87172577 Operation Research 2nd Sem

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MB0048Operation Research

(Book ID: B1137)

Set- 1

Q1. a. Explain how and why Operation Research methods have been valuable in aiding

executive decisions.

b. Discuss the usefulness of Operation Research in decision making process and the role of

computers in this field.

Answer:

Churchman, Aackoff and Aruoff defined Operations Research as: the application of scientificmethods, techniques and tools to operation of a system with optimum solutions to the problems,

where optimumrefers to the best possible alternative.

The objective of Operations Research is to provide a scientific basis to the decision-makers forsolving problems involving interaction of various components of the organisation. You can

achieve this 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 thusobtained is known as an optimal decision.

You can also define Operations Research as The use of scientific methods to provide criteria fordecisions regarding man, machine, and systems involving repetitive operations. OR Operation

Techniques is a bunch of mathematical techniques.

b. Operation Research is an aid for the executive in making his decisions based on

scientific methods analysis. Discuss the above statement in brief.

Answer:

Operation Research is an aid for the executive in making his decisions based on scientific

methods analysis.

Discussion:-

Any problem, simple or complicated, can use OR techniques to find the best possible solution.This section will explain the scope of OR by seeing its application in various fields of everydaylife.

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i) In Defense Operations: In modern warfare, the defense operations are carried out by threemajor independent components namely Air Force, Army and Navy. The activities in each ofthese components can be further divided in four sub-components namely: administration,intelligence, operations and training and supply. The applications of modern warfare techniquesin each of the components of military organisations require expertise knowledge in respective

fields. Furthermore, each component works to drive maximum gains from its operations andthere is always a possibility that the strategy beneficial to one component may be unfeasible foranother component. Thus in defense operations, there is a requirement to co-ordinate theactivities of various components, which gives maximum benefit to the organisation as a whole,having maximum use of the individual components. A team of scientists from various disciplinescome together to study the strategies of different components. After appropriate analysis of thevarious courses of actions, the team selects the best course of action, known as the optimum

strategy.

ii) In Industry: The system of modern industries is so complex that the optimum point ofoperation in its various components cannot be intuitively judged by an individual. The business

environment is always changing and any decision useful at one time may not be so good sometime later. There is always a need to check the validity of decisions continuously against thesituations. The industrial revolution with increased division of labour and introduction ofmanagement responsibilities has made each component an independent unit having their owngoals. For example: production department minimises the cost of production but maximiseoutput. Marketing department maximises the output, but minimises cost of unit sales. Financedepartment tries to optimise the capital investment and personnel department appoints goodpeople at minimum cost. Thus each department plans its own objectives and all these objectivesof various department or components come to conflict with one another and may not agree to theoverall objectives of the organisation. The application of OR techniques helps in overcoming thisdifficulty by integrating the diversified activities of various components to serve the interest ofthe organisation as a whole efficiently. OR methods in industry can be applied in the fields ofproduction, inventory controls and marketing, purchasing, transportation and competitivestrategies.

iii) Planning: In modern times, it has become necessary for every government to have carefulplanning, for economic development of the country. OR techniques can be fruitfully applied tomaximise the per capita income, with minimum sacrifice and time. A government can thus useOR for framing future economic and social policies.

iv) Agriculture: With increase in population, there is a need to increase agriculture output. Butthis cannot be done arbitrarily. There are several restrictions. Hence the need to determine acourse of action serving the best under the given restrictions. You can solve this problem byapplying OR techniques.

v) In Hospitals: OR methods can solve waiting problems in out-patient department of bighospitals and administrative problems of the hospital organisations.

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vi) In Transport: You can apply different OR methods to regulate the arrival of trains andprocessing times minimise the passengers waiting time and reduce congestion, formulate suitabletransportation policy, thereby reducing the costs and time of trans-shipment.

vii) Research and Development: You can apply OR methodologies in the field of R&D for

several purposes, such as to control and plan product introductions.

Q2. Explain how the linear programming technique can be helpful in decision-making in

the areas of Marketing and Finance.

Answer:

Linear programming problems are a special class of mathematical programming problems forwhich the objective functions and all constraints are linear. A classic example of the application

of linear programming is the maximization of profits given various production or costconstraints.

Linear programming can be applied to a variety of business problems, such as marketing mixdetermination, financial decision making, production scheduling, workforce assignment, andresource blending. Such problems are generally solved using the simplex method.

MEDIA SELECTION PROBLEM.

The local Chamber of Commerce periodically sponsors public service seminars and programs.Promotional plans are under way for this years program. Advertising alternatives include

television, radio, and newspaper. Audience estimates, costs, and maximum media usagelimitations are shown in Exhibit 1.

