Jahangirabad Institute of Technology Dr. ZIAUL HASSAN BAKHSHI Associate Professor Semester II, 2016 MASTER SCHEDULE: Operations Research (NMBA-025)

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1 Jahangirabad Institute of Technology Dr. ZIAUL HASSAN BAKHSHI Associate Professor Semester II, 2016 MASTER SCHEDULE: Operations Research (NMBA-025) Week 1 Class 1 Wednesday, 27 January,2016 Introduction of Operations research, uses, Scopes and applications Week 1 Class 3 Friday, 29 January,2016 Decision making under certainty, uncertainty and risk Week 2 Class 5 Monday, February 01,2016 Introduction of Linear Programming problem & Mathematical formulation problem with graphical solution Week 2 Class 7 Wednesday, February 03,2016 Duality of LPP and Sensitivity analysis with examples Week 3 Class 9 Friday, February 05,2016 Transportation problem solved by N-W corner rule Week 4 Class 11 Monday, February 08,2016 Transportation problem solved by vogel s Approximation Methods Week 4 Class 13 Wednesday, February 10,2016 Quiz test on LPP & TP Class 2 Thursday, 28 January,2016 Operations research in managerial decision making Class 4 Saturday, 30 January,2016 Decision tee approaches and its applications Class 6 Tuesday, February 02,2016 Simplex methods to solve LPP Class 8 Thursday, February 04,2016 Transportation problem with examples Class 10 Saturday, February 06,2016 Transportation problem solved by Matrix minima Method Class 12 Tuesday, February 09,2016 Optimal solution of TP Problem Class 14 Thursday, February 11,2016 Introduction to assignment problem Week 4 Class 15 Friday, February 12,2016 Hungarian s Method to solve AP Class 16 Saturday, February 13,2016 Optimal solution of AP Week 5 Class 17 Monday, February 15,2016 Introduction and concept of game theory Week 5 Class 19 Wednesday, February 1,2016 Two-person zero-sum game examples Week 5 Class 21 Friday, February 19,2016 Pue and Mixed Strategy with examples Week 6 22 February to 25 February 2016 Medical Camp Week 6 Class 22 Friday, February 26,2016 Saddle point of games Week 7 Class 24 Monday, February 29,2016 Dominance Methods to solve game Theory Week 7 Class 25 Friday, March 04,2016 Dominance Methods to solve game Theory Class 18 Tuesday, February 16,2016 Two-person zero-sum game Class 20 Thursday, February 18,2016 Pure and Mixed Strategy with examples 20 February to 22 February 2016 Sports Events Class 23 Saturday, February 27,2016 Odds Method fo games Theory 01 March to 03 March,2016 Mid Semester Exam Class 26 Saturday, March 05,2016 Dominance Methods to solve game Theory 1

2 Week 8 07 March Monday, 2016 Mahashivratri Holiday Week 8 Class 28 Wednesday, March 09,2016 Graphical Method for Solving Mixed Strategy Game Week 8 Class 30 Friday, March 11,2016 Introduction to sequencing problem Week 9 Class 32 Monday, March 14,2016 Numerical on n jobs and two machines Week 9 Class 33 Wednesday, March 16,2016 Algorithm for n jobs and three machines Week 9 Class 35 Friday, March 18,2016 Examples on n jobs and two machines Week 10 Week 10 Class 37 Monday, March 21,2016 Description of M/M/1 model with examples 23 to 24 March Holi Festival Holiday Class 27 Tuesday, March 08,2016 Graphical Method for Solving Mixed Strategy Game Class 29 Thursday, March 10,2016 Quiz Test Class 31 Saturday, March 12,2016 Algorithm for n jobs and two machines 15 March, 2016 Kashiram Jayanti Holiday Class 34 Thursday, March 17,2016 Algorithm for 2 jobs and m machines problems Class 36 Saturday, March 19,2016 Introduction of Queueing Theory and applications Class 38 Tuesday, March 22,2016 Numerical on M/M/1 model 25 to 26 March On Leave due to Holi festival Week 11 Class 39 Monday, March 28,2016 Poisson distribution in estimating arrival rate Week 11 Class 41 Wednesday, March 30,2016 Numericals based on the model Week 11 Class 43 Friday, April 01,2016 Quiz Test Week 12 Class 45 Monday, April 04,2016 Replacement of assets that deteriorate with time Week 12 Class 47 Wednesday, April 06,2016 Replacement of assets that deteriorate with time Week 13 Class 48 Monday, April 11,2016 Replacement of assets which fail suddenly Class 40 Tuesday, March 29,2016 Exponential distribution in estimating service rate Class 42 Thursday, March 31,2016 Application of Queue model for better service to the customers Class 44 Saturday, April 02,2016 Introduction of replacement Problem Class 46 Tuesday, April 05,2016 Replacement of assets that deteriorate with time examples 07 April to 09 April, 2016 Second Mid Sem Exam Class 49 Tuesday, April 12,2016 Introduction of PERT & CPM Week 13 Class 50 Wednesday, April 13,2016 Application of PERT & CPM Week April, 2016 Ram Navami Holiday Week 14 Class 52 Monday, April 18,2016 Crashing of Operations 14 April, 2016 Ambedkar Jayanti Holiday Class 51 Saturday, April 16,2016 Application of PERT & CPM in project planning and control Class 53 Tuesday, April 19,2016 Numerical 2

