = -20 x = 5 Check LS = 3[2(5) - 5] = 3(10 - 5) = 3(5) = 15 . RS = —5[7 - 2(5)] = -5(7 - 10) = -5(-3) = 15 Since LS = RS, 5 is the solution. (ii) x - 4(3x - 7) = 3(9 - 5x) - (x - 11) x — 12x + 28 = 27 — 15x — x + 11 — llx + 28 = 38 — 16x — 1 lx + 16x = 38 — 28 5x = 10

- isolate the terms in x

x = 2 Check LS - 2 - 4[3(2) - 7]

RS = 3[9 - 5(2)]

= 2 - 4(6 - 7) = 2 - 4( —1) = 2 + 4 =

6

Since LS = RS, 2 is the solution.

-

(2

-

= 3(9 - 10) - (-9) = 3(— 1) + 9 = -3 + 9 =

6

11)

PART

1:

MATHEMATICS

FUNDAMENTALS

AND

BUSINESS

APPLICATIONS

There is a foolproof method of determining the lowest common denominator (LCD) for a given group of fractions'. STEP 1

Divide the given denominators by integers of 2 or greater until they are all reduced to 1. Make sure the integer divides into at least one of the denominators evenly. all of the resultant integers (divisors) to find the LCD.

STEP 2

4 7 5 5' 9' 6 3 2 13 11 4' 3' 22' 15 Solution:

5

(i)

9

6 -

denominators 2 divides into 6 evenly

-5- 2 =

5

9

3 -

3 =

5

3 1

1 1 -

- 3 divides into 3 evenly

1

1

- 5 divides into 5 evenly

44-

3 =

4-

5 =

5 1

3 divides into 9 and 3 evenly

LCD = 2 x 3x3 x 5 = 90 (ii)

4

3

22

15

11 11

15 5 1

4-

2 =

2

3

4-

2 =

1

4-

3 =

1

3 1

4- 5 = 4- 11 =

1

1

11 11

1

1

1

15 1

LCD = 2 x 2 x 3 x 5 x 11 = 660

B. Solving linear equations containing common fractions The best approach when solving equations containing common fractions is to first create an equivalent equation without common fractions. Multiply each term of the equation by the lowest common denominator (LCD) of the fractions.

(i)

4

3

5X

4

12

+

11

LCD = 60

15

11 60

-60(4)

60( 12) + 60'v15X

12(4x) - 15(3) = 5(7) + 4(1 lx) 48x — 45 = 35 + 44x 48x — 44x = 35 + 45 4x = 80 x = 20

multiply each term by 60 reduce to eliminate the fractions

\y CHAPTER 2: REVIEW OF BASIC ALGEBRA

Check 4 LS = —(20)

3

= 16 - 0.75 = 15.25

7 11 RS = — + —(20) = 0.583333 + 14.666667 = 15.25 Since LS = RS, 20 is the solution. 5

(ii)

5x 6

~

24

( _5x

V 8

24(3) = 24(|

3(5x) - 72 15x - 72 — 5x x

LCD = 24

~

+ 24lf

6(3) + 4(5x) 18 + 20x 90 -18

Check LS = |~( —18) - 3 = -11.25 - 3 = -14.25 3 5 RS = - + —(—18) = 0.75 - 15.00 = -14.25 Since LS = RS, the solution is —18.

C. Solving linear equations involving fractional constants and

multiplication

When solving this type of equation, the best approach is first to eliminate the fractions and then to expand. 3 2 y(x- 2) -y(2x— 1) = 5

(i)

6(|)(x - 2) - 6(|)(2x - 1) = 6(5) 3(3)(x - 2) - 2(2)(2x - 1) 9(x - 2) - 4(2x — 1) 9x - 18 - 8x + 4 x - 14 x

= 30 =30 = 30 = 30 = 44

LCD = 6 multiply each side by 6 reduce to eliminate fractions

Check LS = -|(44 - 2) - j(2 X 44 - 1) = y(42) - j(87) = 63 - 58 = 5 RS = 5 Since LS = RS, 44 is the solution.

PART

1:

MATHEMATICS

(ii)

FUNDAMENTALS

AND

-J-(4x - 1) + -|(4x - 3) = —

BUSINESS

APPLICATIONS

LCD = 40

4o(^)(4x - 1) + 4o(|)(4x - 3) = 40 8( —3)(4x - 1) + 5(5)(4x - 3) = 4(—11) —24(4x - 1) + 25(4x - 3) = -44 — 96x + 24 + lOOx - 75 = -44 4x - 51 = -44

4x = 7 1_ x = 4 Check

41-1-1

LS = -

+t 4

-f(7 - I) +f(7 18 5

36

- 3

3)

25

U_

10 + 10

10

U_

RS = -TTT

10

Since LS = RS, the solution is

4'

When using a lowest common denominator (LCD) to eliminate fractions from an equation involving both fractional constants and multiplication, the LCD must be multiplied by each quantity (on both sides of the equation) that is preceded by a + or - sign outside of brackets. (Some students multiply the LCD by each quantity inside and outside the brackets, which leads to an answer much greater than the correct answer.)

D. Formula rearrangement Formula rearrangement, also known as formula manipulation, is the process of rearranging the terms of an equation. To solve for a particular variable, we want the variable to stand alone on the left side of the equation. If it does not already do so, then we have to rearrange the terms. Developing your skill in rearranging for¬ mulas is very important as it saves a lot of time in memorization. You need only memorize one form of any particular formula. For example, consider the formula I = Prt. Once we have memorized this formula, there is no need to memorize equivalent forms as long as we are skilled in formula rearrangement. Thus, for example, we need not “memorize” the form P = Vrt. The key to formula manipulation is the concept of undoing operations. Addition and subtraction are inverse operations (i.e., they undo each other).



-

o,

[

+1

STEP 2

f FV~] Result: [7640.159341] STEP 3

[ - ] 6000 ( = ] [1640.159341

±1

PV )

(CPT) ( FV ) Result: [1895.409139 j

EXERCISE 9.2 If you choose, you can use Excel’s Future Value (FV) function to calculate the

Excel

future value of an investment subject to compound interest. Refer to FV on the Spreadsheet Template Disk to learn how to use this Excel function.

ND

INTEREST —FUTURE

VALUE

AND

PRESENT

VALUE

Find the future value lor each of the investments in the table below. Nominal

Frequency of

Principal

Rate

Conversion

Time

1.

$ 400.00

7.5%

annually

2.

1000.00

3.5%

semi-annually

12 years

3.

1250.00

6.5%

quarterly

9 years

4.

500.00

12%

monthly

3 years

5.

1700.00

8%

quarterly

14.75 years

6.

840.00

5.5%

semi-annually

8.5 years

7.

2500.00

8%

8.

150.00

10.8%

9.

480.00

10.

1400.00

11.

2500.00

7%

annually

7 years, 6 months

12.

400.00

9%

quarterly

3 years, 8 months

13.

1300.00

5%

semi-annually

9 years, 3 months

14.

4500.00

3.5%

monthly

7.5 months

8 years

monthly

12.25 years

quarterly

27 months

9.4%

semi-annually

42 months

4.8%

monthly

18.75 years

Answer each of the following questions. 1. What is the maturity value of a five-year term deposit of $5000.00 at 6.5% compounded semi-annually? How much interest did the deposit earn? 2. How much will a registered retirement savings deposit of $1500.00 be worth in 15 years at 8% compounded quarterly? How much of the amount is interest? 3. You made a registered retirement savings plan deposit of $1000.00 on December 1, 2000 at a fixed rate of 5.5% compounded monthly. If you with¬ draw the deposit on August 1,2007, how much will you receive? 4. Roy’s parents made a trust deposit of $500.00 on October 31, 1986 to be with¬ drawn on Roy’s eighteenth birthday on July 31,2004. To what will the deposit amount on that date at 7% compounded quarterly? 5. What is the accumulated value of $100.00 invested for eight years at 9% p.a. compounded (a) annually?

(b) semi-annually?

(c) quarterly?

(d) monthly?

6. To what future value will a principal of $500.00 amount in five years at 7.5% p.a. compounded (a) annually?

(b) semi-annually?

(c) quarterly?

(d) monthly?

7. What is the future value of and the amount of compound interest for $100.00 invested at 8% compounded quarterly for (a) 5 years?

(b) 10 years?

(c) 20 years?

8. Find the future value of and the compound interest on $500.00 invested at 4.5% compounded monthly for (a) 3.5 years;

(b) 6 years;

(c) 11.5 years.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT TO

9. A demand loan for $5000.00 with interest at 9.75% compounded semi¬ annually is repaid after five years, ten months. What is the amount of interest paid?

10. Suppose $4000.00 is invested for four years, eight months at 8.5% compound¬ ed annually. What is the compounded amount? 11. Determine the maturity value of a $600.00 promissory note dated August 1, 2000 and due on June 1, 2005, if interest is 5% p.a. compounded semi-annually. 12. Find the maturity value of a promissory note for $3200.00 dated March 31, 1998 and due on August 31, 2004, if interest is 7% compounded quarterly. 13. A debt of $8000.00 is payable in seven years and five months. Determine the accumulated value of the debt at 10.8% p.a. compounded annually. 14. A $6000.00 investment matures in three years, eleven months. Find the matu¬ rity value if interest is 9% p.a. compounded quarterly. 15. The Canadian Consumer Price Index was approximately 98.5 (base year 1992) at the beginning of 1991. If inflation continued at an average annual rate of 3%, what would the index have been at the beginning of 2004? 16. Peel Credit Union expects an average annual growth rate of 20% for the next five years. If the assets of the credit union currently amount to $2.5 million, what will the forecasted assets be in five years? 17. A local bank offers $5000.00 five-year certificates at 6.75% compounded semi¬ annually. Your credit union makes the same type of deposit available at 6.5% compounded monthly.

(a) Which investment gives more interest over the five years? (b) What is the difference in the amount of interest? 18. The Continental Bank advertises capital savings at 7.25% compounded semi¬ annually while TD Canada Trust offers premium savings at 7% compounded monthly. Suppose you have $1000.00 to invest for two years.

(a) Which deposit will earn more interest? (b) What is the difference in the amount of interest? Answer each of the following questions. 1. A deposit of $2000.00 earns interest at 7% p.a. compounded quarterly. After two and a half years, the interest rate is changed to 6.75% compounded monthly. How much is the account worth after six years? 2. An investment of $2500.00 earns interest at 4.5% p.a. compounded monthly for three years. At that time the interest rate is changed to 5% compounded quarterly. How much will the accumulated value be one-and-a-half years after the change? 3. A debt of $800.00 accumulates interest at 10% compounded semi-annually from February 1, 2002 to August 1, 2004, and 11% compounded quarterly thereafter. Determine the accumulated value of the debt on November 1,2007.

CHAPTER

9:

COMPOUND

I N T E R E ST— F U T U R E

VALUE

AND

PRESENT

VALUE

4. Accurruilate $1300.00 at 8.5% p.a. compounded monthly from March 1, 2002 to July 1, 2004, and thereafter at 8% p.a. compounded quarterly. What is the amount on April 1, 2007? 5. Pat opened an RRSP deposit account on December 1, 2002 with a deposit of $1000.00. He added $1000.00 on July 1, 2003 and $1000.00 on November 1, 2004. How much is in his account on January 1, 2006, if the deposit earns 6% p.a. compounded monthly?

6. Terri started an RRSP on March 1, 2000 with a deposit of $2000.00. She added $1800.00 on December 1, 2002 and $1700.00 on September 1, 2004. What is the accumulated value of her account on December 1, 2007, if interest is 7.5% compounded quarterly? 7. A debt of $4000.00 is repaid by payments of $1500.00 in nine months, $2000.00 in 18 months, and a final payment in 27 months. If interest was 10% compounded quarterly, what was the amount of the final payment? (f§)

8. Sheridan Service has a line of credit loan with the bank. The initial loan bal¬ ance was $6000.00. Payments of $2000.00 and $3000.00 were made after four months and nine months respectively. At the end of one year, Sheridan Service borrowed an additional $4000.00. Six months later, the line of credit loan was converted into a collateral mortgage loan. What was the amount of the mort¬ gage if the line of credit interest was 9% compounded monthly?

@

9. A demand loan of $3000.00 is repaid by payments of $1500.00 after two years, $1500.00 after four years, and a final payment after seven years. Interest is 9% compounded quarterly for the first year, 10% compounded semi-annually for the next three years, and 10% compounded monthly thereafter. What is the size of the final payment?

(§§) 10. A variable rate demand loan showed an initial balance of $12 000.00, payments of $5000.00 after eighteen months, $4000.00 after thirty months, and a final payment after five years. Interest was 11% compounded semi-annually for the first two years and 12% compounded monthly for the remaining time. What was the size of the final payment?

9.3 Present Value and Compound Discount A. The present value concept and related terms Find the principal that will amount in six years to $17 715.61 at 10% p.a. com¬ pounded annually. The problem may be graphically represented as shown in Figure 9.5.

PART

FIGURE 9.5

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Graphical Representation of Data End of year Now

1234

5

6 Known

Find the

future

principal

value

This problem is the inverse of the problem used to illustrate the meaning of com¬ pound interest. Instead of knowing the value of the principal and finding its future value, we know that the future value is $17 715.61. What we want to deter¬ mine is the value of the principal. To solve the problem, we use the future value formula FV = PV(1 + i)n and substitute the known values. FV= 17 715.61; I/Y = 10; P/Y = 1; i = —-—= 10% — 0.10; n — 6 17 715.61 = PV(1.10)6

by substituting in FV = PV(1 + i)n

17 715.61 = PV(1.771561)

computing (1.10)6

= i77_l T61

1.771561

solve for PV by dividing both sides by 1.771561

PV = $10 000.00 The principal that will grow to $17715.61 in six years at 10% p.a. com¬ pounded annually is $10 000.00. This principal is called the present value or discounted value or proceeds of the known future amount. The difference between the known future amount of $17 715.61 and the computed present value (principal) of $10 000.00 is the compound discount and represents the compound interest accumulating on the computed present value. The process of computing the present value or discounted value or proceeds is called discounting.

B. The present value formula The present value of an amount at a given time at compound interest is defined as the principal that will grow to the given amount if compounded at a given peri¬ odic rate of interest over a given number of conversion periods. Since the problem of finding the present value is equivalent to finding the prin¬ cipal when the future value, the periodic rate of interest, and the number of con¬ version periods are given, the formula for the future value formula, FV = PV( 1 + i)'\ applies.

CHAPTER

9:

COMPOUND

I N T E R E ST—F U T U R E

VALUE

AND

PRESENT

VALUE

However, because the problem of finding the present value of an amount is frequently encountered in financial analysis, it is useful to solve the future value formula for PV to obtain the present value formula.

FV = PV. (1 + i)n

start with the future value formula. Formula 9.1A

_ pv (i + iy (i + iy (i + iy FV

divide both sides by the compounding factor (1 + i )n

FV

(i + iy

M -f j )n reduce the fraction ^ , . yt to 1

= pv

The present value formula for compound interest is:

PV =

FV

(i +

Formula 9.IB

iy

Find the present value of $6836.56 due in nine years at 6% p.a. compounded quarterly.

6%

FV= 6836.56; I/Y = 6; P/Y = 4; i = —

PV

1.5% = 0.015; n = 36

FV using the present value formula

=(i + iy 6836.56

by substitution

(1 + 0.015)36 6836.56 1.709140 $4000.00

Note: The division of 6836.56 by 1.7091395, like any division, maybe changed to a multiplication by using the reciprocal of the divisor. 6836.56 the division to be changed into a

1.709140

multiplication

— 6836.56

f-

1

\ 1.709140

= 6836.56(0.585090)

the reciprocal of the divisor 1.709140 is found by dividing 1 by 1.709140 computed value of the reciprocal

= $4000.00 For calculators equipped with the reciprocal function key [ llx 1, converting the division into a multiplication is easily accomplished by first computing the compounding factor and then using the

1he

key to obtain the reciprocal.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

What principal will amount to $5000.00 seven years from today if interest is 9% p.a. compounded monthly? Finding the principal that amounts to a future sum of money is equivalent to finding the present value. 9% FV = 5000.00; I/Y = 9; P/Y = 12; i = — = 0.75% = 0.0075; 5000.00

PV

n = 84

using Formula 9.1 B

(1.0075)84 5000.00

computing the factor (1.0075)34

1.873202

using the reciprocal function key

5000.00(0.533845) $2669.23

Using the reciprocal of the divisor to change division into multiplication is reflected in the practice of stating the present value formula with a negative exponent.

negative exponent rule

= a

(m?“Fva + i)Formula 9.IB, the present value formula, can be restated in multiplication form using a negative exponent.

Formula 9.1C

PV = FV(1 + i)'"

The factor (1 + 0“" is called the discount factor and is the reciprocal of the com¬ pounding factor (1 + 0"-

C. Using preprogrammed financial calculators to find present value rjssA

As explained in Section 9.2B, preprogrammed calculators provide quick solutions to compound interest calculations. Three of the four variables are entered and the value of the fourth variable is retrieved. To solve Example 9.3C, in which FV = 5000, i — 0.75%, n — 84, and PV is to be determined, use the following procedure. (Remember that the interest rate for the year, 9% must be entered into the calculator as I/Y.) Key in

Press

Display shows

( 2nd j(CLR TVM)

0

•clears the function key registers

ptad|(P/Y)

0

■checks the P/Y register

12

P/Y = 12

Q

changes the value to "12

C/Y = 12

—changed automatically to match

0-

- returns to the standard calcula¬

the P/Y

2nd j (QUIT)

tion mode 5000

FV

5000 - -enters the future value amount FV

9

I/Y

9-enters the conversion rate /

84

84 ing periods n PV

— 2669.226329- - retrieves the unknown principal (present value) PV, an investment or cash outflow as indicated by the negative sign

The principal is $2669.23. Excel

Excel has a Present Value (PV) function you can use to calculate the present value of an investment subject to compound interest. Refer to PV on the Spreadsheet Template Disk to learn how to use this Excel function.

D. Finding the present value when n is a fractional value Use Formula 9.1A, FV = PV(1 + i)n, or Formula 9.IB, PV =

FV ^ + -y, ,

or Formula 9.1C, PV = FV(1 + z)~" where n is a fractional value representing the entire time period, FV is a known value, and PV is to be deter¬ mined.

Find the present value of $2000.00 due in three years and eight months if money is worth 8% p.a. compounded quarterly. FV= 2000.00; I/Y = 8; P/Y = 4; i =

8% 4

2% = 0.02;

n = ^3^(4) = 14} = 14.666667 PV

FV using Formula 9.IB

(1 + i)n

2000.00 (1 + 0.02)14666667

use as many decimals as are available in your calculator

= 2000.00(0.747936)

multiply by the reciprocal

= $1495.87 Programmed Solution (Set P/Y = 4)

2nd

(CLR TVM) 2000 i PV

8

I/Y

14.666667 ( N ~) fcPT) f~PV~] (^495.870992/

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Determine the principal that will accumulate to $2387.18 from September 1, 2000 to April 1, 2004 at 5% p.a. compounded semi-annually. Finding the principal that will grow to the given amount of $2387.18 is equiva¬ lent to finding the present value or discounted value of this amount. The time period September 1, 2000 to April 1, 2004 contains three years and seven months; that is, it consists of seven whole conversion periods of six months each and a fractional conversion period of one month. FV Use PV =

(1 +

i)n

FV = 2387.18; 18; I/Y = 5; P/Y = 2; i = n = (3 12 "7 \2) = PV

2.5% = 0.025;

= 7.166667

2387.18 (1.025)7-166667 2387.18 ~ 1.193588 = 2387.18(0.837810) = $2000.00

Programmed Solution (Set P/Y = 2) [2nd ) (CLR TVM) 2387.18 ( FV ) 5 fl/F] 7.166667

EXERCISE

N

C.PT

PV j (-2000.003696)

9.3 If you choose, you can use Excel’s Present Value (PV) function to calculate the

Excel

present value of an investment subject to compound interest. Refer to PV on the Spreadsheet Template Disk to learn how to use this Excel function.

fe>©

Find the present value of each of the following amounts.

Amount

Nominal Rate

1

.

$1000.00

8%

2.

1500.00

6.5%

3.

600.00

4. 5.

Frequency of Conversion

Time

quarterly

7 years

semi-annually

10 years

8%

monthly

6 years

350.00

7.5%

8 years

1200.00

9%

annually monthly semi-annually

5 years, 6 months

12 years

6.

3000.00

12.25%

7. 8.

900.00

6.4%

quarterly

9 years, 3 months

500.00

8.4%

monthly

9.

1500.00

4.5%

10.

900.00

8 years, 10 months

11.

6400.00

5.5% 7%

annually semi-annually

15 years 15 years, 9 months

quarterly

5 years, 7 months

12.

7200.00

6%

monthly

21.5 months

CHAPTER

9:

COMPOUND

I N T E R E ST— F U T U R E

VALUE

AND

PRESENT

VALUE

Answer each of the following questions. 1. Find the present value and the compound discount of $ 1600.00 due four-anda-half years from now if money is worth 4% compounded semi-annually. 2. Find the present value and the compound discount of $2500.00 due in six years, three months, if interest is 6% compounded quarterly. 3. Find the principal that will amount to $1250.00 in five years at 10% p.a. com¬ pounded quarterly. 4. What sum of money will grow to $2000.00 in seven years at 9% compound¬ ed monthly? 5. A debt of $5000.00 is due November 1,2011. What is the value of the obliga¬ tion on February 1, 2005, if money is worth 7% compounded quarterly?

6. How much would you have to deposit in an account today to have $3000.00 in a five-year term deposit at maturity if interest is 7.75% compounded annually? 7. What is the principal that will grow to $3000.00 in eight years, eight months at 9% compounded semi-annually?

8. Find the sum of money that accumulates to $1600.00 at 5% compounded quarterly in six years, four months.

9.4 Application—Discounting Negotiable Financial Instruments at Compound Interest A. Discounting long-term promissory notes Long-term promissory notes (written for a term longer than one year) are usually subject to compound interest. As with short-term promissory notes, long-term promissory notes are negotiable and can be bought and sold (discounted) at any time before maturity. The principles involved in discounting long-term promissory notes are simi¬ lar to those used in discounting short-term promissory notes by the simple dis¬ count method except that no requirement exists to add three days of grace in determining the legal due date of a long-term promissory note. The discounted value (or proceeds) of a long-term promissory note is the pre¬ sent value at the date of discount of the maturity value of the note. It is found FV using the present value formula PV = ^ + or PV = FV(1 + ?')”". For non-interest-bearing notes, the maturity value is the face value. However, for interest-bearing promissory notes, the maturity value must be determined first by using the future value formula FV = PV(1 + i)n. Like promissory notes, long-term bonds promise to pay a specific face value at a specified future point in time. In addition, there is a promise to periodically pay a specified amount of interest. Long-term bonds will be covered in detail in Chapter 15.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

B. Discounting non-interest-bearing promissory notes Since the face value of a non-interest-bearing note is also its maturity value, the proceeds of a non-interest-bearing note are the present value of its face value at the date of discount. Determine the proceeds of a non-interest-bearing note for $1500.00 discounted two-and-a-quarter years before its due date at 9% p.a. compounded monthly. The maturity value FV = 1500.00; the rate of discount I/Y = 9; P/Y = 12;i =

9%

— 0.75% — 0.0075;

the number of conversion periods n = (2.25)(12) = 27. PV = FV(1 + i)“" = 1500.00(1 + 0.0075)-27

= 1500-00(nik5) = 1500.00(0.817304) = $1225.96 Programmed Solution (Set P/Y = 12) [2nd] (CLR TVM) 1500 [~FVJ 9 [7/y] 27 flT) [cpt] fpv [-1225.955705]

A three-year, non-interest-bearing promissory note for $6000.00 dated August 31, 2004 was discounted on October 31, 2005 at 6% p.a. compounded quarterly. Determine the proceeds of the note. The due date of the note is August 31,2008; the discount period October 31, 2005 to August 31, 2008 contains 2 years and 10 months.

6%

FV = 6000.00; I/Y = 6; P/Y = 4; i = — = 1.5% = 0.015; n = (2^)(4) = lly = 11.333333 PV = FV(1 + i)~n = 6000.00(1 + 0.015)~11 333333 = 6000.00(0.844731) = $5068.38 Programmed Solution (Set P/Y = 4)f2nd] (CLR TVM) 6000 [ FV ) 6 [ I/Y | 11.333333 [~fT{ fCPT PV

[-5068.383148

CHAPTER 9: COMPOUND I N T E R E S T — F U T U R E VALUE AND PRESENT VALUE

You signed a promissory note at the Continental Bank for $3000.00 due in 27 months. If the bank charges interest at 8% p.a. compounded semi-annually, determine the proceeds of the note. The amount shown on the note is the sum of money due in 27 months, that is, the maturity value of the note. FV = 3000.00; I/Y = 8; P/Y = 2; i = ^ = 4% =# 0.04; PV = = = =

n = (j^)(2) = 4.5

FV(1 + i)~n 3000.00(1 + 0.04)_4-5 3000.00(0.838205) $2514.61

Programmed Solution (Set P/Y = 2)

[2nd]

(CLR TVM) 3000 ;

FV

] 8 (

i/Y

j 4.5

N

CPT

J>V

^2514.613414)

C. Discounting interest-bearing promissory notes The proceeds of an interest-bearing note are equal to the present value at the date of discount of the value of the note at maturity. Therefore, the maturity value of an interest-bearing promissory note must be determined before finding the dis¬ counted value.

Determine the proceeds of a promissory note for $3600.00 with interest at 6% p.a. compounded quarterly, issued September 1, 2004, due on June 1, 2010, and discounted on December 1, 2006 at 8% p.a. compounded semi-annually.

STEP 1

Find the maturity value of the note using Formula 9.1A, FV =PV(1 + i)n. 6% PV — 3600.00; I/Y = 6; P/Y = 4; i = 1.5% = 0.015; the interest period, September 1, 2004 to June 1, 2010, contains 5 years and 9 months: n = (59/i2)(4) = 23. FV = 3600.00(1 + 0.015)23 = 3600.00(1.408377) = $5070.16 Find the present value at the date of discount of the maturity value found in Step 1 using PV = FV(1 + i)~n. FV= 5070.16; I/Y = 8; P/Y = 2; i = ^ = 4% = 0.04; the discount period, December 1, 2006 to June 1, 2010, contains 3 years and 6 months: n = (36/i2)(2) = 7.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

PV = 5070.16(1 + 0.04)~7 = 5070.16(0.759918) = $3852.90 The proceeds of the note on December 1,2006 are $3852.90. The method and the data are represented graphically in Figure 9.6. FIGURE 9.6

Graphical Representation of Method and Data Date of issue

Date of discount

Due date

(Sept. 1, 2004)

(Dec. 1, 2006)

(June 1, 2010)

| ( Face value

FV = ?

Accumulate interest at 6% p.a. compounded quarterly for 5 years and 9 months using Formula 9.1 A. PV = $3600.00; i = 0.015; n = 23

{

•

Step 2 “

Proceeds

FV = ?

Use FV = PV(1 + i)n

Since the due date falls before the focal date, use the future value formula.

PART

3:

MATHEMATICS

OF

FINANCE

PV= 4000.00; I/Y = 9; P/Y = 2; i = — = 0.045;

AND

INVESTMENT

« = 4(2) = 8

FV = 4000.00(1 + 0.045)8 = 4000.00(1.4221006) = $5688.40

The equivalent value of the $4000.00 seven years from now is $5688.40. (ii) Using “now” as the focal date, the method and the data can be repre¬ sented graphically as shown in Figure 9.8. FIGURE 9.8

Graphical Representation of Method and Data Now

3 years

Focal date

Due date

PV = ?


_ 25 000.00 PMT =54.354225 PMT =$459.95

Using Formula 12.2a, which is Formula 12.2 rearranged, PMT

PMT =

FV p _ HILL_ (1 + p)n - 1

25 000.000(0.015075) 1.01507540 - 1

-substituting in Formula 12.2A

PMT = $459.95

Programmed Solution (Set P/Y - 4; C/Y - 12)

2nd

(CLR TVM) 6 J/Y : 0 (jPV 25 000 jjFVj 40 [ N | [cpt] (PMT): [-459.945847]

The required quarterly deposit is $459.95.

B. Finding the periodic payment PMT when the present value of a general annuity is known If the present value of a general annuity PVnc, the number of conversion periods n, and the conversion rate i are known, you can find the periodic payment PMT by substituting the given values in the present value Formula 12.3.

PVnc = PMT

~ (1 +P)

Formula 12.3

P

When using a scientific calculator, you can find PMT by first rearranging the terms of the present value formulas. Then substitute the three known values (PV, n, and i) into the rearranged formula and solve for PMT.

PV nc = PMT

i - (i + prn p

Formula 12.3

PART

3:

MATHEMATICS

OF

FINANCE

PVnc PMT = i - (i +

AND

INVESTMENT

1 ~ (1 + p)~"

prn

- divide both sides by

P

p

PMT

PV ncxp l - (i + p)~n

Formula 12.3A: dividing by a fraction is the same as inverting the fraction and multiplying

When using a preprogrammed financial calculator, you can find PMT by entering the five known values (PV N, I/Y, P/Y, and C/Y) and pressing [cpt j PMT .

Mr. and Mrs. White applied to their credit union for a first mortgage of $60 000.00 to buy a house. The mortgage is to be amortized over 25 years and interest on the mortgage is 8.5% compounded semi-annually. What is the size of the monthly payment if payments are made at the end of each month? PV

2

= 60 000.00; n = 25(12) = 300; P/Y = 12; C/Y = 2; c = —

nc

12

I/Y = 8.5; i =

6

’

= 4.25% = 0.0425

The effective monthly rate of interest p = 1.0425* - 1 = 1.006961 - 1 = 0.006961 = 0.6961% (\ - 1.006961 ~300\ 60 000.00 = PMT I- 006961-) 60 000.00 = PMT(125.72819) PMT = 60 °00-^ 125.72819 PMT = $477.22

Programmed Solution (Set P/Y = 12; C/Y = 2) 8.5 ( I/Y ) 0 ( FV j 60 000 ["±"1(^1 300 ( N

J [ CPT

(PMT) 1477.2181 11 1

The monthly payment due at the end of each month is $477.22.

EXERCISE 12.3 If you choose, you can use Excel’s Payment (PMT) function to answer all of the questions below. Refer to PMT on the Spreadsheet Template Disk to learn how

Excel

to use this Excel function.

4A CHAPTER

12:

ORDINARY

GENERAL

ANNUITIES

For each of the following ten ordinary general annuities, determine the size of the periodic rent. Future Value

Present Value

1.

$15 000.00

2.

$6000.00

3.

$12 000.00

4.

$7 000.00

5.

$8000.00

6. 7.

$20 000.00 $45 000.00

8.

$35 000.00

9. 10.

$20 000.00 $16 500.00

Payment Period

Term of Annuity

Interest Rate

Conversion Period

6 months

7 years, 6 mos

1 quarter

9 years, 9 mos

12 months

15 years

4.5%

6 months

12.5 years

7.5%

quarterly

3 months

6 years

6.8%

semi-annually

3 months

20 years

6 months 1 year 1 month

8 years

3 months

15 years

5.5% 8%

annually monthly semi-annually

12%

monthly

10 years

9%

quarterly

15 years

4%

quarterly

7%

semi-annually

5.75%

annually

Answer each of the following questions. 1. What payment made at the end of each quarter for fifteen years will accu¬ mulate to

$12 000.00

at 6% compounded monthly?

2. What payment is required at the end of each month for five years to repay a loan of $6000.00 at 7% compounded semi-annually? 3. A contract can be fulfilled by making an immediate payment of $9500.00 or equal payments at the end of every six months for eight years. What is the size of the semi-annual payments at 7.4% compounded quarterly? 4. What payment made at the end of each year for eighteen years will amount to $16 000.00 at 4.2% compounded monthly? 5. What payment is required at the end of each month for fifteen years to amor¬ tize a $32 000.00 mortgage if interest is 9.5% compounded semi-annually?

6. How much must be deposited at the end of each quarter for ten years to accu¬ mulate to

$12 000.00

at 6% compounded monthly?

7. What payment made at the end of every three months for twenty years will accumulate to $20 000.00 at 7% compounded semi-annually? 8.

Deragh bought a car priced at $9300.00 for 15% down and equal monthly payments for four years. If interest is 8% compounded semi-annually, what is the size of the monthly payment?

9. Equal payments are to be made at the end of each month for fifteen years with interest at 9% compounded quarterly. After the last payment, the fund is to be invested for seven years at

10%

compounded quarterly and have a

maturity value of $20 000.00. What is the size of the monthly payment? 10. To finance the development of a new product, a company borrowed $30 000.00 at 7% compounded monthly. If the loan is to be repaid in equal quarterly payments over seven years and the first payment is due three months after the date of the loan, what is the size of the quarterly payment?

PART 3: MATHEMATICS OF FINANCE AND INVESTMENT vjJ

12.4 Ordinary General Annuities—Finding the Term n A. Finding the term n when the future value of a general annuity is known If the future value of an annuity FVn(., the periodic payment PMT, and the conver¬ sion rate i are known, you can find the term of the annuity n by substituting the given values in the future value Formula 12.2.

FV nc = PMT

(1 +P)H- i~

P

where p = (1 + i)c — 1

Formula 12.2

When using a scientific calculator, you can find n by first rearranging the terms of the future value Formula 12.2. Then substitute the three known values (FV, PMT, and p) into the rearranged formula and solve for n. FVnc FV nc

PMT

(i +

PMT

(i + pr n ln(l + p)

(1 + P)n ~ 1

pr

-

P -1

Formula 12.2

- divide both sides by PMT

p multiply both sides by p

(!VP) + 1.

and add 1 to both sides

\ PMT ) ln[(FVcp/PMT) + 1]

-solve for n using natural logarithms

ln[FVifp/PMT + 1] n

-Formula 12.2B: divide both sides by

ln(l + p)

ln(1 + p)

When using a preprogrammed financial calculator, you can find n byjentering the five known values (FV, PMT, I/Y, P/Y, and C/Y) and pressing fcPT N J. What period of time is required for $125.00 deposited at the end of each month at 11% compounded quarterly to grow to $15 000.00? 4 1 FVm = 15 000.00; PMT = 125.00; P/Y =12; C/Y = 4; c = — = y; \n = 11; i =

= 2.75% = 0.0275

The effective monthly rate of interest p = 1.0275^ - 1 = 1.009084 - 1 = 0.009084 = 0.9084% /1.009084" - 1\ 15 000.00 = 125.00( 0 0Q9034 ~ j

CHAPTER

12: ORDINARY GENERAL ANNUITIES

(1.009084" - 1)

120.00

0.009084

1.090068 = 1.009084" - 1 1.009084" = 2.090068

n In 1.009084 = In 2.090068 n(0.009043) = 0.737197 0.737197

n —

0.009043

n = 81.522227 n = 82 months approximately With a preprogrammed calculator, the procedure is (Set P/Y = 12; C/Y = 4) 11 ( I/Y ) 0 ( PV ) 15 000

FV

125 f~*n [PMT1 fcPT] (

N 1 [ 81.522240 ~]

It will take about six years and ten months to accumulate $15 000.00.

B. Finding the term rt when the present value of a general annuity

is known If the present value PVnc, the periodic payment PMT, and the conversion rate i are known, you can find the term of the annuity n by substituting the given values in the present value Formula 12.3.

PVnc

PMT

~ 1 — (1 + p)-"~ P

Formula 12.3

Alternatively, you can first rearrange the terms of Formula 12.3 to solve for n. PV = PMT

~i-(i + pyn

p PVnc

1 - (1 + PY" _

PMT PVncrp PMT

p

1 = -(1 +p)-»-

/PV p\ (1 + p) " - 1 - (pMT )

(ELc?)

-n ln(l + p) = In

\PMT )

In 1 n =

-

Formula 12.3 divide both sides by PMT multiply both sides by / and subtract 1 from both sides multiply both sides by —1 solve for n using natural logarithms

(WncP \PMT

-In (1 + p)

Formula 12.3B: divide both sides by -ln(1 + p)

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT ^

When using a scientific calculator, you can find n by first rearranging the terms of the present value formulas above. Then substitute the three known values (PV, PMT, and i) into the rearranged formula and solve for n. When using a preprogrammed financial calculator, you can find n by entering the five known values (PV, PMT, I/Y, P/Y, and C/Y) and pressing

$3000.00(1.06)2 -> $3000.00(1.06)3 -> $3000.00(1.06)4 -+ $3000.00(1.06)5

As shown in Figure 13.1, the first deposit is located at the beginning of Year 1, which is the same as “now”; the second deposit is located at the beginning of Year 2, which is the same as the end of Year 1; the third at the beginning of Year 3; the fourth at the beginning of Year 4; and the fifth and last deposit at the begin¬ ning of Year 5, which is also the beginning of the last payment period. The focal date, however, is located at the end of the last payment period. The accumulated values of the individual deposits are obtained by using Formula 9.1A, FV = PV(1 + z)". Finding the combined total of the five accumu¬ lated values is made easier by taking out the common factors 3000.00 and 1.06 as follows:

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Deposit 5 3000.00(1.06)' =

(1.0)

(1.0)

Deposit 4 3000.00(1.06)2 =

(1.06)

(1.06)

Deposit 3 3000.00(1.06)3 = 3000.00(1 j

(1.06)2 = 3000.00(1.06) (1.1236)

Deposit 2 3000.00(1.06)4 =

(1.06)3

(1.191016)

Deposit 1 3000.00(1.06)5 =

(1.06)4

(1.262477) = 3000.00(1.06) (5.637093) = 16 911.28(1.06) = $17 925.96

Note: This example is the same as Example 11.2B except that the deposits are made at the beginning of each payment period rather than at the end. The answer to Example 11.2B was $16 911.28. We could have obtained the answer to Example 13.1 A simply by multiplying $16 911.28 by 1.06. It appears that the future value of the annuity due can be obtained by multiplying the future value of the ordinary annuity by the factor (1 + i). The general notation for simple annuities due is the same as for ordinary sim¬ ple annuities except that the accumulated value (future value) of the annuity due is represented by the symbol FVn(due). The formula for the future value of a simple annuity due is:

S„(due) = R(1 + i)

(1 + i)" - 1 i

restated as FV„(due) = PMT(1 + i)

(1 + i)" - 1

— Formula 13.1

i

Note: Formula 13.1, the future value of a simple annuity due, differs from Formula 11.1, the future value of an ordinary simple annuity, only by the factor

(1

+ i). FUTURE VALUE OF A SIMPLE ANNUITY DUE

FUTURE VALUE OF THE

= (1 + i) X

ORDINARY SIMPLE ANNUITY

The relationship between an annuity due and the corresponding ordinary annuity is graphically illustrated in the comparison of the line diagrams. Ordinary

n

Now

—

2

n

— 1

n

annuity

Annuity due

PMT

PMT

PMT

PMT

PMT

PMT

PMT

PMT

PMT

Now

\y PMT

The two line graphs show the shift of the payments by one period. In an annuity due, every payment earns interest for one more period than in an ordinary annu¬ ity and this explains the factor (1 + i).

CHAPTER

13:

ANNUITIES

DUE,

DEFERRED

ANNUITIES,

AND

PERPETUITIES

Find the accumulated value at the end of the last payment period of quarterly payments of $50.00 made at the beginning of each quarter for ten years if inter¬ est is 6% compounded quarterly. Since the payments are of equal size made at the beginning of each period, the payment series is an annuity due and since the focal date is the end of the last payment period, the future value of the annuity due is to be found.

6%

PMT = 50.00; P/Y = 4; C/Y = 4; I/Y = 6; i = ~ = 0.015; n = 10(4) = 40 / 1.01540 — 1 \ FV„(due) = 50.00(1.015)1-J = 50.00(1.015) (54.267894) = 2713.39(1.015) = $2754.10 You deposit $100.00 at the beginning of each month for five years in an account paying 4.2% compounded monthly. (i) What will the balance in your account be at the end of five years? (ii) How much of the balance will you have contributed? (iii) How much of the balance will be interest? (i) PMT = 100.00; P/Y = 12; C/Y = 12; I/Y = 4.2; 4.2% i - —— - 0.35% - 0.0035; n = 5(12) = 60 (\ 003560 — 1\ FV„(due) = 100.00(1.0035)^ ' 0.0035 / = 100.00(1.0035) (66.635949) - 6663.5949(1.0035) = 6686.92 (ii) Your contribution is (100.00)(60) = $6000.00. (iii) The interest earned = 6686.92 — 6000.00 = $686.92.

To distinguish between problems dealing with ordinary annuities and annuities due, look for key words or phrases that signal one type of annuity or the other. Ordinary annuities:

“payments (or deposits) made at the end of each (or every)...”

Annuities due:

“payments (or deposits) made at the beginning of each (or every)...” “first payment is due on the date of sale (or signing)” “payable in advance”

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

B. Present value of a simple annuity due Find the present value of five payments of $3000.00 each made at the beginning of each of five consecutive years respectively if money is worth 6% compounded annually. As Figure 13.2 shows, the present value of the individual payments is obtained using Formula 9.1C, PV - FV(1 + i)~". The sum of the individual present val¬ ues is easier to find when the common factor 3000.00 is taken out. FIGURE 13.2

Graphical Representation of Method and Data End of yea: Present value

Focal

of each payment

date

using

|

PV = FV(1 + i)~n

Now

$3000.00

$3000.00

$3000.00

$3000.00

$3000.00

3000.00

3000.00(1.06)-'

3000.00(1.06)~2 n = 3 3000.00(1.06)“3

M1 : 7

IV

LHi) [CPT j

N )

28.00020963 :

n — 28 quarters (approximately) It will take seven years to build up the fund. For how long can you withdraw $480.00 at the beginning of every three months from a fund of $9000.00 if interest is 10% compounded quarterly? 4; C/Y = 4; I/Y = 10; PV„(due) - 9000.00; PMT = 480.00; P/Y . 10% = 2.5% = 0.025 1 - 1.025 — substituting in Formula 13.2 9000.00 = 480.00(1.025)l Q^ 9000.00 = 19 680.00(1 - 1.025“") -9000-QQ 19 680.00 0.457317 1.0251.025“" n In 1.025 — «(0.024693)

=1-1 025= = = =

1 - 1.0251 - 0.457317 0.542683 In 0.542683 -0.611230 0.611230 n = 0.024693 n = 24.753561 n = 25 (quarters)

Programmed Solution (Set P/Y = 4; C/Y = 4) (“BGN” mode) 9000 0

1V

480 (PMT) 10

PV I/Y

[CPT

N

[2475355963]

n = 28 quarters (approximately) Withdrawals of $480.00 can be made for six years and three months. (The last withdrawal will be less than $480.00.)

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

A lease contract valued at $7800.00 is to be fulfilled by rental payments of $180.00 due at the beginning of each month. If money is worth 9% compounded monthly, what should the term of the lease be? 12; I/Y = 9; C/Y = 12;

PV (due) = 7800.00; PMT = 180.00; P/Y 9% i = ~ = 0.75% Programmed Solution

(“BGN” mode) (Set P/Y = 12; C/Y = 12) 0 \ FV j 7800 j± 180 PMT

9

I/Y

PV N j [52.12312544

CPT

n — 53 months (approximately) The term of the lease should be four years and five months.

