Operations Research Assignment 2

 

Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

Master of Business Administration-MBA Semester 2

Operation Research – MB0048 Assignment Set - 2 Q1. What are the essential characteristics of Operation Research? Mention different phases in an Operation Research study. Point out some limitations of O.R? A. The basic dominant characteristic feature of operations research is that it employs mathematical

representations or models to analyse problems. This distinct approach represents an adaptation of the scientific methodology used by the physical sciences. The scientific method translates a real given  problem into a ma mathemat thematical ical represe representation ntation w which hich is solved solved and retransformed retransformed into into the original original context. context. The OR approach to problem solving consists of the following steps: Defining the problem, Constructing the model, Solving the model, Validating the model and Implementing the final result. 

Features of Operation Research: Some key features of OR are as follows: 

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

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

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OR makes use of scientific methods to solve problems.

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OR increases effectiveness of the management’s decision-making ability.

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

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OR offers a quantitative solution.

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OR also takes into account the human factors.

 Phasess of O  Phase Operat perations ions R Researc esearch: h:

The scientific method in OR study generally involves the 1

 

Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

following three phases.

 Phases  Phas es of o operati perations ons research   Judg  Judgmen mentt

Pha Phase: se:

This phase includes the following activities: 

 Determination  Determi nation of of the operations operations

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 Establishment  Establi shment of the objectiv objectives es and values values related related to the the operations operations

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 Determination  Determi nation of of the suitable suitable measures measures of of effectiveness effectiveness

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 Formulation  Formula tion of of the problems problems relative relative to to the objective objectivess

  Resea  Research rch

P Phase: hase:

This phase utilises the following methodologies:    

Operations and data collection for a better understanding of the problems  Formulation  Formula tion of of hypothesis hypothesis and and model  Observation and experimentation to test the hypothesis on the basis of additional data  Analysis of the available available information information and and verification verification of of the hypothesis hypothesis using using prepreestablished measure of effectiveness

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 Prediction  Predictio n of various various results results and consideration consideration of alternative alternative methods methods

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Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

  Actio  Action n

Ph Phase: ase: The action phase involves making recommendations for the decision process. The recommendations can be made by those who identified and presented the problem or anyone who influences the operation in which the problem has occurred.  Limitatio  Limi tations ns of OR: The limitations are more related to the problems of model building, time and money factors. 

 Magnitude de   Magnitu

of comput computation: ation:

Modern problems involve a large number of variables. The magnitude of computation makes it difficult to find the interrelationship.  Intangible le   Intangib

fact factors: ors:

 Non – quantitat quantitative ive factors an andd human eemotiona motionall factor cannot cannot be

taken into account. 

Communication gap:

There is a wide gap between the expectations of managers and the aim of research professional professionals. s. 

Time and Money factors:

When you subject the basic data to frequent changes then incorporation of them into OR models becomes a costly affair.   Human Factor:

Implementation of decisions involves human relations and behaviour.

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Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

Q2. What are the common methods to obtain an initial basic feasible solution for a transportation transportation

problem whose cost and requirement table is given? Give a stepwise procedure for one of them? A. The transportation problem involves m sources, each of which has available ai (i = 1, 2… m) units of  homogeneous product and n destinations, each of which requires bj (j = 1, 2…., n) units of products. Here ai and bj are positive integers. The cost cij of transporting one unit of the product from the ith source to the jth destination is given for each i and j. The objective is to develop an integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost.

It is assumed that the total supply and the total demand are equal.

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

The Initial Basic Feasible Solution:

Let us consider a TP involving m-origins and ndestinations. Since the sum of origin capacities equals the sum of destination requirements, a feasible solution always exists. Any feasible solution satisfying m + n – 1 of the m + n constraints is a redundant one and hence it can be deleted. This also means that a feasible solution to a TP can have only m + n – 1  positive component component;; otherw otherwise ise the sol solution ution wi will ll degenerate. degenerate. It is always possible to assign an initial feasible solution to a TP, satisfying all the rim requirements. This can be achieved either by inspection or by following some simple rules. You can begin by imagining that the transportation table is blank that is initial xij = 0. The simplest procedures for initial allocation are discussed in the following section. The common methods to obtain an initial basic feasible solution for a transportation problem are: 

