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  • ,A model is a simplified representation of a real-world situation that ideally, strips a natural phenomenon of its bewildering complexity and replicates its essential behavior. Models may be represented in a variety of ways.

  • Quantitative Techniques in Management

    Chapter 1 : MANAGEMENT AND DECISION - MAKING

    INTRODUCTION

    Decision-making is an essential and dominating part of the management process. Although authorities sometimes differ in their definitions of the basic functions of management, everybody agrees that one is not a manager unless he has some authority to plan, organise and control the activities of an enterprise and behaviour of the others. Within this context, decision-making may be viewed as the power to determine what plans will be made and how activities will be organized and controlled. The right to make decisions is an internal part of right of authority upon which the entire concept of management rests. Essentially then, decision-making pervades the activities of every business manager. Further, since to carry out the key managerial functions of planning, organizing, directing and controlling, the management is engaged in a continuous process of decision-making pertaining to each of them, we can go to the extent of saying that management may be regarded as equivalent to decision-making.

    Traditionally, decision-making has been considered purely as an art, a talent which is acquired over a period of time through experience. It has been considered so because a variety of individuaI styles can be traced in handling and successfully solving similar type of managerial problems in actual business. However, the environment in which the management has to operate nowadays is complex and fast changing. There is a greater need for supplementing the art of decision-making by systematic and scientific methods. A systematic approach to decision-making is necessary because today's business and the environment in which it functions are far more complex than in the past, and the cost of making errors is becoming graver with time. Most of the business decisions cannot be made simply on the basic of rule of thumb, using commonsense and / or snap judgment. Commonsense may be misleading and snap judgments may have painful implications. For large business, a single wrong decision may not only one ruinous but may also have ramifications in national or even international economies. As such, present day management's cannot rely solely on a trial and error approach and the managers have to be more sophisticated. They should employ scientific methods to help them make proper choices. Thus, the decision makers, in the business world of today must understand scientific methodology for making decisions. This calls for (1) defining the problem in a clear manner, (2) collecting pertinent facts, (3) analyzing the facts thoroughly, and (4) deriving and implementing the solution.

    DECISION - MAKING AND QUANTITATIVE TECHNIQUES.

    ManageriaI decision-making is a process by which management, when faced with a problem, chooses a specific course of action from a set of possible options. In making a decision, a business manager attempts to choose that course of action which is most effective in the given circumstances in attaining the goals of the organization. The various types of decision-making situations that a manager might encounter can be listed as follows.

    1. Decisions under certainly where all facts are known fully and for sure, or uncertainly where the event that would actually occur is not known but probabilities can be assigned to various possible occurrences.

    2. Decisions for one time-period only called static decisions, or a sequence of interrelated decisions made either simultaneously or over several time periods called dynamic decisions.

    3. Decisions where the opponent is nature (digging an oil well, for example) or a national opponent (for instances, setting the advertising strategy when the actions of competitors have to be considered)

    These classes of decisions-making situations are not mutually exclusive and a given situation would exhibit characteristics from each class. Stocking of an item for sale in a certain trade fair, for instance, illustrates a static decision making situation where uncertainly exists and nature is the opponent.

    The elements of any decision are :

    a. a decision-maker who could be an individual, group, organization, or society; b. a set of possible actions that may be taken to solve the decision problem; c. a set of possible states that might occur; d. a set of consequences (pay-offs) associated with various combinations of courses of action and the states that may occur; and e. the relationship between the pay-offs and the values of the decision maker;

    In an actual decision-making situation, definition and identification of the alternatives, the states and the consequences are most difficult, albeit not the most crucial, aspects of the decision problem.

    In real life, some decision-making situations are simple while others are not. Complexities in decision situations arise due to several factors. These include the complicated manner of interaction of the economic, political, technological, environmental and competitive forces in

  • society, the limited resources of an organization; the values, risk attitudes and knowledge of the decision-makers and the like. For example, a company's decision to introduce a new product will be influenced by such considerations as market conditions, labour rates and availability, and investment requirements and availability of funds. The decision will be of multidimensional response, including the production methodology, cost and quality of the product, price, package design, and marketing and advertising strategy. The results of the decision would conceivably affect every segment of the organisation. The essential idea of the quantitative approach to decision-making is that if the factors that influence the decisions can be identified and quantified then it becomes easier to resolve the complexity of the decision-making situations. Thus, in dealing with complex problems, we may use the tools of quantitative analysis. In fact, a large number of business problems have been given a quantitative representation with varying degrees of success and it has led to a general approach which is variably designated as operations research (for operational research), management science, systems analysis, decision analysis, decision science, etc. Quantitative analysis is now extended to several areas of business operations and represents probably the most effective approach to handling of some types of decision problems.

    A significant benefit of attaining some degree of proficiency with quantitative methods is exhibited in the way the problems are perceived and formulated. A problem has to be well defined before it can be formulated into a well-structured framework for solution. This requires an orderly and organised way of thinking.

    Two observations may be made here. First, it should be understood clearly that a decision by itself does not become a good and right decision for adoption merely because it is made within an orderly and mathematically precise framework. Quantification at best is an aid to business judgment and not its substitute. A certain degree of constructive skepticism is as desirable in considering a quantitative analysis of business decisions as it is in any other process of decision-making. Further, some allowances should be made for qualitative factors involving morale, motivation, leadership, etc. which cannot be ignored. But they should not be allowed to dominate to such an extent that the quantitative analysis may look to be an interesting academic exercise, but worthless. In fact, the manager should seek some balance between quantitative and qualitative factors.

    Should, it may be noted that the various names for quantitative analysis; operations research, management science, etc. cannot more or less the same general approach. We shall not attempt to discuss the differences among the various labels as it is prone to create more heat than light, but only state that the basic reason for so many titles is that the field is relatively new and there is not consensus regarding which field of knowledge it includes. We shall now briefly discuss operations research - its historical development, its nature and characteristics, and its methodology. This shall be followed by a discussions of the plan of this book.

    HISTORICAL DEVELOPMENT OF OPERATIONS RESEARCH (OR)

    While it is difficult to mark the 'beginning' of the operations research/ management science, the scientific approach to management can be traced back to the era of Industrial Revolution and even to periods before that. But operations research, as it exists today, was born during the Second World War when the British military management called upon a group of scientists to examine the strategies and tactics of various military operations with the intention of efficient allocation of scarce resources for the war effort.

    The name operational research was derived directly from the context in which it was used - research activity on operational areas of the armed forces. British scientists spurred the American military management to similar research activities (where it came to be known as operations research). Among the investigations carried out by them were the determination of (i) optimum convoy size to minimise losses from submarine attacks, (ii) the optimal way to deploy radar units in order to maximise potential coverage against possible enemy attacks, and (iii) the invention of new flight patterns, and the detemtination of correct color of the aircraft in order to minimise the chance of detection by the submarines.

    After the war, operations research was adopted by the industry and some of the techniques that had been applied to the complex problems of war were successfully transferred and assimilated for use in the industrialized sector.

    The dramatic development and refinement of the techniques of operations research and the advent of digital computers are the two prime factors that have contributed to the growth and application of OR in the post-war period. In the 1950s OR was mainly used to handle management problems that were clear cut, well structured and repetitive in nature Typically, they were of a tactical and operational nature such as inventory control. resources allocation scheduling of construction projects, etc. Since the 1960s, however, formal approaches have been increasingly adopted for the less well-structured planning problems as well. These problems are strategic in nature and are the ones that affect the future of the organization. The development of corporate planning models and those relating to the financial aspects of the business, for example, are such type of problems. Thus, in the field of business and industry, operations research helps the management to determine their tactical and strategic decisions more scientifically.

    Nature and Characteristic Features Of OR.

    In general terms, operations research attempts to provide a systematic and rational approach to the fundamental problems involved in the control of systems by making decisions which, in a sense, achieve the best results considering all the information that can be profitably used. A classical definition of OR is given by Churchman et aI, "....Operations Research is the application of scientific methods, techniques and tools to problems involving the operations of systems so as to provide those in control of operations with optimum solutions to the problems".

  • Thus, it may be regarded as the scientific method employed for problems solving and decision-making by the management.

    The significant features of operations research are given below:

    1. Decision Making Primarily, OR is addressed to managerial decision-making or problem solving a major premise of OR is that decision-making irrespective of the situations involved, can be considered a general systematic process that consists of the following steps.

    a. Define the problem and establish the criterion which will be used The criterion may be the maximization of profits utility and minimization of costs. etc.

    b. Select the alternative courses of action for consideration. c. Determine the model to be used and the values of the parameters of the process. d. Evaluate the alternative and choose the one which is optimum.

