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Paper 1.4: QUANTITATIVE METHODS Unit – 1

Basic Mathematical concepts : Nature of quantitative analysis in the practice of management – problem definition – Models and their development – Concept of trade off – Notion of constants – Variables and function – Linear and Non-linear – Simple examples.

Graphical representation of functions and their application – Concepts of slope and its relevance – Plotting graphs of functions.

Use of functional relationships to understand elasticity of demands. Productive function – Costs of operating a system – Measuring the level of activity of a system in terms of volume – Value and other parameters – Relationship between costs and level of activity – Costs and Profits – Relevance of marginal average and total costs. Importance of “relevant costs” for decisions- making – Break-even analysis and its uses.

Unit – 2

Introduction to the linear programming – Concepts of optimisation – Formulation of different types of linear programming – Duality and Sensitivity analysis for decision-making.

Unit – 3

Solving LP using graphical and simplex method (only simple problems) – Interpreting the solution for decision-making – Other types of linear programming – Transportation – Formulation and solving methods.

Unit – 4

Introduction to the notion of probability – Concepts of events – Probability of events – Joint, conditional and marginal probabilities.

Unit – 5

Introduction to simulation as an aid to decision-making. Illustration through simple examples of discrete event simulation. Emphasis to be on identifying system parameter, variables, measures of performance etc.

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Unit – 6

Introduction to Decision Theory: Pay-off and Loss tables – Expected value of pay-off – Expected value of perfect Formation – Decision Tree approach to choose optimal course of action – Criteria for decision – Mini-max, Maxi-max, Minimising Maximal Regret and their applications.

Course Material Prepared by :

Dr.M.SELVAM, M.Com., M.B.A., Ph.D.,

Professor & Head,

Dept. International Business and Commerce,

Alagappa University, Karaikudi.

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LESSON 1

MATHEMATICAL MODELLING FOR MANAGEMENT DECISION

MAKING Mathematics is the science of magnitude and number, and of all

their relationships. Mathematics is being applied in almost all fields. Management is no exception. In fact there is a kind of mathematicism, that is, everything can be described in terms of mathematical terms i.e., numbers and symbols. True.

1.1 MEASUREMENT AND MATHEMATICS

Yes, the world is moving towards becoming more and more

quantitative. To be quantitative is to be more precise and pointed in what is being said or done. Measurement is fundamental aspect of being quantitative. To measure means, to express the physical, biological, psychological, sociological and other observed phenomena in terms of numbers or symbols such that the phenomena measured bear the same relationship as do the used numbers or symbols have, inter se in terms of magnitude and other relations.

The world of business has many phenomena and all these have to be measured for their management and manipulation. And recourse to mathematics is needed in the pursuit of such measurement. Hence the need for mathematics for management. There is a saying: one's knowledge is one-third if one can only speak on a topic, two-third if one can write in words and 100% if one can express in numbers. Mathematics the queen of sciences, thus has king's size applications in all walks of life.

1.2 MATHEMATICAL OPERATIONS

We know the basic mathematical operations, namely, addition,

subtraction, multiplication and division. But many higher order manipulative mechanisms are available. We have differentiation, integration, programming, simulating, networking, gaming and other techniques which help in business optimisation, i.e., maximising revenue/sales/profit or minimising losses/costs/ inputs, subject to constraints and conditions.

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1.3 MANAGEMENT CONCERNS

Management is concerned with income, expense, assets, liabilities,

productivity, efficiency, profitability, costs, volatility or fluctuation in market, market share, media effectiveness, media reach, brand penetration level, brand equity, effect of pricing strategy on sales and profitability, motivation levels, motivation and productivity nexus, coefficient, leadership styles and productivity nexus coefficient, organisational climate and productivity nexus coefficient, cost of capital, value of firm, capital structure and value of the firm nexus correlation, credit policy and profitability linkage coefficient, social responsibility and profitability linkage coefficient, economic value addition and corporate governance correlation relationship, inventory levels and cost of inventory nexus, competition levels and business prospects nexus, and so on. All these need measurement of the underlying factors to have a greater understanding of individual factors and their relationship, inter se. Management is thus full of measurement and hence draws heavily from the science of numbers, i.e., mathematics.

1.4 QUANTITATIVE ANALYSIS IN THE PRACTICE OF

MANAGEMENT

In the preceding paragraphs a partial list of the concerns of

management was presented. All these concerns require decision making in order to optimise ultimate results for the organisation. Income maximisation, expense minimisation, asset-quality maximisation, liability cost minimisation, productivity maximisation, volatility optimisation, market share maximisation, optimising pricing strategy for sales and profit maximisation and so on are called for. To achieve these diverse objectives of the organisation a scientific approach need to be devised.

There are different alternatives to reach the goals set. The right

alternative has to be chosen. Thus a decision-making situation evolves. This situation can be handled qualitatively as well as quantitatively.

The qualitative approach is heuristic or whimsical or rule-of-thumb

based or intuitive or judgement or trial and error based. The quantitative

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approach is scientific. It is open book. It can be tested for efficiency and improved upon.

Quantitative approach is preferred under the following situation:

- the problem is complex, involving many variables

- the interrelationship is complex and multi-faceted

- the problem is repetitive and that quantitative solutions can be repeated several times thus reducing cost and time.

The quantitative approach is preferred for the following reason:

- the language of quantitative approach to decision making is more concise and precise

- there exists a wealth of mathematical theorems for our use

- forces explicit expression of "ifs & buts"

- enables us to handle any number of variables at a time

- ensured the repeat application through the general n-variable case.

1.5 MODEL OF QUANTITATIVE ANALYSIS

Quantitative analysis is a scientific approach to decision making.

As a first step to decision making, decision model has to be evolved. The decision model depends on two factors, namely the problem and the problem environment.

1.5.1 Problem Definition

The first step in decisions making is defining the problem. The

problem (i.e. the threat, opportunity, etc.) must be fully understood as to its nature, dimensions, intensity and so on. Take labour absenteeism. It becomes a problem when it is rampant affecting work schedules. (one or two absentees, here and there is no problem). What is its nature? Deliberate absenteeism,

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absenteeism due to unavoidable causes, absenteeism as a mark of protest to certain managements attitudes/actions, absenteeism due to ‘take-it-easy’ way of life of employees, absenteeism due to healthy/family/social reasons, etc. are indicative of the nature of the problem of absenteeism. Specific nature must be understood, since each type of absenteeism needs a different approach to solving the same. Dimensions of absenteeism comes next. How does it affect production? How does it affect inter-departmental relations, work flow, co-ordination, morale etc? What is its impact on turnover and customer satisfaction? What is the scale of idle capacity resulting from labour absenteeism? Will motivational/coercive courses work in containing the absenteeism? There are but a few dimensions of the problem. How much grave the problem is? Should it be solved immediately? These indicate the intensity of the problem. The wholesome knowledge of the problems is very essential so that we can solve the problem. Not all problems can be known fully. Some keep change from time to time. So, to the extent possible the decision maker must grasp the problem. This step/function is also called as the ‘intelligence’ function of decision making. All these help to recognize the problem. Problem definition goes a little for.

In the given situation what is your objective? The decision maker

must set the decision objectives. What constitutes an effective solution? The answer to this question is what is dealt by decision objectives. Reduce the level of absenteeism from the present five present of scheduled work-hours to two percent could be one of the objectives. Let us deal with deliberate absenteeism and protest absenteeism at the moment, could be another objective. That is, which part of the problem must be solved now is one of the decision objectives. Let us adopt motivational approaches to solve the deliberate absenteeism could be another objective. Protest absenteeism must be resolved by hook-crook approach, otherwise the management may be weakened-this could be another objective. We must settle things within a span of a year-a time frame or setting is another objective. Objectives give directions to possible courses of actions. And that they lead to developing alternative solutions that satisfied some broad decisions parameters with objective formulation, the problem stands defined.

Take another example. There is an opportunity of diversification.

If the opportunity of diversification is seized, backward and forward integration is possible; competition can be held at bay. But, how to get the necessary resources to take up the opportunity. Acquisition or "startup from scratch",

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internal and external resources, debt and equity capital etc are the decision issues. In what time frame this must be achieved? What if competitors try to copy or thwart our efforts? All those make problem complex.

The problem may be a threat. The launch of new designs by

competitors is eating up your market space leading to declining top line and hence bottom line. How to deal with the problem? Should the firm also launch new models? But that will take some time. Should it gear up promotion? Should it initiate a price war? How to react to competitive price war, if that results eventually?

1.5.2 Problem Situation or Environment

Decision situation or environment refers to situational factors that

contribute to the problem, the situational factors that influence the implementation of solution to the problem and the situational factors that affect the choice of solution and so on.

The problem may be dynamic or static in nature. Dynamic problems keep changing from time to time and place to place. Many variables affect the problem and the decision maker has no control over these. Besides a majority of these variables are external to the organization. Competitive factors, consumer attitudes, technological factors, political and legal environment etc., are continually changing and constantly influencing business decisions. Capital expenditure decision, long-run product and marketing decisions, etc. are typical examples. On the other hand, some decision situations are static. The variables are all known and can be manipulated by the decision maker. Inventory decisions, employee compensation decisions, passenger seat-reservation-cum-cancellation etc., are certain examples where the closed decision system can be used with advantage. Knowledge about outcomes of a course of action is yet another factor constituting the decision situation. The outcome of certain courses of action can be forecast accurately, while that of others cannot be. Forecast of

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return from Government bonds, lease rentals, etc., can be done with 100 per cent attitude, while that of equity investments cannot be made accurately. While this frame work of decision making is common to all decision situations, the practice of decision making varies from case to case, depending on the decision environment. Therefore, decision models have to be so developed that they suit the different decision environment conditions. The decision environment may be complex, volatile and less amendable to manipulation or be simple, static and more amendable to manipulation. Decision situations with certain environment require closed decision model and with volatile environment require open decision model. The two decision models are however, the extremes of a continuum. Mixed decision models are, therefore, more realistic in nature. However, an understanding of the open and closed decision models is very essential in developing a mixed decision system. A closed decision model has a defined environment and known components. Such a model can be modeled and its functioning programmed. The model variables are all known and no random variable enters the model. The model strives to obtain the optimum outcome. As on approach to business decision making, the closed decision model has only limited us, since a closed decision model cannot function effectively in dynamic business environment. However there are certain routine decision situations where a closed decision model can be used. An open decision model has on environment that is generally less defined. The decision maker is not fully aware of the environment he is confronted with. The variables are too many to incorporate in the model. The exact nature of the relationship among variables is not fully known. Random and unpredictable variables may enter into the system and effect the outcome. Under such conditions, the decision maker makes manageable delimitation of the problem. The model goal, variables, alternative courses of actions, etc., are defined by the decision maker to the best of his perception and knowledge. Such limited search is referred to as bounded rationality. The open model has a tendency to adopt to changing environment and, therefore, is increasingly

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relevant today as a business needs to adapt on a continuous basis in this dynamic world.

1.6 DECISION MODEL

A decision model is an idealised representation of the problem. Decision model refers to structured presentation of the problem, solution there to and simulation of working of the solution. The mode's purpose is to enable the decision analyst to forecast the effect of factors crucial to the solution of the problem.

1.6.1 Types of Models

There are different types of models. Iconic and Symbolic models are the prime two types. Iconic Models are concretized. It is a physical representation of any real life object on a different scale. Think of a prototype of a plane/car/machine/globe/idol and so on. Symbolic models are abstract models. A cost curve, a supply curve, a marginal revenue curve, a production possibility curve, etc., is a symbolic model. A forecast profit and loss account is also symbolic model. The statement symbolizes summary of financial effects of commercial activities planned over the next period. An equation is a symbolic model of the variables, their relationship and the effect of independent variables on the dependent variable.

Symbolic models can be classified into: i) quantitative and

qualitative, ii) standard and customized, iii) probabilistic and deterministic, iv) descriptive and optimising, v) static and dynamic simulative and realistic models.

In a quantitative model all variables are expressed in numbers,

while in a qualitative model some or all of the variables are given only verbal expression.

Standard model is universalized. Computer languages and

operating systems are standardized, while application programmes are customized. Nowadays customized-standardisation is the order, powered by enormous computing powers of computers. Mass customization is preferred.

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Probabilistic model deals with situations where the outcomes of current action are not known, but their probability distribution is known. Deterministic model deals with decision situations where certainty of outcomes is taken for granted.

Descriptive model describes the basic relationship between

variables. If Rs. 10 and Rs. 12 are the contribution per unit of Product A and Product B, respectively and if a units of A and b units of B are produced, total contribution is described by: 10a + 12b. In an optimising model, ways to maximise the total contribution, given the resources and consumption pattern of resources by A and B.

Static model assumes, the set of conditions affecting the problem

remains unchanged in a given time frame. Linear programming is a static model. Dynamic model assumes that the conditions keep changing with time. Dynamic programming is a dynamic model.

Simulative model tries to replicate the real world situation either

through a computer programme or through Monte Carlo Simulation System or through such other methods. A realistic model is action play. An advertisement campaign is full scale on. The firm tries to evaluate the sales profit, awareness and brand equity effectiveness of the programme.

1.6.2 Model Construction

The construct a model, one needs to know the variables, constants and constraints. A verbal representation of the relationships among variables is then attempted. Next, a model is constructed by developing a symbolic statement of the relationships among variables and then these symbolic relationships are turned into a mathematical model which is tested with real-world data and evaluated. But all decision situations do not lend themselves to modeling. In the case of decision situations that requires an open system approach, accurate modeling is utopian, because the decision components are not fully known. Hence, open decision systems are not perfectly modeled or structured. Depending on individual situations, models vary from partly structured to totally unstructured. In its partly structured from, the decision

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model is an approximation of the real world situation. Consider the case of sales forecasting, which is a typical dynamic decision exercise. Sales response constant – ‘r’ advertisement effort as a function of time – A(t), sales saturation level - ‘M’, sales rate per time – ‘S’, and sales decay constant - ‘Z’, are some of the relevant variables identified. A simple model is developed using these variables which runs as follows: ds/dt = r.A(t).(M-S) M-ZS, where ds/dt is the rate of change in sales per unit time. Many variables would affect the rate of sales, but a few only are considered important here. Thus, the above model is only an approximation.

In certain cases of open decisions models, no modeling, however much an approximation, is possible. In such cases the system remains unstructured. On the contrary, closed decision models are fully structured for reasons best known. Even here, constructing a workable model may prove difficult in some cases. The simulation technique is adopted then to develop the model.

In the case of open decision making models, the environment is

risky and stochastic and in such an environment as outcomes of decisions cannot be known with certainty, the objective criterion is one of satisfying. In regard to closed systems, on the other hand, the functioning of the system is highly structured and that optimization is the ultimate goal.

1.6.3 Constructing a Mathematical Decision Model

Mathematical model is an idealized representation expressed in mathematical symbols and expressions. A mathematical model of a business problem might be in the form of a set of equations and related mathematical expressions that describe the essence of the problem. An economic order

quantity model is given by: EOQ = √2AO/C where A - annual requirement, O - ordering cost and C - carrying cost. A linear programming model is given by objective function: Say Maximize Z = 10a + 12b, subject to 2a + b < = 60, 3a + 4b < = 120, a, b > = 0, where a and b are units of products A and B, respectively to be produced to maximise total contribution given individual contribution of Rs. 10 per unit of A and Rs. 12 per unit of B, the resource constraints being that resource 1 and 2 are available respectively to the extent of 60 and 120 units only.

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An internal rate of return (IRR) model is given by: -I + CFt/(1+k)t = 0, where `k' is the IRR to be found, while `I' is the initial investment, CFt refers to periodic cash flows. To construct a mathematical model objective of the firm, variables, constants and constraints must be known, besides the relationship amongst the variables. The relationship may be linear or non-linear. The EOQ and linear programming models given above are linear relationship based, while the IRR model is non-linear relationship based as it in exponential form.

1.6.4 Variables in Mathematical Models

Variables are something whose magnitude can change. These can take different values. Price, profit, revenue, cost, quantity produced, quantity sold, imports, exports, income tax, excise duty, national income, savings, consumption, investment, inflation rate, cost of living, etc., are all variables and can take different values entity to entity, time to time, place to place and so on.

While variables can take any value, we may give them particular

values and `freeze' them in a given context. Variables may be i) dependent or independent, ii) stochastic,

probabilistic or deterministic, iii) endogenous or exogenous, iv) continuous or discrete, v) slack or surplus, vi) choice or random, vii) real or artificial, viii) non-negative or unrestricted, ix) integer or non-integer variables.

Let E = (P-V)Q - F - D - I, where, E - earnings, P - price per unit,

V - variable cost per unit, Q - quantity produced and sold, F - fixed cost for the period, D - depreciation for the period and I - interest on borrowed capital. Here, `E' is the dependent variable and all those on the right side of the equation are independent variables. This is a profit model. The set of all permissible values that the independent variables can take in a given context is known as domain of the profit function and the values of E, i.e., profit for the domain values are called the range of the function.

A stochastic variable takes any value and its value cannot be

predicted at all. The rate of change in sales per time is a stochastic variable. A probabilistic variable takes values according to a given probability distribution. The number of hits that a web-site gets per unit time is perhaps Poisson

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distribution based. A deterministic variable gets a value that is known beforehand.

Endogenous variables originate from within, say the organisation.

The `Q', `D' F and V in the profit function are mostly endogenous, i.e., emanating from inside the organisation. But, P (assumed to be competition triggered) and I (assumed to be money supply triggered) are exogenous, emanating from outside the environment.

Continuous variables take any value in a continuum, fractional,

integer in the continuum. In normal probability distribution the variable is assumed to be continuous, while in binomial distribution the variable is discrete, i.e., the variable `jumps', not `moves'.

Slack variable is introduced to an inequality of the "< =" type on

the left hand side to convert the same into an equality. A stack thus represents a shortfall being met. Against this, a surplus variable is the excess of actual over minimum expected. In respect of a cheaper input, a surplus usage and in respect of a dearer output a surplus achievement are generally worked out, if possible.

Choice variables, alternatively referred to decision or policy

variables are the variables whose magnitude the organisation can pick and choose. Against this, the magnitude of random variables is not in the hands of the organisation. Choice variables can be also called controllable as to magnitude, while random variables are uncontrollable.

Real variables are variable that enter the solution set and exist

there. Artificial variables are mathematical artifice and are involved to find a solution, but cannot be part of the solution.

Non-negative variables can take only positive values or zero. They

can't take negative magnitude. These are referred to a, b > = 0. There cannot be a situation of -5 units of product A or -3 units of product B being produced. Unrestricted variables can take positive and/or negative values.

Integer variables can take only whole number values while non-

integer variables can take any value. Number of chairs/ships produced cannot

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take fractional value, but amount of time, consumed can take fractional days/hours.

In constructing a mathematical model all the variables and their

nature must be known.

1.6.5 Constants in Mathematical Models

A constant can take only a specified magnitude. Hence a constant is the antithesis of a variable. A constant when added to a variable, it is called coefficient of that variable. However, a coefficient may be symbolic instead of numeric. Suppose let the symbol ‘C’ stand for a given constant and the use of the expression C in lieu of 6C in a model is perfectly all right and this expression permits greater level of generality. The symbol is a rather a peculiar case. It is supposed to represent a given constant. Yet since it is not assigned a specific numeric value, it can virtually take any value. In other words, it is a `constant' that is `variable'!! Such constants are known as parametric constants or parameters.

Take the famous equation of Einstein: E = MC2. Here, E - energy

produced, M - mass of the object and C - velocity of light. The velocity of light, C, is a constant. In the Poisson distribution we use. `e' is a constant, whose value is corrected to four decimals at 2.7183. The above two constants are classical constants, never change. These are non-changing constants. But, the parametric constants are constants at the given time and space. If any of these is changed, the parameter also changes. Thus, a parametric constant changes with time and space.

Total cost of production for a given level of output, can be

expressed as sum of fixed cost + Total variable cost. Let the output level be, Q, fixed cost, F, Rs. 10,00,000 and variable cost, V, Rs. 5000 per unit.

Then Total Cost = TC = F + VQ or TC(Q) = 10,00,000 + 5000Q Here, total cost varies as `Q' changes. The fixed cost in total is a

constant and the unit variable cost is also a constant. But, these constants are not universal constants. For the time being these are constants. Fixed cost may

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change as rent, administrative and depreciation change and unit variable cost will change if costs of direct material and labour change. Thus there are two constants. Universal constants and temporal constants.

1.6.6 Enrichment and Evaluation of Model

The model must be enriched and evaluated. Usually, to begin with a mathematical model is a simple version or an evolutionary design. Step by step the model is enriched through incorporation of additional relevant elements of (variables and constants) and conditions and constraints. But too many variables, constants, conditions and constraints might make model intractable (i.e., incapable of being solved) and imprecise. Enrichment must be attempted to the extent of achieving a valid representation of the problem.

The validity of the model must be evaluated. Content, context,

construct, concurrent, predictive, convergent, divergent, nomological validities of the model must be evaluated. Criterion for judging the validity of the model is whether or not the model measures what it is supposed to measure, predicts the effects of the alternative courses of action with sufficient degree of accuracy and so on.

Finally, reliability of the model must be evaluated. A retrospective test may be performed here. The test involves using historical data to reconstruct the past and then determining how will the model and resulting solution would have performed if it had been used. If the results of the model mostly conform with the ex-post results, the model can be taken to possess reliability. Further, with repeated applications, if the same results are turned out, the reliability of the model can be taken as established.

1.7 TRADE-OFF

A mathematical model of a business problem aims to achieve the maximum gain with minimum pain. This is essentially an exercise of trade-off. Risk-Return trade off, Cost-benefit trade-off, time-cost trade-off and so on are involved.

In Risk-Return trade-off, risk is traded off for return. High risk and

high return and low risk and low return are the order. But a business man wants

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high return with low risk. How is it possible? Impossible the businessman opts for certain level of risk and tries to maximise return for that level of risk or targets a return and attempts to minimize risk as far as possible for that level of return. This process is called risk-return trade-off.

Similarly, higher benefit involves higher cost and vice versa.

Everyone wants high benefit, but low cost. This is impossible. So, one must decide how much benefit is wanted and try to reach that level with the least possible cost.

In the case of time-cost trade-off, a project leader wants to

complete a project ahead of normal time. Thus he wants a benefit. But this is accompanied by a cost, cost of spending up the work. Thus, every gain has a price-tag. The decision maker has to decide how much cost per unit or utility is optimum. And he settles for that optimum level of risk-return trade-off, cost-benefit trade-off and time-cost trade off and so on.

Below are given diagrams regarding trade-off. B C Total

Risk direct Cost

A D Return Duration time

In the risk-return trade-off, curve AB shows the locus of risk and

return. For a unit rise in return, how much rise in risk takes place? This answer is given by the slope of the curve. And the slope of the curve is not same throughout the length AB, as AB is not a straight line, but a curve. But, successive unit rise in return trade-off with increasing level of risk. The decision maker has to decide the maximum risk per unit rise in return he is prepared to

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trade-off. When that level of return is reached, the choice of risk-return combination is made.

In the time-cost trade-off, initially, for a unit time saved, addition

to cost is small. For successive units of time saved further and further, a higher and further higher addition to total cost is involved. The decision-maker has to decide how much maximum additional cost he can afford per unit time saved. Until such level is reached, project duration is crashed.

Questions

1. Present the application of quantitative analysis in the practice of management.

2. Explain problem definition and problem environment and synthesizing the two.

3. What do you mean by decision model? Explain the types of models and their info needs.

4. How do you construct a model? How do variables and constants figure in the model?

5. Explain the concept and types of variables and constants.

6. What is trade-off? How is the same relevant to decision making? How is it effected?

� � �

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LESSON 2

MATHEMATICAL FUNCTIONS AND THEIR APPLICATIONS

A function is called a mapping, or transformation; both words

connote the action of associating one thing with another. A function is a set of ordered pairs with the property that any particular magnitude of independent variable, say x, uniquely determines the value of the dependent variable, say y,. Thus, a function implies a unique value y for each x. The converse may or may not be true.

2.1 FUNCTION IN THE FORM OF EQUATION

Function can be presented in the form an equation as below: y = f(x), where y is the dependent variable and x is the independent variable. [Note, y = f(x), when y is a function of x, and it does not mean f times x].

x is referred to as the argument of the function and y is called the

value of the function. The set of all permissible values that x can take is known as the domain of the function. The y value with which an x value is mapped is called the image of that x value. The set of all images is called the range of the function, which is the set of all values that y can take.

Let the total cost (C) function of a firm per day is associated with

daily output Q; C = 1500 + 80Q. The firm has a capacity limit of 100 units of output per day. Then the domain of the function is the set of values: 0 = < Q = < 100. The range is lowest at 1500 when Q = 0 and highest at 9500 when Q = 100; or 1500 = < C = < 9500.

By placing the domain on the x-axis and the range on the y-axis,

we get the function in the form of a two-dimensional graph in which the association between x values and y values is specified by a set of ordered pairs such as, (x1, y1), (x2, y2) ... (xn, yn). C = 1500 when Q = 0 and is 9500 when Q = 100, or 1500 = < C = < 9500.

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2.2 DIFFERENT FUNCTIONS AND THEIR GRAPHIC EXPRESSIONS

The expression y = f(x) is a general statement of a function. The actual mapping is not explicit here.

Constant functions, polynomial functions and relation functions

are one class of function.

2.2.1 Constant Function takes only one value as its range. y = f(x) = 7; or y = 7; or f(x) = 7. Regardless the value of x, value of y = 7. Value of y is, perhaps exogenously determined. Such a function, in the coordinate plane, will appear as a horizontal straight line. Graph 2.1 gives the constant function.

Graph 2.1

y7 y = f(x) = 7 x 2.2.2 Polynomial Function is a multi-term function. The general form of a single variable, x, polynomial function is: y = a0x

0 + a1x1 + a2x

2 + a3x3 + .... anx

n. Each form contains a coefficient as well as a non-negative integer power of variables. [The first two terms can be written as a0 + a1x, since, x0 = 1 and x1 is commonly written as x].

Depending on the value of the integer N (which specifies the highest power of x), several subclasses of polynomial function emerge:

Case of n = 0: y = a0 [Constant function]

Case of n = 1: y = a0 + a1x1 [Linear function]

Case of n = 2: y = a0 + a1x1 + a2x

2 [Quadratic function]

Case of n = 3: y = a0 + a1x1 + a2x

2 + a3x3 [Cubic function]

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We can proceed further assigning other values to n. The powers of x are called exponents. The highest power involved i.e., the value of n, is often called the degree of the polynomial function. A cubic function, with n = 3, is a third-degree polynomial constant function. The constant function is also called the degenerate or polynomial of the zero degree polynomial.

Graphs 2.2, 2.3 and 2.4, respectively give the linear, quadratic and

cubic functions:

Graph 2.2 Graph 2.3 Graph 2.4

Y = ao + a1x1 Y = ao + a1x

1+a2x2 Y=ao+a1x

1+a2x2+a3x

3 or or or Y = a + bx Y = a + bx + cx2 Y = a+bx+cx2+dx3 Y y y ao ao ao x x x ao = Y – intercept ao = Y – Intercept ao = Y – Intercept a1 = Slope

2.2.3 Relational functions

A relational function is one which is expressed as a ratio of two

polynomials in the variable x. A function such as

is a polynomial in which y is expressed as a ratio of x – 5 to

x2 + 2x + 20 (both are polynomials), is a relational function. Any polynomial function is a relational function, because it can be always expressed as a ratio to 1, which is a constant function.

,202

52 ++

−=

xx

xY

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A special relational function that has quite an interesting application in business is the function: y = a/x or xy = a. This function plots as a rectangular hyperbola (Graph 2.5).

Graph 2.5

Y = a/x or XY = a a > 0 y x

Since the product of the two terms is always a given constant, this

function may be used to represent average fixed cost curve, a special demand curve where total expenditure (i.e., price x quantity) is always the same.

The rectangular hyperbola drawn for xy = 1, never meets the axes,

for whatever levels of upward and sideward extensions.

2.2.4 Algebraic and non-algebraic functions

Algebraic and non-algebraic functions are another classification of

functions. Any function expressed in terms of polynomials and/or roots of polynomials is an algebraic function. The functions so far dealt are all algebraic.

Non-algebraic functions are exponential, logarithmic,

trigonometric or so. In an exponential function, the independent variable appears as the exponent as in the function: Y = bx. A logarithmic function may be as: y = x log b. Non-algebraic functions are also known as transcendental functions.

Graphs 2.6 and 2.7 give the functions.

22

Graph 2.6 Graph 2.7

Y = bx y = x log b Y Y b > 1 o x x

2.2.5 Linear Functions

Functions can be classified on the basis of number of independent

variables. So far only one independent variable, x, was dealt by us. Instead two, three or any number of independent variables may be involved. We know production is a function of labour (L) and capital (K). So, a production function may be presented as: Q = f(K, L). Consumer utility can be given as a function of 3 different commodities and the function is y = f(u,v,w).

Functions of more than one variable can be constant, linear or

nonlinear. Look at these forms: Y = a1 + a2 + a3+ a4 ………..an (constant function) Y = a1x1+ a2 x2+ a3 x3+ a4 x4 ………..an xn (linear function)

Y = a1x12+ a2 x1x2+ a3 x2

2 (quadratic function)

Y = a1x13+ a2 x1

2x2+ a3 x1x22+ a4 x1x2

3 (cubic function)

As presented already, linear functions are 1st degree polynomial. These could have single or more independent variables. A single variable linear function runs like this. Y = a + bX. A two variable linear function is:

Y = a + bx1 + cx2.

23

2.2.5.1 Concept of slope of a line

Slope refers to the inclination of a line to the x-axis. It gives a measure of the rate of change in dependent variable, y, for a unit change in independent variable, x. Both the direction and quantity of change produced are indicated by slope.

Slope of a line can be studied in five ways. (i) Slope is the tangent of the angle made by the line with the x-

axis when we move anti clockwise from the x-axis to the line. If θ is the angle that the line makes with the positive direction of x-axis, than slope of the line = tangent = opposite side/Adjacent side. See the graphs 2.8(i), 2.8(ii).

Graph 2.8 (i) Graph 2.8 (ii)

θ

θ

(ii) Slope of a line can be measured through ordinates and coordinates. If (x1,y1) and (x2,y2) are any two points on the line, then slope is given by: (y2-y1)/(x2-x1).

(iii) When an equation of a line is given as, y=a+bx, then slope is

given by `b' in that equation. (iv) Slope is nothing but the regression coefficient of y on x or byx

= Σxiyi / Σxi2 where xi = xi – x and yi = yi – y

24

(v) Slope is also studied by amount of change in y, ∆y divided by

amount of change in x, ∆x. Slope = ∆y/ ∆x.

Illustrations:

1) Suppose three lines make each an angle of i) 30 , ii) 45 and 60 with the positive direction of x-axis. Then the slope of the lines are:

i) tan 30° = 1/√3 ii) tan 45 = 1; tan 60°= √3

2) Suppose two points on a line are: 2. -4 and 1, 7. Its slope = (y2-y1)/(x2-x1) = 7 - (-4)/1-2 = 11/-1 = -11. The line has negative slope.

3) A machine costs Rs. 1,00,000. 5 years after, its value falls to Rs. 60000. If value is a linear function of time, find the depreciation function.

Solution: Let the value of the machine at year `t' be: v = a + bt. Put v = 1,00,000 at t = 0 and v = 60000 at t = 5. We get the following two equations:

(i) 1,00,000 = a + b(0) or

a = 1,00,000 …. (1)

(ii) 60000 = a + b(5) or

a + 5b = 60000 …. (2)

Solving (1) & (2) we get, 5b = -40000 or

b = -8000

The annual rate of depreciation is 8000 and the value of the machine falls by Rs. 8000 p.a.

2.2.5.2 Linear Cost function and slope thereof

Total cost (TC) = 1500 + 400Q, where `Q' is the quantity produced, and the constant term is 1500 being the fixed cost and coefficient of Q is 400, which is the slope of the curve indicating the change in total cost, if output rises or falls by one unit. In the case of a linear function, the slope is constant at all points on the curve.

25

Slope is a highly relevant concept in our analysis, slope measures the rate of change in the dependent variable for a unit change in the independent variable. This is given by the tangent of the curve at a point.

Opposite Side ∆Y Tangent A = =

Adjacent side ∆X In graph 2.8 (iii) the slope is presented. Slope = ∆Y/∆X

Graph 2.8 (iii)

Cost curve and slope

Y ∆y

∆x x

We have standardized, ∆x as 1 unit. Then ∆y, the vertical side of

the shaded triangle, is the slope. And it is constant, for all the four triangles depicted.

Actually slope is the regression coefficient in the regression line. Suppose x as the quantity and y, the total cost. Let the details be as follows.

Total Maximum

X : 2 3 5 8 12 30 6

Y : 2300 2700 3500 4700 6300 17500 39500

X – X : -4 -3 -1 2 6 0-

Y – Y : -1600 -1200 -400 800 2400 0 -

XY : 6400 3600 400 1600 14400 26400 -

X2 : 16 9 1 4 36 66 -

26

The regression of X on Y: Y = a + bX The value of `b' = Σxy / Σx2 = 26400/66 = 400. This is the slope of the curve.

a = Y – bX = 3900 - 400(6) = 1500 So, the regression equation is: Y = 1500 + 400X. Actually this

equation is the same as the TC = 1500 + 400Q, with which we started. Graph 2.9 (i) gives this.

Slope of a curve can also be found by differentiation. We know,

TC = 1500 + 400Q. Differentiate TC with respect to Q and we will get the slope. dTC/dQ = 400. A linear function of the type Y = bX is also possible with no

intercept. Suppose the following pattern of X and Y: X: 1 2 3 4 5 Y: 100 200 300 400 500 This pattern also depicts a straight line, but its Y intercept is zero.

In other words, the line passes through the origin. Graph 2.9 (ii) gives an account of the same.

