Quantitative Techniques

QUANTITATIVE TECHNIQUES

Contents Section I

UNIT 1 ROLE OF MATHEMATICS AND STATISTICS IN BUSINESS DECISIONS Basics of Decision Making Decision Contaminants Managerial Decision Making System Managerial Decision Making Environment Managerial Decision Models and Algorithms Quantitative Models The Decision Summary Keywords Review Questions Further Readings

UNIT 2 T H E O R Y OF SETS

Introduction The Concept of a Set Notations Representation of a Set Some Basic Definitions Theorem on Subsets Venn Diagram Set Operations Laws of Union of Sets Laws of Intersection of Sets Law of Complement of a Set Theorem (On Symmetric Difference) De-Morgan's Laws Applications of Venn Diagrams Summary Keywords Review Questions Further Readings

UNIT 3 L O G A R I T H M S & PROGRESSIONS

Introduction Logarithms Laws of Operations Compound Interest Arithmetic Progressions Geometric Progressions Annuities, Loans and Mortgages

Methods of Investment Evaluation Perpetual Annuities and Infinite Series Depreciation Summary Keywords Review Questions Further Readings

UNIT 4 EQUATIONS Introduction Equations Applications of Linear Equations in Business Supply and Demand Functions Irregular, Unequal and Discontinuous Functions Quadratic Equations Fitting a Quadratic Cost Curve Summary Keywords Review Questions Further Readings

UNIT 5 MATRIX ALGEBRA

Introduction Vectors Multiplication of Vectors Matrices Use of Matrices for Production Planning Solving Linear Equations Determinants Cramer's Rule Applications in Management Summary Keywords Review Questions Further Readings

UNIT 6 MATHEMATICAL INDUCTION

Introduction Induction and Deduction Principle of Mathematical Induction Summary Keywords Review Questions Further Readings

57

73

112

Section II

UNIT 7 DATA ANALYSIS

Introduction Data Collection and Presentation Frequency Distribution Measure of Central Tendency Mathematical Averages Positional Averages Commercial Averages Measure of Dispersion Skewness Kurtosis Summary Keywords Review Questions Further Readings

UNIT 8 CORRELATION A N D REGRESSION

Introduction Correlation Analysis Scatter Diagram Covariance Karl Pearson Coefficient of Linear Correlation Spearman's Rank Correlation Regression Analysis Fitting Regression Lines Summary Keywords Review Questions

Further Readings

Section III

UNIT 9 TIME SERIES ANALYSIS AND INDEX NUMBERS Introduction Lime Series Analysis Graphical Method Method of Averages Nonlinear Analysis Measuring Periodic Variations Index Numbers Construction of Index Numbers Price Index Numbers Nature of Weights Laspeyres Index Paasche Index Fisher Index

Dorbish and Bowley Index Marshall and Edgeworth Index Walsh Index Summary Keywords Review Questions Further Readings

UNIT 10 PROBABILITY THEORY Introduction Probability Concepts Permutations Combinations Objective and Subjective Probabilities Revised Probabilities Random Variables and Probability Distribution Discrete Random Variables Continuous Random Variables Binomial Distribution Poisson Distribution Normal Distribution Summary Keywords Review Questions Further Readings

UNIT 11 THEORY OF ESTIMATION AND T E S T OF HYPOTHESIS

Introduction Theory of Estimation Point Fstimation (Properties of Good Estimators) Methods of Point Estimation Interval Estimation Sampling Distributions Sampling Theory The Quantitative Models of Inferential Decisions Statistical Approaches to Inferential Decision-making > The General Inferential Decision Algorithm Specific Decision Areas Chi-square Distribution The Z and t Distributions The One-Sample Mean Problem The F-Distribution Concluding Comments Summary Keywords Review Questions Further Readings

S E C T I O N - I

Unit 1 Role of Mathematics and Statistics in Business Decisions

Unit 2 Theory of Sets

Unit 3 Logari thms & Progressions

Unit 4 * Equations ^

Unit 5 Matrix Algebra

Unit 6 Mathematical Induction

Quantitative Techniques

R o l e o f M a t h e m a t i c s and S t a t i s t i c s in B u s i n e s s D e c i s i o n s

Notes

Unit Structure Basics of Decision Making Decision Contaminants Managerial Decision Making System Managerial Decision Making Environment Managerial Decision Models and Algorithms Quantitative Models The Decision Summary Keywords Review Questions Further Readings

Learning Objectives After reading this unit you should be able So: Define decision contaminants Describe managerial decision making system Explain managerial decision making environment Apply and interpret managerial decision models and algorithms Apply and interpret quantitative models

B a s i c s o f D e c i s i o n M a k i n g

Decision-making comes into play, sometimes voluntarily and other times involuntarily, when one needs to take an action or a stand in a situation and when one does not know what that would be. It implies therefore, that decision-making is primarily a reasoning process. Reasoning is subjective by nature, which can he rational or irrational. Moreover it is almost always based on assumptions - explicit or tacit.

All types of human decision-making are essentially intellectual processes. This process has its roots in both the conscious as well as the subconscious mind and always involves three stages described below:

1. Cogtiition stage: It is the starting point for the mind that has searched for facts in the environment in order to make a decision. Cognition means discovery or recognition of data that are assembled into an information system.

2. Assembly stage: The assembly of recognized facts obtained in the first stage into usable information systems represents the second stage. The mind may employ convergent or divergent thinking properties in the assembly process.

3. Convergent means the conventional grouping of data into a system, whereas divergent signifies unusual or new ways of relating the data.

4. Testing stage: At this point the decision maker evaluates the cognites in terms of their relevancy to a given problem. Either a decision is made or not and any number of managerial action programs are the outcomes of this intellectual process.

There might be one or more subconscious intellectual components that affect the decision-making process negatively. While discussing about managerial decision controls, especially when they are of a quantitative nature, we usually do not address ourselves to this problem. The subtle yet powerful ability of these components to redirect the decision-making process should be borne in mind by the decision-maker and decision-analvst alike.

The two most common flaws in decision making are inertia and impatience. This is a paradoxical situation. Inertia is often due to a fear of change. Impatience, if regarded superficially, may appear to be somewhat of an opposite to inertia. But it has the same roots.

D e c i s i o n C o n t a m i n a n t s

In the preceding section we have discussed decision making in terms of three major intellectual stages, that is, cognition, assembly, and test. We have also discussed the effect of certain subconscious contaminants

A checklist of subconscious contaminants is given belovv to help you avoid them. Of course, the list is not exhaustive. N

Note that this list is not exclusive and you may experience different or additional symptoms in different situations.

Dishonesty : trying to obtain someone else's decision; trying to anticipate the outcome without actually going through the three intellectual stages,....

Inertia : skipping the study of facts because it is "not important" promising oneself to come back to it later; being "confused" because certain things mav not be clear,...


Notes

Student Activity

1. Cite one example of decision-making where falsification makes the decision unfit for action.

2. What are the effects of acquiescence and semantics on a decision making situation?

Impatience ; Skipping the process of analysis and reaching a conclusion without any backup reasoning,...

Acquiescence : Doing everything as asked without questioning the logic,...

Gambling : Trying to fit decision variable configurations in the process of decision making by trial and error; assuming that the problem cannot be solved,....

Semantics : Calling the decision making situation totally absurd, very tough, too much time-consuming,...

Falsification : not being able to solve the problem, illegible recording of calculations and final outcomes,...

Analogy, tabloid thinking, over generalization, etc. can also be included in the above list. Guarding against these containments is one of the major tasks of a good decision maker.

M a n a g e r i a l Dec is ion M a k i n g S y s t e m

Reasoning consisting of logic and contaminants is part of any human decision, making process. After becoming aware of the contaminents let us now concentrate on the quantitative aspects of managerial decision making and the peculiar environment in which the managerial decision maker operates. The decision making task may be conceptualised as an input-output system as shown in Figure 1.1.

Figure 1.1: Decision Making System

Every decision making task results in an output which is the evidence of the decision taken. In industry it is ultimately some kind of product, that is, a good service or on idea. The reasoning takes place in the Decision Making rectangle which is sometimes referred to as, quite appropriately, the black box. Here a transformation of the inputs takes place that results in the output. The transformation process has both physical and mental properties. On the input side a large number of variables may be listed. These variables can be classified in terms of the traditional factors of production, i.e., land, labor and

Selj-Instructional Material ^^^^ l^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

capital as well as the more recently emerged complex variables related to systems, technology and entrepreneurship. Underlying this input-output system is a feed-back loop identified as managerial control systems. Its function is to optimize the transformation of inputs into the desired output, Seen in a nutshell, in industry optimization means the minimization of costs and the maximization of profits subject td legal, social and ideological constraints. The computer has forced the decision maker to very carefully delineate and quantify the variables that makeup the building blocks of the decision task. What is needed and how much is needed for decision optimization have become the important questions. In addition, the proper time sequencing of the decision variables within the decision process had to be understood. And all answers had to be unequivocally quantified. It soon became apparent to every decision maker that quantified variables had different properties and specific quantitative control mechanisms had to be designed. Not only was the decision maker confronted with variable-inherent properties, the decision tasks themselves have such peculiar quantitative properties.

A variable, the building block of the decision task, may be seen as a small piece of a complex behavior. Buying a house, manufacturing a product, spending money on a show are examples of variables. Each variable represents a distinct dimension of the decision making task. So the decision space is always multidimensional, and it is a major task for the decision maker to find out which variables make up that space. If an important variable is overlooked, obviously the decision will be less than optimal, furthermore, the quantitative impact of the variable must be ascertained. And here the special variable-inherent properties come into play. The following illustrations may show the differences among the three types of variables.