3 Week 14 Class 54 Wednesday, April 20,2016 Quiz Test Week April, 2016 Earth Day Celebration Week 15 Class 56 Monday, April 25,2016 Numerical Problems Week 15 Week 15 Semester Exam Preparation and Query Classes Semester Exam Preparation and Query Classes 21 April, 2016 Hazrat Ali Birthday Holiday Class 55 Saturday, April 23,2016 Numerical Problems Class 57 Tuesday, April 26,2016 Numerical Problems Semester Exam Preparation and Query Classes Semester Exam Preparation and Query Classes Details are found in the following sections: General Information, Class Schedule, and Project Schedule. GENERAL INFORMATION Teaching Staff: Dr. ZIAUL HASSAN BAKHSHI JITM, NMBA-025 Course Web Site: Class Meetings: JIT HoD Room MBA Goals : The main course of goal is to provide students with a complete concept of each and every chapter methods. At the completion of the course students should be able to solve OR problems easily step by step and know how to used it in real life situations. Course Objectives: A. By the time of Exam No. 1 (after approximately 24 (Fifty five minutes) of lectures), the students should be able to do the following: 1. Understand the concepts of operations research and its uses in managerial decision-making. 2. Solve decision making problem under certainty, uncertainty and risk 3. Mathematical formulation of LP Model and solve LPP by graphical method & simplex method 4. Solve sensitivity analysis and duality problem 5. To find initial basic feasible solution and obtain optimal solution of transportation problem 6. To find initial basic feasible solution and obtain optimal solution of assignment problem 7. Understand the concepts of games theory, pure and mixed strategy, saddle point. 8. To find the solution of games problem by dominance Method 9. Be able to solve Mixed strategy game by Graphical Method. B. By the time of Exam No. 2 (after approximately 30 lectures), the students should be able to do all the items listed the following: 3

4 10. Understand the concepts of sequencing problem and solve n jobs and two machines. 11. solve n jobs and three machines, and solve two jobs and m-machines. 12. Understand the concepts of queuing theory and it application 13. Be able to Understand the concepts of M/M/1 Model and find its solution 14. Understand the concepts of replacement problem and its application 15. Be able to understand replacement of assets that deteriorate with time and problems. 16. Understand the concept of replacement of assets which fail suddenly and based on problems. 17. Understand the concept of project management CPM/PERT and its application. 18. calculate solution and rules for drawing the network diagram 19. Understand the crashing of operations and its solutions. Course Outcome: 1. Able to understand OR & LPP problems and solve it by different methods 2. Able to understand transportation, assignment problems and find optimal solution. 3. Able to understand games theory and solve it by dominance and graphical method 4. Able to understand queuing theory, concepts of M/M/1 Model and find its solution 5. Able to understand replacement problem and find solution in different cases. 6. Able to understand project management CPM/PERT and its application and solutions. Grading: Based On Four Points: 1. Sessional Marks (2 hours paper/ 30 Marks) 2. Attendance (10 Marks) 3. Assignments (Total 5/5 Marks) 4. Class Test (Total 5 / Marks) Class Preparation and Participation: Reading assignments are given in the Class Schedule for each class session. You are expected to come to class prepared to discuss the readings and the suggested questions. Your individual class participation grade will be based upon your in-class remarks during discussions. Classes and Topics Class 1 Wednesday, 27 January,2016 Introduction of Operations research, uses, Scopes and applications OPERATIONS RESEARCH is a branch of mathematics - specially applied mathematics, used to provide a scientific base for management to take timely and effective decisions to their problems. The objective of Operations Research is to provide a scientific basis to the decision maker for solving the problems involving the interaction of various components of an organization by employing a team of scientists from various disciplines, all working together for finding a solution which is in the best interest of the organisaton as a whole. The best solution thus obtained is known as optimal decision. we can use operations research techniques to get best solution. (i) In Defense Operations 4

5 (ii)in Industry (iii)in Planning For Economic Growth (iv)in Agriculture (v) In Traffic control (vi) In Hospitals Class 2 ` Thursday, 28 January,2016 Operations research in managerial decision making A decision is the conclusion of a process designed to weigh the relative uses or utilities of a set of alternatives on hand, so that decision maker selects the best alternative which is best to his problem or situation and implement it. Decision Making involves all activities and thinking that are necessary to identify the most optimal or preferred choice among the available alternatives. The basic requirements of decision-making are (i) A set of goals or objectives, (ii) Methods of evaluating alternatives in an objective manner, (iii) A system of choice criteria and a method of projecting the repercussions of alternative choices of courses of action. The evaluation of consequences of each course of action is important due to sequential nature of decisions. Class 3 Friday, 29 January,2016 Decision making under certainty, uncertainty and risk The decisions are classified according to the degree of certainty as deterministic models, where the manager assumes complete certainty and each strategy results in a unique payoff, and Probabilistic models, where each strategy leads to more than one payofs and the manager attaches a probability measure to these payoffs. The scale of assumed certainty can range from complete certainty to complete uncertainty hence one can think of decision making under certainty (DMUC) and decision making under uncertainty (DMUU) on the two extreme points on a scale. The region that falls between these extreme points corresponds to the concept of probabilistic models, and referred as decision-making under risk (DMUR). Class 4 Saturday, 30 January,2016 STEPS IN DECISION THEORY APPROACH 1.List the viable alternatives (strategies) that can be considered in the decision. 2. List all future events that can occur. These future events (not in the control of decision maker) are called as states of nature. 3. Construct a payoff table for each possible combination of alternative course of action and state of nature. 4. Choose the criterion that results in the largest payoff. Class 5 Monday, February 01,,2016 Decision tee approaches and its applications A decision tree consists of nodes, branches, probability estimates, and payoffs. There are two types of nodes, one is decision node and other is chance node. A decision node is generally represented by a square, requires that a conscious decision be made to choose one of the branches that emanate from the node (i.e. one of the availed strategies must be chosen). The branches emanate from and connect various nodes. We shall identify two types of branches: decision branch and chance branch. A decision branch denoted by parallel lines ( ) represents a strategy or course of action. Another type of branch is chance branch, represented by single line ( ) represents a chance determined event. Indicated alongside the chance branches are their respective probabilities. When a branch marks the end of a decision tree i.e. it is not followed by a decision or chance node will be called as terminal branch. A terminal branch can represent a decision alternative or chance outcome. The payoffs can be positive (profit or sales) or negative (expenditure or cost) and they can be associated with a decision branch or a chance branch. The payoffs are placed alongside appropriate 5