F. Finding the periodic rate of interest of a simple annuity due Preprogrammed financial calculators are especially helpful when solving for the conversion rate i. Determining i without a financial calculator is extremely timeconsuming. However, it can be done by hand, as illustrated in Appendix B on the CD-ROM. When the future value or present value, the periodic payment PMT, and the term n of an annuity due are known, the periodic rate of interest i can be found by entering the five known values into a preprogrammed financial calculator. (Remember, if both PV and PMT are non-zero, enter PV only as a negative amount.) For simplejmnuities due, retrieve the answer by being in “BGN mode and pressing ( cpt > fi/Y j . This is the nominal annual rate of interest. Compute the nominal annual rate of interest at which $100.00 deposited at the beginning of each month for ten years will amount to $15 000.00. FV„(due) = 15 000; PMT = 100.00; P/Y = 12; C/Y =12; n = 120; m = 12 (“BGN” mode) (Set P/Y = 12; C/Y = 12) 0 I PV j 15 000 ( JVJ 100 flTj [pMt] 120 ["nH [cpt] [7/y] [ 4.282801 ] t

allow several seconds for the computation The nominal annual rate of interest is 4.28% compounded monthly. The monthly conversion rate is 4.282801%/12 = 0.356900%. A lease agreement valued at $7500.00 requires payment of $450.00 at the beginning of every quarter for five years. What is the nominal annual rate of interest charged?

CHAPTER

13: ANNUITIES DUE,

DEFERRED ANNUITIES, AND PERPETUITIES

PV (due) - 7500.00; PMT = 450.00; P/Y = 4; C/Y = 4; n = 20; m = 4 (“BGN” mode) (Set P/Y 5 4; C/Y 5 4) 0 f~Fvl 7500

450.00 PMT 20

N

*

[pv CPT

8.032647

I/Y

J

allow several seconds for the computation The nominal annual rate of interest is 8.03% compounded quarterly. The quarterly compounding rate is 8.032647/4 = 2.008162%.

EXERCISE 13.1 If you choose, you can use Excel’s Present Value (PV) function or Future Value

Excel

(FV) function to answer the questions indicated below. Refer to PV and FV on the Spreadsheet Template Disk to learn how to use these Excel functions. Find the future value and the present value of each of the following six simple annuities due. Periodic Payment 1.

$3000.00

2.

$750.00

Payment Interval

Interest Rate

Term

Conversion Period quarterly

8 years

8%

1 month

5 years

7.2%

monthly

12 years

5.6%

semi-annually quarterly

3 months

3.

$2000.00

6 months

4.

$450.00

3 months

15 years

4.4%

5.

$65.00

1 month

20 years

9%

monthly

6.

$160.00

1 month

15 years

6%

monthly

Find the periodic payment for each of the following four simple annuities due. Future Value

Present Value

Payment Period 3 months

1. $20 000.00

Term

6%

quarterly

8 years

7%

annually

3%

semi-annually

$12 000.00

1 year

3.

$18 500.00

6 months

12 years

1 month

5 years

$9 400.00

Conversion Period

15 years

2. 4.

Interest Rate

12%

monthly

Find the length of the term for each of the following four simple annuities due. Future Value 1.

Present Value

$5 300.00

Periodic Payment

Payment Period

$35.00

1 month 3 months

2.

$8400.00

$440.00

3.

$6450.00

$1120.00

4. $15 400.00

$396.00

1 year 6 months

Interest Rate

Conversion Period

6%

monthly

7%

quarterly

10%

annually

5%

semi-annually

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Compute the nominal annual rate of interest for each of the following four simple annuities due. Future Value

Present Value

o

Payment Interval

Term

Compounding Period

$1014.73

1 year

25 years

annually

2.

$42 000.00

$528.00

1 month

10 years

monthly

3. 4. $36 000.00

$28 700.00

$2015.00

6 months

15 years

semi-annually

$584.10

3 months

12 years

quarterly

1. $70 000.00

%t

Periodic Rent

Answer each of the following questions. 1. Find the amount of an annuity due of $300.00 payable at the beginning of every month for seven years at 6% compounded monthly. 2. Determine the accumulated value after twelve years of deposits of $360.00 made at the beginning of every three months and earning interest at 7% compounded quarterly. 3. Until he retires sixteen years from now, Mr. Lait plans to deposit $300.00 at the beginning of every three months in an account paying interest at 5% compounded quarterly. (a) What will be the balance in his account when he retires?

(b) How much of the balance will be interest? 4. Joanna contributes $750.00 at the beginning of every six months into an RRSP paying interest at 8% compounded semi-annually. (a) How much will her RRSP deposits amount to in twenty years?

(b) How much of the amount will be interest? 5. Find the present value of payments of $2500.00 made at the beginning ot every six months for ten years if money is worth 9.5% compounded semi¬ annually. 6. What is the discounted value of deposits of $240.00 made at the beginning of every three months for seven years if money is worth 8.8% compounded quarterly? 7. A washer-dryer combination can be purchased from a department store by making monthly credit card payments of $52.50 for two-and-a-half years. The first payment is due on the date of sale and interest is 21% compounded monthly. (a) What is the purchase price?

(b) How much will be paid in installments? (c) How much is the cost of financing? 8. Diane Wallace bought a living-room suite on credit, signing an installment contract with a finance company that requires monthly payments of $62.25 for three years. The first payment is made on the date of signing and interest is 24% compounded monthly.

S' CHAPTER

13:

ANNUITIES

DUE,

DEFERRED

ANNUITIES,

AND

PERPETUITIES

(a) What was the cash price? (b) How much will Diane pay in total? (c) How much of what she pays will be interest? 9. The monthly premium on a three-year insurance policy is $64.00 payable in advance. What is the cash value of the policy if money is worth 4.8% com¬ pounded monthly?

10. The monthly rent payment on office space is $535.00 payable in advance. What yearly payment made in advance would satisfy the lease if interest is 6.6% compounded monthly?

11. Elspeth McNab bought a boat valued at $12 500.00 on the installment plan requiring equal monthly payments for four years. If the first payment is due on the date of purchase and interest is 7.5% compounded monthly, what is the size of the monthly payment?

12. How much does a depositor have to save at the beginning of every three months for nine years to accumulate $35 000.00 if interest is 8% compounded quarterly?

13. Payments on a seven-year lease valued at $12 200.00 are to be made at the beginning of each month during the last five years of the lease. If interest is 9% compounded monthly, what is the size of the monthly payments? >14. Julia deposited $1500.00 in an RRSP at the beginning of every six months for twenty years. The money earned interest at 6.25% compounded semi¬ annually. After twenty years, she converted the RRSP into an RRIF from which she wants to withdraw equal amounts at the beginning of each month for fifteen years. If interest on the RRIF is 6.6% compounded monthly, how much does she receive each month?

15. Mr. Clark wants to receive payments of $900.00 at the beginning of every three months for twenty years starting on the date of his retirement. If he retires in twenty-five years, how much must he deposit in an account at the beginning of every three months if interest on the account is 5.25% com¬ pounded quarterly?

16. Quarterly payments of $1445.00 are to be made at the beginning of every three months on a lease valued at $25 000.00. What should the term of the lease be if money is worth 8% compounded quarterly?

17. Tom is saving $600.00 at the beginning of each month. How soon can he retire if he wants to have a retirement fund of $120 000.00 and interest is 5.4% compounded monthly?

18. If you save $75.00 at the beginning of every month for ten years, for how long can you withdraw $260.00 at the beginning of each month starting ten years from now, assuming that interest is 6% compounded monthly?

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

19. Ali deposits $450.00 at the beginning of every three months. He wants to build up his account so that he can withdraw $1000.00 every three months starting three months after the last deposit. If he wants to make the with¬ drawals for fifteen years and interest is 10% compounded quarterly, for how long must Ali make the quarterly deposits?

20. What nominal annual rate of interest was paid if contributions of $250.00 made into an RRSP at the beginning of every three months amounted to $14 559.00 after ten years?

21. A vacation property valued at $25 000.00 was bought for fifteen payments of $2200.00 due at the beginning of every six months. What nominal annual rate of interest was charged?

22. An insurance policy provides a benefit of $250 000.00 twenty years from now. Alternatively, the policy pays $4220.00 at the beginning of each year for twenty years. What is the effective annual rate of interest paid?

13.2 General Annuities Due A. Future value of a general annuity due As with a simple annuity due, the future value of a general annuity due is greater than the future value of the corresponding ordinary general annuity by the inter¬ est on it for one payment period. Since the interest on a general annuity for one payment period is (1 + i)c, or

(1 + p)> PRESENT VALUE OF THE CORRESPONDING

PRESENT VALUE OF A GENERAL = (1

ANNUITY DUE

+p)

x

ORDINARY GENERAL ANNUITY

Thus, for the future value of a general annuity due use Formula 13.3: (1 + p)'1 - 1 Sgdue) = R(l +p)

P

restated as FV„f(due) = PMT(1 + p)

(i + py -1

Formula 13.3

p

where p — (1 + i Y — 1 What is the accumulated value after five years of payments of $20 000 made at the beginning of each year if interest is 7% compounded quarterly?

CHAPTER

13:

ANNUITIES

DUE,

DEFERRED

ANNUITIES,

AND

PERPETUITIES

PMT = 20 000.00; n = 5; c = 4; P/Y = 1; C/Y = 4; I/Y = 7; 7% i = = 1.75% - 0.0175 4 The effective annual rate of interest p = 1.01754 - 1 = 1.071859 - 1 = 0.071859 = 7.1859% /1 071859s i\ FV„c(due) = 20 000.00(1.071859)^ ' Q Q71859 j

substituting in Formula 13.3

= 20 000.00(1.071859)(5.772109) = 20 000.00(6.186888) = $123 737.75 Programmed Solution (“BGN” mode) (Set P/Y 5 1; C/Y = 4) 7

20 000

fl/Y

j 0 f PV

;

f™T] 5 flT} fcpf];

IV

[|2373^75351

The accumulated value after five years is $123 737.75.

B. Present value of a general annuity due For a general annuity due, the present value is greater than the present value of the corresponding ordinary general annuity by the interest on it for one payment period. THE PRESENT VALUE OF A GENERAL ANNUITY DUE

THE PRESENT VALUE OF THE ~

^

CORRESPONDING ORDINARY GENERAL ANNUITY

Thus, for the present value of a general annuity due use Formula 13.4. Using the effective rate of interest per payment period,

A,,/due) = R(1 + p)

1" 1 - (1 + p)-"l P

restated as rvjauej — r.vi 1 yi i- p)

ri - (1 + 1s)~"l p

Formula 13.4

J

where p = (1 + z)c — 1

A three-year lease requires payments of $1600.00 at the beginning of every three months. If money is worth 9.0% compounded monthly, what is the cash value of the lease? PMT = 1600.00; n = 3(4) = 12; P/Y = 4; C/Y =12; c = ^ = 3; I/Y = 9; 9.0% 4 1 = = 0.75% = 0.0075 The effective quarterly rate of interest p = 1.00753 - 1 = 1.022669 - 1 = 0.022669 = 2.26692%

PART

3:

MATHEMATICS

OF

FINANCE

/I — 1.022669 12\ pV„c(due) = 1600.00(1.022669)^ q.022669 /

AND

INVESTMENT

substituting in

= 1600.00(1.022669) (10.404043) = 1600.00(10.639894) = $17 023.83 Programmed Solution (“BGN” mode) (Set P/Y = 4; C/Y = 12) 9

1600.00

I/V

0 [.FV j

[PMT] 12 HnI fcPT

17 023.83049

The cash value of the lease is $17 023.83.

C. Finding the periodic payment PMT of a general annuity due If the future value of an annuity FVnc(due), the number of conversion periods n, and the conversion rate i are known, you can find the periodic payment PM1 by substituting the given values in the future value formula. (1 + p)n - 1 FV„c(due) = PMT(1 + p)

P

Formula 13.3

where p = (1 + i)c — 1 If the present value of an annuity PVJdue), the number of conversion periods n, and the conversion rate i are known, you can find the periodic payment PMT by sub¬ stituting the given values in the present value formula.

PV„c(due) = PMT(1 + p)

ri

-

(i

+ P)~ni

p

- Formula 13.4

where p = (1 + i)c — 1 When using a scientific calculator, you can find PMT by first rearranging the terms of the appropriate formulas above. Then substitute the three known values (FVjdue), PVjdue), n, and i) into the rearranged formula and solve for PMT. When using a preprogrammed financial calculator, you can find PM 1 by entering the five known values (FV„f(due) or PV„c(due), N, I/Y, P/Y, and C/Y) and pressing [

CPT

j

PM T).

CHAPTER

13:

ANNUITIES

DUE,

DEFERRED

ANNUITIES,

AND

PERPETUITIES

What semi-annual payment must be made into a fund at the beginning of every six months to accumulate to $9600.00 in ten years at 7% compounded annually? FV„(due) = 9600.00; P/Y = 2; C/Y = 1; c = 1/2; I/Y = 7; i = 7%/l = 0.07; p = 1.071/2 - 1 = 0.034408 9600.00 = PMT( 1.034408)

[1.034408 20

_ 1]

substituting in Formula 13.3

0.034408

9600.00 = PMT( 1.034408) (28.108280) 9600.00 = PMT(29.075430) 9600.00 PMT =29.075430 PMT = $330.18

Programmed Solution (“BGN” mode) (Set P/Y = 2; C/Y =1)0 1 PV

9600 [ FV

7 ( I/Y } 20 (

N ) (CPT) [pMt) [-330.1755229

The semi-annual payment is $330.18. What deposit made at the beginning of each month will accumulate to $18 000.00 at 5% compounded quarterly at the end of eight years? FV (due) = 18 000.00; n = 8(12) = 96; P/Y = 12; C/Y = 4; c I/Y = 5; i =

5%

1 12

= 1.25% = 0.0125

The effective monthly rate of interest p = 1.0125 - - 1 = 1.004149 - 1 = 0.004149 = 0.4149% 18 000.00 = PMT(1.004149)(

1.004149s6 - r 0 0Mm

substituting in Formula 13.3

18 000.00 = PMT( 1.004149)(117.638106) 18 000.00 = PMT(118.126236) M 18 00.00 PMT = 118.126236 PMT = $152.38

Programmed Solution (“BGN” mode) (Set P/Y = 12; C/Y = 4) 5 !j7yJ o >[

18 000 [ FV ) 96 ( The monthly deposit is $152.38.

N

] [cPt] (pMt)

[-152.3793585

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

What monthly payment must be made at the beginning of each month on a fiveyear lease valued at $100 000.00 if interest is 10% compounded semi-annually?

2

PVjdue) = 100 000.00; n = 5(12) = 60; P/Y - 12; C/Y = 2; c = — I/Y - 10; i =

10%

1

= 5% = 0.05

The effective monthly rate of interest 0.008165 = 0.8165% 1 - 1.008165~6° 100 000.00 =PMT(1.008165) 0.008165

p = 1.05^ - 1 = 1.008165 - 1

substituting in Formula 13.4

100 000.00 = PMT( 1.008165) (47.286470) 100 000.00 = PMT(47.672557) PMT =

100 000.00 47.672557

PMT = $2097.64

Programmed Solution (“BGN”mode) (Set P/Y = 12; C/Y = 2) 10 [j/Yj 0 IJY 100 000 C±J

PV

] 60

N

CPT

PMT

[2097.642904)

The monthly payment due at the beginning of each month is $2097.64.

D. Finding the term n of a general annuity due If the future value of an annuity FVn(. (due), the periodic payment PMT, and the conversion rate i are known, you can find the term of the annuity n by substitut¬ ing the given values in the future value Formula 13.3.

FVjdue) = PMT(1 + p)

rd

+pY-

i]

p

Formula 13.3

where p = (1 + i)c — 1 If the present value PV„C (due), the periodic payment PMT, and the conversion rate i are known, you can find the term of the annuity n by substituting the given values in the present value Formula 13.4. 1 - (1 + pyn PVjdue) = PMT(1 + p) where p = (1 + i)c ~ 1

P

Formula 13.4

CHAPTER

13:

ANNUITIES

DUE,

DEFERRED

ANNUITIES,

AND

PERPETUITIES

When using a scientific calculator, you can find n by first rearranging the terms of the appropriate formula. Then substitute the three known values (FVnc(due) or PV)I PP

discount

$34.37

3%

3.5%

b < i

15.IB

$5 000.00

$5 487.79

PP > FV

premium

$487.79

5.25%

4.5%

b> i

15.1C

$10 600.00

$8 226.73

FV > PP

discount

$2 373.27

4%

5%

b < i

15.ID

$26 750.00

$27 270.74

PP > FV

premium

$520.74

2.75%

2.5%

b> i

15.IE

$1 000 000.00

$915 535.44

FV > PP

discount

$84 464.56

10%

11.46%

b < i

15.IF

$103 000.00

$98 222.68

FV > PP

discount

$4 777.32

3.75%

4.04%

b FV

premium

$41.26

5%

4.5%

b>i

15.1H

$5 000.00

$4 776.08

FV > PP

discount

$223.92

3.25%

3.75%

b < i

$10 800.00

$11 364.07

PP > FV

premium

$564.07

5.5%

5%

b > i

15.11

If the purchase price of a bond is greater than the redemption price, the bond is said to be bought at a premium and the difference between the purchase price and the redemption price is called the premium. PREMIUM = PURCHASE PRICE — REDEMPTION PRICE

where purchase price > redemption price If the purchase price of a bond is less than the redemption price, the bond is said to be bought at a discount and the difference between the redemption price and the purchase price is called the discount. DISCOUNT = REDEMPTION PRICE — PURCHASE PRICE

where redemption price > purchase price An examination of the size of the bond rate b relative to the size of the market rate i shows that this relationship determines whether there is a premium or discount. The bond rate (or coupon rate) stated on the bond is the percent of the face value of the bond that will be paid at the end of each interest period to the bond¬ holder. This rate is established at the time of issue of the bonds and remains the same throughout the term of the bond. On the other hand, the rate at which lenders are willing to provide money fluc¬ tuates in response to economic conditions. The combination of factors at work in the capital market at any given time in conjunction with the perceived risk associ¬ ated with a particular bond determines the yield rate (or market rate) for a bond and thus the price at which a bond will be bought or sold.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

The bond rate and the market rate are usually not equal. However, if the two rates happen to be equal, then bonds that are redeemable at par will sell at their face value. If the bond rate is less than the market rate, the bond will sell at a price less than the face value, that is, at a discount. If the bond rate is greater than the market rate, the bond will sell at a price above its face value, that is, at a premium. Conversely, if a bond is redeemable at par (that is, at 100), purchasers will real¬ ize the bond rate if they pay 100. They will realize less than the bond rate if they buy at a premium and more than the bond rate if they buy at a discount. At any time, one of three possible situations exists for any given bond: (1) Bond rate = Market rate {b — i)

The bond sells at par.

(2) Bond rate < Market rate (b < i)

The bond sells at a discount.

(3) Bond rate > Market rate (b > i)

The bond sells at a premium.

A $10 000 bond is redeemable at par and bears interest at 10% compounded semi-annually. (i) What is the purchase price ten years before maturity if the market rate com¬ pounded semi-annually is (a) 10%;

(b) 12%;

(c) 8%?

(ii) What is the purchase price five years before maturity if the market rate com¬ pounded semi-annually is (a) 10%;

(b) 12%;

(c) 8%?

(i) FV = 10 000.00; PMT = 10 000.00(0.05) = 500.00; n = 10(2) = 20; b = 5% 10% (a) i = —-— = 5% = 0.05;

(b = i)

( 1 - 1.05-20\ Purchase price = 10000.00(1.05 20) + 500.00( J = 10 000.00(0.376890) + 500.00(12.46221) - 3768.90 + 6231.11 = $10000.01

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2)

2nd j (CLRTVM) 10 000 (jvj 10 IjTW

20 (

FV

500

(PMT)

10

(J7y)

20 (

N ) fcPT] fpvj [-3768.894829]

N j jCPT] pPV~] f-6231.10517l]

The purchase price is 3768.89 + 6231.11 = $10 000.00. The bond sells at par.

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

12% (b) i =~^-= 6% = 0.06; Purchase price

(b < i)

= 10 000.00(1.06_2°) + 500.00

1 - 1.06 20

0.06

= 10 000.00(0.311805) + 500.00(11.469921) = 3118.05 + 5734.96 = $8853.01 Programmed Solution (“END”mode) (SetP/Y = 2;C/Y = 2) [ 2nd"] (CLRTVM) 10000 J;vj 12 20

0

( FV |

500

(PMT)

12

20

( I/Y )

(

'17Y

;

f~N~l [CPT j ( PV ) [-3118.047269]

N ) [CPT] (~PV~) [-5734.960609]

The purchase price is 3118.05 + 5734.96 = $8853.01. The bond sells below par. The discount is 10 000.00 — 8853.01 = $1146.99. (c) i = ~ = 4% = 0.04; Purchase price

(b > i)

= 10 000.00(1.04-20) + 500.00

1 - 1.04~20 0.04

= 10 000.00(0.456387) + 500.00(13.590326) = 4563.87 + 6795.16 = $11 359.03 Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) [2nd] (CLR TVM) 10 000 ( FV ] 8 ( I/Y ] 20 ( N ] [CPT] f py"] [-4561869462] 0 [ FV ] 500

[PMT] 8

[ I/Y ] 20 f~N~|

(CPT) [ PV

] [-6795.163172

The purchase price is 4563.87 + 6795.16 = $11 359.03. The bond sells above par. The premium is 11 359.03 — 10 000.00 = $1359.03. (ii) FV = 10 000.00;

PMT = 500.00;

n = 5(2) = 10;

b = 5%

(a) i = 5%; (b = i) Purchase price

= 10 000.00(1.05~10) + 500.00( = 6139.13 + 3860.87 = $10000.00

1 - 1.05 -10 0.05

PART 3: MATHEMATICS OF FINANCE AND INVESTMENT

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) 2nd

(CLR TVM) 10 000 IQ

FV

500

PMT

10( I/Y

10

N

( CPT

N

CPT

PV

IV

PV

10 fl/Y [-6139.132535

(^3860.867465)

The purchase price is 6139.13 + 3860.87 = $10000.00. The bond sells at par. (b) i = 6%;

(b < i) (\ - 1.06 10 = 10000.00(1.06 10) + 500.00( = 5583.95 + 3680.04 = $9263.99

Purchase price

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y =

2) [2nd

I (CLR TVM) IQ (

o ( FV

| 500

[pMt] 12

[7/y]: 10 (

10 000

FV

12! 1/Y

N ( (CPT[ [ PV } [-5583.947769

N ) (err) [ PV | (-3680.043526J

The purchase price is 5583.95 + 3680.04 = $9263.99. The bond sells below par. The discount is 10 000.00 - 9263.99 = $736.01. It is smaller than in part (i) because the time to maturity is shorter. (c) x — 4%;

(b > i)

Purchase price

= 10 000.00(1.04~10) + 500.001

1 - 1.04~10\ 0.04 )

- 6755.64 + 4055.45 = $10811.09 Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) (jndjl (CLRTVM) 10 000 JFV_, 8 ip[

p[ FV

j

500

(

PMT 8 I/Y

1

IQ

r N j ( CPT | [

I/Y

N~~| [ CPT| ( PV )i [-6755.641688]!

PV

j (■-4055.44789]

The purchase price is 6755.64 + 4055.45 = $10 811.09. The bond sells above par. The premium is 10811.09 - 10000.00 = $811.09. It is smaller than in part (i) because the time to maturity is shorter.

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

B. Direct method of computing the premium or discount— alternative method for finding the purchase price A $5000, 12% bond with semi-annual coupons is bought six years before matu¬ rity to yield 10% compounded semi-annually. Determine the premium. 12% FV = 5000.00; P/Y = 2; C/Y = 2; I/Y = 10; b = -y = 6% = 0.06; PMT = 5000.00(0.06) = 300.00;

10% i = —r— = 5% = 0.05;

n = 6(2) = 12

Since b > i, the bond will sell at a premium. Purchase price

1 - 1.05' 0.05

= 5000.00(1.05 12) + 300.00

= 5000.00(0.556837) + 300.00(8.863252) = 2784.19 + 2658.97 = $5443.16

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) [2nd] (CLR TVM) 5000 !

12 0 ( FV ) 300

[PMT]

10

( I/Y ] 12 (

N

CPT

FV~]

10

fTTF

( PV | [-2784.18709?]

N ] (CPT ] ( PV ) [-2658.975491

The purchase price is 2784.19 + 2658.97 = $5443.16. The premium is 5443.16 — 5000.00 = $443.16. While you can always determine the premium by the basic method using Formula 15.1, it is more convenient to determine the premium directly by con¬ sidering the relationship between the bond rate b and the yield rate i. As previously discussed, a premium results when b > i. When this is the case, the premium is paid because the periodic interest payments received exceed the periodic interest required according to the yield rate.

In Example 15.2B The semi-annual interest payment

5000.00(0.06) = $300.00

The required semi-annual interest based on the yield rate

5000.00(0.05) = $250.00

The excess of the actual interest received over the required interest to make the yield rate

= $ 50.00

This excess is received at the end of every payment interval; thus it forms an ordi¬ nary annuity whose present value can be computed at the yield rate i. PMT (the excess interest) = 50.00;

i = 5%;

n = 12

PART

PV„ = 50.00

1 - 1.05

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

-12

0.05

50.00(8.863252) $443.16

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) 0

FV

50

PMT

12 [

10

I/Y

N ) (cPT] ( PV ) (-443.1625818

The premium is $443.16. The purchase price is 5000.00 + 443.16 = $5443.16. The premium is the present value of the ordinary annuity formed by the excess of the actual bond interest over the required interest based on the yield rate. We can obtain the purchase price by adding the premium to the redemption price. 1

-

(1

PREMIUM = (PERIODIC BOND INTEREST - REQUIRED INTEREST)

—

(FACE VALUE X

b-

+ /)“"

i

ri - (1 + Q-"

REDEMPTION PRICE X

A $5000, 6% bond with semi-annual coupons is bought six years before maturi¬ ty to yield 8% compounded semi-annually. Determine the discount. FV = 5000.00; P/Y = 2; C/Y = 2;

b = -r~ = 3% = 0.03; 8%

PMT = 5000.00(0.03) - 150.00; I/Y = 8;

i=—=

40/0 = °-04;

n

= 6(2) = 12

Since b < i, the bond will sell at a discount. Purchase price

(\ - 1.04 ~12\ = 5000.00(1.04 u) + 150.00(--j = 5000.00(0.624597) + 150.00(9.385074) = 3122.99 + 1407.76 = $4530.75

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) [2nd] (CLR TVM) 5000 JVj 8 f I/Y ) 12 (

0 [ FV ) 150 [PMT] 8 [ I/Y ) 12 (

N ) [cPT

PV

N ) (CPT) j PV ) (-3122.985248)

E

The purchase price is 3122.99 + 1407.76 = $4530.75. The discount is 5000.00 - 4530.75 = $469.25.

1407.761064

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

As in the case of a premium, while you can always determine the discount by the basic method using Formula 15.1, it is more convenient to determine the discount directly. When b < i, a discount results. The discount on a bond is received because the periodic interest payments are less than the periodic interest required to earn the yield rate. In Example 15.2C The semi-annual interest payment

5000.00(0.03) = $150.00

The required semi-annual interest based on the yield rate

5000.00(0.04) = $200.00

The shortage of the actual interest received

=

$50.00

compared to the required interest based on the yield rate This shortage occurs at the end of every interest payment interval; it forms an ordinary annuity whose present value can be computed at the yield rate i. PMT = 50.00;

i = 4%;

n = 12

(\ - 1.04~12\ pv„= 50.00(—) = 50.00(9.385074) = $469.25

Programmed Solution (“END” mode) (Set PA2 = 2; C/Y = 2) 0

50 (pmt) g [ I/Y ) 12 (

N I (CPT| ( PV I f-469.253688}

The discount is $469.25. The purchase price is 5000.00 — 469.25 = $4530.75. The discount is the present value of the ordinary annuity formed by the shortage of the actual bond interest received as compared to the required interest based on the yield rate. We can obtain the purchase price by subtracting the discount from the redemption price.

~ 1 - (1 + t)~" DISCOUNT = (REQUIRED INTEREST - PERIODIC BOND INTEREST)

i

1 - (1 + i)~n = -(PERIODIC BOND INTEREST — REQUIRED INTEREST)

= — (FACE VALUE

X b —

REDEMPTION PRICE

X l)

1

-

(1

+

i)~n

Since in both cases the difference between the periodic bond interest and the required interest is involved, the premium or discount on the purchase of a bond can be obtained using the same relationship.

PART

3:

MATHEMATICS

PREMIUM

or

OF

1

[b X FACE VALUE - 2 X REDEMPTION PRICE)

—

FINANCE

(1

AND

INVESTMENT

+

Formula 15.3

2

DISCOUNT

A $1000, 8.5% bond with semi-annual coupons redeemable at par in fifteen years is bought to yield 7% compounded semi-annually. Determine (i) the premium or discount; (ii) the purchase price. 8.5% (l) FV = 1000.00; P/Y = 2; C/Y = 2; b = ~^r~ = 4.25% = 0.0425; , 7% PMT = 1000.00(0.0425) = 42.50; I/Y = 7; i = — = 3.5% = 0.035; n = 15(2) - 30 Since b > i, the bond will sell at a premium. The required interest based on the yield rate is 1000.00(0.035) = 35.00; the excess interest is 42.50 — 35.00 = 7.50. /I - 1 035“30\ The premium is 7.50(-()-(^r:—j = 7.50(18.392045) = $137.94.

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) 0

FV

30

7.50

PMT

fN

[ CPT

7

I/Y

( PV

[-137.9403406

(ii) The purchase price is 1000.00 + 137.94 = $1137.94. A $50 000, 10% bond with quarterly coupons redeemable at par in ten years is purchased to yield 11% compounded quarterly. (i) What is the premium or discount? (ii) What is the purchase price? (i) FV = 50 000.00; P/Y = 4; C/Y = 4; b = n = 10(4) = 40; I/Y =11; i

11%

10%

= 2.5% = 0.025;

= 2.75% = 0.0275

Since b < i, the bond will sell at a discount. Discount = (0.025 X 50 000.00 - 0.0275 X 50 000.00)

1_

using Formula 15.3

= (1250.00 - 1375.00)(24.078101) = ( — 125.00X24.078101) = —$3009.76

1 - 1.0275 0.0275

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Programmed Solution (“END” mode) First compute PMT = (0.025 X 50 000 - 0.0275 X 50000) = -125.00 (Set P/Y = 4; C/Y = 4) 0 (jFv) 125 [pmt] 11 ( I/Y ) 40 (

N ) (CPT) ( PV ) [-3009.762633]

(ii) The purchase price is 50 000.00 — 3009.76 = $46 990.24. Bonds with a face value of $15 000 redeemable at 108 with interest at 9% payable semi-annually are bought twelve years before maturity to yield 11% compound¬ ed semi-annually. (i) What is the premium or discount? (ii) What is the purchase price? (i) FV = 15 000.00(1.08) = 16 200.00; P/Y = 2; C/Y = 2; 9% b = i =

2 11%

= 4.5% = 0.045; n = 12(2) = 24; I/Y = 11; = 5.5% = 0.055

Since b < i and the redemption price is greater than par, the bond will sell at a discount. Discount - (0.045 X 15 000.00 - 0.055 X 16 200.00)1

1 - 1 055-24' 0.055

= (675.00 - 891.00)(13.151699) = (-216.00X13.151699) = -$2840.77

Programmed Solution (“END” mode) PMT = (0.045 X 15 000 - 0.055 X 16 200) = -216.00 (Set P/Y = 2;C/Y = 2) 0

FV

216

I PMT;

l l

I/Y

24

N

[ CPT) [ PV ] [-2840.766974]

(ii) The purchase price is 16 200.00 - 2840.77 = $13 359.23. A $10 000, 8% bond with quarterly coupons redeemable at 106 in seven years is purchased to yield 6% compounded quarterly. (i) What is the premium or discount? (ii) What is the purchase price?

8% (i) FV= 10000.00(1.06) = 10 600.00; P/Y = 4; C/Y = 4; I/Y = 6; b = —p = 2% = 0.02;

6% 4

n = 7(4)= 28; 6 i =- = 1.5% = 0.015

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENTS

Since b > i, the bond is expected to sell at a premium. Premium

(0.02 X 10 000.00

0.015 X 10 600.00)

/I - 0.015~28\

V

0.015

)

= (200.00 - 159.00)(22.726717) = (41.00)(22.726717) = $931.80

Programmed Solution (“END” mode) PMT = (0.02 X 10 000 - 0.015 X 10 600) = 200.00 - 159.00 = 41.00 (Set P/Y = 4; C/Y = 4) 0

f

FV

j 41

I’M I

: 6 [m 28 (

N ) (CPT) ( PV ) [-931.795385

(ii) The purchase price is 10 600.00 + 931.80 = $11 531.80. A $1000, 7% bond redeemable at 110 with interest payable annually is bought nine years before maturity to yield 6.5% compounded annually. Determine (i) the premium or discount; (ii) the purchase price. (i) FV = 1000.00(1.10) = 1100.00; P/Y - 1; C/Y = 1; b = 7% - 0.07; n = 9; 1 !Y= 6.5; i = 6.5% = 0.065 Since b > i, the bond is expected to be sold at a premium. While this is always true for bonds redeemable at par, it does not necessarily follow for bonds redeemable above par. In this particular case: the actual bond interest per year

1000.00(0.07)

=$70.00

the interest required to make the yield rate

1100.00(0.065) = $71.50

Because of the redemption premium, the required interest exceeds the actu¬ al interest: the bond will, in fact, sell at a discount. This conclusion is borne out when using Formula 15.3. Premium/Discount = (0.07 X 1000.00 — 0.065 X 1100.00)

/ 1 ~ 1.065~9 \ V

0.065

= (70.00 - 71.50)(6.656104) = (-1.50)(6.656104) = -$9.98

Programmed Solution (“END” mode) PMT = (0.07 X 1000 - 0.065 X 1100) = -1.50 (Set P/Y = 1; C/Y = 1) 0 f FV~) 1.50 >MT 6.5 9[

1/Y , N

;

CPT

PV

[—9.984156281

)

CHAPTER 15: BOND VALUATION AND SINKING FUNDS

(ii) Since the answer (PMT) is negative, the bond sells at a discount of $9.98. The purchase price is 1100.00 — 9.98 = $1090.02. A $25 000 bond, redeemable at 104 on July 1, 2015 with 11% coupons payable quarterly, is bought on May 20, 2006 to yield 10% compounded quarterly. What is (i) the premium or discount? (ii) the purchase price? (iii) the quoted price? FV = 25 000.00(1.04) = 26 000.00; P/Y = 4; C/Y = 4; 1/Y = 10

11%

10%

b = —= 2.75% = 0.0275; 1/Y = 10; i = —p = 2.5% = 0.025 The interest payment dates are October 1, January 1, April 1, and July 1. The inter¬ est payment date preceding the date of purchase is April 1, 2006. The time period April 1, 2006 to July 1, 2015 contains 9 years and 3 months; n = 9.25(4) = 37. The premium on April 1, 2006, / 1 — 1 025“37 = (25 000.00 X 0.0275 - 26 000.00 X 0.025)(-pp= (687.50 - 650.00) (23.957318) = (37.50)(23.957318) = $898.40

Programmed Solution (“END” mode) PMT = (25 000 X 0.0275 - 26 000 X 0.025) - 37.50 (SetP/Y = 4;C/Y = 4) 0 ( Fv j 37.50 (PMTj |0

l/Y 37 [ N

| [CPT ] ( PV J f-898.3994293]

The purchase price on April 1, 2006 is 26 000.00 + 898.40 = $26 898.40. The time period April 1 to May 20 contains 49 days; the number of days in the interest payment interval April 1 to July 1 is 91. PV = 26 898.40;

i = 0.025;

49 t = p

The accumulated value (flat price) on May 20, 2006, 26 898.40 1 + 0.025

49 Y 91 J

= 26 898.40(1.013462) = $27 260.49 The accrued interest to May 20 is 25 000.00(0.0275)

49 91

The quoted price is 27 260.49 - 370.19 = $26 890.30. Thus, on May 20, 2006,

$370.19.

PART

3:

MATHEMATICS

OF

FINANCE

AND

HI

INVESTMENT TlVpF

(i) the premium is 26 890.30 — 26 000.00 = $890.30; (ii) the purchase price is $27 260.49; (iii) the quoted price is $26 890.30.

EXERCISE 15.2 If you choose, you can use these Excel functions to answer the questions indicated

Excel

below: Bond Purchase Price (PRICE), Days Since Last Interest Date (COUPDAYBS), Total Number of Days in the Coupon Period (COUPDAYS), and Total Number of Remaining Coupon Periods (COUPNUM). Refer to PRICE,

COUPDAYBS, COUPDAYS, and COUPNUM on the Spreadsheet Template Disk to learn how to use these Excel functions. For each of the six bonds in the table below, use Formula 15.3 to determine (a) the premium or discount;

(b) the purchase price. Bond Rate Payable Semi¬ annually

Par Value

Redeemed At:

$25 000

par

6%

2.

$5 000

par

8.5%

3.

$10 000

104

9%

4.

$7 000

110

9%

5.

$1 000

105

6.5%

6.

$50 000

108

1.

Time Before Redemption

Yield Rate Compounded Semi-annually 9%

10 years 8 years

7%

15 years

11%

5 years

8.5%

6 years, 10 months

12%

4 years, 5 months

8% 11.5%

Answer each of the following questions. 1. A $100 000, 8% bond redeemable at par with quarterly coupons is purchased to yield 6.5% compounded quarterly. Find the premium or discount and the purchase price if the bond is purchased (a) fifteen years before maturity;

(b) five years before maturity.

2. A $5000, 7.5% bond redeemable at 104 with semi-annual coupons is pur¬ chased to yield 6% compounded semi-annually. What is the premium or dis¬ count and the purchase price if the bond is bought (a) ten years before maturity?

(b) six years before maturity?

3. A $25 000, 9% bond redeemable at par with interest payable annually is bought six years before maturity. Determine the premium or discount and the purchase price if the bond is purchased to yield (a) 13.5% compounded annually;

(b) 6% compounded annually.

4. A $1000, 8% bond redeemable at 108 in seven years bears coupons payable

annually. Compute the premium or discount and the purchase price if the yield, compounded annually, is (a) 6.5%;

(b) 7.5%;

(c) 8.5%.

CHAPTER

15:

BOND

VALUATION

4

AND

SINKING

FUNDS

5. Twelve $1000 bonds redeemable at par bearing interest at 10% payable semi¬ annually and maturing on September 1, 2006 are bought on June 18, 2001 to yield 7% compounded semi-annually. Determine (a) the premium or discount on the preceding interest payment date; (b) the purchase price;

4

(c) the quoted price.

6. Bonds with a face value of $30 000 redeemable at 107 on June 1, 2012 are offered for sale to yield 9.2% compounded quarterly. If interest is 7% payable quarterly and the bonds are bought on January 24, 2003, what is (a) the premium or discount on the interest payment date preceding the date of sale? (b) the purchase price? (c) the quoted price? 7. A $5 000 000 issue of ten-year bonds redeemable at par offers 7.25% coupons payable semi-annually. What is the issue price of the bonds to yield 8.4% compounded monthly? 8. A $3000 issue of nine-year bonds redeemable at 107 offers 7.5% coupons paid semi-annually. What is the issue price of the bonds to yield 10.5% semi-annually?

9.

Twenty $5000 bonds redeemable at 110 bearing 12% coupons payable quar¬ terly are sold eight years before maturity to yield 11.5% compounded annu¬ ally. What is the purchase price of the bonds?

10. Sixty

$1000 bonds redeemable at 108 bearing 7% coupons payable semi¬

annually are sold seven years before maturity to yield 9.5% compounded semi-annually. What is the purchase price of the bonds?

The Price of Liquidity Investors now have a choice when they invest in the Government of Canada. The traditional Canada Savings Bonds that many Canadians have invested in for many years have been complemented by a new series of bonds called Canada Premium Bonds. The Government of Canada guarantees the principal and interest on both series of bonds. The difference lies in the flexibility to withdraw your money for other investments or uses. Canada Savings Bonds are cashable at any time, with interest being paid up to the first day of the month in which redemption takes place. Canada Premium Bonds, however, are cashable without penalty only on the anniversary of their issue date or during the 30 days thereafter. Canada Premium Bonds, however, carry a higher rate of interest. The interest rate is fixed and interest is compounded if the Canada Premium Bonds are held for more than one year. Canada Savings Bonds have a lower minimum guaranteed interest rate than Canada Premium Bonds for series on sale at the same time with the same issue date, and interest is also com¬ pounded if Canada Savings Bonds are held for more than one year. Both Canada Premium Bonds and Canada Savings Bonds are available in Regular, non-compounding-interest versions as well to which inter¬ est is paid annually on the anniversary of issue.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

QUESTIONS

1. Assume the annual rate of interest on Canada Savings Bonds is set at 2.00% on April 1, 2004, and 2.50% on April 1, 2005, and 3.00% on April, 2004. What is the accumulated value of a $1000 Canada Savings Bond purchased on April 1, 2004, on April 1, 2007? 2. Assume the annual rate of interest on Canada Premium Bonds is set at 2.50% on April 1, 2004, and

3.00% on April 1,2005, and 3.50% on April 1, 2006. What is the accumulated value of a $1000 Canada Premium Bond purchased on April 1, 2004, on April 1, 2007? 3. What is the rate of interest compounded annually for a Canada Premium Bond purchased on April 1, 2004, on April 1, 2007? 4. Why would an investor purchase Canada Savings Bonds instead of Canada Premium Bonds? Why

would an investor purchase Canada Premium Bonds instead of Canada Savings Bonds?

15.3 Bond Schedules A. Amortization of premium If a bond is purchased for more than the redemption price, the resulting premium is not recovered when the bond is redeemed at maturity, so it becomes a capital loss. To avoid the capital loss at maturity, the premium is written down gradually over the period from the date of purchase to the maturity date. The writing down of the premium gradually reduces the bond’s book value until it equals the redemption price at the date of maturity. The process of writing down the premium is called amortization of the

premium. The most direct method of amortizing a premium assigns the difference between the interest received (coupon) and the interest required according to the yield rate to write down the premium. The details of writing down the premium are often shown in a tabulation referred to as a schedule of amortization of premium. A $1000, 12% bond redeemable at par matures in three years. The coupons are payable semi-annually and the bond is bought to yield 10% compounded semi¬ annually. (i) Compute the purchase price. (ii) Construct a schedule of amortization of premium. (i) FV = 1000.00; b =

12% 2

n = 3(2) = 6;

6% - 0.06;

P/Y = 2;

I/Y =10;

C/Y = 2;

10%

i = —= 5% = 0.05

Since b > i, the bond sells at a premium. /1 - 1.05'6 Premium = (0.06 X 1000.00 - 0.05 X 1000.00)1= (60.00 - 50.00)(5.075692) = (10.00)(5.075692) = $50.76

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Programmed Solution PMT = (0.06 X 1000.00 - 0.05 X 1000.00) = 10.00 (“END” mode) (Set P/Y = 2; C/Y = 2) 0 [

FV 1 ]()

PMT

|()

I/Y

6 f N ) (CPT) ( PV ) (-50.75692067^

The purchase price is 1000.00 + 50.76 = $1050.76. (ii) Schedule of amortization of premium End of Interest Payment Interval

Bond Interest Interest on Received Book Value (Coupon) at Yield Rate b = 6% i = 5%

Amount of Premium Amortized

0

Book Value of Bond

Remaining Premium

1050.76

50.76

1

60.00

52.54

7.46

1043.30

43.30

2

60.00

52.17

7.83

1035.47

35.47

3

60.00

51.77

8.23

1027.24

27.24

4

60.00

51.36

8.64

1018.60

18.60

5

60.00

50.93

9.07

1009.53

9.53

6

60.00

50.47

9.53

1000.00

0.00

360.00

309.24

50.76

TOTAL

Explanations of schedule 1. The original book value shown is the purchase price of $1050.76. 2. At the end of the first interest payment interval, the interest received (coupon) is 1000.00(0.06) = $60.00; the interest required according to the yield rate is 1050.76(0.05) = $52.54; the difference 60.00 — 52.54 = 7.46 is used to write down the premium to $43.30 and reduces the book value from $1050.76 to $1043.30. 3. The coupon at the end of the second interest payment interval is again $60.00. The interest required according to the yield rate is 1043.30(0.05) = $52.17; the difference 60.00 — 52.17 = 7.83 reduces the premium to $35.47 and the book value of the bond to $1035.47. 4. Continue in a similar manner until the maturity date when the redemption price is reached. If a rounding error becomes apparent at the end of the final interest payment interval, adjust the final interest on the book value at the yield rate to make the premium zero and to obtain the exact redemption price as the book value. 5. The totals provide useful accounting information showing the total interest received ($360.00) and the net income realized ($309.24).