Deterministic variables can be measured with certainty. Thus, equal measures have equal cumulative impact, or, to use a simple illustration, a+a = 2a.

Stochastic variables are characterized by uncertainty. Thus, a+a=2a+X, where X is a value that comes about because of the uncertainty that is associated with the variable.

Heuristic variables are those that exist in highly complex, unstructured, perhaps unknown decision making situations. The impact of each variable may be explained contingent upon the existence of a certain environment. For example, a+a=3a but only if certain conditions hold. Actual industrial decision making situations in each case may involve the number of gallons of aviation fuel obtained by cracking a barrel of crude oil (deterministic), projected product sales given amount spent on advertising the product (stochastic) and the construction of a platform in outer space (heuristic).

Role of Mathematics and Statistics in

Business Decisions

Student Activity

1. Differentiate clearly between deterministic, stochastic and heuristic decision variables using appropriate examples.

2. Give two examples of decision-making situations where uncertainty is inherent.

t e e l s i o n M a k i n g E n v i r o n m e n t

The reason for the existence of a managerial hierarchy, that is, lower, middle and top management, finds itself in different parameters in which an organization operates. There are industry-wide and market-wide decisions that have to be made. Often these decisions must transcend domestic

Punjab Technical University 5

Notes

considerations to incorporate international aspects. Such decisionsusually made by top managementoccur in a broad-based, complex, ill-defined and non-repetitive problem situation. Middle management usually addresses itself to company-wide problems. It sees to it that the objectives and policies of the organization are properly implemented and that operations are conducted in such a way that optimization may occur.

Lower management is responsible for the conduct of operationsthe firing line so to speakbe this in production, marketing, finance or any of the staff functions like personnel or research. This decision environment is usually well-defined and repetitive. Obviously, with reference to a given decision making situation, the distinction between top, middle and lower management may become blurred. In other words, in any on-going business there is always a certain overlapping of the managerial decision making parameters.

The study and analysis of the existence and interaction of these parameters is of great importance to the management systems designer or communication expert. From the quantitative managerial decision making point of view, their importance lies in recognizing their peculiar constraints and then to build the appropriate decision models and to select the best suited quantitative decision tools. A brief discussion of each environment in this light may enhance the understanding of the tools that are discussed later on.

Figure L2: Top Management Decision Environment

The top management decision environment is shown in Figure 1.2. The company's approach to the domestic or international market is filtered through industry-wide considerations. What does the market want, what does the competition already supply? Where is our field of attack? Do we have the know-how, do we have the resources? What is the impact of our actions upon the market, our own industry and other industries? These are some of the questions that have to be asked, defined and answered. The problems are unstructured and complex. Thus, often a heuristic decision making process can be utilized to good advantage. Forecasting is of major importance and hence stochastic decision making is widely employed in this uncertain decision environment. But even a deterministic toolusually intended for decision making situations that assume certaintyinput-output analysis, can be effectively used in this environment.


Notes

6 Seq-lmmn usual Materia!

Middle management decisions are primarily company-wide in nature. As mentioned before and shown in Figure 1.3, these decisions steer the organization through its life cycle.


Business Decisions

-J oejfCTivE r r DISSOLUTION

I

Figure 1.3: Middle Management Decision Environment

Major features of a firm's life are objectives, planning, operation and the ultimate dissolution. The objectives are general and specific in nature. Obviously top management establishes the objectives, but middle management functions as their guardian. Indeed, as Figure 1.3 shows, every decision at this level must provide feed-back control for each of the other components.

Planning refers to both policy execution as well as policy development. Scale of production, pricing of the product, product mix, in short the orderly and efficient arrangement of the input factors shown in Figure 1.2 is to be decided at this point. Making these factors into a product is the job of operations. It may appear somewhat odd that the decision environment includes attention being paid to the dissolution of the firm. The life cycle concept has been mentioned, and it will be encountered again as one of the major underlying conceptual aids in forecasting. It is well known that business organizations are born, live and die like natural organisms.

Therefore decision making should always be cognizant of the possibility of dissolution. The lower management decision making environment represents a specialized, narrowly defined area within a company's total decision or operational field. Supervisory personnel of all types are operating in this environment. The decision tasks are normally well defined and repetitive.

Student Activity

1. Compare and contrast between different levels of decision making environment.

2. Is it possible to develop a single algorithm to suit all the decision making situations? Give reasons.

M a n a g e r i a l Dec i s ion M o d e l s a n d A l g o r i t h m s

It is highly important that every decision maker, has a firm understanding of the philosophy upon which quantitative decision making is based. Under no circumstances is it sufficient to just know how to perform a certain quantitative analysis and to obtain a solution to be able to make a decision

njab Technical University 7

Notes

Quamimtiif 1 edmiques

Notes

To turn to the specific aspects of the quantitative decision making process, it is possible to recognize three distinct phases in every decision situation. Given a carefully defined problem, a conceptual model is generated first. This is followed by the selection of the appropriate quantitative model that may lead to a solution. Lastly, a specific algorithm is selected. Algorithms are the

orderly delineated sequences of mathematical operations that lead to a solution given the quantitative model that is to be used. The algorithms generate the decision which is subsequently implemented by managerial action programs. The entire process is shown in Figure 1.4.

DEFINED * CONCEPTUAL > QilMiTIWM ; ) ALGORITHMS PROBLEM MODE! WJDEt DECISION

P R O G R A

N M S

Figure 1.4: Phases of Decision Making

Problem Definition

Problem definition is a cultural artifact which is especially visible in a society's economic and industrial decision making process.

Obviously if such cultural determinants are operative in the first phase of managerial decision making, their effect can be noticed at various stages in the process irrespective of the quantitative, thus hopefully objective, methods that are used in the design of the models and algorithms as well as the decision itself.

In the private sectors of free enterprise economies, however, a manager's ability to recognize problems and even to anticipate problems that may emerge at some future time is vital to the survival of the firm. Those managers that make effective decision concerning a known problem are good administrators; those that in addition can recognize and anticipate problems are creative. It is known that creativity is partially inborn and partially acquired. Thus, the quantitative decision maker will not only try to master the methodology but also attempt to sharpen his or .her problem identification skillshis or her creativity. -

The Design of Conceptual Models The conceptual model represents the logic that underlies a decision. Based on this logic the quantitative model, and specific algorithms are constructed. The logic ma)' be a priori or empirical in nature. When shooting craps in a casino, a gambler has pre-established a conceptual model concerning the odds of the game. On a priori groundufing only his or her intellectin determining the odds of every roll of the dice, the concept dictates that the win of a seven or eleven on the first roll has likelihoods of V 3 6 and 2 / 3 6 , respectively. (There are 6 possible combinations of spots showing on 2 dice thai yield a seven and 2

combinations that yield an eleven with 36 combinations for all spots from two through twelve.) Given this conceptional model, quantitative models and algorithms can be designed that facilitate the betting decision.

Now suppose that our gambler stumbles across a floating craps game in some dark alley. After observing the action on the pavement for a while, he notices that seven's and eleven's do not occur on the first roll with the likelihood dictated by his conceptual model. Rather there seems to be a preponderance of two's, three's or twelve'swhich he knows are losses. Crooked dice he may very quietly think to himself. For crooked dice an a priori logic which is based on the ideal situation in which every spot on a dice has an equal probability of occurring ( ] / 6 ) and any spot on two dice as well (' / 6 x 1 / 6 = V 3 6 ) according to the multiplication theorem) is unsuitable. Rather, he will now ascertain by observation (by experiment) the empirical probabilities which are determined by the weights that have been cleverly or crudely (it is a dark alley) concealed in or on the dice. Once this empirical conceptual model has been generated, our gambler may continue the betting decision process in terms of the amount of the bets at each roll, etc. He may also redefine the problem and leave.

In the design of the conceptional model, it is important to observe that the decision maker clearly delineates the interrelationships that make up the realityor the systemsin which the problem occurs. But in the model building process it is virtually impossible to include all variables that have a bearing on the decision. The model includes only the major variables (endogenous variables) as seen from the decision maker's vantage point. Once the conceptual model has been designed and its logic expressed in terms of some systems configuration such as the graph or matrix or perhaps network or flow diagram, the quantitative models are simply superimposed by quantifying the logic. Once that has been accomplished a relatively minor task remains in the selection of the algorithms and the computerization of the process. This is shown in detail for every type of quantitative decision that is discussed in the chapters that follow. It may be surmised that in the "model" that is shown in Figure 1.4, the Defined Pre components are the most important ones. Indee process has been needlessly and most of tim commenced because of faulty problem definiti building. Then there is no optimal or even saris:

Q u a n t i t a t i v e Mode l s

and Conceptual Model IS true. Many a decision

uiously to some extent, poor conceptual model

mfceome.

7%

Once the conceptual model has been properly designed, the quantitative model and its algorithms should almost "flow" out of it. The transition is natural, smooth, almost automatic. The quantitative model is selected from the many such models that have been designed by mathematicians. So while the decision maker will always build a conceptual model, the quantitative model is typically selected from an available pool of such decision making tools. The selection is made on the basis of !

stochastic, deterministic or heuristic nature of available quantitative models for each kind as d chapters, and the decision makers task is to select given decision situation. "Know thy tools" shou

either predominantly variables. There are

tssed in the following appropriate one for a

be inscribed on every


Business Decisions

S t u d e n t A c t i v i t y

Given that 3 out of 10 electric bulb? fuse within the last month, would vou buy 3 new electric bulbs for the current month in advance? Explain and justify your decision.

