Operations Research Assignment 2

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Name : Revathi S Murthy Course : MBA ID : 521033347 Semester : 2nd Center Code : 02908 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 mathematical representation which is solved and retransformed into the original 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 part of the system. 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. 1

Transcript of Operations Research Assignment 2

Page 1: Operations Research Assignment 2

Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 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 mathematical representation which is solved and retransformed into the original 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 part of the system.

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.

OR makes use of scientific methods to solve problems.

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

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

OR offers a quantitative solution.

OR also takes into account the human factors.

Phases of Operations Research:

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

The scientific method in OR study generally involves the following three phases.

Phases of operations research

Judgment Phase:This phase includes the following activities:

Determination of the operations

Establishment of the objectives and values related to the operations

Determination of the suitable measures of effectiveness

Formulation of the problems relative to the objectives

Research Phase:This phase utilises the following methodologies:

Operations 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

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Action Phase: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.

Limitations of OR:The limitations are more related to the problems of model building, time

and money factors.

Magnitude of computation:Modern problems involve a large number of variables. The

magnitude of computation makes it difficult to find the interrelationship.

Intangible factors:Non – quantitative factors and human emotional factor cannot be

taken into account.

Communication gap:There is a wide gap between the expectations of managers and the

aim of research professionals.

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 Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

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

destinations. 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; otherwise the solution will 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 corner rule

Matrix minimum method

Vogel‟s approximation method

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

North West Corner rule:

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

x11 = 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:

x21 = min (a2, b1 – x21) in the cell (2, 1)

This either exhausts the capacity of origin O2 or satisfies the remaining demand at destination D1.

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:

x12 = min (a1 – x11, b2) in the cell (1, 2)

This either exhausts the remaining capacity of origin O1 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

x12 = min (a1 – a1, b2) = 0 in the cell (1, 2)OR

x21 = min (a2, b1 – b2) = 0 in the 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.

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Q3. a. What are the properties of a game? Explain the “best strategy” on the basis of minmax criterion of optimality.b. State the assumptions underlying game theory. Discuss its importance to business decisions.

A. Characteristics of a Competitive Game:

A competitive game has the following characteristics:

The number of players or competitors is finite.

Each player has finite number of courses of action or moves.

A game is played when each player adopts any one course of action.

Every time a game is played, the corresponding combination of courses of action leads

to a transaction or payment to each player. The payment is called pay-off or gain. The

pay-off 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 criterion of optimality: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 plays A1,

Then, his gain would be a11, a12 ,...a1n accordingly B’s choice would be B1 ,B2 ,...Bn.

Let 1 = {a11, a12, ..., a1n} .

Then, 1 is the minimum gain of A when he plays A1 (1 is the minimum pay-off in the first row.)

Similarly, if A plays A2, his minimum gain is 2 , the least pay-off in the second row.

You will find corresponding to A’s play A1, A2,......Am, the minimum gains are the row minimums 1, 2,.. m.

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Suppose A chooses the course of action where 1 is maximum.

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

= max min I j (aij)

Similarly, when B plays, he would minimise his maximum loss.

The maximum loss to b is when Bj is j = max i (aij).

This is the maximum pay-off in the jth column.

The minimum of the column maximums in the pay-off matrix is called minimax.

The minimax is

= max min I j (aij)

If ==v (Say), the maximin and the minimax are equal and the game is said to have saddle point. If <, then the game does not have a saddle point.

Note: cannot be greater than

b. Some assumptions used in some parts of game theory have been challenged in philosophy; psychological egoism states that rationality reduces to self-interest—a claim debated among philosophers.

The first known use is to describe and model how human populations behave. Some scholars believe that by finding the equilibria of games they can predict how actual human populations will behave when confronted with situations analogous to the game being studied. This particular view of game theory has come under recent criticism. First, it is criticized because the assumptions made by game theorists are often violated. Game theorists may assume players always act in a way to directly maximize their wins (the Homo economicus model), but in practice, human behavior often deviates from this model. Explanations of this phenomenon are many; irrationality, new models of deliberation, or even different motives (like that of altruism). Game theorists respond by comparing their assumptions to those used in physics. Thus while their assumptions do not always hold, they can treat game theory as a reasonable scientific ideal akin to the models used by physicists. However, additional criticism of this use of game theory has been levied because some experiments have demonstrated that

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

individuals do not play equilibrium strategies. For instance, in the centipede game, guess 2/3 of the average game, and the dictator game, people regularly do not play Nash equilibria. There is an ongoing debate regarding the importance of these experiments.

Alternatively, some authors claim that Nash equilibria do not provide predictions for human populations, but rather provide an explanation for why populations that play Nash equilibria remain in that state. However, the question of how populations reach those points remains open.

Some game theorists have turned to evolutionary game theory in order to resolve these worries. These models presume either no rationality or bounded rationality on the part of players. Despite the name, evolutionary game theory does not necessarily presume natural selection in the biological sense. Evolutionary game theory includes both biological as well as cultural evolution and also models of individual learning (for example, fictitious play dynamics).

Mathematical game theory had beginnings with some publications by Émile Borel, which led to his 1938 book Applications aux Jeux de Hasard. However, Borel's results were limited, and his conjecture about the non-existence of a mixed-strategy equilibria in two-person zero-sum games was wrong. The modern epoch of game theory began with the statement of the theorem on the existence of mixed-strategy equilibria in two-person zero-sum games and its proof by John von Neumann. Von Neumann's original proof used Brouwer's fixed-point theorem on continuous mappings into compact convex sets, which became a standard method in game theory and mathematical economics. His paper was followed by his 1944 book Theory of Games and Economic Behavior, with Oskar Morgenstern, which considered cooperative games of several players. The second edition of this book provided an axiomatic theory of expected utility, which allowed mathematical statisticians and economists to treat decision-making under uncertainty.