If the promotional budget is limited to $18,200, how many commercial messages should be runon each medium to maximize total audience contact? Linear programming can find the answer.

Q3. a. How do you recognise optimality in the simplex method?

b. Write the role of pivot element in simplex table?

Answer:

Simplex method is used for solving Linear programming problem especially when more thantwo variables are involvedSIMPLEX METHOD

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1. Set up the problem.

That is, write the objective function and the constraints.

2. Convert the inequalities into equations.

This is done by adding one slack variable for each inequality.

3. Construct the initial simplex tableau.

Write the objective function as the bottom row.

4. The most negative entry in the bottom row identifies a column.

5. Calculate the quotients. The smallest quotient identifies a row. The element in the

intersection of the column identified in step 4 and the row identified in this step is identified

as the pivot element.

The quotients are computed by dividing the far right column by the identified column in step 4.A quotient that is a zero, or a negative number, or that has a zero in the denominator, is ignored.

6. Perform pivoting to make all other entries in this column zero.

This is done the same way as we did with the Gauss-Jordan method.

7. When there are no more negative entries in the bottom row, we are finished;

otherwise, we start again from step 4.

8. Read off your answers.

Get the variables using the columns with 1 and 0s. All other variables are zero. The maximumvalue you are looking for appears in the bottom right hand corner.

Example

Niki holds two part-time jobs, Job I and Job II. She never wants to work more than a total of 12hours a week. She has determined that for every hour she works at Job I, she needs 2 hours ofpreparation time, and for every hour she works at Job II, she needs one hour of preparation time,

and she cannot spend more than 16 hours for preparation. If she makes $40 an hour at Job I, and$30 an hour at Job II, how many hours should she work per week at each job to maximize herincome?

Solution: In solving this problem, we will follow the algorithm listed above.

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1. Set up the problem. That is, write the objective function and the constraints. Since thesimplex method is used for problems that consist of many variables, it is not practical to use thevariables x, y, z etc. We use the symbols x1, x2, x3, and so on.

Let x1 = The number of hours per week Niki will work at Job I.

and x2 = The number of hours per week Niki will work at Job II.

It is customary to choose the variable that is to be maximized as Z.

The problem is formulated the same way as we did in the last chapter.

Maximize Z = 401 + 302

Subject to: x1 + x2 12

21 + x2 16

x1 0; x2 0

2. Convert the inequalities into equations. This is done by adding one slack variable for eachinequality.

For example to convert the inequality x1 + x2 12 into an equation, we add a non-negativevariable y1, and we get

x1 + x2 + y1 = 12

Here the variable y1 picks up the slack, and it represents the amount by which x1 + x2 falls shortof 12. In this problem, if Niki works fewer that 12 hours, say 10, then y1 is 2. Later when weread off the final solution from the simplex table, the values of the slack variables will identifythe unused amounts.

We can even rewrite the objective function Z = 401 + 302 as401302 + Z = 0.

After adding the slack variables, our problem reads

Objective function: 401302 + Z = 0

Subject to constraints: x1 + x2 + y1 = 12

21 + x2 + y2 = 16

x1 0;x2 0

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3. Construct the initial simplex tableau. Write the objective function as the bottom row.

Now that the inequalities are converted into equations, we can represent the problem into anaugmented matrix called the initial simplex tableau as follows.

x1 x2 y1 y2 Z C1 1 1 0 0 122 1 0 1 0 16

40 30 0 0 1 0

Here the vertical line separates the left hand side of the equations from the right side. Thehorizontal line separates the constraints from the objective function. The right side of theequation is represented by the column C.

The reader needs to observe that the last four columns of this matrix look like the final matrix forthe solution of a system of equations. If we arbitrarily choose x1 = 0 and x2 = 0, we get

Which reads

y1 = 12

y2 = 16

Z = 0

The solution obtained by arbitrarily assigning values to some variables and then solving for theremaining variables is called the basic solution associated with the tableau. So the above

solution is the basic solution associated with the initial simplex tableau. We can label the basicsolution variable in the right of the last column as shown in the table below.

x1 x2 y1 y2 Z1 1 1 0 0 12 y12 1 0 1 0 16 y2

40 30 0 0 1 0 Z

4. The most negative entry in the bottom row identifies a column.

The most negative entry in the bottom row is40; therefore the column 1 is identified.

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x1 x2 y1 y2 Z1 1 1 0 0 12 y12 1 0 1 0 16 y2

40 30 0 0 1 0 Z


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Q4. What is the significance of duality theory of linear programming? Describe the general

rules for writing the dual of a linear programming problem.