Describe a decision-making situation and identify various endogenous and exogenous variables for decision making.

Notes


N o t e s


1. Lis! the various tools available (or quantitative decision making.

2. What are the bases on which you will setecl a particular model for decision making?

decision maker's desk. As it is possible to build a wall with a spade when the trowel would be the more appropriate tool, decision makers may sometimes misuse quantitative tools.

The scope of quantitative managerial decision is vast, indeed. Any industrialized society's economic and industrial decisions are the most complex and important ones made by that society. The reader already has a good understanding of what these decisions entail. In order to provide a very brief overview of decision tasks, some examples may be cited of specific criteria for selecting quantitative models and algorithms.

With respect to the former, the decision maker needs to establish a time frame. There are static and dynamic models. Static models are used when the decision process focuses on a single time period f. When there are several time periods (f, tu+p tn+2,...) over a planning horizon, dynamic models should be selected. Secondly, the decision maker must analyze the degree of certainty vs uncertainty in the decision environment. Remember that uncertainty calls for stochastic models. Certainty calls for deterministic models. Any type of inferential decision, forecasting studv, quality control problem, waiting line or network analysis and simulation involves a high degree of uncertainty. Therefore, stochastic models are used. On the other hand automated production processes, allocation or inventory or transportation problems, and returns on investment or input-output analysis involve lesser degrees of uncertainty or sometimes, although rarely, complete certainty. In these decision situations, deterministic models may be employed. In some situations the decision maker does not know the variables or does not understand their characteristics. Deep sea mining or outer-space flight may come to mind. So, here heuristic models can be used.

The criteria for algorithm selection, once the quantitative model has been decided upon, rest with the methodological efficiency of the algorithm but also, at times, with the decision maker's training and preference for one over another method. In general, it may be said that the algorithm must render itself applicable to computerization; for, the period in time is very quickly disappearing when the managerial decision process is not based on the computer in some form. Furthermore, the algorithm must result in optimization.

As an illustration for these three criteria of algorithm selection, the familiar, widely used linear programming quantitative model may be mentioned. This deterministic decision tool is used in manufacturing operations where the input iactors (see Figure 1.1) are transformed into a product, subject to well-known and well-quantified constraints.

T h e Dec is ion

The decision is the end product of a sequence of mental activities as illustrated in the preceding pages. To m a k e a decision does not necessarily mean that it gets carried out. In order to accomplish that, numerous managerial action programs are necessary They represent the physical extension to the decision

making process. This book stops at the point when the decision is rendered. The action programs, the physical component, cannot be discussed because they must be specifically designed for each situation. A good decision maker, however, will try to place the seeds for proper implementation into the decision.


Business Decisions

S u m m a r y

Decision making occurs in all fields of human endeavor.

Human decision making are intellectual processes involving both conscious and subconscious efforts comprising of three stages -cognition, assembly and testing stage.

Inertia and impatience are the two most common flaws in decision making.

9 Dishonesty, inertia, impatience, acquiescence, gambling, semantics and falsification are some of the most prevalent contaminents in a good decision making.

The conceptual model represents the logic that underlies a decision.

A decision involves choice among several alternatives. In the most basic sense a decision always involves the answer to the question "to do or not to do?" Not to do (inaction) determines that decision.

To make a decision does not necessarily mean that it gets carried out.

K e y w o r d s Cognition stage: It is the starting point for the mind that has searched for facts in the environment in order to make a decision.

Assembly stage: The assembly of recognized facts obtained in the first stage into usable information systems represents the second stage.

Convergent: Convergent means the conventional grouping of data into a system.

Divergent: divergent signifies unusual or new ways of relating the data.

Testing stage: At this point the decision maker evaluates the cognites in harms of their relevancy to a given problem. Impatience: Skipping the process of analysis and reaching a conclusion without any backup reasoning. Acquiescence: Doing everything as asked without questioning the logic.

Gambling: Trying to fit decision variable configurations in the process of decision making by trial and error; assuming that the problem cannot be solved.

Semantics: Calling the decision making situation totally absurd, very tough, too much time-consuming.

Notes

Quantitative Technique*

Notes

Falsification: not being able to solve the problem, illegible recording of calculations and final outcomes.

R e v i e w Q u e s t i o n s 1. How does a manager use mathematics and statistics in decision

making?

2. Explain with the help of an illustration the decision making system.

3. How do environmental factors play a role in decision making?

4. Draw a step by step decision making algorithm taking upon illustration of any business decision.

F u r t h e r R e a d i n g s

E R Tufte, The Visual Display of Quantitative Information, Graphics Press Anderson , D.R., D.J. Sweeney, and T.A. Williams, Quantitative Methods for Business, 5th edition, West Publishing Company

R.S.Bhardwaj, Business Mathematics, Excel Books

Unit 2 Theory of S e t s Theory of Sets

Notes

Unit Structure Introduction

The Concept of a Set

Notations

Representation of a Se t

Some Basic Def ini t ions

Theorem on Subse t s

Venn Diagram

e Set Operations

Laws of Union of Se t s

Laws of Intersection of S e t s

Law of Complemen t of a S e t

Theorem (On S y m m e t r i c Di f fe rence) De-Morgan's L a w s

Applications o f V e n n D i a g r a m s

Summary

* Keywords

Review Quest ions Further Readings

Learning Object ives After reading this uni t y o u s h o u l d be a b l e to:

W Define a set

e Represent a set in different no ta t ions

c Identify different types of se ts

" Use Venn d iagrams to represen t and manipulate sets

* Apply various set o p e r a t i o n s

s Use set theory to so lve prac t ica l p rob lems

In troduct ion

The theory of sets i fundamental notion. It has a great contribution in the study of different branches of mathematics and has, a great importance in modern mathematics. The beginning of set theory was laid down by a German mathematician George Cantor (1845 -1918 ) .

T h e C o n c e p t o f & S e t

A collection of well-defined objects is called a set. The 'objects' are called elements. The elements are definite and distinct. By the term 'u'ell-defined', we mean that it must be possible to tell beyond doubt, whether or not a given object belongs to the collection (set) under consideration. The term 'distinct', means that no element should be repeated.

Following are the

(ij The set of habet = ta, e, i, o, u}

S t u d e n t Ac t iv i t y -

How many elements are there in each of the following sets?

a. Set of the presidents of free India.

b. Set of all the numbers which are both odd and even.

c. Set of all the numbers which are prime and even.

d. Set of all the numbers greater than 1.

e. Set of all the MB As who are under-graduates.

'uniai> Technical University 13

Quantitative Technique (ii) T h e set of all straight lines in a plane. (iii) The set of odd numbers between 3 & 19 is |5, 7, 9 ,11 ,13 ,15 ,19}

Notes N o t a t i o n s

Sets are usually denoted by capital letters

A, B, C, D, ...

and their elements are denoted by corresponding small letters

a, b, c, d, ....

Note that it is not necessary that the elements of a set A are denoted by a.

If a is an element of set A, then this fact is denoted by the symbol a e A and read as "a belongs to A".

If a is not an element of A, then we write a i A and read it as "a does not belong to A."

R e p r e s e n t a t i o n of a S e t

A set can be represented in the following two ways:

(i) Tabular form (or Roster method) and the tabular form is also called the listing method. If all the elements of a set are kept within {} and elements are separated from one another by comma (,), then this form of the set is called tabular form.

E.g.= (1, 2, 3,...} = the set of natural numbers.

(ii) Set builder form (rule method)

Set builder form is also called the rule method. In this form, we specify the defining property of the elements of the sets, e.g., if A is the set of all prime numbers, we use a letter, usually x, to represent the elements and we write

A = |x : x is a prime number) Note: When the number of elements in a set is smail, we use listing method, but when the number of elements in the set is large or infinite, we use the set builder form.

Example 1

Write the set of the letters in the word CALCUTTA.

Solution

Since no element enters the set more than once. Therefore A, C and T will occur only once. Hence the required set is

(C, A, L, U, T)

Answer

Example 2 Theory ofSets Write the following sets into the tabular form:

(i) A = |x I 0 < x < 5, x is the set of integers). Notes

(ii) A = {x I 3 x : - 12x = 0, x is a natural number). Solution

(i) The integers satisfying the inequality 0 < x < 5 are 1, 2, 3, 4 A = (1, 2, 3, 4}

(ii) 3x2-11% = 0 3x(x-4) = 0 =>x = 0&x = 4

Since x is a natural number. Therefore x - 4

Hence A = \4) S o m e B a s i c Def ini t ions

1. Singleton set: A set containing only one element is called a singleton set. E.g. The sets (0), fx), {} all consist of only one element.

2. Empty set: The set having no element is called the empty set or the null set or the void set. It is denoted by or {}. E.g.: A - (x: x2 + 1 = 0, x is a real number). A is a null set. Since there is no real number satisfying the equation x1 + 1 = 0. So the set A is null set.

3. Subset: A set A is called a subset of B if every element of the set A is also an element of the set B. We write it as

A c B ,

and read as 'A is a subset of B or A is contained in B'. In symbols, if A & B are two sets such that

x e A = > x e B .

Then A is a subset of B.

Note: for every set A , A c A, i.e., A is itself a subset of A.

Also note that the empty set is always subset of every set. If A is not a subset of B. We denote A eg B.

E.g.: If A = (a, b, c) and B - |a, b, c, e). Then clearly A c B. 4. Proper subset: If A is a subset of B and if there is at least one element in

B which is not in A, then A is called the proper subset of B.