    2. Scientific Approach OR employs scientific methods for the purpose of solving problems. It is a formalized process of reasoning and consists of the following steps.

    a. The problem to be analyzed is defined clearly and the conditions for observations are determined b. Observations are made under varying conditions to determine the behaviour of the system. c. On the basis of the observations, a hypothesis describing how the various factors involved are believed to interact and the

    best solution to the problem is formulated. d. To test the hypothesis, an experiment is designed and executed. Observations are made and measurements are recorded. e. Finally, the results of the experiments are analyzed and the hypothesis is either accepted or rejected. If the hypothesis is

    accepted the best solution to the problem is obtained. 3. Objective: OR attempts to locate the best or optimal solution to the problem under consideration. For this purpose, it is necessary

    that a measure of effectiveness is defined which is based on the goals of the organization. This measure is then used as the basis to compare the alternative courses of action.

    4. Inter-disciplinary Team Approach: OR is inter-disciplinary in nature and requires a team approach to a solution of the problem. So

    single individual can have a thorough knowledge of the myriad aspects of operations research and how the problems may be addressed. Managerial problems have economic physical, psychological, biological, sociological and engineering aspects. This requires a bind of people with expertise in the areas of mathematics, statistics, engineering, economics, management, computer science and so on. Of course, it is not always so. Some problem situations may be adequately handled even by one individual.

    5. Digital Computer: Use of a digital computer has become an integral part of the operations research approach to decision-making.

    The computer may be required due to the complexity of the model, volume of data required or the computations to be made. Many OR techniques are available in the form of 'canned' programmes.

    Methodology of Operations Research

    The basic and dominant characteristic feature of operations research is that it employs mathematical representations or models to analyse problems. This distinctive approach represents an adaptation of the scientific methodology used by other physical sciences. The scientific method translates a real given problem into a mathematical representation which is solved and retransformed into the original context. The OR approach to problem solving consists of the following steps.

    1. Formulate the problem. 2. Determine the assumptions (model building) and formulate the problem in a mathematical framework. 3. Solve the model formulated and interpret the results 4. Validate the model. 5. Implement the solution obtained.

    We shall now discuss each of the steps one by one.

    1. Problem Formulation: The first and the most important requirement is that the root problem should be identified and understood

    Logically speaking we can not expect to get the right answer if the problem is identified incorrectly. In that case, the solution derived from it is apt to be useless and all the efforts in that direction shall be a waste. The problem would be identified properly because often what is described as a problem may only be its symptom. For example, excessive costs per se do not constitute a problem. They are only an indication of some problem which may, for instance, be improper inventory levels, excessive wastage and the like. Often the symptoms of a problem may extend beyond a single manager's control to other personnel and other departments in an organization. Thus, it is necessary for an operations researcher to understand that the formulation of a problem develops from a complicated interaction that involves the selection and interpretation of the data between himself and the manager. Once the problem has been identified, it is categorized as being standard or special. The standard problems are also known as programmed problems. As has already been mentioned, they are the well- structured problems characterized by routine, repetitive decisions which utilise specific decision-making techniques in their solution strategy. Standard solution procedures have been developed to handle such prototype problems. Examples of these problems include the assigmnent of workers to jobs, fixing the product-mix for a mouth and determination of the quantity of materials to be brought. On the other hand, there arc special or non-programmed problems. They are unique and non-recurrent in nature and, therefore, ill-structured. Undertaking of a research and

  • development project and the merger and consolidation decisions illustrate such type of decision situations.

    2. Model Building: Once the problem is defined, the next step is to build a suitable model. As has already been mentioned, the

    concepts of models and model building lie at the very heart of the operations research approach to problem solving. A model is a theoretical abstraction of a real-life problem. In fact, many real-life situations tend to be very complex because there are literally innumerable inherent factors in any given situation. Thus, the decision maker has to abstract from the empirical situation those factors which are most relevant to the problem. Having selected the critical factors, he combines them in some logical manner so that they form a counterpart or a model of the actual problem. Thus, a model is a simplified representation of a real-world situation that ideally, strips a natural phenomenon of its bewildering complexity and replicates its essential behavior. Models may be represented in a variety of ways. They can be classified as physical and symbolic models.

    a. Physical Models: a physical model is a physical or a schematic representation of the real thing. There are two types of physical models Iconic and analogue.

    i. Iconic Models: They are essentially the scaled-up / down versions of the particular thing they represent a model airplane in a

    wind tunnel, a model of a proposed building provided by an architect, models of the sun and its planets housed in a planetarium, a model of a particular molecular structure of a chemical - are examples of iconic models because they look like what they represent (except size). Maps, pictures or drawings may also be categorised as Iconic models since they represent essentially the images of certain things. The chief merit of an iconic model is that it is cooperate and specific. It resembles visually the thing it represents and, therefore, there are likely to be fewer problems in translating any finding from the model into the real-life situation. However the disadvantage of such models is that they often do not lend themselves to manipulation for experimental purposes.

    ii. Analogue Models: The analogue models use one set of properties to represent another set. To illustrate, an electrical network

    model may be used as in analogue model to study the flows in a transportation system. Similarly, a barometer which indicates changes in atmospheric pressure through movements of a needle represents an examples of analogue model and the contour lines on a map are analogues of elevation. In general, the analogue models are less specific and less concrete but they are easier to manipulate as compared to the iconic models.

    b. Symbolic Models: Many real-life problems can be described by symbolic models or mathematical forms. These are the most

    general and abstract type of models. They employ letters, numbers and other types of symbols to represent the variables and their inter-relationships. As such, they are capable of experimental manipulation most easily. The symbolic models are capable of experimental manipulation most easily. The symbolic models can be verbal or mathematical. Whereas the verbal models describe a situation in spoken language or written works, the mathematical models employ mathematical notion to represent, in a precise manner, the variables of the real situations. The mathematical models take the form of mathematical relationships that portray the structure of what they are intended to represent. The use of a verbal versus mathematical model could be shown by the formula for finding the perimeter of a rectangle. A verbal model would express this problem as follows. The perimeter (P) of a rectangle is equal to the sum total of two times the length (L) and two times the width (W) of the rectangle. In contrast, the advantage of the mathematical model is demonstrated by the following statement: P=2L+2W. If applied to the same rectangle; both models would yield identical results. However. a mathematical model is more precise. Symbolic models are used in operations research because they are easier to manipulate and they yield more accurate results under manipulation than do either the iconic or the analogue models. Use of Mathematical Models. Various types of mathematical models are used in modern operations research. Two broad

    categories of these are deterministic and probabilistic models. A deterministic model is the one in which all parameters in the mathematical formulation are fixed at predetermined values so that no uncertainty exists. In a probabilistic model, on the other hand, some or all the basic characteristics may be random variables (capable of assuming different values with given probabilities). In such models, uncertainty and errors are required to be given explicit consideration. Probabilistic models are also termed as stochastic or chance models. The mathematical models comprise three basic components: decision variables, result variables and uncontrollable variables. The decision variables represent those factors where a choice could be made. These variables can be manipulated and, therefore, are controllable by the decision-maker. The result variables indicate the level of effectiveness of a system. They represent output of the system and are also termed as dependent variables. Finally, the uncontrollable variables are those which influence the result variables but are beyond the control of the decision-maker. To illustrate in the area of marketing, the decision variables may be the advertising budget, the number of regional salesmen employed, the number of products, etc. results variables may be the market share for the company, level of customer satisfaction, etc. while the uncontrollable variables may be the competitors strategies, consumer incomes, etc. As mentioned earlier, the different components of a mathematical model are tied together with the relationships in the form of equations, inequalities etc. .Such a model consists of an objective function and the constraints under which a given system functions. The objective function describes how a dependent (result) variable is related to independent (decision) variables. For example, the profit function of a firm making two products can be stated as follows.

  • p = p1x1 - p2x 2

    in which p indicates the total profit of the firm, x1 and x2 are the number of units (independent variables) of the two products produced and sold, while p1 and p2 represent the profit per unit on the two products respectively (the uncontrollable variables) The objective function is called for to maximise (or minimise), subject to certain constraints (representing the uncontrollable variables) for example, in this case of production, the firm might be able to sell no more than a certain number of units, say 80. Then the marketing constraint (an uncontrolIable variable) can be expressed as follows:

    x1+ x 2 80

    Similarly, other constraints, if any, of the system can be expressed.

    3. Solution of Model: Once an appropriate model has been formulated, the next stage in the analysis calls for its solution and the

    interpretation of the solution in the context of the given problem. A solution to a model implies determination of a specific set of decision variables that would yield a desired level of output. The desired level of output, in turn, is determined by the principle of choice adopted and represents the level which optimises. Optimisation might mean maximising the level of goal attainment to cost.