Graph 2.9 (i) Graph 2.9 (ii)

Cost curve and slope Cost curve and slope

Y = a + bx Y = bx

500 8000 400 6000 300 cost 4000 200 2000 100 1500 0 2 4 6 8 10 12 0 1 2 3 4 5

Quantity

27

2.2.5.3 Linear Demand Function and Slope thereof

Suppose the quantity demanded and price are linearly related and the demand function is as follows:

Unit price P : 4 5 6 7 8 Quantity Q 1400 1200 1000 800 600 We are interested in finding the rate of change in Q, for a unit

change in P. You know we have made this rate as 200. We can work out it through the regression equation, as well as the differentiation route.

Illustration :

First the regression model is attempted.

Total Mean

P: 4 5 6 7 8 30 6

Q: 1400 1200 1000 800 600 5000 1000

P-P=p: -2 -1 0 1 2 0 -

Q-Q=q 400 200 0 -200 -400 0 -

pq -800 -200 0 -200 -800 -2000 -

p2 4 1 0 1 4 10 -

The slope or repression coefficient = b =Σpq/Σp2 = -2000/10 = -200.

The constant = Q – bp = 1000 – (-200x6) = 1000 + 1200 = 2200 The regression equation is: Q = 2200 - 200P. The negative slope

indicates that, quantity demanded in indirectly proportional with price, i.e., as P rises Q falls and vice versa.

Now that we know: Q = 2200 - 200P, Differentiate Q w.r.t. P, we get: dQ/dP = -200 This is the same as regression coefficient we got earlier.

28

Graph 2.10 depicts the demand curve. The curve is downward sloping, indicating that its slope is negative.

Graph 2.10

Dividend curve and slope Quantity 1800 1200

600 ∆Q

∆P 0 1 2 3 4 5 6 7 8 Price

The slope = Change in Q / Change in P

= ∆Q / ∆P

2.2.5.4 Linear Supply Function and Slope thereof

Let the supply of goods at different prices be as per the schedule

given: Unit price ; P : 1 2 3 4 5 Quantity : Q : 600 800 1000 1200 1400

We can show that the slope is positive and is equal to 200. The equation can be worked out as before. P – P =p -2 -1 0 1 2 0 Q – Q = q -400 -200 0 200 400 0 pq 800 200 0 200 800 2000 p2 4 1 0 1 4 10

29

Let the equation be: Q = a + bp

b = slope = Σpq / Σp2 = 2000 / 10 = 200

a = Q – bp = 1000 – 200(3) = 400

The regression equation : Q = 400 + 200p

Graph 2.11 gives the curve

The slope = ∆Q / ∆P = 200

Graph 2.11

Supply curve and slope 1400 1000 800

∆Q 400

∆P 0 1 2 3 4 5 6 Price

2.2.5.5 Linear Function with plural independent variables

2.2.5.6

So far we dealt with linear functions involving only one independent variable. We can have linear functions with more than one independent variable.

Suppose a firm produces two products A and B. A's contribution

per unit is Rs. 100 and B's contribution per unit is Rs. 80. Assuming `a' units of A and `b' units of B being produced. The total contribution function is a linear function as follows: 100a + 80b. The contribution function can be expressed in a graph in the form of combinations of A and B, giving a particular level of contribution.

30

Suppose Rs. 2000 and Rs. 4000 levels of contribution are needed. We can plot in the graph combinations of A and B giving Rs. 2000 and Rs. 4000 levels of contribution.

Rs. 4000 contribution can be got through 40 units of A or through

50 units of B or combinations of A and B. In the graph 2.12, `A' is taken on x-axis and `B' is taken on y-axis.

The line joining 40 units of A and 50 units of B, represents combinations of A and B yielding a total contribution of Rs. 4000. Similarly the line joining 20 units of A and 25 units of B represents combinations of A and B yielding a total contribution of Rs. 2000. All lines parallel to these lines represent combinations of A and B giving certain levels of total contribution. For higher total contribution the curve shall be farther from origin and vice versa.

If 3 independent variables are involved, we can still have graphical

version, a 3 dimensional graph. Beyond 3 independent variables, the equation form is the only method of presentation.

Graph 2.12 Bi-variate linear functions

50

40

30

B 20

10

0 10 20 30 40 50

A

31

Suppose there are `n' independent variables. The general order of linear equation here is: Y = a0 + a1X1 + a2X2 + a3X3 + .... + anXn, where X1 ... Xn are independent variables and a0, a1, .... an are constants or coefficients.

2.2.6 Non-linear Functions

Non-linear functions are simply the 2nd or further higher degree polynomials. We know the 2nd degree polynomial is a quadratic function, the 3rd degree polynomial is a cubic function and so on. You may refer back graphs 2.3 and 2.4 for quadratic and cubic functions. In graph 2.3 a2 < 0 and the graph appears like a `hill'. If a2 > 0, then the graph will form a `Valley' as shown in graph 2.13.

2.2.6.1 Quadratic Function

Let the demand for a product be given by the quadratic function: Q = P2 - 7P + 10, where P is price.

We can get the graph plotted by first mapping the function in the

form of a table as below. As P cannot take negative value, we take it as zero first and allow it to rise a unit at every step and relevant Q values got thus: P: 0 1 2 3 4 5 Q: 10 4 7 12 19 28

Graph 2.13 Quadratic Function

Y=ao+a1x+a2x2

Case: a2 > 0 Y x

32

Graph 2.14

Quadratic Function

Q = P2 + 7P + 10

0

7

14

21

28

0 1 2 3 4 5Price

Graph 2.14 depicts the function Q = P2 - 7P + 10. The curve does

not conform to normal demand curve, generally speaking. Perhaps this is vanity demand curve. A quadratic function is different from a quadratic equation. A quadratic equation results, when a quadratic function is set to zero. That is, the quadratic function, P2 - 7P + 10 if made as equal to zero, i.e., P2 - 7P + 10 = 0, a quadratic equation results. For the quadratic equation the roots can be worked out.

2.2.6.2 Cubic Function

Let us continue with the quadratic function of Q = quantity

demanded as a function of P = price, as: Q = P2 - 7P + 10. From this we can get the total revenue function, TR = PQ = P(P2 - 7P + 10) = P3 - 7P2 + 10P. This is a cubic function, a non-linear form of 3rd order polynomial.

We can plot the curve by assigning values to `P' and getting those

of TR. The same is tabled below:

Price per unit = P = 0 1 2 3 4 5 6 7 Total revenue TR = 0 4 0 -6 -8 0 24 70 (P3 – 7P2 + 10P)

Graph 2.15 presents the revenue function.

33

The first derivative of the total revenue function is the marginal

revenue function. We know. That is the rate of change in total revenue for a given instantaneous

change in price is 3P2 - 14P. This is the slope of the curve. It varies from place to place on the curve.

2.2.6.3 Slope in the case of non-linear functions

Slope in the case of non-linear functions is not constant for all points on the curve, unlike the case with linear functions with same slope for all points on the curve.

Graph 2.15 Cubic function

70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 -10 Price

34

Let the cost curve and revenue curve be 500 + 13Q + 2Q2 and 125Q - 2Q2.

The cost curve can be derived as follows: Q 0 5 10 15 20 TC: 500 615 830 1145 1560

The revenue curve can be derived as follows: Q: 0 5 10 15 20 TR: 0 575 1050 1425 1700

Graphs 2.16 and 1.17 depict the cost and revenue curves: The slope of the curves are not same at all points. For different The slope of the curves are not same at all points. For different

magnitude of `Q', different slope levels exist. The first derivative of cost function w.r.t. Q gives the slope. It is:

13 + 4Q. So, slope for different `Q' values are as tabled below, obtained by putting the value of Q in the above slope function:

Q : 0 5 10 15 20 Slope: 13 33 53 73 93 (13 + 4Q)

Graph 2.16 Cost curve

2000

1500 1000 500 0 5 10 15 20

Graph 2.17 Revenue curve

1800 1200 600 0 5 10 15 20

35

The first derivative of revenue function w.r.t. Q gives its slope. It is: 125 + 4Q. So, slope for diff. `Q' values are as tabled below:

Q: 0 5 10 15 20 Slope: 125 105 85 65 45 (125 + 4Q)

2.2.6.4 Profit Function

E = Profit = Total Revenue - Total Cost. For the example dealt above, Profit = 125Q - 2Q2 - (500 + 13Q + 2Q2) = 125Q - 2Q2 - 500 - 13Q - 2Q2 = -4Q2 + 112Q - 500. Profit for 0, 5, 10, 15 & 20 units are: -500, 40, 220, 280 and 140.

The slope of the profit curve: i.e. first order derivative with respect

to e = dE/dQ of – 4Q2 + 112Q – 500 = -8Q + 112. At Q = 0, 5, 10, 15 and 20 the slope is: 112, 72, 32, -8 and -48. Graphs 2.18 gives the total profit and marginal profit curves.

What is the profit optimizing output level?

The profit function is: E = -4Q2 + 112Q - 500. First derivative dE/dQ = -8Q + 112. Profit maximizing output is

given by letting the first derivative equal to zero and solve for Q. So, -8Q + 112 = 0; 8Q = 112: Q = 14. At 14 units profit = 284. To ensure the correctness of the answer, the 2nd derivative must be done and if its value is negative, the answer is correct. And d2E/dQ2 = -8. So our answer is correct.

When marginal profit (i.e., slope of total profit), reaches zero, total

profit is maximum. The quantity at this level of profit is profit maximizing quantity. From the graph 2.18 we read it as 14 units, the same as we got earlier algebraically.

36

2.3 ELASTICITY OF DEMAND THROUGH FUNCTIONS

Elasticity is a concept very much used in business. Elasticity of demand, supply, etc are common parlance terms in business studies.

Elasticity of demand to price of the product, to increase in the

income of the customer, to price of competitor product (known as cross elasticity of demand), to advertisement and promotion campaigns, etc are very useful concepts in businesses. All these aim to measure the rate of change of quantity demanded to a given change in price, income, price of competitor product or advertisement and proportional spending.

Graph 2.18 400 300 Peak 200 Total Profit 100 5 10 15 20 0 Marginal -100 profit (Rate of change -200 in profit) -300 -400 -500

37

2.3.1 Price elasticity of demand

Let us now consider price elasticity of demand. Let it be denoted by ‘E’. By definition E is given by:

-∆Q/Q

= , Where Q – is the quantity, P – is the price and ∆Q and

∆P/P

∆P are the changes in Q & P

-∆Q P P -∆Q = x = x

Q ∆P Q ∆P

You must remember ∆Q/∆P is a slope function, slope of quantity w.r.t. price. So, elasticity of demand is slope of demand curve multiplied by –P/Q. [Note the negative sign is used as elasticity in both directional and dimensional and as ‘Q’ varies opposite to the price, the ‘-‘ sign is attenuated to the ∆Q.] For all normal goods and luxury goods the price elasticity is negative, but conventionally the –ve sign ignored.

Let the demand function be Q = 280 – 7P. its elasticity is given by

–(-P/Q) x Slope. The slope is simply the first derivative w.r.t. P. So, dQ/dP = -7. The ‘E’ for the different values of P are computed and given below; and the graph 2.19 gives the demand curve.

pricein change of Rate

demanded

in quantity change of Rate =

38

P : 10 20 30

Corresponding Q : 210 140 70

Slope : -7 -7 -7

Elasticity: -(P/Q) *(Slope) : 0.333 1 3

Total revenue(PxQ) : 2100 2800 2100

When elasticity is >1, the demand is said to be price elastic and

reducing price, more volume can be sold. When elasticity is <1, the demand is inelastic. By reducing price you can not sell more, at the same time by raising the price, your sales volume is not going to be severely pruned.

When price is reduced from Rs.30 to Rs.20, i.e., by 33%, Q i.e.,

the sales volume has increased by 100% from 70 to 140. that is, when elasticity was high, a reduction in P enabled the firm to enhance sales. Also, the sales revenue increased from Rs.2100 to Rs.2800. When price elasticity was less than 1 an increase in price will get good revenue. See, when price was increased from Rs.10 to Rs.20, by 100% volume of sales declined only by 33% from 210 to 140. at the same time revenue increased from Rs.2100 to RS.2800. Thus for the seller, it pays to rise prices when he has price inelastic demand curve. Similarly, when he forces an elastic demand curve, it pays to reduce price.

Graph 2.19 Demand curve

210 E = 0.333 Quantity 140 e<1 E = 1 70 e>1 E = 3 0 10 20 30 Price

39

The price elasticity is > 1, beyond the price level of Rs.20. So, any price rise from Rs.20 will have the seller with low sales. But price reduction to Rs.20 from higher price level will benefit him. The price elasticity is <1, below the price level of Rs.20. So, any price reduction from Rs.20 will not benefit the seller, but any price rise, movement in price towards the unit elasticity point from any previous position of elasticity pays off well the seller. You can notice the use of slope in computing price elasticity. Price elasticity is simply slope times P/Q. you can also note, given the slope price elasticity is directly proportional to price and inversely proportional to quantity.

2.3.2 Income elasticity of demand

Income elasticity of demand process light on the ratio of rate of change in quantity demand to rate of change in income of the consumers.

Income Elasticity = ∆I

∆Q

Q

I

∆I

I

Q

∆Q

I∆I

Q∆Q×=×=

Suppose the following demand income function is present: Q = 500+0.05

I, where Q is quantity demanded and I is income. Then the following table of Q and I ordered pairs can be worked out

I : 1000 1500 2000 2500

Q : 550 975 600 625

Slope: ∆Q/∆I : 0.05 0.05 0.05 0.05

Elasticity: (I/Q)x(∆Q/∆I) : 0.091 0.130 0.167 0.2

[Slope of the Income-demand function is the 1st order derivative of

that function. We know the function is :Q = 500 + 0.05 I. Its 1st order derivative:

2.3.3 Cross elasticity of demand

]05.0dI

dQ=

40

The ratio of rate of change in quantity demanded of a product to the rate of change in price of a related product;

If X any Y are complementary goods [that is, use of one needs use

of the other too, like car and car insurance], when the price of Y rises demand for X falls and vice-versa.

Cross Elasticity = y

x

x

y

yy

x

∆P

∆Q

Q

P

P∆P

Q∆Q×=X

If X and Y are competitive goods [that is use of one results in non-

use of the other, like coffee and tea), when the price of Y rises demand for X rises and vice versa.

2.3.4 Advertisement and Promotion Elasticity of Demand

The ratio of rate of change in quantity demanded to rate of change in amount spent on advertisement and promotion.

Advt. & Promotion Elasticity of demand = (A/Q) (∆Q/∆A), where `A' is the amount spent on advertisement. Here the slope is normally positive and that demand is elastic to advertisement and promotion. That is why big companies spend lot on advertisement promotion.

2.4 MAXIMA AND MINIMA OF A FUNCTION

A function, especially a higher degree function, might be highest for some value of the independent variable and lowest for some value of the independent variable and there may be many `highs' and `lows'. These `highs' are called `maxima' and the `lows' are called the `minima'. Hence the need to study the maxima and minima of functions.

Look at the graph 2.20 given below for the function P = f(Q)

41

Consider the graph in the interval Q = 1 to 4, f(Q) is increasing to

the left of x = 2 and decreasing to the right of Q = 2 and is highest at P, when Q = 2. Consider the interval, Q = 5 to 7. To the left of Q = 6, f(Q) is decreasing and to the right of Q = 6, f(Q) is increasing and is lowest when Q = 6. When f(Q) is rising, the slope is positive and when decreasing the slope is negative. And, when slope is zero, either f(x) is maximum or minimum. So, if the slope is zero, how can we say definitely that f(x) is at its maximum or at its minimum? Here comes to our rescue the 2nd order derivative or d2P / dQ2. If the value of d2P/dQ2 < 0, i.e., negative, f(Q) is at its maximum point and if d2P/dQ2 > 0, i.e., positive, f(Q) is at its minimum point.

Suppose the profit function is, P = Q

3/3 – 4Q

2 + 12Q.

Find the ‘Q’ for which profit is maximum and the‘Q’ for which

profit is minimum?

Graph 2.20 Profit function : P=f(Q)

11

10 P

9

8

7 P 6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 Q

42

Solution

.072 14472 )6 (12)6 (43

6 profit the And 6. Qat

minimum isProfit 0, value theSince, 4.8 -1282(6) Q2d

Pd 6,Qput weIf

. 3

210 2416

3

8)2(12 )2(4

3

2 profit theAnd 2. Qat

maximum isProfit 0, value theSince, -4.8-4 82(2)Q2d

Pd 2,Qput weIf

Q2d

Pd

6Qor 2Q i.e.,

0 6)-(Q 2)-(Q i.e.,

0 128QQdQ

dP

12Q4Q3

Q Profit

2 3

2

2

3

2

2

2

2

3

= +−=+−==

>= =− ==

=+ −=+−==

<==−= =

of Q2 + 8Q + 12 = 2Q - 8 ==

=

=+− =

+ −=

43

We can get ordered pair points for profit for the different ‘Q’ levels as below: Q : 0 1 2 3 4 5 6 7 Profit = P : 0 9 10.67 9 5.33 1.67 0 2.33 These points are plotted in the graph 2.20. A function can have many maxima and corresponding minima.

Questions

1. Give the concept and the types of functions. Illustrate with graphs.

2. What is relational function? Explain the specialty of the function y = a/x.

3. What are linear functions? Comment on the slope of linear functions.

4. Explain linear supply and demand functions.

5. What is a non-linear function? Explain the types of the same with algebraic and graphic illustrations.

6. Comment on the nature of slope in the case of non-linear functions.

7. What is elasticity of demand? Explain measures of computing the same.

8. Explain income and price elasticity of demand?

9. What do you mean by maxima and minima of a function. Explain rules regarding the same.

10. Explain concepts of domain, image, map, argument, value, dependent variable and independent variable of a function.

11. Graph the functions: a) y = -x2 + 5x - 2 and b) y = x2 + 5x - 2 with set of values -5 < x < 5 as the domain. Which of the functions will have a `hill' and which a `valley'. Which factor determine the formation of `hill' or `valley'. [Hint: It is the sign of x2 in the quadratic function decide the formation of `hill' or `valley'].

44

12. A survey shows that there is a linear function between population of a city and time. In 1986 the population was 68 lakhs, and in 2000 its population was 75 lakhs. Estimate the population of city in 2005 and in 2010. (Answer: P = 68 + 0.56T)

13. Find the average rate of change in the supply function, Q = 3p - 4 over the interval p[1,5]. (answer 3).

14. A trader earns Rs. 380 when price = Re 1 per unit, Rs. 660 when price = Rs. 2 per unit, Rs. 860 when price = Rs. 3 per unit. On plotting the points (1,380), (2,660) and (3,860) he finds that a quadratic equation may fit the data. Construct the quadratic equation and estimate his earnings for price = Rs. 4 per unit. [Hint: Construct 3 set of equations find the coefficients of a, b and c. Earnings at p = 4 = Rs. 980].

15. The total profits of a firm is given by: -2x2 + 400x - 1700, where x = number of units sold. Find the maximum profit giving sales volume and profit per unit at that level. (answer: 100 and 183).

16. Cost of production is given by: x2 - 120x + 15000, where x is the units produced. Find the cost minimizing volume and cost per unit at that level.

17. A firm produces x tons of a product at a total cost given by TC = x3-4x2+7x. Find the average cost minimizing output level.

18. Find the maxima and minima of the function: x3/3 - 4x2 + 12x.

19. The demand function of a product is: p = 100 - 0.5q2. If price p = 10, find the elasticity of demand. (Answer: 0.05).

20. The demand function is: q = 100 - 2p. Find income elasticity of demand between p = Rs. 25 and Rs. 36.

� � �

45

LESSON 3

COST-VOLUME-PROFIT RELATIONSHIP

Cost, volume and profit are interrelated. Cost refers to the money value of material, human and other resources expended in the production of a product or rendering a service. Cost of production depends on the volume of production. Cost of production is proportionately, linearly or curvi-linearly related to volume. Hence study of the pattern of cost-volume relationship assumes importance. Profit refers to the excess of revenue over expenses. Profit is a function of volume, cost and price. If the term volume is taken to mean sales revenue, then profit is influenced by volume and cost. Further as cost itself is influenced by volume a direct relationship between volume and profit could be thought. In this lesson varied aspects of the cost-volume profit relationship are analyzed.

3.1 COST OF OPERATING SYSTEM

There are different costs in the operation of a system. These are: variable cost, fixed cost, total cost, average cost, average fixed cost, marginal cost, differential cost, historical cost, future cost, controllable cost, non-controllable cost, opportunity cost, actual cost, out-of-pocket cost, imputed cost, avoidable cost, sunk cost, book cost, replacement cost, direct cost, indirect cost and so on. These are briefly explained below.

Variable Cost varies with volume of activity in direct proportion. Fixed cost remains constant irrespective of volume of activity up to certain range. Total cost is the sum of variable cost and fixed cost. Average cost is the average of total cost or total cost divided by number of units. Average fixed cost is the total fixed cost divided by number of units. It forms a rectangular hyperbola. Marginal cost is the change in total cost caused by a unit change in output. Normally, this is equal to variable cost. Differential cost is the difference in total cost when different alternative productions methods are considered. Both change in fixed cost and in variable cost are involved in differential cost. Historical cost is past cost or cost of events already happened, while future cost is cost of events to happen. Controllable cost is one which can be modified by

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executive actions. Uncontrollable cost is one not lending itself for manipulation. Opportunity cost is the benefit of next best alternative foregone. Actual cost is cost duly and really incurred. Out-of-pocket cost is money actually spent while imputed cost is cost attributed to a job, though not incurred. Avoidable cost is one that can be avoided. Sunk cost is one already incurred and nothing can be done about the same now. Replacement cost is the cost involved in replacing a part or segment at current prices, while book cost is cost historically incurred. Direct cost is cost of direct material, direct labour and direct expenses incurred in a job, while indirect cost is overhead costs.

It is better to throw some light on the nature of costs graphically:

Graphs 3.1 to 3.4 give the graphs for certain costs.

Graph 3.1 : Total Variable Cost Graph 3.2 : Total Fixed Cost

Cost Cost TVC TFC Output Output

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Graph 3.3 : Total Cost Graph 3.4 : Average Fixed Cost

TC Cost TVC TFC Output

3.2 PRODUCTION FUNCTION AND COST ANALYSIS

Production function expresses the relationship between input and

output. A production function can be expressed in the form a table, graph or equation.

Production functions are of several types. One-process production

function meaning only one technically efficient production process, two-process production function meaning the existence of two efficient processes of production and so on.

Production function in the form of an equation is: Q = f(L,K),

where, Q = units of output (dependent variable), L = units of labour (independent variable), K = units of capital (independent variable).

Production function varies with time. There are short run and long run production functions.

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3.2.1 Short-run Production Function

A short run production implies that of the independent variables, a

few only can be changed and the rest cannot be changed. So, keeping the unchangeable factors of production at given levels, how the variable factors be changed to optimize the output? The answer to the question depends on the trend in total product (TP), average product (AP) and marginal product. As long as additional inputs of variable factor of production results in addition to TP, i.e., MP > 0, we can go on introducing additional units of that factor into production.

Table 3.1 gives TP, AP and MP of labour with fixed capital

Units of L TP AP MP Units of L TP AP MP

1 10 10.00 10 7 67 9.57 7

2 21 10.50 11 8 72 9.00 5

3 33 11.00 12 9 75 8.33 3

4 43 10.75 10 10 75 7.50 0

5 52 10.40 9 11 73 6.64 -2

6 60 10.00 8 12 70 5.88 -3

From the table it is found that TP is rising till the 9th unit of labour is added, remains stable between 9

th & 10

th units of labour and falls when 11

th

unit of labour is added. That is, MP is positive till 9th unit, zero at 10th unit and turns negative at 11th unit of labour onwards. When MP is at its peak, 11 when 3 units of labour are employed, AP is at its peak too. TP is highest when MP = 0. TP rises as long as MP > 0, remains constant at previous level when MP = 0, falls when MP < 0.

As long as MP is rising, it is greater than AP. As MP starts

receding, AP is greater than MP. As MP turns negative, AP starts falling bit steeply. The first stage of production, i.e., MP > AP, is increasing returns stage, the second stage, i.e., AP > MP > 0, is decreasing returns stage and the third

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stage, i.e., MP < 0, is negative returns stage as far as the factor of production that is being varied.

Graph 3.5 : TP, AP, MP

TP Output Stage I Stage II Stage III Output AP Units of varying Input, i.e., labour MP

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Graph 3.5, in two periods, gives a vivid picture of the three stages. When MP falls below AP, inflexion takes place in TP, marking the end of stage I. When MP = 0, TP is maximum, marking the end of stage II.

3.2.2 Cost of Operation: Relation between cost and activities in the short

run

So far we studied the behaviour of output with respect to input in physical units. If the physical units of inputs are converted into cost, the behaviour of cost and output.

In the short-run, total cost has two components. A fixed

component attributed to factors of production that are kept constant, here capital and a variable component which is allowed to change, here labour. So, total cost = TC = TFC + TVC, where TFC is total fixed cost and TVC is total variable cost. Economists consider TVC a non-linear costs while accountants take it linear cost. The accountants' version was given earlier in graphs 3.1 & 3.4. And the economists' version is given now.

Graph 3.6 gives, in two panels, the short-run cost-output or cost-

volume relationship.

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Graph 3.6 : Quadratic Cost Function

TC TVC Cost (Rs) TFC Output MC ATC AVC Cost (Rs) AFC Output

Average total cost, ATC = TFC/Q + TVC/Q = AFC + AVC. Marginal Cost (MC) is the rate of change in total cost for a unit

change in output. So, MC is measured through the slope of TC. The slope of TC

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is different at different points. Since, TC and TVC are similar in pattern, MC is also measured through the slope of TVC. This is so, because TFC is constant. Also, using differential calculus, MC can be studied. This we did in Lesson 2.

3.3 LEVEL OF ACTIVITY AND ITS PARAMETERS

Level of activity, i.e., production is influenced by different

parameters. The relationship may be linear, quadratic, etc. These are presented now.

i) Q = a + bL (Linear relationship) ii) Q = a + bL – cL2 (quadratic relationship) iii) Q = a + bL + cL2 - dL3 (cubic relationship) iv) Q = a + bLb (exponential relationship)

3.3.1 Linear Function : Illustration : 1

Suppose production function is linear, for 10 units of labour Q =

45 and for 12 units of labour Q = 33. Establish relationship between activity and its parameters.

Solution : 45 = a + 10b -------------- (1) 53 = a + 12b -------------- (2)

(1)-(2), -8 = 0 - 2b or, b = 4.

Putting b = 4, we get a = 5. So, Q = 5 + 4L

MP is the first derivative of Q with respect to L. So, dQ/dL = 4. This linear relationship has a limitation. It assumes that unlimited

introduction of the variable factor of production will yield a continuously rising output. It is not so in reality.

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3.3.2 Quadratic Relationship: Illustration: 2

Suppose the activity-parameter relationship is quadratic form given

by the equation: Q = 5 + 4L - L2. Find the MP, output maximizing level of L etc. Solution Q = 5 + 4L - L2

To find output maximizing level of L, set MP = 0 and get the value

of L. So, 4 - 2L = 0, or 4 = 2L or L = 2. When, L = 2, Q = 5 + 4(2) + 22 = 9.

To confirm whether this is the output maximizing level, we have to get d

2Q/dL

2 and see whether it is < 0. Here, d

2Q/dC

2, is less than zero. So,

output is maximized at L = 2.

3.3.3 Cubic Relationship: Illustration : 3 Let the activity-parameter relationship be cubic. In a particular

case it is found to be: Q = 25 –18L + 4.5L2 - L3/3. Find the MP, maxima, minima, etc.

Solution : Q = 25 –18L + 4.5L2 - L3/3

So, MP = dQ/dL = - 18 + 9L -L2 By setting dQ/dL = 0, we get, -L2+ 9L + 18 = 0 or,

L2- 9L - 18 = 0 Or, (L - 6)(L - 3) = 0

Or, L = 6 or 3

If we put L = 6, in the second derivative, we get 9 - 2L = 9 - 12 = -3. Since it is negative, the function is at one of its maxima when L = 6. At L = 6, total Q = 25 - 18(6) + 4.5 (6x6) - (6x6x6)/3 = 25 - 108 + 162 - 72 = 7. If we put L = 3, in the second derivative, we get, 9 - 2L = 9 - 6 = 3. Since its positive, the function is at one of its minimum. And at L = 3, Q = 25 - 18(3) + 4.5(3x3) - (3x3x3)/3 = 25 - 54 + 40.5 - 9 = 2.5.

2L4dL

dQMP −==

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At the maxima, MP should be higher. And it is: 18 + 9(6) - (6x6) = 18 + 54 - 36 = 36. At the minima MP should not be higher than the above figure of 36. Could be less or equal. And it is: 18 + 9(3) - 3x3 = 18 + 27 - 9 = 36. It is equal here.

3.3.4 Exponential relationship: Illustration : 4

Let the activity – parameter relationship be exponential. Say it is: Q = aLb if Q = 4 when L = 4 and Q = 8 when L = 16, find the exponential function. Solution: a4b = 4 …. (1) a16b = 8 …. (2) Taking logarithm on both sides we get, log a + b log 4 = log 4 …. (3) log a + b log 16 = log 8 …. (4) (4)-(3) b log16 – b log 4 = log8 – log 4 b (log16 – log4) = log 8 – log 4 b = (log8-log4) / (log16-log4) = (0.9031-0.6021) / (1.2041-0.6021) = 0.301 / 0.602 = 0.5

Putting the value of b = 0.5 in equation (1), we get a4(0.5) = 4 or, 2a = 4 or a = 2. Therefore, the activity-parameter is Q = aL(0.5) = 2L(0.5)

3.4 EMPIRICAL PRODUCTION FUNCTIONS

Certain empirical models of production functions are dealt here.

3.4.1 Cobb-Douglas Production Function

One of the important production functions is given by C.W.Cobb

and Douglas. The function is of the exponential form, with sum of powers of the inputs equal to 1. The function is as follows:

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Q = A Lb K1-b , (so, b + 1-b=1), where, Q = output, A = a constant, L = Labour, K = Capital and b =

Power, a parameter. Actually, the `b' and `1-b', gives the factor-wise shares in total output.

MP of Labour = b(Q/L) and MP of Capital = 1-b(Q/K)

a) Linear Form of Cobb-Douglas Production Function

This is obtained by taking log on both sides of the equation: Q = A Lb K1-b So, log Q = log A b log L + (1-b) log K

b) Constant Returns to scale of C-D Function

We know: Q = A Lb K1-b. Let K & L are increased by ‘λ ' proportion. Then `Q' increases, say to Q . We can prove Q* = λ Q, rise in Q is proportionate to rise in inputs. By original equation,

Q * = A (λ L)b (λ K)1-b = A λb Lb λ1-b K1-b = A λb+1-b Lb K1-b = λ1 A Lb K1-b = λ(Q)

c) Improved Version of C-D Production Function

The original version of Cobb-Douglas production function is of

constant returns to scale type. Later, this was changed to represent both increasing returns and decreasing returns to scale as well. The power of L & K

were taken as α & β, if α + β = 1, there is constant return α + β > 1, there is

increasing returns and α + β < 1, there is decreasing returns to scale. To revised

format is therefore, Q = A Lα Kβ.

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d) Marginal Product of L and K

Marginal product of L and K are derived through partial derivatives of Q w.r.t. L and Q w.r.t. K.

MPL = dQ/dL

= α d.AL α-1 Kβ

= α (AL α Kβ).L-1

= α (Q)L-1

= α (Q/L)

= α (APL), APL (average product w.r.t labour)

similarly, MPK = β(Q/K) or β.APK

e) Value of αααα & ββββ

MPL = α (APL), α = MPL/APL and

as MPK = β.APK β =MPK/ APK

f) Elasticity of Production (EQ)

Let L be the changing variable.

αAPMPL

Q

∆L

∆QLL =÷=÷=

So, the exponent of an input is its elasticity of production.

= = Q

Q

L .

∆L

∆Q

L ∆L

Q ∆Q

E

ninput%changeina

utput%changeinoEQ =

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3.4.2 Fixed Proportion or Leon tiff Production Function

This production function assumes that the factors of production

must be used in fixed ratio. Excess quantity of either of the input, in relation to other as per the fixed ratio, is useless. The function is: Q = minimum (K/a, L/b), when, K and L refer to capital and labour, a and b are constants and Q is output. Output depends on smaller of the two ratios, K/a and L/b.

3.4.3 Linear Programming Production Function

In practice, a firm produces many products and uses many

resources, which are limited by supply, capable of multiple uses and consumed in definite proportion for a given product. The mix of products must maximize revenue or profit. In such cases, linear programming production model is used. [Please refer lesson 4 for LP Production Function]

3.5 LONG-RUN PRODUCTION FUNCTION

In the long run all factors of production can be altered, enhanced,

improved. As a result, all combinations of inputs giving some level of output can be thought of. Isoquants and Isocost lines are used in this context.

3.5.1 Isoquants

An isoquant depicts all combinations of inputs, L & K, giving a

particular level of output. Different isoquants indicate different levels of output by various sets combinations of inputs.

Graph 3.7 gives isoquant map consisting of different isoquants,

each giving a particular level of output, with different combination of inputs.

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Graph 3.7 : ISO quantity

C A P I K1 300 units T

A K2 200 units L 100 units

L1 L2 Labour

100 units of output can be got through different combinations of K and L. L1 and K1 units of L and K and L2 and K2 units of L and K give some output level. As one input is decreased, here K from K1 to K2, other input, here L is increased from L1 to L2.

3.5.2 ISO cost Lines

An ISO Cost line is a line depicting several combinations of the inputs involving same cost. Graph 3.8 gives a set of ISO Cost lines.

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Graph 3.8 : ISO cost lines

C A P I T A L Labour

The line C = 300 indicates several combinations of K and L, having same total cost of Rs. 300. Similarly, C = 400 indicates, combinations of K & L having same total cost of Rs. 400.

3.5.3 Cost Minimization given Output

Cost minimizing given output is graphically shown in graph 3.9.

Graph 3.9 : Minimum cost output

C A P E I

T K A L 400 Units 600 500 400 O L Labour

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*

First the isoquant of a particular level of output is drawn. Then isocost lines for different cost levels are drawn. The isocost line that is tangential to the isoquant gives the minimum cost level for a given output. The isocost = Rs. 500 is tangential to isoquant at point E. OK level of capital and OL level of labour will be used producing 400 units of output at a total cost of Rs. 500.