E.g.: if A = |1, 2, 3) and B = (1, 2, 3, 4). Then A is a proper subset of B, and is denoted by A c B( read as 'A is a proper subset of B'.

We can also say, B is a super set of A, and write it as BD A.

Student Activity Write the following sets into the tabular form:

i. X = |t I 0 > t < 5, t is the set of integers].

ii. X = (t I 3 x 2 - 12x < 18, t is a natural number|.

Quantitatii* Techniques 5.

Notes

6.

7.

8.

9

to

Comparability: Two sets, A & B, are said to be comparable if either of the following conditions is satisfied:

A c B o r B c A .

E.g.:

I. If A = {1, 2, 3}, B = (1, 2,3, 4, 5}. Then A c B s o that A and B are comparable.

II. If A = {a, b, c} and B = (a}. Then B c A, and hence A & B are comparable.

Equal sets: Two sets are said to be equal if they contain the same elements, i.e., if every element of A is an element of B, and every element of B is also an element of A. It is denoted by A = B.

Two sets are equal if and only if A c B and B c A. => A = B.

E.g.: If A = (1, 2}, B = {1, 2} and C = {x: x 2 - 3x + 2 = 0). Then A = B = C.

Equivalent sets: Two sets are called equivalent sets, if and only if there is one to one correspondences between their elements.

IfA = {a,b, cj and B = { 1 , 2 , 3 } . Then correspondence in the elements of A and of B is one to one, A is equivalent to B, we write it as A - B.

Note: If two sets are equal, they are equivalent but two equivalent sets are not necessarily equal.

Finite set: A set is said to be finite set if in counting its different elements, the counting process comes to an end. Thus a set with finite number of elements is a finite set.

E.g.:

I. The set of vowels = {a, e, i, o, u}. II. The set of people living in Delhi.

Infinite set: A set, which is neither a null set nor a finite set is called an infinite set. The counting process can never come to an end in counting the elements of this set.

E.g.: The set of natural numbers^ {1, 2, 3, 4 . . . } . Set of sets: If the elements of a set are sets, then the set is called a set of sets.

E.g.: [[a), |a ,bl, {a, b, e} | Power set: The set of all possible subsets of a given set A, is called the power set A. The power set A is denoted by P (A). If the number of a set is n then No. of elements in power set is 2 n .

E.g.: If A - {a, b. cj. 1 hen its subsets are

i {ah fb}, {c}, {a, b),{a, d {b, c), {a, b, c}

P (A) = {*, {a}, (b), (c), {a, b}, {a, c), {b, c}, {a, b, c}}

12. Universal set: A set, which contains all sets under consideration as subsets, is called the universal set. It is denoted by U.

Note: Different universal sets are used in different contexts. The choice of the universal set is not unique.

E.g.: In the study of different set of letters of English alphabet, the universal set is the set of all letters of English alphabet.

13. Index set and Indexed set: Let Ar be non-empty set for each r in a set A. In this case the sets A,, A2, A y . . . A n, are called indexed sets and the set A = (1,2,3...n} is called index set. Here the suffix r e A of Ar is called index. Such a family of set is denoted by {A ) r e A.

e.g.: Let A, = (1, 2,3) , A2 = {3 ,4 ,5 , 6}, A, = {6, 7, 8), A4 = {1 ,4 ,5 ,12}, A = {a, b, c, d, e}.

Here we find that E a non-empty set Aa, cce A. Hence Ais called index set and the sets A, A2 A3 A4 A, are called indexed sets.

T h e o r e m o n S u b s e t s

Theorem 1

The empty set is the subset of every set.

Proof

Let A be any set and be the empty set. It is clear that has no element of A. Thus t is a subset of A.

Proved

Theorem 2

Every set is the subset of itself.

Proof

Let A be any set. Therefore everv element of A is an element of the set A.

By definition of the subset, the set A is the subset of the set A. Hence

A c A

Theory of Sets

Proved

Theorem 3

If A c B and B c A . then A = B.

Proof

NOTES

STUDENT ACTIVITY

1. If N(A) = m THEN WHAT IS N(P(P(A)))?

2. If A, B AND C are THREE SETS SUCH THAT: A. A C B

B. FIEF AND

C. A n B = THEN WHAT CAN YOU SAY ABOUT N(C)?

* T e c h n i q u e s

Notes

From (1) and (2), x e A o x e B. => A = B.

Proved

Theorem 4

If A c B and B c C . then A c C .

P r o o f

A c B

.. x e A => x e B (1) Again B c C .

.-. x e B => x e C (2)

From (1) and (2), it is clear

x e A =>x e C=> A c C

Proved

Theorem 5

If a set A contains n elements. Then P (A) contains 2" elements. Proof

Let av av ay ... an, be n elements of a set A. Then the number of subsets of A having one element each of the type

{ , } , {

Solution

Subset A = {1,2 , 3,10,14). - v . U ,

f A \

\ 1 4 /

2. Universal set U = {1, 2, 3...25)

A = {1 ,2 , 3 ,10, 14),

13 = {2, 3)

Clearly B c A.

3. Universal set U - { 1 , 2 , 3...25)

No common elements

A = {1 ,2 , 3)

B = {4, 5, 6)

4. Universal set U = 11,2, 3...25)

A = {3, 4, 6, 7}

B = {4, 6, 10, 12)

The elements 4, 6 of the sets A & B are common.

S e t O p e r a t i o n s

1. Union of Sets: The union or join of two sets A and B, written as A u B (read as A cup B), is the set of all elements, which are either in A, or in B, or in both.

/ i A /

i A / 4 \ B \

I 3 1 6 1 0 1 V7 yL y

Theory of Sets

Notes


1. If X is the set of all the letters in ABRACADABRA, what is the total number of proper subsets of X?

2. Enumerate the power set of the set A where,

Punjab Technical Unr;:ers;


" X / u

Notes B \

)

( A O B )

Thus A u B = (x: x e A or x e B). Here 'or' means that x is in atleast one of A or B, and may lie in both. The common elements of A and B are taken only once in A u B.

The Venn-diagram for A u B is given by the shaded area in the above figure. Clearly A and B are each a subset of A u B.

E.g. If A = |1,2, 3,4, 5} and B = {2, 4, 5, 6,11,12) Then A u B = |1,2, 3, 4, 5, 6 ,11,12) . Remark

i When Be A, then A u B = A

ii. A and B are both subsets of A u B, i.e., A c ( A u B ) and B c ( A u B )

iii. A u A = A

iv. If n,be the number of elements in A and n 2be the number of elements in B, then the number of elements of A u B cannot exceed n, -t n2; for the elements common to A and B are to be counted only once in AuB.

Intersection of Sets: The intersection or meet of two sets A and B written as A n B (read as A cap B), is the set of all elements that belong to both A and B.

Thus A n B = l x : x e A and x e B).

The Venn-diagram for A rTf54's^given shaded area. Clearly A n B is a subset o

>ove figure by the id B. Also An A =

E.9.: If A = (1 ,2 ,3 ,4 ,5 )

20 Self Instructional Material

B = (2, 4, 5, 6 ,11 , 12}

T h e n A n B = {2, 4 , 5 } .

Disjoint Sets: Two sets A and B are said to be d i s j o i n t , if they have no element in common, i.e., A n B =

U

(A n B = *f>) E.g.: The set of all even integers and the set of all odd integers are disjoint sets. Remarks

i. When A and B are two disjoint sets, then A n B = f ii. Each of the sets A and B contains A n B as a subset, i.e., (A n B)

c A and ( A n B ) c B .

Difference of two sets: The difference of two sets A and B denoted by A - B (read as A minus B), is the set of all elements of A which are not inB.

Thus A - B = (x : x e A, x B} Similarly B - A = {x : x e B, x ( A}

The shaded area in the above Venn - diagram represents the sets A - B and B - A

E.g.: If A = {a, b, c, d, e} & B = (d, e, p, qj T h e n A - B - |a, b , c j .

Complement of a set: The complement of a set is defined as another set consisting of all elements of the universal set which are not elements of the original set.

Quantitative Technique} 7 h e complement of the set A for the universal set U is generally denoted by A', and thus

A' = {x : x

S t u d e n t A c t i v i t y Theory of Sets 1. Write a set expression for each of the following Venn-diagrams,

(i) I I

(iii) (iv)

Notes

(v)

2. If

A = | 1 , 2 , 3 | A u B = | l , 5 , 2 , 3 , 6 , 7 , 8 ) B n C = 11,6,81

Can you determine the set C?

As a human resource manager of your company you deal with marketing executives and accountants. What inferences can you deduce from the following scenarios?

(i)

(iii)

fi) A M

Q 0 . (iv)

Where A is the set of accountants and M that of the marketing executives.


we Techniques

N o t e s

L a w s o f Union o f S e t s

If A, B, C are any three sets, (|> the empty set and U the universal set. Then their union follows the following laws:

1. Idempotent law: A u A - A

2 . Commutative law: A u B = B u A

3. Associative law: ( A u B ) u C = A u ( B u C )

4. Identity law: (a)Au x e A o r x e A

= > x e A

i.e., A u A c A (1) Let y e A. Then

y e A => y e A or y e A

=> y e A u A

i.e., A c A u A (2) From (1) & (2), we have A u A = A.

2 . To prove A u B = B u A

Let x e A u B = > x e A o r x e B

=> x e B or x e A

= > x e B u A

i.e., A u B c B u A (1) Let y e B u A => y e B or y e A

=> y e A o r y e B S

=> y e A u B

i.e., B u A c A u B (2) From (1) & O, we have A u B = B u A .

