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    McGILL UNIVERSITY

    FACULTY OF SCIENCE

    DEPARTMENT OF

    MATHEMATICS AND STATISTICS

    MATH 329 2004 01

    THEORY OF INTEREST

    Information for Students(Winter Term, 2003/2004)

    Pages 1 - 8 of these notes may be considered theCourse Outline for this course.

    W. G. Brown

    April 29, 2004

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    Information for Students in MATH 329 2004 01

    Contents

    1 General Information 1

    1.1 Instructor and Times . . . . . . 11.2 Course Description . . . . . . . 1

    1.2.1 Calendar Description . . 11.2.2 Syllabus (in terms of sec-

    tions of the text-book) . 11.2.3 Verbal arguments . . 4

    1.3 Evaluation of Your Progress . . 41.3.1 Term Mark . . . . . . . 41.3.2 Assignments. . . . . . . 4

    1.3.3 Class Test . . . . . . . . 41.3.4 Final Examination . . . 51.3.5 Supplemental Assessments 51.3.6 Machine Scoring . . . . 51.3.7 Plagiarism . . . . . . . . 5

    1.4 Published Materials . . . . . . 61.4.1 Required Text-Book . . 61.4.2 Website . . . . . . . . . 61.4.3 Reference Books . . . . 6

    1.5 Other information . . . . . . . 71.5.1 Prerequisites . . . . . . 7

    1.5.2 Calculators . . . . . . . 71.5.3 Self-Supervision . . . . . 71.5.4 Escape Routes . . . . . 71.5.5 Showing your work; good

    mathematical form; sim-plifying answers . . . . . 7

    2 Timetable 9

    3 First Problem Assignment 11

    4 Second Problem Assignment 12

    5 Solutions, First Problem Assign-

    ment 14

    6 Third Problem Assignment 18

    7 Solutions, Second Problem Assign-

    ment 20

    8 Solutions, Third Problem Assign-

    ment 30

    9 Class Tests 41

    9.1 Class Test, Version 1 . . . . . . 419.2 Class Test, Version 2 . . . . . . 469.3 Class Test, Version 3 . . . . . . 519.4 Class Test, Version 4 . . . . . . 56

    10 Fourth Problem Assignment 61

    11 Solutions to Problems on the Class

    Tests 63

    12 Fifth Problem Assignment 73

    13 Solutions, Fourth Problem Assign-

    ment 75

    14 Solutions, Fifth Problem Assign-

    ment 80

    15 References 901

    A Supplementary Lecture Notes 2001

    A.1 Supplementary Notes for the Lec-tures of January 5th, 7th, and9th, 2004 . . . . . . . . . . . . 2001A.1.1 1.2 The accumulation and

    amount functions . . . . 2001A.1.2 1.3 The effective rate of

    interest . . . . . . . . . 2002A.1.3 1.4 Simple interest . . 2003A.1.4 1.5 Compound interest 2003A.1.5 1.6 Present value . . . 2003A.1.6

    1.7 The effective rate of

    discount . . . . . . . . . 2004A.1.7 1.8 Nominal rates of in-

    terest and discount . . . 2005A.2 Supplementary Notes for the Lec-

    ture of January 12th, 2004 . . . 2006A.2.1 1.9 Forces of interest and

    discount . . . . . . . . . 2006A.2.2 1.10 Varying interest . 2008

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    Information for Students in MATH 329 2004 01

    A.2.3 1.11 Summary of results 2008A.3 Supplementary Notes for the Lec-

    ture of January 16th, 2004 . . . 2009A.3.1 2.1 Introduction . . . . 2009A.3.2 2.2 Obtaining numeri-

    cal results . . . . . . . . 2009A.3.3 2.3 Determining time pe-

    riods . . . . . . . . . . . 2011A.3.4 2.4 The basic problem 2011

    A.4 Supplementary Notes for the Lec-ture of January 19th, 2004 . . . 2012A.4.1

    2.4 The basic problem

    (continued) . . . . . . . 2012A.4.2 2.5 Equations of value 2012A.4.3 Unknown principal . . . 2012A.4.4 2.6 Unknown time . . . 2013A.4.5 2.7 Unknown rate of in-

    terest . . . . . . . . . . 2014A.4.6 2.8 Practical examples 2014

    A.5 Supplementary Notes for the Lec-ture of January 21st, 2004 . . . 2016

    A.6 Supplementary Notes for the Lec-ture of January 23rd, 2004 . . . 2018

    A.6.1 3.1 Introduction . . . . 2018A.6.2 3.2 Annuity-immediate 2018

    A.7 Supplementary Notes for the Lec-ture of January 26th, 2004 . . . 2020A.7.1 3.2 Annuity-immediate

    (continued) . . . . . . . 2020A.7.2 3.3 Annuity-due . . . . 2022

    A.8 Supplementary Notes for the Lec-ture of January 28th, 2004 . . . 2023A.8.1 3.3 Annuity-due (contin-

    ued) . . . . . . . . . . . 2023

    A.9 Supplementary Notes for the Lec-ture of January 30th, 2004 . . . 2025A.9.1 3.4 Annuity values on

    any date . . . . . . . . . 2025A.10 Supplementary Notes for the Lec-

    ture of February 2nd, 2004 . . . 2029A.10.1 3.4 Annuity values on

    any date (continued) . . 2029

    A.11 Supplementary Notes for the Lec-

    ture of February 4th, 2004 . . . 2033A.11.1 3.5 Perpetuities . . . . 2033A.11.2 3.6 Nonstandard terms

    and interest rates . . . . 2034A.12 Supplementary Notes for the Lec-

    ture of February 6th, 2004 . . . 2035A.12.1 3.6 Nonstandard terms

    and interest rates . . . . 2035A.12.2 3.7 Unknown time . . . 2035

    A.13 Supplementary Notes for the Lec-ture of February 9th, 2004 . . . 2038

    A.13.1 3.7 Unknown time (con-tinued) . . . . . . . . . 2038

    A.14 Supplementary Notes for the Lec-ture of February 11th, 2004 . . 2040A.14.1 3.8 Unknown rate of in-

    terest . . . . . . . . . . 2040A.14.2 3.9 Varying interest . . 2047A.14.3 3.10 Annuities not in-

    volving compound interest2047A.15 Supplementary Notes for the Lec-

    ture of February 13th, 2004 . . 2048

    A.15.1 4.1 Introduction . . . . 2048A.15.2 4.2 Annuities payable at

    a different frequency thaninterest is convertible . 2048

    A.16 Supplementary Notes for the Lec-ture of February 16th, 2004 . . 2051A.16.1 4.3 Further analysis of

    annuities payable less fre-quently than interest isconvertible . . . . . . . 2051

    A.16.2

    4.4 Further analysis of

    annuities payable more fre-quently than interest isconvertible . . . . . . . 2055

    A.16.3 4.5 Continuous annuities 2057A.16.4 4.6 Basic varying annu-

    ities . . . . . . . . . . . 2058A.17 Supplementary Notes for the Lec-

    ture of February 18th, 2004 . . 2059

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    A.17.1 4.6 Basic varying annu-ities (continued) . . . . 2059

    A.17.2 4.7 More general vary-ing annuities . . . . . . 2064

    A.17.3 4.8 Continuous varyingannuities . . . . . . . . 2064

    A.17.4 4.9 Summary of results. 2064A.18 Supplementary Notes for the Lec-

    ture of February 20th, 2004 . . 2065A.18.1 5.1 Introduction . . . . 2065A.18.2 5.2 Discounted cash flow

    analysis . . . . . . . . . 2065

    A.18.3 5.3 Uniqueness of the yieldrate . . . . . . . . . . . 2065

    A.18.4 5.4 Reinvestment rates 2066A.19 Supplementary Notes for the Lec-

    ture of March 1st, 2004 . . . . 2067A.19.1 5.4 Reinvestment rates

    (continued) . . . . . . . 2067A.19.2 5.5 Interest measurement

    of a fund . . . . . . . . 2069A.19.3 5.6 Time-weighted rates

    of interest . . . . . . . . 2069

    A.19.4 5.7 Portfolio methods andinvestment year methods 2069

    A.19.5 5.8 Capital budgeting . 2069A.19.6 5.9 More general borrow-

    ing/lending models . . . 2069A.20 Supplementary Notes for the Lec-

    ture of March 3rd, 2004 . . . . 2070A.20.1 6.1 Introduction . . . . 2070A.20.2 6.2 Finding the outstand-

    ing loan balance . . . . 2070A.21 Supplementary Notes for the Lec-

    ture of March 5th, 2004 . . . . 2073A.21.1 6.2 Finding the outstand-

    ing loan balance (contin-ued) . . . . . . . . . . . 2073

    A.21.2 6.3 Amortization sched-ules . . . . . . . . . . . 2075

    A.22 Supplementary Notes for the Lec-ture of March 8th, 2004 . . . . 2078

    A.22.1 6.3 Amortization sched-ules (continued) . . . . 2078

    A.22.2 6.4 Sinking funds . . . 2082A.23 Supplementary Notes for the Lec-

    ture of March 12th, 2004 . . . . 2083A.23.1 6.4 Sinking funds (con-

    tinued) . . . . . . . . . 2083A.24 Supplementary Notes for the Lec-

    ture of March 15th, 2004 . . . . 2086A.24.1 6.4 Sinking funds (con-

    cluded) . . . . . . . . . 2086A.24.2

    6.5 Differing payment pe-

    riods and interest conver-sion periods . . . . . . . 2089

    A.24.3 6.6 Varying series of pay-ments . . . . . . . . . . 2089

    A.24.4 6.7 Amortization withcontinuous payments . . 2089

    A.24.5 6.8 Step-rate amounts ofprincipal . . . . . . . . . 2089

    A.25 Supplementary Notes for the Lec-ture of March 17th, 2004 . . . . 2090A.25.1

    7.1 Introduction . . . . 2090

    A.25.2 7.2 Types of securities 2090A.26 Supplementary Notes for the Lec-

    ture of March 19th, 2004 . . . . 2093A.26.1 7.3 Price of a bond . . 2093

    A.27 Supplementary Notes for the Lec-ture of March 22nd, 2004 . . . 2099A.27.1 7.4 Premium and dis-

    count . . . . . . . . . . 2099A.28 Supplementary Notes for the Lec-

    ture of March 24th, 2004 . . . . 2102A.28.1

    7.4 Premium and dis-

    count (concluded) . . . 2102A.29 Supplementary Notes for the Lec-

    ture of March 26th, 2004 . . . . 2104A.29.1 7.5 Valuation between

    coupon payment dates . 2104A.29.2 7.6 Determination of yield

    rates . . . . . . . . . . . 2104A.29.3 7.7 Callable bonds . . . 2104

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    Information for Students in MATH 329 2004 01

