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OPERATION RESEARCH Hemal Rajyaguru
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Introduction One has to take decisions for himself and for others.
(E.g.-students which specialization they should take.) Everyone has to develop their talents in such a way so that they can take correct decision at a proper time. (E.g. selection of proper engg. branch,Investments in proper property,etc.)
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Decision Dependency An effective decision depends on following factors: Economic, Social, Political
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Starting of a new factory
Economic factors would be construction cost, labour cost, raw material cost, transportation cost, taxes, energy, etc. Social factors would be number of persons available for labour, engineers, experts, etc. Political factors would be government policies, rules & regulations (E.g. Gujarat is the best)
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Decision maker has to faced with…..
(i)large number of interacting variables as discussed in factors (ii)actions of the other competitors, over which we don’t have control
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People make decisions…..
Based on their past experience and intuitions(understanding) but they are not always enough. Some formal system is needed to determine an effective course of actions.
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Operation Research Quantitative methods, Management science,
Decision science provides scientific approach to the executives for making better decisions for operation under their control.
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Operation research is not decision making
It hopefully improve the decisions made by the management in rational, logical, systematic and scientific way.
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Operations Research Operations research is “a scientific approach to decision making which seeks to determine how best to design and operate a system, under conditions requiring the allocation of scarce resources.” Provides a set of algorithms that act as tools for effective problem solving and decision making Extensive applications in engineering, business and public systems. Used extensively by manufacturing and service industries in decision making that’s why also useful in IT industry.
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Operations Research Origin during World War II when the British military asked scientists to analyze military problems The application of mathematics and scientific method to military applications was called Operations Research. Operational analysis System analysis Cost benefit analysis Management science Decision Science
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Purpose Helps in determining the best(optimum) solution(course of action) to problems where decision has to be taken under the restriction of limited resources. It is possible to convert any real life problem into a mathematical model. Basic feature of OR is to formulate a real world problem as a mathematical model.
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Any industry want to lower their labour or production costs to achieve higher profits and OR can implied that. OR=problem solving technique =science+art where, science is using mathematical techniques for solving decision problems Art is the ability to develop good report with good supplying information
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Quantitative approach to decision-making: or/ms
Decision under certainty or uncertainty Static decision: Decisions for one time period only come under this. E.g.: (1)Toss result in match (II)Match result Dynamic decision: Decisions over several time periods comes under this. E.g.: Bowling, Batting, Fielding strategies at different stage of match.
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Methodology of Operations Research
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Methodology of Operations Research
Problem Formulation Model Building Physical Models Mathematical Models Obtaining input Data Solution of Model Feasible and Infeasible Solutions Optimal and Non-optimal Solutions Unique and Multiple Solutions
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Fundamentals of Operations Research
Linear Programming – Formulations Linear Programming – Solution Duality and Sensitivity Analysis Transportation Problem Assignment Problem Dynamic Programming Deterministic Inventory Models
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Linear Programming First conceived by George B Dantzig around 1947.
The work of Kantorovich (1939) was published in 1959 Dantzig’s first paper was titled “Programming in Linear Structure” Koopmans coined the term “Linear Programming” in 1948 Simplex method was published in 1949 by Dantzig
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LINEAR PROGRAMMING FORMULATIONS
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Example 1 – Maximisation Problem (Product Mix Problem)
Consider a small manufacturer making two products A and B. Two resources R1 and R2 are require to make these products. Each unit of product A requires 1 unit of R1 and 3 units of R2. Each unit of product B requires 1 unit of R1 and 2 units of R2. The manufacturer has 5 units of R1 and 12 units of R2 available, The manufacturer also makes a profit of Rs. 6 per unit of product A sold and Rs. 5 per unit of product B sold.
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Product Resource 1 Resource 2 Profit A 1 3 Rs. 6 B 2 Rs. 5 TOTAL 5 12
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Solution The manufacturer has to decide on the number of units of products A and B to produce. It is acceptable that the manufacturer would like to make as much profit as possible and would decide on the production quantities accordingly. The manufacturer has to ensure that the resources needed to make the products are available. Before we attempt to find out the decisions of the manufacturer, let us redefine the problem in an algebraic form. The manufacturer has to decide on the production quantities. Let us call them x and y and define Let x be the number of units of product A made Let y be the number of units of product B made.