3.5.4 Given Cost, Output Maximization

Maximizing out, given cost is explained below in graph 3.10.

Graph 3.10 : Maximum output at given cost 400 300 200 100 K E O L

Here, given the isoquant line, C = Rs. 500, we should reach the

highest possible isoquant. Draw isoquants. That isoquant, that is farthest from origin, `O', and tangential to the isocost line gives, the maximum output level given the cost.

Isoquant 400, is farthest from origin and is tangential to isocost

line at E. So, given cost Rs. 500, maximum output is 400 units, produced by OK units of capital and OL units of labour.

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3.6 COST OF OPERATION: RELATION BETWEEN COST AND

ACTIVITY IN THE LONG-RUN

The long-run cost-activity relationship is different from short-run

cost activity relationship. In the long-run there is no fixed cost; so, long-run total cost (LRTC) starts from origin `O', while short-run total cost (SRTC) starts from the level of fixed cost. Similarly short-run average cost, long-run average cost, short-run marginal cost and long-run marginal cost are different.

3.6.1 SRTC and LRTC

Graph 3.11 gives the SRTC and LRTC.

Graph 3.11 : SRTC and LRTC

SRTC LRTC C T O S

T TPC O Q1 Q2

Output or Activity SRTC and LRTC are tangential at point corresponding to output Q1. Except that output level for all other output levels, LRTC < SRTC, because

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in the long run no fixed cost is involved and cost minimizing opportunities abound.

3.6.2. Short-run average cost (SAC) and Long-run average cost (LAC) Graph 3.12 gives the SACs and LAC.

Graph 3.12 gives the SACs and LAC

C SAC1 SAC7 O SAC2 SAC6 S SAC3 SAC5 T SAC4 LAC Output or activity LAC curve is tangential to the various SAC curves. LAC curve is flatter ‘U’ shaped, while SAC curves are sharper ‘U’ shaped. LAC < SAC at any level of output. The minimum point of LAC curve shows the optimum level of activity.

3.6.3 Short-run Marginal Cost (SMC) and Long run Marginal Cost

Graph 3.13 gives the SMC Curves and LMC Curve.

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Graph 3.13 gives the SMC curves and LMC curve

SMC1 LMC SAC1 SAC4 SMC2 SAC2 SMC4 SMC3 SAC3 LAC

At each level of activity, a particular SAC is tangent to LAC, the relevance SMC = LMC. LMC Curve intersects LAC curve at the latter’s minimum point. Some short run plants have SAC coinciding with the minimum of LAC. At such level of activity SMC = LMC. So at that point, SAC = LAC = SMC = LMC.

3.7 RELATIONSHIP BETWEEN COST AND LEVEL OF ACTIVITY

The relationship between cost and level of activity may be: i)

linear total cost relationship, ii) quadratic total cost relationship, iii) cubic total cost relationship or some other higher polynomial farms. The first three are presented below.

3.7.1 Linear Total Cost (TC) Relationship

In linear form, TC = a + bQ, where, TC = total cost, a = a constant

or the fixed cost, b = rate of change in total cost for a unit change in activity or output level. Graph 3.3 gives a vivid graphical version of linear TC function.

Illustration 5

Suppose for 100 units of output the total cost is Rs.1,02,000 while for 120 units the total cost is Rs.1,18,000. If linear cost model exists, get the parameters of the model.

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Solution:

Given a + b(100) = Rs.1,02,000 …. (1) a + b(120) = Rs.1,18,000 …. (2) (2) – (1) we get, 20b = 16000 b = Rs.800 Putting this value in equation 1, we get, a = 22,000 The linear model is: TC = 22000 + 800Q The average cost is : a/Q + 800. The marginal cost is : b = Rs.800

3.7.2 Quadratic TC relationship A quadratic equation between total cost and output or activity

means TC is disproportionately changing with output. The rate of rise/fall changes with output level.

Quadratic TC form may be of two types. (i) TC = a + bQ + cQ2 or (ii) TC = a + bQ - cQ2. The first means an increasingly rising total cost with output. The second type means a decreasingly rising total cost with output.

Graphs 3.14 and 3.15 picture these models of Total cost.

Graph 3.14 : Graph 3.15

Increasing rising TC Decreasing rising TC

TC TC C O S T FC FC Activity Activity

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In the case of 3.14 model, AC falls at a smaller rate with rising output, while in 3.15, average cost falls at a higher rate with rising output.

AC in the case of TC = a + bQ + cQ2, is: a/Q + b + cQ. AC in the

case of TC = 1 + bQ – cQ2 i.e., a/Q + b – cQ. The change in AC in the former case is: -a/Q2 + c and in the latter is –a/Q2 – C.

Illustration 6:

You are given two total cost functions namely : i) TC = 10000 + 500Q + 0.3 Q2 and ii) TC = 10000 + 500Q – 0.3 Q2

Computer, MC, AC, ∆MC and ∆AC at Q = 100.

Solution

i) First case: TC : 10000 + 500Q + 0.3Q2 MC = dTC/dQ = 500 + 0.6Q At Q = 100, MC = 500 + 60 = 560

∆MC = dMC/dQ = 0.6 AC = TC/Q = 10000/Q + 500 + 0.3Q At Q = 100. AC = 100 + 500 + 30 = 630

∆AC = dAC/dQ = 10000 / Q2 + 3 = -1 + 0.3 = -0.7

ii) Second case: TC : 10000 + 500Q - 0.3Q2 MC = dTC/dQ = 500 – 0.6Q At Q = 100, MC = 500 - 60 = 440

∆MC = dMC/dQ = - 0.6 AC = TC/Q = 10000/Q + 500 - 0.3Q At Q = 100, AC = 100 + 500 - 30 = 570

∆AC = dAC/dQ = 10000 / Q2 + 3 = -1 - 0.3 = -1.3

As ∆MC is +ve in first case and negative in second case, AC falls steeper in second case (at 1.3 per unit) than in first case (at 0.7 per unit).

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3.7.3 Cubic TC Relationship

Suppose the cost function is of the cubic type. The general model is: TC = a + bQ - cQ2 + dQ3. Graph 3.6 gives a model of the cubic total cost function.

Illustration 7

Let total cost be: TC = 10000 + 500Q – 0.3Q2 + 0.01Q3

Compute TC, AC, MC, ∆AC and ∆MC for output = 100.

Solution:

TC = 10000 + 500x100 – 0.3(100x100) + 0.01 (100x100x100) = 10000 + 50000 – 3000 + 10000 = 67000 AC = TC/Q = 10000/Q + 500 – 0.3Q + 0.01Q2 = 10000/100 + 500 – 0.3x100 + 0.01 (100x100) = 100 + 500 – 30 + 100 = 670

∆AC = dAC/dQ = - 10000/Q2 – 0.3 + 0.02Q = - 1 – 0.3 + 2 = 0.7 MC = dTC/dQ = 500 - 0.6Q + 0.03Q2 = 500 – 0.6(100) + 0.03(100x100) = 500 – 60 + 300 = 740

∆MC = dMC/dQ = -0.6 + 0.06Q = -0.6 + 6 = 5.4

We can also find the maximum total cost and minimum total cost output levels.

3.8 COSTS AND PROFITS

The relationship between costs and profits is inverse. As cost rises,

profit falls and vice versa. Profit = Revenue - Cost. We need to bring in revenue function here, to study relationship between costs and revenues.

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Illustration 8

Let `p' be price per unit and `x' be units. A firm's price function is: P = 300 - 4x and its cost function is: TC = 500 + 28x. Find profit maximizing output.

Solution: The total revenue function is: P*(x) = (300 – 4x) x

= 300x – 4x2

π = Profit = Total Revenue - Total Cost = (300x – 4x2) – (500 – 28X) = 4x2 + 272x – 500 Profit maximizing output is computed as follows: First get the 1st

derivative of π w.r.t. x.

dπ / dx = -8x + 272

If we set it to zero, we get, -8x = 272 or x = 34. d2π/dπ2 = - 8 and it

is -ve. So, profit is maximized at x = 34. Alternatively, since, the coefficient of x2 is -ve, the maximum

point of x is = (-) Coefficient of x divided by 2 (Coefficient of x2) i.e., -272/2(-4) = 272/8 = 34. The general rule of maximum/minimum point is: -b/2a, in the quadratic form, ax

2 + bx + c = 0. Whether the function is at maximum or

minimum is given by the sign `a'. If a > 0, minimum point and if a < 0, maximum point results.

3.9 RELEVANCE OF MARGINAL COST AND TOTAL COST

Here in this section marginal cost, fixed cost, total cost, absorption

cost, etc. are dealt.

Marginal Cost

Marginal cost is the change in aggregate costs when output is changed by a unit. Total cost is the sum of total fixed cost plus the total variable cost.

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The costing terminology defines marginal cost as ‘the amount at any given volume of output by which the aggregate costs are changed if the volume of output is changed by one unit and it is measured by the total of variable cost’. Analyzing this definition, one can find that with the increase in one unit of output the total cost is increased and this increase in total cost from the existing level to the new level in known as marginal cost. For example if a company produces 100 units at a total cost of Rs.10,000 and if the production is increased by 1 more unit, say the total cost is increased to Rs.10,080. The increase of Rs.80 is the marginal cost at that level of output.

The later part of the definition says that marginal cost is measured

by the total of variable cost. Variable cost is always equal to prime cost plus variable overheads hence marginal cost is also known as variable cost. If in the above example, if the variable cost per unit is Rs.80 and fixed cost for the period Rs.2000, the total cost will be as follows for 100 units:

Variable cost 100 units @ Rs.80 = 8,000 Fixed cost = 2,000 Total cost Rs. = 10,000

If the output is increased by one more unit to 101 units the following emerge: Variable cost for 101 units @ Rs.80 = 8.080 Fixed cost = 2,000 Total cost Rs. = 10,080

The difference in the total cost as well as in total variable cost when the production is increased by one more unit is Rs.80 which is the marginal cost. The institute of Chartered Accountants of England defines marginal as follows: ‘Marginal cost is the very expense (Whether of production, selling or distribution) incurred by taking of a particular decision’. Thus, it implies that Marginal cost is nothing but the cost incurred as a result of a particular decision. Economists define Marginal cost as the additional to total cost or reduction in the total cost by the production of one more or one unit less.

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If we compare the Accountant’s view of marginal cost with that of the economist’s view we can find out the following differences.

i) Economists take total cost at two different levels of production whereas the Accountants take the total variable costs only at two different levels.

ii) As viewed by the economists, marginal cost curve would fall in the beginning and rise rapidly afterwards as the Marginal cost curve will be always saucer shaped. As per the Accountant’s view, the marginal cost curve would be a horizontal straight line as the Marginal cost i.e. variable cost per unit is assumed to be constant.

iii) Economist’s concept of marginal cost is a long-run concept; whereas, accountant’s view of marginal cost is a short-run concept, because the total fixed cost and variable cost per unit remaining constant would be possible only in the short run.

3.9.2 Marginal Costing : Definition

The costing terminology defines marginal costing as “the

ascertainment of Marginal costs and of the effect on profit of changes in the volume or type of output by differentiating between fixed and variable costs”.

The Institute of Cost and Management Accountants of England

defines Marginal costing as follows “Marginal costing is the technique where only the variable costs are charged to cost units, the fixed cost attributable being written off in full against the contribution for that period”.

If one analyze the above two definitions, are can find out:

i) Marginal costing is neither a method nor a system, but it is only a technique i.e. a technique of a decision making.

ii) It involves ascertainment of marginal cost.

iii) Through the ascertainment of marginal cost, we can find out the effect on profit due to changes in the volume or type of output.

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iv) The ascertainment of marginal cost requires differentiating between fixed and variable costs.

v) The variable costs are charged to cost units as they only are the product cost, and

vi) The fixed cost, being the period cost should be written off in full against the contribution for that period. Contribution refers to the difference between sales and the variable cost of sales. It can be presented as: Sales – Variable cost = Contribution. Contribution – Fixed cost = Profit.

3.9.3 Fixed Vs Variable Costs

The whole edifice of marginal costing is built upon one important behaviour of the cost i.e. variability. On the basis of variability, cost can be classified as (i) Variable (ii) Fixed and (iii) Semi-Variable.

i) Variable cost: Variable cost is a cost which in aggregate varies directly with variations in volume of production. Such cost are supposed to vary in the same direction and in the same proportion as the variation in the output. Hence, such costs are also known as ‘product costs’ as additional production involves only the variable cost. Total variable cost will be proportionately increasing or decreasing with the volume of production and variable cost per unit will be constant. ii) Fixed cost: Fixed cost is that cost which, in aggregate, tends to be unaffected by variations in the volume of output. The amount of fixed cost tends to remain constant for all volumes of production within the installed capacity of the plant. For example, salary to Manager, Rent for factory etc. shall remain the same even if production goes up or comes down in a period. It is to be remembered that this distinction of variable or fixed cost is made in the short run only. In the log run all costs are variable. iii) Semi – Variable or Semi – Fixed cost: This is a cost partly fixed and partly variable. It has the characteristics of both the fixed and variable cost. This cost varies with production, but not in the same proportion. E.g. Maintenance of

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plant cost, telephone charge, etc., In fact most of the overhead cost will be in the nature of semi-variable cost only.

3.9.4 Segregation of fixed and variable cost

As the whole marginal costing technique is based upon the differentiation between variable and fixed cost and as most of the overhead costs are only semi-variable in nature it becomes necessary to segregate semi-variable expenses into their fixed and variable elements. The following are the methods employed for this purpose.

a) High and Low Method

This method is based on the analysis of the past records of expenses. The overheads at the highest level of activity and the lowest level of activity are analyzed and the variability rate of expenditure is determined as under: Machine Hours Maintenance Cost (Rs.)

High 2,000 9,000

Low 1,200 7,800

Difference 800 1,200

Variable expenses per machine hour = ∆cost / ∆Hrs = 1200 / 800 = Rs.1.50 Substituting the variable cost in the high or low activity, we can get the fixed cost: High 2,000 Machine hours @ Rs.1.50 = 3,000; Fixed cost Rs.6,000 (i.e., 9000-3000). Low 1,200 Machine hours ~ Rs.1.50 = 1,800; Fixed cost Rs. 6,000 (i.e., 7800-1800). With this information one can project the estimated cost for the future period. For example, if the estimated machine hours in the future is 3,000 hours, then, Variable cost 3,000 x 1.50 = Rs. 4,500 Fixed cost = Rs. 6,000 Total cost = Rs. 10,500

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This method presumes that the variable portion of the semi-variable cost has linear relationship i.e. variable cost per unit is assumed to be constant. This may not be true always.

b) Scatter graph

This method is also known as regression line or the line of best fit. The cost at all lends are plotted in a graph; for example, the machine hours at different levels or periods are plotted in the ‘X’ axis and the semi-variable cost in the ‘Y’ axis. After plotting all the points, a straight line is drawn through the points plotted in such a way that equal distance is maintained as far as possible from these points with almost equal number of points on each side. The point in the ‘Y’ axis where the straight line intersects determines the fixed portion of the cost”. Let the monthly maintenance expenses for 6 months be as follows:

Month Machine Hour Semi Variable cost

(Rs.)

January 4,000 5,600

February 3,000 5,200

March 2,000 4,800

April 6,000 6,400

May 5,000 6,000

June 8,000 7,200

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8000

7000

6000

5000

4000

3000

2000

1000

0 1000 2000 3000 4000 5000 6000 7000 8000

The least square method uses a mathematical approach to find out the variable and fixed expenses. The method of least sequences is based on the regression equation y = a+bx where ‘a’ is the fixed element and ‘b’ is the degree of variability and y the total cost. Applying this basic equation and given a set of observations, n, two simultaneous equations can be developed that will fit a regression line to a linear array of data. The equations are

ΣY = Na + bΣX …. (1)

ΣXY = aΣX + bΣX2 …. (2)

Where a = fixed cost b = variable rate n = number of observations X = activity measure (hours etc) Y = total mixed cost observed

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Solving the equations for a & b will get us to find the fixed cost and variable cost.

c) Least Square Computation

Let x be the hrs and y be the amount. The figures are taken here is Rs.100

omitting the two zeros.

S.No. X Y XY X2

1

2

3

4

5

6

40

30

20

60

50

80

56

52

48

64

60

72

2240

1560

960

3840

3000

5760

1600

900

400

3600

2500

6400

N = 6

ΣX = 280

ΣY = 352

ΣXY = 17360

ΣX2 = 15400

Total 280 352 17360 15400

Putting these values in the equations, we get: 6a + 280b = 352 …. (1) 280a + 15400b = 17360 …. (2) (1) x 14 84a + 3920b = 4928 …. (3) (2) x 0.3 84a + 4620b = 5208 …. (4) (4) – (3) 700b = 280 b = 0.4

Putting this value in equation (1), we get: 6a + 280 (.4) = 352 or 6a = 352 – 112 6a = 240 a = 40

∴Y intercept = 40 hundred or Rs.4000

∴ The equation is : Y = 4000 + 0.4X

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3.9.5 Marginal cost vs. Absorption cost

Marginal costing has to be differentiated from absorption costing

which is also known as total costing or full costing. Absorption costing is ‘the practice of charging all costs to operations, products or processes’. Under this approach, the total cost is the sum total of all costs, variable or fixed, the total cost per unit will remain constant only when the level of output or mixture remains the same from period to period. As these factors are varying, the cost also would be varying from period to period and hence creates difficulties in making decisions regarding price fixing, activity level planning, profit estimating etc. There are certain specific difficulties in absorption costing as discussed below.

1) It involves apportionment of fixed overheads between products, processes etc. to find at the total cost and apportionment can be done in any of the various methods available and as such an element of arbitrariness is introduced in cost determination resulting in over or under-absorption of costs.

2) In absorption costing as the closing stock of finished goods and work-in-progress are valued at total cost which includes variable and fixed cost, the cost of one period is carried forward to the other period, through the value of closing stock. Fixed cost, strictly speaking is a period cost and should be only recovered from the revenue of the same period in which it is incurred.

3) Carrying forward of the cost of one period to another as above distorts the results of both the years, vitiating the very purpose of costing and financial reporting.

4) As a result, cost ascertainment and profit planning become difficult under absorption costing and this necessitates the application of marginal costing for managerial decision – making. Hence the significance of marginal costing.

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3.9 RELEVANT COSTS FOR DECISION MAKING

There are different cost concepts but all are not relevant in all decision situations. Table I gives pairs of relevant and irrelevant costs for decision making.

TABLE I : Relevant and Irrelevant Costs

Relevant Costs Irrelevant Costs

Future cost

Variable cost

Differential cost

Controllable cost

Opportunity cost

Out-of-pocket cost

Avoidable cost

Replacement cost

Historical cost

Fixed cost

Total cost

Non-controllable cost

Actual cost

Imputed cost

Unavoidable cost

Book cost

Relevant cost is one that is significant in a given decision making situation. It influences the choice of the decision maker. All other costs which are insignificant in the given contest are irrelevant costs. For instance, a concern interested in achieving cost efficiency would have to initiate actions to control and contain costs that are to be incurred. On the other hand, costs that were already incurred cannot be controlled or contained now. In other words, future costs are relevant for decision making while past costs are not. In a similar way, variable cost, differential cost, opportunity cost, etc., are relevant to decision making while fixed cost, total cost, etc. are not. Managerial decisions are varied, complex and conflicting in nature. This requires the manager to adopt a wholesome approach in a given situation. That is, as the decisions are varied various cost concepts have to be used in different circumstances. Again, as the decisions are complex, more than one cost concept may have to be depended on in one particular situation.

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Further, as some decisions are mutually conflicting, non-cost considerations, including the so-called irrelevant cost concepts, have to be taken into account as well. In other words, the above cost classification is not absolute, but only relative. Thus a particular cost concept or concepts would be relevant in one situation and irrelevant in another. But the above classification mostly holds good.

3.10.1 Future vs. Historical Cost

Planning and control are two important managerial functions. Both are forward looking as management decisions relate to the future. One can plan for the future only. Similarly, one can control what is to happen rather than what has already happened. It follows that it is the future cost and not the historical cost that is relevant for decision – making. Future cost is associated with future events and course of actions. It includes, among others, standard cost, budgeted cost and costs to be incurred. Standard cost and budgeted cost serve as yardsticks against which the incurred cost is compared to assess performance. The deviation between the standard and actual is analyzed for its cause and effect control. Thus, future cost is relevant to management decision making. Historical cost, on the other hand, cannot be influenced or controlled since it relates to the past. However, it should not be concluded that historical cost is totally irrelevant. In fact, it serves as a guide to planning, it helps constitutes the planning to some extent.

3.10.2 Variable vs. Fixed Cost Level of activity and profit planning, price determination, product mix planning, make or buy decisions, sell or process further, etc, are some of the important managerial decision areas. In these decisions, variable cost plays a key role and hence it is a relevant cost. Fixed cost is irrelevant in this context. Variable cost changes with the level and mode of operation. It may vary in linear or non-linear proportion. The management has to decide the particular level of operation or mix of products where there is cost advantage. In

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quoting special prices to special consumers and in fixing minimum prices for products, variable cost is an essential factor. Thus it is relevant to managerial decision making. On the contrary, fixed cost, for the total cost is not affected by fixed cost. However, in the long run all costs are variable and long run decisions are taken without highlighting the variability of costs with volume.

3.10.3 Differential vs. Total Cost To decide is to choose from between or from among alternatives. This involves evaluation of each alternative against the other. But management decision is influenced by the difference in the costs of the alternatives rather than by the total costs of the alternatives. This means the differential cost is relevant for decision making, while the total cost is irrelevant. When we consider the differential cost the distinction between variable cost and fixed cost is not entertained. That is the differential variable cost and differential fixed cost together influence decision making.

3.10.4 Controllable vs. Non-controllable Cost Controllable cost refers to a cost whose incurrence and its scale

and time are subject to executive authority. Non-controllable cost is thrust on the management by external factors. Litigation costs, cost of meeting statutory requirements etc. are examples of non-controllable cost. Non-controllable cost is irrelevant decision making as management cannot control it. But controllable cost is relevant to decision making as it can be manipulated by executive decisions. Such costs can be planned and also controlled.

3.10.5 Opportunity vs. Actual Cost

Opportunity cost refers to the value of benefits of the next best foregone alternative. Simply, it means the income from the next best missed alternative. It follows that opportunity cost would arise only when there are alternatives and decision making itself arises only when there are alternatives. Hence in all decision situations the opportunity cost concept emanates. It is specially used in resource allocation problems, when the resource in question can be put to plural uses, but one at a time.

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3.10.6 Out-of-pocket vs. Imputed Cost

Out-of-pocket cost refers to cost that involves an actual outlay of cash immediately or in the near future. Material cost, labour cost and interest on borrowed funds are examples of out of pocket cost. On the other hand, imputed costs does not involve actual outlay. It refers to assumed costs of using owned resources for the business carried on by the owner himself. Assumed interest on owner’s capital, assumed remuneration for owner’s service, assumed rental changes on owner’s building used for the business, etc. are some typical examples. For decision making the imputed cost is irrelevant as it does not affect the business funds and the cost of operation of he business in not altered by the imputed cost. On the other hand, cost involving actual outlay is relevant as the manager can plan it and exercise control over it.

3.10.7 Avoidable vs. Sunk Cost

If a manager decides to raise additional finance by issuing equity shares instead of debentures, he can avoid payment of interest. Even if the debentures are already issued the manager can raise additional equity capital and redeem the debentures, thereby avoiding interest. Avoidable cost can be of two types – that which can be avoided once and for all and that which can be avoided for the time being. Both types are relevant to decision making. It is a future cost. Sunk cost results from past decisions of actions. It is already incurred and that it is a type of historical cost. cost of production and storage of stale or unrealizable stock, investment in fixed assets are typical examples of sunk cost. But it is a past cost and future decision would not be influenced by it. Thus sunk cost is irrelevant to decision making, being unavoidable and unalterable.

3.10.8 Replacement vs. Book Cost

Replacement cost is a hypothetical cost but is relevant to decision making. It refers to the cost of replacing an asset by a new one. It is obviously a future cost. asset replacements often taken place in businesses due to technological changes, wear and tear and other factors. In this areas of decision making, the replacement cost of existing assets is an important factor. In the

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factor market, different brands and varieties of assets would be available, each involving a different cost. Hence, with the right decision the management can control the scale and time of incurrence of this cost which is relevant to decision making. On the other hand, book cost – as recorded in the accounting records – is irrelevant in replacement decisions. It is useful only for tax and financial accounting. From the decision making point of view it is of no significance, as it is only a past cost. Further, whether the book cost is the original cost or the ‘written down cost’ makes no difference. To manager deciding to replaced a particular asset its book cost, whether adjusted or unadjusted for depreciation, is irrelevant. It is the replacement cost that is relevant because it is a future cost. Thus, with the availability of a number of cost concepts, the manager has to first ascertain which costs are relevant and which irrelevant in a given situation. It can be observed from the analysis that all relevant costs have one common feature – they relate to the ‘future’. It can be stated that future costs are relevant costs. But this has to be qualified, as all future costs cannot be planned and controlled – for instance, the non-controllable and unavoidable costs which are beyond the manger’s authority, comprehension and hence control. One can conclude that all controllable future costs are relevant costs and all other costs are irrelevant for decision making.

3.11 BREAK-EVEN ANALYSIS AND ITS USES

The relationship among the volume of business, cost and profit is very significant. The profit depends upon volume of production and the type of production. Volume is the single largest factor which influences cost and profit. Break-even analysis goes deep into analyzing the relationship among cost, volume and profit. Hence it is also called C-V-P analysis. The objective of break-even analysis (otherwise called cost-volume-profit analysis) is to assist profit planning, cost control and decision making. With a view to maximize profit, the management has to expand output which will affect the variable cost, contribution and profit. Hence the management has to know the effect of an increase in output on its profit. This

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information will help the management in its budgeting and budgetary control. Decisions relating to the volume of production, type of production and pricing are influenced by a knowledge of the cost behaviour and hence the cost-volume-profit relationship becomes vital for decision-making.

3.11.1 Break-even Point

In the C-V-P analysis, the determination of break-even point is an important step and break-even analysis becomes the key component of C-V-P analysis. The break-even point is that level of production at which there is neither a profit nor a loss. In other words this is the level of production and sales at which the sales value equals to its total cost. At this point of sales, contribution is equal to fixed cost. Any sales above the break even sales only will yield profit and if the actual sales is less than the break-even sales, it would result in a loss because of the fact that the contribution is not sufficient to cover the fixed cost. That portion of the fixed cost that could not be covered is the loss. The Break-even point of sales can be computed by any one of the formula given below : Fixed cost x Sales F x S Break-Even point (in Rs.) = = Sales – Variable cost S – V

Fixed cost x Sales F x S

or = = Contribution C

Fixed cost x Sales F x S

or = = Fixed cost + profit F + P

Fixed cost F F

or = = or Variable cost 1-V/S P/V ratio 1 – Sales Break-even point (in units ) = Total Fixed cost / Contribution per unit

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Illustration 9 Calculate Break-even sales from the following figures.

Sales Rs. 1,50,000

Fixed cost Rs. 37,500

Direct Material Rs. 50,000

Direct Labour Rs. 30,000

Direct Expenses Rs. 20,000

Solution

Break-even sales = F*S / S-V 37,500 X 1,50,000 = = Rs. 1,12,500 1,50,000 – 1,00,000 Verification B.E Sales = 1,12,500 Less V.C at that level 75,000 Contribution 37,500 Less: Fixed cost 37,500 Profit Nil

3.11.2 Profit Volume Ratio (Contribution – Sales Ratios) The profit volume ratio (P/V ratio) is the ratio of contribution to sales and indicates the relative profitability of different processes, departments etc. It reveals the effect on profit of charges in the volume of production and sales. It indicates that an increase in the volume without an increase in the fixed cost would yield a higher profit and vice-versa. It is computed as follows: Contribution C i) P/V Ratio = x 100 = x 100 Sales S If selling price and variable cost are constant, then

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Change in Contribution ∆C P/V Ratio = x 100 = x 100

Change in Sales ∆S If selling price, variable cost and fixed cost are constant, then

Change in Profit ∆P P/V Ratio = x 100 = x 100

Change in Sales ∆S Using the same relationship it can be derived that : ii) Sales = Contribution / P/V Ratio iii) Contribution = Sales x P/V ratio Fixed cost + Desired profit iv) Sales required to make a desired profit = P/V ratio Fixed cost v) Sales at Break-even profit = P/V ratio

3.11.3 Properties of P/V Ratio:

i) It is the result of linking contribution to sales.

ii) It remains constant so long as selling price and variable cost per unit remain constant or fluctuate proportionately.

iii) It is unaffected by any change in level of activity.

iv) It is unaffected by any fluctuation in fixed cost.

3.11.4 Uses of P/V ratio

i) It helps in determining the Break-even point. BEP = FC / P/V ratio

ii) It helps in determining contribution / profit at any volume of sales

iii) (Contribution = S x P/V ratio; profit = C – F)

iv) It helps in determining the sales volume to earn the desired profit required.

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v) Sales = (F+P) / P/V, Where P is desired profit

vi) Helps to determine the profitability of products/product mix.

Any improvement in P/V ratio would result in increased profit and profitability. To improve the P/V ratio, either you have to reduce marginal cost, increase the selling price or by selling more profitable product or product mix. Illustration 10: A factory produces 600 units per month. Selling price is Rs.240 per unit; variable cost is Rs.160 per unit. Fixed cost per month is Rs.16,000. i) Calculate the estimated profit in a month wherein 480 units are produced and ii) find out the sales to make a profit of Rs.14,000 per month.

Solution:

Selling price per unit = Rs.240 Variable cost per unit = Rs.160 Contribution unit = Rs. 80 P/V ratio = C/S x 100 = 80 / 240 x 100 = 33 1/3 % ii) Profit on 480 units: Rs. Sales (480 x 240) = 1,15,200 Contribution 33 1/3% on sales = 38,400 Profit = C – F = 38400 – 16000 = 22400 iii) Sales to make a profit of Rs.14,000 Required profit = 14,000 Add: Fixed cost = 16,000 Required Contribution = 30,000 Required Contribution 30000 Sales required = = P/V ratio 33 1/3%

= Rs.90,000

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3.11.5 Margin of Safety The excess of sales over break-even sales is margin of safety. In other words, it is the excess of actual sales over break-even sales. Margin of safety is calculated using a formula as shown below: M.S = Profit / P/V M.S.S Margin of safety ratio = x 100 Actual Sales

Or

Actual Sales – BES

X 100 Actual Sales

Properties

i) Margin of safety indicates the strength of business.

ii) High margin indicates that concern would make profits even if there should be a fall in production or sales.

iii) Low margin of safety indicates high fixed cost and profits cannot be made unless volume of sales is increased or selling price is raised or costs reduced or a more profitable product is substituted.

Illustration 11

M.Ltd manufactures one uniform product X. The following figures are available for two successive years. Year I Year II Rs. Rs.

Sales 1,00,000 1,20,000

Fixed cost 30,000 40,000

Variable cost 50,000 65,000

The directors want information about the BEP, P/V ratio and margin of safety.

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Solution: I Year: Sales Rs. 1,00,000 Less: Variable cost Rs. 50,000 Contribution 50,000 Fixed cost 30,000 Profit 20,000 F x S 30,000 x 1,00,000 B.E. Sales = = = Rs.60,000 C 50,000 C 50,000 P/V ratio = x 100 = x 100 = 50% S 1,00,000 Margin of safety = Actual sales – BE sales = 1,00,000 – 60,000 = Rs.40,000 Margin of Safety sales Margin of Safety Ratio = x 100 Actual sales 40,000 = x 100 = 40% 1,00,000 Profit 20,000 x 100 Alternatively M.S = = = 40,000 P/V ratio 50 II Year: Sales Rs. 1,20,000 Less: Variable cost Rs. 65,000 Contribution 55,000 Fixed cost 40,000 Profit 15,000

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F x S 40,000 x 1,20,000 B.E. Sales = = = Rs.87,273 C 55,000 M.S = 1,20,000 – 87,273 = Rs.32,727. 32,727 M.S ratio = x 100 = 27.27% 1,20,000

Illustration 12

MV Ltd. a multi product company furnishes you the following data relating to the year 2001. First half of the year Second half of the year Rs. Rs. Sales 45,000 50,000 Total cost 40,000 43,000 Assuming that there is no change in prices and variable costs and that the fixed expenses are incurred equally in the two half-year periods. Calculate for the year 2001: I) The /PV ratio, ii) Fixed expenses, iii) Break-even sales and iv) Percentage of margin of safety.

Solution i) P/V ratio

Sales Total Cost Profit Rs. Rs. Rs. Second half year 50,000 43,000 7,000 First half year 45,000 40,000 5,000 Differences 5,000 3,000 2,000 Difference in Profit P/V ratio = x 100 Difference in Sales

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2,000 = x 100 = 40% 5,000 ii) Fixed expenses Sales for the I half year 45,000 40 Contribution 1 half year 45,000 x = 18,000 100 Less: profit for the I Year 5,000 Fixed expenses 13,000 Therefore fixed expenses for 1 year : 13,000 x 2 = Rs.26,000 iii) BE Sales

FC 26,000 26,000 x 100

BE Sales = = = P/V ratio 40/100 40 = Rs.65,000 iv) Margin of Safety percentage MS = AS – BES = 95,000 – 65,000 = Rs.30,000 % of M.S = (30,000 / 95,000) X 100 = 31.6%

Illustration 13

Two business units X Ltd. and Y ltd. sell the same type of product in the same type of market. Their budgeted profit and loss accounts for the coming year are as follows: X Ltd. Y Ltd. Rs. Rs. Sales 1,50,000 1,50,000 Less: Variable costs 1,20,000 1,00,000

Fixed cost 15,000 1,35,000 35,000 1,35,000 15,000 15,000

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You are required to:

a) Calculate the break even point of each business.

b) Calculate the sales volume at which the business will earn Rs.5000/- profit

c) State which business is likely to earn greater profits in conditions of

i) Heavy demand for the product

ii) Low demand for the product

Solution

FC + Profit a) i) P/V ratio = x 100 Sales 15,000 + 15,000 For X Ltd. = x 100 = 20% 1,50,000 35,000 + 15,000 For Y Ltd. = x 100 = 33 1/3% 1,50,000 FC ii) Break-even sales = PV Ratio 15,000 15,000 x 100 For X Ltd. = = = Rs.75,000 20/100 20 35,000 35,000 x 300 For Y Ltd. = = = Rs.1,05,000 100/300 100

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iii) Sales required to make a profit of Rs.5,000 FC + Profit 15,000 + 5,000 For X Ltd. = = P/V 20/100 = 20,000 X 100 / 20 = Rs.1,00,000 FC + Profit 35,000 + 5,000 For Y Ltd. = = P/V 100/300 = 40,000 X 300 / 100 = Rs.1,20,000

iv) In conditions of heavy demand a concern with larger P/V ratio can earn greater profit because of greater contribution. Thus Y Ltd. is likely to earn greater profit.

v) In conditions of low demand, a concern with lower breakeven point is likely to earn more profit because it will start earning profit form a lower level of sales. X Ltd. will start earning its profit from Rs.75,000 sales level whereas Y Ltd. will start earning profit only from Rs.1,05,000 sales level.

