1 1 ■ 1 i ( , 1 ■ ■'1 Bil 1 2— -f ■ 0— cl 2 §m ■ J H HF ^73 Mathenatic, accountant SfcV-^lSHH^iJPSjR Southern Branch of the University of California Los Angeles Form L-1 This book is DUE on tne last aate siampea oeiow. ;.^A/ 4J y 1923 OCT 2 19^4 DEC 1 5 1924 r;. 1 WAR 4 OCT 2 1 1946 MAR ^ 1949 JAN 3 1957 APR 2 2 I960 MAR 7 iQ?c i nr"^ "» r* t r- r-^ % '*^ i^l{?60 SEP15 1962 MATHEMATICS FOR THE ACCOUNTANT BY EUGENE R. VINAL, A.M. Professor of Actuarial Mathematics and Accounting School of Commerce and Finance Northeastern College, Boston SECOND EDITION NEW YORK BIDDLE BUSINESS PUBLICATIONS, Inc. 1921 '3 7-/6/ Copyright, 1920, by The Biddle Publishing Company. Copyright, 1921, by BiDOLE Business Publications, Inc. PREFACE This book is the outgrowth of a course which has been given for several years in the School of Commerce and Finance of Northeastern College. It has always been the policy of the School to give its students a thorough train- ing in all phases of accountancy, and annuity studies have been considered a part of the training for that profession. While the course is forming in the early part of the year, it has seemed well to treat some elementary subjects which frequently appear in C. P. A. examinations, such as averaging of accounts and foreign exchange, and at the end of the year it has seemed practical to include a few lectures on the slide rule A new impetus was given to tne work by an editorial which appeared in the ''Journal of Accountancy" for August, 1918. By that editorial the American Institute of Accountants committed itself definitely to the policy of requiring certain annuity studies of all candidates for its examinations. The editorial read in part as follows: "The scope of the examination in Actuarial Science is to include certain problems relative to interest and annuities, certain sinking funds, loans repayable by instalments, and so forth, and also the construction and use of tables relating thereto. In other words, the candidate will be expected to answer questions based upon a knowledge which will have been obtained in the study of algebra. PREFACE Any candidate who has an intelligent conception of algebra should have no difficulty in answering the questions which will be propounded in actuarial science. . . . "There is a great deal to be said in favor of the inclusion of actuarial problems and we believe that the need for knowledge of this kind will increase as time goes on. Heretofore there have been many accountants who have had practically no need to exercise any knowledge which they have possessed of actuarial matters, but with the growth of accounting work and the broadening of its scope, there must be many problems which can be solved at a great saving of time and effort if the accountant be able to deal with them with the advantage of actuarial knowledge." This requirement seems to imply a general study of com- pound interest and its application to those problems which are commonly solved by simple interest. It is axiomatic that every dollar in the business world is at work; it is earning other dollars either for its owner or for someone else. At the end of every fiscal period all these earnings, after deduction of expenses, should become capital, and the same or a larger income rate should be earned on this increased capital. Such studies are compound interest and present worth, with the various aspects of interest rates; annuities, immediate, due or deferred; their amount and present worth; sinking funds and various related prob- lems in valuation of assets ; amortization ; bond valuation in its many aspects. It is obvious that the accountant has no time for alge- braic studies, unless he has attended a good high school or college. To prepare a book for those who have not had these advantages and who nevertheless are promising students of accountancy, all the work must be reduced to PREFACE a basis of arithmetic and common sense. Formulae there must be, and they are numerous and sometimes com- plicated; yet there is no reason why any man who is fitted mentally for high grade accounting practise should have difficulty with any formulae which are germane to his profession. The interest rates are so varied and so seldom common enough to be included in annuity tables that a study of logarithms is absolutely necessary. This study should be pursued no further than is demanded by the formulae for compound amount and the simpler formulae for the time and rate involved in a transaction. In short, the difficulty has been to take several highly technical books on actuarial science and reduce them to a content and method which will give the accountant a maximum of training in a minimum of time. Some actuarial works are suitable for the student who knows some algebra and likes to read mathematics. Such are Todhunter's ''Textbook of the Institute of Actuaries, Book I", King's 'Theory of Finance", and ]\Iackenzie's "Interest and Bond Values". For the student who knew algebra once and desires to renew his knowledge there are the Went worth-Smith "Commercial Algebra, Book 2", and Skinner's "Mathematical Theory of Investment". For the student who has no algebra there is only the Sprague-Perrine "Accountancy of Investment", which is written from the standpoint of the savings bank man or trustee, and does not treat some important phases of general aecounting knowledge. These are all excellent books, and have been of great service to the author. The most convenient tables for the student are those in Skinner's "Mathematical Theory of Investment", and reprinted separately by the publishers, Ginn and Com- pany. PREFACE This book, then, is for the use of the accountant who desires to prepare for the examinations of the American Institute oi Accountants or for the Ceitified Pubhc Accountants' examinations of the various states; for the accountant who will be required to handle sinking fund or bond accounts in a scientific way; for the accountant who may be called upon to audit the accounts of an insurance company, savings bank, brokerage house, or any concern which operates its bond accounts scien- tifically ; and finally for the accountant who believes that the broadest training is best for the professional man. The book ends with a chapter in which the fundamentals of actuarial science are treated in a different manner, as a quiz for candidates for examinations. In this chapter the treatment is wholly arithmetical, and no logarithmic or algebraic knowledge is required. Problems from the recent examinations of the American Institute of Account- ants and from some recent C. P. A. examinations are solved and discussed. Since no books or tables are allowed in any of these examinations, the rules are presented in such form as makes them easy to remember and apply. Inasmuch as this book is written for the benefit of accountants, their co-operation is requested and will be gratefully accepted. Any suggestions as to content, method, or problems will be acknowledged and con- sidered seriously. Any formulae which accountants have found useful are particularly welcome. It is planned that if the book goes through a revision a collection of formulae shall appear as a separate chapter, a reference list such as engineers find so useful. Any C. P. A. problems will also be especially welcome, because the aspects of the subject which appear important to examiners are so varied that the most complete collection which can be put to- PREFACE gether will be none too complete for the prospective candidate to study. In closing, I wish to express my sincere appreciation of the help and inspiration I have received from Professor Charles F. Rittenhouse, C. P. A. He first brought the subject to my attention as suitable matter for a college course, and has in many ways inspired the writing of this book. EUGENE R. VINAL. Boston, Massachusetts. ix CONTENTS Chapter Page I Preliminary Suggestions 3 Table of Multiples. Contracted Multiplication. Contracted Division. The Number of Decimal Places Neces- sary for Accuracy. Problems. II Simple Interest 9 Principal. Rate. Frequency. Time. Common Time. Exact Time. Bankers' Time. Exact Interest. One Per Cent Method. Discounting Notes. Interest on Daily Balances. Problems. Ill Accounts Current 18 Averaging an Account. Finding the Equated Date of an Ac- count. Rule for Reckoning from the Focal Date to the Equated Date. Problems. xii CONTENTS Chapter Page IV Foreign Exchange 25 Par of Exchange between the United States and Other Commercial Na- tions. The Rate of Exchange on London. The Value of a Bill on London. Exporting Gold. The Rate of Exchange on France, Belgium, Switzerland, and Italy. South American Exchange. Arbitrage of Exchange. Problems. V Powers and Roots : Logarithms 33 The Four Laws of Exponents. Multiplication. Division. Raising to a Power. Finding a Root. Fractional Exponents. Negative Exponents. The Zero Exponent. Logarithms. Determining the Characteristic. Logarithmic Tables. Finding a Logarithm. The Cologarithm of a Number. Finding the Antilogarithm. Illustrative Problems. VI Compound Interest AND Present Worth 44 Actuarial Science. Compound Interest — Its Significance, for the Accountant. CONTENTS xiii Chapteb Page Computation of Compound Amount. Nominal and Effective Rates. Changing from Nominal to Effective Rates. Nominal Rate when the Effective Rate Is Given. Formula for Finding the Nominal Rate when the Effective Rate Is Given. Present Worth and Compound Dis- count. Summary of the Processes Used in Compound Interest Computation. Problems. VII Annuities : Amount and Present Worth 60 Annuity Defined. The Amount of an Ordinary Annuity. Formula for the Amount of an Annuity : Effective Rates. The Present Worth of an Annuity: Effective Rates. Three Special Types of Annuity. An Annuity Due. Amount of an Annuity Due. The Present Worth of an Annuity Due. A Deferred Annuity. A Perpetuity. The Interest Rate on a Given Annuity. Problems. VIII Sinking Funds 75 Principles of Sinking Fund Mathe- matics. xiv CONTENTS Chapter Page Effective Rates. Problem Involving Logarithms. Problem Involving Annuity Due. Formula to Determine the Time re- quired to Accumulate a Stated Amount. Problems. IX Valuation of Assets 81 Valuation of a Plant as a Whole. The Rate of Depreciation. The Composite Life of a Plant. The Wearing Value of an Asset at any Time. The Condition Per Cent of a Plant at any Time. Other Methods of Reckoning De- preciation. Fixed Percentage on a Diminishing Book Value. Provision Against the Total Exhaus- tion of Wasting Assets. Capitalization of Assets. Summary of Formulae Used in Ac- counting for Assets. Problems. X Amortization 93 Relation between Sinking Fund and Amortization Schedules. Logarithmic Solution. •• Effective Rates. Annuity Due. CONTENTS XV Chapter Page The Amount Due After Any Number of Payments. The Time Required to Amortize a Given Debt. The Increased Cost of Lengthening the Life of an Asset. Problems. XI Valuation of Bonds 101 Bonds and Their Value. Formula for Finding the Correct Value of a Bond. Schedule of Amortization. Redemption at a Premium or Discount. Valuation with Allowance for Income Tax. Valuation between Interest Dates. Serial Bonds. Redemption of a Series. Instalment Bonds. Determining the Income Rate on a Bond. Problems. XII The Slide Rule 123 The Logarithmic Scale, Rules for Determining the Char- acteristic. Illustrations in Multiplication and Division. How to Square a Number. •• Square Root. Proportion. Problems. xvi CONTENTS Chapter Page XIII Review Problems 131 Problems. Two Examination Papers. XIV Problems from the Examinations of THE American Institute of Ac- countants 143 On Averaging an Account with an English Correspondent. On Compound Interest. On Present Worth. On Compound Discount. On Annuities. On Deferred Annuity. On Sinking Funds. On Depreciation. CHAPTER I PRELIMINARY SUGGESTIONS This chapter is a collection of processes which will prove to be timesavers in the use of tables. It is not in any sense a collection of short cuts such as is found in the texts on arithmetic, but simply a brief consideration of methods of shortening and simplifying the numerous and otherwise laborious operations which are necessary when tables of annuities or logarithms have to be used exten- sively. It is obvious that if an operation can be performed one way and lead to a ten-place answer, and by another method will give only a four-place answer, that the second method is preferable for the accountant. These processes are called reversed or contracted multiplication, and con- tracted division. A brief consideration of the number of decimal places desirable in any given computation will also be found helpful. Table of Multiples : In any case where the same num- ber must be used often as multiplicand or divisor, a table of multiples is a time saver. It is formed as follows : Suppose we are finding interest at exact time; that is we are using 365 as a divisor: 1 365 2 730 3 1095 4 1460 5 1825 6 2190 7 2555 8 2920 9 3285 10 3650 4 MATHEMATICS FOR THE ACCOUNTANT On the first line set down the number. The next line is twice the number. The third line is the sum of the first and second. The fourth line is the sum of the first and third; and so on. The tenth, which is the sum of the first and ninth, is also ten times the original number, and so is a proof line. Now in multiplying or dividing we can read the desired multiples and quotient figures from this table Contracted multiplication, sometimes called reversed multiplication. Suppose we are to receive $459.73 at the end of five years, and we wish to "discount" this amount to the present time, at 5% compounding annually. SI. 00 due in five years at 5% compounding annually is worth $.7835262 now; this is called the present worth. So the present worth of $459.73 is the product of these amounts. Form a table of multiples : 1 45973 2 91946 3 137919 4 183892 6 229865 6 275838 7 321811 8 367784 9 413757 10 459730 The product will run to nine places of decimals and we need only four. We can do away with the five superfluous digits by beginning with the left-hand digit of the mul- tiplier and proceeding as follows: PRELIMINARY SUGGESTIONS 459.73 .7835262 321.811 36 . 7784 1.3792 2299 91 27 360.2103 Locate the decimal point in the first partial product. This is usually a matter of common sense, rather than rule. In this example, 450 X .8 = 360; that is, there are three digits at the left of the point. In the second partial product we have four places and are ready to reject all digits at the right of the fourth. But it is advisable to take account of all ''carrying" digits. So the third partial product ought to be 13792 rather than 13791. In all such multipUcation as this the right-hand figure is inaccurate, but the error is negligible if we are keeping a sufficient number of decimal places. Contracted division: The above problem could have been solved by division. The compound amount of $1.00 for five years at 5% compounding annually is $1.2762816. Therefore the number of dollars which will amount to 459.73 will be the quotient of 459.73 divided by 1.2762816. We will not show the table of multiples here. MATHEMATICS FOR THE ACCOUNTANT 12762816)4597300000(360 . 21047 38288448 76845520 76576896 2686240 2552563 133677 127628 6049 5105 944 893 The division proceeds as usual uDtil we reach the decimal point, after which we do not bring down any more digits from the dividend, but at each operation cut off the right hand digit of the divisor. The number of decimal places necessary for accuracy varies. In general, the more the better; for the accuracy of any result is somewhat leps than the accuracy of the least accurate item. In ordinary dollars and cents work, as in schedules of annuities and bonds, all values ought to be carried to four places, that is to hundredths of a cent. In logarithmic work, there ought to be not less than six places, and seven are better. Logarithms of the com- pound interest ratios must run to at least ten places, because the rates are so close together and so near the beginning of the number scale that less than ten places will not show the differences accurately. Furthermore there are so many cases in which the results of these PRELIMINARY SUGGESTIONS 7 operations are multiplied by 100,000 or more, that less than ten digits will not give even approximately correct results. Even then it is best to test out all results by reversing the operation or setting up a schedule, then if an error appeal's it can be spread over the entire life of the transaction by methods which will be shown later. PROBLEMS ON CHAPTER I 1. Required the present worth of $2,283.67 due in 10 years at 4% .compounding annually : (a) Solve by multiplying by $ . 6755642, the present worth of $1.00 due in 10 years at 4% annually. (b) Solve by dividing by $1.4802443, the compound amount of $1.00 for the same time and rate. 2. Out of 89,032 persons living at age 25, 74,173 are alive at age 45. Find by division the probability that a person now aged 25 will live to age 45. 3. An asset costing $750.00 has an estimated life of five years and scrap value of $100.00. By one method of reckoning depreciation it is estimated that this asset depreciates each year 33.17% (or as a decimal .3317) of its value at the beginning of that year. Construct a schedule showing the net value of the asset year by year, as follows: Cost $750.00 Less depreciation, first year — 750 X .3317 248.775 Value, beginning of second year $501 . 225 Less depreciation— 501.225 X .3317... 166.2563 And so on. MATHEMATICS FOR THE ACCOUNTANT Value at end of fifth year should be almost exactly $100.00. A Profit and Loss Statement shows: Net Sales $167,834.29 Cost of Sales $82,387.49 Selling Expenses 31,093.77 G. &: A. Expenses 22,798.44 Net Profit xx,xxx.xx What per cent of the Net Sales is each of these items? Total of per cents should of course be 100%. (Form a table of multiples of the divisor; why?) CHAPTER II SIMPLE INTEREST "Interest is the increase of indebtedness through the lapse of time." — Sprague. Any interest contract must take account of four things: Principal, the number of units originally invested. Rate, the part of a unit of value, usually a few hundredths, which is added to each unit of principal by the lapse of one period of time. Frequency, the length of a unit of time, measured in years, months, or days; weeks are not used, nor are parts of days, Time, the number of these units of time during which the indebtedness continues. Some of the above are usually represented by letters. The principal is P, the rate is i, the time is n. More common than i is the symbol for the amount at the end of one period, 1 + i; this symbol is constantly used in investment mathematics. For instance, on one dollar for one year at 3% 1 + i is 1.03, which means that at the end of each period the debt has increased to 1.03 of its size at the beginning of the period. Since each dollar increases at the same rate as every other dollar, it is correct and usual to reckon the interest or amount of one dollar, and multiply the result by the number of dollars stated in the problem. There are three common methods of reckoning simple interest. They differ according to the way in which the time is reckoned. By common time the year is divided into 12 months of 30 days each, regardless of the calendar. This method 9 10 MATHEMATICS FOR THE ACCOUNTANT is unsatisfactory, and ought never to be used by any person who claims to aim at accuracy. By exact time the year is divided into 365 days, and no account is made of months. This is the just method of reckoning interest, and is simple enough, especially when exact interest tables can be used. By bankers' time the methods are combined. The exact number of days is computed, and the result is divided into groups which are multiples or fractions of 60 days. If a bill is timed three months from June 19 it is due September 19; but if it is timed 90 days from June 19 it is due September 17. There are many variants of this method, and only one of them will be treated here. Since most interest calculations are for short periods, the results obtained under the various methods do not difTer greatly. Moreover, it is a simple matter to change from common to exact time or from exact time to common time. If we let I represent the interest at common time and I' the interest at exact time, we have two equations: To change from common to exact P = I 73. Rule: From the ordinary interest subtract 1/73 of itself. To change from exact to common 1 = 1'+ :z- 72 Rule : To the exact interest add 1 /72 of itself. Exact interest: To find the exact interest on $382.00 at 5% from May 11 to October 1. First count the days, 143. Then reckon one yeai''s interest, $19.10. We have now to divide 19.10 by 366 and multiply by 143; but it is easier and more accurate to multiply first. One year's interest 19.10 times 143 2,731.30 divided by 366 7.48 SIMPLE INTEREST 11 In such computation a table of the multiples of 365 is useful, and also a table of the time between dates, or a calendar showing the number of each day in the year. One percent method: Interest by bankers' time is usually figured by some variation of the one percent method; when the rate is 6% it is usually called the 60-day or 200-month method. At 6% one year's interest on $1.00 is six cents; therefore an interest of one cent will be earned in 1/6 of a year, which by this method is 60 days. In general, any prin- cipal at 6% will earn 1% of itself in 60 days. And since 6 days is 1/10 of 60 days, any principal will earn l/lO of 1% of itself in six days. Rule : At 6%, pointing off two places in the principal gives the interest for 60 days; pointing off three places gives the interest for six days. Any period of time can be divided into multiples and fractions of 60 days, as follows: 120 days is twice 60 days. 180 days is 3 times 60 days, and so on. 30 days is 1/2 of 60 days. 20 days is l/3 of 60 days. 15 days is l/4 of 60 days. 12 days is l/5 of 60 days. 10 days is l/6 of 60 days. 6 days is l/ 10 of 60 days. 3 days is l/2 of 6 days. 2 days is 1/3 of 6 days. 1 day is l/6 of 6 days. It is often convenient to build in other ways: 15 days is l/2 of 30 days. 2 days is 1/10 of 20 days, and so on. 12 MATHEMATICS FOR THE ACCOUNTANT Problem: Find the interest on $125.00 from April 18 to August 3 at 6%, bankers' time. The number of days is: April 12 days May 31 days June 30 days July 31 days August 3 days Total 107 days Divide the time as follows: 60 days' interest $1.25 (point off 2 places) 30 " " .625 (1/2 of 60) 10 " " .2083 (1/3 of 30) 6 " " .125 (1/10 of 60) 1 « « .0208 (1/10 of 10) Total S2.2291 or, rounding up . 2.23 If the rate is not 6% it is usually easier to find at 6% and adjust. For instance, to find at 5%, after finding at 6% divide by 6 and subtract. It is possible, however, to work out special methods for some rates : At 5%, pointing off 2 places gives interest for 72 days At 4%, " " " " 90 " At4H%, " " " '' 80 " At 3%, " " " '' 120 " No such device can be used for odd rates, such as 6)^%, 4/^%, and so forth. Problem: What is 119 days' interest on $147.35 at 5%, bankers' time? SIMPLE INTEREST 13 72 days' interest $1.4735 (point off 2 places) 24 « " .4913 (1/3 of 72) 18 " « .3683 (1/4 of 72) 4 « « .0818 (1/6 of 24) 1 '' " . 0204 (1/4 of 4) 119 $2'4353 or $2.44 Discounting notes : Find the net proceeds of a 90-day note for $350.00, dated May 19, interest at 5%, discounted June 2 at 6%. Face $350.00 Add 90 days' interest at 5% 4.38 Value at maturity $354.38 Due date, 90 days from May 19 is Aug. 17 Term of discount, from June 2 to Aug. 17 is 76 days Discount, 76 days at 6% on $354.38 4.49 Net proceeds $349.87 Interest on daily balances: Many banks advertise in- terest on daily balances of checking accounts at a nominal rate. One method of computing this interest is the follow- ing: Each day's balance is made out, and at the end of the month these balances are totalled and one day's interest is allowed on the total. The following problem will illus- trate the use of exact interest : (The rate used here is 2 %) Deposits Withdrawals Balances March 1 600 ... 600 2 ... ... 600 3 200 300 500 4 100 ... 600 5 300 200 700 B 300 150 860 14 MATHEMATICS FOR THE ACCOUNTANT Deposits Withdrawals Balances 7 . • • 200 650 8 . . . • • . 650 9 . . . • • • 650 10 200 . . . 850 11 300 550 12 . . . 550 13 600 450 700 14 300 300 700 15 500 350 850 16 ... 850 17 400 300 950 18 300 150 1100 19 200 550 750 20 1200 1350 600 21 300 600 300 22 500 100 700 23 . . . 700 24 400 250 850 25 300 100 1050 26 300 200 1150 27 850 300 1700 28 400 150 1950 29 200 . . . 2150 30 2150 31 300 650 1800 28750 /■'s interest on $28,750 is $1.5753, or $1.58. If there are not many items in the month, the use of what may be termed * 'day-numbers" will shorten the work. In this case a "day-number" is the number of days during which a balance remains unchanged. For instance, if th« balance on the seventh of the month is $500.00, and SIMPLE INTEREST 15 there is no change until the fifteenth, there has been an un- changed balance of $500.00 for eight days, which is obviously equivalent to a balance of eight times $500.00 for one day. Problem : Find the interest at 2%, exact time, on the following account: Debits: March 2, $500.00; March 11, $450.00; March 19, $600.00; March 27, $300.00. Credits : March 8, $300.00; March 14, $200.00; March 28, $500.00. Day- Date Debit Credit Balance Numbers" Products 2 500 ... 500 8 ... 300 200 11 450 ... 650 14 ... 200 450 19 600 ... 1050 27 300 . . . 1350 28 ... 500 850 Total 20100 One day's interest on $20,100 is $1,1014, or $1.16. A better method, involving a different use of 'May- numbers," will be demonstrated in the next chapter. PROBLEMS ON CHAPTER II 5. Find the interest on a 3H% Liberty Bond for $50.00 from June 16 to December 15. 6. Find the interest at 6%, bankers' time, on $386.55 from June 17 to October 12. 7. Find the interest at 4 3^ %, bankers' time, on $1 ,200.00 from July 4 to December 25. 8. A note for $500.00, three months, is dated March 9, interest at 5%. It is discounted May 17 at 6%. Find the net proceeds. 6 3000 3 600 3 1950 5 2250 8 8400 1 1350 3 2550 16 MATHEMATICS FOR THE ACCOUNTANT 9. Find the interest on daily balances for the month on the following checking account at 2%, exact time. Deposits Checks 1 3000 2000 2 800 500 3 1200 800 4 560 210 5 . . . . . . 6 380 160 7 650 200 8 1340 620 9 730 400 10 520 260 11 1630 1250 12 . . . 13 870 1420 14 1130 825 15 760 385 16 920 440 17 1420 675 IS 1430 1260 19 .... 20 1165 1530 21 680 445 22 1480 1260 23 260 420 24 650 725 25 1180 855 26 .... 27 380 1560 28 640 210 29 360 620 30 1245 1135 31 680 320 SIMPLE INTERESI^ 17 10. Find the interest due January 31, at 2)/^%, exact time, on these daily balances for the month of January : Debits: 3rd, $6,800; 7th, $4,500; 22nd, $5,250; 29th, $1,850. Credits: 6th, $4,250; 11th, $5,300; 20th, $200; 30th. $1,200. CHAPTER III ACCOUNTS CURRENT Under the title of this chapter are included two types of problem. The first is the averaging of an account; that is, the statement of the balance as of a certain date, with interest charged and credited at a certain rate. The second is the finding of the equated date; that is, the date on which the balance of the account is at a minimum: when the account can be settled with a minimum payment of interest. Averaging an account: Problem: Find the balance of the following account as of July 1, interest at 5% bankers' time: Debits: April 27, 30-day note without interest, $500.00; May 18, 60-day note without interest, $600.00; June 22, cash, $1,200.00. Credits: June 4, cash, $500.00; June 18, cash, $200.00. Of course the 30-day note must be dated from maturity. May 27; and the 60-day note from July 17, which means that when we reckon interest we must deduct 16 days' ''discount" for this item. There are two methods of solution which give the same result. We may reckon interest on each item and balance the account as usual. Or we may simplify the work by using day-numbers. 18 ACCOUNTS CURRENT 19 First solution: Debit Amount Due Time Interest $500.00 May 27 35 2.4304 600.00 July 17 16d 1.3333d 1,200.00 June 22 9 1.50 $2,300.00 2.5971 Credit Amount Due Time Interest $500.00 June 4 27 1.875 200.00 June 18 13 .3611 $700.00 2.2361 Balance of principal due $1,600.00 Balance of interest due .3611 Total $1,600.36 Second solution : In this solution the day-number of each item is the number of days to the balancing date; in other words, the time as in the first solution. In short form the solution is: 500 X 35 17,500 500 X 27 13,500 600 X 16d 9,600 deduct 200 X 13 2,600 1,200 X 9 10,800 2,300 18,700 700 16,100 Debit balance of day-numbers, 2,600; one day's interest at 5% is .3611, agreeing with the result of the first solution. When such a statement is made out on the exact interest basis, the advantage of using day-numbers is obvious. Under such conditions the reckoning of a large number of interest items would be a waste of time. By the day- number method we find only one interest item, yet the result is the same to the last decimal place. 20 MATHEMATICS FOR THE ACCOUNTANT Finding the equated date of an account: This has been defined as the date on which the balance of the account is at a minimum. Evidently that is the date on which the interest is at a minimum. Consider any series of trans- actions. As goods are purchased, interest on the in- debtedness begins; as payments are made, the indebted- ness is lessened. Between the purchase and the payment there is a date on which the interest on the debt and the discount on the payment equalize each other. A purchase of $500.00 on July 11 and a payment of $500.00 on August 6 would thus equalize each other on July 24, the date half way between. Generally, however, there are many items both debit and credit, and the process of finding this neutral date is complicated. It is necessary to assume some date as a balancing date. This date may be any date whatever, but for convenience it is usual to select either the earliest or the latest date in the problem. This is in order that we may have to perform only one kind of operation ; if we use the earliest date, we discount all items to that date; if we use the latest date, we accumu- late all items to that date. It seems easier to use the latest date, and the solutions given will all be in that form. After we have balanced the account as of this date, which is called the "focal date," we can find how many days out of the way our focal date is, and reckon backward or forward to the equated date. An account may be one-sided or two-sided ; it may be all debits or all credits, or it may be composed of both debits and credits. We will study a one-sided account first. Problem: Find the equated date of this account at 6%, bankers' time: Debits: July 18, $600.00; July 20, $200.00; July 28, 30- day note, $2,800.00. ACCOUNTS CURRENT 21 The latest date is the due date of the note, August 27; we will use that as our focal date. Amount Due Time Interest $600.00 July 18 40 $4.00 200.00 July 20 38 1.27 2,800.00 August 27 $3,600.00 $5.27 Since both principal and interest are debits, it is evident that the longer the account remains open the larger the balance will be; that is, the equated date must be earlier than the focal date. A balance of $3,600.00 is accruing interest at the rate of 60 cents a dsij. It has already accrued $5.27. This indicates that interest has been accruing for practically nine days (5.27 -=- 60). Reckoning backward nine days from August 27, we find the equated date to be August 18. This result can be tested by balanc- ing the account as of August 18. Another problem will show the solution of a two-sided account, and also the use of day-numbers: Debits: May 2, 30 days, $1,200.00, interest at 5%. May 21, $840.00. June 1, 1,200.00. Credits: May 15, $600.00. May 20, $1,900.00. Interest at 6%, common time. Focal date, June 1. Note that the 30-day note is due June 1 ; but tnat the interest on it cannot appear on the ledger account, and consequently that the amount appearing in the statement is the face of the note, and its date is May 2. Assume June 1 as focal date. 22 MATHEMATICS FOR THE ACCOUNTANT Due Amount Days Products May 2 $1,200.00 30 30,000 May 21 840.00 11 9,240 June 1 1,200.00 $3,240.00 Total 45,240 Paid Amount Days Products May 15 $600.00 17 10,200 May 20 1,900.00 12 22,800 Total $2,500.00 33,000 Balance of day-numbers is 12,240; one day's interest $2.04. Balance of principal is $740.00; one day's interest 123/^ cents. Quotient 16.7 or 17 days. Again we have both balances on the same side of the account, so we must reckon backward from the focal date. Seventeen days from June 1 is May 15, the equated date. Rule for reckoning from the focal date to the equated date: If balances of both principal and interest are on the same side of the account, reckon backward. This is because, as was explained, the interest on the principal is continually increasing the interest already accrued. But if one balance is debit and the other balance is credit, reckon forward. This is because the interest on a debit principal will tend to extinguish a credit balance of interest, and vice versa. PROBLEMS ON CHAPTER III Find the cash balance of each of these accounts: 11: Jan. 1, 6%, bankers' time: Debits: Oct. 11, $350.00, 30 days; Oct. 22, 30 days, $662.50; Nov. 19, $2,258.00; Dec. 11, 30 days, $350.00. ACCOUNTS CURRENT 23 Credits: Nov. 18, 30-day note without interest, $600.00; Nov. 22, cash, $2,000.00 12: July 1, 5H%, exact time: Debits: May 11, $252.87 at 30 days; May 13, $350.00 at 30 days; May 18, $250.00 at 20 days; June 11, $353.83 at 10 days. Credits: May 15, 30-day note for $500.00, interest at 5%; June 1, cash, $200.00; June 16, cash, $200.00; June 28, cash, $200.00. 13: May 1, 5%, exact time: Debits: Feb. 27, $300.00; March 15, $655.35, 30 days; March 19, $225.75, 30 days; March 24, $225.75; April 3, $625.38, 30 days. Credits: March 1, 60-day note, with interest at 5%, $300.00; April 1, 30-day note, without in- terest, $1,000.00. 14: Jan. 1, 6%, exact time: Debits: Oct. 5, 60 days, $500.00; Oct. 13, 30 days, $500.00; Oct. 18, 30 days, $350.00; Nov. 8, 30 days, $425.00. Credits: Nov. 18, $750.00; Dec. 8, 30-day note, with interest at 63^%, $500.00; Dec. 29, $200.00. Find the equated date of each of these accounts: 15: 6%, bankers' time: Debits: May 17, $350.00; May 29, $255.38; June 19, $125.66; July 5, $264.87. 16: 6%, exact time: Credits: Aug. 15, $325.00; Aug. 19, $2,703.59; Sept. 14, $665.32; Oct. 1, $225.75. 24 MATHEMATICS FOR THE ACCOUNTANT 17: 5%, exact time: Debits: June 25, 30 days, $1,600.00; July 2, 45 days, $800.00; Aug. 14, cash, $500.00. Credits: July 1, 60 days, $1,200.00; July 5, cash, $1,000.00. 18: 5}4%, exact time: Debits: Aug. 14, 30 days, $300.00; Aug. 22, 45 days, $500.00; Sept. 17, 30 days, $225.83; Sept. 29, 30 days, $246.73. Credits: Aug. 20, 30 days, $600.00; Sept. 19, cash, $250.00. CHAPTER IV FOREIGN EXCHANGE The treatment of foreign exchange in this chapter is wholly from the standpoint of arithmetic. The financial and economic phases of the subject are fully treated in many books on the subject as well as incidentally in most books on banking and foreign trade. WTien a sum of money is to be sent from one country to another, it is necessary to change the value from the currency of the sender's country to that of the recipient. This process is the arithmetic of foreign exchange, and the ratio used in making the change is called the rate of exchange. In all the great commercial nations the monetary sys- tem for international trade is based on a gold standard. So theoretically the rate of exchange between any two coun- tries ought to be the ratio between the amounts of gold in their standard monetary units. This theoretical ratio is called par of exchange. For instance, a United States dollar weighs 25.8 grains, and is .9 fine; it therefore con- tains 23.22 grains of pure gold. The English pound sterling is 113.0016 grains of pure gold. Par of exchange between the two countries is therefore the quotient of these num- bers, which is 4.8665. This par of exchange practically never is the actual rate. For the last few years the rate has fluctuated very rapidly and violently, depending on the ravages of the submarines, the varying fortunes of the battlefields all over the earth, the purchases made in this country by England, and many other circumstances. 25 26 MATHEMATICS FOR THE ACCOUNTANT The following table gives par of exchange between the United States and the other more important com- mercial nations: Monet ar}^ Value in Country- Unit U. S. Money Great Britain pound sterling $4.8665 Germany mark .2380 France franc .1930 Belgium franc .1930 Switzerland franc .1930 Italy lira .1930 Greece drachma .1930 Spain peseta .1930 Russia ruble . 5150 Japan yen .4980 Holland guilder .4000 This theoretical par of exchange is almost never the actual rate, as was said on the previous page. The actual rate is dependent on economic conditions, such as the balance of trade, the supply of gold and its location, labor or other disturbances, and so on. As these vary from day to day, and from hour to hour, so the exchange rate varies. Rates of exchange are also dependent on the kind of paper to be bought or sold. That is, on the time between the transaction and the day when the cash can actually be collected. The usual kinds of paper are cables, demand, and time: 30, 60, 90 or 120-day paper. Naturally, the longer the time of the paper, the lower the rate; just as in any discount transaction. In the case of a time draft on London, there are also three days of grace to be con- sidered ; and the discount or interest is reckoned on these days. FOREIGN EXCHANGE 27 In addition to the actual time of the paper, bankers figure "transit." For instance, a 60-days' sight draft on London made on September 15, may not be presented in London until early in October. It may be held in this country several days waiting for a steamer; the voyage to England occupies several more days; and there may be some delay in securing acceptance. During all this time money is tied up in the paper, and the banker who sold it is entitled to his interest. In this chapter, however, transit will be ignored. England has always been the leader in the world's trade, and has controlled the gold market. So London has be- come the center of the world of exchange, and sterling paper has come to be an acceptable medium of exchange in most civilized countries. This means that the great bulk of international settlements are effected in this way. So we will study London exchange first, and the applica- tion of the principles demonstrated to other monetary systems will be obvious. The rate of exchange on London is stated in this country as the dollars and cents value of the pound sterling. It fluctuates by .0005 or fractions or multiples thereof. The smaller English coins are the shilling, 1/20 or .05 of a pound; and the penny, 1/12 of a shilling, which is usually regarded as two cents. The penny is ordinarily indicated by the letter d. Thus, £35 7s. 4d. means 35 pounds, 7 shillings, 4 pence. Problem: Change £43 Us. 7d. to U.S. money, exchange at $4,855: £43 Us. = £43.55 43.55 X 4.855 = $211.44 7 X .02 .14 $211.58 28 MATHEMATICS FOR THE ACCOUNTANT Problem: Change $368.74 to English money, exchange at $4,878: Dividing 368.74 by 4.878 gives £75.592 .592 X 20 = 11.84s. .84 X 12 = lOd. Answer: £75 lis. lOd. The value of a bill on London depends more or less directly on: (a) The current rate of demand exchange; (b) The cost of revenue stamps to be affixed in London ; (c) Interest on the money tied up in the bill; (d) Commission of the London correspondent. Problem: What can a banker afford to pay for a bill on London for £355 16s. 3d., at 60 days' sight, demand rate 4.87 H, interest 4%, stamp 1/20%, commission 1/4%? One method of solution is to reckon the price per unit of currency, in this case per pound sterling: Quotation .A. 875 From which must be deducted Interest for 63 days, exact time 03366 Stamp 00244 Commission 01219 .04829 He can afford to buy the bill at 4 . 