MB0032 - Operations Research - Completed

8/9/2019 MB0032 - Operations Research - Completed

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OPERATIONS RESEARCH

MB0032

SET2MBA 2ndSEM

Name Mohammed Roohul Ameen

Roll Number

Learning Center SMU Riyadh (02543)

Subject Operations Research

Date of Submission 28 Feb 2010

Assignment Number MB0032


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Mohammed Roohul Ameen Roll Number:Assignment MBA 2nd Semester Subject: MB0032

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1.Describe in details the OR approach of problem solving. What are the limitationsof the Operations Research?

Operations Research encompasses a wide range of problem-solving techniques and methods

applied in the pursuit of improved decision-making and efficiency. Some of the tools used by

operations researchers are statistics, optimization, probability theory, queuing theory, game

theory, graph theory, decision analysis, mathematical modeling and simulation. Because of the

computational nature of these fields, OR also has strong ties to computer science. Operations

researchers faced with a new problem must determine which of these techniques are most

appropriate given the nature of the system, the goals for improvement, and constraints on time

and computing power.

Work in operations research and management science may be characterized as one of three

categories:

Fundamental or foundational work takes place in three mathematical disciplines:Probability, Optimization, and Dynamical systems theory.

Modeling work is concerned with the construction of models, analyzing themmathematically, gathering and analyzing data, implementing models on computers, solving

them, playing with them. This level is mainly instrumental, and driven mainly by statistics

and econometrics.

Application work in operations research, like other engineering and economics' disciplines,attempts to use models to make a practical impact on real-world problems.

The basic dominant characteristic feature of operations research is that it employs

mathematical representations or models to analyze and solve problems. This distinctive

approach represents an adaption 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 transformed into the original context.

The OR approach of problem solving consists of the following steps:

a. Definition of the problem:The first and the most important requirement is that the root problem should be identified

and understood. The problem should be identified properly, this indicates three major

aspects:

1. A description of the goal or the objective of the study,2. An identification of the decision alternative to the system, and3. Recognize the limitations, restrictions and requirements of the system.


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b. Construction of the model:Depending on the definition of the problem, the operations research team should decide on

the most suitable model for representing the system. Such a model should specify

quantitative expressions for the objective and the constraints of the problem in terms of its

decision variables. A model gives a perspective picture of the whole system and helps

tackling it in a well organized manner. If the resulting model fits into one of the common

mathematical techniques; and if the mathematical relationships of the model are toocomplex to allow analytic solutions, a simulation model may be more appropriate.

c. Solution of the model:Once an appropriate model has been formulated, the next step in the analysis calls for its

solution and the interpretation of the solution in the context of given problem. A solution to

a model implies determination of optimum solution is one which maximizes or minimizes

the performance of any measure in a model subject to the conditions and constrain

imposed on the model.

d.

Validation of the model:A model is a good representative of a system, and then the optimal solution must improve

the systems performance. A common method for testing validity of a model is to compare

its performance with some past data available for the actual system. The model will be valid

if under similar conditions of inputs, it can reproduce the past performance of the system.

There is no assurance that the future performance ill replicate past performance. Also, since

the models based on the past performance data, the comparison always reveals favorable

results. In some instances, this problem may be overcome by using data from the trail runs

of the system. It must be noted that such a validation method is not appropriate for

nonexistent systems, since data will not be available for comparison.

e. Implementation of the final result:The optimal solution obtained from a model should be applied to improve the performance

of the system and the validity of the solution should be verified under changing conditions.

It involves the translation of these results into detailed operating instructions issued in an

understandable form to the individuals who will administer and operate the recommended

system. The interaction between the operations research team and the operating personnel

is at its peak in this phase.

Limitations of Operations Research:

The limitations are more related to the problems of model building, time and money factors.

Magnitude of computation: Modern problems involve large number of variables; and henceto find the interrelationship, among makes it difficult.

Non Quantitative factors and human emotional factor cannot be taken into account. There is a wide gap between the managers and the operation researchers. Time and money factors when the basic data is subjected to frequent changes then

incorporation of them into OP models is a costlier affair.

Implementation of discussions involves human relations and behavior.


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2. What are the characteristics of the standard form of L.P.P.? What is the standard

form of L.P.P.? State the fundamental theorem of L.P.P..

Characteristics of standard form of L.P.P.:

The characteristics of the standard form are:

All constraints are equations except for the non-negativity condition which remaininequalities [>0] only.