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

A $25 000, 7.5% bond redeemable at 106 with coupons payable annually matures in seven years. The bond is bought to yield 6% compounded annually. (i) Compute the premium and the purchase price. (ii) Construct a schedule of amortization of premium. (i) FV = 25 000.00(1.06) = 26 500.00; n = 7; P/Y = 1; C/Y = 1; b = 7.5% = 0.075; 1/Y = 6; z = 6% = 0.06 Since b > i, the bond is expected to sell at a premium. 1 - 1.06~7\ Premium = (0.075 X 25 000.00 - 0.06 X 26 500.00)(

0.06

)

= (1875.00 - 1590.00)(5.582381) = (285.00)(5.582381) = $1590.98

Programmed Solution PMT = (0.075 X 25 000 - 0.06 X 26 500) - 285 (“END” mode) (Set P/Y - 1; C/Y - 1)0 (jv j 285

PMT

6 j 1/Y

7

(~N~] (CPT) f~PV~) [-1590.97871]

The purchase price is 26 500.00 + 1590.98 = $28 090.98. (ii) Schedule of amortization of premium End of Interest Payment Interval

Coupon b = 7.5%

Interest on Book Value at Yield Rate / = 6%

Amount of Premium Amortized

0

Book Value of Bond

Remaining Premium

28 090.98

1590.98

27 901.44

1401.44

1

1 875.00

1 685.46

189.54

2

1 875.00

1 674.09

200.91

27 700.53

1200.53

3

1 875.00

1 662.03

212.97

27 487.56

987.56

4

1 875.00

1 649.25

225.75

27 261.81

761.81

5

1 875.00

1 635.71

239.29

27 022.52

522.52

6 7

1 875.00

1 621.35

253.65

26 768.87

268.87

1 875.00

1 606.13

268.87

26 500.00

0.00

TOTAL

13 125.00

11 534.02

1590.98

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

B. Accumulation of discount If a bond is bought at less than the redemption price, there will be a gain at the time of redemption equal to the amount of discount. It is generally accepted accounting practice that this gain does not accrue in total to the accounting peri¬ od in which the bond is redeemed. Instead, some of the gain accrues to each of the accounting periods from the date of purchase to the date of redemption. To adhere to this practice, the discount is decreased gradually so that the book value of the bond increases gradually until, at the date of redemption, the discount is reduced to zero while the book value equals the redemption price. The process of reducing the discount so as to increase the book value is called accumulation of discount. In the case of discount, the interest required according to the yield rate is greater than the actual interest received (the coupon). Similar to amortization of a premium, the most direct method of accumulating a discount assigns the difference between the interest required by the yield rate and the coupon to reduce the dis¬ count. The details of decreasing the discount while increasing the book value of a bond are often shown in a tabulation called a schedule of accumulation of discount. A $10 000 bond, redeemable at par in four years with 5.5% coupons payable semi-annually, is bought to yield 7% compounded semi-annually. (i) Determine the discount and the purchase price. (ii) Construct a schedule of accumulation of discount. (i) FV = 10 000.00; n = 4(2) = 8; P/Y = 2; C/Y = 2; , 5.5% _ 7% b = —f~ = 2-75% = 0.0275; I/Y = 7; i = — = 3.5% = 0.035 Since b < i, the bond sells at a discount. Discount = (0.0275 X 10 000.00 - 0.035 X 10 000.00)

1 - 1.035" 0.035

= (275.00 - 350.00)(6.873956) = ( — 75.00) (6.873956) = -$515.55

Programmed Solution PMT = (0.0275X 10 000 - 0.035 X 10 000) = -75.00 (“END” mode) (Set P/Y = 2; C/Y = 2) 0 (FV

75

j PMT

7

I/Y 8 ( N j (cPT) ( PV ) [-515,5466653

The purchase price is 10 000.00 - 515.55 = $9484.45.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

(ii) Schedule of accumulation of discount

Coupon b = 2.75%

Interest on Book Value at Yield Rate l- 3.5%

Amount of Discount Accumulated

Book Value of Bond

Remaining Discount

1

275.00

331.96

56.96

9484.55 9541.41

458.59

2

275.00

333.95

58.95

9600.36

399.64

3

275.00

336.01

61.01

9661.37

338.63

4

275.00

338.15

63.15

9724.52

275.48

5

275.00

340.36

65.36

9789.88

210.12

6

275.00

342.65

67.65

9857.53

142.47

7

275.00

345.01

70.01

9927.54

72.46

10 000.00

0.00

End of Interest Payment Interval 0

8

275.00

347.46

72.46

TOTAL

2200.00

2715.55

515.55

515.55

Explanations of schedule 1. The original book value shown is the purchase price of $9484.55. 2. At the end of the first interest payment interval, the coupon is 10 000.00(0.0275) = $275.00; the interest required according to the yield rate is 9484.55(0.035) = $331.96; the difference used to reduce the discount and to increase the book value is 331.96 — 275.00 = $56.96; the book value is 9484.55+ 56.96 = $9541.41; and the remaining discount is 515.55 — 56.96 = $458.59. 3. The coupon at the end of the second interest payment interval is again $275.00; the interest required on the book value is 9541.41(0.035) = $333.95; the difference is 333.95 - 275.00 = $58.95; the book value is 9541.41 + 58.95 = $9600.36; and the remaining discount is 458.59 — 58.95 = $399.64. 4. Continue in a similar manner until the maturity date when the redemption price is reached. If a rounding error becomes apparent at the end of the final interest payment interval, adjust the final interest on the book value at the yield rate to make the remaining discount equal to zero and obtain the exact redemption price as the book value. 5. The totals provide useful accounting information showing the total interest received ($2200.00) and the net income realized ($2715.55). A $5000, 10% bond redeemable at 102 on April 1, 2008 with coupons payable quarterly is bought on October 1, 2006 to yield 13% compounded quarterly. (i) Compute the discount and the purchase price. (ii) Construct a schedule showing the accumulation of the discount. (i) FV = 5000.00(1.02) = 5100.00; P/Y - 4; C/Y = 4; the time period October 1, 2006 to April 1,2008 contains 18 months

CHAPTER

5:

BOND

VALUATION

AND

SINKING

FUNDS

n = —r~ = 6 (quarters); , 10% b = —— = 2.5% - 0.025; 4

13% i = —- = 3.25% = 0.0325 4

Since b < i, the bond sells at a discount. (1 - l 0325-6 Discount - (*0.025 X 5000.00 - 0.0325 X 5100.00)(-= (125.00 - 165.75X5.372590) = ( — 40.75) (5.372590) = -$218.93

Programmed Solution PMT = (0.025X 5000 - 0.0325 X 5100) = -40.75 (“END” mode) (Set P/Y = 4; C/Y = 4) 0 (tv] 40.75 [pmt] 13 fl/Y

6

LiL 1 (CPT j [ pV ) !

-218.93304 ]

The purchase price is 5100.00 — 218.93 = $4881.07. (ii) Schedule of accumulation of discount End of Interest Payment Interval

Coupon b = 2.5%

Interest on Book Value at Yield Rate / = 3.25%

Amount of Discount Accumulated

Oct. 1, 2006

Book Value of Remaining Bond Discount 4881.07

218.93

Jan. 1, 2007

125.00

158.63

33.63

4914.70

185.30

Apr. 1, 2007

125.00

159.73

34.73

4949.43

150.57

July 1, 2007

125.00

160.86

35.86

4985.29

114.71

Oct. 1, 2007

125.00

162.02

37.02

5022.31

77.69

Jan. 1, 2008

125.00

163.23

38.23

5060.54

39.46

Apr. 1, 2008

125.00

164.46

39.46

5100.00

0.00

750.00

968.93

218.93

TOTAL

C. Book value of a bond—finding the gain or loss on the sale of a bond A $10 000, 8% bond redeemable at par with semi-annual coupons was purchased fifteen years before maturity to yield 6% compounded semi-annually. The bond was sold three years later at 10114. Find the gain or loss on the sale of the bond. The market quotation of 10114 indicates that the bond was sold at 101.25% of its face value. The proceeds from the sale of the bond are 10 000.00(1.0125)

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

= $10 125.00. To find the gain or loss on the sale of the bond, we need to know the book value of the bond at the date of sale. This we can do by determining the orig¬ inal purchase price, constructing a bond schedule, and reading the book value at the time of sale from the schedule.

FV= 10 000.00;

n = 15(2) = 30

P/Y = 2; C/Y = 2;

I/Y = 6;

8% b = ~Y =

.

4% = 0.04;

i

6% = -y = 3% = 0.03

Since b > i, the bond was bought at a premium. Premium = (0.04

X

(\ - 1.03“30\ 10 000.00 - 0.03 X 10 000.00)1-^-1

= (400.00 - 300.00)(19.600441) = (100.00) (19.600441) = $1960.04 Programmed Solution PMT = (0.04 X 10 000 - 0.03 X 10 000) = 100.00 (“END” mode) (SetP/Y = 2; C/Y = 2) 0 f^v"1 100 (pMT) 6 (T/y] 30 [n] ( CPT i

PV j 0960.044135)

The purchase price is 10 000.00 + 1960.04 = $11 960.04. Schedule of amortization of premium End of Interest Payment Interval

Coupon b = 4%

Interest on Book Value at Yield Rate /= 3%

Amount of Premium Amortized

Book Value of Bond

Remaining Premium

11 960.04

1960.04

0 1

400.00

358.80

41.20

11 918.84

1918.84

2

400.00

357.57

42.43

11 876.41

1876.41 1832.70

3

400.00

356.29

43.71

11 832.70

4

400.00

354.98

45.02

11 787.68

1787.68 1741.31 1693.55

5 6

400.00

353.63

46.37

11 741.31

400.00

352.24

47.76

11 693.55

etc. The book value after three years (six semi-annual periods) is $11 693.55. Since the book value is greater than the proceeds, the loss on the sale of the bond is 11 693.55 - 10 125.00 = $1568.55. We can solve this problem more quickly by computing the book value directly. The book value of a bond at a given time is the purchase price of the bond on that date. We can determine the book value of a bond without constructing a bond schedule by using Formula 15.1 or 15.3. This approach can also be used to verify book values in a bond schedule.

VALUATION

AND

SINKING

FV = 10 000.00;

FUNDS

n = (15 - 3)(2) = 24;

b = 4%;

i = 3%

1 - 1.03“24

Premium = (0.04 X 10 000.00 - 0.03 X 10 000.00)

0.03

= (100.00) (16.935542) = $1693.55

Programmed Solution PMT = (0.04 X 10000 - 0.03 X 10 000) = 100.00 (“END” mode) (Set P/Y = 2; C/Y = 2) 0 ( FV ) 100 [pmt] 6 ( I/Y } 24

N

CPT

PV

Jfc 1693.55421213

The purchase price is 10 000.00 + 1693.55 = $11 693.55. The loss on the sale is 11 693.55 — 10 125.00 = $1568.55. A $5000,11% bond redeemable at 106 with semi-annual coupons was purchased twelve years before maturity to yield 10.5% compounded semi-annually. The bond is sold five years later at 98 7/s. Find the gain or loss on the sale of the bond. Proceeds from the sale of the bond are 5000.00(0.98875) = $4943.75. FV = 5000.00(1.06) = 5300.00; n = (12 - 5)(2) = 14; P/Y = 2; C/Y = 2; b =

= 5.5% = 0.055; I/Y = 10.5; i =

= 5.25% = 0.0525 / l — i

Premium/Discount = (0.055 X 5000.00 - 0.0525 X 5300.00)

14

0.0525

= (275.00 - 278.25)(9.742301)

= ( — 3.25X9.742301) = —$31.66

discount

Programmed Solution PMT = (0.055 X 5000 - 0.0525 X 5300) = -3.25 (“END” mode) (Set P/Y = 2; C/Y = 2) 0

f FV

] 3.25 [pmt| 10.5 [ I/Y 14

QD i CPT1 HpvI

The purchase price or book value is 5300.00 — 31.66 = $5268.34. The loss on the sale is 5268.34 — 4943.75 = $324.59.

[-31.66247742)

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

A $1000, 10% bond with quarterly coupons redeemable at 104 on May 1, 2012 was purchased on August 1, 2002 to yield 12% compounded quarterly. If the bond is sold at 95 '/2 on December 11, 2005, what is the gain or loss on the sale of the bond? The interest payment dates are August 1, November 1, February 1, and May 1. The interest date preceding the date of sale is November 1, 2005. The proceeds from the sale of the bond on December 11, 2005, = 1000.00(0.955) + accrued interest from November 1 to December 11 = 955.00 + 1000.00(0.025)(||j = 955.00 + 10.87 = $965.87 FV = 1000.00(1.04) = 1040.00; P/Y = 4; C/Y = 4; b =

10%

= 2.5% = 0.025; I/Y =12; i =

12%

= 3%

= 0.03 The time interval November 1, 2005 to May 1, 2012 contains six years and six months: n = 6.5(4) = 26. Since b < i, the bond will sell at a discount. Discount = (0.025 X 1000.00 - 0.03 X 1040.00) = (25.00 - 31.20)(17.876842) = (—6.20) (17.876842) = -$110.84

Programmed Solution PMT = (0.025 X 1000 - 0.03 X 1040) = -6.20 (“END” mode) (Set P/Y = 4; C/Y = 4) 0

FV

6.20

PMT

\2 JAY 26 Qn

CPT

PV

110.836423

The purchase price on November 1, 2005 is 1040.00 — 110.84 — $929.16. The accumulated value on December 11,2005, = 929.16 1 + 0.03

40 92

= 929.16(1.0130435) = $941.28 The gain from the sale of the bond is 965.87 — 941.28 = $24.59.

[yjjv CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

EXERCISE 15.3 If you choose, you can use these Excel functions to answer the questions indicated

Excel

below: Bond Purchase Price (PRICE), Days Since Last Interest Date (COUPDAYBS), Total Number of Days in the Coupon Period (COUPDAYS), and Total Number of Remaining Coupon Periods (COUPNUM). Refer to PRICE, COUP-

DAYBS, COUPDAYS, and COUPNUM on the Spreadsheet Template Disk to learn how to use these Excel functions. For each of the following bonds, compute the premium or discount and the pur¬ chase price, and construct the appropriate bond schedule. 1. A $5000, 6% bond redeemable at par in three-and-a-half years with semi¬ annual coupons is purchased to yield 6.5% compounded semi-annually. 2. A $25 000 bond with interest at 12.5% payable quarterly redeemable at par is bought two years before maturity to yield 11% compounded quarterly. 3. A $1000, 12% bond with semi-annual coupons redeemable at 103 on September 1, 2005 is bought on March 1, 2002 to yield 10% compounded semi-annually. 4. A $10 000, 7.75% bond with annual coupons redeemable at 110 in seven years is bought to yield 7.25% compounded annually. Find the gain or loss on the sale of each of the following bonds without construct¬ ing a bond schedule. 1. A $25 000, 10.5% bond redeemable at par with semi-annual coupons bought ten years before maturity to yield 12% compounded semi¬ annually is sold four years before maturity at 99‘A. 2. Four $5000,8.5% bonds with interest payable semi-annually redeemable at par were bought twenty years before maturity to yield 7.5% compounded semi¬ annually. The bonds were sold three years later at 103%. 3. Seven $1000, 9.25% bonds with annual coupons redeemable at 107 were bought nine years before maturity to yield 8.25% compounded annually. The bonds are sold three years before maturity at 94A. 4. A $100 000, 7% bond with semi-annual coupons redeemable at 102 was pur¬ chased eleven-and-a-half years before maturity to yield 6% compounded semi¬ annually. The bond was sold five years later at 99'/s. 5. A $5000 bond with 8% interest payable semi-annually redeemable at par on June 1, 2012 was bought on December 1,1998 to yield 9% compounded semi¬ annually. The bond was sold on September 22, 2002 at 101%. 6. Three $10 000, 10.5% bonds with quarterly coupons redeemable at 109 on August 1, 2009 were bought on May 1, 1995 to yield 12% compounded quar¬ terly. The bonds were sold on January 16, 2003 at 93 A.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

15.4 Finding the Yield Rate A. Quoted price of a bond—buying bonds on the market Bonds are usually bought or sold through a bond exchange where agents trade bonds on behalf of their clients. To allow for the different denominations, bonds are offered at a quoted price stated as a percent of their face value. It is understood that, if the bond is bought between interest dates, such a quot¬ ed price does not include any accrued interest. As explained in Section 15.1, the seller of a bond is entitled to the interest earned by the bond to the date of sale and the interest is added to the quoted price to obtain the purchase price (flat price). A $5000, 8% bond with semi-annual coupons payable April 1 and October 1 is purchased on August 25 at

1043/4.

What is the purchase price of the bond?

The quoted price is 5000.00(1.0475) = $5237.50. The time period April 1 to August 25 contains 146 days; the number of days in the interest payment interval April 1 to October 1 is 183. 8% PV = 5000.00;

4% = 0.04;

t

146

183 146 The accrued interest is 5000.00(0.04)y^“ = $159.56. The purchase price (flat price) is 5237.50 + 159.56 = $5397.06.

B. Finding the yield rate—the average investment method When bonds are bought on the market, the yield rate is not directly available; it needs to be determined. The simplest method in use is the so-called method of averages, which gives a reasonable approximation of the yield rate as the ratio of the average income per interest payment interval to the average book value.

AVERAGE INCOME PER INTEREST PAYMENT INTERVAL APPROXIMATE VALUE OF i

AVERAGE BOOK VALUE

where = — (QUOTED PRICE + REDEMPTION PRICE)

AVERAGE BOOK VALUE

and

AVERAGE INCOME PER INTEREST PAYMENT INTERVAL

TOTAL INTEREST PAYMENTS =

— PREMIUM , + DISCOUNT

NUMBER OF INTEREST PAYMENT INTERVALS

Formula 15.4

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

A $25 000, 7.5% bond with semi-annual coupons redeemable at par in ten years is purchased at 103V2. What is the approximate yield rate? The quoted price (initial book value) is 25 000.00(1.035) = $25 875.00; the redemption price is $25 000.00. The average book.-value is y(25 875.00 + 25 000.00) = $25 437.50. The semi-annual interest payment is 25 000.00

0.075 = $937.50;

the number of interest payments to maturity is 10(2) = 20; the total interest payments are 20(937.50) = $18 750.00; the premium is 25 875.00 - 25 000.00 = $875.00. Average income per interest payment interval =

18 750.00 - 875.00

20

= $893.75 a • , r ■ 893.75 Approximate value of z = ^~43 5Q = 0.035135 = 3.51% The yield rate is 2(3.51) = 7.02%. Eight $1000, 6% bonds with semi-annual coupons redeemable at 105 in seven¬ teen years are purchased at 973/8. What is the approximate yield rate? The quoted price is 8000.00(0.97375) = $7790.00; the redemption price is 8000.00(1.05) = $8400.00; uii- (7790.00 + 8400.00) „ the average book value is ---- = $8095.00.

The semi-annual interest payment is 8000.00^-~j = $240.00; the number of interest payments to maturity is 17(2) = 34; the total interest payments are 34(240.00) = $8160.00; the bond discount is 8400.00 - 7790.00 = $610.00. Average income per interest payment interval = _(8160.00^+ 610.00) = $257.94 The approximate value of z is

257 94 QQ = 0.031864 = 3.19%.

The approximate yield rate is 2(3.19%) = 6.38%. A $5000, 10% bond with semi-annual coupons redeemable at par on July 15, 2018 is quoted on December 2, 2006 at 103y4. What is the approximate yield rate? lo find the approximate yield rate for a bond purchased between interest dates,

PART 3: MATHEMATICS OF FINANCE AND INVESTMENT V

assume that the price was quoted on the nearest interest date. Since the interest dates are January 15 and July 15, the nearest interest date is January 15, 2007, which is 11.5 years before maturity. The quoted price is 5000.00(1.0375) = $5187.50; the redemption price is 5000.00; , , , , . (5187.50 + 5000.00) the average book value is---

The semi-annual interest is 5000.00'

$5093.75.

$250.00;

the number of interest payments to maturity is 11.5(2) = 23; the total interest payments are 23(250.00) = $5750.00; the premium is 5187.50 — 5000.00 = $187.50. , (5750.00 - 187.50) The average income per interest payment interval =-^ = $241.85 The approximate value of i is

j— = 0.047480 = 4.75%.

The approximate yield rate is 2(4.75%) = 9.50%.

C. Finding the accurate yield rate by trial and error A method of trial and error similar to the one used to find the nominal rate of interest may be used to obtain as precise an approximation to the yield rate as desired. The method is illustrated in Appendix B on the CD-ROM to this book.

EXERCISE 1 5.4 Use the method of averages to find the approximate yield rate for each of the six bonds shown in the table below.

Face Value

Bond Rate Payable Semiannually

Time Before Redemption

Redeemed At:

Market Quotation

1.

$10 000

6%

15 years

par

3 1018

2.

$5 000

10.5%

7 years

par

944

3.

$25 000

7.5%

10 years

104

97s

4.

$1 000

8.5%

8 years

109

5.

$50 000

9%

5 years, 4 months

par

98s

6.

$20 000

7%

9 years, 8 months

106

1094

101

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

15.5 Sinking Funds A. Finding the size of the periodic payment Sinking funds are interest-bearing accounts into which payments are made at periodic intervals to provide a desired sum of money at a specified future time. Such funds usually involve large sums of money used by both the private and the public sector to repay loans, redeem bonds, finance future capital acquisitions, provide for the replacement of depreciable plant and equipment, and recover investments in depletable natural resources. The basic problem in dealing with sinking funds is to determine the size of the periodic payments that will accumulate to a known future amount. These payments form an annuity in which the accumulated value is known. Depending on whether the periodic payments are made at the end or at the beginning of each payment period, the annuity formed is an ordinary annuity or an annuity due. Depending on whether or not the payment interval is equal in length to the interest conversion period, the annuity formed is a simple annuity or a general annuity. However, since sinking funds are normally set up so that the payment interval and the interest conversion period are equal in length, only the simple annuity cases are considered in this text. (a) For sinking funds with payments at the end of each payment interval, FV„ = PMT

(1 + i)n i

-Formula 11.1

(b) For sinking funds with payments at the beginning of each payment interval, FV„(due)= PMT(1 + i)

(1 + i)n - 1 -Formula 13.1

Western Oil plans to create a sinking fund of $20 000.00 by making equal deposits at the end of every six months for four years. Interest is 6% com¬ pounded semi-annually. (i) What is the size of the semi-annual deposit into the fund? (ii) What is the total amount deposited into the fund? (iii) How much of the fund will be interest? (i) FVn = 20 000.00; P/Y = 2; C/Y = 2; n = 4(2) = 8; I/Y = 6; i = 3% / 1.03s-1 ' 20 000.00 = PMTf 20 000.00 = PMT(8.892336) PMT = $2249.13

6% 2

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Programmed Solution (“END” mode) (Set IVY - 2; C/Y - 2) 0 fpv ] 20 000 ; FV j 6 ( I/Y

8f

N ) ( cPt] [pMt] [-2249.127777)

The size of the semi-annual payment is $2249.13. (ii) The total deposited into the sinking fund is 8(2249.13) = $17 993.04. (iii) The amount of interest in the fund is 20 000.00 — 17 993.04 = $2006.96. Ace Machinery wants to provide for replacement of equipment seven years from now estimated to cost $60 000.00. To do so, the company set up a sinking fund into which it will pay equal sums of money at the beginning of each of the next seven years. Interest paid by the fund is 11.5% compounded annually. (i) What is the size of the annual payment into the fund? (ii) What is the total paid into the fund by Ace Machinery? (iii) How much of the fund will be interest? (i) FV„(due) = 60 000.00; P/Y = 1; C/Y = 1; I/Y =11.5; n = 7; i = 11.5% , ,/ 1.1157 - 1\ 60000.00 = PMT(1.115)(—^5-) 60 000.00 = PMT(1.115)(9.934922) 60 000.00 = PMT( 11.077438) PMT = $5416.42

Programmed Solution (“BGN” mode) (Set P/Y = 1; C/Y = 1) 0 [_pV

60 000 (jFV

11.5 7(

Q/Y

N ) [cPt] (pMt| [-5416.415057}

The size of the annual payment is $5416.42. (ii) The total paid into the fund by Ace Machinery will be 7(5416.42) = $37 914.94. (iii) The interest earned by the fund will be 60 000.00 — 37 914.94 = $22 085.06.

B. Constructing sinking fund schedules The details of a sinking fund can be presented in the form of a schedule. Sinking fund schedules normally show the payment number (or payment date), the peri¬ odic payment into the fund, the interest earned by the fund, the increase in the fund, and the accumulated balance.

^“CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Construct a sinking fund schedule for Example 15.5A. PMT = 2249.13;

n = 8;

i = 3% = 0.03

Sinking fund schedule Payment Interval Number

Periodic Payment

Interest for Payment Interval / = 0.03

Increase in Fund

Balance in Fund at End of Payment Interval

0

0.00

1

2 249.13

0.00

2 249.13

2 249.13

2

2 249.13

67.47

2 316.60

4 565.73

3

2 249.13

136.97

2 386.10

6 951.83

4

2 249.13

208.55

2 457.68

9 409.51

282.29

2 531.42

11 940.93

5

2 249.13

6

2 249.13

358.23

2 607.36

14 548.29

7

2 249.13

436.45

2 685.58

17 233.87

8

2 249.13

517.02

2 766.15

20 000.02

TOTAL

17 993.04

2006.98

20 000.02

Explanations regarding the construction of the sinking fund schedule 1. The payment number 0 is used to introduce the beginning balance. 2. The first deposit is made at the end of the first payment interval. The inter¬

est earned by the fund during the first payment interval is $0, the increase in the fund is $2249.13 and the balance is $2249.13. 3. The second deposit is added at the end of the second payment interval. The

interest for the interval is 0.03(2249.13) = $67.47. The increase in the fund is 2249.13 + 67.47 = $2316.60 and the new balance in the fund is 2249.13 + 2316.60 = $4565.73. 4. The third deposit is made at the end of the third payment interval. The inter¬

est for the interval is 0.03(4565.73) = $136.97, the increase in the fund is 2249.13 + 136.97 = $2386.10, and the new balance in the fund is 2386.10 + 4565.73 = $6951.83. 5. Calculations for the remaining payment intervals are made in a similar manner. 6. The final balance in the sinking fund will probably be slightly different from the expected value. This difference is a result of rounding. The balance may be left as shown ($20 000.02) or the exact balance of $20 000.00 may be obtained by adjusting the last payment to $2249.11. 7. The three totals shown are useful and should be obtained for each schedule.

The total increase in the fund must be the same as the final balance. The total periodic payments are 8(2249.13) = 17 993.04. The total interest is the differ¬ ence: 20 000.02 - 17 993.04 = $2006.98.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Construct a sinking fund schedule for Example 15.5B (an annuity due with pay¬ ments at the beginning of each payment interval). PMT = 5416.42 (made at the beginning); n = 7; i — 11.5% = 0.115 Sinking fund schedule

Periodic Payment

Interest for Payment Interval / = 0.115

1

5 416.42

622.89

6 039.31

2

5 416.42

1 317.41

6 733.83

12 773.14

3

5 416.42

2 091.80

7 508.22

20 281.36

4

5 416.42

2 955.24

8 371.66

28 653.02

5

5 416.42

3 917.99

9 334.41

37 987.43

5 416.42

4 991.44

10 407.86

48 395.29 60 000.06

Payment Interval Number

Increase in Fund

Balance in Fund at End of Payment Interval 0.00

0

6 7 TOTAL

5 416.42

6 188.35

11 604.77

37 914.94

22 085.12

60 000.06

6 039.31

Explanations regarding the construction of the sinking fund schedule

1. The starting balance is $0.00. 2. The first deposit is made at the beginning of the first payment interval and the interest earned by the fund during the first payment interval is 0.115(5416.42) = $622.89. The increase in the fund is 5416.42 + 622.89 = $6039.31 and the balance is $6039.31. 3. The second deposit is made at the beginning of the second payment interval,

the interest earned is 0.115(6039.31 + 5416.42) = $1317.41, the increase is 5416.42 + 1317.41 = $6733.83, and the balance is 6039.31 + 6733.83 = $12 773.14. 4. The third deposit is made at the beginning of the third payment interval, the interest earned is 0.115(12 773.14 + 5416.42) = $2091.80, the increase is 5416.42 + 2091.80 = $7508.22, and the balance is 12 773.14 + 7508.22 = $20 281.36. 5. Calculations for the remaining payment intervals are made in a similar man¬

ner. Be careful to add the deposit to the previous balance when computing the interest earned. 6. The final balance of $60 000.06 is slightly different from the expected balance of $60 000.00 due to rounding. The exact balance may be obtained by adjust¬ ing the last payment to $5416.36. 7. The total increase in the fund must equal the final balance of $60000.06. The total periodic payments are 7(5416.42) = $37 914.94. The total interest is 60000.06 - 37 914.94 = $22 085.12.

CHAPTER

15: BOND VALUATION AND SINKING FUNDS

C. Finding the accumulated balance and interest earned or increase in a sinking fund for a payment interval; constructing partial sinking fund schedules

For Examples 15.5A and 15.5B, compute (i) the accumulated value in the fund at the end of the third payment interval; (ii) the interest earned by the fund in the fifth payment interval; (iii) the increase in the fund in the fifth interval. (i) The balance in a sinking fund at any time is the accumulated value of the payments made into the fund. For Example 15.5A (when payments are made at the end of each payment interval) PMT = 2249.13;

n = 3;

i = 3%

r,

/ 1.033 - 1 \ FV„ = 2249.13(^^) = 2249.13(3.0909) = $6951.83

-see sinking fund schedule. Example 15.5C

Programmed Solution (“END” mode) (Set P/Y = 2; C/Y = 2) 0 ( PV ] 2249.13 [~±~] (pmt) 6 jj/Y

3

CPT

FV

[6951,835917

For Example 15.5B (when payments are made at the beginning of each pay¬ ment interval) PMT = 5416.42; n = 3; i= 11.5% FV„(due) = 5416.42(1.115)(i^^J-) = 5416.42(1.115)(3.358225) = $20281.36

see sinking fund schedule. Example 15.5D

Programmed Solution (“BGN” mode) (Set P/Y = 1; C/Y =1)0 [_PV

J 5416.42 f ± J

PMT

)

11.5 n^t 3 ( N~~] (CPT) ( FV ) [:20 281.35612

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

(ii) The interest earned during any given payment interval is based on the balance in the fund at the beginning of the interval. This figure is the same as the balance at the end of the previous payment interval. For Example 15.5A The balance at the end of the fourth payment interval 1.034 — 1 FV4= 2249.13

0.03

= 2249.13(4.183627) = $9409.52

Programmed Solution (“END” mode) 2249.13 6 [ I/Y

PMT

± 4

N

CPT

( FV

The interest earned by the fund in the fifth payment interval is 0.03(9409.52) = $282.29. For Example 15.5B The balance in the fund at the end of the fourth payment interval FV4 = 5416.42(1.115) 4 ——v— '\

1.1154 - 1 0.115

= 5416.42(1. 5416.42(1.115X4.744421) = $28 653.02

Programmed Solution (“BGN” mode) (Set P/Y = 1; C/Y = 1) 0 1 PV j 5416.42 [[±_

PMT

11.5 ( 1/Y ) 4 ( N ) [cfF] ( FV ] [28 653.02037] The interest earned by the fund in the fifth payment interval is 0.115(28 653.02 + 5416.42) = 0.115(34 069.44) = $3917.99. (iii) The increase in the sinking fund in any given payment interval is the interest earned by the fund during the payment interval plus the periodic payment. For Example 15.5A The increase in the fund during the fifth payment interval is 282.29 + 2249.13 = $2531.42. For Example 15.5B The increase in the fund during the fifth payment interval is 3917.99 + 5416.42 = $9334.41.

32*' CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

The board of directors of National Credit Union decided to establish a building fund of $130 000.00 by making equal deposits into a sinking fund at the end of every three months for seven years. Interest is 12% compounded quarterly. (i) Compute the increase in the fund during the twelfth payment interval. (ii) Construct a partial sinking fund schedule showing details of the first three deposits, the twelfth deposit, the last three deposits, and totals. Size of the quarterly deposit: 12% FV„ = 130 000.00; P/Y = 4; C/Y = 4; I/Y = 12; n = 7(4) = 28; i = —— = 3% 4 / 1 0328 — 1 130 000.00 = PMTf-—— 130 000.00 = PMT(42.930922) PMT = 3028.12

Programmed Solution (“END” mode) (Set P/Y = 4; C/Y = 4) 0 f PV

130 000 (

FV

) 12

I/Y

28 ( N )

(CPT) (PMT) ('-3028.12034)

(i) Balance in the fund at the end of the eleventh payment interval / 1 0311 - 1 FV„ = 3028.12' 0.03 = 3028.12(12.807796) = $38 783.54

Programmed Solution (“END” mode) (Set P/Y = 4; C/Y = 4) 0 i_PV_j 3028.12 [' ± J PMT 12 ( I/Y I 11 ( N | ( CPT | ( FV ) [38 783.54229 The interest earned by the fund during the twelfth payment interval is 0.03(38 783.54) = $1163.51. The increase in the fund during the twelfth payment interval is 1163.51 + 3028.12 = $4191.63. (ii) The last three payments are Payments 26, 27, and 28. To show details of these, we must know the accumulated value after 25 payment intervals. FV25 = 3028.12(-—-J = 3028.12(36.459264) = $110403.03

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Programmed Solution (“END” mode) 3028.12

(Set P/Y = 4; C/Y = 4) 0 ( PV

12

±

I/Y

25

PMT N

CPT

FV

( 110 403.0275

Partial sinking fund schedule

Payment Interval Number

Periodic Payment Made at End

Interest for Payment Interval / = 0.03

Increase in Fund

Balance in Fund at End of Payment Interval 0.00

0 0.00

3 028.12

3 028.12

1

3 028.12

2

3 028.12

90.84

3 118.96

6 147.08

3 •

3 028.12 •

184.41 •

3 212.53 •

9 359.61 •

•

•

•

•

•

11

•

•

•

38 783.54

12 •

3 028.12 •

1 163.51 •

4 191.63 •

42 975.17 •

•

•

•

•

•

25

•

•

•

110 403.03

26

3 028.12

3 312.09

6 340.21

116 743.24

27

3 028.12

3 502.30

6 530.42

123 273.66

28

3 028.12

3 698.21

6 726.33

129 999.99

TOTAL

84 787.36

45 212.63

129 999.99

Laurin and Company want to build up a fund of $75 000.00 by making payments of $2000.00 at the beginning of every six months into a sinking fund earning 11% compounded semi-annually. Construct a partial sinking fund schedule showing details of the first three payments, the last three payments, and totals. To show details of the last three payments, we need to know the number of payments. FV„(due) = 75 000.00; PMT = 2000.00; P/Y = 2; C/Y = 2; I/Y - 11;

11%

= 5.5% 1.055"- 1

75 000.00 = 2000.00(1.055) 1.055" = 2.954976 n In 1.055 = In 2.954976 n(0.053541) = 1.0834906 n = 20.236741

0.055

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Programmed Solution (“BGN” mode) (Set P/Y = 2; C/Y = 2) 0 ( PV ] 75 000 fiy") 2000 f~±~] [pmT 11 ( I/Y ) (CPT) [

N~~) [2023674097]

Twenty-one payments are needed. The last three payments are Payments 19, 20, and 21. The balance in the fund at the end of the 18th payment interval / 1 05518 — 1 \ FVI8(due) = 2000.00(1.055)^ qQ55 ] = 2000.00(1.055) (29.481205) = $62 205.34

Programmed Solution (“BGN” mode) (Set P/Y = 2; C/Y = 2) 0 (pv] 2000 ( ± ) ; PMT .

I/Y

18

N

CPT

FV

K

62 205.3422

J

Partial sinking fund schedule Periodic Payment Made at Beginning

Interest for Payment Interval / = 0.055

Increase in Fund

Balance in Fund at End of Payment Interval

1

2 000.00

110.00

2 110.00

2 110.00

2

2 000.00

226.05

2 226.05

4 336.05

3 •

2 000.00 •

348.48 •

2 348.48 •

6 684.53 •

•

•

•

•

•

18

•

•

•

62 205.34

19

2 000.00

3 531.29

5 531.29

67 736.63

20

2 000.00

3 835.51

5 835.51

73 572.14

21

1 427.86

0.00

1 427.86

75 000.00

41 427.86

33 572.14

75 000.00

Payment Interval Number 0

TOTAL

0.00

Note: The desired balance in the sinking fund will be reached at the beginning of the 21st payment interval by depositing $1427.86.

D Computer application—sinking fund schedule The schedule in Example 15.5C displays the manual calculations for a sinking fund with an interest rate of 3% and eight periodic payments. Microsoft Excel and other

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

spreadsheet programs can be used to create a file that will immediately display the results of a change in the interest rate, the increase in the fund and the accumulat¬ ed balance. The following steps are general instructions for creating a file to calculate the

Excel

sinking fund schedule in Example 15.5C. The formulas in the spreadsheet file were created using Excel; however, most spreadsheets work similarly. Although this exercise assumes a basic understanding of spreadsheet applica¬ tions, someone without previous experience with spreadsheets will be able to com¬ plete it.

STEP 1

Enter the labels shown in Figure 15.2 in row 1 and in column A.

STEP 2

Enter the numbers shown in cells E2, F2 and F4. Do not type in a dollar sign or a comma.

step 3

Enter only the formulas shown in cells B3, C3, D3, and E3. Make sure that the for¬ mula entry includes the dollar ($) sign as shown in the figure.

STEP 4

The formulas entered in Step 3 can be copied through the remaining cells. (a) Select and Copy the formulas in cells B3, C3, D3, and E3. (b) Select cells B4 to BIO and then Paste. (c) The formulas are now active in all the cells.

STEP 5

Enter the formula shown in cell B11, and then use Copy and Paste to enter the for¬ mula in cells C11 and D11.

sTEP 6

To ensure readability of the spreadsheet, format the numbers to display in two dec¬ imal places, and widen the columns to display the full labels. This spreadsheet can now be used to reflect changes in aspects of the sinking fund and create a new schedule. Use cell F2 for new payment amounts and cell F4 for new interest rates.

FIGURE 15.2 — Workbook

p= A 1

B

Payment Interval Periodic Payment

1 =

c

D

Interest

Increase in Fund

E Balance at End 0

2

0

3

1

= $F$2

= $F$4*E2

= B3+C3

= E2+D3

4

2

= $F$2

= $F$4*E3

= B4+C4

= E3+D4

5

3

= $F$2

= $F$4*E4

= B5+C5

= E4+D5

6

4

= $F$2

= $F$4'E5

= B6+C6

= E5+D6

7

5

= $F$2

= $F$4*E6

= B7+C7

= E6+D7

8

6

= $F$2

= $F$4‘E7

= B8+C8

= E7+D8

9

7

= $F$2

= $F$4*E8

= B9+C9

= E8+D9

= B10+C10

= E9+D10

= SUM(D3:D10)

1 0

8

= $F$2

= $F$4*E9

1 1

Totals

= SUM(B3:B10)

= SUM(C3:C10)

F

2249.13

= 0.03

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

E. Debt retirement by the sinking fund method When a sinking fund is created to retire a debt, the debt principal is repaid in total at the due date from the proceeds of the sinking fund while interest on the princi¬ pal is paid periodically. The payments into the sinking fund are usually made at the same time as the interest payments are made. The sum of the two payments (debt interest payment plus payment into the sinking fund) is called the periodic cost of the debt. The difference between the debt principal and the sinking fund balance at any point is called the book value of the debt.

When a debt is retired by the sinking fund method, the borrower is, in effect, paying two separate annu¬ ities. Sinking fund installments are made to the fund’s trustee, while periodic interest payments are made to the lender. Because of these two payment streams, two interest rates must be quoted in questions involving sinking fund debt retirement. One rate determines the interest revenue generated by the sinking fund, and the other rate determines the interest penalty paid by the borrower to the lender.

The City Board of Education borrowed $750 000.00 for twenty years at 13% compounded annually to finance construction of Hillview Elementary School. The board created a sinking fund to repay the debt at the end of twenty years. Equal payments are made into the sinking fund at the end of each year and inter¬ est earned by the fund is 10.5% compounded annually. Rounding all computa¬ tions to the nearest dollar, (i) determine the annual cost of the debt; (ii) compute the book value of the debt at the end of ten years; (iii) construct a partial sinking fund schedule showing the book value of the debt, the three first payments, the three last payments, and the totals. (i) The annual interest cost on the principal: PV = 750 000; P/Y = 1; C/Y = 1; I/Y = 13; i = 13% = 0.13; I = 750 000(0.13) = $97 500 The annual payment into the sinking fund: FV„ = 750 000; n = 20; I/Y = 10.5; i = 10.5% / 1.10520 - 1 750 000 = PMT -—-V 0.105 750 000 = PMT(60.630808) PMT = $12370

Programmed Solution (“END” mode) (Set P/Y = 1; C/Y = 1) 0

I’vJ 750 000 [

FV~] 10.5

20 (

N

I/Y

)[cPT

PMT

The annual cost of the debt is 97 500 + 12 370 = $109 870.

[-12 369.94895)

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

(ii) The balance in the sinking fund after the tenth payment FV10 =

12 370

1.10510 - 1 0.105

= 12 370(16.324579) = $201 935

Programmed Solution (“END” mode) (Set P/Y = 1; C/Y = 1)0

PV

12 370 ( ± ) (PMT| 10.5 fl/Y

10 [ N

CPT

FV

[201935.0483

The book value of the debt at the end of the tenth year is 750 000 - 201 935 = $548 065. (iii) The last three payments are Payments 18, 19, and 20. The balance in the sinking fund at the end of Year 17 FV17 = 12 370

1.10517 - 1

12 370(42.47213) = $525 380

0.105

Programmed Solution (“END” mode) (Set P/Y = 1; C/Y = 1)0 fpv] 12 370 f±~l [pmt] 10.5

l/Y j 17 f~N

CPT

FV ] [525 380.2441 j

Partial sinking fund schedule

Payment Interval Number

Periodic Payment Made at End

Interest for Payment Interval i= 0.105

Increase in Fund

0

Balance in Fund at End of Payment Interval 0

Book Value of Debt 750 000

1

12 370

12 370

737 630

12 370

0 1 299

12 370

2

13 669

26 039

723 961

3 •

12 370 •

2 734 •

15 104 •

41 143 •

708 857 •

•

•

•

•

•

•

17

•

•

•

525 380

224 620

18

12 370

55 165

67 535

592 915

157 085

19

12 370

62 256

74 626

667 541

82 459

12 367

70 092

82 459

750 000

0

247 397

502 603

750 000

20 TOTAL

Note: The last payment has been adjusted to create a fund of exactly $750 000.

