ASSIGNMENT

SUBJECT &CODE NAME :- MB0048 OPERATIONS RESEARCH

SEMESTER :- 2ND

Q 1 :- Explain the types of Operations Research Models. Briefly explain the phases of Operations Research.

Ans :-Meaning of Operations Research: - Personal Churchman, Aackoff, and Aruoff defined

operations research as “the application of scientific methods, techniques and tools to the operation of a system with optimum solutions to the problems” where 'optimum' refers to the best possible alternative.The objective of OR is to provide a scientific basis to the decision-makers for solving problems involving interaction with various components of the organisation. This can be achieved by employing a team of scientists from different disciplines to work together for finding the best possible solution in the interest of the organisation as a whole. The solution thus obtained is known as an optimal decision..

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

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

Iconic models Analogue models

Let us now study the two types of physical models in detail.Iconic models are primarily images of objects or systems, represented on a smaller scale. These models can simulate the actual performance of a product. Analogue models are small physical systems having characteristics similar to the objects they represent, such as toys.

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

Phases of Operations ResearchThe scientific method in OR study generally involves three phases

Judgment phase This phase includes the following activities: Determination of the operations Establishment of objectives and values related to the operations Determination of suitable measures of effectiveness Formulation of problems relative to the objectives

Research phase This phase utilizes the following methodologies: Operation and data collection for a better understanding of the problems Formulation of hypothesis and model 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 Prediction of various results and consideration of alternative methods

Action phase This phase involves making recommendations for the decision process. Therecommendations can be made by those who identify and present the problem or by anyone who influences the operation in which the problem has occurred..

Q 2 :- a. Explain the graphical method of solving Linear Programming Problem.b. A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper in a week. There are 160 production hours in a week. It requires 0.20 and 0.40 hours to produce a ton of grade X and Y papers. The mill earns a profit of Rs. 200 and Rs. 500 per ton of grade X and Y paper respectively. Formulate this as a Linear Programming Problem?

Ans :- Meaning of Linear programming problem:- Linear Programming (LP) is a mathematical technique designed to help managers in their planning and decision-making. It is usually used in an organisation that is trying to make the most effective use of its resources. Resources typically include machinery, manpower, money, time, warehouse space, and raw materials. A few examples of problems in which LP has been successfully applied are: Developments of a production schedule that will satisfy future demands for a firm’s product and at the same time minimise total production and inventory costs. Establishment of an investment portfolio from a variety of stocks or bonds that will maximise a company’s return on investment. Allocation of a limited advertising budget among radio, TV, and newspaper spots in order to maximise advertising effectiveness. Determination of a distribution system that will minimisess total shipping cost from several warehouses to various market locations

Explanation of graphical method of solving Linear Programming Problem The collection of all feasible solutions to an LPP constitutes a convex set whose extreme points correspond to the basic feasible solutions.

There are a finite number of basic feasible regions within the feasible solution space. If the convex set of the feasible solutions of the system of simultaneous equation is a convex polyhedron, then at least one of the extreme points

gives an optimal solution. If the optimal solution occurs at more than one extreme point, the value of the objective function will be the same for all convex combination of

these extreme points.Working ruleThe method of solving an LPP on the basis of the above analysis is known as the graphical method. The working rule for the method is as follows.Step 1: Formulate the problem in terms of a series of mathematical equations representing objective function and constraints of LPP.Step 2: Plot each of the constraints equation graphically. Replace the inequality constraint equation to form a linear equation. Plot the equations on the planar graph with each axis representing respective variables.Step 3: Identify the convex polygon region relevant to the problem. The area which satisfies all the constraints simultaneously will be the feasible region. This is determined by the inequality constraints.Step 4: Determine the vertices of the polygon and find the values of the given objective function Z at each of these vertices. Identify the greatest and the least of these values. These are respectively the maximum and minimum value of Z.Step 5: Identify the values of (x1, x2) which correspond to the desired extreme value of Z. This is an optimal solution of the problem..

Q-3:- a. Explain how to solve the degeneracy in transportation problems. b. Explain the procedure of MODI method of finding solution through optimality test .

Ans : Degeneracy in transportation problem

A basic solution to an m-origin, n destination transportation problem can have at the most m+n-1 positive basic variables (non-zero), otherwise the basic solution degenerates. It follows that whenever the number of basic cells is less than m + n – 1, the transportation problem is a degenerate one. The degeneracy can develop in two ways:

1. Case 1 - The degeneracy develops while determining an initial assignment via any one of the initial assignment methods discussed earlier.

To resolve degeneracy, you must augment the positive variables by as many zero-valued variables as is necessary to complete the required m + n – 1 basic variable. These zero-valued variables are selected in such a manner that the resulting m + n – 1 variable constitutes a basic solution. The selected zero valued variables are designated by allocating an extremely small positive value ε to each one of them. The cells containing these extremely small allocations are then treated like any other basic cells. The ε’s are kept in the transportation table until temporary degeneracy is removed or until the optimum solution is attained, whichever occurs first. At that point, we set each ε = 0.