3.11.6 Break-even Chart The relationship between costs, sales and profits can be shown in the form of a chart. Such a chart not only depicts the level of activity, where there will be neither loss or profit but also shows the profit or loss at various levels of activity. On the X axis of the graph is plotted the volume of production or the quantities of sales and on the Y axis costs and sales revenues are plotted. The fixed cost line is drawn parallel to the X-axis and the variable cost line is depicted above the fixed cost line which shows that the cost is increasing with the increase in the volume of output. The line can also be regarded as the total

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cost line because it starts form the point where the variable cost is zero and certain fixed cost has been incurred. Thereafter the sales revenue are plotted from the origin and a line is drawn up which goes in the upward direction with the increase in production/sales. The two lines – total cost line and total sales line shall intersect each other at one point and a perpendicular can be drawn from this point to find out the level of output where the business is at no profit no loss position, called break even sales. Sales below this point will result in loss and above this point will produce profit.

Illustration 14:

From the following data compute the break-even point by a break-even chart.

Selling price per unit Rs.10

Variable cost per unit Rs.6

Total fixed costs Rs.2000

Solution

Take cost and revenue on Y axis and output on X axis. At Rs.2000 land draw the horizontal line for fixed cost. The BE point = FC/contribution per unit = 2000 / 4 = 500 units. To draw total revenue line we can take origin as one point and another at 200 units yielding a revenue of Rs.2000. To draw the TC line we start at fixed cost point at Rs.2000, which is total cost for zero output. At 200 units the TC is 2000 + 200(6) = 3200. So the two points are (0, 2000) and (200, 3200). The line is drawn. TR and TC intersect at ‘ Q’ which gives BE position. A perpendicular to x-axis gives BE production (500 units) and to y-axis gives BE sales (Rs.5000).

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C 6000 TR O TC S 5000 T Q / 4000 R E V 2000 E TFC N U E 0 200 400 500 1000 Units

3.11.7 Assumptions Of Break-Even Charts

i) Fixed cost remain constant at all levels and they do not vary with the production.

ii) Variable cost varies in the same proportion and in the same direction.

iii) All costs are capable of being bifurcated into fixed and variable elements.

iv) Selling price remains constant at all levels of production and sales

v) Cost and revenue depend only on volume and not on any other factor.

vi) Production and sales figures are identical

vii) Either only one product is produced or the product mix is constant.

3.11.8 Advantages of Break-Even Charts

i) Provides detailed and clearly understandable information regarding the break-even point, margin of safety, profit at different sales level, etc.

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ii) Profitability of the product can be known

iii) Effect of changes in cost and selling price can be demonstrated

iv) Through budget break even chart, planning and control of cost is possible.

v) Economy and efficiency in operation can be achieved.

3.11.9 Limitations of Break-Even Charts

i) Break even charts are based on certain assumptions as shown above but they are not valid in all circumstances. Fixed costs and variable costs can not be correctly segregated, fixed costs do not always remain constant; variable costs do not always vary proportionately; production and sales are always not equal and so on. Hence, the concept is based on unrealistic assumptions.

ii) Break even charts provide only limited information. If we have to study the effect of changes of fixed cost, variable cost and selling prices a number of charts will have to be drawn up.

iii) There is no necessity for a break-even chart because (a) simple tabulation of data is sufficient (b) conclusive decisions are not possible (c) difficult for a layman to understand and (d) there is no basis for comparative efficiency of different units.

In spite of all these limitations, it is a useful tool for the management for analyzing its different day to day problems in decision making.

Questions

1) Explain different costs of operating a system.

2) Explain production function and behaviour of TP, MP, AP with an example.

3) Graph the behaviour of TP, MP and AP.

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4) Graph the behaviour of TC, TFC, TVC, AC, AFC and AVC.

5) With 15 units of labour 45 units of output and obtained while 20 units of labour yield 55 units of output. Find the linear production function between output and labour.

6) A quadratic production function is as follows: Q = -20x2 + 160x + 400, where Q = output and x = an input. Find the marginal productivity of x and output optimizing level of x.

7) A cubic production function present is as follows: Q = 20x + 4x2 - x3/3, where Q = output and x = an input. Find units of x that gives highest Q and maximum value of marginal productivity of x.

8) A production function is as follows: Q = 2HLK - AL2 - BK2. Find the MP of L and MP of K and from that marginal rate of substitution of L for K at any point on the iso-product curve.

9) Explain the different forms of production function.

10) Explain the features of Cobb-Douglas production function.

11) Explain the difference in behaviour of SRTC and LRTC.

12) Explain the derivation of LAC, LMC and LTC.

13) Total cost function is: TC = x2 - 120x + 15000. Find the MC and cost minimizing total output.

14) Comment on the behaviour of cost of the two functions: C1 = -x2 + 60x + 6000 and C2 = x2 - 60x + 6000.

15) The profit function is as follows: P = -2x2 + 400x - 1700, where profit =

P, and x = units of input. Find the profit maximizing output level.

16) The cost function for a firm producing x units is: TC = 5x + 350 and its revenue function is: TR = 5x - x2. Find the demand function, break-even-point, profit maximizing output and sales maximizing output. [Hint: i) For break-even point, solve for x, by equating TC = TR].

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17) A review, made by the top management of Sweat and Struggle Ltd. which makes only one product, of the result of the first quarter of the year revealed the following:

Sales in unit 10,000 Loss in Rs. 10,000 Fixed cost (for the year Rs.1,20,000) in Rs.30,000 Variable cost per unit in Rs. 8 The finance manger, who feels perturbed suggests, that the company should atleast break even in the second quarter with a drive for increases sales. Towards this, the company should introduce a better packing which will increase the cost by Rs.0.50 per unit. The sales manager has an alternate proposal. For the second quarter additional sales promotion expenses can be increased to the extent of Rs.5,000 and a profit of Rs.5,000 can be aimed at the period with increased sales. The production manager feels otherwise. To improve the demand, the selling price per unit has to be reduced by 3%. As a result the sales volume can be increased to attain a profit level of Rs.4,000 for the quarter. The Managing Director asks you as a Cost Accountant to evaluate these three proposals and calculate the additional sales volume that would be required in each cash, in order to help him take a decision.

18) Details about the single product marketed by a company are as under: Per unit Rs. Selling price 100 Directory material 60 Direct labour 10 Variable overheads 10

No. of units sold in the year 5035. Pursuant to an agreement reached with the Employees union there would be next year a 10% increase in wages, across the board for all those directly engaged in production. Work out:

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i) How many more unit have to be sold next year to maintain the same quantum of profit.

ii) In else, by what percentage the selling price has to be raised to maintain the same P/V ratio.

19) Ram Dass Pvt. Ltd. Nasik is currently operating at 80 per cent capacity. The profit and loss account shows the following:

Rupees in lakhs Sales 640 Cost of sales : Direct materials 200 Direct Expenses 80 Variable Overheads 40 Fixed Overheads 260 580 60 The managing Director has been discussing an offer from Middle East of a quantity which will require 50 percent capacity of the factory. The price is 10 percent less than the current price in the local market. Order cannot be split. You are asked by him to find out the most profitable alternative. The factory can be augmented by 10 percent adding facilities at an increase of Rs.40 lakhs in fixed cost.

� � �

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LESSON – 4

LINEAR PROGRAMMING

Linear programming (LP) is an optimization technique. In business there are scores of problems requiring optimal solutions. Generally the resources of a business are limited. These resources can be put to multiple was. Minimum of maximum levels for certain uses are fixed, adding further constraints. Under the situations, the business wants to maximize return, revenue or profit or minimize cost or consumption of resources. These situations can be effectively handled through the LP technique. LP technique helps in finding the optimal solution. The minimization / maximization objective is fulfilled subject to the different constraints. Linear programming has its origin in the input-output analysis developed by W.W.Leontiff, a reputed economist. LP assumes that the variables are in linear, i.e., straight line, relationship with one another. It assumes that the problem can be stated in mathematical equations. The term ‘programming’ refers to the use of certain mathematical techniques to arrive at the optimum solution. ‘Programming’ also means a sort of routine, or doing things as per predetermined rules and procedures. In the LP technique a linear objective of the firm and a set of linear equations/inequalities being the constraints are involved. We have to optimize the objective function satisfying the constraints.

4.1 CONCEPT OF OPTIMISATION Optimization is maximization subject to constraints and minimization subject to constraints. Business wants maximization of sales revenue, profit, market share, etc. But limited resources, multiple uses of resources , price differences of resources, etc. put difficulties on maximizing sales revenue, profit, etc. So, constrained maximization i.e., maximization subject to constraints is sought. This is optimization. In the same way minimization of cost, time or risk subject to constraints is called constrained optimization.

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4.2 USES OF LP TECHNIQUE IN DECISION MAKING AND CONTROL

There are diverse situations where the LP technique can be adopted. Prominent of them are enumerated below:

i) In the field of production management the distribution of limited resources (raw material, labour hours, machine capacity, etc.) to different uses (products, divisions, markets) so as to maximize contribution is decided using LP technique.

ii) In production management, determining the optimum mix of inputs so that total cost of production is minimized, subject to some constraints as to minimum or maximum limits on the use of one or more inputs.

iii) Instead using the available resources in production of certain goods or services, the resources themselves can be dispose, if so decided. In such cases the minimum price at which these resources may be sold can be determined through LP technique.

iv) In marketing management, sales-mix can be decided to maximize profit/ revenue given the capacity constraint for each product and the demand constraint of each market for the different products.

v) In distribution management to decide the least cost transportation of goods from certain number of supply centers, each with certain quantity of the goods, to certain number of demand centers, each requiring certain quantity, given the origin-destination-wise transportation cost per unit. (A better method is the transportation model to be dealt with in the next lesson).

vi) In the case of a banking concern to decide the sector-wise credit-mix so as to maximize return, subject to overall sectoral limits and region-wise sectoral credit plans.

vii) In financial management to decide the least cost capitalization, subject to ceiling on the use of on one or more forms of capital.

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viii) In farm management to decide the acreage under each crop to maximize return given the return per acre for each crop and constraints as to amount available for land preparation and the total acreage.

We can go on giving examples like the above. But this is not the intention. You should note whenever a gain is to be maximized or an outgo is to be minimized and there are many linear constraints, LP technique can be used.

PRESENTATION OF PROBLEMS IN LP MODE

In presenting business problems in LP mode certain terminologies are involved. These are: objective function, constraints, inequality and non-negativity constraints, etc. These you will know as we present the problems in LP mode.

Illustration 1:

Consider this. A firm makes a contribution to Rs.10 and Rs.12 from the two products A and B, respectively, it manufactures. One unit of A requires 2 units of raw material-I and 3 units of material-II, and one unit of B needs 1 unit of material I and 4 units of material II. The firm has 60 units of material I and 120 units of material II. Find the optimum number of units of A and B. In this case the objective is to maximize total contribution. Total contribution is a function of quantity and per unit contribution. If we decide that ‘a’ units of product A and units of product B, be produced, then total contribution is given by 10a + 12b. One unit of A gives Rs.10; therefore ‘a’ units will give 10a. similarly one of B gives Rs.12; so ‘b’ units will give 12b amount of contribution. So total contribution = 10a + 12b. For a particular value of a and b, 10a + 12b would be the highest. Our objective is to find those values of a and b. Hence the objective is: Maximize 10a + 12b. This is what we call as objective function, i.e., the function representing our objective. If there were no constraints, as unit contribution is maximum in the case of B, to maximize total contribution we may simply product ‘B’ only. But there are constraints. One unit of A requires 2 units of material I and one unit of

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B requires 1 unit of material I. We have 60 units of material I and decide to produce ‘a’ units of A. So, 2a units of material I are required for production of the given level of product A; and the number of units of material I required to produce ‘b; units of B is 1b. Total of 2a + 1b should be equal to or less than the total of material I available, viz. 60 units. That is, 2a+b<60. Similarly for material II, 3a+4b<120. The symbol, < means “equal to or less than”. Further the number of units of A and ‘B’ cannot be less than zero. In other words, a and b, can be equal to or more than zero. This we write as a, b>0. So, the constraints are 2a + b < 60 …. (1) 3a + 4b < 120 …. (2) and a, b > 0 …. (3) The last one is referred to as the non-negativity condition, since it stresses that the variable, a and b, cannot take negative values. Items (1) and (2) as well as (3) are not equations form. The R.H.S. is not equal to L.H.S. Hence these are not equations or otherwise put these are inequalities. So far the terminologies are explained. Now we can put the problem as below: Objective function : Maximize Profit = P = 10a + 12b Subject to : 2a + b < 60 | Constraints 3a + 4b < 120 | a, b > 0 Non-negativity condition. You may now do some exercises on the mathematical formulation of an LPP.

Illustration 2:

A firm has to produce 400 lbs mixture containing three ingredients A, B and C, a pound of each costs Rs.18, 8 and 10 respectively. The total volume of A and C should not be less than 250 lb, that of B should not be more than 100 lbs and that of C not more than 125. Find the optimal mix.

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Solution:

Here you need to produce a mixture of A, B and C measuring in all 400 lbs. In so doing you have to minimize cost. And there are other constraints as well. Let us decide that “a” units of ingredient ‘A’, “b” units of ‘B’ and “c” units of ‘C’ are mixed. Then the Objective function: minimize Cost = C = 18a + 8b + 10c Subject to : a + b + c = 400 a + c > 250 b < 100 Constraints c < 125 a,b,c > 0 Non-negativity condition.

Illustration 3 :

Let us taken another example. A firm produces 2 products, M and N, 1 unit of M requires 15 and 20 minutes to grind and assemble respectively, and 1 unit of N needs 15 and 10 minutes respectively. The production run calls for at least 7.5 hours of grinding time and at least 8 hours of assembly time. If M costs Rs.60 and N Rs. 90 to manufacture find the optimum number of units of M and N.

Solution:

This is a minimization problem. We have to minimize cost. If we take that ‘m’ units of M and ‘n’ units of N are to be produced the objective function would be. Minimize 60m + 90n This is to be done subject to: 30m + 30n > 450 minutes | 24m + 48 n > 480 minutes | constraints and non-negatively condition m,n > 0 | (Note 7-5 hours – 450 minutes and 80 hours = 480 minutes).

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Since the grinding and assembling hours have to be at least 7.5 hours and 80 hours, 30m + 30n be ‘equal to or more than 15 hours or 900 minutes and 24m + 4bn be equal to or more than 8 hours or 480 minutes.

Illustration 4 :

Take another example. Old hens cost Rs.2 each. Young ones cost Rs.5 each. The old lays 3 eggs a week and the young 5 a week. Each egg is worth a 30 paise of Re.0.3, Re.1 is needed to feed a hen a week. If Rs.80 are available and not more than 20 hens be maintained at a time, find the optimal mix of old and young hens. This is a maximization problem. Let us take ‘a’ number of old and ‘b; number of young hens. Total egg production per week is therefore 3a+5b. The value of this at the rate of 30 paise per egg is: Re.0.3 (3a + 5b). This is gross earnings. We have to feed the hens which cost Re.1 per hen per week. So, the weekly feeding costs Rs.1 (a+b) or a+b. So, net earnings per week = 0.3 (3a+5b) – (a+b). So our objective function is:

Maximize: 0.3(3a+5b) – (a+b) = 0.9a + 1.5b – a – b = 0.5b – 0.1a

Subject to: 2a + 5b < 80

(the purchase cost be equal to or less than Rs.80)

: a+b < 20

(number of hens at any time be restricted to 20 or less)

: a, b > 0 (non-negativity condition).

4.3 USE OF GRAPHICAL METHOD OF SOLVE LPP So far we just formulated LPP in a mathematical way. We have to solve them. Graphical method can be used to solve LPP, whenever the number of unknown variables is 2. We use certain terms like, feasible solution, bounded and unbounded area here. These will be explained as we solve the problems.

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Illustration 5 : Maximization problem through graphic solution

Consider our illustration 1 given earlier. We may solve the problem using the graphical method. Objective function: Maximize P = 10a+12b Subject to : 2a + b < 60 3a + 4b < 120 a, b > 0 To solve that problem, first the inequalities are, for the time being, taken as equalities. That is, 2a + b < 60, becomes 2a + b = 60 and 3a + 4b < 120, becomes 3a + 4b = 120

On the X-axis product `A' and on the Y-axis product `B' are taken. Say, all the available material I is used to produce product A only. Then 60/2 = 30 units of A can be produced. If all of material I is used to produce B, 60/1 = 60 units of B can be produced. We obtain a straight line connecting these two points: 30 on X-axis and 60 on Y-axis. This line represents,

2a + b = 60. Similarly, if all of material II is used to produce A, 120/3 = 40

units can be produced. If all of material II is used to produce B, 120/4 = 30 units can be produced; join these points: 40 on X-axis and 30 on Y-axis. This line represents 3a + 4b = 120. The area OPQR is the bounded area, bounded by both the equations. Graph 4.1 gives the picture. Any point in the bounded are would meet all the constraints we set in Example 1. So, the bounded area represent all feasible solutions. Feasible Solution thus refers to any solution meeting all constraints. Outer the bounded area, either or both of the capacity constraints is/are not met.

The solution to the problem lies at the corner points of the bounded

area, viz., O, P and R. Point `O' represents nil production of both A & B. Well that is also a feasible solution as it satisfies all the constraints. But the contribution is zero. Point P means a production of 30 units of A and no unit of B. The contribution is 10 x 30 + 12 x 0 = 300. Point R represents 30 units of B

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GRAPH 4.1

60 Maximization

50

B 40 2a+b = 60

30 R

20 Q

10 3a+4b = 120

0 10 20 30 P 40 50 60

A

and no unit of A. The contribution is 10 x 0 + 12 x 30 = Rs. 360. Point Q can be solved by solving the two equations:

2a + b = 60 - (1) 3a + 4b = 120 - (2) Multiply (1) by 4 and we get, 8a + 4b = 240 - (3) (2) - (3), -5a + 0 = -120 or, a = 24. Then b = 12. This means 24 units of A and 12 units of B

are represented by point Q. The contribution is (10 x 24) + (12 x 12) = Rs. 384 Of the 4 corner points, point `Q' produces maximum contribution.

Hence 24 units of A and 12 units of B would have to be produced to maximize contribution.

Alternatively, we choose a minimum contribution level, say Rs.

120. To meet this, 12 units of A and 10 units of B are needed. Get a straight line corresponding to 12 units of A and 10 units of B. Draw parallel lines to the above line. That parallel line farthest from origin and touching any one of the points indicates the optimal solution.

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Illustration 6

Illustration 3 given earlier may be taken up now for solution. Given: Minimize cost C = 60m + 90n Subject to: 30m + 30n >50 min. 24m + 48n>480 min. m, n >0 Now we convert inequalities into equalities. So we get: 30m + 30n = 450 24m + 48n = 480 If all grinding time, viz., 450 minutes, is used to produce M, 30

units of M can be produced. If this time were used to produce N, then a maximum of 360 units can be produced. By joining these two points on X and Y axes, respectively, we get the line representing 30m + 30n = 450. Similarly if all available assembly time is used to produce M, 24 units of M, or if diverted to produce N, 48 units of N can be produced. So we get the line, 24m + 48n = 480. You may note from the inequality that 30m + 30n > 450 and 24m + 48n > 480. Graph 4.2 gives the picture.

So, the solution lies in the area outside the bounded area of the two

lines on the graph, 4.2. In other words, the area lying beyond PQR (the shaded area) contains all feasible solutions. As we want to minimize cost, and also satisfy the inequality constraints, we would get the solution if we solve for P, Q and R. Corner `P' satisfies all constraints, i.e., assembling time is 480 mts., and grinding time well over 450 mts. Here the production cost is Rs. 48 x 90 units of N = Rs. 4320. If we consider corner R, representing 30 units of M, the cost would be Rs. 30 x 60 = Rs. 1800. This point also satisfies all constraints. Point Q is solved by solving the two equations.

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GRAPH 4.2 : MINIMISATION

800

48 P

Less

30 N

20

10

M R

5 10 15 20 25 30 35

30m + 30 = 450 - (1) 24m + 48n = 480 - (2) (1) x 4 we get 120m + 120n = 1800 - (3) (2) x 5 we get 120m + 240n = 2400 - (4) (3) - (4), -120n = -600 n = 5; m = 10 Then the cost of 10 M and 5 N = (60 x 10) + (90 x 5) = Rs. 1050. Of the three options, production of 10 units of M and 5 units of N

would cost the least, viz., Rs. 1050. So, this is the optimum production mix.

4.5 USING SIMPLEX METHOD TO SOLVE LPP

LPP can be solved through another method called simplex method. It was mentioned earlier that graphical method is effective only when 2 variables are there. If there are more number of variables, different method is called for. Simplex method is the other method.

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Simplex method is an interactive (repeat) procedure which either solves a LPP in a finite number of steps or gives an indication that there is an unbounded solution to the given LPP. Here we use some terms like slack

variable or slack resource, artificial variable or resource basic feasible

solution, improvement index, optimal column, intersectional elements, etc. These will be explained as we take up solutions to LPPs.

4.5.1 Maximization through simplex method

The simplex method, the iterative method, is now adopted to solve a maximization problem.

Illustration: 7

We may now take the illustration 1 we discussed earlier for adopting the simplex method. You may recollect that the:

Objective function: Maximize 10a + 12b Subject to: 2a + b < 60 - (1) 3a + 4b < 120 - (2) a, b > 0

Solution:

To adopt the simple x method, the inequalities (1) and (2) have to be converted into equalities by introducing slack variables. You may note that in 2a + b < 60, the LHS is less than the RHS.

To equate the LHS to RHS, on the lower side we add few quantity,

which we call as slack resource. This may be referred to by S1, S2, ... Sn. So the inequality (1) is converted into an equality as follows: 2a + b + S1 = 60. The S1 simply tells the unutilized material 1. If all 60 units of I are used S1 will be zero. If some units are left S1 will be a positive quantity. Whether S1 is zero or some positive quantity, the equation 2a + b + S1 = 60 will hold good. Similarly the inequality 3a + 4b < 120 is converted into an equation, 3a + 4b + S2 = 120, S2 being the slack variable, whose value would be zero or more than zero. In other words, the non-negativity condition would cover both real and slack variables. While products A and B would give us profits, unutilized material I, i.e., the

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slack resource S1 and utilized material II, i.e., the slack resource S2 would not give any. So, the return from S1 and S2 is zero, each.

In the simplex method the slack variables are also incorporated in

the objective function. So, the objective function is: Maximize P = 10a + 12b + 0S1 + 0S2 Subject to: 2a + b + S1 = 60 - (1) 3a + 4b + S2 = 120 - (2) a, b, S1, S2 > 0 Equations (1) and (2) are modified in such a way that both contains

the other slack variable also. The coefficient of the `other slack variable' would be put as zero. This means `other slack variable' is included only to satisfy some structural requirements, but no value is attached.

So we get, 2a + b + S1 + 0S2 = 60

3a + 4b + 0S1 + S2 = 120

Now, we can proceed with the solution process. Below, initial solution table is presented. Tableau-1 gives the same.

Tableau 1: Initial Solution

Cj Column

Product mix

Quantity Cj Row: 10 12 0 0

Variable Row: a b S1 S2

0

0

S1

S2

60

120

2 (1) 1 0

2 (4) 0 1

Zj 0 0 0 0 0

Cj-Zj -0 10 12 0 0

(Optimum column)

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Explanation of the Table :

Cj row gives the contribution per unit for the real and slack variables. Variables row gives the variables - real and slack. Cj column gives contribution per unit for those variable found in the product mix column. Product mix column gives the products (real and in slack variables) that enter into the solution. Quantity column gives the quantity produced for each of the product found in the product-mix column, Zj row gives the loss that would be incurred if one unit of each of the variables in the variables row is produced. From Cj if we deduct Zj - the net loss / net gain that would be incurred by producing unit each of the variables is obtained.

Now coming to the table, Cj row gives the contribution per unit for

a, b, S1, S2 which are Rs. 10, Rs. 12, Rs. 0 and Rs. 0 respectively. The variables row gives the variables a & b (real) and S1 and S2 (slack).

The initial solution in a simplex method is a situation where you

do not produce any real variable. So, a and b become zero, giving S1 and S2 values respectively 60 and 120. These S1 and S2 are, therefore, in the product mix and 60 against S1 and 120 against S2 are placed in the quantity column. S1 and S2 simply represent unutilized resource and, hence they do not give any return. So, in the column against S1 and S2, their contributions viz., Rs. 0 and Rs. 0 respectively are entered. In the S1 row, below each of the variables, are written the coefficients of the constraint equation, (1). In the S2 row, below each of the variables, are written the coefficients of the constraint equation (2). The figures under each variable in the Zj row are the loss they would result if a unit of the respective variable is produced. Take variable a, i.e., product A. To produce one unit of A, 2 units of S1 and 3 units of S2 (that is 2 units of material - 1 and 3 units of material - II) are required. But as S1 and S2 are worth nil, by producing one unit of `A' no loss is incurred, i.e., (2 X 0 + 3 X 0 = 0). To produce a unit of B, that is product B, 1 unit of S1 and 4 units of S2 are required. But the value of these is only 0. Similarly to produce one unit of S1, one unit of S1 is required and the value is 0 and similarly for S2 it also the value is zero. In the quantity column of Zj row the total profit by producing the products in the product-mix shown in the quantity column is given. This is simply the sum of respective quantity times respective contribution. Here it is 60 X 0 + 120 X 0 = 0. Now the Cj-Zj row giving figures for each of the variable is worked out. It is simply got by

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subtraction of the figures in the Zj row from the corresponding figures in the Cj row given at the top of the table. Cj-Zj is the net profit by making one unit of each of the variables in the variables row. So, Cj-Zj for variable a is, 10-0=10: for b is 12-0=12: for S1 it is 0-0=0: and for S2: 0-0=0. Obviously, where the net gain is more, you would be attracted there. Here, variable `b' has a maximum of Rs. 12 net contribution per unit. This `b' column with the highest positive Cj-Zj figure is called as the optimum column.

The figures in the optimum column against each of the variables in

the product mix column are referred to as inter-sectional element. These are bracketed for easy identification.

Now we have to decide which of the products, A or B, to be

produced, for tableau 1 simply gives the initial and not the final solution. Final solution is reached only when all figures in the Cj-Zj row are either negative or zero. When we decide to produce a product, that variable will enter into the product mix column and consequently one variable already there will have to be removed. Which one to enter and which to leave the product mix column are decided in the following manner. The variable of the optimum column enters the product mix column. We have to calculate improvement index (I.I) values for each product of the present product mix by dividing the corresponding quantity column figure by the corresponding inter-sectional element.

Improvement Index = I.I = elementsectionalInter

valuecolumnQuantity

−

In so doing, if for a row the inter-sectional element is zero or a negative figure, for that row I.I will not be calculated and the product of that row will not be disturbed. That is, that will continue to exist.

Now, I.I for S1 row = 60/1 = 60 I.I for S2 row = 120/4 = 30 � (Leaving Variable) The variable with the lowest (or lower) I.I will be removed from

the product and mix the variable of the optimum column be introduced in the

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product-mix. In this case `b' will be introduced and S2 will be removed. The S2 row is called leaving or replaced row. Now the simplex tableau-2 may be worked out.

In the tableau-2 `b' is entered into product-mix column first. The

contribution of Rs. 12 per unit is entered in the Cj column.

Cj Column

Product mix

Quantity Cj Row: 10 12 0 0

Variable Row: a b S1 S2

12

0

b

S1

30

30

(3/4) 1 0 1/4

(5/4) 0 1 (-1/4)

Zj 360 9 12 0 3

Cj-Zj -- 1 0 0 -3

(Optimum column)

As per second solution 30 units of b are produced. This will leave

30 units of S1 unutilized. Hence this appears in the product-mix and quantity columns. `b' row is the entering row It is introduced in the place of S2 row of the previous tableau. The figures for each variable of the entering row is obtained simply by dividing value of each column, right from quantity column, of the leaving S2 row of previous tableau by the intersectional element of that row. So, for `b' row in the tableau 2, the quantity column figure is obtained by dividing corresponding figure of the leaving S2 row of previous tableau i.e., 120/4 = 30. Similarly for `a' variable the figure is 3/4; for `b' variable the figure is 4/4 or 1; for S1, 0/4 or 0; for S2 it is 1/4. Now `b' row - the entering row, is completed. The S1 row continues in the 2nd tableau. But its values are different from those of tableau 1. Value of each variable of the S1 row in the tableau 2 is to be obtained as follows:

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Value of the Corresponding Inter-sectional Variable of value of the element of the the row in – variable of the X row in the old the previous replacing row tableau tableau in new tableau So, quantity Quantity col. Quantitycol. Intersectional Column figure value of S1 row value of the element of the for S1 row in = in the old – entering row X S1 row in the the new tableau (b row) in the old tableau tableau new tableau = 60 - (30 X 1) = 30

Similarly value: for `a' = 2 - (3/4 X 1) = 5/4 for `b' = 1 - (1 X 1) = 0 for `S1' = 1 - (0 X 1) = 1 for `S2' = 0 - (1/4 X 1) = -1/4 These, 30, 5/4, 0, 1 and -1/4 are written in S1 row of the tableau 2.

Now the Zj and Cj-Zj are worked out in the usual way. In the Cj-Zj row, we find that only one positive value in the `a' column is found. That is the optimal solution is not still achieved. So we have to continue our iteration. The optimum column is `a' column now. Therefore in the next solution tableau variable `a' will enter the product mix. To find which of the existing items in the product-mix will have to be removed, we calculate I.I values for tableau 2. The intersectional elements are: 3/4 and 5/4, which are bracketed for easy identification.

So, I.I for `b' row = 30/(3/4) = 40 I.I for `S1' row = 30/(5/4) = 24 The lower of the two I.I values is 24 and it is learnt that S1 will

have to be removed. So in the tableau 3, given below `a' variable enters as the entering or replacing row. Of course `b' row is also there as it is retained.

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Simplex tableau 3: Third solution

Cj Column

Product mix

Quantity Cj Row: 10 12 0 0

Variable Row: a b S1 S2

10

12

a

b

24

12

1 0 4/5 -1/5

0 1 -3/5 8/20

Zj 384 10 12 4/5 14/5

Cj-Zj -- 0 0 -4/5 -14/5

The values for row `a' are obtained by dividing S2 row figures of

the previous tableau by the intersectional element of that row. The `a' row is the entering row. The `b' row figures in tableau 3 are obtained in the same way as we obtained the values for S1 row of tableau 2. Zj and Cj-Zj are obtained as usual. Since all values in the Cj-Zj are either zero or negative, no more further improvement is possible. In other words, the optimal solution is reached. So, we produce 24 units of A and 12 units of B and we get Rs. 384 as contribution. For all these you may refer to the product mix and quantity columns. You may further note, the same results had been obtained under our graphical solution to the above problem.

4.5.2 Minimization through simplex method

Illustration 8

We may take up the following example. Objective function : Minimize: 60P + 120Q subject to : 2P + 3Q >10 P + 4Q > 12 P, Q > 0

Solution :

The inequalities have to be converted into equations. The LHS is greater. So we subtract slack variables. We get:

2P + 3Q - S1 = 10 P + 4Q - S2 = 12

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If P and Q were to be equal to Zero, then, S1 = -10 and S2 = -12.

But this will not be permitted by the non-negativity condition which stress that no variable can take negative value. So, the above equations have to be modified. We introduce artificial resources called A1, A2, ... An. The artificial variable (resource) is one which is of very high cost, taken as, great M, and is a substitute for the various resources. If we add these A's to LHS, we can get over the problem of not fulfilling non-negativity condition. We thus get:

2P + 3Q - S1 + A1 = 10 P + 4Q - S2 + A2 = 12 Now the objective function can be rewritten as: Minimize: 60P + 120Q + 0S1 + 0S2 + MA1 + MA2 Subject to: 2P + 3Q – S1 + S2 + A1 + 0A2 = 10 P + 4Q + 0S1 - S2 + 0A1 + A2 = 12 P, Q, S1, S2, A1, A2 > 0 Now the initial solution is obtained in tableau 1. We start with the

artificial variables, if there are any. That is we take up the high cost option first and move to least cost options thro' iterations. Since in both the constraints, we have artificial resources, these will enter our product mix. If in any equation there is no artificial resource, the slack resource of that constraint will enter the solution. Note both A1 & A2 cost M, assumed very high figure.