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

EXERCISE 15.5 Excel

■■■■■

If you choose, you can use Excel’s Periodic Payment Size (PMT) functions to answer the questions indicated below. Refer to PMT on the Spreadsheet Template Disk to learn how to use these Excel functions. For each of the four sinking funds listed in the table below, compute (a) the size of the periodic payment; (b) the accumulated balance at the time indicated. Accumulated Balance Required After:

Amount of Sinking Fund

Payment Interval

1.

$15 000.00

6 months

end

2.

$9 600.00

1 month

end

3.

$8 400.00

1 month

beginning

15 years

9%

monthly

96th payment

4.

$21 000.00

3 months

beginning

20 years

8%

quarterly

28th payment

Payments Made At:

Interest Rate

Conversion Period

10 years

6%

semi-annually

10th payment

8 years

12%

monthly

36th payment

Term

Each of the four debts listed in the table below is retired by the sinking fund method. Interest payments on the debt are made at the end of each payment inter¬ val and the payments into the sinking fund are made at the same time. Determine (a) the size of the periodic interest expense of the debt; (b) the size of the periodic payment into the sinking fund; (c) the periodic cost of the debt; (d) the book value of the debt at the time indicated. Interest Rate Term of Debt

Payment Interval

On Debt

On Fund

Conversion Period

Book Value Required After:

1. $20 000.00

10 years

3 months

10%

12%

quarterly

6 years

2. $14 500.00

8 years

6 months

7%

8.5%

semi-annually

5 years

3. $10 000.00

5 years

1 month

7.5%

6.0%

monthly

4 years

4. $40 000.00

15 years

3 months

8.0%

7.0%

quarterly

10 years

Debt Principal

Answer each of the following questions. 1. Hein Engineering expects to expand its plant facilities in six years at an esti¬ mated cost of $75 000.00. To provide for the expansion, a sinking fund has been established into which equal payments are made at the end of every three months. Interest is 5% compounded quarterly. (a) What is the size of the quarterly payments? (b) How much of the maturity value will be payments? (c) How much interest will the fund contain? 2. To redeem a $100 000.00 promissory note due in ten years, Cobblestone Enter¬ prises has set up a sinking fund earning 7.5% compounded semi-annually. Equal deposits are made at the beginning of every six months. (a) What is the size of the semi-annual deposits? (b) How much of the maturity value of the fund is deposits? (c) How much is interest?

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

3. Equal deposits are made into a sinking fund at the end of each year for seven years. Interest is 5.5% compounded annually and the maturity value of the fund is $20 000.00. Find the size of the annual deposits and construct a sink¬ ing fund schedule showing totals. 4. A sinking fund amounting to $15 000.00 is to be created by making payments

at the beginning of every six months for four years. Interest earned by the fund is 12.5% compounded semi-annually. Determine the size of the semi-annual payments and prepare a sinking fund schedule showing totals. 5. For Question 3, calculate the increase in the fund for the fourth year. Verify your answer by checking the sinking fund schedule. 6. For Question 4, compute the interest earned during the fifth payment interval. Verify your answer by checking the sinking fund schedule. 7. HY Industries Ltd. plans to replace a warehouse in twelve years at an antici¬ pated cost of $45 000.00. To pay for the replacement, a sinking fund has been established into which equal payments are made at the end of every quarter. Interest is 10% compounded quarterly.

(a) What is the size of the quarterly payments? (b) What is the accumulated balance just after the sixteenth payment? 8. To provide for expansion, Champlain Company has established a sinking fund earning 7% semi-annually. The fund is anticipated to reach a balance of $72 000.00 in fifteen years. Payments are made at the beginning of every six months.

(a) What is the size of the semi-annual payment? (b) What is the accumulated balance at the end of the twentieth payment period? _p

@)

9. Winooski & Co. has borrowed $95 000.00 for capital expansion. The company must pay the interest on the loan at the end of every six months and make equal payments at the time of the interest payments into a sinking fund until the loan is retired in twenty years. Interest on the loan is 9% compounded semi-annually and interest on the sinking fund is 7% compounded semi¬ annually. (Round all answers to the nearest dollar.)

(a) Determine the size of the periodic interest expense of the debt. (b) Determine the size of the periodic payment into the sinking fund. (c) What is the periodic cost of the debt? (d) What is the book value of the debt after fifteen years? 10. The City of Chatham has borrowed $80 000.00 to expand a community cen¬ tre. The city must pay the interest on the loan at the end of every month and make equal payments at the time of the interest payments into a sinking fund until the loan is retired in twelve years. Interest on the loan is 6% compound¬ ed monthly and interest on the sinking fund is 7.5% compounded monthly. (Round all answers to the nearest dollar.)

(a) Determine the size of the periodic interest expense of the debt. (b) Determine the size of the periodic payment into the sinking fund. (c) What is the periodic cost of the debt? (d) What is the book value of the debt after eight years? _p

@ 11. Kirk, Klein & Co. requires $100 000.00 fifteen years from now to retire a debt. A sinking fund is established into which equal payments are made at the end of every month. Interest is 7.5% compounded monthly.

CHAPTER

5:

BOND

VALUATION

AND

SINKING

FUNDS

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

(b) What is the balance in the sinking fund after five years? (c) How much interest will be earned by the fund in the 100th payment interval?

(d) By how much will the fund increase during the 150th payment interval? (e) Construct a partial sinking fund schedule showing details of the first three payments, the last three payments, and totals. 12. The Town of ICeewatin issued debentures worth $120000.00 maturing in ten years to finance construction of water and sewer facilities. To redeem the debentures, the town council decided to make equal deposits into a sinking fund at the beginning of every three months. Interest earned by the sinking fund is 6% compounded quarterly. (a) What is the size of the quarterly payment into the sinking fund?

(b) What is the balance in the fund after six years? (c) How much interest is earned by the fund in the 28th payment interval?

(d) By how much will the fund increase in the 33rd payment interval? (e) Prepare a partial sinking fund schedule showing details of the first three payments, the last three payments, and totals. The Township of Langley borrowed $300 000.00 for road improvements. The debt agreement requires that the township pay the interest on the loan at the end of each year and make equal deposits at the time of the interest payments into a sinking fund until the loan is retired in twenty years. Interest on the loan is 8.25% compounded annually and interest earned by the sinking fund is 5.5% compounded annually. (Round all answers to the nearest dollar.) (a) What is the annual interest expense?

(b) What is the size of the annual deposit into the sinking fund? (c) What is the total annual cost of the debt?

(d) How much is the increase in the sinking fund in the tenth year? (e) What is the book value of the debt after fifteen years?

(f) Construct a partial sinking fund schedule showing details, including the book value of the debt, for the first three years, the last three years, and totals. _p

(f§) 14. Ontario Credit Union borrowed $225 000.00 at 13% compounded semi¬ annually from League Central to build an office complex. The loan agree¬ ment requires payment of interest at the end of every six months. In addition, the credit union is to make equal payments into a sinking fund so that the principal can be retired in total after fifteen years. Interest earned by the fund is 11% compounded semi-annually. (Round all answers to the nearest dollar.) (a) What is the semi-annual interest payment on the debt?

(b) What is the size of the semi-annual deposits into the sinking fund? (c) What is the total annual cost of the debt?

(d) What is the interest earned by the fund in the twentieth payment interval? (e) What is the book value of the debt after twelve years?

(f) Prepare a partial sinking fund schedule showing details, including the book value of the debt, for the first three years, the last three years, and totals.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT V

^ Review Exercise 1. A $5000, 11.5% bond with interest payable semi-annually is redeemable at par in twelve

8. Four $10 000 bonds bearing interest at 6% payable quarterly and redeemable at 106 on

years. What is the purchase price to yield

September 1, 2014 were purchased on January

(a) 10.5% compounded semi-annually?

23, 2002 to yield 5% compounded quarterly.

(b) 13% compounded semi-annually?

Determine (a) the premium or discount;

2. A $10 000, 6% bond with semi-annual coupons is redeemable at 108. What is the purchase price to yield 7.5% compounded semi-annually

(b) the purchase price; (c) the quoted price. 9. A $5000, 8% bond with semi-annual coupons

(a) nine years before maturity?

redeemable at 108 in ten years is purchased to

(b) fifteen years before maturity?

yield 10% compounded semi-annually. What is the purchase price?

3. A $25 000, 9% bond with interest payable quarterly is redeemable at 104 in six years. What is the purchase price to yield 8.25% compounded annually?

10. A $1000 bond bearing interest at 8% payable semi-annually redeemable at par on February 1, 2009 was purchased on October 12, 2002 to yield 7% compounded semi-annually.

4. A $1000, 9.5% bond with semi-annual coupons redeemable at par on March 1, 2010

Determine the purchase price. 11. A $25 000, 13% bond with semi-annual

was purchased on September 19, 2001 to yield

coupons redeemable at 107 on June 15, 2014

7% compounded semi-annually. What was the

was purchased on May 9, 2003 to yield 14.5%

purchase price?

compounded semi-annually. Determine

5. Four $5000, 7% bonds with semi-annual coupons are bought seven years before matu¬ rity to yield 6% compounded semi-annually. Find the premium or discount and the pur¬ chase price if the bonds are redeemable

(b) at 107.

(a) at par;

(a) the premium or discount; (b) the purchase price; (c) the quoted price. 12. A $50 000, 11% bond with semi-annual coupons redeemable at par on April 15, 2007 was purchased on June 25, 2000 at 923/&. What

6. Nine $1000, 8% bonds with interest payable semi-annually and redeemable at par are pur¬ chased ten years before maturity. Find the pre¬ mium or discount and the purchase price if the bonds are bought to yield (a) 6%;

(b) 8%;

(c) 10%.

7. A $100 000, 5% bond with interest payable semi-annually redeemable at par on July 15,

was the approximate yield rate? 13. A $1000, 8.5% bond with interest payable annually is purchased six years before matur¬ ity to yield 10.5% compounded annually. Compute the premium or discount and the purchase price and construct the appropriate bond schedule. 14. A $5000, 12.25% bond with interest payable

2012 was purchased on April 18, 2001 to yield

annually redeemable at par in seven years is

7% compounded semi-annually. Determine

purchased to yield 13.5% compounded annu¬

(a) the premium or discount;

ally. Find the premium or discount and the

(b) the purchase price; (c) the quoted price.

purchase price and construct the appropriate bond schedule.

mV

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

24. A $25 000, 8% bond with semi-annual coupons,

15. A $20 000, 9.5% bond with semi-annual coupons redeemable at 105 in three years is

redeemable at 104 in fifteen years, is purchased

purchased to yield 8% compounded semi¬

to yield 6% compounded semi-annually.

annually. Find the premium or discount and

Determine the gain or loss if the bond is sold

purchase price and construct the appropriate

three years later at 107'A.

bond schedule.

25. To provide for the purchase of heavy construc¬ tion equipment estimated to cost $110 000.00,

16. Three $25 000, 11% bonds with semi-annual coupons redeemable at par were bought eight

Valmar Construction is paying equal sums of

years before maturity to yield 12% compounded

money at the end of every six months for five

semi-annually. Determine the gain or loss if the

years into a sinking fund earning 7.5% com¬

bonds are sold at 89 % five years later.

pounded semi-annually. (a) What is the size of the semi-annual payment

17. A $10 000 bond with 5% interest payable quar¬

into the sinking fund?

terly redeemable at 106 on November 15, 2012 was bought on July 2, 1996 to yield 9% com¬

(b) Compute the balance in the fund after the

pounded quarterly. If the bond was sold at 923A

third payment.

on September 10, 2002, what was the gain or

(c) Compute the amount of interest earned dur¬

loss on the sale?

ing the sixth payment interval.

18. A $25 000, 9.5% bond with semi-annual

(d) Construct a sinking fund schedule showing

coupons redeemable at par is bought sixteen

totals. Check your answers to parts (b) and

years before maturity at 78 A. What was the

(c) with the values in the schedule.

approximate yield rate? 26. Alpha Corporation is depositing equal sums of

19. A $10 000, 7.5% bond with quarterly coupons

money at the beginning of every three months

redeemable at 102 on October 15, 2013 was

into a sinking fund to redeem a $65 000.00

purchased on May 5, 2001 at 98 A. What is the

promissory note due eight years from now.

approximate yield rate?

Interest earned by the fund is 12% compounded 20. What is the approximate yield realized if the

quarterly.

bond in Question 19 was sold on August 7, (a) Determine the size of the quarterly pay¬

2006 at 92? P

ments into the sinking fund.

(§§) 21. A 6.5% bond of $50 000 with interest payable

(b) Compute the balance in the fund after three

quarterly is to be redeemable at par in twelve

years.

years.

(c) Compute the increase in the fund during the

(a) What is the purchase price to yield 8%

24th payment interval.

compounded quarterly?

(d) Construct a partial sinking fund schedule

(b) What is the book value after nine years?

showing details of the first three deposits, (c) What is the gain or loss if the bond is sold nine years after the date of purchase at 995A? 22. A $100 000, 10.75% bond with interest payable annually is redeemable at 103 in eight years. What is the purchase price to yield 12% com¬ pounded quarterly? 23. A $5000, 14.5% bond with semi-annual coupons redeemable at par on August 1, 2014 was purchased on March 5, 2003 at 95lh. What was the approximate yield rate?

the last three deposits, and totals. p

(§§) 27. The municipality of Kirkfield borrowed $100 000.00 to build a recreation centre. The debt principal is to be repaid in eight years and interest at 13.75% compounded annually is to be paid annually. To provide for the retirement of the debt, the municipal council set up a sink¬ ing fund into which equal payments are made at the time of the annual interest payments. Interest earned by the fund is 11.5% com¬ pounded annually.

PART

(a) What is the annual interest payment? (b) What is the size of the annual payment into the sinking fund? (c) What is the total annual cost of the debt? (d) Compute the book value of the debt after three years. (e) Compute the interest earned by the fund in Year 6. (f) Construct a sinking fund schedule showing

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

(b) How much of the maturity value of the fund will be interest? (c) What is the accumulated value of the fund after two years? (d) How much interest will the fund earn in the 15th payment interval? 30. A sinking fund of $10 000.00 is to be created by equal annual payments at the beginning of each year for seven years. Interest earned by the fund

the book value of the debt and totals. Verify

is 7.5% compounded annually.

your computations in parts (d) and (e)

(a) Compute the annual deposit into the fund.

against the schedule.

(b) Construct a sinking fund schedule showing

# 28. The Harrow Board of Education financed the acquisition of a building site through a

totals. 31. Joe Ngosa bought a retirement fund for

$300 000.00 long-term promissory note due in

$15 000.00. Beginning twenty-five years from

fifteen years. Interest on the promissory note is

the date of purchase, he will receive payments of

9.25% compounded semi-annually and is

$17 500.00 at the beginning of every six months.

payable at the end of every six months. To pro¬

Interest earned by the fund is 12% compounded

vide for the redemption of the note, the board

semi-annually.

agreed to make equal payments at the end of every six months into a sinking fund paying 8% compounded semi-annually. (Round all answers to the nearest dollar.)

(a) How many payments will Joe receive?

(b) What is the size of the last payment? 32. The town of Kildare bought firefighting equip¬

(a) What is the semi-annual interest payment?

ment for $96 000.00. The financing agreement

(b) What is the size of the semi-annual payment

provides for annual interest payments and equal

into the sinking fund?

payments into a sinking fund for ten years. After ten years the proceeds of the sinking fund will

(c) What is the annual cost of the debt? (d) Compute the book value of the debt after five years. (e) Compute the increase in the sinking fund in the 20th payment interval. (f) Construct a partial sinking fund schedule showing details, including the book value of the debt, for the first three years, the last

be used to retire the principal. Interest on the debt is 14.5% compounded annually and inter¬ est earned by the sinking fund is 13% com¬ pounded annually. (a) What is the annual interest payment?

(b) What is the size of the annual payment into the sinking fund? (c) What is the total annual cost of the debt?

three years, and totals. (d) What is the book value of the debt after four # 29. Northern Flying Service is preparing to buy an aircraft estimated to cost $60 000.00 by making equal payments at the end of every three months into a sinking fund for five years. Interest earned by the fund is 8% compounded quarterly. (a) What is the size of the quarterly payment into the sinking fund?

years? (e) Construct a partial sinking fund schedule showing details, including the book value of the debt, for the last three years and totals.

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Self-Test 1. A $10 000,10% bond with quarterly coupons redeemable at par in fifteen years is purchased to yield 11% compounded quarterly. Determine the purchase price of the bond. 2. What is the purchase price of a $1000, 7.5% bond with semi-annual coupons redeemable at 108 in ten years if the bond is bought to yield 6% compounded semi-annually? 3. A $5000, 8% bond with semi-annual coupons redeemable at 104 is bought six years before maturity to yield 6.5% compounded semi-annually. Determine the premium or discount. 4. A $20 000,10% bond with semi-annual coupons redeemable at par March 1,2009 was purchased on November 15, 2002 to yield 9% compounded semi-annually. What was the purchase price of the bond? 5. A $5000, 7% bond with semi-annual coupons redeemable at 102 on Decem¬ ber 15, 2012 was purchased on November 9, 2001 to yield 8.5% compounded semi-annually. Determine the quoted price. 6. A $5000, 11.5% bond with semi-annual coupons redeemable at 105 is bought four years before maturity to yield 13% compounded semi-annually. Construct a bond schedule. 7. A $100 000, 13% bond with semi-annual interest payments redeemable at par on July 15, 2010 is bought on September 10, 2003 at 1025/s. What was the approximate yield rate? 8. A $25 000, 6% bond with semi-annual coupons redeemable at 106 in twenty years is purchased to yield 8% compounded semi-annually. Determine the gain or loss if the bond is sold seven years after the date of purchase at 98 ‘A. 9. A $10 000, 12% bond with semi-annual coupons redeemable at par on December 1, 2012 was purchased on July 20, 2001 at 93%. Compute the approximate yield rate. 10. Cottingham Pies made semi-annual payments into a sinking fund for ten years. If the fund had a balance of $100 000.00 after ten years and interest is 11% compounded semi-annually, what was the accumulated balance in the fund after seven years? 11. A fund of $165 000.00 is to be accumulated in six years by making equal pay¬ ments at the beginning of each month. If interest is 7.5% compounded monthly, how much interest is earned by the fund in the twentieth payment interval? 12. Gillian Armes invested $10 000.00 in an income fund at 13% compounded semi-annually for twenty years. After twenty years, she is to receive semi¬ annual payments of $10 000.00 at the end of every six-month period until the fund is exhausted. What is the size of the final payment? 13. A company financed a plant expansion of $750 000.00 at 9% compounded annually. The financing agreement requires annual payments of interest and the funding of the debt through equal annual payments for fifteen years into a sinking fund earning 7% compounded annually. What is the book value of the debt after five years?

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

14. Annual sinking fund payments made at the beginning of every year for six years earning 11.5% compounded annually amount to $25 000.00 at the end of six years. Construct a sinking fund schedule showing totals.

Challenge Problems A $2000 bond with annual coupons is redeemable at par in five years. If the first coupon is $400, and subsequent annual coupons are worth 75% of the previous year’s coupon, find the purchase price of the bond that would yield an interest rate of 10% compounded annually. 2. An issue of bonds, redeemable at par in n years, is to bear coupons at 9% com¬ pounded semi-annually. An investor offers to buy the entire issue at a premi¬ um of 15%. At the same time, the investor advises that if the coupon rate were raised to 10% compounded semi-annually, he would offer to buy the whole issue at a premium of 25%. At what yield rate compounded semi-annually are these two offers equivalent?

Case Study 15.1 Investing in Bonds » Recently Ruja attended a personal financial planning seminar. The speaker men¬ tioned that bonds should be a part of everyone’s balanced investment portfolio, even if they are only a small part. Ruja’s RRSP contains mutual funds and a guar¬ anteed investment certificate (GIC), but no bonds. She has decided to invest up to $4500.00 of her RRSP funds in bonds and has narrowed her choices to three. Bond A is a $1000, 6.8% bond with semi-annual coupons redeemable in five years. Ruja can purchase up to four of these bonds at 104.25. Bond B is a $1000, 7.1% bond with semi-annual coupons redeemable in four years. She can purchase up to four of these bonds at 103.10. Bond C is a $1000, 6.2% bond with semi-annual coupons redeemable in seven years. Ruja can purchase up to four of these bonds at 101.85.

QUESTIONS 1. Suppose Ruja wants to invest in only one bond. Use the average investment method to answer the following questions. (a) What is the approximate yield rate of Bond A?

(b) What is the approximate yield rate of Bond B? (c) What is the approximate yield rate of Bond C?

(d) Assume Ruja is willing to hold the bond she chooses until it matures. Which bond has the highest yield? 2. Suppose Ruja decides to buy two $1000 denominations of Bond A. Bond A’s semi-annual coupons are payable on January 1 and July 1. Suppose Ruja pur¬ chases these bonds on February 27. (a) What is the accrued interest on these two bonds up to the date of Ruja’s purchase? (b) What is Ruja’s purchase price (or flat price) for these two bonds?

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

3. Suppose Ruja decides to buy two $1000 denominations of Bond B on February 27. Bond B’s semi-annual coupons are payable on February 1 and August 1.

(a) What is the accrued interest on these two bonds up to the date of Ruja’s purchase? (b) What is Ruja’s purchase price (or flat price) for these two bonds?

Case Study 15.2 The Business of Bonds » Beaucage Development Company is developing a new process to manufacture com¬ pact discs. The development costs were higher than expected, so Beaucage required an immediate cash inflow of $4 800 000.00. To raise this money, the company decid¬ ed to issue bonds. Since Beaucage had no expertise in issuing and selling bonds, the company decided to work with an investment dealer. The investment dealer bought the company’s entire bond issue at a discount, then sold the bonds to the public at face value or the current market value. To ensure it would raise the $4 800 000.00 it required, Beaucage issued 5000 bonds with a face value of $1000 each on January 20, 2004. Interest is paid semi-annually on July 20 and January 20, beginning July 20, 2004. The bonds pay interest at 7.5% compounded semi-annually. Beaucage directors realize that when the bonds mature on January 20, 2024, there must be $5 000 000.00 available to repay the bondholders. To have enough money on hand to meet this obligation, the directors set up a sinking fund using a specially designated savings account. The company earns interest of 5.5% com¬ pounded semi-annually on this sinking fund account. The directors began making semi-annual payments to the sinking fund on July 20, 2004. Beaucage Development Company issued the bonds, sold them all to the invest¬ ment dealer, and used the money raised to continue its research and development.

QUESTIONS 1. How much would an investor have to pay for one of these bonds to earn 8% compounded semi-annually?

2. (a) What is the size of the sinking fund payment? (b) What will be the total amount deposited into the sinking fund account? (c) How much of the sinking fund will be interest? 3. Suppose Beaucage discovers on January 20, 2014 that it can earn 8% interest compounded semi-annually on its sinking fund account.

(a) What is the balance in the sinking fund after the January 20, 2014 sink¬ ing fund payment? (b) What is the new sinking fund payment if the fund begins to earn 8% on January 21, 2014? (c) What will be the total amount deposited into the sinking fund account over the life of the bonds? (d) How much of the sinking fund will then be interest? (e) How does the amount of sinking fund interest calculated in part (d) com¬ pare to the amount of interest calculated in Question 2(c)?

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

SUMMARY OF FORMULAS Formula 15.1 i - (i + PP = FV(1 +

PMT

Basic formula for finding the purchase price of a bond

iy

when the interest payment interval and the yield rate conversion period are equal

Formula 15.2 PP = FV(1 + pyn+ PMT

i - (i + prn p

Basic formula for finding the purchase price of a bond when the interest payment interval and the yield rate conversion period are different

where p = (1 + i)c — 1

Formula 15.3 Direct formula for finding the premium or discount of a bond (a negative answer indicates a discount) PREMIUM OR DISCOUNT

1 - (1 + z)~n

= (b X FACE VALUE — i X REDEMPTION PRICE)

i

Formula 15.4 Basic formula for finding the yield rate using the method of averages. AVERAGE INCOME PER INTEREST PAYMENT INTERVAL APPROXIMATE VALUE OF

i

AVERAGE BOOK VALUE

where AVERAGE BOOK VALUE

= y (QUOTED PRICE + REDEMPTION PRICE)

and TOTAL INTEREST PAYMENTS

AVERAGE INCOME PER INTEREST PAYMENT INTERVAL

— PREMIUM , + DISCOUNT

= NUMBER OF INTEREST PAYMENT INTERVALS

In addition, Formulas 9.1C, 11.1, 11.2, 12.3, and 13.1 were used in this chapter.

GLOSSARY Accumulation of discount

the process of reduc¬

Bond rate

the rate of interest paid by a bond,

ing a bond discount (p. 664)

stated as a percent of the face value (p. 635)

Amortization of premium

Book value of a debt

the process of writing

down a bond premium (p. 661)

the difference at any time

between the debt principal and the associated sinking fund balance (p. 684)

CHAPTER

15:

BOND

VALUATION

AND

SINKING

FUNDS

Coupon

a voucher attached to a bond to facilitate

the difference between the purchase

the collection of interest by the bondholder (p. 635)

price of a bond and its redemption price when the

Coupon rate

purchase price is greater than the redemption

see

Debentures

Bond rate

bonds for which no security is

Quoted price

offered (p. 635) Denomination Discount

see

sale (p. 643)

the difference between the purchase price

chase price is less than the redemption price (p. 648) see

Face value

Redemption date

(including any accrued interest) (p. 643)

Maturity date

see

see

bonds that are redeemed at

their face value (p. 635) the date at which the bond

principal is repaid (p. 635) the amount that the issuer of

the bond pays to the bondholder upon surrender

Redemption date

Method of averages

redemption price is greater than the face

Redemption price

Quoted price

bonds whose

value (p. 635)

Redemption date

the total purchase price of a bond

Market price

Redeemable at a premium

Redeemable at par

the amount owed by the issuer of the

bond to the bondholder (p. 635) Flat price

the net price of a bond (without

accrued interest) at which a bond is offered for

Face value

of a bond and its redemption price when the pur¬

Due date

price (p. 648)

of the bond on or after the date of maturity (p. 635)

a method for finding the

Sinking fund

a fund into which payments are

approximate yield rate (p. 671)

made to provide a specific sum of money at a

Nominal rate

see Bond rate

future time; usually set up for the purpose of

Par value

Face value

see

Periodic cost of a debt

meeting some future obligation (p. 674)

the sum of the interest

paid and the payment into the sinking fund

field rate

the rate of interest that an investor

earns on his or her investment in a bond (p. 637)

when a debt is retired by the sinking fund method (p. 684)

USEFUL INTERNET SITES www.cis-pec.gc.ca Canada Investment and Savings (CIS)

CIS is a special operating agency of the Department of Finance

that markets and manages savings and investment products for Canadians. This Website provides information on products, including Canada Savings Bonds and Canada Premium Bonds. www.bankofcanada.ca/en Bond Market Rates

Click on Bond Securities. The current Government of Canada bond yields and marketable bond average yields are found at this site. The site also provides links to selected historical interest rates.

www.carswell.com/payroll/index.asp The Payroll Community

Hosted by Carswell publishers, this site provides information about payroll,

FAQs, new products, payroll publications, and links to relevant sites. www.benefitscanada.com Benefits Canada

This magazine deals with employee benefits and pension investments.

Investment Decision Applications

OBJECTIVES Upon completing this chapter, you will be able to do the following: 1.

Determine the discounted value of cash flows and choose between alter¬ native investments on the basis of the discounted cash flow criterion.

2.

Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible or not.

3. Compute the rate of return on investment.

Choices among different investment opportunities must be made often by both indi¬ viduals and companies. Whether one is deciding between buying and leasing a car or how to increase plant capacity, an understanding of the time value of money is critical. When comparing different ways of achieving the same goal, we should always examine cash flows at the same point in the time—usually at the beginning when a decision must be made. Only then can we know whether it is better to buy or lease that car, or to expand the plant now or to wait.

INTRODUCTION When making investment decisions, all decision makers must consider the com¬ parative effects of alternative courses of action on the cash flows of a business or of an individual. Since cash flow analysis needs to take into account the time value of money (interest), present-value concepts are useful. When only cash inflows are considered, the value of the discounted cash flows is helpful in guiding management toward a rational decision. If outlays as well as inflows are considered, the net present value concept is applicable in evaluating projects. The net present value method indicates whether or not a project will yield a specified rate of return. To be a worthwhile investment, the rate of return on a cap¬ ital project must be attractive. Required rates of return tend to be high and may even reach or exceed the 20% level. Knowing the actual rate of return provides use¬ ful information to the decision maker. It may be computed using the net present value concept.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

16.1 Discounted Cash Flow A. Evaluation of capital expenditures—basic concepts Projects expected to generate benefits over a period of time longer than one year are called capital investment projects, and they result in capital expenditures. The benefits resulting from these projects may be in either monetary or non-monetary form. The methods of analysis considered in this chapter will deal only with invest¬ ment projects generating cash flows in monetary form. While capital expenditures normally result in the acquisition of assets, the pri¬ mary purpose of investing in capital expenditures is to acquire a future stream of benefits in the form of an inflow of cash. When an investment is being considered, analysis of the anticipated future cash flows aids in the decision to acquire or replace assets, and whether to buy or lease them. The analysis generally uses the technique of discounted cash flow. This tech¬ nique involves estimating all anticipated future cash flows flowing from an invest¬ ment or project, projecting an interest rate, and calculating the present value of these cash flows. It is important that the present value be calculated, not the future value, due to the need to make a decision on the investment in the beginning, or in the present, before any cash or other resources are invested. The decision to be made might involve determining whether to invest in a project, or choosing which project to invest in. In some situations, the cash flows estimated may not be certain, or there may be other, non-fmancial concerns. The analysis techniques considered in this text are concerned only with the amount and the timing of cash receipts and cash pay¬ ments under the assumption that the amount and timing of the cash flow are certain. From the mathematical point of view, the major issue in evaluating capital expenditure projects is the time value of money. This value prevents direct com¬ parison of cash received and cash payments made at different times. The concept of present value, as introduced in Chapter 9 and subsequent chapters, provides the vehicle for making sums of money received or paid at different times comparable at a given time.

B, Discounted cash flow Discounted cash flow is the present value of all cash payments. When using the dis¬ counting technique to evaluate alternatives, two fundamental principles serve as decision criteria. 1. The bird-in-the-hand principle—Given that all other factors are equal, earlier benefits are preferable to later benefits. 2. The-bigger-the-better principle—Given that all other factors are equal, bigger benefits are preferable to smaller benefits.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT V

Suppose you are offered a choice of receiving $1000.00 today or receiving $1000.00 three years from now. What is the preferred choice? Accepting the bird-in-the-hand principle, you should prefer to receive $1000.00 today rather than three years from now. The rationale is that $1000.00 can be invested to earn interest and will accumulate in three years to a sum of money greater than $1000.00. Stated another way, the present value of $1000.00 to be received in three years is less than $1000.00 today. Consider a choice of $2000.00 today or $3221.00 five years from now. Which alternative is preferable? No definite answer is possible without considering interest. A rational choice must consider the time value of money; that is, we need to know the rate of inter¬ est. Once a rate of interest is established, we can make the proper choice by con¬ sidering the present value of the two sums of money and applying the the-bigger-the-better principle. If you choose “now” as the focal date, three outcomes are possible. 1. The present value of $3221.00 is greater than $2000.00. In this case, the preferred choice is $3221.00 five years from now. 2. The present value of $3221.00 is less than $2000.00. In this case, the pre¬ ferred choice is $2000.00 now. 3. The present value of $3221.00 equals $2000.00. In this case, either choice is equally acceptable. (a) Suppose the rate of interest is 8%. FV = 3221.00; i = 8% = 0.08; n = 5 PV = 3221.00(1.08-5) = 3221.00(0.680583) = $2192.16 Since at 8% the discounted value of $3221.00 is greater than $2000.00, the preferred choice at 8% is $3221.00 five years from now. (b) Suppose the rate of interest is 12%. FV = 3221.00; i = 12% = 0.12; n = 5 PV = 3221.00(1.12"5) = 3221.00(0.567427) = $1827.68 Since at 12% the discounted value is less than $2000.00, the preferred choice is $2000.00 now. (c) Suppose the rate of interest is 10%. FV = 3221.00; i = 10% = 0.10; n = 5 PV = 3221.00(1.10 5) = 3221.00(0.620921) = $1999.99 Since at 10% the discounted value is equal to $2000.00, the two choices are equally acceptable.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

Programmed Solution (“END” mode) (Set P/Y = 1; C/Y = 1) (a)

( FV

(b)

: 2nd : (CLR TVM) 3221 ( FV

(c)

8 [ I/Y

51

12 | I/Y

2nd ] (CLR TVM) 3221 ! FV j i()

I/Y

N

(CPT

N

_

( CPT

N

CPT

5

PV | [-2192.158478] [ PV ] [-1827.681902] PV j

-1999.987582

Two investment alternatives are available. Alternative A yields a return of $6000 in two years and $10 000 in five years. Alternative B yields a return of $7000 now and $7000 in seven years. Which alternative is preferable if money is worth (i) 11%?

(ii) 15%?

To determine which alternative is preferable, we need to compute the present value of each alternative and choose the alternative with the higher present value. Since the decision is to be made immediately, choose a focal point of “now.” (i) For i — 11% Alternative A The present value of Alternative A is the sum of the present values of $6000 in two years and $10 000 in five years. Present value of $6000 in two years = 6000(1.11“2) = 6000(0.811622)

=

$ 4 870

= 10 000(1. ll'-5) = 10 000(0.593451)

=

5 935

The present value of Alternative A

=

$10 805

Present value of $10 000 in five years

Alternative B The present value of Alternative B is the sum of the present values of $7000 now and $7000 in seven years. Present value of $7000 now

$ 7 000

Present value of $7000 in seven years = 7000(1.11“7) = 7000(0.481658)

3 372

The present value of Alternative B

$10 372

Programmed Solution Alternative A (“END” mode) (Set P/Y = 1; C/Y = 1) [2nd] (CLR TVM) 6000 [fT] 11 [ I/Y \ 2 f~N~l [CPT7] fpy~) [-4869.734599] [ 2nd ] (CLR TVM) 10 000 I FV $4870 + 5935 = $10 805

11

I/Y

5 (

N ) [CPT] fpV~] [-5934.513281]

PART 3: MATHEMATICS OF FINANCE AND INVESTMENT

Alternative B 2nd]

(CLRTVM)

7000

fivl

11

[17y~]

( CPT| [" PV ) (-337L6088761

N

7

$3372 + 7000 = $10 372 Since at 11% the present value of Alternative A is greater than the present value of Alternative B, Alternative A is preferable. (ii) For i = 15% Alternative A

Present value of $6000 in two years = 6000(1.15“2) = 6000(0.756144) Present value of $10 000 in five years = 10 000(1.15~5) = 10 000(0.497177)

=

$4537

=

4972

=

$9509

Present value of $7000 now Present value of $7000 in seven years = 7000(1.15“7) = 7000(0.375937)

=

$7000

=

2632

The present value of Alternative B

=

$9632

The present value of Alternative A Alternative B

Programmed Solution Alternative A

(“END” mode) (Set P/Y = 1; C/Y 2nd]

1)

(CLR TVM) 6000 [

FV )

15

[ I/Y

I 2

[

N

) ( CPT ) ( PV ) [-4536.862004

[ 2nd 1 (CLRTVM) lOOPofFy'] 15fT^1 5pN~

CPT

PV

[-4971.767353

$4537 + 4972 = $9509 Alternative B 2nd] (CLR

TVM)

7000 ( FV ] 15

[1/y]

7[

N ) ( CPT) [ PV ) [-2631.559279]

$2632 + 7000 = $9632 Since at 15% the present value of Alternative B is greater than the present value of Alternative A, Alternative B is preferable. Note: Applying present value techniques to capital investment problems usually involves estimates. For this reason, dollar amounts in the preceding example and all following examples may be rounded to the nearest dollar. We suggest that you do the same when working on problems of this nature.

CHAPTER

16: INVESTMENT DECISION APPLICATIONS

An insurance company offers to settle a claim either by making a payment of $50 000 immediately or by making payments of $8000 at the end of each year for ten years. What offer is preferable if interest is 8% compounded annually? Present value of $8000 at the end of each year for ten years is the present value of an ordinary annuity in which PMT = 8000, n — 10, and i = 8%. PV„ = 8000i

1 - 1.08 l0\ 0.08

)

8000(6.710081)

$53 681

Programmed Solution (“END” mode) (Set P/Y = 1; C/Y = 1) 0 ( FV ) 8000 [PMT) 10 ( N ) 8 ( I/Y ) (CPT ) fPV~) [-53680.65119] Since the immediate payment is smaller than the present value of the annual pay¬ ments of $8000, the annual payments of $8000 are preferable.

There are several numerical measures that are used by corporations, financial analysts, and investors to determine whether a particular common stock's trading price is realistic: (i) Book Value: Book value is calculated as (Total assets - Total liabilities) h- Total number of shares of com¬ mon stock. Because the dollar amount of assets is sometimes overstated or understated in the financial records of a company compared to their market value, the book value may be misleading. (ii) Earnings per Share: Earnings per share is calculated as After-tax earnings Total number of shares of common stock. As you might expect, any increase in earnings per share is generally viewed as a healthy sign for a corporation and its shareholders. (iii) Price-Earnings (P/E) Ratio: The price-earnings (P/E) ratio is calculated as the Price per common share -e Earnings per share. The P/E ratio is a key factor used by serious investors to evaluate stocks. A low P/E ratio indicates that a particular stock may be a good investment because its stock price is lower than it could be given its level of earnings. A high P/E ratio suggests that it could be a poor investment. For most corporations, P/E ratios range between 5 and 25. Some dot-com or Internet-related companies have P/E ratios of 60 and higher or have negative P/E ratios because they have not yet earned income. Negative P/E ratios are not normally reported. (iv) Beta: Beta is an index reported in many financial publications that compares the risk associated with a particular stock to the risk of the stock market in general. The beta for the stock market in general is 1.0, while the majority of stocks have beta values between 0.5 and 2.0. Typically, low-risk stocks have low beta values and high-risk stocks have high beta values.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

National Credit Union needs to decide whether to buy a high-speed scanner for $6000 and enter a service contract requiring the payment of $45 at the end of every three months for five years, or to enter a five-year lease requiring the pay¬ ment of $435 at the beginning of every three months. If leased, the scanner can be bought after five years for $600. At 9% compounded quarterly, should the credit union buy or lease? To make a rational decision, the credit union should compare the present value of the cash outlays if buying the scanner with the present value of the cash out¬ lays if leasing the scanner. Present value of the decision to buy Present value of cash payment for the scanner

=

$6000

Present value of the service contract involves an ordinary annuity in which PMT = 45, n — 20, P/Y = 4; C/Y = 4; I/Y = 9; i — 2.25% - 45^1 V^-) - 45(15.963712)

=

718

Present value of decision to buy

=

$6718

Present value of the decision to lease Present value of the quarterly lease payments involves an annuity due: PMT = 435, n = 20, i = 2.25% / 1 — 1 0225_20\ = 435(1.0225)^ 1 0Q^ j = 435(1.0225)(15.963712)

$7100

Present value of purchase price after five years 384

= 600(1.0225^20) = 600(0.640816)

$7484

Present value of decision to lease

Programmed Solution Present value of the decision to buy $6000

Present value of cash payment for the scanner Present value of the service contract (“END” mode) (Set P/Y = 4; C/Y = 4) 0

J V ) 45 [PMT] 20 (

N ) 9 ( I/Y ) (cPT) ( PV ) [-718.3670566

Present value of decision to buy is $6000 + 718 = $6718. Present value of the decision to lease Present value of the quarterly lease payments (“BGN” mode) PV

(^7100.459716)

0 (pmt| 600 f FV ) 20 pN~] 9 fl/vl [CPT) f~PV

( -384,489883 J

0 fl^l 435 |PMt] 20 CiD 9 fuD fcPT

Present value of purchase price after five years

Present value of the decision to lease is $7100 + 384 — $7484.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

In the case of costs, the selection criterion follows the the-smaller-the-better prin¬ ciple. Since the present value of the decision to buy is smaller than the present value of the decision to lease, the credit union should buy the scanner. Hans Machine Service needs a brake machine. The machine can be purchased for $4600 and after five years will have a salvage value of $490, or the machine can be leased for five years by making monthly payments of $111 at the beginning of each month. If money is worth 10% compounded annually, should Hans Machine Service buy or lease? Alternative 1: Buy machine Present value of cash price Less:

$4600

Present value of salvage value

= 490(1.10“5) = 490(0.620921)

304

Present value of decision to buy

$4296

Alternative 2: Lease machine The monthly lease payments form a general annuity due in which PMT = 111; P/Y = 12; C/Y - 1; I/Y =10; c = -jy; n = 60; i = 10%; p = l.lO1^ - 1 = 1.007974 - 1 = 0.7974%

^ , . ,(

Present value of the monthly lease payments

1 10 007974 1^1

= 111(1.007974) (47.538500) = $5319 The present value of the decision to lease is $5319.

Programmed Solution Alternative 1: Buy machine Present value of cash price Less:

=

$4600

Present value of salvage value

(“END” mode) (Set P/Y = 12; C/Y = 1) [2nd] (CLR TVM) 490 fFV~) 60 f~N~]: 10 ( I/Y ^

CPT

PV

-304.2514483]

Present value of decision to buy is $4600 — 304 = $4296. Alternative 2: Lease machine Present value of the monthly lease payments (“BGN” mode) (Set P/Y = 12; C/Y = 1) 0 60

FV

j 11

PMT

N~~] IQ ( I/Y ) (CPT) ( PV ] (-5318.851263]

Since the present value of the decision to buy is smaller than the present value of the decision to lease, Hans Machine Service should buy the machine.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

EXERCISE 16. You can use Excel’s Net Present Value (NPV) function to answer the questions below.

Excel

%*Q

Refer to NPV on the Spreadsheet Template Disk to learn how to use this function. For each of the following, compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion. 1. The D Company must make a choice between two investment alternatives. Alternative 1 will return the company $20 000 at the end of three years and $60 000 at the end of six years. Alternative 2 will return the company $13 000 at the end of each of the next six years. The D Company normally expects to earn a rate of return of 12% on funds invested. 2. The B Company has a policy of requiring a rate of return on investment of 16%. Two investment alternatives are available but the company may choose only one. Alternative 1 offers a return of $50 000 after four years, $40 000 after seven years, and $30 000 after ten years. Alternative 2 will return the company $750 at the end of each month for ten years. 3. An obligation can be settled by making a payment of $10 000 now and a final

payment of $20 000 in five years. Alternatively, the obligation can be settled by payments of $1500 at the end of every three months for five years. Interest is 10% compounded quarterly. 4. An unavoidable cost may be met by outlays of $10 000 now and $2000 at the

end of every six months for seven years or by making monthly payments of $500 in advance for seven years. Interest is 7% compounded annually. 5. A company must purchase new equipment costing $2000. The company can pay cash on the basis of the purchase price or make payments of $108 per month for 24 months. Interest is 7.8% compounded monthly. Should the company purchase the new equipment with cash or make payments on the installment plan? 6. For less than a dollar a day, Jerri can join a fitness club. She would have to pay $24.99 per month for 30 months, or she can pay a lump sum of $549 at the beginning. Interest is 16.2% compounded monthly. Should Jerri pay a lump sum or use the monthly payment feature?