82671 times 355.8 and adding .06 $1717.3434 Exporting gold is sometimes necessary ; the gold may be either bars or coin. Such an occasion occurs whenever the exchange market is so deranged that rates are exces- sively high; that is, whenever the balance of trade is seriously disturbed. Problem: Find the market quotation on demand ex- change at which it is advisable to export gold bars to Lon- don if London pays 77s. lO^^d. per ounce for gold 11/12 fine (the English standard). The U. S. Treasury values gold .995 fine at $20.67183 per ounce. FOREIGN EXCHANGE 29 First change 77s. lO^d. to pence = 934Md. This is at 11/12 fine; now convert to the U. S. standard, which is prac- tically 100%, by multiplying by 12/11 = 1019.72727d. And 1019.72727 -^ 240 = £4.248863 the sterUng value of one ounce of pure gold. Then 20.67183 ^ 4.248863 = 4.865263 the dollar value of a pound sterling under those conditions. To this add charges: Freight, about 1/8% 006082 Insurance, 1/20% 0024326 Other charges, 1/20% 0024326 Interest, 6%, 20 days 016217 U. S. bar charge, 40 cents per $1,000 .001946 .0291102 Cost per £ sterling 4.8943732 That is, if demand exchange is selling at more than 4.894^/8 it will probably be cheaper to pay London in gold bars rather than in sterling drafts. If gold coin is exported the problem is much the same except that there is no bar charge, but an allowance of about 1/10% must be made for abrasion. We must also keep in mind that the United States gold dollar is .900 fine. The rate of exchange on France, Belgium, Switzerland, and Italy is quoted by giving the exchange value of one dollar in francs or lire. Thus, when exchange on Paris is quoted at 5.18, a dollar will buy 5.18 francs, or 5 francs 18 centimes. Exchange on Paris fluctuates by 5/8 of a centime, about 1 /8 of a cent. These rates are sometimes further modified by adding or subtracting 1/32% or even 30 MATHEMATICS FOR THE ACCOUNTANT 1/64%. Note that the greater the numerical value of the quotation, the lower the value of French currency. A market change from 5.20 to 5.19^ is a rise; from 5.20 to 5.205^ is a fall. The processes for changing from French to American currency and vice versa are the exact opposites of those demonstrated for English exchange. Problem: What is the cost of a bill on Paris for f 1,500 exchange at 5.97^4? 1,500 divided by 5.9775 = $250.94 Problem : What number of francs can be purchased for $250.00 at the same quotation? 250 multiplied by 5.9775 = f 1,494.375 The rate of exchange on Germany is quoted bj giving the value in cents of four marks. For instance, a quotation on Berlin of .96^ means that each mark is worth 24)^ cents. The following problems will illustrate the pro- cesses: Problem: How many marks can be purchased for $500.00, exchange at 73 1^? 73 J4 divided by 4 = .183125 500 divided by .183125 = M2,730.38 Problem: What is the proceeds of a German bill for M 1,250, at the same quotation? 1,250 X .183125 = $228.91 South American exchange has always been uncertain, and now that it is coming to be based more on the United States gold dollar no separate treatment is necessary. Arbitrage of Exchange is the process of making remit- tances to a country by way of a third country where rates are more favorable. Exchange rates are now so uniform that arbitrage is useful only rarely. FOREIGN EXCHANGE 31 Suppose a banker wishes to remit f25,250 to Paris, ex- change at 5.17 1^. The cost of the draft is then $4,879.23. But if steHing drafts cost 4.84, and London quotes francs at 25.25 per £ sterling, the cost via London is $4,840.00. This is a saving of $39.23. PROBLEMS ON CHAPTER IV 19. A Boston importer buys goods from a Dresden manu- facturer to the value of M2 1,320. Find the cost of a bill of exchange at 95^^. 20. A bill on London for £342 12s. 6d. is offered at 4.86 1^. What is the value in dollars? 21. A bill on Paris for f33,250 cost $6,412.72. What was the rate of exchange? 22. A Liverpool merchant draws on an American importer for £540 10s. 6d. What is the value, exchange at 4.85 3^? 23. What are the proceeds of a documentary bill of ex- change on London for £528 8s. 6d. at 60 days' sight, demand exchange at 4.77 3^^, interest 4%, stamp and commission as usual. Allow 15 days for transit. 24. Tiffany and Co. import an invoice of statuettes from Florence amounting to 14,725.35 lire. They buy a 3-days' sight draft on a Florence banker at 7.3514, discount 3%, stamp 1/20%, commission 1/8%. Find the cost of the draft. 25. J. P. Morgan and Co. remit to London via Paris when the rates are: New York on London, 4.8675; New York on Paris, 5.19^; Paris on London, 25.04. What is the gain on a remittance of £50,000.00? 26. An importer can purchase certain goods in Amsterdam for 15,000 guilders, exchange at 38^. He can pur- chase the same goods in Paris for f35,000, exchange at 5.85. Which is cheaper, and how much in United States currency? 32 MATHEMATICS FOR THE ACCOUNTANT 27. Find the cost of exporting enough gold to settle a debt of £20,000 if gold bars are shipped to London at 77s. 93^d. per ounce. Charges as on preceding page. 28. A banking concern dealing in foreign exchange has the following transactions on its account with its London correspondent: Debits: Sept. 1, Remittance, 30-day bill, £400 @4.86 10, Remittance, sight bill, £200 10s. @ 4.87 15, Remittance, demand bill, £200 6d. @ 4.8675 Credits : Sept. 2, Draft, sight, £300 @ 4.87 ^ 12, Draft, demand, £200 12s. 5d. @ 4.87 20, Cable, demand, £100 @ 4.88 Ascertain the profit or loss on the account for the month of September, and state the balance as of September 30, in both dollars and sterling, the current rate on that date being 4.89. (Mass. C. P. A., 1914.) CHAPTER V POWERS AND ROOTS: LOGARITHMS If a number is to be multiplied by itself a certain number of times, it is possible to indicate the operation in very brief form. For instance, if it is necessary to multiply 5 together four times, the operation may be indicated thus: 5^ This expression is then read, "5 to the fourth power." 5X5X5X5 is 625; and 625 is called the fourth power of 5. In general, 5 is called the base, and 4 is called the exponent, showing how many times 5 is to be multiplied together. Certain operations of compound interest work can be very much shortened by the use of four elementary laws of exponents. Consider the powers of 10. 10 X 1 = 101 = 10 10 X 10 = 10* = 100 10 X 10 X 10 = 10» = 1,000 10 X 10 X 10 X 10 = 10^ = 10,000 10 X 10 X 10 X 10 X 10 = 10^ = 100,000 10 X 10 X 10 X 10 X 10 X 10 = 10« = 1,000,000 We can now demonstrate the four laws of exponents. Multiplication: 10» x 10* = 100 X 1,000 = 100,000, which is 10^ I. Rule: To multiply powers of the same base, add the exponents and keep the base unchanged. Division: 10« ^ 10^ = 1,000,000 ^ 100 =10,000, which is 10^ II. Rule: To find the quotient of powers of the same base, subtract the exponent of the divisor from the ex- ponent of the dividend and leave the base unchanged. 33 34 MATHEMATICS FOR THE ACCOUNTANT Raising to a power: (lO^)^ = 100 X 100 X 100 = 1,000,000, which is 10^ Evidently the exponent of the result could be obtained by multiplying 2X3. III. Rule : To raise a base having an exponent to a given power, multiply the exponents and leave the base un- changed. Finding a root: The reverse of raising to a power is called finding a root; instead of asking what is the result if we multiply 100 together three times, we ask what was the number which was multiplied together three times so that the result was 1,000,000 — in other words, what is the third root of 1,000,000. This question is indicated by the sign v^l, 000,000. The sign is called a radical sign, and the 3 is called an index. The answer to our question is evidently 100, or 10^. Since 1,000,000 is 10«, the result could have been obtained by dividing 6 by 3. IV. Rule: To find a root of a base which has an ex- ponent, divide the exponent by the index and leave the base unchanged. Exponents may be fractional, negative, or zero. Fractional exponents; It is evident that in most cases where a root is to be found, the result will have a fractional exponent. If we have to find the 5th root of the 7th power the result will have the exponent 7/5. In general, the numerator of a fractional exponent indicates a power and the denominator indicates a root. Most of the ex- ponents we deal with are fractional, and for convenience the fractions are reduced to decimals. Negative exponents : It is also evident that in division the exponent of the divisor might be larger than the ex- ponent of the dividend. This would result in a negative exponent. 10* -^ 10^ is a good illustration. 10,000 -^ 1,000,000 = 1/100. Subtracting exponents, we have as POWERS AND ROOTS: LOGARITHMS 35 the exponent of the result — 2. So lO"^ must mean the same thing as 1/100. In general, a negative exponent indicates that the base has become the denominator of a fraction whose numerator is 1. This idea is important in computing present worth. The zero exponent: 10^ ^ 10^ = 10° because 2—2 = 0. Since 100 -^ 100 = 1 it is evident that when the exponent is the base disappears and the value is 1. Logarithms are an ingenious device for using the rules of exponents. If all the numbers, both whole, mixed and fractional, can be expressed as powers of one and the same base, evidently these rules can be applied in the use of such exponents. Since our number system is decimal, it is natural to select 10 as the base of this system of logar- ithms. Only a few of these exponents are whole numbers, because only a few numbers are exact powers of 10. The following are a few: 106 = 1,000,000 nl 6 105 = 100,000 nl 5 10^ = 10,000 nl 4 103 = 1,000 nl 3 102 = 100 nl 2 101 = 10 nl 1 10° = 1 nl 10-1 = .1 nl -1 10-2 = .01 nl -2 and so on. The sign nl is used to separate a number from its logarithm, and is read 'Hhe number whose logarithm is". The opposite sign hi ''the logarithm of the number" is used in changing from logarithm to number, or anti- logarithm, as it is called. 36 MATHEMATICS FOR THE ACCOUNTANT In the table of powers of 10 note that the logarithm of each of the positive powers is one less than the number of figures at the left of the decimal point, while in the nega- tive powers the numerical value of the logarithm is one more than the number of zeros between the point and the first significant digit (a significant digit is any of the digits from 1 to 9). So far we have spoken only of integral powers: powers whose exponents, or logarithms, are whole numbers. But between 10 and 100, for example, there are 89 whole numbers and an infinite number of mixed numbers. These must all be expressed as powers of 10, and all these powers must lie between 1, the exponent of 10, and 2, the exponent of 100. Evidently these exponents must all be 1 + a decimal. The logarithm of 84, for example, is composed of two parts; a whole number 1, to indicate that it is more than 10, and a decimal to indicate the part of the interval from 10 to 100. The whole number, which indicates the power of 10 just below, is called the characteristic; the decimal is called the mantissa, or "modifier". Tables of logarithms never give character- istics, because they can be determined by inspection, keeping in mind the table of powers of 10. Rule for determining the characteristic: If the number has digits to the left of the decimal point, the characteristic is 1 less than the number of such digits. If the number is a decimal fraction, with no digits to the left of the point, the characteristic is negative, and is 1 more than the number of zeros between the point and the first significant digit. For instance, the characteristic of 317.8895432 is 2. The characteristic of .000895 is — 4. The characteristic of 7.65 is 0. This last perplexes some students, and it is POWERS AND ROOTS: LOGARITHMS 37 important that it be thoroughly understood ; it is a direct appHcation of the first part of the rule. Logarithmic tables, then, are wholly tables of mantissae, each one corresponding to a certain series of digits. There are no decimal points, because the characteristic enables us to locate them. It follows that any series of digits always has the same mantissa. In the tables, for instance, the mantissa corresponding to 8427 is 925673. That is S427 nl 3.925673 842.7 nl 2.925673 84.27 nl 1.925673 8.427 nl 0.925673 .8427 nl 1.925673 and so on. This last logarithm is also written 1.925673, and 9.925673—10. This last method will be followed in this book, as it simplifies certain processes which are quite common in annuity work. The logarithm of .000895 on the preceding page would be written 6.951823—10. As was stated in the first chapter, logarithms should run to at least six places, and seven are better. The first two digits of the mantissa change slowly, and are indi- cated only when they change. This is an important fact to remember in using tables, because the first two digits are apt to be several lines above the others, and must be found and used in order to get correct results. Finding a logarithm in the tables is a simple matter. The digits of the antilogarithm are in a column at the left of the page and across the top. To find the logarithm or rather the mantissa of 357.4 turn to the page or pages containing the 3,000 series, and follow down the left margin until 357 is reached. Across the top are the 38 MATHEMATICS FOR THE ACCOUNTANT digits from to 9, and Id the 4 column, opposite 357 we read 3,155. Remembering that the first two digits must be taken also, we read just above 55. The mantissa is then 553,155. Now determine the characteristic, which in this case is 2, and the entire logarithm is 2.553155. If there are less than four digits in the antilogarithm, affix zeros. For the mantissa of 37 look for 3,700; the logarithm is 1.568202. Usually the problem is to find the mantissa of a number having more than four digits. This is known as inter- polation. To find the logarithm of 231 .24. The character- istic is 2. The mantissa is between the mantissae of 2,312 and 2,313, and evidently is 4/10 of the interval. 2,313 nl 364,176 2,312 nl 363,988 difference 188 4/10 of 188 is 75.2, which, added to 363,988, gives 3,640,632. Prefixing the characteristic we have 231.24 nl 2.3640632. The cologarithm of a number is the logarithm of its reciprocal. We met a leciprocal is discussing negative exponents, and remarked in passing that the com- monest occurrence is in finding present worths. A recip- rocal is 1 divided by a number. Now the logarithm of 1 is 0, as was explained on page 21. To find the cologarithm of 387.2, or the logarithm of 1 divided by 387.2: 1 nl 10.000000—10 387.2 nl 2.587935 Subtracting 7.412065—10 One other very common use for cologarithms is in exam- ples where multiplication and division must be performed POWERS AND ROOTS: LOGARITHMS 39 together. The division can be performed by adding the cologarithm of the divisor, as will be demonstrated presently. Finding the antilogarithm is somewhat more difficult, because it almost always requires interpolation. To find the number whose logarithm is 2.7851. The characteristic 2 indicates that there are three digits at the left of the decimal point. Looking in the tables for mantissae nearest 7,851 we find: 785,116 In 6,097 7851 the given mantissa 785,045 In 6,096 The difference between the logarithms in the table is 71; the given logarithm is 55 more than 785,045. The required antilogarithm is therefore 55/71 of the interval from 6,096 to 6,097. 55/71 of 1 is .774; affixing this to 6,096 we have 6,096,774, and placing the decimal point after the third digit, we have the antilogarithm 609.6774. The following problems will illustrate all the logarithmic processes which are necessary for this book. Multiplication: 387.2 X .04752: 387.2 nl 2.587935 .04752 nl 8.676876—10 Adding 11.264811-10 or 1.264811 In 18.3997 It is very commonly necessary to add 10 — 10 or multiples of it, or to subtract them. Since 10—10 is we evidently do not change the value of any expression by this device. The addition of the logarithms is in accordance with the first law of exponents, in multiplication, page 33. 40 MATHEMATICS FOR THE ACCOUNTANT Division; 69.45 -^ 3.894 69.45 3.894 Subtracting In nl nl 1.841672 .590396 1.251276 17.835123 The subtraction is in accordance with the second law of exponents, in division. Multiplication and division in the same example: 68.39 X 31,750 243.6 68.39 nl 1.834993 31750 nl 4.501744 243.6 n — colog 7.613323—10 Adding 13.950060—10 or 3.950060 In 8913.7347 Raising to a power: a) (2.798)'* 2.798 nl .446848 times 8 3.574784 In 3756.504 b) to illustrate treatment of a negat ive characteristic (.0693)5 .0693 nl 8.840733—10 times 5 44.203665—50 or 4.203665—10 In .0000015983 Multiply by the exponent; see rule III - Finding a root: a) W968.4 968.4 nl 2.980055 divided by 7 .4265793 In 2.67042 POWERS AND ROOTS: LOGARITHMS 41 b) to illustrate treatment of a negative characteristic >J/.08355 .08355 nl after we divide by 4 the charac- teristic must be — 10; so before division it must be — 40 ; so it is necessary to add 30 — 30, mak- ing the log now divide by 4 In 8.921946-10 38.921946^0 9.7304865-10 .537634 Divide by the index as in rule IV, page 34. Combinations of these processes are rarely necessary, a) (3.066)11 ^12.74 3.066 nl .486572 times 11 5.352292 12.74 nl 1.105169 divided by 5 .2210338 Subtracting 5.1312582 In 153287.6 4.925 X >J/16740 (38.25)^ 4.924 nl .692406 16740 nl 4.223755 divided by 3 1.407918 38.25 nl 1.582631 times 4 6.339524 divide by adding ( jolog 3.669476—10 Adding 5.769800-10 hi .0000588573 42 MATHEMATICS FOR THE ACCOUNTANT PROBLEMS ON CHAPTER V A. Problems to give facility in use of tables: 29. Find the logs of 432.5 4.325 .004325 30. Find the logs of 5700 5.7 570000 31. Find the logs of 78.9342 05.4386 7762.47 32. Find the cologs of 13.97 .04953 620000 33. Find the cologs of 6.91843 100.764 .000015 34. Find the antilogs of 3.270679 1.284656 8.817962—10 35. Find the antilogs of 1.316274 2.967777 4.830034—10 36. Find the antilogs of .00317255 6.779999—10 3.84506234 B. Perform the following by the use of logs: 37. Multiplication: 378.2 X 56.43 .06925 X 34700 616500 X 3.9483 X .00745 38. 39. Division : 930.07 - - 1.05 .58023 - - 7.7438 58.023 - - .077438 8.9094 - - .89094 Multiplicatic >n and division: 9.1932 > : .5307 .0076 6800000 X .0038902 1.025 40. Raising to a power; (3.4684) « (1.01225)14 41. Raising to a power: (.6307)9 (.1213)6 POWERS AND ROOTS: LOGARITHMS 43 42. Finding a root: >^810.5 ^1.0575 43. Finding a root: V.0069183 ^.0003301 Miscellaneous : 44. 4.6173 X (2.837)^ .10765 X 48.27 45. /3 X 7.5 y (1.015)4 46. Ay392.7 X 61.3455 X 33.9 (1.0175)« ~ 47. r2.94635 X ^671 e L 1.023 48. ^300.75 X .000045 (.8512)8 CHAPTER VI COMPOUND INTEREST AND PRESENT WORTH At this point we enter on the study of what the Ameri- can Institute of Accountants calls Elementary Actuarial Science. An actuary is the mathematical expert of a life insurance company. It is his business to calculate the income and expenditures of the company and see that the company is on a safe financial basis: to see that it has sufficient funds on hand to meet all ordinary calls and an additional amount to meet emergencies. These expendi- tures are concerned mostly with payments on account of policies, either life or endowment. Such expenditures therefore can be estimated pretty closely if there are sufficient data regarding the number of persons who will live or die during the year. In other words, the actuary is concerned with the probabilities, as they are called. By the study of many thousand lives actuaries have been enabled to formulate laws stating that out of a certain number of persons alive on a given date a stated propor- tion will live through an entire year and the rest will die within the year. These contingencies, so called, affecting sometimes a single life and sometimes a group of lives, make the mathematics of the actuary exceedingly diffi- cult. Moreover, these calculations must be based on a long period of years. The mortality tables for straight life policies run to the ninety-sixth year. Most endowment insurance is for periods of ten to twenty-five years. Under these circumstances all calculations must be on a com- pound interest basis. The series of payments on an en- 44 COMPOUND INTEREST AND PRESENT WORTH 45 dowment insurance policy are invested by the company and earn interest. It is reasonable that the income on the payments of each policyholder should be credited to his policy. The mathematics which have been outlined in the last two paragraphs are Actuarial Science. The effect of probability on annuities means nothing to the accountant, however, because he assumes that the financial arrange- ments which he supervises will be permanent in most cases. To describe this situation, mathematicians speak of annuities certain; annuities which will not be affected by the death or mishap of any person or persons, but will continue to the limit of the time assigned them. This is Elementary Actuarial Science. A very common way of referring to compound interest is "interest on interest". This definition is not true, and is misleading. It is misleading because it implies usury. No such taint inheres in compound interest; every dollar of capital is always at work, and inasmuch as interest rates are stated for definite periods, a year, six months, three months, at the end of the period the interest which has been accruing becomes due and is no longer interest, but an addition to capital. So again it is not interest on interest, but interest on a periodically increasing capital. The chief argument for compound interest is this. If a sum of money is invested, the interest is supposed to be paid when due. It is loaned fcr the sake of this in- come. If the interest is not paid promptly , part of the income is lost and the interest rate is lowered. But this interest is earning money for the party with whom the money is invested. So he and not the rightful owner is reaping the benefit of the investment. What is the significance for the accountant? In the first place, many of the transactions of the business world are 46 MATHEMATICS FOR THE ACCOUNTANT already on a compound interest basis. This applies es- pecially to investment accounting; to bond accounts and businesses, savings banks, insurance companies, and even the interest on checking accounts in commercial banks. And many accountants in supervising bond issues of public and private corporations say that they build the sinking fund schedules on a compound interest basis whenever the accounting staff of the concern are capable of handling schedules built in that way. Evidently, if an accountant is to audit the accounts of such a concern he must know his annuities thoroughly. Again, as has been said already, every business has its fiscal period, and at the end of that period, the net earnings are transferred to capital in any one of a number of ways. If the investment in the business is already adequate, any overplus of capital will be invested in other businesses or in securities, or at least in some banking house which offers a fair rate of interest. In some way this money will earn an income at not less than the savings bank rate. Usually an additional investment in the business is de- sirable, which may take the form of a direct increase in surplus, or may be apportioned among the various re- serves, for sinking funds, depreciation, or so on. Such additional investments of course earn an additional in- come, just as the previous investment did. If the business man expected an 8% return on his original investment he will hope for an 8% return on the increase. And it seems reasonable that such income should be credited to the surplus or reserve on which it was earned. Because if such reserves were represented by funds in the hands of a trustee, they would earn an income at not less than the savings bank rate, and this income would not be trans- ferred to the free surplus of the company, but would be added to the fund in the hands of the trustee and re- invested by him. COMPOUND INTEREST AND PRESENT WORTH 47 These are the arguments for compound interest. They are the arguments from reason and consistency and from the best custom. Failure to recognize them causes mis- statement of income and failure to allocate it properly between the various parts of free or appropriated capital on which it was earned. Compound interest is actually at work whether realized or unrealized; acknowledgment of it as a fact is the duty of the progressive accountant. Computation of compound amount : We have been dis- cussing compound interest. As a matter of fact we usually find the compound amount first, and then deduct the principal, the result being the interest. The reason for this is that there is a very simple process for finding the compound amount of $1.00 for any time at any rate; and since all dollars in a given transaction increase at the same rate, the compound amount of any principal is merely the compound amount of $1.00 multiplied by the given principal. We have perhaps been accustomed to find compound amount by a laborious process, thus: To find the compound amount of $1.00 at 5% for five years : 1st year $1.00 add 5% .05 2nd 3'ear, principal 1.05 add 5% .0525 3rd year, principal 1.1025 add 5% .055125 4th year, principal 1.157625 add 5% .05788125 5th year, principal 1.21550625 add 5% .06077531 Compound amount $1.27628156 48 MATHEMATICS FOR THE ACCOUNTANT Of course no savings bank or savings department of a commercial bank carries its interest to eight places of decimals. But this must be done in investment mathe- matics, where the items which are being accumulated may run into milUons of dollars. A much simpler way to solve the problem is given by an elementary principle of algebra. The principal is 1 ; the rate of interest, hereafter called i, is .05. The amount at the end of one period is, therefore, 1.05. We are compounding annually, so the mmaber of conversion in- tervals, as they are called (the number of times that the interest becomes an addition to principal) is 5; this time element is indicated by n. Now if we multiply 1.05 together 5 times, or, as the mathematicians say, raise it to the fifth power, the product mil be $1.27628156, the very same result as we obtained by the other method. In general, to find the compound amount of 1 for n periods at rate i, raise 1 + i to the n-th power. Expressed as a formula this is (1 + i)°. This is indicated conmionly by the symbol s"^. It is important to remember that n is the number of conversion intervals, not the number of years. For in- stance, if a savings bank states that its rate is 4%, it usually means 2% every six months. A dollar left in this bank five years will amount to the tenth power of 1.02. For ordinary times and rates there are good tables of compound amount. Problem: To find the compound amount of $150.00 for 17 years at 5% compounding semiannually. The rate is 23^% and the time is 34 periods. In the 2^/2% column of the tables opposite 34 of the time column read, 2.31532213. Multiplying by 150 gives the compound amount $347.2983. The compound interest is $197.2983. COMPOUND INTEREST AND PRESENT WORTH 49 For unusual rates, which often occur, logarithms are necessary. And they should be at least ten-place logar- ithms, as was stated in the first chapter. Problem: Find the compound amount of $246.75 for 12 years at 2^% compounding annually. i is .02375 1 + i is 1.02375 n is 12. We have to find (1.02375)'2 1.02375 nl .0101939148 times 12 .1223269776 In 1.325338 tunes 246.75 = $327,027 Nominal and effective rates: At the beginning of the chapter on interest one element of the interest problem mentioned was "frequency", the length of the "con- version interval". For instance, savings banks and de- partments state a yearly rate, but compound semiannually or even monthly. Bonds usually pay semiannual interest, and this is considered when the price is quoted. These are only two common instances out of many. When a bank advertises compound interest at 4%, convertible semiannually, it usually means to pay 2% every six months; the actual yearly rate is 4.04%. In this case 4% is called the nominal rate, indicated by j, and 4.04% is called the effective rate, indicated by i. If the com- pound amount of $1.00 for five years at 4% nominal, convertible semiannually, is desired, i is .02, 1 + i is 1.02, and T) is 10. In the table of s ° in the 2% column opposite 10 read 1.21899442. Similarly, the compound amount of $1.00 for 15 years at 6% nominal, convertible quarterly, is Ihe compound amount for 60 periods at 1^%, or 2.44321978. Such problems usually lead to unusual rates and require the use of logarithms in solution. 50 MATHEMATICS FOR THE ACCOUNTANT Evidently, given a nominal rate, the shorter the con- \'ersion interval, the higher the effective rate. At 6% compounding yearly (1 + i) ° is 1.06 " semi-annually 1.0609 « quarterly 1.061364 " monthly 1.061678 daily 1.061826 There is a limit to this increase, however. For the nominal rate 6%, the highest possible effective rate, for instan- taneous compounding, is slightly less than 6.184%. This idea of instantaneous compounding, or continuous com- pounding, is treated at length by the books on Actuarial Science. The rate 6.184% is called "force of interest". It has no significance for the accountant. To change from nominal to effective rates: [i + i] L mJ The formula is i = 1 1 + - 1 —1 m- where i is the effective rate, j is the nominal rate, and m is the number of conversion intervals in one year. Rule: Find the compound amount of the periodic rate for m periods and subtract 1. Problem: What is the effective rate when the yearly nominal rate is 3%, and compoundings are semiannual? j is .03 m is 2 J- is .015 m .1 -h ^ is 1.015 m Then the effective rate is (1.015)2—1, or .030225. Stated as a percent it is 3.0225%. COMPOUND INTEREST AND PRESENT WORTH ol Nominal rate when the effective rate is given: Insurance companies and some other concerns which have a great variety of investments having varying interest dates and periods find it advisable to annuahze all interest rates. This leads to a situation which is rather difficult to grasp at first. They assume a yearly rate of income, say 6%. Now if a certain investment pays semiannual interest and is to be valued at 6% annually, it is evident that 6% is the effective rate. What is the semiannual rate? Not 3%, for that would yield an annual rate of 6.09%, as was stated on the previous page. Referring to the discussion of roots on page 20, the rate which after two conversions yields 1.06 must be the square root (or second root) of 1.06. If (1 + i)2 = 1.06, then 1 + i = >yi.06. This solution of such a problem is logarithmic and based on the fourth rule of exponents. 1.06 nl .0253058653 dividing by 2 .01265293265 In (10-place table) 1.029563 or 2.9563% per period that is 5.9126% is the nominal rate. Such a nominai rate is indicated by the symbol j p, where p indicates the number of compoundings per year. A table of the values of j p for the usual rates and for two, four and twelve compoundings yearly is given in all good collections of tables. To illustrate a little differently, the value of j 4 for 5% is given in that table as .0490889. This is a nominal rate, which means that the periodic rate is 1/4 of the annual rate. Now 1/4 of .0490889 is .0122722. Then, according to the definition of a periodic rate given, 1.0122722 raised to the fourth power ought to yield 1.05. By actual multi- plication we find the rate to be 1.0500025. The error is of course due to rounding up whenever digits were discarded. 52 MATHEMATICS FOR THE ACCOUNTANT Formula for finding the nominal rate when the effective rate is given is j P = p (^1+i - 1) Rule: Find the p-th root of 1 + i and subtract 1 ; this is the periodic interest rate; multiply by p to find the annual nominal rate. To sum up, rates are always given on a yearly basis. If compoundings are other than yearly, there may be two ways of stating the rate. Either the rate given is a nominal rate, in which case the periodic rate is the n-th part of the given rate. Or the rate may be effective, in which case the periodic rate is the n-th root of the given rate. Present worth and compound discount: If a sum of money is due at some future date, it can be discounted to the present time by deducting interest from the present to the due date. This is a common process on short time notes, in which case the interest is simple interest and the process a very simple one. But in the case of long time transactions, where compound interest must be reckoned, the process is somewhat different. There is an old prin- ciple of arithmetic that if the product of two numbers is known and also one of the factors, the other factor can be found by division. Since compound interest is found by multiplication, evidently present worth on a com- pound interest basis is found by division. And since all our compound amounts are based on 1, and expressed by the formula (1 + i) °, evidently present worth of 1 is 1 divided by (1 + i) °, or expressed as a formula, present worth of 1 == (1 + i) " Rule: Divide 1 by the compound amount of 1 for the given time and rate. COMPOUND INTEREST AND PRESENT WORTH 53 Present worth is indicated by the symbol v °. Tables for the usual rates and times are given in all good collec- tions of tables. Problem: Find the present worth of S125.00 due in 12 years on a 6% nominal basis, convertible semiannually. 1 is 3% n is 24 In the table of v °, 3 % column, opposite 24 read .49193374 times 125 $61.4917. Problem: (to illustrate use of logarithms when the rate is unusual): Find the present worth of 1 due in two years at 3M% nominal, convertible semiannually, i is .01625 n is 4 1.01625 nl .0070005586 times 4 .0280022344 colog 9.9719977656—10 In .937557 Note that since the (1 + i) ° is in the denominator, the value of V " always requires finding of a cologarithm. SUMMARY OF THE PROCESSES USED IN COM- POUND INTEREST COMPUTATION Compound amount: s ° = (1 + i) °. Problem: Find the compound amount of $1.00 at 4 1^2% nominal, convertible semiannually, for 20 years. 1 = 2M% n = 40 1.0225 nl .0096633167 times 40 .3865332668 In 2.435189 54 MATHEMATICS FOR THE ACCOUNTANT Present worth: v ° = — - (1 + i) " Problem: Find the present worth of 11.00 due in 22 years at 33^% nominal, convertible semiannually. 1 = i'AVo n = 44 1.0175 nl .0075344179 times 44 .3315143876 colog In 9.6684856124-10 .466107 Present worth of an interest bearing debt: Given a debt due in n periods with interest at rate k per period. To discount it to the present time at rate i per period. / There are two cases, one when the conversion intervals at rates k and i are the same, and the other case when the conversion intervals differ. Let m be the number of periods in the term of discount. The formula is v = — (l + i)- Rule: Find the compound amount of 1 at rate k, and divide by the compound amount (or multiply by the present worth) of 1 at rate i. Problem: A debt of 1 bears interest at 5% annually, and is due in 12 years. What is the present worth at 6% annually. (1.05)12 = 1.79585633 -^^ = .49696936 (1.06)12 product .8924856 Problem (to illustrate differing conversion intervals) : Find the present worth of $155.63 due in 5 years with COMPOUND INTEREST AND PRESENT WORTH 55 interest at 5% nominal, convertible semiannually, dis- counted at 6% annually: (1.0125) 1 (1.06)^ product 1.28008454 .74725817 .9565536 times 155.63 = $148.86843 To change from nominal rate to effective: The formula is i = (1 + — ) m -1 Problem : WTiat is the effective rate when the stated rate is 6%, compounded monthly. J = 6% m = ± m 1.005 times 12 In subtract 1 or 12 .005 nl .0021660618 .0259927416 1.0616774 .0616774 6.16774% To change from effective to nominal: The formula is jp = p (1/1 + i— 1) Problem: What is the nominal rate equivalent to an effective rate of 434%, i^ there are four conversion inter- vals in the year. 1.0425 nl divided by 4 In (10-place table) subtract 1 or times 4 .0180760636 .0045190159 1.0104597 .0104597 1.04597% periodic rate 4.18388% yearly 56 MATHEMATICS FOR THE ACCOUNTANT To find the rate, given principal, amount, and time: The formula is i = „/ - —I Rule: Divide the amount by the principal; find n-th root of the quotient; subtract 1. Note that usually the easiest way to divide S by P is to find the difference of the logarithms. Problem: At what annual rate will $1.00 amount to $2.00 in 27 years. S divided by P = 2. 2 nl V.30103 divide by 27 .0111492555 In (10-place tables) 1.026045 subtract 1 and the rate is 2.6045% Problem: At what nominal rate, compounding quar- terly, will $3.00 amount to $5.00 in 10 years. .69897 .477121 .221849 .005546225 1.012853 1.2853% periodic rate 5.1412% annually. To find the time, given principal, amount, and rate: , . log S— log P The formula is n = -; — , ., log (1 + i) Rule: From log of S subtract log of P; divide by log of 1 + i. -• Problem: In what time will $2.00 amount to $3.00 at 5% annually? 3 nl .477121 2 nl .301030 5 nl 3 nl subtract divide by 40 In or times 4 COMPOUND INTEREST AND PRESENT WORTH S7 subtract .176091 1 +i = 1.05 nl .021189 quotient 8.31 years. The quotient should not be changed to months and days, because compound interest is not assumed for parts of periods. In this type of problem, and whenever the logarithm of 1 + i is used as a divisor, the 6-place man- tissa is sufficient; if the 10-place mantissa were used it would be cumbersome and we would begin to cut off digits at once. Note also that although one logarithm is divided by another, the quotient is not a logarithm. This situa- tion always occurs in time-problems. When the rate is nominal, it is necessary to proceed as follows : Problem: In what time will $1.00 amount to $2.00 at 5% nominal, convertible quarterly? periodic rate i is .0125 2 nl .301030 1 nl .000000 subtract .30103 1.0125 nl .005395 quotient 55.8 periods or 13.95 years Compare this w ith the approximation commonly used, n = .693 i + .35 .693 in this case \- .35 = 55.79 periods. .0125 58 MATHEMATICS FOR THE ACCOUNTANT PROBLEMS ON CHAPTER VI A. Simple problems to be solved by use of the tables and multiplication or division : 49. What is the compound amount of $1.00 for 15 years at 6% annually? 50. What is the compound amount of $1.00 for 15 years at 6% nominal, convertible semiannually? 51. What is the compound amount of $383.97 for W}4 years at 5% nominal, convertible quarterly? 52. What is the compound amount of $1,500.00 for 22 years at 6% nominal, convertible quarterly. (Note: If the table does not include values for 88 periods, they can be obtained by multiplying the value for 85 periods by the value for 3 periods. See the first law of exponents.) 53. What is the compound amount of $2,785.94 for 20 years at 3% nominal, convertible semiannually? 54. What is the effective rate equivalent to 3%% nom- inal, convertible quarterly? 55. What is the present worth of $1.00 due in 17 years at 6% annually. 56. What is the present worth of $1.00 due in 17 years at 5% nominal, convertible quarterly? 57. What is the present worth of $1,934.17 due in 11 years at 6% nominal, convertible semiannually? B. More difficult problems, usually requiring the use of logarithms in solution : 58. What is the compound amount of $1.00 for 11 years at 4:}4% nominal, convertible quarterly? 59. What is the compound amount of $1.00 for 16 years at 6% nominal, convertible monthly? 60. What is the compound amount of $1.00 for 11 years at 4^% nominal, convertible semiannually? COMPOUND INTEREST AND PRESENT WORTH 59 61. What is the effective rate equivalent to 3^% nom- inal, convertible monthly? 62. What is the effective rate equivalent to 7^% nom- inal, convertible quarterly? 63. What nominal rate, compounded quarterly, is equiv- alent to 6% effective? 64. What nominal rate, compounded monthly, is equiv- alent to 534% effective? 65. What is the present worth of $1.00 due in 11 years, ^t 434% nominal, convertible semiannually? 66. What is the present worth of $1.00 due in 8 years at 3.9% nominal, convertible monthly? 67. At what annual rate will $15,000 double itself in 25 years? 68. At what nominal rate, convertible quarterly, will $3,000.00 amount to $5,000.00 in 5 years? 69. The British government sold non-interest bearing certificates for 15s. 6d., redeemable after 5 years at £1. What is the equivalent savings bank rate, nominal, convertible semiannually. 70. Benjamin Franklin bequeathed $5,000.00 to the City of Boston. After 104 years it had amounted to $431,000.00. What is the equivalent rate, nominal, convertible quarterly? 71. In what time will $1.00 double itself at 6% annually? Solve by logs and check the solution by the approx- imation formula. 