6 branch except that the payoffs associated with the terminal branches of the decision tree will be shown at the end of these branches. The decision tree can be deterministic or probabilistic (stochastic), and it can he represent a single-stage (one decision) or a multistage (a sequence of decisions) problem. Class 6 Tuesday, February 02,2016 Simplex methods to solve LPP 1. In the given inequalities, there should not be any negative element on right hand side (b i 0). If any b i is negative, multiply the inequality by -1 and change the inequality sign. 2. Sometimes, the objective function may be maximisation type and the inequalities may be type. In such cases, multiply the objective function by -1 and convert it into minimisation type and vice versa. 3. While selecting, the incoming variable, i.e., key column, in maximisation case, we have to select the highest positive opportunities cost and in minimisation case, select the highest element with negative sign (smallest element). While doing so, sometimes you may find the highest positive element in maximisation case or lowest element in minimisation case falls under the slack variable in maximisation case or under surplus variable in minimisation case. Do not worry. As per rule, select that element and take the column containing that element as key column. 4. Some times the columns of non-basis variables (decision variables) may have their net evaluation elements same. That is the net evaluation elements are equal. This is known as a TIE in Linear Programming Problem. To break the time, first select any one column of your choice as the key column. In the next table, everything will go right. 5. While selecting the out going variable i.e., key row, we have to select limiting ratio (lowest ratio) in net evaluation row. In case any element of key column is negative, the replacement ratio will be negative. In case it is negative, do not consider it for operation. Neglect that and consider other rows to select out going variable. 6. Sometimes all the replacement ratios for all the rows or some of the rows may be equal and that element may be limiting ratio. This situation in Linear Programming Problem is known as DEGENERACY. We say that the problem is degenerating. When the problem degenerate, the following precautions are taken to get rid of degeneracy. (a)take any one ratio of your choice to select key row or out going variable. If you do this, there is a possibility that the problem may cycle. Cycling means, after doing many iterations, you will get the first table once again. But it may not be the case all times. (b)select the variable, whose subscript is small. Say S 1 is smaller than S 2 and S 2 is smaller than S 3 or X 1 is smaller than X 2 an so on or x is smaller than y or a is smaller than b and so on. (c)if we do above two courses of action, we may encounter with one problem. That one of the remaining variable in the next table (the one corresponding to the tied variable that was not considered) will be reduced to a magnitude of zero.this causes trouble in selecting key column in the next table. (d)identify the tied variable or rows. For each of the columns in the identity (starting with the extreme left hand column of the identity and proceeding one at a time to the right), compute a ratio by dividing the entry in each tied row by the key column number in that row. Compare these ratios, column by column, proceeding to the right. The first time the ratios are unequal, the tie is broken. Of the tied rows, the one in which the smaller algebraic ratio falls is the key row. (e)if the ratios in the identity do not break the tie, form similar ratios in the columns of the main body and select the key row as described in (d) above. The application of the above we shall see when we deal with degeneracy problems. 7. While solving the linear programming problems, we may come across a situation that the opportunity cost of more than one non- basic variables are zero, then we can say that the problem has got ALTERNATE SOLUTIONS. 8. If in a simplex table only one unfavourable C j - Z j identifying the only incoming variable and if all the elements of that column are either negative elements or zeros, showing that no change in the basis can be made and no current basic variable can be reduced to zero. Actually, as the incoming arable is introduced, we continue to increase, without bounds, those basic variables whose ratios of substitutions are negative. This is the indication of UNBOUND SOLUTION. 9. In a problem where, the set of constraints is inconsistent i.e., mutually exclusive, is the case of NO FEASIBLE SOLUTION. In simplex algorithm, this case will occur if the solution is optimal (i.e., the test of optimality is 6

