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MB0048-Unit-01-Introduction to Operations Research

Unit-01-Introduction to Operations Research

Structure:

1.1 Introduction

Learning objectives

1.2 Historical Background

Definitions of Operations Research

1.3 Scope of Operations Research

1.4 Features of Operations Research

1.5 Phases of Operations Research

1.6 Types of Operations Research models

1.7 Operations Research Methodology

Definition

Construction

Solution

Validation

Implementation

1.8 Operations Research Techniques and Tools

1.9 Structure of the Mathematical Model

1.10 Limitations of Operations Research

1.11 Summary

1.12 Terminal Questions

1.13 Answers to SAQs and TQs

Answers to Self Assessment Questions

Answers to Terminal Questions

1.14 References

1.1 Introduction

Welcome to the unit on Operations Research Management. Operations Research Management focuses on the mathematical scoring of consequences of a decision aiming to optimise the use of time, effort and resources, and avoid blunders. The act of obtaining the best results under any given circumstances is known as optimising. The key purpose of Operations Research (OR) is to do preparative calculations that aid the decision-making process.

Now, you will agree that decision-making is a key part of our daily life. The ultimate goal of all decisions is to maximise benefits and minimise effort and time. OR gives decision makers the power to make effective decisions and improve day-to-day operations. Decision makers consider all the available options, study the outcomes and estimate the risks.

In simple situations, you use your common sense and judgement to take decisions. For example, if you are buying a microwave or washing machine, the decision-making process is not very complicated. You can simply compare the price, quality and durability of the well known brands and models in the market and take a decision based on it.

However, in complex situations, although it is possible to take decisions based on ones common sense, a decision backed by mathematical calculations reduces the risk factor and increases the probability of success. Some such situations, where decision-makers have to reply on mathematical scoring and reasoning, are finding an appropriate product mix amidst competitors products or planning a public transportation network in a city.

Learning Objectives

By the end of this unit, you should be able to:

List the significant features of Operations Research

Describe the methodology of Operations Research

Define the structure of a mathematical model in Operations Research

Describe the significance of the function of Operations Research

1.2 Historical Background

During the World War II, scientists from United Kingdom studied the strategic and tactical problems associated with air and land defense of the country. The aim of this study was to determine the effective utilisation of limited military resources to win the battle. The technique was named Operations Research. After World War II, Operations Research techniques were developed and deployed in the decision making process in complicated situations in various fields, such as industrial, academic and government organisations.

1.2.1 Definitions of operations research

Churchman, Aackoff and Aruoff defined Operations Research as: the application of scientific methods, techniques and tools to operation of a system with optimum solutions to the problems, where optimum refers to the best possible alternative.

The objective of Operations Research is to provide a scientific basis to the decision-makers for solving problems involving interaction of various components of the organisation. You can achieve this by employing a team of scientists from different disciplines, to work together for finding the best possible solution in the interest of the organisation as a whole. The solution thus obtained is known as an optimal decision.

You can also define Operations Research as The use of scientific methods to provide criteria for decisions regarding man, machine, and systems involving repetitive operations.

Self Assessment Questions

Fill in the blanks:

1. The main objective of OR is to provide a _______ ________ to the decision-makers.

2. OR employs a team of _________ from _________ __________.

1.3 Scope of Operations Research

Any problem, simple or complicated, can use OR techniques to find the best possible solution. This section will explain the scope of OR by seeing its application in various fields of everyday life.

i) In Defense Operations: In modern warfare, the defense operations are carried out by three major independent components namely Air Force, Army and Navy. The activities in each of these components can be further divided in four sub-components namely: administration, intelligence, operations and training and supply. The applications of modern warfare techniques in each of the components of military organisations require expertise knowledge in respective fields. Furthermore, each component works to drive maximum gains from its operations and there is always a possibility that the strategy beneficial to one component may be unfeasible for another component. Thus in defense operations, there is a requirement to co-ordinate the activities of various components, which gives maximum benefit to the organisation as a whole, having maximum use of the individual components. A team of scientists from various disciplines come together to study the strategies of different components. After appropriate analysis of the various courses of actions, the team selects the best course of action, known as the optimum strategy.

ii) In Industry: The system of modern industries is so complex that the optimum point of operation in its various components cannot be intuitively judged by an individual. The business environment is always changing and any decision useful at one time may not be so good some time later. There is always a need to check the validity of decisions continuously against the situations. The industrial revolution with increased division of labour and introduction of management responsibilities has made each component an independent unit having their own goals. For example: production department minimises the cost of production but maximises output. Marketing department maximises the output, but minimises cost of unit sales. Finance department tries to optimise the capital investment and personnel department appoints good people at minimum cost. Thus each department plans its own objectives and all these objectives of various department or components come to conflict with one another and may not agree to the overall objectives of the organisation. The application of OR techniques helps in overcoming this difficulty by integrating the diversified activities of various components to serve the interest of the organisation as a whole efficiently. OR methods in industry can be applied in the fields of production, inventory controls and marketing, purchasing, transportation and competitive strategies.

iii) Planning: In modern times, it has become necessary for every government to have careful planning, for economic development of the country. OR techniques can be fruitfully applied to maximise the per capita income, with minimum sacrifice and time. A government can thus use OR for framing future economic and social policies.

iv) Agriculture: With increase in population, there is a need to increase agriculture output. But this cannot be done arbitrarily. There are several restrictions. Hence the need to determine a course of action serving the best under the given restrictions. You can solve this problem by applying OR techniques.

v) In Hospitals: OR methods can solve waiting problems in out-patient department of big hospitals and administrative problems of the hospital organisations.

vi) In Transport: You can apply different OR methods to regulate the arrival of trains and processing times minimise the passengers waiting time and reduce congestion, formulate suitable transportation policy, thereby reducing the costs and time of trans-shipment.

vii) Research and Development: You can apply OR methodologies in the field of R&D for several purposes, such as to control and plan product introductions.


3. Mention two applications of OR.

4. How can a hospital benefit from the application of OR methods?

1.4 Features of Operation Research

Some key features of OR are as follows:

1. OR is system oriented. OR scrutinises the problem from an organisations perspective. The results can be optimal for one part of the system, while the same can be unfavourable for another part of the system.

2. OR imbibes an interdisciplinary team approach. Since no single individual can have a thorough knowledge of all fast developing scientific know-how, personalities from different scientific and managerial cadre form a team to solve the problem.

3. OR makes use of scientific methods to solve problems.

4. OR increases effectiveness of the managements decision-making ability.

5. OR makes use of computers to solve large and complex problems.

6. OR offers a quantitative solution.

7. OR also takes into account the human factors.


Fill in the blanks:

5. OR ________ inter-disciplinary approach.

6. OR increases the effectiveness of ________ ability.

1.5 Phases of Operations Research

The scientific method in OR study generally involves the following three phases.

Figure 1.1: Phases of operations research

1. Judgment Phase: This phase includes the following activities:

a) Determination of the operations

b) Establishment of the objectives and values related to the operations

c) Determination of the suitable measures of effectiveness

d) Formulation of the problems relative to the objectives

2. Research Phase: This phase utilises the following methodologies:

a) Operations and data collection for a better understanding of the problems

b) Formulation of hypothesis and model

c) Observation and experimentation to test the hypothesis on the basis of additional data

d) Analysis of the available information and verification of the hypothesis using pre-established measure of effectiveness

e) Prediction of various results and consideration of alternative methods

3. Action Phase: The action phase involves making recommendations for the decision process. The recommendations can be made by those who identified and presented the problem or anyone who influences the operation in which the problem has occurred.


State True/False:

7. OR gives qualitative solution

8. One of the OR phases is Action phase

1.6 Types of OR Models

A model is an idealised representation or abstraction of a real-life system. The objective of a model is to identify significant factors that affect the real-life system and their interrelationships. A model aids the decision-making process as it provides a simplified description of complexities and uncertainties of a problem in a logical structure. The most significant advantage of a model is that it does not interfere with the real-life system.

1.6.1 A broad classification of OR models

You can broadly classify OR models into the following types.