    A.30 Supplementary Notes for the Lec-

    ture of March 29th, 2004 . . . . 2105A.30.1 7.7 Callable bonds (con-

    tinued) . . . . . . . . . 2105A.31 Supplementary Notes for the Lec-

    ture of March 31st, 2004 . . . . 2107A.31.1 7.5 Valuation between

    coupon payment dates . 2107A.31.2 7.8 Serial bonds . . . . 2108A.31.3 7.9 Some generalizations 2108A.31.4 7.10 Other securities . 2108A.31.5

    7.11 Valuation of secu-

    rities . . . . . . . . . . . 2109A.32 Supplementary Notes for the Lec-

    ture of April 5th, 2004 . . . . . 2110

    B Problem Assignments and Tests from

    Previous Years 3001

    B.1 2002/2003 . . . . . . . . . . . . 3001B.1.1 First 2002/2003 Problem

    Assignment, with Solu-tions . . . . . . . . . . . 3001

    B.1.2 Second 2002/2003 Prob-

    lem Assignment, with So-lutions . . . . . . . . . . 3007

    B.1.3 Third 2002/2003 Prob-lem Assignment, with So-lutions . . . . . . . . . . 3012

    B.1.4 Fourth 2002/2003 Prob-lem Assignment, with So-lutions . . . . . . . . . . 3019

    B.1.5 Fifth 2002/2003 ProblemAssignment, with Solu-tions . . . . . . . . . . . 3029

    B.1.6 2002/2003 Class Tests, withSolutions . . . . . . . . 3037

    B.1.7 Final Examination, 2002/20033042

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    Information for Students in MATH 329 2004 01 1

    1 General Information

    Distribution Date: 0th version, Monday, January 5th, 2004This version, Monday, January 12th, 2004

    (All information is subject to change, either by announcements at lectures,on WebCT, or in print.)

    An updated version may be placed, from time to time, on the Math/Statwebsite (cf. 1.4.2 below), and will also be accessible via a link from WebCT.)

    The Course Outline for MATH 329 2004 01 can be considered to be pages1 through 8 of these notes.

    1.1 Instructor and Times

    INSTRUCTOR: Prof. W. G. BrownOFFICE: BURN 1224OFFICE HRS. W 13:2014:15 h.;

    (subject to F 1011 h.;change) and by appointment

    TELEPHONE: 3983836E-MAIL: [email protected]

    CLASSROOM: BURN 1B24CLASS HOURS: MWF 14:3515:25 h.

    Table 1: Instructor and Times

    1.2 Course Description

    1.2.1 Calendar Description

    THEORY OF INTEREST. (3 credits) (Prerequisite: MATH 141.) Simple and com-pound interest, annuities certain, amortization schedules, bonds, depreciation.

    1.2.2 Syllabus (in terms of sections of the text-book)

    The central part of the course consists of many of the topics in the first nine chapters ofthe textbook [1]1; section numbers, where shown, refer to that book. In the list below

    1[n] refers to item n in the bibliography, page 901.

    UPDATED TO April 29, 2004

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    Information for Students in MATH 329 2004 01 2

    we show the chapters and appendices of the textbook. Following each is a description

    as of the date of this revision, of the sections to be excluded. This list will be updatedduring the semester, as becomes apparent that certain sections are not appropriate tothe level of the course or the lecture time available.

    Chapter 1. The Measurement of Interest 1.1-1.8. Portions of 1.9, will beomitted. For the present 1.10 will also be omitted.

    Chapter 2. Solution of Problems in Interest In 2.6 You may omit the discussion[1, pp. 45-46] of the method of equated time.

    Chapter 3. Elementary Annuities You may omit [1, 3.6 Nonstandard terms andinterest rates], [1, 3.10 Annuities not involving compound interest, pp. 82-88] and corre-sponding exercises. We will also omit [1, 3.9 Varying interest] for the present (possiblyto return).

    Chapter 4. More General Annuities In the following section we shall consider theproblems strictly on an ad hoc basis: students are not expected to derive nor to applythe identities obtained: [1, 4.2 Annuities payable at a different frequency than interestis convertible; 4.3 Further analysis of annuities payable less frequently than interest isconvertible;

    4.4 Further analysis of annuities payable more frequently than interest is

    convertible]. Omit [1, 4.5 Continuous annuities] for the present. In [1, 4.6 Nonstandardterms and interest rates] we shall consider the derivation of formul for (Ia)n, (Is)n,(Da)n, (Ds)n, and their due and perpetual variants, also the question of annuities ingeometric progression. Omit [1, 4.7 More general varying annuities, 4.8 Continuousvarying annuities, 4.9 Summary of results] for the present, together with their exercises.

    Chapter 5. Yield Rates Omit [1, 5.2 Discounted cash flow analysis], except for thedefinition [1, p. 131] ofyield rate. Omit [1, 5.3 Uniqueness of the yield rate] except youshould read and understand the example [1, p. 133] of a problem where the yield rateis not unique. Omit [1,

    5.5

    5.9] and accompanying exercises; but we will study [1,

    5.4 Reinvestment rates] in preparation for Chapter 6.Chapter 6. Amortization Schedules and Sinking Funds In [1, 6.4] omit pages178-179, where the function an i&j is introduced. Omit [1, 6.5 6.8] and accompanyingexercises.

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    Information for Students in MATH 329 2004 01 3

    Chapter 7. Bond and Other Securities Omit [1,

    7.6 Determination of yield rates],

    [1, 7.8 Serial bonds], [1, 7.9 Some generalizations], [1, 7.10 Other securities], [1, 7.11Valuation of securities].

    Chapter 8. Practical Applications Omit this chapter.

    Chapter 9. More Advanced Financial Analysis Omit this chapter.

    Chapter 10. A Stochastic Approach to Interest Omit this chapter.

    Appendix I. Table of compound interest functions While most calculations willbe done using calculators, these tables may prove useful.

    Appendix II. Table numbering the days of the year

    Appendix III. Basic mathematical review Topics that are beyond the requiredprerequisites will be explained if, as, and when they are used.

    Appendix IV. Statistical background Omit this section: no background in prob-ability is prerequisite to Math 329.

    Appendix V. Iteration methods

    Appendix VI. Further analysis of varying annuities Omit this Appendix, whichis concerned with the formula for Summation by Parts, analogous to integration byparts for functions of a continuous variable.

    Appendix VII. Illustrative mortgage loan amortization schedule

    Appendix VIII. Full immunization Omit this Appendix, which is related to [1,

    9.9], which is not in the syllabus.Appendix IX. Derivation of the variance of an annuity Omit this Appendix.

    Appendix X. Derivation of the Black-Scholes formula Omit this Appendix.

    UPDATED TO April 29, 2004

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    Information for Students in MATH 329 2004 01 4

    1.2.3 Verbal arguments

    An essential feature of investment and insurance mathematics is the need to be able tounderstand and to formulate verbal arguments; that is, explanations of the truth ofan identity presented verbally i.e., a proof in words, rather then an algebraic proof. Ina verbal argument we seek more than mathematically correctness: we wish to see anexplanation that could be presented to a layman who is not competent in the mathe-matical bases of this subject, but is still possessed of reason, and needs to be assuredthat he is not being exploited. This facet of the course will be seen, at first, to be quitedifficult. When the skill has been mastered it can be used to verify the correctness ofstatements proved mathematically. Verbal arguments require some care with the under-lying language; students who have difficulty with expression in English are reminded that

    all students have the right to submit any written materials in either English or French.2

    1.3 Evaluation of Your Progress

    1.3.1 Term Mark

    The Term Mark will be computed one-third from the assignment grades, and two-thirdsfrom the class test. The Term Mark will count for 30 of the 100 marks in the finalgrade, but only if it exceeds 30% of the final examination percentage; otherwise the finalexamination will be used exclusively in the computation of the final grade.

    1.3.2 Assignments.

    A total of about 6 assignments will be worth 10 of the 30 marks assigned to Term Work.

    1.3.3 Class Test

    A class test, will be held on Wednesday, March 10th, 2004, at the regular class time,counting for 20 of the 30 marks in the Term Mark. There will be no make-up testfor persons who miss the test. (This date has been changed from the tentative dateannounced earlier, after discussion with the class at the lecture of Friday, February 13th,

    2004.)2For a lexicon of actuarial terms in English/French, see The Canadian Institute of Actuaries English-

    French lexicon [8], at

    http://www.actuaries.ca/publications/lexicon/

    UPDATED TO April 29, 2004

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    Information for Students in MATH 329 2004 01 6

    No student shall, with intent to deceive, represent the work of another person

    as his or her own in any academic writing, essay, thesis, research report,project or assignment submitted in a course or program of study or representas his or her own an entire essay or work of another, whether the material sorepresented constitutes a part or the entirety of the work submitted.

    You are also referred to the following URL:

    http://www.mcgill.ca/integrity/studentguide/

    1.4 Published Materials

    1.4.1 Required Text-BookThe textbook for the course this semester is [1] Stephen G. Kellison, The Theory ofInterest, Second Edition. Irwin/McGraw-Hill, Boston, etc. (1991), ISBN 0-256-09150-1.

    1.4.2 Website

    These notes, and other materials distributed to students in this course, will be accessibleat the following URL:

    http://www.math.mcgill.ca/brown/math329b.html

    The notes will be in pdf (.pdf) form, and can be read using the Adobe Acrobat reader,which many users have on their computers. This free software may be downloaded fromthe following URL:

    http://www.adobe.com/prodindex/acrobat/readstep.html 4

    Where revisions are made to distributed printed materials for example these informa-tion sheets it is expected that the last version will be posted on the Web.