    It may be noted that the solutions can be classified as being feasible or infeasible, optimal or non-optimal and unique or multiple.

    a. Feasible and Infeasible Solution: A solution (as set of values of the decision variables, as already mentioned) which satisfies

    all the constraints of the problem is called a feasible solution, whereas an infeasible solution is the one which does not satisfy all the constraints. Since an infeasible solution falls to meet one or more of the system requirements, it is an unacceptable one. Only feasible solutions are of interest to the decision-maker.

    b. Optimal and Non-optimal Solutions: An optimal solution is one of the feasible solutions to a problem that optimises and is,

    therefore, the best of all of them. For example, for a multiproduct firm working under some given constraints of capacity, marketing, finance, etc. the optimal solution would be that product-mix which meet all the constrains and yield the maximum contribution margin towards profits. The feasible solutions other than the optimal solution are called non-optimal solutions. To continue with the example, several other product-mixes would satisfy the restrictions imposed and hence qualify for acceptance but they would be ignored because lower contribution margins would be associated with them. They would be non-optimal.

    c. Unique and Multiple Solutions. If only one optimal solution to a given problem exists, it is called a unique solution. On the

    other hand, if two or more optimal solutions to a problem exist which are equally efficient then multiple optimal solutions are said to exist. Of course, these are preferable from the management's point of view as they provide a greater flexibility in implementation.

    Once the principle of choice has been specified, the model is solved for optimal solution. For this, the feasible solutions are considered and of them the one (or more) that optimises is chosen. For this purpose, a complete enumeration may be made so that all the possible solutions are checked and evaluated However, this approach is limited to those situations where the number of alternatives is small alternatively, and more commonly, methods involving algorithms may be used to get optimal solutions. It is significant to note that in contrast to complete enumeration, where all solutions are checked and evaluated. However, this approach is limited to those situations where the number of alternatives is small. Alternatively, and more commonly, methods involving algorithms may be used to get optimal solutions. It is significant to note that in contrast to complete enumeration, where all solutions are checked an algorithm represents a trial and error process where only a part of the feasible solutions are considered and the solutions are gradually improved until an optimal solution is obtained.

    While algorithms exist for most of the standardised problems, there are also some numerical techniques which yield solutions that are not necessarily optimal. Heuristics and simulation illustrate those methods. Heuristics are step-by-step logical roles which, in a certain number of steps, yield some acceptable solutions to a given problem. They are applied in those cases where no algorithms exist. Similarly, the technique of simulation is also applied where a given system is sought to be replicated and experimented with solutions using simulation need not be optimal because the technique is only descriptive in nature.

    Sensitivity Analysis. In addition to the solution of the model formulation by any technique, sensitivity analysis should also be

    performed. By sensitivity analysis we imply determination of the behavior of the system to changes in the system inputs and specifications. This is done because the input data and the structural assumptions of the model may not be valid.

    4. Model Validation. The validation of a model requires determining if the model can adequately and reliably predict the behaviour of

    the real system it seeks to represent. Also, it involves testing the structural assumtions of the model to ascertain their validity. Usually, the validity of a model is tested by comparing its performance with the past data available in respect of the actual system. There is, of course, no assurance that the future performance of the system will continue to be in the same manner as in its past.

  • Therefore, one must take cognisance of the change in the system over time and adjust the model as required.

    5. Implementation: No standard prescription can be given which would ensure that the solution obtained would automatically be

    adopted and implemented. This is because the techniques and models used in operations research may sound high and may be detailed in mathematical terms, but they generally do not consider the human aspects which are significant in implementation of a solution. The impact of a decision may cut across various segments of the organization and the factors like resistance to change. desire to be consulted and informed, motivation, etc. may come in the way of implementation. Equally important as the skill and expertise needed in developing a model is the requirement of tackling issues related to the factors which may have a bearing on the implementation of a solution in a given solution. Thus, a model-which secures a moderate theoretical benefit and is implemented is better than a model which ranks very high on obtaining theoretical advantage but cannot be implemented. In fact the importance of having managers in the organisation who would act on the results of the study of the team that analysis the problem can hardly be over-emphasised.

  • Quantitative Techniques in Management

    Chapter 2 : FUNCTIONS AND THEIR APPLICATIONS

    Functions

    In mathematics and its applications we study certain correspondences between two sets of objects. Examples of this type of correspondence are 'price of a commodity and its demand', advertising expenses and sales' 'income and expenditure' etc.. This correspondence is denoted by the concept of a function. It is a rule for relating particular objects in one set to particular objects in another set. For example, one set of objects could be the length of a square and the other set of objects the set of areas corresponding to each length.

    Length (x) in cm of a square 1 2 3 4 5 --

    Area (A) of square in cm2 1 4 9 16 25 --

    A convenient way of expressing this correspondence is through symbols. If x represents the length in cm of a square and A represents the corresponding area, the rule determining the area is written as

    A = x2

    We say that A is a function of x. Symbol x is a variable in the sense that it can assume varying numerical values. It is also conventional to label x as the 'independent' variable and' A as the 'dependent' variable.

    If to each value of a. variable x there corresponds one definite value of another variable y, we say that y is a function of x and denote it by y=f(x). The set of values of x for which the value of the function y is determined is called the domain of the function while the set of values of y is called the range of the function.

    Some examples of functions are given by -

    In (i), the domain of x is the closed interval (0, 1) and the range of y is (- I5, -14). In (ii), the domain of x is the interval (1,2) and the range of y is (5, 14). In (iii), the domain of x is (- 8,1) U ( 1, - 8 ), i.e. the set of all real numbers except the value 1 and the range of y is (0, 8 ). A function whose range is a set of real numbers is called a real valued function. Graphs of functions and Coordinate Geometry

    It is possible to represent real valued functions in graphical form using the rectangular Cartesian coordinate system. This system consists of two perpendicular lines OX and OY (Fig. 2.1 )

    The lines OX and OY are called X-axis and Y-axis respectively. Their intersection is the point 0 called the origin of the system, A point is located giving its direction and distance from each axis.

    The direction is indicated by + or. Distances measured along the direction of the arrows shown in Fig. (2.1) are positive and those in the opposite direction are negative. Any point in the plane detemlined by the axes can be represented by an ordered 2-ple (x.y). For example, (-2, 3) means 2 units from the origin on the negative side of the X-axis and 3 units in the positive direction of the Y-axis. The graph of a function y = f(x) is merely the diagram obtained by plotting {x, f(x)} for all values of x in the domain of f(x). Normally, a few points are plotted

  • and smooth curve is drawn over them. Figure 2.1.2. shows the graph of y = 2x -3 and Fig. 2.1.3. shows the graph of y = x 2. In Fig 2.1.2. we

    have a 'straight line' and in Fig. 2.1.3. we have a ' parabola'. The equation (y-b) 2+ (x- a)

    2 = 4 when plotted, yields a circle with centre at the

    point (a,b) and with radius 4 = 2.

    CONSTRUCTION OF FUNCTIONS

    In business applications we commonly talk of profit functions, loss functions, cost functions and revenue functions. The functions are usually set up following the definition and calculation of the functional values. We will take up a few examples to illustrate the procedure of constructing such functions.

    Example

    A banana seller buys bananas at R1 rupees and sells at R2 rupees (R2> R1). The unsold fruits at the end of the day are sold at R3 rupees (R3 < R1). What is the profit function for the fruit seller?

    The profit depends on how many fruits he is able to sell in relation to the quantity of fruits he brought in the beginning of the day. Let D be the demand of fruits at R2 per fruit and Q the stock of fruits.

    Example

    A factory has 100 items on hand for shipment to a destination at the cost of Rs. 1 a piece to meet a certain demand d. In case the demand d overshoots the supply, it is necessary to meet the unsatisfied demand by purchases on the local market at Rs. 2 a piece. Construct the cost function if x is the number shipped from the factory.

    Solution:

    Example:

    (Quantity Discounts and Price Breaks) A retailer offers the following price breaks on an item.

    Rs. 10 per kg for any amount ordered up to 10 Kg Rs. 10 per kg for any amount ordered up to 10 kg. Rs. 7 per additional kg above 10 kg and up to 100 kg Rs. 5 per additional kg beyond 100 kg.

    Construct the cost function for x kg ordered.

    Solution :

    Let C(x) be the cost function. Then from the definition of the discounts, we have

    The graph of this function is given in Fig. 2.2.1.

    The curve obtained in Fig, 2.2.1 is not a straight line in the entire domain of x, but it is piecewise linear.