72. $5,000.00 is invested at 5% nominal, convertible semi- annually. In what time will it amount to $7,500.00? CHAPTER VII ANNUITIES: AMOUNT AND PRESENT WORTH An annuity is a series of equal payments at equal intervals of time, accumulated at a fixed rate of interest. The amount of the periodic payment is always stated on a yearly basis. A good example of this is the in- terest on a bond. The bond is advertised as a 5% bond; but the issuing or selling concern means that the bond will pay 2y-f/o every six months. Another important fact to remember is that payments occur at the end of the vear or other period. This is reasonable, as will appear if we consider the source of the money with which the payments are made. A corporation pays the interest on its bonds out of income. This income is reckoned at the end of the fiscal period. A person paying interest on a mortgage does so at the end of the interest period. There are numerous other instances which will occur to any accountant. The most common instances of payments at the beginning of the period are premiums on insurance policies and some rental payments. Such an annuity is called an annuity due and calls for special treatment. There is also the deferred annuity, in which no payments are made until after a specified time. These types will be taken up after the ordinary annuity has been studied. The amount of an ordinary annuity is indicated by s n. This must always be carefully distinguished from s ", which is the symbol for compound amount. Problem : Find the amount of an annuity of five annual payments of $1.00 each, accumulated at 4% annually. CO ANNUITIES: AMOUNT AND PRESENT WORTH 61 First we will establish the value by accumulating the payments singly. Then w^e will learn a much simpler way of arriving at the same result. First payment, accumulated 4 periods, $1.16985856 Second " « 3 '^ 1.12486400 Third « « 2 « 1.08160000 Fourth « « 1 « 1.04000000 Fifth « , cash 1.00 Total $5.41632256 The better way is this: The compound interest on $1.00 for 5 years at 4% annually is .2166529. This amount divided by the single interest .04 gives the same result as above, $5.4163225. Remembering that s ° oj (1 + i) ° is the symbol for com- pound amount, it is evident that compound interest is s ° — 1, or (1 + i) "" — 1. So we have the Formula for the amount of an annuity: s°— 1 (l+i)°— 1 S n — — i 1 Rule : Divide the compound interest by the interest for a single period. Tables of Sn for the usual rates and times are found in all good collections of tables. In accounting practise, whenever an annuity occurs, a schedule should be set up at once covering the entire life of the annuity. This schedule serves first as a check on the accuracy of the computation, and second as the source of the periodic Journal entries necessitated by the interest or other elements of the annuity. Problem : To prepare a schedule of a series of 5 annual ■payments of $1,000.00 each, improved at 4% annually. We have just seen that the amount of such an annuity Interest Year 4% 1 2 $ 40.00 3' 81.60 4 124.864 5 169.8586 Totals 416.3226 62 MATHEMATICS FOR THE ACCOUNTANT of $1.00 is $5.4163226. For $1,000.00 the amount is evidently $5,146.3226, as is proved by the following: Schedule of Accumulation of Annuity of Five Annual Payments of $1,000.00 at 4% Annually. Total Annual Addition Amount Payment to Fund in Fund $1,000.00 $1,000.00 $1,000.00 1,000.00 1,040.00 2,040.00 1,000.00 1,081.60 3,121.60 1,000.00 1,124.864 4,246.464 1,000.00 1,169.8586 5,416.3226 5,000.00 5,416.3226 Such a schedule should always be footed down as a proof of the correctness of the additions. Total Interest plus total Payments must equal the footing of Addition to Fund, and also the last item in the Amount in Fund column. Also, each item in Total Addition to Fund multiplied by 1 + i must give the next item. If the schedule is lengthy it is a wise precaution to prove every five or at most ten years. The first method of proof is by testing footings as has just been explained. A better method is to find the amount which should be in the fund at the time. The proper value of Sn multiplied by the periodic payment must equal the amount in the fund at the time. Annuity problems often require logarithmic solution. For this purpose the second formula given is the most easily understood. Problem: Find the amount of an annuity of $1.00 a year for 20 years at 4M% annually. ANNUITIES: AMOUNT AND PRESENT WORTH 63 The compound amount must first be found. 1 + i = 1.0425 nl .0180760636 times 20 .361521272 In = s° = 2.2989062 subtract 1 1.2989062 compound interest divide by single interest .0425 gives $30.5625 Effective rates: So far all our calculations have been based on annual or nominal rates. Sometimes, however, the effective rate is stated, but there are several payments a year, and there are as many conversion intervals as payments. There are two methods of attacking this problem. The first method is to reckon the value^of the periodic payment so that nominal rates can be used. Suppose an annuity of $1.00 at 4% effective, payments to be made quarterly. If we assume that the quarterly rate is 1%, the payment will be slightly less than 25 cents, because four payments of 25 cents improved at 1 % would amount to more than the specified $1.00 at 4%. Tables of the value of such periodic payments are sometimes given. Another method involves the use of j p. Suppose an annuity of $1,000.00 for three years, payments semi- annual, at 3% effective. If we assume that the paj'ment is $500.00, we must accumulate at such a rate as is equiva- lent to 3% annually. The table of jp for 3% semi- annually gives .0297783; that is, the semiannual rate is 1.4889%. It is worth while to show the proof schedule in full. This schedule will be presented in columnar form : End of first period, payment $500.00 * second " interest 7.4445 payment 500.00 Total $1,007.4445 Gl MATHEMATICS FOR THE ACCOUNTANT « third " interest 14.9998 payment 500.00 Total $1,522.4443 « fourth " interest 22.6676 payment 500.00 Total $2,045.1120 « fifth " interest 30.4495 payment 500.00 Total $2,575.5615 " sixth " interest 38.3475 payment 500.00 Total $3,113.9090 Whenever tables can be used, the value found in the tables should be multiphed by the quantity -;-' The effect of this is to cancel i out of the denominator of the formula for s n and put j p in its place. A table of the values of — is given in all good collections of tables, jp In the problem for which the schedule appears on the preceding page, the table value for a 3-year annuity of 1 at 3% annually is 3.0909. The value of — when i is Jp 3% and p is 2 is 1.0074446. The product of 3.0909 X 1.0074446 X 1000 is 3113.91, as before. Whenever logarithms are used, the compound interest should be divided by j p instead of i. In such cases both compound amount and j p have to be found by use of logarithms. Problem: Find the amount of an annuity of $1.00 foi' 5 years at 33^% effective, the annuity being payable semi- annually. ANNUITIES: AMOUNT AND PRESENT WORTH 66 First we will find the compound interest: 1.0325 nl .0138900603 times 5 .0694503015 In = s° 1.1734116 less 1 .1734116 Next we will find jpi 1.0325 nl .0138900603 divided by 2 .00694503015 In 1.01612 less 1 .01612 times 2 = j p .03224 The quotient is $5.3787 Formulae for use of j p in finding the amount of an annuity : With tables : s n times — With logarithms: Jp (1 + i) "-1 Jp In both cases n is the number of years. The present worth of an ordinary annuity is indicated by the symbol an. We will demonstrate the formula for finding the present worth just as we demonstrated the amount. Problem: Find the present worth of an annuity of $1.00 for 5 years at 4% annually. First payment, discounted 1 year .96153846 Second « « 2 « .92455621 Thu-d « " 3 « .88899636 Fourth « « 4 « .85480419 Fifth « « 5 « .821927 11 Total $4.45182233 66 MATHEMATICS FOR THE ACCOUNTANT A quicker way is this: The present worth of $1.00 due in 5 years at 4% annually is .82102711; therefore the com- pound discount is .17807289. This, divided by the single interest ,04, gives 4.45182225 as above, allowing for in- accuracies in the right-hand digits. Remembering that v ° is the symbol for present worth, 1 — V ° is the symbol for compound discount. So we have the formula for present worth of an annuity: 1 — V ° an = — : 1 For logarithmic computation this formula may be better stated 1 1 (1 -f i) -^ 1 Rule : Divide the compound discount for the given rate and time by the single interest. Tables of a n for the usual rates and times are found in all good collections of tables. The schedule in tabular form is necessary whenever such a situation occurs. For the annuity of $1,000.00, 5 years at 4% annually, the value of a n is $4,451.8225. Schedule of Amortization of a Debt of $4,451.8225 Payable in Annual Instalments of $1,000.00, with Interest on Outstandings at 4% Annually Annual Net Balance Interest Instal- Amortiza- Outstanding Year 4% ment tion $4,451.8225 1 $178.0729 $1,000.00 $821.9271 3,629.8954 2 145.1958 1,000.00 854.8042 2,775.0912 3 111.0036 1,000.00 888.9964 1,886.0948 4 75.4437 1,000.00 924.5563 961.5385 5 38.4615 1,000.00 961.5385 Totals $548.1775 $5,000.00 $4,451.8225 ANNUITIES: AMOUNT AND PRESENT WORTH 67 The usual checks should be observed. Total Interest deducted from total Annual Payments leaves total Net Amortization, which is the same as the amount of the debt at the beginning. If the schedule runs to twenty years or more it should be tested every five or ten years, by computing a n as of that date. Also, each item in Net Amortization multiplied by 1 + i gives the next item. The logarithmic solution is difficult, but needs to be clearly understood, as the situation is rather frequent. Problem: Find the present worth of a 15-year annuity of $1.00 at 43^% annually. The compound discount must first be found. 1 + i = 1.0475 nl .0201540316 times 15 .302310474 colog 9.697689526—10 In = v " = .4985282 from 1 = .5014718 compound discount divided by .0475 = 10.5573 Effective rates : The changes in the formulae when j p must be used are the same as in the case of the amount of an annuity. The formulae are: For use with tables, a n times ^ Jp 1 — V ° For use with logarithms. Jp In both cases n is the number of years; but v " is reckon- ed at rate i. The significance of the quantity an is twofold. In the first place it is the present worth of a debt of 1 due periodically for n periods. It is the present payment which is equivalent to such a debt and will extinguish it mth all interest that will become due on the successive dates of payment. 68 MATHEMATICS FOR THE ACCOUNTANT Another significance is this : It is the present investment which will allow the owner to withdraw 1 periodically; in this case the interest due periodically is added to the fund remaining in the hands of the trustee or banker, and serves to lengthen the time required to consume the investment. Three special types of annuity are common : An annuity due is an annuity payable at the beginning of the period. Insurance premiums and rentals were mentioned as common instances. For such an annuity the amount and present worth are indicated by the symbols Sn and an. Suppose an annuity due of $1.00 annually for five years at 4% annually. We will first find the amount of this annuity: First payment, accumulated 5 periods, $1.21665290 Second Third Fourth « « Fifth Total $5.63297546 From the tables we find the amount of an ordinary annuity for six periods to be 6.63297546. This differs from the amount of the annuity due for five periods by one cash instalment. The reason is obvious. The ordinary an- nuity for six periods ends with a cash payment on which no interest is earned. The previous payments have earned interest during one to five periods. The annuity due is exactly the same excepting for the cash payment. Amount of an annuity due, first rule: Find the amount of an ordinary annuity for one more period, and subtract one cash payment. Another aspect of the case: Each paj^ment of the an- nuity due has been improved one period more than each 4 « 1.16985856 3 u 1.124864 2 u 1.0816 1 u 1.04 ANNUITIES: AMOUNT AND PRESENT WORTH 69 payment of an ordinary annuity for the same time. So if we take the amount of an ordinary annuity for the same time and multiply by 1 + i we shall have the amount of an annuity due. 5.41632256 X 1.04 = 5.63297546. Formulae for these two rules: First, s n = s n + 1 — 1 Second: Sn = Sn times 1 -\- i The present worth of an annuity due of 1 per period for five periods at 4% annually is: First payment, cash $1.00 Second " discounted 1 period .96153846 Third « " 2 " .92455621 Fourth " « 3 " .88899636 Fifth « « 4 « .85480419 $4.62989522 Now the present worth of an ordinary annuity for four years is $3.62989522, which differs from that of the annuity due by one cash payment. This is because each payment of the ordinary annuity for four years is dis- counted from one to four periods. The annuity due for five periods has an extra cash payment, made when the annuity was entered on, and this payment is not dis- counted. So one way to find the present worth of an annuity due is to find the present worth of an ordinary annuity for one less period, and add one cash payment. This is the first rule. Another rule corresponding to the second rule for summation of an annuity due is this: Multiply the pre&snt worth of an ordinary annuity for the same time by 1 + i. 4.55182233 X 1.04 = 4.62989522. This is evidently because an annuity due is payable one period nearer the time when it is entered on, and so each payment is subject to discount for one less period than the payments of an annuity made at the end of the period. 70 MATHEMATICS FOR THE ACCOUNTANT Expressed as formulae, these rules are, First, a n = a n- 1 + 1 Second, an = an times 1 + i. A deferred annuity is one which is not entered until after a certain time. The amount of such an annuity is not different from the amount of an ordinary annuity. No payments are made until after the period of deferment and consequently there are no extra interest accumu- lations. The present worth, however, is important is some trans- actions. There are two methods of finding the present worth of such an annuity. Problem : Find the present worth of a five-year annuity of 1 per year, deferred three years, on a 4% annual basis. At the date on which the annuity is entered, its present worth is $4.45182233. This amount must be discounted to the present time. Multiplying by v^ at 5% we find 4.45182233 X .88899636 = $3.95765384. Rule : Multiply the present worth of the annuity by the present worth of 1 for the period of deferment. Another method, useful when table values are obtain- able, is the following: as at 4% = $6.73274487 a3 at 4% = 2.77509103 Subtracting, $3.95765384 Rule: Subtract the present worth of an annuity for the period of deferment from the present worth of an annuity for the total time from the present to the last payment. Formulae for these two rules are: For an n-period annuity deferred m periods: First, a n times v ". Second, a n + m — am. ANNUITIES: AMOUNT AND PRESENT WORTH 71 A peq)etuity is a series of periodic payments which are expected to continue indefinitely. The dividends on a share of preferred stock are one instance ; the income from a farm is another. The amount of such an annuity has no meaning, but the present worth is important in some cases, such as an endowment of a charitable institution, or the financing of permanent improvements. By one of the laws of simple interest, the principal can be found if we know the periodic income and rate: in- come divided by rate equals principal. If a perpetuity of 1 annually is valued on a 5% annual basis, its present worth is $20.00. The symbol for a perpetual series or infinite length of time is m- Rule : Divide the rent by the periodic rate of interest : As a formula, a^ = .— 1 The interest rate on a given annuity can be found ap- proximately by Baily's formula. The only method of find- ing the exact rate is to assume a rate and find the sum or present worth as the case may be, and continue estimating and computing until the correct rate is found. Baily's formula always gives a rate slightly above the correct, but it is accurate enough for most occasions. It is neces- sary to know the periodic payment, the term, and either the amount or the present worth. The first step is to read from the tables the nearest value of Sn or an for the given time. Indicate these by s' and a'. Indicate the interest rate for s' or a' by j. Baily's formula provides for finding a corrective which we will call q. Then the correct rate i = j + q or i —'] — q. The formulae for q are 72 MATHEMATICS FOR THE ACCOUNTANT j(s^-s) s'-n(l+j)n-» j^ (a' — a) a — nv " "*^ ' Required the income rate on an annuity of 1 yearly for 10 years, when the present worth is 8. In the table of an in the 10-year row we find 7.91271818 in the 43^% column. V" at 43^% is .61619874. .045(8 — 7.91271818) ^ ~ 7.91271818 — (10 X .61619874) which gives q = .00224. Since the value of a' chosen was less than a, the rate must have been too high. Sub- tracting .00224 from .045 gives i = .04276. PROBLEMS ON CHAPTER VII Find the amount of these ordinary annuities of $1.00 each: 73. 12 years at 4 3/^% annually. 74. 20 years at 5% nominal quarterly. (That is, $.25 quarterly.) 75. 8 years at 5% effective quarterly. 76. 9 years at 4 3/^% nominal quarterly. Find the present worth of these ordinary annuities of $1.00 each: 77. 8 years at 5% annually. 78. 11 years at 6% nominal semiannually. 79. 14 years at 5% effective semiannually. 80. 6 years at 53^% annually. 81. 24 years at 43^% nominal semiannually. Find the amount of these annuities due of $1.00 each: 82. 6 years at 5% nominal semiannually. ANNUITIES: AMOUNT AND PRESENT WORTH 73 83. 8 years at 43^% nominal quarterly. 84. Find the present worth of an annuity due of $1.00 a year for 15 years at 4)^% annually. 85. Find the present worth of a 30-year annuity of $1.00 on a 43/^% annual basis, deferred 10 years. Solve two ways. 86. If $50.00 is deposited at the end of each six months for 5 years in a bank paying 4% nominal semi- annually, what will be the amount credited to the depositor at the end of the term. Set up a schedule. 87. A man buys a house for $1,000.00 cash and $1,000.00 at the end of each year for three years. On a 6% annual basis what is the cash equivalent of the pur- chase price? 88. The present worth of an ordinary annuity of $1.00 a year for 8 years on a 5% annual basis is given in the tables as $6.4632. Prove by setting up a schedule. 89. A man on his 40th birthday wishes to purchase a 10-year annuity of $1,000.00 a year, the first payment to be mjade on his 65th birthday. The insurance company bases on 3% annually. Omitting prob- ability and loading, what is the net single premium? (If the first payment is made on his 65th birthday, the annuity is entered on the 64th birthday. Re- member also that insurance premiums are prepaid). 90. A concern wishes to move their manufacturing plant from its present location on leased property. The lease has ten years to run, and the rent is payable annually in advance. The annual rental is $15,000.00 a year for the first five years and $16,000.00 a year for the last five yeare. What is the cash value of the lease on a 5% annual basis. 91. The sum of a 15-year annuity of $1.00 a year is $18.00. What is the annual interest rate? 74 MATHEMATICS FOR THE ACCOUNTANT 92. A debt of $5,000.00 is to be paid in 20 annual pay- ments of $400.00 each. What is the interest rate? (First reduce this to a unit basis. The present worth of this annuity is $12.50 per dollar of annual pay- ment.) CHAPTER VIII SINKING FUNDS At the beginning of Chapter VII we discussed the amount that could be accumulated by periodic payments of 1 improved at a given rate of interest. In that case we knew the periodic payment and computed the amount. A more conmion type of problem is that in which we know the amount which is to be accumulated and the time and rate, and must find the periodic payment. Such an amount to be accumulated within a given time is called a sinking fund, and the periodic payment is called the sinking fund payment. Such a sinking fund payment can be found by division from the table of s n. If annual pay- ments of 1 will amount in five years at 4% annually, to 5.41632256, what is the annual payment necessary to amount to 1 in five years at 4% annually? By the principle of division which was mentioned some time ago — that if a product and one factor are known, the other factor can be found by division — the sinking fund pay- ment to amount to 1 is evidently 1 divided by 5.41632256, or .18462711. In general, if periodic payments of 1 accumulated for n periods amount to s n, then to accumulate 1 the periodic payment must be 1 divided by Sn, or, expresssed as a fraction, — Sn Tables of this quantity are given in most collections. In some collections, however, tables of another quantity, — an 75 76 MATHEMATICS FOR THE ACCOUNTANT are given. In such cases, as will be explained in Chapter X, — can be found by subtracting i from the table value of-i aa Since — is the reciprocal of Sn, the formula must be the reciprocal of the formula for sn. The formula is 1 i i - or Sn S°— 1 (l+i)°— 1 This is for use in logarithmic computation. Rule: Divide the single interest by the compound interest. The following problem and schedule will illustrate the principles of sinking fund mathematics: Problem : What is the annual sinking fund contribution necessary to provide a sum at the end of five years to retire $10,000.00 bonds, on a 4% annual basis. The table value is .18462711; multiphed by 10,000 gives the annual payment, $1,846.2711. This amount can be proved. The amount of five annual payments of 1 improved at 4% annually is given in the tables as 5.41832256; and multiphed by the amount of the annual payment, $1,846.2711, the product is about 10,000. When such a sinking fund is begun it is usual and ad- visable to build a schedule covering the entire life of the fund. The schedule for the above bonds is: Year 1 2 3 4 5 Interest at 4% $ 73,8508 150.6557 230.5328 313.6050 Totals 768.6443 SINKING FUNDS 77 Schedule for Sinking Fund to Redeem $10,000.00 5% Bonds, Accumulated at 4% Annually Annual Total Amount Instal- Accumula- in ment tion Fund U,846.2711 $1,846.2711 $1,846.2711 1,846.2712 1,920.1220 3,766.3931 1,846.2811 1,996.9268 5,763.3199 1,846.2712 2,076.8040 7,840.1239 1,846.2711 2,159.8761 10,000.0000 9,231.3557 10,000.0000 Note that on any date the total of Annual Instalments plus Interest is the total of Total Additions and also the Amount in the Fund as of that date. Also that each Total Addition multiplied by 1 + i gives the next Total Addi- tion. If the schedule runs for a long time, the amount in the fund ought to be proved just as in the other schedules which have been given. We have shown that the amount in the fund on any date can be computed in advance by multiplying the sinking fund factor by the amount of an annuity for the given number of periods. In the above case the amount at the end of the third year will be $1,846.2711 multiplied by Sn for three years: $1,846.2711 X 3.1216 = $5,763.3198. This bond problem is unusual in that we have assumed annual interest payments, whereas the usual bond pays interest semiannually. The general subject of bond re- demption will be reserved for the chapter on bonds, at which time the method of handling semiannual interest payments will be fully discussed. The computation of the sinking fund payment for effec- tive rates, unusual rates which are computed by logar- ithms, and annuities due will now be considered. The 78 MATHEMATICS FOR THE ACCOUNTANT situations are in general the exact reverse of those treated in the chapter on s n. Effective rates: The formula is Sn Jp Rule : Divide the sinking fund payment for the time in years by the effective rate factor. Problem: What is the monthly payment required to accumulate SI, 000. 00 in 5 years at 5% effective, con- vertible monthly? .1809748 ^ 1.0227148 = .1769563 the yearly charge per dollar. .1769563 X 1,000 - 12 = $14.7463. Problem involving logarithms: The formula has already been explained: R= ^— Sn — 1 Problem : What is the semiannual payment into a sink- ing fund to accumulate $200,000.00 in 25 years at 4^% nominal, convertible semiannually. IH- i = 1.02125 nl .0091320695 times 50 .456603475 In = s° = 2.8615623 less 1 1.8615623 divided into .02125 .01141514 times 200,000 = 2283.028 Annuity due: The formula is R = -- ^ (1 + i) Sn Rule: Divide the sinking fund factor for an ordinary annuity by the compound interest ratio. Problem: What payment beginning now and continuing through five years from date will accumulate $1,000.00 at 4>^% annually. SINKING FUNDS 79 This is an annuity due of six payments. — is given in the tables as .14887839 divided by 1.045 = .1424673 times 1,000 = $142.4673 To determine the time required to accumulate a stated amount: let K be the amount and R the periodic pay- ment. The formula is Ki log [l + ^] R n = — ; ; r; — log (1 + i) Rule: Multiply the amount to be accumulated by the periodic rate and divide by the rent; add the result to 1 and find the log. Divide this log by the log of the com- pound interest ratio. (Six-place log is sufficient.) Problem: In what time can a fund of S10,000.00 be accumulated by semiannual payments of $500.00 improv- ed at 5% nominal, semiannually? K = 10,000 i = .025 product 250 divided by 500 .5 add 1 1.5 nl .176091 1 -1- i = 1.025 nl .010724 quotient 16.42 semiannual periods. PROBLEMS ON CHAPTER VIII 93. What is the semiannual payment into a sinking fund which is to be accumulated at 5% nominal, con- vertible semiannually, to redeem $50,000.00 of bonds at maturity, 25 years from date? From " Mathematical Theory of Investment," by Professor Ernest Brown Skinner. 80 MATHEMATICS FOR THE ACCOUNTANT 94. A debt of $250,000.00 is to be paid in one sum at the end of 20 years. The sinking fund is to be accumu- lated at 4% annually. How large is the fund at the end of ten years? (Schedule is not necessary.) 95. What sum must be deposited quarterly in a bank which pays 4% nominal quarterly to amount to $1,000.00 in 10 years? 96. In what time will $50.00 deposited semiannually in a bank which pays 4% nominal semiannually amount to $1,000.00? 97. What sum paid at the end of each month to an in- surance company which accumulates at 4% effective will amount to $1,000.00 in 10 years? 98. What would be the monthly payment in problem 97 if made in advance? 99. What will be the annual payment into a sinking fund accumulated at 4% annually to extinguish a debt of $225,000.00 at the end of 25 years. Build a schedule for the first ten years and prove the amount in the fund at the end of the tenth year. CHAPTER IX VALQATION OF ASSETS One of the commonest purposes for the building up of a sinking fund is the replacement of an asset. Such a fund is called a reserve fund for depreciation. When an asset is purchased, its probable effective life and replacement cost must be estimated. The phrase effective life is used because in any modern plant an asset is scrapped long before it has reached the limit of its work- ing power. This is due to obsolescence, the loss of effectiveness in the face of more modern methods and machines. This aspect of the replacement problem must be kept in mind when the term of the sinking fund is set, so that the reserve may be sufficient in amount at a fairly early date. Also the replacement cost will be much higher than the original cost ; how much higher is usually difficult to estimate. It is customary also to set an arbitrary value on the wornout machine as scrap. The cost, less scrap value, if any, is called the wearing value ; and it is this wearing value which must be accumulated during the effective life of the machine. The reserve fund then is a sinking fund which must be accumulated during the life of the machine. The amount of the fund is replacement cost less scrap value, and the term is the effective life of the machine. The replace- ment cost is indicated by C; the scrap value by S; the life by n ; and the wearing value by W. W = C — S, and the formula for the annual depreciation charge is D = WX — Sd 81 82 MATHEMATICS FOR THE ACCOUNTANT Rule: Multiply the wearing value by the sinking fund factor for the given time and rate. Problem: Determine the annual depreciation charge against an asset which will cost $1,000.00 to replace less a scrap allowance of $50.00 at the end of 5 years, on a 33^% annual basis. W = $950.00 n = 5 -i- = .18648137 Sn D = $177.1573 A schedule should be set up at the time the asset is acquired; one form is as follows: Interest Annual Total Total Year on Fund Payment Addition Fund 1 $177.1573 $177.1573 $177.1573 2 $ 6.2005 177.1573 183.3578 360.5151 3 12.6181 177.1573 189.7754 550.2905 4 19.2602 177.1573 196.4175 746.7080 5 26.1347 177.1573 203.2920 950.0000 $ 64.2135 $885.7865 $950.0000 Notice that the total Annual Charge plus the total Interest is the Wearing Value. Also that the Total Addition for any year multiplied by 1 + i gives the Total Addition for the next year. From the schedule the periodic Journal entry is made, and the amount in the reserve at any time can be read from the Total Fund column. This amount can be verified at any time by multiplying the annual charge by Sn. The total of the a})ove fund at the end of the third year is $177.1573 X 3.106225 = $550.2903. VALUATION OF ASSETS 83 Valuation of a plant as a whole: In a manufacturing plant the wearing values and terms of effective life of the machines vary widely, yet it is important to arrive at some method of considering the plant as a whole. The depreciation charge on each machine or class of machines must be computed separately, and the results combined, as in this problem: Problem : On a 4% annual basis find the annual charge for depreciation against the following electrical power plant : Wearing Annual Part Life Value Charge Building 40 $ 8,000.00 $ 84.1879 Engine 20 3,500.00 117.5361 Boiler 16 2,000.00 91,6400 Dynamo 18 7,000.00 272.9533 $20,500.00 $566.3173 The depreciation charge against a plant as a whole can conveniently be indicated by D'. Starting with this annual charge on the whole plant, four important factors can be worked out :j The rate of depreciation, d, is the ratio of the periodic charge to the wearing value at the beginning; for this plant it is 566.3173 -^ 20,500 - 2.7625%. The composite life of a plant is the time within which the total of the additions to the depreciation fund or funds will accumulate to the original wearing values. This is exactly the time problem of page 79: n = log (1 + i) 84 MATHEMATICS FOR THE ACCOUNTANT Rule : Divide i by the rate of depreciation, add one, and find the log; divide by the log of 1 + i- This notion is important when a bond issue is secured by a mortgage on a plant and equipment. Such bonds should never be issued for term longer than the composite life of the plant, and to provide a proper margin of safety the time should be considerably shorter. For the above plant — = 1.448 d add 1 and find log .388811 1.04 nl .017033 quotient 22.83 years The wearing value of an asset at any time is the original wearing value less the amount in the reserve at that time. The condition percent of a plant at any time is the ratio of the wearing value at that time to the original wearing value. Other methods of reckoning depreciation: By allowing interest on the wearing value of the asset at the time. The formula is D = (Cs"-S) X - Sn Rule: Multiply the cost by the compound amount factor, and subtract the scrap value; multiply the result by the sinking fund factor. The composition of this formula is easily understood. The original cost multiplied by the compound amount shows the capital tied up in the machine during its entire life. Of this a small part will be returned as a scrap VALUATION OP ASSETS 85 allowance. Deducting scrap, we have a loss of capital which must be spread over the life of the machine by the sinking fund method. The objection to this method is that interest ought to be reckoned on the original cost if it is to be reckoned at all. For the asset con- sidered at the beginning of this chapter, the annual charge by this method is 1,000 X 1.18768631 less 50 X .18648137 = $212.1573 The schedule is Net Reduction ear Book Interest Annual in Value 33^% Charge Book Value 1 11,000.00 $35,0000 $212.1573 $177.1573 2 822.8427 28.7995 212.1573 183.3578 3 639.4849 22.3819 212.1573 189.7754 4 449.7095 15.7398 212.1573 196.4175 5 253.2920 8.8653 212.1573 203.2920 $50.0000 $110.7865 $1,060.7865 $950.0000 The phrase net book value indicates the difference be- tween the cost and the amount in the reserve. It is difficult to make proper Journal entries from such a schedule because the asset should always be carried on the books at cost. Yet the interest must be reckoned on the net book value, else this method of charging for depreciation cannot be used. Fixed percentage on a diminishing book value: The formula is r = 1 ^ c 86 MATHEMATICS FOR THE ACCOUNTANT Rule : Subtract the log of C from the log of S; divide the result by n; find the antilog and subtract from 1. For the asset we have been considering S nl 1.698970 C nl 3.000000 subtracting 8.698970—10 divided by 5 9.739794—10 In .54928 from 1 .45072 or 45.072% each year of the net book value at the beginning of that year. The schedule is Net Book Depreciation Amount in Year Value 45.072% Reserve 1 $1,000.00 $450.72 $450.72 2 549.28 247.5714 698.2914 3 301.7086 135.9861 834.2775 4 165.7225 74.6944 908.9719 5 9 1.0281 41.02 81 95 0.0000 $50.0000 $950.0000 13,842.2608 Hatfield, in his "Modern Accounting," discusses these two methods in detail. He says that the best arguments for this last method are, first that the depreciation charge ought to be heaviest when the machine is new and less in need of repairs, so that year by year as repair charges increase they may be compensated by lessening deprecia- tion charges; and second, that investigations show that a machine depreciates most rapidly in its first year and that the rate of actual physical depreciation decreases after that time. Any reader who is interested in the more complicated methods of reckoning depreciation, such as the equal VALUATION OF ASSETS 87 annual payment method and the unit cost method, should read Saliers' "Principles of Depreciation", in uhich they are treated fully, with schedules and diagrams. Miscellaneous formulae covering certain problems in valuation and accounting for assets: Provision against the total exhaustion of wasting assets such as mines, timber lands, oil properties, and so on. If such assets were permanent or renewable, the provision would take the form of a perpetuity, which is valued by the formula — when R is the annual income, i But since the asset has a limited life it is necessary to provide also against the time when the rent will cease. Such provision should take the form of a sinking fund payment. This payment, at a rate which we may indicate by i', must be added to i as an extra expense. The formula for the value of the asset is then V = 5 — (at rate i ) + i Sn Rule: Add the sinking fund charge at rate V to the income rate i; divide the result into the periodic rent. Problem: Find the value of a mine w^hich will net $15,000.00 a year for 30 years, if the annual dividend rate is to be 6% payable annually, and the sinking fund is to be accumulated at 33^2% annually. — 30 years at 3^% = .01937133 s i = .06 adding .07937133 dividing into 15,000 = $188,985,110 88 MATHEMATICS FOR THE ACCOUNTANT Viewed from another standpoint this solution gives the amount of stock which can safely be issued. This can be proved. Basing on 1,900 shares of stock at 100, annual dividends, 6% on $190,000.00.. . .$11,400.00 sinking fund, $188,985,116 times .01937133 3,660.897 $15,060,897 which is slightly more than the annual income estimated at $15,000.00. Of course it would be conservative to allow for contingencies by issuing a smaller amount of stock. Capitalization of assets : This is defined as the first cost plus cost of perpetual renewals. The cost of a perpetuity is - ; the rent of a sinking fund is — . So the product of 1 Sn these two factors is the cost of an indefinite series of re- newals occurring every n years. The capitalized cost is the first cost plus this cost of indefinite renewals. This problem occurs especially when endowments for hospitals, colleges, and so on have to be valued. There are usually items of upkeep which occur every year, and these must be valued as perpetuities. Problem : A philanthropist desires to present a building to a university. The building will cost $200,000.00 and will have a life of 50 years. The annual upkeep will cost not over $16,000.00. What must be the amount of his gift? First cost $200,000.00 Renewals 200,000 X - X .0065502 32751.00 .04 16000 400,000.00 ^ ^^^ .04 $632,751.00 VALUATION OF ASSETS 89 It is necessary to distinguish carefully between — which is the value of a series of annual payments; 1 and - X — which is the value of a series of payments at i Sd intervals of n years. SUMMARY OF FORMULAE USED IN ACCOUNTING FOR ASSETS Let C be the cost, original or replacement; S the estimated scrap value; W the wearing value; n the life of a single asset; n' the composite life of an entire plant; D the depreciation charge on a single asset; D' the depreciation charge on an entire plant; d the rate of depreciation of an entire plant, being the ratio of the total periodic charge to the wearing value at the begining. Annual depreciation D = W X — Sn Amount in fund at end of r-th year Sr = D X s n. Rate of depreciation d = — logFi+l] Composite life of a plant n' = log(H-i) Wearing value at end of r-th period W, = W — D X s n. Wr Condition percent = -— ' W 90 MATHEMATICS FOR THE ACCOUNTANT Allowing interest on book value D = (CXs"— S)X — Vc Fixed rate on diminishing book value t = 1 —y - Value of a wasting asset to pay dividends at rate i' when R is the annual output: R V = — (at rate i') + i Sn Value of a perpetuity, payable annually = R i R 1 Value of a perpetuity, payable every n years = — X — 1 Sn Capitalized cost of an asset (this will be treated in the next chapter on amortization, but the formula is given here for the sake of completeness) : X = C X- X an+m an For most of thn=o formulae I am indebted to The Mathematical Theory of Invest- ment, by Professor Ernest Brown Skinner, of the University of Wisconsin. PROBLEMS ON CHAPTER IX 100. An asset having an estimated hfe of 10 years and a scrap value of at least $500.00 will cost $12,000.00 to replace. Set up a schedule of the reserve by the sinking fund method, basing on 3i^% annually. 101. An asset having an estimated life of at least 8 years will cost $8,000.00 to replace less an allowance of $200.00 on the old machine. Set up a schedule of the reserve on the fixed percentage method. 102. An asset having an estimated life of 5 years and estimated replacement cost of $15,000.00, less scrap Part Life A 15 B 25 C 45 D 30 E 8 VALUATION OF ASSETS 91 allowance of $1,200.00 is to be valued by the method of allowing interest on the book value at the begin- ning of each year. Set up a schedule on a 4% annual basis. 103. In an investigation of a public utility, the various parts were appraised as follows: Cost Scrap $150,000.00 $ 8,000.00 65,000.00 3,800.00 122,000.00 6,750.00 85,500.00 15,625.00 8,300.00 650.00 On a 3% annual basis compute a) annual depreciation charge, sinking fund method; b) rate of depreciation; c) composite life. 104. What is the wearing value and condition percent at the end of the 7th year of the first asset of problem 103? 105. Certain timber lands will yield a net revenue of $8,000.00 a year for 35 years. 6H% stock, dividends annual, is to be issued, and provision is to be made for annual payments to a sinking fund at 4% annually. Supposing that the land has no residual value, what is a reasonable purchase price? 106. It is estimated that a mine will yield $12,000.00 per year and will be exhausted in 20 years. The stock is to pay 7% annually and the sinking fund is to be accumulated at 43/2% annually. Find the amount of stock that can be safely issued. 107. How much can a railroad afford to spend in elimi- nating a grade crossing which is guarded by two 92 MATHEMATICS FOR THE ACCOUNTANT watchmen at $650.00 a year each. Assume 4% annually. 108. What sum must be set aside annually to provide for rebuilding every 25 years of a bridge costing $30,000.00? Assume that money will be worth 4% annually. 109. On a certain railroad grade a helper engine and two crews are necessary at an annual cost of $12,000.00; a new engine must be provided every 20 years at a cost of $14,000.00. How much can the railroad afford to spend to level the grade if money will be worth 4J^% steadily? 