    The notes will also be available via a link from the WebCT URL:

    http://webct.mcgill.ca

    but not all features of WebCT will be implemented.

    1.4.3 Reference Books

    The textbook used for 2001-2003 may be used as a reference: [5] Michael M. Parmen-tier, Theory of Interest and Life Contingencies, with Pension Applications: A Problem-Solving Approach, 3rd edition. ACTEX Publications, Winstead, Conn. (1999), ISBN0-56698-333-9.

    4At the time of this writing the current version is 5.1.

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    Information for Students in MATH 329 2004 01 7

    1.5 Other information

    1.5.1 Prerequisites

    It is your responsibility as a student to verify that you have the necessary calculusprerequisites. It would be foolish to attempt to take the course without them.

    1.5.2 Calculators

    The use ofnon-programmable, non-graphing calculators only will be permitted in home-work, tests, or the final examination in this course. Students may be required to convinceexaminers and invigilators that all memories have been cleared. The use of calculatorsthat are either graphing or programmable will not be permitted during test or examina-tions, in order to level the playing field.

    1.5.3 Self-Supervision

    This is not a high-school course, and McGill is not a high school. The monitoring ofyour progress before the final examination is largely your own responsibility. While theinstructor is available to help you, he cannot do so unless and until you identify theneed for help. While the significance of the homework assignments and class test in thecomputation of your grade is minimal, these are important learning experiences, andcan assist you in gauging your progress in the course. This is not a course that can

    be crammed for: you must work steadily through the term if you wish to develop thefacilities needed for a strong performance on the final examination.

    Working Problems on Your Own. You are advised to work large numbers of prob-lems from your textbook. The skills you acquire in solving textbook problems could havemuch more influence on your final grade than either the homework or the class test.

    1.5.4 Escape Routes

    At any time, even after the last date for dropping the course, students who are experi-encing medical or personal difficulties should not hesitate to consult their advisors or theStudent Affairs office of their faculty. Dont allow yourself to be overwhelmed by suchproblems; the University has resource persons who may be able to help you.

    1.5.5 Showing your work; good mathematical form; simplifying answers

    When, in a quiz or examination problem, you are explicitly instructed to show all yourwork, failure to do so could result in a substantial loss of marks possibly even allof the marks; this is the default. The guiding principle should be that you want to be

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    Information for Students in MATH 329 2004 01 8

    able to communicate your precise reasoning to others and to yourself. You are always

    expected to simplify any algebraic or numerical expressions that arise in your solutionsor calculations. Verbal proofs are expected to be convincing: it will not be sufficientto simply describe mathematical expressions verbally.

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    Information for Students in MATH 329 2004 01 9

    2 Timetable

    Distribution Date: 0th version: Monday, January 5th, 2004This version, Monday, January 12th, 2004

    (Subject to correction and change.)Section numbers refer to the text-book. 5

    MONDAY WEDNESDAY FRIDAY

    JANUARY05 1.1 1.8 07 1.1 1.8 09 1.1 1.812 Problems from 1.1

    1.8

    14 1.9-1.10, prob-lems

    16 Chapter 2

    Course changes must be completed on MINERVA by Jan. 18, 200419 Chapter 2 21 Chapter 2 23 3.1, 3.2

    Deadline for withdrawal with fee refund = Jan. 25, 200426 3.2, 3.3 1 28 3.3 30 3.4

    FEBRUARYVerification Period: February 26, 2004

    02 3.4 04 3.5 06 3.7Deadline for withdrawal (with W) from course via MINERVA = Feb. 15, 2004

    09 3.7, 3.8 2 11 3.8 13 4.1-4.4

    164.3, 4.4

    184.6

    204.6, 5.4

    Study Break: February 2327, 2004No lectures, no regular office hours

    23 NO LECTURE 25 NO LECTURE 27 NO LECTURE

    (Page 10 of the timetable will not be circulated; however, a version is available in theonline version of these notes.)

    5

    Notation: n = Assignment #n due todayR = Read OnlyX = reserved for eXpansion or review

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    Information for Students in MATH 329 2004 01 10

    MONDAY WEDNESDAY FRIDAY

    MARCH01 5.3, 5.4 3 03 Chapter 6 05 Chapter 608 Chapter 6 10 CLASS TEST 12 Chapter15 Chapter 17 Chapter 4 19 Chapter22 Chapter 24 Chapter 26 Chapter29 Chapter 31 Chapter 5

    APRIL02 Chapter

    05 Chapter 07 09 NO LECTURE12 NO LECTURE 13 (TUESDAY)

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    Information for Students in MATH 329 2004 01 11

    3 First Problem Assignment

    Distribution Date: Thursday, January 15th, 2005 (mounted on the Web)Hard copy distributed on Monday, January 19th, 2004

    Solutions are to be submitted by Monday, January 26th, 2004

    (This is a short assignment. Subsequent assignments can be expected to belonger.)

    1. It is known that the accumulation function a(t) is of the form b (1.1)t + ct2, whereb and c are constants to be determined.

    (a) If $100 invested at time t = 0 accumulates to $170 at time t = 3, find theaccumulated value at time t = 12 of $100 invested at time t = 1.

    (b) Show that this function satisfies the requirement [1, p. 2, #2] that it be non-decreasing.

    (c) Determine a general formula for in, and show that limn

    in = 10%. (Use

    LHopitals Rule.)

    2. It is known that 1000 invested for 4 years will earn 250.61 in interest, i.e., that thevalue of the fund after 4 years will be 1250.61. Determine the accumulated valueof 3500 invested at the same rate of compound interest for 13 years.

    3. It is known that an investment of 750 will increase to 2097.75 at the end of 25years. Find the sum of the present values of payments of 5000 each which willoccur at the ends of 10, 15, and 25 years.

    4. Find the accumulated value of 1000 at the end of 10 years:

    (a) if the nominal annual rate of interest is 6% convertible monthly;

    (b) if the nominal annual rate of discount is 5% convertible every 2 years.

    5. Given that i(m) = 5

    66

    2 and d(m) = 2

    8

    0.06, find m, the equivalent annual

    compound interest rate, and the equivalent annual compound discount rate.

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    Information for Students in MATH 329 2004 01 12

    4 Second Problem Assignment

    Distribution Date: Mounted on the Web on Sunday, January 18th, 2004Distributed in hard copy on Wednesday, January 28th, 2004

    Solutions are to be submitted by Monday, February 9th, 2004

    1. Find the present value of 1000, to be paid at the end of 37 months under each ofthe following scenarios:

    (a) Assume compound interest throughout, and a (nominal) rate of discount of6% payable quarterly.

    (b) Assume compound interest for whole years only at a (nominal) rate of discountof 6% payable quarterly, and simple discount at the rate of 1.5% per 3 monthsduring the final fractional period.

    (c) Assume compound interest throughout, and a nominal rate of interest of 8%payable semi-annually.

    2. The sum of 5,000 is invested for the months of April, May, and June at 7% simpleinterest. Find the amount of interest earned

    (a) assuming exact simple interest in a non-leap year

    (b) assuming exact simple interest in a leap year (with 366 days);

    (c) assuming ordinary simple interest;

    (d) assuming the Bankers Rule.

    3. Find how long 4,000 should be left to accumulate at 5% effective in order that itwill amount to 1.25 times the accumulated value of another 4,000 deposited at thesame time at a nominal interest rate of 4% compounded quarterly.

    4. The present value of two payments of 100 each, to be made at the end of n yearsand 2n years is 63.57. If i = 6.25%, find n.

    5. (a) Find the nominal rate of interest convertible quarterly at which the accumu-lated value of 1000 at the end of 12 years is 3000.

    (b) Find the nominal rate of discount convertible semi-annually at which a pay-ment of 3000 12 years from now is presently worth 1000.

    (c) Find the effective annual rate of interest at which the accumulated value of1000 at the end of 12 years is 3000.

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    6. An investor deposits 20,000 in a bank. During the first 4 years the bank credits an

    annual effective rate of interest of i. During the next 4 years the bank credits anannual effective rate of interest ofi 0.02. At the end of 8 years the balance in theaccount is 22,081.10. What would the account balance have been at the end of 10years if the annual effective rate of interest were i + 0.01 for each of the 10 years?

    7. A bill for 1000 is purchased for 950 4 months before it is due. Find

    (a) the nominal rate of discount convertible monthly earned by the purchaser;

    (b) the annual effective rate of interest earned by the purchaser.

    8. A signs a 2-year note for 4000, and receives 3168.40 from the bank. At the end

    of 6 months, a year, and 18 months A makes a payment of 1000. If interest iscompounded semi-annually, what is the amount outstanding on the note at thetime if falls due?

    9. The Intermediate Value Theorem for continuous functions tells us that such afunction f(x) whose value at x = a has the opposite sign from its value at x = bwill assume the value 0 somewhere between a and b. By computing the value off at the point 1

    2(a + b), we can infer that there is a 0 of f in an interval half as

    long as [a, b], and this procedure may be repeated indefinitely to determine a zeroof f to any desired accuracy. Assuming that polynomials are continuous, use this

    idea to determine the nominal quarterly compound interest rate under which thefollowing payments will accumulate to 1000 at the end of 4 years:

    300 today

    200 at the end of 1 year

    300 at the end of 2 yearsYour answer should be accurate to 3 decimal places, i.e., expressed as a percentageto 1 decimal place.

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    5 Solutions, First Problem Assignment

    Distribution Date: Mounted on the Web on Wednesday, 4 February, 2004Assignment was mounted on the Web on Thursday, January 15th, 2005

    Hard copy was distributed on Monday, January 19th, 2004Solutions were to be submitted by Monday, January 26th, 2004

    (This is a short assignment. Subsequent assignments can be expected to belonger.)

    1. It is known that the accumulation function a(t) is of the form b (1.1)t + ct2, whereb and c are constants to be determined.

    (a) If $100 invested at time t = 0 accumulates to $170 at time t = 3, find theaccumulated value at time t = 12 of $100 invested at time t = 1.