%SkO

Answer each of the following questions. 1. A contract offers $25 000 immediately and $50 000 in five years or $10 000 at the end of each year for ten years. If money is worth 6%, which offer is preferable? 2. A professional sports contract offers $400 000 per year paid at the end of each of six years or $100 000 paid now, $200 000 paid at the end of each of the sec¬ ond and third years and 800 000 paid at the end of each of the last three years. If money is worth 7.3%, which offer is preferable? 3. Bruce and Carol want to sell their business. They have received two offers. If

they accept Offer A, they will receive $15 000 immediately and $20 000 in three years. If they accept Offer B, they will receive $3000 now and $3000 at the end of every six months for six years. If interest is 10%, which offer is preferable?

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

4. When Peter decided to sell his farm, he received two offers. If he accepts the first offer, he would receive $250 000 now, $750 000 one year from now, and $500 000 two years from now. If he accepts the second offer, he would receive $600 000 now, $300 000 one year from now, and $600 000 two years from now. If money is worth 9.8%, which offer should he accept? 5. A warehouse can be purchased for $90 000. After twenty years the property will have a residual value of $30 000. Alternatively, the warehouse can be leased for twenty years at an annual rent of $10 000 payable in advance. If money is worth 8%, should the warehouse be purchased or leased? 6. A car costs $9500. Alternatively, the car can be leased for three years by mak¬ ing payments of $240 at the beginning of each month and can be bought at the end of the lease for $4750. If interest is 9% compounded semi-annually, which alternative is preferable?

16.2 Net Present Value Method A. Introductory examples Net cash inflows from two ventures are as follows:

End of Year

1

2

3

4

5

Total

Venture A

12 000

14 400

17 280

20 736

24 883

89 299

Venture B

17 000

17 000

17 000

17 000

17 000

85 000

Which venture is preferable if the required yield is 20%? Present value of Venture A = 12 000(1.20-') + 14 400( 1.20-2) + 17 280(1.20“3) + 20 736(1.20“4) + 24 883(1.20“5) = 12 000(0.833333) + 14 400(0.694444) + 17 280(0.578704) + 20 736(0.482253) + 24 883(0.401878)

=

10 000

+

10 000

+

10 000

+

10 000

+

10 000

= $50 000 Present value of Venture B = 17 000

1

1.20"5 0.20

-

17 000(2.990612) = $50 840

Programmed Solution Present value of Venture A (“END” mode) (Set P/Y = 1; C/Y = 1) r2nd j (CLRTVM) 12 000 [fv] 20 ( I/Y

~N~| [cpt] ( PV ] [

-10000

PART

2nd i (CLR TVM)

3:

MATHEMATICS

( FV

20

I/Y

2

N

20

I/Y

3

20

I/Y

20 ( I/Y

2nd

(CLR TVM)

FV

I 2nd

(CLR TVM)

( FV

2nd

(CLR TVM)

( FV

j

OF

FINANCE

AND

INVESTMENT

CPT

( PV

[

-10000

N

(CPT

PV

[

-10000

4

N

CPT

PV

(

-10000

5

N

CPT

PV 1 (

-10000

]

)

)

The present value of Venture A is $10 000 + 10 000 + 10 000 + 10 000 + 10 000 = $50 000. Present value of Venture B

0 fFvl 17 000 f±~]

(PMT)

20 (j/yJ 5

N

CPT

PV

( 50840.40638

The present value of Venture B is $50 840. Since at 20% the present value of Venture B is greater than the present value of Venture A, Venture B is preferable to Venture A. Assume for Example 16.2A that Venture A requires an immediate nonrecoverable outlay of $9000 while Venture B requires a non-recoverable outlay of $11 000. At 20%, which venture is preferable?

Present value of cash inflows

Venture A

Venture B

$50 000

$50 840

9000

11000

$41000

$39 840

Present value of immediate outlay Net present value

Since the net present value of Venture A is greater than the net present value of Venture B, Venture A is preferable.

B. The net present value concept When the present value of the cash outlays is subtracted from the present value of cash inflows, the resulting difference is called the net present value. In Example 16.2A, Venture B was preferable, where only the cash inflows were considered. In Example 16.2B, the present value of the outlays as well as the present value of the cash inflows resulted in the net present value being calculated. This approach is necessary when the cash outlays are different. NET PRESENT VALUE (npv)

PRESENT VALUE

PRESENT VALUE

OF INFLOWS

OF OUTLAYS

Formula 16.1

Since the net present value involves the difference between the present value of the inflows and the present value of the outlays, three outcomes are possible: 1. If the present value of the inflows is greater than the present value of the out¬ lays, then the net present value is greater than zero. 2. If the present value of the inflows is smaller than the present value of the out¬ lays, then the net present value is smaller than zero.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

3. If the present value of the inflows equals the present value of the outlays, then the net present value is zero. PVIN > PV0UT

NPV > 0 (positive)

PVIN = PV0UT — NPV = 0 PV)n < PV0UT -

NPV < 0 (negative)

Criterion rule At the organization’s required rate of return, accept those capital investment pro¬ jects that have a positive or zero net present value and reject those projects that have a negative net present value. For a given rate of return: ACCEPT if NPV > 0 or NPV = 0; REJECT if NPV < 0.

To distinguish between a negative and a positive net present value, use NPV = PVIN - PV0UT. If a company is considering more than one project but can choose only one, the project with the greatest positive net present value is preferable. Assumptions about the timing of inflows and outlays The net present value method of evaluating capital investment projects is particu¬ larly useful when cash outlays are made and cash inflows received at various times. Since the timing of the cash flows is of prime importance, follow these assumptions regarding the timing of cash inflows and cash outlays. Unless otherwise stated: 1. All cash inflows (benefits) are assumed to be received at the end of a period. 2. All cash outlays (costs) are assumed to be made at the beginning of a period.

C. Applications A company is offered a contract promising annual net returns of $36 000 for seven years. If it accepts the contract, the company must spend $150 000 imme¬ diately to expand its plant. After seven years, no further benefits are available from the contract and the plant expansion undertaken will have no residual value. Should the company accept the contract if the required rate of return is (i)

12%?

(ii) 18%?

(iii) 15%?

The net inflows and outlays can be represented on a time graph. End of period (year) Now

1

2

3

4

5

6

7

36 000

36 000

36 000

36 000

36 000

36 000

36 000

Note: Cash outlays (costs) are identified in such diagrams by a minus sign or by using accounting brackets.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENTS

(i) For i — 12% Since we assume the annual net returns (benefits) are received at the end of a period unless otherwise stated, they form an ordinary annuity in which PMT = 36 000; PVIN = 36 000

n = 7;

1

i = 12%

1.12-7' = 36 000(4.563756) 0.12

-

PVout — Present value of 150 000 now

=

$164 295

=

150 000

=

$ 14 295

The net present value (NPV) = 164 295 - 150 000

Since at 12% the net present value is greater than zero, the contract should be accepted. The fact that the net present value at 12% is positive means that the contract offers a return on investment of more than 12%. (ii) For i = 18% PV,IN

36 000

1 - 1.18“7 0.18

= 36 000(3.811528)

PVrOUT NPV = 137 215

150 000

=

$137 215

=

150 000

=

-$12 785

Since at 18% the net present value is less than zero, the contract should not be accepted. The contract does not offer the required rate of return on investment of 18%. (iii) For i = 15% PVIN = 36 000

1 - 1.15~7 0.15

= 36 000(4.16042)

PV,OUT

=

$149 775

=

$150 000 -$225

NPV = 149 775 - 150 000

The net present value is slightly negative, which means that the net present value method does not provide a clear signal as to whether to accept or reject the contract. The rate of return offered by the contract is almost 15%.

Programmed Solution (i) For i = 12% (“END” mode) PVIN: (Set P/Y = 1; C/Y = 1) 0

I

FV ] 36 000 pmt} 7 12 [ I/Y

j

CPI

f ~N j

( PVj [-764295.2354]

PVOUT: = Present value of 150 000 now = $150 000 The net present value (NPV) = $164 295 — 150 000 = $14 295. (ii) Fori = 18% (“END” mode) PVIN: (Set P/Y = 1; C/Y =1)0 iJVJ 36 000 PMT

7;

N }

18 ( I/Y | ( CPT| [ PV | [-137214.9934]

^CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

PVOUT: = $150 000 NPV

= $137 215 - 150 000 = —$12 785

(iii) For i = 15% (“END” mode) PVIN: (Set P/Y = 1; C/Y = 1) 0 [ FV~) 36 000 (pmt) 7 f N~' 15 ( I/Y ) (CPT] ( PV ] [-149775.1104]

PV0UT: = $150 000 NPV

= $149 775 - 150 000 = -$225

A project requires an initial investment of $80 000 with a residual value of $15 000 after six years. It is estimated to yield annual net returns of $21 000 for six years. Should the project be undertaken at 16%? The cash flows are represented in the diagram below. End of period(year)

i = 16%

Now

Out

1

2

3

4

5

21

21

21

21

21

In

15 21

Note: The residual value of $15 000 is considered to be a reduction in outlays. Its present value should be subtracted from the present value of other outlays.

PVIN = 21 OOO^1

Q^6' ' j = 21 000(3.684736)

PV0UT = 80 000 - 15 000(1.16"6) = 80 000 - 15 000(0.410442) = 80 000 - 6157 = Net present value (NPV)

=

$77 379

=

73 843

=

$ 3 536

Programmed Solution (“END” mode) PVIN: (Set P/Y = 1; C/Y = 1) 0 1 FV

21 000

PMT

16 [ PV0UT: 0 I™! 15 000 (~ivH NPV

6

flTt 16 fT7T|

6 !

N

I/Y ) (CPT) ( PV ) [-77379.45407

[CPT] fFV~] [-6156.63382

= $77 379 - (80 000 - 6157) = $3536

Since the net present value is positive (the present value of the benefits is greater than the present value of the costs), the rate of return on the investment is greater than 16%. The project should be undertaken.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

The UBA Corporation is considering developing a new product. If undertaken, the project requires the outlay of $100 000 per year for three years. Net returns beginning in Year 4 are estimated at $65 000 per year for twelve years. The resid¬ ual value of the outlays after fifteen years is $30 000. If the corporation requires a return on investment of 14%, should it develop the new product? End of year Now

i = 14% 1

2

3

4

5

6

7

13

14

65

65

65

65

65

65

30

Out In

15

65

The net returns, due at the end of Year 4 to Year 15 respectively, form an ordinary annuity deferred for three years in which PMT = 65 000, n — 12, d = 3, i — 14%. / l — l 14-12\ PVIN= 65 000^-^-j(1.14-3) = 65 000(5.660292)(0.674972) = $248 335 The outlays, assumed to be made at the beginning of each year, form an annuity due in which PMT = 100 000, n = 3, i = 14%. PV0UT = 100 000(1.14)^ 1 "q1^4 3) - 30 000(1.14-15) = 100 000(1.14)(2.321632) - 30 000(0.140096) = 264 666 - 4203 = $260 463 NPV = 248 335 - 260 463 =

Programmed Solution (“END” mode) PVjN: (Set P/Y = 1; C/Y = 1) 0 fjvj 65 000 (pmt) 12 f~N j 14 [ I/Y ) (cPT) ( PV ) [-367918.9882]

14

I/Y

CPT

N ] 14

( I/Y

j CPT

N ] 14

I/Y

CPT

N

PV

[-248334.8453]

[ PV

[-264666.0511]

( PV

[-4202.894462

(“BGN” mode) vOUT’

0 PMT

30 000

j

NPV = $248 335 - (264 666 - 4203) = —$12 128 Since the net present value is negative, the investment does not offer a 14% return. The corporation should not develop the product.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

A feasibility study concerning a contemplated venture yielded the following estimates: Initial cost outlay: $1 300 000; further outlays in Years 2 to 5: $225 000 per year; residual value after 20 years: $625 000; net returns: Years 5 to 10:

$600 000 per year;

Years 11 to 20: $500 000 per year. Should the venture be undertaken if the required return on investment is 15%? End of year Now

i = 15% 1

2

3

4

Out In

5

6

9

10

11

600

500

12

19

20

625 600

600... 600

500 ... 500

500

/ i — 1 l S“6\ / 1 — 1 15~10\ PV1N— 600 000^-^-j(1.15-4) + 500 000^-^-J(1.15-10) = 600 000(3.784483)(0.571753) + 500 000(5.018769)(0.247185) = 1 298 274 + 620 281 = $1 918 555 PV0UT= 1 300 000 + 225 OOo(^-~r—) - 625 000(1.15“20) = 1 300 000 + 225 000(2.854978) - 625 000(0.061100) = 1 300 000 + 642 370 - 38 188 = $1 904 182 NPV = 1 918 555 - 1 904 182 = $14 373

Programmed Solution (“END” mode) PVIN: (SetP/Y = 1; C/Y = 1) 0

FV

600i 000 [pmt] 15 fDYj 6 f~N~] [cpt] [pv] £-2270689.616

4(

o [pmt] 2 270 689 [ FV ] 15 (j/Y

N ] [cpt] [ PV ] [-1298273.805]

PV~) [[-2509384.313] 0 [FVJ 500 000 [PMT] 15 finPi 10 f~N~l (CPT) [pvj 2509384.313}

0 [pmt] 2 509 384 {jvJ 15 (7TP

10 ‘

N

CPT

[ PV ] [-620281.3466] r.

PVIN = $1 298 274 + 620 281 = $1 918 555 0

FV

0 PMT

225 000 PMT

15

I/Y

625 000 ( FV } 15 [ I/Y

4

N

20

N

CPT ICPT

PV

(y

PV

PV0UX = $1 300 000 + 642 370 - 38 188 = $1 904 182 NPV

= $1 918 555 - 1 904 182 = $14 373

Since the net present value is positive, the rate of return on investment is greater than 15%. The venture should be undertaken.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Net present value can be determined by using a preprogrammed financial calcula¬ tor. For instructions on using this function, refer to the CD-ROM accompanying this book.

EXERCISE 16.2 If you choose, you can use Excel’s Net Present Value (NPV) function to answer the questions in Part A and Part B below. Refer to NPV on the Spreadsheet Template Disk to learn how to use this Excel function. For each of the following six investment choices, compute the net present value. Determine which investment should be accepted or rejected according to the net present value criterion. 1. A contract is estimated to yield net returns of $3500 quarterly for seven years. To secure the contract, an immediate outlay of $50 000 and a further outlay of $30 000 three years from now are required. Interest is 12% compounded quarterly. 2. Replacing old equipment at an immediate cost of $50 000 and an additional outlay of $30 000 six years from now will result in savings of $3000 per quarter for twelve years. The required rate of return is 10% compounded annually. 3. A business has two investment choices. Alternative 1 requires an immediate outlay of $2000 and offers a return of $7000 after seven years. Alternative 2 requires an immediate outlay of $1800 in return for which $250 will be received at the end of every six months for the next seven years. The required rate of return on investment is 17% compounded semi-annually. 4. Suppose you are offered two investment alternatives. If you choose Alternative 1, you will have to make an immediate outlay of $9000. In return, you will receive $500 at the end of every three months for the next ten years. If you choose Alternative 2, you will have to make an outlay of $4000 now and $5000 in two years. In return, you will receive $30 000 ten years from now. Interest is 12% compounded semi-annually. 5. You have two investment alternatives. Alternative 1 requires an immediate out¬ lay of $8000. In return, you will receive $900 at the end of every quarter for the next three years. Alternative 2 requires an immediate outlay of $2000, and an outlay of $1000 in two years. In return, you will receive $300 at the end of every quarter for the next three years. Interest is 7% compounded quarterly. Which alternative would you choose? Why?

6. Your old car cost you $300 per month in gas and repairs. If you replace it, you could sell the old car immediately for $2000. To buy a new car that would last five years, you need to pay out $10 000 immediately. Gas and repairs would cost you only $120 per month on the new car. Interest is 9% compounded monthly. Should you buy a new car? Why?

S' CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

Answer each of the following questions. 1. Teck Engineering normally expects a rate of return of 12% on investments. Two projects are available but only one can be chosen. Project A requires an immediate investment of $4000. In return, a revenue payment of $4000 will be received in four years and a payment of $9000 in nine years. Project B requires an investment of $4000 now and another $2000 in three years. In return, rev¬ enue payments will be received in the amount of $1500 per year for nine years. Which project is preferable? 2. The owner of a business is presented with two alternative projects. The first project involves the investment of $5000 now. In return the business will receive a payment of $8000 in four years and a payment of $8000 in ten years. The second project involves an investment of $5000 now and another $5000 three years from now. The returns will be semi-annual payments of $950 for ten years. Which project is preferable if the required rate of return is 14% com¬ pounded annually? 3. Northern Track is developing a special vehicle for Arctic exploration. The development requires investments of $60 000, $50 000, and $40 000 for the next three years respectively. Net returns beginning in Year 4 are expected to be $33 000 per year for twelve years. If the company requires a rate of return of 14%, compute the net present value of the project and determine whether the company should undertake the project. 4. The Kellog Company has to make a decision about expanding its production facilities. Research indicates that the desired expansion would require an immediate outlay of $60 000 and an outlay of a further $60 000 in five years. Net returns are estimated to be $15 000 per year for the first five years and $10 000 per year for the following ten years. Find the net present value of the project. Should the expansion project be undertaken if the required rate of return is 12%? 5. Agate Marketing Inc. intends to distribute a new product. It is expected to pro¬ duce net returns of $15 000 per year for the first four years and $10 000 per year for the following three years. The facilities required to distribute the prod¬ uct will cost $36 000 with a disposal value of $9000 after seven years. The facil¬ ities will require a major facelift costing $10 000 each after three and after five years respectively. If Agate requires a return on investment of 20%, should the company distribute the new product?

6. A company is considering a project that will require a cost outlay of $15 000 per year for four years. At the end of the project the salvage value will be $10 000. The project will yield returns of $60 000 in Year 4 and $20 000 in Year 5. There are no returns after Year 5. Alternative investments are available that will yield a return of 16%. Should the company undertake the project? 7. Demand for a product manufactured by Eagle Company is expected to be 15 000 units per year during the next ten years. The net return per unit is $2. The manufacturing process requires the purchase of a machine costing $140 000. The machine has an economic life of ten years and a salvage value of $20 000 after ten years. Major overhauls of the machine require outlays of

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

$20 000 after four years and $40 000 after seven years. Should Eagle invest in the p

(§)

machine if it requires a return of 12% on its investments? 8. Magnum Electronics Company expects a demand of 20 000 units per year for a special purpose component during the next six years. Net return per unit is $4.00. To produce the component, Magnum must buy a machine costing $250 000 with a life of six years and a salvage value of $40 000 after six years. The company estimates that repair costs will be $20 000 per year during Years 2 to 6. If Magnum requires a return on investment of 18%, should it market the component?

BUSINESS MATH NEWS BOX Yamaha recently ran the following advertisement: Lease a New Vstar 1100 Classic

Kodiak Ultramatic 4 X 4

for $222.95*

for $176.88*

per month for

per month for

36 months with

36 months with

$1500.00 down.

$1200.00 down.

*Plus applicable taxes, freight, and PDI. Does not include insurance, licence, or reg¬ istration fees. Limited time offer. Consumer may be required to purchase leased goods at the end of the lease term for $5399.50 for Vstar™ and $4274.50 for Kodiak™ plus applicable taxes. See your dealer for return or refinancing options. Source: Used with permission from Yamaha Motor Canada Ltd.

QUESTIONS

To answer these questions, assume the combined PST and GST tax rate is 15%. Assume freight, PDI, insurance, licence, and registration costs are the same whether you lease or buy a vehicle. 1. Suppose the Vstar 1100 Classic has a manufacturer’s suggested retail price (MSRP) of $10 799.00 plus taxes. (a) If you can earn 10% on your money, is it cheaper to lease or buy this motorcycle? (b) If you can earn 20% on your money, is it cheaper to lease or buy this motorcycle? 2, Suppose the Kodiak Ultramatic 4 X 4 has an MSRP of $8549.00 plus taxes. (a) If you can earn 10% on your money, is it cheaper to lease or buy this motorcycle? (b) If you can earn 20% on your money, is it cheaper to lease or buy this motorcycle? (c) Based on your calculations in parts (a) and (b), about how much interest would you have to earn on your money to make leasing a Kodiak Ultramatic 4X4 cost the same as buying it?

¥ CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

16.3 Finding the Rate of Return on Investment A. Net present value, profitability index, rate of return The rate of return on investment (R.O.I.) is widely used to measure the value of an investment. Since it takes interest into account, knowing the rate of return that results from a capital investment project provides useful information when evalu¬ ating a project. The method of finding the rate of return that is explained and illustrated in this section uses the net present value concept introduced in Section 16.2. However, instead of being primarily concerned with a specific discount rate and with com¬ paring the present value of the cash inflows and the present value of the cash out¬ lays, this method is designed to determine the rate of return on the investment. As explained in Section 16.2, three outcomes are possible when using Formula 16.1. These three outcomes indicate whether the rate of return is greater than, less than, or equal to the discount rate used in finding the net present value. 1. If the net present value is greater than zero (positive), then the rate of return is greater than the discount rate used to determine the net present value. 2. If the net present value is less than zero (negative), then the rate of return is less than the discount rate used. 3. If the net present value is equal to zero, then the rate of return is equal to the rate of discount used.

If NPV > 0 (positive)

R.O.I. > i

If NPV < 0 (negative)

R.O.I. < i

If NPV = 0

R.O.I. = i

It follows, then, that the rate of return on investment (R.O.I.) is that rate of dis¬ count for which the NPV = 0, that is, for which PV,N = PV0UT. The above definition of the rate of return and the relationship between the net present value, the rate of discount used to compute the net present value, and the rate of return are useful in developing a method of finding the rate of return. However, before computing the rate of return, it is useful to consider a ratio known as the profitability index or discounted benefit-cost ratio. It is defined as the ratio that results when comparing the present value of the cash inflows with the present value of the cash outlays. PROFITABILITY INDEX (or DISCOUNTED BENEFIT-COST RATIO)

PV IN PV,OUT

Formula 16.2

Since a division is involved, three outcomes are possible when computing this ratio. 1. If the numerator (PVIN) is greater than the denominator (PV0UT), then the profitability index is greater than one.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

2. If the numerator (PVIN) is less than the denominator (PVOUT), then the prof¬ itability index is less than one. 3. If the numerator (PVIN) is equal to the denominator (PVOUT), then the prof¬ itability index is equal to one. The three outcomes give an indication of the rate of return. 1. If the profitability index is greater than 1, then the rate of return is greater than the discount rate used. 2. If the profitability index is less than 1, then the rate of return is less than the discount rate used. 3. If the profitability index is equal to 1, then the rate of return equals the dis¬ count rate used. The relationship between the present value of the inflows, the present value of the outlays, the net present value, the profitability index, and the rate of return at a given rate of discount i is summarized below. Net Present Value (NPV)

Profitability Index

Rate of Return (R.O.I.)

PVIN > PV0UT

NPV > 0

> 1

>

PV,N = PV0UT

NPV = 0

= 1

= i

PVIN < PV0UT

NPV < 0

< 1

< i

IN Versus PV0UT

i

B. Procedure for finding the rate of return by trial and error From the relationships noted above, the rate of return on investment can be defined as the rate of discount for which the present value of the inflows (benefits) equals the present value of the outlays (costs). This definition implies that the rate of return is the rate of discount for which the net present value equals zero or for which the profitability index (benefit-cost ratio) equals 1. This conclusion permits us to determine the rate of return by trial and error. STEP 1

Arbitrarily select a discount rate and compute the net present value at that rate.

STEP 2

From the outcome of Step 1, draw one of the three conclusions. (a) If NPV = 0, infer that the R.O.I. = i. (b) If NPV > 0, infer that the R.O.I. > i. (c) If NPV < 0, infer that the R.O.I. < i.

STEP 3

(a) If, in Step 1, NPV = 0, then R.O.I. = i and the problem is solved. (b) If, in Step 1, NPV > 0 (positive), then we know that R.O.I. > i. A second attempt is needed. This second try requires choosing a discount rate greater than the rate used in Step 1 and computing the net present value using the higher rate.

S’ CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

If the resulting net present value is still positive, choose a still higher rate of discount and compute the net present value for that rate. Repeat this proce¬ dure until the selected rate of discount yields a negative net present value. (c) If, in Step 1, NPV < 0 (negative), then we know that R.O.I. < i. The second try requires choosing a discount rate less than the rate used in Step 1 and com¬ puting the net present value using the lower rate. If the resulting net present value is still negative, choose a still lower rate of discount and compute the net present value for that rate. Repeat this proce¬ dure until the selected rate of discount yields a positive net present value. step 4

The basic aim of Step 3 is to find one rate of discount for which the net present value is positive and a second rate for which the net present value is negative. Once this has been accomplished, the rate of return must be a rate between the two rates used to generate a positive and a negative net present value. You can now obtain a reasonably accurate value of the rate of return by using linear interpolation. To ensure sufficient accuracy in the answer, we recommend that the two rates of discount used when interpolating be no more than two per¬ centage points apart. The worked examples in this section have been solved using successive even rates of discounts when interpolating.

STEP 5

(Optional) You can check the accuracy of the method of interpolation when using an electronic calculator by computing the net present value. Use as the discount rate the rate of return determined in Step 4. Expect the rate in Step 4 to be slight¬ ly too high. You can obtain a still more precise answer by further trials.

C. Selecting the rate of discount—using the profitability index While the selection of a discount rate in Step 1 of the procedure is arbitrary, a sen¬ sible choice is one that is neither too high nor too low. Since the negative net pre¬ sent value immediately establishes a range between zero and the rate used, it is preferable to be on the high side. Choosing a rate of discount within the range 12% to 24% usually leads to quick solutions. While the initial choice of rate is a shot in the dark, the resulting knowledge about the size of the rate of return combined with the use of the profitability index should ensure the selection of a second rate that is fairly close to the actual rate of return. In making the second choice, use the profitability index. 1. Compute the index for the first rate chosen and convert the index into a percent. 2. Deduct 100% from the index and divide the difference by 4. 3. If the index is greater than 1, add the above result to obtain the rate that you should use for the second attempt. If, however, the index is smaller than 1, deduct the above result from the rate of discount initially used.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Assume that the rate of discount initially selected is 16%. The resulting PV[N = 150 and the PVOUT = 120. 1. The profitability index is -jyy = 1.25 = 125%. 2. The difference (125% — 100%) divided by 4 = 6.25%. 3. Since the index is greater than 1, add 6.25% to the initial rate of 16%; the recommended choice is 22%. Assume that the rate of discount initially selected is 20%. The resulting PV1N = 200 and the PV0UT = 250. 1. The profitability index is yyy = 0.80 = 80%. 2. The difference (80% — 100%) divided by 4 = —5%. 3. Since the index is less than 1, subtract 5% from the initial rate of 20%; the rec¬ ommended choice is 15%. (If, as in this text, you are using only even rates, try either 14% or 16%.)

D. Using linear interpolation The method of linear interpolation used in Step 4 of the suggested procedure is illustrated in Example 16.3A below. Assume that the net present value of a project is $420 at 14% and —$280 at 16%. Use linear interpolation to compute the rate of return correct to the nearest tenth of a percent. The data can be represented on a line diagram. NPV

$420

$0

-$280

A

X

B

H-b ii

d — 16%

14%

The line segment AB represents the distance between the two rates of discount that are associated with a positive and a negative net present value respectively. At Point A,

where i = 14%, the NPV = 420;

at Point B,

where i = 16%, the NPV = —280;

at Point X,

where i is unknown, the NPV = 0.

By definition, the rate of return is that rate of discount for which the net present value is zero. Since 0 is a number between 420 and —280, the NPV = 0 is located at a point on AB. This point is marked X.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

We can obtain two useful ratios by considering the line segment from the two points of view shown in the diagram. (i) In terms of the discount rate i, AB = 2% -16% - 14% AX = d% added to 14% to obtain the rate of discount at which the NPV = 0 AX _ d%_ AB 2% (ii) In terms of the net present value figures, AB - 700 AX = 420 AX = 420 AB

700

Since the ratio AX:AB is written twice, we can derive a proportion statement. d% _ 420

2%

"

d% =

700 420 700

X 2%

d% = 1.2% Therefore, the rate at which the net present value is equal to zero is 14% + 1.2% = 15.2%. The rate of return on investment is 15.2%.

E, Computing the rate of return A project requires an initial outlay of $25 000. The estimated returns are $7000 per year for seven years. Compute the rate of return (correct to the nearest tenth of a percent). The cash flows (in thousands) are represented in the diagram below. The inflows form an ordinary annuity since inflows are assumed to be received at the end of each year. End of year Now

Out In

1

2

3

4

5

6

7

7

7

7

7

7

7

7

PV,IN

7000

1 - (1 + i)~ i

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

The outlays consist of an immediate payment. PV0UT = 25 000 To determine the rate of return, we will choose a rate of discount, compute the net present value, and try further rates until we find two successive even rates. For one, the NPV > 0 (positive) and, for the other, NPV < 0 (negative). STEP 1

Try i = 12%. PVIN = 7000^ 1 ~

= 7000(4.563757)

=

$31 946

PV0UT

=

25 000

NPV at 12%

=

$ 6 946

Since the NPV > 0, R.O.I. > 12%. STEP 2

Compute the profitability index to estimate what rate should be used next. PVIN

31 946

MDEX = FV^ = TSOOO = L278 = 127'8% Since at i = 12%, the profitability index is 27.8% more than 100%, the rate of dis¬ count should be increased by 27,8% = 7% approximately. To obtain another even 4

rate, the increase should be either 6% or 8%. In line with the suggestion that it is better to go too high, increase the previous rate by 8% and try i = 20%. STEP 3

Try i = 20%. / l - l 20~7\ PVIN = 7000(-—-J = 7000(3.604592)

=

$25 232

PVOUT

=

25 000

NPV at 20%

=

$

232

Since the NPV > 0, R.O.I. > 20%. STEP 4

Since the net present value is still positive, a rate higher than 20% is needed. The profitability index at 20% is 25 232 = 1.009 = 100.9%. The index exceeds 25 000

100% by 0.9%; division by 4 suggests an increase of 0.2%. For interpolation, the recommended minimum increase or decrease is 2%. The next try should use i = 22%. STEP 5

Try i — 22%. PV1N = 7000^ 1' ~o2Y~ ) = 7000(3.415506)

=

PV0UT

= -$25 000

NPV at 22% Since the NPV < 0, R.O.I. < 22%. Therefore, 20% < R.O.I. < 22%.

$23 909

- ~$ 1091

CHAPTER

16:

INVESTMENT

STEP 6

DECISION

APPLICATIONS

Now that the rate of return has been located between two sufficiently close rates of discount, linear interpolation can be used as illustrated in Example 16.3A. NPV

$232

$0

A

X

B

*

d i d

2

=

22%

20%

232 232 + 1091 232(2) 1323

0.35

The rate of discount for which the NPV = 0 is approximately 20% + 0.35% = 20.35%. The rate of return is approximately 20.3%. (A more precisely computed value is 20.3382%.) Note: Three attempts were needed to locate the R.O.I. between 20% and 22%. Three is the usual number of tries necessary. The minimum number is two attempts. Occasionally four attempts may be needed. To produce a more concise solution, organize the computation as shown below. Since estimates are involved, it is sufficient to use present value factors with only three decimal positions. In the fol¬ lowing examples, all factors are rounded to three decimals. Attempts

Present Value of Amounts in General Form

/ = 22%

i = 20%

/ = 12%

PV,N

Factor

$

Factor

$

Factor

$

1 - (1 + i)~71 7000 --—

4.563757

31 946

3.604592

25 232

3.415506

23 909

PVout 25 000 now

NPV

25 000

25 000

6 946

232

25 000

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Programmed Solution

(Set P/Y = 1; C/Y = 1) Attempts

Present Value of Amounts in General Form

/=

12%

0

FV

PVlr 7000

1

-

(1

+

i)1

7000 PMT

7

N

12

I/Y

20%

/=

0

0 fFVJ -7000 (PMT)

7

20

CPT

-7000 PMT

22 (1/Y

I/Y

20.5

CPT

fcPT ’

0

-7000 PMT

fcPT

FV

-7000 PMT N

LL

7

I/Y

20.34 ; I/Y

( CPT

CPT

PV

PV

20.3

I/Y

PV I

PV

/ = 20.34%

FV

■r'N~"j

7LJLJ

PV PV„

o j;v;

FV

7000 PMT

N

/ = 20.3%

/ = 20.5%

/ = 22%

31 946

25 232

23 909

24 890

25 026

24 999

25 000

25 000

25 000

25 000

25 000

25 000

6946

232

-1091

-110

26

>12%

>20%

0; R.O.I. > 12% 12% < R.O.I. < 14%

STEP 4

d_ = 2

9708 9708 + 17 100

=

9708 26 808

0.362131

d = 2(0.362131) = 0.724 The rate of discount at which the net present value is zero is 12% + 0.72% = 12.72%. The rate of return is 12.7%.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Programmed Solution (SetP/Y = 1;C/Y= 1) Attempts

Present Value of Amounts in General Form

PVIN

-64 000 pMT ]

-64 000 PMT

-64 000 PMTi

-64 OOP!PMT;

/ = 12.68% oCtlJ -64 000 PMT 10

N j

N ]

iof._N j

14©

12 i I/Y j

12.7 [ I/Y ]

12.68 1 I/Y j

CPT ]

| CPT j

fcPTj

i.CPT)

[cpt]

(W)

fPV )

(PV1

10 .£j

10'

20! I/Yj

N

i

10

> u-

i

/ = 12.7% 0

1 - (1 + i)~'°

oLixJ

0

64 000

oO

a

PV of benefits

/= 12%

/= 14%

/ = 20%

QpyJ

1 PV 1

268 318

333 831

361 614

351 484

351 767

320 000

320 000

320 000

320 000

320 000

PV of costs 320 000 now

ofPMTj

96 000(1 + /)"5

96 000 f FV 1

0;

R.O.I. > 20%

42 203 Index = S' TV = 1-310 = 131.0% 32 216 31.0% Increase in rate = —-— = 7.75% or 8% 4 STEP 3

For i = 28%, Index =

NPV < 0;

32 636

= 0.970 = 97.0%

Decrease in rate

step 4

For z = 26%,

R.O.I. < 28%

3% 4

NPV > 0;

0.75% or 2% (rounded up)

R.O.I. > 26%

26% < R.O.I. < 28% STEP 5

1303

2606

1.14499 2276 1303 + 973 The rate of discount for which the net present value is zero is approximately d =

X 2

26% + 1.14% = 27.14%. The rate of return, correct to the nearest tenth of a percent, is 27.1%.

Investors should always be aware of the fact and that higher rates of return (yield) are accompanied typically by higher levels of investment risk. For the most daring investors, high risk/high yield invest¬ ments include derivatives, commodities, precious metals, gemstones, collectible items, common stocks, and growth stocks. For investors more comfortable with moderate risk/moderate yield, investment choices include mutual funds, real estate, corporate bonds, and preferred stocks. Low risk/low yield options such as savings accounts, term deposits, guaranteed investment certificates (GICs), and Canada Savings Bonds (CSBs) are designed to appeal to the most conservative investors.

Excel

Internal rate of return can be determined by using a preprogrammed financial cal¬ culator. For instructions on using this function, refer to the CD-ROM accompany¬ ing this book.

If you choose, you can use Excel’s Internal Rate of Return (IRR) function to

Excel

answer the questions in Part B below. Refer to IRR on the Spreadsheet Template Disk to learn how to use this Excel function. Use linear interpolation to find the approximate value of the rate of return for each of the projects below. State your answer correct to the nearest tenth of a percent. Positive NPV at /

Negative NPV at i

$2350 at 24%

-$1270 at 26%

1. 2.

$850 at

8%

-$370 at 10%

3.

$135 at 20%

-$240 at 22%

4.

$56 at 16%

— $70 at 18%

Find the rate of return for each of the six situations below (correct to the nearest tenth of a percent). 1. The proposed expansion of CIV Electronics’ plant facilities requires the immediate outlay of $100 000. Expected net returns are Year 1: Nil

Year 2: $30 000

Year 3: $40 000

Year 4: $60 000

Year 5: $50 000

Year 6: $20 000

2. The introduction of a new product requires an initial outlay of $60 000. The

anticipated net returns from the marketing of the product are expected to be $12 000 per year for ten years. 3. Your firm is considering introducing a new product for which net returns are expected to be Year 1 to Year 3 inclusive:

$2000 per year;

Year 4 to Year 8 inclusive:

$5000 per year;

Year 9 to Year 12 inclusive:

$3000 per year.

The introduction of the product requires an immediate outlay of $15 000 for equipment estimated to have a salvage value of $2000 after twelve years. @

4. A project requiring an immediate investment of $150 000 and a further outlay of $40 000 after four years has a residual value of $30 000 after nine years. The project yields a negative net return of $10 000 in Year 1, a zero net return in Year 2, $50 000 per year for the following four years, and $70 000 per year for the last three years.

P

(§)

5. You are thinking of starting a hot dog business that requires an initial invest¬ ment of $16 000 and a major replacement of equipment after ten years amounting to $8000. From competitive experience, you expect to have a net loss of $2000 the first year, a net profit of $2000 the second year, and, for the remaining years of the first fifteen years of operations, net returns of $6000 per year. After fifteen years, the net returns will gradually decline and will be zero at the end of twenty-five years (assume returns of $3000 per year for that peri¬ od). After twenty-five years, your lease will expire. The salvage value of equip¬ ment at that time is expected to be just sufficient to cover the cost of closing the business.

PART

@

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

6. The Blue Sky Ski Resort plans to install a new chair lift. Construction is esti¬ mated to require an immediate outlay of $220 000. The life of the lift is esti¬ mated to be fifteen years with a salvage value of $80 000. Cost of clearing and grooming the new area is expected to be $30 000 for each of the first three years of operation. Net cash inflows from the lift are expected to be $40 000 for each of the first five years and $70 000 for each of the following ten years.

Review Exercise 1. Wells Inc. has to choose between two invest¬

5. A real estate development project requires

ment alternatives. Alternative A will return the

annual outlays of $75 000 for eight years. Net

company $20 000 after three years, $60 000

cash inflows beginning in Year 9 are expected

after six years, and $40 000 after ten years.

to be $250 000 per year for fifteen years. If the

Alternative B will bring returns of $10 000 per

developer requires a rate of return of 18%,

year for ten years. If the company expects a

compute the net present value of the project.

return of 14% on investments, which alterna¬ tive should it choose?

6. A company is considering a project that will require a cost outlay of $30 000 per year for

2. A piece of property may be acquired by mak¬

four years. At the end of the project, the com¬

ing an immediate payment of $25 000 and

pany expects to salvage the physical assets for

payments of $37 500 and $50 000 three and

$30 000. The project is estimated to yield net

five years from now respectively. Alternatively,

returns of $60 000 in Year 4, $40 000 in Year 5,

the property may be purchased by making

and $20 000 for each of the following five

quarterly payments of $5150 in advance for

years. Alternative investments are available

five years. Which alternative is preferable if

yielding a rate of return of 14%. Compute the

money is worth 15% compounded semi¬

net present value of the project.

annually? 3. An investor has two investment alternatives.

7. An investment requires an initial outlay of

$45 000. Net returns are estimated to be

If he chooses Alternative 1, he will have to

$14 000 per year for eight years. Determine the

make an immediate outlay of $7000 and will

rate of return.

receive $500 every three months for the next

8. A project requires an initial outlay of $10 000

nine years. If he chooses Alternative 2, he will

and promises net returns of $2000 per year

have to make an immediate outlay of $6500

over a twelve-year period. If the project has a

and will receive $26 000 after eight years. If

residual value of $4000 after twelve years, what

interest is 12% compounded quarterly, which

is the rate of return?

alternative should the investor choose on the basis of the net present value criterion? 4. Replacing old equipment at an immediate cost

of $65 000 and $40 000 five years from now will result in a savings of $8000 semi-annually for ten years. At 14% compounded annually, should the old equipment be replaced?

9. Compute the rate of return for Question 5.

10. Compute the rate of return for Question 6. 11. Superior Jig Co. has developed a new jig for which it expects net returns as follows. Year 1:

$8000

Year 2 to 6 inclusive:

$12 000 per year

Year 7 to 10 inclusive: $6000 per year The initial investment of $36 000 has a resid¬ ual value of $9000 after ten years. Compute the rate of return.

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

12. The owner of a sporting goods store is consid- (|§) 16, The introduction of a new product requires an ering remodelling the store in order to carry a immediate outlay of $45 000. Anticipated net larger inventory. The cost of remodelling and

returns from the marketing of the product are

additional inventory is $60 000. The expected

expected to be $12 500 per year for ten years.

increase in net profit is $8000 per year for the

What is the rate of return on the investment

next four years and $10 000 each year for the following six years. After ten years, the owner plans to retire and sell the business. She expects to recover the additional $40 000

4

(correct to the nearest tenth of a percent)? 17. Games Inc. has developed a new electronic game and compiled the following product information.

invested in inventory but not the $20 000 invested in remodelling. Compute the rate of return.

Production Cost

Promotion Cost

Sales Revenue

$32 000

—

—

Outway Ventures evaluates potential investment

Year 1

projects at 20%. Two alternative projects are

Year 2

32 000

$64 000

$ 64 000

available. Project A will return the company

Year 3

32 000

96 000

256 000

$5800 per year for eight years. Project B will

Year 4

32 000

32 000

128 000

return the company $13 600 after one year,

Year 5

32 000

32 000

$17 000 after five years, and $20 400 after eight

Should the product be marketed if the company

years. Which alternative should the company

requires a return of 16%?

choose according to the discounted cash flow criterion? 14. Project A requires an immediate investment of

18. Farmer Jones wants to convert his farm into a golf course. He asked you to determine his rate of return based on the following

$8000 and another $6000 in three years. Net

estimates.

returns are $4000 after two years, $ 12 000 after

Development cost for each of the first three years, $80 000.

four years, and $8000 after six years. Project B requires an immediate investment of $4000, another $6000 after two years, and $4000 after four years. Net returns are $3400 per year for

Construction of a clubhouse in Year 4, $240 000.

seven years. Determine the net present value at

Upon his retirement in fifteen years, improve¬

10%. Which project is preferable according to

ments in the property will yield him $200 000.

the net present value criterion?

Net returns from the operation of the golf

15. Net returns from an investment are estimated

course will be nil for the first three years

to be $13 000 per year for twelve years. The

and $100 000 per year afterwards until his

investment involves an immediate outlay of

retirement.

$50 000 and a further outlay of $30 000 after six years. The investments are estimated to have a residual value of $10 000 after twelve years. Find the net present value at 20%.

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

Self-Test 1. Opportunities Inc. requires a minimum rate of return of 15% on investment proposals. Two proposals are under consideration but only one may be chosen. Alternative A offers a net return of $2500 per year for twelve years. Alternative B offers a net return of $10 000 each year after four, eight, and twelve years respectively. Determine the preferred alternative according to the discounted cash flow criterion. 2. A natural resources development project requires an immediate outlay of $100 000 and $50 000 at the end of each year for four years. Net returns are nil for the first two years and $60 000 per year thereafter for fourteen years. What is the net present value of the project at 16%? 3. An investment of $100 000 yields annual net returns of $20 000 for ten years. If

the residual value of the investment after ten years is $30 000, what is the rate of return on the investment (correct to the nearest tenth of a percent)? ((d):

4. A telephone system with a disposable value of $1200 after five years can be purchased for $6600. Alternatively, a leasing agreement is available that requires an immediate payment of $1500 plus payments of $100.00 at the beginning of each month for five years. If money is worth 12% compounded

_p @

monthly, should the telephone system be leased or purchased? 5. A choice has to be made between two investment proposals. Proposal A requires an immediate outlay of $60 000 and a further outlay of $40 000 after three years. Net returns are $20 000 per year for ten years. The investment has no residual value after ten years. Proposal B requires outlays of $29 000 in each of the first four years. Net returns starting in Year 4 are $40 000 per year. The residual value of the investment after ten years is $50 000. Which proposal is preferable at 20%?