3. T o p r j v e ( A u B ) u C = Au ( B u C ) L e t x e ( A u B ) u C =c> x e (A u B) or x e C

=5> (x e A or x B) or x e C => x e A or (x e B or x e C) => x e A or (x e B u C)

24 Self-Instructional Material

i.e., x e A u ( B u C ) (1) Let y A u (B u C)

=> y e A or (y F B u C) y e A or (y e B or y e C)

=> (y e A or y e B) or (y e C) ^ ( y e A u B ) o r ( y e C ) = > y e ( A u B ) u C

i.e., A u ( B u C ) c ( A u B ) u C (2) From (1) & (2), we have ( A u B ) u C = A u ( B u C )

4. To prove A u x e A o r x e < ) ) => x e A

i.e., Au< j>cA (1) But A c A u < t > (2) From (1) &(2), we have A u i j ) = A Again let y e A u U

=> y e A or y U

= j y e U i.e., A u U c U (1) But U c A u U (2) From (1) and (2), we have A u U - U

L a w s o f I n t e r s e c t i o n o f S e t s

If A, B, C are any three sets, then their intersection follows the following laws:

1. Idempotent law: A n A = A

2. Commutative law: A n B = B n A

3. Associative law: ( A n B ) n C = A n ( B n C ) 4. Identity law: A n U = A, A n= 5. Distributive law: A u (B n C) = (A u B) n (A u C)

A n ( B u C ) = ( A n B ) u ( A n C )

antitative Techniques Proof

Notes => x e A and x eA

= > X 6 A

i.e. A N A c A (1) Let y A

=> y e A

y G A and y e A

=> y G A N A

i.e., A c A n A (2) From (1) and (2), we have

A N A = A

2. To prove A N B = B N A

Let X G A N B

=> X G A and X G B

=> X G B and X G A

=> X G B n A

i.e., A n B c B n A (1) Let y G B n A

=> y G B and y e A

=> y G A and y E B

=> y G A n B

i.e. B n A c A n B (2) From (1) and (2), we have

A n B = B n A

3. To prove (A n B) n C = A n ( B n C ) Let X e (A n B) n C

=> X G (A n B) and X G C => (X G A and X E B) and X e C

X e A and (X E B and X G C) X G A and (X e B n C)

=> X e A n (B n C)

1. To prove A N A = A

Let x A N A

i.e., ( A o B ) n C c A n ( B n C ) (1) Let y e A n (B n C)

=> y E A and y G (B n C) => y e A and (y e B and y e C) = > ( y e A and y G B) and y e C =>y G ( A n B ) n C

i.e. A n ( B n C ) c ( A n B ) n C (2) From (1) and (2), we have (A n B) n C = A n (B n C)

To prove A n U = A, An x A [Because A c U]

i.e., A n U c A (1) Again, let y eA

=> y G A and y e U [Because A c U]

i.e. A c A n U (2)

from (1) and (2), we have

A n U = A

Again to prove A n $ = (x e A u B) and (x G A u C) => x e ( A u B ) n ( A u C ) i.e., A u ( B n C ) c ( A u B ) n ( A u C ) (1) Again, let y e ( A u B ) n ( A u C )


Techniques ( y e A u B ) a n d (y e/A u C )

=> (y e A or y eB) and (y e A or y e C)

N , ; e . => y e A or (y eB and y e C) => y e A or (y e/B n C) = > y e A u ^ n C )

i.e., ( A u B ) n ( A a Q c A u ( B n C ) (2) From ( 1 ) and (2), we have A u (B njC) = (A u B) n (A u C).

(ii) To prove A n ( B u C ) = ( A n B ) u ( A n C). Let x e A n (B u C)

x e A and (x e B u C) => x e A and (x e B or x e C) => (x e A and x e B) or (x e A and x e C) => (x e A n B) or (x e A n C)

x e (A n B) u (A n C) i.e., A n ( B u C ) c ( A n B ) u ( A n C ) ( 1 ) Similarly, let y e ( A n B ) u ( A n C ) => y e (A n B) or y e (A n C) => ( y e A and y eB) or (y e A and y e C ) => y e A and (y eB or y e C) => y e A a n d ( y e B u C ) => y e A n ( B u C )

i.e., ( A n B ) u ( A n Q c A n ( B u C ) (2) From (1) and (2), we have A n ( B u C ) = ( A n B ) u ( A n C).

L a w o f C o m p l e m e n t o f a S e t

If is the empty set and U the universal set and A be any one of its subsets. Then the component of A, i.e., A' follows the following laws:

1. A u A ' = U

2. A n A' =

Since every set is a subset of a universal set,

therefore the set A u A' c U

Again, let x e U, then

x e U => x e A or x A'.

=> x e (A u A') i . e . , U c A u A '

From (1) & (2), we have A u A' = U

Proof 2

To prove A n A' = Since the null set is a subset of every set, therefore it follows that

$ c A n A' Again, let x e A n A'

=> x e A and x e A'

^ x e A and x i A

=> x e i.e. A n A' = $ From (1) and (2), we have

A n A' = Proof 3

To prove (A')' = A.

Let x e (A')'

Then x e (A')' ^ x f ? A '

=> x e A

i.e. ( A ' ) ' c A

Again, let x e A

Then x G A => x A'

=> X G (A')'

i.e. A c (A')'

From (1) and (2), we have

(A')' = A.

Theory of Sets

(1)

Notes

(2)

(1)

(2)

Student Activity

1. Prove that ((A') ') ' = A' 2. Prove that AnA' = ty

(1)

(2)


Quantitative Techniques T h e o r e m (On S y m m e t r i c D i f f erence ) If A and B are any two sets. Then

Notes A A B = (A u B) - (A n B).

Proof

Let x A A B.

Thenx e ( A - B ) u ( B - A ) o (x e A and x t B) or (x e B and x ^ A ) (x e A and x e B) or x e B and [x e A and x e B] or x e A o [x G B or (x G A and x B)] and [x A or (x e A and x G B)] [(x e B or x e A) and (x G B or x g B)] and [(x x t A u B

=> x A or x $. B. L

=> x G A' and x G B'

= > x e A ' o B '

i.e. ( A u B ) t A ' n B ' (1) Again, Let y e A ' n B '

y e A ' n B ' => y 6 A' and y e B'.

= > V A o r y g B

= > y s ? ( A u B )

=> y E (A U B ) ' i.e. A ' n B ' c (A u B ) ' ( 2 ) From (i) and (ii), we have (A u B) '= A' n B'.

2. To prove (A n B ) ' = A ' u B ' . Let x e (A n B ) ' => x t A h B

=> x A or x 0 B.

=> x E A' and x e- B'

=> x E A' U B'

i.e. ( A n 8 ) ' c A ' u B ' ( 1 )

Similarly, we can show

A' u B ' c (A n B ) ' ( 2 ) From (i) and (ii), we have

(A n B) '= A' U B'. 3. To prove A - (B u C) = (A - B) n (A - C).

Let x e A - (B u C) => x 6 A and x ( B u C ) => x e A and ( x ? B and x ? C ) => (x e A and x ? B ) and ( x e A and x e C) => x e (A - B) n x e (A - C)

x e ( A - B ) n ( A - C ) . i.e., A - ( B u C ) c ( A - B ) n ( A - C ) ( 1 ) Similarly, we can show

( A - B ) n ( A - Q c A - ( B u C ) (2) From (i) and (ii), we have

A - ( B u C ) = ( A - B ) n ( A - C )

A P P L I C A T I O N S O F V E N N D I A G R A M S

T h e o r y o f S e t s

NOTES

STUDENT ACTIVITY

1. Prove that A - (B n C ) = ( A - B ) o ( A - C )

2. Use Venn diagram to prove the above identity.

Some important results from Venn diagrams are, if A and B are two non empty intersecting sets, then

(1) n ( A u B ) = n(A) + n ( B ) - n ( A n B ) (2) n ( A u B ) = n ( A - B ) + n ( A n B ) + n ( B - A ) (3) n (A - B) + n (A n B) = n (A) (4) n (B - A) + n (A n B) = n (B).

P - i i n i n h T e r h n i r a t University 3 1

Quantitative Techniques Example 4

If A = { 1 , 2 , 3 , 4 } , B = {2 ,3 ,5 , 6} and C = { 4 , 5 , 6 , 7 } . Then verify that

N o t e s i ) A u ( B n C ) = ( A u B ) r > ( A u C ) . ii) A n ( B u C ) = ( A n B ) u ( A n C ) Solution

(i) B n C = {2 ,3 ,5 ,6} n {4, 5, 6, 7} = {5, 6}. A u ( B n C ) = {1 ,2 ,3 , 4} u {5, 6} = | 1 , 2 , 3 , 4 , 5 , 6 ) (1) A u B = {1 ,2 ,3 ,4} u {2, 3, 5, 6} = {1, 2 , 3 , 4 , 5 , 6} A O C = |1 ,2 ,3 ,4} U {4, 5, 6, 7} = {1, 2 , 3 , 4 , 5 , 6 , 7 } .

(A u B) n (A u C) = (1, 2, 3, 4, 5,6} n (1,2, 3,4,5,6, 7} = {1,2,3,4,5,6} (2)

From (1) & (2), we have A u ( B n C ) = ( A u B ) n ( A u C).