Answer:

Linear programming (LP) is a mathematical method for determining a way to achieve the bestoutcome (such as maximum profit or lowest cost) in a given mathematical model for some list ofrequirements represented as linear relationships. Linear programming is a specific case ofmathematical programming.

More formally, linear programming is a technique for the optimization of a linear objectivefunction, subject to linear equality and linear inequality constraints. Given a polytope and a real-valued affine function defined on this polytope, a linear programming method will find a pointon the polytope where this function has the smallest (or largest) value if such point exists, by

searching through the polytope vertices.

Linear programs are problems that can be expressed in canonical form:where x represents the vector of variables (to be determined), c and b are vectors of (known)coefficients andA is a (known) matrix of coefficients. The expression to be maximized orminimized is called the objective function (cTx in this case). The equationsAxb are theconstraints which specify a convex polytope over which the objective function is to beoptimized. (In this context, two vectors are comparable when every entry in one is less-than orequal-to the corresponding entry in the other. Otherwise, they are incomparable.)

Linear programming can be applied to various fields of study. It is used most extensively in

business and economics, but can also be utilized for some engineering problems. Industries thatuse linear programming models include transportation, energy, telecommunications, andmanufacturing. It has proved useful in modeling diverse types of problems in planning, routing,scheduling, assignment, and design.

Duality: Every linear programming problem, referred to as aprimal problem, can be convertedinto a dual problem, which provides an upper bound to the optimal value of the primal problem.In matrix form, we can express theprimal problem as:

Maximize cTx subject toAx b, x 0;with the corresponding symmetric dual problem,

Minimize b

T

y subject toA

T

y c, y 0.An alternative primal formulation is:Maximize cTx subject toAx b;with the corresponding asymmetric dual problem,Minimize bTy subject toATy = c, y 0.

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http://en.wikipedia.org/wiki/Mathematical_modelhttp://en.wikipedia.org/wiki/Mathematical_programminghttp://en.wikipedia.org/wiki/Optimization_%28mathematics%29http://en.wikipedia.org/wiki/Linearhttp://en.wikipedia.org/wiki/Objective_functionhttp://en.wikipedia.org/wiki/Objective_functionhttp://en.wikipedia.org/wiki/Linear_equalityhttp://en.wikipedia.org/wiki/Linear_inequalityhttp://en.wikipedia.org/wiki/Constraint_%28mathematics%29http://en.wikipedia.org/wiki/Polytopehttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Affine_functionhttp://en.wikipedia.org/wiki/Canonical_formhttp://en.wikipedia.org/wiki/Vector_spacehttp://en.wikipedia.org/wiki/Matrix_%28mathematics%29http://en.wikipedia.org/wiki/Convex_polytopehttp://en.wikipedia.org/wiki/Comparabilityhttp://en.wikipedia.org/wiki/Dual_problemhttp://en.wikipedia.org/wiki/Dual_problemhttp://en.wikipedia.org/wiki/Comparabilityhttp://en.wikipedia.org/wiki/Convex_polytopehttp://en.wikipedia.org/wiki/Matrix_%28mathematics%29http://en.wikipedia.org/wiki/Vector_spacehttp://en.wikipedia.org/wiki/Canonical_formhttp://en.wikipedia.org/wiki/Affine_functionhttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Polytopehttp://en.wikipedia.org/wiki/Constraint_%28mathematics%29http://en.wikipedia.org/wiki/Linear_inequalityhttp://en.wikipedia.org/wiki/Linear_equalityhttp://en.wikipedia.org/wiki/Objective_functionhttp://en.wikipedia.org/wiki/Objective_functionhttp://en.wikipedia.org/wiki/Linearhttp://en.wikipedia.org/wiki/Optimization_%28mathematics%29http://en.wikipedia.org/wiki/Mathematical_programminghttp://en.wikipedia.org/wiki/Mathematical_model


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There are two ideas fundamental to duality theory. One is the fact that (for the symmetric dual)the dual of a dual linear program is the original primal linear program. Additionally, everyfeasible solution for a linear program gives a bound on the optimal value of the objectivefunction of its dual. The weak duality theorem states that the objective function value of the dualat any feasible solution is always greater than or equal to the objective function value of the

primal at any feasible solution. The strong duality theorem states that if the primal has an optimalsolution, x*, then the dual also has an optimal solution, y*, such that cTx*=bTy*.

A linear program can also be unbounded or infeasible. Duality theory tells us that if the primal isunbounded then the dual is infeasible by the weak duality theorem. Likewise, if the dual isunbounded, then the primal must be infeasible. However, it is possible for both the dual and theprimal to be infeasible.

