Operation Research

Operation Research (MBA) Module 1 Unit 1 Origin of Operations Research Concept and Definition of OR Characteristics of OR Management Applications of OR Phases of OR OR Models Principles of Modeling Simplifications of OR Models General Methods for Solving OR Models Scope of OR Role of Operations Research in Decision-Making Scientific Method in Operations Research Development of Operations Research in India Role of Computers in OR Unit 2 2.1 Introduction to Linear Programming 2.2 General Form of LPP 2.3 Assumptions in LPP 2.4 Applications of Linear Programming 2.5 Advantages of Linear Programming Techniques 2.6 Limitations of Linear Programming 2.7 Formulation of LP Problems Unit 3 3.1 Graphical solution Procedure 3.2 Definitions 3.3 Example Problems 3.4 Special cases of Graphical method 3.4.1 Multiple optimal solutions 3.4.2 No optimal solution 3.4.3 Unbounded solution Module 2 Unit 1 Introduction Steps to convert GLPP to SLPP Some Basic Definitions Introduction to Simplex Method Computational procedure of Simplex Method Worked Examples Unit 2 2.1 Computational Procedure of Big – M Method (Charne’s Penalty Method) 2.2 Worked Examples 2.3 Steps for Two-Phase Method 2.4 Worked Examples Unit 3 3.1 Special cases in Simplex Method 3.1.1 Degenaracy 3.1.2 Non-existing Feasible Solution 3.1.3 Unbounded Solution 3.1.4 Multiple Optimal Solutions Module 3 Unit 1 The Revised Simplex Method Steps for solving Revised Simplex Method in Standard Form-I Worked Examples Unit 2 2.1 Computational Procedure of Revised Simplex Table in Standard Form-II 2.2 Worked Examples 2.3 Advantages and Disadvantages Unit 3 3.1 Duality in LPP 3.2 Important characteristics of Duality 3.3 Advantages and Applications of Duality 3.4 Steps for Standard Primal Form 3.5 Rules for Converting any Primal into its Dual 3.6 Example Problems 3.7 Primal-Dual Relationship 3.8 Duality and Simplex Method Module 4 Unit 1 Introduction Computational Procedure of Dual Simplex Method Worked Examples Advantage of Dual Simplex over Simplex Method Introduction to Transportation Problem Mathematical Formulation Tabular Representation Some Basic Definitions Unit 2 2.1 Methods for Initial Basic Feasible Solution 2.1.1 North-West Corner Rule 2.1.2 Row Minima Method 2.1.3 Column Minima Method 2.1.4 Lowest Cost Entry Method (Matrix Minima Method) 2.1.5 Vogel’s Approximation Method (Unit Cost Penalty Method) Unit 3 3.1 Examining the Initial Basic Feasible Solution for Non-Degeneracy 3.2 Transportation Algorithm for Minimization Problem 3.3 Worked Examples Module 5 Unit 1 Introduction to Assignment Problem Algorithm for Assignment Problem Worked Examples Unbalanced Assignment Problem Maximal Assignment Problem Unit 2 2.1 Introduction to Game Theory 2.2 Properties of a Game 2.3 Characteristics of Game Theory 2.4 Classification of Games 2.5 Solving Two-Person and Zero-Sum Game Unit 3 3.1 Games with Mixed Strategies 3.1.1 Analytical Method 3.1.2 Graphical Method 3.1.3 Simplex Method Module 6 Unit 1 Shortest Route Problem Minimal Spanning Tree Problem Maximal Flow Problem Unit 2 2.1 Introduction to CPM / PERT Techniques 2.2 Applications of CPM / PERT 2.3 Basic Steps in PERT / CPM 2.4 Frame work of PERT/CPM 2.5 Network Diagram Representation 2.6 Rules for Drawing Network Diagrams 2.7 Common Errors in Drawing Networks 2.8 Advantages and Disadvantages 2.9 Critical Path in Network Analysis Unit 3 3.1 Worked Examples on CPM 3.2 PERT 3.3 Worked Examples Module 1 Unit 1 Origin of Operations Research Concept and Definition of OR Characteristics of OR Management Applications of OR Phases of OR OR Models Principles of Modeling Simplifications of OR Models General Methods for Solving OR Models Scope of OR Role of Operations Research in Decision-Making Scientific Method in Operations Research Development of Operations Research in India Role of Computers in OR 1.1 Origin of Operations Research The term Operations Research (OR) was first coined by MC Closky and Trefthen in 1940 in a small town, Bowdsey of UK. The main origin of OR was during the second world war – The military commands of UK and USA engaged several inter-disciplinary teams of scientists to undertake scientific research into strategic and tactical military operations. Their mission was to formulate specific proposals and to arrive at the decision on optimal utilization of scarce military resources and also to implement the decisions effectively. In simple words, it was to uncover the methods that can yield greatest results with little efforts. Thus it had gained popularity and was called “An art of winning the war without actually fighting it” The name Operations Research (OR) was invented because the team was dealing with research on military operations. The encouraging results obtained by British OR teams motivated US military management to start with similar activities. The work of OR team was given various names in US: Operational Analysis, Operations Evaluation, Operations Research, System Analysis, System Research, Systems Evaluation and so on. The first method in this direction was simplex method of linear programming developed in 1947 by G.B Dantzig, USA. Since then, new techniques and applications have been developed to yield high profit from least costs. Now OR activities has become universally applicable to any area such as transportation, hospital management, agriculture, libraries, city planning, financial institutions, construction management and so forth. In India many of the industries like Delhi cloth mills, Indian Airlines, Indian Railway, etc are making use of OR activity. 1.2 Concept and Definition of OR Operations research signifies research on operations. It is the organized application of modern science, mathematics and computer techniques to complex military, government, business or industrial problems arising in the direction and management of large systems of men, material, money and machines. The purpose is to provide the management with explicit quantitative understanding and assessment of complex situations to have sound basics for arriving at best decisions. Operations research seeks the optimum state in all conditions and thus provides optimum solution to organizational problems. Definition: OR is a scientific methodology – analytical, experimental and quantitative – which by assessing the overall implications of various alternative courses of action in a management system provides an improved basis for management decisions. 1.3 Characteristics of OR (Features) The essential characteristics of OR are Inter-disciplinary team approach – The optimum solution is found by a team of scientists selected from various disciplines. Wholistic approach to the system – OR takes into account all significant factors and finds the best optimum solution to the total organization. Imperfectness of solutions – Improves the quality of solution. Use of scientific research – Uses scientific research to reach optimum solution. To optimize the total output – It tries to optimize by maximizing the profit and minimizing the loss. 1.4 Management Applications of OR Some areas of applications are Finance, Budgeting and Investment Cash flow analysis , investment portfolios Credit polices, account procedures Purchasing, Procurement and Exploration Rules for buying, supplies Quantities and timing of purchase Replacement policies Production management Physical distribution Facilities planning Manufacturing Maintenance and project scheduling Marketing Product selection, timing Number of salesman, advertising Personnel management Selection of suitable personnel on minimum salary Mixes of age and skills Research and development Project selection Determination of area of research and development Reliability and alternative design 1.5 Phases of OR OR study generally involves the following major phases Defining the problem and gathering data Formulating a mathematical model Deriving solutions from the model Testing the model and its solutions Preparing to apply the model Implementation 1. Defining the problem and gathering data The first task is to study the relevant system and develop a well-defined statement of the problem. This includes determining appropriate objectives, constraints, interrelationships and alternative course of action. The OR team normally works in an advisory capacity. The team performs a detailed technical analysis of the problem and then presents recommendations to the management. Ascertaining the appropriate objectives is very important aspect of problem definition. Some of the objectives include maintaining stable price, profits, increasing the share in market, improving work morale etc. OR team typically spends huge amount of time in gathering relevant data. To gain accurate understanding of problem To provide input for next phase. OR teams uses Data mining methods to search large databases for interesting patterns that may lead to useful decisions. 2. Formulating a mathematical model This phase is to reformulate the problem in terms of mathematical symbols and expressions. The mathematical model of a business problem is described as the system of equations and related mathematical expressions. Thus Decision variables (x1, x2 … xn) – ‘n’ related quantifiable decisions to be made. Objective function – measure of performance (profit) expressed as mathematical function of decision variables. For example P=3x1 +5x2 + … + 4xn Constraints – any restriction on values that can be assigned to decision variables in terms of inequalities or equations. For example x1 +2x2 ≥ 20 Parameters – the constant in the constraints (right hand side values) The alternatives of the decision problem is in the form of unknown variables Example – consider a company producing chairs and tables with the aim of getting maximum profit, then the decision variables are number of chairs and tables to be produced (say mathematically x and y). Decision variables are used to construct the objective function and restrictions in mathematical functions. The end result of OR model is a mathematical form relating the objective function, constraints with variables. The mathematical function is to optimize (maximize/ minimize) the magnitude of the objective function, simultaneously satisfying all the facility constraints. The resulting solution in the form of magnitude of decision variables value of objective function is known as “optimum feasible solution”. A mathematical model of OR is organized as Maximize or Minimize (Objective Function) Subject to (Constraints) Example Maximize Z = 45x + 80y Subject to 5x+ 20y ≤ 400 10x + 15y ≤ 450 x ≥ 0 , y ≥ 0 Here x and y are decision variables say x = number of chairs y = number of tables x and y should always be nonnegative values z= objective function Linear Programming (LP) It is a mathematical technique which optimizes the available resources. Optimization The solution of the model yields the values of the decision variables that maximize or minimize the value of the objective function while satisfying all the constraints of that system. Hence optimization may be maximization or minimization. Example Maximize the profit of the production oriented company. Minimize the loss of the trading company. The advantages of using mathematical models are Describe the problem more concisely Makes overall structure of problem comprehensible Helps to reveal important cause-and-effect relationships Indicates clearly what additional data are relevant for analysis Forms a bridge to use mathematical technique in computers to analyze 3. Deriving solutions from the model This phase is to develop a procedure for deriving solutions to the problem. A common theme is to search for an optimal or best solution. The main goal of OR team is to obtain an optimal solution which minimizes the cost and time and maximizes the profit. Herbert Simon says that “Satisficing is more prevalent than optimizing in actual practice”. Where satisficing = satisfactory + optimizing Samuel Eilon says that “Optimizing is the science of the ultimate; Satisficing is the art of the feasible”. To obtain the solution, the OR team uses Heuristic procedure (designed procedure that does not guarantee an optimal solution) is used to find a good suboptimal solution. Metaheuristics provides both general structure and strategy guidelines for designing a specific heuristic procedure to fit a particular kind of problem. Post-Optimality analysis is the analysis done after finding an optimal solution. It is also referred as what-if analysis. It involves conducting sensitivity analysis to determine which parameters of the model are most critical in determining the solution. 4. Testing the model After deriving the solution, it is tested as a whole for errors if any. The process of testing and improving a model to increase its validity is commonly referred as Model validation. The OR group doing this review should preferably include at least one individual who did not participate in the formulation of model to reveal mistakes. A systematic approach to test the model is to use Retrospective test. This test uses historical data to reconstruct the past and then determine the model and the resulting solution. Comparing the effectiveness of this hypothetical performance with what actually happened, indicates whether the model tends to yield a significant improvement over current practice. 5. Preparing to apply the model After the completion of testing phase, the next step is to install a well-documented system for applying the model. This system will include the model, solution procedure and operating procedures for implementation. The system usually is computer-based. Databases and Management Information System may provide up-to-date input for the model. An interactive computer based system called Decision Support System is installed to help the manager to use data and models to support their decision making as needed. A managerial report interprets output of the model and its implications for applications. 6. Implementation The last phase of an OR study is to implement the system as prescribed by the management. The success of this phase depends on the support of both top management and operating management. The implementation phase involves several steps OR team provides a detailed explanation to the operating management If the solution is satisfied, then operating management will provide the explanation to the personnel, the new course of action. The OR team monitors the functioning of the new system Feedback is obtained Documentation 1.6 OR Models The OR models are Allocation models Waiting line models Game theory Inventory models Replacement models Job sequencing models Network models Simulation models Markovian models Allocation models (Distribution models) These models are concerned with the allotment of available resources so as to maximize profit or minimize loss subject to available and predicted restrictions. Methods for solving allocation models are Linear programming problems Transportation problems Assignment problems Waiting line models (Queueing) This model is an attempt to made to predict How much average time will be spent by the customer in a queue? What will be an average length of the queue? What will be the utilization factor of a queue system? Etc. This model provides to minimize the sum of costs of providing service and cost of obtaining service, associated with the value of time spent by the customer in a queue. Game theory (Competitive strategy models) These models are used to determine the behavior of decision-making under competition or conflict. Methods for solving such models have not been found suitable for industrial applications mainly because they are referred to an idealistic world neglecting many essential feature of reality. Inventory (Production) models These models are concerned with the determination of the optimal order quantity and ordering production intervals considering the factors such as demand per unit time, cost of placing orders, costs associated with goods held up in the inventory and the cost due to shortage of goods, etc. Replacement models These models deals with determination of best time to replace an equipment in situations that arise when some items or machinery need replacement by a new one or scientific advance or deterioration due to wear and tear, accidents etc. Individual and group replacement policies can be used in the case of such equipments that fail completely and instantaneously. Job sequencing models These models involve the selection of such a sequence of performing a series of jobs to be done on machines that optimizes the efficiency measure of performance of the system. Network models These models are applicable in large projects involving complexities and interdependencies of activities. CPM (Critical Path Method) and PERT (Project Evaluation and Review Technique) are used for planning, scheduling and controlling activities of complex project which can be characterized as network diagram. Simulation models This model is used much for solving when the number of variables and constrained relationships may be too large. Markovian models These models are applicable in such situations where the state of the system can be defined by some descriptive measure of numerical value and where the system moves from one state to another on a probability basis. 1.7 Principles of Modeling The model building and their uses both should be consciously aware of the following ten principles Do not build up a complicated model when simple one will suffice Beware of molding the problem to fit the technique The deduction phase of modeling must be conducted rigorously Models should be validated prior to implementation A model should never be taken too literally A model should neither be pressed to do nor criticized for failing to do that for which it was never intended Beware of over-selling a model Some of the primary benefits of modeling are associated with the process of developing the model A model cannot be any better than the information that goes into it Models cannot replace decision makers 1.8 Simplifications of OR Models While constructing a model, two conflicting objectives usually strike in our mind The model should be as accurate as possible It should be as easy as possible in solving Besides, the management must be able to understand the solution of the model and must be capable of using it. So the reality of the problem under study should be simplified to the extent when there is no loss of accuracy. The model can be simplified by Omitting certain variable Changing the nature of variables Aggregating the variables Changing the relationship between variables Modifying the constraints, etc 1.9 General Methods for Solving OR Models Generally three types of methods are used for solving OR models Analytic method Iterative method Monte- Corlo method Analytic method If the OR model is solved by using all the tools of classical mathematics such as differential calculus and finite differences available for this task, then such type of solutions are called analytic solutions. Solutions of various inventory models are obtained by adopting the so called analytic procedure. Iterative method If classical methods fail because of complexity of the constraints or the number of variables, then we are usually forced to adopt an iterative method. Such a procedure starts with a trial solution and a set of rules for improving it. The trial solution is then replaced by the improved solution and the process is repeated until either no further improvement is possible or the cost of further calculation cannot be justified. Iterative method can be divided into three groups After a finite number of repetitions, no further improvement will be possible Although successive iterations improve the solutions, we are only guaranteed the solution as a limit of an infinite process Finally, include the trial and error method which however is likely to be lengthy, tedious and costly even if electronic computers are used. The Monte-Corlo method The basis of so called Monte-Corlo technique is random sampling of variables values from a distribution of that variable. Monte-Corlo refers to the use of sampling methods to estimate the value of non-stochastic variables. The following are the main steps of this method Step 1 – In order to have a general idea of the system, first draw a flow diagram of the system Step 2 – Then, take correct sample observations to select some suitable model for the system. In this step, compute the probability distributions for the variables of our interest. Step 3 – Then, convert the probability distribution to cumulative distribution function Step 4 – A sequence of random numbers is now selected with the help of random number tables Step 5 – Determine the sequence of values of variables of interest with the sequence of random numbers obtained in step 4. Step 6 – Finally construct standard mathematical function to the values obtained in step 5 1.10 Scope of OR In recent years of organized development, OR has entered successfully in many different areas of research. It is useful in the following various important fields In agriculture With the explosion of population and consequent shortage of food, every country is facing the problem of Optimum allocation of land to various crops in accordance with the climatic conditions Optimum distribution of water from various resources like canal for irrigation purposes Thus there is a need of determining best policies under the prescribed restrictions. Hence a good amount of work can be done in this direction. In finance In these modern times of economic crisis, it has become very necessary for every government to have a careful planning for the economic development of the country. OR techniques can be fruitfully applied To maximize the per capita income with minimum resources To find out the profit plan for the company To determine the best replacement policies, etc In industry If the industry manager decides his policies only on the basis of his past experience and a day comes when he gets retirement, then a heavy loss is encountered before the industry. This heavy loss can be immediately be compensated by newly appointing a young specialist of OR techniques in business management. Thus OR is useful to the industry director in deciding optimum allocation of various limited resources such as men, machines, material, etc to arrive at the optimum decision. In marketing With the help of OR techniques a marketing administrator can decide Where to distribute the products for sale so that the total cost of transportation is minimum The minimum per unit sale price The size of the stock to meet the future demand How to select the best advertising media with respect to time, cost, etc? How, when and what to purchase at the minimum possible cost? In personnel management A personnel manager can use OR techniques To appoint the most suitable person on minimum salary To determine the best age of retirement for the employees To find out the number of persons to be appointed in full time basis when the workload is seasonal In production management A production manager can use OR techniques To find out the number and size of the items to be produced In scheduling and sequencing the production run by proper allocation of machines In calculating the optimum product mix To select, locate and design the sites for the production plans In L.I.C OR approach is also applicable to enable the L.I.C offices to decide What should be the premium rates for various modes of policies? How best the profits could be distributed in the cases of with profit policies? 1.11 Role of Operations Research in Decision-Making The Operation Research may be regarded as a tool which is utilized to increase the effectiveness of management decisions. OR is the objective supplement to the subjective feeling of the administrator (decision maker). Scientific method of OR is used to understand and describe the phenomena of operating system. The advantages of OR study approach in business and management decision making may be classified as follows Better control The management of big concerns finds it much costly to provide continuous executive supervisions over routine decisions. An OR approach directs the executives to devote their attention to more pressing matters. For example, OR approach deals with production scheduling and inventory control. Better coordination Sometimes OR has been very useful in maintaining the law and order situation out of chaos. For example, an OR based planning model becomes a vehicle for coordinating marketing decisions with the limitations imposed on manufacturing capabilities. Better system OR study is also initiated to analyze a particular problem of decision making such as establishing a new warehouse. Later OR approach can be further developed into a system to be employed repeatedly. Consequently the cost of undertaking the first application may improve the profits. Better decisions OR models frequently yield actions that do improve an intuitive decision making. Sometimes a situation may be so complicated that the human mind can never hope to assimilate all the important factors without the help of OR and computer analysis. 1.12 Scientific Method in Operations Research The scientific method in OR study generally involves three phases The judgment phase The research phase The action phase The judgment phase includes A determination of the operation The establishment of the objectives and values related to the operation The determination of the suitable measure of effectiveness Lastly, the formulation of the problems relative to the objectives The research phase utilizes Observations and data collection for a better understanding of what the problem is Formulation of hypothesis and models Observation and experimentation to test the hypothesis on the basis of additional data Analysis of the available information and verification of the hypothesis using pre-established measure of effectiveness. Predictions of various results from the hypothesis, generalization of the result and consideration of alternative methods The action phase OR consists of making recommendations for decision process by those who first posed the problem for consideration or by anyone in a position to make a decision influencing the operation in which the problem occurred. 