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The profit associated with x units of product A and y units of product B is 6x + 5y. The manufacturer would determine x and y such that this function has a maximum value. The requirements of two resources are x + y for R1 and 3x + 2y for R2 and the manufacturer has to ensure that these are available. The problem is to find x and y such that 6x + 5y is maximized and x + y ≤ 5 and 3x + 2y ≤ 12 are satisfied. Also it is necessary that x, y ≥ 0 so that only nonnegative quantities are produced. If we redefine the production quantities as x1 and x2 (for reasons of uniformity and consistency) then problem is to Maximize Z = 6x1 + 5x2 Subject to x1 + x2 ≤ 5 3x1 + 2x2 ≤ 12 x1, x2 ≥ 0
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Terminology The problem variables x1 and x2 are called decision variables and they represent the solution or the output decision from the problem. The profit function that the manufacturer wishes to increase, represents the objective of making the decisions on the production quantities and is called the objective function. The conditions matching the resource availability and resource requirement are called constraints. These usually limit (or restrict) the values the decision variables can take. We have also explicitly stated that the decision variable should take non negative values. This is true for all Linear Programming problems. This is called non negativity restriction The problem that we have written down in algebraic form represents the mathematical model of the given system and is called the Problem Formulation.
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The problem formulation has the following steps.
Identifying the decision variables. Writing the objective function Writing the constraints. Writing the non negativity restrictions. In the above formulation, the objective function and the constraints are linear. Therefore the model that we formulated is a linear programming problem. A Linear Programming problem has A linear objective function, linear constraints and the non negativity constraints on all the decision variables.
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Example – 2 Minimization Problem
The Agricultural Research institute suggested to a farmer to spread out at least 4800 kg of a special phosphate fertilizer and not less than 7200 kg of a special nitrogen fertilizer to raise productivity of crops in his fields. There are two sources two obtaining these-mixtures A and B. Both of these are available in bag weighing 100 kg each and they cost Rs 40 and Rs 24 respectively. Mixture A contains phosphate and nitrogen equivalent of 20 kg and 80 kg respectively. While mixture B contains these ingredients equivalent of 50 kg each. Write this as a linear programming problem to determine how many bags of each type the farmer should buy in order to obtain the required fertilizer at minimum cost.
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Bag Type (100 Kg) Price per Bag Phosphate/Bag Nitrogen / Bag A 40 20 80 B 24 50 Min Requirement 4800 7200
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Solution The objective function The constraints
Non-negativity condition Minimise Z = 40x1 + 24x2 Cost Subject to 20x1 + 50x2 ≥ Phosphate requirement 80x1 + 50x2 ≥ Nitrogen requirement x1, x2 ≥ 0 Non-negativity restriction
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Examples A firm is engaged in producing two products, A and B. Each unit of product A requires 2 kg of raw material and 4 labour hours for processing, whereas each unit of product B requires 3 kg of raw material and 3 hours of labour, of the same type. Every week, the firm has an availability of 60 kg of raw material and 96 labour hours. One unit of product A sold yields Rs 40 and one unit of product B sold gives Rs 35 as profit. Formulate this problem as a linear programming problem to determine as to how many units of each of the products should be produced per week so that the firm can earn the maximum profit. Assume that there is no marketing constraint so that all that is produced can be sold.
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Product Raw Material Labour Profit A 2 4 40 B 3 35 Total 60 96
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A company is manufacturing two different types of products, A and B
A company is manufacturing two different types of products, A and B. Each product has to be processed in 3 different departments – casting, machining and finally inspection. The capacity of 3 departments is limited to 35 hrs., 32 hrs and 24 hrs. per week respectively. Product A requires 7 hrs. in casting department, 8 hrs. in machining shop and 4 hrs. in inspection, whereas product B requires 5 hrs., 4 hrs. and 6 hrs. respectively in respective shop. The profit contribution for a unit product of A and B is Rs. 30 and Rs. 40 respectively. Formulate the problem.
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Department Product Capacity Per week A B Casting 7 5 35 Machining 8 4 32 Inspection 6 24 Profit contribution Per unit Rs. 30 /- Rs. 40 /-
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A24-hour supermarket has the following minimal requirements for cashiers:
Period 1 follows immediately after period 6. A cashier works eight consecutive hours, starting at the beginning of one of the six time periods. Determine a daily employee Worksheet which satisfies the requirements with the least number of personnel. Formulate the problem as a linear programming problem. Period: 1 2 3 4 5 6 Time of day (24-hour clock): 3-7 7-11 11-15 15-19 19-23 23-3 Minimum number required: 7 20 14 10
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