Figure 1.2: Classification of models

a. Physical Models include all form of diagrams, graphs and charts. They are designed to tackle specific problems. They bring out significant factors and interrelationships in pictorial form to facilitate analysis. There are two types of physical models:

I. Iconic models

II. Analog models

Iconic models are primarily images of objects or systems, represented on a smaller scale. These models can simulate the actual performance of a product. Analog models are small physical systems having characteristics similar to the objects they represent, such as toys.

b. Mathematical or Symbolic Models employ a set of mathematical symbols to represent the decision variable of the system. The variables are related by mathematical systems. Some examples of mathematical models are allocation, sequencing, and replacement models.

c. By nature of Environment: Models can be further classified as follows:

I. Deterministic model in which everything is defined and the results are certain, such as an EOQ model.

II. Probabilistic Models in which the input and output variables follow a defined probability distribution, such as the Games Theory.

d. By the extent of Generality Models can be further classified as follows:

I. General Models are the models which you can apply in general to any problem. For example: Linear programming.

II. Specific Models on the other hand are models that you can apply only under specific conditions. For example: You can use the sales response curve or equation as a function of only in the marketing function.


State True/False

9. Diagram belongs to the physical model

10. Allocation problems are represented by iconic model

1.7 OR Methodology

The basic dominant characteristic feature of operations research is that it employs mathematical representations or models to analyse problems. This distinct approach represents an adaptation of the scientific methodology used by the physical sciences. The scientific method translates a real given problem into a mathematical representation which is solved and retransformed into the original context. The OR approach to problem solving consists of the following steps: Defining the problem, Constructing the model, Solving the model, Validating the model and Implementing the final result.

Figure 1.3: Steps in the OR methodology

1.7.1 Definition

The first and the most important step in the OR approach of problem solving is to define the problem. You need to ensure that the problem is identified properly because this problem statement will indicate three major aspects:

1) A description of the goal or the objective of the study

2) An identification of the decision alternative to the system

3) The recognition of the limitations, restrictions and requirements of the system.

1.7.2 Construction

Based on the problem definition, you need to identify and select the most appropriate model to represent the system. While selecting a model, you need to ensure that the model specifies quantitative expressions for the objective and the constraints of the problem in terms of its decision variables. A model gives a perspective picture of the whole problem and helps tackling it in a well-organised manner. Therefore, if the resulting model fits into one of the common mathematical models, you can obtain a convenient solution by using mathematical techniques. If the mathematical relationships of the model are too complex to allow analytic solutions, a simulation model may be more appropriate. There are various types of models which you can construct under different conditions.

1.7.3 Solution

After deciding on an appropriate model you need to develop a solution for the model and interpret the solution in the context of the given problem. A solution to a model implies determination of a specific set of decision variables that would yield an optimum solution. An optimum solution is one which maximises or minimises the performance of any measure in a model subject to the conditions and constraints imposed on the model.

1.7.4 Validation

A model is a good representation of a system. However, the optimal solution must work towards improving the systems performance. You can test the validity of a model by comparing its performance with some past data available from the actual system. If under similar conditions of inputs, your model can reproduce the past performance of the system, then you can be sure that your model is valid. However, you will still have no assurance that future performance will continue to duplicate the past behaviour. Secondly, since the model is based on careful examination of past data, the comparison should always reveal favourable results. In some instances, this problem may be overcome by using data from trial runs of the system. Note that such validation methods are not appropriate for non-existent systems, since data will not be available for comparison.

1.7.5 Implementation

You need to apply the optimal solution obtained from the model to the system and note the improvement in the performance of the system. You need to validate this performance check under changing conditions. To do so, you need to translate these results into detailed operating instructions issued in an understandable form to the individuals who will administer and operate the recommended system. The interaction between the operations research team and the operating personnel reaches its peak in this phase.

1.8 OR Techniques and Tools

The different techniques and tools used in OR are as follows:

1. Linear programming: You can use linear programming to find a solution for optimising a given objective. The objective may be to maximise profit or to minimise cost. You need to ensure that both the objective function and the constraints can be expressed as linear expressions of decision variables. You will learn about the various uses of linear programming in Chapter-2.

2. Inventory control methods: The production, purchasing and material managers are always confronted with questions, such as when to buy, how much to buy and how much to keep in stock. The inventory model aims at optimising these inventory levels.

3. Goal programming: In linear programming, you take a single objective function and consider all other factors as constraints. However, in real life there may be number of important objective functions. Goal programming has several objective functions, each having a target value Programme models are developed to minimise deviations from these targets.

4. Queuing model: The queuing theory is based on the concept of probability. It indicates the capability of a given system and the changes possible in the system when you modify the system. In formulating a queuing model you need not take into account all the constraints. There is no maximisation or minimisation of an objective function. Therefore, the application of queuing theory cannot be viewed as an optimisation process. You can use the queuing theory to estimate the required balance between customer waiting time and the service capability of the system. You need to first consider several alternatives, evaluate them through queuing models, study their effect on the system, and then make a choice. The criteria for evaluation will be measures of efficiency of the system, such as the average length of a queue, expected waiting time of a customer and the average time spent by the customer in the system. In this approach, your success primarily depends on the alternatives considered and not so much on the queuing models developed.

5. Transportation model: The transportation model is an important class of linear programs. The model studies the minimisation of the cost of transporting a commodity from a number of sources to several destinations. The supply at each source and the demand at each destination are known. The objective of the model is to develop an integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost.

The transportation problem involves m sources, each of which has available ai (i = 1, 2, ..,m) units of homogeneous product and n destinations, each of which requires bj (j = 1, 2., n) units of products. Here ai and bj are positive integers. The cost cij of transporting one unit of the product from the ith source to the jth destination is given for each i and j. It is assumed that the total supply and the total demand are equal.

(1)

The condition (1) is guaranteed by creating either a fictitious destination with a demand equal to the surplus if total demand is less than the total supply or a (dummy) source with a supply equal to the shortage if total demand exceeds total supply. The cost of transportation from the fictitious destination to all sources and from all destinations to the fictitious sources are assumed to be zero so that total cost of transportation will remain the same.

6. In addition to the above there are tools, such as the sequence model, the assignment model, and network analysis which you will learn in detail in later units.


State True/False

11. OR methodology consists of definition, solution and validation only.

12. The interaction between OR team and Management reaches peak level in implementation phase.

1.9 The Structure of the Mathematical Model

Many industrial and business situations are concerned with planning activities. In each case of planning, there are limited sources, such as men, machines, material and capital at the disposal of the planner. One has to take decision regarding these resources to either maximise production, or minimise the cost of production or maximise the profit. These problems are referred as the problems of constrained optimisation.

Linear programming is a technique for determining an optimal schedule of interdependent activities, for the given resources. Therefore, you can say that programming refers to planning and the process of decision-making about a particular plan of action from a given set of alternatives.

Any business activity or production activity to be formulated as a mathematical model can best be discussed through its parts which are as follows:

1. Decision variables

2. Objective function

3. Constraints

4. Diet problem

Decision variables

Decision variables are the unknowns, which you need to determine from the solution of the model. The parameters represent the controlled variables of the system.

Objective function

The objective function defines the measure of effectiveness of the system as a mathematical function of its decision variables. The optimal solution to the model is obtained when the corresponding values of the decision variable yield the best value of the objective function whilst satisfying all constraints. Therefore, you can say that the objective function acts as an indicator for the achievement of the optimal solution.

While formulating a problem, the desire of the decision-maker is expressed as a function of n decision variables. This function is a linear programming problem that is each of its items will have only one variable raised to the power one). Some of the objective functions in practice are:

Maximisation of contribution or profit

Minimisation of cost

Maximisation of production rate or minimisation of production time

Minimisation of labour turnover

Minimisation of overtime

Maximisation of resource utilisation

Minimisation of risk to environment or factory

Constraints

To account for the physical limitations of the system, you need to ensure that the model includes constraints, which limit the decision variables to their feasible range or permissible values. These are expressed as constraining mathematical functions.

For example, in chemical industries, restrictions come from the government about throwing gases in the environment. Restrictions from sales department about the marketability of some products are also treated as constraints. A linear programming problem then has a set of constraints in practice.

The mathematical models in OR may be viewed generally as determining the values of the decision variables x J, J = 1, 2, 3, n, which will optimize Z = f (x 1, x 2, - x n).

Subject to the constraints:

g i (x 1, x 2 x n) ~ b i, i = 1, 2, - m

And xJ 0 j = 1, 2, 3 - n where ~ is , or =.

The function f is called the objective function, where xj ~ bi, represent the ith constraint for i = 1, 2, 3 - m where b i is a known constant. The constraints x j 0 are called the non-negativity condition, which restrict the variables to zero or positive values only.

Diet problem

Formulate the mathematical model for the following:

Vitamin A and Vitamin B are found in food 1 and food 2.

One unit of food 1 contains 5 units of vitamin A and 2 units of vitaminB.

