21757331 MB0032 Operations Research Assignments

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    ASSIGNMENT

    Subject code: MB0032

    (3 credits)

    Set 2

    SUBJECT NAME: OPERATION

    RESEARCH

    NAME : PRANAY SRIVASTAVA

    MBA 2ND

    SEMROLL NO: 510933855

    L C CODE: 02782

    SIKKIM MANIPAL UNIVERSITY

    DDE

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    ASSIGNMENTSMB 0032

    (3 or 4 credits)Set 2

    Marks 60

    OPERATIONS RESEARCH

    Note: Each Question carries 10 marks

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

    Answer:

    OR approach of problem solvingOptimization is the act of obtaining the best result under any given circumstance. In variouspractical problems we may have to take many technical or managerial decisions at severalstages. The ultimate goal of all such decisions is to either maximize the desired benefit orminimize the effort required. We make decisions in our every day life without even noticingthem. Decision-making is one of the main activity of a manager or executive. In simplesituations decisions are taken simply by common sense, sound judgment and expertisewithout using any mathematics. But here the decisions we are concerned with are rathercomplex and heavily loaded with responsibility. Examples of such decision are finding the

    appropriate product mix when there are large numbers of products with different profitcontributions and productional requirement or planning public transportation network in atown having its own layout of factories, apartments, blocks etc. Certainly in such situationsalso decision may be arrived at intuitively from experience and common sense, yet they aremore judicious if backed up by mathematical reasoning. The search of a decision may alsobe done by trial and error but such a search may be cumbersome and costly. Preparativecalculations may avoid long and costly research. Doing preparative calculations is thepurpose of Operations research. Operations research does mathematical scoring ofconsequences of a decision with the aim of optimizing the use of time, efforts and resourcesand avoiding blunders.

    The application of Operations research methods helps in making decisions in such

    complicated situations. Evidently the main objective of Operations research is to provide ascientific basis to the decision-makers for solving the problems involving the interaction ofvarious components of organization, by employing a team of scientists from differentdisciplines, all working together for finding a solution which is the best in the interest of theorganization as a whole.

    The solution thus obtained is known as optimal decision. The main features of ORare:

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    It is System oriented: OR studies the problem from over all points of view of organizationsor 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 athorough knowledge of all fast developing scientific know-how, personalities from differentscientific 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 identifiedand 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, and(3) A 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.Magnitude of computation: Modern problem involve large number of variables and hence tofind 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 researches.

    Time and Money factors when the basic data is subjected to frequent changes thenincorporation of them into OR models are a costly affair.Implementation of decisions involves human relations and behaviour

    2. What are the characteristics of the standard form of L.P.P.? What is the standard formof L.P.P.? State the fundamental theorem of L.P.P.

    Answer:

    The characteristics of the standard form are:

    1.All constraints are equations except for the non-negativity condition which remaininequalities (, 0) only.2.The right-hand side element of each constraint equation is non-negative.3.All variables are non-negative.4.The objective function is of the maximization or minimization type. The inequalityconstraints can be changed to equations by adding or subtracting the left-hand side of eachsuch constraint by a non-negative variable. The non-negative variable that has to be addedto a constraint inequality of the form to change it to an equation is called aslack variable. The non-negative variable that has to be subtracted from a constraintinequality of the form to change it to an equation is called a surplus variable. The right handside of a constraint equation can be made positive by multiplying both sides of the resulting

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    equation by (-1) wherever necessary. The remaining characteristics are achieved by usingthe elementary transformations introduced with the canonical form.

    The Standard Form of the LPPAny standard form of the L.P.P. is given by

    Fundamental Theorem of L.P.P.

    Given a set of m 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.

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

    example.

    Answer:Two Phase MethodThe drawback of the penalty cost method is the possible computational error that could resultfrom assigning a very large value to the constant M. To overcome this difficulty, a new methodis considered, where the use of M is eliminated by solving the problem in two phases. They arePhase I:Formulate the new problem by eliminating the original objective function by the sum of theartificial variables for a minimization problem and the negative of the sum of the artificialvariables for a maximization problem. The resulting objective function is optimized by thesimplex 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 artificialvariables are 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 original problem. Thenthe objective function is taken without the artificial variables and is solved by simplex method.Examples:Use the two phase method toMaximize 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 withthe initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplex table is

    4. What do you understand by the transportation problem? What is the basicassumption behind the transportation problem? Describe the MODI method ofsolving transportation problem.

    Answer:Transportation Problem & its basic assumptionThis 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 eachdestination are known. The transportation problem involves m sources, each of which hasavailable a

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    i (i = 1, 2, ..,m) units of homogeneous product andn destinations, each of which requires

    bj (j = 1, 2., n) units of products. Here ai and bj are positiveintegers. The cost cij of transporting one unit of the product from theith source to the

    jth destination is given for eachi and j. The objective is to develop an integral transportation schedule that meets all demandsfrom the inventory at a minimum total transportation cost.It is assumed that the total supplyand the total demand are equal.i.e.

    Condition (1)The condition (1) is guaranteed by creating either a fictitious destination with ademand 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 thefictitious sources are assumed to be zero so that total cost of transportation will remain thesame.Formulation of Transportation ProblemThe 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 thedestination jThen 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 feasible solution.Rather than determining a first approximation by a direct application of the simplex method itis more efficient to work with the table given below called the transportation table. Thetransportation 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. improvingthe solution, when it is not optimal iv. repeating steps (ii) and (iii) until the optimal solution isobtained. The solution to T.P is obtained in two stages. In the first stage we find Basicfeasible solution by any one of the following methods a) North-west corner rule b) MatrixMinima Method or least cost method c) Vogels approximation method. In the second stagewe test the B.Fs for its optimality either by MODI method or by stepping stone method.

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

    Answer:

    The Initial basic Feasible solution using North-West corner ruleLet us consider a T.P involving m-origins and n-destinations. Since the sum of origincapacities 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 andhence 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 isalways possible to assign an initial feasible solution to a T.P. in such a manner that the rimrequirements are satisfied. This can be achieved either by inspection or by following somesimple 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:The first assignment is made in the cell occupying the upper left hand (north west) corner ofthe 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 destinationD1 is satisfied or both. This value of x11 is entered in the upper left hand corner (smallsquare) of cell (1, 1) in the transportation table.

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    Step 2: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 ona positive value. Move down vertically to the second row and make the second allocation ofmagnitude x21 = min (a2, b1 x21) in the cell (2,1). This either exhausts the capacity oforigin O2 or satisfies the remaining demand at destination D1.If a1 > b1 the requirement atdestination D1 is satisfied but the capacity of origin O1 is not completely exhausted. Move tothe right horizontally to the second column and make the second allocation of magnitudex12 = min (a1 x11, b2) in the cell (1, 2) . This either exhausts the remaining capacity oforigin O1 or satisfies the demand at destination D2 .If b1 = a1, the origin capacity of O1 iscompletely exhausted as well as the requirement at destination is completely satisfied.There is a tie for second allocation, An arbitrary tie breaking choice is made. Make thesecond 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 destinationrequirements and exhausting the origin capacities one at a time, move down towards thelower right corner of the transportation table until all the rim requirements are satisfied.

    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

    2 x1 U1

    in one problem and

    L1 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

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    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 )