Final Assignment on Or

8/8/2019 Final Assignment on Or

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ASSIGNMENT ON OPERATIONS RESEARCH

Q.1: Describe in details the OR approach of problem solving. What

are the limitations of the Operations Research?Answer:

OR approach of problem solving:

Optimization is the act of obtaining the best result under any given circumstance. In

various practical problems we may have to take many technical or managerial decisions

at several stages. The ultimate goal of all such decisions is to either maximize the

desired benefit or minimize the effort required. We make decisions in our every day life

without even noticing them. Decision-making is one of the main activities of a manageror executive.

In simple situations decisions are taken simply by common sense, sound judgment

and expertise without using any mathematics. But here the decisions we are concerned

with are rather complex and heavily loaded with responsibility. Examples of such

decision are finding the appropriate product mix when there are large numbers of

products with different profit contributions and production requirement or planning

public transportation network in a town having its own layout of factories, apartments,

blocks etc. Certainly in such situations also decision may be arrived at intuitively from

experience and common sense, yet they are more judicious if backed up by

mathematical reasoning.

The search of a decision may also be done by trial and error but such a search may

be cumbersome and costly. Preparative calculations may avoid long and costly research.

Doing preparative calculations is the purpose of Operations research. Operations

research does mathematical scoring of consequences of a decision with the aim of

optimizing the use of time, efforts and resources and avoiding blunders.

The application of Operations research methods helps in making decisions in suchcomplicated situations. Evidently the main objective of Operations research is to

provide a scientific basis to the decision-makers for solving the problems involving the

interaction of various components of organization, by employing a team of scientists

from different disciplines, all working together for finding a solution which is the best

in the interest of the organization as a whole. The solution thus obtained is known as

optimal decision.

RAHUL GUPTA, MBAHCS (2ND SEM), SUBJECT CODE-MB0032, SET-2 Page 1


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Features of Operations Research: It is System oriented:

OR studies the problem from over all points of view of organizations or

situations since optimum result of one part of the system may not be optimum for some

other part.

It imbibes 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.

It makes use of Scientific methods to solve problems.

OR increases the effectiveness of a management Decision making ability.

It makes use of computer to solve large and complex problems.

It gives Quantitative solution.

It considers the human factors also.

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:

A description of the goal or the objective of the study.

An identification of the decision alternative to the system.

Recognition of the limitations, restrictions and requirements of the system.

Limitations of OR:

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



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Linear programming can be applied to various fields of study. Most extensively it is

used in business and economic situations, but can also be utilized for some engineering

problems. Some industries that use linear programming models include transportation, energy,telecommunications, and manufacturing. It has proved useful in modeling diverse types of

problems in planning, routing, scheduling, assignment, and design.

Uses:

Linear programming is a considerable field of optimization for several reasons. Many

practical problems in operations research can be expressed as linear programming problems.Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on

specialized algorithms for their solution. A number of algorithms for other types ofoptimization problems work by solving LP problems as sub-problems. Historically, ideas from

linear programming have inspired many of the central concepts of optimization theory, such as

duality, decomposition, and the importance of convexity and its generalizations. Likewise,linear programming is heavily used in microeconomics and company management, such as

planning, production, transportation, technology and other issues. Although the modern

management issues are ever-changing, most companies would like to maximize profits orminimize costs with limited resources. Therefore, many issues can boil down to linear

programming problems.

Standard form:

Standard form is the usual and most intuitive form of describing a linear programming

problem. It consists of the following three parts:

A linear function to be maximized

E.g. maximize

Problem constraints of the following form

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e.g.

Non-negative variables

e.g.

The problem is usually expressed in matrix form, and then becomes:

Maximize

Subject to

Other forms, such as minimization problems, problems with constraints on alternative

forms, as well as problems involving negative variables can always be rewritten into an

equivalent problem in standard form.