110. A man desires to endow an art gallery. The building will cost $75,000.00. The life is 100 years. The annual cost for purchase and rental of paintings and so forth will be $50,000.00. Repairs and renewals will cost $2,000.00 yearly. The lighting system will need renovating every 5 years at a cost of $1,200.00. Upkeep and service will cost $4,500.00 a year. Value the endowment on a 33^% annual basis. CHAPTER X AMORTIZATION In Chapter VII we discussed amortization somewhat. It can be defined as the series of equal payments which will extinguish a given debt, the periodic payments being used first to meet interest accrued on balances outstand- ing. In that chapter we learned that five equal payments of 1 will extinguish a debt of 4.45182233 with interest at 4%. The more common situation is that in which the amount of the debt is known and the amount of the periodic payment must be found. What, for instance, is the periodic payment which will extinguish a debt of 1 in 5 years, interest on outstandings at 4% annually? Evidently, if payments of 1 will extinguish a debt of 4.45182233, then the payment required to extinguish a debt of 1 is 1 divided by 4.45182233, or .22462711. And the payment required to extinguish a debt of $10,000.00 is $2,246.2711. A schedule will prove this: Schedule of Amortization of a Debt of $10,000.00 in 5 Equal Annual Instalments, with Interest on Outstandings at 4% Annually Balance Net Out- Interest Annual Amortiza- Year standing at 4% Payment tion 1 $10,000.0000 $400.0000 $2,246.2711 $1,846.2711 2 8,153.7289 326.1492 2,246.2711 1,920.1219 3 6,233.6070 249.3443 2,246.2711 1,996.9268 4 4,236.6802 169.4672 2,246.1712 2,076.8040 5 2,159.8762 86.3950 2,246.2712 2,159.8762 1,231.3557 11,231.3557 10,000.0000 93 94 MATHEMATICS FOR THE ACCOUNTANT The usual checks should be observed. Total Annual Payments less total Interest is total Net Amortization, which is the same as the Balance Outstanding at the beginning. Each Net Amortization times 1 + i gives the next amount. The balance outstanding on any interest date is the present worth of the annual payments still due. In the above schedule, the amount due just after the third payment is the present worth of the two remain- ing payments; $2,246.2711 X 1.88609467 = S4,236.678. If the schedule is lengthy it should be proved every few years with this last check. "To pass to algebra, since a debt of a n can be amortized by a series of n payments of 1, a debt of 1 can be amortized by a series of n payments of 1 divided by a n, or, expressed 1 as a fraction, — an. For logarithmic computation, the formula is the recip- rocal of the formula for a n, J^ ^ _J i an 1 — V° - 1 (1 + i) " and in the usual circumstances, when the given rate is efifective, if we are using table values, we divide : ■ — ■ -^ — 1 Jp where n is the number of years ; and if we are using logar- ithms, we use j p instead of i in the numerator of the formula, while n in the denominator is the number of years. These will be illustrated later. Now, if we compare the interest columns of the sinking fund and amortization schedules, we shall find an in- teresting relation between the sinking fund and amortiza- tion factors. In the sinking fund schedule the interest is a credit ; in the amortization schedule it is a debit. AMORTIZATION Amortization Sinking Year Debit Fund Credit 1 $400.00 2 326.1492 $ 73.8508 3 249.3443 150.6557 4 169.4672 230.5328 5 86.3950 313.6050 95 Id each year the sum of the two items is 4% of $10,000.00. Or, a debit of $400.00 set against each credit to the sinking fund results in the corresponding debit to amortization. This shows why the tables do not include any tab'e of — an' for that quantity is always greater than -^ by the annual Sn interest. This fact, stated on the basis of a unit, as usual, gives the formula: . . — = - + i an Sn Rule : To find the amortization factor for any time and rate, read the table value for the corresponding sinking fund factor and add the interest rate. In the sinking fund table for 4% when n = 5 we find .18462711; adding .04 we have the amortization factor .22462711, the same amount we found on the previous page by division. Logarithmic solution : Find the rent of an amortization of a debt of 1 in 10 semiannual payments at 5}/2% nominal, convertible semiannually : 1 + i = 1.0275 nl .0117818305 times 10 .117818305 colog = log of v" 9.882181695—10 antilog = v° .7623977 from 1 .2376023 the compound discount divide into .0275 = .11574 96 MATHEMATICS FOR THE ACCOUNTANT Rule: Divide the interest rate by the compound dis- count. Effective rates: What is the monthly payment required to amortize a debt of $500.00, beginning at the end of one month and continuing for 5 years, interest at 5% effective. 1 an i .2309748 at 5%, p = 12, = 1.0227148 Jp quotient = .2258448 times 500 = 112.9224 divided by 12 = $9.41 Annuity due: Since the formula for the present worth of an annuity due is En X (1 + i), it is evident that the amortization factor payable in advance is — ^ (1 + i). an Problem: A debt of 1 is to be amortized in 5 annual payments, the first payment to be made immediately. What is the annual payment on a 6% annual basis? — = .2373964 dividing by 1.06 = .223959 Four special problems in amortization are important: The amount due after any number of payments have been made: This has been mentioned already, and was explained as the present worth of the payments still due. This is true inmaediately after a payment has been made. If, however, the amount due immediately before a date of payment is in question, the payments still due con- stitute an annuity due and must be valued according to the formula for such an annuity. Problem: A debt of $10,000.00 is to be amortized over a period of 10 years by annual payments, interest at 5% AMORTIZATION 97 annually. On the date of the seventh payment it is desired to liquidate the balance in one amount. What is the balance? .12950458 _1 aio times 10,000 1,295.0458 annual payment as 2.72324803 product 3,526.7309 add 7th payment 1,295.0458 Amount due $4,921.7767 Or, we might have multiplied 3.549505 X 1.05 X 1,295.0458, since 3.5459505 is the present worth of an ordinary annuity for 4 periods at 5%. The time required to amortize a given debt: Let A be the debt and R the rent, or periodic payment. The formula is* colog[l-|] "" - log (1 + i) Rule : Multiply the amount by the interest rate, divide by the rent, subtract from 1, find the colog; divide by the log of the interest ratio. Problem: In what time can a debt of $1,000.00 be liquidated by annual payments of $80.00, interest at 5% annually? A = $1,000.00 R = 80.00 i = .05 Ai = 50 divided by 80 .62J from 1 .37i From " Mathematical Theory of Investment," by Professor Ernest Brown Skinner. 98 MATHEMATICS FOR THE ACCOUNTANT nl 9.574031—10 colog .425969 1.05 nl .021189 quotient 20.1 When the rate is effective, j p is used in place of i in the numerator (multipHed by A), but R is the annual pay- ment, and i is the annual rate. Problem : In what time will a debt of $300.00 be amor- tized by quarterly payments of $10.00, beginning 3 months from date, interest at 33^2% effective? A = $300.00 R = 40.00 j p when p = 4 .0345498 A times j p 10.36494 divided by 40 .2591235 from 1 .7408765 nl 9.869746—10 colog .130254 1.035 nl .01494 quotient 8.72 or 8^ years and a small balance due. The increased cost of lengthening the life of an asset Let C be the cost of the present asset, andC the cost of the longer-lived asset. Let n be the present life, and m the added life, so that n + m is the total life of the new asset. Now, if we multiply the present cost by the amortization factor, we shall have the annual cost of the present asset. And if we multiply that by the present worth of an annuity of 1 for the lengthened term of the new asset we shall have the amount which we can afford to pay for the new asset. Expressed as a formula* From "Mathematical Theory of Investment," by Professor Ernest Brown Skinner. AMORTIZATION 99 1 C = C X X an + m an Problem : If a piece of equipment costs'^! 100.00 and has a life of 3 years, how much can be paid for a better piece of equipment which will last 5 years, on a 4% annual basis? C = $100.00 — = .36034854 aa product 36.034854 annual cost Ef = 4.45182233 product S160.42 That is, the equipment costs us .36034854 a year, for any term of years. The present worth of a 5-year article is that amount times the present worth of an annuity of 1 for 5 years. The amordzation of a series of bonds will be discussed in the next chapter. PROBLEMS ON CHAPTER X 111. A man buys an estate valued at S8,000.00, agreeing to pay cash S2,000.00 and the balance in 5 equal annual payments with interest on outstandings at 5% annually. Set up a schedule of the payments. 112. A man borrows $3,000.00 at 5% annually, agreeing to pay $300.00 annually. By the use of a formula determine the life of the debt. What is the amount due after the last full period. Solve without schedule. 113. What is the annual rent to amortize a debt of $1,000.00 in 10 years at 4^% annually. 1 14. What quarterly payment continued three years will pay a debt of $200.00 with interest at 4% effective. 100 MATHEMATICS FOR THE ACCOUNTANT 115. What is the annuity required to amortize a debt of SI, 500.00 with interest at 3^% annually, in 5 pay- ments, the first payment to be made at once. 116. Set up a schedule for problem 115. 117. A debt of $3,000.00 with interest at 43^% annually is to be amortized in 10 annual payments, beginning one year from date. If the payments are made promptly for five years, what amount will liquidate the balance due on the date of the sixth payment. 118. A city incurs a debt of $150,000.00. From a mathe- matical standpoint which is better: to amortise in 20 equal annual payments at Q}-i% annually, or to accumulate a sinking fund by annual payments im- proved Sit 4:% annually, paying interest on the face of the debt meanwhile at 5% annually. 119. A man is offered $12,000.00 a year payable in ad- vance on a 99-year lease or cash $225,000.00 for a piece of land. Which is better for him on a 5% annual basis? 120. On a 43^% annual basis what amount can be spent in improving a piece of apparatus which now costs $25.00 and has a life of 3 years so that it may last 5 years? 121. On a 5% annual basis what can a company afford to spend in improving a machine which now costs $85.00 and has a life of 6 years so that it may last 9 years? CHAPTER XI VALUATION OF BONDS A bond is a promise to pay a certain sum, usually $1,000.00 or $100.00, after a certain number of years from the date of issue. The commonest exception to this statement is the issue of a series redeemable by drawings. It also promises, during the life of the bond, to pay interest at a fixed rate, called the coupon rate; such interest is usually paid semiannually. In this book this coupon rate will be indicated by c; actuarial writers usually use j, because, as we shall see, this coupon rate is really a nominal rate and does not indicate the income rate realized by the holder. Bonds are commonly pur- chased either above par, ''at a premium"; or below par, "at a discount". The amount of premium or discount may depend on a number of things; but if the bond is "gilt-edged" and purchased for investment lather than for speculation the price depends chiefly on the income rate expected, and after that on the value at which the bond will be redeemed and the time to elapse before redemption. The income rate, which we will call i, is independent of c; in fact it is usually not calculated until after purchase and when the bond is brought on the books. If the coupon and income rates are the same the bond is bought at par. If the purchaser is satisfied that his income shall be less than the nominal coupon rate, the bond will be purchased at a premium. If the pur- chaser wishes his income rate to be above the coupon rate, he will purchase at a discount. The value of a bond depends then on the redemption value and the relation between the coupon and income 101 102 MATHEMATICS FOR THE ACCOUNTANT rates. Since coupons are usually payable semiannually it is customary to value bonds on a semiannual basis, unless the bond is stated to pay annual or quarterly coupons. Let us value a bond by valuing a) the present value of the redemption price, and b) the present value of the coupons. Problem: What is the price of a 7% 25-year bond, redeemable at par $1,000.00, to yield 6%. Since the valuation is to be on a semiannual basis: n = 50 c = .035 i = .03 Present worth of the redemption value is 1,000 X v^o at 3%, which is 1,000 X .22821708 = $228.1071 Present worth of 50 coupons of $35 each is 35 X 25.729764 = 900.5417 Total $1,128.6488 This establishes the correct value of the bond. The best method of valuation is the formula of Mr. Makeham, as follows : Let C be the redemption value (cash) ; K the present worth of C ; c the coupon rate (Makeham uses j) ; i the income rate; A the present value of the bond. The formula is A = K -^ X (C - K) 1 Rule: Find the present worth of the redemption value. Subtract this from the redemption value and multiply the result by the coupon rate and divide by the income rate. Add these results. VALUATION OF BONDS 103 For the bond we have just been considering: K = 1,000 X .22810708 = S228.1071 C — K = 771.8929 .035 .03 X 771.8929 = 900.5417 $1,128.6488 The reason for this formula is this. If the bond were purchased so that coupon and yield rates were the same, the price would be the same as the redemption value. Since the present worth of the redemption value is $228.1071, the value of the coupons would be the re- mainder of the $1,000.00, or $771.8929. But the yield rate is lower than the coupon rate, which shows that the bond cost more than par. It cost as much above par as the coupon rate is above the yield rate, which in this . -035. , case IS — r— In every case the coupon rate is the numer- ator of this fraction. It is usual to set up a schedule at the time the bond is brought on the books ; the first ten periods of this schedule are: Schedule of Amortization of 7% 25- Year Bond of the Co., Redeemable at Par $1,000.00 on , Income Rate 6% Net Coupons Income Amortiza- Book Period 33^% 3% tion Value $1,128.6488 1 $35.00 $33.8595 $1.1405 1,127.5083 2 35.00 33.8252 1.1748 1,126.3335 3 35.00 33.7900 1.2100 1,125.1235 4 35.00 33.7537 1.2463 1,123.8772 104 MATHEMATICS FOR THE ACCOUNTANT 5 35.00 33.7163 1.2837 1,122.5935 () 35.00 33.6778 1.3222 1,121.2713 7 35.00 33.6381 1.3619 1,119.9094 8 35.00 33.5973 1.4027 1,118.5067 9 35.00 33.5552 1.4448 1,117.0619 35.00 33.5119 1.4881 1,115.5738 350.00 336.9250 13.0750 There are several checks to such a schedule, the first of which is by the usual additions. If the bond was bought at a premium, total Coupons less total Income equals total Net Amortization; each Net Amortization multi- plied by 1 + i gives the next. If the bond was bought at a discount, total Income less total Coupons gives total Net Accumulation; and each Net Accumulation times 1 -1- i gives the next. The next check is by valuing the bond again: K = 1,000 X v^o = 1,000 X .3065568 = $306.5568 C-K = 693.4432 .035 — - X 693.4432 = 809.0171 $1,115.5739 Still another check is by difTerencing. This has the merit that it is not in any way similar to the methods used in valuation, and so is less likely to repeat any mistake that may have been made. Set down in a column the successive Net Amortizations (in the table they have been set down in reverse order to facilitate subtraction). Subtract each one from the one directly over it; this is called the first order of differences. Repeat the operation, securing a second order of differences. If desirable subtract once more. In the accompanying schedule this is unnecessary. It is unnecessary to pro- ceed beyond the point where the differences are nearly VALUATION OF BONDS 105 equal, some being slightly more and some slightly less. If any wide variation should appear it would indicate an error in that region of the schedule. 1.4881 .0433 .0012 1.4448 .0421 .0013 1.4027 .0408 .0011 1.3619 .0397 .0012 1.3222 .0385 .0011 1.2837 .0374 .0011 1.2463 .0363 .0011 1.2100 .0352 , .0009 1.1748 .0343 1.1405 In the second column the differences are progressive; that is, they decrease successively. In the third column the differences waver. In a short-term schedule the differences are larger than in a long-term schedule. If this were a five-year bond it would probably be neces- sary to obtain a fourth column to reach the point where the differences were nearly equal. When the income rate is not in the tables the use of logarithms is necessary in finding v °. The process has been demonstrated, but will be shown here again. Problem: Value a 5% 20-year bond redeemable at par $1,000.00, to yield 3.6%. c = .025 i = .018 n = 40 1 -1- i = 1.018 nl .0047747778 times 40 .30991112 colog 9.69008888-10 In = v ° = . .4898794 106 MATHEMATICS FOR THE ACCOUNTANT K = 1,000 X .4898794 = $489.8794 C— K = 510.1206 — X 510.1206 = 708.5008 ^•^ $1,198.3802 Redemption at a premium or discount: Bonds are often issued subject to an agreement that they may be redeemed above or below par. This is especially common when the issuing company reserves the right to redeem before maturity. Suppose a 5% bond can be redeemed at 110, when par is 100. The income rate is based on the pur- chase price, so the redemption value has no direct effect on the income rate. But 5% on 100 is not 5% on 110. The coupon rate must be converted to this new value at maturity. X .05 = .045454 annually, or .022727 semiannually. Suppose this bond is redeemable in 15 years, that is n = 30. Income rate 4%: K = 1,100 X v30at2% = 1,100 X .55207089 = $607.2780 C— K = 492.7220 .022727 .02 X 492.7220 = 559.9113 $1,167.1893 Valuation with allowance for income tax: The writers on actuarial science are forced to treat this subject at length and to give complicated formulae because the English income tax law bases taxation on the net in- come after accumulation of discount or amortization of premium. But in this country where the tax is levied on the coupons themselves, the problem is simple. Consider a 20-year 5% bond redeemable at par $1,000.00. The purchaser wishes to net 4% after taxation, and esti- VALUATION OF BONDS 107 mates that the tax will consume 10% of all his income. Of every 2.5% of income one- tenth will satisfy the tax. That means that he will net 2.25%. The problem then is to value a 40-period 23^% bond on a 2% yield: K = 1,000 X .4528904 = $452.8904 C— K = 547.1096 .0225 .02 X 547.1096 = 615.4983 $1,068.3887 Without the allowance for tax the cost of the bond would have been $1,136.7774, the difference being 68.3887. The proof of this is very interesting. The holder of this bond expects to pay $2.50 of every coupon in tax. The present worth of a 40-period annuity on a 2% basis is 27.35547924. The product of 27.35547924 X 2.50 is $68.3887, which was exactly the decrease in purchase price due to the allowance for taxation. Valuation between interest dates: So far we have been valuing bonds on interest dates or at least ex-interest. In practise, bonds are usually bought between interest dates, and if the price is accurately fixed there must be not only an allowance for interest due the former holder but also adjustment for discount or premium for the part of a period already past. It is usual to reckon the interest and premium or discount by simple interest, as follows: Problem: A 7% 25-year bond, redeemable at par $1,000.00, is bought on a 6% basis. The coupons are due January and July 1. The bond is bought March 1. What is the price? We found that the price January 1 v/as $1,128.6488. The seller is entitled to the third of the coupon which has accrued and must expect to lose the third of the premium to be amortized: 108 MATHEMATICS FOR THE ACCOUNTANT Value January 1 $1,128.6488 1/3 of $35.00 $11.6667 less 1/3 of 3% of book value 11.2865 .3802 Subtracting, value March 1 $1,128.2686 Also, it often happens that the coupon dates do not coincide with the dates for closing the books. In this case it is desirable to write a schedule which will show the value of the bond on the dates of closing. Problem: A bond redeemable for $1,000.00 on October 15, 1916, bears interest at 5% payable April and October 15. The bond is purchased April 15, 1913, to yield 6%. Value the bond and write a schedule which can be used as a source of entries on a set of books which are closed June 30 and December 31 : Value April 15 $968.8486 5/12 of 3% of book value $12.1106 5/12 of coupon 10.4167 1.6939 Adding, value June 30 $970.5425 The schedule is : Net Income Accumu- Book Date 3% Coupons lation Value 4/15/'13 $968.8486 6/30/'13 $12.1106 $10.4167 $1.6939 970.5425 12/31/'13 29.1163 25.00 4.1163 974.6588 6/30/'14 29.2398 25.00 4.2398 978.8986 12/31/'14 29.3670 25.00 4.3670 983.2656 6/30/'15 29.4980 25.00 4.4980 987.7636 12/31/'15 29.6329 25.00 4.6329 992.3965 6/30/'16 29.7719 25.00 4.7719 997.1684 10/15/'16 17.4149 14.5833 2.8316 1,000.0000 $206.1514 $175.0000 $31.1514 VALUATION OF BONDS 109 The usual checks should be observed. A slight discrep- ancy will usually appear in the net accumulation or amortization for the last partial period. In the above schedule this has all been taken up by adjusting the income. There are various methods of eliminating this residue, as it is called. It can all be taken care of during the first period or during the last period. It can be spread equally over all tne periods. Or it can be spread over all the periods proportionately to the net accumu- lations or amortizations. By this last method an adjust- ment per dollar is found, basing on the total of the net column. This adjustment is then multiplied by each accumulation in turn, and the result is added to or sub- tracted from the proper item. For a detailed discussion with tables see the Sprague-Perrine "Accountancy of Investment". A last method which may be suggested is to multiply the error by — at rate i, and add the result Sn to the Income items or subtract from them as may be necessary. Serial bonds: Bonds are often issued under an agree- ment that they may be redeemed at various dates rather than all at once. These periodic redemptions may take place semiannually, annually, or at intervals of several years. The bonds may be redeemable at par or at a premium or discount; in practise they are oft^n redeem- able at a premium which decreases as the final redemption date approaches. They may be redeemable at the pleasure of the trustee of the sinking fund or in stated amounts, regular or otherwise. If the trustee is instructed merely to purchase bonds in open market to the amount of the fund in his hands at any time, it is obviously impossible to value the series 110 MATHEMATICS FOR THE ACCOUNTANT with any mathematical accuracy. For such redemptions will vary with varying market conditions, both price and amount being variables. If the schedule of redemptions is known, dates, amounts and prices being stated, the series can be valued as a whole. No single bond can ever be valued except at a so-called average price, which is only an arbitrary valua- tion, unless the redemption date of that particular bond is stated. The valuation of an irregular series proceeds as follows: Problem: A series of $100,000.00 of bonds paying 5% annually is issued on January 1, 1920, redeemable on and after January 1, 1926, in these amounts: Jan. 1, 1926 $10,000.00 Jan. 1, 1927 15,000.00 Jan. 1, 1928 20,000.00 Jan. 1, 1929 20,000.00 Jan. 1, 1930 35,000.00 Value this series on a 4% annual basis. It is first necessary to schedule the interest payments: Jan. 1, 1921, to Jan. 1, 1926 $5,000.00 each year Jan. 1, 1927 4,500.00 Jan. 1, 1928 3,750.00 Jan. 1, 1929 2,750.00 Jan. 1, 1930 1,750.00 The value of the series on a 4 % annual basis is : Jan. 1, 1921, to Jan. 1, 1925, 5,000 X as $22,259.1117 Jan. 1, 1926, 15,000 X v«. . . . 11,854.7180 Jan. 1, 1927, 19,500 X v^. . . . 14,818.3973 Jan. 1, 1928, 23,750 X v^. . . . 17,353.8924 Jan. 1, 1929, 22,750 X v*. . . . 15,983.8483 Jan. 1, 1930, 36,750 X v'".. . . 24,826.9832 Total $107,096.8909 VALUATION OF BONDS 111 If a series is regular in every way, redeemable in equal amounts at a fixed price and at equal intervals of time, it can be valued by Makeham's formula. Consider a series of ten $1,000.00 bonds, 5% J J, redeemable one each six months beginning July 1, 1922, at par. This series is to be valued as of Jan. 1, 1917, ex interest on a 6% basis. The method of stating the coupon rate is a common method, and means that the bonds pay 23^% on the first of January and July. This is a 10-period series deferred 10 periods. So in Makeham's formula the present worth factor, instead of v, is the factor for a deferred annuity, which is an+m — SLm, where m is the term of deferment. In this case it is a 2 — Si 1 C = $10,000.00 K = 1,000 X a2o— aio at 3% or 6,347.2720 C-K = 3,652.7280 -^^^ X 3,652.7280 = 3,043.9400 no — l , $9,391.2120 The problem is usually complicated by the fact that redemptions are annual, while interest payments are semi- annual. In such a case the yield is also semiannual, and the best method is to annualize both coupon and income rates. A series of five $10,000.00 bonds, 5% JJ, issued Jan. 1, 1920, is redeemable at par, $10,000.00 at the end of each year for five years. Value the series on a 43^% basis. The rates can be changed to an annual effective basis by the rule on page 35. (1.025)2 1.050625 annualized c = 5.0625% (1.0225)2 1.0455 annualized i = 4.55% 112 MATHEMATICS FOR THE ACCOUNTANT Next we have to find a 5 at 4.55% 1.0455 nl .01932398745 times 5 .09661993725 colog 9.90338006275—10 In = V ° .80053466 from 1 .19946534 divided by .0455 = 4.3838536 = a n Now, using Makeham's formula K = 10,000 X an = $43,838.54 C-K = 6161.464 5.0625 4.55 X 6161.464 = 6,855.47 $50,694.01 If the bonds are to be redeemed by an amortization scheme the valuation is no more difficult. Consider a series of five $10,000.00 bonds, paying 5% annually, to be amortized over a period of ten years, the first amor- tization to take place at the end of one year. To value the series on a 4% annual basis. — at 5% = .12950548 aio 50,000 X .12950548 = $6,475,229 This is the annual amortization payment, disregarding the necessity of redeeming whole bonds. If this annuity is purchased on a 4% annual basis, aioat4% = 8.11089578 times 6.475.229 = $52,519.9075 Redemption of a series: As has been said, a series may be redeemed at the discretion of the trustee, or the arrangements as to amounts and dates may be stated in VALUATION OF BONDS 113 the deed of trust under which the bonds were issued. Two methods which lead to some mathematical diffi- culties will now be discussed. It may be desirable to retire the series by an amorti- zation scheme, so that periodic payments on account of principal and coupons combined may be as nearly equal as possible. Of course the amount of principal repaid at each redemption date must be an exact multiple of the face of the bonds. In such a case the periodic payments cannot be exactly equal, and the schedule requires con- siderable adjustment. Also this situation is not to be confused with the sinking fund method, by which no bonds are redeemed until the entire series matures, and periodic payments are allowed to accumulate interest in the hands of the trustee. Suppose an issue of 50 bonds of par SI, 000.00 is to be retired over a period of ten years, beginning one year from date. Interest on bonds outstanding is to be at the rate of 5% annually. The amortization factor is .12950458, which, multipHed by 50,000, gives the average annual cost $6,475,229, The first year's coupons are $2,500.00, leaving $3,975,229 for redemptions. This will almost redeem four bonds, and it will be good policy to appropriate the necessary $24.77. The annual payment then is $6,500.00, and the par of bonds outstanding at the beginning of the second year is $46,000.00. 5% interest for the second year is $2,300.00, leaving $4,175,229 for redemptions. It will be best to appropriate only $4,000.00. The extra $175,229 may be appropriated if desired and left to draw interest and later it can be used to make up deficiencies such as occurred in the first year. This is ultra conservative, however. The schedule can now be given without comment: 114 MATHEMATICS FOR THE ACCOUNTANT Schedule of Amortization of $50,000.00 Bonds at Par $1,000.00, with Interest at 5% Annually, in Ten Equal Annual Instalments Par of Number Bonds Out- Interest Total Net of Bonds Year standing 5% Payment Payment Retired 1 $50,000.00 $2,500.00 $6,500.00 $4,000.00 4 2 46,000.00 2,300.00 6,300.00 4,000.00 4 3 42,000.00 2,100.00 6,100.00 4,000.00 4 4 38,000.00 1,900.00 6,900.00 5,000.00 5 6 33,000.00 1,650.00 6,650.00 5,000.00 5 6 28,000.00 1,400.00 6,400.00 5,000.00 5 7 23,000.00 1,150.00 6,150.00 5,000.00 5 8 18,000.00 900.00 6,900.00 6,000.00 6 9 12,000.00 600.00 6,600.00 6,000.00 6 10 6,000.00 300.00 6,300.00 6,000.00 _6 14,800.00 64,800.00 50,000.00 50 The usual checks should be observed. If there were a question regarding the payment on account of redemp- tions in any year, it would be wise to redeem too few bonds, rather than too many. This satisfies the bond- holders and allows the issuing company the use of the money for one year more. In general the number of bonds redeemed is progressive, increasing slowly from the first date to the last. The other method which we ought to consider is the method which requires what we may term a redemption fund. By this method annual costs are equalized regard- less of amount of redemptions. This is done by appro- priating a fixed amount and allowing any excess to remain in the bank until it is needed. In this way interest is earned on all unused balances, yet they are available without any strain on the finances of the issuing com- pany. VALUATION OF BONDS 115 A series of $100,000.00 of 5% bonds such as we valued on page 110 is a good example. Suppose we appoint a trustee and agree to pay him equal amounts at intervals of six months from six months after the date of issue to the date of the last redemption. This arrangement does away with the necessity of annualizing rates an^ simplifies the problem. The trustee is to meet all coupons and redemp- tions and keep the balance of the fund invested at a rate which we assume will be not less than 4% annually. The balance thus invested must be available in any year when total redemptions and coupons exceed the annual payment into the fund. First it is necessary to value the series at the date of issue. This value has already been found to be $107,096.8909. The annual payment to meet this must be 107,096.9809 times — at 4%, which is .12329094; the aio product is the annual payment, $13,204.0763. The schedule is given on the following page. Instalment bonds: Sometimes a bond is payable, principal and interest, at regular intervals and in equal amounts. This is the usual amortization problem. Con- sider a bond for $5,000.00 to be repaid in ten semiannual instalments with interest on outstandings at 4% nominal, convertible semiannually, payments to be equal in amount. The bond is to be priced on a 3% basis. The periodic payment is 5,000 X .11132653 = $556.6327. The present worth of these ten payments at 1^% is 556.6327 X 9.22218455 = $5133.3695. The two operations may be expressed in one formula: A = C ( — at rate c) X (an at rate i). an 116 MATHEMATICS FOR THE ACCOUNTANT Determining the income rate on a bond: This is a sub- ject into which the accountant need not go as deeply as the actuarial writers, but a fe^; of the simpler methods may be useful at times. For most purposes the income rate can be read accur- ately enough from bond tables. If the rate obtained from the tables is not acceptable, a closer rate may be obtained by interpolation between two table values. Selecting the nearest values above and below the given value proceed as in finding logarithms. A $1,000.00 bond redeemable at par after 40 years, bearing interest at 4%, is bought at 93.50, that is, $935.00. In the tables for a 40-year 4% bond we find yield 4.35 cost $942,955 4.30 952.116 differences .05 9.161 partial differences X 2.116 x: .05 = 2.116:9.101 X = .1155 rate = 4.26155% VALUATION OF BONDS 117 d O fl S c o o p a; O O O § 'S '^ : : : : ' '=''^^ lis MATHEMATICS FOR THE ACCOUNTANT After finding table values it is possible to proceed by the method of trial and error. That is, in the bond just discussed, after finding the 4.26155%, and ascertaining that the corresponding price is $949.99, the yield rate can be slightly increased and the bond valued logarithmically. Of course this will eventually give as correct a value as the number of places used can offer. It is veiy laborious, and generally unnecessary. It should be borne in mind that too high a yield rate results in too low a price; and too low a yield rate results in too high a price. The next method is given by Sprague, and is perhaps the best of all, in that the result of each step checks up the assumed rate. It is based on a $100.00 bond, and may be expressed in the following rule: 1) Assume an income rate; half of this will be i; 2) find an for $1.00 at rate i; 3) take hdf the result of 2) ; 4) divide the result of 3) into the known premium or discount; 5) if the bond was bought at a premium, subtract the result of 4) from c ; if bought at a discount, add ; 6) compare the result of 5) wuth 1); modify 1) and repeat until the result of 5) is near enough to 1) to be acceptable. Problem : To determine the income rate on a 5 % bond bought at 110, redeemable at par after 20 years. The successive steps may be: Trial Resulting Rate a n Rate 4 27.35548 4.268885 4.1 27.11714 4.262459 4.2 26.88179 4.25341 4.25 26.76526 4.252763 I 4.252 26.76254 4.252687 4.2524 26.75967 4.25261 VALUATION OF BONDS 119 The income rate is apparently about 4.2525. Valuing the bond at that rate we find $1,100.0134. Two of the many methods given by Todhunter are simple and fairly accurate if used with care. First, Let k be the premium per dollar of redemption A— C- value ;:that is k = — — — ' then k 1 = '-^x*^ Problem: Determine the income rate on a 4^% bond redeemable at $1,125.00 after 25 years, bought at 120. Adjusting c, the periodic coupon rate is 2%. A = $1,200.00 C = 1,125.00 A-C = 75.00 divided by C = .06666 this is k divided by 50 = .0013333 from c .0186666 This is the numerator of the formula. n + 1 =51 2n = 100 quotient .51 times k .034 add 1 1.036 This is the denominator. The quotient is 1.8053 semiannually, or 3.6106% nominal. The correct rate is 3.5936%, an error of .017 annually. If the bond is bought at a premium this result will be too high; if bought at a discount the result will be too low. 120 MATHEMATICS FOR THE ACCOUNTANT Second: Assume a rate i' and compute K' as for Make- ham's formula. If K' is not near enough to the given A, modify the rate i'. This part of the solution cnn be shortened by use of bond tables. After a satisfactory K' has been found, use this formula: "C-A i = c + i' Lc-K'J For the bond just considered the solution is: Assume i' = 1.8% semiannually. To find K' at this rate: yn = 1125 .007747778 .3873889 9.6126111—10 .409837 461.0666 this is K' -1200 1 1.018 nl times 50 colog In times 1,125 i = 2 + 1.8, L1125-461.0666J Note that the numerator of the parenthesis is negative, so that the value of the fraction must be subtracted from 2. i = 1.796664% or 3.5933%, annually. PROBLEMS ON CHAPTER XI Value these bonds (rates c and i are quoted for the whole year) : 122. par $1,000.00 c ' 4% i3% n 40 years 123. 1,000.00 33^% 23^% 5 build a schedule 124. 1,300.00 6% 3.6% 15 125. 1,000.00 4% 5% 45 126. 1,000.00 4>^% 33^% 333^ 127. 1,000.00 4% 3% 12 VALUATION OF BONDS 121 128. 1,000.00 7% 5% 10 redeemable at 108 129. 1,000.00 3H% 43^% 20 redeemable at 105 130. 1,000.00 5%JJ 3% due Jan. 1, 1918; purchased April 1, 1915 131. 1,000.00 7% 5% interest June 15 and Dec. 15, due June 15, 1916, purchased Oct. 1, 1898 132. 1,000.00 5% 6% 20 income tax 15% 133. 1,000.00 43^% 5% 30 redeemable at 102; income tax 25% 134. 1,000.00 33^% 3% 20 income tax 5% 135. A 4% bond AO, redeemable at $1,000.00 on Oct. 1, 1916, is purchased April 1, 1914, on a 6% basis. Value the bond and set up a schedule to be used as the source of entries on a set of books which are closed June 30 and December 31. 136. Find the value on a 4% basis of a series of bonds amounting to $100,000.00 issued on January 1, 1885, paying 5% J J, redeemable $2,000.00 each six months at par on and after July 1, 1910. 137. A series of bonds amounting to $50,000.00, bearing 53^% annually is to be redeemed $8,000.00 at the end of one year at 105, 9,000.00 at the end of two years at 103, 10,000.00 at the end of three years at 102, 11,000.00 at the end of four years at 101, 12,000.00 at the end of five years at 100. Value the series on a 43^% annual basis. 138. A series of bonds to the value of $250,000.00 bearing 5% annually is to be amortized over a period of ten years. Value the series on a 6% annual basis. 122 MATHEMATICS FOR THE ACCOUNTANT 139. Set up a schedule for the company issuing the above bonds showing how the annual payments on account of coupons and redemptions combined may be kept as nearly equal as possible. Redemptions will be at par S1,000.00. 140. A series of bonds to the amount of $500,000.00, bearing 6% annually, is issued January 1, 1908. Par of the bonds is $1,000.00. For 10 years they pay interest only, and beginning January 1, 1919, the series is to be amortized over a period of 20 years. What must be the annual payment into a redemption fund which can be accumulated at 43^2% annually for the full thirty years from January 1, 1908, to January 1, 1938, to meet the interest charges and retire the bonds? 141. A series of bonds to the value of $50,000.00 is issued under an agreement that they shall be redeemed $10,000.00 each year for five years. They pay 6% JJ, and are purchased on a 5% annual basis. What is the price? Find the income rate on the following bonds; if possible use the method given by Sprague (seepage 118), and check by one of the methods taken from Todhunter, or else by valuing the bonds logarithmically. Bond tables may be used for first approximations, but answers must be accurate to hundredths of a percent. 142. A 5% 50-year bond, redeemable at par, bought at 110. 143. A 4% 20-year bond, costing 115, redeemable at 108. 