Simplex Tableau –I : Initial Solution

Cj Mix Qty 60 120 0 0 M M

P Q S1 S2 A1 A2

M A1 10 2 (3) -1 0 1 0

M A2 12 1 (4) 0 -1 0 1

Zj 22M 3M 7M -M -M M M

Cj-Zj - 60-3M 120-7M M M 0 0

Optimum column

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In the case of minimization problem the column with the highest negative Cj-Zj value is taken as the optimum column. (You may remember in the case of maximization problem the optimum column is given by the highest positive Cj-Zj value). Variable `Q' has the highest negative value with 120-7 M. This enters into solution. The intersectional elements are: for A1 row, it is 3 and for A2 row, it is 4. The respective I.I values are: 10/3 and 12/4 or 3. The lower of these is 3. So in our next solution A2 is replaced by Q. Simplex tableau 2 is given below: `Q' row figures in tableau 2 are obtained by dividing outgoing A2 row figures by the intersectional element: 12/4 = 3, 1/4, 4/4 = 1. 0, -1/4, 0 and ¼.

Simplex Tableau – 2

Cj Mix Qty 60 120 0 0 M M

P Q S1 S2 A1 A2

120 Q 3 ¼ 1 0 -1/4 0 1/4

M A1 1 5/4 0 -1 3/4 1 -3/4

Zj M+360 5/4 M+30 120 -M 3/4M-30

M 30-3/4M

Cj-Zj - 30-5/4M 0 M 30-3/4M

0 1/4M-30

OPTIMUM COLUMN A1 row figures in tableau 2 are obtained as follows: (refer formula

given under maximization): Qty Column: 10 - (3 X 3) = 1 P Column: 2 - (1/4 X 3) = 5/4 Q Column: 3 - (1 X 3) = 0 S1 Column: -1 - (0 X 3) = -1 S2 Column: 0 - (-1/4 X 3) = 3/4 A1 Column: 1 - (0 X 3) = 1 A2 Column = 0 - (1/4 X 3) = -3/4

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The optimum column is the `P' column with highest Cj-Zj for value of 30 - 5/4M. The I.I value for Q row: 3 / (1/4) = 12 and for A1 row = 1 / (5/4) or 4/5. 4/5 is the lower of the two. So, A1 is replaced by P. The next solution is found in tableau 3. `P' row figures are obtained by dividing outgoing A1 row figures by its intersectional element. The figures are: 1 / (5/4) = 4/5, (5/4) / (5/4) = 1, 0, -1 / (5/4) = -4/5, (3/4)/(5/4) = 3/5, 1/(5/4) = 4/5 and (- 3/4) / 5/4 = -3/5

Simplex Tableau – 3

Cj Mix Qty 60 120 0 0 M M

P Q S1 S2 A1 A2

60 P 4/5 1 0 -4/5 3/5 4/5 -3/5

120 Q 14/5 0 1 1/5 -2/5 -1/5 2/5

Zj 384 60 120 -24 -12 24 12

Cj-Zj - 0 0 24 12 M-24 M-12

The figures for Q row in tableau 3 are obtained as follows: Qty col: 3 - (4/5 X 1/4) = 14/5 P col: 1/4 - (1 X 1/4) = 0 Q col: 1 - (0 X 1/4) = 1 S1 col: 0 - (-4/5 X 1/4) = 1/5 S2 col: -1/4 - (3/5 X 1/4) = -8/20 or -2/5 A1 col: 0 - (4/5 X 1/4) = -1/5 A2 col: 1/4 - (-3/5 X 1/4) = 8/20 or 2/5 In the Cj-Zj row all values are non-negative. M being very high

figure M-24 and M-12 are all positive which you should note. Since no value is negative in the Cj-Mj row the optimum solution is reached.

So the value of P is 4/5 and that of Q is 14/5. The value of the

objective function is: 60 X 4/5 + 120 X 14/5 = 48 + 336 = 384

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4.6 DUALITY

Every linear programming problem has two sides: the primal and

the dual. That is, in maximization problem there is a minimization problem and vice versa. The LPP as given taken as the primal. Its other side, is the dual. So for a maximization problem. maximization is primal and its minimization is its dual. You can solve the primal thro' its dual. From primal, the dual's answers can be obtained and from the dual solution, the solution to its primal can be obtained. So why do you present the dual at all? the answer is: firstly a problem can be easily solved by its dual than thro' the primal. The primal may have more constraints, whereas its dual may not. Then solving the dual and finding answers to the primal thro' the dual is good and easy. The opposite situation will prevail if the dual has many constraints.

4.6.1 What is duality?

Consider our illustration 1 where we produce products A & B with

the help of material I and II. We solved that LPP for maximization of contribution. It can be solved for its dual. If done so, our question would be what minimum price would be acceptable to us for a unit of material I and material II? It is not always necessary to convert materials into goods and sell the produced goods. The materials themselves can be disposed off, it is applicable to all resources. So, what minimum price would be acceptable to us is the question.

There we say that 2 units of material I and 3 units of material II are

required to produce one unit of ‘a’ which gives Rs. 10 contribution. Suppose, one unit of material I is worth P rupees and that of material II worth Q rupees. If you sell the materials, that would be committed to a unit of A, then you would expect that 2P + 3Q>= 10. 2P + 3Q is the value of both of the materials spent to produce one unit of A which gives Rs. 10 contribution. So to expect that 2P + 3Q is at least equal to Rs. 10 is very reasonable. Similarly, the value of materials spent on one unit of B is, P + 4Q. This should be at least equal to the contribution from B, viz., Rs. 12. That is P + 4Q >= 12. You would be happy to get the acceptable minimum for both the materials. You have 60 units of I and 120 units of material II. So their value is 60P + 120Q. So, the objective function is:

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Minimize 60P + 120Q Subject to: 2P + 3Q >= 10 P + 4Q >= 12 P, Q >= 0 This we have solved in our illustration 8. So, illustration 8 is the dual of illustration 7.

4.6.2 Solving dual through primal I mentioned that from the primal you can solve for the dual. Take

our primal-maximization done earlier. As per its dual, the minimum acceptable price for a unit each of material I and material II are needed. Look at the maximization tableau 3 in Illustration 7. Under S1 and S2 columns, in the Cj-Zj row we find values -4/5 and -14/5. Just convert the negative signs into +ve signs and the values are 4/5 and 14/5. Now look at the minimization tableau 3 in Illustration 8 under quantity column. Against P there is 4/5 and against Q there is 14/5. Thus Rs. 4/5 is the value of a unit of material I and 14/5 is the value of material II.

4.6.3 Solving Primal through dual From the dual (minimization) the solution for the primal

(maximization) can be obtained. Look at the minimization tableau 3 of Illustration 8, S1 and S2 columns. In the Cj-Zj row we get the figures 24 and 12. Now look at the maximization tableau 3 of Illustration 7, quantity column. Against A there is 24 and against B there is 12. The same values are found in the dual's Cj-Zj row under the corresponding slack resources columns. Hence the primal is solved through dual.

You need to know how to write the dual under some situation. Some cases are presented.

Illustration 9: You are given the primal:

Maximize: 15A + 20B Subject to: 2A + 4B = 100 5A + 4B =< 160 A, B >= 0 Write the dual.

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Solution:

Before we attempt to write dual, the constraints should be converted such that all of them are of the same form i.e., "equal to or greater than" or "Equal to or less than". The equal to type constraint, 2A + 4B = 100 can be dealt by two inequalities as follows:

2A + 4B >= 100 and - (1) 2A + 4B =< 100 - (2) Then (1) can be written as -2A - 4B < - 100

Now all the constraints are of the "equal to or less than" form. That is: -2A - 4B < - 100 - (4) 2A + 4B < 100 - (5) 5A + 4B < 160 - (6) If we take (4), (5) and (6) as different resources with unit value of

Rs. P , Q and R respectively, the dual (minimization) can be written as follows: Minimize: -100P + 100Q + 160R Subject to: -2P + 2Q + 5R >= 15 -4P + 4Q + 4R >= 20 P, Q, R >= 0 Note that when the constraints of the primal are "Equal to or less

than type", those of the dual are "equal to or more than type" and vice versa. Of course non-negativity condition is always of the "equal to or greater than type".

Illustration 10:

Jim Jones produces inexpensive furniture to students. Currently it produces book cases and tables. Each book-case gives a yield of Rs.60 and each table Rs.50. Each product has to pass through two stages of production, namely, cutting and finishing. Bookcases take 4 hrs a unit in cutting and 4 hours in finishing. Tables take respectively 3 hrs and 5 hrs. Currently , there are available 96 hrs in cutting and 120 hrs in finishing in the factor y per week. Solve the LPP graphically and through simplex.

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Solution

Let B units of book cases and T units of tables be produced. The yield from book cases = Rs.60xB = Rs.60B and yield from table is Rs.50XT = Rs.50T. So, the objective function is: Maximize Y = 60B + 50T Subject to : 4B + 3T < 96 (cutting constraint) 4B + 5T < 120 (finishing constraint) B, T > 0

Graphic Method

First convert the inequalities into equalities as below : 4B + 3T = 96 (cutting equation) 4B + 5T = 120 (finishing equation) If all cutting hrs are used to produce book-cases, 24 cases (96/4) can be produced and if all are used in table production, 32 tables (96/3) can be produced. Similarly using all finishing hrs in bookcases, 30 cases (120/4) and in tables 24 tables (120/5) can be produced. In the graph 3, book cases are taken in X-axis and tables in Y-axis. The production possibilities are shown. The feasible solution are a is bounded by OPQR. The value of Q is got through solving the two equalities: 4B + 3T = 96 …. (1) 4B + 5T = 120 …. (2) (1)-(2) - 2T = -24 T = 12 So, B = 15 We have to evaluate the corner points, O, P, Q and R For ‘O’, the yield is : 60 x 0 + 50 x 0 = 0 For ‘P’, the yield is : 60 x 24 + 50 x 0 = 1440 For ‘Q’, the yield is : 60 x 15 + 50 x 12 = 1500 For ‘R’, the yield is: 60 x 0 + 50 x 24 = 1200

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So, optimum production is 15 book cases and 12 tables gives a total yield of Rs.1500.

Simplex Solution

In simplex solution, we have to introduce slack variables to convert the inequalities into equalities as below: Objective function: Maximize Y = 60B + 50T + 0S1 + 0S2 Subject to : 4B + 3T + S1 + 0S2 = 96 4B + 5T +0S1 + S2 = 120 B, T, S1, S2 > 0 The initial solution table is worked out now.

Y 32 Graph 3 30

T A B L 24 E

20 R 20 Q P 0 10 20 24 30 X O

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Initial Solution : Tableau 1

Cj Column Product Mix

Quantity Column

Cj value: 60 50 0 0

Cj row : B T S1 S2

Improvement Index I I

O

O

S1

S2

96

120

(4) 3 1 0

(4) 5 0 1

96/4 = 24

120/4 = 30

Zj

Cj – Zj

0

-

0 0 0 0

60 50 0 0

-

-

Optimum column The optimum column is ‘B’ (i.e. book cases) and that it enters into production. So it will be the first now in Tableau – 2. Since, improvement index is low for S1, S1 will be replaced by B in tableau – 2.

Improvement Solution : Tableau – 2

Cj Column

Product Mix

Quantity Column

Cj value: 60 50 0 0

Cj row : B T S1 S2

Improvement Index I I

60

0

B

S2

24

24

1 (¾) (¼) 0

0 (2) -1 1

32

12

Zj

Cj – Zj

1440

60 45 15 0

0 5 -15 0

-

-

Optimum column The values of ‘B’ row are obtained by dividing the values of S1 row (outgoing row) by its intersectional element 4. We get: 96/4 = 24, 4/4 = 1, 3/4, 1/4 and 0.

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The values of ‘S2’ row in the tableau - 2 are obtained as follows: Old values in tableau 1 – (Incoming B row values in tableau 2 x S2 row’s inter-sectional element in tableau 1)

120 - ( 24 x 4 ) 24 (Qty column) 4 - ( 1 x 4 ) = 0 (‘B’ Column) 5 - (3/4 x 4 ) = 2 (‘T’ column) 0 - (1/4 x 4 ) = -1 (S1 column) 1 - ( 0 x 4 ) = 1 (S2 column)

Now the optimum column, i.e.; the column with highest positive number in the Cj – Zj row, is that of T. So, in our next improved solution, T, i.e., Tables , will enter production mix. When ‘T’ enters, something must be removed from existing product mix, either B or S2 which has to be removed? This is decided by the lowest value of the improvement index found in the last column of tableau – 2. It is least for S2. So, S2 will be outgoing and B will be incoming in tableau 3.

Improvement Solution : Tableau – 3

Cj Column

Product Mix

Quantity Column

Cj value: 60 50 0 0

B T S1 S2

Improvement Index (I I)

50

60

T

B

12

15

0 1 -1/2 1/2

1 0 5/8 -3/8

Zj

Cj – Zj

1500

60 50 12.5 2.5

0 0 -12.5 -2.5

Optimum column The value of the incoming ‘T’ now in tableau – 3, are obtained by dividing the values of S2 row in tableau-2 by its intersectional element. So, the values for ‘T’ row are : 24/2 = 12,0, 2/2 = 1, -1/2 and 1/2. The values of ‘B’ row in the tableau-3 are obtained as follows:

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Old values in tableau 2 – (Income T row values in tableau 3 x Inter sectional element of ‘B’ row in tableau 2)

24 - ( 12 x 3/4 ) = 15 1 - ( 0 x 3/4 ) = 1 3/4 - (1 x 3/4 ) = 0 1/4 - (-1/2 x 3/4 ) = 5/8 0 - ( 1/2 x 3/4 ) = -3/8

Since the values in Cj – Zj row in tableau –3 are negative or zero, further improvement is not possible. So, 12 units of ‘T’ (tables) and 15 units of B (Book cases) will be produced giving a total contribution of Rs.1500. All these values are found in quantity column of tableau –3. The same result was obtained through graphic solution as well, as you are aware of.

Duality of the above problem:

In the duality of this problem, we have to find the solution as to what minimum price for each hour of cutting facility and from each hour of finishing facility will make the firm earn at least the yield it is getting by manufacturing book cases and tables using these facilities. Let an hour of cutting can be hired out for Rs.C and finishing for Rs.F. We have to find the minimum of ‘C’ and ‘F’. We now attempt to set the objective function and constraints. We have 96 hrs of cutting and 120 hrs of finishing. So, 96C + 120F give the objective function. Minimize : 96C + 120F Subject to : 4C + 4F > 60 and 3C + 5T > 50 C, T > 0 That is 4 hrs of cutting and 4 hrs of finishing used in producing 1 book case, must be hired out such that the value got, i.e.; 4C + 4F, is equal to or greater than the yield of Rs.60, obtained by selling a book-case. Similarly for

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table usage. 3C + 5T must be equal to or greater than or equal to yield of Rs.50 obtained from producing one table. The answer to the minimization duality can be got from the simplex of the primal maximization solved already. In the final tableau of primal, in the Cj–Zj row, under S1 and S2 columns we find values –12.5 and –2.5. These the amounts of by which profit will fall if a unit of S1 and S2, respectively, are created. Now that all units of cutting and finishing hours are used, creating slack hours will mean, diverting them from production. So, profit, will have to fall. Indirectly, these figures tell the price per hour of cutting facility and per hour of finishing facility. So, C = Rs.12.5 and F = Rs.2.5. With 96 cutting hours and 120 finishing hours, we get: 96 x 125 + 12.0 x 2.5 = Rs.1500. This is the same value of objective function obtained for primal as well.

4.7 SENSITIVITY ANALYSIS Let us take an example. A publishing firm can publish technology book, children book and business book. Each of these respectively give a net profit of Rs.2, Rs.4 and Rs.3. Three resources, authorship, filming and printing are needed. The available capacities are 60, 40 and 80 hours each, respectively in a week. Technology book requires 3, 2 and 1 hours of each of the resources Children book requires 4, 1 and 3 hours of each of the resources Business book requires 2, 2 and 2 hours of each of the resources If we take ‘T’ units of Tech. Book, ‘C’ units of Children book and ‘B’ units of Business book be produced, the objective function is : Maximize : 2T + 4C + 3B Subject to : 3T + 4C + 2B < 60 2T + 1C + 2B < 40 1T + 3C + 2B < 80 T, C, B > 0

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The final solution to the problem is given in tableau – I. (Initial and improvement solutions are not given).

Tableau – I : Final solution

Cj Product

Mix

Qty 2

T

4

C

3

B

0

SA

0

SF

0

SP

4

3

0

C

B

SP

Zj

6 2/3

16 2/3

26 2/3

76 2/3

1/3

5/6

-5/3

23/6

1

0

0

4

0

1

0

3

1/3

-1/6

-2/3

5/6

-1/3

2/3

-1/3

2/3

0

0

1

0

Cj – Zj - -11/6 0 0 -5/6 -2/3 0

SA, SF and SP are slack resources as to authorship, filming and printing. The final solution tells that, 6 2/3 children books and 16 2/3 business books be produced, giving a total profit of Rs.76 2/3. No slack capacity in authorship and filming is available. But, 26 2/3 hrs of printing is available. As a result printing capacity has no value, (see ‘0’ in Cj – Zj row), while authorship has a value of Rs.5/6 per hour and filming Rs. 2/3 per hour. These are known as ”Shadow prices”. Sensitivity analysis is done with respect to right hard side range (RHS range) and objective function coefficients.

4.7.1 RHS Range Since, excess printing capacity is there, by adding authorship, and filming capacity we can make additional profit. But we cannot add unlimited authorship and/or filming capacities as this would affect the constraint equations beyond a point. In other words, for how large addition of these resources, the ‘shadow price’ will hold good? This is sensitivity analysis, analyzing the sensitivity of shadow prices up to certain level of addition of the resources, shadow prices remain same, but after that level these prices become sensitivity and change.

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Sensitivity analysis is done as follows: Let us do the sensitivity w.r.t authorship. For that we reproduce quantity and SA columns of final solution below and divide the former by the latter row-wise and get the values as follows in tableau 2.

Tableau – 2 : Sensitivity w.r.t authorship

Quantity column SA column Quotient = Quantity / SA

6 2/3

16 2/3

26 2/3

1/3

-1/6

-2/3

( 6 2/3 ) / (1/3) = 20/3 x 3/1 = 20

(16 2/3) / (-1/6) = 50/3 x –6/1 = -100

(26 2/3) / (-2/3) = 80/3 x –3/2 = -40

The fastest positive quotient, here 20, is the level by which authorship resource can be reduced without altering shadow prices and least negative quotient, here 40, is the quantity by which authorship resource can be increased without altering shadow prices. So, the shadow price for authorship resource, at Rs.5/6, is valid for a range of (-20) to (+40) hours of authorship hours. Since we have started with 60 hrs of authorship resource, the right hand side range for it is 40 (i.e. 60 – 20) to 100 (i.e. 60 + 40).

Sensitivity of shadow price of SF : Tableau – 3

Quantity column SA column Quotient = Quantity / SA

6 2/3

16 2/3

26 2/3

-1/3

-2/3

-1/3

(6 2/3) / (-1/3) = 20/3 x -3/1 = -20

(16 2/3) / (-2/3) = 50/3 x 3/2 = 25

(26 2/3) / (-1/3) = 80/3 x –3/1 = -80

We conclude that for filming capacity the range for which shadow price holds at the present level is 15, (40-25) to 60 (40+20). With regard to the surplus resource, here printing, the shadow of Re.0, will get altered only when its availability is reduced by quantity more than

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the slack quantity, i.e. below 53 1/3 hrs (80-26 2/3). For quantity > 53 1/3 the shadow price will be same as ‘0’ as now.

4.7.2 Sensitivity w.r.t. objective function coefficients

This analysis is done in two types. For variables already in solution and for variables not in the solution.

4.7.2.1 Variable in solution

Here children book is in the solution. Sensitivity analysis here shows, how large or how small that the coefficient of children book (i.e. how large or how small the profit per unit of children book) could become without altering the optimal solution. To get the answer, we present the Cj – Zj row and Cj row from the final solution and divide the former by the latter.

Tableau 4 : Sensitivity Analysis

Cj – Zj -11/6 0 0 -5/6 -2/3 0

Cj 1/3 1 0 1/3 -1/3 0

(Cj – Zj) / Cj -11/6 x 3/1 0 0 -5/6x3/1 -2/3x-3/1 0*

Quotient -5.5 0 0 -2.5 2 0

(* Instead of division by Cj, multiplication by increase of Cj is done)

The least positive quotient sets the limit by which profit can rise without

altering optimal solution. Here it Rs.2. The least negative quotient is the amount by which profit per children book go down without altering optical solution. Here it is Rs.2.5, So, range is : Rs.1.5 to 6 [(4-2.5) and (4+2)]. Similarly we can find that the range for business book is Rs.2 to Rs.8. That is a decrease up to Re.1 and increase up to Rs.5 will not change optimal solution.

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Variable not in solution

In the final solution, technology book is not in solution. The reason is loss from it is higher at Rs.23/6 than its profit of Rs.2 per book. To come into solution, profit per technology book must exceed Rs.23/6. From current profit of Rs.2, it must increase by Rs.11/6 (i.e. Rs.23/6 – Rs.2), so that it can enter the solution.

4.8 OTHER TYPES OF LINEAR PROGRAMMING Production smoothening application, portfolio selection ingredient mixing, diet combination, etc. are certain special applications of LP. Transportation and assignment problems are special types of linear programming. These are dealt in the next lesson.

4.9 LIMITATIONS OF LP

i) Assumption of linearity as to the relationship among variables is a mathematical convenience, but devoid of practical relevance.

ii) The graphic solution seems easy, but can not be adopted when there are more than two variables.

iii) The simplex method involves some complications. Those with lack of a mathematical bent of mind would find it difficult to understand.

iv) LP would become more and more involved when more constraints are there. Better special purpose algorithms have been developed, thus relegating simplex method.

v) Degeneracy might result. That is some +ve value in the max. LPP or -ve value in the min-LPP might continue to appear in the Cj-Zj row, yet the Qty. Col. total remains same, indicating no improvement after successive iterations.

vi) Unboundness is another problem. There is no definite solution to the LPP.

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vii) Redundancy is a problem, wherein constraints that not at all arise are stipulated.

The last few are however, strictly not limitations. The LPP is not well formulated. So, it is futile to blame the technique as such.

4.10 CERTAIN CONCEPTS Infeasibility, un-boundedness, redundancy, multi-optimal, degeneracy, etc. are certain concepts associated with LPPs. Infeasibility means there is no solution to the LPP meeting all constrains. In graphic solution, infeasibility is expressed when there is no area of feasible solution. In simplex solution when one or more artificial resources are found in final solution there is infeasibility. Un-boundedness means that in an LPP infinite solutions exist. Practical world is beset with constraints. So when an LPP produces unlimited solutions, there is something wrong in its formulation. In simplex method, unboundedness is expressed when the improvement indices are all negative or zero. This means an infinite results. Redundancy means, that in the LPP one or more constraints are really no constraints. When maximum production possible is say only 15 table and 12 chairs, a marketing constraint of no more than 20 tables and 15 chairs could be sold is a redundant constraint not affecting the production feasibility at all. Degeneracy in LPP arises when improvement index values for two or more variables are equal. Then, one of them is randomly replaced and this leads to degeneracy. Whenever a variable in the solution has the value 0 in the quantity column, that variable and that solution are said to be degenerate. When there is a degeneracy, the value of objective function does not improve with the next solution. Multiple optima in an LPP means that the LPP has many solutions giving the same objective function value.

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SELF-ASSESSMENT QUESTIONS

1. Explain the LP technique as an optimization technique.

2. What are the managerial applications of LP?

3. Enumerate the limitations of LP.

4. A company produces 3 products A, B and C giving a profit of Rs. 1000, Rs. 4000 and Rs. 5000 per unit respectively. O1, O2 and O3 are the materials required. The requirement is as follows per unit of output of A, B and C:

- A B C

O1 3 - 3

O2 1 2 3

O3 3 2 -

The available quantity of O1, O2 and O3 is 22, 14 and 14 units respectively. To maximize profit, find the optimum product mix.

5. Write the dual for the above problem, solve the same. 6. Maximize: 6X + 4Y

Subject to: 2X + 3Y < 30 3X + 2Y < 24 X + Y > 3 X, Y > 0

7. Write notes on: a) Initial solution; b) Objective function; c) Non-negativity condition; d) Constraints; e) Feasible region; f) Optimum column; g) Intersectional elements; h) Artificial resource

8. What are degeneracy, redundancy, unboundedness and infeasibility.

9. Explain the significance and process of sensitivity analysis.

10. A Toffee company mixes 3 types of toffees to form one kilogram of Toffee gift pack sold at Rs. 300. The three toffees cost Rs. 200, Rs. 250 and Rs. 280 per kg each. The mixture must contain at least 0.3 kg of first type of Toffee and the weight of the first two must at least be equal to that of the third. Find the optimal mix.

� � �

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LESSON 5

TRANSPORTATION AND ASSIGNMENT MODELS (Special Purpose Linear Programming)

Transportation and assignment models are special purpose

algorithms of the linear programming. The simplex method of LPP seen in the previous lesson proves to be inefficient is certain situations like determining optimum assignment of jobs to persons, supply of materials from several supply points to several destinations and the like. More effective solution models have been evolved. These are called assignment and transportation models which are dealt with in this chapter.

5.1 MEANING:

The transportation model is concerned with selecting the routes between supply and demand points in order to minimize costs of transportation subject to constraints of supply at any supply point and demand at any demand point. Assume a company has 4 manufacturing plants with different capacity levels, and 5 regional distribution centres. 4 x 5 = 20 routes are possible. Given the transportation costs per load of each of 20 routes between the manufacturing (supply) plants and the regional distribution (demand) centres, and supply and demand constraints, how many loads be transported through different routs so as to minimize transportation costs? The answer to this question is obtained easily through the transportation algorithm. Similarly, how are we to assign different jobs to different persons/machines, given cost of job completion for each pair of job machine/person? The objective is minimizing total cost. This is best solved thro’ assignment algorithm.

5.2 USES OF TRANSPORTATION AND ASSIGNMENT MODELS IN

DECISION MAKING:

The broad purposes of these models are just mentioned above.

Now we are just enumerate the different situations where we can make use of these models.

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i) To decide the transportation of new materials from various centres to different manufacturing plants. In the case of multi-plant company this is highly useful.

ii) To decide the transportation of finished goods from different manufacturing plants to the different distribution centres. For a multi-plant-multi-market company this is useful. These two are the uses of transportation model. The objective is minimizing transportation cost.

Assignment model is used in the following:

i) To decide the assignment of jobs to persons/machines, the assignment model is used.

ii) To decide the route a traveling executive has to adopt (dealing with the order in which he/she has to visit different places).

iii) To decide the order in which different activities performed on one and the same facility be taken up.

In the case of transportation model, the supply quantity may be

less or more than the demand. Similarly the assignment model, the number of jobs may be equal to, less or more than the number of machines/persons available. In all cases, solutions are optimizing the objective set. In all these cases the simplex method of LPP can be adopted, but transportation and assignment models are more effective, less time consuming and easier than the LPP. Further, transportation and assignment models are useful only in these specific instances dealt within this lesson.

Solution to transportation problems consist of initial solution and

improved solutions. To find initial solution North West Corner Rule Methods, Vogel’s approximation, or row cost minima, etc, are used. For improved solution Stepping stone and ‘MODI’ methods are also used.

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5.3 MECHANISM OF TRANSPORTATION MODEL:

The mechanism of transportation model is explained with an example.

Illustration 1:

Let us take an example. A construction company has 3 projects,

A,B and C. Truck load of sands required per week for these 3 projects are: 80, 100 and 120. Three supply points of sand are there namely, X,Y and Z with a capacity per week of 120,80 and 100 loads respectively. The delivery cost per load from each supply site to each demand point are as follows:

Costs per load

To A To B To C

From X From Y From Z

12 17 19

8 10 5

18 12 8

Decide how many truck loads of sand from the supply points to the

different projects in order to reduce transportation cost.

Solution

The above problem can be presented in a table form.

Origin Destination Total supply

X Y Z

A B C 120 80

100

12 17 19

8 10 5

18 12 8

Total demand 80 100 120 300

Initial solution: North West corner Rule:

Here supply = demand. So this is a balanced problem. To solve the problem an initial solution has to be worked out. Later

this solution is improved. Initial solution is the starting point. There are several methods to find initial solution. North west corner method, Vogel’s

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Approximation method, Row Minima Method, Column Minima method and inspection method are some of the methods. Let us take up the North West Corner method to develop the initial solution. Consider table 1 given below.

Table 1: Initial Solution

Project A Project B Project C Total Supply

Supply X XA (12 (80)

XB ( 8

(40)

XC (18

120

Supply Y YA (17

YB (10 (60)

YC (12 (20)

80

Supply Z ZA (19

ZB ( 5

ZC (8 (100)

100

Total Demand 80 100 120 300

The table simply reproduces the problem with additional

information. The cost figures are placed in sub-cells within each cell. The XA, XB, ….XC simply indicate the origin - destination pair.

The north west corner rule is logical and systematic method to find

the initial solution. You start with north west corner cell, viz, XA cell. It means sending supplies from X to project A. X has a total capacity of 120 loads. A needs only 80 loads. So, ‘A’s demand is fully met from X. This allotment is written in a bracket within XA cell. Since A’s need is fully met we can move to project B. Now balance available with ‘X’ needs to be exhausted first. So we move horizontally from XA cell to XB cell.. The balance 40 loads of X is allotted to B. This is indicated in the bracketed figure in XB cell. B requires a total of 100 loads. So, it still needs 60 more load of sands. We have to use the supply with the next source viz. Y. So we move vertically down to YB from XB. From Y, 60 loads are to be sent B. This is indicated by the figure in the bracket in YB. Now that B’s requirement is fully met, you move to C and still some capacity is available with Y. Our move is horizontal to YC. 20 units of available balance with ‘Y’ allotted to project C. Still project ‘C’ needs 100 more loads. We move down to ZC and the available stock with Z is just sufficient to meet the balance needs of C. This is a case of supply being equal to demand type

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problem. Initial solution as obtained under north west corner method is as follows: From X to A 80 loads, X to B 40 loads, Y to B 60 loads, from Y to C 20 loads and from Z to C 100 loads are transported. This is only an initial solution. A better solution you have to find from the above. Arrow marks are made to present flow of allotments made.

To proceed further we have to consider the number of stone cells.

Stone cells are those with allotment. XA, XB, YB, YC and ZC are stone cells. The number of stone cells must be equal to m+n-1, where ‘m’ indicates the total number of supply sources (here 3) and ‘n’ indicates the total number of demand points (here the project numbering 3). 3+3-1= 5= No. of stone cells. This equality is referred to as rim-requirement. The satisfaction of rim-requirement, i.e., m+n-1 = Number of stone cells, is a must for further attempt on improving the solution at any stage. Now the rim-requirement is met, we can proceed further.

To proceed hence forth, we have to find what would happen to cost

if one load is allotted to each of the unoccupied cells, or non-stone cells. There are two methods here, viz. stepping stone method or modified or MODI method. Now we are adopting the stepping stone method.

Stepping Stone method for optimization:

If one load is allotted to XC, one load must be subtracted from XB, one load must be added to YB and one load must be subtracted from YC. This is necessary so that the supply constraints of X and Y and the demand levels of B and C are not affected. What would be the effect on cost of this more? X to C costs Rs.18 and Y to B costs Rs.10. Withdrawal of a load from X to B saves RS.8 and form Y to C saves Rs.12. The overall effect of this : 18-8+10-12=Rs.8. Sending a load of sand to XC results in extra cost of Rs.8. this has to be avoided. Similarly for other unoccupied cells evaluation is to be done. To send a load to YA: YA-YB+XB-XA. The effect on cost would be :17-10+8-12=5. This is also no good.

To send a load A to Z, would require one load less for XA, one

load more for XB, one load less for YB, one load more for YC and one load less for ZC. That is: ZA – ZC + YC + YB + XB – XA. You might note the

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mechanism. You start from the empty cell (giving a+). You move either vertically or horizontally, never diagonally to a stone cell. You take turn only at the stone cell. And so on. Ultimately you get back to the empty cell you started from. These routes are explained in table 2.

Table 2. Routes to unoccupied cells in table 1.

A B C Total supply

X (80) (40)

+ 120

Y (60)

(20) 80

Z (+)

(100) (-)

100

Total demand 80 100 120 300

The process is one of making ‘loop’, a closed loop. Successive turn

points given ‘-‘ and ’+’ signs alternatively. The dash-dash route in table 2 gives this, make sure you never halt or turn in any intervening cells. The effect on cost of the above more is: 19-12+8-10+12-8=9. Again this move is not beneficial as cost goes up by 9 for every load thus transported. The only remaining empty cell is ZB. For this a simple loop is possible and the same is: ZB-ZC+YC-YB. The cost effect is: 5-8+12-10=-1. So sending one load to ZB reduces the cost Re.1. This route has to be utilized to the maximum extent, so that maximum reduction in cost is possible. So we have to do reallocations among the involved cells here. There are four cells involved. Of these 4 two are ‘+’, i.e., receiving cells and two are ‘-’, i.e., sending cells. How much to send? Of the sending cells, find the one with least allocation. That much will be moved. Here ZC and YB are sending cells. ZC has 100 and YB has 60. So, 60 loads will be moved from YB to ZB, 60 from ZC to YC. The relevant reallocation would be: ZB: 60 loads: YB: nil; YC 80 loads and ZC 40 loads.(refer table 3) Note allocations to XA and XB are not changed. Because these cells, are not involved in the ‘loop’ for ZB. Now also the rim-requirement is fulfilled as, m+n-1= stone cells. So, we can think of further improvements, if any. Now the empty cells are: XC,YA,YB and ZA. The routes to these empty cells and additional cost of allocations empty cells would be as follows:

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XC: XC-XB+ZB-ZC = 18-8+5-8 =3 YA: YA-XA-XB+ZB-ZC+YC =17-12+8-5+8-12 =4 (loop again) YB: YB-YC+ZC-ZB =10-12+8-5 =1 ZA: ZA-XA+XB-ZB =19-12+8-5 =10

Table 3: Second Solution

A B C Total supply

X

12 (80)

8 (40)

18 120

Y

17 10

12 (80)

80

Z

19

5 (60)

8 (40)

100

Total demand 80 100 120 300

Allocation to any of the empty cells prove to be inefficient. That is,

allocation to them involve additional cost. So, a final solution is already obtained. As per table 3, to XA 80 loads, XB 40 loads, YC 80 loads, to ZB 60 loads and to ZC 40 loads allocations are made. The total transportation cost would be:

XA : 80 x 12 = 960 XB : 40 x 8 = 320 YC : 80 x 12 = 960 ZB : 60 x 5 = 300 ZC : 40 x 8 = 320

Total Rs. 2860

Any other allocation would be costlier than the above.