5. Use Two-Phase simplex method to solve:

Solution:

Rewriting in the standard form,

Manimize Z = X1 + X2 + X3 + MA1 + 0S1 + 0S2

Subject to

X13X2 +4X3 +MA1 = 5

X1 -2X2 + S1 = 3

2X2 + X3 - S2 = 4,

X1, X2, X3, S1, S2, A1 > 0.

Phase 1: Consider the new objective,

Maximize Z* = A1

Subject to 2 1 + x2S1 + A1 = 2

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X1 X2 X3 S1 S2 A1

Cj 0 0 0 0 0 1 Ratio

A1 - 1 1 -3* 4 0 0 1 5 -5/3

S1 0 1 -2 0 1 0 0 3 -1

S2 0 0 2 1 0 -1 0 4 -4/3

Zj 1 -3 4 0 0 1

Zj-Cj 1 -3 4 0 0 0

most

ve Work

Column

*Pivot elementX1 enters the basic set replacing A1.

The first iteration gives the following table:

X1 X2 X3 S1 S2 A1

Cj 0 0 0 0 0 1

X1 0 -1/3 1 -4/3 0 0 -1/3 -5/3

S2 0 -1/2 -7/2 -2 1 0 -1/2

S3 0 0 2 1 0 -1 0 4

Zj 0 0 0 0 0 0 0

Zj-Cj 0 0 0 0 0 -1

Phase I is complete, since there are one negative elements in the last row.

Phase II: Consider the original objective function,

Maximize z = x1 + x2+x3 + 0S1 + 0S2

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Subject to -3x1/3 + x24x3/3 =-5/3

-x1/27x2/22x3 + S1 =1/2

2x2 + x3S2 = 4

x1, x2, S1, S2 = 0

with the initial solution X1 = 1, S1 = 0, S2 = 0, the corresponding simplex table is

X1 X2 X3 S1 S2 A1 Ratio

Cj 1 1

X1 1 -1/3 1 1/2 1/2/2=1

S1 0 -1/2 4

S2 0 0

Zj -1/3

Zj-Cj -4/3

Work column *Pivot element

S1 S2 S3

X1 1 0 0S2 0 1 0 0S3 0 7/4 1 1/2 0Zj-Cj 0 0 0 0

Since all elements of the last row are negative, the current solution is optimal.

The maximum value of the objective function Z = 6 which is attained for x1 = 2, x2 = 0.

6. Use Branch and Bound method to solve the following L.P.P:

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Solution:

At the starting iteration we can consider z (0) = 0 to be the lower bound, forx, since all xj = 0 isfeasible. The master list contains only the L.P.P. (1) (2) and (3). which is designated as problem1. Choose it in step 0, and in step 1 determine the optimum solution.

Z0=63 X1= 9 /2, X2= 7/2 (Solution to problem 1)

Since the solution is not integer valued, proceed from step 2 to step 3, and selectx1. Then since

[X1*]= [9 /2] = 4, place on the master list the following two additional problems.

Problem 2: (1) (2) and 5 x1 7 0 x2 7Problem 3: (1) (2) and 0 x1 4 0 x2 7Returning to step 0: with z(1) = z(0) = 0,

We choose problem 2, Step 1 establishes that problem 2 has the feasible solutionz0= 35, x1 = 5 x2= 0[solution to problem (2)] (5)

Since this satisfies the integer constraints, therefore at step 2 we record it by enclosing in arectangle and let z (2) = 35.

Returning to step 0 with z (2) = 35, we find that problem 3 is available. Step 1 determines thefollowing optimum feasible solution to it

Z0 = 58, x1 = 4, X2 = 10 / 3 (Problem 3)

Since the solution is not integer valued, proceed from step 2 to step 3 and select x2. Then

[X1*]= [10/3] = 3. We add the following additional problems on the master list:

Problem 4: (1) (2) and 0 < x1 < 4; 4 < x2


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Returning to step 0 with z(3) = z(2) = 35, choose problem 4 from step 1 we know that problem 4has no feasible solution and so we again return to step 0 with z (4) = z (3) = 35.Only problem 5 isavailable in the master list. In step 1 we determine the following optimum solution to thisproblem.z0= 55 x1 = 4 x2= 3(solution to problems) -------- (6)

Since this satisfies the integer constraints, therefore at step 2. We record it by enclosing inside arectangle and let z(5) = 55.

Returning to step 0, we find that the master list is empty; and thus the algorithm terminates.