S’

@)

6. Introducing a new product requires an immediate investment in plant facili¬ ties of $180 000 with a disposal value of $45 000 after seven years. The facilities will require additional capital outlays of $50 000 each after three and five years respectively. Net returns on the investment are estimated to be $75 000 per year for each of the first four years and $50 000 per year for the remaining three years. Determine the rate of return on investment (correct to the nearest tenth of a percent).

Challenge Problems 1. The owners of a vegetable processing plant can buy a new conveyor system for $85 000. The owners estimate they can save $17 000 per year on labour and maintenance costs. They can purchase the same conveyor system with an automatic loader for $114 000. They estimate they can save $22 000 per year with this system. If the owners expect both systems to last ten years and they require at least 14% return per year, should the owners buy the system with the automatic loader?

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

2. CheeseWorks owns four dairies in your province and has planned upgrades for all locations. The owners are considering four projects, each of which is inde¬ pendent of the other three projects. The details of each project—A, B, C, and D—are shown below. _ . . _ Revenues, and Cost Savings at End of: Cost at Beginning _~_ Project

of Year 1

Year 1

Year 2

A

300 000

150 000

120 000

B

360 000

0

40 000

C

210 000

10 000

10 000

D

125 000

30 000

40 000

Year 3

Year 4

Year 5

120 000

0

0

200 000

200 000

200 000

100 000

120 000

120 000

40 000

40 000

40 000

The owners of CheeseWorks have $700 000 to invest in these projects. They expect at least 12% return on all of their projects. In which projects should the owners of a CheeseWorks invest to maximize the return on their investment?

Case Study 16.1 To Lease or Mot to Lease? » Jonathon was just hired for a new job. To travel to his new job, he needed a new car. Reading the newspaper, he noticed an ad for a 2003 Jeep Liberty Sport. It was just the car he wanted. The ad quoted both a cash purchase price of $33 650 and a monthly lease payment option. Since he did not have enough money to pay for a car, he would have to borrow the money from the bank, paying interest of 5.3% compounded annually on the loan. The lease option showed payments of $339 a month for 48 months with a $3300 down payment or equivalent trade. Freight and air tax was included. Jonathon did not currently have a vehicle to offer as a trade-in. If the vehicle was leased, after 48 months it could be purchased for $20 242. The lease was based on a finance interest rate of 5.3%. During the term of the lease, kilometres were limited to 81 600, with a charge of $0.15 per kilometre for excess kilometres. The costs included freight, but excluded taxes, registration, licence, and dealer admin¬ istration charges. Jonathon was particularly impressed with the “seven-years or 115 000-kilometre” warranty on the engine and transmission. They also offered 24-hour roadside assistance. Jonathon must decide whether to buy or lease this car. He lives in a province with a combined PST and GST tax rate of 15%. He realizes that the costs of licence and insurance must be paid, but he will ignore these in his calculations.

QUESTIONS 1. If Jonathon buys the car, what is the total purchase price, including taxes? Assume tax is charged on any additional costs. 2. Since Jonathon has no down payment, he must finance the car if he purchases it. (a) Is it cheaper to borrow the money from the bank or to lease?

PART

3:

MATHEMATICS

OF

FINANCE

AND

INVESTMENT

(b) The dealer is offering a vehicle loan rate of 6% compounded annually. Is it better to buy or lease the car at this rate? (c) Find the rate of interest at which the cost of buying is the same as the cost

of leasing. Calculate your answer to two decimal places 3. Suppose Jonathon has a $2000 down payment for this car.

(a) What is the purchase price of the car if he pays cash for it? Assume the down payment is subtracted from the price of the car including tax.

(b) If the monthly lease payment is $279.00, is it cheaper to lease or buy the car if Jonathon can get the special dealer rate of 5%? (c) Is there an advantage to having a $2000 down payment if you want to

lease this car? Why or why not?

Case Study 16.2 Building a Business » MAS Manufacturing has demolished an old warehouse to make room for addi¬ tional manufacturing capacity. The company has decided to construct a new build¬ ing, but must decide how to proceed. It has two alternatives for the new building, both of which will create a building with an expected life of fifty years. The resid¬ ual value is unknown but will be the same for either alternative. Alternative A is to construct a new building that would have 180 000 square feet. Construction costs will total $2 100 000 at the end of Year 1. Maintenance costs are expected to be $15 000 per year. The building will need to be repainted every ten years (starting in ten years) at an estimated cost of $10 000. Alternative B is to construct the building in two stages: build 100 000 square feet now; and add 80 000 square feet in ten years. Construction costs for the first stage will be $1 600 000 at the end of Year 1. Construction costs for the second stage will be $900 000 when the addition is completed at the end of Year 10. Maintenance costs are expected to be $10 000 per year for the first ten years, then $17 000 per year after that. The building and the addition will have to be painted every ten years, beginning in Year 20, at an estimated cost of $10 000.

QUESTIONS 1. Suppose the company’s required rate of return is 15%.

(a) What is the present value of Alternative A? (b) What is the present value of Alternative B? (c) Which alternative would you recommend on the basis of your discount¬

ed cash flow analysis? 2. Which alternative would you recommend if the company’s required rate of return was 20%? Show all calculations. 3. Suppose the company could rent a portion of its building for $48 000 per year for the first ten years if it chose Alternative A. (a) If the company’s required rate of return is 15%, what is the net present value of Alternative A?

(b) On the basis of the new information, would you recommend Alternative A or Alternative B if the company’s required rate of return is 15%?

CHAPTER

16:

INVESTMENT

DECISION

APPLICATIONS

SUMMARY OF FORMULAS Formula 16.1 Formula for finding the difference between the present value of cash

NET present VALUE _ PRESENT VALUE _ PRESENT VALUE (npv)

OF INFLOWS

OF OUTLAYS

inflows and the present value of cash outflows, known as the net pre¬ sent value.

Formula 16.2 Formula for finding the relationship PROFITABILITY INDEX =

PRESENT VALUE OF INFLOWS

PVlN

PRESENT VALUE OF OUTLAYS

PVqut

by dividing the present value of cash inflows by the present value of cash outflows, known as the profitability index.

In addition, Formulas 9.IB, 9.1C, 11.2, 12.3, 13.2, and 13.4 were used in this chapter.

GLOSSARY Discounted benefit-cost ratio

see Profitability

Profitability index

the ratio of the present value

of the inflows (benefits) to the present value of the

index Discounted cash flow

the present value of cash

outlays (costs) of a capital investment project (p. 716)

payments (p. 698) Net present value

the difference between the

Rate of return

the rate of discount for which the

present value of the inflows (benefits) and the pre¬

net present value of a capital investment project is

sent value of the outlays (costs) of a capital invest¬

equal to zero (p. 716)

ment project (p. 707)

USEFUL INTERNET SITES www.tse.com TSX Group

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A comprehensive financial glossary with definitions and related terms

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A full education site that also contains quotes, charts, historical analysis, and reports.

Review of Basic Algebra 1.1 Basic Laws, Rules, and Definitions A. The fundamental operations The fundamental operations of algebra are addition, subtraction, multiplication, and division. The symbols used to show these operations are the same as the symbols used in arithmetic. For any two numbers ‘a’and ‘b’, the fundamental operations are as follows. 1. Addition is denoted by‘a + Farid referred to as the sum of ‘a’ and ‘b If a = 7 and b = 4, then a + b = 7 + 4= 11. . Subtraction is denoted by ‘a — F and referred to as the difference between V

2

and ‘bl If a = 7 and b — 4, then a — b = 7 — 4 = 3. 3. Multiplication is denoted by‘a X F or‘(a){bY or ‘ab! ‘a’ and ‘F are called factors and ‘aF is referred to as the product of‘a’and ‘b.’ If a = 7 and b = 4, then ab — (7)(4) = 28. . Division is denoted by ‘a:F or ‘-7-’ or ‘a/fo.’ V is the dividend, ‘F is the divisor, .ca,. . . ' 0 and y 1S ^e quotient.

4

If a = 7 and b = 4, then -7- = -7. b 4

B. Basic laws The basic laws governing algebraic operations are the same as those used for arithmetic operations.

1. The Commutative Laws for Addition and Multiplication (a) When adding two numbers, the two numbers (addends) may be interchanged. a + b = b + a

Formula 1.1

If a = 7 and b — 4, then 7 + 4 = 4 + 7 = 11.

(b) When multiplying two numbers, the two factors may be interchanged. ab = ba

-Formula 1.2

If a = 7 and b = 4, then (7)(4) = (4)(7) = 28.

APPENDIX

I

2. The Associative Laws for Addition and Multiplication (a) When adding three or more numbers, the numbers (addends) may be combined in any order. a + b + c

(a + b) + c = a + (b + c) — b + (a + c) — Formula 1.3

If a = 7, b = 4, and c = 2, then 7 + 4 + 2 = (7 + 4)+ 2 = 7+ (4 + 2) = 4 + (7 + 2) = 13.

(b) When multiplying three or more numbers, the numbers (factors) may be combined in any order. Formula 1.4

abc = (ab)c = a(bc) = b(ac)

If a = 7, b = 4, andc = 2, then 7 X 4 X 2 = (7 X 4) X 2 = 7 X (4 X2) = 4 X (7 X 2) = 56.

3. The Distributive Law of Multiplication over Addition The product of‘a’times the sum of‘b’and V is equal to the sum of the prod¬ ucts ‘ah' and ‘ac. Formula 1.5

a{b + c) = ab + ac

If a = 7, b = 4, and c = 2, then 7(4 + 2) = 7 X 4 + 7 X 2 = 42.

4. Special Properties of '1' (a)

a X 1 = 1 X a = a

When any number ‘a’ is multiplied by ‘1,’ the product is the number ‘a.’

If a = 5, then 5 X 1 = 1 X 5 (b)

f = a

When any number V is divided by£l,’ the quotient is the number ‘a.

If a = 5, then y = 5. (c)

" = 1 a

When any number V is divided by itself, the quotient is ‘I.’

If a = 5, then y = 1.

5. Special Properties of '0' (a) Addition with ‘O' a + 0 — 0 + a = a

When ‘0’ is added to any number ‘a, the sum is the number ‘a'.

REVIEW

OF

BASIC

ALGEBRA

If a = 5, then 5 + 0 = 0 + 5 = 5.

(b) Subtraction with ‘O' a — 0 = a

When ‘0’ is subtracted from any number ‘a’ the difference is the number ‘a'.

— 5,then 5 — 0 When any number ‘a’ is subtracted from

0 — a = —a

‘0,’ the difference is the inverse value of ‘a’, that is, ‘a’with the sign changed. If a — 5, then 0 — 5 = —5. (c) Multiplication with ‘O'

flX0=0Xa=0

When ‘0’ is multiplied by any number ‘a,’the product is ‘O’

If a = 5, then 5X0 = 0X5 = 0. (d) Division with ‘O’ (i)

®

When ‘0’ is divided by any number ‘a’

°=0 a

other than ‘0,’ the quotient is ‘0.’

q = undefined

Division by‘O’ has no meaning.

5

If a = 5, then — = undefined.

C. Definitions 1.

An algebraic expression is a combination of numbers, variables representing numbers, and symbols indicating an algebraic operation. 3 1 7ab, 3a - 5b, x2 3 - 3x + 4, ^x - ~y

2.

are algebraic expressions.

A term is a part of an algebraic expression separated from other parts by a positive ( + ) sign or by a negative ( —) sign. The preceding ( + ) sign or ( —) sign is part of the term. The terms for the algebraic expressions listed in part (1) are 7ab;

3.

3a and —5b\

3 1 x2, ~3x, and +4; —x and ——y

A monomial is an algebraic expression consisting of one term, such as lab. A binomial is an algebraic expression consisting of two terms, such as . 3 1 3a — 5b or — x — —y. A trinomial is an algebraic expression consisting of three terms, such as x2 — 3x + 4. A polynomial is an algebraic expression consisting of more than one term.

#

APPENDIX

I

4.

A factor is one of the numbers that when multiplied by another number or numbers yields a given product. The factors of the term lab are 7, a, and b.

5. A factor of a term is called the coefficient of the rest of the term. In the term lab,

7 is the coefficient of ab, la is the coefficient of b, lb is the coefficient of a.

6. The numerical coefficient is the part of a term formed by numerals. In the term lab, the numerical coefficient is 7; in the term x2, the numerical coefficient is understood to be 1 (1 is usually not written); in the term —yy, the numerical coefficient is —— (the sign is considered to be part of the numerical coefficient). 7. The literal coefficient of a term is the part of the term formed with letter symbols. In the term lab, ab is the literal coefficient; in the term 3x2, x2 is the literal coefficient. 8. Like terms are terms having the same literal coefficients. la, — 3a, a, —~a are like terms; 1 f x2, —2x1, --x2, 5x2 are like terms. 9. Combining like terms or collecting like terms means adding like terms. Only like terms can be added. 10. Signed numbers are numbers preceded by a positive ( + ) or a negative ( —) sign. Numbers preceded by a positive ( + ) sign are called positive numbers, while numbers preceded by a negative ( —) sign are called negative numbers. 11. Like signed numbers are numbers that have the same sign while numbers with different signs are called unlike signed numbers. + 7 and +8 are like signed numbers; — 7 and —8 are like signed numbers; + 7 and —8 are unlike signed numbers; — 7 and 8 are unlike signed numbers. Note: If no sign is written in front of a number, a plus ( + ) sign is understood to precede the number. ‘6’ means ‘ + 6.’ 12. The absolute value of a signed number is the value of the number without the sign and is denoted by the symbol | |. The absolute value of +5 = | + 5| = 5; the absolute value of —5 = | —5| = 5.

REVIEW

OF

BASIC

EXERCISE 1.1 Answer each of the following questions. 1.

2.

List the terms contained in each of the following expressions, (a) —3xy

(b) 4a — 5c — 2d

(c) x2 — yx ~ 2

(d) 1.2x — 0.5xy + 0.9y — 0.3

Name the numerical coefficient of each of the following terms, (a) —3b

(b) 7c

(c) —a

(d) x

(e) 12a2b

(f) —3ax

(g) —yx2

(h) y

3. Name the literal coefficient of each of the following. (a) 3x

(b) ab

(c) —4y

(d) —xy

(e) -15X2/

(f) 3.5abx

(g) yx3

(h) yr-

1.2 Fundamental Operations with Signed Numbers A. Additions with signed numbers 1. Addition of like signed numbers To add like signed numbers, (i) add their absolute values, and (ii) prefix the common sign. Add each of the following. (i)

—6

and

—8

The absolute values are the sum of 6 and

8

6

and 8;

is 14;

the common sign is ( —). (-6) + (-8) = -6 - 8 = -14 (ii) +6, +5, and + 12 The absolute values are 6, 5, and 12; the sum of 6, 5, and 12 is 23; the common sign is ( + ). ( + 6) + ( + 5) + ( + 12) =

+6

(iii) —9, —3, — 1, and -15

+ 5 + 12 = +23, or 23

ALGEBRA

APPENDIX

I

The absolute values are 9, 3, 1, and 15; the sum of the four numbers is 28; the common sign is (—). (-9) + (-3) + (-1) + (-15) = - 9 - 3 - 1 - 15 = -28 2. Addition of unlike signed numbers To add unlike signed numbers, (i) subtract the smaller absolute value from the larger absolute value, and (ii) prefix the sign of the larger absolute value. Add each of the following. (i)

8

and —5

The absolute values are

8

and 5;

the difference between the absolute values is 3; the sign of the larger absolute value is ( + ). ( + 8) + (-5) =

+8

- 5 = +3, or 3

(ii) 4 and —9 The absolute values are 4 and 9; the difference between the absolute values is 5; the sign of the larger absolute value is (—). (+4) + (-9) = 4 - 9 = -5 (iii) —6,+8,+3,—4, and—5 When more than two numbers are involved and unlike signs appear, two approaches are available. METHOD

1

Add the first two numbers and then add the sum to the next number and so on. (-6) + ( + 8) + ( + 3) + (-4) + (-5) = —6 + 8 + 3 — 4 — 5 = +2 + 3 — 4 — 5 = +5 — 4 — 5 -add +2 and +3, which equals +5 — +1 ~ 5-

METHOD 2

-add +5 and —4, which equals +1

= —4 -— add +1 and —5, which equals —4 First add the numbers having like signs and then add the two resulting unlike signed numbers. (-6) + ( + 8) + ( + 3) + (-4) + (-5) = —6 + 8 + 3 — 4 — 5 = (—6 — 4 — 5) + (+ 8 + 3)

= (-15) + ( + 11) = -15 + 11 = -4

REVIEW

OF

BASIC

ALGEBRA

B. Subtraction with signed numbers The subtraction of signed numbers is changed to addition by using the inverse of the subtrahend. Thus, to subtract with signed numbers, change the sign of the subtrahend and add.

Perform each of the following subtractions, (i) ( + 6) from (4) (+4) - (+6) = (+4) + (-6) = +4 - 6 =

-2

change the subtrahend (+6) to (—6) and change the subtraction to an addition use the rules of addition to add +4 and — 6

(ii) ( — 12) from ( + 7) ( + 7) - (-12) = ( + 7) + ( + 12) = +7 + 12

and add

= 19 (iii) ( + 9) from (—6) (-6) - ( + 9)

= (-6) + (-9) = -6 - 9 = -15

C. Multiplication with signed numbers The product of two signed numbers is positive or negative according to the following rules. (a) If the signs of the two numbers are like, the product is positive.

( + )( + ) = ( + ) (-)(-) = (+) (b) If the signs of the two numbers are unlike, the product is negative.

( + )(-) = (-)

(-)(+) = (-) (i) ( + 7)(+4) - 28

the signs are like (both positive); the product is positive

(ii) (—9)(—3) = 27

the signs are like (both negative); the product is positive

(iii) (-8)(3) = -24

the signs are unlike; the product is negative

(iv) (7)( —1) — —7 negative

(v) ( — 8)(0) = 0 —=— (vi) (

7)(3)( — 4)

— the product of any number and 0 is 0

= (— 21)(—4) = 84

(vii) (—2)( —1)(—4)(3) = (2)(—4)(3) = (

)(3) = -24

8

Note: Brackets around one or both numbers indicate multiplication.

D. Division with signed numbers The quotient of two signed numbers is positive or negative according to the following rules. (a) If the signs are like, the quotient is positive.

(+)-( + ) = ( + ) (-)-(-) = ( + ) (b) If the signs are unlike, the quotient is negative.

( + )-(-) = (-) (-)-( + ) = (-) 15.

-tr

(+5) = 3--the signs are like; the quotient is positive

(—24) -^ (—4) = 6

-the signs are like; the quotient is positive

( —18)

2 = — 9 -the signs are unlike; the quotient is negative

(12)

s-

( — 1) — —12 --the signs are unlike; the quotient is negative

(v) 0

(-10) = 0

0 divided by any number is 0

(vi) ( —16) -T- 0 = undefined

division by 0 has no meaning

-5-

REVIEW

OF

BASIC

ALGEBRA

E. Absolute value of signed numbers The absolute value of signed numbers, denoted by | |, is the value of the numbers without the signs.

(i)|-7| = 7 (ii) |— 3 +

| = | + 5| = 5

8

(hi) |4-9| = |

5| = 5

(iv) | —9 - 4| = | — 13| - 13 (v) |4(

7)| = |-28| = 28

(vi) |(

9)(

3)| = |27| = 27

(vii) |(

12) = (4)| = | — 3| = 3

(viii) |(

30) = (-5)| = |+6| =

6

EXERCISE 1.2 Simplify. 1. ( + 3) + ( + 7)

2. (+12) + ( + 6)

3. (-5) + (-9)

4. (-15) + (-12)

5. 4 + ( + 5)

6. ( + 6) + 8

7. -8 + (-7)

8. (-18) - 7

9. +3 + 14

10. +12 13.

-8

+ 1

+ 3

16. 0-9

11. -6-9

12. -14 - 3

14. -12 + 16

15. 8-12

17. 1 - 0.6

18.

19. (-4) + (6) + (-3) + ( + 2) . -3 - 7 + 9 +

21

-

0.02

20. 12 + (- 15) + ( + 8) + (-10)

- 5

6

1

22. 10 -

8

- 12 + 3 - 7

Simplify. 1. ( + 9) - ( + 8)

2. ( + 11) - ( + 14)

3. ( + 6) - (-6)

4. ( + 11) - (-12)

5. (-8)-(-7)

6

7. (-4)-(+6)

8. ( — 15) — (+3)

9. 0 - (-9)

10. 1-(-0.4)

11. 1-(-0.03)

. (-9) - (-13)

12. 0 - ( + 15)

13. 6 - (-5) + (-8) - ( + 3) + (-2) 14. -12 - (-6) - ( + 9) + (-4) - 7 Simplify. 1

2. 11( + 3)

5. ( + 7)( — 1)

6

. ( + 5)(+4)

9. 0( —6)

. 10( — 5)

10. -12(0)

3. ( —4)( —6)

4. —7( —3)

7. -3(12)

8

. 6(—4)( —3)(2)

11

. -9(1)

12. —3(5)(—2)( —1)

APPENDIX

I

Simplify. 1. ( + 18) - ( + 3)

2. (32) + ( + 4)

3. ( + 45) + (-9)

4. (63) - (-3)

5. (-28)

( + 7)

6.

7. (-16) 4- (-1)

8. (-48)

(-8)

9. 0 + (-5)

11. ( + 4) -i- 0

10. 0 + 10;

(-36)

-5-

( + 12)

12. (-12) - 0

Simplify. 1. | — 9|

2. | + 4|

5. |—7 - 8|

6. |o “3|

9. |20 -r ( — 5)|

1.3

I

3. 6 — io| 7. l(--3) X 3|

4. |—5 + 121

8. |4 X (-5)|

10. l(--35) -- (7)|

Common Factoring

A. Basic concept In arithmetic, certain computations, such as multiplication and division involving common fractions, are helped by factoring. Similarly, algebraic manipulation can be made easier by the process of finding the factors that make up an algebraic expression. Factoring an algebraic expression means writing the expression as a product in component form. Depending on the type of factors contained in the expression, the process of factoring takes a variety of forms. Only the simplest type of factor¬ ing applies to the subject matter dealt with in this text. Accordingly only this type, called common factoring, is explained in this section. A common factor is one that is divisible without remainder into each term of an algebraic expression. The factor that is common to each term is usually found by inspection; the remaining factor is then obtained by dividing the expression by the common factor.

B. Examples Factor 14a + 21 b. By inspection, recognize that the two terms 14a and 21 b are both divisible by 7. The common factor is 7. The second factor is now found by dividing the expression by 7. 14a + 21b 7

14a , 7

21 b 7

— 2a + 3b

Thus the factors of 14a + 21b are 7 and 2a + 3b. 14 a + 21b = 7(2a + 3b)

REVIEW

OF

BASIC

ALGEBRA

Factor 18a — 45. By inspection, the highest common factor is 9; , . c 18a — 45 the second factor is---= 2a — 5. 18a - 45 * 9(2a - 5) Note: If 3 is used as the common factor, the second factor, 6a — 15, contains a common factor 3 and can be factored into 3(2a — 5). Thus, 18a — 45 = 3[6a — 15] = 3[3(2a - 5)] = 9(2a - 5) When factoring, the accepted procedure is to always take out the highest common factor.

Factor mx — my. The common factor is m; , , . mx — my the second factor is-= x — y. m mx — my = m(x — y)

Factor 15X3 — 25X2 — 20x. The common factor is 5x. The second factor is •

15x3 - 25X2 - 2Ox — v2 3x2 — 5x — 4.

5x

15X3 — 25x2 — 20x = 5x (3x2 — 5x — 4) Factor P + Prt. The common factor is P. , r . P + Prt The second factor is--P + P rt= P(1 + rt)

^P

P

P rt

1 + rt.

APPENDIX

I

Factor a(x + y) — b(x + y). The common factor is (x + y). b{x + y)

a(x + y)

The second factor is ^x + ^+ /) x + y

x + y

x

a(x + y) — b(x + ’y) = (x + y){a — b)

Factor (1 + i) + (1 + i)2 + The common factor is The second factor is

_ (1 + 0 (1

+ z)

,

(1 (1

(1

(1

(1

+ z')3.

+ i).

+ 0 + (1 + 02 + (1 + i)3 (1 + z)

+ j)2 + z)

(1 (1

= 1 + (1 + z) + (1 + z)2 + i) + (1 + i)2 + (l +

(1

z')3

+ z')J + z) =

+ z)

(1

[1

+

(1

+ i) +

(1 + z)2]

EXERCISE L3

O

Factor each of the following. .

1

o

x -

8

12

2. 27 - 36a

3. 4n2 — 8n

4. 9x2 — 2lx

5. 5ax — 10ay — 20a

6. 4ma — 12 mb + 24mab

Factor each of the following. . mx + my

1

2. xa — xb

3. m(a — b) + n(a — b)

4. k(x

5. P + Pz

6. A — Adt

7. r — r2 — r3

8

.

(1

—

1) — 3(x — 1)

+ z')4 +

(1

+ z’)3 +

(1

+ i)2

- b.

REVIEW

1.4

OF

BASIC

ALGEBRA

Graphing Inequalities

A. Basic concepts and method A straight line drawn in a plane divides the plane into two regions: Y axis 1t

y/rL

Region to the left of the line L

/

Region to the right of the line L

/ ’

(a) the region to the left of the line drawn in the plane; (b) the region to the right of the line drawn in the plane. When a system of axes is introduced into the plane, each region consists of a set of points that may be represented by ordered pairs (x, y). Relative to the divid¬ ing line, the two sets of ordered pairs (x, y) that represent the points in the regions are defined by the two inequalities associated with the equation of the dividing line. For the equation x = 5, the associated inequalities are x < 5 (x is less than 5) and x > 5 (x is greater than 5). For the equation y — —3, the associated inequali¬ ties are y < -3 and y> -3. For the equation 2x + 3y= 6, the associated inequal¬ ities are 2x + 3y < 6 and 2x + 3y > 6. Graphing an inequality means identifying the region that consists of the set of points whose coordinates satisfy the given inequality. To identify this region, use the following method. 1. Draw the graph of the equation associated with the inequality. 2. Test an arbitrarily selected point that is not a point on the line by substituting its coordinates in the inequality. The preferred point for testing is (0, 0). If (0, 0) is not available because the line passes through the origin, try the points (0, 1) or (1, 0). 3. (a) If substituting the coordinates of the selected point in the inequality yields a mathematical statement that is true, the selected point is a point in the region defined by the inequality. Thus, the region is identified as the area containing the selected point. (b) If substituting the coordinates of the selected point in the inequality yields a mathematical statement that is false, the selected point is not a point in the region defined by the inequality. Thus, the region defined by the inequality is the area that does not contain the point tested.

APPENDIX

I

B. Graphing inequalities of the form ax+ by> c and ax+ by< c

Graph each of the following inequalities. (i) x — y> — 3

(ii) 3x + 2y < —8

(i) The equation associated with the inequality x — y> —3 is x — y = —3. Table of values X

0

-3

2

y

3

0

5

Note: To show that the coordinates of the points on the line x — y = —3 do not satisfy the inequality, the graph of the equation is drawn as a broken line. Since the line does not pass through the origin, the point (0, 0) may be used for testing. Substituting x — 0 and y = 0 in the inequality x — y > — 3 yields the statement 0 - 0 > -3 0 > -3 Since the statement 0 > —3 is true, the point (0, 0) is a point in the region. The region defined by the inequality x — y > — 3 is the area to the right of the line as shown in the diagram below. Y axis

(ii) The equation associated with the inequality 3x + 2y < — 8 is 3x + 2y — —8. Table of values X

0

-2

-4

y

-4

-1

2

REVIEW

OF

BASIC

ALGEBRA

Y axis

Testing the point (0, 0) 3(0) + 2(0) < -8

0 + 0 < -8 0 < -8 Since the statement 0 < —8 is false, the point (0, 0) is not a point in the region defined by 3x + 2y < —8. The region defined by the inequality is the area to the left of the line as shown.

C. Graphing inequalities of the form ax> by or ax< by

Graph each of the following inequalities. (i) y < —x

(ii) 3x < 2y

(i) The equation associated with the inequality / ^ — x is y = —x. Table of values X

0

3

-3

y

0

-3

3

Note: The inequality includes y = — x. Because the points on the line meet the condition stated, the graph of the equation is drawn as a solid line.

Since the line passes through the origin, the point (0, 0) cannot be used for testing. Instead, we test (1,0). Substituting x = 1, y = 0 in the inequality y < — x yields the statement

0 < -1 Since the statement 0 < — 1 is false, the point (1,0) is not a point in the region defined byy < —x. The region defined by the inequality is the area to the left of the line including the line. (ii) The equation associated with the inequality 3x < 2y is 3x = 2y. Table of values X

0

2

-2

y

0

3

-3 Y axis

Since (0, 0) is on the line, test (0, 1). Substituting x = 0, y = 1 in the inequality 3x < 2y yields the statement

0 < 2 Since the statement 0 < 2 is true, the point (0, 1) is a point in the region defined by 3x < 2y. The region defined by the inequality is the area to the left of the line as shown.

REVIEW OF BASIC ALGEBRA

D. Graphing inequalities involving lines parallel to the axes Graph each of the following inequalities. (i) x < 3

(ii) / > — 3

(i) The equation associated with the inequality x < 3 is x = 3. The graph of x 3 is a line parallel to the Y axis three units to the right. Y axis 4--

Region defined

3--

by x < 3

2--

x= 3

1--

H-1-1-I-5 -4 -3 -2 -1 0

H-1-1— 1 2 3 4

"► X; 5

-r

-2‘ 3-

-

-4- -

Test (0, 0). Substituting x = 0 in the inequality x < 3 yields the statement 0 < 3 Since the statement 0 < 3 is true, (0,0) is a point in the region defined by the inequality x < 3. The region defined by the inequality is the region to the left of the line as shown. (ii) The equation associated with the inequality/ 2: — 3 is y = —3. The graph of y = —3 is a line parallel to the X axis three units below it. Test (0, 0). Substituting / = 0 in the inequality / > — 3 yields the statement 0 > -3 Since the statement 0 > -3 is true, (0, 0) is a point in the region defined by / > —3. The region defined by the inequality is the area above the line includ¬ ing the line as shown. Y axis 4

Region defined by y 3= —3

3 2

1 - 0 is y — 0. The graph of y = 0 is the X axis. The region defined by the inequality / > 0 is the area above the X axis and includes the points forming the X axis. Y axis

The equation associated with the inequality 4x + 5y^ 20 is 4x + 5y= 20. The graph of the equation is the line passing through the points A(0, 4) and B(5, 0). The true statement 0 < 20 shows that the origin is a point in the region defined by the inequality. The region defined by 4x + 5y^ 20 is the area to the left of the line and includes the line itself. The equation associated with the inequality 4x — 3y S —12 is 4x — 3y — —12. The graph of this equation is the line passing through the points A(0, 4) and C( —3, 0). The true statement 0 > -12 shows that the origin is a point in the region defined by the inequality. The region defined by 4x — 3y a —12 is the area to the right of the line, including the line itself. The region defined by the three inequalities is the area formed by the intersec¬ tion of the three regions. It is the triangle ABC shown in the diagram.

EXERCISE 1.4 Graph each of the following inequalities. 1. x + y > 4

2. x~y< -2

3. x - 2y < 4

4. 3x- 2y> -10

5. 2x< -3y

6. 4y a 3x

7. x>-2

8. y < 5

Graph the region defined by each of the following linear systems. 1. y < 3 and x + y > 2

2. x - 2y < 4 and x > -3

3. 3x — y < 6 and x + 2y > 8

4. 5x > -3y and 2x - 5y > 10

5. 2y - 3x < 9, x < 3, and y a 0

6. 2x + y < 6, x > 0, and y a 0

7. y a — 3x, y < 3, and 2x - y < 6

8. 2x < y,x > -3y, and x - 2y > -6

APPENDIX

I

SUMMARY OF FORMULAS (LAWS) Formula 1.1 a + b = b + a

The commutative law for addition that permits the addition of two numbers in any order

Formula 1.2 ab = ba

The commutative law for multiplication that permits the multiplication of two numbers in any order

Formula 1.3 a + b + c = (a + b) + c = a + (b + c)

The associative law for addition that permits the addition of three or more numbers in any order

= b + (a + c) Formula 1.4 The associative law for multiplication that permits

abc = (ab)c = a(bc)

the multiplication of three or more numbers in any

= b(ac)

order

Formula 1.5 a(b + c) = ab + ac

The distributive law of multiplication over addition that provides the basis for the multiplication of algebraic expressions

GLOSSARY Absolute value its sign

the value of a number without

Algebraic expression

a combination of numbers,

variables representing numbers, and symbols indi¬ cating an algebraic operation

(p. 737)

(p. 737)

Collecting like terms Combining like terms Common factor

Monomial

(p. 738)

an algebraic expression consisting

of one term

(p. 737)

Negative numbers

an algebraic expression consisting of

two terms

the part of a term formed

with letter symbols

(p.738)

Binomial

Literal coefficient

signed numbers preceded

by a minus ( —) sign

(p. 738)

Numerical coefficient adding like terms

(p.738)

see Collecting like terms

a factor that is divisible without

with numerals Polynomial

the part of a term formed

(p. 738)

an algebraic expression consisting

of more than one term

(p. 737)

remainder into each term of an algebraic expression

Positive numbers

signed numbers preceded

(p. 744)

by a plus ( + ) sign

(p. 738)

one of the numbers that when multiplied

Signed numbers

numbers preceded by a plus

with the other number or numbers yields a given

( + ) or by a minus ( —) sign

product

Term

(p. 738)

equality

a mathematical statement involving

a part of an algebraic expression separated

from other parts by a plus ( + ) sign or by a minus

relationships between variables described as

(-)sign

“greater than” or “less than”

Trinomial

Like signed numbers sign

(p. 747)

numbers having the same

(p. 738)

Like terms coefficient

terms having the same literal (p. 738)

(p.738)

(p.737) an algebraic expression consisting

of three terms

(p. 737)

Unlike signed numbers different signs

(p.738)

numbers having

Instructions and Tips for Four Preprogrammed Financial Calculator Models Different models of financial calculators vary in their operation and labelling of the function keys and face plate. This appendix provides you with instructions and tips for solving compound interest and annuity problems with these financial cal¬ culators: Texas Instruments BAII Plus and BA-35 Solar, Sharp EL-733A, and Hewlett-Packard 10B. The specific operational details for each of these calculators are given using the following framework: A. Basic Operations 1. Turning the calculator on and off 2. Operating modes 3. Using the Second function 4. Clearing operations 5. Displaying numbers and display formats 6. Order of operations 7. Memory capacity and operations 8. Operating errors and calculator dysfunction B. Pre-Calculation Phase (Initial Set-up) 1. Setting to the financial mode, if required 2. Adjusting the calculator’s interest key to match the text presentation, if required 3. Setting to the floating-decimal-point format, if required 4. Setting up order of operations, if required C. Calculation Phase 1. Clearing preprogrammed registers 2. Adjusting for annuities (beginning and end of period), if required 3. Entering data using cash flow sign conventions and correcting entry errors 4. Calculating the unknown variable D. Example Calculations 1. Compound interest 2. Annuities E. Checklist for Resolving Common Errors Go to the section of this appendix that pertains to your calculator. You may want to flag those pages for easy future reference. II. 1. Texas Instruments BAII Plus Advanced Business Analyst, page 756 II. 2. Texas Instruments BA-35 Solar Business Analyst, page 760 II. 3. Sharp Business/Financial Calculator EL-733A, page 764 II. 4. Hewlett-Packard 10B Business, page 767

ft* APPENDIX

II

II. 1. Texas Instruments BAII Plus Advanced Business Analyst A. Basic operations 1. Turning the calculator on and off The calculator is turned on by pressing foN/QFF i . If the calculator was turned off using this key, the calculator returns in the standard-calculator mode. If the Automatic Power Down (APD) feature turned the calculator off, the calculator will return exactly as you left it—errors and all, if that was the case. The calculator can be turned off either by pressing i ON/OFF; again or by not pressing any key for approximately 10 minutes, which will activate the APD feature. 2. Operating modes The calculator has two modes: the standard-calculation mode and the promptedworksheet mode. In the standard-calculation mode, you can perform standard math operations and all of the financial calculations presented in this text. This is the default mode for your calculator. Refer to your calculator’s Guidebook to learn more about the worksheet mode, since it is not addressed in this appendix. 3. Using the Second function The primary function of a key is indicated by a symbol on its face. Second func¬ tions are marked on the face plate directly above the keys. To access the Second function of a key, press ( 2nd ) (“2nd” will appear in the upper left corner of the dis¬ play) and then press the key directly under the symbol on the face plate (“2nd” will then disappear from the display). 4. Clearing operations

—*J clears one character at a time from the display, including decimal points. (ce/c| clears an incorrect entry, an error condition, or an error message from the display. | 2nd j (QUIT) clears all pending operations in the standard-calculation mode and returns the display to 0.

ICE/C - fcE/c] clears any calculation you have started but not yet completed. j 2nd ; (CLR TVM) sets the financial function registers to 0 and returns to stan¬

dard-calculation mode. 5. Displaying numbers and display formats The display shows entries and results up to 10 digits but internally stores numeric values to an accuracy of 13 digits. The default setting in the calculator is 2 decimal places. To change the number of fixed decimal places, press QZndJ (Format) along with a number key for the decimal places desired. Then press

lENTERj

to complete

INSTRUCTIONS

AND

TIPS

FOR

FOUR

PREPROGRAMMED

FINANCIAL

CALCULATOR

the installation. For a floating-decimal-point format, press f2ncl ENTER . Return to standard-calculation mode by pressing [ 2nd

MODELS

(Format) Q*

(QUIT).

6. Order of operations The default for the BAII Plus is Chn. To change to AOS, which will have the calcu¬ lator do all mathematical calculations in the proper order according to the rules of mathematics, press (2nd Chn press [_2yid

[Formatl, arrow down four times, and with display on

( SET ) to change display to AOS, ( 2nd ]

Qud

to go back to the

standard-calculation mode. 7. Memory capacity and operations The calculator has ten memory addresses available, numbered 0 through 9. To store a displayed value in a memory address (0 through 9), press ( STO j and a digit key ( 0

through [ 9

and a digit key [ o

. To recall a value from memory and display it, press ( RCI. through ( 9

. The numeric value is displayed but is also

retained in that memory address. To clear each memory address individually, store “0” in each selected memory. To clear all of the addresses at the same time, press [^>d

(MEM) f2nd

(CLR

Work). Memory arithmetic allows you to perform a calculation with a stored value and then store the result with a single operation. You may add, subtract, multiply, divide, or apply an exponent to the value in the memory. Use this key sequence:

(number in display) (STo] ( + } (or ( - } or [ digit key Toy to

CD

or

CD or 1 y*l

) and a

for the memory address.

8. Operating errors and calculator dysfunction The calculator reports error conditions by displaying the message “Error n,” where n is a number that corresponds to a particular error discussed in the calculator’s Guidebook on pages 80-82. Errors 4, 5, 7, and 8 are the most common financial calculation errors. A list of possible solutions to calculator dysfunction is given on page 87 of the Guidebook. Generally, if you experience difficulties operating the calculator, press [ 2nd calculations.

(Reset) [ENTER to clear the calculator, and repeat your

B. Compound interest and annuity calculations The BAII Plus calculator can be used for virtually all compound interest calcula¬ tions using the third row of the calculator after the payment and interest schedules have been set up in the Second function area of the calculator. Each key represents one of the variables in the formula. The variables are: •

N—Represents time. The value is arrived at by taking the number of years involved in the transaction and multiplying it by the value set up in P/Y.

•

I/Y—The stated or nominal yearly interest rate.

•

PV—The amount of money one has at the beginning of the transaction.

•

PMT—The amount of money paid on a regular basis.

•

FV—The amount of money one has at the end of the transaction.

To perform compound interest or annuity calculations, the process will be to input the variables that are known and to compute the unknown variable. For compound interest, the process will be: 1. Set up the payment and interest schedules in the Second function of the cal¬ culator. This is done by going (

2nd

) f P/Y) and inputting the payment and

interest schedules as prompted. Since the transaction will not have any pay¬ ments, simply make the payment and interest schedules the same. For exam¬ ple, if there are no payments and interest is compounded quarterly, the process would be

2nd

P/Y

j,

4, (enter ;

2nd

]

Quit]. This will set up the proper

schedules in both P/Y and C/Y and take you back to the calculator mode. 2. Clear out any old information with [2nd J (CLR TVM). 3.

Input the variables you know.

4. Compute the variable you need to find. For annuity calculations, the process will be: 1. Set up the calculator for either an ordinary annuity (payments made at the end) or an annuity due (payments made at the beginning). This is done by hit¬ ting [

2nd

) (BGN) and then setting up the display to END or BGN.

will allow you to switch between the two options. (jindJ

2nd

j (Set)

Quit) will take you

back to the calculcator. Note if the calculator is in END mode, the display will be clear in the upper right-hand corner of the display; if it is in BGN mode, the letters BGN will appear in the upper right-hand corner. 2. Set up the payment and interest schedules in the Second function of the cal¬ culator. This is done by going

2nd

(P/Y ) and inputting the payment and

interest schedules as prompted. For example, if the transaction had monthly payments with quarterly compounding, the process would be j 12, [enter] , ( i

], 4,

[enter] ;

2nd

J (P/Y ),

] Quit ]. This would set up monthly payments

with interest compounded quarterly. 3.

2nd

Clear out the old information with

(~2nd

4. Input the variables you know. 5. Compute the variable you need to find.

| (CLR

TVM).

INSTRUCTIONS

AND

TIPS

FOR

FOUR

PREPROGRAMMED

FINANCIAL

CALCULATOR

MODELS

C. Calculation phase 1, 2. The steps required to perform calculations and an example calculation appear on pages 357-358 in Chapter 9. The steps required for annuities and sam¬ ple annuity calculations appear on pages 491-492 in Chapter 11. 3. Entering data using cash flow sign conventions and correcting entry errors Data can be entered in any order, but you must observe the cash flow sign conven¬ tions. For compound interest calculations, always enter PV as a negative number and all other values (N, I/Y, FV) as positive numbers. An error message will be dis¬ played when calculating I/Y or N if both FV and PV are entered using the same sign. For annuity calculations, enter either PV or PMT as negative numbers and all other values (N, I/Y, FV) as positive numbers. When PV = 0, designate PMT as the negative number and all other values (N, I/Y, FV) as positive numbers. Failure to observe this sign convention will result in either an error message in the display when calculating I/Y values or an incorrect negative number when calculating val¬ ues of N.

--

Data entry errors can be corrected one character at a time by using l-or the entry and error mes¬ sage can be cleared from the display by using (CE/Cj.

4. Calculating the unknown variable Press i CPT

and the financial key representing the unknown variable after all the

known variable data are entered (including 0 for PV or FV if required). Successive calculations are possible because numerical values stored in the function key reg¬ isters remain there until cleared or replaced. The value stored in any of the func¬ tion key registers can be determined without altering its value by pressing RCL) and the function key.