Vogel’s Approximation Method (VAM) for Initial solution:

Vogel’s Approximation Method can also be used, instead of north west corner rule method to find the initial solution. Under this method the following steps are involved.

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i) Calculate the difference between lowest of two transportation costs for each column and each row. These differences are known as column and row penalties and are written by respective row and column by providing additional column and additional rows in the transportation table.

ii) Suppose in any row or column the smallest cost figures repeat. These repeat values are taken for penalty computation and the penalty is case in such cases.

iii) Select the row or column with the largest penalty and circle that value. In case of a tie, select the column or row that permits the greatest movement of materials.

iv) Assign the largest possible allocation within the restrictions of the row and column requirements to the lowest cost cell for the row or column selected in step (iii) above.

v) Cross out the column or row whose demand/supply is completely met with/exhausted, as the case may be, as per our assignment made in the previous step.

vi) With a column or row thus deleted, (as per step in above) with the reduced number of columns/rows repeat the process (i) to (ii) until all assignments are made.

Let us consider our problem in question.

Illustration2:

Illustration 1 is solved through VAM and MODI method. Table 1 gives the problem and application of VAM.

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Table 1 : VAM : PENALTY

A B C Supply Row penalty

X 12 (80)

(8) (18) 120 12 – 8 = 4

Y (17) (10) (12) 80 12 – 10 = 2

Z (19) (5) (8) 100 8 – 5 = 3

Demand 80 100 120 300

Column penalty

17 – 12 = 5 8 – 5 = 3 12 – 8 = 4

Look at table 1, the largest penalty is 5 in column A. The least cost cell in that column is XA. X has 120, but A needs only 80. So, 80 units given to A from X. This reduces supply with X to 40 and A gets deleted Now Column A is eliminated as its need is fully met. So, the reduced matrix is in table 2 below.

B C Supply Penalty

X (40) (8) (18) 40 18 – 8 = 10

Y (10) (12) 80 12 – 10 = 2

Z (5) (8) 100 8 – 5 = 3

Demand 100 120 220

Penalty 8 – 5 = 3 12 – 8 = 4

Now the highest penalty is 10 in row X. The least cost cell in row X is XB. 40 units with X are given to B and B > read reduces to 60 from 100. new penalty is now constructed in table 3 below.

B C Supply Penalty

Y (60) (10)

(20) (12)

80 12 - 10= 2

Z (5) (100) (3) 100 5 – 3 = 2

Demand 60 120 180

Penalty 10 – 5 = 5 12 – 3 = 9

The highest penalty is 9 in column C. The least cost cell is ZC. So, 100 units from Z are sent to C. Now Z row gets eliminated. This leaves us with

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only Y row. So, whatever is available with Y has to be given to ‘B’ and ‘C’ B gets 60 from Y and C gets 20. And this ends all initial distribution.

The rim-requirement = m + n - 1 = 5 = Number of stone cells, is fulfilled; so we can proceed further.

Note this initial solution is same as the one obtained under the

North west corner rule. But of need not be so. Improved solutions have to be found. Table 4 gives the picture.

MODI method for optimization

At this stage we may adopt the modified or MODI method for improved solution. Under this method the columns and rows are represented by K's and R's respectively.

So, K1, K2 and K3 and R1, R2 and R3 are the notations for the three

columns and three rows respectively. The cost of a cell can be represented by Cij-ith row and jth column. C1,1 refers of cost of the first row - first column cell.

Cost at Stone (i.e. occupied) cell i, j = Cij = Ri + Kj. In our case

under the Vogel's approximation method the stone cells are: 1.1, 1.2, 2.2, 2.3 & 3.3. So

C1,1 = R1 + K1 = 12 - (1) C1,2 = R1 + K2 = 8 - (2) C2,2 = R2 + K2 = 10 - (3) C2,3 = R2 + K3 = 12 - (4) C3,3 = R3 + K3 = 8 - (5) We have 5 equations, but 6 unknowns. So, one of the unknowns be

given value zero. Let any one of Ri's or Kj's be equal to zero. Say K1 = 0. Then, as

per equation (1) 12 = R1 + 0 or R1 = 12 (2) 8 = 12 + K2 or K2 = -4

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(3) 10 = R2 + (-4) or R2 = 14 (4) 12 = 14 + K3 or K3 = -2 (5) 8 = R3 + (-2) or R3 = 10 Now, the cost effect of sending one load to each of the unoccupied

cells viz., 1,3; 2,1; 3,1; and 3,2; have to be found. This is given by Cij - Ri - Rj So The cost effect for XC = C13 - R1 - K3 = 18 - 12 - (-2) = 8 So The cost effect for YB = C32 - R2 - K2 = 17 - 14 - (0) =3 So The cost effect for ZA = C31 - R3 - K1 = 19 - 10 – 0 = 9 So The cost effect for ZB = C32 - R3 - K2 = 5 - 10 - (-4) = -1

Now you look for the cell with the largest -ve cost effect. It is ZB. We have to send loads from Z to B. The closed loop would be: ZB - YB + YC - ZC. Now, from YB to YC 60 units are transferred making the total to 80. From ZC to ZB, 60 units are transferred. The new position is given in table 4.

A B C Total supply

X

12 (80)

8 (40)

18 120

Y

17

10 (60)

12 (20)

80

Z

19

5

8

(100)

100

Total demand

80

100

120

300

143

Again the cost effect of sending a load to the empty cells need to be worked out. For this purpose we must know the values of Kj's and Ri's.

Cost at a stone cell: Cij = Ri + Kj So for, XA: C11 = 12 = R1 + K1 - (1) XB: C12 = 8 = R1 + K2 - (2) YC: C23 = 12 = R2 + K3 - (3) ZB: C32 = 5 = R3 + K2 - (4) ZC: C33 = 8 = R3 + K3 - (5) Setting any of the Rj's or Kj's as equal to zero, we can solve the

remaining Ki's and Kj's. Let R1 = 0. Then 12 = 0 + K1; or K1 = 12 (based on equation. 1) 8 = 0 + K2; or K2 = 8 (based on equation. 2) 5 = R3 + 8; or R3 = -3 (based on equation. 4) 8 = -3 + K3; or K3 = 11 (based on equation. 5) 12 = R2 + 11; or R2 = 1 (based on equation. 3) The cost effect for non-stone cells be worked out using the

formula: Cij – Ri – kj

Cost effect for XC: 18 – R1 – K3 = 18 – 0 – 11 = 7 Cost effect for YA: 17 – R2– K1 = 17 – 1 – 12 = 4 Cost effect for YB: 10 – R2 – K2 = 10 – 1 – 8 = 1 Cost effect for ZA: 19 – R3– K1 = 19 – (-3) – 12 = 10

Now all figures are +ve indicating no improvement is possible hence form. Table 5

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So, we have reached the final solution as in table 5. This solution is the same as the one obtained earlier.

5.4 TRANSPORTATION MODEL WHEN DEMAND IS LESS THAN

SUPPLY

This is called unbalanced problem. To solve the problem, first balancing is needed.

Here you add a `Dummy' demand absorbing the excess supply.

The cost figures for this `dummy' demand column would be zero for all rows. Any supply centre supplying to this `Dummy' demand does not, in fact supply to this destination as the latter is non-existent. A problem on this pattern is worked out now.

Illustration 3

Given in the table are the supply and demand factors and the transportation cost matrix. Find the optimal distribution.

A B C Total supply

X

12 (80)

8 (40)

18 120

Y

17

10

12 (80)

80

Z

19

5 (60)

8 (40)

100

Total demand

80

100

120

300

145

Deport

Factory

A B C D Total

Supply

P

Q

R

4

3

3

6

5

9

8

2

6

6

5

5

700

550

550

Total

Need

400

450

350

500

1800

1700

Solution

Here the supply exceeds demand. So a Dummy Depot is to be added with a need or demand of 100 units. The initial solution under North West Corner Rule is given in table 1.

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Table 1: Initial Solution

A B C D Dummy Total supply

P 4 (400)

6 (300)

8 6 0 700

Q 3 5

(150)

2

(350)

5

(50)

0

550

R 3 9 6 5 (450)

0 100

550

Total Demand

400 450

350

500

100

1800

Now the rim requirement is m + n - 1 = 3 + 5 - 1 = 7 = number of

stone cells. We can proceed further. The effect on cost for transportation of one unit to each of the non-stone cells may be worked out now.

For PC: PC - PB + QB – QC = 8 - 6 + 5 - 2 = 5 For PD: PD = PB – QB – QD = 6 - 6 + 5 - 5 = 0 For P Dummy: P-Dummy - PB + QB - QD + RD - R-Dummy : 0 - 6 + 5 - 5 + 5 - 0 = -1 For QA: QA - PA + PB - QB = 3 - 4 + 6 - 5 = 0 For RA: RA - PA + PB - QB + QD - RD : 3 - 4 + 6 - 5 + 5 - 5 = 0 For Q-Dummy: Q-Dummy - R-Dummy + RD - QD = 0 - 0 + 5 - 5

= 0 For RB: RB - QB + QD - RD = 9 - 5 + 5 –5 = 4 For RC: RC - QC + QD - RD = 6 - 2 + 5 –5 = 4 From the above cost effect computations, cost minimization is

possible only if distribution from P to Dummy is made.

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Note that the movement in any improvement is limited to the minimum quantity of the -ve stone cells in the loop. In this case the loop is P-Dummy – PB + QB - QD + PD -R-Dummy. QD in old assignment has the minimum of 50 units. So the movement is limited to successive add/less of 50 units. Any higher movement would result in -ve figure at the minimum -ve stone cell which is not realistic. Table 2 gives the 2nd solution.

Table 2: 2nd

Solution

A B C D Dummy Total supply

P

4 (400)

6 (250)

8 6 0 (50)

700

Q

3

5

200

2

(350)

5

0

550

R

3

9 6 5 (400)

0 (50)

550

Total Demand

400

450

350

500

100

1800

Now improvement may be worked out for the empty cells in the

table 2.

For PC: PC - PB + QB - QC = 8 - 6 + 5 - 2 = 5

For PD: PD - P-Dummy + R-Dummy - RD = 6 - 0 + 0 - 5 = 1

For QD: QD - QB + PB - P-Dummy + R-Dummy - RD = 5 - 5 + 6 - 0 + 0 - 5 = 1

For Q-Dummy: Q-Dummy - QB + PB - P-Dummy = 0 - 5 + 6 - 0 = 1

For RA: RA - R-Dummy + P-Dummy - PA = 3 - 0 + 0 - 4 = -1

For RB: RB - R-Dummy + P-Dummy - PB = 9 - 0 + 0 - 6 = 3

For RC: RC - R-Dummy + P-Dummy - PB + QB - QC = 6 - 0 + 0 - 6 + 5 - 2 = 3

For QA: QA - PA + PB - QB = 3 - 4 + 6 - 5 = 0

For RA cell there is –ve cost impact figure.

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So, distribution from R to A is advantageous. Table 3 gives the new distribution.

Table 3: 3rd Solution

A B C D Dummy Total supply

P

4 (350)

6 (250)

8 6 0 (100)

700

Q

3

5

200

2

(350)

5

0

550

R

3 (50)

9 6 5 (500)

0

550

Total Demand

400

450

350

500

100

1800

Now, all the empty cells have (computations not given) positive

cost effects. So, table 3 gives the optimum solution. As per this, the distribution is:

P to A: 350 units; cost: 350 x 4 = 1400

P to B: 250 units; cost: 250 x 6 = 1500

(P to Dummy is in effect no distribution) = 0

Q to B: 200 units; cost: 200 x 5 = 1000

Q to C: 350 units; cost: 350 x 2 = 700

R to A: 50 units; cost: 50 x 3 = 150

R to D: 500 units; cost: 500 x 5 = 2500

Total Cost Rs. 7250

You may note the transportation as per table 1 involves a cost of

Rs. 7350 and table 2 involves a cost of Rs. 7300. So, successive iterations lead to reduced cost. The optimum is reached when no further reduction in cost is possible.

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5.5 TRANSPORTATION MODEL WITH SUPPLY LESS THAN

DEMAND

When supply falls short of demand a dummy is added to the

supply source and the cost of transportation from this dummy source to any of the demand centres is taken as 0. Dummy's supply is taken as the excess of demand over available supply. The problem is solved in the usual way. Any destination getting allotment from `dummy' in effect does not get real allotment to that extent.

5.6 MAXIMISATION PROBLEM IN TRANSPORTATION

Transportation model is designed to solve cost minimization problems. However it can be used to solve maximization problems too. Consider the problem.

Illustration 4

A firm has 3 factories, viz., A, B & C. It has 3 sales territories, viz., X, Y & Z. Excluding transportation cost the per unit returns, i.e., the ex-factory returns are, Rs. 10, 9 and 11 at these three factories. The transportation costs and capacities factors are as under.

From To X To Y To Z Total supply

A 1 2 3 1000

B 5 3 4 800

C 3 5 3 1200

800 1200 1000 3000

You are required to distribute in such a way, that maximum return

is obtained.

Solution

Step I

The above cost figures are not to be considered in isolation. First we have to ascertain the per unit return net of transportation costs. This is as

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follows: Net return = Gross ex-factory return - Relevant transport cost. This is given below in the table 1.

Table 1 : Net Return Matrix

X Y Z

A 10-1=9 10-2=8 10-3=7

B 9-5=4 9-3=6 9-4=5

C 11-3=8 11-5=6 11-3=8

Step II :

This return per unit matrix has to be converted into a cost matrix. This is done as follows. Find the highest net return figure. It is Rs. 9. Subtract all net return figures from 9 and construct the matrix. This is done in the table 2 that follows, which also gives the quantity factors and initial solution.

Table – 2 – Initial Solution: NWCR Method

X Y Z Total supply

A

0 (800)

1 (200)

2 1000

B 5

3 (800)

4

800

C

1

3 (200)

1 (1000)

1200

Total Demand

800

1200

1000

3000

Table – II – Final Solution

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Table 2 solution is the optimum one. Detailed workings are not given. So, from A 600 units to X and 400 units to Y, from B 800 units to Y, from C 200 units to X and 1000 units to Z are to be sent. The profit would be = units times net return.

= 600 x 9 + 400 x 8 + 800 x 6 + 200 x 8 + 1000 x 8 = Rs. 23000

5.7 ASSIGNMENT MODEL

Assignment model is a special case of transportation problem.

Assignment of jobs to machines/persons, sequence in which jobs be undertaken, sequence in which places have to be visited etc., are better determined using assignment model. In all these cases cost minimization is the root objective. Then one-to-one basis underlies the assignment. That is, one machine to one job, is the basis. Similarly, in sequence determination, only once a job or place be taken up/visited and no retreat is permitted. Let us examine the use of the assignment model now. The method adopted here is called Flood's technique or the Hungarian method of assignment.

JOB-MACHINE ASSIGNMENT

Illustration 5:

The cost matrix for each job-machine combination is as follows:

X Y Z Total supply

A

0 (600)

1 (400)

8 1000

B 5

3 (800)

4

800

C

1 (200)

3 1 (1000)

1200

Total Demand

800

1200

1000

3000

152

Job Mac. X Mac. Y Mac. Z

A 30 30 25

B 45 18 25

C 21 17 15

Find the optimum assignment of jobs to machines. Step 1: Column-wise subtract the least cost figure from the

respective column figures. This is done as found in table 1. Table 1

Job Mac. X Mac. Y Mac. Z

A 30-21=9 30-17=13 25-15=10

B 45-21=24 18-17=1 25-15=10

C 21-21=0 17-17=0 15-15=0

Step2: For resulting figures, go row-wise subtracting each row

minimum from respective row figures. We get table 2 as follows: TABLE 2

Job Mac. X Mac. Y Mac. Z

A 9-9=0 13-4=4 10-9=1

B 24-1=23 1-1=0 10-1=9

C 0 0 0

Step 3: Draw straight lines, horizontally and vertically, to connect the zeros in the above table. Minimum number of lines must be used. If that minimum number of lines equals the jobs/machines, an assignment is possible. Here it is possible.

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Step 4 : Match jobs with machines wherever the cell figures are zero. Give priority to row/column with only one zero. Z column has only one zero. So, job C be assigned to Z; And the zero is CZ is boxed. Any other zero in C row to Z column will be crossed out. This reduces the number of zeros in other rows and columns too. Row A has only one zero; in X column. So, job A be assigned to machine X. This leaves job B be assigned to machine Y. The assignment is therefore A to X, B to Y and C to Z. The cost = 30 + 18 + 15 = 63.

Illustration 6:

The job-machine cost matrix is given below:

Job Mac. X Mac. Y Mac. Z

A 25 31 35

B 15 20 24

C 22 19 17

Determine the optimum assignment

Solution:

Step 1: Row-wise, subtract minimum value from respective row figures. (instead column wise, you can do this way also). 25 is subtracted from row 1 figures, 15 subtracted from row 2 figures and 17 from row 3 figures.

We get figures as found in table 1.

Table 1

Job Mac. X Mac. Y Mac. Z

A 0 6 10

B 0 5 9

C 5 2 0

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Step 2: For the resulting figures we go column-wise deducting each column minimum value from respective column figures and we get figures as found in table 2.

Table 2

Job Mac. X Mac. Y Mac. Z

A 0 4 10

B 0 3 9

C 5 0 0

Step 3: Draw St. lines to connect the zeros. We need only a

minimum of two st. lines for the purpose. Since the number of st. lines is less than the number of jobs/machines, optimum assignment is not decided at this stage. In such case we proceed further.

Step 4: Find the lowest value number not covered by any st. line.

Subtract this number from each of the numbers not covered by any st. line. Add that number to the figure found at the intersection of any two st. lines. Other numbers are left untouched. Accordingly, the lowest number is 3. This is subtracted from 4, 10, 3 and 9 and added to 5 at the intersection. We get:

Job Mac. X Mac. Y Mac. Z

A 0 1 7

B 0 0 6

C 8 0 0

Now draw lines to connect the zeros. We need a minimum of 3

lines which is equal to the number of rows/columns. So an optimum solution is possible at this stage. Allot machine Z to job C, Machine Y to job B and Machine X to job A. The total cost would be: 25 + 20 + 17 = 62. No other assignment would produce a lower cost than the above.

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5.8 TRAVELLING EXECUTIVE/SALESPERSON PROBLEM:

An executive has to visit a number of places. With distance matrix, we can find the route he has to follow to reduce total distance traveled. The condition is that he should get back to the place he started first and he should not visit the same place twice. This is a special case of assignment.

Illustration 7

The following distance matrix among 5 places are given. To places (figs. In kms)

From A B C D E

A 0 44 82 91 65

B 44 0 26 71 36

C 82 26 0 49 80

D 91 71 49 0 59

E 65 36 80 59 0

Find the least cost route if the person has to start from A and get

back to A visiting all places.

Solution

Follow steps 1 and 2 of the assignment model, which diagonal figures A-A, B-B, C-C, D-D & E-E set to infinity.

With α (infinity) for diagonal moves such moves will go

impossible. Tables 1 and 2 give the resulting figure.

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TABLE1 Row-wise subtraction

A B C D E

A ∞ 0 38 47 21

B 18 ∞ 0 45 10

C 56 0 ∞ 23 64

D 42 22 0 ∞ 10

E 29 0 44 23 (

TABLE 2

Column wise subtraction

A B C D E

A ( (0) 38 24 11

B � � (0) ( (0) 21 � � (0)

C 38 � � (0) ( (0) 54

D 24 22 � � (0) ( (0)

E (11) 0 44 (0) (

Step 3: The zero cells are our solution cells. Let him go to B first from A. The zero at A B is bracketed. From B

he can go to A, C or E as all has zeros. B to A is no good as he has to visit other places before getting back to A. So B to C and B to E are the possible solution. B to C is the best choice. (The reason will be known later). The zeros at B-A and B-E are cross-marked as these are not used for solution. Now he is in C. Then from C to D. Then from D to E. And finally from E to A. Of course this last

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move costs 11 kms. Still it is the available best, the unused zeros are cross marked. The route is A-B-C-D-E-A. The total distance is: 44 + 26 + 49 + 59 + 65 = 243 kms.

If instead of B to C, B to E is chosen, then E to D would have been

chosen, then D to C and finally C to A, with extra 38 kms. So this is not good. Here the distance would be: 270 kms. The last move, with non-zero element, is also bracketed as this is in solution.

Questions

1. Explain the meaning and uses of transportation and assignment models.

2. Explain the methods of North West corner rule and Vogel's approximation method for finding the initial solution.

3. Explain the MODI and stepping stone methods with an example of your own.

4. Explain the mechanism of assignment model.

5. How would you deal with supply-demand inequalities in transportation?

6. Discuss the procedure of assignment model for traveling executive problem.

7. T.C.Mellott trucking company has a contract to move 115 truckloads of sand per week between three sand-washing plants, W, X and Y, and three destinations, A, B and C. Cost and volume information is given below. Compute the optimal transportation cost using the stepping – stone method.

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Project Requirement per week, truckloads

Plant Available per week, truckloads

A

B

C

45

50

20

W

X

Y

35

40

40

COST INFORMATION

From To Project A To Project B To Project C

Plant W

Plant X

Plant Y

Rs. 5

20

5

Rs. 10

30

8

Rs.10

20

12

8. Sid Lane hauls oranges between Florida groves and citrus packing plants. His schedule this week calls for 520 boxes with locations and costs as follows:

Grove Available per week Packing plant Requirement per week

A

B

C

170

250

100

W

X

Y

130

200

190

COST INFORMATION

From To plant W To plant X To plant Y

Grove W

Grove X

Grove Y

Rs.12

11

2

Rs. 8

15

7

Rs.5

10

6

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9. Jack Evans Owns several trucks used to haul crushed stone to road projects in the country. The road contractor for whom Jack hauls, N, Teer, has given Jack this schedule for next week.

Project Requirement per week Plant Available per week

A

B

C

50

75

50

W

X

Y

45

60

60

Jack figures his costs from the crushing plant to each of the road projects to be these:

COST INFORMATION

From To Project A To Project B To Project C

Plant W

Plant X

Plant Y

Rs.4

6

8

Rs.8

7

2

Rs.3

9

5

10. Coley’s Machine shop has four machines on which to do three jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following table. What are the job assignments which will minimize cost.

Job

Machine

W X Y Z

A

B

C

Rs. 18

8

10

Rs.24

13

15

Rs.28

17

19

Rs.32

19

22

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LESSON 6

PROBABILITY : CONCEPTS AND USES

One of the statistical tools that is much used in the world of uncertainty by business managers is probability Jacob Bernoulli, Abraham de Moirre, Rev. Thomas Bayes and Joseph Lagrange developed probability concepts Pierre Simon and Laplace unified these concepts. Concept of probability is a part of every-day life.

6.1 CONCEPT OF TYPES OF PROBABILITY Probability means chance. What is the “probability” that the BSE SENSEX will appreciate by 50 points today over the yesterday’s close? You can replace the term “probability” by “chance” and represent the above interrogative sentence as follows: What is the “Chance” that the BSE SENSEX will appreciate, 50 points over yesterday’s close? There is absolutely no difference in meaning between the two sentences. Probability is thus a science dealing will chance or uncertain outcomes. It deals with prediction of future uncertain outcomes. Every walk of our life is now beset with uncertainties and that probability science has great value to us to predict the future. Business managers use probability in a wide variety of situations. There are different concepts of probability.

6.1.1 Mathematical or Prior or Classical Probability Mathematically probability is the ratio of number of favourable outcomes to the number all possible outcomes of an action. Take the action of tossing of a coin. What is the probability of getting head (H)? There are two possible outcomes – getting head and getting tail (T). Therefore number of favourable outcomes (i.e.; getting head, in this case is one and total number of outcomes is two. So, probability of getting head = No. of favourable outcome/ Total No. of possible outcomes = ½ = 0.5.

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Similar to the above, the probability of getting an even number side up of a dice is = No. of even number sides of a dice / total no. of sides of the dice = 3/6 = 0.5. Similarly, the probability of drawing an ACE card from a pack of the usual card game is, number of ACE cards / total number of cards =4/52=0.0769. And probability of drawing a SPADE = 13/52 = 0.25. And probability of drawing SPADE KING = 1/52 = 0.0192. The above probability approach cannot be applied to problems other than experiments with coins, cards and dice. Hence the limitation of priori or mathematical probability.

6.1.2 Statistical or Empirical Probability or Relative Frequency Approach to Probability

Statistical probability is the ratio of the number of times a considered favourable outcome is obtained from an experiment to total number of outcomes obtained from the experiment when the experiment is repeated number of times under stable conditions. What is the probability of the prices of scrips rising when higher divided is declared? Say of the 300 scrips for which higher (than previous dividend) dividend has been declared, 240 scrips have reported initial gains. Then the probability = 240/300 = 0.8. This approach is also known as relative frequency approach.

6.1.3 Subjective or Intuitive Probability Subjective probability is defined as the probability assigned to an event by an individual based on whatever evidence is available. Intuition, educated guess or relative frequency of past occurrences could be assumed as evidence. This approach is generally adopted when events occur only once or at most a very few times only.

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6.2 BASIC TERMS In probability some concepts are used. We must know each of these concepts. “Experiment” is a term used, which means the activity performed, like tossing a coin or playing a dice or drawing a card and so on. “Outcome” is a term used to depict the results of an experiment. In tossing a coin head or tail is got as result. “Event” is a term which refers to one or more of the possible outcomes emanating from the experiments. “Sample space” is a term used which refers to the set of all possible outcomes of an experiment. In a dice game the same space = S = [1,2,3,4,5 and 6]. “Mutually exclusive” events are those that cannot occur together but only one at a time. Head or tail and not both can occur when a coin is tossed. So, the events are mutually exclusive. Sample space of mutually exclusive events is said to contain the list of mutually exclusive but collectively exhaustive events. ‘Ace’ and ‘King’ are mutually exclusive events. “Mutually inclusive” events are those that can occur together. “Ace and Spade” are mutually inclusive events. Families having CTVs and washing machines – mutually inclusive. “Statistical independence” of events means, the outcome of an event has no effect on the chance of the occurrence of any other event. “Statistical dependence” of events means, the probability of some event is dependent upon or affected by the occurrence of some other event.

6.3 MARGINAL, JOINT AND CONDITIONAL PROBABILITIES The concepts of marginal, joint and conditional probabilities are dealt in two sections.

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6.3.1 Statistical Independence Under conditions of statistical independence the formulas for marginal, joint and conditional probabilities are attempted below:

i) Marginal Probability or unconditional probability is the simple probability of the occurrence of an event. It is given by, say P(H) = 1/2 (1/(1+1) = 1/2 or 0.5) where P(H) is probability of getting head in a coin-toss experiment. The symbol for marginal probability is P(A) read as probability of event A happening. Each toss is statistically independence.

ii) Joint Probability is the probability of two or more events happening together or in succession. It is given by: P(AB) = P(A)*P(B), where P(AB)-is probability two events A and B happening together or in succession and P(A) and P(B) are marginal probability of events (A) and (B). In a fair coin tossing experiment P(H1, H2) = P(H1)*P(H2) = 0.5 * 0.5 = 0.25, where P(H1,H2) – is P of getting heads in successive two losses or getting heads up on the two coins tossed simultaneously. In a dice game P(2 & 2) = P(2)*P(2) = 1/6 * 1/6 = 1/36.

iii) Conditional Probability is probability of an event happening given that another event has happened. It is written as P(B/A) and read as probability of event B happening, given that event A has already happened. Under statistical independence. P(B/A) = P(B), for no event has influence over other.

6.3.2 Statistical Dependence

Under conditions of statistical dependence marginal and joint probabilities are some as in statistical independence. But conditional probability differs. Conditional probability = P(B/A) = P(AB)/P(A), where P(AB) joint probability and P(A) is marginal probability. Conditional probability of P(A/B) = P(AB) / P (B).

6.4 RULES FOR ADDITION & MULTIPLICATION Addition and multiplication rules of probability are dealt now.

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6.4.1 Addition Rule The addition rule of probability is about probability of getting either this or that. For mutually exclusive events, the addition rule is given by: P(A or B) = P of (A) or (B) happening = P(A) + P(B) Note carefully, that addition rule deals with either or situation. What is the probability that a card drawn is either diamond or spade? P(diamond or spade) = P(Diamond) + P(Spade) = ¼ + ¼ = ½. For mutually inclusive events, the addition rule is given by: P(A or B) = P(A) + P(B) – P(AB), where P(AB) is probability of events A and B happening together. What is the probability that a card drawn is a spade or king? P(Spade or king) = P(Spade) + P(King) = 1/4 + 1/13 – 1/52 = 16/52 = 4/13

6.4.2 Multiplication Rule For multiplication rule of probability is concerned about joint occurrence of events. For statistically independent events, the joint prob = P(AB) = P(A) P(B). Similarly, P(ABC) = P(A) P(B) P(C). For statistically dependent events, P(AB) = P(B/A)

P(A) and P(AB) = P(A/B) – P(B).

Note carefully that multiplication rule is concerned with this and that or both, or all or joint happening.

Illustration 1

The frequency distribution of brokerage commission earned in a month from a survey of 300 stock brokers is as follows: Rs. 0-5000 5000-10000 10000-15000 15000-20000 20000-25000 25000+ frequency 15 25 35 125 70 30 (%) 5 8.33 11.67 41.67 23.33 10

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What is probability that a randomly selected stock broker earns say (i) Rs.5000 to Rs.10,000 and (ii) more than Rs.15,000.

Solution

The probability that a broker earns between $ 5000 and $ 10000 is 0.0833 or 8.33% and more than $ 150000 is, 100% minus (P of earning less than $ 5000 + P of earning $ 5000 to 10000 + P earning $ 10000 to $ 15000) = 100% - (5% + 8.33% + 11.67%) = 75% or 0.75. This is a goal example for statistical probability.

Illustration 6.2

Out of every 1000 investors 20 complain non-receipt of share certificates and 18 have their names wrongly spelled. Of these, 5 have both these complaints. What is the P of any randomly chosen investor has any complaint?

Solution:

This is a problem of mutually inclusive type involving addition rule. Let ‘A’ refers to complaint of non-receipt of share certificate, ‘B’ refers to wrong spelling of name and therefore ‘AB’ refers to both complaints happening together. P(A or B) = P(A) + P(B) – P(AB) = 20/1000 + 18/1000 – 5/1000 = 33/1000 = 0.033 or about 3%.

Illustration 6.3

A research group says, illiquidity of scrips is occurring only one-third as often as quoting below par. The P of both illiquidity and below par is 0.05. If 80 per cent of scrips have none of these problems, how low must the illiquidity problem probability be?

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Solution

Let ‘A’ denote illiquidity, ‘B’ denote below par and ‘AB’ denote both. P(A or B) = P(A) + P(B) – P(AB) P(A or B) = 1 – % of nil problem = 1 - .8 = .2 So, 0.2 = P(A) + P(B) – P(AB) So, P(A) + P(B) = 0.2 + 0.05 = 0.25 i.e., P(A) + 3[P(A)] = 0.25 (Since, ‘A’ is a third ‘B’) So, P(A) = 0.25 / 4 = 0.0625.

Illustration 6.4

An inventor is setting 3 independent criteria for an ideal investment. First it must be liquid. Second, it must not fall below issue price. Third it must pay commensurate return. By experience, 85% of investments are found to be liquid. Out of liquid investments 80% are found to be holding their value. Out of these that hold their value 82% are giving good return. What is the probability that a randomly selected investment will prove to be ideal.

Solution

This is a case of mutually independent, joint probability problem. The required is the type P(ABC) given, P(A), P(B) and P(C). So, P(ABC) = P(A)*P(B)*P(C) = (.85)(.8)(.82) = 0.5576. So there is about 56% chance.

Illustration 6.5

A stock broker says out of his clients 63% give “purchase” orders and 32% have over 3 years association with him. 21% of clients are with 3 years association placing “purchase” orders. What is the P that a client placing “purchase” order has more than 3 years association?

Solution

This is a case of conditioned probability. Let ‘A’ denote “purchase”, ‘B’ denote over 3 year standing and ‘AB’ denote those with 3 year

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placing a “purchase” order. What is required is P(B/A) = P(BA)/P(A) = 0.21/0.63 = 0.33.

Illustration 6.6

In a survey, the P that a family makes equity investments if its annual income exceeds Rs.3,50,000 is 0.75. Of the surveyed families, 60% have income exceeding Rs.3,50,000 and 52% have equity investments. What is the P that a family has equity investments and income over Rs.3,50,000.

Solution

Let ‘A’ denote equity investment and ‘B’ denote income exceeding Rs.3,50,000. P(A) = .52; P(B) = .6 and P(A/B) = 0.75. Required is P(AB). We know that, P(A/B) = P(BA)/P(B). So, 0.75 = P(BA) / 0.6. So, P(BA) = 0.45.

Illustration 6.7

When a sample of 250 investors, are distributed as to sex and first time or experienced investors, the following distribution is obtained. First time Experienced Male 65 65 Female 44 76 109 141

i) What is P of a randomly selected investor is male? ii) What is P that the investor is first-time investor given that the person

is male? iii) What the P that the person is female given that she is an experienced

investor.

Solution

The P (male) = Total no. of male / Total no. of investors = 130 / 250 = 0.52

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P(First-time/male) = P(First time and male) / P(Male) = (65/250) / (130/250) = 65 / 130 = 0.5 P(Female/experienced) = P(Female and experienced) / P(Experienced) = (76/250) / (141/250) = 76 / 141 = 0.539

Illustration 6.8

A study of a stock exchange reveals that in the fortnight beginning new-moon, 10 out of 15 days the market grew and in the fortnight beginning full moon on 8 out of 15 days it declined as indicated by the market index. On no day it remained at same level as at the previous day’s value. i) Find the P of growth on any randomly selected day, ii) Find the P of growth given that the day followed full moon day.