    (b) Show that this function satisfies the requirement [1, p. 2, #2] that it be non-decreasing.

    (c) Determine a general formula for in, and show that limn

    in = 10%. (Use

    LHopitals Rule.)

    Solution: [1, Exercise 4, p. 30] Denote the corresponding amount function by A(t).

    (a) An accumulation function must have the property that a(0) = 1; this impliesthat 1 = a(0) = b + 0, so b = 1.

    The given data imply that

    170 = 100(a(3)) = 100(1(1.331) + c 32) (1)

    which implies that c = 0.041. We conclude that

    A(t)

    A(0)= a(t) = (1.1)t + 0.041t2 , (2)

    implying that a(1) = 1.141, a(12) = 9.042428377. Then

    A(12) = A(1) A(12)A(1)

    = A(1) a(12)a(1)

    = A(1) 9.0424283771.141

    = A(1)(7.925002959) = 792.5002959

    so $100 at time t = 1 grows to $792.50 at time t = 12.

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    (b) It follows from (2) that a(t) = (1.1)t ln 1.1 + 0.082t, which is positive for

    positive t; thus a(t) is an increasing function of t for positive t.(This property may also be proved from first principles. Let t1 t2. Then

    a(t2) a(t1) = (1.1)t2 + 0.041t22 (1.1)t1 + 0.041t21= (1.1)t1

    (1.1)t2t1 1+ (0.041)(t2 t1) (t2 + t1)

    where both of the summands are non-negative for 0 t1 t1.(c)

    in =a(n) a(n 1)

    a(n 1)=

    (0.1)(1.1)n1 + (0.041)(2n 1)

    (1.1)n

    1

    (0.041)(n 1)2

    =0.1 + 0.041

    2n1

    (1.1)n1

    1 0.041

    (n1)2

    (1.1)n1

    By LHopitals Rule

    limx

    2x 1(1.1)x1

    = limx

    2

    (1.1)x1 ln 1.1= 0

    limx

    (n 1)2(1.1)n1

    = limx

    2(x 1)(1.1)x1 ln 1.1

    = limx

    2(1.1)x1(ln1.1)2

    = 0

    Hence

    limn

    in =0.1 + 0.041 (0)

    1 0.041 (0) = 0.1 = 10%.

    2. It is known that 1000 invested for 4 years will earn 250.61 in interest, i.e., that thevalue of the fund after 4 years will be 1250.61. Determine the accumulated valueof 3500 invested at the same rate of compound interest for 13 years.

    Solution: [1, Exercise 14, p. 30] Let i be the rate of compound interest. Then

    1000(1 + i)4 = 1250.61. The accumulated value of 3500 after 13 years will be

    3500(1 + i)13 = 3500

    1250.61

    1000

    134

    = 7239.57 .

    3. It is known that an investment of 750 will increase to 2097.75 at the end of 25years. Find the sum of the present values of payments of 5000 each which willoccur at the ends of 10, 15, and 25 years.

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    Solution: [1, Exercise 21, p. 31] Let i be the interest rate. The known fact is that

    750(1+ i)25 = 2097.75. Hence (1+ i)25 = 2.797 , so v25 = 0.357535924. The presentvalue of three payments of 5000 after 10, 15, and 25 years will, therefore, be

    5000(v10 + v15 + v25)

    = 5000

    (0.357535924)1025 + (0.357535924)

    1525 + (0.357535924)

    2525

    = 5000(0.662709221 + 0.539500449 + 0.357535924)

    = 7798.73.

    4. Find the accumulated value of 1000 at the end of 10 years:

    (a) if the nominal annual rate of interest is 6% convertible monthly;(b) if the nominal annual rate of discount is 5% convertible every 2 years.

    Solution: [1, Exercise 32, p. 31]

    (a) The accumulation factor for each month is 1 + 6%12

    = 1.005. After 10 years1000 grows to

    1000(1.005)1012 = 1819.40.

    (b) The discount factor for each 2 years is 125% = 0.09 (moving backwards),corresponding to an accumulation factor of 1

    0.9. After 10 years 1000 grows to

    1000(0.09)102 = 1693.51 .

    5. Given that i(m) = 5

    66

    2 and d(m) = 2 80.06, find m, the equivalent annualcompound interest rate, and the equivalent annual compound discount rate.

    Solution: [1, Exercise 30, p. 32] For an mth of a year the relationship between i(m)

    and d(m) is given by 1 +

    i(m)

    m

    1 d

    (m)

    m

    = 1

    which is equivalent to

    (m + i(m))(m d(m)) = m2or

    m =i(m)d(m)

    i(m) d(m) .Substituting the given values

    i(m) = 0.041241452

    d(m) = 0.040408205

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    gives m = 2. It follows that

    i =

    1 +

    i(2)

    2

    2 1

    = (1.020620726)2 1 = 4.1666667% = 416

    % =1

    24

    d = 1

    1 d(2)

    2

    2

    = 1 0.96 = 4% = 125

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    6. Give an algebraic proof and a verbal explanation for the formula

    m |an = a am vm+na .

    7. A level perpetuity-immediate is to be shared by A, B, C, and D. A receives thefirst n payments, B the next 2n payments, C payments ##3n + 1, . . . 5n, and Dthe payments thereafter. It is known that the present values of Bs and Ds sharesare equal. Find the ratio of the present value of the shares of A, B, C, D.

    8. (a) Find an the present value of an annuity which pays 4,000 at the beginning ofeach 3-month period for 12 years, assuming an effective rate of 2% interestper 4-month period.

    (b) Suppose that the owner of the annuity wishes to pay now so that paymentsunder his annuity will continue for an additional 10 years. How much shouldhe pay?

    (c) How much should he pay now to extend the annuity from the present 12 yearsto a perpetuity?

    (It is intended that you solve this problem from first principles, not by substitu-tion into formul in [1, Chapter 4].)

    9. (No time diagram is needed for the solution to this problem.) In Problem 9 ofAssignment 2 you were asked to apply the Bisection Method to determine the

    solution to an interest problem to 3 decimal places. The equation in question was:

    300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 .

    and the solution given began with the values of

    f(x) = 3x4 + 2x3 + 3x2 10 .at x = 0 (f(0) = 10), x = 2 (f(2) = 66 > 0), and x = 2 (f(2) = 34 >0), and we were interested in the solution between 0 and 2 a solution thatis unique because f is positive in this interval. Apply Linear Interpolation 4times in an attempt to determine the solution we seek. (You are not expectedto know the general theory of error estimation.) The intention is that you applylinear interpolation unintelligently, using it to determine a point where you find thefunction value and thereby confine the zero to a smaller subinterval: the point thatyou find will replace the midpoint in the bisection method. In some situations, asin the present one, the procedure may not be better than the bisection method.Indeed, in the present example, it could take many more applications than thebisection method to obtain the accuracy you obtained with that method.

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    7 Solutions, Second Problem Assignment

    Distribution Date: Mounted on the Web on Friday, February 20th, 2004Assignment was mounted on the Web on January 19th, 2004.Hard copy was distributed on Wednesday, January 28th, 2004

    Solutions were due by Monday, February 9th, 2004(Solutions presented subject to correction of errors and omissions.)

    1. Find the present value of 1000, to be paid at the end of 37 months under each ofthe following scenarios:

    (a) Assume compound interest throughout, and a (nominal) rate of discount of

    6% payable quarterly.(b) Assume compound interest for whole years only at a (nominal) rate of discount

    of 6% payable quarterly, and simple discount at the rate of 1.5% per 3 monthsduring the final fractional period.

    (c) Assume compound interest throughout, and a nominal rate of interest of 8%payable semi-annually.

    Solution: (cf. [1, Exercise 2, p. 53])

    (a) Present value = 10001 0.06

    4

    373

    = 829.94.

    (b) Present value = 1000

    1 0.06

    4

    363

    1 0.0153

    = 10000.8728230.9995 =

    829.96.

    (c) Present value = (1.04)376 = 1000 0.785165257 = 785.17.

    2. The sum of 5,000 is invested for the months of April, May, and June at 7% simpleinterest. Find the amount of interest earned

    (a) assuming exact simple interest in a non-leap year

    (b) assuming exact simple interest in a leap year (with 366 days);

    (c) assuming ordinary simple interest;

    (d) assuming the Bankers Rule.

    Solution: (cf. [1, Exercise 6, p. 54])

    (a) The number of days is 30 + 31 + 30 = 91; exact simple interest is 91365

    5000 (0.07) = 87.26.

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    (b) Exact simple interest is 91366

    5000

    (0.07) = 87.02

    (c) Ordinary simple interest is 30+30+30360 5000 (0.07) = 87.50(d) Interest under the Bankers Rule is 91

    360 5000 (0.07) = 88.47.

    3. Find how long 4,000 should be left to accumulate at 5% effective in order that itwill amount to 1.25 times the accumulated value of another 4,000 deposited at thesame time at a nominal interest rate of 4% compounded quarterly.

    Solution: (cf. [1, Exercise 13, p. 55]) The equation of value at n years is

    4000(1.05)n = (1.25)(4000)(1.01)4n

    so

    n = ln 1.25ln105 4ln101 = 24.82450822 .

    4. The present value of two payments of 100 each, to be made at the end of n yearsand 2n years is 63.57. If i = 6.25%, find n.

    Solution: (cf. [1, Exercise 14, p. 55]) Solving the equation of value, 100v2n+100vn =63.57, we obtain

    vn =1 3.5428

    2,

    in which only the + sign is acceptable, since vn > 0. Taking logarithms gives

    n =0.818446587

    ln 1.0625 = 13.50023411.

    We conclude, to the precision of the problem, that n = 13.5 years.

    5. (a) Find the nominal rate of interest convertible quarterly at which the accumu-lated value of 1000 at the end of 12 years is 3000.

    (b) Find the nominal rate of discount convertible semi-annually at which a pay-ment of 3000 12 years from now is presently worth 1000.

    (c) Find the effective annual rate of interest at which the accumulated value of1000 at the end of 12 years is 3000.

    Solution:(a) (cf. [1, Exercise 19, p. 55]) The equation of value at time t = 12 is

    1000

    1 +

    i(4)

    4

    412= 3000 ,

    implying that

    i(4) = 4

    3148 1

    = 9.260676%.