1.13 Development of Operations Research in India In 1949 OR came into picture when an OR unit was established at the regional research lab, Hyderabad. At the same time, Prof. R. S. Verma (Delhi University) set up an OR team in the Defense Science laboratory to solve the problems of store, purchase and planning. In 1953, Prof. P.C. Mahalanobis established an OR team in the Indian Statistical Institute, Calcutta for solving the problem of national planning and survey. In 1957, OR society of India was formed and this society became a member of the International Federation of OR societies in 1960. Presently India is publishing a number of research journals, namely OPSEARCH Industrial Engineering and Management Materials Management Journal of India Defense Science Journal SCIMA Journal of Engineering Production, etc As far as the OR education in India is concerned, University of Delhi was the first to introduce a complete M.Sc course in OR in 1963. The industries have engaged OR teams and are using OR techniques. Some of them are Hindustan Lever Ltd Union Carbide TELCO Hindustan Steel Imperial Chemical Industries Tata Iron and Steel Company Sarabhai Group FCI Kirloskar Company, etc 1.14 Role of Computers in OR Computers have played a vital role in the development of OR. But OR would not have achieved its present position for the use of computers. The reason is that – in most of the OR techniques computations are so complex and involved that these techniques would be of no practical use without computers. Many large scale applications of OR techniques which require only few minutes on the computer may take weeks, months and sometimes years even to yield the same results manually. So the computers has become essential and integral part of OR. Now a days, OR methodology and computer methodology are growing up simultaneously. It seems that in the near future the line dividing the two methodologies will disappear and the two sciences will combine to form a more general and comprehensive science. Exercise What is Operations Research? Explain the phases of OR. Explain the different models used in OR. What are the significance and applications of OR? What are the general methods of solving OR models? Mention the principles of modeling. Give the essential characteristics of the following types of process Allocation Competitive games Inventory Waiting line Write short notes on the following Role of constraints and objectives in the construction of mathematical models Area of applications of OR OR as an decision making science Write a note on scope of OR. “Much of success of OR in past few decades is due to computers”. Discuss. Unit 2 2.1 Introduction to Linear Programming 2.2 General Form of LPP 2.3 Assumptions in LPP 2.4 Applications of Linear Programming 2.5 Advantages of Linear Programming Techniques 2.6 Limitations of Linear Programming 2.7 Formulation of LP Problems 2.1 Introduction to Linear Programming A linear form is meant a mathematical expression of the type a1x1 + a2x2 + …. + anxn, where a1, a2, …, an are constants and x1, x2 … xn are variables. The term Programming refers to the process of determining a particular program or plan of action. So Linear Programming (LP) is one of the most important optimization (maximization / minimization) techniques developed in the field of Operations Research (OR). The methods applied for solving a linear programming problem are basically simple problems; a solution can be obtained by a set of simultaneous equations. However a unique solution for a set of simultaneous equations in n-variables (x1, x2 … xn), at least one of them is non-zero, can be obtained if there are exactly n relations. When the number of relations is greater than or less than n, a unique solution does not exist but a number of trial solutions can be found. In various practical situations, the problems are seen in which the number of relations is not equal to the number of the number of variables and many of the relations are in the form of inequalities (≤ or ≥) to maximize or minimize a linear function of the variables subject to such conditions. Such problems are known as Linear Programming Problem (LPP). Definition – The general LPP calls for optimizing (maximizing / minimizing) a linear function of variables called the ‘Objective function’ subject to a set of linear equations and / or inequalities called the ‘Constraints’ or ‘Restrictions’. 2.2 General form of LPP We formulate a mathematical model for general problem of allocating resources to activities. In particular, this model is to select the values for x1, x2 … xn so as to maximize or minimize Z = c1x1 + c2x2 +………….+cnxn subject to restrictions a11x1 + a12x2 + …..........+a1nxn (≤ or ≥) b1 a21x1 + a22x2 + ………..+a2nxn (≤ or ≥) b2 . . . am1x1 + am2x2 + ……….+amnxn (≤ or ≥) bm and x1 ≥ 0, x2 ≥ 0,…, xn ≥ 0 Where Z = value of overall measure of performance xj = level of activity (for j = 1, 2, ..., n) cj = increase in Z that would result from each unit increase in level of activity j bi = amount of resource i that is available for allocation to activities (for i = 1,2, …, m) aij = amount of resource i consumed by each unit of activity j Resource Resource usage per unit of activity Amount of resource available Activity 1 2 …………………….. n 1 2 . . . m a11 a12 …………………….a1n a21 a22 …………………….a2n . . . am1 am2 …………………….amn b1 b2 . . . bm Contribution to Z per unit of activity c1 c2 ………………………..cn Data needed for LP model The level of activities x1, x2………xn are called decision variables. The values of the cj, bi, aij (for i=1, 2 … m and j=1, 2 … n) are the input constants for the model. They are called as parameters of the model. The function being maximized or minimized Z = c1x1 + c2x2 +…. +cnxn is called objective function. The restrictions are normally called as constraints. The constraint ai1x1 + ai2x2 … ainxn are sometimes called as functional constraint (L.H.S constraint). xj ≥ 0 restrictions are called non-negativity constraint. 2.3 Assumptions in LPP Proportionality The contribution of each variable in the objective function or its usage of the resources is directly proportional to the value of the variable i.e. if resource availability increases by some percentage, then the output shall also increase by same percentage. Additivity Sum of the resources used by different activities must be equal to the total quantity of resources used by each activity for all resources individually or collectively. Divisibility The variables are not restricted to integer values Deterministic Coefficients in the objective function and constraints are completely known and do not change during the period under study in all the problems considered. Finiteness Variables and constraints are finite in number. Optimality In LPP, we determine the decision variables so as to optimize the objective function of the LPP. The problem involves only one objective, profit maximization or cost minimization. 2.4 Applications of Linear Programming Personnel Assignment Problem Suppose we are given ‘m’ persons, ‘n’ jobs and the expected productivity cij of ith person on the jth job. We want to find an assignment of person’s xij ≥ 0 for all i and j, to ‘n’ jobs so that the average productivity of person assigned is maximum, subject to the conditions Where ai is the number of persons in personnel category i bj is the number of jobs in personnel category j Transportation Problem Suppose that ‘m’ factories (sources) supply ‘n’ warehouses (destinations) with certain product. Factory Fi (i=1, 2 … m) produces ai units and warehouse Wj (j=1, 2, 3 … n) requires bj units. Suppose that the cost of shipping from factory Fi to warehouse Wj is directly proportional to the amount shipped and that the unit cost is cij. Let the decision variables xij be the amount shipped from factory Fi to warehouse Wj. The objective is to determine the number of units transported from factory Fi to warehouse Wj so that the total transportation cost The supply and demand must be satisfied exactly. Mathematically, this problem is to find xij (i=1, 2 … m; j=1, 2 … n) in order to minimize the total transportation cost Subject to constraints Efficiency on Operation of system of Dams In this problem, we determine variations in water storage of dams which generate power so as to maximize the energy obtained from the entire system. The physical limitations of storage appear as inequalities. Optimum Estimation of Executive Compensation The objective here is to determine a consistent plan of executive compensation in an industrial concern. Salary, job ranking and the amounts of each factor required on the ranked job level are taken into consideration by the constraints of linear programming. Agriculture Applications Linear programming can be applied in agricultural planning for allocating the limited resources such as labour, water supply and working capital etc, so as to maximize the net revenue. Military Applications These applications involve the problem of selecting an air weapon system against gurillas so as to keep them pinned down and simultaneously minimize the amount of aviation gasoline used, a variation of transportation problem that maximizes the total tonnage of bomb dropped on a set of targets and the problem of community defense against disaster to find the number of defense units that should be used in the attack in order to provide the required level of protection at the lowest possible cost. Production Management Linear programming can be applied in production management for determining product mix, product smoothing and assembly time-balancing. Marketing Management Linear programming helps in analyzing the effectiveness of advertising campaign and time based on the available advertising media. It also helps in travelling salesman in finding the shortest route for his tour. Manpower Management Linear programming allows the personnel manager to analyze personnel policy combinations in terms of their appropriateness for maintaining a steady-state flow of people into through and out of the firm. Physical distribution Linear programming determines the most economic and efficient manner of locating manufacturing plants and distribution centers for physical distribution. 2.5 Advantages of Linear Programming Techniques It helps us in making the optimum utilization of productive resources. The quality of decisions may also be improved by linear programming techniques. Provides practically solutions. In production processes, high lighting of bottlenecks is the most significant advantage of this technique. 2.6 Limitations of Linear Programming Some limitations are associated with linear programming techniques In some problems, objective functions and constraints are not linear. Generally, in real life situations concerning business and industrial problems constraints are not linearly treated to variables. There is no guarantee of getting integer valued solutions. For example, in finding out how many men and machines would be required to perform a particular job, rounding off the solution to the nearest integer will not give an optimal solution. Integer programming deals with such problems. Linear programming model does not take into consideration the effect of time and uncertainty. Thus the model should be defined in such a way that any change due to internal as well as external factors can be incorporated. Sometimes large scale problems cannot be solved with linear programming techniques even when the computer facility is available. Such difficulty may be removed by decomposing the main problem into several small problems and then solving them separately. Parameters appearing in the model are assumed to be constant. But, in real life situations they are neither constant nor deterministic. Linear programming deals with only single objective, whereas in real life situation problems come across with multi objectives. Goal programming and multi-objective programming deals with such problems. 2.7 Formulation of LP Problems Example 1 A firm manufactures two types of products A and B and sells them at a profit of Rs. 2 on type A and Rs. 3 on type B. Each product is processed on two machines G and H. Type A requires 1 minute of processing time on G and 2 minutes on H; type B requires 1 minute on G and 1 minute on H. The machine G is available for not more than 6 hours 40 minutes while machine H is available for 10 hours during any working day. Formulate the problem as a linear programming problem. Solution Let x1 be the number of products of type A x2 be the number of products of type B After understanding the problem, the given information can be systematically arranged in the form of the following table. Type of products (minutes) Machine Type A (x1 units) Type B (x2 units) Available time (mins) G 1 1 400 H 2 1 600 Profit per unit Rs. 2 Rs. 3 Since the profit on type A is Rs. 2 per product, 2 x1 will be the profit on selling x1 units of type A. similarly, 3x2 will be the profit on selling x2 units of type B. Therefore, total profit on selling x1 units of A and x2 units of type B is given by Maximize Z = 2 x1+3 x2 (objective function) Since machine G takes 1 minute time on type A and 1 minute time on type B, the total number of minutes required on machine G is given by x1+ x2. Similarly, the total number of minutes required on machine H is given by 2x1 + 3x2. But, machine G is not available for more than 6 hours 40 minutes (400 minutes). Therefore, x1+ x2 ≤ 400 (first constraint) Also, the machine H is available for 10 hours (600 minutes) only, therefore, 2 x1 + 3x2 ≤ 600 (second constraint) Since it is not possible to produce negative quantities x1 ≥ 0 and x2 ≥ 0 (non-negative restrictions) Hence Maximize Z = 2 x1 + 3 x2 Subject to restrictions x1 + x2 ≤ 400 2x1 + 3x2 ≤ 600 and non-negativity constraints x1 ≥ 0 , x2 ≥ 0 Example 2 A company produces two products A and B which possess raw materials 400 quintals and 450 labour hours. It is known that 1 unit of product A requires 5 quintals of raw materials and 10 man hours and yields a profit of Rs 45. Product B requires 20 quintals of raw materials, 15 man hours and yields a profit of Rs 80. Formulate the LPP. Solution Let x1 be the number of units of product A x2 be the number of units of product B Product A Product B Availability Raw materials 5 20 400 Man hours 10 15 450 Profit Rs 45 Rs 80 Hence Maximize Z = 45x1 + 80x2 Subject to 5x1+ 20 x2 ≤ 400 10x1 + 15x2 ≤ 450 x1 ≥ 0 , x2 ≥ 0 Example 3 A firm manufactures 3 products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respectively. The firm has 2 machines and below is given the required processing time in minutes for each machine on each product. Products Machine A B C X 4 3 5 Y 2 2 4 Machine X and Y have 2000 and 2500 machine minutes. The firm must manufacture 100 A’s, 200 B’s and 50 C’s type, but not more than 150 A’s. Solution Let x1 be the number of units of product A x2 be the number of units of product B x3 be the number of units of product C Products Machine A B C Availability X 4 3 5 2000 Y 2 2 4 2500 Profit 3 2 4 Max Z = 3x1 + 2x2 + 4x3 Subject to 4x1 + 3x2 + 5x3 ≤ 2000 2x1 + 2x2 + 4x3 ≤ 2500 100 ≤ x1 ≤ 150 x2 ≥ 200 x3 ≥ 50 Example 4 A company owns 2 oil mills A and B which have different production capacities for low, high and medium grade oil. The company enters into a contract to supply oil to a firm every week with 12, 8, 24 barrels of each grade respectively. It costs the company Rs 1000 and Rs 800 per day to run the mills A and B. On a day A produces 6, 2, 4 barrels of each grade and B produces 2, 2, 12 barrels of each grade. Formulate an LPP to determine number of days per week each mill will be operated in order to meet the contract economically. Solution Let x1 be the no. of days a week the mill A has to work x2 be the no. of days per week the mill B has to work Grade A B Minimum requirement Low 6 2 12 High 2 2 8 Medium 4 12 24 Cost per day Rs 1000 Rs 800 Minimize Z = 1000x1 + 800 x2 Subject to 6x1 + 2x2 ≥ 12 2x1 + 2x2 ≥ 8 4x1 +12x2 ≥ 24 x1 ≥ 0 , x2 ≥ 0 Example 5 A company has 3 operational departments weaving, processing and packing with the capacity to produce 3 different types of clothes that are suiting, shirting and woolen yielding with the profit of Rs. 2, Rs. 4 and Rs. 3 per meters respectively. 1m suiting requires 3mins in weaving 2 mins in processing and 1 min in packing. Similarly 1m of shirting requires 4 mins in weaving 1 min in processing and 3 mins in packing while 1m of woolen requires 3 mins in each department. In a week total run time of each department is 60, 40 and 80 hours for weaving, processing and packing department respectively. Formulate a LPP to find the product to maximize the profit. Solution Let x1 be the number of units of suiting x2 be the number of units of shirting x3 be the number of units of woolen Suiting Shirting Woolen Available time Weaving 3 4 3 60 Processing 2 1 3 40 Packing 1 3 3 80 Profit 2 4 3 Maximize Z = 2x1 + 4x2 + 3x3 Subject to 3x1 + 4x2 + 3x3 ≤ 60 2x1 + 1x2 + 3x3 ≤ 40 x1 + 3x2 + 3x3 ≤ 80 x1≥0, x2 ≥0, x3≥0 Example 6 ABC Company produces both interior and exterior paints from 2 raw materials m1 and m2. The following table produces basic data of problem. Exterior paint Interior paint Availability M1 6 4 24 M2 1 2 6 Profit per ton 5 4 A market survey indicates that daily demand for interior paint cannot exceed that for exterior paint by more than 1 ton. Also maximum daily demand for interior paint is 2 tons. Formulate LPP to determine the best product mix of interior and exterior paints that maximizes the daily total profit. Solution Let x1 be the number of units of exterior paint x2 be the number of units of interior paint Maximize Z = 5x1 + 4x2 Subject to 6x1 + 4x2 ≤ 24 x1 + 2x2 ≤ 6 x2 – x1≤ 1 x2≤ 2 x1≥0, x2 ≥0 b) The maximum daily demand for exterior paint is atmost 2.5 tons x1≤ 2.5 c) Daily demand for interior paint is atleast 2 tons x2 ≥ 2 d) Daily demand for interior paint is exactly 1 ton higher than that for exterior paint. x2 > x1 + 1 Example 7 A company produces 2 types of hats. Each hat of the I type requires twice as much as labour time as the II type. The company can produce a total of 500 hats a day. The market limits daily sales of I and II types to 150 and 250 hats. Assuming that the profit per hat are Rs.8 for type A and Rs. 5 for type B. Formulate a LPP models in order to determine the number of hats to be produced of each type so as to maximize the profit. Solution Let x1 be the number of hats produced by type A Let x2 be the number of hats produced by type B Maximize Z = 8x1 + 5x2 Subject to 2x1 + x2 ≤ 500 (labour time) x1 ≤ 150 x2 ≤ 250 x1≥0, x2 ≥0 Example 8 A manufacturer produces 3 models (I, II and III) of a certain product. He uses 2 raw materials A and B of which 4000 and 6000 units respectively are available. The raw materials per unit of 3 models are given below. Raw materials I II III A 2 3 5 B 4 2 7 The labour time for each unit of model I is twice that of model II and thrice that of model III. The entire labour force of factory can produce an equivalent of 2500 units of model I. A model survey indicates that the minimum demand of 3 models is 500, 500 and 375 units respectively. However the ratio of number of units produced must be equal to 3:2:5. Assume that profits per unit of model are 60, 40 and 100 respectively. Formulate a LPP. Solution Let x1 be the number of units of model I x2 be the number of units of model II x3 be the number of units of model III Raw materials I II III Availability A 2 3 5 4000 B 4 2 7 6000 Profit 60 40 100 x1 + 1/2x2 + 1/3x3 ≤ 2500 [ Labour time ] x1 ≥ 500, x2 ≥ 500, x3 ≥ 375 [ Minimum demand ] The given ratio is x1: x2: x3 = 3: 2: 5 x1 / 3 = x2 / 2 = x3 / 5 = k x1 = 3k; x2 = 2k; x3 = 5k x2 = 2k → k = x2 / 2 Therefore x1 = 3 x2 / 2 → 2x1 = 3x2 Similarly 2x3 = 5x2 Maximize Z= 60x1 + 40x2 + 100x3 Subject to 2x1 + 3x2 + 5x3 ≤ 4000 4x1 + 2x2 + 7x3 ≤ 6000 x1 + 1/2x2 + 1/3x3 ≤ 2500 2 x1 = 3x2 2 x3 = 5x2 and x1 ≥ 500, x2 ≥ 500, x3 ≥ 375 Example 9 A person wants to decide the constituents of a diet which will fulfill his daily requirements of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods. The yields per unit of these foods are given in the table. Food Type Yield/unit Cost/Unit Rs Proteins Fats Carbohydrates 1 3 2 6 45 2 4 2 4 40 3 8 7 7 85 4 6 5 4 65 Minimum Requirement 800 200 700 Formulate the LP for the problem. Solution Let x1 be the number of units of food type l x2 be the number of units of food type 2 x3 be the number of units of food type 3 x4 be the number of units of food type 4 Minimize Z = 45x1 + 40x2 + 85x3 + 65x4 Subject to 3x1 + 4x2 + 8x3 + 6x4 ≥ 800 2x1 + 2x2 + 7x3 + 5x4 ≥ 200 6x1 + 4x2 + 7x3 + 4x4 ≥ 700 x1≥0, x2 ≥0, x3≥0, x4≥0 Exercise Define the terms used in LPP. Mention the advantages of LPP. What are the assumptions and limitations of LPP? A firm produces three products. These products are processed on three different machines. The time required manufacturing one unit of each of the three products and the daily capacity of the three machines are given in the table. Machine Time per unit (mins) Machine capacity Min /day Product 1 Product 2 Product 3 M1 2 3 2 440 M2 4 - 3 470 M3 2 5 - 430 It is required to determine the daily number of units to be manufactured for each product. The profit per unit for product 1, 2 and 3 is Rs. 4, Rs. 3 and Rs. 6 respectively. It is assumed that all the amounts produced are consumed in the market. Formulate the mathematical model for the model. A chemical firm produces automobiles cleaner X and polisher Y and realizes Rs. 10 profit on each batch of X and Rs. 30 on Y. Both products require processing through the same machines, A and B but X requires 4 hours in A and 8 hours in B, whereas Y requires 6 hours in A and 4 hours in B. during the fourth coming week machines A and B have 12 and 16 hours of available capacity, respectively. Assuming that demand exists for both products, how many batches of each should be produce to realize the optimal profit Z? A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine. Size B contains 1 grain of aspirin, 8 grains of bicarbonate and 6 grains of codeine. It is formed by users that it requires at least 12 grains of aspirin, 74 grains of bicarbonate and 24 grains of codeine fro providing immediate effect. It is required to determine the least number of pills a patient should take to get immediate relief. Formulate the problem as a standard LPP. Unit 3 3.1 Graphical solution Procedure 3.2 Definitions 3.3 Example Problems 3.4 Special cases of Graphical method 3.4.1 Multiple optimal solutions 3.4.2 No optimal solution 3.4.3 Unbounded solution 3.1 Graphical Solution Procedure The graphical solution procedure Consider each inequality constraint as equation. Plot each equation on the graph as each one will geometrically represent a straight line. Shade the feasible region. Every point on the line will satisfy the equation of the line. If the inequality constraint corresponding to that line is ‘≤’ then the region below the line lying in the first quadrant is shaded. Similarly for ‘≥’ the region above the line is shaded. The points lying in the common region will satisfy the constraints. This common region is called feasible region. Choose the convenient value of Z and plot the objective function line. Pull the objective function line until the extreme points of feasible region. In the maximization case this line will stop far from the origin and passing through at least one corner of the feasible region. In the minimization case, this line will stop near to the origin and passing through at least one corner of the feasible region. Read the co-ordinates of the extreme points selected in step 5 and find the maximum or minimum value of Z. 3.2 Definitions Solution – Any specification of the values for decision variable among (x1, x2… xn) is called a solution. Feasible solution is a solution for which all constraints are satisfied. Infeasible solution is a solution for which atleast one constraint is not satisfied. Feasible region is a collection of all feasible solutions. Optimal solution is a feasible solution that has the most favorable value of the objective function. Most favorable value is the largest value if the objective function is to be maximized, whereas it is the smallest value if the objective function is to be minimized. Multiple optimal solution – More than one solution with the same optimal value of the objective function. Unbounded solution – If the value of the objective function can be increased or decreased indefinitely such solutions are called unbounded solution. Feasible region – The region containing all the solutions of an inequality Corner point feasible solution is a solution that lies at the corner of the feasible region. 3.3 Example problems Example 1 Solve 3x + 5y < 15 graphically Solution Write the given constraint in the form of equation i.e. 3x + 5y = 15 Put x=0 then the value y=3 Put y=0 then the value x=5 Therefore the coordinates are (0, 3) and (5, 0). Thus these points are joined to form a straight line as shown in the graph. Put x=0, y=0 in the given constraint then 0<15, the condition is true. (0, 0) is solution nearer to origin. So shade the region below the line, which is the feasible region. Example 2 Solve 3x + 5y >15 Solution Write the given constraint in the form of equation i.e. 3x + 5y = 15 Put x=0, then y=3 Put y=0, then x=5 So the coordinates are (0, 3) and (5, 0) Put x =0, y =0 in the given constraint, the condition turns out to be false i.e. 0 > 15 is false. So the region does not contain (0, 0) as solution. The feasible region lies on the outer part of the line as shown in the graph. Example 3 Max Z = 80x1 + 55x2 Subject to 4x1+ 2x2 ≤ 40 2x1 + 4x2 ≤ 32 x1 ≥ 0 , x2 ≥ 0 Solution The first constraint 4x1+ 2 x2 ≤ 40, written in a form of equation 4x1+ 2 x2 = 40 Put x1 =0, then x2 = 20 Put x2 =0, then x1 = 10 The coordinates are (0, 20) and (10, 0) The second constraint 2x1 + 4x2 ≤ 32, written in a form of equation 2x1 + 4x2 =32 Put x1 =0, then x2 = 8 Put x2 =0, then x1 = 16 The coordinates are (0, 8) and (16, 0) The graphical representation is The corner points of feasible region are A, B and C. So the coordinates for the corner points are A (0, 8) B (8, 4) (Solve the two equations 4x1+ 2 x2 = 40 and 2x1 + 4x2 =32 to get the coordinates) C (10, 0) We know that Max Z = 80x1 + 55x2 At A (0, 8) Z = 80(0) + 55(8) = 440 At B (8, 4) Z = 80(8) + 55(4) = 860 At C (10, 0) Z = 80(10) + 55(0) = 800 The maximum value is obtained at the point B. Therefore Max Z = 860 and x1 = 8, x2 = 4 Example 4 Minimize Z = 10x1 + 4x2 Subject to 3x1 + 2x2 ≥ 60 7x1 + 2x2 ≥ 84 3x1 +6x2 ≥ 72 x1 ≥ 0 , x2 ≥ 0 Solution The first constraint 3x1 + 2x2 ≥ 60, written in a form of equation 3x1 + 2x2 = 60 Put x1 =0, then x2 = 30 Put x2 =0, then x1 = 20 The coordinates are (0, 30) and (20, 0) The second constraint 7x1 + 2x2 ≥ 84, written in a form of equation 7x1 + 2x2 = 84 Put x1 =0, then x2 = 42 Put x2 =0, then x1 = 12 The coordinates are (0, 42) and (12, 0) The third constraint 3x1 +6x2 ≥ 72, written in a form of equation 3x1 +6x2 = 72 Put x1 =0, then x2 = 12 Put x2 =0, then x1 = 24 The coordinates are (0, 12) and (24, 0) The graphical representation is The corner points of feasible region are A, B, C and D. So the coordinates for the corner points are A (0, 42) B (6, 21) (Solve the two equations 7x1 + 2x2 = 84 and 3x1 + 2x2 = 60 to get the coordinates) C (18, 3) Solve the two equations 3x1 +6x2 = 72 and 3x1 + 2x2 = 60 to get the coordinates) D (24, 0) We know that Min Z = 10x1 + 4x2 At A (0, 42) Z = 10(0) + 4(42) = 168 At B (6, 21) Z = 10(6) + 4(21) = 144 At C (18, 3) Z = 10(18) + 4(3) = 192 At D (24, 0) Z = 10(24) + 4(0) = 240 The minimum value is obtained at the point B. Therefore Min Z = 144 and x1 = 6, x2 = 21 Example 5 A manufacturer of furniture makes two products – chairs and tables. Processing of this product is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours of time per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs 2 and Rs 10 respectively. What should be the daily production of each of two products? Solution Let x1 denotes the number of chairs Let x2 denotes the number of tables Chairs Tables Availability Machine A Machine B 2 6 5 0 16 30 Profit Rs 2 Rs 10 LPP Max Z = 2x1 + 10x2 Subject to 2x1+ 5x2 ≤ 16 6x1 + 0x2 ≤ 30 x1 ≥ 0 , x2 ≥ 0 Solving graphically The first constraint 2x1+ 5x2 ≤ 16, written in a form of equation 2x1+ 5x2 = 16 Put x1 = 0, then x2 = 16/5 = 3.2 Put x2 = 0, then x1 = 8 The coordinates are (0, 3.2) and (8, 0) The second constraint 6x1 + 0x2 ≤ 30, written in a form of equation 6x1 = 30 → x1 =5 The corner points of feasible region are A, B and C. So the coordinates for the corner points are A (0, 3.2) B (5, 1.2) (Solve the two equations 2x1+ 5x2 = 16 and x1 =5 to get the coordinates) C (5, 0) We know that Max Z = 2x1 + 10x2 At A (0, 3.2) Z = 2(0) + 10(3.2) = 32 At B (5, 1.2) Z = 2(5) + 10(1.2) = 22 At C (5, 0) Z = 2(5) + 10(0) = 10 Max Z = 32 and x1 = 0, x2 = 3.2 The manufacturer should produce approximately 3 tables and no chairs to get the max profit. 3.4 Special Cases in Graphical Method 3.4.1 Multiple Optimal Solution Example 1 Solve by using graphical method Max Z = 4x1 + 3x2 Subject to 4x1+ 3x2 ≤ 24 x1 ≤ 4.5 x2 ≤ 6 x1 ≥ 0 , x2 ≥ 0 Solution The first constraint 4x1+ 3x2 ≤ 24, written in a form of equation 4x1+ 3x2 = 24 Put x1 =0, then x2 = 8 Put x2 =0, then x1 = 6 The coordinates are (0, 8) and (6, 0) The second constraint x1 ≤ 4.5, written in a form of equation x1 = 4.5 The third constraint x2 ≤ 6, written in a form of equation x2 = 6 The corner points of feasible region are A, B, C and D. So the coordinates for the corner points are A (0, 6) B (1.5, 6) (Solve the two equations 4x1+ 3x2 = 24 and x2 = 6 to get the coordinates) C (4.5, 2) (Solve the two equations 4x1+ 3x2 = 24 and x1 = 4.5 to get the coordinates) D (4.5, 0) We know that Max Z = 4x1 + 3x2 At A (0, 6) Z = 4(0) + 3(6) = 18 At B (1.5, 6) Z = 4(1.5) + 3(6) = 24 At C (4.5, 2) Z = 4(4.5) + 3(2) = 24 At D (4.5, 0) Z = 4(4.5) + 3(0) = 18 Max Z = 24, which is achieved at both B and C corner points. It can be achieved not only at B and C but every point between B and C. Hence the given problem has multiple optimal solutions. 