(ii) B u C = {2 ,3 ,5 ,6} u {4, 5, 6, 7} = {2, 3 , 4 ,5 ,6 , 7}. A n ( B u C ) = { 1 , 2 , 3 , 4 } n {2, 3, 4, 5, 6,7} = {2 ,3 ,4} (1) Again A n B = {1, 2, 3, 4} n {2, 3, 5, 6} = {2,3} (2)

A n C = { l , 2 , 3 , 4 } n { 4 , 5, 6, 7} = {4} (3) (A n B) u (A n C) = {2,3} U {4}

= {2,3,4} (4) From (1) & (4), we observe

A n (B u C) = (A n B) u (A n C) Example 5

If is the set of complex numbers and A = {x e c : x4 - 1 = 0),

B = {x e c : x3 - 1 = 0). Find A - B and A u B. Solution

We have x4 - 1 = 0

=>(x2 + l ) ( x 2 - l ) = 0

=> x = 1, i

Since, x e C => x = i

A = {- i,i} Again x3 - 1 = 0

^ ( x - l ) ( x 2 + x + 1) = 0


=> x = 1 and x = (-1 Vl-4) / 2

=>x= 1 andx = (-l + ^ 3 i ) / 2

Since, x e C => x = (-1 W3) / 2

.-. B = {(-1 + iV3) / 2, (-1 - W3) / 2

A - B ={-i , i] - {(-1 + iV3) / 2, (-1 - W3) / 2}

= Ki .

A u B = {-i, i} u {(-1 .+ bb) I 2, (-1 - W3) / 2} = K i / ( - l + W 3 ) / 2 , ( - l - W 3 ) / 2 } .

Example 6

Theory of Sets

Answer

Answer

Sets A and B have 3 and 6 elements respectively. What is the least number of elements in A u B?

Solution

The number of elements in the set A u B is least, when A c B . Then all the 3 elements of A are in B. Since B has 6 elements. Therefore, the least number of elements is 6.

It is clear from the fact that A c B

= ) A u B c B .-. n (A u B) = n (B) = 6

Answer

Example 7

In a group of athletic team in a certain institute, 21 are in the basket ball team, 26 in the hockey team, 29 in the football team. If 14 play hockey and basketball, 12 play football and basket ball, 15 play hockey and foot ball, 8 play all the three games.

(i) How many players are there in all?

(ii) How many play only football? Solution

Given:

No. of basket ball players B = 21,

No. of hockey players H = 26,

No. of football players F = 29,

No. of players playing hockey & basket ball both H n B = 14

Notes

Student Activity

1. Shade the area represented by the following pairs of expressions and verify that they are equal.

(i) (AuB)nC and

(AnQUBnQ

(ii) A u A a n d U

2. In a group of athletic team in a certain institute, 21 are in the basket ball team, 26 in the hockey team, 29 in the football team. If 14 play hockey and basketball, 12 play football and basketball. 15 play hockey and football. 8 play all the three games.



Notes

No. of players playing football & basket ball both F n B = 12

No. of players playing hockey & football both F n H = 15.

Therefore

(i) The no. of total players = n ( B U H U F) = n (B) + n(H) + n(F) - n(B n H) - n(H n F) - n(B n F) + n(F n B n H) = 21 + 26 + 2 9 - 1 4 - 1 2 - 1 5 + 8

= 43

No of players playing football, basket ball but not hockey = 1 2 - 8 = 4.

No of players playing foot ball, hockey but not basket ball = 15-8=7

No of players playing football only

= No. playing foot ball - (no. Playing football and hockey + No. playing foot ball, basket ball & hockey)

= 2 9 - ( 7 + 4 + 8) = 10.

Answer

S u m m a r y

The beginning of set theory was laid down by a German mathematician George Cantor (1845-1918). A collection of well-defined objects is called a set. The 'objects' are called elements. The elements are definite and distinct.

A set can be represented in the following two ways - tabular form (or Roster method) and set builder form. A set containing only one element is called a singleton set.

The set having no element is called the empty set or the null set or the void set. v

A set A is called a subset of B if every element of the set A is also an element of the set B.

If A is a subset of B and if there is at least one element in B which is not in A, then A is called the proper subset of B.

Two sets are said to be equal if they contain the same elements

Two sets are called equivalent sets, if and only if there is one to one correspondences between their elements.

A set is said to be finite set if in counting its different elements, the counting process comes to an end.

H)

iswer

' = 4 .

ry +

swer

'man

are

h ( o r

'r the

yo an

Is not

) one

the

The set of all possible subsets of a given set A, is called the power set A.

The difference of two sets A and B denoted by A - B (read as A minus B), is the set of all elements of A which are not in B.

The complement of a set is defined as another set consisting of all elements of the universal set which are not elements of the original set.

K e y w o r d s

Tabular form (or Roster method): If all the elements of a set are kept within {J and elements are separated from one another by comma (,), then this form of the set is called tabular form.

Set builder form (rule method): Set builder form is also called the rule method. In this form, we specify the defining property of the elements of the sets.

Singleton set: A set containing only one element is called a singleton set. E.g. The sets {0}, {x}, jf} all consist of only one element. Empty set: The set having no element is called the empty set or the null set or the void set. It is denoted by f or {}. Subset: A set A is called a subset of B if every element of the set A is also an element of the set B. We write it as

A e B

Equal sets: Two sets are said to be equal if they contain the same elements

Equivalent sets: Two sets are called equivalent sets, if and only if there is one to one correspondences between their elements.

Finite set: A set with finite number of elements is a finite set.

Infinite set: A set, which is neither a null set nor a finite set is called an infinite set.

Set of sets: If the elements of a set are sets, then the set is called a set of sets. Power set: The set of all possible subsets of a given set A, is called the power set A.

Universal set: A set, which contains all sets under consideration as subsets, is called the universal set. It is denoted by U.

Intersection of Sets: The intersection or meet of two sets A and B written as A n B (read as A cap B) , is the set of all elements that belong to both A and B. Complement of a set: The complement of a set is defined as another set consisting of all elements of the universal set which are not elements of the original set.

R e v i e w Q u e s t i o n s 1. Is the set A = )x: x + 5 = 5} null? 2. Write down all the subsets of the set |1, 2, 3j.

Theory of Sets

Notes



Notes

3. How many subsets of the letters of the word ALLAHABAD will be formed?

4. Are the following sets equal?

(i) A = (x: x is a letter in the word WOLF}, (ii) B = {x: x is a letter in the word FOLLOW}.

5. If A c B, B c C and C c A. show that B = A.

6. If A = {1, 2, 3, 4}, B = {2 ,3 ,4 ,5} & C = {4, 5, 6, 7}, find A - (B - C). 7. If A = (I , 3, 6 ,10,15,21}, & B = {15,3,6}, find (A - B) n (B - A). 8. If X = {1, 2 , 3 , 4 , 5 } & Y = { 1 , 3 , 5 , 7 , 9 } , find the values of X n Y and

( X - Y ) u ( Y - X ) . 9. If A = {1, 2, 3 ,4 , 5} & B = { 1 , 3 , 5 , 7 , 9 ) , find the symmetric difference of

A & B .

10. If A = {a, b, c, d), & B = {e, f, c, d}, find A A B. 11. If A = A u B, show B = A n B.

12. If A & B are two sets, find the value of A n (A u B ) . 13. If A, B are subsets of a set S, and A', B' are the complements of A& B

respectively. Prove that A c B B' c A'.

14. Prove that for any two sets A & B ,

(A - B) u (B - A) = (A u B) - (A n B).

F u r t h e r R e a d i n g s

P. N. Mishra, Quantitative Techniques for Managers, Excel Books D.R., D.J. Sweeney, and T.A. Williams Anderson, Quantitative Methods for Business, 5th edition, West Publishing Company

E R Tufte, The Visual Display of Quantitative Information, Graphics Press

36 Self-lnstnutional M n t e r i n l

Unit 3 Logar i thms & P r o g r e s s i o n s

Unit Structure Introduction Logarithms Laws of Operations Compound Interest Arithmetic Progressions Geometric Progressions Annuities, Loans and Mortgages Methods of Investment Evaluation Perpetual Annuities and Infinite Series Depreciation Summary Keywords Review Questions Further Readings

Learning Objectives After reading this unit you should be able to:

Define logarithms Prove and use logarithmic operations Compute compound interest Define and use arithmetic progressions Define and use geometric progressions Calculate annuities, loans and mortgages Evaluate investments by different methods Compute perpetual annuities using infinite series

In troduct ion Mathematical tools have been developed time to time to simplify mathematical computations. Logarithms are mathematical tools that convert multiplication, division, exponentiation and root operations into addition, subtraction, multiplication and division operations respectively.

L o g a r i t h m s When we have two numbers such as 4 and 16, which can be related to each Dther by the equation 42 = 16, we define the exponent 2 to be the logarithm of 16 to f he base of 4, and write

log4 16 = 2

Logarithms & Progressions

Notes


it is clear from this example that the logarithm is nothing but the power to which a base (4) must be raised to attain a particular number (16). In general, we may state that

y = b' o f = ^ y

which indicates that the log of y to the base b (denoted by logb y) is the power to which the base b must be raised in order to attain the value y. For this reason, it is correct, to write

This implies that any positive number y must posses a unique logarithm f to a base b > 1 such that the larger the y, the larger its logarithm. As y is necessarily positive in the exponential function y = bl; negative number or zero cannot a logarithm.

The base of the logarithm, b > 1, does not have to be restricted to any particular number, but in actual log applications two numbers are widely chosen as bases - the number 10 and the number e. When 10 is the base, the logarithm is known as common logarithm, symbolized by logio (or if the context is clear, simply by log). With e as the base, on the other hand, the logarithm is referred to as natural logarithm and is denoted either by log e or by In (for natural log). We may also use the symbol log (without subscript e) if it is not ambiguous in the particular context.