Now, on terminating we find that only two feasible integer solution namely (5) and (6) have beenrecorded. The best of these gives the optimum solution to the given L.P.P. Hence the optimuminteger solution to the given L.P.P. is

Z0= 55, x1 = 4, x2= 3

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Tree diagram of above example

Z*=63

1 1[X1= 4 / , X2 = 3 / ]

2 2

x1 < 4 x1 > 5Node (3) Node (2)

Z*=58

1

[X1 = 4 , X2 = 3 / ]3

Z*=35

(x1 = 5, x2= 0)

x2< 3 x2> 4

Node (5) Node (4)

Z* = 55(x1 = 4, x2= 3)

Optimal solution

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No solution


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NodeSolution Additional

ConstraintsType of solution

x1 x2 z*

(1) 9 / 2 7 / 2 63 __Non-integer(Original problem)

(2) 5 0 35 x1 > 5 Integer z* (1)(3) 4 10 /3 58 x1 < 4 Non-integer(4) x1 < 4 ,x2 > 4 No solution

(5) 4 3 55x1 < 4 , x2 < 3

Integer z* (2) (Optimal)

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MB0048Operation Research

(Book ID: B1137)

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

Answer:

Characteristics of Operations Research

Operations research, an interdisciplinary division of mathematics and science, uses statistics,

algorithms and mathematical modeling techniques to solve complex problems for the

best possible solutions. This science is basically concerned with optimizing maxima and minima

of the objective functions involved. Examples of maxima could be profit, performance and yield.

Minima could be loss and risk. The management of various companies has benefited immensely

from operations research.

Operations research is also known as OR. It has basic characteristics such as systems orientation,

using interdisciplinary groups, applying scientific methodology, providing quantitative answers,

revelation of newer problems and the consideration of human factors in relation to the state

under which research is being conducted.

Systems Orientation

o This approach recognizes the fact that the behavior of any part of the system has an effect onthe system as a whole. This stresses the idea that the interaction between parts of the system iswhat determines the functioning of the system. No single part of the system can have a bearingeffect on the whole. OR attempts appraise the effect the changes of any single part would haveon the performance of the system as a whole. It then searches for the causes of the problem thathas arisen either in one part of the system or in the interrelation parts.

Interdisciplinary groups

o The team performing the operational research is drawn from different disciplines. The

disciplines could include mathematics, psychology, statistics, physics, economics andengineering. The knowledge of all the people involved aids the research and preparation of thescientific model.

Application of Scientific Methodology

o OR extensively uses scientific means and methods to solve problems. Most OR studies cannot

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be conducted in laboratories, and the findings cannot be applied to natural environments.Therefore, scientific and mathematical models are used for studies. Simulation of these models iscarried out, and the findings are then studied with respect to the real environment.

New Problems Revealedo Finding a solution to a problem in OR uncovers additional problems. To obtain

maximum benefits from the study, ongoing and continuous research is necessary. New problemsmust be pursued immediately to be resolved. A company looking to reduce costs inmanufacturing might discover in the process that it needs to buy one more component tomanufacture the end product. Such a scenario would result in unexpected costs and budgetoverruns. Ensuring flexibility for such contingencies is a key characteristic of OR.

Provides Quantitative Answers

o The solutions found by using operations research are always quantitative. OR considers two ormore options and emphasizes the best one. The company must decide which option is the bestalternative for it.

Human Factors

o In other forms of quantitative research, human factors are not considered, but in OR, humanfactors are a prime consideration. People involved in the process may become sick, which wouldaffect the companys output.

PHASES OPERATIONS RESEARCH

Formulate the problem:

This is the most important process, it is generally lengthy and time consuming. The activities thatconstitute this step are visits, observations, research, etc. With the help of such activities, theO.R. scientist gets sufficient information and support to proceed and is better prepared toformulate the problem. This process starts with understanding of the organizational climate, itsobjectives and expectations. Further, the alternative courses of action are discovered in this step.

Develop a model:

Once a problem is formulated, the next step is to express the problem into a mathematical modelthat represents systems, processes or environment in the form of equations, relationships orformulas. We have to identify both the static and dynamic structural elements, and devicemathematical formulas to represent the interrelationships among elements. The proposed model

may be field tested and modified in order to work under stated environmental constraints. Amodel may also be modified if the management is not satisfied with the answer that it gives.

Select appropriate data input:Garbage in and garbage out is a famous saying. No model will work appropriately if data input is

not appropriate. The purpose of this step is to have sufficient input to operate and test the model.

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Solution of the model:

After selecting the appropriate data input, the next step is to find a solution. If the model is notbehaving properly, then updating and modification is considered at this stage.

Validation of the model:

A model is said to be valid if it can provide a reliableprediction of the systems performance. Amodel must be applicable for a longer time and can be updated from time to time taking intoconsideration the past, present and future aspects of the problem.

Implement the solution:

The implementation of the solution involves so many behavioural issues and the implementingauthority is responsible for resolving these issues. The gap between one who provides a solutionand one who wishes to use it should be eliminated. To achieve this, O.R. scientist as well asmanagement should play a positive role. A properly implemented solution obtained through O.R.techniques results in improved working and wins the management support.