Example-

Suppose that a farmer has a piece of farm land, say A square kilometers large, to be planted

with either wheat or barley or some combination of the two. The farmer has a limited

permissible amountF of fertilizer and P of insecticide which can be used, each of which isrequired in different amounts per unit area for wheat (F1,P1) and barley (F2,P2). Let S1 be the

selling price of wheat, and S2 the price of barley. If we denote the area planted with wheat andbarley by x1 and x2 respectively, then the optimal number of square kilometers to plant with

wheat vs. barley can be expressed as a linear programming problem:

maximize (maximize the revenue revenue is the "objective function")

subject to (limit on total area)

(limit on fertilizer)

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(limit on insecticide)

(cannot plant a negative area)

Which in matrix form becomes?

Maximize

Subject to

Augmented form (slack form):

Linear programming problems must be converted into augmented form before being solved

by the simplex algorithm. This form introduces non-negative slack variables to replace

inequalities with equalities in the constraints. The problem can then be written in the followingblock matrix form.

MaximizeZin:

Where the newly introduced slack variables, and Z are is the variable to be maximized.

Example-

The example above is converted into the following augmented form:

maximize (objective function)

subject to (augmented constraint)

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(augmented constraint)

(augmented constraint)

Where are (non-negative) slack variables, representing in this example the unused

area, the amount of unused fertilizer, and the amount of unused insecticide.

In matrix form this becomes:

MaximizeZ in:

Duality:

Every linear programming problem, referred to as a primal problem, can be converted into a

dual problem, which provides an upper bound to the optimal value of the primal problem. Inmatrix form, we can express theprimal problem as:

Maximize

Subject to

The corresponding dual problem is:

Minimize

Subject to

Wherey is used instead ofx as variable vector.

There are two ideas fundamental to duality theory. One is the fact that the dual of a dual

linear program is the original primal linear program. Additionally, every feasible solution for a

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linear program gives a bound on the optimal value of the objective function of its dual. The

weak duality theorem states that the objective function value of the dual at any feasible solution

is always greater than or equal to the objective function value of the primal at any feasiblesolution. The strong duality theorem states that if the primal has an optimal solution, 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 is unbounded then the dual is infeasible by the weak duality theorem. Likewise, if the

dual is unbounded, then the primal must be infeasible. However, it is possible for both the dual

and the primal to be infeasible (See also Farkas' lemma).

Example-

Revisit the above example of the farmer who may grow wheat and barley with the setprovision of someA land, Ffertilizer andPinsecticide. Assume now that unit prices for each

of these means of production (inputs) are set by a planning board. The planning board's job is tominimize the total cost of procuring the set amounts of inputs while providing the farmer with a

floor on the unit price of each of his crops (outputs), S1 for wheat and S2 for barley. This

corresponds to the following linear programming problem:

minimize(minimize the total cost of the means of production as the"objective function")

subject

to(the farmer must receive no less than S1 for his wheat)

(the farmer must receive no less than S2 for his barley)

(prices cannot be negative)

Which in matrix form becomes?

Minimize

Subject to

The primal problem deals with physical quantities. With all inputs available in limited

quantities, and assuming the unit prices of all outputs is known, what quantities of outputs to

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produce so as to maximize total revenue? The dual problem deals with economic values. With

floor guarantees on all output unit prices, and assuming the available quantity of all inputs is

known, what input unit pricing scheme to set so as to minimize total expenditure? To eachvariable in the primal space corresponds an inequality to satisfy in the dual space, both indexed

by output type? To each inequality to satisfy in the primal space corresponds a variable in thedual space, both indexed by input type? The coefficients that bound the inequalities in the

primal space are used to compute the objective in the dual space, input quantities in this

example. The coefficients used to compute the objective in the primal space bound theinequalities in the dual space, output unit prices in this example. Both the primal and the dual

problems make use of the same matrix. In the primal space, this matrix expresses the

consumption of physical quantities of inputs necessary to produce set quantities of outputs. Inthe dual space, it expresses the creation of the economic values associated with the outputs

from set input unit prices. Since each inequality can be replaced by equality and a slack

variable, this means each primal variable corresponds to a dual slack variable, and each dualvariable corresponds to a primal slack variable. This relation allows us to complementaryslackness.

Covering-Packing Dualities:

A covering LP is a linear program of the

form

Minimize

Subject to

Such that the matrix and the vectors and are non-negative.