Common logarithms, used frequently is computational work, are exemplified by the following:

logio 1000 = 3 [because 10 3 = 1000] logio 100 = 2 [because 10 2 = 100] logio 10 = 1 [because 10 1 = 10] logio 1 = 0 [because 10 = 1] logio 0.01 = -1 [because 1 0 - 1 = = 0.01] logio 0.01 = -2 [because 10" 2 = = 0.01]

There is a close relation between the set of numbers immediately to the left of the equals signs and the set of numbers immediately to the right. From these, it should be apparent that the common logarithm of^ a number between 10 and 100 must be between 1 and 2 and that the common logarithm of a 1 and 10 must be a positive fraction, etc. The exact logarithms can easily be obtained from a table of common logarithms or electronic calculators with log capabilities.

In analytical work, however, natural logarithms prove vastly more convenient to use than common logarithms. Since, by the definition of logarithm, we have the relationship.

y = c! &t=logey(ort = lny)

it is easy to see that the analytical convenience of e in exponential function Logarithms & Progressions will automatically extend into the realm of logarithms with e as the base.

The following example will serve to illustrate natural logarithms:

l n e 3 = l o g e e 3 = 3 N o t e s

In e2 = loge e2 = 2

In e1 = loge e1 = 1

In 1 = loge e = 0

I n - = l o g e e _ 1 = -1 e

The general principle emerging from these examples is that, given an expression e", where n is any real number, we can automatically read the exponent n as the natural log of e". In general, therefore, we have the result that In e" = n.

Common log and natural log are convertible into each other; i.e., the base of a logarithm can be changed, just as the base of an exponential expression can. A pair of conversion formulas will be developed after we have studied the basic rules of logarithms.

L a w s o f O p e r a t i o n s

Logarithms are in the nature of exponents; therefore, they obey certain rules closely related to the rules of exponents. These can be of the great help in simplifying mathematical operations. The first three rules are stated only in terms of natural log, but they are also valid when the symbol In is replaced by logb.

R u l e I ( l o g o f a p r o d u c t ) in{uv)-lnu + lnv (u,v>0\

Example 1

/ ( e V ) = / c 6 +lne6 +lne4 =6 + 4 = 10 Example 2

In(Ae7 )= In A + In e7 + In A + 7

R u l e I I ( l o g of a q u o t i e n t ) > ln(u/v)=lnu-lnv (u,v>0)

Example 3

ln[e2 /c)=lne2 -lnc = 2-lnc

Example 4

' ln(e2/e5)=lne2-Ine5 = 2 - 5 = -3

Student Activity

1. If log x 64=3, what is the value of x?

2. If log3X = 21og9X, what is the value of x?

humtitative Techniques

Notes

Student Activity

1. What is the value of

lne 1 5? 2. Show that

log^ x.log ; v.log t

3. What are the values of the following logarithms?

(a) log 1 0 10.1000 (b) logio 0.0001 (c) logjSl (d) logs 3125

4. Evaluate the following:

(a) l n e 2

( b ) l 0 g e e - 4 (c) lnf l /e 3 ) ( d ) l 0 g e ( l / e 2 ) (e) ( e l n 3 ) ! (f) l n e " - e l n *

5. Evaluate the following by application of the rules of logarithms:

(a) log 1 0 (100) 1 4

I (b) log, 0

100 (c) ln(3/B) (d) In Ae 2

W lnABe- 1

(f) (log4e)(log,64) 5. Which of the following

are valid?

w (a) In u ~ 2 In

(t>) 3 + In v = In v

(c) In u + In v u v

In vv = In w

(d) In 3 + In 5 = In 8 Prove that

In ( / V) = In u - In u .

Rule III (log of a power) Inu" =alnu (u>0)

Example 5

InA3 =3/nA

These three rules are useful devices for simplifying the mathematical operations in certain types of problems. Rule I serves to convert, via logarithms, a multiplicative operation (uv) into an additive one (In u + In v); Rule II turns a division (u/v) into a subtraction (In u - In v); and III enables us to reduce a power to a multiplicative constant. Moreover, these rules can be used in combination.

Example 6

ln(uv")-lnu + lnv" -lnu + alnv

Your are warned, however, that when we have additive expressions to begin with, logarithms may be of no help at all. In particular, it should be remembered that

ln(uv)*lnu*lnv

Let us now introduce two additional rules concerned with changes in the base of a logarithm.

Rule IV (conversion of log base)

The rule, which resembles the chain rule in spirit (witness the "chain" e

~~^ b ~ ^ 7 1 1 ) , enables us to derive a logarithm log e u (to base e) from the logarithm log e u (to base b), or vice versa. Rule IV can readily be generalized to

ty w)

where c is some base other than b.

Rule V (inversion of base) 1

logbe = logeb

This rule, which resembles the inverse-function rule of differentiation, enables us to obtain the log of b to the base e immediaLely upon being given the log of e to the base b, and vice versa. (This rule can also be generalized to the form logbc = yiogcb).

From the last two rules, it is easy to derive the following pair conversion formulas between common log and natural log:

Iog10 N = (log10 eXk)ge N) = 0.4343loge N logio N = (loge 1 0 X l o g l 0 N) = 2.3026log10 N

for N a positive real number. The first equals sign in each formula is easily justified by Rule IV. In the first formula, the value 0.4343 (the common log of 2.71828) can be found from a table of common logarithms or an electronic calculator; in the second, the value 2.3026 (the natural log of 10) is merely the reciprocal of 0.4343, so calculated because of Rule V.

Example 7

log, 100 = 2.3026(/o 1 0100)=2.3026(2)=4.6052. Conversely, we have logw 100 = 0.4343(Jo c 100)=0.4343(4.6052)=2.

C o m p o u n d I n t e r e s t If we are getting a return of 10 % in one year what is the return we are going to get in two years? 20 %, right. What about the return on 10 % that you are going to get at the end of one year?; If we also take that into consideration the interest that we get on this 10 % then We get a return of 10 + 1 = 11 % in the second year making for a total return of 21 %. This is the same as the compound value calculations that you must have learned earlier.

Future Value = (Investment or Present Value) * (1 + Interest)"" The compound values can be calculated on a yearly basis, or on a half-yearly basis, or on a monthly basis or on continuous basis or on any other basis you may so desire. This is because the formula takes into consideration a specific time period and the interest rate for that time period only.

To calculate these values would be very tedious and would require scientific calculators. To ease our jobs there are tables developed which can take care of the interest factor calculations so that our formulas can be written as:

Future Value = (Investment or Present, Value) * (Future Value Interest Factor ) where n = no of time periods and i = interest rate.

A r i t h m e t i c P r o g r e s s i o n s

A series of quantities form an arithmetic progression if each subsequent term is obtained by adding to the previous term a constant amount, whicji is called the common difference. An arithmetic progression always has the form:

a, a+ d, a+ 2d, , a+ (n-1) d here a is the first term, d is the common difference and n is the number of terms.

The first application would be a regular increase in salary, say by Rs X per year. Another application of arithmetic progressions is in the depreciation of machinery and other fixed assets. In the financial accounts of manufacturing firms, it is necessary to make a deduction from gross profits to allow for the decrease in value of the machinery due to wear and tear and perhaps also


Notes

Student Activity 1. At what rate Rs 5,000

must be invested compounded annually so that amount receivable after 20 years is Rs. 20,000?

2. Solve the following equation.

20.5*-13.8 = 0

Student Activity ABC Transformers has produced 780 transformers in 199S and is decreasing the annual production by 40 transformers per year because of the competition by Class Transformers. Class Transformers has produced 100 transformers in 1998 and is increasing the annual production by 30 transformers. In which year Class Transformers will become the larger producer?

A plant measures 3 inches at present. What will be its height after 10 years if the height increases at a steady rate of 1.5 inches a year?



Notes


ABC Transformers has produced 780 transformers in 1998 and is decreasing the annual

40 year

the Class Class

has 100

production by transformers per because of competition by Transformers. Transformers produced transformers in 1998 and is increasing the annual production by 30 transformers. In which year Class Transformers will become the larger producer?

A plant measures 3 inches at present. What will be its height after 10 years if the height increases at a steady rate of 1.5 inches a year?


Find the annual salary increment of a man who

retires after his 3 7 m year, having tamed Rs 61,700 in his fifth year and an average of Rs 1,00,900 over his whole career.

What is the sum of first 100 even numbers?

obsolescence. This reduces the claim of the owners on the profit, so that cash is made available for the eventual replacement of the machinery. The increase in cash is matched in the balance sheet by a decrease in the 'book value' of the machinery. If the depreciation is assessed as a fixed amount each year, then the model for the changing value of the machinery is an arithmetic progression. The common difference as in this case is always negative.

Illustration 3.1

A machine is bought for Rs 90,000 and the depreciation on it is assessed at Rs 7,200 per year. What will be its book value at the end of eight years?

Solution

This method of depreciation is called the straight line method and the depreciation value is assessed by dividing the machine cost less the cost of the scrap by the number of useful years it is in operation.