Limitations

Dependence on an Electronic Computer:

O.R. techniques try to find out an optimal solution taking into account all the factors. In themodern society, these factors are enormous and expressing them in quantity and establishingrelationships among these require voluminous calculations that can only be handled bycomputers.

Non-Quantifiable Factors:

O.R. techniques provide a solution only when all the elements related to a problem can bequantified. All relevant variables do not lend themselves to quantification. Factors that cannot be

quantified find no place in O.R. models.

Distance between Manager and Operations Researcher:

O.R. being specialists job requires a mathematician or a statistician, who might not be aware of

the business problems. Similarly, a manager fails to understand the complex working of O.R.Thus, there is a gap between the two.

Money and Time Costs:

When the basic data are subjected to frequent changes, incorporating them into the O.R. modelsis a costly affair. Moreover, a fairly good solution at present may be more desirable than aperfect O.R. solution available after sometime.

Implementation:

Implementation of decisions is a delicate task. It must take into account the complexities ofhuman relations and behaviour.

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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?

Answer:

Transportation Problem & its basic assumption

This model studies the minimization of the cost of transporting a commodity from a number of

sources to several destinations. The supply at each source and the demand at each destination are

known. The transportation problem involves m sources, each of which has available. i (i = 1, 2,

.., m) units of homogeneous product and n destinations, each of which requires bj (j = 1, 2.,

n) units of products. Here a i 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 areequal .i.e. Condition (1) 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.

Formulation of Transportation Problem

The standard mathematical model for the transportation problem is as follows. Let xij benumber of units of the homogenous product to be transported from source i to the destination j

Then objective is to Theorem:

A necessary and sufficient condition for the existence of a feasible solution to the transportation

problem (2) is that

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.

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Answer:

a) Minimax (sometimes minmax) is a decision rule used indecision theory, game theory,

statistics and philosophy for minimizing the possible loss while maximizing the potential gain.

Alternatively, it can be thought of as maximizing the minimum gain (maxim in). Originally

formulated for two-player zero-sum game theory, covering both the cases where players takealternate moves and those where they make simultaneous moves, it has also been extended to

more complex games and to general decision making in the presence of uncertainty.

Game theory

In the theory of simultaneous games, a minima strategy is a mixed strategy which is part of the

solution to a zero-sum game. In zero-sum games, the minima solution is the same as the Nash

equilibrium. Minimax theorem The minimax theorem states: For every two-person, zero-sum

game with finitely many strategies, there exists a value V and a mixed strategy for each player,

such that (a) Givenplayer 2s strategy, the best payoff possible for player 1 is V, and (b) Givenplayer 1s strategy, the best payoff possible for player 2 is V. Equivalently, Player 1s strategy

guarantees him a payoff of V regardless of Player 2s strategy, and similarly Player 2 can

guarantee himself a payoff of V. The name minimax arises because each player minimizes the

maximum payoff possible for the othersince the game is zero-sum, he also maximizes his own

minimum payoff. This theorem was established by John von Neumann,

[1] Who is quoted as saying As far as I can see, there could be no theory of games without

that theorem I thought there was nothing worth publishing until the Minimax Theorem was

proved. [2]See Scions minimax theorem and Parthasarathys theorem for generalizations; see

also example of a game without a value. Example The following example of a zero-sum game,where A and B make simultaneous moves, illustrates minimax solutions. Suppose each player

has three choices and consider the payoff matrix for A displayed at right. Assume the payoff

matrix for B is the same matrix with the signs reversed (i.e. if the choices are A1 and B1 then B

pays 3 to A). Then, the minimax choice for A is A2 since the worst possible result is then having

to pay 1, while the simple minimax choice for B is B2 since the worst possible result is then no

payment. However, this solution is not stable, since if B believes A will choose A2 then will

choose B1 to gain 1; then if A believes B will choose B1 then will choose A1 to gain 3; and then

will choose B2; and eventually both players will realize the difficulty of making a choice. So a

more stable strategy is needed. Some choices are dominated by others and can be eliminated:

will not choose A3 since either A1 or A2 will produce a better result, no matter what B chooses;B will not choose B3 since some mixtures of B1 and B2 will produce a better result, no matter

what A chooses. A can avoid having to make an expected payment of more than 1/3 by choosing

A1 with probability 1/6 and A2 with probability 5/6, no matter what B chooses. B can ensure an

expected B chooses B1B chooses B2B chooses B3A chooses A1+3 2 +2A chooses A21 0

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+4A chooses A34 3 +1 wouldnt have to pay as much to license these characters. Changing

the rules is another way in which companies can benefit. The authors introduce the idea of judo

economics, where a large company may be willing to allow a smaller company to capture a small

market share rather than compete by lowering its prices. As long as it does not become too

powerful or greedy, a small company can often participate in the same market without having to

compete with larger companies on unfavorable terms. Kiwi International Air Lines introduced

services on its carriers that were of lower prices to get market share, but made sure that the

competitors understood that they had no intention of capturing more than 10% of any market.