3.4.2 No Optimal Solution Example 1 Solve graphically Max Z = 3x1 + 2x2 Subject to x1+ x2 ≤ 1 x1+ x2 ≥ 3 x1 ≥ 0 , x2 ≥ 0 Solution The first constraint x1+ x2 ≤ 1, written in a form of equation x1+ x2 = 1 Put x1 =0, then x2 = 1 Put x2 =0, then x1 = 1 The coordinates are (0, 1) and (1, 0) The first constraint x1+ x2 ≥ 3, written in a form of equation x1+ x2 = 3 Put x1 =0, then x2 = 3 Put x2 =0, then x1 = 3 The coordinates are (0, 3) and (3, 0) There is no common feasible region generated by two constraints together i.e. we cannot identify even a single point satisfying the constraints. Hence there is no optimal solution. 3.4.3 Unbounded Solution Example Solve by graphical method Max Z = 3x1 + 5x2 Subject to 2x1+ x2 ≥ 7 x1+ x2 ≥ 6 x1+ 3x2 ≥ 9 x1 ≥ 0 , x2 ≥ 0 Solution The first constraint 2x1+ x2 ≥ 7, written in a form of equation 2x1+ x2 = 7 Put x1 =0, then x2 = 7 Put x2 =0, then x1 = 3.5 The coordinates are (0, 7) and (3.5, 0) The second constraint x1+ x2 ≥ 6, written in a form of equation x1+ x2 = 6 Put x1 =0, then x2 = 6 Put x2 =0, then x1 = 6 The coordinates are (0, 6) and (6, 0) The third constraint x1+ 3x2 ≥ 9, written in a form of equation x1+ 3x2 = 9 Put x1 =0, then x2 = 3 Put x2 =0, then x1 = 9 The coordinates are (0, 3) and (9, 0) The corner points of feasible region are A, B, C and D. So the coordinates for the corner points are A (0, 7) B (1, 5) (Solve the two equations 2x1+ x2 = 7 and x1+ x2 = 6 to get the coordinates) C (4.5, 1.5) (Solve the two equations x1+ x2 = 6 and x1+ 3x2 = 9 to get the coordinates) D (9, 0) We know that Max Z = 3x1 + 5x2 At A (0, 7) Z = 3(0) + 5(7) = 35 At B (1, 5) Z = 3(1) + 5(5) = 28 At C (4.5, 1.5) Z = 3(4.5) + 5(1.5) = 21 At D (9, 0) Z = 3(9) + 5(0) = 27 The values of objective function at corner points are 35, 28, 21 and 27. But there exists infinite number of points in the feasible region which is unbounded. The value of objective function will be more than the value of these four corner points i.e. the maximum value of the objective function occurs at a point at ∞. Hence the given problem has unbounded solution. Exercise 1. A company manufactures two types of printed circuits. The requirements of transistors, resistors and capacitor for each type of printed circuits along with other data are given in table. Circuit Stock available (units) A B Transistor 15 10 180 Resistor 10 20 200 Capacitor 15 20 210 Profit Rs.5 Rs.8 How many circuits of each type should the company produce from the stock to earn maximum profit. [Ans. Max Z = 82, 2 units of type A circuit and 9 units of type B circuit] 2. A company making cool drinks has 2 bottling plants located at towns T1 and T2. Each plant produces 3 drinks A, B and C and their production capacity per day is given in the table. Cool drinks Plant at T1 T2 A 6000 2000 B 1000 2500 C 3000 3000 The marketing department of the company forecasts a demand of 80000 bottles of A, 22000 bottles of B and 40000 bottles of C during the month of June. The operating cost per day of plants at T1 and T2 are Rs. 6000 and Rs. 4000 respectively. Find graphically the number of days for which each plants must be run in June so as to minimize the operating cost while meeting the market demand. [Ans. Min Z = Rs. 88000, 12 days for the plant T1 and 4 days for plant T2] Solve the following LPP by graphical method Max Z = 3x1 + 4x2 Subject to x1 - x2 ≤ -1 -x1+ x2 ≤ 0 x1 ≥ 0 , x2 ≥ 0 [Ans. The problem has no solution] Max Z = 3x1 + 2x2 Subject to -2x1 + 3x2 ≤ 9 x1- 5x2 ≥ -20 x1 ≥ 0 , x2 ≥ 0 [Ans. The problem has unbounded solution] Max Z = 45x1 + 80x2 Subject to 5x1 + 20x2 ≤ 400 10x1+ 15x2 ≤ 450 x1 ≥ 0 , x2 ≥ 0 [Ans. Max Z = 2200, x1 = 24, x2 = 14] Module 2 Unit 1 Introduction Steps to convert GLPP to SLPP Some Basic Definitions Introduction to Simplex Method Computational procedure of Simplex Method Worked Examples 1.1 Introduction General Linear Programming Problem (GLPP) Maximize / Minimize Z = c1x1 + c2x2 + c3x3 +……………..+ cnxn Subject to constraints a11x1 + a12x2 + …..........+a1nxn (≤ or ≥) b1 a21x1 + a22x2 + ………..+a2nxn (≤ or ≥) b2 . . . am1x1 + am2x2 + ……….+amnxn (≤ or ≥) bm and x1 ≥ 0, x2 ≥ 0,…, xn ≥ 0 Where constraints may be in the form of any inequality (≤ or ≥) or even in the form of an equation (=) and finally satisfy the non-negativity restrictions. 1.2 Steps to convert GLPP to SLPP (Standard LPP) Step 1 – Write the objective function in the maximization form. If the given objective function is of minimization form then multiply throughout by -1 and write Max z׳ = Min (-z) Step 2 – Convert all inequalities as equations. If an equality of ‘≤’ appears then by adding a variable called Slack variable. We can convert it to an equation. For example x1 +2x2 ≤ 12, we can write as x1 +2x2 + s1 = 12. If the constraint is of ‘≥’ type, we subtract a variable called Surplus variable and convert it to an equation. For example 2x1 +x2 ≥ 15 2x1 +x2 – s2 = 15 Step 3 – The right side element of each constraint should be made non-negative 2x1 +x2 – s2 = -15 -2x1 - x2 + s2 = 15 (That is multiplying throughout by -1) Step 4 – All variables must have non-negative values. For example: x1 +x2 ≤ 3 x1 > 0, x2 is unrestricted in sign Then x2 is written as x2 = x2׳ – x2׳׳ where x2׳, x2׳׳ ≥ 0 Therefore the inequality takes the form of equation as x1 + (x2׳ – x2׳׳) + s1 = 3 Using the above steps, we can write the GLPP in the form of SLPP. Write the Standard LPP (SLPP) of the following Example 1 Maximize Z = 3x1 + x2 Subject to 2 x1 + x2 ≤ 2 3 x1 + 4 x2 ≥ 12 and x1 ≥ 0, x2 ≥ 0 SLPP Maximize Z = 3x1 + x2 Subject to 2 x1 + x2 + s1 = 2 3 x1 + 4 x2 – s2 = 12 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0 Example 2 Minimize Z = 4x1 + 2 x2 Subject to 3x1 + x2 ≥ 2 x1 + x2 ≥ 21 x1 + 2x2 ≥ 30 and x1 ≥ 0, x2 ≥ 0 SLPP Maximize Z׳ = – 4x1 – 2 x2 Subject to 3x1 + x2 – s1 = 2 x1 + x2 – s2 = 21 x1 + 2x2 – s3 = 30 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Example 3 Minimize Z = x1 + 2 x2 + 3x3 Subject to 2x1 + 3x2 + 3x3 ≥ – 4 3x1 + 5x2 + 2x3 ≤ 7 and x1 ≥ 0, x2 ≥ 0, x3 is unrestricted in sign SLPP Maximize Z׳ = – x1 – 2 x2 – 3(x3׳ – x3׳׳) Subject to –2x1 – 3x2 – 3(x3׳ – x3׳׳) + s1= 4 3x1 + 5x2 + 2(x3׳ – x3׳׳) + s2 = 7 x1 ≥ 0, x2 ≥ 0, x3׳ ≥ 0, x3׳׳ ≥ 0, s1 ≥ 0, s2 ≥ 0 1.3 Some Basic Definitions Solution of LPP Any set of variable (x1, x2… xn) which satisfies the given constraint is called solution of LPP. Basic solution It is a solution obtained by setting any ‘n’ variable equal to zero and solving remaining ‘m’ variables. Such ‘m’ variables are called Basic variables and ‘n’ variables are called Non-basic variables. Basic feasible solution A basic solution that is feasible (all basic variables are non negative) is called basic feasible solution. There are two types of basic feasible solution. Degenerate basic feasible solution If any of the basic variable of a basic feasible solution is zero than it is said to be degenerate basic feasible solution. Non-degenerate basic feasible solution It is a basic feasible solution which has exactly ‘m’ positive xi, where i=1, 2, … m. In other words all ‘m’ basic variables are positive and remaining ‘n’ variables are zero. Optimum basic feasible solution A basic feasible solution is said to be optimum if it optimizes (max / min) the objective function. 1.4 Introduction to Simplex Method It was developed by G. Danztig in 1947. The simplex method provides an algorithm (a rule of procedure usually involving repetitive application of a prescribed operation) which is based on the fundamental theorem of linear programming. The Simplex algorithm is an iterative procedure for solving LP problems in a finite number of steps. It consists of Having a trial basic feasible solution to constraint-equations Testing whether it is an optimal solution Improving the first trial solution by a set of rules and repeating the process till an optimal solution is obtained Advantages Simple to solve the problems The solution of LPP of more than two variables can be obtained. 1.5 Computational Procedure of Simplex Method Consider an example Maximize Z = 3x1 + 2x2 Subject to x1 + x2 ≤ 4 x1 – x2 ≤ 2 and x1 ≥ 0, x2 ≥ 0 Solution Step 1 – Write the given GLPP in the form of SLPP Maximize Z = 3x1 + 2x2 + 0s1 + 0s2 Subject to x1 + x2+ s1= 4 x1 – x2 + s2= 2 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0 Step 2 – Present the constraints in the matrix form x1 + x2+ s1= 4 x1 – x2 + s2= 2 Step 3 – Construct the starting simplex table using the notations Cj → 3 2 0 0 Basic Variables CB XB X1 X2 S1 S2 Min ratio XB /Xk s1 s2 0 4 0 2 1 1 1 0 1 -1 0 1 Z= CB XB Δj Step 4 – Calculation of Z and Δj and test the basic feasible solution for optimality by the rules given. Z= CB XB = 0 *4 + 0 * 2 = 0 Δj = Zj – Cj = CB Xj – Cj Δ1 = CB X1 – Cj = 0 * 1 + 0 * 1 – 3 = -3 Δ2 = CB X2 – Cj = 0 * 1 + 0 * -1 – 2 = -2 Δ3 = CB X3 – Cj = 0 * 1 + 0 * 0 – 0 = 0 Δ4 = CB X4 – Cj = 0 * 0 + 0 * 1 – 0 = 0 Procedure to test the basic feasible solution for optimality by the rules given Rule 1 – If all Δj ≥ 0, the solution under the test will be optimal. Alternate optimal solution will exist if any non-basic Δj is also zero. Rule 2 – If atleast one Δj is negative, the solution is not optimal and then proceeds to improve the solution in the next step. Rule 3 – If corresponding to any negative Δj, all elements of the column Xj are negative or zero, then the solution under test will be unbounded. In this problem it is observed that Δ1 and Δ2 are negative. Hence proceed to improve this solution Step 5 – To improve the basic feasible solution, the vector entering the basis matrix and the vector to be removed from the basis matrix are determined. Incoming vector The incoming vector Xk is always selected corresponding to the most negative value of Δj. It is indicated by (↑). Outgoing vector The outgoing vector is selected corresponding to the least positive value of minimum ratio. It is indicated by (→). Step 6 – Mark the key element or pivot element by ‘1’‘.The element at the intersection of outgoing vector and incoming vector is the pivot element. Cj → 3 2 0 0 Basic Variables CB XB X1 X2 S1 S2 (Xk) Min ratio XB /Xk s1 s2 0 4 0 2 1 1 1 0 1 -1 0 1 4 / 1 = 4 2 / 1 = 2 → outgoing Z= CB XB = 0 ↑incoming Δ1= -3 Δ2= -2 Δ3=0 Δ4=0 If the number in the marked position is other than unity, divide all the elements of that row by the key element. Then subtract appropriate multiples of this new row from the remaining rows, so as to obtain zeroes in the remaining position of the column Xk. Basic Variables CB XB X1 X2 S1 S2 (Xk) Min ratio XB /Xk s1 x1 0 2 3 2 (R1=R1 – R2) 0 2 1 -1 1 -1 0 1 2 / 2 = 1 → outgoing 2 / -1 = -2 (neglect in case of negative) Z=0*2+3*2= 6 ↑incoming Δ1=0 Δ2= -5 Δ3=0 Δ4=3 Step 7 – Now repeat step 4 through step 6 until an optimal solution is obtained. Basic Variables CB XB X1 X2 S1 S2 Min ratio XB /Xk x2 x1 2 1 3 3 (R1=R1 / 2) 0 1 1/2 -1/2 (R2=R2 + R1) 1 0 1/2 1/2 Z = 11 Δ1=0 Δ2=0 Δ3=5/2 Δ4=1/2 Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z = 11, x1 = 3 and x2 = 1 1.6 Worked Examples Solve by simplex method Example 1 Maximize Z = 80x1 + 55x2 Subject to 4x1 + 2x2 ≤ 40 2x1 + 4x2 ≤ 32 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Maximize Z = 80x1 + 55x2 + 0s1 + 0s2 Subject to 4x1 + 2x2+ s1= 40 2x1 + 4x2 + s2= 32 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0 Cj → 80 55 0 0 Basic Variables CB XB X1 X2 S1 S2 Min ratio XB /Xk s1 s2 0 40 0 32 4 2 1 0 2 4 0 1 40 / 4 = 10→ outgoing 32 / 2 = 16 Z= CB XB = 0 ↑incoming Δ1= -80 Δ2= -55 Δ3=0 Δ4=0 x1 s2 80 10 0 12 (R1=R1 / 4) 1 1/2 1/4 0 (R2=R2– 2R1) 0 3 -1/2 1 10/1/2 = 20 12/3 = 4→ outgoing Z = 800 ↑incoming Δ1=0 Δ2= -15 Δ3=40 Δ4=0 x1 x2 80 8 55 4 (R1=R1– 1/2R2) 1 0 1/3 -1/6 (R2=R2 / 3) 0 1 -1/6 1/3 Z = 860 Δ1=0 Δ2=0 Δ3=35/2 Δ4=5 Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 860, x1 = 8 and x2 = 4 Example 2 Maximize Z = 5x1 + 3x2 Subject to 3x1 + 5x2 ≤ 15 5x1 + 2x2 ≤ 10 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Maximize Z = 5x1 + 3x2 + 0s1 + 0s2 Subject to 3x1 + 5x2+ s1= 15 5x1 + 2x2 + s2= 10 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0 Cj → 5 3 0 0 Basic Variables CB XB X1 X2 S1 S2 Min ratio XB /Xk s1 s2 0 15 0 10 3 5 1 0 5 2 0 1 15 / 3 = 5 10 / 5 = 2 → outgoing Z= CB XB = 0 ↑incoming Δ1= -5 Δ2= -3 Δ3=0 Δ4=0 s1 x1 0 9 5 2 (R1=R1– 3R2) 0 19/5 1 -3/5 (R2=R2 /5) 1 2/5 0 1/5 9/19/5 = 45/19 → 2/2/5 = 5 Z = 10 ↑ Δ1=0 Δ2= -1 Δ3=0 Δ4=1 x2 x1 3 45/19 5 20/19 (R1=R1 / 19/5) 0 1 5/19 -3/19 (R2=R2 –2/5 R1) 1 0 -2/19 5/19 Z = 235/19 Δ1=0 Δ2=0 Δ3=5/19 Δ4=16/19 Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 235/19, x1 = 20/19 and x2 = 45/19 Example 3 Maximize Z = 5x1 + 7x2 Subject to x1 + x2 ≤ 4 3x1 – 8x2 ≤ 24 10x1 + 7x2 ≤ 35 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Maximize Z = 5x1 + 7x2 + 0s1 + 0s2 + 0s3 Subject to x1 + x2 + s1= 4 3x1 – 8x2 + s2= 24 10x1 + 7x2 + s3= 35 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Cj → 5 7 0 0 0 Basic Variables CB XB X1 X2 S1 S2 S3 Min ratio XB /Xk s1 s2 s3 0 4 24 0 35 1 1 1 0 0 3 -8 0 1 0 10 7 0 0 1 4 /1 = 4→outgoing – 35 / 7 = 5 Z= CB XB = 0 ↑incoming -5 -7 0 0 0 ←Δj x2 s2 s3 4 0 56 0 7 1 1 1 0 0 (R2 = R2 + 8R1) 11 0 8 1 0 (R3 = R3 – 7R1) 3 0 -7 0 1 Z = 28 2 0 7 0 0 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z = 28, x1 = 0 and x2 = 4 Example 4 Maximize Z = 2x – 3y + z Subject to 3x + 6y + z ≤ 6 4x + 2y + z ≤ 4 x – y + z ≤ 3 and x ≥ 0, y ≥ 0, z ≥ 0 Solution SLPP Maximize Z = 2x – 3y + z + 0s1 + 0s2 + 0s3 Subject to 3x + 6y + z + s1= 6 4x + 2y + z + s2= 4 x – y + z + s3= 3 x ≥ 0, y ≥ 0, z ≥ 0 s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Cj → 2 -3 1 0 0 0 Basic Variables CB XB X Y Z S1 S2 S3 Min ratio XB /Xk s1 s2 s3 0 6 4 0 3 3 6 1 1 0 0 4 2 1 0 1 0 1 -1 1 0 0 1 6 / 3 = 2 4 / 4 =1→ outgoing 3 / 1 = 3 Z = 0 ↑incoming -2 3 -1 0 0 0 ←Δj s1 x s3 0 3 2 1 0 2 0 9/2 1/4 1 -3/4 0 1 1/2 1/4 0 1/4 0 0 -3/2 3/4 0 -1/4 1 3/1/4=12 1/1/4=4 8/3 = 2.6→ Z = 2 ↑incoming 0 4 1/2 0 1/2 0 ←Δj s1 x z 7/3 2 1/3 1 8/3 0 5 0 1 -2/3 -1/3 1 1 0 0 1/3 -1/3 0 -2 1 0 -1/3 4/3 Z = 10/3 0 3 0 0 1/3 2/3 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 10/3, x = 1/3, y = 0 and z = 8/3 Example 5 Maximize Z = 3x1 + 5x2 Subject to 3x1 + 2x2 ≤ 18 x1 ≤ 4 x2 ≤ 6 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Maximize Z = 3x1 + 5x2 + 0s1 + 0s2 + 0s3 Subject to 3x1 + 2x2 + s1= 18 x1 + s2= 4 x2 + s3= 6 x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Cj → 3 5 0 0 0 Basic Variables CB XB X1 X2 S1 S2 S3 Min ratio XB /Xk s1 0 18 3 2 1 0 0 18 / 2 = 9 s2 0 4 1 0 0 1 0 4 / 0 = ∞ (neglect) s3 0 6 0 1 0 0 1 6 / 1 = 6→ Z = 0 -3 ↑ -5 0 0 0 ←Δj (R1=R1-2R3) s1 0 6 3 0 1 0 -2 6 / 3 = 2 → s2 0 4 1 0 0 1 0 4 / 1 = 4 x2 5 6 0 1 0 0 1 -- Z = 30 ↑ -3 0 0 0 5 ←Δj (R1=R1 / 3) x1 3 2 1 0 1/3 0 -2/3 (R2=R2 - R1) s2 0 2 0 0 -1/3 1 2/3 x2 5 6 0 1 0 0 1 Z = 36 0 0 1 0 3 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 36, x1 = 2, x2 = 6 Example 6 Minimize Z = x1 – 3x2 + 2x3 Subject to 3x1 – x2 + 3x3 ≤ 7 -2x1 + 4x2 ≤ 12 -4x1 + 3x2 + 8x3 ≤ 10 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution SLPP Min (-Z) = Max Z׳ = -x1 + 3x2 - 2x3 + 0s1 + 0s2 + 0s3 Subject to 3x1 – x2 + 3x3 + s1 = 7 -2x1 + 4x2 + s2 = 12 -4x1 + 3x2 + 8x3 + s3 = 10 x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Cj → -1 3 -2 0 0 0 Basic Variables CB XB X1 X2 X3 S1 S2 S3 Min ratio XB /Xk s1 0 7 3 -1 3 1 0 0 - s2 0 12 -2 4 0 0 1 0 3→ s3 0 10 -4 3 8 0 0 1 10/3 Z' = 0 1 ↑ -3 2 0 0 0 ←Δj (R1 = R1 + R2) s1 0 10 5/2 0 3 1 1/4 0 4→ (R2 = R2 / 4) x2 3 3 -1/2 1 0 0 1/4 0 - (R3 = R3 – 3R2) s3 0 1 -5/2 0 8 0 -3/4 1 - Z' = 9 ↑ -5/2 0 0 0 3/4 0 ←Δj (R1 = R1 / 5/2) x1 -1 4 1 0 6/5 2/5 1/10 0 (R2 = R2 + 1/2 R1) x2 3 5 0 1 3/5 1/5 3/10 0 (R3 = R3 + 5/2R1) s3 0 11 0 1 11 1 -1/2 1 Z' = 11 0 0 3/5 1/5 1/5 0 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Z' =11 which implies Z = -11, x1 = 4, x2 = 5, x3 = 0 Example 7 Max Z = 2x + 5y x + y ≤ 600 0 ≤ x ≤ 400 0 ≤ y ≤ 300 Solution SLPP Max Z = 2x + 5y + 0s1 + 0s2 + 0s3 x + y + s1 = 600 x + s2 = 400 y + s3 = 300 x1 ≥ 0, y ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0 Cj → 2 5 0 0 0 Basic Variables CB XB X Y S1 S2 S3 Min ratio XB /Xk s1 0 600 1 1 1 0 0 600 / 1 = 600 s2 0 400 1 0 0 1 0 - s3 0 300 0 1 0 0 1 300 /1 = 300→ Z = 0 -2 ↑ -5 0 0 0 ←Δj (R1 = R1 – R3) s1 0 300 1 0 1 0 -1 300 /1 = 300→ s2 0 400 1 0 0 1 0 400 / 1 = 400 y 5 300 0 1 0 0 1 - Z = 1500 ↑ -2 0 0 0 5 ←Δj x 2 300 1 0 1 0 -1 (R2 = R2 – R1) s2 0 100 0 0 -1 1 1 y 5 300 0 1 0 0 1 Z = 2100 0 0 2 0 3 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Z = 2100, x = 300, y = 300 Exercise Solve by simplex method Maximize Z = 5x1 + 3x2 Subject to 3x1 + 5x2 ≤ 15 5x1 + 2x2 ≤ 10 and x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 235/19, x1= 20/19, x2= 45/19] Maximize Z = 5x1 + 7x2 Subject to x1 + x2 ≤ 4 3x1 - 8x2 ≤ 24 10x1 + 7x2 ≤ 35 and x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 28, x1= 0, x2= 4] Maximize Z = 2x1 + 4x2 + x3+ x4 Subject to x1 + 3x2 + x4 ≤ 4 2x1 + x2 ≤ 3 x2 + 4x3 + x4 ≤ 3 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 [Ans. Max Z = 13/2, x1= 1, x2= 1, x3= 1/2, x4= 0] Maximize Z = 7x1 + 5x2 Subject to -x1 - 2x2 ≥ -6 4x1 + 3x2 ≤ 12 and x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 21, x1= 3, x2= 0] Maximize Z = 3x1 + 2x2 Subject to 2x1 + x2 ≤ 10 x1 + 3x2 ≤ 6 x1 + x2 ≤ 21 and x1 ≥ 0, x2 ≥ 0 Unit 2 2.1 Computational Procedure of Big – M Method (Charne’s Penalty Method) 2.2 Worked Examples 2.3 Steps for Two-Phase Method 2.4 Worked Examples 2.1 Computational Procedure of Big – M Method (Charne’s Penalty Method) Step 1 – Express the problem in the standard form. Step 2 – Add non-negative artificial variable to the left side of each of the equations corresponding to the constraints of the type ‘≥’ or ‘=’. When artificial variables are added, it causes violation of the corresponding constraints. This difficulty is removed by introducing a condition which ensures that artificial variables will be zero in the final solution (provided the solution of the problem exists). On the other hand, if the problem does not have a solution, at least one of the artificial variables will appear in the final solution with positive value. This is achieved by assigning a very large price (per unit penalty) to these variables in the objective function. Such large price will be designated by –M for maximization problems (+M for minimizing problem), where M > 0. Step 3 – In the last, use the artificial variables for the starting solution and proceed with the usual simplex routine until the optimal solution is obtained. 2.2 Worked Examples Example 1 Max Z = -2x1 - x2 Subject to 3x1 + x2 = 3 4x1 + 3x2 ≥ 6 x1 + 2x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Max Z = -2x1 - x2 + 0s1 + 0s2 - M a1 - M a2 Subject to 3x1 + x2 + a1= 3 4x1 + 3x2 – s1 + a2 = 6 x1 + 2x2 + s2 = 4 x1 , x2 , s1, s2, a1, a2 ≥ 0 Cj → -2 -1 0 0 -M -M Basic Variables CB XB X1 X2 S1 S2 A1 A2 Min ratio XB /Xk a1 -M 3 3 1 0 0 1 0 3 /3 = 1→ a2 -M 6 4 3 -1 0 0 1 6 / 4 =1.5 s2 0 4 1 2 0 1 0 0 4 / 1 = 4 Z = -9M ↑ 2 – 7M 1 – 4M M 0 0 0 ←Δj x1 -2 1 1 1/3 0 0 X 0 1/1/3 =3 a2 -M 2 0 5/3 -1 0 X 1 6/5/3 =1.2→ s2 0 3 0 5/3 0 1 X 0 4/5/3=1.8 Z = -2 – 2M 0 0 0 X 0 ←Δj x1 -2 3/5 1 0 1/5 0 X X x2 -1 6/5 0 1 -3/5 0 X X s2 0 1 0 0 1 1 X X Z = -12/5 0 0 1/5 0 X X Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z = -12/5, x1 = 3/5, x2 = 6/5 Example 2 Max Z = 3x1 - x2 Subject to 2x1 + x2 ≥ 2 x1 + 3x2 ≤ 3 x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Max Z = 3x1 - x2 + 0s1 + 0s2 + 0s3 - M a1 Subject to 2x1 + x2 – s1+ a1= 2 x1 + 3x2 + s2 = 3 x2 + s3 = 4 x1 , x2 , s1, s2, s3, a1 ≥ 0 Cj → 3 -1 0 0 0 -M Basic Variables CB XB X1 X2 S1 S2 S3 A1 Min ratio XB /Xk a1 -M 2 2 1 -1 0 0 1 2 / 2 = 1→ s2 0 3 1 3 0 1 0 0 3 / 1 = 3 s3 0 4 0 1 0 0 1 0 - Z = -2M ↑ -2M-3 -M+1 M 0 0 0 ←Δj x1 3 1 1 1/2 -1/2 0 0 X - s2 0 2 0 5/2 1/2 1 0 X 2/1/2 = 4→ s3 0 4 0 1 0 0 1 X - Z = 3 0 5/2 ↑ -3/2 0 0 X ←Δj x1 3 3 1 3 0 1/2 0 X s1 0 4 0 5 1 2 0 X s3 0 4 0 1 0 0 1 X Z = 9 0 10 0 3/2 0 X Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 9, x1 = 3, x2 = 0 Example 3 Min Z = 2x1 + 3x2 Subject to x1 + x2 ≥ 5 x1 + 2x2 ≥ 6 and x1 ≥ 0, x2 ≥ 0 Solution SLPP Min Z = Max Z׳ = -2x1 - 3x2 + 0s1 + 0s2 - M a1 - M a2 Subject to x1 + x2 – s1+ a1= 5 x1 + 2x2 – s2+ a2= 6 x1 , x2 , s1, s2, a1, a2 ≥ 0 Cj → -2 -3 0 0 -M -M Basic Variables CB XB X1 X2 S1 S2 A1 A2 Min ratio XB /Xk a1 -M 5 1 1 -1 0 1 0 5 /1 = 5 a2 -M 6 1 2 0 -1 0 1 6 / 2 = 3→ Z׳ = -11M -2M + 2 ↑ -3M+3 M M 0 0 ←Δj a1 -M 2 1/2 0 -1 1/2 1 X 2/1/2 = 4→ x2 -3 3 1/2 1 0 -1/2 0 X 3/1/2 =6 Z׳ = -2M-9 ↑ (-M+1) / 2 0 M (-M+3)/2 0 X ←Δj x1 -2 4 1 0 -2 1 X X x2 -3 1 0 1 1 -1 X X Z׳ = -11 0 0 1 1 X X Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Z' = -11 which implies Max Z = 11, x1 = 4, x2 = 1 Example 4 Max Z =3x1 + 2x2 + x3 Subject to 2x1 + x2 + x3 = 12 3x1 + 4x2 = 11 and x1 is unrestricted x2 ≥ 0, x3 ≥ 0 Solution SLPP Max Z = 3(x1' - x1'') + 2x2 + x3 - M a1 - M a2 Subject to 2(x1' - x1'') + x2 + x3 + a1= 12 3(x1' - x1'') + 4x2 + a2 = 11 x1', x1'', x2 , x3, a1, a2 ≥ 0 Max Z = 3x1' - 3x1'' + 2x2 + x3 - M a1 - M a2 Subject to 2x1' - 2x1'' + x2 + x3 + a1= 12 3x1' - 3x1'' + 4x2 + a2 = 11 x1', x1'', x2 , x3, a1, a2 ≥ 0 Cj → 3 -3 2 1 -M -M Basic Variables CB XB X1' X1'' X2 X3 A1 A2 Min ratio XB /Xk a1 -M 12 2 -2 1 1 1 0 12 /2 = 6 a2 -M 11 3 -3 4 0 0 1 11/3 =3.6→ Z = -23M ↑ -5M-3 5M+3 -5M-2 -M-1 0 0 ←Δj a1 -M 14/3 0 0 -5/3 1 1 X 14/3/1 = 14/3→ x1' 3 11/3 1 -1 4/3 0 0 X - 0 -6 5/3M+1 ↑ -M-1 0 X ←Δj x3 1 14/3 0 0 -5/3 1 X X x1' 3 11/3 1 -1 4/3 0 X X Z = 47/3 0 0 1/3 0 X X Since all Δj ≥ 0, optimal basic feasible solution is obtained x1' = 11/3, x1'' = 0 x1 = x1' - x1'' = 11/3 – 0 = 11/3 Therefore the solution is Max Z = 47/3, x1 = 11/3, x2 = 0, x3 = 14/3 Example 5 Max Z = 8x2 Subject to x1 - x2 ≥ 0 2x1 + 3x2 ≤ -6 and x1 , x2 unrestricted Solution SLPP Max Z = 8 (x2' – x2'') + 0s1 + 0s2 - M a1 - M a2 Subject to (x1' - x1'') - (x2' – x2'') – s1+ a1= 0 -2(x1' - x1'') - 3(x2' – x2'') - s2 + a2 = 6 x1', x1'', x2', x2'', s1, a1, a2 ≥ 0 Max Z = 8x2' – 8x2'' + 0s1 + 0s2 - M a1 - M a2 Subject to x1' - x1'' - x2' + x2''– s1+ a1= 0 -2x1' + 2x1'' - 3x2' + 3x2'' - s2 + a2 = 6 x1', x1'', x2', x2'', s1, a1, a2 ≥ 0 Cj → 0 0 8 -8 0 0 -M -M Basic Variables CB XB X1' X1'' X2' X2'' S1 S2 A1 A2 Min ratio XB /Xk a1 -M 0 1 -1 -1 1 -1 0 1 0 0→ a2 -M 6 -2 2 -3 3 0 -1 0 1 2 Z = -6M M -M 4M-8 ↑ -4M+8 M M 0 0 ←Δj x2'' -8 0 1 -1 -1 1 -1 0 X 0 - a2 -M 6 -5 5 0 0 3 -1 X 1 6/5→ Z = -6M 5M-8 ↑ -5M+8 0 0 -3M+8 M X 0 ←Δj x2'' -8 6/5 0 0 -1 1 -2/5 -1/5 X X x1'' 0 6/5 -1 1 0 0 3/5 -1/5 X X Z = -48/5 0 0 0 0 16/5 8/5 X X Since all Δj ≥ 0, optimal basic feasible solution is obtained x1' = 0, x1'' = 6/5 x1 = x1' - x1'' = 0 – 6/5 = -6/5 x2' = 0, x2'' = 6/5 x2 = x2' – x2'' = 0 – 6/5 = -6/5 Therefore the solution is Max Z = -48/5, x1 = -6/5, x2 = -6/5 2.3 Steps for Two-Phase Method The process of eliminating artificial variables is performed in phase-I of the solution and phase-II is used to get an optimal solution. Since the solution of LPP is computed in two phases, it is called as Two-Phase Simplex Method. Phase I – In this phase, the simplex method is applied to a specially constructed auxiliary linear programming problem leading to a final simplex table containing a basic feasible solution to the original problem. Step 1 – Assign a cost -1 to each artificial variable and a cost 0 to all other variables in the objective function. Step 2 – Construct the Auxiliary LPP in which the new objective function Z* is to be maximized subject to the given set of constraints. Step 3 – Solve the auxiliary problem by simplex method until either of the following three possibilities do arise Max Z* < 0 and atleast one artificial vector appear in the optimum basis at a positive level (Δj ≥ 0). In this case, given problem does not possess any feasible solution. Max Z* = 0 and at least one artificial vector appears in the optimum basis at a zero level. In this case proceed to phase-II. Max Z* = 0 and no one artificial vector appears in the optimum basis. In this case also proceed to phase-II. Phase II – Now assign the actual cost to the variables in the objective function and a zero cost to every artificial variable that appears in the basis at the zero level. This new objective function is now maximized by simplex method subject to the given constraints. Simplex method is applied to the modified simplex table obtained at the end of phase-I, until an optimum basic feasible solution has been attained. The artificial variables which are non-basic at the end of phase-I are removed. 