D. Example calculations 1. Compound interest See page 357 in Chapter 9 for an example of a compound interest calculation using this calculator. 2. Annuities See pages 492 and 495 in Chapter 11 for examples of annuity calculations using this calculator.

APPENDIX

II

E. Checklist for resolving common errors 1. Confirm that the P/Y and C/Y are properly set. 2. Confirm that the decimal place format is set to a floating decimal point. 3. If attempting annuity calculations, check to see that the calculator is in the appropriate payment mode (“END” or “BGN”). 4. Clear all function key registers before entering your data. 5. Be sure to enter a numerical value, using the cash flow sign convention, for all known variables before solving for the unknown variable, even if one of the variables is 0.

II. 2. Texas Instruments BA-35 Solar Business Analyst A. Basic operations 1. Turning the calculator on and off Turn on the calculator by pressing [ac/On] . The calculator turns off automatically when the solar cell panel is no longer exposed to light. 2. Operating modes The available operational modes are financial (FIN), statistical (STAT), and Profit Margin (no message). The message FIN, STAT, or no message appears in the lower left corner of the display to indicate the current mode. Change the mode by pressing (mode) until the desired mode is displayed. Changing to a new mode clears the contents from the mode registers. 3. Using the Second function The primary function of a key is indicated by a symbol on the face of the key. Second functions are marked on the face plate directly above the keys. To access the Second function of a key, press (j^dj (“2nd” will appear in the upper left corner of the display) and then press the key directly under the symbol on the face plate (“2nd” will then disappear from the display). 4. Clearing operations (AC/ON;; jn addition to turning the calculator on, clears the display, all pending operations, the memory, and the mode registers. Furthermore, the key also sets the calculator to floating-decimal format and to the financial mode. CE/Cj clears incorrect entries, error conditions, the display, or pending operations. It does not clear the memory, the mode registers, or the display format. [2nd J

(cE/cj clears values stored in the mode registers. clears the last digit entered.

INSTRUCTIONS

AND

TIPS

FOR

FOUR

PREPROGRAMMED

FINANCIAL

CALCULATOR

MODELS

5. Displaying numbers and display formats The display shows entries and results up to 10 digits but internally stores numeric values to an accuracy of 13 digits. The calculator normally displays numbers in the floating-decimal-point format unless the result of a calculation is too large or too small to be displayed; then scientific notation is used. The floating-decimal-point format appears when the calculator is turned on. To change the format to any number of decimal places up to 9, press [2nd I (Fix) then press the appropriate digit key from (_o ;

to ( 9j . The calculator can be returned to the floating-

decimal-point format by pressing (~2nd

(Fix) f~* I or by pressing [AC/ON ,

6. Memory capacity and operations STO j stores the displayed numeric value in the memory, replacing any value previously stored. 1SUM; or ( ±

(sum] adds or subtracts the displayed numeric value to the

contents of the memory. ( RCL j f~2nd j (MEM) recalls and displays the number stored in memory, without affecting the memory’s contents. 7. Operating errors and calculator dysfunction “Error” appears in the display when there is an error condition. General error con¬ ditions normally occur when the function is out of range of the calculator or not defined, or a key or key sequence is pressed that cannot be performed in the cur¬ rent mode. In the financial mode, errors can occur when insufficient data are entered or when no solution exists. The display can appear blank (digits do not appear) for a number of reasons. The most common are: • The calculator has powered down due to an inadequate light source. (The rem¬ edy is to turn the calculator back on.) • A long calculation is in progress. (The remedy is to wait for at least 20 seconds, then investigate further.) • A function does not appear to work. (The remedy is to check that the calculator is set to the correct operating mode.) • The number of decimal digits expected is not displayed. (The remedy is to reset the decimal format and repeat the calculation.) For other, less common dysfunctions, refer to pages 35-36 of the Guidebook.

APPENDIX

II

B. Pre-calculation phase Financial mode is the default mode when the calculator is turned on. If the calcu¬ lator is in one of the other modes, press I MODE - unqj “fin” appears in the lower left corner of the display. No adjustment is required to match this calculator’s per¬ formance to the text presentation. Floating-decimal-point format is the default decimal setting when the calculator is turned on.

C. Calculation phase 1. Clearing preprogrammed registers [ac/On] clears the financial modes and registers. 2. Adjusting for annuities (beginning and end of period) The default mode for annuity calculations is “end of period.” If “beginning of period” calculations are required, press (_2nd j (BGN). “Begin” will appear in the lower middle portion of the display. To return to “end of period” mode, press I 2nd (BGN) again or press (AC/ON; .

j

3. Entering data using cash flow sign conventions and correcting entry errors This calculator does not require data to be entered using the cash flow sign con¬ vention for simple compound interest calculations. In other words, both PV and FV can be entered as positive values. However, for problems involving annuity cal¬ culations where PV = 0, enter the PMT amount as a negative number. Otherwise, anomalies will occur in the display when calculating FV(it will appear as a nega¬ tive value), N (an incorrect negative answer will appear), or i (“Error” appears in the display). When FV = 0 (in problems involving paying off mortgages and loans), all values can be entered as positive values. Furthermore, when calculating PMT, all other values can be entered as positive. The resulting absolute value for PMT will be correct but the sign will be positive when FV = 0 and negative when PV = 0. Refer to “Clearing Options” above for correcting data entry errors. 4. Calculating the unknown variable Press j CPTj and the financial key representing the unknown variable after all the known variable data are entered (including 0 for PV or FV if required). Successive calculations are possible because numerical values stored in the function key reg¬ isters remain there until cleared or replaced. The value stored in any of the func¬ tion key registers can be determined without altering its value by pressing j RCL and the function key.

INSTRUCTIONS

AND

TIPS

FOR

FOUR

PREPROGRAMMED

FINANCIAL

CALCULATOR

MODELS

D. Example calculations 1. Compound interest (Example 9.2A, page 354)

Key in

Display shows

Press AC/ON) or [2nd] (CMR)

clears the mode registers

( PV

6000

6000

value P (principal)

I/Y cz:

2.5

(

20

2.5 -

|

j CPT

FV

• this enters the present this enters the periodic interest rate / as a percent

20

this enters the number of compounding periods n

9831.698641

this computes and displays the unknown future value S

2. Annuities (Example 11.2D, page 445)

Press

/ in 0

Display shows

{ PV

0

a precaution to avoid incorrect answers

10

)

-

PMT

-10 -

- this enters the periodic payment R

0.5

[777

0.5

this enters the conversion rate i as a percent

60

(

60

N ]

this enters the number of payments n

CPT

FV

697.7003047

this computes the unknown future value S„

E. Checklist for resolving common errors 1. Check to see that your calculator is in the financial (FIN) mode. 2. Check to see that the calculator is in the appropriate payment mode (“Begin” or end mode). 3. Clear all registers before entering your variable data by pressing AC/ON or

2nd

(CMR).

4. Be sure to enter values for all variables except the unknown variable, before solving for the unknown variable, even if one of variables is 0. 5. Observe the cash flow sign conventions (discussed above) when entering the data to avoid unwanted negative signs, display errors, or incorrect answers.

II. 3. Sharp EL-733A Business/Financial Calculator A. Basic operations 1. Turning the calculator on and off PCE turns the calculator on. ( OFF j turns the calculator off. To conserve battery

life, the calculator will turn itself off automatically 9 to 13 minutes after the last key operation.

2. Operating modes The available operational modes are financial (FIN), statistical (STAT), and Normal (no message). The message FIN, STAT, or no message appears in the upper right corner of the display to indicate the current mode. Change the mode by pressing (2ndf| (MODE) until the desired mode is displayed.

3. Using the Second function The primary function of a key is indicated by a symbol on the face of the key. Second functions are marked on the face plate directly above the keys. To access the Second function of a key, press '(2nd Fj (“2ndF” will appear in the upper left corner of the display) and then press the key directly under the symbol on the face plate (“2ndF” will then disappear from the display).

4. Clearing operations 2nd Fj (CA) clears the numerical values and calculation commands including data for financial calculations. The contents of memory register storage are not affected. iOCEj (x-»m) clears the memory.

C*CE clears the last entry. (pce'} }pce) clears the calculator of all data except the data for financial calculations,

clears the last digit entered.

5. Displaying numbers and display formats The display shows entries and results up to 10 digits. The default setting in the cal¬ culator is the floating decimal. To change the number of fixed decimal places, press 2nd F

(TAB) along with a number key for the decimal places desired. For a

floating-decimal-point format, press [2nd Fj (TAB) ( * ). The number of decimal places is retained even when the power is turned off.

INSTRUCTIONS

AND

TIPS

FOR

FOUR

PREPROGRAMMED

FINANCIAL

CALCULATOR

MODELS

Various messages can appear in the display from time to time. Refer to page 73 of the Operation Manual and Application Manual for a complete list. 6. Memory capacity and operations This calculator has one memory address. To store a displayed value in memory, press

.

To clear the memory of values other than zero, press

(oce) [x-»Mj.

To recall a value from memory and display it, press

RM

To add a displayed amount to the value in the memory, press ! M+ . Tc> subtract a displayed amount to the value in the memory, press ( ± ) I

M+).

7. Operating errors and calculator dysfunction Operational errors are indicated by the symbol E in the lower left corner of the display. See pages 74-77 of the Operation Manual and Application Manual for a complete description of errors and error conditions that may affect the operation and functioning of your calculator. The error symbol is cleared from the display by pressing }C»Ce],

B. Pre-calculation phase With the calculator on, set the financial mode by pressing

(2nd~F)

(MODE) until the

FIN message appears in the upper right corner of the display. The calculator requires no change to a register or mode in order to match the text presentation. To set the calculator to the floating-decimal-point format, press

[2nd F|

(TAB) ( • ).

C. Calculation phase 1. Clearing preprogrammed registers [2nd f)

(CA) clears the preprogrammed registers of numerical values and sets them

to 0 for financial calculations. 2. Adjusting for annuities (beginning and end of period) The default mode for annuity calculations is “end of period.” If “beginning of period” calculations are required, press fBGN) ■ “BGN” will appear in the upper right corner of the display. To return to “end of period” mode, press !, BGN] again. “BGN” will disappear from the display.

APPENDIX

II

3. Entering data using cash flow sign conventions and correcting entry errors Data can be entered in any order but you must observe the cash flow sign conven¬ tions to avoid operational errors and incorrect answers. For compound interest calculations, always designate PV as a negative number and all other values (N, i, FV) as positive numbers. If you do not observe this sign convention when you enter data, your answer will be the same numerical value but the opposite sign of the answer in the text. An error message will be displayed when calculating i or N if both FV and PV are entered using the same sign. For annuity calculations, when FV = 0, designate PV as a negative number and all other values (N, i, PMT) as pos¬ itive numbers. When PV = 0, designate PMT as the negative number and all other values (N, z, FV) as positive numbers. Failure to observe this sign convention will result in either an error message in the display when calculating i values or an incorrect negative number when calculating values of N.

4. Calculating the unknown variable Press (comp) and the financial key representing the unknown variable after all the known variable data are entered (including 0 for PV or FV if required). Successive calculations are possible because numerical values stored in the function key reg¬ isters remain there until cleared or replaced. The value stored in any of the Junc¬ tion key registers can be determined without altering its value by pressing 2nd F (RCL) and the function key.

D. Example calculations 1. Compound interest (Example 9.2A, page 354) Display shows

(2nd f] (CA)

no change

-

clears all registers

-6000

-

this enters the present

1+

--

6000


11. 1

1. (a) -2 - 8x (b) —2x - 9 (c) —16 a — 7 (d) ~6x + 6x + 12

5. 88 B.

B.

1. 21 months

(b) $0.90

5. (a) $3600 (b) $9000

3. (a) -8 (b) f

A.

5. 2% C.

5. 36

(d) 2187

Exercise 3.3 A.

(f) -x15 6

(b) n = 5

B.

D.

1. (a) —1.4%; (b) -2.4% 3. -3.2%

7. 30

Review Exercise

1. 32

1. (a) 5:6

3. $130

(b) 6:1

5. 18 7. 6

5. $4.40

(c) 9:40

9. 0.5

Exercise 3.5 A.

1. 27

(d) 6:1 (e) 240:20:1 (f) 15:4:3

13. 7.5

3. 12.5%

3. (a) 210

15. 17.5

5. $4320

(b) 7.2

1. $16 3. $1950

Exercise 3.1

5. $9

A.

7. $200

1. (a) 3:8

Exercise 3.9

3. 300

11. 2

7. (a) P = A

CHAPTER 3

1. 36

1. $8978.06 3. $7035.46

A.

1. 20 3. 440

7. $100 800

(c) 1

Exercise 3.8

1. 50% 3. 200%

3. 600 km

1. (a) $0.92, $0.83 3. $39 173.21

1. 168 3. $1140

1.102.08,110.64, 98.33

7. $60 000 9. $6

5. 7.4

Self-Test

n =

1. 60% 3. 115%

5. $45 000

(i) 164

SELF-TESTS

Exercise 3.6

D.

Exercise 3.2

lb) superlight, 27; ordinary, 45

5. (a)

AND

1. $200

,

3. S5250; $2800; S1400

(g) Machine C, 35 minutes

EXERCISES,

(d) 23:14:5

(g) x = -A

water = S341

REVIEW

3- W Te (b) f ✓ (c) 2:7:11

LS = 25 = RS

(e) heat = S814 power = S1056

PROBLEMS,

9. $600

B.

7. (a) $180 000

(c) 195

(b) $225 000

(d) 3.6

1. $14.52 3. $130 5. $85.10

(b) 3:2 (c) 5:8:13

11. $49

7. $5000

(d) 3:6:13

13. $135

9. 83i%

15. $60

3

5. (a) $6.66 (b) $8.30 (c) $90.00 (d) $27.72

ANSWERS TO ODD NUMBERED PROBLEMS. REVIEW EXERCISES

17. U) -033% fb) -0.53%

(c) 16|% (d) 550%

(e) S56 (f) $340

7.

15. S0.88

7. (a) SI8 (b) S1935

AD 5E---ESTS

CHAPTER 4 Exercse 4.*

(g) S140 9. S2400; S4200;

A.

l.x = —8,}’ = — 1 3. x = 10,7 = 12

S4800 11. S75 000; S50 000;

3-

5. x = —3,7 = 3

x

0-2

t

0

~

S60 000

3. x = —I,/ = 3

15. 540 17. Bonds S56 250

5. x = 4, y = 3

5. x = 6,7 = 10 7. x = \,y = |

19. (a) S400 000

1

--

1. For7=3x—20

C

Forx = 0 tox = 40

F (0,3) G (—4,4i

X

0

20

40

r

20

so

140

H -5,0

31. S141.18U.S. 33. S17 261.96

or m = 3 b= 20

3. a X 1r

—5 1 -31

-4 -2

—3

o 1 l

7

3 J

-i i 0

7 i 3 ; 4

3 i

i

Y

(b)

1. (a) S350 X

(b) S76.05

(c) S 145.00 (d) SI3.20 3. 45%

Australian dollars

-3

E 4,3)

(b) S250 000 (c) 312.5%

Australian dollars (b) 554-2

r

-4

D (2,0)

25. SI65 27. SI5 000 29. (a) $80 000

7. S16 875;S11 250; S6 750; S5 625 9. $3730 11. 180 13. (a) 1.1084

4

B (0, -4) C (3, -4)

(b) 266=%

5. S10 000

o

A. LA(-4,-3)

23. (a) 166|%

Self-Test

X

'Zi

Exercse 4.2

©

(b) 16|%

slope, m = 2 •.-intercept, h = — 3

3. x = 13,7 = 23

S9375

21. (a) 7.5%

9. For 7= 2x- 3

C 1. x = 12,7 = 8

S84 375 Preffered Shares

(b) S280 000

2 -2

1.x =-4,>-=3

B.

13. S182 000

Common stock

|

2

•y

3

J

| 3 j 3

X

1 3 j

,

i

0

-l — 1 ; -3 —5 '

(O

| r

i 4

6

0

■» 1 0

-1

-2 i -3'

_y | -4 -6

(d) X ; —5 i -4 ! -3 .1 -2 -i ‘

T

B. j

5

4

3

1.

X

0

3

2

J

-3

0

-1

■y

0

i 1 0

I -•

3

4

5

_7 j -3 -4 -5

;• ANSWERS

TO

ODD-NUMBERED

3. For 3x + 4/ = 1200

PROBLEMS,

REVIEW

EXERCISES,

5. 3x-4y= 18

AND

SELF-TESTS

9. 55 quarters;

Y

72 loonies

Forx = 0 to x - 40 0

X

300 150

y

Review Exercise

200 400

1. (a) m = —b = 2

0

(b) m =

or m = — |

b= 0

(c) m = |, b = 4

b = 300 (2,-3) is the solution.

Y

Y \

=

3x + 3/ = 2400

4-

\ -3x

(d) m = —6, b = 10

3. For For x = 0 to x = 800

3' 2-

3x - 4y = 18

V

-2 -i 0 -1-

1

2

3

0

800 400

7

4

400 800 0

(e) m is undefined, there is no /-intercept (f) m = A, 6 = -3 (g) m = i, b = -1

-2-

-3-

(h) m = 0, b — 5

Sx (2,_3)

-4-

3. (a) 3x + / = 6 and / -6

x—/= 2

A.

1.

X

0

2

4

y

6

0

-6

X

0

2

4

y

-2

0

2

o

y

II

II

Exercise 4.3

CN

7. 5x

4

2

6

0

-5

5

(6,5) is the solution. Exercise 4.4 A.

1. .-. 30.v + 40v = 6000

(0,4) is the solution. B.

1. For / — 4x = 0 For x = 0 to x = 10 0000 0

y (1,1) is the solution.

3.

(b) x + 4/ = —8 and v = 200 + 4a

5000 10 000

0

20 000 40 000

For y- 2x — 10 000 = 0 For x = 0 to x = 10 0000 X

y

0

5000 10 000 B.

10 000 20 000 30 000

1.15; 9 3. Brand X, 90; No-Name, 50 5. Kaya, $31 500; Fred, $23 500 7. Type A, 42; Type B, 18

3x + 4/ = 0 X

'0

y

-2 -3

4

-4 -1

X

0

4

-4

y

0

-3

3

ANSWERS

(c) 5x = 3y and

TO

ODD-NUMBERED

(f) y = —2x and

y=- s

PROBLEMS,

REVIEW

EXERCISES,

AND

SELF-TESTS

Y

5. (a)

x= 4

v = 1000 + 75-t

X

0

3

-3

X

0

4

-2

y

0

5

-5

y

0

-8

4

5. $8000; $4000 (b)

CHAPTER 5 5x + 2v = 200

Exercise 5.1 A.

1. (a) (i) Revenue = 120x (ii) Cost = 2800 + 50x

where x represents the number of units per period Self-Test

(b) (i) 40 (ii) 40%

1. (a) m = 0, b = —

(iii) $4800

(b) m = 6, b = —9 (c) m = — i, b = 0

(c) (i) $70.00 (ii) 58.33%

(d) m = 0,b = —3 (e) m is undefined,

(d)

there is no y-intercept

3. (a) For — x = — 55 + y For x = 0 to x = 55

y

X

0

5 3

-1

y

-2

3

-5

0

25

55

55

30

0

(h) y = —2 and 5x + Revenue = x

3y = 15 X

y

3

0

0

5

(ii) Cost =

4.2

220 000 + 0.45x -2

(b) (i) Not applicable (ii) 50% (iii) $400 000 (c)

(i) $0.55 (ii) 55%

(b) For

(d)

3x + 2 y+ 600 = 0 for x = 0 to x = 200 X

0

too

200

y -300 -450 -600

Output m Site* Dollar*

ANSWERS

B.

TO

ODD-NUMBERED

1. 45

PROBLEMS,

REVIEW

EXERCISES,

Total cost = 3136 + \57x

(c) $103.20

5. Original cost = $45 000; residual value = $2500; n = 10

(b) 246 (c) 250 (d) $13 Exercise 5.3 1. (a) $3500

= $45

Depreciation in year 1: 20% X 45 000.00

=

9 OOQ.f,

Book value at end of year 1

=

Depreciation in year 2: 20% X 36 000.00

=

36 OOO.C 7 200.C

Book value at end of year 2 Depreciation in year 3: 20% X 28 800.00

= =

28 800.0 5 750.0

Book value at end of year 3

= $23 040.0 3. $2507.19

(c) Wearing value

(iii) 70%

636.3636

3. (a) (i) $0.35

20%

(b) (i) 60%

1. (a) Yearly

A.

(c) Total cost =

(d) $335 000 5. 802

year 16 = $12 300

7. (a) 265

Review Exercise 1. (a) (i) $28.00

1. 15.9%

A.

(c) $2.16

(d) 25% (e) 20% 3. (a) $35.00 (b) $31.50

1. $30.24; 32.5%

(c) $3.50

3. $137.89; 48.55%

(b) Rate of

(d) 66|%

5. $1583.33; 61%

depreciation 25% B.

9. 330

(e) 40%

1. (a) $127.68

5. (a) $10.50

(b) $112.32 Self-Test

(b) $12.75

(c) $46.8%

7. $180.00 9. 15%

(iii) 40%

Exercise 6.4

Cost = 18 000 + 7.00x A.

1. $640.00 3. $776.11

(b) (i) 112

5. $1136.80

(ii) 35%

7. $4581.50

(iii) $20 720 B.

1. $582.00; $850.00 3. $564.50; $536.28 5. $810.00; $810.00

8

10

12

(e) 28%

5. 15%

(b) (i) 6000 CDs

(ii) 15.135%

(d) 38f%

(b) 48.26%

(ii) 30%

(c) Revenue = 10x

(c) ($2.25)

3. (a) 38.75%

1. (a) (i) $3.00

1. (a) $6.00 (b) $3.84

Exercise 6.3

(ii) $60 000

$12 972.15

Exercise 6.5

7. $426.00

(ii) Depreciation charge in

(ii) Depreciation in Year 16

(b) $3400.00

5. $74.10

Book value $375 500

(i) $259 443.07

9. (a) $1925.00

5. 15%; $252.60

3. $2436.00

15 years $184 500;

(c) 5%

(c) $600.00

3. 33±%; $25.65

A. B.

depreciation after

(ii) $15 000

(b) $733.54

9. $134.96; $50.61

(i) Accumulated

(i) $263 000

7. (a) $1164.00

1. $11.07; $13.53

7. $133.36; $50.01

10 5000 + 0.65x

depreciation $12 300

(b) $600

5. $2184.00

Exercise 6.2

(ii) $300 000

(e) Class 8 rate 20% B.

CHAPTER 6

(ii) 35%

depreciation

ooo.c

Original book value

(b) $0.35

(d) Rate of

1. (a) Sept. 10 (b) $5276.85

3. $35 000

1. (a) 164

A.

C.

(e) 5500

185x;

Exercise 5.2

SELF-TESTS

(d) 5600

(c) Revenue =

3. $19.75 per unit

AND

B.

1. $6.25; 25%; 20% 3. $102.40; 60%; 37.5% 5. $75.95; $21.70; 28.6% 7. 44.24; $22.12; 33±% 9. $78.10; $46.86; 150% 11. $111.30; $133.56; 20%

ANSWERS

1. $13 .60

13. (a) $1645.00 (b) $1500.00 15. (a) $90.00

5. $102.08 7. (a) 150% (b) 60% 9. (a) $36.50 (b) $29.93 (c) 21.95% 11. (a) $1.65 (b) 66|%

1. $51.00; $59.00; 3. $96.40; $30.65; 5. $160.00; $19.20; 12%

Self-Test

(b) -$8.00 1. $295.77

(c) 10.3%

(b) $957.00 (c) 34.5% 9. (a) $72.00

3. 45.66% 5. 15% 7. $30.00 9. September 12; $25 117.40

7. $1599.35

1. $1224.00 Review Exercise 1. (a) 172 (b) 214 3. (a) $1160.00

1. $3296.00

(b) $601.77

3. 9.5%

5. (a) $750.00

5. 11 months

(b) $5709.97

7. $876.00

7. $3000.17

9. $400 000

9. 8.25%

11. 9% 13. November 18, 2006

11. 196 days

15. (a) $51 975.00

13. $1601.89 15. $3200.00

(b) $51 943.53

17. $1736.47

(c) 3.887% Exercise 7.4

19. $2664.00 21. $3404.32

A.

23. $1614.74

1. $490.13 3. $768.75

25. $961.50

5. $849.21

27. $1587.06

7. $1298.00 B.

Self-Test 1. $21.40

13. 180%

5. $26 954.84

15. 23.40%

7. $13 864.50

7. $4306.81 9. 359 days 11. $7432.80 13. $1163.85

Exercise 7.5 A.

1. $266.00; $13.30 3. $517.50; $547.17

1. 112 days

B.

5. $1600.04

1. December 1, 2007

7. $9755.22

3. April 5, 2007 Exercise 7.2 A.

1. 0.035; 1.25 3. 0.0825; iff

1. $1222.00

CH A PTER 8

3. $1704.60

3. 166 days B.

15. $1799.23

5. $2025.00; 292 days

Exercise 7.1

(b) $24.08 (c) $43%

5. $1408.21

3. 6.5% 5. $6187.50

CHAPTER 7

A.

1. $1156.80 3. $519.17

3. $13 800.00

17. -$660.45

Review Exercise 1. (a) $31.92

B.

11. $348.36

(b) $54.00

15. -$82.50

1. $10.87

1. $2542.53

9. $240.00

$6.60

11. $1070.39

7. 144 days

7. $1450.00

7. (a) $14.85; $9.90;

5. $75.34

5. 14 months

5. $1635.04

5. -$2.23

9. $777.81

3. 10.75%

3. 50.5%

(d) 9|%

SELF-TESTS

7. $856.47

A.

B.

AND

1. $1096.88

(d) $74.88 (e) -$6.48

(c) 40.98%

(b) 54%

EXERCISES,

3. $95.21

Exercise 7.3

27. (a) $2152.40 (b) 69.43%

1. (a) $228.69

REVIEW

3. $21.76

(b) 21.25% (c) 133±%

$7.91

11. -$6.50 13. -$97.50

C.

(b) 25% 25. (a) $189.00

-$8.00

(c) 25%

B.

(b) $58.50 (c) 53.85%

21. (a) -$0.60 (b) 26.98% 23. (a) - $13.20

Exercise 6.6

3. (a) 40%

PROBLEMS,

17. (a) $77.50 (b) 42.86% 19. $240.00

13. (a) $234.20 (b) 47.37%

B.

ODD-NUMBERED

11. (a) $1940.00 (b) $2813.00

3. $3.00

A.

TO

Exercise 8.1 A.

9. $9933.50

3. $530.00 5. 154 days 7. $544.54

Exercise 7.6 A.

1. $829.33 3. $617.50 5. $1103.37

1. December 30, 2005

B.

1. (a) March 3, 2008 (b) 155 days

ANSWERS

TO

ODD-NUMBERED

PROBLEMS,

(c) $21.40

7. $814.17

(d) $861.40

9. $1269.57

3. (a) April 3, 2004

REVIEW

EXERCISES,

B.

11. 6.1493%

9. $1972.80 11. $1492.15

$70.75

7. (a) $148.59; $48.59

Exercise 9.5

(b) $220.80; 120.80

1. $631.24

(c) $487.54; 387.54

Self-Test

A.

1. $5983.40 3. $2537.13

9. $3712.50 1. $19.79

11. $761.75

5. $1673.49

3. $1160.00

13. $17116.96

7. $641.36

1. $500.00

5. $1664.66

15. 144.65

1. $1471.28

7. (a) $98 116.43

17. (a) Bank; $6968.27;

3. $1615.56

(b) 3.680%

1. $98 627.91

$4070.41; $70.41;

C.

$4000.00

A.

1.1; 0.12; 5

1. $3786.27

1. 1.7623417 3. 1.6349754

(b) $4.05

5. 4.5244954

(c) $1.24

7. 2.7092894

(d) $10.00

9. 1.7074946

9. $749.91 11. $4344.21 B.

C.

3. $762.84

(c) $1635.62 3. $1500.00 5. $5125.75

9. (a) $3113.90 (b) $1847.95 Review Exercise 1. (a) $1198.28 (b) $1221.61 (c) $1227.05 3. (a) $2890.09 (b) $890.09

7. $1398.85

5. (a) $6144.45; $4344.45

A.

1. $1767.71; $232.29

(b) $3305.27; $2055.27

(c) 1.02548

3. $3981.87; $1018.13

7. $10 681.77

(d) 3.2714896

5. $2642.50; $557.50

9. $11 102.50

$1233.69; $33.69;

7. $1012.96; $671.00

11. $4194.33

9. $2310.82; $720.62

13. $9791.31

1. $713.39

11. $2036.86; $540.42

15. $2830.68;

3. $2233.21

13. $1332.82; $261.78

Exercise 9.2 A.

5. $655.02

5. $3129.97

1. (a) 48 (b) 2.5%

1. Totals are

(b) $35.62

1. $1338.81; $261.19

Exercise 9.4

Exercise 8.6

1. (a) November 2

1. $574.37

7. $500.24

9. 2; 0.06125; 9 B.

(d) $5610.21 3. $6805.31

5. $409.16

150

1. (a) $0.16

Review Exercise

(b) $3265.19 (c) $4000.00

3. $371.86

7. 12; z = 0.0066667;

5. $178.66

1. (a) $2851.94

7. $2423.97 A.

5. 2; 0.0575; 27 3. $1825.63

C.

5. $3436.38

Exercise 9.3

3. 4; 0.01375; 36

5. $307.56

$1200.00

7. $987.93

3. $1452.79

9. $1444.24 Exercise 9.1

(e) -$956.34

1. $3010.85

7. $1102.13

CHAPTER 9

1. $37.50

Exercise 8.5

5. $1536.03

(b) $54.18

Exercise 8.4

3. $22.50

1. $3426.73 3. $2464.35

$6914.09

(b) $99 636.18 (c) 2.706%

B.

11. Totals are

5. (a) 2.72%

A.

7. $1074.71

(c) $203.81

3. 2.73%

A.

5. $3488.29

(b) $202.24

(d) $204.89

9. $340.26

B.

5. (a) $199.26

$70.82; $69.86;

Exercise 8.3

A.

1. $8452.52 3. $2346.36

15. (a) $54.57; $62.49;

3. $837.19

C.

C.

(c) $14.02

(b) -$10 623.49

B.

1. $6884.47; $1884.47 3. $1441.71

13. $449.12

Exercise 8.2

A.

SELF-TESTS

(b) 63 days

(d) $1264.02

A.

AND

5. $5468.38 7. $6639.51 9. $662.02 11. $4152.58 13. $2052.74

B.

1. $4589.47 3. $1345.06 5. $3800.24 7. $1561.49

$3190.63 17. $9294.85 19. $1035.70 21. $2838.62 23. $2079.94 25. $110 440.03

ANSWERS

27. $26 048.42 29. (a) $2742.41

TO

ODD-NUMBERED

Exercise 10.3

A.

(b) $2911.55

PROBLEMS,

31. $4857.56

(c) 7.397%

3. 5.535675%

(d) 10.691%

5. 7.352%

(c) 7.383%

(d) 4.273%

I. $2129.97 3. 2.7118780 5. $6919.05 7. $14 711.80 9. $4504.29 11. $10 138.19 13. $2661.85 15. $848.88

CHAPTER 10 Exercise 10.1 A.

3. 4.656%

A.

7. 6.2196%

(b) deferred annuity

(b) $114.57

(c) general annuity

(c) 3.5567%

annuity due

(c) 4.3182%

(c) simple annuity

(e) 37.313

5. 22.517085 half-

5. $17 915.08

5. 5.622 years 7. November 1, 2001 9. 21 months II. 2 years, 298 days 13. 3 years, 230 days

A.

years (11 years, B.

7. 6.03%

A.

(c) 3.5462%

1. 4.5% 3. 9.6% 5. 9.778%

(d) 3.2989% 17. 1.831663 years (1 year, 304 days)

B.

1. 9.237% 3. (a) 10.402%

(b) 7.585% 5. 3.5% 7. 4.771%

19. 1.5677878 years (1 year, 208 days) 21. 6.8287904 years

A.

.

1 14.6023573 years (14 years, 8 months) 3. 93.0687373

months (7 years, 10 months) 5. 15.5775778 half-

years (7 years, 10 months) 7. 15.395937 semi-

(c) $4031.12

annual periods

7. $62 177.25

(7 years, 9 months) 9. 71.517450 months

Exercise 11.3 A.

(6 years)

.

1 $9515.19 3. $30 941.11

15. (a) 4.0% (b) 3.7852%

Exercise 11.5

(b) $4500.00

(d) 8.24%

Exercise 10.2

19. $1984.62

1. $13 045.68 3. $32 434.02

(c) 8.833%

13. 8.44%

17. $114.89

1. $54 193.60

(b) 14.35%

(b) 5.96%

15. $3141.41

5. (a) $8531.12

9. (a) 5.92%

11. (a) 4.59%

11. $557.65 13. $229.33

Exercise 1 1.2

3. $59 185.19

3. 17.501 years

7. $410.00 9. $290.55

(b) deferred

(b) $631.40

1. 5.592%

1. 9.329 years

5. $320.61

5. (a) annuity certain

13. (a) $1831.40

.

1 $207.87 3. $591.66

3. (a) perpetuity

11. (a) $714.57

95 days)

9. $1653.70

(c) general annuity

9. 7.385%

3. 1 year, 231 days

years

7. $232.54

(b) annuity due

(d) 8 half-years

quarters

5. $272.73

1. (a) annuity certain

5. 8.945%

.

1 $821.39 3. $1117.37

B. Exercise 11.1

(c) 117.8 months

(f) 37.167 half-

A.

CHAPTER 1 1

1. 6.991%

Review Exercise

Exercise 11.4

9. 7.5%

1. (a) 13.4 years (b) 28 quarters

B.

B.

(e) $21.28

7. 10.0%

(b) 6.465%

5. $10 544.91 B.

SELF-TESTS

(d) $500.03

(63 months)

3. (a) 8.9%

Self-Test

AND

1. 10.50 half-years

1. (a) 12.891%

33. $1820.32 35. $3574.57

EXERCISES,

Self-Test

(b) 6.168%

(c) $3281.79

REVIEW

B.

.

1 74.4983312 months (6 years, 3 months)

.

1 $6897.02 3. (a) $9906.20 (b) $2093.80 5. (a) $2523.82 (b) $372.06 7. $12 710.96 9. (a) $13 683.13

(6 years, 303 days-

(b) $1428.88

2007-09-30)

(c) $14 183.16

3. 69.835347 months (5 years, 10 months) 5. 8.380061 quarters (2 years, 1.14 months) 7. 40.080206 months (3 years, 4 months)

ANSWERS

TO

ODD-NUMBERED

9. 26.241960 months

(2 years, 3 months)

B.

Exercise 11.6 A.

REVIEW

1. $23 268.52

5. 13 years,

7. $31 293.63

7. 11.37%

Exercise 12.2

.

1 12.5%

A.

3. $42 505.51

5. 9.50%

5. $5106.97

7. 3.04%

7. $34 627.97

9. 7.685%

B.

1. $10 041.88 3. $31 736.57

Review Exercise

5. (a) $29 829.03

1. (a) $26 734.60

(b) $5170.97

(b) $17 280.00

7. $80 000.02

(c) $9454.60 3. $722.62 5. 12.575297 years (12 years,

9. $33 393.84 Exercise 12.3 A.

7 months)

3. $1121.26

7. $411.57

5. $273.20

9. $101 517.64

7. $1426.84

11. 10.524175 half-

9. $271.68

years (6 years) B.

13. 7.25%

.

1 $823.60

.

1 $124.41

15. $34 031.63

3. $799.39

17. 8.5838838 half-

5. $330.63 7. $117.26

years (9 semi¬ annual payments) 19. $16 102.46 Self-Test

9. $26.63 Exercise 12.4

A.

1. $35 786.08

1. 14.51 years (14 years, 6 months)

3. 10.9%

3. 92.97 months (7 years, 9 months)

5. 28.06 quarters

5. 15.64 semi-annual

(84 months)

periods (7 years,

7. $40 385.39

10 months)

9. $3268.62

7. 15.43 half-years 12

(7 years, 9 months) 9. 71.57 months

Exercise 12.1

A. 1. $45 855.46

(6 years) B.

1. 44.76 months

3. $16 317.77

(3 years,

5. $63 686.72

9 months)

1. $55 246.47 3. 17.34 quarters (4 years, 4 months)

9. 25 years

5. $3003.81

Exercise 12.5

7. 32 quarters (8 years)

A.

9. $13 355.86; $96.29

1. 6.01% 3. 6.86%

1. $47 583.23

3. 11.7%

Self-Test

7. 40 deposits

(b) $3265.50

5. 7.6%

SELF-TESTS

7.3 months

5. (a) $13 265.50

.

AND

3. 133 months

3. $2326.65

1 5.0%

CHAPTER

EXERCISES,

7. $32 876.06

3. 9.0%

B.

PROBLEMS,

5. 12.41% 7. 4.40% B.

1.7.31%

CHAPTER 13 Exercise 13.1

A.

3. 8.82%

$71 813.10

5. 10.778%

3. $69 033.69;

7. 9.24%

$35 581.67 5. $43 738.24

Review Exercise

$7278.61

1. $13 509.62 3. $64 125.87

B.

1. $204.80 3. $909.95

5. $75 962.59 7. (a) $101.81

1. $135 334.71;

C.

(b) $1714.37

1. n = 113 months (9 years, 5 months)

9. (a) 30.62 quarters

3. n = 8 years

(7 years, 8 months) (b) 27.84 semi¬

D.

rate = 7.125% 3. nominal annual

annual periods

rate = 12.498%

(13 years, 11 months) 11. (a) 12.46 months (6 years, 3 months) (b) 23.24 months

1. nominal annual

E.

1. $31 378.29 3. (a) $29 513.14 (b) $10 313.14 5. $33 338.44 7. (a) $1238.56

(1 year,

(b) $1575.00

11 months)

(c) $336.44

13. 16.33 quarters

(4 years, 1 month) 15. 8.30 quarters (2 years, 1 month) 17. (a) $13 719.16 (b) $3951.16

(c) $382.25 (d) $8580.00 19. 96.82 months

(8 years, 1 month)

9. $2150.38 11. $300.36 13. $300.74 15. $217.19 17. n = 142.48 months (11 years, 11 months) 19. n = 41 quarterly deposits 21. nominal annual rate = 8.58%

ANSWERS

Exercise 13.2

A.

1. $39 342.74

1. n = 35 half-years

A.

1. nominal annual rate = 11.28% 3. nominal annual rate = 4.89%

(f) $1037.88

29. n

29 years

=

.

.

5. $459.47 7. n = 37 quarters (9 years, 3 months) 9. nominal annual rate = 9.18% Exercise 13.3

1. $1403.20

Exercise 14.1

A.

(b) $5633.77

7.51 years

(c) $140.84

(c) n

=

10.111884

(d) $408.38

quarters =

3. (a) $1103.73

8.86133

(b) $4913.61

half-years

(c) $196.54

33. (a) $16 102.46

(d) $907.19

(b) $3120.21 (c) $17 860.97

B.

(b) $2493.97

35. $301.80

3. (a) n = 14.53

37. $910.15 39.

n =

(b) $8035.94

26 quarters

(6 years,

1. (a) n = 19.48 payments

(d) $120.21

7. $41 854.67

Review Exercise

1. (a) $549.22

=

(d) n

3. $850.87 5. $210 240.91 9. $166.70

SELF-TESTS

(b) n

7. $96 992.34

1 $214.60

AND

CHAPTER 14

years

5. $44 762.78

B.

EXERCISES,

31. (a) n = 6.78 half-

1 $73 529.41

1. $59 113.10 3. $24 111.08

6 months)

1. (a) $920.57 (b) $15 052.64

(b) S„(due)= $11 728.44

41. $20 964.79

(c) $59 234.20

A„(due)= $6060.32

45. $8908.36

3. $19 153.93

47. n = 28.605079

1. (a) S„ = $11 223.39 A„ = $5799.35

5. (a) $2638.84

(b) $3444.00 (c) $805.16

43. $34 543.53

quarters 49. (a) $60 229.26

(b) $49 312.99

C.

(d) $23 234.20 3. (a) $136.87 (b) $4199.54 (c) $31.50

(d) $105.37 5. $2054.05; totals are $14 378.41;

3. $6427.70

7. n = 109 months

(c) $29 620.76

5. $21 748.37

9. n = 28 months

(d) $36 229.73

$4378.41;

11. nominal annual

(e) $86 633.66

$10 000.00

7. $10 272.72 9. $962.88 B.

27. n = 11 years

3. $91 027.01

(4.5 years)

A.

(e) $89.52

Exercise 13.4

3. n = 18 quarters

E.

REVIEW

25. $27 246.85

years

(17.5 years)

D.

PROBLEMS,

1. $1316.98 3. $903.61

C.

ODD-NUMBERED

29. n = 41.52 half-

3. $11 304.97

B.

TO

1. $126 738.19 3. (a) -$147 329.91 (b) $272 000.00 (c) $124 670.09

rate = 5.51% 13. (a) $411.57

(b) $403.50 15. (a) $15 749.42

(b) $3149.42

(f) $90 133.66 51. $318 181.82 53. $105.30 55. $28 089.24 Self-Test

7. totals are $14 943.16; $5743.16; $9200.00 9. $651.11

11. $1160.09

5. $576.10

(c) $432.91

7. $2820.21

(d) $8179.68

9. $2752.22

17. n = 51 months 19. n = 49 months

3. $1258.38 5. $357.34

(c) $2799.85

(3 years) 13. n = 26.712 months

21. (a) $28 435.38

7. Nominal annual

(d) $10 445.14;

15. $21 831.32

23. $42 092.99

17. $89 397.79 19. n = 27.65 half-

11. n = 36 months

(b) $28 552.64

1. $84 209.75

rate = 8.185%

9. $43 246.07

13. (a) $3621.90 (b) $1035.27

totals are $115 900.80;

25. $2736.49

11. $1170.69

$30 900.80;

27. (a) $503.87

13. $6056.04

$8500.00

years

(b) $1437.96

15. $21.97

21. $147.55

(c) -$275.12

17. $122 263.83

23. $165.11

(d) $1782.29

15. (a) n = 14.022508 (b) $957.79 (c) $1910.53

ANSWERS

TO

ODD-NUMBERED

PROBLEMS,

(d) $4664.98;

EXERCISES,

AND

SELF-TESTS

(c) $23 981.92

(e)

totals are

(d) $1000.88

Partial Amortization Schedule

$35 057.75;

Payment

Amount

Interest

Principal

$11 057.75;

number

paid

paid

repaid

$24 000.00

0

Exercise 14.2

A.

REVIEW

1. (a) $1829.69

Outstanding

(e) $15 900.11;

principal

totals are

40 000.00

$24 843.68;

1

318.15

278.44

39.71

39 960.29

$11 087.74;

2

318115

278.17

39.98

39 920.31

$13 755.94

3

318.15

277.89

40.26

39 880.05

7. (a) n = 9.3198366

(b) $20 286.42

(b) $1139.88

(c) $819.57

9. (a) n = 11.744464

(d) $1010.12

60

3. (a) $164.04

B.

(b) $3211.27 11. (a) $1091.28

61

364.44

317.37

47.07

37 009.21

(b) $4563.53

62

364.44

316.97

47.47

36 961.74

(b) $125 324.69

(c) $22.54

63

364.44

316.56

47.88

36 913.86

(c) $1023.15

(d) $141.50

13. (a) n = 117.33202

1. (a) n = 17.13

months (b) $332.79

(b) $2685.85 3. (a) n = 14.894762 (b) $1584.96 C.