  • Example:

    The simplest model relating revenue (R) in '000s of Rs and advertising expenditure (A) in '000 of Rs can be given as

    The difference between revenue and advertising expendinture may be written as

    D is a function of A. Suppose, we have three alternative decisions of advertising expenditure.

    Which is the best decision?

    We calculate D for each of the three decisions and we have

    Decision D(in thousands of rupees)

    d1 126

    d2 150

    d3 150

    The best decision is clearly either d2 or d3. Both these decisions yield the same value for D. Suppose, we increase the advertising expenditure to 256, then D = 14 which shows that it does not always mean that an increase in advertising expenditure leads to an increase in net revenue, D.

    LINEAR AND QUADRATIC FUNCTIONS

    A linear function is defmed as y = f(x) = ax + b, where a and b are given real numbers and x is a variable taking all numerical values in an interval. It is called linear in x because the graph of such a function is a straight line (Fig. 2.3.1.).

    Also, note that in the definition of y = ax + b, the power of x is 1. The straight line cuts the X-axis at a distance -b/a units from the origin and the Y -axis at a distance of b units from the origin. In Fig. 2.1.2., the straight line y = 2x -3 has an intercept equal to -3 on the Y-axis an 3/2 on the Y-axis and 3/2 on the X-axis.

    A quadratic function is defined by

    where a, b and c are any real numbers. Here the maximum power of x = 2. Figure 2.1.3. gives the quadratic function y = x2. In general, a

    function of the form

    y = a1xn + a2x

    n-1 + ... +anx+an+1

    where aI's are real numbers, aI 0. and n is a positive integer is called polynominal of degree n. Thus, a polynominal of degree 1 is a linear function and that of degree n.

    Thus, a polynominal of degree 1 is a linear function and that of degree 2 is a quadratic function.

  • Fig 2.3.1

    Example

    Assume that for certain values of the advertising expenditure, the sale is a linear function of the expenditure. It is known that for an advertising expenditure of Rs.50,000, the sales would be Rs.4,00,000 and if no advertising expenditure is incurred the sales would be Rs.2,00,000. Construct the sales function.

    Since y is assumed to be linear in x, let y = ax + b

    When x = 0, we are given that y = 200, y = ax + b, for x = 0 give y = b. Hence b = 200.

    The sales function is therefore given by

    Example

    The demand for a certain item is given by

    where q denotes the amount demanded and p the price per unit. It costs Rs. 4 to produce each unit. What is the profit function of the firm for this item?

    which is a quadratic function in p.

    Example:

    (Depreciation calculation)

    Consider an asset costing C. At the end of n years it has a scrap value of S. The difference C -S is the depreciation of the asset in n years. In financial planning, a firm has to provide for depreciation and has to set apart a certain amount every year to account for depreciation There are several methods of determining this amount. In the straight-line method of calculating the annual contribution to the depreciation found, we assume that the fund earns no interest. Further, equal contributions are to be made at the end of each year throughout the life of the asset. The formula for the annual connibution is then

    The depreciation fund F at the end of the kth year is F = kR. This is a linear function of k and hence is a straight line though we calculate it only for integral k. The book value of an asset on a given date is the differcnce between the original cost and the amount in the depreciation fund at that time. Denoting the book value at k

    th year by BV(k), then

  • This is also a straight line as a function of k. Suppose for example c = 25.000, S = 5,000, n = 10. Then

    The depreciation schedule and the book value are given in Table 2.3.1.

    Figure 2.3.2. shows the graphic representation of the depreciation by the straight-line method

    Other methods of depreciation calculations are given in inter chapters of the book.

    SOME SPECIAL FUNCTIONS

    Absolute Value Functions

    First, we define the absolute value of a real numbers, denoted by | x |.

    Thus I x I is the magnitude of the number x without caring for its sign.. Thus, I 8 I = 8 and |-3| = 3. The function f(x) = I x l is called the absolute value function. The graph of the absolute value function is shown in Fig. 2.4.1.

    Step Functios

    A function f(x) that takes a constant valuc for values of x within an interval but possibly takes djfferent values in djfferent intervals is called a step functjon.

    Fig. 2.4.1

    Example 2.4.1.

    The unit price of a commodity is shown below which depends on the quantity ordered.

    The graph of the function giving the unit price in terms of quantity ordered is shown in Fig.2.4.2.

    Convex Sets and Convex Functions

    Convex Set Consider a Set S of points in the tWo -dimensional plane (Figs. 2.4.3. and 2.4.4.).

  • Take any two points (x1, y1) and (x2, y2) falling within the sets. The line segment joining (x1, y1) and (x2, y2) is the set of all points lying on the straight line joining these two points. If the line segment is also wholly contained in the set S, we call the set S a convex set. In Fig. 2,.4. 4, S is not a convex set since the line segment joining the points within the set as shown in the figure is not wholly contained in the set Loosely speaking, a convex set cannot have holes in it.

    Given two points (x1, y1) and (x2, y2) how does one represent the co-ordinates of a point lying on the line segment joining (x1, y1) and (x2,y2)? A typical point s given by the co-ordinates.

    where is a number with 0 = = 1. Note that when = 0, this representation gives the point (x2, y2) and when = 1, we get the point (x1, y1). In particular when =1/2, the corresponding point on line segment is

    which is the mid point of the line segment joining (x1, y1) and (x2, y2). Examples of convex sets are a circle and a triangle. There are other examples which are important in operations research and economics and these will be taken up later in the book.

    Convex Function. In some areas, e.g. operations research and optimization problems, a class of functions called convex functions assumes a great importance. We give the definition and a few basic properties.

    Definition: A function f(x) defined over a convex set S is said to be a convex function if for any two points x1, x2 lying in S and for any 0 1

    Geometrically, this means that the curve f(x) lies below the chord joining the two points x1and x2 (Fig 2.4.5.). .

  • Quantitative Techniques in Management

    Chapter 3 : CALCULUS AND MANAGERIAL APPLICATION

    Many of the decisions facing managers fall into the category of optimisation problems. For example decisions relating to maximizing profit or minimizing cost clearly involve optimization. Often such problems can be solved graphically or by using algebra. In other cases, however, the solution requires the use of calculus. More importantly, in many cases calculus can be used to solve such problems more easily and with greater insight into the economic principles underlying the solution.

    Consider a firm whose total revenues from sales are given by the function

    Where Q represents the rate of output. Assume that the total cost of producing any rate of output is given by the equation

    Given the firm's objective of maximizing profits (i.e. total revenue minus total cost) how much output should be produced? One approach would be to graph both the total revenue and total cost functions. The vertical distance between the two functions is profit. By identifying that point where this vertical distance is the largest the profit-maximized by producing eight units of output (Q = 8).

    An alternative approach is to develop a table showing revenue, cost and profit at each rate of output. Such data are provided in Table 2A-1. The tabular method has the advantage of being somewhat more precise. That is. at an output rate of 8, total revenue is 96, total cost is 82, and profit is 14. From the graph in Figure 2A-1 these values can only be approximated.

    Fig. 2A-1

    The problem is that neither method is very efficient. What if the profit-maximizing output level is 8,000 or 80,00,000 instead of 8? The answer could be found using either approach. but fInding that answer could have taken considerable time. What if there had been two outputs (Q1 and Q2) and three inputs (land, labour and capital)? In this case, there would be no practical way to determine profit- maximizing output rates for Q1 and Q2 using these methods.

    A more powerful technique is needed so that the solution process can be both precise and straightforward. Elementary calculus is easily adapted to optimization problems in economics. Indeed, a few basic principles can be used in many different kinds of problems. The profit-maximization problem just outlined could have been solved quickly using the most elementary calculus. In the following pages, some basic principles of calculus are outlined and their application to economic problems is demonstrated.

    The Derivative of a Function

    From algebra recall that for the function y = f(x), the slope of that function is the change in y (denoted by y) divided by a change in x (i.e. x). The slope sometimes is referred to as the rise (the change in the variable measured on the vertical axis) over the run (the change in Ihe variable measured on the horizontal axis). The slope is positive if the curve slopes upward from left to righl and negative if the function slopes downward from left to right. A horizontal line has a zero slope and a vertical line is said to have an infinite slope. For a positive change in x (i.e. x > 0), a positive slope implies that y is positive and a negative slope implies that y is negative.

    The function y= 10 + x2

    is graphed in Figure 2A -2. To deternine the average slope of this function over the range x=1 to x = 2, first find the couesponding y values. If x1 = 1 then y1= 11, and if x2 = 2 then y2= 14. Then the slope is found by using the formula.

  • In reality, this meiliod determines the slope of a straight line through the points a and b in Figure 2A -2.