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Q4. a. Compare CPM and PERT explaining similarities and mentioning where they mainly differ.b. What is meant by graphing in Network Analysis?

A. Basic Difference between PERT and CPM:

Though there are no essential differences between PERT and CPM as both of them share in common the determination of a critical path. Both are based on the network representation of activities and their scheduling that determines the most critical activities to be controlled so as to meet the completion date of the project.

PERT:Some key points about PERT are as follows:

PERT was developed in connection with an 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 so as to characterise it.

It is an event-oriented network because in the analysis of a network, emphasis is

given on the important stages of completion of a task rather than the activities

required to be performed to reach a particular event or task.

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.

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

CPM is used for projects involving activities of repetitive nature.

When the plan is complete and the project ready to kick off, the project network is fixed and the buffers size is "locked" (i.e. their planned duration may not be altered during the project), because they are used to monitor project schedule and financial performance.

With no slack in the duration of individual tasks, the resources are encouraged to focus on the task at hand to complete it and hand it off to the next person or group. The object here is to eliminate bad multitasking, and this is done by providing priority information to all resources. An analogy is drawn in the literature with a relay race. Each element on the project is encouraged to move as quickly as they can: when they are running their "leg" of the project, they should be focused on completing the assigned task as quickly as possible, with no distractions or multitasking. In some case studies, actual batons are reportedly hung by the desks of people when they are working on critical chain tasks so that others know not to interrupt. The goal, here, is to overcome the tendency to delay work or to do extra work when there seems to be time. The CCPM literature contrasts this with "traditional" project management that monitors task start and completion dates. CCPM encourages people to move as quickly as possible, regardless of dates.

Because task durations have been planned at the 50% probability duration, there is pressure on the resources to complete critical chain tasks as quickly as possible, overcoming student's syndrome and Parkinson's Law.

Monitoring is, in some ways, the greatest advantage of the Critical Chain method. Because individual tasks will vary in duration from the 50% estimate, there is no point in trying to force every task to complete "on time;" estimates can never be perfect. Instead, we monitor the buffers that were created during the planning stage. A fever chart or similar graph can be easily created and posted to show the consumption of buffer as a function of project completion. If the rate of buffer consumption is low, the project is on target. If the rate of consumption is such that there is likely to be little or no buffer at the end of the project, then corrective actions or recovery plans must be developed to recover the loss. When the buffer consumption rate exceeds some critical value (roughly: the rate where all of the buffer may be expected to be consumed before the end of the project, resulting in late completion), then those alternative plans need to be implemented.

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Q5. Consider the following transportation problem:

Godowns

Factory 1 2 3 4 5 6Stock

availableA 7 5 7 7 5 3 60B 9 11 6 11 - 5 20C 11 10 6 2 2 8 90D 9 10 9 6 9 12 50

Demand 60 20 40 20 40 40

It is not possible to transport any quantity from factory B to Godown 5.Determine: (a) Initial solution by Vogel’s approximation method. (b) Optimum basic feasible solution.

A. The transportation table of the given TP has 23 cells. The differences between the smallest and next to smallest costs in each row and each column are computed and displayed inside the parenthesis against the respective columns and rows. The largest of these differences is (3) and is associated with the fourth row of the transportation table.

The minimum cost in the fourth row is C42 = –1. Accordingly, you allocate x42 = min (19, 25) = 19 in the cell (4, 2). This exhausts the availability at warehouse D. Cross the fourth row. The row and column differences are now computed for the resulting reduced transportation table (ii) as shown below:

Resulting reduced transportation table5 1 3 3 34(2)3 3 5 5 15(0)6 4 4 3 12(1)4 19-1 4 2 19(3)21(1)

25(2)

17(1)

17(1)

(i)

The largest of this is (2) and is associated with the first row as well as the first and second column. Arbitrarily, select the first row whose minimum cost is C12 = 1. Thus the second allocation of magnitude x12 = min (34, 6) = 6 is made in the cell (1, 2). Cross the second column from the table. Continuing in this way, the subsequent reduced transportation tables and the differences for the surviving rows and columns

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Surviving rows and columns

Eventually, the basic feasible solution is obtained as shown in table below:

The transportation cost according to the above route is given by:

Z = 6 x 5 + 6 x 1 + 17 x 3 + 5 x 3 + 15 x 3 + 12 x 3 + 19 x (– 1) = 164

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Name : Revathi S Murthy Course : MBAID : 521033347 Semester : 2ndCenter Code : 02908 Subject Code : MB0048

Q6. A machine operator processes five types of items on his machine each week, and must choose a sequence for them. The set-up cost per change depends on the item presently on the machine and the set-up to be made according to the following table:

From Item To itemA B C D EA ∞ 4 7 3 4B 4 ∞ 6 3 4C 7 6 ∞ 7 5D 3 3 7 ∞ 7E 4 4 5 7 ∞

If he processes each type of item once and only once each week, how should he sequence the items on his machine in order to minimize the total set-up cost? (Hint: A-> E -> C-> B-> D-> A Cost: 20)

A. Let x1 and x2 be the number of days the units A and B be operated per week respectively. Then the objective of the manager is to: Minimise the cost function

Z = 20 x1 + 20 x2 + 20 x3 + 20 x4

Subject to the constraints ∞

x1+ 4x2 + 3 x3 + 7 x4 ≥ 20 (Super fine)

2x1+2x2 ≥ 8 (medium)

4x1+12x2 ≥ 24 (low grade)and x1 ≥ 0, x2 ≥ 0

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