The right hand side element of each constraint equation is nonnegative. All variables are nonnegative. The objective function is of the maximization or minimization type.The inequality constraints can be changed to equations by adding or subtracting the left handside of each such constraint by a non negative variable. The non negative variable that has to be

added to a constraint inequality of the form < to change it to an equation is called a slack

variable. The non negative variable that has to be subtracted from a constraint inequality is

called a surplus variable. The right hand side of a constraint equation can be made positive by

multiplying both sides of the resulting equation with -1 wherever necessary. The remaining

characteristics are achieved by using the elementary transformations introduced with the

canonical form.

The standard form of L.P.P.:

Any standard form of LPP is given by

Fundamental theorem of L.P.P.:

Given a set of simultaneous linear equations in n unknowns/variables,

n >m, AX=b, with r(A) =m.

If there is a feasible solution X>0, then there exists a basic feasible solution.erce,

Trevelyan


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3. Describe the Two-Phase method of solving a linear programming problem with an

example?

Two phase method:

The drawback of the penalty cost method is the possible computational errors that could result from

assigning a very large value to the constant M. to overcome this difficulty, a new method is considered,

where the use of M is eliminated by solving the problem in two phases. They are:

Phase I: Formulate the new problem by eliminating the original objective function by the sum of

artificial variables for a minimization problem and the negative of the sum of the artificial variables for

a maximization problem. The resulting objective function is optimized by the simplex method with the

constraints of the original problem. If the problem has a feasible solution, the optimal value of the new

objective function is zero. Then we proceed to Phase II. Otherwise, if the optimal value of the newobjective function is non zero, the problem has no solution and the method terminates.

Phase II: Use the optimum solution of the phase I as the starting solution of the variables and is solved

by simplex method.

Example:

Use the two phased method to

Maximise Z=3x1-x2

Subject to 2x1 +x2 >2; x1+3x2 0

Phase I:

Consider the new objective,

Maximise Z* =-A1

Subject to

2x1 +x2S1+A1=2; x1+3x2+S2=2;

x2+S3 =4;

x1,x2,S1,S2,S3,A1>0


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Solving by simplex method, the initial simplex table is given by

x1 x2 S1 A1 S2 S3

0 0 0 -1 0 0 Ratio

A1-1 2* 1 -1 1 0 0 2 2/2 =1

S2 0 1 3 0 0 1 0 2 2/1 = 2

S3 0 0 1 0 0 0 1 4

-2 -1 1 0 0 0 -2

Work column * Pivot element

x1 enters the basic set replacing A1The first iteration gives the following table:

x1 x2 x1 A1 S2 S3

0 0 0 -1 0 0

x1 0 1 - 0 0 1

S2 0 0 5/2 - 1 0 1

S3 0 0 1 0 0 0 1 4

0 0 0 1 0 0 0Phase I is complete, since there are no negative elements in the last row.

The optimal solution of the new objective is Z* =0.

Phase II:

Consider the original objective function,

Maximise Z=3x1-x2+0S1-MA1+0S2+0S3

Subject to

x1 +(x2/2)(S1/2)=1;

(5/2)x2 + S1 /2 + S2=1;

x2+S3 =4;x1,x2,S1,S2,S3,A1>0

With the initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplex table is

x1 x2 S1 S2 S3

3 -1 0 0 0 Ratio

x1 3 1 -1/2 0 0 1

S2 0 0 5/2 * 1 0 1 1/(1/2) = 2

S3 0 0 1 0 0 1 4

0 5/2 -3/2 0 0 3

Work column, * Pivot table

Proceeding to the next iteration

x1 x2 S1 S2 S3

3 -1 0 0 0

x1 3 1 3 0 1 0 1 2

S2 0 0 5 1 2 0 1 2

S3 0 0 1 0 0 1 4 4

0 10 0 3 0 3 6

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


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4. What do you understand by the transportation problem? What is the basic

assumption behind the transportation problem? Describe the MODI method of solving

transportation problem.

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

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 i th source to the j th 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 thefictitious 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 Transportation Algorithm (MODI Method)

The first approximation to (2) is always integral and therefore always a feasible solution. Rather

than determining a first approximation by a direct application of the simplex method it is more

efficient to work with the table given below called the transportation table. The transportation

algorithm is the simplex method specialized to the format of table it involves:

i) Finding an integral basic feasible solution

ii) Testing the solution for optimality

iii) Improving the solution, when it is not optimal

iv) Repeating steps (ii) and (iii) until the optimal solution is obtained.

The solution to T.P is obtained in two stages. In the first stage we find Basic feasible solution by

any one of the following methods a) Northwest corner rule b) Matrix Minima Method or least

cost method c) Vogels approximation method. In the second stage we test the B.Fs for its

optimality either by MODI method or by stepping stone method.


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Modified Distribution Method / Modi Method / U V Method.

Step 1: Under this method we construct penalties for rows and columns by subtracting the least

value of row / column from the next least value.