    Thus, it is only a rough approximation of the slope of the function y = 10 + x2, which actually changes at every point on that function. By

    making the interval smaller, a better estimate of the slope is determined. For example, consider the slope over the interval x = 1 to x = 1.1. If x1 = 1, then y1 = 11 and x1 = 1.1 implies that y2 = 11.21. Thus the slope is

    By using calculus, the exact slope at any point on the function can be determined. The first derivative of a function (denoted as dy/dx) is simply the slope of a function when the interval along the horizontal axis is infinitesimally small. Technically, the derivative is the limit of the ratio as y / x as x approaches zero, that is,

    Thus the calculus term dy/dx is analogous to is analogous to y / x, but dy/dx is the precise slope at a point, whereas y / x is the average slope over an interval of the function. The derivative can be thought of as the slope of a straight line drawn tangent to the funcrion at that point. For example, the slope of the function y = 10 + x

    2 at point a in Figure 2A -2 is the slope of the straight line drawn tangent to that

    function at A. the derivative of y = f(x) is sometimes written as f'(x). What is the significance of this concept for management economics? Recall from the discussion of total and marginal relationships that the marginal function is simply the slope of the total function. Calculus offers an easy way to find the marginal function by taking the first derivative of the total function. Calculus also offers a set of rules for using these derivatives to make optimizing decisions such as minimizing cost or maximizing profit.

    Standard calculus texts present numerous formulas for the derivative of various functions. In the hands of the skilled mathematician, these formulas allow the derivative of virtually any function to be found. However, only a few of these rules are necessary to solve most of the relevant problems in managerial economics. In this section each of these basic rules is explained and its use demonstrated.

    The Derivative of a Constant

    The derivative of any constant is zero. When plotted, the equation of a constant (such as y = 5) is a horizontal line. For any x, the change in y is always zero. Thus for any equation y = a, where a is a constant.

    The Derivative of a constant Times a Function

    The derivative of a constant times a function is that constant times the derivative of the function. Thus the derivative of y = af(x), where a is constant, is

    For example, if y = 3x, the derivative is

    The Derivative of a Power Function

  • For the general power function of the form y = axb, the derivative is

    The function y =x2 is a specific case of a power function where a =1 and b = 2. Hence the derivative of this function is

    The interpretation of this derivative is that the slope of the function y =x2 at any point x is 2x. For example, the slope at x = 4 would be found

    by substituting x = 4 into the derivative. That is

    Thus when x = 4, the change in y is 8 times a small change in x. The function y = x

    2 is shown in Figure 2A-3. Note that the slope changes continually. The slope at x = 4 is 8. As x increases, the slope

    becomes steeper. For negative values of x, the slope is negative. For example, if x = -3, the slope is -6. The Derivative of a sum or Difference

    The derivative of a function that is a sum or difference of several terms is the sum for difference of the derivatives of each of the terms. That js, if y = f(x) + g(x), then

    For example, the derivative of the function

    Y = 10 + 5x + 6x2

    Is equal to the sum of the derivatives of each of the three terms of the righthand side. Note that the rules for the derivative of a constant, a content times a function. and a power function must be used. This

    The Derivative of the Product of Two Functions

    Given a function of the form

    That is, the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second.

    For example, the derivative of the function

    The Derivative of a Quotient of Two Functions

    For a function of the form y = f(x) / g(x), the derivative is

  • Given the function

    The derivative would be

    The Derivative of a Function of a Function

    The function

    Is really two functions combined. That is, by writing

    It is seen that y is a function of a function. That is

    This derivative is the derivatjve of y with respect to u and multiplied by the derivative of u with respect to x, or

    Now using the rule for the derivative of a power function yields

    Consider another example

    Which can be rewritten as

    In this case

    So the solution is

    Substituting x5 + 2x + 6 for u and multiplying the two derivatives just given yields.

    These seven rules of differentiation are sufficient to determine the derivatives of all the functions used in this book. However, sometimes two or more of the rules must be used at the same time.

  • Example

    Finding the Marginal Function

    Given a total revenue function

    And a total cost function

    Find the marginal revenue and marginal cost functions.

    Solution:

    Recall that a marginal function is simply the slope of the corresponding total functions. For example, marginal revenue is the slope of total revenue. This, by finding the derivative of the total revenue function, the marginal revenue function will be obtained.

    Similarly, the marginal cost function will be found by taking the rust derivative of the total cost function:

    KEY CONCEPTS

    The slope of a function y = f(x) is the change in y (i.e., y) divided by the corresponding change in x (i.e . x)

    For a function y = f(x), the derivative, written as dy/dx or f (x). is the slope of the function at a particular point on the function. Equivalently, the derivative is the slope of a straight line drawn tangent to the function at that point.

    By using one or more of the seven formulas outlined in this appendix, the derivative of most functions encountered in managerial economics can be found.

    Higher -Order Derivatives

    The derivative of a function sometimes is called the first derivative to indicate that ther are higher -order derivatives. The second derivative of a function is simply the first derivative of the first derivative, it is written d

    2y / dx

    2 or f ". In the context of economics, the first derivative of a

    total function is the marginal function. The second derivative of the total function is the slope of the first derivative or the rate at which the marginal function is changing.

    Higher-order derivatives are easy to find. One simply keeps taking the first derivative against. Given the function.

    The second derivative has an important application in finding the maximum and / or minimum of a function. This concept is explained in the following section.

  • Quantitative Techniques in Management

    Chapter 4 : CALCULUS AND OPTIMIZATION

    Recall from the discussion of total and marginal relationships that if the marginal function is positive, the total function must be increasing, if the marginal function is negative, the total function must be decreasing. It was shown that if the marginal function is zero, then the total function must be at either a maximum or a minimum. In Figure 2A-4, a total function and its associated marginal function are shown. At point a, which corresponds to x = x1, the total function is at a maximum and the marginal function is zero. A point , corresponding to x= x2, the total function is minimized and the marginal function is again zero. Thus the marginaI curve is zero at both x1 and x2. However, note the difference in the slope of the marginal function at these points. At x1, the marginal curve has a negative slope whereas at x2, the slope is positive.

    Fig 2A-4

    Because the total function is at a maximum or a minimum (i.e., an extremism) when its slope is zero, one way to find the value of x that results in a maximum or a minimum is to set the first derivative of the total function equal to zero and solve for x. This is a better approach than the triaI-and-error method used earlier. In that example, a total revenue function,

    And a total cost function

    were given. The problem was to find the rate of output, Q, that maximized profit. The total profit function () is found by subtracting total cost from total revenue.

    The profit function will have a slope of zero where that function is at a maximum and also at its minimum point. To fund the profit-maximizing output, take the fIrst derivative equal to zero, and solve for output. That is, find the rate (or rates) of output where the function is zero.

    But will this output rate result in a profit maximum or a profit minimum? Remember, setting the first derivative equal to zero and solving results in an extremism, but it could be a maximum or a minimum. For Figure 2A-4 shows that if the total function is maximised, the marginal function has a negative slope. Conversely, a minimum point on a total function is associated with a positive slope of the marginal function.

    Because the slope of the marginal function is the first derivative of the marginal function, a simple test to determine if a point is a maximum or minimum if suggested. Find the second derivative of the total function and evaluate it at the suggested. Find the second derivative of the total function, and evaluate it at the point where the slope of the total function is zero. If the second derivative is negative (i.e. the marginal function is decreasing) the total function is at a maximum, if the second derivative is positive at that point (i.e. the marginal function is increasing), the total function is at a minimum point.

    In the profit-maximization problem just discussed, the second derivative of the profit function is -2, which is negative. Therefore, profit is maximized at Q = 8. When finding the extremum of any function, setting the first derivarive equal to zero is called the first-order condition. meaning that this condition is necessary for an extremum but not sufficient to determine if the function is at a maximum or a minimum. The test for a maximum or a minimum using the second derivative is called the second-order condition. The first- and second-order conditions together are said to be sufficient to test for either a maximum or a minimum point. These conditions are summarized as follows:

  • Fig 2A-5 : Graph of a function having Several Extrema

    In some problems there will be two or possibly more points where the first derivative is zero. Therefore, all these points will have to be evaluated using the second order condition to test for a minimum or a maximum. As shown in figure 2A-5, a function could have several points, such as a,b,c, and d. where the slope is zero. Points a and c are relative maximun and b and d are relative minimum. The point a is a maximum relative to other points on the function around it. It is not the maximun point for the entire function because the value of y at point c exceeds that at point for the entire function because the value of y at point c exceeds that at point a. To find the absolute maximum the value of the equation must be determined for all relative maxima within the range that the function is defined and also at each of the end points.