2.4 Worked Examples Example 1 Max Z = 3x1 - x2 Subject to 2x1 + x2 ≥ 2 x1 + 3x2 ≤ 2 x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 3x1 - x2 Subject to 2x1 + x2 – s1+ a1= 2 x1 + 3x2 + s2 = 2 x2 + s3 = 4 x1 , x2 , s1, s2, s3,a1 ≥ 0 Auxiliary LPP Max Z* = 0x1 - 0x2 + 0s1 + 0s2 + 0s3 -1a1 Subject to 2x1 + x2 – s1+ a1= 2 x1 + 3x2 + s2 = 2 x2 + s3 = 4 x1 , x2 , s1, s2, s3,a1 ≥ 0 Phase I Cj → 0 0 0 0 0 -1 Basic Variables CB XB X1 X2 S1 S2 S3 A1 Min ratio XB /Xk a1 -1 2 2 1 -1 0 0 1 1→ s2 0 2 1 3 0 1 0 0 2 s3 0 4 0 1 0 0 1 0 - Z* = -2 ↑ -2 -1 1 0 0 0 ←Δj x1 0 1 1 1/2 -1/2 0 0 X s2 0 1 0 5/2 1/2 1 0 X s3 0 4 0 1 0 0 1 X Z* = 0 0 0 0 0 0 X ←Δj Since all Δj ≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we proceed to phase II. Phase II Cj → 3 -1 0 0 0 Basic Variables CB XB X1 X2 S1 S2 S3 Min ratio XB /Xk x1 3 1 1 1/2 -1/2 0 0 - s2 0 1 0 5/2 1/2 1 0 2→ s3 0 4 0 1 0 0 1 - Z = 3 0 5/2 ↑ -3/2 0 0 ←Δj x1 3 2 1 3 0 1 0 s1 0 2 0 5 1 2 0 s3 0 4 0 1 0 0 1 Z = 6 0 10 0 3 0 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z = 6, x1 = 2, x2 = 0 Example 2 Max Z = 5x1 + 8x2 Subject to 3x1 + 2x2 ≥ 3 x1 + 4x2 ≥ 4 x1 + x2 ≤ 5 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 5x1 + 8x2 Subject to 3x1 + 2x2 – s1+ a1 = 3 x1 + 4x2 – s2+ a2 = 4 x1 + x2 + s3 = 5 x1 , x2 , s1, s2, s3, a1, a2 ≥ 0 Auxiliary LPP Max Z* = 0x1 + 0x2 + 0s1 + 0s2 + 0s3 -1a1 -1a2 Subject to 3x1 + 2x2 – s1+ a1 = 3 x1 + 4x2 – s2+ a2 = 4 x1 + x2 + s3 = 5 x1 , x2 , s1, s2, s3, a1, a2 ≥ 0 Phase I Cj → 0 0 0 0 0 -1 -1 Basic Variables CB XB X1 X2 S1 S2 S3 A1 A2 Min ratio XB /Xk a1 -1 3 3 2 -1 0 0 1 0 3/2 a2 -1 4 1 4 0 -1 0 0 1 1→ s3 0 5 1 1 0 0 1 0 0 5 Z* = -7 -4 ↑ -6 1 1 0 0 0 ←Δj a1 -1 1 5/2 0 -1 1/2 0 1 X 2/5→ x2 0 1 1/4 1 0 -1/4 0 0 X 4 s3 0 4 3/4 0 0 1/4 1 0 X 16/3 Z* = -1 ↑ -5/2 0 1 -1/2 0 0 X ←Δj x1 0 2/5 1 0 -2/5 1/5 0 X X x2 0 9/10 0 1 1/10 -3/10 0 X X s3 0 37/10 0 0 3/10 1/10 1 X X Z* = 0 0 0 0 0 0 X X ←Δj Since all Δj ≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we proceed to phase II. Phase II Cj → 5 8 0 0 0 Basic Variables CB XB X1 X2 S1 S2 S3 Min ratio XB /Xk x1 5 2/5 1 0 -2/5 1/5 0 2→ x2 8 9/10 0 1 1/10 -3/10 0 - s3 0 37/10 0 0 3/10 1/10 1 37 Z = 46/5 0 0 -6/5 ↑ -7/5 0 ←Δj s2 0 2 5 0 -2 1 0 - x2 8 3/2 3/2 1 -1/2 0 0 - s3 0 7/2 -1/2 0 1/2 0 1 7→ Z = 12 7 0 ↑ -4 0 0 ←Δj s2 0 16 3 0 0 1 2 x2 8 5 1 1 0 0 1/2 s1 0 7 -1 0 1 0 2 Z = 40 3 0 0 0 4 Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 40, x1 = 0, x2 = 5 Example 3 Max Z = -4x1 - 3x2 - 9x3 Subject to 2x1 + 4x2 + 6x3 ≥ 15 6x1 + x2 + 6x3 ≥ 12 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution Standard LPP Max Z = -4x1 - 3x2 - 9x3 Subject to 2x1 + 4x2 + 6x3 - s1+ a1= 15 6x1 + x2 + 6x3 - s2 + a2 = 12 x1 , x2 , s1, s2, a1, a2 ≥ 0 Auxiliary LPP Max Z* = 0x1 - 0x2 - 0x3 + 0s1 + 0s2 -1a1 -1a2 Subject to 2x1 + 4x2 + 6x3 - s1+ a1= 15 6x1 + x2 + 6x3 - s2 + a2 = 12 x1 , x2 , s1, s2, a1, a2 ≥ 0 Phase I Cj → 0 0 0 0 0 -1 -1 Basic Variables CB XB X1 X2 X3 S1 S2 A1 A2 Min ratio XB /Xk a1 -1 15 2 4 6 -1 0 1 0 15/6 a2 -1 12 6 1 6 0 -1 0 1 2→ Z* = -27 -8 -5 ↑ -12 1 1 0 0 ←Δj a1 -1 3 -4 3 0 -1 1 1 X 1→ x3 0 2 1 1/6 1 0 -1/6 0 X 12 Z* = -3 4 ↑ -3 0 1 -1 0 X ←Δj x2 0 1 -4/3 1 0 -1/3 1/3 X X x3 0 11/6 22/18 0 1 1/18 -4/18 X X Z* = 0 0 0 0 0 0 X X Since all Δj ≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we proceed to phase II. Phase II Cj → -4 -3 -9 0 0 Basic Variables CB XB X1 X2 X3 S1 S2 Min ratio XB /Xk x2 -3 1 -4/3 1 0 -1/3 1/3 - x3 -9 11/6 22/18 0 1 1/18 -4/18 3/2→ Z = -39/2 ↑ -3 0 0 1/2 1 ←Δj x2 -3 3 0 1 12/11 -3/11 1/11 x1 -4 3/2 1 0 18/22 1/22 -4/22 Z = -15 0 0 27/11 7/11 5/11 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z = -15, x1 = 3/2, x2 = 3, x3 = 0 Example 4 Min Z = 4x1 + x2 Subject to 3x1 + x2 = 3 4x1 + 3x2 ≥ 6 x1 + 2x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Min Z = Max Z' = – 4x1 – x2 Subject to 3x1 + x2 + a1 = 3 4x1 + 3x2 – s1+ a2 = 6 x1 + 2x2 + s2 = 4 x1 , x2 , s1, s2, a1, a2 ≥ 0 Auxiliary LPP Max Z* = 0x1 – 0x2 + 0s1 + 0s2 –1a1 –1a2 Subject to 3x1 + x2 + a1 = 3 4x1 + 3x2 – s1+ a2 = 6 x1 + 2x2 + s2 = 4 x1 , x2 , s1, s2, a1, a2 ≥ 0 Phase I Cj → 0 0 0 0 -1 -1 Basic Variables CB XB X1 X2 S1 S2 A1 A2 Min ratio XB /Xk a1 -1 3 3 1 0 0 1 0 1→ a2 -1 6 4 3 -1 0 0 1 6/4 s2 0 4 1 2 0 1 0 0 4 Z* = -9 ↑ -7 -4 1 0 0 0 x1 0 1 1 1/3 0 0 X 0 3 a2 -1 2 0 5/3 -1 0 X 1 6/5→ s2 0 3 0 5/3 0 1 X 0 9/5 Z* = -2 0 ↑ -5/3 1 0 X 0 x1 0 3/5 1 0 1/5 0 X X x2 0 6/5 0 1 -3/5 0 X X s2 0 1 0 0 1 1 X X Z* = 0 0 0 0 0 X X Since all Δj ≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we proceed to phase II. Phase II Cj → -4 -1 0 0 Basic Variables CB XB X1 X2 S1 S2 Min ratio XB /Xk x1 -4 3/5 1 0 1/5 0 3 x2 -1 6/5 0 1 -3/5 0 - s2 0 1 0 0 1 1 1→ Z' = -18/5 0 0 ↑ -1/5 0 ←Δj x1 -4 2/5 1 0 0 -1/5 x2 -1 9/5 0 1 0 3/5 s1 0 1 0 0 1 1 Z' = -17/5 0 0 0 1/5 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained Therefore the solution is Max Z' = -17/5 Min Z = 17/5, x1 = 2/5, x2 = 9/5 Exercise Solve by Big-M method Min Z = 4x1 + 2x2 Subject to 3x1 + x2 ≥ 27 x1 + x2 ≥ 21 and x1 ≥ 0, x2 ≥ 0 [Ans. Min Z = 48, x1 = 3, x2 =18] Min Z = x1 + x2 + 3x3 Subject to 3x1 + 2x2 + x3 ≤ 3 2x1 + x2+ 2x3 ≥ 3 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 [Ans. Min Z = 3, x1 = 3/4, x2 =0, x3 = 3/4] Solve by Two-phase method Max Z = 3x1 - x2 Subject to 2x1 + x2 ≥ 2 x1 + 3x2 ≤ 2 x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 6, x1 = 2, x2 =0] Max Z = 5x1 - 2x2 +3x3 Subject to 2x1 + 2x2 - x3 ≥ 2 3x1 - 4x2 ≤ 3 x2 + 3x3 ≤ 5 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 [Ans. Max Z = 85/3, x1 = 23/3, x2 =5, x3 =0] Unit 3 3.1 Special cases in Simplex Method 3.1.1 Degenaracy 3.1.2 Non-existing Feasible Solution 3.1.3 Unbounded Solution 3.1.4 Multiple Optimal Solutions 3.1.1 Degeneracy The concept of obtaining a degenerate basic feasible solution in a LPP is known as degeneracy. The degeneracy in a LPP may arise At the initial stage when at least one basic variable is zero in the initial basic feasible solution. At any subsequent iteration when more than one basic variable is eligible to leave the basic and hence one or more variables becoming zero in the next iteration and the problem is said to degenerate. There is no assurance that the value of the objective function will improve, since the new solutions may remain degenerate. As a result, it is possible to repeat the same sequence of simplex iterations endlessly without improving the solutions. This concept is known as cycling or circling. Rules to avoid cycling Divide each element in the tied rows by the positive coefficients of the key column in that row. Compare the resulting ratios, column by column, first in the identity and then in the body, from left to right. The row which first contains the smallest algebraic ratio contains the leaving variable. Example 1 Max Z = 3x1 + 9x2 Subject to x1 + 4x2 ≤ 8 x1 + 2x2 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 3x1 + 9x2 + 0s1 + 0s2 Subject to x1 + 4x2 + s1 = 8 x1 + 2x2 + s2 = 4 x1 , x2 , s1, s2 ≥ 0 Cj→ 3 9 0 0 Basic Variables CB XB X1 X2 S1 S2 XB / XK S1 / X2 s1 0 8 1 4 1 0 1/4 s2 0 4 1 2 0 1 0/2→ Z = 0 -3 ↑ -9 0 0 ←Δj s1 0 0 -1 0 1 -1 x2 9 2 1/2 1 0 1/2 Z =18 3/2 0 0 9/2 Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 18, x1 = 0, x2 = 2 Note – Since a tie in minimum ratio (degeneracy), we find minimum of s1 /xk for these rows for which the tie exists. Example 2 Max Z = 2x1 + x2 Subject to 4x1 + 3x2 ≤ 12 4x1 + x2 ≤ 8 4x1 - x2 ≤ 8 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 2x1 + x2 + 0s1 + 0s2 + 0s3 Subject to 4x1 + 3x2 + s1 = 12 4x1 + x2 + s2 = 8 4x1 - x2 + s3 = 8 x1 , x2 , s1, s2, s3 ≥ 0 Cj→ 2 1 0 0 0 Basic Varibles CB XB X1 X2 S1 S2 S3 XB / XK S1 / X1 S2 / X1 s1 0 12 4 3 1 0 0 12/4=3 s2 0 8 4 1 0 1 0 8/4=2 4/0=0 1/4 s3 0 8 4 -1 0 0 1 8/4=2 4/0=0 0/4=0→ Z = 0 ↑ -2 -1 0 0 0 ←Δj s1 0 4 0 4 1 0 -1 4/4=1 s2 0 0 0 2 0 1 -1 0→ x1 2 2 1 -1/4 0 0 1/4 - Z = 4 0 ↑ -3/2 0 0 1/2 ←Δj s1 0 4 0 0 1 -2 1 0→ x2 1 0 0 1 0 1/2 -1/2 - x1 2 2 1 0 0 1/8 1/8 16 Z = 4 0 0 0 3/4 ↑ -1/4 ←Δj s3 0 4 0 0 1 -2 1 x2 1 2 0 1 1/2 -1/2 0 x1 2 3/2 1 0 -1/8 3/8 0 Z = 5 0 0 1/4 1/4 0 ←Δj Since all Δj ≥ 0, optimal basic feasible solution is obtained. Therefore the solution is Max Z = 5, x1 = 3/2, x2 = 2 3.1.2 Non-existing Feasible Solution The feasible region is found to be empty which indicates that the problem has no feasible solution. Example 1 Max Z = 3x1 +2x2 Subject to 2x1 + x2 ≤ 2 3x1 + 4x2 ≥ 12 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 3x1 +2 x2 + 0s1 + 0s2 – Ma1 Subject to 2x1 + x2 + s1 = 2 3x1 + 4x2 - s2 + a1 = 12 x1 , x2 , s1, s2, s3 ≥ 0 Cj→ 3 2 0 0 -M Basic Variables CB XB X1 X2 S1 S2 A1 Min Ratio XB / XK s1 0 2 2 1 1 0 0 2/1=2→ a1 -M 12 3 4 0 -1 1 12/4=3 Z= -12M -3M-3 ↑ -4M-2 0 M 0 ←Δj x2 2 2 2 1 1 0 0 a1 -M 4 -5 0 -4 -1 1 Z= 4-4M 1+5M 0 2+4M M 0 Δj ≥ 0 so according to optimality condition the solution is optimal but the solution is called pseudo optimal solution since it does not satisfy all the constraints but satisfies the optimality condition. The artificial variable has a positive value which indicates there is no feasible solution. Example 2 Min Z = x1 –2x2– 3x3 Subject to –2x1 + x2 + 3x3= 2 2x1 + 3x2 + 4x3= 1 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution Standard LPP Min Z = Max Z' = –x1 +2x2+ 3x3 Subject to –2x1 + x2 + 3x3 + a1 = 2 2x1 + 3x2 + 4x3+ a2 = 1 x1 , x2 , a1, a2 ≥ 0 Auxiliary LPP Max Z* = 0x1 + 0x2 + 0x3 –1a1 –1a2 Subject to –2x1 + x2 + 3x3 + a1 = 2 2x1 + 3x2 + 4x3+ a2 = 1 x1 , x2 , a1, a2 ≥ 0 Phase I Cj→ 0 0 0 -1 -1 Basic Variables CB XB X1 X2 X3 A1 A2 Min Ratio XB / XK a1 -1 2 -2 1 3 1 0 2/3 a2 -1 1 2 3 4 0 1 1/4→ Z* = -3 0 -4 ↑ -7 0 0 ←Δj a1 -1 5/4 -7/4 -5/4 0 1 X x3 0 1/4 1/2 3/4 1 0 X Z* = -5/4 7/4 5/4 0 1 X ←Δj Since for all Δj ≥ 0, optimum level is achieved. At the end of phase-I Max Z* < 0 and one of the artificial variable a1 appears at the positive optimum level. Hence the given problem does not posses any feasible solution. 3.1.3 Unbounded Solution In some cases if the value of a variable is increased indefinitely, the constraints are not violated. This indicates that the feasible region is unbounded at least in one direction. Therefore, the objective function value can be increased indefinitely. This means that the problem has been poorly formulated or conceived. In simplex method, this can be noticed if Δj value is negative to a variable (entering) which is notified as key column and the ratio of solution value to key column value is either negative or infinity (both are to be ignored) to all the variables. This indicates that no variable is ready to leave the basis, though a variable is ready to enter. We cannot proceed further and the solution is unbounded or not finite. Example 1 Max Z = 6x1 - 2x2 Subject to 2x1 - x2 ≤ 2 x1 ≤ 4 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 6x1 - 2x2 + 0s1 + 0s2 Subject to 2x1 - x2 + s1 = 2 x1 + s2 = 4 x1 , x2 , s1, s2 ≥ 0 Cj→ 6 -2 0 0 Basic Variables CB XB X1 X2 S1 S2 Min Ratio XB / XK s1 0 2 2 -1 1 0 1→ s2 0 4 1 0 0 1 4 Z = 0 ↑ -6 2 0 0 ←Δj x1 6 1 1 -1/2 1/2 0 - s2 0 3 0 1/2 -1/2 1 6→ Z = 6 0 ↑ -1 3 0 ←Δj x1 6 4 1 0 0 1 x2 -2 6 0 1 -1 2 Z = 12 0 0 2 2 ←Δj The optimal solution is x1 = 4, x2 = 6 and Z =12 In the starting table, the elements of x2 are negative and zero. This is an indication that the feasible region is not bounded. From this we conclude the problem has unbounded feasible region but still the optimal solution is bounded. Example 2 Max Z = -3x1 + 2x2 Subject to x1 ≤ 3 x1 - x2 ≤ 0 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = -3x1 + 2 x2 + 0s1 + 0s2 Subject to x1 + s1 = 3 x1 - x2 + s2 = 0 x1 , x2 , s1, s2 ≥ 0 Cj→ -3 2 0 0 Basic Variables CB XB X1 X2 S1 S2 Min Ratio XB / XK s1 0 3 1 0 1 0 s2 0 0 1 -1 0 1 Z = 0 3 ↑ -2 0 0 ←Δj Corresponding to the incoming vector (column x2), all elements are negative or zero. So x2 cannot enter the basis and the outgoing vector cannot be found. This is an indication that there exists unbounded solution for the given problem. Example 3 Max Z = 107x1 + x2 +2x3 Subject to 14/3x1 + 1/3x2 - 2x3 ≤ 7/3 16x1 + 1/2x2 - 6x3 ≤ 5 3x1 - x2 - x3 ≤ 0 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution Standard LPP Max Z = 107x1 + x2 +2x3 Subject to 14/3x1 + 1/3x2 - 2x3 + s1 = 7/3 16x1 + 1/2x2 - 6x3 + s2 = 5 3x1 - x2 - x3 + s3 = 0 x1 , x2 , s1, s2, s3 ≥ 0 Cj→ 107 1 2 0 0 0 Basic Variables CB XB X1 X2 X3 S1 S2 S3 Min Ratio XB / XK s1 0 7/3 14/3 1/3 -2 1 0 0 0.5 s2 0 5 16 1/2 -6 0 1 0 0.8 s3 0 0 3 -1 -1 0 0 1 0→ Z = 0 ↑ -107 -1 -2 0 0 0 ←Δj s1 0 7/3 0 17/9 -4/9 1 0 -14/9 - s2 0 5 0 35/6 -2/3 0 1 -16/3 - x1 107 0 1 -1/3 -1/3 0 0 1/3 - Z = 0 0 -110/3 -113/3 0 0 107/3 ←Δj Corresponding to negative Δ3, all the elements of x3 column are negative. So x3 cannot enter into the basis matrix. This is an indication that there exists an unbounded solution to the given problem. 3.1.4 Multiple Optimal Solution When the objective function is parallel to one of the constraints, the multiple optimal solutions may exist. After reaching optimality, if at least one of the non-basic variables possess a zero value in Δj, the multiple optimal solution exist. Example Max Z = 6x1 + 4x2 Subject to 2x1 + 3x2 ≤ 30 3x1 + 2x2 ≤ 24 x1 + x2 ≥ 3 and x1 ≥ 0, x2 ≥ 0 Solution Standard LPP Max Z = 6x1 + 4x2 + 0s1 + 0s2 + 0s3 - Ma1 Subject to 2x1 + 3x2 + s1 = 30 3x1 + 2x2 + s2 = 24 x1 + x2 – s3 + a1= 3 x1 , x2 , s1, s2, s3, a1 ≥ 0 Cj→ 6 4 0 0 0 -M Basic Variables CB XB X1 X2 S1 S2 S3 A1 Min Ratio XB / XK s1 0 30 2 3 1 0 0 0 15 s2 0 24 3 2 0 1 0 0 8 a1 -M 3 1 1 0 0 -1 1 3→ Z = -3M ↑ -M-6 -M-4 0 0 M 0 ←Δj s1 0 24 0 1 1 0 2 X 12 s2 0 15 0 -1 0 1 3 X 5→ x1 6 3 1 1 0 0 -1 X - Z = 18 0 2 0 0 ↑ -6 X ←Δj s1 0 14 0 5/3 1 -2/3 0 X 42/5→ s3 0 5 0 -1/3 0 1/3 1 X - x1 6 8 1 2/3 0 1/3 0 X 12 Z = 48 0 ↑ 0 0 2 0 X ←Δj Since all Δj ≥ 0, optimum solution is obtained as x1 = 8, x2 = 0, Max Z = 48 Since Δ2 corresponding to non-basic variable x2 is obtained zero, this indicates that alternate solution or multiple optimal solution also exist. Therefore the solution as obtained above is not unique. Thus we can bring x2 into the basis in place of s1. The new optimum simplex table is obtained as follows Cj→ 6 4 0 0 0 -M Basic Variables CB XB X1 X2 S1 S2 S3 A1 Min Ratio XB / XK x2 4 42/5 0 1 3/5 -2/5 0 X s3 0 39/5 0 0 1/5 1/5 1 X x1 6 12/5 1 0 -2/5 3/5 0 X Z = 48 0 0 0 2 0 X ←Δj Exercise Solve Max Z = 3x1 + 2.5x2 Subject to 2x1 + 4x2 ≥ 40 3x1 + 2x2 ≥ 50 and x1 ≥ 0, x2 ≥ 0 [Ans. Unbounded solution] Max Z = 3x1 + 2x2 Subject to 2x1 + x2 ≤ 2 3x1 + 4x2 ≥ 12 and x1 ≥ 0, x2 ≥ 0 [Ans. Pseudo-optimum solution] Min Z = x1 - 2x2 - 3x3 Subject to -2x1 + x2 + 3x3 = 2 2x1 + 3x2+ 4x3 = 1 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 [Ans. No feasible solution] Max Z = 3x1 + 2x2 + x3 + 4x4 Subject to 4x1 + 5x2 + x3 - 3x4 = 5 2x1- 3x2 - 4x3 + 5x4 = 7 x1 + 4x2 + 2.5x3 - 4x4 = 6 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 [Ans. No solution] Max Z = 3x1 + 9x2 Subject to 4x1 + 4x2 ≥ 8 x1 + 2x2 ≥ 4 and x1 ≥ 0, x2 ≥ 0 [Degeneracy exists] In the course of simplex table calculations, describe how u will detect a Degenerate An unbounded Non-existing feasible solution What is degeneracy? Write the role of pivot element in a simplex table. Module 3 Unit 1 The Revised Simplex Method Steps for solving Revised Simplex Method in Standard Form-I Worked Examples 1.1 The Revised Simplex Method While solving linear programming problem on a digital computer by regular simplex method, it requires storing the entire simplex table in the memory of the computer table, which may not be feasible for very large problem. But it is necessary to calculate each table during each iteration. The revised simplex method which is a modification of the original method is more economical on the computer, as it computes and stores only the relevant information needed currently for testing and / or improving the current solution. i.e. it needs only The net evaluation row Δj to determine the non-basic variable that enters the basis. The pivot column The current basis variables and their values (XB column) to determine the minimum positive ratio and then identify the basis variable to leave the basis. The above information is directly obtained from the original equations by making use of the inverse of the current basis matrix at any iteration. There are two standard forms for revised simplex method Standard form-I – In this form, it is assumed that an identity matrix is obtained after introducing slack variables only. Standard form-II – If artificial variables are needed for an identity matrix, then two-phase method of ordinary simplex method is used in a slightly different way to handle artificial variables. 1.2 Steps for solving Revised Simplex Method in Standard Form-I Solve by Revised simplex method Max Z = 2x1 + x2 Subject to 3 x1 + 4 x2 ≤ 6 6 x1 + x2 ≤ 3 and x1, x2 ≥ 0 SLPP Max Z = 2x1 + x2+ 0s1+ 0s2 Subject to 3 x1 + 4 x2 + s1 = 6 6 x1 + x2 + s2 = 3 and x1, x2, s1, s2 ≥ 0 Step 1 – Express the given problem in standard form – I Ensure all bi ≥ 0 The objective function should be of maximization Use of non-negative slack variables to convert inequalities to equations The objective function is also treated as first constraint equation Z - 2x1 - x2 + 0s1 + 0s2 = 0 3 x1 + 4 x2 + s1 + 0s2= 6 -- (1) 6 x1 + x2 + 0s1 + s2= 3 and x1, x2, s1, s2 ≥ 0 Step 2 – Construct the starting table in the revised simplex form Express (1) in the matrix form with suitable notation Column vector corresponding to Z is usually denoted by e1. It is the first column of the basis matrix B1, which is usually denoted as B1 = [β0(1), β1(1), β2(1) … βn(1)] Hence the column β0(1), β1(1), β2(1) constitutes the basis matrix B1 (whose inverse B1-1 is also B1) Basic variables B1-1 XB Xk XB / Xk a1 (1) a2 (1) e1 (Z) β1(1) β2(1) Z 1 0 0 0 -2 -1 s1 0 1 0 6 3 4 s2 0 0 1 3 6 1 Step 3 – Computation of Δj for a1 (1) and a2 (1) Δ1 = first row of B1-1 * a1 (1) = 1 * -2 + 0 * 3 + 0 *6 = -2 Δ2 = first row of B1-1 * a2 (1) = 1 * -1 + 0 * 4 + 0 *1 = -1 Step 4 – Apply the test of optimality Both Δ1 and Δ2 are negative. So find the most negative value and determine the incoming vector. Therefore most negative value is Δ1 = -2. This indicates a1 (1) (x1) is incoming vector. Step 5 – Compute the column vector Xk Xk = B1-1 * a1 (1) Stp 6 – Determine the outgoing vector. We are not supposed to calculate for Z row. Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 0 0 0 -2 - s1 0 1 0 6 3 2 s2 0 0 1 3 6 ↑ incoming 1/2→outgoing Step 7 – Determination of improved solution Column e1 will never change, x1 is incoming so place it outside the rectangular boundary β1(1) β2(1) XB X1 R1 0 0 0 -2 R2 1 0 6 3 R3 0 1 3 6 Make the pivot element as 1 and the respective column elements to zero. β1(1) β2(1) XB X1 R1 0 1/3 1 0 R2 1 -1/2 9/2 0 R3 0 1/6 1/2 1 Construct the table to start with second iteration Basic variables B1-1 XB Xk XB / Xk a4 (1) a2 (1) e1 (Z) β1(1) β2(1) Z 1 0 1/3 1 0 -1 s1 0 1 -1/2 9/2 0 4 x1 0 0 1/6 1/2 1 1 Δ4 = 1 * 0 + 0 * 0 + 1/3 *1 = 1/3 Δ2 = 1 * -1 + 0 * 4 + 1/3 *1 = -2/3 Δ2 is most negative. Therefore a2 (1) is incoming vector. Compute the column vector Determine the outgoing vector Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 0 1/3 1 -2/3 - s1 0 1 -1/2 9/2 7/2 9/7→outgoing x1 0 0 1/6 1/2 1/6 ↑ incoming 3 Determination of improved solution β1(1) β2(1) XB X2 R1 0 1/3 1 -2/3 R2 1 -1/2 9/2 7/2 R3 0 1/6 1/2 1/6 β1(1) β2(1) XB X2 R1 4/21 5/21 13/7 0 R2 2/7 -1/7 9/7 1 R3 -1/21 8/42 2/7 0 Basic variables B1-1 XB Xk XB / Xk a4 (1) a3 (1) e1 (Z) β1(1) β2(1) Z 1 4/21 5/21 13/7 0 0 x2 0 2/7 -1/7 9/7 0 1 x1 0 -1/21 8/42 2/7 1 0 Δ4 = 1 * 0 + 4/21 * 0 + 5/21 *1 = 5/21 Δ3 = 1 * 0 + 4/21 * 1 + 5/21 *0 = 4/21 Δ4 and Δ3 are positive. Therefore optimal solution is Max Z = 13/7, x1= 2/7, x2 = 9/7 1.3 Worked Examples Example 1 Max Z = x1 + 2x2 Subject to x1 + x2 ≤ 3 x1 + 2x2 ≤ 5 3x1 + x2 ≤ 6 and x1, x2 ≥ 0 Solution SLPP Max Z = x1 + 2x2+ 0s1+ 0s2+ 0s3 Subject to x1 + x2 + s1 = 3 x1 + 2x2 + s2 = 5 3x1 + x2 + s3 = 6 and x1, x2, s1, s2, s3 ≥ 0 Standard Form-I Z - x1 - 2x2 - 0s1 - 0s2 - 0s3= 0 x1 + x2 + s1 + 0s2 + 0s3= 3 x1 + 2x2 + 0s1 + s2 + 0s3 = 5 3x1 + x2 + 0s1 + 0s2 + s3 = 6 and x1, x2, s1, s2 , s3 ≥ 0 Matrix form Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a2 (1) e1 (Z) β1(1) β2(1) β3(1) Z 1 0 0 0 0 -1 -2 s1 0 1 0 0 3 1 1 s2 0 0 1 0 5 1 2 s3 0 0 0 1 6 3 1 Computation of Δj for a1 (1) and a2 (1) Δ1 = first row of B1-1 * a1 (1) = 1 * -1 + 0 * 1 + 0 *1 + 0 *3= -1 Δ2 = first row of B1-1 * a2 (1) = 1 * -2 + 0 * 1 + 0 *2+ 0 *1 = -2 Δ2 = -2 is most negative. So a2 (1) (x2) is incoming vector. Compute the column vector Xk Xk = B1-1 * a2 (1) Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) β3(1) Z 1 0 0 0 0 -2 - s1 0 1 0 0 3 1 3 s2 0 0 1 0 5 2 5/2→ s3 0 0 0 1 6 1 ↑ 6 Improved Solution β1(1) β2(1) β3(1) XB Xk R1 0 0 0 0 -2 R2 1 0 0 3 1 R3 0 1 0 5 2 R4 0 0 1 6 1 β1(1) β2(1) β3(1) XB Xk R1 0 1 0 5 0 R2 1 -1/2 0 1/2 0 R3 0 1/2 0 5/2 1 R4 0 -1/2 1 7/2 0 Revised simplex table for II iteration Basic variables B1-1 XB Xk XB / Xk a1 (1) a4 (1) e1 (Z) β1(1) β2(1) β3(1) Z 1 0 1 0 5 -1 0 s1 0 1 -1/2 0 1/2 1 0 x2 0 0 1/2 0 5/2 1 1 s3 0 0 -1/2 1 7/2 3 0 Δ1 = 1 * -1 + 0 * 1 + 1 *1 + 0 *3= 0 Δ4 = 1 * 0 + 0 * 0 + 1 *1+ 0 *0 = 1 Δ1 and Δ4 are positive. Therefore optimal solution is Max Z = 5, x1= 0, x2 = 5/2 Example 2 Max Z = 80x1 + 55x2 Subject to 4x1 +2x2 ≤ 40 2x1 + 4x2 ≤ 32 and x1, x2 ≥ 0 Solution Max Z = 80x1 + 55x2 Subject to 2x1 +x2 ≤ 20 (divide by 2) x1 + 2x2 ≤ 16 (divide by 2) and x1, x2 ≥ 0 SLPP Max Z = 80x1 + 55x2+ 0s1+ 0s2 Subject to 2x1 +x2+ s1 = 20 x1 + 2x2 + s2 = 16 and x1, x2, s1, s2 ≥ 0 Standard form-I Z - 80x1 - 55x2 - 0s1 - 0s2 = 0 2x1 +x2+ s1 + 0s2= 20 x1 + 2x2 + 0s1 + s2 = 16 and x1, x2, s1, s2 ≥ 0 Matrix form Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a2 (1) e1 (Z) β1(1) β2(1) Z 1 0 0 0 -80 -55 s1 0 1 0 20 2 1 s2 0 0 1 16 1 2 Computation of Δj for a1 (1) and a2 (1) Δ1 = first row of B1-1 * a1 (1) = 1 * -80 + 0 * 2 + 0 *1 = -80 Δ2 = first row of B1-1 * a2 (1) = 1 * -55 + 0 * 1 + 0 *2 = -55 Δ1 = -80 is most negative. So a1 (1), (x1) is incoming vector. Compute the column vector Xk Xk = B1-1 * a1 (1) Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 0 0 0 -80 - s1 0 1 0 20 2 10→ s2 0 0 1 16 1 ↑ 16 Improved solution β1(1) β2(1) XB Xk R1 0 0 0 -80 R2 1 0 20 2 R3 0 1 16 1 β1(1) β2(1) XB Xk R1 40 0 800 0 R2 1/2 0 10 1 R3 -1/2 1 6 0 Revised simplex table for II iteration Basic variables B1-1 XB Xk XB / Xk a3 (1) a2 (1) e1 (Z) β1(1) β2(1) Z 1 40 0 800 0 -55 x1 0 1/2 0 10 1 1 s2 0 -1/2 1 6 0 2 Computation of Δj for a3 (1) and a2 (1) Δ3 = first row of B1-1 * a3 (1) = 1 * 0 + 40 * 1 + 0 *0 = 40 Δ2 = first row of B1-1 * a2 (1) = 1 * -55 + 40 * 1 + 0 *2 = -15 Δ2 = -15 is most negative. So a2 (1) (x2) is incoming vector. Compute the column vector Xk Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 40 0 800 -15 - x1 0 1/2 0 10 1/2 20 s2 0 -1/2 1 6 3/2 ↑ 4→ Improved solution β1(1) β2(1) XB Xk R1 40 0 800 -15 R2 1/2 0 10 1/2 R3 -1/2 1 6 3/2 β1(1) β2(1) XB Xk R1 35 10 860 0 R2 2/3 -1/3 8 0 R3 -1/3 2/3 4 1 Revised simplex table for III iteration Basic variables B1-1 XB Xk XB / Xk a3 (1) a4 (1) e1 (Z) β1(1) β2(1) Z 1 35 10 860 0 0 x1 0 2/3 -1/3 8 1 0 x2 0 -1/3 2/3 4 0 1 Computation of Δ3 and Δ4 Δ3 = 1 * 0 + 35 * 1 + 10 *0 = 35 Δ4 = 1 * 0 + 35 * 0 + 10 *1 = 10 Δ3 and Δ4 are positive. Therefore optimal solution is Max Z = 860, x1= 8, x2 = 4 Example 3 Max Z = x1 + x2+ x3 Subject to 4x1 + 5x2 + 3x3≤ 15 10x1 + 7x2+ x3 ≤ 12 and x1, x2, x3 ≥ 0 Solution SLPP Max Z = x1 + x2+ x3+ 0s1+ 0s2 Subject to 4x1 + 5x2 + 3x3+ s1 = 15 10x1 + 7x2+ x3 + s2 = 12 and x1, x2, x3, s1, s2 ≥ 0 Standard form-I Z - x1 - x2 - x3 - 0s1 - 0s2 = 0 4x1 +5x2 + 3x3+ s1 + 0s2= 15 10x1 + 7x2+ x3 + 0s1+ s2 = 12 and x1, x2, x3, s1, s2 ≥ 0 Matrix form Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a2 (1) a3 (1) e1 (Z) β1(1) β2(1) Z 1 0 0 0 -1 -1 -1 s1 0 1 0 15 4 5 3 s2 0 0 1 12 10 7 1 Computation of Δj for a1 (1), a2 (1) and a3 (1) Δ1 = first row of B1-1 * a1 (1) = 1 * -1 + 0 * 4 + 0 *10 = -1 Δ2 = first row of B1-1 * a2 (1) = 1 * -1 + 0 * 5 + 0 *7 = -1 Δ3 = first row of B1-1 * a3 (1) = 1 * -1 + 0 * 3 + 0 *1 = -1 There is a tie, so perform the computation of Δj with second row Δ1 = second row of B1-1 * a1 (1) = 0 * -1 + 1 * 4 + 0 *10 = 4 Δ2 = second row of B1-1 * a2 (1) = 0 * -1 + 1 * 5 + 0 *7 = 5 Δ3 = second row of B1-1 * a3 (1) = 0 * -1 + 1 * 3 + 0 *1 = 3 Since Δj ≥ 0, we obtain pure optimum solution where Max Z = 0, x1= 0, x2= 0 Example 4 Max Z = 5x1 + 8x2 + 7x3 + 4x4 + 6x5 Subject to 2x1 + 3x2 + 3x3+ 2x4 + 2x5≤ 20 3x1 + 5x2+ 4x3 + 2x4 + 4x5≤ 30 and x1, x2, x3, x4, x5 ≥ 0 Solution SLPP Max Z = 5x1 + 8x2 + 7x3 + 4x4 + 6x5+ 0s1+ 0s2 Subject to 2x1 + 3x2 + 3x3+ 2x4 + 2x5+ s1 = 20 3x1 + 5x2+ 4x3 + 2x4 + 4x5+ s2 = 30 and x1, x2, x3, x4, x5, s1, s2 ≥ 0 Standard form-I Z - 5x1 - 8x2 - 7x3 - 4x4 - 6x5 - 0s1 - 0s2 = 0 2x1 + 3x2 + 3x3+ 2x4 + 2x5+ s1 + 0s2= 20 3x1 + 5x2+ 4x3 + 2x4 + 4x5+ 0s1+ s2 = 30 and x1, x2, x3, x4, x5, s1, s2 ≥ 0 Matrix form Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a2 (1) a3 (1) a4 (1) a5 (1) e1 (Z) β1(1) β2(1) Z 1 0 0 0 -5 -8 -7 -4 -6 s1 0 1 0 20 2 3 3 2 2 s2 0 0 1 30 3 5 4 2 4 Computation of Δj for a1 (1) , a2 (1) , a3 (1) , a4 (1) , a5 (1) Δ1 = first row of B1-1 * a1 (1) = 1 * -5 + 0 * 2 + 0 *3 = -5 Δ2 = first row of B1-1 * a2 (1) = 1 * -8 + 0 * 3 + 0 *5 = -8 Δ3 = first row of B1-1 * a3 (1) = 1 * -7 + 0 * 3 + 0 *4 = -7 Δ4 = first row of B1-1 * a4 (1) = 1 * -4 + 0 * 2 + 0 *2 = -4 Δ5 = first row of B1-1 * a5 (1) = 1 * -6 + 0 * 2 + 0 *4 = -6 Δ2 = -8 is most negative. So a2 (1), (x2) is incoming vector. Compute the column vector Xk Xk = B1-1 * a2 (1) Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 0 0 0 -8 - s1 0 1 0 20 3 20/3 s2 0 0 1 30 5 ↑ 6→ Improved solution β1(1) β2(1) XB Xk R1 0 0 0 -8 R2 1 0 20 3 R3 0 1 30 5 β1(1) β2(1) XB Xk R1 0 8/5 48 0 R2 1 -3/5 2 0 R3 0 1/5 6 1 Revised simplex table for II iteration Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a7 (1) a3 (1) a4 (1) a5 (1) e1 (Z) β1(1) β2(1) Z 1 0 8/5 48 -5 0 -7 -4 -6 s1 0 1 -3/5 2 2 0 3 2 2 x2 0 0 1/5 6 3 1 4 2 4 Computation of Δj for a1 (1) , a2 (1) , a3 (1) , a4 (1) , a5 (1) Δ1 = -1/5, Δ7 = 8/5, Δ3 = -3/5, Δ4 = -4/5, Δ5 = 2/5 Δ4 = -4/5 is most negative. So a4(1), (x4) is incoming vector. Compute the column vector Xk Xk = B1-1 * a4 (1) Basic variables B1-1 XB Xk XB / Xk e1 (Z) β1(1) β2(1) Z 1 0 8/5 48 -4/5 - s1 0 1 -3/5 2 4/5 10/4→ x2 0 0 1/5 6 2/5 ↑ 15 Improved solution β1(1) β2(1) XB Xk R1 0 8/5 48 -4/5 R2 1 -3/5 2 4/5 R3 0 1/5 6 2/5 β1(1) β2(1) XB Xk R1 1 1 50 0 R2 5/4 -3/4 5/2 1 R3 -1/2 1/2 5 0 Revised simplex table for III iteration Revised simplex table Additional table Basic variables B1-1 XB Xk XB / Xk a1 (1) a7 (1) a3 (1) a6 (1) a5 (1) e1 (Z) β1(1) β2(1) Z 1 1 1 50 -5 0 -7 0 -6 x4 0 5/4 -3/4 5/2 2 0 3 1 2 x2 0 -1/2 1/2 5 3 1 4 0 4 Computation of Δj for a1 (1) , a2 (1) , a3 (1) , a4 (1) , a5 (1) Δ1 = 0, Δ7 = 1, Δ3 = 0, Δ6 = 1, Δ5 = 0 Δj ≥ 0, Therefore optimal solution is Max Z = 50, x1= 0, x2 = 5, x3= 0, x4 = 5/2, x5= 0 Exercise Solve by Revised Simplex method Max Z = x1 + x2 Subject to 3x1 + 3x2 ≤ 6 x1 + 4x2 ≤ 4 x1, x2 ≥ 0 [Ans. Max Z = 11/5, x1 = 8/5, x2 = 3/5] Max Z = x1 + 2x2 Subject to x1 + 2x2 ≤ 3 x1 + 3x2 ≤ 1 x1, x2 ≥ 0 [Ans. Max Z = 1, x1 = 1, x2 = 0] Max Z = 5x1 + 3x2 Subject to 3x1 + 5x2 ≤ 15 3x1 + 2x2 ≤ 10 x1, x2 ≥ 0 [Ans. Max Z = 285/19, x1 = 22/19, x2 = 45/19] Max Z = x1 + x2 Subject to x1 + 2x2 ≤ 2 4x1 + x2 ≤ 4 x1, x2 ≥ 0 [Ans. Max Z = 10/7, x1 = 6/7, x2 = 4/7] Max Z = 3x1 + x2+ 2x3+ 7x4 Subject to 2x1 + 3x2 - x3+ 4x4 ≤ 40 -2x1 + 2x2 + 5x3 - x4 ≤ 35 x1 + x2 - 2x3+ 3x4 ≤ 100 x1, x2, x3, x4 ≥ 0 [Ans. Max Z = 445/4, x1 = 71/4, x2 = 1, x3 = 29/2, x4 = 4] Unit 2 2.1 Computational Procedure of Revised Simplex Table in Standard Form-II 2.2 Worked Examples 2.3 Advantages and Disadvantages 2.1 Computational Procedure of Revised Simplex Table in Standard Form-II Phase I – When the artificial variables are present in the initial solution with positive values Step 1 – First construct the simplex table in the following form Variables in the basis e1 e2 β1(2) β2(2) … βm(2) XB(2) Xk(2) x0 1 0 0 0 … 0 x'n +1 0 1 0 0 … 0 xn +1 0 0 1 0 … 0 xn +2 0 0 0 1 … 0 . . . . . . . . . . . . xn +m 0 0 0 0 … 1 Step 2 – If x'n +1 < 0, compute Δj = second row of B2-1 * aj(2) and continue to step 3. If max x'n +1 = 0 then go to phase II. Step 3 – To find the vector to be introduced into the basis If Δj ≥ 0, x'n +1 is at its maximum and hence no feasible solution exists for the problem If at least one Δj < 0, the vector to be introduced in the basis, Xk(2), corresponds to such value of k which is obtained by Δk = min Δj If more than one value of Δj are equal to the maximum, we select Δk such that k is the smallest index. Step 4 – To compute Xk(2) by using the formula Xk(2) = B2-1 ak(2) Step 5 – To find the vector to be removed from the basis. The vector to be removed from the basis is obtained by using the minimum ratio rule. Step 6 – After determining the incoming and outgoing vector, next revised simplex table can be easily obtained Repeat the procedure of phase I to get max x'n +1 = 0 or all Δj for phase I are ≥ 0. If max x'n +1 comes out of zero in phase I, all artificial variables must have the value zero. It should be noted carefully that max x'n +1 will always come out to be zero at the end of phase I if the feasible solution to the problem exists. Proceed to phase II Phase II - x'n +1 is considered like any other artificial variable; it can be removed from the basic solution. Only x0 must always remain in the basic solution. However there will always be at least one artificial vector in B2, otherwise it is not possible to have an m+2 dimensional bases. The procedure in phase II will be the same as described in standard form-I 2.1 Worked Examples Solve by revised simplex method Example 1 Min Z = x1 + 2x2 Subject to 2x1 + 5x2 ≥ 6 x1 + x2 ≥ 2 and x1, x2 ≥ 0 Solution SLPP Min Z = Max Z' = -x1 - 2x2+ 0s1+ 0s2 Subject to 2x1 + 5x2 - s1 + a1= 6 x1 + x2 - s2 + a2 = 2 and x1, x2, s1, s2 ≥ 0 Standard form-II Z' + x1 + 2x2 = 0 -3x1 - 6x2 + s1 + s2 + av = -8 where av = - (a1 + a2) 2x1 + 5x2 - s1 + a1= 6 x1 + x2 – s2 + a2 = 2 and x1, x2, s1, s2 ≥ 0 The second constraint equation is formed by taking the negative sum of two constraints. Matrix form Phase -I I Iteration Basic variables B2-1 a1(2) a2(2) a3(2) a4(2) e1 e2 β1(2) β2(2) XB Xk XB/Xk e1 1 0 0 0 0 1 2 0 0 av 0 1 0 0 -8 -3 -6 1 1 a1 0 0 1 0 6 2 5 -1 0 a2 0 0 0 1 2 1 1 0 -1 Calculation of Δj Δ1 = second row of B2-1 * a1(2) = -3 Δ2 = second row of B2-1 * a2(2) = -6 Δ3 = second row of B2-1 * a3(2) = 1 Δ4 = second row of B2-1 * a4(2) = 1 Δ2 is most negative. Therefore a2(2) (x2) is incoming vector Compute the column vector Xk Xk = B2-1 * a2 (2) Basic variables B2-1 e1 e2 β1(2) β2(2) XB Xk XB/Xk e1 1 0 0 0 0 2 av 0 1 0 0 -8 -6 a1 0 0 1 0 6 5 6/5→ a2 0 0 0 1 2 1 ↑ 2 Improved Solution β1(1) β2(1) XB Xk R1 0 0 0 2 R2 0 0 -8 -6 R3 1 0 6 5 R4 0 1 2 1 β1(1) β2(1) XB Xk R1 -2/5 0 -12/5 0 R2 6/5 0 -4/5 0 R3 1/5 0 6/5 1 R4 -1/5 1 4/5 0 II iteration Basic variables B2-1 a1(2) a5(2) a3(2) a4(2) e1 e2 β1(2) β2(2) XB Xk XB/Xk z' 1 0 -2/5 0 -12/5 1 0 0 0 av 0 1 6/5 0 -4/5 -3 0 1 1 x2 0 0 1/5 0 6/5 2 1 -1 0 a2 0 0 -1/5 1 4/5 1 0 0 -1 Calculation of Δj Δ1 = -3/5, Δ5 = 6/5, Δ3 = -1/5, Δ4 = 1 Δ1 is most negative. Therefore a1(2) (x1) is incoming vector Compute the column vector Xk Xk = B2-1 * a1 (2) Basic variables B2-1 e1 e2 β1(2) β2(2) XB Xk XB/Xk z' 1 0 -2/5 0 -12/5 1/5 av 0 1 6/5 0 -4/5 -3/5 x2 0 0 1/5 0 6/5 2/5 3 a2 0 0 -1/5 1 4/5 3/5 ↑ 4/3→ Improved Solution β1(1) β2(1) XB Xk R1 -2/5 0 -12/5 1/5 R2 6/5 0 -4/5 -3/5 R3 1/5 0 6/5 2/5 R4 -1/5 1 4/5 3/5 β1(1) β2(1) XB Xk R1 -1/3 -1/3 -8/3 0 R2 1 1 0 0 R3 1/3 -2/3 2/3 0 R4 -1/3 5/3 4/3 1 III iteration Basic variables B2-1 a6(2) a5(2) a3(2) a4(2) e1 e2 β1(2) β2(2) XB Xk XB/Xk z' 1 0 -1/3 -1/3 -8/3 