7 satisfied) but some artificial variable remains in the optimal solution with a non zero value. Class 7 Wednesday, February 03,2016 Duality of LPP and Sensitivity analysis with examples Most important finding in the development of Linear Programming Problems is the existence of duality in linear programming problems. Linear programming problems exist in pairs. That is in linear programming problem, every maximization problem is associated with a minimization problem. Conversely, associated with every minimization problem is a maximization problem. Once we have a problem with its objective function as maximization, we can write by using duality relationship of linear programming problems, its minimization version. The original linear programming problem is known as primal problem, and the derived problem is known as dual problem. The concept of the dual problem is important for several reasons. Most important are (i) the variables of dual problem can convey important information to managers in terms of formulating their future plans and (ii) in some cases the dual problem can be instrumental in arriving at the optimal solution to the original problem in many fewer iterations, which reduces the labour of computation. Whenever, we solve the primal problem, may be maximization or minimization, we get the solution for the dual automatically. That is, the solution of the dual can be read from the final table of the primal and vice versa. Let us try to understand the concept of dual problem by means of an example. Let us consider the diet problem, which we have discussed while discussing the minimization case of the linear programming problem. While solving a linear programming problem for optimal solution, we assume that: (a). Technology is fixed, (b). Fixed prices, (c). Fixed levels of resources or requirements, (d). The coefficients of variables in structural constraints (i.e. time required by a product on particular resource) are fixed, and (e) profit contribution of the product will not vary during the planning period. These assumptions, implying certainty, complete knowledge, and static conditions, permit us to design an optimal programme. The condition in the real world however, might be different from those that are assumed by the model. It is, therefore, desirable to determine how sensitive the optimal solution is to different types of changes in the problem data and parameters. The changes, which have effect on the optimal solution are: (a) Change in objective function coefficients (a ij ), (b) Resource or requirement levels (b i ), (c) Possible addition or deletion of products or methods of production. The process of checking the sensitivity of the optimal solution for changes in resources and other components of the problem, is given various names such as: Sensitivity Analysis, Parametric Programming and Post optimality analysis or what if analysis. Class 8 Thursday, February 04,2016 Transportation problem with examples The transportation model deals with a special class of linear programming problem in which the objective is to transport a homogeneous commodity from various origins or factories to different destinations or markets at a total minimum cost. To understand the problem more clearly, let us take an example and discuss the rationale of transportation problem. Three factories A, B and C manufactures sugar and are located in different regions. Factory A manufactures, b 1 tons of sugar per year and B manufactures b 2 tons of sugar per year and C manufactures b 3 tons of sugar. The sugar is required by four markets W, X, Y and Z. The requirement of the four markets is as follows: Demand for sugar in Markets W, X, Yand Z is d 1, d 2, d 3 and d 4 tons respectively. The transportation cost of one ton of sugar from each factory to market is given in the matrix below. The objective is to transport sugar from factories to the markets at a minimum total transportation cost. Class 9 Friday, February 05,2016 Transportation problem solved by N-W corner rule (i) Balance the problem. That is see whether Σbi = Σ dj. If not open a dummy column or 7

8 dummy row as the case may be and balance the problem. (ii) Start from the left hand side top corner or cell and make allocations depending on the availability and requirement constraint. If the availability constraint is less than the requirement constraint, then for that cell make allocation in units which is equal to the availability constraint. In general, verify which is the smallest among the availability and requirement and allocate the smallest one to the cell under question. Then proceed allocating either sidewise or downward to satisfy the rim requirement. Continue this until all the allocations are over. (iii) Once all the allocations are over, i.e., both rim requirement (column and row i.e., availability and requirement constraints) are satisfied, write allocations and calculate the cost of transportation. Class 10 Saturday, February 06,2016 Transportation problem solved by Matrix minima Method Identify the lowest cost cell in the given matrix. In this particular example it is = 0. Four cells of dummy column are having zero. When more than one cell has the same cost, then both the cells are competing for allocation. This situation in transportation problem is known as tie. To break the tie, select any one cell of your choice for allocation. Make allocations to this cell either to satisfy availability constraint or requirement constraint. Once one of these is satisfied, then mark crosses ( ) in all the cells in the row or column which ever has completely allocated. Next search for lowest cost cell. Class 11 Monday, February 08,2016 Transportation problem solved by vogel s Approximation Methods In this method, we use concept of opportunity cost. Opportunity cost is the penalty for not taking correct decision. To find the row opportunity cost in the given matrix deduct the smallest element in the row from the next highest element. Similarly to calculate the column opportunity cost, deduct smallest element in the column from the next highest element. Write row opportunity costs of each row just by the side of availability constraint and similarly write the column opportunity cost of each column just below the requirement constraints. These are known as penalty column and penalty row. Class 12 Tuesday, February 09,2016 Optimal solution of TP Problem we get the basic feasible solution for a transportation problem, the next duty is to test whether the solution got is an optimal one or not? This can be done by two methods. (a) By Stepping Stone Method, and (b) By Modified Distribution Method, or MODI method. (a) Stepping stone method of optimality test To give an optimality test to the solution obtained, we have to find the opportunity cost of empty cells. As the transportation problem involves decision making under certainty, we know that an optimal solution must not incur any positive opportunity cost. Thus, we have to determine whether any positive opportunity cost is associated with a given progarmme, i.e., for empty cells. Once the opportunity cost of all empty cells are negative, the solution is said to be optimal. In case any one cell has got positive opportunity cost, then the solution is to be modified. The Stepping stone method is used for finding the opportunity costs of empty cells. Every empty cell is to be evaluated for its opportunity cost. Class 13 Wednesday, February 10,2016 Quiz test on LPP & TP 8