    The Partial Derivative

    Many economic phenomena are described by multivariate functions (i.e. equations that have two or more independent variables). Given a general multivariate function such as

    The first partial derivative of y with respect to x, denoted as dy/dx or f(x) indicates the slope relationship between y and x earn z is held constant. This partial derivative is found by considering z to be fixed and taking the derivative of y with respect to x in the usual way. Similarly, the partial derivative of y with respect to z (i.e. dy/dz or fx) is found by considering x to be a constant and taking the first derivative of y with respect to z.

    For example, consider the function

    To find the partial derivative dy/dx or fx, consider z as a fixed and take the derivative, Thus

    This partial derivative means that a small change in x is associated with y changing at the rate 2x + 3z when z is held constant at a specified level. For example, if z = 2, the slope associated with y and x is 2x + 3(2) or 2x - 6. If z = 5, the slope associated with y and x would be 2x + 3(5), or 2x + 15.

    Similarly, the partial derivative of y with respect to z, would be

    This means that a small change in z is associated with y changing at the rate 3x + 2z when x is held constant.

    Optimization and Multivariate functions

    The approach to finding the maximum or minimum value of a multivariate function involves three steps. First find the partial derivative of the function with respect to each independent variable. Second, set all the partial derivatives equal to zero. Finally, solve the system of equations determined in the second step for the values of each of the independent variables. That is, if

    Then the partial derivatives are

    Setting these derivatives equal to zero

  • -2x + Z = 0

    -2 + x + 4z = 0

    and solving these two equations simultaneously for x and z yields

    x = 2/9

    z = 4/9

    These values of x and z minimize the value of the function. The approach to testing whether the optimizing solution results in a maximizing solution results in a maximum or minimum for a multivariate function is complex and beyond the scope of this book. In this text, the context of the problem will indicate whether a maximum or minimum has been determined.

    KEY CONCEPTS

    Higher-order derivatives are found by repeatedly taking the first derivative of each resultant derivative.

    The maximum or minimum point of a function y = f(x) can be found by setting the first derivative of the function equal to zero and solving for the value or values of x.

    When the first derivative of a function is zero, the function is at a maximum if the second derivative is negative or at a minimum if the second derivative is positive.

    For a function having two or more independent variables (e.g. y = f(x) = -z). the partial derivative dy/dx is the slope relationship between y and x, assuming z to be held constant.

    Optimizing a multivariate function requires setting each partial derivative equal to zero and then solving the resulting system of equations simultaneously for the values of each independent variable.

    PROBLEMS

    2A - 1. Determine the first and second derivatives of each of the following functions

    2A - 2. Determine all the first order partial derivatives for each of the following functions.

    2A - 3. Given the multivariate function

    Determine the values of X and Z that maximize the function.

    2A - 4. The total revenue (TR) function for a firm is given by

    Where Q is the rate of output per period. Determine the rate of output that results in maximum total revenue. (Be sure that you have maximised not minimised total revenue).

    2A - 5. Smith and Wesson have written a new managerial economics book for which they receive royalty payments of 15 percent of total revenue from sales of the book. Because their royalty income is tied to revenue, not profit they want the publisher to set the price so that total revenue is maximised. However, the publisher's objective is maximum profit. If the total revenue functions is

  • and the total cost function is determine

    a. The output rate that will maximize total royalty revenue and also Ihe amount of royalty income that Smith and Wesson would receive.

    b. The output rate that would maximize profit to the publisher. Based on this rate of output what is the amount of royalty income that Smith and Wesson would receive? Compare the royalty income of Smith and Wesson to that determined in part (a) (HINT: first determine a function for total profit by subtracting the cost function from the total revenue function).

    2A-6. A firm had determined that its anual profits depend on the number or salespersons it employees and the amount spent on advertising. Specifically the relationship between profits, (in millions), salespersons S (in thousands) and advertising expenditures, A (in millions), is

    Determine the number of salespersons and the amount of advertising expenditures that would maximize the firm's profits.

  • Quantitative Techniques in Management

    Chapter 5 : MATRICES AND DETERMINANTS

    MATRICES: DEFINITION AND NOTATIONS

    A matrix is an array of m x n numbers arranged into m rows and n columns.

    Let these numbers be denoted by a ij (I= 1,2,------m) (J = 1,2, ------n) The matrix with m rows and n columns can be written as

    A matrix having m rows and n columns is called a matrix of order m x n.

    (read: m by n ) The individual entries of the array, aij, is, are termed as the elements of the matrix A.

    A matrix is indicated by enclosing an array of numbers by tbe parentheses [ ] or () or II.

    In order to locate an element of a matrix, one has to specify the raw and the colwnn to which it belongs. For example, the element a34 lies in the third row and fourth column.

    Representation of Data in Matrix form

    Matrices can be used to present a given set of data in a compact form as shown below:

    1. The processing time (in hours) of the two products that pass through three processes:

    2. The following matrix gives the transportation cost (in Rs. Per unit) from each of the three warehouses to each of the five distribution points:

    Distribution Point

    3. The following matrix gives the productive capacity, the maximum units that can be produced per week, of a manufacuring company producing two goods A and B, each of which requires stamping, assembly and painting operations.

    Productive Capacity (Units / Week)

    4. The following matrix gives the distance ( in kms) by train, berween four metropolitan cities of India.

    5. The following matrix gives the input requirements (in rupee) of each industry from other industries (including itself) to produce a rupee worth of output

  • Input Receiving Industry

    SPECIAL TYPES OF MATRICES

    (i) Rectangular Matrix

    A matrix consisting of m rows and n columns, where m n. is called a rectangular matrix, For example,

    (ii) Square Matrix

    When the number or rows of a matrix is equal to its number of columns, it is said to be a square matrix.

    (iii) Row matrix

    A matrix having only one raw is called a row matrix or row vector. For example. [4 1 2 7] is a 1 x 4 matrix or row matrix having 4 elements.

    (iv) Column Matrix

    A matrix ha\wg only one column is called a column matrix or column vector.

    For Example,

    is 3 x 1 matrix or column or column matrix having 3 elements

    (v) Diagonal Matrix

    A square matrix a = (aij) n x n is said to be diagonal matrix if aij= 0 for I=j;

    The elements aij of matrix A. for I = j are called the diagonal elements and the line along which they lie is called the principal diagonal.

    (vi) Scalar Matrix

    A diagonal matrix in which all its diagonal elements are equal, is called a scalar matrix.

    The matrix T = is a 3 x 3 scalar matrix

    (vii) Identity (or Unit) Matrix

    A scalar matrix in which all its diagonal elements are unity, is called an Identify matrix,

    The Matrix I =

  • is 3 x 3 identiy matrix

    (viii) Null (or Zero) Matrix

    A Matrix (square or rectungular) having all its elements equal to zero, is called a null matrix.

    The matrix 0 =

    is a 2 x 3 null matrix

    (ix) Triangular Matrix

    A triangular matrix can be : (a) Upper triangular or (b) Lower triangular.

    (a) A square matrix A = (aij) is said to be upper triangular if aij = 0 for I > j.

    (b) A square matrix A = (aij) is said to be lower triangular if aij = 0 for I < j.

    Remarks: 1. A diagonal matrix is both, an upper and lower triangular.

    2. If the diagonal elements of a triangular matrix arc all zero, it is said to be strictly triangular.

    MATRIX OPERATIONS

    Equality of Matrices

    Two matrices, say A and B, are said to be equal if (a) they are of same order, and (b) the elements in corresponding positions of the two are same. Thus two matrices A = (aij) and B = (bij) can be equal if aij= bij for all values of I= 1 to m and j = 1 to n.

    Remarks: 1.

    cannot be considered for equality because their dimensions are different.

    2. If A = a b and X = 3 4 and it is given that A = x. then it implies that a = 3, b = 4, c = 2 and d = 5

    3. Two matrices A and B are said to be comparable if they are of same order.

    Scalar Multiplication

    The multiplication of a matrix A by a scalar k implies the multiplication of every element of A by k.

    EXAMPLE 1

    SOLUTION

    Remarks: if k = -1, then (-) A = -A (-2 - 7) is known as negative Matrix of A

  • EXAMPLE 2

    Cars and Jeeps are produced in two manufacturing units, M1 and M2 of a company. It is known that unit M1 manufacnlres 10 cars and 5 jeeps per day and unit M2 manufactures 8 cars and 9 jeeps per day. Write the above in formation in a matrix form. Multiply this by 2 and explain its meaning.

    SOLUTION

    Let A denote the required matrix. Let first row of A denote the output of M1 and the second row denote the output of M2. Further, let first column represent the number of cars and the second column the number of jeeps.