One unit of food 2 contains 6 units of vitamin A and 3 units of vitaminB.

The minimum daily requirement of a person is 60 units of vitamin A and 80 units of Vitamin B.

The cost per one unit of food 1 is Rs. 5/- and one unit of food2 is Rs. 6/-. Assume that any excess units of vitamins are not harmful. Find the minimum cost of the mixture (of food1 and food2) which meets the daily minimum requirements of vitamins.

Mathematical Model of the Diet Problem: Suppose x1 = the number of units of food1 in the mixture and x2 = the number of units of food2 in the mixture.

Lets formulate the constraint related to vitamin-A. Since each unit of food1 contains 5 units of vitamin A, we have that x1 units of food1 contains 5x1 units of vitamin A. Since each unit of food 2 contains 6 units of vitaminA, we have that x2 units of food2 contains 6x2 units of vitaminA. Therefore, the mixture contains 5x1 + 6x2 units of vitamin-A. Since the minimum requirement of vitamin A is 60 units, you can say that 5x1 + 6x2 60.

Now lets formulate the constraint related to vitaminB. Since each unit of food1 contains 2 units of vitaminB we have that x1 units of food1 contains 2x1 units of vitamin-B. Since each unit of food2 contains 3 units of vitaminB, we have that x2 units of food2 contains 3x2 units of vitaminB.

Therefore the mixture contains 2x1 + 3x2 units of vitaminB. Since the minimum requirement of vitaminB is 80 units, you can say that

2x2 + 3x2 80

Next lets formulate the cost function. Given that the cost of one unit of food1 is Rs. 5/- and one unit of food 2 is Rs. 6/-. Therefore, x1 units of food1 costs Rs. 5x1, and x2 units of food 2 costs Rs. 6x2.

Therefore, the cost of the mixture is given by Cost = 5x1 + 6x2.

If we write z for the cost function, then you can write z = 5x1 + 6x2.

Since cost is to be minimised, you can write min z = 5x1 + 6x2.

Since the number of units (x1 or x2) are always non-negative, therefore, you have x1 0, x2 0.

Therefore, the mathematical model is:

5x1 + 6x2 60

2x1 + 3x2 80

x1 0, x2 0, min z = 5x1 + 6x2.

1.10 Limitations of OR

The limitations are more related to the problems of model building, time and money factors.

i. Magnitude of computation: Modern problems involve a large number of variables. The magnitude of computation makes it difficult to find the interrelationship.

ii. Intangible factors: Non quantitative factors and human emotional factor cannot be taken into account.

iii. Communication gap: There is a wide gap between the expectations of managers and the aim of research professionals.

iv. Time and Money factors: When you subject the basic data to frequent changes then incorporation of them into OR models becomes a costly affair.

v. Human Factor: Implementation of decisions involves human relations and behaviour.


Fill in the blanks:

13. OR imbibes _________ team approach.

14. Linear programming is tool of _______.

15. The three phases of OR are ________.

16. To solve any problem through OR approach the first step is _______.

17. _________ represents a real life system.

18. _________ represents the controlled variables of the system

1.11 Summary

The OR approach needs to be equally developed in various agricultural problems on a regional or international basis. With the explosion of population and consequent shortage of food, every country faces the problem of optimum allocation of land in various crops in accordance with climate conditions and available facilities. The problem of optimal distribution of water from a resource like a reservoir for irrigation purposes is faced by each developing country, and a good amount of scientific work can be done in this direction.


1. Define OR.

2. What are the characteristic features of OR?

3. What is a model in OR? Discuss different models available in OR.

4. Write short notes are different phases of OR.

5. What are the limitations of OR?



1. Scientific basis

2. Scientists, different disciplines

3. Industry Planning

4. To solve waiting problems

5. Imbibes

6. Decision making

7. True

8. True

9. True

10. False

11. False

12. False

13. Inter-disciplinary

14. OR

15. Judgement phase, Research phase & Action phase

16. Define the problem

17. Model

18. Parameters


1. Refer to 1.2.1

2. Refer to 1.4

3. Refer to 1.6

4. Refer to 1.5

5. Refer to 1.10

1.14 References

No external sources have been referred for this unit.

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MB0048-Unit-02-Linear Programming

Unit 2 Linear Programming

Structure:

2.1 Introduction

Learning objectives

2.2 Requirements

Basic assumptions of linear programming problems

2.3 Linear Programming

Canonical forms

Case studies of linear programming problems

2.4 Graphical Analysis

Some basic definitions

2.5 Graphical Methods to Solve Linear Programming Problems

Working rule

Solved problems on mixed constraints LP problem

Solved problem for unbounded solution

Solved problem for inconsistent solution

Solved problem for redundant constraint

2.6 Summary





2.9 References

2.1 Introduction

Welcome to the unit of Operations Research on Linear Programming. Linear programming focuses on obtaining the best possible output (or a set of outputs) from a given set of limited resources.

Minimal time and effort and maximum benefit coupled with the best possible output or a set of outputs is the mantra of any decision-maker. Today, decision-makers or managements have to tackle the issue of allocating limited and scarce resources at various levels in an organisation in the best possible manner. Man, money, machine, time and technology are some of these common resources. The managements task is to obtain the best possible output (or a set of outputs) from these given resources.

You can measure the output from factors, such as the profits, the costs, the social welfare, and the overall effectiveness. In several situations, you can express the output (or a set of outputs) as a linear relationship among several variables. You can also express the amount of available resources as a linear relationship among various system variables. The managements dilemma is to optimise (maximise or minimise) the output or the objective function subject to the set of constraints. Optimisation of resources in which both the objective function and the constraints are represented by a linear form is known as a linear programming problem (LPP).

Learning objectives


Construct linear programming problem and analyse a feasible region

Evaluate and solve linear programming problems graphically

2.2 Requirements of LPP

The common requirements of a LPP are as follows.

i. Decision variables and their relationship

ii. Well-defined objective function

iii. Existence of alternative courses of action

iv. Non-negative conditions on decision variables

2.2.1 Basic Assumptions of LPP

1. Linearity: You need to express both the objective function and constraints as linear inequalities.

2. Deterministic: All co-efficient of decision variables in the objective and constraints expressions are known and finite.

3. Additivity: The value of the objective function and the total sum of resources used must be equal to the sum of the contributions earned from each decision variable and the sum of resources used by decision variables respectively.

4. Divisibility: The solution of decision variables and resources can be non-negative values including fractions.


Fill in the blanks

1. Both the objective function and constraints are expressed in _____ forms.

2. LPP requires existence of _______, _______, ____ and _______.

3. Solution of decision variables can also be ____________.

2.3 Linear Programming

The LPP is a class of mathematical programming where the functions representing the objectives and the constraints are linear. Optimisation refers to the maximisation or minimisation of the objective functions.

You can define the general linear programming model as follows:

Maximise or Minimise:

Z = c1 x1 + c2 x2 + - - - - + cn xn

Subject to the constraints,

a11 x1 + a12 x2 + + a1n xn ~ b1

a21 x1 + a22 x2 + + a2n xn ~ b2

-

am1 x1 + am2 x2 + - + amn xn ~ bm

and x1 0, x2 0, xn 0

Where cj, bi and aij (i = 1, 2, 3, .. m, j = 1, 2, 3 - n) are constants determined from the technology of the problem and xj (j = 1, 2, 3 - n) are the decision variables. Here ~ is either (less than), (greater than) or = (equal). Note that, in terms of the above formulation the coefficients cj, bi aij are interpreted physically as follows. If bi is the available amount of resources i, where aij is the amount of resource i that must be allocated to each unit of activity j, the worth per unit of activity is equal to cj.

2.3.1 Canonical forms

You can represent the general Linear Programming Problem (LPP) mentioned above in the canonical form as follows:

Maximise Z = c1 x1+c2 x2 + + cn

Subject to,

a11 x1 + a12 x2 + + a1n xn b1

a21 x1 + a22 x2 + + a2n xn b2

am1 x1+am2 x2 + + amn xn bm

x1, x2, x3, xn 0.

The following are the characteristics of this form.