The dual of a covering LP is apacking LP, a linear program of the form

Maximize

Subject to


Covering-Packing Dualities

Covering problems Packing problems

Minimum Set Cover Maximum Set Packing

Minimum Vertex CoverMaximum Matching

Minimum Edge Cover Maximum Independent Set
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Such that the matrix and the vectors and are non-negative.

Examples-

Covering and packing LPs commonly arise as a linear programming relaxation of acombinatorial problem and are important in the study of approximation algorithms.[1] For

example, the LP relaxation ofset packing problem, independent set problem, ormatching is a

packing LP. The LP relaxation ofset cover problem, vertex cover problem, ordominating setproblem is a covering LP.

Finding a fractional coloring of a graph is another example of a covering LP. In this

case, there is one constraint for each vertex of the graph and one variable for each independent

set of the graph.

Complementary slackness:

It is possible to obtain an optimal solution to the dual when only an optimal solution to the

primal is known using the complementary slackness theorem. The theorem states:

Suppose that x = (x1, x2, xn) is primal feasible and that y = (y1, y2, . . . , ym) is dual feasible.Let (w1, w2, . . ., wm) denote the corresponding primal slack variables, and let (z1, z2, . . . , zn)

denote the corresponding dual slack variables. Then x and y are optimal for their respectiveproblems if and only if xjzj = 0, for j = 1, 2, . . . , n, wiyi = 0, for i = 1, 2, . . . , m.

So if the ith slack variable of the primal is not zero, then the ith variable of the dual is equalzero. Likewise, if the jth slack variable of the dual is not zero, then the jth variable of the primal

is equal to zero.

This necessary condition for optimality conveys a fairly simple economic principle. In

standard form (when maximizing), if there is slack in a constrained primal resource (i.e., thereare "leftovers"), then additional quantities of that resource must have no value. Likewise, if

there is slack in the dual (shadow) price non-negativity constraint requirement, i.e., the price is

not zero, and then there must scarce supplies (no "leftovers").

Theory:

Geometrically, the linear constraints define a convexpolytope, which is called the feasible

region. It is not hard to see that every local optimum (a pointx such that for every unit direction

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vectordwith positive objective value any every > 0 it holds that x + dis infeasible) is also a

global optimum. This holds more generally for convex programs: see the KKT theorem.

There are two situations in which no optimal solution can be found. First, if the constraints

contradict each other (for instance,x 2 andx 1) then the feasible region is empty and there

can be no optimal solution, since there are no solutions at all. In this case, the LP is said to beinfeasible. Alternatively, the polyhedron can be unbounded in the direction of the objective

function (for example: maximizex1 + 3x2 subject tox1 0, x2 0, x1 +x2 10), in which case

there is no optimal solution since solutions with arbitrarily high values of the objective function

can be constructed. Barring these two conditions (which can often be ruled out when dealingwith specific LPs), the optimum is always attained at a vertex of the polyhedron (unless the

polyhedron has no vertices, for example in the feasible bounded linear program

; polyhedral with at least one vertex are called pointed). However, the

optimum is not necessarily unique: it is possible to have a set of optimal solutions covering anedge or face of the polyhedron, or even the entire polyhedron (this last situation would occur if

the objective function were constant on the polyhedron).

The vertices of the polyhedron are also called basic feasible solutions. The reason for thischoice of name is as follows. Let ddenote the dimension, i.e. the number of variables. Then the

following theorem holds: for every vertex x* of the LP feasible region, there exists a set of d

inequality constraints from the LP such that, when we treat those dconstraints as equalities, theunique solution is x*. Thereby we can study these vertices by means of looking at certain

subsets of the set of all constraints (a discrete universe), rather than the continuous universe of

LP solutions. This principle underlies the simplex algorithm for solving linear programs.

Q 3: Describe the Two-Phase method of solving a linear

programming problem with an example?

Answer:

Two Phase Method:

The drawback of the penalty cost method is the possible computational error that couldresult 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:

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Formulate the new problem by eliminating the original objective function by the sum ofthe 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

(which indicates that all artificial variables are zero). Then we proceed to phase II.

Otherwise, if the optimal value of the new objective 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 original problem.

Then the objective function is taken without the artificial variables and is solved by simplexmethod.