0 0 0 0 av 0 1 1 1 0 0 0 1 1 x2 0 0 1/3 -2/3 2/3 0 1 -1 0 x1 0 0 -1/3 5/3 4/3 1 0 0 -1 Since av =0 in XB column. We proceed to phase II Phase II Basic variables B2-1 a3(2) a4(2) e1 e2 β1(2) β2(2) XB Xk XB/Xk z' 1 0 -1/3 -1/3 -8/3 0 0 av 0 1 1 1 0 1 1 x2 0 0 1/3 -2/3 2/3 -1 0 x1 0 0 -1/3 5/3 4/3 0 -1 Δ3 = first row of B2-1 * a3(2) = 1/3 Δ4 = first row of B2-1 * a4(2) = 1/3 Δ3 and Δ4 are positive. Therefore optimal solution is Z' = -8/3→ Z =8/3, x1= 4/3, x2 = 2/3 Example 2 Max Z = x1 + 2x2 + 3x3 - x4 Subject to x1 + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 x1 + 2x2 + x3 + x4 = 10 and x1, x2, x3 ≥ 0 Solution SLPP Max Z = x1 + 2x2 + 3x3 - x4 Subject to x1 + 2x2 + 3x3 + a1= 15 2x1 + x2 + 5x3 + a2 = 20 x1 + 2x2 + x3 + x4 + a3 = 10 and x1, x2, a1, a2 ≥ 0 Standard form-II Z - x1 - 2x2 - 3x3 + x4 = 0 -4x1 - 5x2 - 9x3 - x4 + av = -45 where av = - (a1 + a2+ a3) x1 + 2x2 + 3x3 + a1= 15 2x1 + x2 + 5x3 + a2 = 20 x1 + 2x2 + x3 + x4 + a3 = 10 x1, x2, a1, a2, a3 ≥ 0 Matrix form Phase I I Iteration Basic variables B2-1 a1(2) a2(2) a3(2) a4(2) e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk e1 1 0 0 0 0 0 -1 -2 -3 1 av 0 1 0 0 0 -45 -4 -5 -9 -1 a1 0 0 1 0 0 15 1 2 3 0 a2 0 0 0 1 0 20 2 1 5 0 a3 0 0 0 0 1 10 1 2 1 1 Calculation of Δj Δ1 = second row of B2-1 * a1(2) = -4 Δ2 = second row of B2-1 * a2(2) = -5 Δ3 = second row of B2-1 * a3(2) = -9 Δ4 = second row of B2-1 * a4(2) = -1 Δ3 is most negative. Therefore a3(2) (x3) is incoming vector Compute the column vector Xk Xk = B2-1 * a3 (2) Basic variables B2-1 e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk e1 1 0 0 0 0 0 -3 av 0 1 0 0 0 -45 -9 a1 0 0 1 0 0 15 3 5 a2 0 0 0 1 0 20 5 4→ a3 0 0 0 0 1 10 1 ↑ 10 Improved Solution β1(2) β2(2) β3(2) XB Xk R1 0 0 0 0 -3 R2 0 0 0 -45 -9 R3 1 0 0 15 3 R4 0 1 0 20 5 R5 0 0 1 10 1 β1(2) β2(2) β3(2) XB Xk R1 0 3/5 0 12 0 R2 0 9/5 0 -9 0 R3 1 -3/5 0 3 0 R4 0 1/5 0 4 1 R5 0 -1/5 1 6 0 II Iteration Basic variables B2-1 a1(2) a2(2) a6(2) a4(2) e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk z 1 0 0 3/5 0 12 -1 -2 0 1 av 0 1 0 9/5 0 -9 -4 -5 0 -1 a1 0 0 1 -3/5 0 3 1 2 0 0 x3 0 0 0 1/5 0 4 2 1 1 0 a3 0 0 0 -1/5 1 6 1 2 0 1 Calculation of Δj Δ1 = -2/5, Δ2 = -16/5, Δ6 = 9/5, Δ4 = -1 Δ4 is most negative. Therefore a4(2) (x4) is incoming vector Compute the column vector Xk Basic variables B2-1 e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk Z 1 0 0 3/5 0 12 1 av 0 1 0 9/5 0 -9 -1 a1 0 0 1 -3/5 0 3 0 x3 0 0 0 1/5 0 4 0 a3 0 0 0 -1/5 1 6 1 ↑ 6→ Improved Solution β1(2) β2(2) β3(2) XB Xk R1 0 3/5 0 12 1 R2 0 9/5 0 -9 -1 R3 1 -3/5 0 3 0 R4 0 1/5 0 4 0 R5 0 -1/5 1 6 1 β1(2) β2(2) β3(2) XB Xk R1 0 4/5 -1 6 0 R2 0 8/5 1 -3 0 R3 1 -3/5 0 3 0 R4 0 1/5 0 4 0 R5 0 -1/5 1 6 1 III Iteration Basic variables B2-1 a1(2) a2(2) a6(2) a7(2) e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk Z 1 0 0 4/5 -1 6 -1 -2 0 0 av 0 1 0 8/5 1 -3 -4 -5 0 0 a1 0 0 1 -3/5 0 3 1 2 0 0 x3 0 0 0 1/5 0 4 2 1 1 0 x4 0 0 0 -1/5 1 6 1 2 0 1 Calculation of Δj Δ1 = 1/5, Δ2 = -7/5, Δ6 = 8/5, Δ7 = 1 Δ2 is most negative. Therefore a2(2) (x2) is incoming vector Compute the column vector Xk Basic variables B2-1 e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk z 1 0 0 4/5 -1 6 -16/5 av 0 1 0 8/5 1 -3 -7/5 a1 0 0 1 -3/5 0 3 7/5 15/7→ x3 0 0 0 1/5 0 4 1/5 20 x4 0 0 0 -1/5 1 6 9/5 ↑ 30/9 Improved Solution β1(2) β2(2) β3(2) XB Xk R1 0 4/5 -1 6 -16/5 R2 0 8/5 1 -3 -7/5 R3 1 -3/5 0 3 7/5 R4 0 1/5 0 4 1/5 R5 0 -1/5 1 6 9/5 β1(2) β2(2) β3(2) XB Xk R1 16/7 4/7 -1 90/7 0 R2 1 1 1 0 0 R3 5/7 -3/7 0 15/7 1 R4 -1/7 2/7 0 25/7 0 R5 -9/7 4/7 1 15/7 0 IV Iteration Basic variables B2-1 a1(2) a5(2) a6(2) a7(2) e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk z 1 0 16/7 4/7 -1 90/7 -1 0 0 0 av 0 1 1 1 1 0 -4 0 0 0 x2 0 0 5/7 -3/7 0 15/7 1 1 0 0 x3 0 0 -1/7 2/7 0 25/7 2 0 1 0 x4 0 0 -9/7 4/7 1 15/7 1 0 0 1 Since av =0 in XB column. We proceed to phase II Phase II Basic variables B2-1 a1(2) e1 e2 β1(2) β2(2) β3(2) XB Xk XB/Xk z 1 0 16/7 4/7 -1 90/7 -1 av 0 1 1 1 1 0 -4 x2 0 0 5/7 -3/7 0 15/7 1 x3 0 0 -1/7 2/7 0 25/7 2 x4 0 0 -9/7 4/7 1 15/7 1 Δ1 = 0, Δ1 is positive. Therefore optimal solution is Z =90/7, x1= 0, x2 = 15/7, x3 = 25/7, x4 = 15/7 2.3 Advantages and Disadvantages Advantages The method automatically generates the inverse of the current basis matrix and the new basic feasible solution as well. It provides more information at lesser computational effort It requires lesser computations than the ordinary simplex method A less number of entries are needed in each table of revised simplex table The control of rounding-off-errors occurs when a digital computer is used Disadvantages In solving the numerical problems side computations are also requires, therefore more computational mistakes may occur in comparison to original simplex method. Exercise Solve by revised simplex method Max Z = 3x1 + 5x2 Subject to x1 ≤ 4 x2 ≤ 6 3x1 + 2x2 ≤ 18 x1, x2 ≥ 0 [Ans. Max Z = 36, x1 = 2, x2 = 6] Max Z = 5x1 + 3x2 Subject to 4x1 + 5x2 ≥ 10 5x1 + 2x2 ≤ 10 3x1 + 8x2 ≤ 12 x1, x2 ≥ 0 [Ans. Max Z = 185/17, x1 = 28/17, x2 = 15/17] Max Z = x1 + x2 + 3x3 Subject to 3x1 + 2x2 + x3 ≤ 3 2x1 + x2 + 2x3 ≤ 2 x1, x2, x3 ≥ 0 [Ans. Max Z = 3, x1 = 0, x2 = 0, x3 = 1] Min Z = 3x1 + x2 Subject to x1 + x2 ≥ 1 2x1 + x2 ≥ 0 x1, x2 ≥ 0 [Ans. Min Z = 1, x1 = 0, x2 = 1] Min Z = 4x1 + 2x2 + 3x3 Subject to 2x1 + 4x3 ≥ 5 2x1 + 4x2 + x3 ≥ 4 x1, x2, x3 ≥ 0 [Ans. Min Z = 67/12, x1 = 0, x2 = 11/12, x3 = 5/4] Unit 3 3.1 Duality in LPP 3.2 Important characteristics of Duality 3.3 Advantages and Applications of Duality 3.4 Steps for Standard Primal Form 3.5 Rules for Converting any Primal into its Dual 3.6 Example Problems 3.7 Primal-Dual Relationship 3.8 Duality and Simplex Method 3.1 Duality in LPP Every LPP called the primal is associated with another LPP called dual. Either of the problems is primal with the other one as dual. The optimal solution of either problem reveals the information about the optimal solution of the other. Let the primal problem be Max Zx = c1x1 + c2x2 + … +cnxn Subject to restrictions a11x1 + a12x2 + … + a1nxn ≤ b1 a21x1 + a22x2 + … + a2nxn ≤ b2 . . . am1x1 + am2x2 + … + amnxn ≤ bn and x1 ≥ 0, x2 ≥ 0,…, xn ≥ 0 The corresponding dual is defined as Min Zw = b1w1 + b2w2 + … + bmwm Subject to restrictions a11w1 + a21w2 + … + am1wm ≥ c1 a12w1 + a22w2 + … + am2wm ≥ c2 . . . a1nw1 + a2nw2 + ……….+amnwm ≥ cn and w1, w2, …, wm ≥ 0 Matrix Notation Primal Max Zx = CX Subject to AX ≤ b and X ≥ 0 Dual Min Zw = bT W Subject to AT W ≥ CT and W ≥ 0 3.2 Important characteristics of Duality Dual of dual is primal If either the primal or dual problem has a solution then the other also has a solution and their optimum values are equal. If any of the two problems has an infeasible solution, then the value of the objective function of the other is unbounded. The value of the objective function for any feasible solution of the primal is less than the value of the objective function for any feasible solution of the dual. If either the primal or dual has an unbounded solution, then the solution to the other problem is infeasible. If the primal has a feasible solution, but the dual does not have then the primal will not have a finite optimum solution and vice versa. 3.3 Advantages and Applications of Duality Sometimes dual problem solution may be easier than primal solution, particularly when the number of decision variables is considerably less than slack / surplus variables. In the areas like economics, it is highly helpful in obtaining future decision in the activities being programmed. In physics, it is used in parallel circuit and series circuit theory. In game theory, dual is employed by column player who wishes to minimize his maximum loss while his opponent i.e. Row player applies primal to maximize his minimum gains. However, if one problem is solved, the solution for other also can be obtained from the simplex tableau. When a problem does not yield any solution in primal, it can be verified with dual. Economic interpretations can be made and shadow prices can be determined enabling the managers to take further decisions. 3.4 Steps for a Standard Primal Form Step 1 – Change the objective function to Maximization form Step 2 – If the constraints have an inequality sign ‘≥’ then multiply both sides by -1 and convert the inequality sign to ‘≤’. Step 3 – If the constraint has an ‘=’ sign then replace it by two constraints involving the inequalities going in opposite directions. For example x1+ 2x2 = 4 is written as x1+2x2 ≤ 4 x1+2x2 ≥ 4 (using step2) → - x1-2x2 ≤ - 4 Step 4 – Every unrestricted variable is replaced by the difference of two non-negative variables. Step5 – We get the standard primal form of the given LPP in which. All constraints have ‘≤’ sign, where the objective function is of maximization form. All constraints have ‘≥’ sign, where the objective function is of minimization from. 3.5 Rules for Converting any Primal into its Dual Transpose the rows and columns of the constraint co-efficient. Transpose the co-efficient (c1,c2,…cn) of the objective function and the right side constants (b1,b2,…bn) Change the inequalities from ‘≤’ to ‘≥’ sign. Minimize the objective function instead of maximizing it. 3.6 Example Problems Write the dual of the given problems Example 1 Min Zx = 2x2 + 5x3 Subject to x1+x2 ≥ 2 2x1+x2+6x3 ≤ 6 x1 - x2 +3x3 = 4 x1, x2 , x3 ≥ 0 Solution Primal Max Zx' = -2x2 – 5x3 Subject to -x1-x2 ≤ -2 2x1+x2+6x3 ≤ 6 x1 - x2 +3x3 ≤ 4 -x1 + x2 -3x3 ≤ -4 x1, x2 , x3 ≥ 0 Dual Min Zw = -2w1 + 6w2 + 4w3 – 4w4 Subject to -w1 + 2w2 +w3 –w4 ≥ 0 -w1 + w2 - w3 +w4 ≥ -2 6w2 + 3w3 –3w4 ≥ -5 w1, w2, w3, w4 ≥ 0 Example 2 Min Zx = 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 Primal Max Zx' = -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 Dual Min Zw = -7w1 - 4w2 - 10w3 – 3w4 -2w5 Subject to -3w1 - 6w2 - 7w3 –w4 – 4w5 ≥ -3 -5w1 - w2 + 2w3 + 2w4 – 7w5 ≥ 2 -4w1 - 3w2 + w3 - 5w4 + 2w5 ≥ -4 w1, w2, w3, w4, w5 ≥ 0 Example 3 Max Z = 2x1+ 3x2 + x3 Subject to 4x1+ 3x2 + x3 = 6 x1+ 2x2 + 5x3 = 4 x1, x2 ≥ 0 Solution Primal Max Zx = 2x1+ 3x2 + x3 Subject to 4x1+ 3x2 + x3 ≤ 6 -4x1 - 3x2 - x3 ≤ -6 x1 + 2x2 + 5x3 ≤ 4 -x1 - 2x2 - 5x3 ≤ -4 x1, x2 ≥ 0 Dual Min Zw = 6w1 - 6w2 + 4w3 –4w4 Subject to 4w1 - 4w2 + w3 –w4 ≥ 2 3w1 - 3w2 + 2w3 - 2w4 ≥ 3 w1 - w2 + 5w3 - 5w4 ≥ 1 w1, w2, w3, w4≥ 0 Example 4 Min Zx = x1+ x2 + x3 Subject to x1 - 3x2 + 4x3 = 5 x1 - 2x2 ≤ 3 2x2 - x3 ≥ 4 x1, x2 ≥ 0 ,x3 is unrestricted in sign Solution Primal Max Z' = - x1- x2 – x3' + x3'' Subject to x1 - 3x2 + 4(x3' - x3'') ≤ 5 -x1+ 3x2 - 4(x3' - x3'') ≤ -5 x1 - 2x2 ≤ 3 -2x2 + x3' - x3'' ≤ -4 x1, x2 , x3', x3'' ≥ 0 Dual Min Zw = 5w1 - 5w2 + 3w3 – 4w4 Subject to w1 - w2 + w3 ≥ -1 -3w1 + 3w2 - 2w3 - 2w4 ≥ -1 4w1 - 4w2 + w4 ≥ -1 -4w1 + 4w2 - w4 ≥ 1 w1, w2, w3, w4, ≥ 0 Example 5 Max Z = x1 - x2 + 3x3 Subject to x1 + x2 + x3 ≤ 10 2x1 – x3 ≤ 2 2x1 - 2x2 + 3x3 ≤ 6 x1, x2, x3 ≥ 0 Solution Primal Max Zx = x1 - x2 + 3x3 Subject to x1 + x2 + x3 ≤ 10 2x1 - x3 ≤ 2 2x1 - 2x2 + 3x3 ≤ 6 x1, x2, x3 ≥ 0 Dual Min Zw = 10w1 + 2w2 + 6w3 Subject to w1 + 2w2 +2w3 ≥ 1 w1 - 2w3 ≥ -1 w1 - w2 + 3w3 ≥ 3 w1, w2, w3 ≥ 0 3.7 Primal –Dual Relationship Weak duality property If x is any feasible solution to the primal problem and w is any feasible solution to the dual problem then CX ≤ bT W. i.e. ZX ≤ ZW Strong duality property If x* is an optimal solution for the primal problem and w* is the optimal solution for the dual problem then CX* = bT W* i.e. ZX = ZW Complementary optimal solutions property At the final iteration, the simplex method simultaneously identifies an optimal solution x* for primal problem and a complementary optimal solution w* for the dual problem where ZX = ZW. Symmetry property For any primal problem and its dual problem, all relationships between them must be symmetric because dual of dual is primal. Fundamental duality theorem If one problem has feasible solution and a bounded objective function (optimal solution) then the other problem has a finite optimal solution. If one problem has feasible solution and an unbounded optimal solution then the other problem has no feasible solution If one problem has no feasible solution then the other problem has either no feasible solution or an unbounded solution. If kth constraint of primal is equality then the dual variable wk is unrestricted in sign If pth variable of primal is unrestricted in sign then pth constraint of dual is an equality. Complementary basic solutions property Each basic solution in the primal problem has a complementary basic solution in the dual problem where ZX = ZW. Complementary slackness property The variables in the primal basic solution and the complementary dual basic solution satisfy the complementary slackness relationship as shown in the table. Primal variable Associated dual variable Decision variable (xj) Zj –Cj (surplus variable) j = 1, 2, ..n Slack variable (Si) Wi (decision variable) i = 1, 2, .. n 3.8 Duality and Simplex Method 1. Solve the given primal problem using simplex method. Hence write the solution of its dual Max Z = 30x1 + 23x2 + 29x3 Subject to 6x1 + 5x2 + 3x3 ≤ 26 4x1 + 2x2 + 6x3 ≤ 7 x1 ≥ 0, x2 ≥ 0 Solution Primal form Max Z = 30x1 + 23x2 + 29x3 Subject to 6x1 + 5x2 + 3x3 ≤ 26 4x1 + 2x2 + 6x3 ≤ 7 x1 ≥ 0, x2 ≥ 0 SLPP Max Z = 30x1 + 23x2 + 29x3+ 0s1+ 0s2 Subject to 6x1 + 5x2 + 3x3 + s1 = 26 4x1 + 2x2 + 6x3 + s2 = 7 x1, x2, s1, s2 ≥ 0 Cj→ 30 23 29 0 0 Basic Variables CB XB X1 X2 X3 S1 S2 Min Ratio XB / XK s1 0 26 6 5 3 1 0 26/6 s2 0 7 4 2 6 0 1 7/4→ Z = 0 ↑ -30 -23 -29 0 0 ←Δj s1 0 31/2 0 2 -6 1 -3/2 31/4 x1 30 7/4 1 1/2 3/2 0 1/4 7/2→ Z = 105/2 0 ↑ -8 16 0 15/2 ←Δj s1 0 17/2 -4 0 -12 1 -5/2 x2 23 7/2 2 1 3 0 1/2 Z =161/2 16 0 40 0 23/2 ←Δj Δj ≥ 0 so the optimal solution is Z = 161/2, x1 = 0, x2 = 7/2, x3 = 0. The optimal solution to the dual of the above problem will be Zw* = 161/2, w1 = Δ4 = 0, w2 = Δ5 = 23/2 In this way we can find the solution to the dual without actually solving it. 2. Use duality to solve the given problem. Also read the solution of its primal. Min Z = 3x1 + x2 Subject to x1 + x2 ≥ 1 2x1 + 3x2 ≥ 2 x1 ≥ 0 , x2 ≥ 0 Solution Primal Min Z =Max Z' = -3x1 - x2 Subject to - x1 - x2 ≤ -1 -2x1 - 3x2 ≤ - 2 x1 ≥ 0 , x2 ≥ 0 Dual Min Zw = -w1 - 2w2 Subject to -w1 - 2w2 ≥ -3 -w1 - 3w2 ≥ -1 w1, w2 ≥ 0 Changing the dual form to SLPP Max Zw' = w1 + 2w2 + 0s1+ 0s2 Subject to w1 + 2w2 + s1= 3 w1 + 3w2 + s2 = 1 w1, w2, s1, s2 ≥ 0 Cj→ 1 2 0 0 Basic Variables CB WB W1 W2 S1 S2 Min Ratio WB / WK s1 0 3 1 2 1 0 3/2 s2 0 1 1 3 0 1 1/3← Zw' = 0 -1 ↑ -2 0 0 ←Δj s1 0 7/3 1/3 0 1 -2/3 7 w2 2 1/3 1/3 1 0 1/3 1→ Zw' = 2/3 ↑ -1/3 0 0 2/3 ←Δj s1 0 2 0 -1 1 -1 w1 1 1 1 3 0 1 Zw' = 1 0 1 0 1 ←Δj Δj ≥ 0 so the optimal solution is Zw' = 1, w1 = 1, w2 = 0 The optimal solution to the primal of the above problem will be Zx* = 1, x1 = Δ3 = 0, x2 = Δ4 = 1 3. Write down the dual of the problem and solve it. Max Z = 4x1 + 2x2 Subject to - x1 - x2 ≤ -3 -x1 + x2 ≤ -2 x1 ≥ 0, x2 ≥ 0 Solution Primal Max Z = 4x1 + 2x2 Subject to - x1 - x2 ≤ -3 -x1 + x2 ≤ -2 x1 ≥ 0, x2 ≥ 0 Dual Min Zw = -3w1 - 2w2 Subject to -w1 - w2 ≥ 4 -w1 + w2 ≥ 2 w1, w2 ≥ 0 Changing the dual form to SLPP Max Zw' = 3w1 + 2w2 + 0s1+ 0s2 - Ma1- Ma2 Subject to -w1 - w2 - s1 + a1= 4 -w1 + w2 - s2 + a2= 2 w1, w2, s1, s2, a1, a2 ≥ 0 Cj→ 3 2 0 0 -M -M Basic Variables CB WB W1 W2 S1 S2 A1 A2 Min Ratio WB / WK a1 -M 4 -1 -1 -1 0 1 0 - a2 -M 2 -1 1 0 -1 0 1 2→ Zw' = -6M 2M - 3 ↑ -2 M M 0 0 ←Δj a1 -M 6 -2 0 -1 -1 1 X w2 2 2 -1 1 0 -1 0 X Zw' = -6M+4 2M-5 0 M M-2 0 X ←Δj Δj ≥ 0 and at the positive level an artificial vector (a1) appears in the basis. Therefore the dual problem does not posses any optimal solution. Consequently there exists no finite optimum solution to the given problem. 4. Use duality to solve the given problem. Min Z = x1 - x2 Subject to 2x1 + x2 ≥ 2 -x1 - x2 ≥ 1 x1 ≥ 0 , x2 ≥ 0 Solution Primal Min Z =Max Z' = -x1 + x2 Subject to - 2x1 - x2 ≤ -2 x1 + x2 ≤ -1 x1 ≥ 0 , x2 ≥ 0 Dual Min Zw = -2w1 - w2 Subject to -2w1 + w2 ≥ -1 -w1 + w2 ≥ 1 w1, w2 ≥ 0 Changing the dual form to SLPP Max Zw' = 2w1 + w2 + 0s1+ 0s2 - Ma1 Subject to 2w1 - w2 + s1= 1 -w1 + w2 - s2 + a1 = 1 w1, w2, s1, s2 ≥ 0 Auxiliary LPP Max Zw' = 0w1 + 0w2 + 0s1+ 0s2 - 1a1 Subject to 2w1 - w2 + s1= 1 -w1 + w2 - s2 + a1 = 1 w1, w2, s1, s2, a1 ≥ 0 Phase I Cj→ 0 0 0 0 -1 Basic Variables CB WB W1 W2 S1 S2 A1 Min Ratio XB / XK s1 0 1 2 -1 1 0 0 - a1 -1 1 -1 1 0 -1 1 1→ Zw' = -1 1 ↑ -1 0 1 0 ←Δj s1 0 2 1 0 1 -1 X w2 0 1 -1 1 0 -1 X Zw' = 0 0 0 0 0 X ←Δj Δj ≥ 0 and no artificial vector appear at the positive level of the basis. Hence proceed to phase II Phase II Cj→ 2 1 0 0 Basic Variables CB WB W1 W2 S1 S2 Min Ratio XB / XK s1 0 2 1 0 1 -1 2→ w2 1 1 -1 1 0 -1 - Zw' = 1 ↑ -3 0 0 -1 ←Δj w1 2 2 1 0 1 -1 - w2 1 3 0 1 1 -2 - Zw' = 7 0 0 3 ↑ -4 ←Δj Δj = -4 and all the elements of s2 are negative; hence we cannot find the outgoing vector. This indicates there is an unbounded solution. Consequently by duality theorem the original primal problem will have no feasible solution. 5. Solve the given primal problem using simplex method. Hence write the solution of its dual Max Z = 40x1 + 50x2 Subject to 2x1 + 3x2 ≤ 3 8x1 + 4x2 ≤ 5 x1 ≥ 0, x2 ≥ 0 Solution Primal form Max Z = 40x1 + 50x2 Subject to 2x1 + 3x2 ≤ 3 8x1 + 4x2 ≤ 5 x1 ≥ 0, x2 ≥ 0 SLPP Max Zx = 40x1 + 50x2 + 0s1+ 0s2 Subject to 2x1 + 3x2 + s1 = 3 8x1 + 4x2 + s2 = 5 x1, x2, s1, s2 ≥ 0 Cj → 40 50 0 0 Basic Variables CB XB X1 X2 S1 S2 Min Ratio XB / XK s1 0 3 2 3 1 0 1→ s2 0 5 8 4 0 1 5/4 Zx = 0 -40 ↑ -50 0 0 ←Δj x2 50 1 2/3 1 1/3 0 3/2 s2 0 1 16/3 0 -4/3 1 3/16→ Zx = 50 ↑ -20/3 0 50/3 0 ←Δj x2 50 7/8 0 1 1/2 -1/8 x1 40 3/16 1 0 -1/4 3/16 Zx = 205/4 0 0 15 5/4 ←Δj Δj ≥ 0 so the optimal solution is Z = 205/4, x1 = 3/16, x2 = 7/8 The optimal solution to the dual of the above problem will be Zw* = 205/4, w1 = Δ4 = 15, w2 = Δ5 = 5/4 Exercise Explain the concept of duality in LPP. Explain the characteristics of duality. Mention the advantages and application of duality Write the steps for converting LPP into its dual. Explain the concept of primal- dual relationship Obtain the dual of the following linear programming problems Max Z = 3x1 + 4x2 Subject to 2x1 + 6x2 ≤ 16 5x1 + 2x2 ≥ 20 x1 ≥ 0, x2 ≥ 0 Min Z = 7x1 + 3x2 + 8x3 Subject to 8x1 + 2x2 + x3 ≥ 3 3x1 + 6x2 + 4x3 ≥ 4 4x1 + x2 + 5x3 ≥ 1 x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Max Z = 6x1 + 4x2 + 6x3 + x4 Subject to 4x1 + 4x2 + 4x3 + 8x4 = 21 3x1 + 17x2 + 80x3 + 2x4 ≤ 48 x1 ≥ 0, x2 ≥ 0, x3, x4 are unrestricted Use duality to solve the following LPP Max Z = 3x1 + 2x2 Subject to 2x1 + x2 ≤ 5 x1 + x2 ≤ 3 x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 8, x1 =2, x2 = 1, w1 = 1, w2 =2, Min Zw =8] Min Z = 2x1 + 2x2 Subject to 2x1 + 4x2 ≥ 1 x1 + 2x2 ≥ 1 2x1 + x2 ≥ 1 x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = 4/3, x1 = 1/3, x2 = 1/3] Max Z = 8x1 + 6x2 Subject to x1 - x2 ≤ 3/5 x1 - x2 ≥ 2 x1 ≥ 0, x2 ≥ 0 [Ans. Dual problem does not possess feasible solution] Module 4 Unit 1 Introduction Computational Procedure of Dual Simplex Method Worked Examples Advantage of Dual Simplex over Simplex Method Introduction to Transportation Problem Mathematical Formulation Tabular Representation Some Basic Definitions 1.1 Introduction Any LPP for which it is possible to find infeasible but better than optimal initial basic solution can be solved by using dual simplex method. Such a situation can be recognized by first expressing the constraints in ‘≤’ form and the objective function in the maximization form. After adding slack variables, if any right hand side element is negative and the optimality condition is satisfied then the problem can be solved by dual simplex method. Negative element on the right hand side suggests that the corresponding slack variable is negative. This means that the problem starts with optimal but infeasible basic solution and we proceed towards its feasibility. The dual simplex method is similar to the standard simplex method except that in the latter the starting initial basic solution is feasible but not optimum while in the former it is infeasible but optimum or better than optimum. The dual simplex method works towards feasibility while simplex method works towards optimality. Computational Procedure of Dual Simplex Method The iterative procedure is as follows Step 1 - First convert the minimization LPP into maximization form, if it is given in the minimization form. Step 2 - Convert the ‘≥’ type inequalities of given LPP, if any, into those of ‘≤’ type by multiplying the corresponding constraints by -1. Step 3 – Introduce slack variables in the constraints of the given problem and obtain an initial basic solution. Step 4 – Test the nature of Δj in the starting table If all Δj and XB are non-negative, then an optimum basic feasible solution has been attained. If all Δj are non-negative and at least one basic variable XB is negative, then go to step 5. If at least Δj one is negative, the method is not appropriate. Step 5 – Select the most negative XB. The corresponding basis vector then leaves the basis set B. Let Xr be the most negative basic variable. Step 6 – Test the nature of Xr If all Xr are non-negative, then there does not exist any feasible solution to the given problem. If at least one Xr is negative, then compute Max (Δj / Xr ) and determine the least negative for incoming vector. Step 7 – Test the new iterated dual simplex table for optimality. Repeat the entire procedure until either an optimum feasible solution has been attained in a finite number of steps. Worked Examples Example 1 Minimize Z = 2x1 + x2 Subject to 3x1 + x2 ≥ 3 4x1 + 3x2 ≥ 6 x1 + 2x2 ≥ 3 and x1 ≥ 0, x2 ≥ 0 Solution Step 1 – Rewrite the given problem in the form Maximize Z׳ = – 2x1 – x2 Subject to –3x1 – x2 ≤ –3 –4x1 – 3x2 ≤ –6 –x1 – 2x2 ≤ –3 x1, x2 ≥ 0 Step 2 – Adding slack variables to each constraint Maximize Z׳ = – 2x1 – x2 Subject to –3x1 – x2 + s1 = –3 –4x1 – 3x2 + s2 = –6 –x1 – 2x2 + s3 = –3 x1, x2, s1,s2, s3 ≥ 0 Step 3 – Construct the simplex table Cj → -2 -1 0 0 0 Basic variables CB XB X1 X2 S1 S2 S3 s1 0 -3 -3 -1 1 0 0 s2 0 -6 -4 -3 0 1 0 → outgoing s3 0 -3 -1 -2 0 0 1 Z׳ = 0 2 ↑ 1 0 0 0 ←Δj Step 4 – To find the leaving vector Min (-3, -6, -3) = -6. Hence s2 is outgoing vector Step 5 – To find the incoming vector Max (Δ1 / x21, Δ2 / x22) = (2/-4, 1/-3) = -1/3. So x2 is incoming vector Step 6 –The key element is -3. Proceed to next iteration Cj → -2 -1 0 0 0 Basic variables CB XB X1 X2 S1 S2 S3 s1 0 -1 -5/3 0 1 -1/3 0 → outgoing x2 -1 2 4/3 1 0 -1/3 0 s3 0 1 5/3 0 0 -2/3 1 Z׳ = -2 ↑ 2/3 0 0 1/3 0 ←Δj Step 7 – To find the leaving vector Min (-1, 2, 1) = -1. Hence s1 is outgoing vector Step 8 – To find the incoming vector Max (Δ1 / x11, Δ4 / x14) = (-2/5, -1) = -2/5. So x1 is incoming vector Step 9 –The key element is -5/3. Proceed to next iteration Cj → -2 -1 0 0 0 Basic variables CB XB X1 X2 S1 S2 S3 x1 -2 3/5 1 0 -3/5 1/5 0 x2 -1 6/5 0 1 4/5 -3/5 0 s3 0 0 0 0 1 -1 1 Z׳ = -12/5 0 0 2/5 1/5 0 ←Δj Step 10 – Δj ≥ 0 and XB ≥ 0, therefore the optimal solution is Max Z׳ = -12/5, Z = 12/5, and x1=3/5, x2 = 6/5 Example 2 Minimize Z = 3x1 + x2 Subject to x1 + x2 ≥ 1 2x1 + 3x2 ≥ 2 and x1 ≥ 0, x2 ≥ 0 Solution Maximize Z׳ = – 3x1 – x2 Subject to –x1 – x2 ≤ –1 –2x1 – 3x2 ≤ –2 x1, x2 ≥ 0 SLPP Maximize Z׳ = – 3x1 – x2 Subject to –x1 – x2 + s1 = –1 –2x1 – 3x2 + s2 = –2 x1, x2, s1,s2 ≥ 0 Cj → -3 -1 0 0 Basic variables CB XB X1 X2 S1 S2 s1 0 -1 -1 -1 1 0 s2 0 -2 -2 -3 0 1 → Z׳ = 0 3 ↑ 1 0 0 ←Δj s1 0 -1/3 -1/3 0 1 -1/3 → x2 -1 2/3 2/3 1 0 -1/3 Z׳ = -2/3 7/3 0 0 ↑ 1/3 ←Δj s2 0 1 1 0 -3 1 x2 -1 1 1 1 -1 0 Z׳ = -1 2 0 1 0 ←Δj Δj ≥ 0 and XB ≥ 0, therefore the optimal solution is Max Z׳ = -1, Z = 1, and x1= 0, x2 = 1 Example 3 Max Z = –2x1 – x3 Subject to x1 + x2 – x3 ≥ 5 x1 – 2x2 + 4x3 ≥ 8 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution Max Z = –2x1 – x3 Subject to –x1 – x2 + x3 ≤ –5 –x1 + 2x2 – 4x3 ≤ –8 x1, x2, x3 ≥ 0 SLPP Max Z = –2x1 – x3 Subject to –x1 – x2 + x3 + s1 = –5 –x1 + 2x2 – 4x3 + s2 = –8 x1, x2, x3, s1, s2 ≥ 0 Cj → -2 0 -1 0 0 Basic variables CB XB X1 X2 X3 S1 S2 s1 0 -5 -1 -1 1 1 0 s2 0 -8 -1 2 -4 0 1 → Z = 0 2 0 ↑ 1 0 0 ←Δj s1 0 -7 -5/4 -1/2 0 1 1/4 → x3 -1 2 1/4 -1/2 1 0 -1/4 Z = -2 7/4 ↑ 1/2 0 0 1/4 ←Δj x2 0 14 5/2 1 0 -2 -1/2 x3 -1 9 3/2 0 1 -1 -1/2 Z = -9 1/2 0 0 1 1/2 ←Δj Δj ≥ 0 and XB ≥ 0, therefore the optimal solution is Z = -9, and x1= 0, x2 = 14, x3 = 9 Example 4 Find the optimum solution of the given problem without using artificial variable. Max Z = –4x1 –6x2 – 18x3 Subject to x1 + 3x3 ≥ 3 x2 + 2x3 ≥ 5 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 Solution Max Z = –4x1 –6x2 – 18x3 Subject to –x1 – 3x3 ≤ –3 –x2 – 2x3 ≤ –5 x1, x2, x3 ≥ 0 SLPP Max Z = –4x1 –6x2 – 18x3 Subject to –x1 – 3x3 + s1 = –3 –x2 – 2x3 + s2 = –5 x1, x2, x3, s1, s2 ≥ 0 Cj → -4 -6 -18 0 0 Basic variables CB XB X1 X2 X3 S1 S2 s1 0 -3 -1 0 -3 1 0 s2 0 -5 0 -1 -2 0 1 → Z = 0 4 ↑ 6 18 0 0 ←Δj s1 0 -3 -1 0 -3 1 0 → x2 -6 5 0 1 2 0 -1 Z = -30 4 0 ↑ 6 0 6 ←Δj x3 -18 1 1/3 0 1 -1/3 0 x2 -6 3 -2/3 1 0 2/3 -1 Z = -36 2 0 0 2 6 ←Δj Δj ≥ 0 and XB ≥ 0, therefore the optimal solution is Z = -36, and x1= 0, x2 = 3, x3 = 1 Example 5 Min Z = 6x1 + 7x2 + 3x3 + 5x4 Subject to 5x1 + 6x2 - 3x3 + 4x4 ≥ 12 x2 + 5x3 - 6x4 ≥ 10 2x1 + 5x2 + x3 + x4 ≥ 8 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 Solution Max Z' = - 6x1 - 7x2 - 3x3 - 5x4 Subject to -5x1 - 6x2 + 3x3 - 4x4 ≤ -12 -x2 - 5x3 + 6x4 ≤ -10 -2x1 - 5x2 - x3 - x4 ≤ -8 x1, x2, x3, x4 ≥ 0 SLPP Max Z' = - 6x1 - 7x2 - 3x3 - 5x4 Subject to -5x1 - 6x2 + 3x3 - 4x4 + s1 = -12 -x2 - 5x3 + 6x4 + s2 = -10 -2x1 - 5x2 - x3 - x4 + s3 = -8 x1, x2, x3, x4, s1, s2, s3 ≥ 0 Cj → -6 -7 -3 -5 0 0 0 Basic variables CB XB X1 X2 X3 X4 S1 S2 S3 s1 0 -12 -5 -6 3 -4 1 0 0 → s2 0 -10 0 -1 -5 6 0 1 0 s3 0 -8 -2 -5 -1 1 0 0 1 Z'= 0 6 ↑ 7 3 5 0 0 0 x2 -7 2 5/6 1 -1/2 2/3 -1/6 0 0 s2 0 -8 5/6 0 -11/2 20/3 -1/6 1 0 → s3 0 2 13/6 0 -7/2 7/3 -5/6 0 1 Z'= -14 1/6 0 ↑ 13/2 1/3 7/6 0 0 x2 -7 30/11 25/33 1 0 2/33 -5/33 -1/11 0 x3 -3 16/11 -5/33 0 1 -40/33 1/33 -2/11 0 s3 0 78/11 18/11 0 0 -21/11 -8/11 -7/11 1 Z'= -258/11 38/33 0 0 271/33 32/33 13/11 0 Δj ≥ 0 and XB ≥ 0, therefore the optimal solution is Z= 258/11 and x1= 0, x2 = 30/11, x3 = 16/11, x4= 0 Advantage of Dual Simplex over Simplex Method The main advantage of dual simplex over the usual simplex method is that we do not require any artificial variables in the dual simplex method. Hence a lot of labor is saved whenever this method is applicable. 1.5 Introduction to Transportation Problem The Transportation problem is to transport various amounts of a single homogeneous commodity that are initially stored at various origins, to different destinations in such a way that the total transportation cost is a minimum. It can also be defined as to ship goods from various origins to various destinations in such a manner that the transportation cost is a minimum. The availability as well as the requirements is finite. It is assumed that the cost of shipping is linear. 1.6 Mathematical Formulation Let there be m origins, ith origin possessing ai units of a certain product Let there be n destinations, with destination j requiring bj units of a certain product Let cij be the cost of shipping one unit from ith source to jth destination Let xij be the amount to be shipped from ith source to jth destination It is assumed that the total availabilities Σai satisfy the total requirements Σbj i.e. Σai = Σbj (i = 1, 2, 3 … m and j = 1, 2, 3 …n) The problem now, is to determine non-negative of xij satisfying both the availability constraints as well as requirement constraints and the minimizing cost of transportation (shipping) This special type of LPP is called as Transportation Problem. 