Solution

Fortnight Beginning Full moon New moon Total Growth 7 10 17 Decline 8 5 13 15 15 30 P(G) = 17 / 30 = 0.567 P(G/F) = P(GF) / P(F) = (7/30) / (15/30) = 7 / 15 = 0.467

Illustration 6.9

In a sample of 100 people surveyed with debt or equity investments only, 60 were equity investors and 40 debt investors. Of the 60 equity investors 40 were metro-based investors and the 20 were non-metro-based. Of the 40 debt investors, 10 were metro and 30 non-metro. What is the P that a randomly selected investor is metro, given that he is an equity investor?

Solution

The required answer is : P(M/E) = P(ME) / P(E)

= (40/100) / (60/100) = 40 / 60 = 0.67 Where M – denotes “metro” and ‘E’ denotes “equity”.

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Illustration 6.10

Two cards are drawn from a pack of 52 cards with replacement, i.e.., the second card is drawn after replacing the first card in the pack. Find the probability that, i) both are king, ii) first is king and second is queen and iii) one is king and other is queen.

Solution

i) P(both cards are king) = P(1st king).P(2nd King) = 4/52 x 4/52 = 1/169

ii) P(first king & second queen) = P(1st king) P(2nd queen) = 4/52 x 4/52 = 1/169

iii) P(one king and other queen) = P(first is king & 2nd queen).P(first is queen & 2nd is king)

= (4/52 x 4/52) + (4/52 x 4/52) = 2/169

Illustration 6.11

A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of i) 3 success, ii) at least 2 successes and iii) at most 2 successes.

Solution

Doubles could be : 1,1; 2,2; 3,3; 4,4; 5,5; 6,6 Let p = P(1,1; 2,2; 3,3; 4,4; 5,5 or 6,6) = 6/36 = 1/6 q = 1 – p = 1 – 1/6 = 5/6 n = 3 (Number of trials) i) 3 success: with n = 3, p = 1/6, q = 5/6 and r = 3

P(r=3) = ncr pr qn-r = 3c3 (1/6)3 (5/6)(3-3) = 1 x 1/216 x 1 = 1/216

ii) P (r>2) = P(2 or 3 success here) We know P(r=3) = 1/216

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P(r=2) = nc2 p2 q3-2

= 3 x 1/6 x 1/6 x 5/6 = 15/216 P(r>2) = 1/216 + 15/216 = 16/216 = 4/54 = 2/27

iii) P(at most 2 success) = P(0) + P(1) + P(2) = 1 – P(3) = 1 – 1/216 = 215/216

Illustration 6.12

An urn contains 3 while balls, 4 red balls and 5 black balls. Two balls are drawn: what is the probability that i) both and red? Ii) both are white? And iii) one red and one white?

Solution

Total balls = 3 + 4 + 5 = 12

i) P(2 red) = 4C2 / I2C2 = [(4X3)/2] / [(12X11)/2] = 1/11

ii) P(2 white) = 3C2 / I2C2 = 3 / (12x11 / 2) = 1/22

iii) P(1R & 1W) = [(4C1X3C1) / I2C2 = (4X3) / (12X11 / 2) = 2/11

Illustration 6.13

A problem in statistics is given to 3 students. The marginal probability of solving the problem for each of the students is 0.8, 0.6 and 0.9. What is the probability that the problem will be solved by one or other?

Solution

The P(Solving) = 1 – P (none solving) P(None solving)= Joint prob. Of none solving = (0.2) (0.4) (0.1) = 0.008 P(Solving) = 1 – P(none solving) = 1 – 0.008 = 0.992

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6.5 BAYE’S THEOREM (REVISING PRIOR ESTIMATES OF

PROBABILITY

Baye’s theorem is the basis for conditional probability under statistical dependence. We know that P(A/B) = P(AB) / P(B) and P(B/A) = P(AB) / P(A) Baye’s theorem gives a superior method of evaluating new information and our earlier estimates (based on limited information only) of the probability revising that things are in one or other state. Decision making is immensely helped by Baye’s theorem.

Illustration 6.2.1

In a factory 3 machines A, B & C produce respectively 25%, 35% and 40%. Of total output of their outputs, 2%, 4% and 5% are defective. A product is randomly chosen and it is found to be defective. What is the probability that it was produced by i) A, ii) A or B, iii) C?

Solution

Given P(A) = 0.25, P(B) = 0.35 and P(C) = 0.4 P(D/A) = 0.02, P(D/B) = 0.04 and P(D/C) = 0.05 P(A) x P(D/A) i) P(A/D) = P(A)xP(D/A) + P(B)xP(D/B) + P(C)xP(D/C) 0.25 x 0.02 = (0.25x0.02) + (0.35x0.04) + (0.4x0.05) 0.005 0.005 = = 0.005 + 0.014 + 0.020 0.039 = 0.128

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P(A) x P(D/A) + P(B) x P(D/B) ii) P(A/D or B/D) = P(A)xP(D/A) + P(B)xP(D/B) + P(C)xP(D/C) 0.005 + 0.014 0.029 = = 0.005 + 0.014 + 0.020 0.039 = 0.487 iii) P(C/D) = 1 – P(A/D or B/D) = 1 – 0.487 = 0.513 (or) 0.02 / 0.039 = 0.513

Illustration 6.2.2

In a competitive exam, multiple choice questions are used. 4 choices of possible answers to each question are given, of which 1 answer is correct. An intelligent candidate knows 90% of answers and below-average candidate knows only 20% of answers. But guess work also help the candidates gets correct answer. What is the probability that the intelligent candidate gets it through guess? What is the probability that a below average candidate got correct answer through guess?

Solution

Let ‘A’ denote knowing the correct answer and ‘α’ not knowing the correct answer. Let ‘B’ denote getting the correct answer either knowing the answer or through guess. So, P(B/A) = 1;

P(B/α) = ¼ = 0.25, since one of the 4 answers is correct.

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i) Intelligent candidate

P(A) = 0.9 and P(α) = 1 – 0.9 = 0.1

P(α) x P(B/α)

P(α/B) =

P(A) x P(B/A) + P(α) x P(B/α) 0.1 x 0.25 = (0.9 x 1) + (0.1 x 0.25) 0.025 0.025 = = = 0.027 0.9+0.025 0.925

ii) Below average candidate

P(A) = 0.2 and P(α) = 1 – 0.2 = 0.8

P(α) x P(B/α)

P(α/B) =

P(A) x P(B/A) + P(α) x P(B/α) 0.8 x 0.25 = (0.2 x 1) + (0.8 x 0.25) 0.2 0.2 = = = 0.5 0.2+0.2 0.4

6.6 PROBABILITY AND EXPECTED VALUE Probability theory helps in computing expected values. Expected return, expected EPS or DPS, expected risk, expected price, etc. can all be computed using probability theory. Expected value of anything is the weighted value, the weights being respective probability value.

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Illustration 6.3.1

Compute expected return on a mutual fund scheme and its risk given the following: Return = Ri = 12% 13% 14% 15% 16% Probability = Pi = 0.1 0.2 0.4 0.2 0.1

Solution

Σ(R) = ΣRi Pi

Σ(R) = 12%(0.1) + 13%(0.2) + 14%(0.4) + 15%(0.2) + 16%(0.1) = 1.2 + 2.6 + 5.6 + 3 + 1.6 = 14%.

Σ(Risk) = [ΣPi (Ri – E(R))]1/2 = [ .1(12-14)2 + .2(13-14)2 + .4(14-14)2 + .2(15-14)2 + .1(16-14)2 ]

= [.4 + .2 + 0 + .2 + 4]0.5 = √1.2 = 1.095%

Illustration 6.3.2

The absenteeism on Monday has the following distribution in a Govt. office. % Absenteeism : 5% 10% 15% 20% 25% Probability : 40% 30% 20% 6% 4% Find the expected level of absenteeism.

Solution

Expected absenteeism = ΣPi Ai = 0.4(5%) + 0.3(10%) + 0.2(15%) + 0.06(20%) + 0.04(25%) = 2% + 3% + 3% + 1.2% + 1% = 10.2% of total staff members.

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QUESTIONS

1. Explain the concept and types of probability.

2. Present the uses of probability to investment decision making

3. Compute expected return and risk of a security whose probability distribution of returns are:

Ri = 12% 14% 16% 18% 20% Pi = 0.1 0.15 0.5 0.15 0.1

4. In a survey of 400 investors 50% hold equity investment, 80% hold debt investments. Find those who hold both debt and equity investments.

5. In an investment game, the choice of investment is made randomly by picking up applications from a bag which contains 4 forms for share investments, 10 for debenture investments, 6 for mutual funds and 3 for fixed deposits. Find the probability of (i) an equity, a debenture, a mutual fund and an FD investment, (ii) the joint probability of (a) two successive mutual fund investment, (b) one equity and one FD investment.

6. Explain basic concepts of probability theory.

7. A company uses 3 machines, A, B and C. The proportion of products produced by the machines are 0.3, 0.48 and 0.22. The percent defective for each machine is 1%, 2% and 3% of their individual output. A defective output is drawn from a day’s production. What is the probability that it was not produced by A?

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LESSON – 7

SIMULATION FOR DECISION MAKING

Simulation means replicating the reality. The reality here refers to a situation. In business management, decision situations differ. As such, decision making is difficult. In such cases, a replica i.e.; model, duplicate, or copy of the decision situation is developed. Solution under the model, duplicate or copy situation is tried over a number of times. These trial solutions are noted down. From these solutions the most acceptable one is chosen.

7.1 A CASE OF SIMULATION IN PRACTICAL LIFE Case of practical life of situation were simulation was adopted to evolve guidelines for operations are given now.

7.1.1 Carpet Cutting Carpet cutting is a case here. A company produces carpets in 175 foot rolls, all in 12 feet wide in different styles and colours. When orders are received, the carpets to the order size are cut and delivered. The carpet cutting division is staffed by very intelligent people with long experience. With all these, the remnants (i.e.. tail pieces) are sizable. These remnants under 3-feet lengths have to be thrown out and a foot costs Rs.720 and remnants between 3 to 6 feet are sold at a third of normal price. In a regular year the total loss due to remnants is a whopping sum of Rs.1 crore. To reduce remnants several cutting rules are adopted in such a way that unusable piece left at the end of the roll is reduced to the minimum. Some of the rules adopted are: when an order is received, i) cut it from the longest roll, ii) cut it from the roll which would leave the shortest left over piece, iii) combine two orders/ three orders and so on so that the left over piece is shortest, iv) to practice rule 3, want for 1 day / 2 days / 3 days. So that combinable orders are obtained, v) to practice rule 4, delivery time has to vary and some customers’ wrath has to be swallowed.

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To optimize on the situation a simulation model was thought of with the components:

i) the size distribution of orders for carpets

ii) the spatial distribution of orders for carpets

iii) the style distribution of orders for carpets

iv) the shade distribution of orders for carpets

v) the production operation by style

vi) the production operation by shade

vii) the inventory of carpets by size

viii) the inventory of carpets by shade

ix) the inventory of carpets by style

x) the cutting process time

xi) the cutting process personnel capabilities

xii) the price of sold carpet and remnants

Simulated cutting operations under a varied possible “cutting rules” were run for 1000 days each. Such long period trial is necessary to avoid sampling fluctuations and to capture the true patterns of incoming orders and production. For every different simulation, the effect on inventory, remnants, cost of labour, revenue from carpets sold, revenue from sale of remnants, maximum, average and minimum waiting times for customers, etc were noted down. The results showed that, accumulation of orders beyond 2 days had no appreciable effect on reducing remnants. A set of 7 cutting rules was found superior to any other set of cutting rules. Application of the 7-set rules reduced remnants by 21% and savings resulted to the tune of Rs.20 lakhs.

7.1.2 Service Level Problem A bank wants to provide optimum service to customers. The level of service depends on pattern of arrivals of customers. At times very few customers and at times vary large number of customers arrive leading to idle

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times and peak loads. A simulation of pattern of arrivals of customers, say every half-an-hour from 10 am to 2 pm, would help the management in deciding the allocation of staff members to handle customers. Simulation is a good exercise here.

7.1.3 Parking Facility Problem A drive-in-restaurant wants to provide adequate parking facilities for customers. How much facility is needed. To park as many as 10 cars, or 15 cars or 17 cars? How many parking slots? You cannot go for vast facility as it costs you heavily nor meagre facility as customers are put to hardship. The number of parking slots needed depends on arrival patterns and service time for customers. By simulating these two, one can find, how many cars will have to be provided parking facilities on the maximum, minimum and the optimum service levels.

7.1.4 Enrolment Problems What will be our college enrolment per year over the nest 5 years? This depends on i) Number of schools (contributing to our enrolment) and colleges (competing with us for enrolment) within a radius of say 50 kms, the fee structure in the different colleges, the demand for different courses, the launching of new courses, the population in the area, transformation facility, etc. A simulation model can be thought of.

7.1.5 Inventory Problem

How much inventory of stock or cash be held? Outflow and inflow patterns influence this. Outflow can be simulated so also the inflow. As a result inventory pattern can be worked out. Based on the simulated inventory, optimum inventory size can be thought of.

7.1.6 Capital Budgeting Problem

A capital budgeting problem involves among other things, finding the net present value of a long term investment. This is influenced by a host of factors like, price per unit, variable cost per unit, units saleable, original

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investment, life of the project in years, cost of capital, tax rate etc. Some of these factors may not lend themselves for forecast. These factors may be simulated and at their simulated values, the net present value of the capital investment may be worked out.

7.1.7 New Product Launching Problem In the case of a new product launching problem the number of months needed to launch a product may have to be found through simulation. Two set of factors influence this i) the different activities in product launch and ii) the activity time. By simulating the activities and their times over a large number of trials, the product launch time can be found.

7.2 SIMULATION AND MATHEMATICAL SOLUTION Mathematical solutions are possible only for text-book cases of business problems. Practical business problems are so complex that mathematical solutions could not be arrived at without simplifying the problem situation at the cost of realism. Simulation is a better course than mathematical approach in complex business cases since reality is kept as such. Simulation does not provide exact result. It is not exact as mathematics. But it is superior to mathematics as the problem situation is close to realism, while mathematical model does simplify the situation and get on exact result which is devoid of realism. Do you want on inexact but realistic solution or an exact but un-realistic solution? The former is better. So, most business executives use simulation, of all the quantitative techniques adopted by business executives, simulation technique is the single most widely used technical (by 29% of 1000 surveyed in US), followed by LPP (by 21%) and by inventory modeling (by 12%).

7.3 REASONS FOR USING SIMULATION Many use simulation technique. What are the reasons? The reasons are:

180

i) For most business problem cases simulation is the only technique to be used, because a long period experimentation in real life is difficult, but the same can be done through simulation taking each trial as a day’s experiment.

ii) Total reality can be replicated, something not possible in mathematical models.

iii) No mathematical model even after simplification of the problem case is possible in certain cases.

iv) Actual experiment as opposed to simulation, involves destruction of property. (eg: testing life of bulbs, tubes, etc.)

v) Actual observation can’t be done just like that as in the case of accidents.

7.4 SHORT-COMINGS OF SIMULATION Simulation is imprecise. No definite answer does it give. It does give a set of results under different situations. Simulation model may be expensive. To simulate itself huge investment is called for. Simulation is useless when there is no uncertainty in the business environment.

7.5 STEPS IN SIMULATION

The steps involved in simulation are as follows:

• Define the problem to be simulated

• Design the model to be used

• Test the model with relevant data

• Try the model and run the simulation

• Analyze the results of simulation

• Validate the simulation if results are closer to reality.

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7.6 TYPES OF SIMULATION Computer simulation and make Carlo simulation are certain types of simulation. Computer simulation uses the competing power of computers to develop solutions. Monte Carlo simulation uses random numbers to deal with uncertainty. Hence forth, using Monte Carlo method, problems are presented and solved.

7.7 ILLUSTRATIONS Using Monte Carlo method same problems are take up. Using this method involves using random numbers.

7.7.1 Capital Budgeting Problem When uncertainly haunts in the estimation of variables in a capital budgeting exercise, simulation technique may be used with respect to a few of the variables, taking the other variables at their best estimates. We know that P, V, F, Q, T, K, I, D and N are the important variables. (P – Price per unit of output, V – Variable cost per unit of output, F – Fixed cost of operation, Q – Quantity of output, T – Tax rate, K – Discount rate or cost of capital, I – Original investment, D – Annual depreciation and N – Number of years of the project’s life). Suppose in a project, P, V, F, Q, N and I are fairly predictable but ‘K’ and ‘T’ are playing truant. In such cases, the K and T will be dealt through simulation while others take given values. Suppose that P = Rs.300/unit, V = Rs.150/unit, F = Rs.15,00,000/p.a, Q = 20,000/p.a, N = 3 years and I = Rs.18,00,000. Then annual profit before tax = [(P-V)Q] – F – D = [(300-150)*20000] – 15,00,000 – 6,00,000 = Rs. 9,00,000/p.a.

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The profit after tax and hence cash flow cannot be computed as tax rate, T is not predictable. Further as ‘k’ is not predictable, present value cannot be computed as well. So, we use simulation here. Simulation process gives a probability distribution to each of the truant playing variables. Let the probability distribution for ‘T’ and ‘K’ be as follows:

T K

Probability Value Probability Value

0.20

0.50

0.30

30%

35%

40%

0.30

0.50

0.20

10%

11%

12%

Next, we construct cumulative probability and assign random number ranges, as follows separately for T and K. Two digit random number ranges are used. We start with 00 and end with 99, thus using 100 random numbers. For the different values of the variable in question, as many number of random number as are equal by the probability values of respective values are used. Thus, for variable T, 20% of random numbers aggregated for its first value 30% and 50% of random number for its next value 35% and 40%. Table 4.6 : Cumulative Probability and Random Number range

T K

Value Profit-ability

Cumulative profitability

Random no.

range

Value Profit-ability

Cumulative profitability

Random no.

range

30%

35%

40%

0.20

0.50

0.30

0.20

0.70

1.00

00-19

20-69

70-99

10%

11%

12%

0.30

0.50

0.20

0.30

0.80

1.00

00-29

30-79

80-99

183

For the first value of the unpredictable variable, we assign random number 00 to 19. For the second value was assign random numbers 20 – 69 and for the third value 70 – 99 are assigned. Similarly for the variable ‘K’ random numbers are assigned. These are given in the above table 4.6. Simulation process now involves reading from random number table, random number pairs (one for ‘T’ and another for ‘K’). The values of ‘T’ and ‘K’ corresponding to the random numbers read are taken from the above table. Suppose the random numbers read are: 48 and 80. Then ‘T’ is 35% as the random number 48 falls in the random number range 20-69 corresponding to 35% and ‘K’ is 12% as the random number 80 falls in the random number range 80-99 corresponding to 12%. Now taking the T = 35% and K = 12%, the NPV of the project can be worked out. We know that the project gives a PBT of Rs.9,00,000 p/a for 3 years. So, the PAT = 9,00,000 – Tax @ 35% = Rs.9,00,000 – 3,15,000 = Rs.5,85,000 p.a. To this we have to add depreciation Rs.6,00,000 (i.e. Rs.18,00,000 / 3 years) to get the cash flow. So, the cash flow = 5,85,000 + 6,00,000 = Rs.11,85,000 p.a.

n

NPV = Σ CFt / (1+k)t - I t=1 = (11,85,000/1.12+11,85,000/1.122 + 11,85,000/1.123) – 18,00,000 = 11,85,000 [1/1.12 + 1/1.122 + 1/1.123] – 18,00,000 = 11,85,000 X 2.4018 – 18,00,000 = 28,56,798 – 18,00,000 = Rs. 10,56,798 We have just taken one pair of random numbers from the table and calculated the NPV is Rs.10,56,798. This process must be repeated at least 20 times, reading 20 pairs of random numbers and getting the NPV for values of T and K corresponding to each pair of random numbers read. Suppose the next pair of random numbers is 28 and 49. Corresponding ‘T’ = 35% and ‘K’ = 11%. Then the PAT = PBT – T = 9,00,000 – 3,15,000 = 5,85,000. The cash flow = 5,85,000 + 6,00,000 = Rs.11,85,000.

184

n

NPV = Σ CFt / (1+k)t - I

t=1 = (11,85,000/1.11+11,85,000/1.112 + 11,85,000/1.113) – 18,00,000 = (10,67,598 + 9,61,773 + 8,66,462) – 18,00,000 = 28,95,803 – 18,00,000 = Rs. 10,95,803 Similarly the NPV for other simulations be obtained. Thus computed NPVs may be averaged and if the same is positive the project may be selected.

7.7.2 New Product Launch Problem Headen Ltd. plans to introduce a new device for automobiles to warn the driver of the proximity of the car in front. Two different engineering strategies to develop the product are there. These two have different probability distributions of development time as follows:

Development time in months Strategy I Strategy II

6

9

12

0.21

0.38

0.41

0.42

0.35

0.23

Strategy I will require Rs.6,00,000 in investment and will result in a variable cost of Rs.7.5 per unit. For strategy II the respective figures are Rs.15,00,000 and Rs.6.75. The product will sell for Rs.10. The sales volume depends on development time, with the following probability distribution:

Unit sales volume Development time

6 9 12

10,00,000

15,00,000

0.21

0.79

0.38

0.62

0.51

0.49

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Simulate 5 trials for introducing the new product for each engineering strategy taking the following random numbers: 49, 67, 06, 30 and 95 for Strategy I and 01, 10, 70, 80 and 66 for Strategy II for both development time and sales volume and find the worth of the strategies.

Solution :

First we have to construct cumulative probability distribution and random number range for development time strategy wise, for the two strategies. Cumulative Probability and R. No. Range (two digits)

Development Time

Strategy I Strategy II

P Cu.P R.Nos P Cu.P R.Nos

6

9

12

0.21

0.38

0.41

0.21

0.59

1.00

00-20

21-58

59-99

0.42

0.35

0.23

0.42

0.77

1.00

00-41

42-86

77-99

Similarly, cumulative probability and random number range for sales volume is needed. The same is given below. Development time

Sales volume 6 months 9 months 12 months

P Cu.P R.Nos P Cu.P R.Nos P Cu.P R.Nos

10,00,000

15,00,000

0.21

0.79

0.21

1.00

00-20

21-99

0.38

0.62

0.38

1.00

00-37

38-99

0.51

0.49

0.51

1.00

00-50

51-99

Now simulation runs need to be done. We have to read R.Nos. from the table if these are not given. But we are given the random numbers, strategy-wise. So, we take them. Strategy I : R.Nos. 49, 67, 06, 30 & 95

Development time and corresponding sales volume for given R.Nos. are as follows:

186

R.Nos. Development time Sales Volume

49

67

06

30

95

9

12

6

9

12

15,00,000

15,00,000

10,00,000

10,00,000

15,00,000

Similarly for strategy II we can get the simulated development time and corresponding sales volume given the R.Nos. Strategy II: R.Nos. 01, 10, 70, 80 and 66.

R.Nos. Development time Sales Volume

01

10

70

80

66

6

6

9

12

9

10,00,000

10,00,000

15,00,000

15,00,000

15,00,000

Now the worth of the two strategies can be evaluated. The following index is used. Worth = [(Selling price – Variable cost)*Volume] / Investment The average of worth index figures can be taken as measures of performance.

Strategy I Selling price – Variable cost = Rs.10 – 7.5 = Rs.2.5. Investment = Rs. 6,00,000 Volume varies from run to run. So, worth also varies run to run. Worth Run 1 = (2.5 x 15,00,000) / 6,00,000 = 6.25 Run 2 = (2.5 x 15,00,000) / 6,00,000 = 6.25 Run 3 = (2.5 x 10,00,000) / 6,00,000 = 4.16 Run 4 = (2.5 x 10,00,000) / 6,00,000 = 4.17 Run 5 = (2.5 x 15,00,000) / 6,00,000 = 6.25

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Total = 27.08 Average = 5.416

Strategy II

Selling price – Variable cost = Rs.10 – 6.75 = Rs.3.25. Investment = Rs. 15,00,000 Volume varies from run to run. Worth Run 1 = (3.75 x 10,00,000) / 15,00,000 = 2.5 Run 2 = (3.75 x 10,00,000) / 15,00,000 = 2.5 Run 3 = (3.75 x 15,00,000) / 15,00,000 = 3.75 Run 4 = (3.75 x 15,00,000) / 15,00,000 = 3.75 Run 5 = (3.75 x 15,00,000) / 15,00,000 = 3.75 Total = 16.25 Average = 3.25 Strategy I has worthier than II.

7.7.3 Service Level Problem At a public facility the inter arrival time between two customers and service time by the officer in-change of the facility have the following frequency distribution. Time between arrivals Service time for customer Minutes Frequencies Minutes Frequencies 2 12 1 48 3 27 2 66 4 60 3 36 5 33 6 18 150 150 Using simulation, find the waiting times on the basis of 5 arrivals with the R.Nos. for inter arrival times being 28, 16, 06, 67 & 88 and R.Nos. for service time being 14, 66, 49, 95 and 04.

188

Solution

First of all from the frequencies given, probability (frequency of a value / total frequency of all) be obtained. From that the cumulative probability distributions and random number range assignments (two digits) can be done, as follows. Probability, cumulative probability and R.Nos. ranges

Time between arrivals Service time

Minutes P Cu.P R.No. Range

Minutes P Cu.P R.No. Range

2

3

4

5

6

0.08

0.18

0.40

0.22

0.12

0.08

0.26

0.66

0.88

1.00

00-07

08-25

26-65

66-87

88-99

1

2

3

0.32

0.44

0.24

0.32

0.76

1.00

00-31

32-75

76-99

Now simulation runs be done. Simulation work sheet taking office opening hour as 10.00 am

R.No. for

Arrival

Inter Arrival

time

Arrival time

Service begins

Service Waiting time*

R.No. Time Ends User Officer

28

16

06

67

88

4

3

2

5

6

10.04

10.07

10.09

10.14

10.20

10.04

10.07

10.10

10.14

10.20

14

86

49

95

04

1

3

2

3

1

10.05

10.10

10.12

10.17

10.21

-

-

1

-

-

4

2

-

2

3

189

Officer waits more than the customer. Only one customer has to wait for one minute. Service level is being high as waiting time is very less and queue is also very small, just 1. (* The officer has to wait 4 minutes beginning 10-00 his office start time and till the arrival of the first customer. Again he has to wait 2 minutes from 10.05 to 10.07. The third customer has to wait for one minute as he arrived at 10.09, but the service for 2nd customer had to go till 10.10. And so on.)

7.7.4 Inventory Problem A merchant has found the following probability distribution about the daily sale of telephone receivers. Daily sale 4 5 6 7 8 9 10 11 12 Probability 0.06 0.14 0.18 0.17 0.16 0.12 0.08 0.06 0.03 His lead time is 5 days. His inventory policy is governed by reorder level and order quantity. He orders 50 units, whenever day-end stock falls to 40 or below. Assuming his beginning inventory as 65, simulate for 15 runs and find inventory position. The R.Nos. are 03, 38, 17, 32, 68, 24, 61, 99, 16, 48, 03, 71, 27, 80 & 33.

Solution:

First random number assignment based on cumulative probability distribution is called for. Cumulative probability and Random No. range

Level of sale P Cu.P R.No. Range

4

5

6

7

0.06

0.14

0.18

0.17

0.06

0.20

0.38

0.55

00-05

06-19

20-37

38-54

190

8

9

10

11

12

0.16

0.12

0.08

0.06

0.03

0.71

0.83

0.91

0.97

1.00

55-70

71-82

83-90

91-96

97-99

Now the simulation runs have to be made as follows:

Day R.No. (Given)

Day beginning inventory

Day’s Sales

Day End Inventory

Lost sale

Stock ordered

Stock received

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

03

38

17

32

69

24

61

99

16

48

03

71

27

80

33

65

61

54

49

43

35

29

21

9

4

50

46

37

31

22

4

7

5

6

8

6

8

12

5

7

4

9

6

9

6

61

54

49

43

35

29

21

9

4

0

46

37

31

22

16

-

-

-

-

-

-

-

-

-

3

-

-

-

-

-

-

-

-

-

50

-

-

-

-

-

-

50

-

-

-

-

-

-

-

-

-

-

-

-

50

-

-

-

-

-

191

Notes: on the fifth day-end order is placed for 50 units as stock falls below reorder level. The stock is received on 10th day end. Mean while on 10th day, the opening stock was only 4 units. Demand 7 units; so lost sales (i.e.. sales lost for want of stock) 3 units. Next days, the opening stock is 50 units. The inventory policy of 50 units of reader quantity and 40 units of re-order level has lead to one stock-cut situation during the period of 15 days.

7.7.5 Cash management problem A trader has the following probability distribution of monthly cash receipts and monthly cash expenses.

Receipts (Rs.lakhs) : 15 16 17 18 19 20

Probability : 0.30 0.25 0.15 0.15 0.10 0.05

Expenses (Rs.lakhs): 13 14 15 16 17 18

Probability : 0.15 0.20 0.25 0.20 0.15 0.05

The trader has an opening balance of Rs.2 lakhs. Assume over draft facility as available which has to be cleared at the next opportunity. Find the lowest level, highest level and closing level of cash during the year.

Solution

Step 1: Since monthly figures are given and annual position is asked at least 12 runs must be made. For these 12 runs, let the R.Nos. pairs for receipts – payment combination be: 15:48, 20:98, 73:06, 60:45, 44:15, 18:19, 58:15, 61:67, 18:90, 00:58, 32:68 and 65:73.

192

Step 2: Cumulative probability and R.No. assignment

Receipts Payments

Value P Cu.P R.No Value P Cu.P R.No

15

16

17

18

19

20

0.30

0.25

0.15

0.15

0.10

0.05

0.30

0.55

0.70

0.85

0.95

1.00

00-29

30-54

55-69

70-84

85-94

95-99

13

14

15

16

17

18

0.15

0.20

0.25

0.20

0.15

0.05

0.15

0.35

0.60

0.80

0.95

1.00

00-14

15-34

35-59

60-79

80-94

95-99

Now the simulation runs have to be made as follows:

Month R.No. for

Receipts

Corresp

-onding Receipts

Receipts plus previous closing

R.No. for

payment

Pay-ments

Closing Balance

Over-draft

1

2

3

4

5

6

7

8

9

10

11

12

15

20

73

60

44

18

58

61

18

00

32

65

15

15

18

17

16

15

17

17

15

15

16

17

15+2=17

15+2=17

18+0=18

17+4=21

16+6=22

15+8=23

17+9=26

17+12=29

15+13=28

15+11=26

16+11=27

17+11=28

48

98

06

45

15

19

15

67

90

58

68

73

15

18

13+1

15

14

14

14

16

17

15

16

16

17-15=2

0

18-14=4

21-15=6

22-14=8

23-14=9

26-14=12

29-16=13

28-17=11

26-15=11

27-16=11

28-16=12

0

1

0

0

0

0

0

0

0

0

0

0

193

At year end a cash balance of Rs.12 lakhs remains. The peak point is Rs.13 lakhs on hand during the 8th month. The lowest point is in the 2nd month with overdraft of Rs.1 lakhs. The overdraft is returned in the next month itself.

QUESTIONS

1. Explain the simulation and its practical uses.

2. How does simulation compare with mathematics? State the merits and limitations of simulation.

3. Explain the process of Monte Carlo simulation with two examples.

4. The cashier of a bank maintain Rs.10 lakhs as opening cash on a particular day of heavy deposits and withdrawals. The deposits have the following probability distribution for that day:

Rs. (Lakhs) : 82 87 92 97 102 107 Prob : 0.12 0.13 0.25 0.25 0.12 0.13 The withdrawals have the following probability: Rs. (Lakhs) : 77 84 91 98 105 112 Prob : 0.10 0.15 0.25 0.25 0.15 0.10 On the basis of 5 runs find the cash position of the bank given the random number pairs: 18, 28; 42, 31; 81,09; 92,91; and 07,90 for deposits and withdrawals respectively.

5. A new product has the following distribution of number of pre-launch activities and mean-time per activity.

No. of activities Probability Activity Mean-time (Months)

Probability

5 6 7

0.25 0.45 0.30

4 5 6

0.15 0.65 0.20

194

On the basis of 10 runs. Find the mean launch time. You may use the first to R.Nos. pairs given in 7.7.5 Illustration in respective order.

6. The top management of a company is considering the problem of marketing a new product. The investment or the fixed cost, required in the project is Rs.150,000. The three factors that are uncertain are the selling price, variable cost and the annual sales volume. The product has a life of only one year. The management has collected the following data regarding the possible levels of these three factors:

Unit selling price

Prob. Unit variable cost

Prob. Sales volume (units)

Prob.

Rs.14 Rs.15 Rs.16

0.35 0.50 0.15

Rs.2 Rs.3 Rs.4

0.30 0.50 0.20

3.000 4.000 5.000

0.25 0.40 0.35

Using Monte Carlo simulation technique, determine the expected profit from the above investment on the basis of 50 trials. What is the variance of this expected profit?

7. Analysis of arrival and service time by the attendant at a tool room (between 8 noon and 10 a.m) in a factory gave the following results:

Time between arrival, (minutes)

Frequency Service Time (minutes)

Frequency

2.0 3.0 4.0 5.0 6.0 7.0 8.0

5 22 25 30 10 5 3

1.0 2.0 3.0 4.0 5.0

15 35 28 14 8

Using the simulation procedure, determine the waiting times on the basis of 50 arrivals.

195

8. A firm adopts the model for calculation of profit : Profit = (selling price – variable cost) Number of units sold

- Fixed factory cost – Marketing cost – Administrative cost Using simulation, prepare the profit budget.

a) Sales in units Probability of Achievement

80,000

0.10

90,000

0.10

1,00,000

0.60

1,20,000

0.10

1,30,000

0.10

b) Selling price(Rs./unit) Probability

9.00 0.05

9.50 0.15

10.00 0.50

10.50 0.25

11.00 0.05

c) Variable cost (Rs./unit) Probability

4.75 0.10

5.00 0.80

5.25 0.10

d) Fixed factory cost(Rs.) Probability

1,75,000 2,00,000 2,25,000 0.25 0.50 0.25

e) Marketing Cost (Rs.) Probability

40,000 50,000 60,000 0.20 0.60 0.20

f) Administrative Cost (Rs.) Probability

75,000 1,00,000 1,25,000 0.10 0.80 0.10

9. A trader in a town has studied his varying monthly sales and monthly

expenses (including the value of goods) and has arrived at the following empirical distributions.