9 Class 14 Thursday, February 11,2016 Introduction to assignment problem Another model comes under the class of linear programming model, which looks alike with transportation model with an objective function of minimizing the time or cost of manufacturing the products by allocating one job to one machine or one machine to one job or one destination to one origin or one origin to one destination only. This type of problem is given the name ASSIGNMENT MODEL. Basically assignment model is a minimization model. If we want to maximize the objective function, then there are two methods. One is to subtract all the elements of the matrix from the highest element in the matrix or to multiply the entire matrix by -1 and continue with the procedure. For solving the assignment problem we use Assignment technique or Hungarian method or Flood's technique. Class 15 Friday, February 12,2016 Hungaian s Method to solve AP Deduct the smallest element in each row from the other elements of the row. The matrix thus got is known as Row opportunity cost matrix (ROCM). The logic here is if we assign the job to any machine having higher cost or time, then we have to bear the penalty. If we subtract smallest element in the row or from all other element of the row, there will be at least one cell having zero, i.e zero opportunity cost or zero penalty. Hence that cell is more competent one for assignment. Class 16 Saturday, February 13,2016 Optimal solution of AP Deduct the smallest element in each column from other elements of the column. The matrix thus got is known as Column opportunity cost matrix (COCM). Here also by creating a zero by subtracting smallest element from all other elements we can see the penalty that one has to bear. Zero opportunity cell is more competent for assignment. The property of total opportunity cost matrix is that it will have at least one zero in every row and column. All the cells, which have zero as the opportunity cost, are eligible for assignment. Class 17 Monday, February 15,2016 Introduction and concept of game theory Every business manager is interested in capturing the larger share in the market. To do this they have to use different strategies (course of action) to motivate the consumers to prefer their product. For example you might have seen in newspapers certain company is advertising for its product by giving a number of (say 10) eyes and names of 10 cine stars and identify the eyes of the stars and match the name with the eyes. After doing this the reader has to write why he likes the product of the company. For right entry they get a prize. This way they motivate the readers to prefer the product of the company. When the opponent company sees this, they also use similar strategy to motivate the potential market to prefer the product of their company. Like this the companies advertise in series and measure the growth in their market share. This type of game is known as business game. Managers competing for share of the market, army chief planning or execution of war, union leaders and management involved in collective bargaining uses different strategies to fulfill their objective or to win over the opponent. All these are known as business games or competitive situation. In business, competitive situations arise in advertising and marketing campaigns by competing business firms. Hence, Game theory is a body of knowledge that deals with making decisions when two or more intelligent and rational opponents are involved under conditions of conflict or competition. The competitors in the game are called players.. 9

10 Class 18 Tuesday, February 16,2016 Two-person zero-sum game A game where two persons are playing the game and the sum of gains and losses is equal to zero, the game is known as Two-Person Zero-Sum Game (TPZSG). A good example of two-person game is the game of chess. A good example of n-person game is the situation when several companies are engaged in an intensive advertising campaign to capture a larger share of the market. Class 19 Wednesday, February 1,2016 Two-person zero-sum game examples Let us consider a game by name Two-finger morra, where two players (persons) namely A and B play the game. A is the winner and B is the loser. The matrix shown below is the matrix of A, the winner. The elements of the matrix show the gains of A. Any positive element in the matrix shows the gain of A and the negative element in the matrix show the loss (negative gain) of A. The game is as follows: Both the players A and B sit at a table and simultaneously raise their hand with one or two fingers open. In case the fingers shown by both the players is same, then A will gain Rs.2/-. In case the number of fingers shown is different (i.e. A shows one finger and B shows two fingers or vice versa) then A has to give B Rs. 2/- i.e. A is losing Rs.2/-. In the above matrix, strategy I refer to finger one and strategy II refers to two fingers. The above given matrix is the pay of matrix of A. The negative entries in the matrix denote the payments from A to B. The pay of matrix of B is the negative version of A s pay of matrix; because in two person zeros sum game the gains of one player are the losses of the other player. Always we have to write the matrix of the winner, who is represented on the left side of the matrix. The winner is the maximizing player, who wants to maximize his minimum gains. The loser is the minimizing player, who wants to minimize his maximum losses Class 20 Thursday, February 18,2016 Pure and Mixed Strategy In pure strategy game one knows, in advance of all plays that he will always choose only one particular course of action. Thus pure strategy is a decision rule always to select the same course of action. Every course of action is pure strategy. A mixed strategy is that in which a player decides, in advance to choose on of his course of action in accordance with some fixed probability distribution. This in case of mixed strategy we associate probability to each course of action (each pure strategy). The pure strategies, which are used in mixed strategy game with non-zero probabilities, are termed as supporting strategies. Mathematically, a mixed strategy to any player is an ordered set of m non-negative real numbers, which add to a sum unity (m is the number of pure strategies available to a player). Class 21 Friday, February 19,2016 Pure and Mixed Strategy with examples Mixed Strategies: Consider the matching pennies game: Player 2 Heads Tails Player 1 Heads 1,-1-1,1 Tails -1,1 1,-1. There is no (pure strategy) Nash equilibrium in this game. If we play this game, we should be unpredictable. That is, we should randomize (or mix) between strategies so that we do not get exploited. But not any randomness will do: Suppose Player 1 plays.75 Heads and.25 Tails (that is, Heads with 75% chance and Tails with 25% chance). Then Player 2 by choosing Tails (with 100% chance) can get an expected payoff of (-1) = 0.5. But that cannot happen at equilibrium since Player 1 then wants to play Tails (with 100% chance) deviating from the original mixed stategy. Since this game is completely symmetric it is easy to see that at mixed strategy Nash equilibrium both players will choose Heads with 50% chance and Tails with 50% chance. In this case the expected payoff to both players is (-1) = 0 and neither can do better by deviating to another strategy (regardless it is a mixed strategy or not). In general there is no guarantee that mixing will be at equilibrium. 10