1.7 Tabular Representation Let ‘m’ denote number of factories (F1, F2 … Fm) Let ‘n’ denote number of warehouse (W1, W2 … Wn) W→ F ↓ W1 W2 … Wn Capacities (Availability) F1 c11 c12 … c1n a1 F2 c21 c22 … c2n a2 . . . . . . . . . . . . Fm cm1 cm2 … cmn am Required b1 b2 … bn Σai = Σbj W→ F ↓ W1 W2 … Wn Capacities (Availability) F1 x11 x12 … x1n a1 F2 x21 x22 … x2n a2 . . . . . . . . . . . . Fm xm1 xm2 … xmn am Required b1 b2 … bn Σai = Σbj In general these two tables are combined by inserting each unit cost cij with the corresponding amount xij in the cell (i, j). The product cij xij gives the net cost of shipping units from the factory Fi to warehouse Wj. 1.8 Some Basic Definitions Feasible Solution A set of non-negative individual allocations (xij ≥ 0) which simultaneously removes deficiencies is called as feasible solution. Basic Feasible Solution A feasible solution to ‘m’ origin, ‘n’ destination problem is said to be basic if the number of positive allocations are m+n-1. If the number of allocations is less than m+n-1 then it is called as Degenerate Basic Feasible Solution. Otherwise it is called as Non- Degenerate Basic Feasible Solution. Optimum Solution A feasible solution is said to be optimal if it minimizes the total transportation cost. Exercise Solve by dual simplex method Max Z = -3x1 - 2x2 Subject to x1 + x2 ≥ 1 x1 + x2 ≤ 7 x1 + 2x2 ≤ 10 x2 ≤ 3 and x1 ≥ 0, x2 ≥ 0 [Ans. Max Z = -2, x1 = 0, x2 = 1] Max Z = –2x1 –2x2 – 4x3 Subject to 2x1 + 3x2 + 5x3 ≥ 2 3x1 + x2 + 7x3 ≤ 3 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 [Ans. Max Z = 4/3, x1 = 0, x2 = 2/3, x3 = 0] Min Z = x1 + 2x2 + 3x3 Subject to 2x1 - x2 + x3 ≥ 4 x1 + x2 + 2x3 ≥ 8 x2 - x3 ≥ 2 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 [Ans. Min Z = 10, x1 = 6, x2 = 2, x3 = 0] Min Z = 3x1 + 2x2 + x3 + 4x4 Subject to 2x1 + 4x2 + 5x3 + x4 ≥ 10 3x1 - x2 + 7x3 - 2x4 ≥ 2 5x1 + 2x2 + x3 + 6x4 ≥ 15 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 [Ans. Min Z = 215/23, x1 = 65/23, x2 = 0, x3 = 20, x4 = 0] Min Z = x1 + x2 Subject to 2x1 + x2 ≥ 2 -x1 - x2 ≥ 1 and x1 ≥ 0, x2 ≥ 0 [Ans. Pseudo Optimum basic feasible solution] Unit 2 2.1 Methods for Initial Basic Feasible Solution 2.1.1 North-West Corner Rule 2.1.2 Row Minima Method 2.1.3 Column Minima Method 2.1.4 Lowest Cost Entry Method (Matrix Minima Method) 2.1.5 Vogel’s Approximation Method (Unit Cost Penalty Method) 2.1 Methods for Initial Basic Feasible Solution Some simple methods to obtain the initial basic feasible solution are North-West Corner Rule Row Minima Method Column Minima Method Lowest Cost Entry Method (Matrix Minima Method) Vogel’s Approximation Method (Unit Cost Penalty Method) 2.1.1 North-West Corner Rule Step 1 The first assignment is made in the cell occupying the upper left-hand (north-west) corner of the table. The maximum possible amount is allocated here i.e. x11 = min (a1, b1). This value of x11 is then entered in the cell (1,1) of the transportation table. Step 2 If b1 > a1, move vertically downwards to the second row and make the second allocation of amount x21 = min (a2, b1 - x11) in the cell (2, 1). If b1 < a1, move horizontally right side to the second column and make the second allocation of amount x12 = min (a1 - x11, b2) in the cell (1, 2). If b1 = a1, there is tie for the second allocation. One can make a second allocation of magnitude x12 = min (a1 - a1, b2) in the cell (1, 2) or x21 = min (a2, b1 - b1) in the cell (2, 1) Step 3 Start from the new north-west corner of the transportation table and repeat steps 1 and 2 until all the requirements are satisfied. Find the initial basic feasible solution by using North-West Corner Rule W→ F ↓ W1 W2 W3 W4 Factory Capacity F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Warehouse Requirement 5 8 7 14 34 1. Solution W1 W2 W3 W5 Availability F1 5 2 7 2 0 (19) (30) F2 6 3 9 3 0 (30) (40) F3 4 14 18 14 0 (70) (20) Requirement 5 0 8 6 0 7 4 0 14 0 Initial Basic Feasible Solution x11 = 5, x12 = 2, x22 = 6, x23 = 3, x33 = 4, x34 = 14 The transportation cost is 5 (19) + 2 (30) + 6 (30) + 3 (40) + 4 (70) + 14 (20) = Rs. 1015 2. D1 D2 D3 D4 Supply O1 1 5 3 3 34 O2 3 3 1 2 15 O3 0 2 2 3 12 O4 2 7 2 4 19 Demand 21 25 17 17 80 Solution D1 D2 D3 D4 Supply 21 13 O1 (1) (5) 34 13 0 12 3 O2 (3) (1) 15 3 0 12 O3 (2) 12 0 2 17 O4 (2) (4) 19 17 Demand 21 0 25 12 0 17 14 2 0 17 0 Initial Basic Feasible Solution x11 = 21, x12 = 13, x22 = 12, x23 = 3, x33 = 12, x43 = 2, x44 = 17 The transportation cost is 21 (1) + 13 (5) + 12 (3) + 3 (1) + 12 (2) + 2 (2) + 17 (4) = Rs. 221 3. From To Supply 2 11 10 3 7 4 1 4 7 2 1 8 3 1 4 8 12 9 Demand 3 3 4 5 6  Solution From To Supply 3 1 (2) (11) 4 1 0 2 4 2 (4) (7) (2) 8 6 2 0 3 6 (8) (12) 9 6 0 Demand 3 0 3 2 0 4 0 5 3 0 6 0 Initial Basic Feasible Solution x11 = 3, x12 = 1, x22 = 2, x23 = 4, x24 = 2, x34 = 3, x35 = 6 The transportation cost is 3 (2) + 1 (11) + 2 (4) + 4 (7) + 2 (2) + 3 (8) + 6 (12) = Rs. 153 2.1.2 Row Minima Method Step 1 The smallest cost in the first row of the transportation table is determined. Allocate as much as possible amount xij = min (a1, bj) in the cell (1, j) so that the capacity of the origin or the destination is satisfied. Step 2 If x1j = a1, so that the availability at origin O1 is completely exhausted, cross out the first row of the table and move to second row. If x1j = bj, so that the requirement at destination Dj is satisfied, cross out the jth column and reconsider the first row with the remaining availability of origin O1. If x1j = a1 = bj, the origin capacity a1 is completely exhausted as well as the requirement at destination Dj is satisfied. An arbitrary tie-breaking choice is made. Cross out the jth column and make the second allocation x1k = 0 in the cell (1, k) with c1k being the new minimum cost in the first row. Cross out the first row and move to second row. Step 3 Repeat steps 1 and 2 for the reduced transportation table until all the requirements are satisfied Determine the initial basic feasible solution using Row Minima Method 1. W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 80 70 20 18 Requirement 5 8 7 14 Solution W1 W2 W3 W4 7 X F1 (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 (40) (80) (70) (20) 18 5 8 7 7 W1 W2 W3 W4 7 X F1 (19) (30) (50) (10) 8 F2 (70) (30) (40) (60) 1 F3 (40) (80) (70) (20) 18 5 X 7 7 W1 W2 W3 W4 7 X F1 (19) (30) (50) (10) 8 1 X F2 (70) (30) (40) (60) F3 (40) (80) (70) (20) 18 5 X 6 7 W1 W2 W3 W4 7 X F1 (19) (30) (50) (10) 8 1 X F2 (70) (30) (40) (60) 5 6 7 X F3 (40) (80) (70) (20) X X X X Initial Basic Feasible Solution x14 = 7, x22 = 8, x23 = 1, x31 = 5, x33 = 6, x34 = 7 The transportation cost is 7 (10) + 8 (30) + 1 (40) + 5 (40) + 6 (70) + 7 (20) = Rs. 1110 2. A B C Availability I 50 30 220 1 II 90 45 170 4 III 250 200 50 4 Requirement 4 2 3 Solution A B C Availability I 1 1 0 (30) II 3 1 4 3 0 (90) (45) III 1 3 4 1 0 (250) (50) Requirement 4 1 0 2 1 0 3 0 Initial Basic Feasible Solution x12 = 1, x21 = 3, x22 = 1, x31 = 1, x33 = 3 The transportation cost is 1 (30) + 3 (90) + 1 (45) + 1 (250) + 3 (50) = Rs. 745 2.1.3 Column Minima Method Step 1 Determine the smallest cost in the first column of the transportation table. Allocate xi1 = min (ai, b1) in the cell (i, 1). Step 2 If xi1 = b1, cross out the first column of the table and move towards right to the second column If xi1 = ai, cross out the ith row of the table and reconsider the first column with the remaining demand. If xi1 = b1= ai, cross out the ith row and make the second allocation xk1 = 0 in the cell (k, 1) with ck1 being the new minimum cost in the first column, cross out the column and move towards right to the second column. Step 3 Repeat steps 1 and 2 for the reduced transportation table until all the requirements are satisfied. Use Column Minima method to determine an initial basic feasible solution 1. W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 80 70 20 18 Requirement 5 8 7 14 Solution W1 W2 W3 W4 F1 5 2 (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 (40) (80) (70) (20) 18 X 8 7 14 W1 W2 W3 W4 F1 5 2 X (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 (40) (80) (70) (20) 18 X 6 7 14 W1 W2 W3 W4 F1 5 2 X (19) (30) (50) (10) F2 6 3 (70) (30) (40) (60) F3 (40) (80) (70) (20) 18 X X 7 14 W1 W2 W3 W4 F1 5 2 X (19) (30) (50) (10) F2 6 3 X (70) (30) (40) (60) F3 (40) (80) (70) (20) 18 X X 4 14 W1 W2 W3 W4 F1 5 2 X (19) (30) (50) (10) F2 6 3 X (70) (30) (40) (60) F3 4 14 (40) (80) (70) (20) X X X 14 W1 W2 W3 W4 F1 5 2 X (19) (30) (50) (10) F2 6 3 X (70) (30) (40) (60) F3 4 14 X (40) (80) (70) (20) X X X X Initial Basic Feasible Solution x11 = 5, x12 = 2, x22 = 6, x23 = 3, x33 = 4, x34 = 14 The transportation cost is 5 (19) + 2 (30) + 6 (30) + 3 (40) + 4 (70) + 14 (20) = Rs. 1015 2. D1 D2 D3 D4 Availability S1 11 13 17 14 250 S2 16 18 14 10 300 S3 21 24 13 10 400 Requirement 200 225 275 250 Solution D1 D2 D3 D4 S1 200 50 250 50 0 (11) (13) S2 175 125 300 125 0 (18) (10) S3 275 125 400 125 0 (13) (10) 200 0 225 175 0 275 0 250 0 Initial Basic Feasible Solution x11 = 200, x12 = 50, x22 = 175, x24 = 125, x33 = 275, x34 = 125 The transportation cost is 200 (11) + 50 (13) + 175 (18) + 125 (10) + 275 (13) + 125 (10) = Rs. 12075 2.1.4 Lowest Cost Entry Method (Matrix Minima Method) Step 1 Determine the smallest cost in the cost matrix of the transportation table. Allocate xij = min (ai, bj) in the cell (i, j) Step 2 If xij = ai, cross out the ith row of the table and decrease bj by ai. Go to step 3. If xij = bj, cross out the jth column of the table and decrease ai by bj. Go to step 3. If xij = ai = bj, cross out the ith row or jth column but not both. Step 3 Repeat steps 1 and 2 for the resulting reduced transportation table until all the requirements are satisfied. Whenever the minimum cost is not unique, make an arbitrary choice among the minima. Find the initial basic feasible solution using Matrix Minima method 1. W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Requirement 5 8 7 14 Solution W1 W2 W3 W4 F1 (19) (30) (50) (10) 7 F2 (70) (30) (40) (60) 9 F3 8 10 (40) (8) (70) (20) 5 X 7 14  W1 W2 W3 W4 F1 7 X (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 8 10 (40) (8) (70) (20) 5 X 7 7  W1 W2 W3 W4 F1 7 X (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 8 7 3 (40) (8) (70) (20) 5 X 7 X  W1 W2 W3 W4 F1 7 X (19) (30) (50) (10) F2 (70) (30) (40) (60) 9 F3 3 8 7 X (40) (8) (70) (20) 2 X 7 X W1 W2 W3 W4 F1 7 X (19) (30) (50) (10) F2 2 7 X (70) (30) (40) (60) F3 3 8 7 X (40) (8) (70) (20) X X X X Initial Basic Feasible Solution x14 = 7, x21 = 2, x23 = 7, x31 = 3, x32 = 8, x34 = 7 The transportation cost is 7 (10) + 2 (70) + 7 (40) + 3 (40) + 8 (8) + 7 (20) = Rs. 814 2. To Availability From 2 11 10 3 7 4 1 4 7 2 1 8 3 9 4 8 12 9 Requirement 3 3 4 5 6 Solution To From 4 4 0 (3) 3 5 8 5 0 (1) (1) 3 4 1 1 9 5 4 1 0 (9) (4) (8) (12) 3 0 3 0 4 0 5 1 0 6 1 0 Initial Basic Feasible Solution x14 = 4, x21 = 3, x25 = 5, x32 = 3, x33 = 4, x34 = 1, x35 = 1 The transportation cost is 4 (3) + 3 (1) + 5(1) + 3 (9) + 4 (4) + 1 (8) + 1 (12) = Rs. 78 2.1.5 Vogel’s Approximation Method (Unit Cost Penalty Method) Step1 For each row of the table, identify the smallest and the next to smallest cost. Determine the difference between them for each row. These are called penalties. Put them aside by enclosing them in the parenthesis against the respective rows. Similarly compute penalties for each column. Step 2 Identify the row or column with the largest penalty. If a tie occurs then use an arbitrary choice. Let the largest penalty corresponding to the ith row have the cost cij. Allocate the largest possible amount xij = min (ai, bj) in the cell (i, j) and cross out either ith row or jth column in the usual manner. Step 3 Again compute the row and column penalties for the reduced table and then go to step 2. Repeat the procedure until all the requirements are satisfied. Find the initial basic feasible solution using vogel’s approximation method 1. W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Requirement 5 8 7 14 Solution W1 W2 W3 W4 Availability Penalty F1 19 30 50 10 7 19-10=9 F2 70 30 40 60 9 40-30=10 F3 40 8 70 20 18 20-8=12 Requirement 5 8 7 14 Penalty 40-19=21 30-8=22 50-40=10 20-10=10 W1 W2 W3 W4 Availability Penalty F1 (19) (30) (50) (10) 7 9 F2 (70) (30) (40) (60) 9 10 F3 (40) 8(8) (70) (20) 18/10 12 Requirement 5 8/0 7 14 Penalty 21 22 10 10 W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) (10) 7/2 9 F2 (70) (30) (40) (60) 9 20 F3 (40) 8(8) (70) (20) 18/10 20 Requirement 5/0 X 7 14 Penalty 21 X 10 10 W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) (10) 7/2 40 F2 (70) (30) (40) (60) 9 20 F3 (40) 8(8) (70) 10(20) 18/10/0 50 Requirement X X 7 14/4 Penalty X X 10 10 W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) 2(10) 7/2/0 40 F2 (70) (30) (40) (60) 9 20 F3 (40) 8(8) (70) 10(20) X X Requirement X X 7 14/4/2 Penalty X X 10 50 W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) 2(10) X X F2 (70) (30) 7(40) 2(60) X X F3 (40) 8(8) (70) 10(20) X X Requirement X X X X Penalty X X X X Initial Basic Feasible Solution x11 = 5, x14 = 2, x23 = 7, x24 = 2, x32 = 8, x34 = 10 The transportation cost is 5 (19) + 2 (10) + 7 (40) + 2 (60) + 8 (8) + 10 (20) = Rs. 779 2. Stores Availability I II III IV Warehouse A 21 16 15 13 11 B 17 18 14 23 13 C 32 27 18 41 19 Requirement 6 10 12 15 Solution Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) (13) 11 2 B (17) (18) (14) (23) 13 3 C (32) (27) (18) (41) 19 9 Requirement 6 10 12 15 Penalty 4 2 1 10 Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) 11(13) 11/0 2 B (17) (18) (14) (23) 13 3 C (32) (27) (18) (41) 19 9 Requirement 6 10 12 15/4 Penalty 4 2 1 10 Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) 11(13) X X B (17) (18) (14) 4(23) 13/9 3 C (32) (27) (18) (41) 19 9 Requirement 6 10 12 15/4/0 Penalty 15 9 4 18 Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) 11(13) X X B 6(17) (18) (14) 4(23) 13/9/3 3 C (32) (27) (18) (41) 19 9 Requirement 6/0 10 12 X Penalty 15 9 4 X Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) 11(13) X X B 6(17) 3(18) (14) 4(23) 13/9/3/0 4 C (32) (27) (18) (41) 19 9 Requirement X 10/7 12 X Penalty X 9 4 X Stores Availability Penalty I II III IV Warehouse A (21) (16) (15) 11(13) X X B 6(17) 3(18) (14) 4(23) X X C (32) 7(27) 12(18) (41) X X Requirement X X X X Penalty X X X X Initial Basic Feasible Solution x14 = 11, x21 = 6, x22 = 3, x24 = 4, x32 = 7, x33 = 12 The transportation cost is 11 (13) + 6 (17) + 3 (18) + 4 (23) + 7 (27) + 12 (18) = Rs. 796 Exercise Determine an initial basic feasible solution to the following transportation problem using north-west corner rule. From I II III IV Supply A 13 11 15 20 2000 B 17 14 12 13 6000 C 18 18 15 12 7000 Demand 3000 3000 4000 5000 [Ans. x11 = 2, x21 = 1, x22 = 3, x23 = 2, x34 = 5] Determine an initial basic feasible solution to the following transportation problem using row/column minima method. To Supply From 6 3 5 4 22 5 9 2 7 15 5 7 8 6 8 Demand 7 12 17 9 [Ans. x12 = 12, x13 = 1, x14 = 9, x23 = 15, x31 = 7, x33 = 1] Obtain an initial basic feasible solution to the following transportation problem using matrix minima method. From D1 D2 D3 D4 Capacity O1 1 2 3 4 6 O2 4 3 2 0 8 O3 0 2 2 1 10 Demand 4 6 8 6 [Ans. x12 = 6, x23 = 2, x24 = 6, x31 = 4, x32 = 0, x33 = 6] Determine the minimum cost to the following transportation problem using Vogel’s method. From D1 D2 D3 D4 D5 Capacity O1 2 11 10 3 7 4 O2 1 4 7 2 1 8 O3 3 9 4 8 12 9 Demand 3 3 4 5 6 21 [Ans. Min cost = Rs 68] Determine the minimum cost to the following transportation problem using matrix minima method and vogel’s method From D1 D2 D3 D4 Capacity O1 1 2 1 4 30 O2 3 3 2 1 50 O3 4 2 5 9 20 Demand 20 40 30 10 [Ans. Min cost = Rs 180] Unit 3 3.1 Examining the Initial Basic Feasible Solution for Non-Degeneracy 3.2 Transportation Algorithm for Minimization Problem 3.3 Worked Examples 3.1 Examining the Initial Basic Feasible Solution for Non-Degeneracy Examine the initial basic feasible solution for non-degeneracy. If it is said to be non-degenerate then it has the following two properties Initial basic feasible solution must contain exactly m + n – 1 number of individual allocations. These allocations must be in independent positions Independent Positions Non-Independent Positions 3.2 Transportation Algorithm for Minimization Problem (MODI Method) Step 1 Construct the transportation table entering the origin capacities ai, the destination requirement bj and the cost cij Step 2 Find an initial basic feasible solution by vogel’s method or by any of the given method. Step 3 For all the basic variables xij, solve the system of equations ui + vj = cij, for all i, j for which cell (i, j) is in the basis, starting initially with some ui = 0, calculate the values of ui and vj on the transportation table Step 4 Compute the cost differences dij = cij – ( ui + vj ) for all the non-basic cells Step 5 Apply optimality test by examining the sign of each dij If all dij ≥ 0, the current basic feasible solution is optimal If at least one dij < 0, select the variable xrs (most negative) to enter the basis. Solution under test is not optimal if any dij is negative and further improvement is required by repeating the above process. Step 6 Let the variable xrs enter the basis. Allocate an unknown quantity Ө to the cell (r, s). Then construct a loop that starts and ends at the cell (r, s) and connects some of the basic cells. The amount Ө is added to and subtracted from the transition cells of the loop in such a manner that the availabilities and requirements remain satisfied. Step 7 Assign the largest possible value to the Ө in such a way that the value of at least one basic variable becomes zero and the other basic variables remain non-negative. The basic cell whose allocation has been made zero will leave the basis. Step 8 Now, return to step 3 and repeat the process until an optimal solution is obtained. 3.3 Worked Examples Example 1 Find an optimal solution W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Requirement 5 8 7 14 Solution 1. Applying vogel’s approximation method for finding the initial basic feasible solution W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) 2(10) X X F2 (70) (30) 7(40) 2(60) X X F3 (40) 8(8) (70) 10(20) X X Requirement X X X X Penalty X X X X Minimum transportation cost is 5 (19) + 2 (10) + 7 (40) + 2 (60) + 8 (8) + 10 (20) = Rs. 779 2. Check for Non-degeneracy The initial basic feasible solution has m + n – 1 i.e. 3 + 4 – 1 = 6 allocations in independent positions. Hence optimality test is satisfied. 3. Calculation of ui and vj : - ui + vj = cij ui (19) (10) u1= -10 (40) (60) u2 = 40 (8) (20) u3 = 0 vj v1 = 29 v2 = 8 v3 = 0 v4 = 20 Assign a ‘u’ value to zero. (Convenient rule is to select the ui, which has the largest number of allocations in its row) Let u3 = 0, then u3 + v4= 20 which implies 0 + v4 = 20, so v4 = 20 u2 + v4= 60 which implies u2 + 20 = 60, so u2 = 40 u1 + v4= 10 which implies u1 + 20 = 10, so u1 = -10 u2 + v3= 40 which implies 40 + v3 = 40, so v3 = 0 u3 + v2= 8 which implies 0 + v2 = 8, so v2 = 8 u1 + v1= 19 which implies -10 + v1= 19, so v1 = 29 4. Calculation of cost differences for non basic cells dij = cij – ( ui + vj ) cij ui + vj (30) (50) -2 -10 (70) (30) 69 48 (40) (70) 29 0 dij = cij – ( ui + vj ) 32 60 1 -18 11 70 5. Optimality test dij < 0 i.e. d22 = -18 so x22 is entering the basis 6. Construction of loop and allocation of unknown quantity Ө We allocate Ө to the cell (2, 2). Reallocation is done by transferring the maximum possible amount Ө in the marked cell. The value of Ө is obtained by equating to zero to the corners of the closed loop. i.e. min(8-Ө, 2-Ө) = 0 which gives Ө = 2. Therefore x24 is outgoing as it becomes zero. 5 (19) 2 (10) 2 (30) 7 (40) 6 (8) 12 (20) Minimum transportation cost is 5 (19) + 2 (10) + 2 (30) + 7 (40) + 6 (8) + 12 (20) = Rs. 743 7. Improved Solution ui (19) (10) u1= -10 (30) (40) u2 = 22 (8) (20) u3 = 0 vj v1 = 29 v2 = 8 v3 = 18 v4 = 20 cij ui + vj (30) (50) -2 8 (70) (60) 51 42 (40) (70) 29 18 dij = cij – ( ui + vj ) 32 42 19 18 11 52 Since dij > 0, an optimal solution is obtained with minimal cost Rs.743 Example 2 Solve by lowest cost entry method and obtain an optimal solution for the following problem Available From 50 30 220 1 90 45 170 3 250 200 50 4 Required 4 2 2 Solution By lowest cost entry method Available From 1(30) 1/0 2(90) 1(45) 3/2/0 2(250) 2(50) 4/2/0 Required 4/2/2 2/1/0 2/0 Minimum transportation cost is 1 (30) + 2 (90) + 1 (45) + 2 (250) + 2 (50) = Rs. 855 Check for Non-degeneracy The initial basic feasible solution has m + n – 1 i.e. 3 + 3 – 1 = 5 allocations in independent positions. Hence optimality test is satisfied. Calculation of ui and vj : - ui + vj = cij ui (30) u1= -15 (90) (45) u2 = 0 (250) (50) u3 = 160 vj v1 = 90 v2 = 45 v3 = -110 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 50 220 75 -125 170 -110 200 205 dij = cij – ( ui + vj ) -25 345 280 -5 Optimality test dij < 0 i.e. d11 = -25 is most negative So x11 is entering the basis Construction of loop and allocation of unknown quantity Ө min(2-Ө, 1-Ө) = 0 which gives Ө = 1. Therefore x12 is outgoing as it becomes zero. 1(50) 1(90) 2(45) 2(250) 2(50) Minimum transportation cost is 1 (50) + 1 (90) + 2 (45) + 2 (250) + 2 (50) = Rs. 830 II Iteration Calculation of ui and vj : - ui + vj = cij ui (50) u1= -40 (90) (45) u2 = 0 (250) (50) u3 = 160 vj v1 = 90 v2 = 45 v3 = -110 Calculation of dij = cij – ( ui + vj ) cij ui + vj 30 220 5 -150 170 -110 200 205 dij = cij – ( ui + vj ) 25 370 280 -5 Optimality test dij < 0 i.e. d32 = -5 So x32 is entering the basis Construction of loop and allocation of unknown quantity Ө 2 – Ө = 0 which gives Ө = 2. Therefore x22 and x31 is outgoing as it becomes zero. 1(50) 3(90) 0(45) 2(200) 2(50) Minimum transportation cost is 1 (50) + 3 (90) + 2 (200) + 2 (50) = Rs. 820 III Iteration Calculation of ui and vj : - ui + vj = cij ui (50) u1= -40 (90) (45) u2 = 0 (200) (50) u3 = 155 vj v1 = 90 v2 = 45 v3 = -105 Calculation of dij = cij – ( ui + vj ) cij ui + vj 30 220 5 -145 170 -105 250 245 dij = cij – ( ui + vj ) 25 365 275 5 Since dij > 0, an optimal solution is obtained with minimal cost Rs.820 Example 3 Is x13 = 50, x14 = 20, x21 = 55, x31 = 30, x32 = 35, x34 = 25 an optimal solution to the transportation problem. Available From 6 1 9 3 70 11 5 2 8 55 10 12 4 7 90 Required 85 35 50 45 Solution Available From 50(9) 20(3) X 55(11) X 30(10) 35(12) 25(7) X Required X X X X Minimum transportation cost is 50 (9) + 20 (3) + 55 (11) + 30 (10) + 35 (12) + 25 (7) = Rs. 2010 Check for Non-degeneracy The initial basic feasible solution has m + n – 1 i.e. 3 + 4 – 1 = 6 allocations in independent positions. Hence optimality test is satisfied. Calculation of ui and vj : - ui + vj = cij ui (9) (3) u1= -4 (11) u2 = 1 (10) (12) (7) u3 = 0 vj v1 = 10 v2 = 12 v3 = 13 v4 = 7 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 6 1 6 8 5 2 8 13 14 8 4 13 dij = cij – ( ui + vj ) 0 -7 -8 -12 0 -9 Optimality test dij < 0 i.e. d23 = -12 is most negative So x23 is entering the basis Construction of loop and allocation of unknown quantity Ө min(50-Ө, 55-Ө, 25-Ө) = 25 which gives Ө = 25. Therefore x34 is outgoing as it becomes zero. 25(9) 45(3) 30(11) 25(2) 55(10) 35(12) Minimum transportation cost is 25 (9) + 45 (3) + 30 (11) + 25 (2) + 55 (10) + 35 (12) = Rs. 1710 II iteration Calculation of ui and vj : - ui + vj = cij ui (9) (3) u1= 8 (11) (2) u2 = 1 (10) (12) u3 = 0 vj v1 = 10 v2 = 12 v3 = 1 v4 = -5 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 6 1 18 20 5 8 13 -4 4 7 1 -5 dij = cij – ( ui + vj ) -12 -19 -8 12 3 12 Optimality test dij < 0 i.e. d12 = -19 is most negative So x12 is entering the basis Construction of loop and allocation of unknown quantity Ө min(25-Ө, 30-Ө, 35-Ө) = 25 which gives Ө = 25. Therefore x13 is outgoing as it becomes zero. 25(1) 45(3) 5(11) 50(2) 80(10) 10(12) Minimum transportation cost is 25 (1) + 45 (3) + 5 (11) + 50 (2) + 80 (10) + 10 (12) = Rs. 1235 III Iteration Calculation of ui and vj : - ui + vj = cij ui (1) (3) u1= -11 (11) (2) u2 = 1 (10) (12) u3 = 0 vj v1 = 10 v2 = 12 v3 = 1 v4 = 14 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 6 9 -1 -10 5 8 13 15 4 7 1 14 dij = cij – ( ui + vj ) 7 19 -8 -7 3 -7 Optimality test dij < 0 i.e. d22 = -8 is most negative So x22 is entering the basis Construction of loop and allocation of unknown quantity Ө min(5-Ө, 10-Ө) = 5 which gives Ө = 5. Therefore x21 is outgoing as it becomes zero. 25(1) 45(3) 5(5) 50(2) 85(10) 5(12) Minimum transportation cost is 25 (1) + 45 (3) + 5 (5) + 50 (2) + 85 (10) + 5 (12) = Rs. 1195 IV Iteration Calculation of ui and vj : - ui + vj = cij ui (1) (3) u1= -11 (5) (2) u2 = -7 (10) (12) u3 = 0 vj v1 = 10 v2 = 12 v3 = 9 v4 = 14 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 6 9 -1 -2 11 8 3 7 4 7 9 14 dij = cij – ( ui + vj ) 7 11 8 1 -5 -7 Optimality test dij < 0 i.e. d34 = -7 is most negative So x34 is entering the basis Construction of loop and allocation of unknown quantity Ө min(5-Ө, 45-Ө) = 5 which gives Ө = 5. Therefore x32 is outgoing as it becomes zero. 30(1) 40(3) 5(5) 50(2) 85(10) 5(7) Minimum transportation cost is 30 (1) + 40 (3) + 5 (5) + 50 (2) + 85 (10) + 5 (7) = Rs. 1160 V Iteration Calculation of ui and vj : - ui + vj = cij ui (1) (3) u1= -4 (5) (2) u2 = 0 (10) (7) u3 = 0 vj v1 = 10 v2 = 5 v3 = 2 v4 = 7 Calculation of cost differences for non-basic cells dij = cij – ( ui + vj ) cij ui + vj 6 9 6 -2 11 8 10 7 12 4 5 2 dij = cij – ( ui + vj ) 0 11 1 1 7 2 Since dij > 0, an optimal solution is obtained with minimal cost Rs.1160. Further more d11 = 0 which indicates that alternative optimal solution also exists. Exercise Determine the optimal solution of the given transportation problem To Supply From 2 3 11 7 6 1 0 6 1 1 5 8 15 10 10 Demand 7 5 3 2 17 [Ans. x12 = 5, x13 = 1, x24 = 1, x31 = 7, x33 = 2, x34 = 1 Min cost = Rs 102] Using North-West Corner rule for initial basic feasible solution, obtain an optimum basic feasible solution to the following problem To Available From 7 3 4 2 2 1 3 3 3 4 6 5 Demand 4 1 5 10 [Ans. x13 = 2, x22 = 1, x23 = 2, x31 = 4, x33 = 1 Min cost = Rs 33] Determine the optimal solution of the given transportation problem To Supply From 10 7 3 6 3 1 6 7 3 5 7 4 5 3 7 Demand 3 2 6 4 [Ans. x13 = 3, x21 = 3, x24 = 2, x32 = 2, x33 = 3, x34 = 2, Min cost = Rs 47] Module 5 Unit 1 Introduction to Assignment Problem Algorithm for Assignment Problem Worked Examples Unbalanced Assignment Problem Maximal Assignment Problem 1.1 Introduction to Assignment Problem In assignment problems, the objective is to assign a number of jobs to the equal number of persons at a minimum cost of maximum profit. Suppose there are ‘n’ jobs to be performed and ‘n’ persons are available for doing these jobs. Assume each person can do each job at a time with a varying degree of efficiency. Let cij be the cost of ith person assigned to jth job. Then the problem is to find an assignment so that the total cost for performing all jobs is minimum. Such problems are known as assignment problems. These problems may consist of assigning men to offices, classes to the rooms or problems to the research team etc. Mathematical formulation Cost matrix: cij= c11 c12 c13 … c1n c21 c22 c23 … c2n . . . cn1 cn2 cn3 … cnn Subject to restrictions of the form Where xij denotes that jth job is to be assigned to the ith person. This special structure of assignment problem allows a more convenient method of solution in comparison to simplex method. 1.2 Algorithm for Assignment Problem (Hungarian Method) Step 1 Subtract the minimum of each row of the effectiveness matrix, from all the elements of the respective rows (Row reduced matrix). Step 2 Further modify the resulting matrix by subtracting the minimum element of each column from all the elements of the respective columns. Thus first modified matrix is obtained. Step 3 Draw the minimum number of horizontal and vertical lines to cover all the zeroes in the resulting matrix. Let the minimum number of lines be N. Now there may be two possibilities If N = n, the number of rows (columns) of the given matrix then an optimal assignment can be made. So make the zero assignment to get the required solution. If N < n then proceed to step 4 Step 4 Determine the smallest element in the matrix, not covered by N lines. Subtract this minimum element from all uncovered elements and add the same element at the intersection of horizontal and vertical lines. Thus the second modified matrix is obtained. Step 5 Repeat step 3 and step 4 until minimum number of lines become equal to number of rows (columns) of the given matrix i.e. N = n. Step 6 To make zero assignment - examine the rows successively until a row-wise exactly single zero is found; mark this zero by ‘1’‘to make the assignment. Then, mark a ‘X’ over all zeroes if lying in the column of the marked zero, showing that they cannot be considered for further assignment. Continue in this manner until all the rows have been examined. Repeat the same procedure for the columns also. Step 7 Repeat the step 6 successively until one of the following situations arise If no unmarked zero is left, then process ends If there lies more than one of the unmarked zeroes in any column or row, then mark ‘1’‘one of the unmarked zeroes arbitrarily and mark a cross in the cells of remaining zeroes in its row and column. Repeat the process until no unmarked zero is left in the matrix. Step 8 Exactly one marked zero in each row and each column of the matrix is obtained. The assignment corresponding to these marked zeroes will give the optimal assignment. 