    Addition of Matrices

    Two matrices A and B can be added only if they are comparable (i.e. of same order). Their sum is a matrix C, defined as C = A + B = ( aaij + bij ) m x n

    Remarks

    1. The matrix C = A + B is obtained by adding the elements in corresponding places of A and B. 2. The order of C is same as that of A and B. 3. The subtraction of B from A and be defined as A+ (-1) B

    EXAMPLE 3

    SOLUTION

    Example 4

    SOLUTION

    EXAMPLE 5

    SOLUTION

  • Thus X11 = -7, X12 = -12, X21 = -7, etc Properties of Matrix Addition

    (a) Matrix Addition is Commutative

    If A and 8 are two matrices of the same order, then A + B = B + A. As in the case of scalars, this property implies that it is immaterial whether matrix B is added to A or A is added to B.

    (b) Matrix Addition is Associative

    If A, B and C are three comparable matrices, then A -(B+C) = (A + B) + C = B + (A + C).

    According to this property, to add three matrices, we can first add any two of them and then add this result to the third matrix.

    (c) Scalar Multiplication is Distributive over Matrix addition

    Given a scalar k and matrices A and B of the same order, we can write k(A + B) = kA + kB

    (d) Existence of Additive Identity

    The null matrix 0 is said to be the additive identity of any matrix A. of the same order, because A +) = A, i.e. the identity of A does not change when 0 is added to it.

    (e) Existence of Additive lnverse

    If A and B are two comparable matrices such that A + B = 0, where 0 is a null matrix of the same order as that of A or B, then B is said to be additive inverse of a and vice-versa. Obviously, B = -A, i.e. the negative of a matrix is its additive inverse.

    (f) Matrix Addition admits Cancellation Law

    Matrix Addition, like scalars, admits cancellation law. If A, B and C are three comparable matrices such that A+ C = B + C, then A = B.

    Multiplication of Two Matrices

    Two matrices A and B are said to be conformable for the product A B if the number of columns of A is equal to the number of rows of B. If A = (aij)mxn and B = (bjk)nxp then the product AB is a matrix C of order m x p such that

    Note: If AB is defined then it is not necessary that BA will also be defined.

    EXAMPLE 6

  • SOLUTION

    EXAMPLE 7

    and B = (pqr) can you find A B and B A ? if yes, find the two products two products

    SOLUTION

    Since the number of columns of A = number of rows of B, therefore AB is defined. On the similar argument, BA is also defined.

    Inner Product

    The inner product (or dot product) of these vectors, denoted by X.Y (read as X dot Y) is given by

    The essential requirement for the inner product is that the two vectors must have same dirnentions. However, both can be column (or row) vectors or one a row vector and the other a column vector.

    EXAMPLE 8

    SOLUTION

    The inner product is CD = 3 x 1 + 4 x 0 - 2 x 5 = -7

    EXAMPLE 9

    A firm produces three products A, B and C which it sells in two markets. Annual sales in units are given below:

    In the prices per unit of A, B and C are Rs.2.50 Rs. 1.25 and Rs.1.50 and the costs per unit are Rs. 1.70 Rs.1.20 and Rs.0.80 respectively. Find the totaIprofit in each market by using matrix algebra.

    SOLUTION

    Further, let P and C be the matrices of prices and costs respectively, Thus We can write

    Note : P and C can also be written as row matrices.

    The total revenue (TR) and Total cost (TC) matrices are given by

  • Hence profits from markets I and II are Rs. 17.800 and 12,800 respectively.

    Note: Alternatively, we can write profit matrix = Q (P-C)

    Properties of Matrix Multiplication.

    (a) Matrix Multiplication is not, in general, Commutative

    If A and B are two matrices confomable for the products AB and BA, then in general AB BA.

    In view of this fact, the terms premultiply and postmultiply are often used to specify the order of multiplication. For example, B A can be obtained either by premultiplication of A by B or by post multiplication of B by A. Matrix multiplication is, however, commutative under the following exceptional circumstances:

    i. When one matrix is square and the other an identity matrix of the same order. ii. When one matrix is inverse ( see section 3.6) of the other.

    (b) Matrix Multiplication is Associative

    If A, B and C are three matrices such that conformability conditions for the product ABC are satisfied, then

    ABC = A(BC) = (AB)C

    (c) Matrix Multiplication is Distributive over Addition

    The distributive law is A(B + C) = AB = AC, provided that the conformability conditions for adilition as well as for multiplication are satisfied.

    (d) Existence of Multiplicative identity Matrix

    The identity matrix I is said to be the multiplicative identity matrix or simply the identity matrix of any matrix, say A of the same order, because IA = AI = A i.e. the identity of A remains unchanged as a result of its multiplication of I.

    (e) Existence of Multiplicative Inverse.

    If a and b are two scalars, then they are said to be inverse of each other if ab = ba = 1. In a similar way. two square matrices, say A and B. of the same order. are said to be multiplicative inverse of each other if AB = BA = I.

    (f) Matrix Multiplication does Dot always admit Cancellation Law:

    According to this law, if AB = AC, then it does not necessarily mean that B = C. To illustrate this, consider

    Thus AB = AC while B C

    (g) The product of two Matrices can be a Null Matrix with none of them being Null Matrix.

    Note

    i. If AB = 0, then it is not necessary that BA will also be a null matrix.

  • ii. Properties (f) and (g) hold true only for singular matrices (See section 3,4)

    (g) Positive Integral Powers of Matrices

    If A is any square matrix, then the product A.A is denoted by A2. Further, we can write A

    2.A = A

    3 etc.

    We should note that, like real numbers, we have An.A

    m = A

    m.A

    n = A

    m+n for matrices

    Note: (i) Using the above equation it is possible to find some matrices which commute with respect to multiplication. If we write B = An and C = Am then obviously, BC = CB

    (ii) Matrix A is said to be idempotent if A= A2 =------A

    m

    Transpose of a Matrix

    The transpose of a matrix A is a matrix, denoted by A (or A'). obtained by the interchange of its rows and columns. Symbolically,

    if A = (aij)mxn then A = (aji)nxm

    EXAMPLE 10

    Properties of Transpose

    i. The transpose of the transposed matrix is the original matrix i.e. (A1)1=A

    ii. The transpose of the sum of two ( or more ) matrices is equal to the sum of transposed matrices i.e (A+B)' = A' + B'

    Thus (A + B)' = A' + B'

    iii. The transpose of the product of two ( or more) matrices is equal to the product of the transposed matrices in revised order i.e (AB)' = B'A'

    Note (I) A matrix A with the property that a = A', is blown as a symmetric matrix. (ii) A matrix A with the property that A = -A. is known as skew symmetric matrix. DETERMINANTS AND NON -SINGULARITY

    Let X, Y and Z be in -dimensional (row or column) vectors. If we can write W = aX + bY + cZ, where a, b and c are scalars. then W is said to be a linear combinabon of the vector X, Y and Z or in other words the set of vectors X, Y, Z and W are said to be linearly dependent. Contrary to this, if none of these vectors can be expressed as a linear combination of the others. the vectors are said to be linearly independent. The condition for Linear Independence.

    A set of vectors V1,V2, ---Vm is said to be linearly independent if for scalars k1,k2,--- km3 the linear combination k1 V1 + k2 V2 + ---- + km Vm = 0 only when all k1 ( I = 1 to m) = 0.

    Non-Singular Matrix

    A square matrix consisting of linearly independent rows ( or columns) is said to be non- singular.

  • Let us examine the linear independence of the rows of matrix

    To examine the linear independence, we have to find the values of two scalars k1 and k2 such that

    This gives the following system of simultaneous equations: 3k1 - k2 = 0 -2k1 = 3k1 = 0 On solving these equations simultaneously, we get k1 - k2 = 0 This, the rows of matrix A are linearly independent. Further, since A is a square matrix, with linearly independent rows, it is said to be non-singular. Note : A square matrix with linearly dependent rows (or columns) is said to be a singular matrix. The above method of examining linear independence of rows (or columns) becomes very cumbersome when the number of rows (or colunms) becomes large and each rows (or column) contains several elements. Alternatively, an easier method is provided by determinants. The concept of a determinant is discussed below. Determinant

    A determinant is a uniquely defined scalar associated with a square matrix. The determinant of a square matrix A = (a ij)nxn is denoted as det A or (a') or

    Determinants of Order one

    If a matrix A consists of only one element i.e. A = (a11)' its determinant is defined as (a11). It should be noted here that the concept of determinant is entirely different from the concept of absolute value although the same symbol is used to denote both of them. For example det (-4) = (-4) = -4, however, absolute value of -4, also denoted as (- 4) is 4. Determinants of Order two

    If A = is a 2 x 2 matrix, its determinant is defined as |A| = = a11.a22.a12.a21

    Determinants of Order Three or Higher

    Before giving a rule for the evaluation of this determinant, we introduce the concept of Minors and Cofactors

    Minors

    A minor of an element aij denoted by Mij is a sub-determinant of (A) obtained by determine its ith row and jth column .