 North west west corner corner rule

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 Matrix minimum minimum method  method 

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  Nort  North h

Vogel ‟ s approximat approximation ion method  method 

Wes Westt Corn Corner er rul rule: e:

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Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

 Step1:

The first assignment is made in the cell occupying the upper left hand (north-west) corner of the transportation table. The maximum feasible amount is allocated here is  x 1111 = min (a1 ,  , b1  ))

Either the capacity of origin O1 is used up or the requirement at destination D1 is satisfied or both. This value of x11 is entered in the upper left hand corner (small square) of cell (1, 1) in the transportation table.  Step 2:

If b1 > a1, the capacity of origin O is exhausted and the requirement at destination D1 is still not satisfied. Then at least one variable in the first column will have to take on a positive value. Move down vertically to the second row and make the second allocation of magnitude:  x 2211 = min (a2 ,  , b1 – x 2211 )  ) in the cell cell (2, 1)

This either exhausts the capacity of origin O 2 or satisfies the remaining demand at destination D 1. If a1 > b1, the requirement at destination D1 is satisfied, but the capacity of origin O1 is not completely exhausted. Move to the right in a horizontal position to the second column to make the second allocation of magnitude:  x 1122 = min (a1 – x 1111 ,  , b2 )  ) in the cell cell (1, 2)

This either exhausts the remaining capacity of origin O 1 or satisfies the demand at destination D2. If b1 = a1, the origin capacity of O1 is completely exhausted as well as the requirement at destination is completely satisfied, then there is a tie at the second allocation. An arbitrary tie breaking choice is made. Make the second allocation of magnitude  x 1122 = min (a1 – a1 ,  , b2 )  ) = 0 in the cell cell (1, 2) OR  x 2211 = min (a2 ,  , b1 – b2 )  ) = 0 in the cell cell (2, 1)

 Step 3:

Start from the new north-west corner of the transportation table satisfying the destination requirements and exhausting the origin capacities one at a time, moving down towards the lower right corner of the transportation table until all the rim requirements are satisfied. Q3. a. What are the properties of a game? Explain the “best strategy” on the basis of minmax criterion of optimality. optimality. b. State the assumptions underlying game theory. Discuss its importance to business decisions.

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Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

A. 

Characteristics of a Competitive Game: A competitive game has the following characteristics: The number of players or competitors is finite.  Each player player has finite finite number number of courses courses of action or or moves. moves.  A game is is played when each player player adopts adopts any one one course of action. action.  Every time time a game game is played, played, the correspondi corresponding ng combinatio combination n of courses courses of of action action leads to to a transaction or payment to each player. The payment is called pay-off or gain. The payoff may be monetary (money) or some benefit, such as increased sales. The players do not communicate with each other. The players are aware of the rules before starting the game.

 Minmax  Minm ax cri criterion terion of op optimal timality: ity: Player A and Player B are to play a game without knowing the other player’s strategy. However, player A would like to maximize his profit and player B would like to minimize his loss. Also each player would expect his opponent to be calculative. 

Suppose player  A  A plays A1  ,, Then, his gain would be a11 , , a12 ,...a1n accordingly B’s choice would be B1 ,B2 ,...Bn. Let α1 = {a11 , , a12 , , ..., a1n  }} . Then, α1 is the minimum gain of A when he plays A1 ( α1 is the minimum pay-off in the first row.) Similarly, if A plays A2, his minimum gain is α2 , the least pay-off in the second row. You will find co corresponding rresponding to A A’s ’s play A1 , , A2 ,......A  ,......Am, the minimum gains are the row minimums α1 ,  , α2 ,..  ,.. αm. Suppose A chooses the course of action where α1 is maximum. 6

 

Name : Revathi S Murthy ID : 521033347 Center Code : 02908

Course : MBA Semester : 2nd Subject Code : MB0048

Then the maximum of the row minimum in the pay-off matrix is called maximin. α = max

I

min j (aij)

Similarly, when B plays, he would would mi minimise nimise his maximum maximum loss. The maximum loss to b is when Bj is β j = max i (aij).   This is the maximum pay-off in the j th column. The minimum of the column maximums in the pay-off matrix is called minimax. The minimax is β = max min I

j

(aij)

If α=β=v (Say), the maximin and the minimax are equal and the game is said to have saddle point. If  α