    UPDATED TO April 29, 2004

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    (b) The equation of value at time t = 12 is

    3000

    1 d

    (2)

    2

    212= 1000 ,

    implying that

    d(2) = 2

    1 3 124

    = 9.3678763%.

    (c) The equation of value at time t = 12 is

    1000 (1 + i)12 = 3000 ,

    implying thati = 3

    112 1 = 9.5872691%.

    6. An investor deposits 20,000 in a bank. During the first 4 years the bank credits anannual effective rate of interest of i. During the next 4 years the bank credits anannual effective rate of interest ofi 0.02. At the end of 8 years the balance in theaccount is 22,081.10. What would the account balance have been at the end of 10years if the annual effective rate of interest were i + 0.01 for each of the 10 years?

    Solution: (cf. [1, Exercise 32, p. 57]) The equation of value is

    20000(1 + i)4

    (1 + (i 0.02))4

    = 22081.10 ,

    which we interpret as a polynomial equation. The equation is of degree 8, and wedont have a simple algebraic method for solving such equations in general. Butthis equation has the left side a pure 4th power, so we can extract the 4th roots ofboth sides, obtaining

    (1 + i)(1 + (i 0.02)) = (1.104055)14 = 1.025056201 ,

    which may be expressed as a quadratic equation in 1 + i:

    (1 + i)2

    0.02(1 + i) 1.025056 = 0whose only positive solution is

    1 + i =0.02 +

    (0.02)2 + 4(1.025056)

    2= 1.0225

    from which we conclude that i = 2.25%, and that the account balance after 10years would be 20000(1.0225 + 0.01)10 = 20000(1.0325)10 = 27, 737.89.

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    7. A bill for 1000 is purchased for 950 4 months before it is due. Find

    (a) the nominal rate of discount convertible monthly earned by the purchaser;

    (b) the annual effective rate of interest earned by the purchaser.

    Solution: (cf. [1, Exercise 25, p. 56])

    (a) Ifd(12) be the nominal discount rate, then

    1000

    1 d

    (12)

    12

    4= 950

    implying that

    1 d(12)

    12= 0.9872585

    so d(12) = 15.29%.

    (b) Let i be the effective annual interest rate. Then

    950(1 + i)13 = 1000

    implies that

    i =1000

    9503 1 = 16.635% .

    8. A signs a 2-year note for 4000, and receives 3168.40 from the bank. At the endof 6 months, a year, and 18 months A makes a payment of 1000. If interest iscompounded semi-annually, what is the amount outstanding on the note at thetime if falls due?

    Solution: If i be the rate of interest charged semi-annually, then

    3168.40(1 + i)4 = 4000

    so i = 6.00%; that is i(2) = 12.00%. The value of the 3 payments at the time thenote matures is

    1000

    (1.06)3 + (1.06)2 + (1.06)1

    = 3374.62

    so the amount outstanding before the final payment is 625.38.

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    9. The Intermediate Value Theorem for continuous functions tells us that such a

    function f(x) whose value at x = a has the opposite sign from its value at x = bwill assume the value 0 somewhere between a and b. By computing the value off at the point 1

    2(a + b), we can infer that there is a 0 of f in an interval half as

    long as [a, b], and this procedure may be repeated indefinitely to determine a zeroof f to any desired accuracy. Assuming that polynomials are continuous, use thisidea to determine the nominal quarterly compound interest rate under which thefollowing payments will accumulate to 1000 at the end of 4 years:

    300 today

    200 at the end of 1 year 300 at the end of 2 years

    Your answer should be accurate to 3 decimal places, i.e., expressed as a percentageto 1 decimal place.

    Solution: (We will carry the computations to an accuracy greater than requestedin the problem.)

    (a) Let the effective annual interest rate be i. The equation of value at the endof 4 years is

    300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 . (3)

    (b) We need a continuous function to which to apply the Intermediate ValueTheorem. Some choices may be better than others. We will choose

    f(x) = 3x4 + 2x3 + 3x2 10 .We observe that f(0) = 10, that f(2) = 48+16+1210 = 66 > 0, and thatf(2) = 48 16 + 12 10 = 34 > 0. This tells us that there is a solution toequation (3) for 2 x 0, equivalently for 3 i 1: such a solution isof no interest to us, as it does not fit the constraints of this problem. But thefunction is a cubic polynomial, and has 2 other zeros. We see that it also hasa solution in the interval 0 x 2, and we proceed to progressively halveintervals.

    (c) The midpoint of interval [0, 2] is 1;

    f(1) = 3 + 2 + 3 10 = 2 < 0 < 66 = f(2) ,so there must be a root in the interval [1 , 2].

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    (d) The midpoint of [1, 2] is 1.5;

    f(1.5) = 3(1.5)4 + 2(1.5)3 + 3(1.5)2 10= 15.1875 + 6.75 + 6.75 10 = 18.6875> 0

    so there must be a zero in the interval [1, 1.5], whose midpoint is 1.25.

    (e)

    f(1.25) = 3(1.25)4 + 2(1.25)3 + 3(1.25)2 10= 7.3242 + 3.9063 + 4.6875

    10 = 5.918

    > 0

    so there must be a zero in the interval [1, 1.25], whose midpoint is 1.125.

    (f)

    f(1.125) = 3(1.125)4 + 2(1.125)3 + 3(1.125)2 10= 4.8054 + 2.8477 + 3.7969 10 = 1.45> 0

    so there must be a zero in the interval [1, 1.125], whose midpoint is 1.0625.

    (g)

    f(1.0625) = 3(1.0625)4 + 2(1.0625)3 + 3(1.0625)2 10= 3.8233 + 2.3989 + 3.3867 10 = 0.3911< 0

    so there must be a zero in the interval [1.0625, 1.125], whose midpoint is1.09375.

    (h)

    f(1.09375) = 3(1.09375)4 + 2(1.09375)3 + 3(1.09375)2 10= 4.2933 + 2.6169 + 3.5889 10 = 0.4991> 0

    so there must be a zero in the interval [1.0625, 1.09375], whose midpoint is1.078125.

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    (i)

    f(1.078125) = 3(1.078125)4 + 2(1.078125)3 + 3(1.078125)2 10= 4.053197 + 2.506325 + 3.487061 10 = 0.046582> 0

    so there must be a zero in the interval [1 .0625, 1.078125], whose midpoint is1.0703125.

    (j)

    f(1.0703125) = 3(1.0703125)4 + 2(1.0703125)3 + 3(1.0703125)2

    10

    = 3.936984 + 2.452233 + 3.436707 10 = 0.174076< 0

    so there must be a zero in the interval [1.0703125, 1.078125], whose midpointis 1.07421875.

    (k) f(1.07421875) = .064207958 < 0 so there must be a zero in the interval[1.07421875, 1.078125], whose midpoint is 1.076172.

    (l) f(1.076172) = .008925 < 0 so there must be a zero in the interval[1.076172, 1.078125], whose midpoint is 1.077149.

    (m) f(1.077149) = .018814 > 0 so there must be a zero in the interval[1.076172, 1.077149], whose midpoint is 1.076661.

    (n) f(1.076661) = .004952 > 0 so there must be a zero in the interval[1.076172, 1.076661], whose midpoint is 1.076417.

    (o) f(1.076417) = .001974 < 0 so there must be a zero in the interval[1.076417, 1.076661], whose midpoint is 1.076539.

    (p) f(1.076539) = .001488 > 0 so there must be a zero in the interval[1.076417, 1.076539], whose midpoint is 1.076478.

    (q) f(1.076478) =

    .000243 < 0 so there must be a zero in the interval

    [1.076478, 1.076539], whose midpoint is 1.076509.(r) f(1.076509) = .000637 > 0 so there must be a zero in the interval

    [1.076478, 1.076509], whose midpoint is 1.076494.

    (s) f(1.076494) = .000211 > 0 so there must be a zero in the interval[1.076478, 1.076494], whose midpoint is 1.076486.

    (t) f(1.076486) = .000016 > 0 so there must be a zero in the interval[1.076486, 1.076494], whose midpoint is 1.07649.

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    (u) f(1.07649) = .000097 > 0 so there must be a zero in the interval

    [1.076486, 1.07649], whose midpoint is 1.076488.(v) f(1.076488) = .000041 > 0 so there must be a zero in the interval

    [1.076486, 1.076488], whose midpoint is 1.076487.

    (w) f(1.076487) = .000012. One zero will be approximately x = 1.07649. Thusthe effective annual interest rate is approximately 7.649%. This, however, isnot what the problem asked for. The accumulation function for 3-months willthen be (1.0749)

    14 = 1.01822, so the effective interest rate for a 3-month period

    will be 1.822%, and the nominal annual interest rate, compounded quarterly,will be 7.288, or 7.3% to the accuracy requested.

    THE FOLLOWING PROBLEM WAS CONSIDERED FOR INCLUSION IN THE AS-SIGNMENT, BUT WAS (FORTUNATELY) NOT INCLUDED.

    10. [1, Exercise 6, p. 88]

    (a) Show thatamn = am vmsn = (1 + i)nam sn

    where 0 < n < m.

    (b) Show thatsmn = sm (1 + i)man = vnsm an

    where 0 < n < m.(c) Interpret the results in (a) and (b) verbally.