11 Class 22 Friday, February 26,2016 Saddle point of games Maxi min aij=mini max a ij is called a game with saddle point. This makes us to understand ijjithat the players in the game always use pure strategies. The element at the intersection of their pure strategies is known as saddle point. The element at the saddle point is the value of the game. As the players uses the pure optimal strategies, the game is known as strictly determined game. A point to remember is that the saddle point is the smallest element in the row and the greatest element in the column. Not all the rectangular games will have saddle point, but if the game has the saddle point, then the pure strategies corresponding to the saddle point are the best strategies and the number at the point of intersection of pure strategies is the value of the game. Once the game has the saddle point the game is solved. The rules for finding the saddle point are: 1. Select the minimums of each row and encircle them. 2. Select the maximums of each column and square them. 3. A point where both circle and square appears in the matrix at the same point is the saddle point. Class 23 Saturday, February 27,2016 Odds Method for games Theory Once the game matrix is reduced to 2 2 the players has to resort to mixed strategies. We have already seen how using formulae can an algebraic method to find optimal strategies and value of the game. There is one more method available for the same that is the method of oddments. Steps involved in method of dominance are: 1. Subtract the two digits in column 1 and write them under column 2, ignoring sign. 2. Subtract the two digits in column 2 and write them under column 1 ignoring sign. 3. Similarly proceed for the two rows. These values are called oddments. They are the frequencies with which the players must use their courses of action in their optimum strategies. Class 24 Monday, February 29,2016 Dominance Methods to solve game Theory The general rules of dominance can be formulated as below 1. If all the elements of a column (say ith column) are greater than or equal to the corresponding elements of any other column (say jth column), then ith column is dominated by jth column. 2. If all the elements of rth row are less than or equal to the corresponding elements of any other row, say sth row, then rth row is dominated by sth row. 3. A pure strategy of a player may also be dominated if it is inferior to some convex combinations of two or more pure strategies, as a particular case, inferior to the averages of two or more pure strategies. Class 25 Friday, March 04,2016 Dominance Methods to solve game Theory Continued Class 26 Saturday, March 05,2016 Method of Oddments (for 2 2 games) Once the game matrix is reduced to 2 2 the players has to resort to mixed strategies. We have already seen how using formulae can an algebraic method to find optimal strategies and value of the game. There is one more method available for the same that is the method of oddments. Steps involved in method of dominance are: 1. Subtract the two digits in column 1 and write them under column 2, ignoring sign. 2. Subtract the two digits in column 2 and write them under column 1 ignoring sign. 3. Similarly proceed for the two rows. These values are called oddments. They are the frequencies with which the players must use their courses of action in their optimum strategies. Let us take a simple example and get the answer for a game. 11

12 Class 27 Tuesday, March 08,2016 Graphical Method for Solving Mixed Strategy Game When a m n pay of matrix can be reduced to m 2 or n 2 pay off matrix, we can apply the sub game method. But too many sub games will be there it is time consuming. Hence, it is better to go for Graphical method to solve the game when we have m 2 or n 2 matrixes. Class 28 Wednesday, March 09,2016 Graphical Method for Solving Mixed Strategy Game Continued. Class 29 Thursday, March 10,2016 Quiz Test Class 30 Friday, March 11,2016 Introduction to sequencing problem we have dealt with problems where two or more competing candidates are in race for using the same resources and how to decide which candidate (product) is to be selected so as to maximize the returns (or minimize the cost). Now let us look to a problem, where we have to determine the order or sequence in which the jobs are to be processed through machines so as to minimize the total processing time. Here the total effectiveness, which may be the time or cost that is to be minimized is the function of the order of sequence. Such type of problem is known as SEQUENCING PROBLEM. In case there are three or four jobs are to be processed on two machines, it may be done by trial and error method to decide the optimal sequence (i.e. by method of enumeration). In the method of enumeration for each sequence, we calculate the total time or cost and search for that sequence, which consumes the minimum time and select that sequence. This is possible when we have small number of jobs and machines. But if the number of jobs and machines increases, then the problem becomes complicated. It cannot be done by method of enumeration. Consider a problem, where we have n machines and m jobs then we have (n!) m theoretically possible sequences. Class 31 Saturday, March 12,2016 Algorithm for n jobs and two machines If the problem given has two machines and two or three jobs, then it can be solved by using the Gantt chart. But if the numbers of jobs are more, then this method becomes less practical. (For understanding about the Gantt chart, the students are advised to refer to a book on Production and Operations Management (chapter on Scheduling). Gantt chart consists of X -axis on which the time is noted and Y-axis on which jobs or machines are shown. For each machine a horizontal bar is drawn. On these bars the processing of jobs in given sequence is marked. Let us take a small example and see how Gantt chart can be used to solve the same. Class 32 Monday, March 14,2016 Numerical on n jobs and two machines There are two jobs job 1 and job 2. They are to be processed on two machines, machine A and Machine B in the order AB. Job 1 takes 2 hours on machine A and 3 hours on machine B. Job 2 takes 3 hours on machine A and 4 hours on machine B. Find the optimal sequence which minimizes the total elapsed time by using Gantt chart. Class 33 Wednesday, March 16,2016 Algorithm for n jobs and three machines When there are n jobs, which are to be processed on three machines say A, B, and C in the order ABC i.e first on machine A, second on machine B and finally on machine C. We know processing times in time units. As such there is no 12