  • Cofactor

    A Cofactor of an element aij, denotes by cij, is its minor with appropriate sign Thus, we can write Cij = (-1) I + j M ij

    We note that C11 = M 11 but C12 M12 etc.

    Definitions

    The determinant of any matrix of order 3 x 3 or higher is defined as the inner product for dot product of any row (or column) of that matrix by the corresponding row (or column) of that matrix by the corresponding row (or column) of cofactors. It can be shown that the result is independent of the choice of the row (or column)

    Thus we can write

    EXAMPLE 11

    Evaluate the following determinants

    SOLUTION

    1. Writing the inner product of the element of first row and their co-factors we have

    ii. Since the third row contains two elements as zero, the calculation work will be minimum if we write the determinant as the inner product of the elements of third row and their cofactors

    Sarrus Diagram

    An alternative method, that is very convenient for writing the value of a 3 x 3 determinant is provided by Sarrus Diagram

    This diagram consists of 5 Columns in whic first and second columns of (A) are written as the fourth and fifth columns respectively, as shown below

    The elements in the indicated directions are multiplied. From the diagram we can write.

    On the basis of this rule, we can write the value of determinant given in example 11 (I) as

    Properties of Determinants

    1. The value of a determinant is not affected by the interchange of its rows and columns.

  • 2. The interchange of any two rows ( or columns) of a determinant alters its value by a multiple of - 1.

    Interchanging the first and second row of | A |considered in property (1) above we have

    3. If all the elements of a row ( or colwnn) are multiplied by a scalar k, the determinant thus obtained is k times the original determinant.

    by a scalar k, we get

    4. The addition (or subtraction) of a multiple of any row (or column) to another row (or column) does not affect the value of the determinant.

    5. If any row (or column) is a linear combination of one or more rows (or columns ) or if two or more rows (or columns) are identical, the determinant vanishes.

    Consider a determinant in which first and the second columns are identical i.e.

    Remarks

    This property applies that if two or more rows of a matrix are linearly dependent, its detemlinant will be zero. Thus, to determine whether a matrix is singular or not, we simply have to evaluate its determinant. Since determinants are defined only for square matrices, therefore we can say that a matrix is non-singular if its determinant is non-zero.

    Rank of Matrix

    If the determinant of a matrix is zero, it only indicates that its rows (or colwnns) are not linearly independent. When all the rows (or columns) are not linearly independent, we are often interested to know, how many rows (or columns) out of the total number of rows (or columns) are linearly independent? To answer this question, we introduce the concept of rank of a matrix. Rank of a matrix A is the maximum number of linearly independent rows (or columns) in it. If A is m x n matrix, then its maximum rank can be m or n whichever is smaller. The maximwn possible rank of an n x n matrix can be n. We may note here that if the rank of an n x n matrix is n, then it must be nonsingular. Determination of Rank.

    (a) By the use of Determinants The rank of an m x n matrix A is said to be r, if every minor of order greater than r is zero and there is at least one minor of order r which is non-zero.

    EXAMPLE 12

    SOLUTION

  • The highest order of the minor is 3. Therefore, we first compute | A |

    This implies that there is at least one row ( or column) which is linearly dependent and hence rank of A, to be denoted as p (A), is less than 3. Further, we consider minors or order 2 x 2. Let us consider the minor

    Example 13

    SOLUTION

    Note that A is a 3 x 4 matrix, therefore its maximum possible rank can be 3. To determine whether p (A) is 3 or not, we find whether there exists at least one minor of order 3 that is non-zero.

    Further, consider another minor of order 3.

    SOLUTION

    First we examine the minors of order 3

    Since all the four possible minors of order 3 are zero, the rank of A is less than 3. We now consider minors of order 2.

    Hence p (A) = 2. Note 1. The rank of an identity matrix In is n 2. The rank of a null mab-ix is zero 3. The rank of a matrix whose every element is a (= 0) is unity 4. P (A) = P (A')

    (b) By Elementary Row Operations

    It is obvious from example 14 that the method of finding rank of a matrix, by the use of determinants, may become very cumbersome. Alternatively, the rank of a matrix can be obtained by reducing it to an Echelon form (a matrix whose all elements below its diagonal are zero)

  • and then the rank is given by the number of non-zero rows in its echelon form.

    To obtain an echelon form of a matrix, we introduce the concept of Equivalent Matrices.

    Two matrices A and B are said to be equivalent if one can be obtained from the other by a sequence of elementary row (or column) transformations.

    An elementary row ( or column) transformation can be anyone of the following transformations:

    (i) The interchange of ith and jib rows (or columns). This transformation is denoted by Rij (or Cij) (ii) the multiplication of ith row ( or column) by a non-zero scalar k, to be denoted as Rs = kR1 (or C1 = kC1). (iii) The addition (or substraction) of k times the jth row (or column) to ith row (or colunm), to be denoted as Rl = R1 + kRj (or C1 = C1 + kC1) We state, without proof, that Equivalent Matrices have the same rank.

    Using elementary row operations, a matrix can be reduced to an echelon form. The rank of the matrix is then given by the number of non-zero rows In Its echelon

    Example 15

    Solutions

    Since tbe number of non-zero rows is 3, therefore p (A) = 3

    Example 16

    Solutions

    hence p (A) = 2

    Example 17

    Solution

  • (ii) To get a non zeo diagnal element at the intersectionof second row and second column, apply R24 and write

    Thus p(B) = 2

    Consistency of a System of Equations

    Using matrix algebra, it can be written as the following matrix equation:

    The condition for consistency of the above system can be expressed by using the concept of rank of a matrix. The above system of equations is said to be consistent if p (A) = P (A | B). where A I B (read as A augmented B) is a matrix obtained by augmenting A by B as shown below:

    Case 1: If A is a square mabix- with | A | o Then the rank condition for consistency will always be satisfied. Since p (A) will be equal to the number of rows ( or columns), the addition of another column to A wil not alter the rank of A I B. Further, since all the rows are linearly independent when I A | = 0, tbe system is consistent and independent and has a unique solurion ( refer to sec. 2.4.) Case 2 : If p (A) = P (A I B) and I A | = 0, the system is said to be consistent and dependent. Such a system has infinite number of solutions. (refer to Sec 2.4.) .

    Summary

    This blocks attempts to introduce the concepts of matrices and determinants and their importance in solving real world problems of business. While a matrix is an array of numbers arranged into certain number of rows and columns, a determinant is a scalar associated with a square matrix. Unlike scalars, the basic operations such as addition, subtraction and multiplication can be performed only if certain conditions are satisfied by the participating matrices. Like scalars, division of one matrix by another is not defined. Using matrices, we can write a system of linear equations in a compact form, test their consistency and solve them in an efficient manner.

    Review Question

    1. What are the dimensions of the following matrices:

    2. Indicate whether the following statements are true or false. Give reasons if the statement is false.

    3. Perform the indicated operations. State reasons if an operation is not possible.

  • find A - B, A + B, 4A + 2B, 3A - 3B

    7. Find A B and BA, if possible, for the matrices

    8. An automobile company has two manufacr:wing plants located at Delhi and Pune. It manufactures scooters and motorbikes at each plant. Each vehicle is produced in three models A, B and C. The following two matrces give the number of Vehicles (in thousands) of each model produced in the two plants during 1997.

    i. Write a matrix showing the total production for both in 1997. ii. if the production is increased by 20% in Delhi plant and 10% in Pune plant, writethe matrix for total production for the following year.

    9. Show that, for all values of p, q, r and s, the matrices

    10. if possible, find a matrix X such that

    14. A manufacturer produces three products A, B and C which are sold in Delhi and Calcutta. The annual sales of these products are given by the following matrix

    If the sale price of the products A, B and C per unit be Rs. 2, 3 and 4 respectively, calculate total revenue from each centre by using matrices. 15. A firm uses three ingredients 10 manufacture two products. A and B. The cost (in Rs.) per kg of each ingredient is given by C = (5.0 12.5 15.0). The requirement of each ingredient (in kgs) to produce one unit of each product is shown in the following matrix.

    find the cost per unit of each products

  • Quantitative Techniques in Management

    Chapter 6 : COLLECTION OF DATA

    OBJECTIVES After studying this unit, you should be able to:

    appreciate the need and significance of data collection

    distinguish between primary and secondary data

    know different methods of collecting primary data

    design a suitable questionnaire

    edit the primary data and know the source of secondary data and its use

    understand the concept of census vs. sample.

    INTRODUCTION

    To make a decision in any business situatio