1.3 Worked Examples Example 1 A department head has four subordinates and four tasks have to be performed. Subordinates differ in efficiency and tasks differ in their intrinsic difficulty. Time each man would take to perform each task is given in the effectiveness matrix. How the tasks should be allocated to each person so as to minimize the total man-hours? Tasks Subordinates I II III IV A 8 26 17 11 B 13 28 4 26 C 38 19 18 15 D 19 26 24 10 Solution Row Reduced Matrix 0 18 9 3 9 24 0 22 23 4 3 0 9 16 14 0 I Modified Matrix N = 4, n = 4 Since N = n, we move on to zero assignment Zero assignment Total man-hours = 8 + 4 + 19 + 10 = 41 hours Example 2 A car hire company has one car at each of five depots a, b, c, d and e. a customer requires a car in each town namely A, B, C, D and E. Distance (kms) between depots (origins) and towns (destinations) are given in the following distance matrix a b c d e A 160 130 175 190 200 B 135 120 130 160 175 C 140 110 155 170 185 D 50 50 80 80 110 E 55 35 70 80 105 Solution Row Reduced Matrix 30 0 45 60 70 15 0 10 40 55 30 0 45 60 75 0 0 30 30 60 20 0 35 45 70 I Modified Matrix N < n i.e. 3 < 5, so move to next modified matrix II Modified Matrix N = 5, n = 5 Since N = n, we move on to zero assignment Zero assignment Minimum distance travelled = 200 + 130 + 110 + 50 + 80 = 570 kms Example 3 Solve the assignment problem whose effectiveness matrix is given in the table 1 2 3 4 A 49 60 45 61 B 55 63 45 69 C 52 62 49 68 D 55 64 48 66 Solution Row-Reduced Matrix 4 15 0 16 10 18 0 24 3 13 0 19 7 16 0 18 I Modified Matrix N < n i.e 3 < 4, so II modified matrix II Modified Matrix N < n i.e 3 < 4 III Modified matrix Since N = n, we move on to zero assignment Zero assignment Multiple optimal assignments exists Solution - I Total cost = 49 + 45 + 62 + 66 = 222 units Solution – II Minimum cost = 61 + 45 + 52 + 64 = 222 units Example 4 Certain equipment needs 5 repair jobs which have to be assigned to 5 machines. The estimated time (in hours) that a mechanic requires to complete the repair job is given in the table. Assuming that each mechanic can be assigned only one job, determine the minimum time assignment. J1 J2 J3 J4 J5 M1 7 5 9 8 11 M2 9 12 7 11 10 M3 8 5 4 6 9 M4 7 3 6 9 5 M5 4 6 7 5 11 Solution Row Reduced Matrix 2 0 4 3 6 2 5 0 4 3 4 1 0 2 5 4 0 3 6 2 0 2 3 1 7 I Modified Matrix N < n II Modified Matrix N = n Zero assignment Minimum time = 5 + 7 + 6 + 5 + 4 = 27 hours 1.4 Unbalanced Assignment Problems If the number of rows and columns are not equal then such type of problems are called as unbalanced assignment problems. Example 1 A company has 4 machines on which to do 3 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 Machines Jobs W X Y Z A 18 24 28 32 B 8 13 17 19 C 10 15 19 22 Solution 18 24 28 32 8 13 17 19 10 15 19 22 0 0 0 0 Row Reduced matrix 0 6 10 14 0 5 9 11 0 5 9 12 0 0 0 0 I Modified Matrix N < n i.e. 2 < 4 II Modified Matrix N < n i.e. 3 < 4 III Modified Matrix N = n Zero assignment Multiple assignments exists Solution -I Minimum cost = 18 + 13 + 19 = Rs 50 Solution -II Minimum cost = 18 + 17 + 15 = Rs 50 Example 2 Solve the assignment problem whose effectiveness matrix is given in the table R1 R2 R3 R4 C1 9 14 19 15 C2 7 17 20 19 C3 9 18 21 18 C4 10 12 18 19 C5 10 15 21 16 Solution 9 14 19 15 0 7 17 20 19 0 9 18 21 18 0 10 12 18 19 0 10 15 21 16 0 Row Reduced Matrix 9 14 19 15 0 7 17 20 19 0 9 18 21 18 0 10 12 18 19 0 10 15 21 16 0 I Modified Matrix N < n i.e. 4 < 5 II Modified Matrix N < n i.e. 4 < 5 III Modified Matrix N = n Zero assignment Minimum cost = 19 + 7 + 12 + 16 = 54 units 1.5 Maximal Assignment Problem Example 1 A company has 5 jobs to be done. The following matrix shows the return in terms of rupees on assigning ith ( i = 1, 2, 3, 4, 5 ) machine to the jth job ( j = A, B, C, D, E ). Assign the five jobs to the five machines so as to maximize the total expected profit. Jobs Machines A B C D E 1 5 11 10 12 4 2 2 4 6 3 5 3 3 12 5 14 6 4 6 14 4 11 7 5 7 9 8 12 5 Solution Subtract all the elements from the highest element Highest element = 14 9 3 4 2 10 12 10 8 11 9 11 2 9 0 8 8 0 10 3 7 7 5 6 2 9 Row Reduced matrix 7 1 2 0 8 4 2 0 3 1 11 2 9 0 8 8 0 10 3 7 5 3 4 0 7 I Modified Matrix N < n i.e. 3 < 5 II Modified Matrix N < n i.e. 4 < 5 III Modified Matrix N = n Zero assignment Optimal assignment 1 – C 2 – E 3 – D 4 – B 5 – A Maximum profit = 10 + 5 + 14 + 14 + 7 = Rs. 50 Exercise What is assignment problem? Give any two areas of its applications Find the optimal solution for the assignment problem with the following cost matrix I II III IV A 5 3 1 8 B 7 9 2 6 C 6 4 5 7 D 5 7 7 6 [Ans. A - III, B - IV, C - II, D – I, Min cost = Rs.16] Solve the following assignment problem 1 2 3 4 A 10 12 19 11 B 5 10 7 8 C 12 14 13 11 D 8 15 11 9 [Ans. A - 2, B - 3, C - 4, D – 1, Min cost = Rs.38] The jobs A, B, C are to be assigned to three machines X, Y, Z. The processing costs (Rs.) are as given in the matrix below. Find the allocation which will minimize the overall processing cost. X Y Z A 19 28 31 B 11 17 16 C 12 15 13 [Ans. A – X, B - Y, C – Z] A company is faced with the problem of assigning 4 machines to 6 different jobs (one machine to one job only).the profits are estimated as follows A B C D 1 3 6 2 6 2 7 1 4 4 3 3 8 5 8 4 6 4 3 7 5 5 2 4 3 6 5 7 6 4 [Ans. 2 - A, 3 - B, 4 - D, 6 – C, Max profit = Rs.28] Unit 2 2.1 Introduction to Game Theory 2.2 Properties of a Game 2.3 Characteristics of Game Theory 2.4 Classification of Games 2.5 Limitations of Game Theory 2.5 Solving Two-Person and Zero-Sum Game 2.1 Introduction to Game Theory Game theory is a distinct and interdisciplinary approach to the study of human behavior. The disciplines most involved in game theory are mathematics, economics and the other social and behavioral sciences. Game theory (like computational theory and so many other contributions) was founded by the great mathematician John von Neumann. Game theory is a type of decision theory in which one’s choice of action is determined after taking into account all possible alternatives available to an opponent playing the same game, rather than just by the possibilities of several outcome results. Game theory does not insist on how a game should be played but tells the procedure and principles by which action should be selected. Thus it is a decision theory useful in competitive situations. Game is defined as an activity between two or more persons according to a set of rules at the end of which each person receives some benefit or suffers loss. The set of rules defines the game. Going through the set of rules once by the participants defines a play. A Scientific Metaphor Since the work of John von Neumann, "games" have been a scientific metaphor for a much wider range of human interactions in which the outcomes depend on the interactive strategies of two or more persons, who have opposed or at best mixed motives. Among the issues discussed in game theory are 1) What does it mean to choose strategies "rationally" when outcomes depend on the strategies chosen by others and when information is incomplete? 2) In "games" that allow mutual gain (or mutual loss) is it "rational" to cooperate to realize the mutual gain (or avoid the mutual loss) or is it "rational" to act aggressively in seeking individual gain regardless of mutual gain or loss? 3) If the answers to 2) are "sometimes," in what circumstances is aggression rational and in what circumstances is cooperation rational? 4) In particular, do ongoing relationships differ from one-off encounters in this connection? 5) Can moral rules of cooperation emerge spontaneously from the interactions of rational egoists? 6) How does real human behavior correspond to "rational" behavior in these cases? 7) If it differs, in what direction? Are people more cooperative than would be "rational?" More aggressive? Both? Thus, among the "games" studied by game theory are Bankruptcy Barbarians at the Gate Battle of the Networks Caveat Emptor Conscription Coordination Escape and Evasion Frogs Call for Mates Hawk versus Dove Mutually Assured Destruction Majority Rule Market Niche Mutual Defense Prisoner's Dilemma Subsidized Small Business Tragedy of the Commons Ultimatum Video System Coordination Why Do Economists Study Games? Games are a convenient way in which to model the strategic interactions among economic agents. Many economic issues involve strategic interaction. Behavior in imperfectly competitive markets, e.g. Coca-Cola versus Pepsi. Behavior in auctions, e.g. Investment banks bidding on U.S. Treasury bills. Behavior in economic negotiations, e.g. trade. Game theory is not limited to Economics 2.2 Properties of a Game There are finite numbers of competitors called ‘players’ Each player has a finite number of possible courses of action called ‘strategies’ All the strategies and their effects are known to the players but player does not know which strategy is to be chosen. A game is played when each player chooses one of his strategies. The strategies are assumed to be made simultaneously with an outcome such that no player knows his opponents strategy until he decides his own strategy. The game is a combination of the strategies and in certain units which determines the gain or loss. The figures shown as the outcomes of strategies in a matrix form are called ‘pay-off matrix’. The player playing the game always tries to choose the best course of action which results in optimal pay off called ‘optimal strategy’. The expected pay off when all the players of the game follow their optimal strategies is known as ‘value of the game’. The main objective of a problem of a game is to find the value of the game. The game is said to be ‘fair’ game if the value of the game is zero otherwise it s known as ‘unfair’. Characteristics of Game Theory 1. Competitive game A competitive situation is called a competitive game if it has the following four properties There are finite number of competitors such that n ≥ 2. In case n = 2, it is called a two-person game and in case n > 2, it is referred as n-person game. Each player has a list of finite number of possible activities. A play is said to occur when each player chooses one of his activities. The choices are assumed to be made simultaneously i.e. no player knows the choice of the other until he has decided on his own. Every combination of activities determines an outcome which results in a gain of payments to each player, provided each player is playing uncompromisingly to get as much as possible. Negative gain implies the loss of same amount. 2. Strategy The strategy of a player is the predetermined rule by which player decides his course of action from his own list during the game. The two types of strategy are Pure strategy Mixed strategy Pure Strategy If a player knows exactly what the other player is going to do, a deterministic situation is obtained and objective function is to maximize the gain. Therefore, the pure strategy is a decision rule always to select a particular course of action. Mixed Strategy If a player is guessing as to which activity is to be selected by the other on any particular occasion, a probabilistic situation is obtained and objective function is to maximize the expected gain. Thus the mixed strategy is a selection among pure strategies with fixed probabilities. Repeated Game Strategies In repeated games, the sequential nature of the relationship allows for the adoption of strategies that are contingent on the actions chosen in previous plays of the game. Most contingent strategies are of the type known as "trigger" strategies. Example trigger strategies In prisoners' dilemma: Initially play doesn’t confess. If your opponent plays Confess, then play Confess in the next round. If your opponent plays don’t confess, then play doesn’t confess in the next round. This is known as the "tit for tat" strategy. In the investment game, if you are the sender: Initially play Send. Play Send as long as the receiver plays Return. If the receiver plays keep, never play Send again. This is known as the "grim trigger" strategy. 3. Number of persons A game is called ‘n’ person game if the number of persons playing is ‘n’. The person means an individual or a group aiming at a particular objective. Two-person, zero-sum game A game with only two players (player A and player B) is called a ‘two-person, zero-sum game’, if the losses of one player are equivalent to the gains of the other so that the sum of their net gains is zero. Two-person, zero-sum games are also called rectangular games as these are usually represented by a payoff matrix in a rectangular form. 4. Number of activities The activities may be finite or infinite. 5. Payoff The quantitative measure of satisfaction a person gets at the end of each play is called a payoff 6. Payoff matrix Suppose the player A has ‘m’ activities and the player B has ‘n’ activities. Then a payoff matrix can be formed by adopting the following rules Row designations for each matrix are the activities available to player A Column designations for each matrix are the activities available to player B Cell entry Vij is the payment to player A in A’s payoff matrix when A chooses the activity i and B chooses the activity j. With a zero-sum, two-person game, the cell entry in the player B’s payoff matrix will be negative of the corresponding cell entry Vij in the player A’s payoff matrix so that sum of payoff matrices for player A and player B is ultimately zero. 7. Value of the game Value of the game is the maximum guaranteed game to player A (maximizing player) if both the players uses their best strategies. It is generally denoted by ‘V’ and it is unique. 2.4 Classification of Games Simultaneous v. Sequential Move Games Games where players choose actions simultaneously are simultaneous move games. Examples: Prisoners' Dilemma, Sealed-Bid Auctions. Must anticipate what your opponent will do right now, recognizing that your opponent is doing the same. Games where players choose actions in a particular sequence are sequential move games. Examples: Chess, Bargaining/Negotiations. Must look ahead in order to know what action to choose now. Many sequential move games have deadlines/ time limits on moves. Many strategic situations involve both sequential and simultaneous moves. One-Shot versus Repeated Games One-shot: play of the game occurs once. Players likely to not know much about one another. Example - tipping on your vacation Repeated: play of the game is repeated with the same players. Indefinitely versus finitely repeated games Reputational concerns matter; opportunities for cooperative behavior may arise. Advise: If you plan to pursue an aggressive strategy, ask yourself whether you are in a one-shot or in a repeated game. If a repeated game, think again. Generally games are classified into Pure strategy games Mixed strategy games The method for solving these two types varies. By solving a game, we need to find best strategies for both the players and also to find the value of the game. Pure strategy games can be solved by saddle point method. The different methods for solving a mixed strategy game are Analytical method Graphical method Dominance rule Simplex method 2.5 Limitations of game theory The major limitations are The assumption that the players have the knowledge about their own payoffs and others is rather unrealistic. As the number of players increase in the game, the analysis of the gaming strategies become increasingly complex and difficult. The assumptions of maximin and minimax show that the players are risk-averse and have complete knowledge of the strategies. It doesn’t seem practical. Rather than each player in an oligopoly situation working under uncertain conditions, the players will allow each other to share the secrets of business in order to work out collusion. Then the mixed strategies are not very useful. 2.6 Solving Two-Person and Zero-Sum Game Two-person zero-sum games may be deterministic or probabilistic. The deterministic games will have saddle points and pure strategies exist in such games. In contrast, the probabilistic games will have no saddle points and mixed strategies are taken with the help of probabilities. Definition of saddle point A saddle point of a matrix is the position of such an element in the payoff matrix, which is minimum in its row and the maximum in its column. Procedure to find the saddle point Select the minimum element of each row of the payoff matrix and mark them with circles. Select the maximum element of each column of the payoff matrix and mark them with squares. If their appears an element in the payoff matrix with a circle and a square together then that position is called saddle point and the element is the value of the game. Solution of games with saddle point To obtain a solution of a game with a saddle point, it is feasible to find out Best strategy for player A Best strategy for player B The value of the game The best strategies for player A and B will be those which correspond to the row and column respectively through the saddle point. Examples Solve the payoff matrix 1. Player A Player B B1 B2 B3 A1 2 4 5 A2 10 7 9 A3 4 5 6 Solution Strategy of player A – A2 Strategy of player B – B2 Value of the game = 7 2. Player B Player A I II III IV V I -2 0 0 5 3 II 3 2 1 2 2 III -4 -3 0 -2 6 IV 5 3 -4 2 -6 Solution Strategy of player A – II Strategy of player B - III Value of the game = 1 3.. B1 B2 B3 B4 A1 1 7 3 4 A2 5 6 4 5 A3 7 2 0 3 Solution Strategy of player A – A2 Strategy of player B – B3 Value of the game = 4 4. B’s Strategy A’s Strategy B1 B2 B3 B4 B5 A1 8 10 -3 -8 -12 A2 3 6 0 6 12 A3 7 5 -2 -8 17 A4 -11 12 -10 10 20 A5 -7 0 0 6 2 Solution Strategy of player A – A2 Strategy of player B – B3 Value of the game = 0 5. Solution Value of the game = 4 Exercise Explain the concept of game theory. What is a rectangular game? What is a saddle point? Define pure and mixed strategy in a game. What are the characteristics of game theory? Explain two-person zero-sum game giving suitable examples. What are the limitations of game theory? Explain the following terms Competitive Game Strategy Value of the game Pay-off-matrix Optimal strategy Explain Maximin and Minimax used in game theory For the game with payoff matrix Determine the best strategies for player A and B and also the value of the game. Unit 3 3.1 Games with Mixed Strategies 3.1.1 Analytical Method 3.1.2 Graphical Method 3.1.3 Simplex Method 3.1 Games with Mixed Strategies In certain cases, no pure strategy solutions exist for the game. In other words, saddle point does not exist. In all such game, both players may adopt an optimal blend of the strategies called Mixed Strategy to find a saddle point. The optimal mix for each player may be determined by assigning each strategy a probability of it being chosen. Thus these mixed strategies are probabilistic combinations of available better strategies and these games hence called Probabilistic games. The probabilistic mixed strategy games without saddle points are commonly solved by any of the following methods Sl. No. Method Applicable to 1 Analytical Method 2x2 games 2 Graphical Method 2x2, mx2 and 2xn games 3 Simplex Method 2x2, mx2, 2xn and mxn games 3.1.1 Analytical Method A 2 x 2 payoff matrix where there is no saddle point can be solved by analytical method. Given the matrix Value of the game is With the coordinates Alternative procedure to solve the strategy Find the difference of two numbers in column 1 and enter the resultant under column 2. Neglect the negative sign if it occurs. Find the difference of two numbers in column 2 and enter the resultant under column 1. Neglect the negative sign if it occurs. Repeat the same procedure for the two rows. 1. Solve Solution It is a 2 x 2 matrix and no saddle point exists. We can solve by analytical method V = 17 / 5 SA = (x1, x2) = (1/5, 4 /5) SB = (y1, y2) = (3/5, 2 /5) 2. Solve the given matrix Solution V = - 1 / 4 SA = (x1, x2) = (1/4, 3 /4) SB = (y1, y2) = (1/4, 3 /4) 3.1.2 Graphical method The graphical method is used to solve the games whose payoff matrix has Two rows and n columns (2 x n) m rows and two columns (m x 2) Algorithm for solving 2 x n matrix games Draw two vertical axes 1 unit apart. The two lines are x1 = 0, x1 = 1 Take the points of the first row in the payoff matrix on the vertical line x1 = 1 and the points of the second row in the payoff matrix on the vertical line x1 = 0. The point a1j on axis x1 = 1 is then joined to the point a2j on the axis x1 = 0 to give a straight line. Draw ‘n’ straight lines for j=1, 2… n and determine the highest point of the lower envelope obtained. This will be the maximin point. The two or more lines passing through the maximin point determines the required 2 x 2 payoff matrix. This in turn gives the optimum solution by making use of analytical method. Example 1 Solve by graphical method Solution V = 66/13 SA = (4/13, 9 /13) SB = (0, 10/13, 3 /13) Example 2 Solve by graphical method Solution V = 8/7 SA = (3/7, 4 /7) SB = (2/7, 0, 5 /7) Algorithm for solving m x 2 matrix games Draw two vertical axes 1 unit apart. The two lines are x1 =0, x1 = 1 Take the points of the first row in the payoff matrix on the vertical line x1 = 1 and the points of the second row in the payoff matrix on the vertical line x1 = 0. The point a1j on axis x1 = 1 is then joined to the point a2j on the axis x1 = 0 to give a straight line. Draw ‘n’ straight lines for j=1, 2… n and determine the lowest point of the upper envelope obtained. This will be the minimax point. The two or more lines passing through the minimax point determines the required 2 x 2 payoff matrix. This in turn gives the optimum solution by making use of analytical method. Example 1 Solve by graphical method Solution V = 3/9 = 1/3 SA = (0, 5 /9, 4/9, 0) SB = (3/9, 6 /9) Example 2 Solve by graphical method Solution V = 73/17 SA = (0, 16/17, 1/17, 0, 0) SB = (5/17, 12 /17) 3.1.3 Simplex Method Let us consider the 3 x 3 matrix As per the assumptions, A always attempts to choose the set of strategies with the non-zero probabilities say p1, p2, p3 where p1 + p2 + p3 = 1 that maximizes his minimum expected gain. Similarly B would choose the set of strategies with the non-zero probabilities say q1, q2, q3 where q1 + q2 + q3 = 1 that minimizes his maximum expected loss. Step 1 Find the minimax and maximin value from the given matrix Step 2 The objective of A is to maximize the value, which is equivalent to minimizing the value 1/V. The LPP is written as Min 1/V = p1/V + p2/V + p3/V and constraints ≥ 1 It is written as Min 1/V = x1 + x2 + x3 and constraints ≥ 1 Similarly for B, we get the LPP as the dual of the above LPP Max 1/V = Y1 + Y2 + Y3 and constraints ≤ 1 Where Y1 = q1/V, Y2 = q2/V, Y3 = q3/V Step 3 Solve the LPP by using simplex table and obtain the best strategy for the players Example 1 Solve by Simplex method Solution We can infer that 2 ≤ V ≤ 3. Hence it can be concluded that the value of the game lies between 2 and 3 and the V > 0. LPP Max 1/V = Y1 + Y2 + Y3 Subject to 3Y1 – 2Y2 + 4Y3 ≤ 1 -1Y1 + 4Y2 + 2Y3 ≤ 1 2Y1 + 2Y2 + 6Y3 ≤ 1 Y1, Y2, Y3 ≥ 0 SLPP Max 1/V = Y1 + Y2 + Y3 + 0s1 + 0s2 + 0s3 Subject to 3Y1 – 2Y2 + 4Y3 + s1 = 1 -1Y1 + 4Y2 + 2Y3 + s2 =1 2Y1 + 2Y2 + 6Y3 + s3 = 1 Y1, Y2, Y3, s1, s2, s3 ≥ 0 Cj→ 1 1 1 0 0 0 Basic Variables CB YB Y1 Y2 Y3 S1 S2 S3 Min Ratio YB / YK S1 0 1 3 -2 4 1 0 0 1/3→ S2 0 1 -1 4 2 0 1 0 - S3 0 1 2 2 6 0 0 1 1/2 1/V = 0 ↑ -1 -1 -1 0 0 0 Y1 1 1/3 1 -2/3 4/3 1/3 0 0 - S2 0 4/3 0 10/3 10/3 1/3 1 0 2/5 S3 0 1/3 0 10/3 10/3 -2/3 0 1 1/10→ 1/V =1/3 0 ↑ -5/3 1/3 1/3 0 0 Y1 1 2/5 1 0 2 1/5 0 1/5 S2 0 1 0 0 0 1 1 -1 Y2 1 1/10 0 1 1 -1/5 0 3/10 1/V = 1/2 0 0 2 0 0 1/2 1/V =1/2 V = 2 y1 = 2/5 * 2 = 4/5 y2 = 1/10 * 2 = 1/5 y3 = 0 * 2 = 0 x1 = 0*2 = 0 x2 = 0*2 = 0 x3 = 1/2*2 = 1 SA = (0, 0, 1) SB = (4/5, 1/5, 0) Value = 2 Example 2 Solution Maximin = -1 Minimax = 1 We can infer that -1 ≤ V ≤ 1 Since maximin value is -1, it is possible that value of the game may be negative or zero, thus the constant ‘C’ is added to all the elements of matrix which is at least equal to the negative of maximin. Let C = 1, add this value to all the elements of the matrix. The resultant matrix is LPP Max 1/V = Y1 + Y2 + Y3 Subject to 2Y1 + 0Y2 + 0Y3 ≤ 1 0Y1 + 0Y2 + 4Y3 ≤ 1 0Y1 + 3Y2 + 0Y3 ≤ 1 Y1, Y2, Y3 ≥ 0 SLPP Max 1/V = Y1 + Y2 + Y3 + 0s1 + 0s2 + 0s3 Subject to 2Y1 + 0Y2 + 0Y3 + s1 = 1 0Y1 + 0Y2 + 4Y3 + s2 = 1 0Y1 + 3Y2 + 0Y3 + s3 = 1 Y1, Y2, Y3, s1, s2, s3 ≥ 0 Cj→ 1 1 1 0 0 0 Basic Variables CB YB Y1 Y2 Y3 S1 S2 S3 Min Ratio YB / YK S1 0 1 2 0 0 1 0 0 1/2→ S2 0 1 0 0 4 0 1 0 - S3 0 1 0 3 0 0 0 1 - 1/V =0 ↑ -1 -1 -1 0 0 0 Y1 1 1/2 1 0 0 1/2 0 0 - S2 0 1 0 0 4 0 1 0 - S3 0 1 0 3 0 0 0 1 1/3→ 1/V =1/2 0 ↑ -1 -1 1/2 0 0 Y1 1 1/2 1 0 0 1/2 0 0 - S2 0 1 0 0 4 0 1 0 1/4→ Y2 1 1/3 0 1 0 0 0 1/3 - 1/V = 5/6 0 0 ↑ -1 1/2 0 1/3 Y1 1 1/2 1 0 0 1/2 0 0 Y3 1 1/4 0 0 1 0 1/4 0 Y2 1 1/3 0 1 0 0 0 1/3 1/V =13/12 0 0 0 1/2 1/4 1/3 1/V =13/12 V = 12/13 y1 = 1/2 * 12/13 = 6/13 y2 = 1/3 * 12/13 = 4/13 y3 = 1/4 * 12/13 = 3/13 x1 = 1/2*12/13 = 6/13 x2 = 1/4 * 12/13 = 3/13 x3 = 1/3 * 12/13 = 4/13 SA = (6/13, 3/13, 4/13) SB = (6/13, 4/13, 3/13) Value = 12/13 – C =12/13 -1 = -1/13 Exercise 1. Explain the method of solving a problem with mixed strategy using algebraic method. 2. Solve the following game graphically 1. 2. 3. Use simplex to solve the following 1. 2. 