It is important to note that in this type of problem the number of terms in the arithmetic progression is 1 more than the number of years. This is because the first term is the book value at the beginning of the first year and the final term is the book value at the end of the final year. Putting a = 90000, d = -7200, and n = 9, the eight term works out as:

90000 + 8 x (-7200) = 32400

Sum of an Arithmetic Progression S may represent the sum of the n terms of an arithmetic progression, where:

S s a+(a+ d)+(a+ 2d)+ +[a+ ( n - 2)d]+[a+ (n - l)d] The easiest way to find the formula for S is to write the series again in reverse order:

S = [a+ (n-l)d]+[a+ ( n-2)d]+ -.+(a+ 2d)+(a+ d)+ a Then the two equations are then added together to form a third equation. In this equation the left-hand side becomes 2S; on the right-hand side the sum of the first terms of the equations is I2a+ (n - 1) d] and the sum of the second terms and of each succeeding pair of terms is exactly the same. Since there are n pairs in all:

2S = n\2a+ (n - l)d] > \

hence, S = n[a+ (n -l)d]

Illustration 3.2 The salary of a company secretary is increased by a fixed increment each year. If his total earnings over nine years are Rs 23,40,000 and his salary in the final year is Rs 2,95,000, what was his salary in the sixth year?


Solution

r14 -

,14

740 6200

= 0.11935

Now this equation can be easily solved using scientific calculator or logarithmic tables can be used. As the log tables would permit only four


Notes


Rs 7,000 are invested at 5% per annum compound interest. What will be the amount after 20 years?

The number of rabbits becomes three times in 3 days. In how many days it will become 200 times the original number?

In this example, n~9,S = 23,40,000, [a + (n - 1) d] = 2,95,000 and it is required to find the sixth term of the series which is (a + 5d). Using the formula to express 5 in terms of the unknowns, a and d:

23,40,000 = 9[a+4d] a+4d = 2,60,000

a+8d = 2,96,000

d = 87.5

a+5d = 2,68,750 ' r

And so the salary in the sixth year was Rs 2,68,750 per annum.

G e o m e t r i c P r o g r e s s i o n s .

A series of quantities form a geometric progression if each terms is obtained by multiplying the previous term by a constant, which is called the common ratio. A geometric progression always has the form:

a, ar, ar2, ar

3, , ar"'

1

where a is the first term, r is the common ratio and n is the number of terms. Note that in the last term r has the power (n-1) and not (n). This is because in the first term r has the power (0) and therefore the total terms are n. Illustration 3.3

A small water pump costs Rs 6,200 and is expected to last for 14 years and then have a scrap value of Rs 740. If depreciation is to be calculated as a fixed percentage of the current book value at the end of each year, what should the percentage be?

Solution

This illustration deals with the same situation as the illustration 3.1, but the depreciation is now to be calculated as a fixed proportion of the current book value instead of a fixed amount each year. This method of depreciation is called the Written Down Value (WDV) method. The model for the changing value of the water pump is in this case a geometric progression; the book value is reduced each year to r tinges its previous figure, where r is less than 1. The number of terms is 1 more than the number of years, and so n = 15, a = 6200, andar"'1 = 740.

6200r 1 4 = 740


Notes

decimal places, there would be a difference between the value calculated! through it and the value calculated through the calculator. Both the values are| OK for normal purposes.

For solving this equation through log tables, it is necessary to divide the above] equation into a negative logarithm:

Logr = Log 740-log 6200 -0.9231599

14 14 = 0.0656399

The r is 0.8592, which means that depreciation is reducing the value and tr depreciation percentage can be found by subtracting 1 (which stands for] 100%) from the value found. The value comes out to be -0.1408, whic converted back into percentage from results in 14.08%/ which may be rounde to 14.1% or 14% depending on the accounting practice and the requirements! Note that negative sign did not came in this case as depreciation mea reduction itself so the negative sign is understood. In other cases we have use it explicitly.

Sum of Geometric Progression

If S represents the sum of the n terms of a geometric progression, the easie way to find the formula for S is to write out the series and then multiply throughout by r:

S = a+ar+ar2+ar3+ +arn'2+ar"2

Sr - ar+ar2+ar3+ +arn'1+ar"

The first equation is then subtracted from the second and on the right-har side almost all the terms cancel out to leave:

S(r-l) = arn-a

Student Activity

Sum of the first 3 terms of a GP is 8 times sum of the next 3 terms. Find the 7 t h term of the GP.

A man invests Rs 5,000 at the beginning of each year, and compound interest is added at 5W/o per annum. How much money would he have accumulated after 15 years?

hence, S

Illustration 3.5

- "(r"-l) (r-1)

A company sets aside Rs 5,00,000 each year out of its profits to form a reservd fund, which is invested at 4% per annum compound interest. What will be tha value of the fund after ten years?

Solution

It is assumed that the first Rs 5,00,000 earns interest only from the end of tl year in which it is set aside, which is for nine years. Successive annua instalments earn fewer years of interest, and the final sum for the tenth yea earns no interest at all. So this is the sum of a geometric progression with thfj terms written in reverse order so that the last term is written first.

With a = Rs 5,00,000, r - 1.04 and n = 10, the terms being in the reverse of tl usual order. The formula for S may be applied to these values. It will be note


that n is the number of years, since the model represents the number of instalments and not the beginnings and ends of years.

By logarithms, S is 59,87,500. The possible error due to the use of logarithms is about 20,000 which is not a small sum and so the answer is best quoted as about Rs 60,03,052.

A n n u i t i e s , L o a n s a n d M o r t g a g e s

The examples in above sections either involved a single investment, or a series of regular payments. We will now consider problems that involve both a single initial payment and a series of regular equal payments. A payment Rs a, which is to be made in n years in the future has a present

value of Rs where r is the compound interest ratio. Since an annuity

consists of a payment Rs a at the end of each year for n years, the present value Rs P of the annuity is the sum of the present values of the individual payments:

This is a geometric progression with first term a / r , common ratio V r and n terms, and the sum is:

The only difference between this formula and the one which gives the sum of GP is that there is another term r" in the denominator. You can derive this formula yourself by substituting the first term and common ratio given above in the formula which gives the sum of a GP.

Illustration 3.9

Find the cost of an annuity of Rs 25,000 for 15 years, if compound interest is allowed at 3% per annum.

Solution >

Using the above formula and using the values a = 25000, n = 15, and r = 1.03 and using four-figure logarithms gives P = 2,97,800. The cost is the present value, and so is approximately Rs 2,98,000.

Finding the Annuities

When we need to find the annuities and are given the present value P, the only necessity is to rearrange the formula of the present value:

5,00,000(l.04" -l) {104-1)

= 1,25,00,000(1.04w -1)

rn(r-l)

a -Pr"{r~l)

Quantitative Techniques Illustration 3.10

A firm borrows Rs 10,00,000 and repays it by three equal sums at the ends of the following three years. What is the amount of each repayment, if compound interest at 4Vi% is allowed?

Solution

On a loan, equal repayments are nothing but a form of annuities, the amount of the loan being the present value of the annuity. So it is only necessary to directly apply the formula already obtained:

Student Activity

1. What sum should be paid for an annuity of Rs 25,000 per annum, to be paid half-yearly commencing in six months' time, if compound interest is allowed at 14% per annum and compounded half-yearly and the annuity is to last (i) 10 years (ii) 20 years (iii) 30 years

2. What is the amount payable in the above case in perpetual case?

Pr"(r-1)

W,000,000xl.0453 x(l.045-l) = 3,63,800

(l.0453 -l) The company needs to pay Rs 3,63,800 every year to repay me loan and the interest in the three years.

Some people find it surprising that this method really does give the correct annual repayment within the terms of the question. While the proof by algebra ought to be sufficient, the number of years is here small enough to permit the luxury of a numerical check:

Interest 4.5%

Repay

Interest 4.5%

Repay

Interest 4.5%

Repay

10,00,000 45,000

10,45,000 3,63,800 6,81,200 Amount outstanding after 1 year

30,700 7,11,900 3,63,800 3,48,100 Amount outstanding after 2 years

15,700

3,63,800 3,63,800 ^ 0000000 No amount outstanding after 3 years

Methods of I nves tmen t Evaluat ion

Any commercial or industrial investment can be regarded as the purchase of an annuity. The amount of the annuity is the income or the saving in costs directly attributable to the investment, such as the saving in running costs when replacing an old machine by a new one.

Illustration 3.7 Logarithms & Progressions

N o t e s


A new machine is expected to last for eight years and to produce annual savings of Rs 23,000. What is its present value, allowing interest at 7% per annum?

Solution

The present value is obtained in the same way as found earlier in this unit:

' P -

r"(r-l)

= > P s 23.000(1.07-- l) . 107" (107-1)

The same result can be obtained using appendix 2, the figure 5.9713 then being multiplied by Rs 23,000.

The assumptions that are made here are that the savings are effectively obtained at the end of each year and that the rate of interest quoted is the rate at which money is available to buy the machine. The answer indicates that the machine is worth buying if it costs less than Rs 1,37,340 and not worth buying if it costs more than that.

A refinement on this method is to subtract the cost of a machine from its present value to give the net present value. If the machine, just considered, has a cost price of Rs 1,25,000, the net present value is Rs 1,37,340 - Rs 1,25,000 = Rs 12,340. Any investment is worth making only when it has a positive net present value. This net present value or the NPV, as it is called in short form, is used extensively in appraising investment proposals, about which you would read more in your finance textbooks.

If the new machine replaces an old machine, the selling price of this machine (which will usually be its scrap value) must be added to the net present value of the investment. The book value of the old machine may be different from it selling price, but this is irrelevant. Some accountants argue that the 'loss on book value' of the machine to be scrapped must be included, as part of the cost of the proposed investment, but this is completely erroneous.

This raises the question as to the best method of calculating depreciation. It is usually simpler to depreciate by a fixed amount each year, but more>realistic to depreciate by a fixed percentage so that the decreasing amounts written off in successive years helps to balance the increasing costs of repairs and maintenanc

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Quantitative Techniques