Monthly sales (in thousand Rs.)

Probability Monthly expenses (in thousand Rs.)

Probabilities

15.0 16.0 17.0 18.0 19.0 20.0

0.30 0.25 0.15 0.15 0.10 0.05

12.0 13.0 14.0 15.0 16.0 18.0

0.15 0.20 0.25 0.20 0.15 0.05

196

i) The trader at the beginning of the year has Rs.2,000 in the bank. Simulate his sales and expenses over a year. Assume that the trader can avail temporary overdraft facilities to cover any negative balances.

ii) How much money does the trader have at the end of the year? iii) What is the low point in his finances during the year. iv) Use the R.Nos. in illustration 7.7.5 in respective order.

10. How do you simulate inventory problems? Explain with the choice of variables that need to be brought into our model.

� � �

197

LESSON 8

DECISION THEORY

Decision making is the most significant function of management. Every manager’s prime job is decision making. Decision making process includes the following steps:

i) Knowing the business problem that calls for a decision

ii) Analyzing the decision environment – certainty or uncertainty

iii) Evolving alternative solutions

iv) Evaluating the alternative solutions as to cost, benefit, practicability, etc.

v) Choosing the best alternative course.

Every step in decision making process involves confronting with a host of issues and that decision making is the difficult - most task that managers’ cannot, however, evade. To aid the business managers in this difficult task, certain tools are available, which are presented in this lesson. Actually all the quantitative techniques are of great use in decision-making. So, all are decision aids. However, certain techniques are specially grouped under statistical decision theory and these are: Pay-off technique, expected value of perfect information, decision tree technique and criteria for decision making. These techniques alone are dealt in this lesson. Statistical decision theory deals with decision making under risk and uncertainty. Decision making under risk means, probability distribution of outcomes of current decision is available. Then expected value of outcome can be calculated and on that basis decision can be taken. Decision making under uncertainty implies no knowledge of probability of outcomes. So a set of criteria is used for decision making.

198

8.1 FACTORS IN A DECISION FRAME-WORK There are certain factors in any decision context. These are: i) Acts, ii) Events or the state of nature, iii) Outcomes, iv) Payoffs and v) Expected value of each act. Acts refer to alternative courses. In manufacturing a product a firm can go for either say i) hi-tech course or ii) low-tech course. The choice is very much within the managements purview. Events or states of nature refer to external situations over which the management has no influence, but very vital in the choice of act or course the management goes for. The levels of competition and extent of consumer demand are states of nature affecting sales potentials. These vary with the alternative courses. For example, the hi-tech course may promise less or high competition and hence high or less sales volume. Similarly, low-tech course may have high or low competition and so low or more sales volume. Outcomes are also called as conditional values. For each combination of act and state of nature, an outcome exists. The acts events and outcomes can be preferred in the form of a tree as follows: Acts Events Outcomes E1 O1,1 O

A1 E2 O1,2 E1 O2,1 O

A2 E2 O2,2

199

We can also present these acts, events and outcomes in a matrix form, taking acts in rows and events in columns and the outcomes in the cells as follows: Table 8.1 : Matrix of Acts and Events and Outcomes

Acts Events

E1 E2

A1 O1,1 O1,2

A2 O2,1 O2,2

Pay-off: Pay-off refers to the monetary gain or loss of each outcome or it may be in the form of cost saved or time saved. Pay-off is also called as utility, though strictly, speaking utility is different from pay-off. Utility is personal oriented while pay-off is impersonal. Some level of pay-off may have different levels of utility for different people or for the same person at different times. Expected value of each Act : Every Act has different set of outcomes. The expected value of all these outcomes can be computed by attaching marginal profitability to these outcomes of an Act. Similarly, for the outcomes of each of the different acts expected value can be computed. And choice of Act, i.e., course of action, is based on the expected value.

Illustration 1

Let the two acts be i) High-tech process and ii) Low-tech process. Let the total cost of each of the Acts be: Rs.8 lakhs and Rs. 6 lakhs. Let the events be i) High demand and ii) Low demand. Let the sales value for high demand for high-tech be Rs.10 lakhs and for high demand for high-tech be Rs.7 lakhs. Let the sales volume for high demand for low-tech be Rs.8 lakhs and for low demand for low-tech be Rs.5 lakhs. Let the probability for high and low demand levels for high-tech be 50:50 and the same for low-tech be: 60:40. Find the expected values of high-tech and low-tech.

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Solution:

Table 8.2 gives the problem in a nut shell. Table 8.2 : Events & Acts with probability

Acts Events

High Demand Low Demand

High-tech cost:

Rs.8 lakhs

Sales : Rs. 10 lakhs

Probability = 0.5

Sales Rs. 7 lakhs

Probability = 0.5

Low-tech cost

Rs.6 lakhs

Sales : Rs.8 lakhs

Probability = 0.6

Sales: Rs.4.5 lakhs

Probability = 0.4

Expected value (EV) of an Act = Σij (Salesij – Costi) Probabilityij EV for A1 = (10 – 8) (0.5) + (7 – 8) (0.5) = (2) (0.5) + (-1) (0.5) = 1 – 0.5 = 0.5 = Rs.0.5 lakhs EV for A2 = (8 – 6) (0.6) + (4.5 – 6) (0.4) = (2) (0.6) + (-1.5) (0.4) = 1.2 – 0.6 = 0.6 = Rs.0.6 lakhs Act 2, i.e. low-tech course of action has a higher expected value than high-tech alternative course.

PAY-OFF TABLES

Pay-off tables present the monetary gains or losses of outcomes of Act-Event combinations.

Illustration 2

Suppose you have to conduct a function of large gathering. Two alternative courses are there: you can go for a open-air-stadium or an indoor stadium. These are the two strategies. Three states of nature, mainly normal, hot and chill climates are possible. The turn out of guests is forecast as follows under varied circumstances given in table 8.3.

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Table 8.3 : Events and Acts

Strategy Normal Climate Hot Climate Chill Climate

Open-air stadium

Indoor stadium

1,00,000

1,10,000

1,70,000

80,000

40,000

2,00,000

Suppose the rent for open-air stadium is Rs.10,00,000 and for indoor stadium Rs.12,00,000 and per capita contribution by guest is Rs.10 for open-air and Rs.12 for indoor stadium. Then the pay-off (revenue-expenses), for the different combinations of strategy – States of Acts or Events, is as follows in table 8.4. Table 8.4 : Pay-off table

Strategy Pay-off Table (Figures Rs. lakhs)

Normal Hot Chill

Open-air

Indoor

10 – 10 = 0

11 – 12 = 1

17 – 10 = 7

8 – 12 = -4

4 – 10 = -6

20 – 12 = 8

Pay-off table gives both possible gains and losses of different act-event combinations.

8.2 LOSS TABLES Loss tables gives the conditional losses consisting of loss arising out of excess capacity and loss arising out of deficit capacity .Illustration 3 Suppose daily sales in units be any of the following : 10, 11, 12 & 13. Profit per unit sold is Rs.5 and loss per unit unsold is Rs.3, being the cost unsold as these are worthiness. States of nature, demand levels, are therefore 10, 11, 12 & 13. Suppose four acts, i.e., carrying 10, 11, 12 & 13 units in stock are adopted. Then the excess capacity loss (loss arising out of demand less than supply) and under capacity loss (loss arising out demand greater than capacity to supply) are as follows in the table 8.5.

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Table 8.5 : States of Nature or Events

Stock levels (Acts) Demand Levels (States of Nature or Events)

10 11 12 13

10

11

12

13

0

3

6

9

5

0

3

6

10

5

0

3

15

10

5

0

Values in each column below zero, are excess capacity loss arising out of excess stock and values above zero in each column are over capacity loss figures. Zeros indicate neither of the losses. Excess capacity loss may be called obsolescence loss and the under capacity loss may be called opportunity loss.

8.3 EXPECTED VALUE OF PAY-OFF Expected pay-off is pay-off multiplied by probability. For the above problem, cited in 8.3 we may attach here the following profitability distribution of: 0.10, 0.20, 0.30 and 0.40. The consequent pay-off and expected conditional pay-off can be computed as follows. First, pay-off computation is taken up. Table 8.6 : Conditional Pay-off for

Stock levels Demand levels

10 11 12 13

10

11

12

13

50

47

44

41

50

55

52

49

50

55

60

57

50

55

60

65

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The figures of realized profit are arrived at as follows:

i) Stock = 10, Demand > 10 units, profit = 10 x 5 = 50

ii) Stock = 11, Demand > 11 units, profit = 11 x 5 = 55

iii) Stock = 12, Demand > 12 units, profit = 12 x 5 = 60

iv) Stock = 13, Demand > 13 units, profit = 13 x 5 = 65

v) Stock = 11, Demand = 10, profit = (10x5) – (1x3) = 47

vi) Stock = 12, Demand = 10, profit = (10x5) – (2x3) = 44

vii) Stock = 12, Demand = 11, profit = (11x5) – (1x3) = 52

viii) Stock = 13, Demand = 10, profit = (10x5) – (3x3) = 41

ix) Stock = 13, Demand = 11, profit = (11x5) – (2x3) = 49

x) Stock = 13, Demand = 12, profit = (12x5) – (1x3) = 57

We can generalize the computations in terms of ‘P’ for profit, ‘L’ for loss, ‘D’ for demand and ‘S’ for supply. Conditional pay-off = PD – L(S-D), for S > D and PD for S < D The expected profit for each stock level is simply, the sum of relevant profit multiplied by probability for diff. demand levels.

i) Expected profit for stock = 10 is : 50x.1 + 50x.2 + 50x.3 + 50x.4 = 50

ii) Expected profit for stock = 11 is : 47x.1 + 55x.2 + 55x.3 + 55x.4 = 54.2

iii) Expected profit for stock = 12 is : 44x.1 + 52x.2 + 60x.3 + 60x.4 = 56.8

iv) Expected profit for stock = 13 is : 41x.1 + 49x.2 + 57x.3 + 65x.4 = 57

Alternatively, the expected loss can be computed from the loss figures dealt earlier and the probability of diff. demand levels.

i) Expected loss for stock = 10 is : 0x.1 + 5x.2 + 10x.3 + 15x.4 = 10

ii) Expected loss for stock = 11 is : 3x.1 + 0x.2 + 5x.3 + 10x.4 = 5.8

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iii) Expected loss for stock = 12 is : 6x.1 + 3x.2 + 0x.3 + 5x.4 = 3.2

iv) Expected loss for stock = 13 is : 9x.1 + 6x.2 + 3x.3 + 0x.4 = 3.0

Now compare the expected pay-off computed earlier, with the expected loss, for diff stock levels. The stock level that yields maximum expected profit has the lowest expected loss as well. The sum of these is the profit under certainty. The same is done below in table 8.7. Table 8.7 : Expected profit and loss

Stock Profit Loss Certainty profit

10 50.0 10.0 60.0

11 54.2 5.8 60.0

12 56.8 3.2 60.0

13 57.0 3.0 60.0

The optimum stock level is 13 units, with highest profit and hence lowest loss.

8.4 EXPECTED VALUE OF PERFECT INFORMATION (EVPI) Perfect information refers to complete and accurate information about the states of nature. That is knowing the probability distribution of future states of nature. Different states will prevail. And prior knowledge of the same is known. So, the decision marker can respond to the state of nature in an exact way. That is, if demand is 10 units, exactly 10 units are stocked, if it is 12 units, exactly 12 units are stocked, and so on. So, the profit under such a position of knowledge of the state of nature is computable exactly. Let the probability distribution be 0.1, 0.2, 0.3 and 0.4 for demand levels of 10, 11, 12 and 13 units and unit profit is Rs.5. At these levels of

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demand, the certainty based profit would be Rs.50, 55, 60 and 65 (i.e. units x Rs.5). By multiplying these certainty level profits by the probability values, we get: 50x.1 + 57x.2 + 60x.3 + 65x.4 = 5 + 11 + 18 + 26 = Rs.60. Compare this expected profit of Rs.60 with the expected profit under different levels stock given in table 8.7. The best level of profit in table 8.7 itself is only Rs.57. There is thus an increment of at least Rs.3 in expected profit with perfect information. This is called the expected value of perfect information (EVPI). EVPI puts the lid on the price, if any, payable to the providers of perfect information or the predictor. The value of perfect information is ascertained in this way.

Illustration 4

A manufacturer produces a product at a cost Rs.5 and sells at Rs.8. If a product is not sold within a week, it is a dead loss, i.e.; Rs.5 per unit unsold. The weekly sales record in the past is as follows. Demand ‘D’ : 20 25 30 35 No. of weeks : 5 10 20 15 Suggest the optimal act. Construct loss table also. Ascertain the EPVI.

Solution

Act here refers to number of units to be produced (20, 25, 30 or 35). The manufacturer must not produce less than 20 and more than 35. Profit ‘P” = Rs.8-5 = Rs.3, Loss ‘L’ = cost per unit = Rs.5. The probability distribution of ‘D’ can be constructed from the no. of weeks figures given. Out of total 50 weeks, demand was 20 units for 5 weeks, (i.e. a 10 in 100 or 10% or 0.1) 25 units for 10 weeks (i.e. a 20 in 100 or 20% or 0.2) and so on. So, probability distribution is : 10%, 20%, 40% and 30% respectively for 20, 25, 30 and 35 units.

206

(1)Expected conditional profit method

Conditional profit is given by : PD – L(S-D), for S>D and PD for S<D. With P = 3 and L = 5, the above becomes : 3D – 5(S-D) or 8D-5S for S>D and 3D for S<D. Look at figures in table 8.8

Table 8.8 : Conditional profit

Acts (or) production level (Supply)

States of nature or demand level

20 25 30 35

20

25

30

35

60

35

10

-15

60

75

50

25

60

75

90

65

60

75

90

105

Expected conditional profit (ECP) is computed by attaching probability values to the conditional profit values row-wise. The expected

conditional profit for any act = Σ probability x conditional profit for each state. For production level of 20 units, ECP : 60x.1+60x.2+60x.4+60x.3=60

Similarly for 25 units, ECP : 35x.1 + 75x.2 + 75x.4 + 75x.3 = 71

Similarly for 30 units, ECP : 10x.1 + 50x.2 + 90x.4 + 90x.3 = 74

Similarly for 35 units, ECP : 15x.1 + 25x.2 + 65x.4 + 105x.3 = 61

The optimal production level is 30 units per week giving a profit of 74, the highest. (2)Expected conditional loss method

The same result is also obtained otherwise by conditional loss method as well. The table of conditional loss for different act and state combinations is as follows in table 8.9.

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Table 8.9 : Conditional Loss

Acts (or) levels of production

States of nature or demand levels

20 25 30 35

20

25

30

35

0

25

50

75

15

0

25

50

30

15

0

25

45

30

15

0

The Expected conditional loss (ECL) is computed by attaching probability values of states of nature to relevant loss figures. The ECL for diff. Acts, i.e.; production levels is as follows:

For 20 units, ECL : 0x.1+15x.2+30x.4+45x.3=28.5

For 25 units, ECL : 25x.1 + 0x.2 + 15x.4 + 30x.3 = 17.5

For 30 units, ECL : 50x.1 + 25x.2 + 0x.4 + 15x.3 = 14.5

For 35 units, ECL : 75x.1 + 50x.2 + 25x.4 + 0x.3 = 27.5

The lowest ECL is Rs.14.5 and 30 units of weekly production.

Computation of EVPI

If the manufacturer knows exactly the demand, and he produces accordingly, his certainty based profits will be: 20x3 = 60, 25x3 = 75, 30x3 = 90 and 35x3 = 105 for the 4 acts coinciding the 4 states. The expected profit = 60x.1+75x.2+90x.4+105x.3 = Rs.88.5. Under uncertainty conditions, his best strategy yields a profit of Rs.75, and with perfect information his expected profit is Rs.88.5. Therefore, the EVPI = Rs.88.5 – 74 = Rs.14.5

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8.5 INCREMENTAL ANALYSIS. Incremental analysis is an approach to decision making under risk. Whether a unit addition to level of operation adds to profit or otherwise is the question asked. If answer is positive that addition is made. Further whether a further unit addition to the level of operation results in added profit or not, is the question asked. If the result profit, addition to production is effected. The process is continued to the point until one reaches a stage where the unit addition does in fact reduces the profit. Here the unit addition is the act. Decision on the act is needed.

Illustration 5

A newspaper agent can buy newspapers at Rs.1.2, a concession price, and sell at Rs.3. Unsold newspapers cause a dead-loss. The following demand pattern is facing the agent, with minimum 15 and maximum 20. Demand (No. of copies): 15 16 17 18 19 20 Probability 0.04 0.19 0.33 0.26 0.11 0.07 Find the optimal number of copies, that the agent buys.

Solution

Let the incremental profit be IP, when the incremental unit is sold and incremental loss be IL, when the incremental unit is not sold. Let the probability of sale of additional unit be ‘P’ and it not being sold is 1 – P. Therefore expected incremental profit is P(IP) and the expected incremental loss is (1-P)(IL). As long as P (IP) > (1-P)(IL), addition to stock is profitable. We can deduce the value of ‘p’ from the inequality: P(IP) > (1-P)(IL)

I.e., P(IP) > IL – P(IL) I.e., P(IP) + P (IL) > IL I.e.. P(IP + IL) > IL I.e., P > IL/(IP+IL)

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The value of ‘P’ helps ascertaining the optimal stock level. Here, IL = Rs.1.2 and IP = 3 – 1.2 = 1.8. Therefore P > 1.2 / (1.8 + 1.2) > 1.2 / 3 > 0.4 Now, we have to compute greater than cumulative probability for demand. The highest level of demand at which cumulative probability > 0.4 is the optimum stock quantity. Demand : 15 16 17 18 19 20 > 21 Probability : 0.04 0.14 0.33 0.26 0.11 0.07 0 Greater than Cum. Prob : 1.00 0.96 0.77 0.44* 0.18 0.07 0 The (*) marked cumulative probability is > 0.4, at which level, demand is highest. So, 18 copies of newspaper will be bought by the news agent.

8.6 DECISION TREE APPROACH Decision tree approach is a versatile tools used for decision making under conditions of risk. The features of this approach are: (1) it takes into account the results of all expected outcomes, (ii) it is suitable where decisions are to be made in sequential parts - that is, if this has happened already, what will happen next and what decision has to follow, (iii) every possible outcome is weighed using joint probability model and expected outcome worked out, (iv) a tree-form pictorial presentation of all possible outcomes is presented here and hence the term decision-tree is used. An example will make understanding easier.

Illustration 6 An entrepreneur is interested in a project, say introduction of a fashion product for which a 2 year market span is foreseen, after which the product turns fade and that within the two years all money invested must be realized back in full. The project costs Rs. 4,00,000 at the time of inception. During 1st year, three possible market outcomes are foreseen. Low penetration, moderate penetration and high penetration are the three outcomes, whose probability values, respectively, are 0.3, (i.e., 30% chance), 0.4 and 0.3

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and the cash flows after tax under the three possible outcomes are respectively estimated to be Rs.1,60,000, Rs. 2,20,000 and Rs. 3,00,000. The level of penetration during the 2

nd year is influenced by level

of penetration in the first year. The probability values of different penetration levels in the 2nd year given the level of penetration in the 1st year and respective cash flows are estimated as follows:

Level of Penetration in

year 2

If low penetration in first year

If moderate penetration in first

year

If high penetration in

first year

Cash flow year 2 Cash flow year 2 Cash flow year 2

Amount Prob. Amount Prob. Amount Prob.

Low

Moderate

High

80000

200000

300000

0.2

0.6

0.2

260000

300000

320000

0.3

0.4

0.3

320000

400000

480000

0.1

0.8

0.1

How do you read the above table? It is very simple. If low penetration resulted in 1st year, low presentation in 2nd year with probability of 0.2 and cash flow of Rs.80,000, moderate penetration in 2nd year with probability of 0.6 and cash flow of Rs.2,00,000 and high penetration in 2nd year with probability of 0.2 and cash flow of Rs.3,00,000 are possible. Similarly you can follow for other cases. Combining 1st and 2nd year penetration levels together, 9 outcomes are possible. These are:

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S.

No.

1st Year

penetra- tion

2nd year

penetra-tion

1st year cash flow

1st year proba-bility (P1)

2nd year cash flow

2nd year proba-bility (P2)

Joint probability (P1 x P2)

1

2

3

4

5

6

7

8

9

Low

Low

Low

Moderate

Moderate

Moderate

High

High

High

Low

Moderate

High

Low

Moderate

High

Low

Moderate

High

160000

160000

160000

220000

220000

220000

300000

300000

300000

.3

.3

.3

.4

.4

.4

.3

.3

.3

80000

200000

300000

260000

300000

320000

320000

400000

480000

.2

.6

.2

.3

.4

.3

.1

.8

.1

0.06

0.18

0.06

0.12

0.16

0.12

0.03

0.24

0.03

At this stage, we may go for present value evaluation of these set of outcomes. And this is done below. For this we require a discounting rate. Let us take a 10% discount rate. Then the present value of Rs.1 receivable at 1st year end is Rs.0.909 (i.e. 1/1.1) and at 2nd year end is Rs.0.826 (i.e., 1/1.12). Now the present values of the 9 cash flow streams can be worked out. These values, the NPV relevant to each stream (i.e., the aggregate of the present value of the two cash flows of each stream minus investment Rs.4,00,000), joint probability (i.e., product of probabilities of the two cash flows of each stream) and expected value of NPV (i.e., joint probability times NPV of each stream) are given below in table.

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Table

S.No. PV of 1st year flow

PV of 2nd year flow

PV of both year flows

NPV of each stream

Joint Prob.

Expected NPV

(1) (2) (3) (4) =(2)+(3) (5)=(4)-400000

(6) (7)=(5)(6)

1

2

3

4

5

6

7

8

9

145440

145440

145440

199980

199980

199980

272700

272700

272700

50080

165200

247800

214760

247800

264320

264320

330400

346920

195520

310640

393240

414740

447780

464300

537120

603100

619620

-204480

- 89360

- 6760

14740

47780

64300

137130

203100

219620

0.06

0.18

0.06

0.12

0.16

0.12

0.03

0.24

0.03

-12269

-16085

- 403

1709

7645

7716

4111

48744

6889

Total 1.00 47395

The expected NPV of the project is negative at Rs.12269 if low penetration prevailed both in the 1st and 2nd year and this has a probability of 6 out of 100 or .06. The expected NPV is negative at Rs.16085, if low penetration in 1st year and moderate penetration in 2nd year prevailed and the probability of this happening is 18%. S.No.8 tells that NPV of Rs.48744 with probability of 24% is possible when high penetration in first year and moderate penetration in the 2 year result. The expected NPV of the project is the aggregate of the expected NPVs of the different streams = Rs.47395. Since, it is positive, the project may be taken up.

8.7 CRITERIA FOR DECISION MAKING UNDER UNCERTAINITY When probability distribution of outcomes is not know, the decision situation is called, uncertainty decision situation. Here certain decision criteria are used. These are:

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i) Maximin, i.e., maximum of minima or WALD criterion

ii) Maximax, i.e., maximum of maxima or HURWICZ criterion

iii) Minimax regret criterion or SAVAGE criterion

iv) Minimum regret criterion

v) Discarding on inadmissible act

vi) Hurwicz optimism-pessimism coefficient criterion

vii) Jacob Bernoulli method or Laplace criterion or Baye’s postulate.

There are explained below:

MAXIMIN PAYOFF OR ‘WALD CRITERION

Maximin decision criterion involves ascertaining the minimum values of different acts or strategies and selecting that act having highest minimum pay-off. Hence the name, maximum of minimum. This criterion is somewhat pessimistic optimistic. The stress is an avoiding the worst, and also on not aspiring for too much. This method was conceptualized by Wald.

Illustration 8.7

The pay-off matrix for certain act – event combinations is as follows: Events Act E1 E2 E3 A1 25 400 650 A2 -25 440 400 A3 -125 400 750 Find the choice act as per maximin strategy.

Solution

As per maximin criterion, the row minima are 25, -25, and 125. The maximum of these minima is 25. Therefore, act A1 will be chosen.

214

MAXIMAX PAY-OFF CRITERION OR HURWICZ CRITERION

Maximax decision criterion involves ascertaining the maximum values of different acts and selecting that act having highest maximum pay-off. Very obviously the criterion is highly optimistic. This method was developed by Leonid Hurwicz. In the illustration 8.8.1 cited above, the row maxima are: 25, 440 and 750. The maximum of these maxima is 750 and the strategy corresponding to this is A3.

MINIMAX REGRET CRITERION OR SAVAGE CRITERION

Minimax regret criterion involves ascertaining maximum of regrets for each act. Then the act with lowest of maximum regret value is chosen. This criterion was developed by L.J Savage. The regret value is calculated by subtracting pay off values in a column i.e. event from the best of pay-off of that event. In our illustration 8.8.1, the best of pay-off for E1 is 25, for E2 440 and E3 750. From these values the given column-wise pay-off values are subtracted. A regret pay-off matrix is thus obtained. The following is the regret matrix for our illustration in question.

Table : Regret matrix

Events Act E1 E2 E3 A1 25-25=0 440-400=40 750-650=100 A2 25-(-20)=45 440-440=0 750-700 =50 A3 25-(-125)=150 440-400=40 750-750=0 Row-wise maximum of regrets are: 100, 50 and 150. Minimum of these maxi-regret values is 150 and corresponding strategy or Act is A2 is chosen. Minimizing maximum of regret is the thrust here.

MINI-MIN REGRET CRITERION

Mini-min regret criterion involves ascertaining regrets for each act in the same was as in savage criterion. From the regrets, non-zero minimum

215

regret for each act is chosen. In our problem non-zero the minima of regrets row-wise are: 40, 45 and 40. The minimum of the minima is 40 and either of acts A1 and A3 can be chosen, as both have a regret value of 40 each. Minimizing minimum of non-zero regrets is the thrust here. Thus, it is akin to maxi-max pay-off.

Discarding of an inadmissible act

Payoff matrix can be reduced by eliminating an inadmissible or week act. This is done when an act is dominated by other acts. When the payoff of an act for all events is greater than that of another act, the former act is said to be dominating the latter and the latter can be conveniently removed and the payoff matrix gets reduced.

Illustration 8.8

Consider the following pay-off matrix Events Acts E1 E2 E3 E4 A1 14 9 10 5 A2 11 10 12 7 A3 9 12 10 11 A4 15 10 11 13 A1 is dominated by A4 ‘in all events. So, A1 can be removed and with the reduced matrix, decision can be taken as to choice of act using any of the criteria already dealt.

Hurwicz Optimism – Pessimism Coefficient Criterion

To overcome the pessimism of minimax and optimism of maximax pay-off criteria, a compromise is evolved by Leonid Hurwicz. Under this

method, a coefficient of optimism, α < 1 is used to multiply the maximum

payoff of each act and the minimum payoff of each act is multiplied by 1-α. The aggregate payoff of each act given by the sum of these two products is computed. The act with maximum aggregate payoff figure is chosen.

216

Illustration 8.9

The following payoff matrix is given to you. State of nature or events Acts S1 S2 S3 S4

Process A 1 3 8 5 Process B 2 5 4 7 Process C 4 6 6 3 Process D 6 8 3 5

Using an optimism coefficient of α = 0.6, find the best process as per Hurwicz optimism – pessimism coefficient criterion.

Solution

First we get the row maximum and minimum as follows: Maximum Minimum Process A 8 1 Process B 7 2 Process C 6 3 Process D 8 3 Now, the application of optimism – pessimism coefficient is done

as follows: α = 0.6; hence 1 - α = 0.4 Process A : (8x.6) + (1x.4) = 4.8 + .4 = 5.2 Process B : (7x.6) + (2x.4) = 4.2 + .8 = 5.0 Process C : (6x.6) + (3x.4) = 3.6 + 1.2 = 4.8 Process D : (8x.6) + (3x.4) = 4.8 + 1.2 = 6.0 Process D will be chosen as it has maximum aggregate product value.

Jacob Bernoulli Method or Baye’s Postulate or Laplace criterion

Laplace criterion is based on the premises that there is no reason to believe that an event is more probable than the other. So, equal probabilities are assigned to all events and row-wise mean pay-off is computed by simply

217

dividing the aggregate payoff by number of events. The act, i.e.; the row, with highest mean pay-off is selected.

Illustration 8.10

The following table gives the payoff of certain act – event combinations: Events Act E1 E2 E3 A1 -120 200 260 A2 -80 400 -260 A3 100 -300 600 Adopting Jacob Bernoulli method, find the best act.

Solution

Simply add up the pay-off figures row-wise and divide them by 3, viz, the number of events. This amounts to giving equal probability for all events. We get the following figures. Total Average A1 : - 120 + 200 + 260 340 113 A2 : - 80 + 400 – 260 80 27 A3 : 100 – 300 + 600 400 133 The best act is A3 with highest total and average.

QUESTIONS

1. Explain the concept of statistical decision theory and the concepts used here.

2. Distinguish between Acts and Events.

3. Unit is Rs.10 and unit cost is Rs.5 unsold unit is price worth Rs.2 only. For 10, 11, 12 and 13 units of supply (Acts) and Demand (Events) construct conditional loss table.

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4. For the problem (3) construct conditional profit table and compute expected profit given probability of 0.2, 0.25, 0.25 and 0.2 for 10, 11, 12 and 13 units of demand. Also compute expected loss. Which is the right strategy?

5. For the problem (4) above, Find the FVPI.

6. Following matrix is giving cost under three alternate strategies.

Cost Matrix

Strategies State of Nature

S1 S2 S3

a1 a2 a3

12 6 9

8 10 11

5 10 12

Choose the best strategy by using:

i) Minimum criterion ii) Minimax criterion iii) Minimax regret criterion iv) Minimum regret criterion v) Laplace criterion

7. Consider the following payoff table (probability in parenthesis)

Action State of Nature

S1(2) S2(2) S3(5)

a1 a2

50 -20

105 0

175 300

i) Use the expected payoff criterion to select the better action.

ii) Convert the payoff table to an opportunity loss table and use the opportunity loss criterion to select the better action.

8. Following is an opportunity loss table (probability in parenthesis)

219

Strategies State of Nature

S1 (3) S2 (3) S3 (4)

a1 a2 a3 a4

40 80 50 30

50 30 40 40

80 20 30 50

Use the EOL criterion to choose the best action.

9. Given the following pay-off matrix

Act Payoff (in Rs.) State of Nature

Cold Weather Hot Weather

Sell Cold Drinks Sell Hot Drinks

50 120

100 40

Given the probability of weather being hot is 0.8, set up the opportunity loss table and compute opportunity loss of each action. Select the best act.

10. A fruit seller sells strawberries. If not sold on the day of delivery, they are

worthless. One case of strawberries cost Rs.200, and the seller receives Rs.500 for it. The seller cannot specify the number of customers on any one day, but his analysis of past records shows the following results.

No. of cases sold per day Probability of sale

10 11 12 13

0.15 0.20 0.40 0.25

1.00

Find the best stock option for the seller so that he can optimize his profits. Find the EVPI.

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220

MODEL QUESTION PAPER

Paper 1.4: QUANTITATIVE METHODS

Time : 3 Hrs Max : 100 Marks

PART – A (5 x 8 = 40 Marks) Answer any FIVE questions

1. Explain the concept and types of variables and constants.

2. What are linear functions? Comment on the slope of linear functions.

3. Graph the behaviour of TP, MP and AP.

4. A quadratic production function is as follows: Q = -20x2 + 160x + 400, where Q = output and x = an input. Find the marginal productivity of x and output optimizing level of x.

5. Explain the LP technique as an optimization technique.

6. Coley’s Machine shop has four machines on which to do three jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following table. What are the job assignments which will minimize cost.

Job

Machine

W X Y Z

A B C

Rs. 18 8

10

Rs.24 13 15

Rs.28 17 19

Rs.32 19 22

7. Compute expected return and risk of a security whose probability distribution of returns are: Ri = 12% 14% 16% 18% 20% Pi = 0.1 0.15 0.5 0.15 0.1

8. Following is an opportunity loss table (probability in parenthesis)

221

Strategies State of Nature

S1 (3) S2 (3) S3 (4)

a1 a2 a3 a4

40 80 50 30

50 30 40 40

80 20 30 50

Use the EOL criterion to choose the best action.

PART – B (4 x 15 = 60 Marks) Answer any FOUR questions

9. What is trade-off? How is the same relevant to decision making? How is it effected?

10. Graph the functions: a) y = -x2 + 5x - 2 and b) y = x2 + 5x - 2 with set of values -5 < x < 5 as the domain. Which of the functions will have a `hill' and which a `valley'. Which factor determine the formation of `hill' or `valley'. [Hint: It is the sign of x2 in the quadratic function decide the formation of `hill' or `valley'].

11. The cost function for a firm producing x units is: TC = 5x + 350 and its revenue function is: TR = 5x - x2. Find the demand function, break-even-point, profit maximizing output and sales maximizing output. [Hint: i) For break-even point, solve for x, by equating TC = TR].

12. T.C.Mellott trucking company has a contract to move 115 truckloads of sand per week between three sand-washing plants, W, X and Y, and three destinations, A, B and C. Cost and volume information is given below. Compute the optimal transportation cost using the stepping – stone method.

Project Requirement per week, truckloads

Plant Available per week, truckloads

A B C

45 50 20

W X Y

35 40 40

222

COST INFORMATION

From To Project A To Project B To Project C

Plant W Plant X Plant Y

Rs. 5 20

5

Rs. 10 30

8

Rs.10 20 12

13. A company uses 3 machines, A, B and C. The proportion of products produced by the machines are 0.3, 0.48 and 0.22. The percent defective for each machine is 1%, 2% and 3% of their individual output. A defective output is drawn from a day’s production. What is the probability that it was not produced by A?

14. Explain the application of simulation.

15. Present criteria for decision making under uncertainty.

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