2. Case 2 - The degeneracy develops at the iteration stage. This happens when the selection of the entering variable results in the simultaneous drive to zero of two or more current (pre-iteration) basic variables.

To resolve degeneracy, the positive variables are augmented by as many zero-valued variables as it is necessary to complete m+n-1 basic variables. These zero-valued variables are selected from among those current basic variables, which are simultaneously driven to zero. The rest of the procedure is exactly the same as discussed in case 1.Note - The extremely small value ε is infinitely small and it never affects the value it is added to or subtracted from. Introduce ‘’ in unallocated minimum cost cell to avoid forming a loop.b. Procedure of MODI method A feasible solution has to be found always. Rather than determining a first pproximation by a direct application of the simplex method, it is more efficient to work with the transportation table. The transportation algorithm isthe simple method specialised to the format of table involving the following steps:i) Finding an initial basic feasible solutionii) Testing the solution for optimalityiii) Improving the solution, when it is not optimaliv) Repeating steps (ii) and (iii) until the optimal solution is obtainedThe solution to transportation problem is obtained in two stagesIn the first stage, we find the basic feasible solution using any of thefollowing methods: North-west corner rule Matrix minima method or least cost method Vogel’s approximation method.In the second stage, we test the basic feasible solution for its optimality byMODI method Step 1 - For each row of the transportation table, identify the smallest and the next to smallest costs. Determine the difference between them for each row. Display them alongside the transportation table by enclosing them in parenthesis against the respective rows. Similarly, compute the differences for each column.Step 2 - Identify the row or column with the largest difference among all the rows and columns. If a tie occurs, use any arbitrary tie breaking choice. Let the greatest difference correspond to the ith row and let Cij be the smallest cost in the ith row. Allocate the maximum feasible amount xij = min (ai, bj) in the (i, j)th cell and cross off the ith row or the jth column in the usual manner.Step 3 - Recompute the column and row differences for the reduced transportation table and go to step 2. Repeat the procedure until all the rim requirements are satisfied.

Step 4 - You repeat steps 1 to 3 to till all allocations are over.Step 5 - For allocating all forms of equations ui+ vj = cj , set one of the dual variable ui / vjto zero and solve for others.Step 6 - Use this value to find ij = cij - ui – vj . If all ij 0, then it is the optimal solution.Step 7 - If any ij 0 select the most negative cell and form loop. Starting point of the loop is positive and alternative corners of the loop are negative and positive. Examine the quantities allocated at negative places. Select the minimum, add it to the positive places and subtract from the negative places.

Step 8 - Form a new table and repeat steps 5 to 7 till ij 0

Q – 4 : a. Explain the steps involved in Hungarian method of solving Assignment problems. b. What do you mean by unbalanced assignment problem? How do you

overcome it?

Ans :- Steps in Hungarian method:-Hungarian method algorithm is based on the concept of opportunity cost and is more efficient in solving assignment problems. The following steps are adopted to solve an AP using the Hungarian method algorithm.Step 1: Prepare row ruled matrix by selecting the minimum values for each row and subtract it from the other elements of the row.Step 2: Prepare column-reduced matrix by subtracting minimum value of the column from the other values of that column. Step 3: Assign zero row-wise if there is only one zero in the row and cross (X) or cancel other zeros in that column.Step 4: Assign column wise if there is only one zero in that column and cross other zeros in that row.Step 5: Repeat steps 3 and 4 till all zeros are either assigned or crossed. Ifthe number of assignments is equal to number of rows present, you have arrived at an optimal solution, if not, proceed to step 6.Step 6: Mark () the unassigned rows. Look for crossed zero in that row. Mark the column containing the crossed zero. Look for assigned zero in that column. Mark the row containing assigned zero. Repeat this process till all the makings are done.Step 7: Draw a straight line through unmarked rows and marked column. The number of straight line drawn will be equal to the number of assignments made.Step 8: Examine the uncovered elements. Select the minimum. Subtract it from the uncovered elements. Add it at the point of intersection of lines. Leave the rest as is. Prepare a new table.

Step 9: Repeat steps 3 to 7 till optimum assignment is obtained.

Step 10: Repeat steps 5 to 7 till number of allocations = number of rows.

Unbalanced APUnbalanced assignment problem is an assignment where the number of rows is not equal to the number of columns and vice versa. For example, the number of machines may be more than the number of jobs or thenumber of jobs may be more than the number of machines. In such a situation, you have to introduce dummy rows or columns in the matrix. The dummy rows or columns will contain all cost elements as zero. This balances the problem and then you can use Hungarian method to find the optimal assignment.

Maximisation in AP

Some assignment problems are phrased in terms of maximising the profit or effectiveness or payoff of an assignment of people to tasks or of jobs to machines. You cannot apply the Hungarian method to such maximisation problems. Therefore, you need to reduce it to a minimisation problem.It is easy to obtain an equivalent minimisation problem by converting every number in the table to an opportunity loss. To do so, you need to subtract every value from the highest value of the matrix and then proceed as usual.You will notice that minimising the opportunity loss produces the same assignment solution as the original maximisation problem.