1. All decision variables are non-negative.

2. All constraints are of type.

3. The objective function is of the maximisation type.

You can represent any LPP in the canonical form by using five elementary transformations, which are as follows:

1. The minimisation of a function is mathematically equivalent to the maximisation of the negative expression of this function. That is,

Minimise Z = c1 x1 + c2x2 + . + cn xn

is equivalent to

Maximise Z = c1x1 c2x2 cn xn

2. Any inequality in one direction ( or ) may be changed to an inequality in the opposite direction ( or ) by multiplying both sides of the inequality by 1. For example

2x1+3x2 5 is equivalent to 2x13x2 5

3. An equation can be replaced by two inequalities in opposite direction. For example:

2x1+3x2 = 5 can be written as 2x1+3x2 5 and 2x1+3x2 5 or 2x1+3x2 5 and 2x1 3x2 5

4. An inequality constraint with its left hand side in the absolute form can be changed into two regular inequalities. For example:

2x1+3x2 5 is equivalent to 2x1+3x2 5 and 2x1+3x2 5 or 2x1 3x2 5

5. The variable which is unconstrained in sign ( 0, 0 or zero) is equivalent to the difference between 2 non-negative variables. For example:

if x is unconstrained in sign then x = (x+ x) where x+ 0, x 0

Caselet

An automobile company has two units X and Y which manufacture three different models of cars - A, B and C. The company has to supply 1500, 2500, and 3000 cars of A, B and C respectively per week (6 days). It costs the company Rs. 1,00,000 and Rs. 1,20,000 per day to run the units X and Y respectively. On a day unit X manufactures 200, 250 and 400 cars and unit Y manufactures 180, 200 and 300 cars of A, B and C respectively per day. The operations manager has to decide on how many days per week should each unit be operated to meet the current demand at minimum cost.

The operations manager along with his team uses a LPP model to arrive at the minimum cost solution.

2.3.2 Case Studies of linear programming problems


State True/False

4. One of the characteristics of canonical form in the objective function must be of maximisation.

5. 2x 3y 10 can be written as -2x + 3y -10

2.4 Graphical Analysis

You can analyse linear programming with 2 decision variables graphically.

Example

Lets look at the following illustration.

Maximise Z = 700 x1+500 x2

Subject to 4x1+3x2 210

2x1+x2 90

and x1 0, x2 0

Let the horizontal axis represent x1 and the vertical axis x2. First, draw the line 4x1 + 3x2 = 210, (by replacing the inequality symbols by the equality) which meets the x1-axis at the point A (52.50, 0) (put x2 = 0 and solve for x1 in 4x1 + 3x2 = 210) and the x2 axis at the point B (0, 70) (put x1 = 0 in 4x1 + 3x2 = 210 and solve for x2).

Figure 2.1: Linear programming with 2 decision variables

Any point on the line 4x1+3x2 = 210 or inside the shaded portion will satisfy the restriction of the inequality, 4x1+3x2 210. Similarly the line 2x1+x2 = 90 meets the x1-axis at the point C(45, 0) and the x2 axis at the point D(0, 90).

Figure 2.2: Linear programming with 2 decision variables

Combining the two graphs, you can sketch the area as follows:

Figure 2.3: Feasible region

The 3 constraints including non-negativity are satisfied simultaneously in the shaded region OCEB. This region is called feasible region.

2.4.1 Some basic definitions

Note: The objective function is maximised or minimised at one of the extreme points referred to as optimum solution. Extreme points are referred to as vertices or corner points of the convex regions.


Fill in the blanks

6. The collection of all feasible solutions is known as the ________ region.

7. A linear inequality in two variables is known as a _________.

2.5 Graphical Methods to Solve LPP

Solving a LPP with 2 decision variables x1 and x2 through graphical representation is easy. Consider x1 x2 the plane, where you plot the solution space enclosed by the constraints. The solution space is a convex set bounded by a polygon; since a linear function attains extreme (maximum or minimum) values only on the boundary of the region. You can consider the vertices of the polygon and find the value of the objective function in these vertices. Compare the vertices of the objective function at these vertices to obtain the optimal solution of the problem.

2.5.1 Working rule

The method of solving a LPP on the basis of the above analysis is known as the graphical method. The working rule for the method is as follows.

Step 1: Write down the equations by replacing the inequality symbols by the equality symbols in the given constraints.

Step 2: Plot the straight lines represented by the equations obtained in step I.

Step 3: Identify the convex polygon region relevant to the problem. Decide on which side of the line, the half-plane is located.

Step 4: Determine the vertices of the polygon and find the values of the given objective function Z at each of these vertices. Identify the greatest and least of these values. These are respectively the maximum and minimum value of Z.

Step 5: Identify the values of (x1, x2) which correspond to the desired extreme value of Z. This is an optimal solution of the problem

2.5.2 Solved problems on mixed constraints LP problem

In linear programming problems, you may have:

i) a unique optimal solution or

ii) many number of optimal solutions or

iii) an unbounded solution or

iv) no solutions.

2.5.3 Solved problem for unbounded solution

2.5.4 Solved problem for inconsistent solution

2.5.5 Solved problem for redundant constraint


State True/False

8. The feasible region is a convex set

9. The optimum value occurs anywhere in feasible region

2.6 Summary

In a LPP, you first identify the decision variables with economic or physical quantities whose values are of interest to the management. The problems must have a well-defined objective function expressed in terms of the decision variable.

The objective function is to maximise the resources when it expresses profit or contribution. Here, the objective function indicates that cost has to be minimised. The decision variables interact with each other through some constraints. These constraints arise due to limited resources, stipulation on quality, technical, legal or variety of other reasons.

The objective function and the constraints are linear functions of the decision variables. A LPP with two decision variables can be solved graphically. Any non-negative solution satisfying all the constraints is known as a feasible solution of the problem. The collection of all feasible solutions is known as a feasible region. The feasible region of a LPP is a convex set. The value of the decision variables, which maximise or minimise the objectives function is located on the extreme point of the convex set formed by the feasible solutions. Sometimes the problem may be unfeasible indicating that no solution exists for the problem.


1. Use the graphical method to solve the LPP.

Maximise Z= 5x1 + 3x2

Subject to:

3x1 + 5x2 15

5x1 + 2x2 10

x1, x2 0

2. Mathematically formulate the problem.

A firm manufactures two products; the net profit on product 1 is Rs. 3 per unit and the net profit on product 2 is Rs. 5 per unit. The manufacturing process is such that each product has to be processed in two departments D1 and D2. Each unit of product 1 requires processing for 1 minute at D1 and 3 minutes at D2; each unit of product 2 requires processing for 2 minute at D1 and 2 minutes at D2.

Machine time available per day is 860 minutes at D1 and 1200 minutes at D2. How much of products 1 and 2 should be produced every day so that total profit is maximum. Formulate this as a problem in L.P.P.



1. Linear

2. Alternate course of action

3. Fractious

4. True

5. True

6. Feasible

7. Half-plan

8. True

9. False


1.

2. Maximise 3x1 + 5x2, subject to x1 + 2x2 800 (minutes)

3x1 + 2x2 1200 (minutes) x1, x2 0

2.9 References

No external sources have been referred.

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MB0048-Unit-03-Simplex Method

Unit-03-Simplex Method

Structure:

3.1 Introduction

Learning objectives

3.2 Standard Form of LPP

The standard form of LPP

Fundamental theorem of LPP

3.3 Solution of LPP Simplex Method

Initial basic feasible solution of a LPP

To solve problem by Simplex Method

3.4 The Simplex Algorithm

Steps

3.5 Penalty Cost Method or Big M-method

3.6 Two Phase Method

3.7 Solved Problems on Minimisation

3.8 Summary


3.10 Answers to SAQs & TQs



3.11 References

3.1 Introduction

Welcome to the third unit of Operations Research Management on the simplex method. The simplex method focuses on solving LPP of any enormity involving two or more decision variables.

The simplex algorithm is an iterative procedure for finding the optimal solution to a linear programming problem. The objective function controls the development and evaluation of each feasible solution to the problem. If a feasible solution exists, it is located at a corner point of the feasible region determined by the constraints of the system.

The simplex method simply selects the optimal solution amongst the set of feasible solutions of the problem. The efficiency of this algorithm is because it considers only those feasible solutions which are provided by the corner points, and that too not all of them. You can consider obtaining an optimal solution based on a minimum number of feasible solutions.

Learning objectives


Create a standard form of LPP from the given hypothesis

Apply the simplex algorithm to the system of equations

Interpret the big M-technique

Discuss the importance of the two phase method

Construct the dual from the primal (and vice versa)


The characteristics of the standard form of LPP are:

1. All constraints are equations except for the non-negativity condition, which remain inequalities only.

2. The right-hand side element of each constraint equation is non-negative.

3. All the variables are non-negative.

4. The objective function is of maximisation or minimisation type.

You can change the inequality constraints of equations by adding or subtracting the left hand side of each such constraint by a non-negative variable. The non-negative variable that has to be added to a constraint inequality of the form to change it to an equation is called a slack variable. The non-negative variable subtracted from a constraint inequality of the form to change it to an equation is called a surplus variable.