Companies can also change perceptions to make themselves better off. This can be accomplished

either by making things clearer or more uncertain. In 1994, the New York Post attempted to

make radical price changes in order to get the Daily News to raise its price to regain subscribers.

However, the Daily News misunderstood and both newspapers were headed for a price war. The

New York Post had to make its intentions clear, and both papers were able to raise their prices

and not lose revenue. The authors also show an example of how investment banks can maintain

ambiguity to benefit themselves. If the client is more optimistic than the investment bank, thebank can try to charge a higher commission as long as the client does not develop a more

realistic appraisal of the companys value. Finally, companies can change the boundaries within

which they compete. For example, when Sega was unable to gain market share from Nintendos

8-bit systems, it changed the game by introducing a new 16-bit system. It took Nintendo 2 years

to respond with its own 16-bit system, which gave Sega the opportunity to capture market share

and build a strong brand image. This example shows how companies can think outside the box to

change the way competition takes place in their industry. Brandenburger and Nalebuff have

illustrated how companies that recognize they can change the rules of competition can vastly

improve their odds of success, and sometimes respond in a way that benefits both themselves and

the competition. If companies are able to develop a system where they can make both themselvesand their competitors better off, then they do not have to worry so much about their competitors

trying to counter their moves. Also, because companies can easily copy each others ideas, it is

to a firms advantage if they can benefit when their competitors copy their idea, which is not

usually possible under the traditional win-lose structure. This article has some parallels with the

article Competing on Analytics by (). The biggest factor that both of these articles have in

common is how crucial it is for managers to understand everything they can about their business

and the environment in which they work. In Competingon Analytics, the authors say that it is

important to be familiar with this information so that managers can change the way they compete

to improve their chances of success. At the end of The Right Game: Use Game Theory to Shape

Strategy, the authors discuss how in order for companies to be able to change the environment

or rules under which they compete they need to understand everything they can about the

constructs under which they are competing. Whether a manager intends to use analytics or game

theory to be successful, he or she must first have all available information and use that

information to understand how to make the company better off. However, the work shown in

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Competing on Analytics tends to place an emphasis almost exclusively on the use of

quantitative data to improve efficiency or market share of the company. The Right Game,

however focuses more on using information to find creative ways of changing the constructs or

rules applied between companies, often yielding a much broader impact.

Q4. a. Compare CPM and PERT explaining similarities and mentioning where they mainly

differ.

Answer:

The Major Differences and Similarities between CPM and PERTCPM (Critical Path Method) &

PERT (Program Evaluation and Review Technique)

1PERT is a probabilistic tool used with three

CPM is a deterministic tool, with only single Estimating the duration for completion of estimateof duration.

This tool is basically a tool for planningCPM also allows and explicit estimate of and control of time. Costs in addition to time, thereforeCPM can control both time and cost.

PERT is more suitable for R&D relatedCPM is best suited for routine and those projects where the project is performed for projectswhere time and cost estimates can the first time and the estimate of duration be accuratelycalculated are uncertain.

The probability factor I major in PERTThe deterministic factor is more so values or so outcomes may not be exact. Outcomes aregenerally accurate and realistic. Extensions of both PERT and CPM allow the user to manageother resources in addition to time and money, to trade off resources, to analyze different typesof schedules, and to balance the use of resources.

Tensions of both PERT and CPM allow the user to manage other resources in addition to time

and money, to trade off resources, to analyze different types of schedules, and to balance the use

of resources.

Graphs _ In mathematics, networks are called graphs, the entities are nodes, and the links are

edges _ Graph theory starts in the 18th century, with Leonhard Euler _ The problem of

Konigsberg bridges _ Since then graphs have been studied extensively. Graph Theory

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_ Graph G= (V, E) _ V = set of vertices _ E = set of edges 2 _ An edge is defined by the two

vertices which it connects _ optionally: 1 3A direction and/or a weight _ Two vertices are

adjacent if they are connected by an edge 4 5 _ A vertexs degree is the number of its edges

Graph G= (V, E)

2V = set of vertices E = set of edges Each edge is now an 1 3arrow, not just a line ->directionThe in degree of a vertex is the number of 5incoming edges 4The out degree of a vertex is thenumber of outgoing edges.