Q – 5 :- a. Write a short note on Monte Carlo Simulation.

b.A Company produces 150 cars. But the production rate varies with the distribution.Production Rate 147 148 149 150 151 152 153Probability 0.05 0.10 0.15 0.20 0.30 0.15 0.05At present the track will hold 150 cars. Using the following random numbersdetermine the average number of cars waiting for shipment in the company andaverage number of empty space in the truck. Random Numbers 82, 54, 50, 96, 85, 34,30, 02, 64, 47.Ans :- Monte-Carlo SimulationThe Monte-Carlo method is a simulation technique in which statistical distribution functions are

created by using a series of random numbers. This approach has the ability to develop many months or years of data in a matter of few minutes on a digital computer. The method is generally used to solve the problems that cannot be adequately represented by mathematical models or where solution of the model is not possible by analytical method.Let us now describe each step in detail.Step 1: Define the problem:a) Identify the objectives of the problem.b) Identify the main factors that have the greatest effect on the objectives of the problem.Step 2: Construct an appropriate model:a) Specify the variables and parameters of the model.b) Formulate the appropriate decision rules, i.e., state the conditions under which the experiment is to be performed.c) Identity the type of distribution that will be used. Models use either theoretical distributions or empirical distributions to state the patterns of occurrence associated with the variables.d) Specify the manner in which time will change.e) Define the relationship between the variables and parameters.Step 3: Prepare the model for experimentation:a) Define the starting conditions for the simulation.b) Specify the number of runs of simulation to be made.Step 4: Using steps 1 to 3, experiment with the model:a) Define a coding system that will correlate the factors defined in step 1 with the random numbers to be generated for the simulation.b) Select a random number generator and create the random numbers to be used in the simulation.

c) Associate the generated random numbers with the factors identified in step1 and coded in step 4(a).Step 5: Summaries and examine the results obtained in step 4.Step 6: Evaluate the results of the simulation.Step 7: Formulate proposals for advice to management on the course of action to be adopted and modify the model, if necessary

Q – 6 :- a. Explain the dominance principle in game theory. b. Describe the Constituents of a Queuing System. c. Differentiate between PERT and CPM.

Ans Dominance principle in game theory Dominance In a rectangular game, the pay-off matrix of player A is pay-off in one

specific row ( r row )th exceeding the corresponding pay-off in another specific row ( s row )th . This means that whatever course of action is adopted by player B, for A, the course of action Ar yields greater gains than the course of action As . Therefore, Ar is a better strategy than As irrespective of B’s strategy. Hence, you can say that Ar dominates As . Alternatively, if each pay-off in a specific column ( p column )th is less than the corresponding pay-off in another specific column ( q column ) th , it means strategy Bp offers minor loss than strategy Bq irrespective of A’s strategy. Hence, you can say that Bp dominates Bq Therefore, you can say that:a) In the pay-off matrix, if each pay-off in r row th is greater than (or equal to) the corresponding pay-off in the s row th , Ar dominates As.b) In the pay-off matrix, if each pay-off in p column th is less than (or equal to) the corresponding pay-off in the q column th , Bp dominates Bq

Constituents of a Queuing System The constituents of a queuing system include arrival pattern, service facility and queue discipline.

Arrival pattern: It is the average rate at which the customers arrive. Service facility: Examining the number of customers served at a time and the

statistical pattern of time taken for service at the service facility. Queue discipline : The common method of choosing a customer for service

amongst those waiting for service is ‘First Come First Serve’.Arrival patternThe arrival of customers can be regular as in case of an appointment system of a doctor or flow of components on a conveyor belt. The regular pattern of arrivals is neither very common nor very easy to deal with mathematically. The following are the important arrival characteristics: 1. Size of the population: Unlimited (infinite) or limited (finite) 2. Pattern of arrivals (statistical distribution)3. Behaviour of arrivals

Service Facility

In the previous section, you learnt the constituents of a queuing system. You will now learn the service facility. Service Facility is based on three parameters – Availability of service, number of service centers and duration of service.i)Availability of serviceii) Number of service centersiii) Duration of serviceQueue DisciplineIn the previous section, you learnt the service facility. You will now learn about queue discipline. The queue discipline is the order or manner in which customers from the queue are selected for service. There are a number of ways in which customers in the queue are served:

Differences between PERT and CPM:- PERT

Some key points of PERT are as follows

PERT was developed in connection with an Research and Development (R&D) work. Therefore, it had to cope with the uncertainties that are associated with R&D activities. In PERT, the total project duration is regarded as a random variable. Therefore, associated probabilities are calculated in order to characterize it.

PERT is normally used for projects involving activities of non-repetitive nature in which time estimates are uncertain.

It helps in pinpointing critical areas in a project, so that necessary adjustment can be made to meet the scheduled completion date of the project.

CPM

CPM was developed in connection with a construction project, which consisted of routine tasks whose resource requirements and duration were known with certainty. Therefore, it is basically deterministic.

CPM is suitable for establishing a trade-off for optimum balancing between schedule time and cost of the project.

CPM is used for projects involving activities of repetitive nature.