37 056.28

Exercise 14.3

A.

3. $348.26

1. (a) $2790.38

5. $306.39

(b) $34 892.23 (c) $1599.89

1. $1006.24

B.

(d) $1190.49

1. (a) n = 29.434057 (b) $157.46 3. (a) n = 47.779687

3. (a) $252.15 (b) $34 200.11

7. (a) $86 514.51

5. $234.53

(d) $294.24

7. $301.45

5. (a) n = 15.413529

9. $1152.87 11. (a) n = 61.523813

(b) $2911.17

(d) $21.41

= $38 795.58

(e) $478.06; totals $1264.00;

1. (a) $1491.37

(f) $4300.94;

1. (a) $715.83 (b) $83 375.45 (c) $641.41 3. (a) $503.15

1. $6027.52

3. (a) n = 24.830989

104.522813

(e) $5477.40; totals are

(b) $288.05

$49 664.41;

(c) $2889.95

$9664.41;

= 9.50%

5. (a) $1406.95

(b) $28 940.29 (d) $1807.58

semi-annually

Self-Test

3. $12 866.98

$550.00; n =

rate compounded

(c) $25 598.05

(d) $638.94

$35 000.00

(c) $426.66

5. nominal annual

(b) $1651.04

$12 723.84;

rounded to

months

19. (a) $601.20

totals are $47 723.84;

A.

17. 8.38%

(c) $15 771.75

(d) $68 599.48

$16 000.00

$6500.00

(b) $12 723.84

are $22 484.09;

Exercise 14.4

are $7764.00;

Review Exercise

(e) $1247.91

(d) $364.44

(c) $5086.52

11. Balance December 1

(c) $84 599.48

(c) $263.14

(b) $1264.00

$38 794.01

7. $3212.01; totals

(b) $3323.27

15. (a) $161.75

9. $380.00; balance

(d) $338.49

11. (a) $318.15

$941.61

(c) $54 269.19

(b) $724.48

(c) $4311.17

9. $752.68

$117 332.79;

months

(b) $722.00

(c) $7277.51

$6484.09;

(c) $118 274.40;

(b) n = 165.56125

40 000.00 5. (a) $735.80 (b) $280.57

(b) $174 506.12 (c) $1479.66 7. 10.0% CHAPTER 15 Exercise 15.1

A.

1. $97 780.05 3. $24 098.27 5. $52 193.61

ANSWERS

7. $7234.24 B.

1. $466.70 3. $14 891.87

TO

ODD-NUMBERED

A.

AND

SELF-TESTS

$4922.85 (b) Schedule of Accumulation of Discount

9. $4 278 80 1.75

Interest Book

Discount

value

balance

4922.85 4932.84

77.15

9.99

160.32

10.32

4943.16

56.84

160.65

10.65

4953.81

46.19

150.00

161.00

11.00

4964.81

35.19

161.36

11.36

4976.17

23.83

6

150.00 150.00

161.73

11.73

4987.90

12.10

7

150.00

162.10

12.10

5000.00

—

1050.00

1127.15

77.15

coupon

interval

b = 3%

(b) $792.35

0

(c) $29 465.99

1

150.00

159.99

2

150.00

3

150.00

4 5

15. $5408.13 Exercise 15.2 1. (a) $4877.98 (b) $20 122.02 3. (a) $1773.12

Total

Discount

on book

Payment

13. (a) $30 258.34

(b) $8626.88

EXERCISES,

1. (a)-$77.15;

7. $1184.53

A.

REVIEW

Exercise 15.3

5. $10 759.95

11. $43 667.11

PROBLEMS,

i = 3.25% accumulated

67.16

5. (a) $100.35 (b) $962.31 B.

3. a $49.18;

1. (a) $14 304.00;

$1079.18

$114 304.00

(b)

(b) $6359.60;

Schedule of Amortization of Premium

$106 359.60

Interest

3. (a) $-4435.32;

Payment

$20 564.68

interval

(b) $3687.99;

coupon

on book

Premium

b = 6%

i = 5%

amortized

0

$28 687.99 5. (a) $1620.28

Book

Premium

value

balance

1079.18

49.18

1

60.00

53.96

6.04

1073.14

43.14

2

53.66

6.34

1066.80

53.34

6.66

1060.14

36.80 30.14

53.01

6.99

1053.15

23.15

1045.81

(b) $13 902.68

3

60.00 60.00

(c) $13 547.25

4

60.00

7. $4 569 384.48

5

60.00

52.66

7.34

9. $109 270.75

6

60.00

52.29

7.71

1038.10

15.81 8.10

7 Total

60.00

51.90

8.10

1030.00

—

420.00

370.82

49.18

B.

3. (a) $22.03 (b) $3104.14

1. $976.84 3. -$950.88 5. $387.55

B.

Exercise 15.4 A.

1. (a) $500.00 (b) $265.25 (c) $765.25 (d) $10 868.38 3. (a) $62.50 (b) $143.33 (c) $205.83

1. 5.868% 3. 8.14% 5. 9.26%

Exercise 15.5

(d) $2246.16 A.

1. (a) $558.24 (b) $6399.60

C.

1. (a) $2699.00 (b) $64 776.00

(c) $10 224.00 3. $2419.29; totals are $16 935.03; $3064.99; $20 000.02 5. $2840.83 7. (a) $495.27 (b) $9598.44 9. (a) $4275.00 (b) $1124.00 (c) $5399.00 (d) $36 976.00

# ANSWERS

TO

ODD-NUMBERED

11. (a) $302.01

PROBLEMS,

REVIEW

EXERCISES,

23. 15.:24%

(b) $21 903.91

(b) $28 862.02

(d) $764.19

(c) $1873.21

(e) $97 252.61;

(d) Totals are

totals are

$92 687.50;

$54 361.80;

$17 312.54;

$45 637.43;

$110 000.04

$99 999.23

27. (a) $13 750.00

13. (a) $24 750.00

PV $30 026; $29 769 5. Buy; Cost of buying = $83 564; leasing = $106 036 Exercise 16.2 1. Reject; NPV is

-$5367 3. Alternative 1;

(c) $33 354.00

(c) $22 029.90 (d) $72 194.23

(d) $13 931.00

(e) $5989.29

5. Alternative 1;

(e) $107 196.00

(f) Totals are

(f) $232 277.00; totals are

$66 239.23;

NPV $234; $203 NPV $1666; $352 B.

$33 760.77;

$172 073.00;

3. No; NPV is

29. (a) $2469.40

$300 000.00 Review Exercise

-$8561

3. $304.02 (premium) 5. $4507.68

7. (a) $15 620.41

7. 12.46%

(b) $85 897.02

9. 12.92%

(c) $1284.53;

11. $240.24 13. $603 102.20

Exercise 16.3 A.

CHAPTER 16

13. $914.16; totals art

Exercise 16.1 A.

1. Alternative 2;

$510.00; $595.84;

PV $44 634;

$85.84

$53 448

15. $21 576.64; totals

3. Alternative 1;

are $5700.00;

PV $22 205;

$5123.64; $576.64

$23 384

17. $1645.34

5. Alternative 1

19. 7.73% 21. (a) $44 248.79 (b) $48 017.12

(c) $1795.38

3. 20.7% B.

1. at 18%,

NPV = $6102; at 20%, NPV = -$292; R.O.I. == 19.9% 3. at 18%,

NPV = $1286; NPV = -$104; R.O.I. == 19.9%

(b) $24 625.42

(c) $23 330.78

1. 25.3%

at 20%,

9. $4527.64 11. (a) $3469.30

PV $2000; $2393 B.

at 18%, NPV = — $22 226; R.O.I = 17.4% 11. at 24%, NVP = $2035; at 26%, NVP = -$184; R.O.I = 25.8% 13. Project B; PV $22 256, $22 909

1. $9269.44

$84 612.49

R.O.I. = 26.3% 9. at 16%,

15. -$1215

Self-Test

$22 055.17

NPV = -$1939;

7. Yes; NPV $5142

3. $26 795.83

(b) $655.17;

at 28%,

5. Yes; NPV $5696

(b) $8318.12

$21 129.61

NPV = $370;

(c) $21 194.78 31. (a) n = 38.468063

(b) $4550.35

7. at 26%,

(b) $10 612.00 (d) $788.92

1. (a) $5336.73

5. (a) $1129.61;

1. Project B;

NPV $2568; $1787

$100 000.00

$127 929.00;

5. -$22 226

NPV = $47 272; A.

(b) $8279.90

(b) $8604.00

SELF-TESTS

3. Offer A;

25. (a) $9268.75

(c) $257.62

AND

1. Alternative 2;

5. at 22%,

NPV = $148; at 24%, NPV = -$1538; R.O.I. -: 22.2% Review Exercise

1. Alternative B; PV $51 624; $52 161 3. Alternative 1;

PV $62 363;

NPV $3916;

$73 601

$3597

17. Yes; NPV = $28 940 Self-Test

1. Alternative A; PV $13 552; $10 856 3. at 16%, NPV = $3466; at 18%, NPV =-$4386; R.O.I. = 16.9% 5. Proposal B; NPV $701; $1427

Index A accumulated depreciation, 200 accumulated value of one dollar per period, 445 accumulation factor, 351-353 accumulation factor for annuities, 445 accumulation of discount, 664-666 addition simplification with, 45—46 solving equations with, 70 algebraic expressions cost function, 194-196 described, 45 formulas, 91 revenue function, 194-196 simplification of. See algebraic simplification word problem solving, 80 algebraic simplification addition, 45^6 brackets, 46 brackets and multiplication, 47 combining (collecting) like terms, 45 division, 48 equation solving with, 74-79 evaluation, 48^19 multiplication, 46-48 substitution, 48-49 subtraction, 45^16 allocation problems, 97-98 amortization described, 575 final payment, determination of, 605-611 general annuities, 593-602 of premium, 661-663 prospective method, 582 retrospective method, 582,605 simple annuities, 575-591 amortization period, 616 amortization schedule computer application, 336-337, 590-591, 602 construction of, 335-336 general annuities, 593-594, 602 partial amortization schedule, 584-589, 598-602 purpose of, 333-334 simple annuity, 576-579, 590-591 amount. See future value “and,” 83 annual rent, 440 annuities annual rent, 440 annuities certain, 440 annuity due, 441 compounding factor for annuities, 445 consols, 558 contingent annuities, 440 deferred annuity, 441, 540-554 defined, 440 final payment, determination of, 605—611 future value of an annuity, 444 general annuities, 441

general annuity due, 531-536 ordinary annuity, 441 ordinary general annuities. See ordinary general annuities ordinary simple annuities. See ordinary simple annuities payment interval, 440 periodic rent, 440 perpetuity, 440, 557-563 simple annuities, 441 simple annuity due, 516-528 term of an annuity, 440 types of, 440-442 annuities certain, 440 annuity due described, 441 general annuity due, 531-536 general deferred annuity due, 551-554 simple annuity due, 516-528 simple deferred annuity due, 548-551 arithmetic average/mean, 14-15 assessed value (of real estate), 30 average investment method, 671-673 averages applications arithmetic average/mean, 14—15 basic problems, 13-14 depreciation, allocation of, 201-203 weighted arithmetic average, 15-18

B Bank of Montreal, 433 base common fraction or decimal as, 52-53 illustration of, 51 negative integer as, 52 positive integer as, 51-52 solving for, in percentage problems, 109-110 base period, 131 BEDMAS rule, 2 beta, 702 big city living costs, 33 bigger-the-better principle, 698 bird-in-the-hand principle, 698 blended payments, 297, 333 bond rate, 635, 648-649 bond schedules accumulation of discount, 664-666 amortization of premium, 661-663 book value, determination of, 666-669 bonds average investment method, 671-673 basic concepts, 635-636 bond rate, 635,648-649 book value, 666—669 buying on the market, 671 coupon rate, 635,648-649 debentures, 635 discount, 648-659 flat price, 643-645 formulas, 695 gain or loss on sale, 666-669 market price, 643-645

market rate, 648-649 premium, 648-659 problems, 636 promises associated with, 635 purchase price (alternative method), 652-659 purchase price between interest dates, 641-643 purchase price on interest date, 636-641 quoted price, 643-645 redemption of, 635 in stable market, 643 terminology, 635-636 yield rate, 648-649,671-674 book value, 200, 208-209, 666-669, 702 book value of the debt, 684 brackets and multiplication, 47 simplification with, 46 break-even analysis break-even charts, 191-196 break-even point, 185 contribution margin, 190 contribution rate, 190 cost-volume-profit relationships, 184-189 formulas, 215 loss leader, 197 unit contribution, 190-191 break-even chart algebraic approach to, 194 cost function, 194-196 described, 192 graphical approach to, 194-196 notes on, 193-194 revenue function, 194-196 break-even point, 185

c Canada Mortgage and Housing Corporation (CMHC), 614-616,616 Canada Premium Bonds, 660—661 Canada Savings Bonds (CSBs), 642, 660-661 capital expenditure evaluation, 698-704 cash discount basic concepts, 231 end-of-the-month (proximo) dating, 231, 234 ordinary dating, 231,232-234 partial payments, 235-236 purpose of, 231 receipt-of-goods dating, 231, 235 terms of payment, interpretation of, 231 cash value, 458 change, problems involving, 115-116 combining (collecting) like terms, 45 commission, 23-24 common fractions conversion to decimals, 5 described, 3 linear equation with, 75 common logarithms, 63 common stock trading price, 702

INDEX

comparison date, 382 compound discount, and present value, 368-374 compound interest basic procedure for computation, 346-347 compounding (accumulation) factor, 351-353 compounding (conversion) frequency, 349, 354 compounding (conversion) periods (n), 349,351,409-411 defined,346 effective rate of interest, 422-428, 438 equated date, 412-415 equivalent rate of interest, 422-423, 428^31 equivalent single payment, 382-388 equivalent values, 382-397 formulas, 406 future value formula, 348-349, 354-365 growth and, 350,381 nominal rate of interest (j), 417-420,427 periodic rate of interest (i), 349-350, 354, 417-420 preprogrammed financial calculators, 355-359 principal, changes in, 361-365 rate, changes in, 361-365 Rule of 70,411 spreadsheet application, 347-348 two or more equivalent replacement payments, 388-397 use of, 346 compound interest applications compound discount, and present value, 368-374 promissory notes, discounting, 374—380 compounding (accumulation) factor, 351-353 compounding (conversion) frequency, 349, 354 compounding (conversion) periods (n), 349, 351,409—411 compounding factor for annuities, 445 consols, 558 constant-dollar purchase program, 89 constant-percentage method, 207-208 Consumer Price Index (CPI), 131-133 contingent annuities, 440 contribution margin, 190 contribution rate, 190 conversion frequency, 349 conversion periods (n), 349, 351, 409—411 cost, calculation of, 243-246 cost function, 194-196 cost-volume-profit analysis break-even analysis, 184-197 cost-volume-profit relationships, 184-189 effect of changes, 198-199 formulas, 215 sensitivity analysis, 184, 198-199 coupon rate, 635, 648-649 coupons, 635 CPI “basket,” 131 credit cards, 328-331 credit period, 231

cross-multiplication, 101 cross rate tables, 129 currency conversions cross rate tables, 129 exchange rate, 127 with proportions, 127-128

D dated values, 289-298 debentures, 635 debt book value of, 684 periodic cost of, 684 retirement by sinking fund method, 684-685 decimals common fraction, conversion from, 5 mixed number, conversion from, 5 percent, conversion from, 10 percent, conversion to, 11 declining-balance approach, 324 declining-balance method, 205-207 deferred annuity basic concepts, 540-541 described, 441 future value, 540-541 general deferred annuity due, 551-554 ordinary general deferred annuity, 545-547 ordinary simple deferred annuity, 540-545 period of deferment, 540 periodic payment (PMT), 543 present value, 542 simple deferred annuity due, 548-551 term n, 544 deferred perpetuity due, 562 demand loans declining balance approach, 324 defined, 323 examples, 323-324 partial payments, 324—326 demand mortgage, 613 demand note, 323 denomination, 635 denominator, 3 depreciation accounting treatment, 200 accumulated depreciation, 200 basic concepts, 200 book value, without depreciation schedule, 208-209 common aspects, 200 formulas, 215-216 methods. See depreciation methods use of, 184 depreciation methods allocation as diminishing charge per year, 203-208 allocation based on average, 201-203 constant-percentage method, 207-208 declining-balance method, 205—207 development of, 200 straight-line method, 201 Sum-of-the-Years-Digits Depreciation (SYD), 203-205 units-of-product method, 202-203

depreciation schedules, 200 discount (bonds) accumulation of, 664—666 described, 648-649 direct computation method, 652-659 discount factor, 371 discount period, 231,239 discount series, 224-227 discounted benefit-cost ratio, 716-717, 718-719 discounted cash flow bigger-the-better principle, 698 bird-in-the-hand principle, 698 defined, 698 described, 698 discounted value. See present value discounting defined, 369 promissory notes, 374-380 discounts cash discount, 231-236 rate of discount, 231 trade discount. See trade discount division monomials, 48 of polynomial by monomial, 48 of powers, 54 quotient, power of, 55 solving equations with, 72 dot-coms, profitability of, 124 down payment, 458,459 due date, 231, 635

E earnings per share, 702 effective rate of interest described, 422-428 fixed-rate mortgages, 616-618 formula, 438 per payment period, 489—491, 494-^495 end-of-the-month (proximo) dating, 231, 234 equated date, 412-415 equation of values, 294, 382 equation solving addition, 70 with algebraic simplification, 74-79 checking solution, 73-74 division, 72 formula rearrangement, 77-79 linear equations. See linear equation solving linear systems. See linear systems problems multiple operations, 72-73. multiplication, 71 subtraction, 70-71 systems of equations, 175-176 equations basic terms and concepts, 69 defined, 69 equivalent equations, 69 first-degree equation, 69 linear equation, 69 members of an equation, 69 root of an equation, 69 solving the equation, 69 See also equation solving

INDEXW

equity, 613 equivalent equations, 69 equivalent fractions, 4 equivalent rate of interest, 422-423,428^31 equivalent ratios in higher terms, 96-97 equivalent values compound interest, 382-397 defined, 382 simple interest, 289-298 evaluation of algebraic expressions, 48-49 exchange rate, 127 exponent described, 51 fractional exponents, 58-61 integral exponents, 51-58

F face value, 635 factor, 51 “fewer than,” 83 final payment, determination of, 605-611 financial calculators breakeven function, 189 cash flow sign convention, 358-359 compound interest, 355-359 compounding factor, numerical value of, 352-353 conversion rate, 505 date function, 270 decimal places, 60 effective rate of interest, 422 effective rate of interest per payment period, 489-490 equivalent rate of interest, 422-423 future value, 355-359 interest period, 269 intermediate values, and decimal places, 13-14 internal rate of return, 727 natural logarithm function, 64 ordinary general annuities (future value), 491—493 ordinary general annuities (present value), 495 ordinary simple annuity (future value), 446-448 ordinary simple annuity (present value), 457 percent of increase over base, 111 perpetuities due, 563 power function, 53 present value, 371-372 simple annuity due, 521-522 storage of numbers, 15 time value of money (TVM) worksheet, 412 first-degree equation, 69 first mortgage, 613 fixed costs, 185,186 fixed-rate mortgage, 613,614,616-618 flat price, 643-645 focal date, 290,382 formula manipulation, 77-79 formula rearrangement, 77-79 formulas (summary) basic algebra formulas, 91 bonds, 695

break-even analysis, 215 compound interest, 406 depreciation, 215-216 effective rate of interest, 438 future value, 406,438 markup, 266 net present value, 734 ordinary general annuities, 513 ordinary simple annuity, 483 percents, 145 present value, 406 profitability index, 734 promissory notes, 342 proportions, 145 ratios, 145 simple interest, 308 trade discount, 265-266 fractional constants and multiplication, 76-77 fractional exponents radicals, 58-60 representation of, 60-61 fractions common fractions, 3, 5 complex fractions, 6 denominator, 3 equivalent fractions, 4 improper fraction, 3 in linear systems in two variables, 152-153 lowest common denominator (LCD), 75, 77 mixed number conversion to decimals, 5 percent, conversion from, 8-10 percent, conversion to, 11 proper fraction, 3 reducing to lower terms, 4 rounding, 5 terms of the fraction, 3 future value of an annuity, 444 basic concept of, 281 compound interest, 348-349, 354-365 dated values, computation of, 290 deferred annuity, 540-541 defined,281 formula, 282, 406, 438 general annuity due, 531-532 of investment, 354-355 maturity value of promissory note, 314 n as fractional value, 359-360, 438 ordinary general annuities, 486-493 ordinary simple annuity, 443—452 and perpetuity, 557 preprogrammed financial calculators, 355-359 principal, changes in, 361-365 rate, changes in, 361-365 simple annuity due, 516-518

G general annuities, 441 See also ordinary general annuities general annuity due deferred, 551-554 future value, 531-532 nominal interest rate, 537

periodic payment (PMT), 533-535 present value, 532—533 term n, 535-536 Goods and Services Tax (GST), 28-29, 38-39 graduated commission, 23 graphs cost function, 194—196 of linear equations, 158-169 linear systems of equations in two unknowns, 169-171 lines parallel to X axis, 161 lines parallel to Y axis, 162 ordered pair of numbers, 155 origin, 154 plotting, 156 rectangular coordinates, system of, 154-156 revenue function, 194-196 slope-y-intercept form of linear equation, 162-167 straight-line graph, 167 table of values, 157-158 X axis, 154, 167 x coordinate, 155 Yaxis, 154, 167 /coordinate, 155 Gross Debt Service (GDS) Ratio, 41, 614 gross earnings, 20,24-25 gross profit. See markup

H Harmonized Sales Tax (HST), 29 high risk/high yield investments, 727 house purchase, real cost of, 68

I improper fraction, 3 index, 58 index numbers base period, 131 Consumer Price Index (CPI), 131-133 nature of, 130-131 integral exponents basic concept and definition, 51 common fraction or decimal as base, 52-53 negative exponents, 56-57 negative integer as base, 52 operations with powers, 53-55 positive integer as base, 51-52 powers with positive integer, 51-53 zero exponent, 55-56 interest compound interest. See compound interest defined, 271 rate money is worth, 317 rates of. See interest rates simple interest. See simple interest time period of interest, 277-278 interest-bearing promissory note, 311, 376-378 See also promissory notes interest conversion period, 488 interest period counting exact time, 300-301

INDEX

determination of, 269-270 leap years, 303 manual techniques for determination, 300-303 of promissory note, 311 tables, 301-302 interest rates effective rate of interest. See effective rate of interest equivalent rate of interest, 422-423, 428-431 nominal rate of interest. See nominal rate of interest periodic rate of interest. See periodic rate of interest promissory notes, 311 investment, 89-90 investment decision applications common stock trading price, 702 discounted cash flow, 698-704 net present value method, 706-713 rate of return on investment (R.O.I.), 716-728 investment risk, 727

L leap years, 269, 273, 303 least common denominator (LCD), 75, 77 “less than,” 83 like terms, 45 line rise, 162 run, 162 slope, 162 y-intercept, 162 line of credit, 328-331,333 linear equation described, 69 graphing, 158-169 slope-y-intercept form, 162-166 table of values, 157-158 linear equation solving common fraction, 75-76 fractional constants and multiplication, 76-77 integral constant or binomial product, 74-75 systems of, in two variables, 148-154 linear interpolation, 719-720 linear programming, 184 linear systems in two variables basic concept, 148 with fractions, 152-153 graphing, 169-171 non-numerically equal coefficients, 150-151 removal of decimals or fractions, 153 solving by algebraic elimination, 148-150 linear systems problems one equation in two variables, 173-174 systems of equations, 175-176 liquidity, 660-661 list price computation of, where discount known, 221 described, 220 net price, relationship with, 222

literal coefficients, 45 loan repayment schedules, 333-337 See also amortization schedule loan repayments, 297-298 loans credit cards, 328-331 demand loans, 323-327 line of credit, 328-331, 333 statement of disclosure, 334 logarithms common logarithms, 63 concept, 62 defined, 62 natural logarithms, 63-64 useful relationships, 65-67 long-term promissory notes, and discount¬ ing, 374 loss leader, 197,249 low risk/low yield investments, 727 lowest common denominator (LCD), 75, 77

M manufacturer’s suggested retail price (MSRP), 220 margin. See markup marginal tax rate, 134-135 markdown basic concepts and calculations, 249-252 defined,249 integrated problems, 253-255 pricing strategies, 249 market price, 643-645 market rate, 648-649 markup basic concepts, 239-241 calculations, 239-241 cost, calculation of, 243-246 formulas, 266 rate of market, 242 selling price, calculation of, 243-246 statement of, 242 maturity date, 635 maturity value. See future value members of an equation, 69 merchandising terminology, 219-220 method of averages, 671-673 methods of depreciation. See depreciation methods mill rate, 30 millionaires, 464 monomials, 46 mortgage contract, 613 mortgage loan, 458 See also residential mortgages mortgage statement, 622-624 mortgagee, 613 mortgagor, 613 multiple discounts, 224-230 multiplication and brackets, 47 cross-multiplication, 101 monomials, 46 of monomials with polynomials, 47 of polynomial by polynomial, 47 of powers, 53-54 product. See product

simplification, 46-48 solving equations with, 71 mutual fund, 136 mutual fund returns, 136-138

N National Housing Act, 614-616 natural logarithms, 63-67 negative exponents, 56-57 net asset value, 136 net factor approach, 222-223, 226 net income, 185 net present value applications, 708-713 assumptions, 708 concept, 707-708 criterion rule, 708 formula, 734 introductory examples, 706-707 negative vs. positive, 708 rate of return and, 716, 717 net price computation of, where discount known, 221 described, 220 discount series, 225 list price, relationship with, 222 net factor approach, 222-223 net price factor (NPF), 222-223, 226 net sales, 23 net value, 200 net worth, 41 New Canada Savings Bonds Payroll Savings Program, 288 nominal income, 132 nominal rate (bonds), 635 nominal rate of interest general annuity due, 537 nominal annual rate, 417^120,427 in simple interest formula, 271 as stated rate, 349 non-interest-bearing promissory notes, and discounting, 375-376 numerator, 3 numerical coefficients, 45

o 1% method, 108 operations inverse operations, 77-78 order of operations, 2 undoing operations, 77 order of operations, 2 ordered pair of numbers, 155 ordinary annuities, 441 See also ordinary general annuities; ordinary simple annuities ordinary dating, 231, 232-234 ordinary general annuities amortization, 593-602 amortization schedule, 593-594 deferred, 545-547 effective rate of interest per payment period, 489^91, 494-495 financial calculators, and future value, 491-493

INDEX

financial calculators, and present value, 495 formulas, 513 future value, 486—493 interest conversion period, 488 interest paid and principal repaid, 598-602 outstanding principle, determination of, 595-598 partial amortization schedule, 598-602 payment interval, 488 periodic interest rate (i), 505-506 periodic payment (PMT) (known future value), 497—498 periodic payment (PMT) (known present value), 498-499 present value, 494-495 term n (known future value), 501-502 term n (known present value), 502-503 ordinary general perpetuity, 560 ordinary simple annuities amortization, 575-591 amortization schedule, 576-579 compounding factor for annuities, 445 deferred, 540-545 financial calculators, and future value, 446-448 financial calculators, and present value, 457 formulas, 483 future value, 443-452 future value applications, 448-452 future value formula, 445-446 interest paid and principal repaid, 584-589 outstanding principal balance, calculation of, 580-584 partial amortization schedule, 584-589 periodic payment (PMT) (known future value), 465-466 periodic payment (PMT) (known present value), 467-468 periodic rate of interest (i), 477 present value, 454-462 present value applications, 458-462 present value formula, 456—457 restatement of future value formula, 446 restatement of present value formula, 457 series of payments, future value of, 443—444 series of payments, present value of, 454-455 term n (known future value), 471-473 term n (known present value), 474-475 origin, 154 original amount, solving for, 117-118 original cost, 200 overtime, 24 overtime premium, 25

P par value, 635 partial amortization schedule, 584-589 partial payments cash discounts and, 235-236 demand loans, 324-326 partial sinking fund schedules, 678-682

payment interval, 440,488 payment period, 440 payment terms discount period, 231, 239 end-of-the-month dating, 234 interpretation of, 231 ordinary dating, 232 receipt-of-goods dating, 235 payroll applications commission, 23-24 gross earnings, 20, 24—25 salaries, 20-22 wages, 24-26 percent See also percent problems commonly used percents, 106-107 decimals, conversion from, 11 decimals, conversion to, 10 formulas, 145 fractions, conversion from, 11 fractions, conversion to, 8-10 meaning of, 8 percent problems applications, 111-112 base, solving for, 109-110 basic percentage relationship problems, 121-122 commonly used percents, computation with, 106-107 1% method, 108 original amount, solving for, 117-118 percent change problems, 115-116, 122-124 percentages, computation of, 106 rate of increase or decrease, 116-117, 122-124 rate percent, 108-109, 120 useful relationships, summary of, 119 period of deferment, 540 periodic cost of the debt, 684 periodic payment (PMT) applications, 468-469 deferred annuity, 543 general annuity due, 533-535 known future value, 465-466, 497-498 known present value, 467-468,498-499 ordinary general annuities, 497-499 ordinary simple annuities, 465-469 simple annuity due, 523-524 sinking funds, 674—675 periodic rate of interest and compound interest, 349-350,354, 417-420 ordinary general annuities, 505-506 ordinary simple annuities, 477 simple annuity due, 527-528 periodic rent, 440 perpetuities due deferred perpetuity due, 562 financial calculators, 563 present value of, 560-563 perpetuity defined,557 described, 440 and future value, 557 ordinary general perpetuity, 560 perpetuities due, 560-563 and present value, 557,558-560

perpetuity payment, 558 personal income taxes, 134-135 polynomials division, 48 multiplication of, 47 power, 51 power of sale, 613 powers division, 54 multiplication, 53-54 operations with, 53-55 with positive integral exponents, evaluation of, 51-53 of product, 55 of quotient, 55 raising power to a power, 54-55 of the root, 58 premium amortization of, 661-663 described, 648-649 direct computation method, 652-659 present value and compound discount, 368-374 concept of, 285-286, 368-369 deferred annuity, 542 defined, 284-286 formula, 284, 285—286, 369-371,406 and future value formula, 284 general annuity due, 532-533 n as fractional value, 372-373 ordinary general annuities, 494-495 ordinary simple annuity, 454-462 perpetuities, 557,558—560 perpetuities due, 560-563 preprogrammed financial calculators, 371-372 of promissory notes, 316-320 related terms, 368-369 simple annuity due, 519-520 of treasury bills, 320-321 price-earnings (P/E) ratio, 702 principal, 275-276, 284-286 proceeds. See present value product power of, 55 root of, 58 profit, 249 profitability index, 716-717, 718-719, 734 promissory notes amount of interest, 312 computed values, 312-313 date of maturity, 311 defined, 311 discounting, 374-380 due date, 311 face value, 311,316 formulas, 342 interest-bearing, 311 interest-bearing, and discounting, 376-378 interest period, 311 issue date (date of issue), 311 legal due date, 312 long-term, and discounting, 374 maker of note, 311 maturity value, 312,314-315 non-interest-bearing, and discounting, 375-376

INDEX

payee of note, 31J present value, 316-320 rate of interest, 311 related terms, 311-312 term, 311 three days of grace, 312 use of, 312 proper fraction, 3 property taxes, 28, 30-31 proportions currency conversions, 127-128 described, 100 formulas, 145 problems, 102-104 setting up, 102 solving, 100-102 with unknown term, 100 prospective method, 582 Provincial Sales Tax (PST), 28, 29-30 proximo dating, 231, 234 purchasing power of the dollar, 131

Q quota, 23 quoted price, 643-645 quotient, power of, 55

R radicals, 58-60 radicand, 58 rate money is worth, 317 rate of change, 116-117, 122-124 rate of discount, 231 rate of interest, 276-277 See also interest rates rate of markup, 242 rate of return on investment (R.O.I.) computation of, 720-727 described, 716 discounted benefit-cost ratio, 716-717, 718-719 internal rate of return, 727 linear interpolation, 719-720 and net present value concept, 716, 717 profitability index, 716-717, 718-719, 734 trial and error procedure, 717-718 rate percent, 108-109, 120 ratios allocation problems, 97-98 defined,95 discounted benefit-cost ratio, 716-717, 718-719 earnings per share, 702 equivalent ratios in higher terms, 96-97 formulas, 145 Gross Debt Service (GDS) Ratio, 41, 614 price-earnings (P/E) ratio, 702 profitability index, 716-717, 718-719, 734 reducing to lowest terms, 96 rise/run, 163 setting up, 95-96 terms of the ratio, 95 Total Debt Service (TDS) Ratio, 41,615 real income, 132 receipt-of-goods dating, 231, 235 rectangular coordinates, system of, 154-156

redeemable at a premium, 635 redeemable at par, 635 redemption date, 635 redemption price, 635 “reduced by,” 83 “reduced to,” 83 regular selling price, 250 residential mortgages CMEfC mortgages, 614-616, 616 defined, 613 demand mortgage, 613 effective rate of interest, 616-618 first mortgage, 613 fixed-rate mortgage, 613, 614, 616-618 home buyer strategies, 616 mortgage statement, 622-624 open mortgages, 614 payments and balances, computation of, 618-619 rounded payments, 620-621 second mortgage, 613 variable-rate mortgage, 613 residual value, 200 retrospective method, 582, 605 returns mutual funds, 136-138 rate of return. See rate of return on investment (R.O.I.) simple rate of return, 137 revenue function, 194-196 rise, 162 rise/run ratio, 163 risk, and investments, 727 root of an equation, 69 root of the product, 58 rounding, 5 Royal Bank of Canada, 621 Rule of 70,411 run, 162

s salaries, 20-22 salary plus commission, 23 sale price, 250 salvage value, 200 schedule of amortization of premium, 661 scrap value, 200 second mortgage, 613 secured line of credit, 328 selling price, 239, 243-246 sensitivity analysis, 184, 198-199 simple annuities, 441 See also ordinary simple annuities simple annuity due deferred, 548-551 financial calculators, 521-522 future value, 516-518 vs. ordinary annuities, 518 periodic payment (PMT), 523-525 periodic rate of interest (i), 527-528 present value, 519-520 term n, 525-527 simple interest basic concepts, 271-272 blended payments, 297 computation of amount, 272-273 dated values, 289-298

derived formulas, 274—275 equation of values, 294 equivalent single payment, 291-292 equivalent values, computation of, 289-298 focal date, 290 formula, various forms of, 275 formula for, 271-272 formulas, summary of, 308 future value, computation of, 281-283 interest period, 269-270, 300-303 loan repayments, 297-298 matching rate with time, 271-272 present value, 284-286 principal, 275-276 rate of interest, 276-277 time period, 277-278 simple interest applications credit cards, 328-331 demand loans, 323-327 formulas, 342 line of credit, 328-331, 333 loan repayment schedules, 333-337 maturity value of promissory notes, 314-315 present value of promissory notes, 316-320 promissory notes, 311-314 treasury bills, present value of, 320-321 two ore more equivalent payments, 293-294 simple rate of return, 137 single equivalent rates of discount, 227-230 sinking funds accumulated balance and interest earned, 678-682 debt retirement, 684-685 defined,674 increase for payment interval, 678-682 interest rates, 684 partial sinking fund schedules, 678-682 periodic payment size, 674-675 schedules, 675-677 slope, 162 slope-y-intercept form of linear equation described, 162-166 lines parallel to X axis, 166 lines parallel to Y axis, 166 special cases, 166-168 stock trading price, 702 straight commission, 23 straight-line graph, 167 straight-line method, 201 substitution of algebraic expressions, 48-49 subtraction simplification, 45-46 solving equations with, 70-71 Sum-of-the-Years-Digits Depreciation (SYD), 203-205

T T-bills, 320-321 taxes defined, 28 Goods and Services Tax (GST), 28-29 Harmonized Sales Tax (HST), 29 and investments, 539-540

INDEX

marginal tax rate, 134—135 personal income taxes, 134—135 property taxes, 28, 30-31 Provincial Sales Tax (PST), 28, 29-30 term of an annuity', 440 terms of the fraction, 3 terms of the ratio, 95 theatre chains, 172 time period of interest, 277-278 time value of money, 284,285,289 total cost, 185-186, 249 Total Debt Service (TDS) Ratio, 41,615 total revenue, 185-186 total variable cost, 186 trade discount basic concepts and computations, 220 computation of amount of, 221 defined,220 discount series, 224-227 formulas, 263-266 list price, 220,221 multiple discounts, 224-230 net factor approach, 222-223

net price, 220,221 rate of discount, computation of, 221-222 single equivalent rates of discount, 227-230 trade-in value, 200 treasury bills, present value of, 320-321

weighted arithmetic average, 15-18 what-if analysis. See sensitivity analysis word problems key words, identification of, 83 as practical applications, 80 systematic procedure for solving, 80

X

u units-of-product method, 202-203 unsecured line of credit, 328 used cars, values of, 210 useful life, 200

V variable costs, 185, 186 variable-rate mortgage, 613

X axis, 154,167 x coordinate, 155

Y Y axis, 154,167 y coordinate, 155 y-intercept, 162 yield rate, 648-649,671-674

z w wages, 24—26 wearing value, 200

zero exponent, 55—56

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Finding the effective rate of interest f for a nominal

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annual rate compounded

Formula 11.1

S = R

Formula 11.2

A = R

Formula 12.1

P — (1

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Finding the future value (accumulated value) of an ordinary simple annuity

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Finding the present value (discounted value) of an ordinary simple annuity

Finding the effective rate of interest per payment period p for a nominal annual rate of interest compounded c times per payment interval

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TERMS OF USE AND WARRANTY DISCLAIMERS READ BEFORE OPENING THIS PACKAGE. By opening this sealed package, you accept and agree to the terms and conditions printed below. If you do not agree, DO NOT OPEN THIS PACKAGE and immediately return the sealed package with a copy of your receipt for a full refund.

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NET PRICE FACTOR FOR THE DISCOUNT SERIES

Formula 6.6A

NET PRICE =

Formula 6.7

SINGLE EQUIVALENT RATE OF DISCOUNT

Finding the net amount directly when a list price X LIST PRICE

is subject to a series of discounts

Finding the single rate of discount that has the same effect as a given series of discounts

FOR A DISCOUNT SERIES — 1 — NPF FOR THE DISCOUNT SERIES

= 1- [(l-^xi -d2){\- d3) ...(1- d)}J rt'

SELLING

Formula 6.8A

Basic relationship between selling price, cost, expenses

= COST + EXPENSES + PROFIT

PRICE

such as overhead, and profit

or S = C + E + P

Relationship between markup expenses such as

MARKUP = EXPENSES + PROFIT

overhead and profit

or M = E + P

Formula 6.10

RATE OF MARKUP

RATE OF MARKUP BASED ON SELLING PRICE

Formula 7.1A

Finding the rate of markup as a percent of cost

- MARKU- = —

BASED ON COST

COST

_

MARKUP

_ _M

— SELLING PRICE

Finding the rate of markup as a percent of selling price

S

Finding the amount of interest when the principal, the

I = Prt

rate, and the time are known Finding the future value (maturity value) when the

Formula 7.2

S — P + I

Formula 7.3A

S = P(1 + rt)

principal and the amount of interest are known Finding the future value (maturity value) at simple interest directly when the principal, rate of interest, and time are known

Formula 9.1A

S = P(1 + 0"or FV = PV(1 +

Finding the future value of the compound amount (or

0

"

maturity value) when the original principal, the rate of interest, and the time period are known

Formula 9.2

Finding the periodic rate of interest

. = J_

1

m

Formula 10.1

f= (1 + i)m — 1

Formula 11.1

Sn = R

Formula 11.2

An

Formula 12.1

p=

=

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R

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Finding the effective rate of interest f for a nominal annual rate compounded

or FVr, = PMT

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Finding the future value (accumulated value) of an ordinary simple annuity

i 1 - (1 + irn

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Finding the present value (discounted value) of an ordinary simple annuity

Finding the effective rate of interest per payment period p for a nominal annual rate of interest compounded c times per payment interval

Formula 12.2

where p

Formula 12.3

(l+py-r p

FVnc = PMT

Finding the future value of an ordinary general annuity using the equivalent effective rate of interest per payment period

= (1 + i)c — 1

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~i ~ (i

Finding the present value of an ordinary general

V

annuity using the effective rate of interest per

where p = (l + z)c — 1

Formula 13.1

payment period

Finding the future value of an annuity due

or FV (due) = PMT(1 +

Formula 13.2

An(due) = R(1 +

Formula 13.4

1

-

or PV/ic(due) = PMT(1 +

Formula 13.5

Finding the present value of an annuity due

py -1

Finding the future value of a general

(i

p)

annuity due using the effective rate of interest

+ py-1

per payment period

p

i)c — 1

AJdue) = R(1 + /)

where p

0" -1

p

or FVnc(due) = PMT(1 + where p = (1 +

+

i)

(i +

+ p)

S (due) = R(1

i) (1

i) -L

orPVn(due) = PMT(1 +

Formula 13.3

i)n - 1

(1 +

S (due) = R(1 + z)

(1

+/)-"-

/

Finding the present value of a general annuity

p) i - (i + prn p

due using the effective rate of per payment period

= (l + i)c — l

A = ^ or PV = i

PMT Finding the present value of an ordinary simple

i

perpetuity

Formula 13.6

Formula 13.7

R PMT A = -r or PV =/ P where p = (1 + i)c — 1 A(due) = R + ^ or PV(due) = PMT +

Finding the present value of an ordinary general perpetuity

PMT •

Finding the present value of a simple perpetuity due

Formula 13.8

A(due) = R +

~j or PV(due) = PMT +

where p — (1 +

i)c — 1

PMT

P

Finding the present value of a general perpetuity due

Formula 15.1

PP

=

S(1

+

i)~n+ R

1 - (1 +

i)-"

Finding the purchase price of a bond when the interest payment

or PP = FV(1 + i)_n+ PMT

Formula 15.2

PP = S(1 +/)"”+ R

interval and the yield rate conversion

1 - (1 + i)~n

r i - (i

period are equal

+f) f

Finding the purchase price of a bond when the interest payment

or PP = FV(1 + where p

Formula 15.3

i - (i

pyn + PMT

= (1 +

interval and the yield rate conversion

+prn p

period are different

i)c — 1

Direct formula for finding the premium or

PREMIUM or DISCOUNT =

(b X

discount of a bond (a negative answer

? X

FACE VALUE —

- (i + REDEMPTION VALUE)

indicates a discount)

jyn

i AVERAGE INCOME PER INTEREST PAYMENT INTERVAL

Formula 15.4

APPROXIMATE VALUE OF

i

—

AVERAGE BOOK VALUE

Basic formula for finding the

where

yield rate using =

AVERAGE BOOK VALUE

— (QUOTED PRICE

+ REDEMPTION PRICE)

the method of averages.

and TOTAL INTEREST PAYMENTS

AVERAGE INCOME

- PREMIUM +

PER INTEREST NUMBER OF INTEREST PAYMENT INTERVALS

PAYMENT INTERVAL

Formula 16.1

NET PRESENT VALUE _ PRESENT VALUE _ PRESENT VALUE (NPV)

Formula 16.2

_

OF INFLOWS

OF OUTLAYS

PRESENT VALUE OF INFLOWS PROFITABILITY INDEX =

PRESENT VALUE OF OUTLAYS

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ISBN 0-13-120128-X

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9780131201286

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