13 direct method of sequencing of n jobs on three machines. Before solving, a three-machine problem is to be converted into a two-machine problem. The procedure for converting a three-machine problem into two-machine problem is: (a) Identify the smallest time element in the first column, i.e. for machine 1 let it be A r. (b) Identify the smallest time element in the third column, i.e. for machine 3, let it be C s (c) Identify the highest time element in the second column, i.e. for the center machine, say machine 2, let it be B i. (d) Now minimum time on machine 1 i.e. A r must be maximum time element on machine 2, i.e. B i OR Minimum time on third machine i.e. C s must be maximum time element on machine 2 i.e. B i OR Both A r and C s must be B i (e) If the above condition satisfies, then we have to workout the time elements for two hypothetical machines, namely machine G and machine H. The time elements for machine G, G i =A i +B i. The time element for machine H, is H i = B i + C i (f)now the three-machine problem is converted into two-machine problem. We can find sequence by applying Johnson Bellman rule. (g) All the assumption mentioned earlier will hold good in this case also. Class 34 Thursday, March 17,2016 Examples of n jobs and 3 machines problems A machine operator has to perform three operations, namely plane turning, step turning and taper turning on a number of different jobs. The time required to perform these operations in minutes for each operating for each job is given in the matrix given below. Find the optimal sequence, which minimizes the time required. Job. Time for plane turning Time for step turning Time for taper turning. In minutes in minutes in minutes Class 35 Friday, March 18,2016 PROCESSING OF 2 - JOBS ON M MACHINES There are two methods of solving the problem. (a) By enumerative method and (b) Graphical method. Graphical method is most widely used. Let us discuss the graphical method by taking an example. Graphical Method This method is applicable to solve the problems involving 2 jobs to be processed on m machines in the given order of machining for each job. In this method the procedure is: (a) Represent Job 1 on X- axis and Job 2 on Y-axis. We have to layout the jobs in the order of machining showing the processing times. (b) The horizontal line on the graph shows the processing time of Job 1 and idle time of Job 2. Similarly, a vertical line on the graph shows processing time of job 2 and idle time of job 1. Any inclined line shows the processing of two jobs simultaneously. (c) Draw horizontal and vertical lines from points on X- axis and Y- axis to construct the blocks and hatch the blocks. (Pairing of same machines). (d) Our job is to find the minimum time required to finish both the jobs in the given order of machining. Hence we have to follow inclined path, preferably a line inclined at 45 degrees (in 13

14 a square the line joining the opposite coroners will be at 45 degrees). (e) While drawing the inclined line, care must be taken to see that it will not pass through the region indicating the machining of other job. That is the inclined line should not pass through blocks constructed in step (c). (f) After drawing the line, the total time taken is equals to Time required for processing + idle time for the job. The sum of processing time + idle time for both jobs must be same. Class 36 Saturday, March 19,2016 Introduction of Queueing Theory and applications Queues or waiting lines stands for a number of customers waiting to be serviced. Queue does not include the customer being serviced. The process or system that performs the services to the customer is termed as service channel or service facility. Thus from the above we see that waiting lines or not only the lines formed by human beings but also the other things like railway coaches, vehicles, material etc. Entire queuing system can be completely described by: (a) The input (Arrival pattern) (b) The service mechanism or service pattern, (c) The queue discipline and (d) Customer behavior. Class 37 Monday, March 21,2016 Description of M/M/1 model Poisson Arrival / Poisson output / Number of channels / Infinite capacity / FIFO Model M / M / 1 / ( / FIFO): Formulae used 1. Average number of arrivals per unit of time = λ 2. Average number of units served per unit of time = µ 3. Traffic intensity or utility ratio the condition is ( ) 4. Probability that the system is empty 5. Probability that there are n units in the system = 6. Average number of units in the system = 7. Average number of units in the waiting line = 8. Average waiting length (mean time in the system) = = 9. Average length of waiting line with the condition that it is always greater than zero 10. Average time an arrival spends in the system 11. System is busy 12. Idle time = 13. Probability distribution of waiting time = 14. Probability that a consumer has to wait on arrival 14