To make the right hand side of a constraint equation positive, multiply both the sides of the resulting equation by (-1). Use the elementary transformations introduced with the canonical form to achieve the remaining characteristics.

3.2.1 The standard form of the LPP

3.2.2 Fundamental theorem of LPP

A set of m simultaneous linear equations in n unknowns/variables, n m, AX = b, with r (A) = m.

If there is a feasible solution X 0, then there exists a basic feasible solution.


State True or False

1. We add surplus variable for of constraint.

2. The right hand side element of each constraint is non-negative.

3.3 Solution of the Linear Programming Program Simplex Method

Consider a LPP given in the standard form

To optimise z = c1 x1 + c2 x2 + + cn xn

Subject to

a11 x1 + a12 x2 + + an x n S1 = b1

a21 x1 + a22 x2 + -+ a2n xn S2 = b2

.

am1 x1 + am2 x2 + + amn xn Sm = bm

x1, x2, xn, S1, S2 , Sm 0.

To each of the constraint equations, add a new variable called an artificial variable on the left hand side of every equation which does not contain a slack variable. Subsequently every constraint equation will contain either a slack variable or an artificial variable.

The introduction of slack and surplus variables does not alter either the constraints or the objective function. Therefore, you can incorporate such variables in the objective function with zero coefficients. However, the artificial variables do change the constraints as these are added only to the left hand side of the equations.

The newly derived constraint equation is equivalent to the original equation, only if all the artificial variables have value zero. Artificial variables are incorporated into the objective function with very large positive coefficient M in the minimisation program and very large negative coefficientM in the maximisation program guaranteeing optimal solutions. The large positive and negative coefficients represent the penalty incurred for making a unit assignment to the artificial variable.

Thus the standard form of LPP can be given as follows:

Optimise Z = CT X

Subject to AX = B,

And X 0

Where X is a column vector with decision, slack, surplus and artificial variables, C is the vector corresponding to the costs; A is the coefficient matrix of the constraint equations and B is the column vector of the right hand side of the constraint equations.

3.3.1 Initial basic feasible solution of a LPP

Consider a system of m equations in n unknowns x1, x2 - - an,

a11 x1 + a12 x2 + - - + a1n xn = b1

a21 x1 + a22 x2 + - - + a2n xn = b2

am1 x1 + am2 x2 + - - + amn xn = bn

Where m n

To solve this system of equations, you must first assign any of n m variables with value zero. The variables assigned the value zero are called non-basic variables, while the remaining variables are called basic variables. Then, solve the equation to obtain the values of the basic variables. If one or more values of the basic variables are valued at zero, then solution is said to degenerate, whereas if all basic variable have non-zero values, then the solution is called a non-degenerate solution. A basic solution satisfying all constraints is said to be feasible.

3.3.2 To solve problem by simplex method

To solve a problem by the simplex method, follow the steps below.

1. Introduce stack variables (Sis) for type of constraint.

2. Introduce surplus variables (Sis) and artificial variables (Ai) for type of constraint.

3. Introduce only Artificial variable for = type of constraint.

4. Cost (Cj) of slack and surplus variables will be zero and that of artificial variable will be M

5. Find Zj - Cj for each variable.

6. Slack and artificial variables will form basic variable for the first simplex table. Surplus variable will never become basic variable for the first simplex table.

7. Zj = sum of [cost of variable x its coefficients in the constraints Profit or cost coefficient of the variable].

8. Select the most negative value of Zj - Cj. That column is called key column. The variable corresponding to the column will become basic variable for the next table.

9. Divide the quantities by the corresponding values of the key column to get ratios; select the minimum ratio. This becomes the key row. The basic variable corresponding to this row will be replaced by the variable found in step 6.

10. The element that lies both on key column and key row is called Pivotal element.

11. Ratios with negative and value are not considered for determining key row.

12. Once an artificial variable is removed as basic variable, its column will be deleted from next iteration.

13. For maximisation problems, decision variables coefficient will be same as in the objective function. For minimisation problems, decision variables coefficients will have opposite signs as compared to objective function.

14. Values of artificial variables will always is M for both maximisation and minimisation problems.

15. The process is continued till all Zj - Cj 0.


State True or False

3. A basic solution is said to be a feasible solution if it satisfies all constraints.

4. If one or more values of basic variable are zero then solution is said to be degenerate.



To test for optimality of the current basic feasible solution of the LPP, use the following algorithm called simplex algorithm. Lets also assume that there are no artificial variables existing in the program.

3.4.1 Steps

Perform the following steps to solve the simplex algorithm.

Figure 3.1: Steps to solve simple algorithm

1) Locate the negative number in the last row of the simplex table. Do not include the last column. The column that has negative number is called the work column.

2) Next, form ratios by dividing each positive number in the work column, excluding the last row into the element in the same row and last column. Assign that element to the work column to yield the smallest ratio as the pivot element. If more than one element yields the same smallest ratio, choose the elements randomly. The program has no solution, if none of the element in the work column is non-negative.

3) To convert the pivot element to unity (1) and then reduce all other elements in the work column to zero, use elementary row operations.

4) Replace the x -variable in the pivot row and first column by x-variable in the first row pivot column. The variable to be replaced is called the outgoing variable and the variable that replaces it is called the incoming variable. This new first column is the current set of basic variables.

5) Repeat steps 1 through 4 until all the negative numbers in the last row excluding the last column are exhausted.

6) You can obtain the optimal solution by assigning the value to each variable in the first column corresponding to the row and last column. All other variables are considered as non-basic and have assigned value zero. The associated optimal value of the objective function is the number in the last row and last column for a maximisation program, but the negative of this number for a minimisation problem.

Caselet

A manufacturing company discontinued production of an unprofitable product line. This created excess production capacity. The companys management is considering devoting this excess capacity to one or more of the other ongoing products.

Knowing the capacity of the machines, the number of machine hours required to produce one unit of each product and the unit profit per product, the management needs to find out how much of each product the company should produce to maximise profit.

The management can use the simplex method to arrive at the required solution.


State Yes or No

6. The key column is determined by Zj - Cj row.

7. Pivotal element lies on the crossing of key column and key row.

8. The negative and infinite ratios are considered for determining key row.

3.5 Penalty Cost Method or Big-M Method

Consider a LPP when at least one of the constraints is of type or =. While expressing in the standard form, add a non negative variable to each of such constraints. These variables are called artificial variables.

Addition of artificial variables causes violation of the corresponding constraints as they are added to only one side of an equation. The new system is equivalent to the old system of constraints only if the artificial variables are valued at zero. To guarantee such assignments in the optimal solution, you can incorporate artificial variables into the objective function with large positive or large negative coefficients in a minimisation and maximisation programs respectively. You denote these coefficients by M. Whenever artificial variables are part of the initial solution X0, the last row of simplex table will contain the penalty cost M. You can make the following modifications in the simplex method to minimise the error of incorporating the penalty cost in the objective function. This method is called Big M-method or Penalty cost method.

Figure 3.2: Big M-method

1) The last row of the simplex table is decomposed into two rows, the first of which involves those terms not containing M, while the second involves those containing M.

2) The step 1 of the simplex method is applied to the last row created in the above modification and followed by steps 2, 3 and 4 until this row contains no negative elements. Then step 1 of simplex algorithm is applied to those elements next to the last row that are positioned over zero in the last row.

3) Whenever an artificial variable ceases to be basic, it is removed from the first column of the table as a result of step 4; it is also deleted from the top row of the table as is the entire column under it.

4) The last row is removed from the table whenever it contains all zeroes.

5) If non-zero artificial variables are present in the final basic set, then the program has no solution. In contrast, zero valued artificial variables in the final solution may exist when one or more of the original constraint equations are redundant.


State Yes or No

9. The value of artificial value is M.

10. Artificial variables enter as basic variables.


The drawback of the penalty cost method is the possible computational error resulting from assigning a very large value to the constant M. To overcome this difficulty, a new method is considered, where the use of M is eliminated by solving the problem in two phases.

Phase I: Formulate the new problem. Start by eliminating the original objective function by the sum of the artificial variables for a minimisation problem and the negative of the sum of the artificial variables for a maximisation problem. The Simplex method optimises the ensuing objective with the constraints of the original problem. If a feasible solution is arrived, the optimal value of the new objective function is zero (suggestive of all artificial variables being zero). Subsequently proceed to phase II. If the optimal value of the new objective function is non-zero, it means there is no solution to the problem and the method terminates.