144. A 5% 20-year bond is quoted at "95 to yield about 6.4%." Is this accurate? If not, quote a more accurate yield rate. CHAPTER XII THE SLIDE RULE This chapter is a brief explanation of only such uses of the slide rule as are useful in general accounting practise. The real study of the slide rule must be in the form of practise; simple problems should be solved by arithmetic and the solutions checked by the slide rule. Only in this way can proficiency be attained. The slide rule is a mechanical device for performing multiplication, division, raising to a power (by continued multiplication), square roots, and in general the easier operations which can be performed by the use of logar- ithms. Among the commercial operations for which the slide rule is used are payrolls, invoices, discounts, setting prices on goods to be sold, inventories, determining rates of profit and loss, proportion, foreign prices and exchange, simple interest and discount. The mechanical and engineering uses are numerous, and valuable to the cost accountant. Such processes should be taught in engineering courses, however, and will not be mentioned here. There are many sizes and forms of calculating machines built on the logarithmic principle. Beside the usual eight and ten-inch slide rules there are twenty-inch rules for desk use. Special rules have trigonometric scales, scales of equal parts for use with logarithmic tables, cube root scales, and so on. Other machines are circular or cylindrical, thus securing greater accuracy by reason of their longer scales. Our discussion will concern the shorter slide rules, which are most useful to the accountant. Computations 123 124 MATHEMATICS FOR THE ACCOUNTANT with the ten-inch rule are accurate to the third digit, and the fourth place can be estimated with a fair degree of accuracy when the result is at the left end of the scales. Results to more than four places are useful only to check actual computation. The logarithmic scale: The logarithms of the numbers from 1 to 10 are: 1 nl 2 .301030 3 .477121 4 .602060 5 .698970 6 .778151 7 .845098 8 .903090 9 .954243 10 1 If we multiply each of these logarithms by ten and lay the lengths off in sequence on a line ten inches long we have this scale: wmm The usefulness of such a scale consists in the fact that as multiplication can be performed by adding logarithms, so multiplication can be performed by adding lines whose logarithmic measures correspond to the given numbers. For instance, the sum of the logarithms of 2 and 3 is .778151, which is the logarithm of 6. If we have two lines whose lengths are respectively the logarithms of 2 and 3, the sum of the lengths must be the logarithm of 6. This statement can be tested from the above scale THE SLIDE RULE 125 with the edge of a sheet of paper or any other straight edge, whether it has any scale marked on it or not. A slide rule is composed of three parts: stock, slide, and runner. The runner has a line or wire on its glass, to assist in locating or keeping the location of points. On the face of the rule are four scales, lettered A, B, C and D ; A and D are on the stock, B and C are on the slide. A and B are alike, being scaled from 1 to 10 twice; C and D are scaled only once. In other words, A and B are "double scales". Consider first the D scale; it is precisely like the scale just constructed from the logarithms, with the addition of subdivisions. These subdivisions are not on a logar- ithmic scale, but are equal divisions of the section of the scale in which they occur. This corresponds to the manner in which we interpolate in the use of logarithms, by ordinary proportion. Of course such results are not strictly accurate, but they are satisfactory for ordinary use. The meaning of these subdivisions is evident. If we suppose that the 1 at the left is 10, then the subdivision next it marked 1 is 11 ; and the subdivision next after that is 12. In the division beginning with 2, the subdivision marked 1 is 21, and so on. The subdivision which we called 11 is marked off into ten very small parts; the first of these is 111 with the decimal point wherever the problem requires; and so on. If the 1 at the extreme left were 1000, then the subdivision 1 would be 1100, and the sub-subdivision 1 would be 1110. The scale begins at 1 because the logarithm of 1 is 0, and zero length is at the beginning of the scale. This 1 may be 1 or 100 or 1,000,000 or .01; the logarithms of all these numbers differ only in their characteristics, and the slide rule is a scale of mantissae and not of characteristics. The characteristic in any result can 126 MATHEMATICS FOR THE ACCOUNTANT usually be determined by common sense although there are rules which will be given later. The A scale differs from the D scale only in that the logarithmic scale is repeated and the subdivisions are less numerous and shorter. If the 1 at the extreme left is 1, then the 1 in the middle is 10; if the 1 at the left is 100, the 1 in the middle is 1000. If the 1 at the left is .01, the 1 in the middle is .10; and so forth. Any number on the A scale is the square of the number directly below it on the D scale. For instance, move the runner until the wire is over 9 (not 90) on the A scale; directly under the wire on the D scale find 3. The rules for determining the characteristic are : In multiplication: If the slide projects to the right, the characteristic is the sum of the characteristics of the factors plus 1; If the slide projects to the left, sum of the character- istics. In division: If the slide projects to the right, the characteristic is the characteristic of the dividend less that of the divisor; If the sHde projects to the left, this difference less 1. A few simple illustrations : Multiplication: 3 X 4 = 12. On C and D; set 1 of C to 3 of D; below 4 on C read 12 on D. Notice that in order to get 4 of C over the D scale at all, the slide must be moved to the left. The characteristic is determined by the first clause of the rule for multiplication. On A and B; set 1 of B to 3 (not 30) of A; on A over 4 of B read 12. Note that the slide projects to the THE SLIDE RULE 127 right, but the characteristic is 1 as before; this is because the result is in the right hand half of the double scale A. Division: 35 ^5 = 7. On C and D; set the runner to 35 of D; bring 5 of C to the runner; below 1 of C read 7 on D. The difference of the characteristics is 1; but the slide projects to the left, so the characteristic of the quotient is 1 less than that difference. See the last part of the rule for the characteristic in division. On A and B: set the 5 of B to 35 of A; over 1 of B read 7 on A. Notice that the divisor is on the slide, and is set opposite the dividend on the stock. To square a number: 7' = 49. Set the runner to 7 of D; on A under the runner read 49. Square root: to find the square root of 81. Set the runner to 81 of A; under the runner on D read 9. Proportion: Most engineering problems involve the use of "gauge points", which are ratios comparing two quan- tities; for instance, the diameter and circumference of a circle; the speed of two shafts; and such ratios. Com- mercial gauge points are the ratio of a yard to a metre, foreign exchange rates, and many others. There is always a list of such gauge points as are- constant on the back of a slide inile. A few illustrations: What number of yards is equivalent to 350 metres? The gauge point is 75 :82 ; that is, 75 metres are equivalent to 82 yards. Set 75 of C to 82 of D; under 350 of C read 382 on D. 128 MATHEMATICS FOR THE ACCOUNTANT What is the value in francs of $125.00, exchange at 5.26. Proceed as in multipHcation : 125 X 5.26 = 657.5. At the same quotation what is the value in dollars of 2,000 francs? This is division. Set 5.26 of C to 2,000 of D; under 1 of C read 3803. The characteristic is 3—0 — 1 = 2; answer $380.30. Correct value, $380.25. An article costing $2.75 is to be sold to make 75% on the cost. Find the seUing price.. . The selling price is 175% of the cost. Multiply 2.75 by 1.75, product $4.82. An article costing $3.25 is to be sold to gain 55% on the selling price. Find the selling price. Evidently the cost is 45% of the selling price, and the solution requires division. Set .45 of C to 3.25 of D; under 1 of C read $7.22. Sometimes this use of gauge points leads to a puzzling situation. Problem: Find the weight of 10 gallons of water. The gauge point is 3:25; that is, 3 gallons weigh 25 pounds. Set 3 of C to 25 of D; the answer is expected under 105 of C, but this point is outside the stock. The shde must be thrown to the right its whole length ; to assure accuracy, set the runner to the one at the right end of the slide, and then throw the shde over until the one at its left end is under the runner. This multiplies the result by 10 Under 105 of C read 87.5. Successive operations: What is the interest on a $50.00 Liberty Bond at 3H% from November 15 to May 16? The time is 181 days. Expressed as cancellation, the problem is: 50 X .035 X 181 365 THE SLIDE RULE 129 Set 1 of C to 50 of D ; runner to .035 of C, and under it is one year's interest (characteristic 0). Set 1 of C to this result, runner to 181 of C (characteristic 2). Set 365 of C to the runner, and under 1 of C read 87. (Charac- teristic 2-2 — 1 = —1). Answer .87. PROBLEMS ON CHAPTER XII A. Drill problems to train in fundamentals: 36 X 14 82 X 74 875 -^35 4588 H- 74 28 X 36 ^ 63 mxl 23 « 23' >J/1681 ^15376 23* B. Commercial problems : 145. What amount is due to an employee who has worked 46 M hours at $18.00 for a 48-hour week? 146. What amount is due this employee if he works 334 hours' overtime, at "time and a half?" 147. A chair is listed at $16.50 less 10—10-^5%. What is the net cost? 148. Name a selling price on the above chair to gain 75 % on the cost. 149. A 60-day note for $318.00 is dated May 9 with interest at 53^% exact time. What is the value at maturity? 150. The above note is discounted May 11 at 6%, bankers' time. Find the net proceeds. 151. What is the proceeds of a Russian bill for 550 rubles, exchange at .12^? 130 MATHEMATICS FOR THE ACCOUNTANT 152. Wliat is the proceeds of a draft on London for £385 43., exchange at 4.855? 153. What draft on London can be purchased for $1,250.00, exchange at 4.8675? 154. When gold is quoted at $20.60 per ounce, what is the price in francs per gramme, exchange at 5.75? Gauge point, ounces to grammes 6:170. 155. What is the adjusted coupon rate on a 33^% bond redeemable at 108? 156. In taking an inventory in a hardware store 38 clamps were counted. They were billed from the manufacturer at $6.50 per gross less 25 — 10—10%. The dealer priced them to gain 50% on net cost. What is the inventory value, depreciation excluded? (Value one clamp and multiply by 38.) 157. A desk costing $85.00 is to be priced to gain 40% on the selling ^price. State the selling price. 158. A sprinter rah 100 metres in 11 seconds. What is the equivalent speed for 100 yards? 159. An article costing $25.00 is sold at $40.00 less 10 — 10%. What is the percent of profit on cost? 160. The financial report of a business showed the follow- ing facts, in thousands of dollars: Sales 315 Cost of Sales 189 Selling Expenses 68 General Expenses 22 Profit ? What percent of Sales is each of the four items? Total must be approximately 100%. CHAPTER XIII REVIEW PROBLEMS 161. What is the cash balance October 1, at 5% exact time, of the following account: Debits: May 8, $500.00, 2 months; June 25, $1,000.00, 60 days; August 10, $500.00, cash. Credits: July 1, $1,200.00, 30 days; September 15, $200.00, cash. 162. When a boy was born $500.00 was placed to his credit in a bank which pays 5% nominal, com- pounding semiannually. If the account is not dis- turbed until his twenty-first birthday, what will be the balance on that date? 163. A man buys a farm for $10,000.00, paying cash $2,500.00 and agreeing to pay the balance with accrued interest in three equal annual instalments, interest at 6% annually. What is the annual pay- ment? Prove by a schedule. 164. A clerk expects to go into business for himself as soon as he has saved $5,000.00. If he has now $2,500.00, and can invest all funds at 4)^% annually, how much must he save yearly so that he may have the necessary amount at the end of five years? 165. How many years will be necessary to accumulate $1,000.00 by investing ^150.00 per annum at 6% annually']? Solve without schedule. 166. A city issues bonds for $500,000.00 on which it pays interest at 5% annually. It accumulates a sinking fund at 4% annually to retire the bonds at the end of fifteen years. Find the annual payment into the 131 132 MATHEMATICS FOR THE ACCOUNTANT sinking fund and the amount in the fund at the end of five years, without schedule. 167. What sum must be put semiannually into an invest- ment which pays 4% effective in order to accumulate $1,000.00 at the end of five years? 168. A person invests $5,000.00 at 5% annually. Prin- cipal and interest are to provide a fixed income for ten years, at the end of which time the capital will be exhausted. The first payment is to be made one year from date. What is the annuity? Set up a schedule. 169. A mortgage for $5,000.00 was given on January 1, 1906, to be repaid in ten equal semiannual instal- ments with interest, beginning July 1, 1906. Find the periodic payment and set up a schedule, basing on 6% nominal, convertible semiannually. 170. According to the conditions of a will the sum of $20,000.00 is to be held in trust until it has increased to $25,000.00. If the fund can be invested at 5% nominal, compounding semiannually, when will the beneficiary receive the money. Solve without schedule. 171. A father bequeaths his son on his tenth birthday $10,000.00 worth of preferred stock which pays 6% dividends semi-annually. The will further directs that the income shall be invested by the trustee until the son is twenty-one years of age. Assuming that the stock continues to pay dividends regularly, and that these dividends can be invested at 5% nominal, convertible quarterly, find the value of the property at the end of the trustee's term. 172. How large an endowment is necessary for a room in a hospital, costing $5,000.00, to instal and $1,500.00 annually for maintenance? Assume 4% annually. REVIEW PROBLEMS 133 173. Is it more profitable for a city to pay $3.00 per square yard for paving that will last 5 years or $4.00 for paving that will last 7 years? Assume 5% annually. 174. A business man wishes to set aside on his forty-fifth birthday a sum which will give him an income of $1,000.00 a year for ten years, the first payment to be made on his sixtieth birthday. On a 33^% annual basis, what is the sum necessary? 175. An issue of $50,000.00 of bonds of par value'$ 1,000. 00 bearing interest at 5% annually, is to be retired as follows: For the first five years interest only will be paid, after which time the bonds and interest will be amortized over a period of five years. The money to provide for all this is to be raised by equal payments into a fund through the full ten years; the fund can be accumulated at 4% annually. What is the amount of this annual payment? Construct a schedule. 176. A bond redeemable at par $1,000.00 on January 1, 1920, bearing 6% JJ, is bought January 1, 1914, to yield 4%. Find the price? 177. A 6% bond, JJ, redeemable at par $1,000.00 on January 1, 1919, is purchased July 1, 1916, to yield 10%. Find the price. 178. A manufacturing concern contracts for a factory site for $20,000.00 cash and $5,000.00 at the end of each year for five years. If money is worth 5% annually, what would be a fair cash price for the lease? 179. A man borrows $5,000.00, agreeing to pay 6% interest annually, and repay the principal at the end of ten years. He accumulates a sinking fund at 4% annually. Find the annual cost of carrying the debt. 134 MATHEMATICS FOR THE ACCOUNTANT 180. Would it be better to amortize the above debt over the ten years by semiannual payments at 5% nom- inal, semiannually? 181. On his fortieth birthday a man begins to invest $1,000.00 a year. On his seventieth birthday he makes the usual payment and retires from business. (a) If money has been worth 4% annually through- out the period, what is the value of his investment? (b) Assuming that he will live not more than fifteen years, he plans to exhaust the investment in that time. If money will continue to earn 4%, what sum can he withdraw annually? (c) If he desired to have $10,000.00 left at the end of the time, what would be his annual allowance? 182. 300 members of the graduating class of a college plan to present their college with a scholarship fund of principal $100,000.00 on the twentieth anniversary of their graduation. They propose to accumulate this fund by annual payments, the first payment to be made one year after graduation. If the fund can be accumulated at 4% annually, what is the annual payment necessary from each member? 183. What would be the annual payment in the above problem if the first payment were made on the date of graduation, and the last on the twentieth anni- versary? 184. Explain the meaning of the terms cable, demand, and 60 day, used in quoting sterling exchange. If the cable rate is 4.20, and interest is reckoned at 4%, exact time, state the demand and 60 day rates. Allow 15 days for transit. 185. A debt of $200,000.00 is to be paid at the end of twenty-five years by means of a sinking fund into which annual payments are made. If the first pay- REVIEW PROBLEMS 135 ment is made one year from the date when the debt is incurred, and all sums are invested at an average rate of 43^% annually, what is the amount of the annual payment? 186. How much can a man afford to spend on a piece of equipment which must be replaced every five years at a cost of $1,000.00 in order to get better equipment which will last eight years? Assume 4% annually. 187. A plant consists of: X cost $20,000.00 scrap value $2,000.00 life 18years Y 11,000.00 1,500.00 15 Z 30,000.00 none 5 Find the annual depreciation charge on a 4% annual basis, sinking fund method. 188. Find the composite life of the above plant. 189. A bank loans a farmer $3,000.00 to be repaid with interest at 5 3/^% nominal, payments and interest semiannual for fifteen years. Find the semiannual payment. 190. A man was killed in an accident, and an industrial commission awarded his wife compensation to the amount of $5,000.00, but suggested that the amoimt be paid at the rate of $50.00 a month. On a 3% effective basis, how long will the payments con- tinue? j 191. A man begins at the age of twenty-five to save $10.00 a month and invest his savings in a bank which pays 5% nominal, compounding quarterly. How much will he have when he is sixty years old? 192. If the beneficiary of a life insurance policy for $5,000.00 chooses to accept settlement in twenty-five annual payments, the first to be made at once, what is the annual payment at 3% annually? 136 MATHEMATICS FOR THE ACCOUNTANT 193. A 5% JJ bond redeemable at par $1,000.00 twenty- eight years from date is advertised for sale at 92 "to yield about 5.60". How nearly correct is this statement? Is the yield above or below 5.60? 194. A banker wished to remit to his correspondent in Florence, Italy, 55,000 lire. Direct exchange was at 7.05. He bought $7,500 in pesetas, exchange at .203, and instructed his Madrid correspondent to change them to lire after deducting 3^% commis- sion. Madrid quoted lire at 67.50. (That is, 100 lire for 67.50 pesetas). a) Was the remittance sufficient; what was the margin or deficit in lire? b) If the remittance was insufficient, what should the banker have paid in dollars? c) Which was more profitable: direct exchange or arbitration? 195. What sum paid at the end of each year for five years will extinguish three debts, $1,000.00 due in three years, $650.00 due in one year, and $475.00 due in four years. Base on 5% annually. 196. Which is more profitable: to rent an ocean-going steamer for $30,000.00 per year for twenty years, payable in advance, or to buy it for $400,000.00, assuming that it will be worth $5,000.00 as scrap at the end of the twenty years? Base on 4% annually. 197. A debt of $8475.00 bearing interest at 4% annually is being repaid in annual instalments of $1,000.00. In how many years will the debt be discharged? 198. A bond of par $1,000.00, 5% JJ, redeemable at 105 on January 1, 1921, is bought on January 1, 1917, to yield 8%. Find the price and set up a schedule of accumulation of discount. REVIEW PROBLEMS 137 199. How much must be given to endow a motion picture outfit costing $3,000.00, scrapValue $500.00 at the end of eight years, annual operating expense $2,000.00, on a 5% annual basis? 200. A 5% JJ bond redeemable at par $1,000.00 on January 1, 1918, is purchased May 1, 1911, to yield 4%. What is the price? 201. A woman who has funds on deposit in a savings bank which pays 4% and compounds quarterly is con- sidering the purchase of bonds at par, $10,000.00, which will net 4.15% annually. Assuming that the bond interest will be paid promptly and that she will deposit it at once in the bank mentioned, what is the difference on the return for the first ten years on a deposit of $10,000.00? 202. Change f 25,000 to United States money, exchange at 5.165^. 203. If a mechanical piano for a dance hall costs $2,200.00, and saves $100.00 a month in wages of a pianist and repairs to the piano, in what time will the mechanical piano pay for itself on a 6% effective basis. 204. A farm yields an average annual crop of $5,000.00. The annual expenses are $485.00 for equipment, $350.00 for fertihzer, and $2,775.00 for wages, board and so on. Find a fair cash price for the farm at 6% annually. 205. Change $1,250.00 to francs, exchange at 6.57^. 206. Set up a schedule showing the reserve against a truck costing $3,500.00, scrap value $500.00 at the end of five years, by the method of fixed percentage on a diminishing book value. 138 MATHEMATICS FOR THE ACCOUNTANT 207. A manufacturing plant has the following assets: A wearing value $40,000.00 life 50 years B 25,000.00 25 C 5,750.00 18 D 6,750.00 20 E 2,400.00 5 What is the annual depreciation charge on a 4% annual basis, sinking fund method? 208. Change £382 14s. 7d. to dollars, exchange at 4.83875. 209. A corpoiation issues bonds for $200,000.00, par $1,000.00, paying 5% J J, redeemable at par in ten years. Is it better for the company to pay the semiannual interest and accumulate a sinking fund by semiannual payments improved at 4% nominal, compounding semiannually, to retire the bonds at maturity; or to buy bonds in the open market on each interest date at an average price of 105 through- out the ten years, so that the semiannual charges for redemptions and interest on outstandings shall be as nearly equal as possible. 210. $1,000.00 is invested in a 60 day bill on Paris ex- change at 9.50. At maturity the bill is sold at 9.20. If money is worth 6%, banker's time, what is the net profit on the investment. 211. A debt of $10,000.00 is to be amortized by ten equal biennial payments. If money is worth 6% annual, what is the biennial payment? 212. A charitable organization receives a bequest of $5,000.00 with the understanding that it is all to be used within three years. On a ^%% annual basis, what is the annual income? 213. Find the equated date for these transactions at 6% exact time: REVIEW PROBLEMS 139 Debits: May 11, $500.00; May 19, 82,000.00, 60 days: June 2, 90-day note for $500.00. Credits: May 24, $350.00; June 12, $850.00. 214. A 5% bond, AO, par $1,000.00, redeemable October 1, 1920, at par, is purchased October 1, 1915, at $950.00. What is the rate of income? 215. What is the price of the above bond to yield 6%? 216. Prepare a schedule for the bond of problem 215, changing to a December 31-June 30 basis, and carrying to maturity. 217. A man pays an annual premium of $33.38 on an endowment insurance policy. Payments are of course in advance. At the end of twenty years the insurance company pays him $1,000.00. What simi deposited semiannually in a savings bank which pays 43^% nominal, compounding semiannually, would have accumulated to the same amount? Assume payments at the beginning of the period. On this basis how much of each payment was the cost of the endowment as an investment, and how much was cost of insurance and loading? 218. "WTiat is the better yield : A 4% bond redeemable at par, bought at 95; or a 5% bond redeemable at 110, bought at 102. 219. What is the value in English money of a 90-day bill for $1,500.00 when the demand rate is 4.6875, and the discount rate is 33^^%? 220. Which is better for an American importer of English goods: a quotation of 4.65 on London; or 5.173^ on Paris and 25.40 in Paris on London? The face of the invoice is £10,000. 140 MATHEMATICS FOR THE ACCOUNTANT TWO EXAMINATION PAPERS Do any six examples, in any order.: Tables may be used. 1. Find the balance due on the following account at December 1, on a 5% basis, exact time: Debits: July 7, S500.00, 60 days; August 8, S600.00, 60 days; October 5, $500.00, 30 days; November 12, $450.00, 30 days. Credits: August 9, 60-day note without interest, $1,000.00; November 1, cash, $400.00; November 20, cash, $200.00. 2. Balance the following account as of October 1, without interest. State the balance both in dollars and sterling, exchange at 4.85^. State the apparent profit or loss for the month. Debits: Sept. 5, £350 @ 4.86^; Sept. 9, £225 8s. 5d. @ 4.86; Sept. 21, £310 6s. @ 4.86)^. Credits: Sept. 8, £300 @ 4.86 M; Sept. 16, £405 10s. 5d. @ 4.8615. 3. What is the cost of the draft in payment of an invoice amounting to f 15,000, exchange at 7.453^. 4. Find the amount and present worth of a 10-year annuity due of $200.00 a year at 33^% annually. 5. A concern issues $200,000.00 of serial bonds bearing interest at 6% annually. The trust agreement pro- vides that an annual payment of $20,000.00 shall be made to meet interest charges on outstandings and to retire as many bonds as possible. In what time will the entire series be retired.? Solve without schedule. 6. A coi-poration plans to accumulate a sinking fund to retire a bond issue of $250,000.00 maturing at the end of 25 years. If the trustee can keep the fund invested at an average rate of 4% nominal, compounding semi- REVIEW PROBLEMS 141 annually, what amount must the corporation pay him at the end of each six months? 7. In the preceding problem what amount ought the trustee's books to show at the end of the fifth year? Solve without schedule. 8. A machine costing $2,000.00 has an estimated life of 12 years and scrap value $500.00. The reserve is to be built up by the method of fixed percentage on diminishing book value. Show the percentage. 9. A 20-year bond redeemable at par $1,000.00 on January 1, 1939, bearing interest at 6% J J, is pur- chased January 1, 1919, to yield 5%. What is the price? 10. A $1,000.00 bond bearing semiannual coupons at the rate of 53^%, is redeemable at 110 after 18 years. Find the price to yield 4%. 1. On July 1 your account with your Chicago corres- pondent shows: Debits:" March 8, $500.00, 60 days; May 6, $500.00, 60 days; June 5, $250.00, 10 days. Credits: April 8, $500.00, 60 days; June 24, $300.00, cash. Settlements are made semiannually at 6% exact time. State the balance as of July 1. 2. On what date could the above account be settled with a minimum payment of interest; that is, what is the equated date? 3. a) Change $1,200.00 to sterling at 4.6565. b) At the same quotation change £255 13s. 7d. to dollars. 4. You are importing olive oil, and have these quotations per hundred litres: 142 MATHEMATICS FOR THE ACCOUNTANT From Florence, 600 lire, exchange at 7.55. From Marseilles, 425 francs, exchange at 5.70. Which is cheaper, and how much? 5. Find the amount of a 15-year annuity of $1.00 a year accumulated at 4^% annually. 6. A machine costing $3,000.00 has a life of four years and scrap value $200.00. a) Find the annual depreciation charge by the sink- ing fund method, at 4% annually .- b) Find the annual rate of depreciation by the method of fixed rate on diminishing book value. 7. I have $10,000.00 in a Trust Company which pays 4% nominal, compounding semiannually. Shall I draw the money and buy a 6% stock at 125, dividends payable semiannually, and invest the dividends in the same Trust Company. Which is the better in- vestment and how much over a ten-year period. 8. Value a $1,000.00 bond, redeemable after 13 years at par, on a 7% basis. Interest on the bond 6% J J. 9. Value a 4% AO bond, par $1,000.00, redeemable at 108 after 8 years, to yield 6%. 10. A series of $1,000,000.00 of bonds, par $1,000.00, bearing interest at 5% annually, is redeemable at par $50,000.00 each year from January 1, 1921, to 1940. Value this series as of January 1, 1920, on a 4% annual basis. CHAPTER XIV PROBLEMS FROM EXAMINATIONS OF THE AMERICAN INSTITUTE OF ACCOUNTANTS This chapter is intended to summarize briefly such parts of the book as are most useful to accountants who are preparing for the examinations of the American Institute of Accountants. The student should first read the paragraphs on reversed multiplication and division in chapter I. These shortened processes are valuable time savers in the rush of an examination, and the results obtained by them are sufficiently accurate. The chapter on averaging accounts should be read, as there was such a problem on a recent examination, and even the matter of equated date is sometimes required on various C. P. A. papers. The first matter of importance which has not been taken up in the body of the book is the averaging of an account with an English correspondent. This problem occurred on the Institute examination in November, 1918. A dealer in foreign exchange finds from his books that he has had the following transactions in London exchange during a particular month, viz. : Exchange bought in the local market : Jan. 1, 30-day bill payable in London £300 @ 4.75. 15, bill due at sight in London, £2,500 @ 4.76. Exchange sold in the local market: Jan. 5, bill due in London at sight £1,000 @ 4.77. 20, cable transfer, £2,000 @ 4.78. Foreign correspondent's draft honored and paid : Jan. 20, bill at 30 days after sight accepted Dec. 21, £500 @ 4.78. 143 144 MATHEMATICS FOR THE ACCOUNTANT State how the balance stands on the account at the close of the month, and how much profit or loss has been derived from the transactions. (At Jan. 31 the rate for cable transfers was 4.78). Is the profit or loss so stated final? The solution in the Journal of Accountancy for Feb- ruary, 1919, gave the following form as the usual one in the United States. The transactions are listed in both currencies, the sterling being used as an inventory. This inventory is valued at the end of the month, and the profit or loss determined from this. London Correspondent Debit Jan. 1, £ 300 @ 4.75 $ 1,425 15, 2,500 @ 4.76 11,900 20, 500 @ 4.78 2,390 31, Cr. Exchange a/c 4i) £3,300 S15,764~ Credit Jan. 5, £1,000 @ 4.77 $4,770 20, 2,000 @ 4.78 9,560 31, 300 @ 4.78 ba l. 1,434 £3,300 S15,764 The balance of sterling on hand January 31 was £300; the value of this at 4.78 was SI, 434. This item is entered on the credit side. Then the dollar columns are footed and balanced as usual, and there is found to be a profit of S49.00, which is transferred to Exchange account. The following comment is made in answer to the last question: "The profit is not final. The balance may not realize 4.78. Besides there is the interest on the thirty- day bill, and on the overdraft caused by the credit of January 20 which will not be covered until the sight draft of January 15 reaches London. Also the London corres- PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 145 pondent may charge a commission for handling the business." A similar problem occurred on the Massachusetts C. P. A. examination for 1914. The problem will be given without solution. A banking concern dealing in foreign exchange has the following transactions on its account with its London correspondent: Debits : Sept. 1, remittance, 30-day bill £400 @ 4.86. 10, remittance, sight bill £100 10s. @ 4.87. 15, remittance, demand bill £200 Os. 6d. @ 4.8G75 Credits : Sept. 2, sight draft £300 @ 4.873^. 12, demand draft, £200 12s. 5d. @ 4.87. 20, cable, demand, £100 @ 4.88. Ascertain the profit or loss on the account for the month of September and state the balance as of Sept. 30, in foreign and domestic currency, the current rate on that date being 4.89. Compound interest: The simplest method of arriving at the compound interest on a given amount for a given time and rate is to compute the compound amount first and deduct the principal. All such problems are worked on a basis of one unit: one dollar, one pound sterling, one franc, and so on. One unit at 5% for one unit of time, whether a year, six months or other period, earns .05 of itself, and so at the end of the period amounts to 1.05 of itself. This new principal during the next unit of time earns .05 of itself, and at the end of the period amounts to 1.05 of its value at the beginning of the period. The simplest way to arrive at this amount for two periods is therefore to multiply 1.05, the amount at the end of the first period, by 1.05, the ratio of increase during the 146 MATHEMATICS FOR THE ACCOUNTANT second period. Further, note that the ratio of increase during any period is 1.05; that is, the amount at the end of any period is 1.05 of the amount at the beginning. To find the amount at the end of the sixth period, for in- stance, multiply 1.05 together 6 times. Algebraically expressed, this is (1.05) ^ which we read 1.05 to the sixth power. In multiplying, it is of course wise to use reversed multiplication. The 1.05 is called the base, and 6 is called the exponent, showing how many times the base must be used as a fact . Another shortcut: It is not necessary to multiply five times to obtain this result. When we have multiplied 1.05 X 1.05 X 1.05 we have obtained (1.05)3. Now if we multiply this result by itself, we shall be in effect multi- plying by 1.05 X 1.05 X 1.05 all over again, except that we multiply once instead of three times. This leads to an important algebraic truth which was necessary in answering a question which occurred on the 1917 examination. The truth referred to is this: since (1.05)3X(1.05)3=(1.05)«, m general whenever exponents are added, bases are multiplied. The problem was this: You are called upon to state what is the annual sinking fund payment necessary to redeem a principal of $1,000,000.00 due 30 years hence, it being assumed that the annual sums set aside are in- vested at compound interest at 5% annually. State what computation you would make to arrive at the result desired. You need not work out the computation. The solution of this problem requires the computation of (1.05)^". The shortest solution is: Find (1.05)2 That quantity multiplied by itself gives (1.05) rt u u u u « (1,05)8 U u u a u u (1,05)^^ a a u u u (I (1.05)^2 PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 147 This result is two periods more than the required time. Since adding exponents is the same as multiplying bases, evidently, subtracting exponents is the same as dividing bases. So (1.05) =« 2 divided by (1.05) ^ will give (1.05) »°. The finding of the sinking fund payment will be shown later. To return to the matter of compound amount, the simplest method of working out the amount of one unit for any rate and time is to multiply 1 plus the rate of interest by itself, or to multiply multiples of this quantity together, until the desired total of exponents is obtained. This compound amount is often indicated by the symbol s °. This quantity is usually given in problems where it must be used in finding a sinking fund payment, or summing an annuity, or in other cases. Compound interest for any period is, as has been said, amount less principal. In many problems the interest is in question rather than amount, so this fact should be kept in mind. Present worth: One very common use for compound amount is in finding present worth. If a dollar will amount in five years at 5% to $1.27628156, then the principal required to accumulate $1.00 at the end of five years at 5% must be 1 divided by 1,27628156, or $.78352617. In general, if 1 amounts in n years at rate i to (1+i) "^j ors °, then the present investment which will yield 1 is ^ — or — The rule for finding present worth (1+1)°, s°. then is to divide compound amount into 1. The symbol for this quantity is v °. It is usually given when it is necessary in working out an amortization factor. But if s ° happens to be given, v ° can easily be found by division. Contracted division should be used in this computation. 148 MATHEMATICS FOR THE ACCOUNTANT Compound discount is simply the discount found by compound interest methods. We are familiar with dis- count on notes, found by simple interest. It is the difference between present value and the par value at some future date. So compound discount is the difference between 1 and the present worth of 1. Expressed as a rule, subtract the present worth from the par value due at some time in the future. This quantity is important in annuity calculations. The compound discount on 1 for.5 years at 5% is 1— .78352617, or .21647383. Annuities: An annuity is a series of equal payments made at equal intervals of time, improved at a fixed rate of interest. The rule for finding the amount of an annuity is: Divide the compound interest for the given rate and time by the interest for one period, and multiply by the periodic payment. This is demonstrated with schedules at the beginning of chapter VII. For instance, the amount of payments of 1 each at the end of each year for 15 years accumulated at 4% annually, given that the compound amount of 1 for the same time and rate is 1.80094351, is found as follows: compound amount 1.80094351 compound interest .80094351 single interest .04 quotient 20.02358775 Note that the payments are held to be made at the end of the period. This is because in the business world such payments are usually made out of profits, which are reck- oned at the end of the fiscal period. If the payments are made at the beginning of the period, the result is an annuity due, as it is called. The amount of such an annuity can be found from the amount of an ordinary annuity by multiplying by 1 -f- i. In the above annuity. PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 149 if the payments were made at the beginning of each period, the amount would be 20.02358775 X 1.04 = 20.82453126. The present worth, or cash value, of an annuity is a term used in two senses. It may mean first the cash pay- ment which will extinguish a debt which was intended to be repaid in annuity form with inteiest charged on balances outstanding. Second it may mean the present investment which will yield a given income; the interest on the portion remaining invested increases that protion and therefore increases the term during which the income can continue. This last is the annuity in the banking and insurance sense. The rule for valuing the present worth of an ordinary annuity is : Divide the compound discount for the given time and rate by the interest for one period, and multiply by the periodic payment. This is also demonstrated with schedules in chapter VII. For instance, the present worth of the annuity which we summed just above, is found as follows, given that the present worth of 1 for 15 years at 4% annually is .5552645: present worth .5552645 compound discount .4447355 single interest .04 quotient 11.1183875 The present worth of an annuity due is found by multi- plying the present worth of an ordinary annuity by 1 + i. If the above were an annuity due its present worth would by 11.1183875 X 1.04 =11.563123. One more sort of annuity which occurs frequently is the deferred annuity. This is an aiuiuity of n payments which does not begin until after m periods; that is, the first payment is to be made at the end of the first period after m periods have expired. For instance, the above 15-year annuity might be deferred 5 years. This means 150 MATHEMATICS FOR THE ACCOUNTANT that the first payment is made at the end of the 6th year. There are two methods of finding the present worth of such an annuity. We have already found that at the beginning of the sixth year, which is the same as the end of the fifth year, the present worth is 11.1183875. Since this amount is not due for five years, it must be discounted to the present time. The proper method of doing this is to multiply by the present worth of 1 for 5 years at 4%. If the problem does not state this we must proceed as in the model problem. The present worth is 1 divided by the compound amount. By multiplication we find this com- pound amount to be 1.2166529, which, divided into 1, gives as quotient the present worth, .82192711. Now multiply this by 11.1183875, and we have the worth of this 15-year annuity, which will not begin for five years; this value is 9.13850408. The other method is to subtract the present worth of an annuity for the term of deferment from the present worth of an annuity from the present time to the time of the last payment. For the annuity we are discussing, subtract the present worth of a 5-year annuity from that of a 20-year annuity. If present worth of a 20-year annuity is 13.59032634 and of a 5-year annuity 4.45182233 subtracting, we have 9.13850401 the present worth of a 15-year annuity deferred 5 years, interest at_4% annually. The following Institute problem included both amount and present worth of an ordinary annuity: A owns an annuity of $50.00 per annum, the first payment on which falls due one year hence, and continues for a period of 20 years certain. State PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 151 a) the present value of the benefit; b) the amount which he will have accumulated at the end of the period if he invests each moiety as it becomes due. Assume interest at 4% payable annually. In this con- nection it is stated that the value of 1.04 ="> is 2.191123. , , ,, compound discount a) present worth = single interest compound amount 2191123 by division .456387 is present worth compound discount .543613 divided by .04 present worth of an annuity of 1 13.5904 times 50 $679.52 If only the present worth had been required very likely the present worth of 1, namely v^o, would have been given. In that case the first division would have been avoided, and the solution would begin with finding the compound discount. , , , compound interest b) amount = -. — - — : smgle mterest compound amount 2.191123 compound interest 1.191123 single interest .04 quotient 29.778075 times 50 $1,488.90 On the May, 1919, examination occurj-ed this problem: A lease has run five years to run at $1,000.00 a year pay- able at the end of each year, with an extension for a further five 3 ears at $1,200.00 a year. On a 6% basis what sum should be paid now in lieu of the ten year's rent? v^ at 6% = .7473. 