Examples:

Use the two phase method to Maximize z = 3x1 x2



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Phase 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, Maximize z = 3x1 x2 + 0S1 + 0S2 + 0S3Subject to x1 + x2/2 S1/2=1 5/2 x2 + S1/2 + S2=1 x2 + S3 = 4x1, x2, S1, S2, S3 0 with the initial

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



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Q.4: What do you understand by the transportationproblem? What is the basic assumption behind the

transportation problem? Describe the MODI method of

solving transportation problem.

Answer:



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Transportation Problem & its basicassumption:

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 an 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 transportationschedule 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 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 oftransportation 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

be number of units of the homogenous product to be transported from source i to the

destination j. Then objective is to-



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

transportation problem (2) is that

The Transportation Algorithm (MODI

Method):

The first approximation to (2) is always integral and therefore always a feasiblesolution. 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.



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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) North-west 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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Q.5: Describe the North-West Corner rule for finding theinitial basic feasible solution in the transportation problem.

Answer:

The Initial basic Feasible solution usingNorth-West corner rule:

Let us consider a T.P involving m-origins and n-destinations. Since the sum of

origin capacities equals the sum of destination requirements, a feasible solution alwaysexists. Any feasible solution satisfying m + n 1 of the m + n constraints is a redundant

one and hence can be deleted. This also means that a feasible solution to a T.P can have

at the most only m + n 1 strictly positive component, otherwise the solution will

degenerate.

It is always possible to assign an initial feasible solution to a T.P. in such a manner

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

following some simple rules. We begin by imagining that the transportation table is

blank i.e. initially all xij = 0. The simplest procedures for initial allocation discussed in

the following section.

North West Corner Rule:

Step1:a. The first assignment is made in the cell occupying the upper left hand (North

West) corner of the transportation table.

b. The maximum feasible amount is allocated there, that is x11 = min (a1, b1) Sothat either the capacity of origin O1 is used up or the requirement at destination

D1 is satisfied or both.

c. This value of x11 is entered in the upper left hand corner (Small Square) of cell

(1, 1) in the transportation table.



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Step 2:a. If b1 > a1 the capacity of origin O, is exhausted but the requirement at

destination D1 is still not satisfied , so that at least one more other variable in the

first column will have to take on a positive value.

b. 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.

c. If a1 > b1 the requirement at destination D1 is satisfied but the capacity of origin

O1 is not 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).

d. 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.

e. There is a tie for 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:

a. Start from the new North West corner of the transportation table satisfying

destination requirements and exhausting the origin capacities one at a time.

b. Move down towards the lower right corner of the transportation table until all

the rim requirements are satisfied.



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

problem.

Answer:

The Branch And Bound Technique:

Sometimes a few or all the 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 optimization problems is the

branch and bound technique. The technique is applicable to both all IPP as well as mixed I.P.P. the

technique for a maximization problem is discussed below: Let the I.P.P be



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Or the linear constraint xj I ...(7)To explain how this partitioning helps,

let us assume that there were no integer restrictions (3), and suppose that this then yields

an optimal solution to L.P.P. (1), (2), (4) and (5). Indicating

x1 = 1.66 (for example). Then we formulate and solve two L.P.Ps each containing (1),

(2) and (4). But (5) for

j = 1

Is modified to be



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2 x1 U1

In one problem andL1 x1 1

In the other. Further each of these problems process an optimal solution satisfying

integer constraints (3) Then the solution having the larger value for z is clearly optimum

for the given I.P.P. However, it usually happens that one (or both) of these problems has

no optimal solution satisfying (3), and thus some more computations are necessary. We

now discuss step wise the algorithm that specifies how to apply the partitioning (6) and

(7) in a systematic manner to finally arrive at an optimum solution.

We start with an initial lower bound for z, say) 0(Zat the first iteration which is less than

or equal to the optimal value z*, this lower bound may be taken as the starting Lj for

some xj. In addition to the lower bound) 0(Z, we also have a list of L.P.Ps (to be called

master list) differing only in the bounds (5). To start with (the 0th iteration) the master

list contains a single L.P.P. 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 and Bound Algorithm

At the tth iteration (t = 0, 1, 2 )



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