Step 2: We select the highest penalty constructed for both row and column. Enter that row /

column and select the minimum cost and allocate min (ai, bj)

Step 3: Delete the row or column or both if the rim availability / requirements is met.

Step 4: We repeat steps 1 to 2 to till all allocations are over.

Step 5: For allocation all form equation ui + vj = cj set one of the dual variable ui / vj to zero and

solve for others.

Step 6: Use these values to find ij = cij uivj of all ij >, then it is the optimal solution.

Step 7: If any Dij 0, select the most negative cell and form loop. Starting point of the loop is

+ve and alternatively the other corners of the loop areve and +ve. Examine the quantities

allocated atve places. Select the minimum. Add it at +ve places and subtract fromve place.

Step 8: Form new table and repeat steps 5 to 7 till ij > 0


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5. Describe the North-West Corner rule for finding the initial basic feasible solution in

the transportation problem.

Consider a T.P. involving m-origins and n-destinations. Since the sum of origin capacities equalsto the sum of requirements, a feasible solution always exists. Any feasible solution satisfying

m+n-1 of the m+n constraints is a redundant one an hence can be deleted. This also means that

a feasible solution to a transportation problem can have at the most only m+n-1 strictly positive

compliments, otherwise the solution will degenerate.

It is always possible to assign an initial feasible solution to a transportation problem in such a

manner that the rim requirements are satisfied. This can be achieved either by inspection or by

following some simple rules.

North-West corner rule is one of the simplest procedures for initial allocation of feasible

solution.

North-West Corner Rule:

Step 1:

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 there, that is x11 = min (a1,b1)

So that, either the capacity of origin O1 is used up; or the requirement at the destination D 1 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 but the requirement at destination D1 is still not

satisfied, so that one more other 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,). This either exhausts the capacity of origin O 2 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 no

completely exhausted. Move to the right horizontally to the second column and make the

second allocation of magnitude x12=min (a1-x11,b2) in the cell(1,2). This either exhausts theremaining 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. There is a tie for second allocation. An arbitrary the breaking

choice is made. Make the second allocation of magnitude x 12 =min (a1-a2,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 destination

requirements and exhausting the origin capacities one at a time, move down towards the lowerright corner of the transportation table until al the rim requirements are satisfied.


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6. Describe the Branch and Bound Technique to solve an I.P.P. problem.

Branch and bound (BB) is a general algorithm for finding optimal solutions of various

optimization problems, especially in discrete and combinatorial optimization. It consists of a

systematic enumeration of all candidate solutions, where large subsets of fruitless candidates

are discarded en masse, by using upper and lower estimated bounds of the quantity being

optimized.

The method was first proposed by A. H. Land and A. G. Doig in 1960 for linear programming.

Sometimes a few or all variables of an IPP are constrained by their upper or lower bounds or by

both. The most general technique for the solution of such constrained optimisation problems is

the branch and bound technique. The technique is applicable to both all IPP as well as mixedIPP. The technique for a maximisation problem is discussed below:

Let the IPP be

Maximise Z= =1 Cj Xj ---------------------- (1)

Subject to constraints

=1 aijxj < bi,i=1,2,.....m ---------------------(2)

Xj is integer valued j = 1,2,3....r (0 j=r+1,......n--------------------------(4)

Further let us suppose that for each integer valued xj, we can assign lower and upper bounds forthe optimum values of the variable by

Lj


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We start with an initial lower bound for z, say z(0),

we also have a list of LPPs differing only in

the bounds (5). Th start with the master list contains a single LPP, consisting of (1), (2), (4) and

(5). We now discuss below ,the step by step procedure that specifies how the partitioning (6)

and (7) can be applied systematically to eventually get an optimum integer valued solution.

Branch & Bound Algorithm:

At the tth

iteration (t=0,1, 2...)

Step 0:

If the master list is not empty, choose an LPP out of it. Otherwise stop the process, Go the step

1.

Step 2:

Obtain the optimum solution to the objective function z is less than or equal to z

(t)

, then let z

(t+1)

= z(t)

and return to step 0 otherwise go to step 3.

Step 3:

Select any variable xj, j = 1,2,...p. that does not have an integer value in the obtained optimum

solution to the LPP chosen in step 0. Let xj* denote his optimal value of xj. Add two LPPs to the

master list; these LPPs are identical with the LPP chosen in step 0, except that in one, the lower

bound on xj is replaced by [xj*] +1. Let z(t+1)

= z(t)

, return to step 0.

At the termination of algorithm, if feasible integer valued solution yielding z(t)

has been

recorded it is optimum, otherwise no integer valued feasible solution exists.

MB0032 - Operations Research - Completed

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