15 15. Probability that a new arrival stays in the system Class 38 Tuesday, March 22,2016 Numerical on M/M/1 model A T.V. Repairman finds that the time spent on his jobs have an exponential distribution with mean of 30 minutes. If he repairs sets in the order in which they come in, and if the arrival of sets is approximately Poisson with an average rate of 10 per 8 hour day, what is repairmanís expected idle time each day? How many jobs are ahead of the average set just brought in? Solution This problem is Poisson arrival/negative exponential service / single channel /infinite capacity/ FIFO type problem. Data: λ = 10 sets per 8 hour day = 10 / 8 = 5/4 sets per hour. Given 1/µ = 30 minutes, hence µ = (1/30) 60 = 2 sets per hour. Hence, Utility ratio = ρ= (λ/µ) = (5/4) / 2 = = 5 / 8. = This means out of 8 hours 5 hours the system is busy i.e. repairman is busy. Probability that there is no queue = The system is idle = (1 ρ) = 1 ñ (5 / 8) = 3 / 8 = That is out of 8 hours the repairman will be idle for 3 hours. Number of sets ahead of the set just entered = Average number of sets in system = λ / (µ λ) = ρ / (1 ρ) = / (1 ñ 0.625) = 5 / 3 ahead of jobs just came in. Class 39 Monday, March 28,2016 Poisson distribution in estimating arrival rate The common basic waiting line models have been developed on the assumption that arrival rate follows the Poisson distribution and that service times follow the negative exponential distribution. This situation is commonly referred to as the Poisson arrival and Exponential holding time case. These assumptions are often quite valid in operating situations. Unless it is mentioned that arrival and service follow different distribution, it is understood always that arrival follows Poisson distribution and service time follows negative exponential distribution. Research scholars working on queuing models have conducted careful study about various operating conditions like - arrivals of customers at grocery shops, Arrival pattern of customers at ticket windows, Arrival of breakdown machines to maintenance etc. and confirmed almost all arrival pattern follows nearly Poisson distribution. One such curve is shown in figure 9.5. Although we cannot say with finality that distribution of arrival rates are always described adequately by the Poisson, there is much evidence to indicate that this is often the case. We can reason this by saying that always Poisson distribution corresponds to completely random arrivals and it is assumed that arrivals are completely independent of other arrivals as well as any condition of the waiting line. The commonly used symbol for average arrival rate in waiting line models is the Greek letter Lamda (λ ), arrivals per time unit. It can be shown that when the arrival rates follow a Poisson processes with mean arrival rate λ, the time between arrivals follow a negative exponential distribution with mean time between arrivals of 1/λ. Class 40 Tuesday, March 29,2016 Exponential distribution in estimating service rate The commonly used symbol for average service rate in waiting line models is the Greek letter ëmuí ëµí, the number of services completed per time unit. As with arrivals it can be shown that when service rates follow a Poison process with mean service rate µ, the distribution of serviced times follow the negative exponential distribution with mean service time 1 /µ. The reason for the common reference to rates in the discussion of arrivals and to times in the discussion of service is simply a matter of practice. One should hold it clearly in mind, however, for both arrivals and services, that in the general Poison models, rates follow the Poisson distribution and times follow the negative exponential distribution. One must raise a doubt at this point why the interest in establishing the validity of the Poisson and Negative exponential distributions. The answer is that where the assumptions hold, the resulting waiting line formulas are quite simple. The Poison and Negative exponential distributions are single parameters 15

16 distributions; that is, they are completely described by one parameter, the mean. For the Poisson distribution the standard deviation is the square root of the mean, and for the negative exponential distribution the standard deviation is equal to the mean. The result is that the mathematical derivations and resulting formulas are not complex. Where the assumptions do not hold, the mathematical development may be rather complex or we may resort to other techniques for solution, such as simulation. Class 41 Wednesday, March 30,2016 Numerical based on the model. In a departmental store one cashier is there to serve the customers. And the customers pick up their needs by themselves. The arrival rate is 9 customers for every 5 minutes and the cashier can serve 10 customers in 5 minutes. Assuming Poisson arrival rate and exponential distribution for service rate, find: (a) Average number of customers in the system. (b) Average number of customers in the queue or average queue length. (c) Average time a customer spends in the system. (d) Average time a customer waits before being served. Class 42 Thursday, March 31,2016 Application of Queue model for better service to the customers Queuing management has been applied very successfully in many service-oriented industries. L. L. Bean, a large telemarketer and mail-order catalog house for high-quality sporting goods and apparel, used queuing theory to optimize staffing levels resulting in an estimated $500,000 per year savings. The Department of Motor Vehicles in Virginia and Arizona used queuing technology to virtually eliminate long lines and greatly improve customer satisfaction. In addition, they were able to significantly improve employee morale and reduce operating costs. Queuing models have also been used to plan staffing levels in an outpatient hospital laboratory department and a centralized appointment department in Lourdes Hospital in Binghamton, New York. Queuing models were used to identify an optimal configuration of capacity and staffing levels for both departments. The lengthy delays in answering telephone calls in the centralized appointments department were completely eliminated by rearranging work shifts of current employees. Queuing theory has been used extensively in the banking industry to increase business by careful placement of merchandising materials while at the same time alleviating both the actual and perceived amount of time a customer spends waiting in line. Finally, queuing theory has been applied to computer simulation models to help with business decisions and problems. Class 43 Friday, April 01,2016 Quiz Test Class 44 Saturday, April 02,2016 Introduction of replacement Problem The problem of replacement arises when any one of the components of productive resources, such as machinery, building and men deteriorates due to time or usage. The examples are: (a) A machine, which is purchased and installed in a production system, due to usage some of its components wear out and its efficiency is reduced. (b) A building in which production activities are carried out, may leave cracks in walls, roof etc, and needs repair. (c) A worker, when he is young, will work efficiently, as the time passes becomes old and his work efficiency falls down and after some time he will become unable to work. Class 45 Monday, April 04,2016 Replacement of assets that deteriorate with time (a) If time is measured continuously, then the average annual costs will be minimized by 16