    Solution:

    (a) We prove the first of these identities by technical substitutions in sums, anal-ogous to changes of variables in a definite integral. For the second identity wegive a less formal proof.

    amn =mn

    r=1vr

    =mr=1

    vr m

    r=mn+1vr

    =mr=1

    vr vmm

    r=mn+1vrm

    =mr=1

    vr vm1

    s=n

    v1s

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    under the change of variable s = m

    r + 1

    =mr=1

    vr vm ns=1

    v1s

    reversing the order of the 2nd summation

    =mr=1

    vr vmn

    s=1

    (1 + i)s1

    = am vm snamn = v + v

    2 + . . . + vmn

    = vn+1 + vn+2 + . . . + v0 + v1 + v2 + . . . + vmn

    vn+1 + vn+2 + . . . + v0= vn

    v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

    (1 + i)n1 + (1 + i)n2 + . . . + (1 + i)v0

    = (1 + i)n

    v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

    (1 + i)0 + (1 + i)1 + . . . + (1 + i)vn1= (1 + i)nam sn

    (b) These identities could be proved in similar ways to those used above. Instead,we shall show that these identities can be obtained from the preceding simplyby multiplying the equations by (1 + i)mn:

    smn = (1 + i)mnamn [1, (3.5), p. 60]

    = (1 + i)mn ((1 + i)nam sn)= (1 + i)mam (1 + i)m (1 + i)nsn= sm (1 + i)man

    smn = (1 + i)mnamn

    = (1 + i)mn (am vmsn)= (1 + i)n (1 + i)mam (1 + i)nsn= (1 + i)nsm

    an

    (c) i. An (m n)-payment annuity-immediate of 1 has the same present valueas an annuity for a total term of m = (m n) + n years minus a correc-tion paid today equal to the value of the deferred n payments. Those npayments are worth sm at time t = m, which amount can be discountedto the present by multiplying by vm.The preceding explanation was based on values at the commencement ofthe first year of an m-year annuity-immediate. Let us now interpret the

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    m

    n payments as being the last payments of an m-year annuity whose

    mth payment has just been made. That m-payment annuity was wortham, n years ago a year before its first payment; today it is worth(1 + i)nan, including the payments we attached at the beginning. Thosepayments are worth sn today, for a net value as claimed.

    ii. Consider an value of the first m n payments of an m-payment annuity-immediate of 1, just after the (m n)th payment. Since these could beconsidered simply the accumulated value of an (m n)-payment annuity,they are worth smn. But the payments of the m-payment annuity notyet made are worth an, and the entire annuity is worth sm at termination,hence vnsm today; hence v

    msm an is also the value today. This givesthe equality between the extreme members of the alleged inequality.Now lets evaluate the same mn payments, but this time consider themto be the last m n payments of an m-payment annuity-certain; again,the m nth payment has just been made. From first principles, theaccumulated value of the payments actually received is smn , and we areviewing them from the context of an annuity-certain of m payments thatwould be worth sm today: lets determine the amount that would haveto be paid out today to correct for that expanded annuity. The value ofthe n payments we have tacked on in the past was an one year beforethe payments began, and (1 + i)msn today; so we can also view the value

    today as sm (1 + i)m

    an .

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    8 Solutions, Third Problem Assignment

    Distribution Date: Mounted on the Web on Wednesday, March 3rd, 2004.Assignment was mounted on the Web on February 8th, 2004,

    hard copy of assignment was distributed on Wednesday, February 11th, 2004.Solutions were to be submitted by 9 a.m., Monday, March 3rd, 2004

    SUBJECT TO CORRECTION OF TYPOS AND OTHER ERRORS

    Sketch a time diagram to accompany your solution of all problems except the last.

    1. A skier wishes to accumulate 30,000 in a chalet purchase fund by the end of 8years. If she deposits 200 into the fund at the end of each month for the first 4

    years, and 200 + X at the end of each month for the next 4 years, find X if thefund earns a nominal (annual) rate of 6% compounded monthly.

    Solution: The equation of value at the end of 8 12 = 96 months is200s96 + X s48 = 30000 ,

    which we may solve to yield

    X =30000

    s48 0.005 200 s96 0.005

    s48 0.005

    =30000

    s48 0.005 200(1.005)96

    1

    (1.005)48 1= 30000s1

    48 0.005 200((1.005)48 + 1)

    = 30000(0.018485) 200(2.27049) from the tables= 100.452.

    2. A fund of 2500 is to be accumulated by n annual payments of 50, followed by n + 1annual payments of 75, plus a smaller final payment of not more than 75 made 1year after the last regular payment. If the effective annual rate of interest is 5%,find n and the amount of the final irregular payment.

    Solution: We shall interpret the payments to be made under two annuities-due: thefirst, for 2n + 1 years, consists of an annual deposit of 50 in advance; the second,for n + 1 years, deferred n years after the first, consists of an annual deposit of 25in advance. It is at the end of year 2n + 1 that the final drop payment is tobe made, and it is to be under 75. (Note that this is the type of problem where thedrop payment could turn out to be negative.) We seek the smallest n for which

    50 s2n+1 + 25 sn+1 > 2500 75 = 2425

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    2(1.05)((1.05)n)2 + (1.05)n

    3 +0.05(2425)

    25(1.05)

    ((1.05)n)2 + 12.1

    (1.05)n 12.1

    3 +

    0.05(2425)

    25(1.05)

    (1.05)n +1

    4.2

    2>

    1

    (4.2)2+

    1

    2.1

    3 +

    0.05(2425)

    25(1.05)

    (1.05)n + 14.2

    >

    1

    (4.2)2+

    1

    2.1

    3 +

    0.05(2425)

    25(1.05)

    12

    = 1.919585178

    n > ln 1.919585178ln 1.05

    = 10.65133267

    Thus the drop payment will be when t = 11 + 12, i.e., 23 years after the firstpayment under the annuity with payments of 50. Just before the drop paymentthe accumulated value of all previous payments is

    50s23 + 25s12 =1.05

    0.05

    50 (1.05)23 + 25 (1.05)12 75 = 2592.924516

    so the drop payment at time t = 23 is 2500 2592.924516 = 92.92.Thus we have an example here of a negative drop payment. Could this meanthat we should have taken n = 10? No. In that case we would find that the

    final payment would be larger than the permitted 75. (If tables like those in thetextbook were available, one could determine the value ofn by inspecting the valueof 2s2n+1 + sn+1 . We observe from the 5% tables the following values:

    n 2s2n+1 + sn10 84.016511 97.067812 111.3713

    We seek the smallest n such that

    50s2n+1 + 25sn > 2500 75i.e., such that

    2s2n+1 + sn > 97 ,

    equivalently,

    2s2n+1 + sn >97

    1.05= 92.38 ,

    and so can conclude that n = 11.)

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    3. On his 30th birthday, a teacher begins to accumulate a fund for early retirement

    by depositing 5,000 on that day and at the beginnings of the next 24 years aswell. Since he expects that his official pension will begin at age 65, he plans that,starting at age 55 he will make an annual level withdrawal at the beginning of eachof 10 years. Assuming that all payments are certain to be made, find the amountof these annual withdrawals, if the effective rate of interest is 6% during the first25 years, and 7% thereafter.

    Solution: Let the constant amount of the withdrawals beginning at age 55 be X.The equation of value at age 55, just before the first withdrawal, is

    5000 s256% = X a107% Annual Withdrawal X = 5000 s256%

    a107%

    = 5000 1.061.07

    0.070.06

    (1.06)25 1

    1 (1.07)10= 33477.74

    4. At an effective annual interest rate of i it is known that

    (a) The present value of 5 at the end of each year for 2n years, plus an additional3 at the end of each of the first n years, is 64.6720.

    (b) The present value of an n-year deferred annuity-immediate paying 10 per yearfor n years is 34.2642.

    Find i.

    Solution: It is convenient to distinguish two cases.

    Case i = 0: From (4a) we have an equation of value

    64.6720 = 5 a2n + 3 an ; (4)

    from (4b) we have the equation of value34.2642 = vn 10 an = 10

    a2n an

    . (5)

    Solving these equations, we obtain

    a2n = 9.3689 (6)

    an = 5.9425 , (7)

    UPDATED TO April 29, 2004

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    implying that

    1 v2n1 vn =

    9.3689

    5.9425 1 + vn = 1.5766 vn = 0.5766.

    We can substitute in equation (7) to obtain

    i = 0.07125 = 7.125%

    which implies that

    n = ln 0.5766ln 1.07125

    = 8.000 years.

    Case i = 0: Here Equations (4) and (5) become

    64.6720 = 5(2n) + 3n = 13n

    34.2642 = 10n = 10(2n n)

    which are inconsistent. Thus this case is impossible.

    5. (a) Find a12 if the effective rate of discount is 5%.(b) Charles has inherited an annuity-due on which there remain 12 payments of

    10,000 per year at an effective discount rate of 5%; the first payment is dueimmediately. He wishes to convert this to a 25-year annuity-immediate at thesame effective rates of interest or discount, with first payment due one yearfrom now. What will be the size of the payments under the new annuity?

    Solution:

    (a) Since (1 d)(1 + i) = 1, v = 1 d = 0.95 when d = 0.05.

    a12 = 1 (0.95)12

    0.05= 9.19279825.

    (b) i = d(1 + i) = dv

    = 0.050.95

    = 119

    . Let X be the size of the new payments. Wemust solve the equation of value,

    91927.9825 = X a25i

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    where v = 0.95. Hence

    X =91927.9825

    a25i

    =91927.9825i

    1 v25=

    91927.9825

    (1 (0.95)25) 19= 6695.606220

    So the new annuity-immediate will pay 25 annual payments of 6695.61, be-ginning one year from now.

    6. Give an algebraic proof and a verbal explanation for the formula

    m |an = a am vm+na .

    Solution:

    (a)

    a am vm+na = 1i

    1 vm

    i vm+n 1

    i

    = 1 (1 vm) vm+ni

    =vm (1 vn)

    i= vm 1 v

    n

    i= vm an = m |an

    (b) a i is the present value of a perpetuity at rate i of 1 per year, paymentsstarting a year from now. am is the present value of the first m payments ofthat perpetuity; if we subtract this we have the present value of a perpetuity-immediate that starts m years from now, i.e., where the first payment is m+1years from now. vm+na is the value of a perpetuity-immediate of 1 startingm + n years from now, i.e., where the first payment is m + n + 1 years fromnow; if we subtract this term as well, we are left with the present value ofpayments at the ends of years m + 1, m + 2, . . ., m + n, i.e., with the presentvalue of an n-payment annuity-certain of 1, deferred m years, i.e., of m |an

    7. A level perpetuity-immediate is to be shared by A, B, C, and D. A receives thefirst n payments, B the next 2n payments, C payments ##3n + 1, . . . , 5n, and Dthe payments thereafter. It is known that the present values of Bs and Ds sharesare equal. Find the ratio of the present value of the shares of A, B, C, D.