152 MATHEMATICS FOR THE ACCOUNTANT for the first five years: present worth .7473 compound discount .2527 divided by .06 4.211667 times 1,000 $4,211.67 For the last five years, the value at the beginning of the sixth year is, as above 4.211667 discount this 5 years, mul- tiplying by v^ .7473 product 3.1474 times 1,200 3,776.8 5 total $7,988.52 Comment on the solution: The present worth of the rentals for the first five years is found by the rule for present worth: divide the compound discount by the single interest. The last five years' rentals constitute a five-year annuity deferred five years. The present worth at the beginning of the sixth year is the same per dollar as for the first five years. But this must be discounted from the beginning of the sixth year to the beginning of the first year, multiplying by v^ If more digits had been given in the value of v^ the problem might have been solved by the second method. v^° = v^ times v^; .7473 X .7473 = .5584, about. present worth .5584 compound discount .4416 divided by .06 7.36 the present worth of a 10-year annuity, subtract present worth of a 5-year annuity 4.2117 giving 3.1483 present worth of the deferred annuity, times 1,200 = $3,777.96, which is very in- accurate, owing to the lack of sufficient digits- PROBLEMS FROM EXAMINATIONS OF TliE A. I. A. 153 Sinking funds: We saw that if 1 per annum for 15 years was improved at 4% annually, the sum at the end of the 15 years was 20.02358775. Per contra, if it is desired to make an annual payment for 15 years such that at the end of the 15 years the sum shall be 1, this annual payment is 1 divided by 20.02358775, which is .049941 1. Now this 20.02358775 was obtained by dividing the compound interest by the single interest. Evidently the annual payment which will amount to 1 should be obtained by dividing the single interest by the compound interest. This annual payment which is to amount to a given sum is used most commonly in finding payments into the various reserves, whether for sinking funds or for depreciation. This problem occurred on a recent examination: In auditing the books of a corporation you find that in order to provide a sum to redeem a mortgage of $100,000.00 faUing due at the end of ten years, a reserve of $8,000.00 per annum has been set aside for three years, but that contrary to intention the company has failed to accumulate interest thereon. Assuming interest at 4% convertible annually, what should have been the total accumulations to date and what amount should now be set aside for the next seven years in order to complete the sinking fund. 1.04' = 1.31593. The answer to the first question requires the summing of a three-year annuity of $8,000.00 at 4%. First we must find 1.04 3; then compound amount 1.124864 compound interest .124864 single interest .04 quotient 3.1216 times 8,000 $24,972.80 154 MATHEMATICS FOR THE ACCOUNTANT This is the amount which ought to be in the fund. Whether it is advisable to assume that the interest of $972.80 is added to the fund by an adjusting entry is questionable. There is another question which depends on the date at which the audit actually takes place. If it took place well along in the next fiscal year, we might assume that the amount in the fund would be increased by one year's interest and brought on the books at the end of the year. The remainder of the solution given here assumes that no interest was added, but that the fund was left at $24,000.00, and that the remaining payments must supply the balance of $76,000.00. The rule for finding a sinking fund payment has been given: divide the single interest by the compound interest: compound amount 1.31593 compound interest .31593 dividing into .04 quotient .1266 This is the sinking fund payment per dollar. Multiplying by 76,000 we have the annual payment, $9,621.60. On the Michigan C. P. A. examination in 1915 was this problem : A contractor proposes to build a bridge to Belle Isle and accepts the city's 4% 20-year bonds in payment to the amount of $2,000,000.00. He advocates as a means of retiring the bonds the establishment of a toll system on foot passengers and automobiles at the respective rates of one and five cents each. Assuming the ratio of foot passengers to automobiles to be ten to one, how many of each w^ould be necessary to pay the interest annually and create a fund which, placed at the same rate of interest, would be sufficient to retire the bonds at maturity. $1.00 compounded at 4% for 20 years will amount to 2.19112314. PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 155 The solution will not be given, but the number of foot passengers is 9,810,893. A different type of sinking fund problem is the follow- ing: Argument has been strongly urged that, aside from any question of possible mismanagement or of the difiB- culty of making satisfactory investments to yield the same rate of interest as the bonds, a sinking fund for bonds is more expensive than an arrangement for serial repayment of the bonds. This is illustrated by the case of S20,000.00 of 5% bonds. If these are paid off in a series, one each year, the total payment will be principal $20,000.00, interest $10,500.00, total $30,500.00. The annual sinking fund to pay off these bonds would on a 5% annual basis amount to $604.85, making in 20 years $12,097.00, and the interest paid on the bonds would be $20,000.00, total payments $32,097.00. The apparent excess burden is accordingly $1,297.00. Discuss the above argument and show clearly just what the figures mean and in what the apparent saving actually consists. The solution given in the ''Journal of Accountancy" read as follows: "If the bonds are serial, one bond being paid off each year, the average capital of which the company would have the use would be $10,500.00. As the interest is the same amount, the company paid 100% in 20 years or 5% per annum. "By the sinking fund method the bonds were virtually paid off at the rate ot $604.85 per year. This means that the available capital in use was diminished by that amount at the end of each year — that is, that the com- pany had the use of $20,000.00 the first year, $19,395.15 the second year, and so on. As the last payment was made at the end of the twentieth year, they had the use of $20,000.00 less 10 X $604.85, or $8,507.85. This makes 156 MATHEMATICS FOR THE ACCOUNTANT an average capital in use of $14,253.92. For the use of this average capital the company paid interest $20,000.00, against which there is a credit of $7,903.00, the compound inctrest reahzed from the sinking fund. This means that the net interest charge was $12,097.00. $12,097.00 for the use of $14,253.92 means a rate of a trifle less than 84.87% for twenty years, or 4.1435% per annum. The advantage of the sinking fund method is apparent, and is explained by the fact that the fund earns compound interest. "The apparent excess burden mentioned in the problem is not a true excess. It is reached by calling the $20,000.00 paid for coupons all interest, which is not true. Of this amount $7,903.00 was applied to the principal, being the difference between the face of the bonds, $20,000.00, and the actual payments to the sinking fund, $12,097.00. The true amount of interest was therefore $12,097.00, not $20,000.00. The interest paid must be considered in relation to the capital of which the corporation had the use during the 20 years, not in relation to the face of the bonds." This closes the quotation. The solution does not re- quire any computation, but does bring out the idea of average capita), and is an interesting comparison of re- demption oy smking fund and serial redemption in equal amounts. If, however, the redemption had been on an amortization principle, by which the annual total pay- ments of interest and principal together had been as nearly equal as possible, there would be practically no difference in dollars and cents between amortization and the sinking fund redemption. Depreciation: As has been said, the depreciation re- serve is usually built up on the sinking fund principle. PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 157 The determination of the periodic payment under those conditions would be by the process just demonstrated. Another method of reckoning depreciation is by com- puting affixed percentage periodically, based on a dimin- ishing balance. After the percentage has been determined, the annual deduction from the net book value as at the beginning of the year can be easily found. This method is sometimes required in accounting problems, but the percentage is fixed arbitrarily, and has no reference to any estimated life of the asset. There is a formula by which the percentage can be computed, but the com- putation requires the use of logarithms unless the figures in the problem are "doctored". The following problem illustrates this: A machine costing $81.00 is estimated to have a life of four years with a residual value of $16.00. Prepare a statement showing the annual charge for depreciation under each of these methods: a) straight line; b) constant percentage of diminishing value ; c) annuity method. For convenience in arithmetical calculation assume the rate of interest to be 10%. a) The wearing value, as it is called, is $65.00; that is, that amount of capital is sunk in the machine and will never be reahzed. In a Hfe of four years, the annual consumption of capital by the straight-line method is 1/4 of that amount, or $16.25. b) The formula for determining the constant percent- 1 / g age is r = 1 — V — C Expressed as a rule: Divide the scrap value by the cost; find the number which, multiplied together n times, will give that amount; subtract from 1. 158 MATHEMATICS FOR THE ACCOUNTANT In this problem, S is 16; C is 81; n is 4. 16 _^ 2 ^ ^ sT •" 3 ^ 3 ^ 3 ^ 3 16 2 so the fourth root of — is -, or as a decimal. 81 3 .66 V 3, or as a percent, 662/3%. Subtracting from 1, we have the annual rate. 33V3%. It is well to subjoin this schedule: Schedule Showing Depreciation Charged Against Asset , Costing S81.00, Life 4 Years, Estimated Scrap Value, $16.00 Charged at 33V/ 3% Annually Based on Net Book Value at Beginning of Each Year Value at Amount in Beginning of Depreciation Reserve at ear Year 33V/ 3% End of Year i $81.00 $27.00 $27.00 2 54.00 18.00 45.00 3 36.00 12.00 57.00 4 24.00 8.00 65.00 'emair ider $16.00 Total $65.00 The computation by the sinking fund method is: Compound amount, 4 years, at 10% annually 1.4641 compound interest .4641 divide into .10 quotient .21547 times 65 $14.00555 The schedule, omitting title, is: PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 1^9 Interest Total Amount on Fund, Annual Addition in ear 10% Payment to Fund Fund 1 $14.00555 $14.00555 $14.0055,1 2 $1.4006 14.00555 15.40615 29.41170 3 2.9412 14.00555 16.94675 46.35845 4 4.6358 14.00555 18.64135 64.99980 $8.9776 $56.02220 164.99980 The annuity method assumes interest on the net book value. It is developed by the following reasoning: The machine at first cost a certain amount. At the end of the life of the machine the amount of capital that has been tied up is not merely the original cost, but the compound amount of that cost. Of this compound amount a certain part is returned in the form of scrap allowance. The balance is total loss, and should be spread over the life of the machine in the form of a sinking fund charge. Expressed as a rule: Multiply the cost by the compound amount; subtract the scrap value; multiply by the sinking fund factor. For this machine : cost $ 81.00 compound amount factor 1.4641 product 118.5921 less scrap 16.00 loss of capital 102.5921 sinking fund factor .21547 product 22.1055 The schedule is: Net Book Annual Net Reduc- Value at Interest Depreciation tion of Year Beginning 10% Charge Book Value 1 $81.00 $ 8.10 $22.1055 $14.0055 2 66.9945 6.6995 22.1055 15.4060 3 51.5885 5.1588 22.1055 16.9467 4 34.6418 3.4642 22.1055 18.6413 $16.0005 $23.4225 $88.4220 $64.9995 160 MATHEMATICS FOR THE ACCOUNTANT Amortization: Just as the sinking fund factor is the reciprocal of the amount of an annuity, so the amortiza- tion factor is the reciprocal of the present worth of an annuity. We saw that the cash payment which would yield an income of one per annum for 15 years if interest was allowed on balances at the rate of 4% per annum was 11.1183875. If on the other hand we invest 1 now and desire an annual income for 15 years on a 4% annual basis, evidently that annual income is 1 divided by 11.1183875, which is .0899411. The rule for finding the present worth of an annuity was to divide the compound discount by the single interest; so the rule for finding the amortization factor is to divide the single interest by the compound discount. But the finding of compound dis- count may present difficulties, when the compound amount is given, as it was in the problem on the Institute paper of 1918; because the present worth must be found first, with possibilities for mistakes in the work and probability of error in the last digits; then the compound discount must be found by subtraction. But note that the amortization factor is .0899411, while the sinking fund factor is .0499411, a difTerence of .04, the interest rate. This fact can be proved by algebra, and is in- vestigated in a schedule in chapter X. The simplest way to find the amortization factor, given the compound amount, is to find the sinking fund factor first and add the interest rate. That method will be employed in solv- ing the Institute problem referred to: A corporation wants to retire a debt of $105,000.00 bearing 5% interest payable annually. The tenth pay- ment, including interest, is to be $15,000.00. The other nine periodical payments are all to include interest and to be of the same amount. Required the amount of each of the nine payments. 1.05* = 1.551328. PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 161 The payment of S15, 000.00 at the end of the ienth year represented a balance due at the beginning of the year plus 5% interest on that balance accrued during the year. In other words, the $15,000.00 is 105% of balance due at the beginning of the year. 15,000 -5- 1.05 = $14,285.71 from 105,000 = 90,714.29 the amount to be amortized over nine years. To find the amortization factor: compound amount 1.551328 compound interest .551328 divided into .05 sinking fund factor .0906901 add .05 .1406901 multiply by 90,714.29 $12,762.6025 the annual amortization of principal and interest. To' which must be added every year the 5 % interest on $14,285.71 71 4.2855 Total annual cost $13,476.8880 But if the problem had stated the present worth of 1 for nine years at 5%, .644609, it would have been a simple matter to find the amortization factor directly: present worth .644609 compound discount .355391 divided into .05 .1406901 as above In general, whenever the amount of 1, or s °, is given, it is easier to find the amount of an annuity or the sinking fund factor; whenever the present worth of 1, or v°, is given, it is easier to find the present worth of an annuity or the amortization factor. The four rules are : compound interest amount of annuity- single interest 162 MATHEMATICS I OR THE ACCOUNTANT . , . . , , ^ single interest sinking fund factor compound interest ^, , .^ compound discount present worth of annuity — single interest , . , . . , single interest Amortization factor — or compound discount; amortization factor = sinking fund factor plus single interest. Valuation of bonds: The handiest formula to use in bond valuation is that of Mr. Makeham. It requires only the finding of v °, the present worth of 1. It is as follows: Let C be the redemption value; the coupon rate; n the life of the bond; i the desired income rate; K the present worth of C at rate i. The value of this bond consists of the present worth of the redemption value, which is K, and the present worth of the coupons at the desired yield rate i. If the yield rate were the same as the coupon rate, the bond would obviously be purchased at par. The excess over par, called the premium, or the deficiency under par, called the discount, depend on the relation existing between these rates. If the purchaser is satisfied with a yield rate less than the coupon rate, he will be willing to pay more than par for the bond. If he desires a larger yield than the coupon rate, he will purchase at a discount. The measure of this premium or discount is the ratio of coupon to yield. It should also be kept in mind that most bonds pay semiannual coupons, and that consequently the annual coupon and yield rates must be halved, and the time must be doubled, to conform with the half-year periods. The following problem will illustrate this: PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 1G3 Vahie a 20-year 5% bond, redeemable at par $1,000.00, on a 4% basis. Given that v*" at 2% is .45289. C = $1,000.00 c = .025 n = 40 i = .02 K = 1,000 X .45289 = 452.89 C-K = 547.11 If the bond were purchased to yield 5%, that is, to yield the coupon rate, then $547.11 would be the present worth of the coupons. This can be proved by computing the present worth of a 40-period annuity of $25.00 at 23^%. But the yield is only 2%, so that the coupons must have cost not; ^^^ of $547.11 = $ 683.89 .02 adding the above 452.89 cost of the bond $1,136.78 Expressed as a formula, Makeham's rule is. Value of bond = Cv " +— (C— K). 1 Rule : Find the present worth of the redemption value ; call this K. Subtract this from the redemption value, multiply by the coupon rate, and divide by the income rate. Add this last result to K. Another probleni will illustrate this for a bona bought at a discount: Value a 15-year 5% bond redeemable at $1,000.00 on a 6% basis, v^o at 2^% = .476743. C = $1,000.00 c = .025 n = 30 i = .03 164 MATHEMATICS FOR THE ACCOUNTANT K = 1,000 X .476743 = 476.743 C — K = 523.257 .025 .03 X 523.257 = 436.048 value $912,791 Valuation by discounting: Up to the present time the Institute problems have all treated short-term bonds. In such cases the formula is not necessary, it being easier to discount from maturity. For instance, a 5% bond, redeemable at 1,000, is bought on a 4% basis; that is, the semiannual yield is 2%. At maturity the holder will receive par $1,000.00 plus coupon $25.00. So at the beginning of the last half-year the value was 1,025 h- 1.02. At that time a coupon was payable to the amount of $25.00. This increased the value of the bond by that amount. Going back to the beginning of the previous half-year, the value is found by discounting, that is, dividing by 1.02; and another coupon again increases the value of the bond. Presented in a vertical schedule At maturity, par $1,000.00 plus coupon 25.00 $1,025.00 value 6 months before, divide by 1.02 1,004.9019 plus coupon 25.00 $1,029.9019 value 6 months before that date, divide by 1.02 1,009.7078 plus coupon 25.00 $1,034.7078 Continue until the date stated in the problem has been reached. This method is slightly less accurate than the valuation by formula, because of the error in the right PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 165 hand digits caused by the division. The method is useful only in cases where speed is necessary, and only when the term of the bond is short. The above problem occurred on the Institute examina- tion of November, 1918. The redemption value of the bond in that problem was $10,000.00, which gives the value one year before maturity $10,347.0780 value 1 Yi years before maturity, divide by 1.02 10,144.194 if bought ex interest ; otherwise add coupon 250.000 Value $10,394,194 Redemption at a premium or discount: On the May, 1919 examination of the Institute was this problem: A bond bearing interest at 5% payable annually and repayable in 5 years with a bonus of 10% is for sale. What price can a purchaser pay who desires to realize 6% on his investment? v^ at 6% is .7473. The noteworthy feature of this problem is that the bond is redeemable at a 10% premium; that is, the value of C in Makeham's formula is $1,100.00. This has no effect on the yield, for the problem expressly states that the purchaser desires to realize 6% on his investment. But the coupon rate is 5% on the face of the bond, regardless of the redemption value. And $50.00 on 1,000 is not 5% on the investment as purchased. 5% on 1,000 is only — of 5% when adjusted to the redemption value 1,100. And since the purchaser bases his valuation on this redemption value, the coupon rate must also be based on it. We can value this bond by Makeham's formula more quickly than by discounting: 166 MATHEMATICS FOR THE ACCOUNTANT G c = n = $1,100.00 4 6/11% 5 i = K = 1,100 X .7473 = C — K = 277.97 6% 822.03 * «/" X 277.97 = 6 210.58 Value The solution by discounting is: Value at maturity plus coupon $1,032.61 $1,100.00 50.00 $1,150.00 value one year before, divide* by 1.06 1,084.9057 plus coupon 50.00 $1,134.9057 value two years before, divide by 1.06 1,070.6658 plus coupon 50.00 $1,120.6658 value three years before, divide by 1.06 1,057.2319 plus coupon 50.00 $1,107.2319 value four yeaTs before, divide by 1.06 1,044.5584 plus coupon 50.00 $1,094.5584 value five years before, divide by 1.06 1,032.602 if purchased ex interest. Valuation between interest dates is the usual situation in actual practise. In such cases the value as of the last interest date must be first determined. The change in value up to the next interest date consists of two things: that part of the coupon and of the premium or discount proper to the period. The difference between coupon and PROBLEMS FROxM EXAMINATIONS OF THE A. I. A. 167 premium or discount is the net change in the value of the bond. If the bond is below par this difference causes an increase in value ; if the value is above par, this difference causes a decrease in value. The simplest method is to find the difference between coupon and discount or pre- mium and add it to the value of the bond if below par, or subtract it if the bond is above par. If the bond in the last problem had been purchased three months after the date when its value was $1,032,602, the value would have been: Value on last interest date $1,032,602 1/4 of 6% of that value 15.4890 less 1/4 of $50.00 12.50 2.989 adding, because the value is below redemption value $1,035,591 Serial bonds have never been required in Institute examinations. In fact, the valuation of a series is usually a complicated process, because redemptions are usually at a premium which varies as the final date approaches, and the amount redeemed at the various dates is often variable. If a series is regular in every way (that is, if the redemptions are equal in amount, interval, and redemp- tion value), the series may be valued by Makeham's formula. In this case K is the present worth of a series of payments, and the present worth of an annuity must be found and used in place of v °. Problem: Value a series of 20 $1,000.00 bonds, re- deemable at par, one each year, beginning one year from date, on a 4% annual basis. The bonds pay 5% annually. 1.0420 = 2.1911231. We must find the present worth of an annuity: Compound amount 2.1911231 present worth .4563869 single interest .04 divided into .4563869 = 13.5903263 1G8 MATHEMATICS FOR THE ACCOUNTANT Each redemption is $1,000.00; the present worth is 1,000 X 13.5903263 = 13,590.3263 C— K = 6,409.6737 - X 6,409.6737 = 8,012.0921 4 Value $21,602.4184 This closes the discussion from the investor's standpoint. There remains one phase of the bond problem from the standpoint of the issuer: the writing off of premium or discount at which the bonds were originally sold. Such premium or discount ought to be amortized or accumu- lated on the annuity principle, but many bookkeepers are not competent to do this work, and the preparation of the schedule is a tiresome task. An alternative method, called the ''bonds outstanding method," was required in the following problem: The A company issues $100,000.00 bonds, maturing as follows : $ 50,000.00 in 1 year 75,000 " 2 « 100,000.00 « 3 « 150,000.00 " 4 « 125,000.00 " 5 " These bonds were issued at 90. State the amount of discount to be written off annually, using the bonds out- standing method. The solution in the Journal of Accountancy opened with this quotation from Dickinson's "Accounting Prac- tise and Procedure": "There are various methods in use for determining the proper interest charge to be made to income account under the varying conditions which arise. PROBLEMS FROM EXAMINATIONS OF THE A. I. A. 169 "The first and most correct method, which may be called the effective interest method, consists in charging to income account the effective interest calculated from the known conditions of issue upon the whole amount outstanding during the year. "The second and more common method, which may be called the equal instalment method, is to ignore altogether the effective interest rate; to charge to income account each year the interest actually accrued, together with a proportionate part, according to the term of issue, of the discount on issue or premium on redemption. "A third method which may be called the bonds out- standing method, which may be safely adopted where by reason of complication in the terms of issue or redemp- tion it is difficult or impracticable to determine the true interest rate, is to distribute the discount or premium over the period in the proportion that the bonds out- standing for each year bear to the sum of the bonds outstanding for all the years of the currency of the loan. "When a large accumulated surplus is available the practise is. irequently adopted of charging the whole discount on issue to profit and loss account. A great objection to this practise is that thereby the true rate of interest on loans during their currency is entirely lost sight of; current fixed charges are understated; and the discount is charged against surplus arising out of previous operations instead of against income from current opera- tions which should meet it. "The charge to income account by bonds outstanding method is so close to that given by the effective interest method that for all practical purposes it may safely be adopted." 170 MATHEMATICS FOR THE ACCOUNTANT Table Showing Bonds Outstanding Each Year, and Pro- portion to Total Outstanding during the Entire Term of the Loan, and Corresponding Reduction of Discount Par of Bonds Proportion Reduction of Year Outstanding of Total Discount 1 $ 500,000.00 20/69 $14,492.75 2 450,000.00 18/69 13,043.48 3 375,000.00 15/69 10,869.57 4 275,000.00 11/69 7,971.01 5 125,000.00 5/69 3,623.19 Totals $1,725,000.00 69/69 $50,000.00 Since the bonds were issued at 90 there w^as a discount of 10% of $500,000.00, or $50,000.00. During the first year the bonds outstanding amounted to $500,000.00; at the end of that year $50,000.00 were redeemed, leaving $450,000.00 outstanding during the second year, and so on. The footing of this Outstanding column is $1,725,000.00. Of this amount ^^'^ or - were out- ' ' 1,725,000 69 standing during the first year, and so that year must bear 20 —of the total discount, $50,000.00, or $14,492.75. The 69 remainder of the schedule needs no explanation. Of course the various columns must prove : the total of the fractions 69' must be — and the total of the Reduction of Discount 69 column must be $50,000.00. APPENDIX LOGARITHMIC TABLES Table I — The Compound Amount of One Unit. Table II— The Present Worth of One Unit. Tal)le III — The Amount of Periodic Payments of One Unit Each. Table IV — The Present Worth of Periodic Payments of One Unit Each. Table V— The Periodic Payment That Will Amount to One. The Sinking Fund. Table VI— Effective Rate Factors. Table VII — Ten-Place Logarithms of the Interest Ratios. CONSTEUCTION OF TABLES The following brief description of the manner in which the tables in this book have been constructed will be based on the previous sections of the book in which the various functions of s" were discussed. There is nothing difficult or interesting in the construction of tables, and in fact only their usefulness can justify the manual labor required in their construction. It has seemed desirable to include all the values of the various factors for one to one hundred periods, and to eight places of decimals. If more than a hundred periods are specified in any problem, the proper value of s" or v" can be found by multiplying two or more table values ; if a value of some annuity factor is required, it can then be derived from this value of s" or v" by formula. If more than eight places of decimals are necessary for accuracy, it will be necessary either to build a schedule covering the life of the transaction, and adjust errors, or it will be necessary to find the required factor by the use of twelve place loga- rithms, as will be explained below. Table I: The Compound Amount of One Unit. Great care must be used in constructing this table, as it is the basis of two of the other tables, and any error in this table will be especially magnified in the table of s„. The table is developed by ordinary multiplication, the decimals running to not less than twelve places, and fourteen are better. Every ten periods at most there should be an in- dependent check, either by use of some machine, or, as in this case, by use of the fifteen-place logs of 1+i given in the Si^rague-Perrine "Accountancy of Investment", the anti- logs being found by use of the tables of factors in the same book. If any error in the value as given by actual multi- plication is found, the value, and some preceding values should be adjusted before proceeding with further multi- plication. As an instance, the table of s" at 5%, beginning at 50, is made up as follows : 50 11.467399785753 .573369989288 51 12.040769775041 .602038488762 52 12.642808263793 .632140413190 53 13.274948676983 .663747433849 54 13.9.386961108.32 .696934805542 55 14.635630916374 .731781545819 56 15.367412462193 .768370623110 57 16.1.35783085303 .806789154265 58 16.942572239568 .847128611978 59 17.789700851546 .889485042577 60 18.679185894123 Xow the logarithmic value should be very carefully cal- culated, and if any discrepancy is discovered which affects any figure except the two at the right, a suitable adjust- ment should be made. Such adjustment may be on a pro- portional basis; that is, if the error were 16,* add 16 to the last value, 14 to the preceding one, and so on proceeding liackward as far as desired. Table II: The Present Worth of One Unit. This table is formed by multiplication, but is peculiar in that the actual work begins with the value for 100 peri- ods. Of course we might begin with one period, dividing 1 by 1+i, dividing this result by 1+i, and so on. But the multiplication is easier, and gives the same results. The logarithmic value for 100 periods is calculated, and multi- plication proceeds until the value for 90 periods is reached. This value is checked by logarithms, and the multiplication is resumed. The table for 5%, beginning at 60 periods is: 60 .053535523746 .002676776187 59 .056212299933 .002810614996 58 .059022914929 .002951145746 57 .061974060675 .003098703033 56 .065072763708 .003253638185 55 .068326401893 .003416320094 54 .071742721987 .003587136099 53 .075329858086 .003766492904 52 .079096350990 .003954817549 51 .083051168539 .004152558427 50 .087203726966 The logarithmic value for 50 periods must now be cal- culated, and any necessary adjustment made before pro- ceeding with the multiplication. Table III : The Amount of Periodic Payments of One Unit Each. As was shown on page 61, such amounts may be found by addition from values of the compound amount. This table has been so constructed, and checked every ten peri- ods by the formula Sj,= — -. — . For instance, the value of s" at 5% for 50 periods has just been given as 11.467399785753. Deducting 1 and dividing by .05 we have s'»=209.34799571506. To find s'^ for 51 periods we must add to this value s" for 50 periods. The reason for this is evident. In any annuity, the payments are made at the end of the period. A payment therefore does not accumu- late interest until the end of the follomng period. The payment made at the end of the first period mil have ac- cumulated, at the end of the fifty-first period, only fifty periods interest. The schedule is built as follows: 50 209.34799571506 11.467399785753 51 220.815395500813 12.040769775041 52 232.856165275854 12.642808263793 53 245.498973539647 13.274948676983 54 258.773922216630 and so on. At 60 periods the value should be calculated from the value of s" for 60 periods, and any discrepancy should be adjusted over as many periods as may seem necessary. Tablk IV : The Present Worth of Periodic Payments OF One Unit Each. This table is formed by addition from the table of v" in precisely the same manner as Table III is formed from Table I. The checks are made from the corresponding 1 yn value of V" according to the formula a^= . — . The value of V" for 50 periods being .087203726966, we deduct from 1 and divide by .05, and the result is a^ for 50 periods, 18.25592546068. The table proceeds by addition until the value for 60 periods is reached. This value should be checked and adjusted in a mananer similar to that de- scribed for the table of s^. Table V : The Periodic Payment That Will Amount TO One. The Sinking Fund : This table is formed by dividing the successive values of s^ into 1. There is no method of checking such division except by differencing. This method, as applied to bond schedules, was described on pages 104 and 105. The third order of differences should be sufficient to show whether the schedule is accurate. Any considerable variation, particularly in case a difference should be larger than the preceding one, will indicate an error in the schedule. The periodic payment that Avill extinguish a debt of one unit — the amortization factor — is often given in place of the sinking fund factor. But the schedule presented on page 95 should make it clear that the amortization factor can always be found by adding i to the sinking fund factor. It is desirable to give the sinking fund factors because they are more often used in this book, especially in the various problems dealing with the valuation of assets. Table VI : The Effective Rate Factor. This quantity was discussed at length on pages 51, 52, and 54. The process of finding the values is logarithmic, and the antilogs ma}^ be found from the table of factors, or less accurately from the ten-place logs of 1+i at the end of this book. The effective rate factor for use with annuity tables (see pages 63 to 65) is not given because it is seldom re- quired. If it is necessary to change from a table value at an annual rate to an effective rate, the rule is : ^ In case of Sn or a^, multiply the table value by i, and divide by the correct value of j ^ . In case of „ or — mult i pi v by i and divide by L The Compound Amount of One Unit. s" = (l+i)" n 1% iy4% 1M.'% 1%% 2% 1 1.01 1.0125 1.015 1.0175 1.02 2 1.0201 1.02515625 1.030225 1.035:30625 1.0404 3 1.030301 1.03797070 1.045678.38 1.05:342411 1.061208 4 1.04060401 1.05095434 1.06136355 1.0718.5903 1.08243216 5 1.05101005 1.06408215 1.07728400 1.09061656 1.10408080 6 1.06152015 1.07738318 1.09344326 1.10970235 1.12616242 7 1.07213535 1.09085047 1.01984491 1.12912215 1.14868567 8 1.08285671 1.10448610 1.12649259 1.14888178 1.17165938 9 1.09368527 1.11829218 1.14338998 1.16898721 1.19.509257 10 1.10462213 1.13227083 1.16054083 1.18944449 1.21899442 11 1.11566835 1.14642422 1.17794894 1.21025977 1.24337431 12 1.12682503 1.16075452 1.19561817 1.2314:3931 1.26824179 13 1.13809328 1.17526395 1.21355244 1.25298950 1.29360663 14 1.14947421 1.18995475 1.23175573 1.27491682 1.31947876 15 1.1609G896 1.20482918 1.25023207 1.29722786 1.34586834 16 1.17257864 1.21988955 1.26898555 1.31992935 1.37278571 17 1.1 84:30443 1.23513817 1.28802033 1.34302811 1.40024142 18 1.19614748 1.25057739 1.30734064 1.36653111 1.42824625 19 1.20810895 1.26620961 1.32695075 1.39044540 1.45681117 20 1.22019004 1.28203723 1.34685.501 1.41477820 1.48594740 21 1.23239194 1.29806270 1.36705783 1.43953681 1.51566634 22 1.24471586 1.31428848 1.38756370 1.46472871 1.54597967 23 1.25716302 1.33071709 1.40837715 1.49036146 1.57689926 24 1.26973465 1.34735105 1.42950281 1.51644279 1.60843725 25 1.28243200 1.36419294 1.45094535 1.54298054 1.64060599 26 1.29525631 1.38124535 1.47270953 1.56998260 1.67341811 27 1.30820888 1.39851092 1.49480018 1.59745739 1.70688648 28 1.32129097 1.41599230 1.51722218 1.62541290 1.74102421 29 1.33450388 1.43369221 1.53998051 1.65385762 1.77584469 30 1.34784892 1.45161336 1.56308022 1.68280013 1.81136158 31 1.361.32740 1.46975853 1.58652642 1.71224913 1.84758882 32 1.37494068 1.48813051 1.610324:32 1.74221:349 1.88454059 33 1.38K6!)009 1.50673214 1.63447919 1.77270223 1.92223140 34 1.40257699 1.52556629 1.65899637 1.80372452 1.96067603 35 1.4166027G 1.54463587 1.68388132 1.83528970 1.99988955 36 1. -43076878 1.56394382 1.70913954 1.86740727 2.03988734 37 1.44507647 1.58349312 1.73477663 1.90008689 2.08068509 38 1.45952724 1.60328678 1.76079828 1.9:33:33841 2.12229879 39 1.47412251 1.62332787 1.78721025 1.96717184 2.16474477 40 1.48886373 1.64361946 1.81401841 2.00159734 2.20803966 41 1.50375237 1.66416471 1.84122868 2.03662530 2.25220046 42 1.51878989 1.68496677 1.86884712 2.07226624 2.29724447 43 .1.53397779 1 .7()()02885 1 .89687982 2.1085:3090 2.34318936 44 1.54931757 1.72735421 1.9253:3302 2.1454:3019 2.3900.5314 45 1.56481075 1.74894614 1.95421301 2.18297522 2.4:J785421 n 1% iy4% IV2% 1%% 2% 46 1.58045885 1.77080797 1.98352621 2.22117728 2.48661129 47 l.5iHi2(iSU 1.79294:306 2.01327910 2.260047H9 2.53634351 48 I.(jr222(i08 1.81.535485 2.04347829 2.29959H72 2.5870703!) 49 1.628:31^3-1. 1.83804679 2.0741:3046 2.:3:3984170 2.6:3881179 50 l.()H.(i;5182 1.86102237 2.10524242 2.3807889:3 2.69158803 51 l.()()0178M. 1.88428515 2.13682106 2.42246274 2.74541979 52 l.«77()8892 1.9078:3872 2.16887337 2.46484566 2.800:32819 53 l.()914()581 1.93168670 2.20140647 2.50798046 2.85633475 54 1.711U04.7 1.95583279 2.23442757 2.55187012 2.91346144 55 1.72852+57 1.98028070 2.26794398 2.59652785 2.97173067 56 1.71-580982 2.00503420 2.30196314 2.64196708 3.03116529 57 1.7();}2()792 2.0:3009713 2.33649259 2.68820151 :}.091 78859 58 1.78090060 2.05547335 2.37153998 2.73524503 3.15362436 59 1.79870960 2.08116676 2.40711308 2.78311182 3.21669685 60 1.81669670 2.10718135 2.44321978 2.83181628 3.28103079 61 1.83486367 2.13352111 2.47986807 2.88137306 3.34665140 62 1.85321230 2.16019013 2.51706609 2.93179709 8.4135844.3 63 1.87174443 2.18719250 2.55482208 2.98310354 3.48185612 64 1.89046187 2.2145:3241 2.59314442 3.03530785 3.55149:324 65 1.90936649 2.24221407 2.63204158 3.08842574 3.62252311 66 1.92846015 2.27024174 2.67152221 3.14247319 3.69497357 67 1.94774475 2.29861976 2.71159504 3.19746647 3.76887:304 68 1.96722220 2.32735251 2.75226896 3.25342213 3.84425050 69 1.98689442 2.35644442 2.79355300 3.31035702 3.92113551 70 2.00676337 2.38589997 2.83545629 3.36828827 3.99955822 71 2.02683100 2.41572372 2.87798814 3.42723331 407954939 72 2.04709931 2.44592027 2.92115796 3.48720990 4.16114037 73 2.06757031 2.47649427 2.96497533 3.54823607 4.24436318 74 2.08824601 2.50745045 3.00944996 3.61033020 4.32925045 75 2.10912849 2.53879358 3.05459171 3.67351098 4.41583556 76 2.13021975 2.57052845 3.10041058 8.73779742 4.50415126 77 2.15152195 2.60266011 3.14691674 8.80320888 459423521 78 2.17303717 2.6:3519336 3.19412049 3.86976503 4.68611991 79 2.19476754 2.66813328 3.24203230 3.93748592 4.77984231 80 2.21671522 2.70148494 3.29066279 4.00639192 487543916 81 2.23888237 2.73525350 3.34002273 4.07650378 4.97294794 82 2.26127119 2.76944417 3.39012307 4.14784260 5.07240690 83 2.28388390 2.80406222 3.44097492 4.22042984 5.17:385504 84 2.30672274 2.83911300 3.49258954 4.29428737 5.27733214 85 2.32978997 2.87460191 3.54497838 436943740 5.38287878 86 2.85308787 2.91053444 3.59815.306 444590255 5.49053636 87 2.37661875 2.94691612 3.65212535 452370584 5.60034708 88 2.40038494 2.98375257 3.70690723 4.60287070 5.712:35402 89 2.42438879 3.02104948 3.76251084 4.68342093 5.82660110 90 2.44863267 3.05881260 3.81894851 4.76538080 5.94313313 91 2.47311900 3.09704775 3.87623273 4.84877496 6.06199579 92 2.49785019 3.13576085 3.93437622 493362853 6.18:323570 93 2.52282869 3.17495786 3.993.39187 5.01996703 6.30690042 94 2.54805698 3.21464483 4.05329275 5.10781645 6.43303843 95 2.57353755 3.25482789 4.11409214 5.19720324 6.56169920 96 2.59927293 3.29551324 4.17580352 5.28815429 6.6929:3318 97 2.62526565 3.33670716 423844057 5.38069699 6.82679184 98 2.65151831 3.37841600 4.:30201718 5.47485919 6.963:32768 99 2.67803349 3.42064620 4.36654744 6.57066923 7.10259423 100 2.70481383 3,46340427 4.43204566 6.66815694 7.24464612 n 21/4 % 2y2% 2y4% 3% 31/2% 1 1.0225 1.025 1.0275 1.03 1.035 2 1.04550625 1.050625 1.05575625 1.0609 1.071225 3 1. 060930 U 1.07689063 1.08478955 1.092727 1.10871788 4 1.09308332 1.01381289 1.11462126 1.12550881 1.14752300 5 1.11767769 1.13140821 1.14527334 1.15927407 1.18768631 6 1.142825H. 1.15969342 1.17676836 1.19405230 1.22925533 7 1.1685390 1 1.18868575 1.20912949 1.22987387 1.27227926 8 1.191-83114. 1.21840290 1.24238055 1.26677008 1.31680904 9 1.22171484 1.24886297 1.27654602 1.30477318 1.36289735 10 1.24920348 1.28008454 1.31165103 1.34391638 1.41059876 11 1.27731050 1.31208666 1.34772144 1.38423387 1.45996972 12 1.30604999 1.34488882 1.38478378 1.42576089 1.51006866 13 1.3354:3611 1.37851104 1.42286533 1.46853371 1.56395606 14 1.36548343 1.41297382 1.46199413 1.51258972 1.61869452 15 1.39620680 1.44829817 1.50219896 1.55796742 1.67534883 16 1.42762146 1.48450562 1.54350944 1.60470644 1.73398604 17 1.45974294 1.52161826 1.58595595 1.65284763 1.79467555 18 1.49258716 1.55965872 1.62956973 1.70243306 1.85748120 19 1.52617037 1.59865019 1.67438290 1.75350605 1.92250312 20 1.56050920 1.63861644 1.72042843 1.80611123 1.98978887 21 1.59562066 1.67958185 1.76774021 1.86029457 2.05943148 22 1.63152212 1.72157140 1.81635307 1.91610341 2.13151158 23 1.66823137 1.76461068 1.86630278 1.97358651 2.20611148 24 1.70576658 1.80872595 1.91762610 2.03279411 2.28332849 25 1.74414632 1.85394410 1.97036082 2.09377793 2.36324498 26 1.78338962 1.90029270 2.02454575 2.15659127 2.44595856 27 1.82351588 1 .94780002 2.08022075 2.22128901 2.53156711 28 1.86454499 1.99649502 2.13742682 2.28792768 2.62017196 29 1.90649725 2.04640739 2.19620606 2.35656551 2.71187798 30 1.94939344 2.09756758 2.25660173 2.42726247 2.80679370 31 1.99325479 2.15000677 2.31865828 2.50008035 2.90503148 32 2.03810303 2.20375694 2.38242138 2.57508276 3.00670759 33 2.08396034 2.25885086 2.44793797 2.65233523 3.11194235 34 2.13084945 2.31532213 2.51525626 2.73190530 3.22086033 35 2.17879356 2.37320519 2.58442581 2.81386245 3.33359045 36 2.22781642 2.43253532 2.65549752 2.89827833 3.45026611 37 2.27794229 2.49334870 2.72852370 2.98522668 3.57102543 38 2.32919599 2.55568242 2.80355810 3.07478348 3.69601132 39 2.381 G0290 2.61957448 2.88065595 3.16702698 3.82537171 40 2.43518897 2.68506384 2.95987399 3.26203779 3.95925972 41 2.48998072 2.75219043 3.04127052 3.35989893 4.09783381 42 2.54600528 2.82099520 3.12490546 3.46069589 4.24125799 43 2.60329040 2.89152008 3.21084036 3.56451677 4.38970202 44 2.44186444 2.86380808 3.29913847 3.67145227 4.54334160 45 2.72175639 3.03790328 3.38986478 3.78159584 4.70235855 46 2.78299590 3.11385086 3.48308606 3.89504372 4.86694110 47 2.84561331 3.19169713 3.57887093 4.01189503 5.03728404 48 2.90963961 3.27148956 3.67728988 4.13225188 5.21358898 49 2.97510650 3.35327679 3.77841535 4.25621944 5.39606459 50 3.04204640 3.43710872 3.88232177 4.38390602 5.58492686 n 2 1/4 7c 21/2% 2%% 3% 3y2% 51 3.1 10 1.92 14 3.52303644 3.98908562 4.51542320 5.780.399.30 52 3.1801.7852 8.611112.35 4.09878547 4.65008590 5.98271327 53 3.2520:5929 3.701.39016 4.211.50208 4.79041247 6.19210824 54 3.32521017 3.79392491 4.32731838 4.93412485 6.40883202 55 3.4-0002740 3.88877303 4.44631964 5.08214859 6.63;il4114 56 3.4.7652802 3.98599236 4.56859343 5.23461305 6.86.530108 57 3.55474990 4.09564217 4.69422975 5..39165144 7.01558662 58 3.63473177 4.18778.322 4.82.3.32107 5.55.340098 7.