Phase II: Start phase II using the optimum solution of phase I as the base. Then take the objective function without the artificial variables and solve the problem using the Simplex method.

The first iteration gives the following table:

Table 3.20: The first iteration table

x1

x2

S1

S2

A1

0

0

0

0

1

X2 0

2

1

1

0

0

2

A1 1

5

0

4

1

1

4

5

0

4

1

0

4

Since all elements of the last row are non negative, the procedure is complete. But the existence of non-zero artificial variables in the basic set indicates that the problem has no solution.


3.8 Summary

This unit explains how to solve LPP using the simplex method. The unit also explains the constraints for which you need to introduce the slack, surplus and artificial variables. Examples are used to illustrate the method of solving LPP using the simplex method.


1. Maximise z = 3x1 x2

Subject to the constraints

2x1 + x2 2

x1 + 3x2 3

x2 4,

x1, x2 0.

2. Minimise Z = 6x1 + 7x2


x1 + 3x2 12

3x1 + x2 12

x1 + x2 8

x1 + x2 0



1. False

2. True

3. True

4. True

5. Yes

6. Yes

7. No

8. Yes

9. Yes

10. Yes


1. Z = 9 X1 = 3 X2 = 0

2. Z = 5.8 X1 = 8/3 X2 = 6

3.11 References

No external sources of content have been referred for unit 3

About the Author

admin" admin



MB0048-Unit-03-Simplex Method

Unit-03-Simplex Method

Structure:

3.1 Introduction

Learning objectives


The standard form of LPP

Fundamental theorem of LPP

3.3 Solution of LPP Simplex Method

Initial basic feasible solution of a LPP

To solve problem by Simplex Method


Steps

3.5 Penalty Cost Method or Big M-method



3.8 Summary


3.10 Answers to SAQs & TQs



3.11 References

3.1 Introduction

Welcome to the third unit of Operations Research Management on the simplex method. The simplex method focuses on solving LPP of any enormity involving two or more decision variables.

The simplex algorithm is an iterative procedure for finding the optimal solution to a linear programming problem. The objective function controls the development and evaluation of each feasible solution to the problem. If a feasible solution exists, it is located at a corner point of the feasible region determined by the constraints of the system.

The simplex method simply selects the optimal solution amongst the set of feasible solutions of the problem. The efficiency of this algorithm is because it considers only those feasible solutions which are provided by the corner points, and that too not all of them. You can consider obtaining an optimal solution based on a minimum number of feasible solutions.

Learning objectives


Create a standard form of LPP from the given hypothesis

Apply the simplex algorithm to the system of equations

Interpret the big M-technique

Discuss the importance of the two phase method

Construct the dual from the primal (and vice versa)


The characteristics of the standard form of LPP are:

1. All constraints are equations except for the non-negativity condition, which remain inequalities only.

2. The right-hand side element of each constraint equation is non-negative.

3. All the variables are non-negative.

4. The objective function is of maximisation or minimisation type.

You can change the inequality constraints of equations by adding or subtracting the left hand side of each such constraint by a non-negative variable. The non-negative variable that has to be added to a constraint inequality of the form to change it to an equation is called a slack variable. The non-negative variable subtracted from a constraint inequality of the form to change it to an equation is called a surplus variable.

To make the right hand side of a constraint equation positive, multiply both the sides of the resulting equation by (-1). Use the elementary transformations introduced with the canonical form to achieve the remaining characteristics.

3.2.1 The standard form of the LPP

3.2.2 Fundamental theorem of LPP

A set of m simultaneous linear equations in n unknowns/variables, n m, AX = b, with r (A) = m.

If there is a feasible solution X 0, then there exists a basic feasible solution.


State True or False

1. We add surplus variable for of constraint.


3.3 Solution of the Linear Programming Program Simplex Method

Consider a LPP given in the standard form

To optimise z = c1 x1 + c2 x2 + + cn xn

Subject to

a11 x1 + a12 x2 + + an x n S1 = b1

a21 x1 + a22 x2 + -+ a2n xn S2 = b2

.

am1 x1 + am2 x2 + + amn xn Sm = bm

x1, x2, xn, S1, S2 , Sm 0.

To each of the constraint equations, add a new variable called an artificial variable on the left hand side of every equation which does not contain a slack variable. Subsequently every constraint equation will contain either a slack variable or an artificial variable.

The introduction of slack and surplus variables does not alter either the constraints or the objective function. Therefore, you can incorporate such variables in the objective function with zero coefficients. However, the artificial variables do change the constraints as these are added only to the left hand side of the equations.

The newly derived constraint equation is equivalent to the original equation, only if all the artificial variables have value zero. Artificial variables are incorporated into the objective function with very large positive coefficient M in the minimisation program and very large negative coefficientM in the maximisation program guaranteeing optimal solutions. The large positive and negative coefficients represent the penalty incurred for making a unit assignment to the artificial variable.

Thus the standard form of LPP can be given as follows:

Optimise Z = CT X

Subject to AX = B,

And X 0

Where X is a column vector with decision, slack, surplus and artificial variables, C is the vector corresponding to the costs; A is the coefficient matrix of the constraint equations and B is the column vector of the right hand side of the constraint equations.

3.3.1 Initial basic feasible solution of a LPP

Consider a system of m equations in n unknowns x1, x2 - - an,

a11 x1 + a12 x2 + - - + a1n xn = b1

a21 x1 + a22 x2 + - - + a2n xn = b2

am1 x1 + am2 x2 + - - + amn xn = bn

Where m n

To solve this system of equations, you must first assign any of n m variables with value zero. The variables assigned the value zero are called non-basic variables, while the remaining variables are called basic variables. Then, solve the equation to obtain the values of the basic variables. If one or more values of the basic variables are valued at zero, then solution is said to degenerate, whereas if all basic variable have non-zero values, then the solution is called a non-degenerate solution. A basic solution satisfying all constraints is said to be feasible.

3.3.2 To solve problem by simplex method

To solve a problem by the simplex method, follow the steps below.

1. Introduce stack variables (Sis) for type of constraint.

2. Introduce surplus variables (Sis) and artificial variables (Ai) for type of constraint.

3. Introduce only Artificial variable for = type of constraint.

4. Cost (Cj) of slack and surplus variables will be zero and that of artificial variable will be M

5. Find Zj - Cj for each variable.

6. Slack and artificial variables will form basic variable for the first simplex table. Surplus variable will never become basic variable for the first simplex table.

7. Zj = sum of [cost of variable x its coefficients in the constraints Profit or cost coefficient of the variable].

8. Select the most negative value of Zj - Cj. That column is called key column. The variable corresponding to the column will become basic variable for the next table.

9. Divide the quantities by the corresponding values of the key column to get ratios; select the minimum ratio. This becomes the key row. The basic variable corresponding to this row will be replaced by the variable found in step 6.

10. The element that lies both on key column and key row is called Pivotal element.

11. Ratios with negative and value are not considered for determining key row.

12. Once an artificial variable is removed as basic variable, its column will be deleted from next iteration.

13. For maximisation problems, decision variables coefficient will be same as in the objective function. For minimisation problems, decision variables coefficients will have opposite signs as compared to objective function.

14. Values of artificial variables will always is M for both maximisation and minimisation problems.

15. The process is continued till all Zj - Cj 0.


State True or False

3. A basic solution is said to be a feasible solution if it satisfies all constraints.

4. If one or more values of basic variable are zero then solution is said to be degenerate.



To test for optimality of the current basic feasible solution of the LPP, use the following algorithm called simplex algorithm. Lets also assume that there are no artificial variables existing in the program.

3.4.1 Steps

Perform the following steps to solve the simplex algorithm.

Figure 3.1: Steps to solve simple algorithm

1) Locate the negative number in the last row of the simplex table. Do not include the last column. The column that has negative number is called the work column.

2) Next, form ratios by dividing each positive number in the work column, excluding the last row into the element in the same row and last column. Assign that element to the work column to yield the smallest ratio as the pivot element. If more than one element yields the same smallest ratio, choose the elements randomly. The program has no solution, if none of the element in the work column is non-negative.

3) To convert the pivot element to unity (1) and then reduce all other elements in the work column to zero, use elementary row operations.