5. Consider the following transportation problem:

Godowns

Factory 1 2 3 4 5 6Stock

available

A 7 5 7 7 5 3 60B 9 11 6 11 - 5 20

C 11 10 6 2 2 8 90

D 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 Vogels approximation method.

(b) Optimum basic feasible solution.

Solution:

The initial solution is found by VAM below:

Factory Godowns Availability Diff

1 2 3 4 5 6 60/40/0 2/4/0

A 7 20 box 5 7 7 5 40 box 3 20/10/0 1/3

B10 box

911 10 box 6 11 7 5 90/70/30/0 0/4/2/5

C 11 10 30 box 6 20 box 2 40 box 2 8 50/0 3/0

D 50 box 9 10 9 6 9 12

Demand 60/50/0 20/0 40/10/0 20/0 40/0 40/0

Diff 2 5 0/1 4 3 2

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The above initial solution is tested for optimality. Since there are only 8 allocations and werequire 9(m+n-1=9) allocations, we put a small quantity in the least cost independent cell (2, 6)

and apply the optimality test. Let u3=0 and then we calculate remaining ui and vj

Factory Godowns uj

-2000

1 2 3 4 5 6A 7 20 box 5 7 7 5 40 box 3

B 10 box 9 11 10 box 6 11 E box 5

C 11 10 30 box 6 20 box 2 40 box 2 8

D 50 box 9 10 9 6 9 12

vj 9 7 6 2 2 5

Now we calculate ij = Cij(ui + vj) for non basic cells which are given in the table below:

0 3 7 54 9

2 3 3

3 3 4 7 7

Since all ij are positive, the initial solution found by VAM is an optimal solution. The finalallocations are given below:

Factory to Godown Unit Cost Value

A 2 40 5 100

A 6 40 3 120

B 1 10 9 90

B 3 10 6 60

C 3 30 6 180

C 4 20 2 40

C 5 40 2 80

D 1 50 9 450

Total cost Rs. 1,120

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The above solution is not unique because the opportunity cost of cell (1, 2) is zero. Hencealternative solution exists. Students may find that the alternative solution is as given below:

Factory to Godown Unit Cost Value

A 1 10 7 70

A 2 20 5 100

A 6 30 3 90

B 3 10 6 60

B 6 10 5 50

C 3 30 6 180

C 5 40 2 80

C 4 20 2 40

D 1 50 9 450

Total cost (Rs.) 1,120

6. A machine operator processes five types of items on his machine each week, and mustchoose 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 ItemTo item

A B C D E

A 4 7 3 4

B 4 6 3 4

C 7 6 7 5

D 3 3 7 7

E 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?

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Solution:

Step 1: Reduce the cost matrix using step 1 & 2 of Hungarian algorithm and then makeassignments in rows and columns having single zeros as usual.

1 3 0 1

1 2 0 1

2 1 2 0

0 0 3 4

0 0 0 3

Step 2: Note that row 2 is not assigned. So mark to row 2. Since there is a zero in the 4th

column. Further, there is an assignment in the first row of 4th column. So, tick first row. Drawlines through all unmarked rows and marked columns. We can find the number of lines is 4which is less than order of the matrix. So, go to next step (see table)

1 3 0 1

1 2 0 1

--2-- --1-- ---- --2-- --0--

--0-- --0-- --3-- ---- --4--

--0-- --0-- --0-- --3-- ----

Step 3: Subtract the lowest element from all the elements not covered by these lines and add thesame with the elements at the intersection of two lines. Then we get the table as:

0 2 0 0

0 1 0 0

2 1 3 0

0 0 3 4

0 0 0 4

The optimum assignment is 1 4, 2 1, 3 5, 4 2, 5 3 with minimum cost as Rs 20.

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This assignment schedule does not provide us the solution of the travelling salesman problems asit gives 14, 4 2, 2 1, without passing through 3 and 5.

Next we try to find the next best solution which satisfies this restriction. The next minimum(non-zero) element in the cost matrix is 1. So, we bring 1 into the solution. But the element 1

occurs at two places. We consider all cases separately until we get an optimal solution.

We start with making an assignment at (2, 3) instead of zero assignment at (2, 1). The resultingassignment schedule is

1 4, 4 2, 2 3, 3 5, 5 1When an assignment is made at (3, 2) instead of zero assignment at (3, 5), the resultingassignment schedule is

1 5, 5 3, 3 2, 2 4, 4 1

The total set-up cost in both the cases is 21.

So, Optimum Assignment is:

AECBDA

AE = 4EC = 5CB = 6BD = 3DA = 3Total cost = 21

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87172577 Operation Research 2nd Sem

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