-354282 15 59 3.71651324 4.29247780 4.95596239 5.72000.301 7.61168203 60 3.80013479 4.39978975 5.09225136 5.89160310 7.87809090 61 3.88563782 4.50978449 5.23228827 6.06835120 8.15-382408 62 3.97306467 4.62252910 5.37617620 6.25040173 8.43920793 63 4.06245862 4.73809233 5.52402105 6.43791379 8.73458020 64 4.15386394 4.85654464 5.67593162 6.63105120 9.04029051 65 4.24732588 4.97795826 5.8.3201974 6.82998273 9.35670068 66 4.34289071 5.10240721 5.99240029 7.03488222 9.68418.520 67 4.44060576 5.22996739 6.157191.30 7.24592868 10.02313168 68 4.54051939 5.360716.58 6..32651406 7.46.330654 10.37-394129 69 4.64268107 5.49473449 6.50049319 7.68720574 10.73702924 70 4.74714140 5.63210286 6.67925676 7.91782191 11.11282526 71 4.85395208 5.77290543 6.86293632 8.15535657 11.-50177414 72 4.96316560 r 5.91722806 7.05166706 8.40001727 11.9043.3624 73 5.07483723 6.06515877 7.24558791 8.65201778 12.-32098801 74 5.18902107 6.21678774 7.44484158 8.91157832 12.7-5222259 75 5.30577405 6.37220743 7.64957472 9.17892567 13.19855038 76 5.42515396 6.53151261 7.85993802 9.45429344 13.66049964 77 5.54721993 6.69480043 8.07608632 9.73792224 14.1-3861718 78 5.67203237 6.86217044 8.29817869 10.03005991 14.6-3346873 79 5.79965310 7.03372470 8.52637861 10.3.3096171 15.14564014 80 5.93014.530 7.20956782 8.76085402 10.64089056 15.67573754 81 6.06.357357 7.28980701 9.00177751 10.96011727 16.224388-35 82 6.20000397 7.5745.5219 9.24932639 11.28892079 16.79224195 83 6.33950406 7.76391599 9.50368286 11.62758842 17.37997041 84 6.48214920 7.95801389 9.7650-3414 11.97641607 17.988269-38 85 6.62799112 8.15696424 10.0.3357258 12.33570855 18.61785881 86 6.77712092 8.36088834 10.30949583 12.70577981 19.26948387 87 6.92960614 8.56991055 10.59300696 13.08695320 19.94391.580 88 7.08552228 8.7841.5832 10.88431465 13.47956180 20.6419-5285 89 7.24494653 9.00376227 11.18363331 13.88394865 21.36442120 SO 7.40795782 9.22885633 11.49118322 14.30046711 22.11217595 91 7.57463688 9.45957774 11.80719076 14.72948112 22.88610210 92 7.74506621 9.69606718 12.13188851 15.17136556 23.68711568 93 7.919.33020 9.93846886 12.46551544 15.62650652 24.51616478 94 8.09751512 10.1869.3058 12.80831711 16.09530172 25.37423049 95 8.27970921 10.44160385 13.16054584 16.57816077 26.26232856 96 8.46600267 10.70264.395 13.52246085 17.07550559 27.18151006 97 8.65648773 10.97021004 13.89432852 17.58777076 28.13286291 98 8.8512.5871 11.24446.5.30 14.27642255 18.11540388 29.11751311 99 9.05041203 11.52.557693 14.66902417 18.65886600 30.13662607 100 9.25404630 11.81371635 15.07242234 19.21863198 31.19140798 n '1% *y2% 5% 6y2% 6% 1 1.04 1.045 1.05 1.055 1.06 2 1.0816 1.092025 1.1025 1.113025 1.1236 3 1.124^fi4. 1.14116613 1.157625 1.17424138 1.191016 4 1.1 69858.56 1.19251860 1.21550625 1.23882465 1.26247696 5 1.21665290 1.24618194 1.27628156 1.30696001 1.33822558 6 1.26531902 1.30226012 1.. 34009564 1.37884281 1.41851911 7 1.31593178 1.36086183 1.40710042 1.45467916 1.60363026 8 1.36856905 1.42210061 1.47745544 1.53468651 1.59384807 9 1.42331181 1.48609514 1.551.32822 1.61909427 1.68947896 10 1.48024428 1.55296942 1.62889463 1.70814446 1.79084770 11 1.53945406 1.62285305 1.71033936 1.80209240 1.89829856 12 1.6010.3222 1.69588143 1.79585633 1.90120749 2.01219647 13 1.66507351 1.77219610 1.88564914 2.00577390 2.13292826 14 1.73167645 1.85194491 1.97993160 2.11609146 2.26090396 15 1.80094361 1.93528244 2.07892818 2.23247649 2.39655819 16 1.87298125 2.02237015 2.18287459 2.35526270 2.54036168 17 1.94790050 2.113.37681 2.29201832 2.48480215 2.69277279 18 2.02581652 2.20847877 2.40661923 2.62146627 2.85433915 19 2.10684918 2.30786031 2.52695020 2.76564691 3.02569960 20 2.19112314 2.41171402 2.65329771 2.91775749 3.20713647 21 2.27876807 2.52024116 2.78596259 3.07823415 3.39956360 22 2.36991879 2.63365201 2.92526072 8.24753703 3.60353742 23 2.46471554 2.75216635 3.07152376 3.42615157 3.81974966 24 2.563.30416 2.87601.383 3.22509994 3.61458990 4.04893464 25 2.66583633 2.00543446 3.38635494 3.81339235 4.29187072 26 2.77246978 3.14067901 3.55567269 4.02312893 4.54938296 27 2.88336858 3.28200956 3.73345632 4.24440102 4.82234694 28 2.99870332 3.42969999 3.92012914 4.47784307 5.11168670 29 3.11865145 3.58403649 4.1161.3560 4.72412444 6.41838790 30 3.2433975 | 3.74531813 4.32194238 4.98395129 6.74349117 31 3.37313341 3.91385745 4.53803949 5.25806861 6.08810064 32 3.50805875 4.08998104 4.76494147 5.54726238 6.45338668 33 3.64838110 4.27403018 5.00318854 5.85236181 6.84068988 34 3.79431634 4.46636154 5.25334797 6.17424171 7.26102528 35 3.94608899 4.66734781 5.51601537 6.51382501 7.68608679 36 4.10393255 4.87737846 5.79181614 6.87208538 8.14725200 37 4.26808986 5.09686049 6.08140694 7.25005008 8.63608712 38 4.4.3881345 5.32621921 6.38547729 7.64880283 9.15425235 39 4.61636599 5.56589908 6.70475115 8.0G948()99 9.70350749 40 4.80102063 5.81636454 7.03998871 8.51330877 10.28571794 41 4.99306145 6.07810094 7.39198815 8.98154076 10.90286101 42 5.19278391 6.35161548 7.76158756 9.475,52550 11.55703267 43 5.40049527 6.6374.3818 8.14966693 9.99667940 12.25046463 44 5.61651508 6.93612290 8.55716028 10.54649677 12.98548191 45 5.84117568 7.24824843 8.98500779 11.12655409 13.76461083 46 6.07482271 7.57441961 9.43425818 11.73851456 14.69048748 47 6.31781.562 7.91526849 9.90597109 12.38413287 15.46591673 48 6.57052824 8.2714.5-557 10.40126965 13.06526017 16.39387173 49 6.83334937 8.64:367107 10.921.33313 13.78384948 17.37750403 50 7.10666335 9.03263627 11.46739979 14.64196120 18.42015429 10 n 4% 4V2% 5% 5V::7r 6% 51 7.39095068 9.43910490 12.04076978 15.34176907 19.62636353 52 7.68658871 9.86386Ui3 12.64280826 16.18556637 20.69688534 S3 7.994-05256 10.30773853 13.27494868 17.07577252 21.93869846 54 8.31381 4.;35 10.771.'58677 13.93869611 18.01494001 23.25502037 55 8.64636692 11.25630817 14.63563092 19.00576171 24.650.32159 56 8.99222160 11.76284204 15.36741246 20.05107860 26.12934090 57 9.35191046 12.29216993 16.13578309 21.1.5388793 27.69710134 58 9.72598688 12.84531758 16.94257224 22.317.35176 29.35892742 59 10.11502635 13.42335687 17.78970085 23.54480611 31.12046307 60 10.51962741 14.02740793 18.67918589 24.83977045 32.98769085 61 10.94041250 14.65864129 19.61314519 26.20596782 34.96695230 62 11.37802900 15.31828014 20.59380245 27.64728550 37.06496944 63 11.83315016 16.00760275 21.62.349257 29.16788620 39.28886761 64 12.30647617 16.72794487 22.70466720 30.77211994 41.64619967 65 12.79873522 17.48070239 23.83990056 32.46458654 44.14497165 66 13.31068463 18.26733400 25.03189559 34.25013880 46.79366994 67 13.84311201 19.08936403 26.28349037 36.13389643 49.60129014 68 14.39683649 19.94838541 27.59766488 38.12126074 52.57736755 69 14.97270995 20.84606276 28.97754813 40.21793008 55.73200960 70 15.57161835 21.78413558 30.42642554 42.42991623 59.07693018 71 16.19448308 22.76442168 31.94774681 44.76.356163 62.62048699 72 16.84226241 23.78882066 33.54513415 47.22555751 66.37771515 73 1 7.51595290 24.85931758 35.22239086 49.82296318 70.36037806 74 18.21659102 25.97798688 36.98351040 .52.56322615 74.58200074 75 18.94525466 27.14699629 38.83268592 55.45420359 79.05692079 76 19.70306485 28.36861112 40.77432022 58.50418479 83.80033603 77 20.49118744 29.64519862 42.81303623 61.72191495 88.82835619 78 21.31083494 30.97923256 44.95368804 65.11662027 94.15805757 79 22.16326834 32.37329802 47.20137244 68.69803439 99.80754102 80 23.04979907 33.83009643 49.56144107 72.47642628 105.79699348 81 23.97179103 35.35245077 52.03951312 76.46262973 112.14375309 82 24.93066267 36.94331106 54.64148878 80.66807436 118.87237828 83 25.92788918 38.60576006 57.37356322 85.10481845 126.00472097 84 26.96500475 40.34301926 60.24224138 89.78558347 133.56500423 .85 28.04360494 42.15845513 63.25435344 94.72379056 141.57890449 86 29.16534914 44.05558561 66.41707112 99.93359904 150.07363875 87 30.33196311 46.03808696 69.73792467 105.42994698 159.07805708 88 31.54524163 48.10980087 73.22482091 111.22859407 168.62274050 89 32.80705129 50.27474191 76.88606195 117.34616674 178.74010493 90 34.11933334 52..53710530 80.73036505 123.80020591 189.46451123 91 35.48410668 54.90127503 84.76688330 1.30.60921724 200.82328190 92 86.90347094 67.37183241 89.00522747 137.79272419 212.88232482 93 38.37960978 59.95356487 93.45548884 145.37132402 226.65526431 94 39.91479417 62.65147529 98.12826328 153.36674684 239.19458017 95 41.51138594 65.47079168 103.03467645 161.80191791 253.54625498 96 43.17184138 68.41697730 108.18641027 170.70102340 268.75903027 97 44.89871503 71.49574128 113.59573078 180.08957969 284.88457209 98 46.69466363 74.71304964 119.27551732 189.99450657 301.97764642 99 48.56245018 78.07513687 125.23929319 200.44420443 320.09630520 100 50.50494818 81.58851803 131.50126785 21 1 .46863567 339.30208351 11 The Present Worth of One Unit. V — (1+i)" n 1% 1^4% 11/2% 1%% 2% 1 .99009901 .98765432 .98522168 .98280098 .98029216 2 .98029605 .97546106 .97066175 .96589777 .96116878 3 .97059015 .96341833 .95031699 .94928528 .94232233 4 .9()0980.^4 .95152428 .94218423 .93295851 .92384543 5 .9514G5(i9 .93977706 .92826033 .91691524 .90573081 6 .94204524 .92817488 .91454219 .90114254 .88797138 7 .93271805 .91671593 .90102679 .88564378 .87056018 8 .92348322 .90539845 .88771112 .87041157 .85349037 9 .91433982 .89422069 .87459224 .85544135 .83675527 10 .90528694 .88318093 .86166723 .84072860 .82034830 11 .89632372 .87227746 .84893323 .82626889 .80426304 12 .88744923 .86150860 .83638742 .81205788 .78849318 13 .87866260 .85087269 .82402702 .79809128 .77303253 14 .86996297 .84036809 .81184928 .78436490 .75787502 15 .86134947 .82999318 .79985151 .77087459 .74301473 16 .85282126 .81974635 .78803104 .75761631 .72844581 17 .84437749 .80962602 .77638526 .74458605 .71416256 18 .83601731 .79963064 .76491159 .73177990 .70015937 19 .82773992 .78975866 .75360747 .71919401 .68643076 20 .81954447 .78000855 .74247042 .70682458 .67297133 21 .81143017 .77037881 .73149795 .69466789 .65977582 22 .80339621 .76086796 .72068763 .68272028 .64683904 23 .79544179 .75147453 .71003708 .67097817 .63415592 24 .78756613 .74219707 .69954392 .65943800 .62172149 25 .77976844 .73303414 .68920583 .64809632 .60953087 26 .77204796 .72398434 .67902052 .63694970 .59757928 27 .76440392 .71504626 .66898574 .62599479 .58586204 28 .75683557 .70621853 .65909925 .61522829 .57437455 29 .74934125 .69749978 .64935887 .60464697 .56311231 30 .74192292 .68888867 .63976243 .59424764 .55207089 31 .73457715 .68038387 .63030781 .58402716 .54124597 32 .72730411 .67198407 .62099292 .57398247 .53063330 33 .72010307 .66368797 .61181568 .56411053 .52022873 34 .71297734 .65549429 .60277407 .55440839 .51002817 35 .70591420 .64740177 .59386608 .54487311 .50002761 36 .69892495 .63940916 .58508974 .53550183 .49022315 37 .69200490 .63151522 .57644^309 .52629172 .48061093 38 .68515337 .62371873 .56792423 .51724002 .47118719 39 .67836967 .61601850 .55953126 .50834400 .46194822 40 .67165314 .60841334 .55126232 .49960098 .45289042 41 .66500311 .60090206 .54311559 .49100834 .44401021 42 .65841892 .59348352 .53508925 .48256348 .43530413 43 .65189992 .58615656 .52718153 .47426386 .42676875 44 .64544546 .57892006 .51939067 .46610699 .41840074 45 .63905492 .57177290 .51171494 .45809040 .41019680 46 .63272764 .56471397 .50415265 .45021170 .40215373 47 .62646301 .55774219 .49670212 .4424()850 .39426836 48 .62026041 .55085649 .48936170 .43485848 .38653761 49 .61411921 .54405579 .48212975 .42737934 .37895844 50 .60803882 .53733905 .47500470 .42002883 .37152788 12 n 1% l'/4% ^V2% 1%% 2% 51 .60201864 .53070524 .46798491 .41280475 .36424302 52 .59605806 .52415332 .46106887 .40570492 .35710100 53 .59015649 .51768229 .45425505 .39872719 .35009902 54 .584:31336 .51129115 .44754192 .39186947 .34323433 55 .57852808 .50497892 .44092800 .38512970 .33650425 56 .57280008 .49874461 .43441182 .37850585 .32990613 57 .56712879 .49258727 .42799194 .37199592 .32343740 58 .56151365 .48650594 .42166694 .36559796 .31709547 59 .5.5595411 .48049970 .41543541 .35931003 .31087791 60 .55044962 .47456760 .40929597 .3.5313025 .30478227 61 .54499962 .46870874 .40324726 .34705676 .29880614 62 .53960.358 .46292222 .39728794 .34108773 .29294720 63 .53426097 .45720713 .39141669 .335221.35 .28720314 64 .52897126 .45156259 .38563221 .32945587 .28157170 65 .52373392 .44598775 .37993321 .32378956 .27605070 66 .51854844 .44048173 .37431843 .31822069 .27063793 67 .51341429 .43504370 .36878663 .31274761 .26533130 68 .50833099 .42967277 .36333658 .30736866 .26012873 69 .50329801 .42436817 .35796708 .30208222 .25502817 70 .49831486 .41912905 .35267692 .29688670 .25002761 71 .493.38105 .41395461 .34746495 .29178054 .24512511 72 .48849609 .40884406 .34233000 .28676221 .24031874 73 .48365949 .40379661 .33727093 .28183018 .23560661 74 .47887078 .39881147 .33228663 .27698298 .23098687 75 .47412949 .39388787 .32737599 .27221914 .22645771 76 .46943514 .38902506 .32253793 .26753724 .22201737 77 .46478726 .38422228 .31777136 .26293586 .21766408 78 .46018541 .37947879 .31307523 .25841362 .21339616 79 .45562912 .37479387 .30844850 .25396916 .20921192 80 .45111794 .37016679 .30389015 .24960114 .20510973 81 .44665142 .36559683 .29939916 .24530825 .20108797 82 .44222913 .36108329 .29497454 .24108919 .19714507 83 .4378.5063 .35662547 .29061531 .23694269 .19327948 84 .43351547 .35222268 .28632050 .23285761 .18948968 85 .42922324 .34787426 .28208917 .22886242 .18577420 86 .42497350 .34357951 .27792036 .22492621 .18213157 87 .42076585 .33933779 .27381316 .22105770 .17856036 88 .41659985 .33514843 .26976666 .21725572 .17.505918 89 .41247510 .33101080 .26577996 .21351914 .17162665 90 .40839119 .32692425 .2618.5218 .20984682 .16826142 91 .40434771 .32288814 .25798245 .20623766 .16496217 92 .40034427 .31890187 .25416990 .20269057 .16172762 93 .39638046 .31496481 .25041369 .19920450 .15855649 94 .39245590 .31107636 .24671.300 .19577837 .15544754 95 .38857020 .30723591 .24306699 .19241118 .15239955 96 .38472297 .30344287 .23947487 .18910190 .14941132 97 38091383 .29969666 .23593583 .18584953 .14648169 98 .37714241 .29599670 .23244909 .1826.5310 .14360950 99 .37340832 .29234242 .22901389 .17951165 .14079363 [00 .36971121 .28873326 .22562944 .17642422 .13803297 13 n 21/4% 2V2% 2%% 3% 3V2% 1 2 3 4 5 .97799511 .97560976 .97323601 .97087379 .96618357 .95G47U4. .951814-U) .947188:33 .94259591 .93351070 .9351'2732 .92859941 .92183779 .91514166 .90194271 .9148i:3;55 .90595064 .89716573 .88848705 .87144223 .894.71232 .88385429 .87315400 .86260878 .84197317 Q .87502427 .86229687 .84978491 .83748426 .81350064 7 .85576946 .84126524 .82704128 .81309151 .78599096 8 .8;3()93835 .82074657 .80490635 .78940923 .75941156 9 .81852161 .80072836 .78336385 .76641673 .73373097 10 .80051013 .78119840 .76239791 .74409391 .70891881 11 .78289499 .76214478 .74199310 .72242128 .68494571 12 .76566748 .74355589 .72213440 .70137988 .66178330 13 .74881905 .72542038 .70280720 .68095134 .63940415 14 .73234137 .70772720 .68399728 .66111781 .61778179 15 .71622628 .69046556 .66569078 .64186195 .59689062 16 .70046580 .67362493 .64787424 .62316694 .57670591 17 .68505212 .65719506 .63053454 .60501645 .55720378 18 .66997763 .64116591 .61:365892 .58739461 .538:36114 19 .65523484 .62552772 .59723496 .57028603 .52015569 20 .64081647 .61027094 .58125057 .55367575 .50256588 21 .62671538 .59538629 .56569398 .53754928 .48557090 22 .61292457 .58086467 .55055375 .52189250 .46915063 23 .59943723 .66669724 .53581874 .50669175 .45328563 24 .58624668 .55287535 .52147809 .49193374 .4:3795713 25 .57334639 .53939059 .50752126 .47760557 .42314699 26 .56073000 .52623472 .49393796 .46369478 .40883767 27 .54839117 .51339973 .48071821 .45018906 .39501224 28 .53632388 .50087778 .46785227 .43707675 .38165434 29 .52452213 .48866125 .455:33068 .42434636 .36874815 30 .51298008 .47674269 .44314421 .41198676 .35627841 31 .50169201 .46511481 .43128391 .39998714 .34423035 32 .49065233 .45377055 .41974103 .38833703 .33258971 33 .47985558 .44270298 .40850708 .37702625 .32134271 34 .46929641 .4:3190534 .39757380 .36604490 .31047605 35 .45896959 .42137107 .38693314 .35538340 .29997686 36 .44887002 .41109372 .37657727 .34503243 .28983272 37 .4:3899268 .40106705 .36649856 .33498294 .2800:5161 38 .42933270 .39128492 .35668959 .32522615 .27056194 39 .41988530 .38174139 .:34714316 .31575355 .26141250 40 .41064575 .37243062 .33785222 .30655684 .25257247 41 .40160954 .36334695 .32880995 .29762800 .24403137 42 .39277216 .35448483 .32000968 .28895922 .2:3577910 43 .38412925 .34583886 .31144495 .28054294 .22780590 44 .37567653 .33740376 .:3031()944 .27237178 .22010231 45 .36740981 .32917440 .29499702 .26443862 .21265924 46 .35932500 .32114576 .28710172 .25673653 .20546787 47 .35101809 .31:331294 .27941773 .24925876 .19851968 48 .34:368518 .:30567116 .27193940 .24199880 .19180645 49 .:i36 12242 .29821576 .26466122 .23495029 .185:52024 50 .32872608 .29094221 .25757783 .22810708 .17905337 14 n 2y4% 2y2% 2% % 3% ^Vz7c 51 .32149250 .28384606 .25068402 .22146318 .172998-43 52 .31441810 .27692298 .24397471 .21501280 .16714824 53 .30749936 .27016876 .23744197 .20875029 .161-49589 54 .30073287 .26357928 .23109000 .20267019 .15603467 55 .29411528 .25715052 .22490511 .1967671Z .15075814 56 .28764330 .25087855 .21888575 .19103609 .14566004 57 .28131374 .24-475956 .21302749 .18547193 .1-10734:34 58 .27512347 .23878982 .20732603 .18006984 .13597520 59 .26906940 .23296568 .20177716 .17482508 .13137701 60 .26314856 .22728359 .19637679 .16973309 .12693431 61 .25735801 .22174009 .19112097 .16478941 .12264184 62 .25169487 .21633179 .18600581 .15998972 .11849454 63 .24615635 .21105541 .18102755 .15532982 .11448747 64 .24073971 .20590771 .17618253 .15080565 .11061592 65 .23544226 .20088557 .17146718 .14641325 .10687528 66 .23026138 .19598593 .16687804 .14214879 .10326114 67 .22519450 .19120578 .16241172 .13800853 .09976922 68 .22023912 .1865422.3 .15806493 .13398887 .09639538 69 .21539278 .18199241 .15383448 .13008628 .09313563 70 .21065309 .17755358 .14971726 .12629736 .08998612 71 .20601769 .17322300 .14571023 .12261880 .08694311 72 .20148429 .16899805 .14181044 .11904737 .08400300 73 .19705065 .16487615 .13801503 .11557998 .08116232 74 19271458 .16085477 .13432119 .11221357 .07841770 75 .18847391 .15693149 .13072622 .10894521 .07576590 76 .18432657 .15310389 .12722747 .10577205 .07320376 77 .18027048 .14936965 .12382235 .10269131 .07072828 78 .17630365 .14572649 .12050837 .09970030 .06833650 79 .17242411 .14217218 .11728039 .09679641 .06602560 80 .16862993 .13870457 .11414412 .09397710 .06379285 81 .16491725 .1.3532153 .11108917 .09123990 .06163561 82 .16129022 .13202101 .10811598 .08858243 .05955131 83 .15774105 .12880098 .10522237 .08600236 .057537.50 84 .15426997 .12565949 .10240620 .08349743 .05559178 85 15087528 .12259463 .09966540 .08106547 .05371187 86 .14755528 .11960452 .09699795 .07870434 .05189553 87 .144:30835 .11668733 .09440190 .07641198 .05014060 88 .14113286 .11384130 .09187.533 .07418639 .04844503 89 .13802724 .11106468 .08941638 .07202562 .04680679 90 .13498997 .10835579 .08702324 .06992779 .04522395 91 .13201953 .10571296 .08469415 .06789105 .04369464 92 .12911445 .1031.3460 .08242740 .06591364 .04221704 93 .12627331 .10061912 .08022131 .06399383 .0!')78941 94 .12349468 .09816.500 .07807427 .06212993 03941006 95 .12077719 .09577073 .07598469 .06032032 .03807735 96 11811950 .09343486 .07395104 .05856342 .03678971 97 11.552029 .09115.596 .07197181 .05685769 .03554562 98 11297828 .08893264 .07004556 .05.520614 .03434359 99 .11049221 .08676355 .06817086 .05359383 .0:3318222 100 .10806084 .08464737 .06634634 .05203284 .03206011 15 n 4% *V2% 6% 5V2% 6% 1 .96153846 .95693780 .95238095 .94786730 .94339623 2 .9245.'j621 .91572995 .90702948 .89845242 .88999644 3 .88899636 .87629660 .86383760 .85161366 .83961928 4 .85480419 .83856134 .82270427 .80721674 .79209366 5 .82192711 .80245105 .78352617 .765134:35 .74725817 6 .79031453 .76789574 .74621540 .72524583 .70496054 7 .75991781 .73482846 .71068133 .68743681 .66505711 8 .73069021 .70318513 .67683936 .65159887 .62741237 9 .702.')8674 .67290443 .64460892 .61762926 .59189846 10 .67556417 .64392768 .61391325 .58543058 .55839478 11 .64958093 .61619874 .58467929 .55491051 .52678753 12 .62459705 .58966386 .55683742 .52598152 .49696936 13 .60057409 .56427164 .53032135 .49856068 .46883902 14 .57747508 .53997286 .50506795 .47256937 .44230096 15 .55526450 .51672044 .48101710 .44793305 .41726506 16 .53390818 .49446932 .45811152 .42458019 .39364628 17 .51337323 .47317639 .43629669 .40244653 .37136442 18 .49362812 .45280037 .41552065 .38146590 .35034379 19 .47464242 .43330179 .39573396 .36157906 .33051301 20 .45638695 .41464286 .37688948 .34272896 .31180473 21 .43883360 .39678743 .35894236 .32486158 .29415540 22 42195539 .37970089 .34184987 .30792567 .27750510 23 .40572638 .36335013 .32557131 .29187267 .26179726 24 39012147 .34770347 .31006791 .27665656 .24697855 25 .37511680 .33273060 .29530277 .26223370 .23299863 26 .36068923 .31840248 .28124073 .24856275 .21981003 27 .34681657 .30469137 .26784832 .23560450 .20736795 28 .33347747 .29157069 .25509364 .22332181 .19563014 29 .32065141 .27901502 .24294632 .21167944 .18455674 30 .30831867 .26700002 .23137745 .20064402 .17411013 31 .29646026 .25550241 .22035947 .19018390 .16425484 32 .28505794 .24449991 .20986617 .18026910 .15495740 33 .27409417 .23397121 .19987254 .17087119 .14618622 34 .26355209 .22389589 .19035480 .16196321 .13791153 35 .25341547 .21425445 .18129029 .15351963 .13010522 36 .24366872 .20502817 .17265741 .14551624 .12274077 37 23429685 .19619921 .16443563 .13793008 .11579318 38 .2252854;3 .18775044 .15660536 .13073941 .10923885 39 .21662061 .17966549 .14914797 .12392362 .103055.52 40 .20828904 .17192870 .14204.568 .11746314 .09722219 41 .20027793 .164,52507 .13528160 .11133947 .09171905 42 .19257499 .15714026 .12883962 .10553504 .08652740 43 .18516820 .15066054 .12270440 .10003322 .08162962 44 17804635 .14417276 .11686133 .09481822 .07700908 45 .17119841 .13796437 .11129651 .08987509 .07265007 46 .16461386 .13202332 .10599668 .08518965 .06853781 47 .15828256 .12633810 .10094921 .08074849 .06465831 48 .15219476 .12089771 .09614211 .07653885 .06099840 49 .14634112 .11569158 .09156391 .07254867 .05754.566 50 . 14071262 .11070965 .08720373 .06876652 .05428836 16 n 4% *V2% 5% 5V2% 6% 51 .13530059 .10594225 .08305117 .06518153 .05121544 52 .13009072 .10138014 .07909635 .06178344 .04831645 53 .12509300 .09701449 .07532986 .05856250 .04558156 54 .12028273 .09283683 .07174272 .05550948 .0-4300147 55 115G5551 .08883907 .06832640 .05261562 .04056742 56 .11120722 .08501347 .06507276 .04987263 .03827115 57 .10093002 .0813.5260 .06197406 .0472726.3 .03610486 58 .10281733 .07784938 .05902291 .04-480818 .03406119 59 .0988()282 .07449701 .05621230 .04247221 .03213320 60 .09506040 .07128901 .05353552 .04025802 .03031434 61 .0911.04.23 .06821915 .05098621 .03815926 .02859843 62 .08788868 .06528M* .04855830 .03616992 .02697965 63 .08450835 .06247032 .04624600 .03428428 .02545250 64 .08125803 .05978021 .04404381 .03249695 .02401179 65 .07813272 .05720594 .04194648 .03080279 .02265264 66 .07512762 .05474253 .03994903 .02919696 .02137041 67 .07223809 .05238519 .03804670 .02767485 .02016077 68 .06945970 .05012937 .03623495 .02623208 .019019.59 69 .06678818 .04797069 .03450948 .02486453 .01794301 70 .06421940 .04590497 .03286617 .02356828 .01692737 71 .06174942 .04392820 .03130111 .02333960 .01596921 72 .05937445 .04203655 .02981058 .02117498 .01506530 73 .05709081 .04022637 .02839103 .02007107 .01421254 74 .05489501 .03849413 .02703908 .01902471 .01,340806 75 .05278367 .03683649 .02575150 .01803290 .01264911 76 .05075353 .03525023 .02452524 .01709279 .0119.3313 77 .04880147 .03373228 .02335737 .01620170 .01125767 78 .04692449 .03227969 .02224.512 .01535706 .01062044 79 .04511970 .03088965 .02118582 .014.55646 .01001928 80 .04838433 .02955948 .02017698 .01379759 .00945215 81 .04171570 .02828658 .01921617 .01307828 .00891713 82 .04011125 .02706850 .01830111 .01239648 .00841238 83 .03856851 .02590287 .01742963 .01175022 .00793621 84 .03708510 .02478744 .016.59965 .01113765 .00748699 85 .03565875 .02372003 .01580919 .01055701 .00706320 86 .03428726 .02269860 .01595637 .01000664 .00666340 87 .03296852 .02172115 .01433940 .00948497 .00628622 88 .03170050 .02078579 .01365657 .00899049 .00593039 89 .03048125 .01989070 .01.300626 .00852180 .00559472 90 .02930890 .01903417 .01238691 .ooscrycs .00527803 91 .02818163 .01821451 .01179706 .00765643 .00497928 92 .02709772 .01743016 .01123530 .00725728 .00469743 93 .02605550 .01667958 .01070028 .00687894 .00443154 94 .02505337 .01.596132 .01019074 .00652032 .00418070 95 .02408978 .01527399 .00970547 .00618040 .00394405 96 .02316325 .01461626 .00924331 .00585820 .00372081 97 .02227235 .01398685 .00880315 .00555280 .00351019 98 .02141572 .01338454 .00838395 .0052():331 .00331150 99 .02059204 .01280817 .00798471 .00498892 .00312406 100 .01980004 .01225663 .00760449 .00472883 .00294723 17 The Amount of Pekiodic Payments of One Unit Each. s'^— 1 n 1% o — n iy4% i 1%% 1%% 2% 1 1.00 1.00 1.00 1.00 1.00 2 2.01 2.0125 2.015 2.0175 2.02 3 3.0301 3.03765625 3.045225 3.06280626 3.0604 4 4.060401 4.07562695 4.09090338 4.10623036 4.121608 5 5.10100501 5.12657229 5.16226693 6.17808939 6.20404016 6 6.15201506 6.19065444 6.22955093 6.26870596 6.30812096 7 7.21353521 7.26803762 7.32299419 7.37840831 7.413428338 8 8.28567056 8.35888809 8.43283911 8.50753046 8.68296905 9 9.36852727 9.46337420 9.55933169 9.66641224 9.76462843 10 10.46221524 10.58166637 10.70272167 10.82639946 10.94972100 11 11.56684367 11.71393720 11.86326249 12.01484394 12.16871642 12 12.68250301 12.86036142 13.04121143 13.22510371 13.41208973 13 13.80932804 14.02111594 14.23682960 14.45654303 14.68033152 14 14.94742132 15.19637989 15.45038205 15.70953253 16.97393815 15 16.09689554 16.38633463 16.68213778 16.98444935 17.29341692 16 17.25786449 17.59116382 17.93236984 18.28167721 18.63928625 17 18.43044314 18.81105336 19.20135539 19.60160656 20.01207096 18 19.61474757 20.04611953 20.48937572 20.94463468 21.41231238 19 20.81089504 21.29676893 21.79671636 22.31116579 22.84055863 20 22.09100399 22.56297854 23.12366710 23.70161119 24.29736980 21 23.23919403 23.84.501577 24.47052211 25.11638938 26.78331719 22 24.47158598 25.14307847 25.83757994 26.55592620 27.29898364 23 25.71630183 26.45736695 27.22514364 28.02065490 28.84496321 24 26.97346485 27.78808403 28.63352088 29.61101637 30.42186247 25 28.24319950 29.13543508 30.06302361 31.02736915 32.03029972 26 29.52563150 30.49962802 31.51396897 32.67043969 33.67090672 27 30.82087781 31.88087337 32.98667850 34.14042238 35.34432383 28 32.12909669 33.27938428 34.48147868 36.73787977 37.06121031 29 33.45038766 34.69537659 35.99870086 37.36329266 38.79223461 30 34.78489153 36.12906880 37.63868137 39.01716029 40.56897921 31 36.13274045 87.58068216 39.10176159 40.69995042 42.37944079 32 37.49406785 39.05044068 40.68828801 42.41219955 44.22702961 33 38.861f)0()853 40.53857120 42.29861233 44.15441306 46.11157020 34 40.25769863 42.04530334 43.93309152 45.92711527 48.03380160 35 41.66027560 43.57086963 45.59208789 47.73083979 49.99447763 36 43.07687836 45.11550550 47.27596921 49.66612949 51.99436719 37 44.50764714 46.67944932 48.98510875 61.43353675 54.03426453 38 45.95272361 48.26294243 50.71988538 63.33362365 66.11493962 39 47.41225085 49.86622921 52.48068366 56.26696206 68.23723841 40 48.88637336 51.48955708 54.26789392 67.23413390 60.40198318 41 50.37523709 63.13317654 56.08191232 59.23573124 62.61002284 42 51.87898946 54.79734125 57.92314100 61.27236654 64.86222330 43 53.39777935 66.48230802 59.79198812 63.34462278 67.15946777 44 54.93175715 58.18833687 61.68886794 65.45316368 69.50266712 45 56.48107472 59.91569108 63.61420096 67.59858386 71.89271027 46 58.01.588547 61.66463722 65.66841397 69.78955908 74.33056447 47 59.(i26344;32 63.43544518 67.55194018 72.00273637 76.81717576 48 61.22260777 65.22838824 69.66521928 74.26278426 79.35351927 49 62.831.S3385 67.04;374310 71.60869757 76.66238298 81.94058966 50 64.46318218 68.88178989 73.68282804 78.90222468 84.57940145 18 n 1% iy4% iy2% iy4% 2% 51 66.10781401 70.74281227 75.78807046 81.28301361 87.27098948 52 67.76889215 72.62709742 77.92489152 83.705.4()635 90.01640927 53 69.44658107 74.53493613 80.09376489 86.17031201 92.81673745 54 71.14104688 76.46662284 82.29517136 88.67829247 95.67307221 55 72.85245735 78.42245562 84.52959893 91.23016259 98.59653365 56 74.58098193 80.40273633 86.79754292 93.8266904:3 101.558264:32 57 76.32679175 82.40777053 89.09950606 96.4()865752 104.58942960 58 78.09005966 84.43786766 91.43599865 99.15685902 107.68121819 59 79.87096026 86.49334101 93.80753863 101.89210405 110.83484256 60 81.66966986 88.57450777 96.21465171 104.67521586 114.0515:3942 61 83.48636656 90.68168911 98.65787149 107.5070:3215 117.;3:3257021 62 85.32123023 92.81521022 101.13773956 110.38840522 120.67922161 63 87.17444253 94.97540035 103.65480565 113.32020231 124.09280604 64 89.04618695 97.16259285 106.20962774 116.:30330585 127.57466216 65 90.93664882 99.37712526 108.80277215 119.33861370 131.12615541 66 92.84601531 101.61933934 111.43481374 122.42703944 134.74867851 67 94.77447546 103.88958108 114.10633594 125.56951263 138.44:365208 68 96.72222022 106.18820084 116.81793098 128.76697910 142.21252513 69 98.68944242 108.51555355 119.57019995 132.02040124 146.05677563 70 100.67633684 110.87199776 122.36375295 135.33075826 149.97791114 71 102.68310021 113.25789773 125.19920924 138.69904653 153.97746936 72 104.70993121 115.67362145 128.07719738 142.12627984 158.05701875 73 106.75703053 118.11954172 130.99835534 145.61348974 162.21815913 74 108.82460083 120.59603599 133.96333067 149.16172581 166.46252231 75 110.91284684 123.10348644 136.97278063 152.77205601 170.79177276 76 113.02197531 125.64228002 140.02737234 156.44556699 175.20760822 77 115.15219506 128.21280852 14:5.12778293 160.18336441 179.711760:38 78 117.30371701 130.81546862 146.27469967 163.98657:329 184.30599559 79 119.47675418 133.45066199 149.46882017 167.85633832 188.99211550 80 121.67152172 136.11879526 152.71085247 171.79382424 193.77195780 81 123.88823694 138.82028020 156.00151526 175.80021617 198.64739697 82 126.12711931 141.55553371 159.34153799 179.87671995 203.62034490 83 128.38839050 144.32497788 162.73166106 184.02456255 208.69275181 84 130.67227441 147.12904010 166.17263597 188.24499239 213.86660684 85 132.97899715 149.96815310 169.66522552 192.53927976 219.14393898 86 135.30878713 152.84275501 173.21020390 196.90871716 224.52681776 87 137.66187500 155.75328945 176.80835696 201.35461971 230.017:35411 88 140.03849375 158.70020556 180.46048231 205.87832555 235.61770119 89 142.43887868 161.68395813 184.16738955 210.48119625 241.;3:3005521 90 144.86326745 164.70500761 187.92990039 215.16461718 247.15665632 91 147.31190014 167.76382021 191.74884889 219.92999798 253.09978945 92 149.78501914 170.86086796 195.62508162 224.77877295 259.16178523 93 152.28286933 173.99662881 199.55945784 229.71240146 265.34502094 94 154.80569802 177.17158667 203.55284971 234.73236850 271.65192135 95 157.55375501 180.38623150 207.60614246 239.84018495 278.08495978 96 159.92729255 183.64105941 211.72023459 245.03738819 284.64665898 97 162.52656548 186.93657265 215.89603811 250.32554250 291.:3:3959216 98 165.15183111 190.27327980 220.13447868 255.70623947 298.16638400 99 167.80334945 193.65169580 224.4:3649586 261.18109866 305.12971168 100 170.48138294 197.07234200 228.80304330 266.75176789 312.23200591 19 n 2y4% 2y2% 2%% 3% 3y2% 1 1.00 1.00 1.00 1.00 1.00 2 2.022.5 2.025 2.0275 2.03 2.035 3 3.0fiS00f)2.5 3.075625 3.08325625 3.0909 3.106226 4 4..187036.39 4.1.5251563 4.16804580 4.183627 4.2149428S 5 5.23011971 6.25632852 5.28266706 6.30913581 6.3024ti5^>S 6 fi.3.1.779740 6.38773673 6.42794040 6.4091869 59 120.73;392169 131.69911214 143.85317799 157.3334:3379 188.90520085 60 124.45043493 135.99158995 148.80914038 163.05343680 196.51688288 61 128.25056972 140.39137969 153.90139174 168.94503991 204.39497378 62 132.13620753 144.90116419 159.13368002 175.01339110 212.-5487978G 63 136.10927220 149.52369329 164.50985622 181.26379284 220.98800579 64 140.17173083 154.26178563 170.03387726 187.70170662 229.72258599 65 144.32559477 159.11833027 175.70980889 194.33275782 238.76287650 66 148.57292605 164.09628852 181.54182863 201.16274055 248.11957718 67 152.91581137 169.19869574 187.53422892 208.19762277 257.80376238 68 157.35641712 174.42866313 193.69142022 215.44355145 267.82689406 69 161.89693651 179.78937971 200.01793427 222.90685800 278.20083535 70 166.53961758 185.28411421 206.51842746 230.59406374 288.93786459 71 171.28675898 190.91621706 213.19768422 238.51188565 300.05068985 72 176.14071105 196.68912249 220.06062054 246.66724222 311.55246400 73 181.10387705 202.60635056 227.11228760 255.06725949 323.45680023 74 186.17871429 208.67150932 234.35787551 263.71927727 3-35.77778824 75 191.36773538 214.88829706 241.80271709 272.63085559 348.53001083 76 196.67350940 221.26050449 249.45229181 281.80978126 361.72856121 77 202.09866337 227.79201710 257.31222983 291.26407469 375.38906085 78 207.64588329 234.48681752 265.38831615 301.00199693 389.52767798 79 213.31791567 241.34898795 273.68649485 311.03205684 404.16114671 80 219.11756877 248.38271265 282.21287345 321.36301855 419.30678685 81 225.04771497 255.59228047 290.97372747 332.00390910 434.98252439 82 231.11128763 262.98208748 299.97550499 342.96402638 451.20691274 83 237.31129160 270.55663967 309.22483137 354.25294717 467.99915469 84 243.65079567 278.32055566 318.72851423 365.88053558 485.37912510 85 250.13293857 286.27856955 328.49354837 377.85695165 503.36739448 86 256.76092969 294.43553379 338.52712095 390.19266020 521.98525329 87 263.53805060 302.79642214 348.83661678 402.89844001 541.25473715 88 270.46765674 311.36633269 359.42962374 415.98539321 561.19865295 89 277.55317902 320.15041900 370.31393839 429.46495500 581.84060581 90 284.79812555 329.15425328 381.49757170 443.34890365 603.20502701 91 292.20608337 338.38310961 392.98875492 457.64937076 625.31720296 92 299.78072025 347.84268735 404.79594568 472.37885189 648.20330506 93 307.52578644 357.53875453 416.92783418 487.55021745 671.89042074 94 315.44511665 367.47722340 429.39334962 503.17672397 696.40658546 95 323.54263177 377.66415398 442.20166674 519.27202569 721.78081595 96 331.82234099 388.10575783 455.36221257 535.85018646 748.04.314451 97 340.28834366 398.80840178 468.884.67342 552.92569205 775.22465457 98 348.94483139 409.77861182 482.77900194 570.51:546282 803.35751748 99 357.79609010 421.02307712 497.05542449 588.62886670 832.47503059 100 366.84650213 432.54865404 511.72444867 607.28773270 862.61165066 21 n 4% *V2% 5% 51/2% 6% 1 1.00 1.00 1.00 1.00 1.00 2 . 2.04 2.045 2.0o 2.055 2.06 3 3.121G 3.137025 3.1525 3.168025 3.1836 4 4.246164 4.27819113 4.310125 4.34226638 4.374616 5 6.41632256 6.47070973 6..52563125 5..58109103 5.63709296 6 6.63297546 6.71689166 6.80191281 6.88805103 6.97531854 7 7.89829418 8.01915179 8.14200845 8.26689384 8.39383765 8 9.21422626 9.38001362 9.54910888 9.72157300 9.89746791 9 10.58279531 10.80211423 11.02656432 11.2.562,5951 11.49131598 10 12.00610712 12.28820937 12.577892.54 12.87535379 13.18079494 11 13.48635141 13.84117879 14.20678716 14.58349825 14.97164264 12 15.02580546 15.46403184 15.91712652 16.38559065 16.86994120 13 16.62683768 17.1.5991.327 17.71298284 18.29679814 18.88213767 14 18.29191119 18.93210937 19.59863199 20.29257203 21.01506593 15 20.02358764 20.78405429 21.57856359 22.40866250 23.27596988 16 21.82453114 22.71933673 23.65749177 24.64113999 25.67252808 17 23.69751239 24.74170689 25.84036636 26.99640269 28.21287976 18 25.64541288 26.85508370 28.13238467 29.48120483 30.9056,5255 19 27.67122940 29.06356246 30.53900391 32.10267110 33.75999170 20 29.77807858 31.37142277 33.06595410 34.86831801 36.78559120 21 31.96920172 33.78313680 35.71925181 37.78607,550 39.99272668 22 34.24796979 36.30.337795 38.50521440 40.86430965 4:3.39229028 23 36.617888.58 38.93702996 41.43047512 44.11184669 46.99582769 24 39.08260412 41.68919631 44.50199887 47.53799825 50.81557735 25 41,64590829 44.56521015 47.72709882 51.15258816 54.86451200 28 44.31174462 47.57064460 51.11345376 54.96598051 59.15638272 27 47.08421440 50.71132361 54.66912645 58.9891094:3 63.70576568 28 49.96758298 53.99.333319 58.40258277 63.23351045 68.52811162 29 52.96628630 57.42303316 62.32271191 67.711,3,53.53 73.63979832 30 66.08493775 61.00706966 66.43884750 72.43547797 79.05818622 31 59.32833526 64.75238779 70.76078988 77.41942926 84.80167739 32 62.70146867 68.66624524 75.29882937 82.67749787 90.88977803 33 66.20952742 72.75622628 80.06377084 88.22476025 97.34316471 34 69.857.90851 77.03025646 85.06695938 94.07712206 104.18375459 35 73.65222486 81.49661800 90.32030735 100.251.36378 111.43477987 36 77.59831385 86.16396-581 95.83632272 106.76518879 119.12086666 37 81.70224640 91.04134427 101.62813886 113.63727417 127.26811866 38 85.97033626 96.13820476 107.70954580 120.88732425 135.90420578 39 90.40914971 101.46442.398 114.09502309 128.53612708 145.05845813 40 95.02551570 107.03032306 120.79977424 136.60561407 154.76196562 41 99.82653633 112.84668760 127.83976295 145.11892285 165.04768356 42 104.81959778 118.92478854 135.23175110 154.10046360 175.95054457 43 110.012.38169 125.27640402 142.99.3.33866 16.3.57,598910 187.50757724 44 11.5.41287696 131.91384220 151.1^300559 173.572668.50 199.75803188 45 121.02939204 138.84996510 159.70015587 184.11916,527 212.74351:379 46 126.87056772 146.09821353 168.68516366 195.24571936 226.50812462 47 132.94539043 153.67263314 178.11942185 206.98423392 241.09861210 48 139.26320604 161.58790163 188.02539294 219.36836679 256.56452882 49 145.8337;} 1.29 169.85935720 198.42666259 232.43362696 272.95840055 50 152.66708366 178.60302828 209.34799572 246.21747645 290.33590458 22 n 4% *V2% 5% 5%% 6% 51 159.77:370700 187.53560455 220.81539550 260.75943765 308.75605886 52 167.1G1.71768 196.97476940 2:32.85616528 276.10120672 328.28142239 53 174.851306.39 200.83803408 245.49897354 292.28677.309 348.97830773 54 182.84.5358fi5 217.14637261 258.77392222 309.30254501 370.91700620 55 191.15917299 227.91795938 272.71261833 327.37748562 394.17202657 56 199.80553991 239.17426756 287.34824924 346.38.324733 418.82234810 57 208.7977fil51 250.93710959 302.71566171 360.4.343259:} 444.95168905 58 218.149G7197 263.22927953 318.85144479 387.58821:380 472.04879040 59 227.87505885 270.07459711 3:35.7940170:} 409.90.550562 502.00771782 60 237.9900-8520 289.49795398 353.58371788 433.45037173 533.12818089 61 248.51031261 303.52536190 372.26290378 458.29014217 566.11587174 62 259.45072511 318.18400319 391.87604897 484.49609999 601.08282405 63 270.82875412 333.50228333 412.46985141 512.14338549 6.38.14779:349 64 282.00190428 349.50988608 434.09334398 541.31127109 677.43000110 65 294.90838045 366.23783096 456.79801118 572.08339104 719.08286070 66 307.70711567 383.71853335 480.63791174 604.54797818 763.22783241 67 321.07780030 401.98580735 .505.66980733 0.38.79811098 810.021502:50 68 334.92091231 421.07523138 .531.95.329770 074.93201341 859.62279250 69 349.31774880 441.02301079 559.55090258 713.05327415 912.20016005 70 364.29045876 401.80907955 588.52851071 753.27120423 967.93216965 71 379.80207711 483.05381513 018.9549.3025 795.70112040 1027.00809983 72 396.05050019 506.41823681 050.90208300 840.40408208 1089.02858582 73 412.89882200 530.20705747 084.44781721 887.69023960 1150.000.30097 74 430.41477550 555.06037505 719.08020807 937.51320278 1220.36667902 75 448.63130652 581.044:30193 750.05371848 990.07642893 1300.94867977 76 467.57002118 608.19135822 795.48040440 1045..5.306.3252 1380.00560055 77 487.27908003 030.55996934 830.20072402 1104.0.3481731 1463.80593658 78 507.77087347 666.20516797 879.07370085 1105.7507.3220 15.52.6.3429278 79 529.08170841 097.184400.53 924.02744889 1230.87.335254 164^.79235035 80 551.24497675 729.55769854 971.22882134 1299.57138693 1746.59989137 81 574.29477582 763.38779497 1020.79026240 1372.04781321 1852.39588485 82 598.26656085 798.74024574 1072.82977.5.52 1448.51044293 1964.53963794 83 623.19722952 835.08355080 1127.471264.30 1529.17851729 2083.41201622 84 649.12511870 874.28931080 1184.84482752 1614.28.3.3.3574 2209.41673719 85 676.09012345 914.03233012 1245.08700889 1704.06891921 2.342.98174142 86 704.13372839 950.79071924 1308.34142234 1798.79270977 2484.56064590 87 733.29907752 1000.84037085 1374.75849345 1898.72630880 2634.63428466 88 763.03104003 1040.88446381 1444.49041813 2004.15625579 279:3.712:34174 89 795.17028225 1094.99420468 1517.72123903 2115.38484986 2902.33508224 90 827.98333354 1145.26900659 1594.00730098 22.32.73101660 3141.07518718 91 862.10266688 1197.80611188 1075.33760003 2356.53122251 3330.53909841 92 897.58677356 1252.70738692 1700.104.549.34 2487.14043975 3531.37208032 93 934.49024450 1310.07921933 1849.10977080 2624.93316394 3744.25440513 94 972.80985428 1370.03278420 1942.56526564 2770.30448796 3909.90900944 95 1012.78464845 14:32.68425949 2040.69352892 292:3.67123480 4209.10424901 96 1054.29603439 1498.15505117 2143.72820537 3085.47315271 4462.65050459 97 1097.40787577 1566.57202847 2251.91461564 3256.17417611 4731.40953486 98 1142.36659090 1038.00770975 2365.51034042 3430.26375579 5016.29410095 99 1189.06125443 1712.78081938 2484.78586374 3626.25826236 5318.27175.337 100 1237.62370461 1790.85595626 2610.02515693 3826.70246679 5638.36805857 23 The Present Worth of Periodic Payments of One Unit Each. n 1% a 11 i l%% 2% 1 .99009901 .98765432 .98522167 .98280098 .98039216 2 1.97039506 1.96311538 1.95588342 1.94869875 1.94156094 3 2.94098521 2.92653371 2.91220042 2.89798403 2.88388327 4 3.90196555 3.87805798 3.854138465 3.83094254 3.80772870 5 4.85343124 4.81783504 4.78264497 4.74785508 4.71345951 6 5.79547647 5.74600992 5.69718717 5.64899762 5.6014:3089 7 6.72819453 6.66272585 6.59821396 6.53464139 6.47199107 8 7.65167775 7.56812429 7.48592508 7.40505297 7.32548144 9 8.56601758 8.46234498 8.36051732 8.26049432 8.16223671 10 9.47130453 9.34552591 9.22218455 9.10122291 8.98258501 11 10.36762825 10.21780337 10.07111779 9.92719181 9.78684805 12 11.25507747 11.07931197 10.90750521 10.73954969 10.57534122 13 12.13374007 11.93018766 11.73153222 11.53764097 11.34837375 14 13.00370304 12.77055275 12.54338150 12.32200587 12.10624877 15 13.86505252 13.60054592 13.34323301 13.09288046 12.849263.50 16 14.71787378 14.42029227 14.13126405 13.85049677 13.57770931 17 15.56225127 15.22991829 14.90764931 14.59508282 14.29187188 18 16.39826286 16.02954893 15.67256089 15.32686272 14.99203125 19 17.22600850 16.81930759 16.42616837 16.04605673 15.6784*201 20 18.04555297 17.59931613 17.16863878 16.75288130 16.35143334 21 18.85696313 18.36969495 17.90013673 17.44754919 17.01120916 22 19.66037934 19.13056291 18.62082437 18.13026948 17.65804820 23 20.45582113 19.88203744 19.33086145 18.80124764 18.29220411 24 21.24338726 20.62423451 20.03040537 19.46068565 18.91392560 25 22.02315570 21.35726865 20.71961120 20.10878196 19.52345647 26 22.79520366 22.08125299 21.39863172 20.74573166 20.12103576 27 23.55960759 22.79629925 22.06761946 21.37172644 20.70689780 28 24.31644316 23.50251778 22.72671671 21.98695474 21.28127236 29 25.06578530 24.20001756 23.37607558 22.59160171 21.84438466 30 25.80770822 24.88890623 24.01583801 23.18584934 22.39645555 31 26.54228537 25.56929010 24.64614582 23.76987651 22.93770152 32 27.2()958497 26.24129417 25.26713874 24.34385897 23.46833482 33 27.98969255 26.90496215 25.87895442 24.90796951 23.98856355 34 28.70266589 27.56045644 26.48172849 25.46237789 24.49859172 35 29.40858009 28.20785821 27.07559458 26.00725100 24.99861933 36 30.10750504 28.84726737 27.66068431 26.54275283 25.48884248 37 30.79950994 29.47878259 28.23712742 27.06904455 25.96945341 38 31.48466330 30.10250132 28.80505163 27.58628457 26.44064060 39 32.16303297 30.71851982 29.36458288 28.09462857 26.90258883 40 32.83468611 31.32693316 29.91584520 28.59242955 27.35547923 41 33.49968922 31.92783522 30.45896079 29.08523789 27.79948945 42 34.15810814 32.52131874 30.99405004 29.56780136 28.23479358 43 34.81000806 33.10747529 31.52123157 30.04206522 28.661562.33 44 35.45545352 33.68639535 32.04062223 30.50817221 29.07996307 45 36.09-450844 34.25816825 32.55233718 30.96626261 29.49015988 46 36.72723608 34.82288222 33.05648983 31.41647431 29.89231.360 47 37.35369909 35.38062442 33.55319195 31.8.5894281 .30.28658196 48 37.97395949 35.93148096 34.04255365 32.29380129 30.67311957 49 38.58807871 36.47553670 34.52468339 32.72118063 31.05207801 50 39.19611753 37.01287575 34.99968807 24 33.14120946 31.42360589 n 1% IV4% IM.''