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    Solution: The present values of the shares of A, B, C, D are, respectively, an,

    vna2n = a3n an, v3na2n = a5n a3n, and v5na = a a5n. The fact that Bsand Ds shares are equal implies that

    vn

    i (1 v2n) = v

    5n

    i

    which is equivalent to (v2n)2 + v2n 1 = 0, implying that

    v2n =1 5

    2.

    Since v is positive, only the + sign is admissible:

    v2n =1 + 5

    2= 0.618033988...

    sovn = 0.7861513777....

    Then the shares of A, B, C, D will be in the ratio

    1 vn : vn v3n : v3n v5n : 1 v5n

    i.e. 0.2138486221 : 0.3002831059 : 0.1855851657 : 0.6997168937

    8. (a) Find the present value of an annuity which pays 4,000 at the beginning ofeach 3-month period for 12 years, assuming an effective rate of 2% interestper 4-month period.

    (b) Suppose that the owner of the annuity wishes to pay now so that paymentsunder his annuity will continue for an additional 10 years. How much shouldhe pay?

    (c) How much should he pay now to extend the annuity from the present 12 years

    to a perpetuity?(It is intended that you solve this problem from first principles, not by substitu-tion into formul in [1, Chapter 4].)

    Solution:

    (a) If i be the effective interest rate per 4-month period, the effective rate per

    3-month period will be j = (1 + i)34 1. Accordingly the value of the desired

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    annuity is

    4000a48j = 4000(1 + j) 1 (1 + j)48

    j

    = 4000(1 + i)34 1 (1 + i)

    36

    (1 + i)34 1

    = 40001 (1.02)361 (1.02) 34

    = 138, 317.4894.

    (b) Repeating the calculations above for 48 + 40 = 88 payments, we obtain

    4000a88j = 4000(1 + j) 1 (1 + j)88

    j

    = 4000(1 + i)34 1 (1 + i)

    66

    (1 + i)34 1

    = 40001 (1.02)661 (1.02) 34

    = 197, 897.4338,

    so the additional payments will cost 197, 897.4338

    138, 317.4894 = 59, 579.9444

    today.

    (c) The cost of the perpetuity-due today would be

    4000a88j = 4000(1 + j) 1

    j

    = 4000(1 + i)34 1

    (1 + i)34 1

    =4000

    1 (1.02) 34= 271, 329.4837,

    so the additional payments will cost

    271, 329.4837 138, 317.4894 = 133, 011.9943today.

    9. (No time diagram is needed for the solution to this problem.) In Problem 9 ofAssignment 2 you were asked to apply the Bisection Method to determine the

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    solution to an interest problem to 3 decimal places. The equation in question was

    (3):300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 .

    and the solution given began with the values of

    f(x) = 3x4 + 2x3 + 3x2 10 .

    at x = 0 (f(0) = 10), x = 2 (f(2) = 66 > 0), and x = 2 (f(2) = 34 >0), and we were interested in the solution between 0 and 2 a solution thatis unique because f is positive in this interval. Apply Linear Interpolation 4times in an attempt to determine the solution we seek. (You are not expected

    to know the general theory of error estimation.) The intention is that you applylinear interpolation unintelligently, using it to determine a point where you find thefunction value and thereby confine the zero to a smaller subinterval: the point thatyou find will replace the midpoint in the bisection method. In some situations, asin the present one, the procedure may not be better than the bisection method.Indeed, in the present example, it could take many more applications than thebisection method to obtain the accuracy you obtained with that method.

    Solution: We take x1 = 0, x2 = 2. Then

    x3 = 0 + (2

    0)

    10

    10 66= 0.2631578947f(x3) = 9.741407754

    x4 = 0.2631578947 + (2 0.2631578947) 9.7414077549.741407754 66= 0.4865401588

    f(x4) = 8.891376200x5 = 0.4865401588 + (2 0.4865401588) 8.8913762008.891376200 66

    = 0.6662236083

    f(x5) = 7.486007579x6 = 0.6662236083 + (2 0.6662236083) 7.4860075797.486007579 66

    = 0.8020951913

    f(x6) = 5.796139773x7 = 0.8020951913 + (2 0.8020951913) 5.7961397735.796139773 66

    = 0.8988026707

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    f(x7) =

    4.166425912

    x8 = 0.8988026707 + (2 0.8988026707) 4.1664259124.166425912 66= 0.9641908820

    f(x8) = 2.825434839x9 = 0.9641908820 + (2 0.9641908820) 2.8254348392.825434839 66

    = 1.006713115

    f(x9) = 1.837664218x10 = 1.006713115 + (2 1.006713115) 1.837664218

    1.837664218

    66

    = 1.033620406

    f(x10) = 1.162055435x11 = 1.033620406 + (2 1.033620406) 1.1620554351.162055435 66

    = 1.050340958

    f(x11) = 0.721587971x12 = 1.050340958 + (2 1.050340958) 0.7215879710.721587971 66

    = 1.060611435

    f(x12) = 0.442976533x13 = 1.060611435 + (2 1.060611435) 0.4429765330.442976533 66

    = 1.066874356

    f(x13) = 0.270019571x14 = 1.066874356 + (2 1.066874356) 0.2700195710.270019571 66

    = 1.070676410

    f(x14) = 0.163879567x15 = 1.070676410 + (2

    1.070676410)

    0.163879567

    0.163879567 66= 1.072978227f(x15) = 0.099198908

    x16 = 1.072978227 + (2 1.072978227) 0.0991989080.099198908 66= 1.074369462

    f(x16) = 0.059950550

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    x17 = 1.074369462 + (2

    1.074369462)

    0.059950550

    0.059950550 66= 1.075209488f(x17) = 0.036195811

    x18 = 1.075209488 + (2 1.075209488) 0.0361958110.036195811 66= 1.075716385

    f(x18) = 0.021840825x19 = 1.075716385 + (2 1.075716385) 0.0218408250.021840825 66

    = 1.076022149

    f(x19) = 0.013174269x20 = 1.076022149 + (2 1.076022149) 0.0131742690.013174269 66

    = 1.076206548

    f(x20) = 0.007944937x21 = 1.076206548 + (2 1.076206548) 0.0079449370.007944937 66

    = 1.076317739

    f(x21) = 0.004790698

    x22 = 1.076317739 + (2 1.076317739) 0.004790698

    0.004790698 66= 1.076384781

    f(x22) = 0.002888504x23 = 1.076384781 + (2 1.076384781) 0.0028885040.002888504 66

    = 1.076425201

    f(x23) = 0.001741533x24 = 1.076425201 + (2 1.076425201) 0.0017415330.001741533 66

    = 1.076449571

    f(x24) = 0.001049952x25 = 1.076449571 + (2 1.076449571) 0.0010499520.001049952 66

    = 1.076464263

    f(x25) = 0.000632996x26 = 1.076464263 + (2 1.076464263) 0.0006329960.000632996 66

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    = 1.076473120

    f(x26) = 0.000381633x27 = 1.076473120 + (2 1.076473120) 0.0003816330.000381633 66

    = 1.076478460

    f(x27) = 0.000230079x28 = 1.076478460 + (2 1.076478460) 0.0002300790.000230079 66

    = 1.076481679

    f(x28) = 0.000138723

    x29 = 1.076481679 + (2 1.076481679) 0.0001387233

    0.0001387233 66= 1.076483620

    f(x29) = 0.000083635x30 = 1.076483620 + (2 1.076483620) 0.0000836350.000083635 66

    = 1.076484790

    f(x30) = 0.000050430x31 = 1.076484790 + (2 1.076484790) 0.0000504300.000050430 66

    = 1.076485496f(x31) = 0.000030391

    The reason that the method is not efficient here is that the graph of the functionis far from linear in the interval under consideration.6

    6Issues of this type are beyond MATH 329, and are not covered adequately in the current textbook;if you are interested in Numerical Anaylsis, consider taking MATH 317.

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    9 Class Tests

    9.1 Class Test, Version 1

    McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

    EXAMINER: Professor W. G. Brown DATE: Wednesday, 10 March, 2004.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:3515:20FAMILY NAME:

    GIVEN NAMES:

    STUDENT NUMBER:

    Instructions

    The time available for writing this test is about 45 minutes. This test booklet consists of this cover, Pages 42 through 44 containing questions together

    worth 60 marks; and Page 45, which is blank.

    Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you find

    that your booklet is defective.)

    All your writing even rough work must be handed in. Calculators. While you are permitted to use a calculator to perform arithmetic and/or expo-

    nential calculations, you must not use the calculator to calculate such actuarial functions asani, sni, (Ia)ni, (I s)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for the value ofthe function in terms of exponentials and/or polynomials involving n and the interest rate.You must not use your calculator in any programmed calculations. If your calculator hasmemories, you are expected to have cleared them before the test.

    In your solutions to problems on this test you are expected to show all your work. You areexpected to simplify algebraic and numerical answers as much as you can.

    Your neighbours may be writing a version of this test which is different from yours.PLEASE DO NOT WRITE INSIDE THIS BOX

    1(a) 1(b) 1(c) 1(d) 2 3 Total

    /5 /5 /5 /5 /15 /25 /60

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    1. (a) [5 MARKS] Suppose that the nominal annual rate of interest, compounded

    8 times per year, is 5%. Showing all your work, determine the equivalenteffective annual rate of discount.

    (b) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 3times per year, is 7%. Showing all your work, determine the equivalent annualrate of interest.

    (c) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 4%.

    (d) [5 MARKS] Suppose that the effective interest rate for1

    4year is 3%. Deter-

    mine the equivalent nominal interest rate, compounded every 2 years.

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    3. [25 MARKS] A loan of 5000 is to be repaid by annual payments of 250 to commence

    at the end of the 6th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be larger thanthe regular payments. Assume i = 4%.

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    continuation page for problem number

    You must refer to this continuation page on the page where the problem is printed!

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    9.2 Class Test, Version 2

    McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

    EXAMINER: Professor W. G. Brown DATE: Wednesday, 10 March, 2004.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:3515:20FAMILY NAME:

    GIVEN NAMES:

    STUDENT NUMBER:

    Instructions

    The time available for writing this test is about 45 minutes. This test booklet consists of this cover, Pages 47 through 49 containing questions together

    worth 60 marks; and Page 50, which is blank.

    Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you findthat your booklet is defective.)

    All your writing even rough work must be handed in. Calculators. While you are permitted to use a calculator to perform arithmetic and/or expo-

    nential calculations, you must not use the calculator to calculate such actuarial functions asani, sni, (Ia)ni, (I s)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for the value ofthe function in terms of exponentials and/or polynomials involving n and the interest rate.You must not use your calculator in any programmed calculations. If your calculator hasmemories, you are expected to have cleared them before the test.

    In your solutions to problems on this test you are expected to show all your work. You areexpected to simplify algebraic and numerical answers as much as you can.

    Your neighbours may be writing a version of this test which is different from yours.PLEASE DO NOT WRITE INSIDE THIS BOX

    1 2(a) 2(b) 2(c) 2(d) 3 Total

    /25 /5 /5 /5 /5 /15 /60

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    1. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 100 to commence

    at the end of the 5th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be NO largerthan the regular payments. Assume i = 4.5%.

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    2. (a) [5 MARKS] Suppose that the effective interest rate for1

    5

    year is 0.02. Deter-

    mine the equivalent nominal interest rate, compounded every 3 years.

    (b) [5 MARKS] Suppose that the nominal annual rate of interest, compounded9 times per year, is 6%. Showing all your work, determine the equivalenteffective annual rate of discount.

    (c) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 6times per year, is 5%. Showing all your work, determine the equivalent annualrate of interest.

    (d) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 8%.

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    Information for Students in MATH 329 2004 01 49

    3. [15 MARKS] The accumulated value just after the last payment under a 12-year

    annuity of 1000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity of 500 per annum forever, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the effective interest rate of the perpetuity, assuming it comes into effectjust after the last payment under the annuity.

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    Information for Students in MATH 329 2004 01 50

    continuation page for problem number

    You must refer to this continuation page on the page where the problem is printed!

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    Information for Students in MATH 329 2004 01 51

    9.3 Class Test, Version 3

    McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

    EXAMINER: Professor W. G. Brown DATE: Wednesday, 10 March, 2004.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:3515:20FAMILY NAME:

    GIVEN NAMES:

    STUDENT NUMBER:

    Instructions

    The time available for writing this test is about 45 minutes. This test booklet consists of this cover, Pages 52 through 54 containing questions together

    worth 60 marks; and Page 55, which is blank.

    Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you findthat your booklet is defective.)

    All your writing even rough work must be handed in. Calculators. While you are permitted to use a calculator to perform arithmetic and/or expo-

    nential calculations, you must not use the calculator to calculate such actuarial functions asani, sni, (Ia)ni, (I s)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for the value ofthe function in terms of exponentials and/or polynomials involving n and the interest rate.You must not use your calculator in any programmed calculations. If your calculator hasmemories, you are expected to have cleared them before the test.

    In your solutions to problems on this test you are expected to show all your work. You areexpected to simplify algebraic and numerical answers as much as you can.

    Your neighbours may be writing a version of this test which is different from yours.PLEASE DO NOT WRITE INSIDE THIS BOX

    1 2 3(a) 3(b) 3(c) 3(d) Total

    /15 /25 /5 /5 /5 /5 /60

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    Information for Students in MATH 329 2004 01 52

    1. [15 MARKS] The accumulated value just after the last payment under a 9-year

    annuity of 2000 per year, paying interest at the rate of 8% per annum effective,is to be used to purchase a perpetuity at an interest rate of 4%, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the size of the payments under the perpetuity.

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    Information for Students in MATH 329 2004 01 53

    2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commence

    at the end of the 4th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be larger thanthe regular payments. Assume i = 5%.

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    Information for Students in MATH 329 2004 01 54

    3. (a) [5 MARKS] State the nominal annual interest rate, compounded instanta-

    neously, which is equivalent to an effective annual interest rate of 6%.

    (b) [5 MARKS] Suppose that the effective interest rate for1

    6year is 0.015. De-

    termine the equivalent nominal interest rate, compounded every 4 years.

    (c) [5 MARKS] Suppose that the nominal annual rate of interest, compounded3 times per year, is 8%. Showing all your work, determine the equivalenteffective annual rate of discount.

    (d) [5 MARKS] Suppose that the nominal annual rate of discount, compounded12 times per year, is 6%. Showing all your work, determine the equivalentannual rate of interest.

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    Information for Students in MATH 329 2004 01 55

    continuation page for problem number

    You must refer to this continuation page on the page where the problem is printed!

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    Information for Students in MATH 329 2004 01 56

    9.4 Class Test, Version 4

    McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

    EXAMINER: Professor W. G. Brown DATE: Wednesday, 10 March, 2004.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:3515:20FAMILY NAME:

    GIVEN NAMES:

    STUDENT NUMBER:

    Instructions

    The time available for writing this test is about 45 minutes. This test booklet consists of this cover, Pages 57 through 59 containing questions together

    worth 60 marks; and Page 60, which is blank.

    Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you findthat your booklet is defective.)

    All your writing even rough work must be handed in. Calculators. While you are permitted to use a calculator to perform arithmetic and/or expo-

    nential calculations, you must not use the calculator to calculate such actuarial functions asani, sni, (Ia)ni, (I s)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for the value ofthe function in terms of exponentials and/or polynomials involving n and the interest rate.You must not use your calculator in any programmed calculations. If your calculator hasmemories, you are expected to have cleared them before the test.

    In your solutions to problems on this test you are expected to show all your work. You areexpected to simplify algebraic and numerical answers as much as you can.

    Your neighbours may be writing a version of this test which is different from yours.PLEASE DO NOT WRITE INSIDE THIS BOX

    1(a) 1(b) 1(c) 1(d) 2 3 Total

    /5 /5 /5 /5 /25 /15 /60

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    Information for Students in MATH 329 2004 01 57

    1. (a) [5 MARKS] Suppose that the nominal annual rate of discount, compounded

    6 times per year, is 0.5%. Showing all your work, determine the equivalentannual rate of interest.

    (b) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 10%.

    (c) [5 MARKS] Suppose that the effective interest rate for1

    5year is 2%. Deter-

    mine the equivalent nominal interest rate, compounded every 6 years.

    (d) [5 MARKS] Suppose that the nominal annual rate of interest, compounded7 times per year, is 2%. Showing all your work, determine the equivalenteffective annual rate of discount.

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    Information for Students in MATH 329 2004 01 58

    2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commence

    at the end of the 4th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be NO largerthan the regular payments. Assume i = 5%.

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    Information for Students in MATH 329 2004 01 59

    3. [15 MARKS] The accumulated value just after the last payment under a 10-year

    annuity of 1000 per year, paying interest at the rate of 6% per annum effective,is to be used to purchase a perpetuity of 800 per annum forever, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the effective interest rate of the perpetuity, assuming it comes into effectjust after the last payment under the annuity.

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    Information for Students in MATH 329 2004 01 60

    continuation page for problem number

    You must refer to this continuation page on the page where the problem is printed!

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    Information for Students in MATH 329 2004 01 61

    10 Fourth Problem Assignment

    Distribution Date: Mounted on the Web on Wednesday, March 3rd, 2004Hard copy distributed on Friday, March 5th, 2004

    Solutions are due by Wednesday, March 17th, 2004

    1. (a) Find, to the nearest unit, the accumulated value 19 years after the first pay-ment is made of an annuity on which there are 7 payments of 3000 each madeat 1 1

    2-year intervals. The nominal rate of interest convertible semiannually is

    6%.

    (b) Find, to the nearest unit, the present value of a 20-year annuity-due which

    pays 200 at the beginning of each half-year for the first 8 years, increasing to250 per half-year thereafter. The effective annual rate of interest is 6%.

    2. (a) The present value of a perpetuity-immediate paying 1 at the end of every 5years is 1.637975. Find i and d.

    (b) The present value of a perpetuity-due paying 1 at the beginning of every 5years is 1.637975. Find d and i.

    3. Determine the present value, at a nominal interest rate of 6% compounded quar-terly, of the following payments made under an annuity: 120 at the end of the 3rdyear, 110 at the end of the 4th year, decreasing by 10 each year until nothing is

    paid.

    4. Find the present value, at an effective annual interest rate of 5.75%, of a perpetuity-immediate under which a payment of 100 is made at the end of the 1st year, 300 atthe end of the 2nd year, increasing until a payment of 2500 is made, which level ismaintained for exactly a total of 10 payments of 2500 (including the first of themin the count of 10), after which the payments fall by 400 each year until they reacha level of 100, which is maintained in perpetuity. (Note: You are expected to showexplicitly how you decompose the payments; it is not sufficient to simply show afew numbers and a sum.)

    5. Find, to the nearest unit, the present value of a 25-year annuity-due which pays100 immediately, 104 at the end of the 1st year, 108.16 at the end of the 2nd year,where each subsequent payment is obtained from its predecessor by multiplying bya factor of 1.04. The annual effective rate of interest is 8.%.

    6. (a) A loan of 15,000 is being repaid with payments of 1,500 at the end of eachyear for 20 years. If each payment is immediately reinvested at 6% effective,find the effective annual rate of interest earned over the 20-year period.

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    Information for Students in MATH 329 2004 01 62

    (b) A loan of 15,000 is being repaid with payments of 1,500 at the end of each

    year for 10 years. Determine the yield rate to the investor.

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    Information for Students in MATH 329 2004 01 63

    11 Solutions to Problems on the Class Tests

    Distribution Date: Wednesday, March 17th, 2004, in hard copy.The test was administered on Wednesday, March 11th, 2004.

    (Subject to correction.)

    The first four problems listed are concerned with equivalent rates of interest anddiscount (each in 4 parts); the next four concern annuities and perpetuities; and the lastfour are concerned with unknown type and final balloon or drop payments.

    1. (a) [5 MARKS] [VERSION 1 #1(a)] Suppose that the nominal annual rate of in-terest, compounded 8 times per year, is 5%. Showing all your work, determine

    the equivalent effective annual rate of discount.Solution: We are given that i(8) = 0.05, and