4) Replace the x -variable in the pivot row and first column by x-variable in the first row pivot column. The variable to be replaced is called the outgoing variable and the variable that replaces it is called the incoming variable. This new first column is the current set of basic variables.

5) Repeat steps 1 through 4 until all the negative numbers in the last row excluding the last column are exhausted.

6) You can obtain the optimal solution by assigning the value to each variable in the first column corresponding to the row and last column. All other variables are considered as non-basic and have assigned value zero. The associated optimal value of the objective function is the number in the last row and last column for a maximisation program, but the negative of this number for a minimisation problem.

Caselet

A manufacturing company discontinued production of an unprofitable product line. This created excess production capacity. The companys management is considering devoting this excess capacity to one or more of the other ongoing products.

Knowing the capacity of the machines, the number of machine hours required to produce one unit of each product and the unit profit per product, the management needs to find out how much of each product the company should produce to maximise profit.

The management can use the simplex method to arrive at the required solution.


State Yes or No

6. The key column is determined by Zj - Cj row.

7. Pivotal element lies on the crossing of key column and key row.

8. The negative and infinite ratios are considered for determining key row.

3.5 Penalty Cost Method or Big-M Method

Consider a LPP when at least one of the constraints is of type or =. While expressing in the standard form, add a non negative variable to each of such constraints. These variables are called artificial variables.

Addition of artificial variables causes violation of the corresponding constraints as they are added to only one side of an equation. The new system is equivalent to the old system of constraints only if the artificial variables are valued at zero. To guarantee such assignments in the optimal solution, you can incorporate artificial variables into the objective function with large positive or large negative coefficients in a minimisation and maximisation programs respectively. You denote these coefficients by M. Whenever artificial variables are part of the initial solution X0, the last row of simplex table will contain the penalty cost M. You can make the following modifications in the simplex method to minimise the error of incorporating the penalty cost in the objective function. This method is called Big M-method or Penalty cost method.

Figure 3.2: Big M-method

1) The last row of the simplex table is decomposed into two rows, the first of which involves those terms not containing M, while the second involves those containing M.

2) The step 1 of the simplex method is applied to the last row created in the above modification and followed by steps 2, 3 and 4 until this row contains no negative elements. Then step 1 of simplex algorithm is applied to those elements next to the last row that are positioned over zero in the last row.

3) Whenever an artificial variable ceases to be basic, it is removed from the first column of the table as a result of step 4; it is also deleted from the top row of the table as is the entire column under it.

4) The last row is removed from the table whenever it contains all zeroes.

5) If non-zero artificial variables are present in the final basic set, then the program has no solution. In contrast, zero valued artificial variables in the final solution may exist when one or more of the original constraint equations are redundant.


State Yes or No

9. The value of artificial value is M.

10. Artificial variables enter as basic variables.


The drawback of the penalty cost method is the possible computational error resulting from assigning a very large value to the constant M. To overcome this difficulty, a new method is considered, where the use of M is eliminated by solving the problem in two phases.

Phase I: Formulate the new problem. Start by eliminating the original objective function by the sum of the artificial variables for a minimisation problem and the negative of the sum of the artificial variables for a maximisation problem. The Simplex method optimises the ensuing objective with the constraints of the original problem. If a feasible solution is arrived, the optimal value of the new objective function is zero (suggestive of all artificial variables being zero). Subsequently proceed to phase II. If the optimal value of the new objective function is non-zero, it means there is no solution to the problem and the method terminates.

Phase II: Start phase II using the optimum solution of phase I as the base. Then take the objective function without the artificial variables and solve the problem using the Simplex method.

The first iteration gives the following table:

Table 3.20: The first iteration table

x1

x2

S1

S2

A1

0

0

0

0

1

X2 0

2

1

1

0

0

2

A1 1

5

0

4

1

1

4

5

0

4

1

0

4

Since all elements of the last row are non negative, the procedure is complete. But the existence of non-zero artificial variables in the basic set indicates that the problem has no solution.


3.8 Summary

This unit explains how to solve LPP using the simplex method. The unit also explains the constraints for which you need to introduce the slack, surplus and artificial variables. Examples are used to illustrate the method of solving LPP using the simplex method.


1. Maximise z = 3x1 x2


2x1 + x2 2

x1 + 3x2 3

x2 4,

x1, x2 0.

2. Minimise Z = 6x1 + 7x2


x1 + 3x2 12

3x1 + x2 12

x1 + x2 8

x1 + x2 0



1. False

2. True

3. True

4. True

5. Yes

6. Yes

7. No

8. Yes

9. Yes

10. Yes


1. Z = 9 X1 = 3 X2 = 0

2. Z = 5.8 X1 = 8/3 X2 = 6

3.11 References

No external sources of content have been referred for unit 3

About the Author

admin" admin



MB0048-Unit-04-Duality in LPP

Unit-04-Duality in LPP

Structure:

4.1 Introduction

Learning Objectives

4.2 Importance of Duality Concepts

4.3 Formulation of Dual Concepts

4.4 Economic Interpretation of Duality

Economic interpretation of dual variables

4.5 Sensitivity Analysis

Changes in Cj of non-basic variable

Change in Cj of a basic variable

Change in available resources

Calculating the range

4.6 Summary



Answers to self assessment questions

Answers to terminal questions

4.9 References

4.1 Introduction

Every linear programming problem (LPP) is associated with another linear programming problem involving the same data and optimal solutions. Such two problems are said to be duals of each other. One problem is called the primal, while the other problem is called the dual.

The dual formulation is derived from the same data and solved in a manner similar to the original primal formulation. In other words, you can say that dual is the inverse of the primal formulation because of the following reasons.

If the primal objective function is maximisation function, then the dual objective function is minimisation function and vice-versa.

The column co-efficient in the primal constraint is the row co-efficient in the dual constraint.

The co-efficients in the primal objective function are the RHS constraint in the dual constraint.

The RHS column of constants of the primal constraints becomes the row of co-efficient of the dual objective function.

The concept of duality is useful to obtain additional information about the variation in the optimal solution. These changes could be effected in the constraint co-efficient, in resource availabilities and/or objective function co-efficient. This effect is termed as post optimality or sensitivity analysis.

Learning objectives


Describe the duality concept

Solve a dual problem

Describe the economical interpretation

Apply sensitivity analysis

4.2 Importance of Duality Concepts

The importance of duality concept is due to two main reasons:

i. If the primal contains a large number of constraints and a smaller number of variables, the labour of computation can be considerably reduced by converting it into the dual problem and then solving it.

ii. The interpretation of the dual variable from the loss or economic point of view proves extremely useful in programming future decisions in the activities.

Characteristics of dual solutions

If the primal problem possesses a unique non-degenerate, optimal solution, then the optimal solution to the dual is unique. However, dual solutions arise under a number of other conditions. Several of the cases which can arise are:

When the primal problem has a degenerate optimal solution, the dual has multiple optimal solutions.

When the primal problem has multiple optimal solutions, the optimal dual solution is degenerate.

When the primal problem is unbounded, the dual is infeasible.

When the primal problem is infeasible, the dual is unbounded or infeasible.


State Yes or No

1. Dual LPP always reduces the amount of computation.

2. It is possible to reverse the dual LPP to primal LPP

4.3 Formulation of Dual Concepts

Consider the following LPP

Maximise Z = c1x1 +c2x2 + . . .+ cnxn


a11 x1 + a12 x2 + . . . + a1n xn b1

a21 x1 + a22 x2 + . . . + a2n xn b2

am1 x1 + am2 x2 + . . . + amn xn bm

x1, x2, . . ., xn 0

To construct a dual problem, you must adopt the following guidelines:

i. The maximisation problem in the primal becomes a minimisation problem in the dual and vice versa

ii. () type of constraints in the primal become () type of constraints in the dual and vice versa.

iii. The coefficients c1, c2, . . .,cn in the objective function of the primal become b1, b2,,bm in the objective function of the dual.

iv. The constants b1, b2,,bm in the constraints of the primal become c1, c2, . . .,cn in the constraints of the dual

v. If the primal has n variables and m constraints the dual will have m variables and n constraints

vi. The variables in both the primal and dual are non-negative

Thus the dual problem will be

Minimise W = b1 y1 + b2 y2 + . . . +bm ym


a11 y1 + a21 y2 + . . . + am1 ym c1

a12 y1 + a22 y2 + . . . + am2 ym c2

a1n y1 + a2n y2 + . . . + amn ym cn

y1, y2, . . ., ym 0

Formation of dual LPP is easier when the standard form of LPP for maximisation problem must contain type of constraints, while for minimisation problem, it must contain type of constraints.