/^ 1%% 2% 51 39.79813617 37.54358099 35.46767298 33.55401421 31.78784892 52 40.39419423 38.06773431 35.92874185 33.95971913 32.14494992 53 40.98535072 38.58541660 36.38299690 34.3584t632 32.49504894 54 41.5G866408 39.09670776 36.83053882 34.75031579 32.83828327 55 42.14719216 39.60168667 37.27146681 35.13544550 33.17478752 56 42.71999236 40.10043128 37.70587864 35.51395135 ,33.50469365 57 43.38712102 40.59301855 38.13387058 35.88594727 33.82813103 58 43.84863468 41.07952449 38.55553751 36.25154523 34.14522655 59 44.404.58879 41.56002419 38.97097292 36.61085526 34.45610441 60 44.95503841 42.03459179 39.38026889 36.96398552 .34.76088667 61 45.50003803 42.50330054 39.78351615 37.31104228 35.05969282 62 46-.03964161 42.96622275 40.18080408 37.65213000 35..35264002 63 46.57390258 43.42342988 40.57222077 37.98735134 35.63984315 64 47.10287385 43.87499247 40.95785298 38.31680723 35.92141486 65 47.62660777 44.32098022 41.33778619 38.64059678 36.19746555 66 48.14515621 44.76146194 41.71210462 38.95881748 36.46810348 67 48.65857050 45.19650562 42.08089125 39.27156509 36.73343478 68 49.16690149 45.62617839 42.44422783 39.57893375 36.99.356.351 69 49.67019949 46.05054656 42.80219491 39.88101597 37.24859168 70 50.16851435 46.46967561 43.15487183 40.17790267 37.49861929 71 50.66189539 46.88363023 43.50233678 40.46968321 37.74374440 72 51.15039148 47.29247430 43.84466678 40.75644.542 37.984^6314 73 51.63405097 47.69627091 44.18193771 41.0.3827560 .38.21966975 74 52.11292175 48.09508238 44.51422434 41.31.52.5857 .38.4.5065661 75 52.58705124 48.48897026 44.84160034 41.58747771 .38.67711433 76 53.05648638 48.87799532 45.16413826 41.85501495 .38.89913169 77 53.52127364 49.26221760 45.48190962 42.11795081 39.11679578 78 53.98145905 49.64169640 45.79498485 42.37636443 39.33019194 79 54.43708817 50.01649027 46.10343335 42.63033359 39.53940386 80 54.88820611 50.38665706 46.40732349 42.87993474 39.74451359 81 55..3348574;3 50.75225389 46.70672265 43.12524298 39.94560156 82 55.77708666 51.11333717 47.00169720 43..36633217 40.14274662 83 56.21493729 51.43996264 47.29231251 4:3.60327486 40..33602610 84 56.64845276 51.82218532 47.57863301 43.8.36142.37 40.52551579 85 57.07767600 52.17005958 47.86072218 44.06500479 40.71128999 86 57.50264951 52.51363909 48.13864254 44.28993099 40.89342156 87 57.92441535 52.85297688 48.41245571 44.51097869 41.07198192 88 58.34001520 53.18812531 48.68222237 44.72824441 41.24704110 89 58.75249030 53.51913611 48.94800233 44.94176.355 41.41866774 90 59.16088148 53.84606035 49.20985452 45.15161037 41.58692916 91 59.56522919 54.16894850 49.46783696 45.35784803 41.75189133 92 59.96557346 54.48785037 49.72200686 45.56053860 41.91361895 93 60.36195392 54.80281518 49.97242055 45.75974310 42.07217544 94 60.75440982 55.11389154 50.21913355 45.95.5.52147 42.22762299 95 61.14298002 55.42112744 50.46220054 46.14793265 42.38002254 96 61.52770299 55.72457031 50.70167541 46.33703455 42.52943386 97 61.90861682 56.02426698 50.93761124 46.52288408 42.67591555 98 62.28575923 56.32026368 51.17006034 46.70553718 42.819.52.505 99 62.65916755 56.61260610 51.39907422 46.88504882 42.96031867 100 63.02887877 56.90133936 51.62470367 47.06147304 43.09835164 25 n 21/4% 21/2% 2%% 3% 31/2% 1 .97799551 .97560976 .97323601 .97087379 .96618357 2 1.934+6755 1.92742415 1.920424:34 1.91.346970 1.89969428 3 2.86989()87 2.85602356 2.84226213 2.82861135 2.8016:3698 4 3.784.74021 3.76197421 3.73942787 3.71709840 3.67307921 5 4.6794^253 4.64582850 4.61258186 4.57970719 4.51505238 6 5.55447680 5.50812536 5.46236678 5.41719144 5.32855302 7 C.41024626 6.34939060 6.28940806 6.23028296 6.114.54398 8 7.247184-61 7.17013717 7.09431441 7.01969219 6.87395554 9 8.06570622 7.97086553 7.87767826 7.78610892 7.60768651 10 8.86621635 8.75206393 8.64007616 8.53020284 8.31660532 11 9.64911134 9.51420871 9.38206926 9.25262911 9.00155104 12 10.41477882 10.25776460 10.10420.366 9.95400400 9.663334:33 13 11.16359787 10.98318497 10.80701086 10.634955.33 10.30273849 14 11.89593924 11.69091217 11.49100814 11.29607314 10.92052028 15 12.61216551 12.38137773 12.15669892 11.93793509 11.51741089 16 13.31263131 13.05500266 112.80457315 12.56110203 12.09411681 17 13.99768343 13.71219772 13.43510769 13.16611847 12.65132059 18 14.66766106 14.35336363 14.04876661 13.75351308 13.18968173 19 15.32289590 14.97889134 14.64600157 14.32379911 13.70983742 20 15.96371237 15.58916229 15.22725213 14.87747486 14.21240330 21 16.59042775 16.184.54857 15.79294611 15.41502414 14.69797420 22 17.20335232 16.76541324 16.34349987 15.93691664 15.16712483 23 17.80278955 17.33211048 16.87931861 16.44360839 15.62041047 24 18.38903624 17.88498583 17.40079670 16.93554212 16.05836760 25 18.96238263 18.42437642 17.90831795 i 7.41314769 16.48151459 26 19.52311260 18.95061114 18.40225592 17.87684242 16.89036226 27 20.07150376 19.46401087 18.88297413 18.32703147 17.28536451 28 20.60782764 19.96488866 19.35082640 18.76410823 17.66701885 29 21.13234977 20.45354991 19.80615708 19.18845459 18.03576700 30 21.64532985 20.93029259 20.24930130 19.60044135 18.39204541 31 22.14702186 21.39540741 20.68058520 20.00042849 18.73627576 32 22.63767419 21.84917796 21.10032623 20.38876553 19.06886.547 33 23.117.52977 22.29188094 21.508833.32 20.76579177 19.39020818 34 23.58682618 22.72378628 21.90640712 21.13183667 19.70068423 35 24.04579577 23.14515734 22.29334026 21.48722007 20.00066110 36 24.4946(i579 23.55625107 22.66991753 21.83225249 20.29049:381 37 24.9336581.8 2.3.95731811 23.()3()11609 22.1672354;} 2().57()5251-2 38 25.36299118 24.34860304 23..39310568 22.49246158 20.841087:36 39 25.78287646 24.7303444;? 23.74024884 22.80821513 21.10249987 40 26.19352221 25.10277505 24.07810102 2.3.11477197 21.35507234 41 26.59513174 25.46612200 24.40691101 23.412:39997 21. .599 10371 42 26.98790390 25.820()0683 24.72692069 2:3.701:35920 21.83488281 43 27.37203316 2().l(i(iMr)(i9 25.()3S3(i563 23.98190213 22.()()268870 44 27.74770969 2(i.50384.945 35.34147.507 24.251.27392 22.28279102 45 28.11511950 26.83302386 25.63647209 24.51871254 22.49545026 46 2S.471. 1-4 1-50 27.15416962 25.92357.381 24.77544907 22.7()091HI2 47 2K.H2.1H6259 27.46748255 26.20299151. 25.02470783 22.8994:5780 48 2!>. 16954777 27.77315371 26.47493094 25.26670664 2:5.09 1241-25 49 29..50567019 28.07136947 26.73959215 25.50165693 2;}.276561..50 50 29.83439627 28.36231168 26.99716998 25.72976401 23.45561787 26 n 2y4% ^y2% 2%% 3% 3yo% 51 30.15588877 28.64615774 27.24785399 25.95122719 23.628616-30 52 30.47030fi87 28.92308072 27.49182870 26.16623999 23.79576455 53 30.77780()2;} 29.19324948 27.72927368 26.37499028 23.957260U 54 31.07853910 29.45682876 27.96036368 26.57766047 24.11329511 55 31.37265438 29.71397928 28.18526879 26.77442764 24.26405-324 56 31.(56029768 29.96485784 28.40415454 26.96346373 24.40971327 57 31.94.16114.2 30.20961740 28.61718203 27.15093566 24.55044761 58 32.21673489 30.44840722 28.82450806 27.33100549 24.68642281 59 32.48580429 30.68137290 29.02628522 27.50583058 24.81779981 60 32.74895285 30.90865649 29.22266201 27.67556367 24.94473412 61 33.00631086 31.13039657 29.41378298 27.8403.5.307 25.06737596 62 33.25800573 31.34672836 29.59978879 28.00034279 25.18-587050 63 33.50416208 31.55778377 29.78081634 28.15567261 25.-30035797 64 33.74490179 31.76369148 29.95699887 28.30647826 25.41097389 65 33.98034405 31.96457705 30.1284()605 28.45289152 25.51784916 66 34.21060543 32.16056298 30.295.34409 28.59504031 25.62111030 67 34.43579993 32.35176876 30.45775.581 28.7-3:304884 25.72087952 68 34.65603905 32.53831099 30.61582074 28.86703771 25.81727489 69 34.87143183 32.72030340 .30.76965522 28.99712-399 25.91041053 70 35.08208492 32.89795698 30.91937247 29.12:342135 26.00039664 71 35.28810261 33.07107998 31.06508270 29.24604105 26.087.33976 72 35.48958691 33.24007803 31.20689304 29.365087.52 26.17134276 73 35.68663756 33.40495417 31. .344908 16 29.48066750 26.25250508 74 35.87935214 33.56580895 31.47922936 29.59288107 26.33092278 75 36.06782605 33.72274044 31.609955.58 29.70182628 26.40668868 76 36.25215262 33.875844:33 31.73718,304 29.807598.33 26.47989244 77 36.43242310 34.02521398 31.86100540 29.91028964 26.55062072 78 36.60872675 34.17094047 31.98151377 30.00998994 26.61895721 79 36.78115085 34.31311265 .32.09879685 30.10678635 26.68498281 80 36.94978079 34.45181722 32.21294098 30.20076345 26.74877567 81 37.11470004 34.58718752 .32..32403015 30.29200335 26.81041127 82 37.27599026 34.71915976 .32.4:3214613 30.38058577 26.86996258 83 37.43373130 34.84796074 :32..53736850 .30.46658813 26.92750008 84 37.58800127 34.97362023 32.63977469 .30.55008556 26.98:309186 85 37.73887655 35.09621486 32.73944009 30.63115103 27.03680373 86 37.88643183 35.21581938 32.83643804 30.70985537 27.08869926 87 38.03074018 35..33250671 32.93083995 30.78626735 27.1:3883986 88 38.17187304 35.44634891 33.02271528 30.86045374 27.18728489 89 38.30990028 35.55741269 33.11213165 .30.93247935 27.2:3409168 90 38.44489025 35.66576848 .33.19915489 31.00240714 27.27931564 91 38.57690978 35.77148144 33.28384905 31.07029820 27.-32:301028 92 38.70602423 35.87461604 33.36627644 31.1:3621184 27.36522732 93 38.83229754 35.97523516 33.44649776 31.20020567 27.40601673 94 38.9.5579221 36.073400 Ki .33.-52457203 31.262:]35(;0 27.44542630 95 39.07657940 .36.16917089 33.60055672 31.;i22U5592 27.48350415 96 39.19468890 36.26260574 33.67450775 31.-381219-34 27.52029386 97 39.31020920 36.35376170 33.74647957 31.43807703 27.55583948 98 39.42318748 36.41269434 33.81652512 31.49:327867 27.59018307 99 39.5;3;367968 36.52945790 33.88469599 31.54687250 27.62336529 too 39,64174052 36.61410526 33.95104232 31,59890534 27.65542540 27 n *% *V2% 5% 5V2% 6% 1 .96153846 .95693780 .95238095 .94786730 .94339623 2 1.88609467 1.87266775 1.85941043 1.84631971 1.83339267 3 2.77509103 2.74896435 2.72324803 2.69793338 2.67301195 4 3.62989522 3.58752570 3.54595050 3.50515012 3.46510561 S 4.45182233 4.38997674 4.32947667 4.27028448 4.21236379 6 6.24213686 5.1,5787248 5.07569207 4.99553031 4.91732433 7 6.00205467 5.89270094 5.78637340 5.68296712 5.58238144. 8 6.73274487 6.59588607 6.46321276 6.33456599 6.20979381 9 7.43533161 7.26879050 7.10787168 6.95219525 6.80169227 10 8.11089578 7.91271818 7.72173493 7.53762583 7.36008705 11 8.76047671 8.52891692 8.30641422 8.09253633 7.88687458 12 9.38507376 9.11858078 8.86325164 8.61851785 8.38384394 13 9.98564785 9.68285242 9.39357299 9.11707853 8.85268296 14 10.56312293 10.22282528 9.89864094 9.58964790 9.29498393 15 11.11838743 10.73954573 10.37965804 10.03758094 9.71224899 16 11.65229561 11.23401505 10.83776956 10.46216203 10.10589527 17 12.16566885 11.70719143 11.27406625 10.86460856 10.47725969 18 12.65929698 12.15999180 11.68958690 11.24607447 10.82760348 19 13.13393940 12.59329359 12.08532086 11.60765352 11.15811649 20 13.59032634 13.00793645 12.46221034 11.95038248 11.46992122 21 14.02915995 13.40472388 12.82115271 12.27524406 11.76407662 22 14.45111533 13.78442476 13.16300258 12.58316973 12.04158172 23 14.85684167 14.14777489 13.48857388 12.87504239 12.30007898 24 15.24696314 14.49547837 13.79864179 13.15169895 12.55035753 25 15.62207994 14.82820896 14.09394457 13.41393266 12.78335616 26 15.98276918 15.14661145 14.37518530 13.66249541 13.00316619 27 16.32958575 15.45130282 14.64303362 13.89809991 13.21053414 28 16.66306322 15.74287351 14.89812726 14.12142172 13.40616428 29 16.98371463 16.02188853 15.14107358 14.33310116 13.59072102 30 17.29203330 16.28888854 15.37245103 14.53374517 13.76483115 31 17.58849356 16.54439095 15.59281050 14.72392907 13.92908600 32 17.87355150 16.78889086 15.80267667 14.90419817 1408404339 33 18.14764567 17.02286207 16.00254921 15.07506936 14.23022961 34 18.41119776 17.24675796 16.19290401 15.23703257 14.36814114 35 18.66461323 17,46101240 16.37419429 15.39055220 14.49824636 36 18.90828195 17.66604058 16.54685171 15.53606843 14.62098713 37 19.14257880 17.86223979 16.71128734 15.67399851 14.73678031 38 19.36786423 18.04999023 16.86789271 15.80473793 14.84601916 39 19.58448484 18.22965572 17.01704067 15.92866154 14.94907468 40 19.79277388 18.40158442 17.15908635 16.04612469 15.04629687 41 19.99305181 18.56610949 17.29436796 16.15746416 15.13801592 42 20.18562674 18.72354975 17.42320758 16.26299920 15.22454332 43 20.37079494 18.87421029 17.54591198 16.36303242 15.30617294 44 20.54884129 19.01838305 17.66277331 16.45785063 15.38318202 45 20.72003970 19.15634742 17.77406982 16.54772572 15.45583209 46 20.88465356 19.28837074 17.88006650 16.63291537 15.52436990 47 21.04293612 19.41470884 17.98101571 16.71366386 15.58902821 48 21.19513088 19.535()0654 18.07715782 16.79020271 15.65002661 49 21.34147200 19.65129813 18.16872173 16.86275139 15.70757227 50 21.48218462 19.76200778 18.25592546 16.93151790 15.76186064 28 n ■i% •*y.% 5% 5^2% 6% 51 21.6174.8521 19.86795003 18.33897663 16.99669943 15.81306707 52 21.74758193 19.96933017 18.41807298 17.05848287 15.861392.52 53 21.872()-71.;):3 20.06634466 18.493K)284 17.11704.538 15.90697408 54 21.992J)5(i()7 20.15918149 18.56514.556 17.172.55486 15.94997544 55 22.108()1218 20.24802057 18.63347196 17.22517048 15.99054297 56 22.21981940 20.33303404 18.69854473 17.27504.311 16.02881412 57 22.;32G74943 20.41438664 18.76951879 17.32231575 16.06491898 58 22.42956676 20.49223602 18.81954170 17.36712393 16.09898017 59 22.52852957 20.56673303 18.87575400 17.40959614 16.131113.37 60 22.62348977 20.63802204 18.92928953 17.44985416 16.16142771 61 22.71489421 20.70624118 18.98027574 17.48801343 16.19002614 62 22.80278289 20.77152266 19.02883404 17.524183.34 16.21700579 63 22.88729124 20.83399298 19.07508003 17.55846762 16.24245829 64 22.96854927 20.89377319 19.11912384 17.59096457 16.26647009 65 23.04668199 20.95097913 19.16107033 17.621767.37 16.28912272 66 23.12180961 21.00572165 19.20101936 17.65096433 16.31049314 67 23.19404770 21.05810684 19.23906606 17.6786-3917 16. .3306-5390 68 23.26350740 21.10823621 19.27530101 17.70487125 16.34967.349 69 23.33029558 21.15620690 19.30981048 17.72973579 16.-367616-50 70 23.39451498 21.20211187 19.34267665 17.75330406 16.38454.387 71 23.45626440 21.24604007 19.37397776 17.77564366 16.40051308 72 23.51563885 21.28807662 19.40378834 17.79681864 16.41557838 73 23.57272966 21.32830298 19.43217937 17.81688970 16.42979093 74 23.62762468 21.36679711 19.45921845 17.83591441 16.44319899 75 23.68040834 21.40363360 19.48496995 17.85394731 16.45584810 76 23.73116187 21.43888.383 19.50949519 17.87104010 16.46778123 77 23.77996333 21.47261611 19.53285257 17.88724180 16.47903889 78 23.82688782 21.50489579 19.55.509768 17.90259887 16.48965933 79 23.87200752 21.53578545 19.57628351 17.917155.32 16.49967862 80 23.91539185 21.56534493 19.59646048 17.93095291 16.50913077 81 23.95710755 21.59363151 19.61567665 17.94403120 16.51804790 82 23.99721879 21.62070001 19.63397776 17.95642767 16.52646028 83 24.03578730 21.64660288 19.65140739 17.96817789 16.53439649 84 24.07287241 21.671390.32 19.66800704 17.97931554 16.54188348 85 24.10853116 21.69511035 19.68381623 17.98987255 16.54894668 86 24.14281842 21.71780895 19.69887260 17.99987919 16.55561008 87 24.17578694 21.73953009 19.71321200 18.00936416 16.56189630 88 24.20748745 21.76031588 19.72686857 18.01835466 16.56782670 89 24.23796870 21.78020658 19.73987483 18.02687645 16.57342141 90 24.26727759 21.79924075 19.75226174 18.03495398 16.57869944 91 24.29545923 21.8174.5526 19.76405880 18.04261041 16.58367872 92 24.32255695 21.83488542 19.77529410 18.04986769 16.58837615 93 24.34861245 21.85156499 19.785994:38 18.05674662 16.59280769 94 24.37366582 21.86752631 19.79618512 18.06326694 16.59698839 95 24.39775559 21.88280030 19.80589059 18.06944734 16.60093244 96 24.42091884 21.897416.55 19.81513390 18.07530.553 16.60465325 97 24.44319119 21.91140340 19.82393705 18.08085832 16.60816344 98 24.46460692 21.92478794 19.83232100 18.08612164 16.61147494 99 24.48519896 21.93759612 19.840.30571 18.09111055 16.61459900 100 24.50499900 21.94985274 19.84791020 18.09583939 16.61754623 29 The Periodic Payment That Will Amount to One. The Sinking Fund. Sn S°— 1 n 1% lVi% iy2% 1%% 2% 1 1.00 1.00 1.00 1.00 1.00 2 .49751244 .49689441 .49627792 .49566295 .49504950 3 .33002211 .32920117 .32838296 .32756746 .32675467 4 .26428109 .24536102 .24444478 .24353237 .24262375 5 .19G03980 .19506211 .19408932 .19312142 .19215839 6 .16254837 .16153381 .16052521 .15952256 .15852581 7 .13862828 .13758872 .13655616 .13553059 .13451196 8 .12069029 .11963314 .11959402 .11754292 .11650980 9 .10674036 .10567055 .10460982 .10355813 .10251544 10 .09558208 .09450307 .09343418 .09237534 .09132653 11 .08645408 .08536839 .08429384 .08323038 .08217794 12 .07884879 .07775831 .07667999 .07561337 .07455960 13 .07241482 .07132100 .07024036 .06917283 .06811835 14 .06690177 .06580515 .06472332 .06365562 .06260197 15 .06212378 .06102646 .05994436 .05887739 .05782547 16 .05794460 .05684672 .05576508 .05469958 .05365013 17 .05425806 .05316023 .05207966 .05101623 .04996984 18 .05098205 .04988479 .04880578 .04774492 .04670210 19 .04805175 .04695548 .04587847 .04482061 .04378177 20 .04541531 .04432039 .04:324574 .04219122 .04115672 21 .04303075 .04193748 .04086550 .03981464 .03878477 22 .04086372 .03977238 .03870331 .03765638 .03663140 23 .03888584 .03779666 .03673075 .03568796 .03466810 24 .03707347 .03598665 .03492410 .03388565 .03287110 25 .03540675 .03432247 .03326345 .03222952 .03122044 26 .03386888 .03278729 .03173196 .03070269 .02969923 27 .03244553 .03136677 .03031527 .02929079 .02829309 28 .03112444 .03004863 .02900108 .02798151 .02698967 29 .02989502 .02882228 .02777878 .02676424 .02577836 30 .02874811 .02767854 .02663919 .02562975 .02464992 31 .02767573 .02660942 .02557430 .02457005 .02359635 32 .02667089 .02560791 .02457710 .02357812 .02261061 33 .02572744 .02466786 .02364144 .022()4779 .02168653 34 .02483997 .02378387 .02276189 .02177363 .02081867 35 .02400368 .02295111 .02193363 .02095082 .02000221 36 .02321431 .02216533 .02115240 .02017507 .01923285 37 .02246805 .02142270 .02041 4:}7 .01944257 .01850678 38 .02176150 .02071983 .01971613 .01874990 .01782057 39 .02109160 .020053()5 .01905463 .01809399 .01717114 40 .02045560 .01942141 .01842710 .01747209 .01655575 41 .01985102 .01882063 .01783106 .01688170 .01597188 42 .01927563 .01824906 .01726426 .01632057 .01541729 43 .01872737 .01770466 .01672465 .01578666 .01488993 44 .01820441 .01718557 .01621038 .01527810 .01438794 45 .01770505 .01669012 .01571976 .01479321 .01390962 30 n 1% IV4% 1^3%, iy4% 2% 46 .01722775 .01621675 .01525125 .01433043 .01345342 47 .01677111 .01576406 .01480342 .01388836 .01301792 48 .0ie38384 .01533075 .01437500 .01346569 .01260184 49 .01591474. .01491563 .01396478 .01306124 .01220396 50 .01551273 .01451763 .01357168 .01267391 .01182321 51 .01512680 .01413571 .01319469 .01230269 .01145856 52 .01475603 .01376897 .01283287 .01194665 .01110909 53 .01439956 .013416,53 .01248537 .01160492 .01077392 54 .01405658 .01307760 .01215138 .01127672 .01045226 55 .01372637 .01275145 .01183018 .01096129 .01014337 56 .01340824 .01243739 .01152106 .01065795 .00984656 57 .01310156 .01213478 .01122341 .01036606 .00956120 58 .01280573 .01184303 .01093661 .01008503 .00928667 59 .01252020 .01156158 .01066012 .00981430 .00902243 60 .01224445 .01128993 .01039343 .00955336 .00876797 61 .01197800 .01102758 .01013604 .00930172 .00852278 62 .01172041 .01077410 .00988751 .00905892 .00828643 63 .01147125 .01052904 .00964741 .00882455 .00805848 64 .01123013 .01029203 .00941534 .00859821 .00783855 65 .01099667 .01006268 .00919094 .00837952 .00762624 66 .01077052 .00984065 .00897386 .00816813 .00742122 67 01055136 .00962560 .00876376 .00796372 .00722316 68 .01033889 .00941724 .00856033 .00776597 .00703173 69 .01013280 .00921527 .00836329 .00757459 .00684665 70 .00993282 .00901941 .00817235 .00738930 .00666765 71 .00973870 .00882941 .00798727 .00720985 .00649446 72 .00955019 .00864501 .00780779 .00703600 .00632683 73 .00936706 .00846600 .00763368 .00686750 .00616454 74 .00918910 .00829215 .00746473 .00670413 .00600736 75 .00901609 .00812325 .00730072 .00654570 .00585508 76 .00884784 .00795910 .00714146 .00639200 .00570751 77 .00868416 .00779953 .00698676 .00624285 .00556447 78 .00852488 .00764436 .00683645 .00609806 .00542576 79 .00836983 .00749341 .00669036 .00595748 .00529123 80 .00821885 .00734652 .00654832 .00582093 .00516071 81 .00807179 .00720356 .00641019 .00568828 .00503405 82 .00792851 .007064;37 .00627583 .00555936 .00491110 83 .00778887 .00692881 .00614509 .00543406 .00479173 84 .00765273 .00679675 .00601784 .00531223 .00467581 85 .00751988 .00666808 .00589396 .00519375 .00456321 86 .00739050 .00654267 .00577333 .00507850 .00445381 87 .00726418 .00642041 .00565584 .00496636 .00434750 88 .00714089 .00630119 .00554138 .00485724 .00424416 89 .00702056 .00618491 .00542984 .00475102 .00414370 90 .00690306 .00607146 .00532113 .00464760 .00404602 91 .00678832 .00596076 .00521516 .00454690 .00395101 92 .00667624 .00585272 .00511182 .00444882 .00385859 93 .00656673 .00574724 .00501104 .00435327 .00376868 94 .00645971 .00564425 .00491273 .00426017 .00368118 95 .00635511 .00554366 .00481681 .00416944 .00359602 96 .00625284 .00544541 .00472321 .00408101 .00351313 97 .00615284 .00534941 .00463186 .00399480 .00341242 98 .00605503 .00525560 .00454268 .00391074 .00335383 99 .00595936 .00516391 .00455560 .00382876 .00327729 100 .00586574 .00507428 .00437057 .00374880 .00320274 31 n 21/4% 2y3% 2%% 3% 3yo% 1 1.00 1.00 1.00 1.00 1.00 2 .494.«758 .49382716 .49321825 .49261048 .49140049 3 .32591.1-58 .32513717 .3243324:3 .32353036 .32193418 4 .24171893 .24081788 .23992059 .23902705 .23725114 5 .19120021 .19024686 .18929832 .18835457 .18648137 6 .157.53496 .15654997 .15557083 .15459750 .15266821 7 .13350026 .13249543 .13149747 .13050635 .12854449 8 .11548462 .11446735 .11344795 .1124.5639 .11047665 9 .10148170 .10045689 .09944095 .09843386 .09644601 10 .09028768 .08925876 .08823972 .08723051 .08524137 11 .08113649 .08010596 .07908629 .07807745 .07609197 12 .07351740 .07248713 .07146871 .07046209 .06848395 13 .06707686 .06604827 .06503252 .06402954 .06206157 14 .06156230 .06053653 .05952457 .05852634 .05657073 15 .05678852 .05576646 .05475917 .05376658 .05182507 16 .05261663 .05159899 .0.5059710 .04961085 .04768483 17 .04894039 .04792777 .04693186 .04595253 .04404313 18 .04567720 .04467008 .04368063 .04270870 .04081684 19 .04276182 .04176062 .04077802 .03981388 .03794033 20 .04014207 .03914713 .03817178 .03721571 .03536108 21 .03777572 .036787.33 .03581941 .03487178 .03303659 22 .03562821 .03464661 .03368640 .03274739 .03093207 23 .03367097 .03269638 .03174410 .03081390 .02901880 24 .03188023 .03091282 .02996863 .02904742 .02727283 25 .03023599 .02927592 .02833997 .02742787 .02567404 26 .02872134 .02776875 .02684116 .02592829 .02420540 27 .02732188 .02637687 .02545776 .02456421 .02285241 28 .02602525 .02508793 .02417738 .02329323 .02160265 29 .02482081 .02389127 .02298935 .02211467 .02044538 30 .02369934 .02277764 .02188442 .02101926 .01937133 31 .02265280 .02173900 .02085453 .01999893 .01837240 32 .02167415 .02076831 .01989263 .01904662 .01744150 33 .02075722 .019859.38 .01899253 .01815612 .01657242 34 .01989655 .01900675 .01814875 .01732196 .01575966 35 .01908731 .01820558 .01735645 .01653929 .01499835 36 .018.32522 .01745158 .016611.32 .01580379 .01428416 37 .01760643 .01674090 .01590953 .01511162 .01361325 38 .01692753 .01607012 .01524764 .01445934 .01298214 39 .0162H51.;3 .01543615 .01462256 .01384385 .01238775 40 .01567738 .01483623 .01403151 .01326238 .01182728 41 .01510087 .01426786 .01347200 .01271241 .01129822 42 .01455364 .01372876 .01294175 .01219167 .01079828 43 .01403.364 .01321688 .01243871 .01169811 .01032.539 44 .01353901 .01273037 .01196100 .01122985 .00987768 45 .01306805 .01226752 .011,50693 .01078518 .009453443 46 .01261921 .01182676 .01107493 .01036254 .00905108 47 .01219107 .01140669 .01066358 .00996051 .00866916 48 .01178233 .01100599 .01027158 .00957777 .00830646 49 .01139179 .01062348 .00989773 .00921314 .00796167 50 .01101836 .01025806 .00954092 .00886660 .00763371 32 n 2y4% 2^% 2%% 3% ^Vz% 51 .010()6102 .00990870 .00920014 .00853382 .00732156 52 .01031884. .00957446 .00887M4 .00821718 .00702429 53 .00!)99091. .00925449 .00856297 .00791471 .00674100 54 .009G7()51. .00894799 .00826491 .00762558 .00647090 55 .00937489 .00865419 .00797953 .00734907 .00621323 56 .00908530 .00837243 .00770612 .00708447 .00596730 57 .0088071:i .00810204 .00744404 .00683114 .00573245 58 .00853977 .007842U .00719270 .00658848 .00550810 59 .00828268 .00759307 .00695153 .00635593 .00529366 60 .00803533 .00735340 .00672002 .00613296 .00508862 61 .00779724 .00712294 .00649767 .00591908 .00489249 62 .0075()795 .00690126 .00628402 .00571385 .00470480 63 .00734704 .00668790 .00607866 .00551682 .00452513 64 .00713411 .00648249 .00588118 .00532760 .00435308 65 .00692878 .00628463 .00569120 .00514581 .00418826 66 .00673070 .00609398 .00550837 .00497110 .00403031 67 .00653955 .00591021 .00533236 .00480313 .00387892 68 .00635500 .00573300 .00516285 .00464159 .00373375 69 .00617677 .00556206 .00499955 .00448618 .003594.53 70 .00600458 .00539712 .00484218 .00433663 .00346095 71 .00583816 .00523790 .00469048 .00419266 .00333277 72 .00567728 .00508417 .00454420 .00405404 .00320973 73 .00552169 .00493568 .00440311 .00392053 .00309160 74 .00537118 .00479222 .00426698 .00379191 .00297816 75 .00522554 .00465358 .00413560 .00366796 .00286919 76 .00508457 .00451956 .00400878 .00354849 .00276450 77 .00494808 .004438997 .00388633 .00343331 .00266390 78 .00481589 .00426468 .00376806 .00332224 .00256721 79 .00468784 .00414338 .00365382 .00321510 .00247426 80 .00456376 .00402605 .00354342 .00811175 .00238489 81 .00444850 .00391248 .00.343674 .00301201 .00229894 82 .00432692 .00380254 .00333361 .00291576 .00221628 83 .00421387 .00369608 .00323389 .00282284 .00213676 84 .00410423 .00359298 .00313747 .00273313 .00206025 85 .00399787 .00349810 .00304420 .00264650 .00198662 86 .00389467 .00339633 .00295397 .00256284 .00191576 87 .00379452 .00330255 .00286667 .00248202 .00184756 88 .00369730 .00321165 .00278219 .00240393 .00178190 89 .00360291 .00312353 .00270041 .00232848 .00171868 90 .00351126 .00.303809 .00262125 .00225556 .00165781 91 .00342224 .00295523 .00254460 .00218508 .00159919 92 .003.33577 .00287486 .00247038 .00211694 .00154273 93 .00325176 .00279690 .00239850 .00205107 .00148834 94 .00317012 .00272126 .00232887 .00198737 .00143591 95 .00309078 .00264786 .00226141 .00192577 .00138546 96 .00.301366 .00257662 .00219605 .00186619 .00133682 97 .00293868 .00250747 .00213272 .00180856 .00128995 98 .00286578 .00244034 .00207134 .00175281 .00124478 99 .00279489 .00237517 .00201185 .00169886 .00120124 100 .00272594 .00231188 .00195418 .00164667 .00115927 33 n 4% ^y^'/c 5% 5%% 6% 1 1.00 1.00 1.00 1.00 1.00 2 .49019608 .48899756 .487804^8 .48661800 .48543689 3 .32034854 .31877336 .31720856 .31565407 .31410981 4 .23549005 .23374365 .2:3201183 .23029449 .22859149 5 .18462711 .18279164 .18097480 .17917644 .17739010 6 .15076190 .14887839 .14701747 .14517895 .1433026:? 7 .12660961 .12470147 .12281982 .120964'}.2 .11913502 8 .10852783 .10660905 .10472181 .10286401 .1010:3594 9 .0941.9299 .09257447 .09069008 .08883946 .08702224 10 .08329094 ,08137882 .07950458 .07760777 .07586796 11 .07414904 .07224818 .07038889 .06857065 .06689294 12 .06655217 .00406619 .06282541 .06102923 .05927703 13 .06014:373 .05S27535 .0504.5577 .05408426 .0.5296011 14 .054;)6897 .05282032 .05102397 .04927912 .04758491 15 .04994110 .04811381 .04634229 .041.62560 .04296176 16 .04582000 .04401537 .04226991 .04058254 .0389.5214 17 .04219852 .04041758 .03809914 .03704197 .03544480 18 .03899333 .03723690 .03554622 .0.3391992 .03235654 19 .03613862 .03440734 .03274501 .0311.5006 .02962086 20 .03358173 .03187614 .03024259 .02867933 .02718456 21 .03128011 .02960057 .02799611 .02646478 .02500455 22 .02919881 .02754546 .02597051 .02447123 .02:304557 23 .02730906 .02568249 .0241:3682 .02266965 .02127848 24 .02558683 .02398703 .02247090 .0210.3580 .01967900 25 .02401196 .02243903 .02095246 .01954935 .01822672 26 .02256738 .02102137 .01956432 .01819.307 .01690435 27 .02123854 .01971946 .01829186 .01695228 .01509717 28 .02001298 .01852081 .01712253 .01581440 .014592.55 29 .01887993 .01741-461 .01004551 .01470857 .01357961 30 .01783010 .01639154 .01505144 .01380539 .01264891 31 .01685535 .01544345 .01413212 .01291005 .01179222 32 .01594859 .01456:320 .01328042 .01209519 .011002.34 33 .01510357 .01374453 .01249004 .011.3:3469 .01027293 34 .014:31477 .01298191 .01175545 .01002958 .00959843 35 .01357732 .01227045 .01107171 .00997493 .00897386 36 .01288688 .01160578 .01043440 .009.36635 .00839483 37 .0122:3957 .01098402 .00983979 .00879993 .007857-1:3 38 .01163192 .010 W 169 .00928423 .00827217 .007:35812 39 .01100083 .00985567 .00870462 .00777991 .00(i89377 40 .01052-349 .00934315 .00827816 .00732034 .00646154 41 .01001738 .00886158 .00782229 .00689090 .00605886 42 .00954020 .00840868 .00739471 .00648927 .00508342 43 .00908989 .00798235 .00099333 .00611337 .00,5:5:3:512 44 .00966454 .00758071 .00001025 .00570128 .00500606 45 .00820246 .00720202 .00026173 .00543127 .00470050 46 .00788205 .00684471 .00592820 .00512175 .004-tl485 47 .00752189 .00050734 .00501421 .00483129 .00414768 48 .00718005 .00018858 .0053184:3 .00455854 .0():5897()6 49 .00685712 .00588722 .00503965 .004.30230 .00306356 50 .00655020 .00560215 .00477674 .00406145 .00344429 34 n *% *y3% 6% 5V2% 6% 51 .00625885 .00533232 .00452867 .00383495 .00323880 52 .00598212 .00507679 .00429450 .00362186 .00304616 53 .00571914 .00483469 .00407334 .00342130 .00286551 54 .0051.6912 .00460519 .00386438 .00323245 .00269602 55 .00523124. .00438754 .00366686 .00305458 .00253696 56 .00500487 .00418105 .00348010 .00288698 .00238765 57 .00478932 .00398506 .00333032 .00272900 .00224744 58 .00458401 .00379897 .00313626 .00258006 .00211574 59 .00438836 .00362221 .00297802 .00243959 .00199200 60 .00420185 .00345426 .00282818 .00230707 .00187572 61 .00402398 .00329461 .00268627 .00218202 .00176642 62 .00385430 .00314284 .00255183 .00206400 .00166366 63 .00369237 .00299848 .00242442 .0019.5258 .00156703 64 .00353779 .00286115 .00230365 .00184737 .00147615 65 .00339019 .00273047 .00218915 .00174800 .00139066 66 .00324921 .00260608 .00208057 .00165413 .00131022 67 .00311451 .00248765 .00197757 .00156544 .00123453 68 .00298578 .00237492 .00187986 .00148163 .00116330 69 .00286272 .00226745 .00178715 .00140242 .00109625 70 .00274506 .00216511 .00169915 .00132754 .00103313 71 .00263254 .00206759 .00161563 .00125675 .00097370 72 .00252489 .00197465 .00153633 .00118982 .00091774 73 .00242190 .00188606 .00146103 .00112652 .00086504 74 .00232334 .00180159 .00138951 .00106665 .00081542 75 .00222900 .00172104 .00132161 .00101002 .00076867 76 .00213869 .00164422 .00125709 .00095645 .00072463 77 .00205221 .00157094 .00119580 .00090577 .00086315 78 .00196939 .00150101 .00113756 .00085781 .00064407 79 .00189006 .00143434 .00108222 .00081243 .00060724 80 .00181408 .00137069 .00102962 .00076948 .00057254 81 .00174127 .00130995 .00097963 .00072884 .00053984 82 .00167150 .00125196 .00093211 .00069036 .00050902 83 .00160463 .00119663 .00088694 .00065395 .00047998 84 .00154054 .00114379 .00084399 .00061947 .00045261 85 .00147909 .00109334 .00080316 .00058683 .00042681 86 .00142018 .00104516 .00076433 .00055593 .00040241 87 .00136370 .00099915 .00072740 .00052667 .00073956 88 .00130953 .00095522 .00069228 .00049896 .00035795 89 .00125758 .00091235 .00065888 .00047273 .00033757 90 .00120775 .00087316 .00062711 .00044788 .00031836 91 .00115995 .00083486 .00059689 .00042435 .00030025 92 .00111410 .00079827 .00056815 .00040207 .00028318 93 .00107012 .00076331 .00054080 .00038096 .00026708 94 .00102789 .00072991 .00051478 .00036097 .00025189 95 .00098738 .00069799 .00049003 .00034204 .00023758 96 .00094850 .00066749 .00046648 .00032410 .00024408 97 .00091119 .00063834 .00044407 .0003071 1 .00021135 98 .00087538 .00061048 .00042274 .00029101 .00019935 99 .00084100 .00058385 .00040245 .00027577 .00018803 100 .00080800 .00055839 .00038314 .00026132 .00017736 35 TABLE VI Effective Kate Factors. The following are the effective rate factors for a single mit. The formula is jp= ^p(Vl+i- -1) p 2 1% .00998756 .01246118 .01494417 1%% .01742410 2% .01990099 4 .00996272 .01244183 .01491636 .01739631 .01985173 12 .00995446 .01242895 .01489785 .01736119 .01981898 P 2 2>/4% .02237484 2Vb7c .02484567 2%% .02731349 .02977831 31/2% .03469859 4 .02231261 .02476899 .02722087 .02966829 .03454978 12 .02227125 .02471804 .02715936 .02959524 .03445078 P 2 4^^ .03960781 ^V2 7c .04450483 5% .04939015 .05426386 6% .05921603 4 .03941363 .04425996 .04908894 .05390070 .05869538 12 .03928488 .04409771 .04888949 .05366039 .05841061 The effective rate factors for use with annuity tables are seldom of use to the accountant. The factors are in- dicated by the fraction ^-, as explained on pages 63 to 65, 3 p where these factors were discussed. In cases where there is an effective rate and the yearly rate is in the tables, these rules w^ill apply : In case of &„ or a^,: multiply the table value by the yearly rate and divide by th correct value of jp. 1 1 In case of ^- or — multiply by the correct value of j p 6 „ a , and divide by the yearly rate. 36 TABLE VII TEN-PLACE LOGARITHMS OF THE INTEREST RATIOS Rate % 1+i log. Rate % 1 + i log. 1 1/40 1.01025 .0044288591 1/20 1.0105 .0045363179 3/40 1.01075 .0046437500 1/10 1.011 .0047511556 1/8 1.00125 .0005425291 1/8 1.01125 .0048585346 3/20 1.0015 .0006509536 3/20 1.0ia5 .0049658871 7/40 1.00175 .0007593511 7/40 1.01175 .0050732131 1/5 1.002 .0008677215 1/5 1.012 .0051805125 9/40 1.00225 .0009760649 9/40 1.01225 .0052877854 1/4 1.0025 .0010843813 1/4 1.0125 .0053950319 11/40 1.00275 .0011926707 11/40 1.01275 .0055022519 3/10 1.003 .0013009330 3/10 1.013 .0056094454 13/40 1.00325 .0014091684 13/40 1.01325 .0057166124 7/20 1.0035 .0015173768 7/20 1.0135 .0058237530 3/8 1.00375 .0016255583 3/8 1.01375 .0059308672 2/5 1.004 .0017337128 2/5 1.014 .0060379550 17/40 1.00425 .0018418404 17/40 1.01425 .0061450164 9/20 1.0045 .0019499411 9/20 1.0145 .0062520514 19/40 1.00475 .0020580149 19/40 1.01475 .0063590600 1/2 1.005 .0021660618 1/2 1.015 .0064660422 21/40 1.00525 .0022740818 21/40 1.01525 .0065729982 11/20 1.0055 .0023820749 11/20 1.0055 .0066799277 23/40 1.00575 .0024900412 23/40 1.01575 .0067868310 3/5 1.006 .0025979807 3/5 1.016 .0068937079 6/8 1.00625 .0027058934 5/8 1.01625 .0070005586 13/20 1.0065 .0028137792 13/20 1.0165 .0071073830 27/40 1.00675 .0029216383 27/40 1.01675 .0072141811 7/10 1.007 .0030294706 7/10 1.017 .0073209529 29/40 1.00725 .0031372761 29/40 1.01725 .0074276985 3/4 1.0075 .0032450548 3/4 1.0175 .0075344179 31/40 1.00775 .0033528068 31/40 1.01775 .0076411111 4/5 1.008 .0034605321 4/5 1.018 .0077477780 33/40 1.00825 .0035682307 33/40 1.01825 .0078544188 17/20 1.0085 .0036759025 17/20 1.0185 .0079610333 7/8 1.00875 .0037835477 7/8 1.01875 .0080676217 9/10 1.009 .0038911662 9/10 1.019 .0081741840 37/40 1.00925 .0039987581 37/40 1.01925 .0082807201 19/20 1.0095 .0041063233 19/20 1.0195 .0083872301 39/40 1.00975 .0042138618 39/40 1.01975 .0084937140 1 1.01 .0043213738 2 1.02 .0086001718 37 TABLE VII— (Continued) TEN-PLACE LOGARITHMS OF THE INTEREST RATIOS— Continued Rate % 1 +i log. Rate % 1 +i log. 2 1.40 1.02025 .0087066034 3 1/20 1.0305 .0130479961 1/20 1.0205 .0088130091 1/10 1.031 .0132586653 3/40 1.02075 .0089193886 3/20 1.0315 .0134692323 1/10 1.021 .0090257421 1/5 1.032 0136796973 1/8 1,02125 .0091320695 1/4 1.0325 0138900603 3/20 1.0215 .0092383710 3/10 1.033 .0141003215 7/40 1.02175 , 0093446464 7/20 1.0335 .0143104810 1/5 1.022 .0094508958 2/5 1.034 .0145205388 9/40 1.02225 .0095571192 9/20 1.0345 .0147304950 1/4 1.0225 .0096633167 1/2 1.035 .0149403498 11/40 1.02275 .0097694882 3/10 1.023 .0098756337 11/20 1.0355 0151501032 13/40 1.02325 .0099817533 3/5 1.036 .0153597554 7/20 1.0235 .0100878470 3/4 1.0375 .0159881054 3/8 1.02375 .0101939148 4/5 1.038 .0161973535 2/5 1.024 .0102999566 9/10 1.039 0166155476 17/40 1.02425 .0104059726 9/20 1.0245 .0106119627 4 1.04 0170333393 19/40 1.02475 .0106179270 1/10 1.041 .0174507295 1/2 1.025 ' .0107238654 1/4 1.0425 .0180760636 3/10 1.043 .0182843084 21/40 1.02525 .0108297780 2/5 1.044 .0187004987 11/20 1.0255 .0109356647 1/2 1.045 .0191162904 23/40 1.02575 .0110415256 3/5 1.026 .0111473608 3/5 1.046 .0195316845 5/8 1.02625 .0112531701 3/4 1.0475 .0201540316 13/20 1.0265 .0113589537 4/5 1.048 .0203612826 27/40 1.02675 .0114647115 9/10 1.049 .0207754882 7/10 1.027 .0115704436 29/40 1.02725 .0116761499 5 1.05 0211892991 3/4 1.0275 .0117818305 1/2 1.055 .0232524596 31/40 1.02775 .0118874854 4/5 1 . 028 .0119931147 6 1.06 .0253058653 33/40 1.02S25 .0120987182 1/2 1.065 .0273496078 17/20 1.0285 .0122042960 7/8 1.02S75 .0123098482 7 1.07 .0293837777 9/10 1.029 .0124153748 1/2 1.075 .0314084643 37/40 1.02925 .0125208757 19/20 1.0295 .0126263510 8 1.08 .0334237555 39/40 1.02975 .0127318006 3 1.03 .0128372247 9 1.09 .0374264979 10 1.10 .0413926852 38 UNIVERSITY OF CALIFORNIA LIBRARY Los Angeles This book is DUE on the last date stamped below. SEP 1 4 1962 SEP i ? Form L9-50m-9.'60(B3610s4)444 AA 001 020 219