4. Two companies A and B are competing for the same product. Their different strategies are given in the following pay off matrix Module 6 Unit 1 Shortest Route Problem Minimal Spanning Tree Problem Maximal Flow Problem 1.1 Shortest Route Problem The criterion of this method is to find the shortest distance between two nodes with minimal cost. Example 1 Find the shortest path Solution n Solved nodes directly connected to unsolved nodes Closest connected unsolved node Total distance involved nth nearest node Minimum distance Last connection 1 a c 7 c 7 a-c 2 a c b e 13 7+6 =13 b e 13 13 a-b c-e 3 b c e d f h 13+5 =18 7+11 =18 13+8 =21 d f - 18 18 - b-d c-f - 4 e d f h g h 13+8 =21 18+9 =27 18+5 =23 h - - 21 - - e-h - - 5 e h d g i g 13+10 =23 21+10 =31 18+9 =27 g - - 23 - - e-g - - 6 g h i i 23+6 =29 21+10 =31 i - 29 - g-i - The shortest path from a to i is a → c →e →g → i Distance = 7 + 6 + 10 + 6 = 29 units Example 2 Solution n Solved nodes directly connected to unsolved nodes Closest connected unsolved node Total distance involved nth nearest node Minimum distance Last connection 1 1 3 1 3 1 1-3 2 1 3 2 2 5 1+2 =3 - 2 - 3 - 3-2 3 2 3 5 4 3+1 =4 1+6 =7 5 - 4 - 2-5 - 4 2 3 5 6 4 4 3+6 =9 1+6 =7 4+3 =7 - 4 4 - 7 7 - 3-4 5-4 5 2 4 5 6 6 6 3+6 =9 7+4 =11 4+5 =9 6 - 6 9 - 9 2-6 - 5-6 6 4 5 6 7 7 7 7+6 =13 4+9 =13 9+2 =11 - - 7 - - 11 - - 6-7 The shortest path from 1 to 7 can be 1 →3 → 2 → 6 →7 Total distance is 11 units 1 → 3 → 2 →5 → 6 →7 Total distance = 11 units 1.2 Minimal Spanning Tree Problem A tree is defined to be an undirected, acyclic and connected graph. A spanning tree is a subgraph of G (undirected, connected graph), is a tree and contains all the vertices of G. A minimum spanning tree is a spanning tree but has weights or lengths associated with edges and the total weight is at the minimum. Prim’s Algorithm It starts at any vertex (say A) in a graph and finds the least cost vertex (say B) connected to the start vertex. Now either from A or B, it will find the next least costly vertex connection, without creating cycle (say C) Now either from A, B or C find the next least costly vertex connection, without creating a cycle and so on. Eventually all the vertices will be connected without any cycles and a minimum spanning tree will be the result. Example 1 Suppose it is desired to establish a cable communication network that links major cities, which is shown in the figure. Determine how the cities are connected such that the total cable mileage is minimized. Solution C = {LA} C' = {SE, DE, DA, EH, NY, DC} C = {LA, SE} C' = {DE, DA, EH, NY, DC} C = {LA, SE, DE} C' = {DA, EH, NY, DC} C = {LA, SE, DE, DA} C' = {EH, NY, DC} C = {LA, SE, DE, DA, EH} C' = {NY, DC} C = {LA, SE, DE, DA, EH, NY} C' = {DC} C = {LA, SE, DE, DA, EH, NY, DC} C' = { } The resultant network is Thus the total cable mileage is 1100 + 1300 + 780 + 900 + 800 + 200 = 5080 Example 2 For the following graph obtain the minimum spanning tree. The numbers on the branches represent the cost. Solution C = {A} C' = {B, C, D, E, F, G} C = {A, D} C' = {B, C, E, F, G} C = {A, D, B} C' = {C, E, F, G} C = {A, D, B, C} C' = {E, F, G} C = {A, D, B, C, G} C' = {E, F} C = {A, D, B, C, G, F} C' = {E} C = {A, D, B, C, G, F, E} C' = { } The resultant network is Cost = 2 + 1 + 4 + 3 + 3 + 5 = 18 units Example 3 Solve the minimum spanning problem for the given network. The numbers on the branches represent in terms of miles. Solution C = {1} C' = {2, 3, 4, 5, 6} C = {1, 2} C' = {3, 4, 5, 6} C = {1, 2, 5} C' = {3, 4, 6} C = {1, 2, 5, 4} C' = {3, 6} C = {1, 2, 5, 4, 6} C' = {3} C = {1, 2, 5, 4, 6, 3} C' = {} The resultant network is 1 + 4 + 5+ 3 + 3 = 16 miles 1.3 Maximal Flow Problem Algorithm Step1 Find a path from source to sink that can accommodate a positive flow of material. If no path exists go to step 5 Step2 Determine the maximum flow that can be shipped from this path and denote by ‘k’ units. Step3 Decrease the direct capacity (the capacity in the direction of flow of k units) of each branch of this path ‘k’ and increase the reverse capacity k1. Add ‘k’ units to the amount delivered to sink. Step4 Goto step1 Step5 The maximal flow is the amount of material delivered to the sink. The optimal shipping schedule is determined by comparing the original network with the final network. Any reduction in capacity signifies shipment. Example 1 Consider the following network and determine the amount of flow between the networks. Solution Iteration 1: 1 – 3 – 5 Iteration 2: 1 – 2 – 3 – 4 – 5 Iteration 3: 1 – 4 – 5 Iteration 4: 1 – 2 – 5 Iteration 5: 1 – 3 – 2 – 5 Maximum flow is 60 units. Therefore the network can be written as Example 2 Solve the maximal flow problem Solution Iteration 1: O – A – D – T Iteration 2: O – B – E – T Iteration 3: O – A – B – D – T Iteration 4: O – C – E – D – T Iteration 5: O – C – E – T Iteration 6: O – B – D – T Therefore there are no more augmenting paths. So the current flow pattern is optimal. The maximum flow is 13 units. Exercise Find the shortest path Solve the maximal flow problem Explain prim’s algorithm Solve the minimal spanning tree Unit 2 2.1 Introduction to CPM / PERT Techniques 2.2 Applications of CPM / PERT 2.3 Basic Steps in PERT / CPM 2.4 Frame work of PERT/CPM 2.5 Network Diagram Representation 2.6 Rules for Drawing Network Diagrams 2.7 Common Errors in Drawing Networks 2.8 Advantages and Disadvantages 2.9 Critical Path in Network Analysis 2.1 Introduction to CPM / PERT Techniques CPM/PERT or Network Analysis as the technique is sometimes called, developed along two parallel streams, one industrial and the other military. CPM (Critical Path Method) was the discovery of M.R.Walker of E.I.Du Pont de Nemours & Co. and J.E.Kelly of Remington Rand, circa 1957. The computation was designed for the UNIVAC-I computer. The first test was made in 1958, when CPM was applied to the construction of a new chemical plant. In March 1959, the method was applied to maintenance shut-down at the Du Pont works in Louisville, Kentucky. Unproductive time was reduced from 125 to 93 hours. PERT (Project Evaluation and Review Technique) was devised in 1958 for the POLARIS missile program by the Program Evaluation Branch of the Special Projects office of the U.S.Navy, helped by the Lockheed Missile Systems division and the Consultant firm of Booz-Allen & Hamilton. The calculations were so arranged so that they could be carried out on the IBM Naval Ordinance Research Computer (NORC) at Dahlgren, Virginia. The methods are essentially network-oriented techniques using the same principle. PERT and CPM are basically time-oriented methods in the sense that they both lead to determination of a time schedule for the project. The significant difference between two approaches is that the time estimates for the different activities in CPM were assumed to be deterministic while in PERT these are described probabilistically. These techniques are referred as project scheduling techniques. In CPM activities are shown as a network of precedence relationships using activity-on-node network construction Single estimate of activity time Deterministic activity times USED IN: Production management - for the jobs of repetitive in nature where the activity time estimates can be predicted with considerable certainty due to the existence of past experience. In PERT activities are shown as a network of precedence relationships using activity-on-arrow network construction Multiple time estimates Probabilistic activity times USED IN: Project management - for non-repetitive jobs (research and development work), where the time and cost estimates tend to be quite uncertain. This technique uses probabilistic time estimates. Benefits of PERT/CPM Useful at many stages of project management Mathematically simple Give critical path and slack time Provide project documentation Useful in monitoring costs Limitations of PERT/CPM Clearly defined, independent and stable activities Specified precedence relationships Over emphasis on critical paths 2.2 Applications of CPM / PERT These methods have been applied to a wide variety of problems in industries and have found acceptance even in government organizations. These include Construction of a dam or a canal system in a region Construction of a building or highway Maintenance or overhaul of airplanes or oil refinery Space flight Cost control of a project using PERT / COST Designing a prototype of a machine Development of supersonic planes 2.3 Basic Steps in PERT / CPM Project scheduling by PERT / CPM consists of four main steps Planning The planning phase is started by splitting the total project in to small projects. These smaller projects in turn are divided into activities and are analyzed by the department or section. The relationship of each activity with respect to other activities are defined and established and the corresponding responsibilities and the authority are also stated. Thus the possibility of overlooking any task necessary for the completion of the project is reduced substantially. Scheduling The ultimate objective of the scheduling phase is to prepare a time chart showing the start and finish times for each activity as well as its relationship to other activities of the project. Moreover the schedule must pinpoint the critical path activities which require special attention if the project is to be completed in time. For non-critical activities, the schedule must show the amount of slack or float times which can be used advantageously when such activities are delayed or when limited resources are to be utilized effectively. Allocation of resources Allocation of resources is performed to achieve the desired objective. A resource is a physical variable such as labour, finance, equipment and space which will impose a limitation on time for the project. When resources are limited and conflicting, demands are made for the same type of resources a systematic method for allocation of resources become essential. Resource allocation usually incurs a compromise and the choice of this compromise depends on the judgment of managers. Controlling The final phase in project management is controlling. Critical path methods facilitate the application of the principle of management by expectation to identify areas that are critical to the completion of the project. By having progress reports from time to time and updating the network continuously, a better financial as well as technical control over the project is exercised. Arrow diagrams and time charts are used for making periodic progress reports. If required, a new course of action is determined for the remaining portion of the project. 2.4 The Framework for PERT and CPM Essentially, there are six steps which are common to both the techniques. The procedure is listed below: Define the Project and all of its significant activities or tasks. The Project (made up of several tasks) should have only a single start activity and a single finish activity. Develop the relationships among the activities. Decide which activities must precede and which must follow others. Draw the "Network" connecting all the activities. Each Activity should have unique event numbers. Dummy arrows are used where required to avoid giving the same numbering to two activities. Assign time and/or cost estimates to each activity Compute the longest time path through the network. This is called the critical path. Use the Network to help plan, schedule, and monitor and control the project. The Key Concept used by CPM/PERT is that a small set of activities, which make up the longest path through the activity network control the entire project. If these "critical" activities could be identified and assigned to responsible persons, management resources could be optimally used by concentrating on the few activities which determine the fate of the entire project. Non-critical activities can be replanned, rescheduled and resources for them can be reallocated flexibly, without affecting the whole project. Five useful questions to ask when preparing an activity network are: Is this a Start Activity? Is this a Finish Activity? What Activity Precedes this? What Activity Follows this? What Activity is Concurrent with this? 2.5 Network Diagram Representation In a network representation of a project certain definitions are used 1. Activity Any individual operation which utilizes resources and has an end and a beginning is called activity. An arrow is commonly used to represent an activity with its head indicating the direction of progress in the project. These are classified into four categories Predecessor activity – Activities that must be completed immediately prior to the start of another activity are called predecessor activities. Successor activity – Activities that cannot be started until one or more of other activities are completed but immediately succeed them are called successor activities. Concurrent activity – Activities which can be accomplished concurrently are known as concurrent activities. It may be noted that an activity can be a predecessor or a successor to an event or it may be concurrent with one or more of other activities. Dummy activity – An activity which does not consume any kind of resource but merely depicts the technological dependence is called a dummy activity. The dummy activity is inserted in the network to clarify the activity pattern in the following two situations To make activities with common starting and finishing points distinguishable To identify and maintain the proper precedence relationship between activities that is not connected by events. For example, consider a situation where A and B are concurrent activities. C is dependent on A and D is dependent on A and B both. Such a situation can be handled by using a dummy activity as shown in the figure. 2. Event An event represents a point in time signifying the completion of some activities and the beginning of new ones. This is usually represented by a circle in a network which is also called a node or connector. The events are classified in to three categories Merge event – When more than one activity comes and joins an event such an event is known as merge event. Burst event – When more than one activity leaves an event such an event is known as burst event. Merge and Burst event – An activity may be merge and burst event at the same time as with respect to some activities it can be a merge event and with respect to some other activities it may be a burst event. 3. Sequencing The first prerequisite in the development of network is to maintain the precedence relationships. In order to make a network, the following points should be taken into considerations What job or jobs precede it? What job or jobs could run concurrently? What job or jobs follow it? What controls the start and finish of a job? Since all further calculations are based on the network, it is necessary that a network be drawn with full care. 2.6 Rules for Drawing Network Diagram Rule 1 Each activity is represented by one and only one arrow in the network Rule 2 No two activities can be identified by the same end events Rule 3 In order to ensure the correct precedence relationship in the arrow diagram, following questions must be checked whenever any activity is added to the network What activity must be completed immediately before this activity can start? What activities must follow this activity? What activities must occur simultaneously with this activity? In case of large network, it is essential that certain good habits be practiced to draw an easy to follow network Try to avoid arrows which cross each other Use straight arrows Do not attempt to represent duration of activity by its arrow length Use arrows from left to right. Avoid mixing two directions, vertical and standing arrows may be used if necessary. Use dummies freely in rough draft but final network should not have any redundant dummies. The network has only one entry point called start event and one point of emergence called the end event. 2.7 Common Errors in Drawing Networks The three types of errors are most commonly observed in drawing network diagrams 1. Dangling To disconnect an activity before the completion of all activities in a network diagram is known as dangling. As shown in the figure activities (5 – 10) and (6 – 7) are not the last activities in the network. So the diagram is wrong and indicates the error of dangling 2. Looping or Cycling Looping error is also known as cycling error in a network diagram. Drawing an endless loop in a network is known as error of looping as shown in the following figure. 3. Redundancy Unnecessarily inserting the dummy activity in network logic is known as the error of redundancy as shown in the following diagram 2.8 Advantages and Disadvantages PERT/CPM has the following advantages   A PERT/CPM chart explicitly defines and makes visible dependencies (precedence relationships) between the elements, PERT/CPM facilitates identification of the critical path and makes this visible, PERT/CPM facilitates identification of early start, late start, and slack for each activity, PERT/CPM provides for potentially reduced project duration due to better understanding of dependencies leading to improved overlapping of activities and tasks where feasible.   PERT/CPM has the following disadvantages:   There can be potentially hundreds or thousands of activities and individual dependency relationships, The network charts tend to be large and unwieldy requiring several pages to print and requiring special size paper, The lack of a timeframe on most PERT/CPM charts makes it harder to show status although colours can help (e.g., specific colour for completed nodes), When the PERT/CPM charts become unwieldy, they are no longer used to manage the project. 2.9 Critical Path in Network Analysis Basic Scheduling Computations The notations used are (i, j) = Activity with tail event i and head event j Ei = Earliest occurrence time of event i Lj = Latest allowable occurrence time of event j Dij = Estimated completion time of activity (i, j) (Es)ij = Earliest starting time of activity (i, j) (Ef)ij = Earliest finishing time of activity (i, j) (Ls)ij = Latest starting time of activity (i, j) (Lf)ij = Latest finishing time of activity (i, j) The procedure is as follows Determination of Earliest time (Ej): Forward Pass computation Step 1 The computation begins from the start node and move towards the end node. For easiness, the forward pass computation starts by assuming the earliest occurrence time of zero for the initial project event. Step 2 Earliest starting time of activity (i, j) is the earliest event time of the tail end event i.e. (Es)ij = Ei Earliest finish time of activity (i, j) is the earliest starting time + the activity time i.e. (Ef)ij = (Es)ij + Dij or (Ef)ij = Ei + Dij Earliest event time for event j is the maximum of the earliest finish times of all activities ending in to that event i.e. Ej = max [(Ef)ij for all immediate predecessor of (i, j)] or Ej =max [Ei + Dij] Backward Pass computation (for latest allowable time) Step 1 For ending event assume E = L. Remember that all E’s have been computed by forward pass computations. Step 2 Latest finish time for activity (i, j) is equal to the latest event time of event j i.e. (Lf)ij = Lj Step 3 Latest starting time of activity (i, j) = the latest completion time of (i, j) – the activity time or (Ls)ij =(Lf)ij - Dij or (Ls)ij = Lj - Dij Step 4 Latest event time for event ‘i’ is the minimum of the latest start time of all activities originating from that event i.e. Li = min [(Ls)ij for all immediate successor of (i, j)] = min [(Lf)ij - Dij] = min [Lj - Dij] Determination of floats and slack times There are three kinds of floats Total float – The amount of time by which the completion of an activity could be delayed beyond the earliest expected completion time without affecting the overall project duration time. Mathematically (Tf)ij = (Latest start – Earliest start) for activity ( i – j) (Tf)ij = (Ls)ij - (Es)ij or (Tf)ij = (Lj - Dij) - Ei Free float – The time by which the completion of an activity can be delayed beyond the earliest finish time without affecting the earliest start of a subsequent activity. Mathematically (Ff)ij = (Earliest time for event j – Earliest time for event i) – Activity time for ( i, j) (Ff)ij = (Ej - Ei) - Dij Independent float – The amount of time by which the start of an activity can be delayed without effecting the earliest start time of any immediately following activities, assuming that the preceding activity has finished at its latest finish time. Mathematically (If)ij = (Ej - Li) - Dij The negative independent float is always taken as zero. Event slack - It is defined as the difference between the latest event and earliest event times. Mathematically Head event slack = Lj – Ej, Tail event slack = Li - Ei Determination of critical path Critical event – The events with zero slack times are called critical events. In other words the event i is said to be critical if Ei = Li Critical activity – The activities with zero total float are known as critical activities. In other words an activity is said to be critical if a delay in its start will cause a further delay in the completion date of the entire project. Critical path – The sequence of critical activities in a network is called critical path. The critical path is the longest path in the network from the starting event to ending event and defines the minimum time required to complete the project. Exercise What is PERT and CPM? What are the advantages of using PERT/CPM? Mention the applications of PERT/CPM Explain the following terms Earliest time Latest time Total activity slack Event slack Critical path Explain the CPM in network analysis. What are the rules for drawing network diagram? Also mention the common errors that occur in drawing networks. What is the difference between PERT and CPM/ What are the uses of PERT and CPM? Explain the basic steps in PERT/CPM techniques. Write the framework of PERT/CPM. Unit 3 3.1 Worked Examples on CPM 3.2 PERT 3.3 Worked Examples 3.1 Worked Examples on CPM Example 1 Determine the early start and late start in respect of all node points and identify critical path for the following network. Solution Calculation of E and L for each node is shown in the network Activity(i, j) Normal Time (Dij) Earliest Time Latest Time Float Time (Li - Dij ) - Ei Start (Ei) Finish (Ei + Dij ) Start (Li - Dij ) Finish (Li) (1, 2) (1, 3) (1, 4) (2, 5) (4, 6) (3, 7) (5, 7) (6, 7) (5, 8) (6, 9) (7, 10) (8, 10) (9, 10) 10 8 9 8 7 16 7 7 6 5 12 13 15 0 0 0 10 9 8 18 16 18 16 25 24 21 10 8 9 18 16 24 25 23 24 21 37 37 36 0 1 1 10 10 9 18 18 18 17 25 24 22 10 9 10 18 17 25 25 25 24 22 37 37 37 0 1 1 0 1 1 0 2 0 1 0 0 1 Network Analysis Table From the table, the critical nodes are (1, 2), (2, 5), (5, 7), (5, 8), (7, 10) and (8, 10) From the table, there are two possible critical paths 1 → 2 → 5 → 8 → 10 1 → 2 → 5 → 7 → 10 Example 2 Find the critical path and calculate the slack time for the following network Solution The earliest time and the latest time are obtained below Activity(i, j) Normal Time (Dij) Earliest Time Latest Time Float Time (Li - Dij ) - Ei Start (Ei) Finish (Ei + Dij ) Start (Li - Dij ) Finish (Li) (1, 2) (1, 3) (1, 4) (2, 6) (3, 7) (3, 5) (4, 5) (5, 9) (6, 8) (7, 8) (8, 9) 2 2 1 4 5 8 3 5 1 4 3 0 0 0 2 2 2 1 10 6 7 11 2 2 1 6 7 10 4 15 7 11 14 5 0 6 7 3 2 7 10 11 8 12 7 2 7 11 8 10 10 15 12 12 15 5 0 6 5 1 0 6 0 5 1 1 From the above table, the critical nodes are the activities (1, 3), (3, 5) and (5, 9) The critical path is 1 → 3 → 5 → 9 Example 3 A project has the following times schedule Activity Times in weeks Activity Times in weeks (1 – 2) (1 – 3) (2 – 4) (3 – 4) (3 – 5) (4 – 9) (5 – 6) 4 1 1 1 6 5 4 (5 – 7) (6 – 8) (7 – 8) (8 – 9) (8 – 10) (9 – 10) 8 1 2 1 8 7 Construct the network and compute TE and TL for each event Float for each activity Critical path and its duration Solution The network is Event No.: 1 2 3 4 5 6 7 8 9 10 TE: 0 4 1 5 7 11 15 17 18 25 TL: 0 12 1 13 7 16 15 17 18 25 Float = TL (Head event) – TE (Tail event) – Duration Activity Duration TE (Tail event) TL (Head event) Float (1 – 2) (1 – 3) (2 – 4) (3 – 4) (3 – 5) (4 – 9) (5 – 6) (5 – 7) (6 – 8) (7 – 8) (8 – 9) (8 – 10) (9 – 10) 4 1 1 1 6 5 4 8 1 2 1 8 7 0 0 4 1 1 5 7 7 11 15 17 17 18 12 1 13 13 7 18 16 15 17 17 18 25 25 8 0 8 11 0 8 5 0 5 0 0 0 0 The resultant network shows the critical path The two critical paths are 1 → 3 → 5 →7 → 8 → 9 →10 1 → 3 → 5 → 7 → 8 →10 3.2 Project Evaluation and Review Technique (PERT) The main objective in the analysis through PERT is to find out the completion for a particular event within specified date. The PERT approach takes into account the uncertainties. The three time values are associated with each activity Optimistic time – It is the shortest possible time in which the activity can be finished. It assumes that every thing goes very well. This is denoted by t0. Most likely time – It is the estimate of the normal time the activity would take. This assumes normal delays. If a graph is plotted in the time of completion and the frequency of completion in that time period, then most likely time will represent the highest frequency of occurrence. This is denoted by tm. Pessimistic time – It represents the longest time the activity could take if everything goes wrong. As in optimistic estimate, this value may be such that only one in hundred or one in twenty will take time longer than this value. This is denoted by tp. In PERT calculation, all values are used to obtain the percent expected value. Expected time – It is the average time an activity will take if it were to be repeated on large number of times and is based on the assumption that the activity time follows Beta distribution, this is given by te = ( t0 + 4 tm + tp ) / 6 The variance for the activity is given by σ2 = [(tp – to) / 6] 2 3.3 Worked Examples Example 1 For the project Task: A B C D E F G H I J K Least time: 4 5 8 2 4 6 8 5 3 5 6 Greatest time: 8 10 12 7 10 15 16 9 7 11 13 Most likely time: 5 7 11 3 7 9 12 6 5 8 9 Find the earliest and latest expected time to each event and also critical path in the network. Solution Task Least time(t0) Greatest time (tp) Most likely time (tm) Expected time (to + tp + 4tm)/6 A B C D E F G H I J K 4 5 8 2 4 6 8 5 3 5 6 8 10 12 7 10 15 16 9 7 11 13 5 7 11 3 7 9 12 6 5 8 9 5.33 7.17 10.67 3.5 7 9.5 12 6.33 5 8 9.17 Task Expected time (te) Start Finish Total float Earliest Latest Earliest Latest A B C D E F G H I J K 5.33 7.17 10.67 3.5 7 9.5 12 6.33 5 8 9.17 0 0 5.33 0 16 3.5 3.5 23 23 28 29.33 0 8.83 5.33 10 16 13.5 18.5 23 25.5 30.5 29.33 5.33 7.17 16 3.5 23 13 15.5 29.33 28 36 31.5 5.33 16 16 13.5 23 23 30.5 29.33 30.5 38.5 38.5 0 8.83 0 10 0 10 15 0 2.5 2.5 0 The network is The critical path is A →C →E → H → K Example 2 A project has the following characteristics Activity Most optimistic time (a) Most pessimistic time (b) Most likely time (m) (1 – 2) (2 – 3) (2 – 4) (3 – 5) (4 – 5) (4 – 6) (5 – 7) (6 – 7) (7 – 8) (7 – 9) (8 – 10) (9 – 10) 1 1 1 3 2 3 4 6 2 5 1 3 5 3 5 5 4 7 6 8 6 8 3 7 1.5 2 3 4 3 5 5 7 4 6 2 5 Construct a PERT network. Find the critical path and variance for each event. Solution Activity (a) (b) (m) (4m) te (a + b + 4m)/6 v [(b – a) / 6]2 (1 – 2) (2 – 3) (2 – 4) (3 – 5) (4 – 5) (4 – 6) (5 – 7) (6 – 7) (7 – 8) (7 – 9) (8 – 10) (9 – 10) 1 1 1 3 2 3 4 6 2 5 1 3 5 3 5 5 4 7 6 8 6 8 3 7 1.5 2 3 4 3 5 5 7 4 6 2 5 6 8 12 16 12 20 20 28 16 24 8 20 2 2 3 4 3 5 5 7 4 6.17 2 5 4/9 1/9 4/9 1/9 1/9 4/9 1/9 1/9 4/9 1/4 1/9 4/9 The network is constructed as shown below The critical path = 1 → 2 → 4 → 6 → 7 →9 →10 Example 3 Calculate the variance and the expected time for each activity Solution Activity (to) (tm) (tp) te (to + tp + 4tm)/6 v [(tp – to) / 6]2 (1 – 2) (1 – 3) (1 – 4) (2 – 3) (2 – 5) (3 – 6) (4 – 7) (5 – 8) (6 – 7) (6 – 9) (8 – 9) (7 – 10) (9 – 11) (10 – 11) 3 6 7 0 8 10 8 12 8 13 4 10 6 10 6 7 9 0 12 12 13 14 9 16 7 13 8 12 10 12 12 0 17 15 19 15 10 19 10 17 12 14 6.2 7.7 9.2 0.0 12.2 12.2 13.2 13.9 9.0 16.0 7.0 13.2 8.4 12.0 1.36 1.00 0.69 0.00 2.25 0.69 3.36 0.25 0.11 1.00 1.00 1.36 1.00 0.66 Example 4 A project is represented by the network as shown below and has the following data Task: A B C D E F G H I Least time: 5 18 26 16 15 6 7 7 3 Greatest time: 10 22 40 20 25 12 12 9 5 Most likely time: 15 20 33 18 20 9 10 8 4 Determine the following Expected task time and their variance Earliest and latest time Solution 1. Activity Least time (t0) Greatest time (tp) Most likely time (tm) Expected time (to + tp + 4tm)/6 Variance (σ2) (1-2) (1-3) (1-4) (2-5) (2-6) (3-6) (4-7) (5-7) (6-7) 5 18 26 16 15 6 7 7 3 10 22 40 20 25 12 12 9 5 8 20 33 18 20 9 10 8 4 7.8 20.0 33.0 18.0 20.0 9.0 9.8 8.0 4.0 0.69 0.44 5.43 0.44 2.78 1.00 0.69 0.11 0.11 2. Earliest time E1 = 0 E2 = 0 +7.8 = 7.8 E3 = 0 +20 = 20 E4 = 0 +33 = 33 E5 = 7.8 + 18 = 25.8 E6 = max [7.8 + 20, 20 + 9] = 29 E7 = max [33 + 9.8, 25.8 + 8, 29 + 4] = 42.8 Latest time L7 = 42.8 L6 = 42.8 – 4 = 38.8 L5 = 42.8 – 8 = 34.3 L4 = 42.8 – 9.8 = 33 L3 = 38.8 – 9 = 29.8 L2 = min [34.8 – 18, 38.8 – 20] = 16.8 L1 = min [16.8 – 7.8, 29.8 – 20, 33 - 33] = 0 Exercise What is PERT? For the following data, draw network. Find the critical path, slack time after calculating the earliest expected time and the latest allowable time Activity Duration Activity Duration (1 – 2) (1 – 3) (2 – 4) (2 – 5) (2 – 6) (3– 7) (3 – 8) (4 – 9) 5 8 6 4 4 5 3 1 (5 – 9) (6 – 10) (7 – 10) (8 – 11) (9 – 12) (10 – 12) (11 – 13) (12 – 13) 3 5 4 9 2 4 1 7 [Ans. Critical path: 1 → 3 → 7 → 10 → 12 →13] A project schedule has the following characteristics Activity Most optimistic time Most likely time Most pessimistic time (1 – 2) (2 – 3) (2 – 4) (3 – 5) (4 – 5) (4 – 6) (5 – 7) (6 – 7) (7 – 8) (7 – 9) (8 – 10) (9 – 10) 1 1 1 3 2 3 4 6 2 4 1 3 2 2 3 4 5 5 5 7 4 6 2 5 3 3 5 5 4 7 6 8 6 8 3 7 Construct a PERT network and find out The earliest possible time Latest allowable time Slack values Critical path Explain the following terms optimistic time Most likely time Pessimistic time Expected time Variance Calculate the variance and the expected time for each activity