Solved Problem 1

Write dual of

Max Z = 4x1 + 5x2

Subject to 3x1 + x2 15

x1 +2 x2 10

5x1 + 2x2 20

x1, x2, 0

Solution: The given problem is in its standard form. Therefore, its dual is

Mini W = 15y1 + 10 y2 + 20 y3

Subject to 3y1 + y2 + 5 y3 4

y1 + 2y2 + 2y3 5

y1, y2, y3, 0

Solved Problem 2

Write the dual of

Min Z = 10x1 + 12x2

Subject to 2x1 +3 x2 10

5x1 +6 x2 20

x1 + 2x2 15

2x1 + 3x2 12

x1, x2, 0

Solution: The given problem is in its standard form. Therefore, its dual problem is

Max W = 10y1 + 20 y2 + 15 y3 + 12y4

Subject to 2y1 + 5y2 + y3 = 2y4 12

6y1 + 2y2 + 3 y3 10

y1, y2, y3, 0

Solved Problem 3

Write the dual of

Max Z = 100x1 + 200x2

Subject to 3x1 10 x2 15

+4x1 +15 x2 20

x1, x2, 0

Solution: First we convert

3x1 10x2 15

to type as shown here.

-3x1 + 10x2 -15

Therefore, its dual is

Mini Z = -15y1 + 20 y2

Subject to -3y1 + 4y2 100

10y1 + 15y2 200

y1, y2, 0

Solved Problem 4

When the constraints contain = sign, write the dual of

Max Z = 40x1 + 30x2


5x1 +6 x2 -10

x1 + x2 = 9

x1 + x2 10

x1, x2, 0

Solution: Firstly,

5x1 7x2 -10

is rewritten as

-5x1 +7x2 10

Secondly,

x1 + x2 = 9

is written as

x1 + x2 9

Also

x1 + x2 9 this is same as -x1 x2 -9

Therefore, the given problem is

Max Z = 40x1 + 30x2

Subject to 10x1 + 6 x2 15

- 5x1 +7 x2 10

x1 +x2 9

- x1 x2 9

Therefore, its dual is

Mini W = -15y1 + 10 y2 + 9y31 9y311

Subject to 10y1 5y2 + y31 y311 40

6y1 + 7y2 + y31 y311 30

y1, y2, y31 y311 0

Let y31 y311= y3

Then the dual becomes

Max W = 15y1 -5 y2 + y3

Subject to 10y1 5y2 + y3 40

6y1 + 7y2 + y3 30

y1, y2, 0

y3 is unrestricted in sign

Solved Problem 5

Write the dual of

Max Z = 12x1 + 15x2


x1 + x2 5

x1 0

x2 is unrestricted in sign

Solution: Let x2 =x21 x211 0. Therefore, its standard form is

Max Z = 12x1 + 15x21 -15x211

Subject to 5x1 +3(x21- x211) 10

- [x1 +(x21- x211)] -5

Or x1 x21+ x211 -5

x1, x21, x211 0

Therefore, dual is

5y1 y2 12 .(1)

3y1 y2 15 .(2)

-3y1 + y2 15 (3)

No.3 constraint is 3y1 y2 15 (4)

Constraints 2 and 4 give

3y1 y2 = 15

The dual problem is

Max W = 10y1 15 y2

Subject to 5y1 y2 12

3y1 y2 =15

y1 0 y2 unrestricted in sign

Solved Problem 6

Write the dual of the following LPP

Minimise Z = 3x1 2x2 + 4x3

Subject to 3x1 + 5x2 + 4x3 7

6x1 + x2 + 3x3 4

7x1 2x2 x3 10

x1 2x2 + 5x3 3

4x1 + 7x2 2x3 2,

x1, x2, x3 0

Solution: Since the problem is of minimisation, all constraints should be of type. Therefore, you have to multiply the third constraint throughout by -1 so that -7x1 + 2x2 + x3 -10

Let y1, y2, y3, y4 and y5 be the dual variables associated with the above five constraints. Then the dual problem is given by

Maximise W = 7y1 + 4 y2 10 y3 + 3 y4 + 2y5

Subject to 3y1 + 6y2 7 y3 + y4 + 4y5 3

5y1 + y2 + 2y3 2y4 + 7y5 -2

4y1 + 3y2 + y3 + 5y4 2y5 4

y1, y2, y3, y4, y5 0

Solved Problem 7

Find the dual of

Maximise 12x1 + 10x2

Subject to 2x1 + 3x2 18

2x1 + x2 14

x1, x2 0

Solution: The dual formulation for this problem will be

Minimise 18y1 + 14y2

Subject to 2y1 + 2y2 12

3y1 + y2 10

y1 >= 0, y2 0

Points to be noted:

The column coefficients in the primal constraint namely (2, 2) and (3, 1) have become the row co-efficient in the dual constraints.

The coefficient of the primal objective function namely, 12 and 10 have become the constants in the right hand side of the dual constraints.

The constants of the primal constraints, namely 18 and 14, have become the coefficient in the dual objective function.

The direction of the inequalities has been reserved.

While the primal is a maximisation problem the dual is a minimisation problem

Solved Problem 8

Obtain the dual problem of the following primal formulation.

Maximise Z = 2x1 + 5x2 + 6x3

Subject to 5x1 + 6x2 -x3 3

-2x1 + x2 + 4x 4

x1 5x2 + 3x3 1

-3x1 3x2 + 7x3 6

x1, x2, x3 0

Solution: Step 1: Write the objective function of the dual. As there are four constraints in the primal, the objective function of the dual will have 4 variables.

Minimise r* = 3 w 1 + 4 w 2 + w 3 + 6 w 4

Step 2: Write the constraints of the dual. As all the constraints in the primal are . The column COM+ efficient of the primal becomes the row co-efficient of the dual.

Constraints 5w1 2w2 + w3 3w4 2

6w1 + w2 5w3 -w4 5

-w1 + 4w2 + 3w3 + 7w4 6

Step 3 : Therefore the dual of the primal is :

Minimise Z = 3w1 + 4w2 + w3 + 6w4

Subject to: 5w1 2w2 + w3 3w4 2

6w1+ w2 5w3 3w4 5

- w1 + 4w2 + 3w3 + 7w4 6

w1, w2, w3, w4 >= 0

Solved Problem 9

Obtain the dual of the following linear programming problem.

Minimise Z = 5x1 -6x2 + 4x3

Subject to the constraints: 3x1 + 4x2 + 6x3 9

x1 + 3x2 + 2x3 5

7x1 2x2 x3 10

x1 2x2 + 4x3 4

2x1 + 5x2 3x3 3

x1, x2, x3 0

Solution: As you can see, one of the primal constraints is a " " constraint while the others are all " constraints. The dual cannot be worked out unless all the constraints are in the same direction. To convert this into " constraint, multiply both the sides of the equation by "-" sign. After multiplying the constraint by "-" sign, it will become 7x1 + 2x2 + x3 -10. Now that all the constraints are in the same direction and the dual can be worked out, the dual formulation is:

Maximise Z = 9w1 + 5w2 10w3 + 4w4 + 3w5

Subject to: 3w1 + w2 7w3 + w4 + 2w5 5

4w1 + 3w2 + 2w3 2w4 + 5w5 -6

6w1 + 2w2 + w3 + 4w4 + -3w5 4

w1, w2, w3, w4, w5 0


Fill in the blanks

3. The coefficients of decision variables in the objective function become quantities on the right hand side of _________ __________.

4. constraints changes to ________ type in dual LP.

5. For every LPP, there exists a unique ________ problem.

4.4 Economic Interpretation of Duality

The linear programming problem is thought of as a resource allocation model where the objective is to maximise revenue or profit subject to limited resources. The associated dual problem offers interesting economic interpretations of the LP resource allocation model. Consider a representation of the general primal and dual problems where primal takes the role of a resource allocation model.

In the above resource allocation model, the primal problem has n economic activities and m resources. The coefficient cj in the primal represents the profit per unit of activity j. Resource i, whose maximum availability is bi, is consumed at the rate aij units per unit of activity j.

4.4.1 Economic interpretation of dual variables:

For any pair of feasible primal and dual solutions,

(Objective value in the maximisation problem) (Objective value in the minimisation problem).

At the optimum, the relationship holds as a strict equation.

Note: Here, the sense of optimisation is very important.

Hence, for any two primal and dual feasible solutions, the values of the objective functions, when finite, must satisfy the following ineq

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