MB0032 Operation Research Set1

download MB0032 Operation Research Set1

of 31

Transcript of MB0032 Operation Research Set1

  • 8/8/2019 MB0032 Operation Research Set1

    1/31

    ASSIGNMENTS

    Subject code: MB

    0032

    (2 credits)

    Set 1

    Marks 60

    SUBJECT NAME: OPERATIONS RESEARCH

    Note: Each Question carries 10 marks

    1. Describe in details the different scopes of application of Operations Research.

    Answer: Operations Research (OR) in the USA, South Africa and Australia, and Operational

    Research in Europe and Canada, is an interdisciplinary branch of applied mathematics and formal

    science that uses methods such as mathematical modeling, statistics, and algorithms to arrive at

    optimal or near optimal solutions to complex problems. It is typically concerned with optimizing

    the maxima (profit, assembly line performance, crop yield, bandwidth, etc) or minima (loss, risk,

    etc.) of some objective function. Operations research helps management achieve its goals using

    scientific methods. The terms operations research and management science are often used

    synonymously.

    When a distinction is drawn, management science generally implies a closer relationship to the

    problems of business management. The field of operations research is closely related to Industrial

    engineering. Industrial engineers typically consider Operations Research (OR) techniques to be a

    major part of their toolset. Some of the primary tools used by operations researchers are statistics,

    optimization, probability theory, queuing theory, game theory, graph theory, decision analysis,and simulation. Because of the computational nature of these fields, OR also has ties to computer

    science, and operations researchers use custom-written and off-the-shelf software.

    Operations research is distinguished by its frequent use to examine an entire management

    information system, rather than concentrating only on specific elements (though this is often done

    as well). An operations researcher faced with a new problem is expected to determine which

  • 8/8/2019 MB0032 Operation Research Set1

    2/31

    techniques are most appropriate given the nature of the system, the goals for improvement, and

    constraints on time and computing power. For this and other reasons, the human element of OR is vital.

    Like any other tools, OR techniques cannot solve problems by themselves.

    Scope of operation Research

    Examples of applications in which operations research is currently used include:

    Critical path analysis or project planning: identifying those processes in a complex project which affect

    the overall duration of the project

    1. Designing the layout of a factory for efficient flow of materials

    2. Constructing a telecommunications network at low cost while still guaranteeing QoS

    (quality of service) or QoS (Quality of Experience) if particular connections become very busy

    or get damaged

    3. Road traffic management and 'one way' street allocations i.e. allocation problems.

    4. Determining the routes of school buses (or city buses) so that as few buses are needed as

    possible

    5. Designing the layout of a computer chip to reduce manufacturing time (therefore reducing

    cost)

    6. Managing the flow of raw materials and products in a supply chain based on uncertain

    demand for the finished products

    7. Efficient messaging and customer response tactics

    8. Robotizing or automating human-driven operations processes

    9. Globalizing operations processes in order to take advantage of cheaper materials, labor, land

    or other productivity inputs

    10. Managing freight transportation and delivery systems (Examples: LTL Shipping,

    intermodal freight transport)

    11. Scheduling:

    Personnel staffing

    Manufacturing steps

    Project tasks

    Network data traffic: these are known as queuing models or queueing systems.

  • 8/8/2019 MB0032 Operation Research Set1

    3/31

    sports events and their television coverage

    12. blending of raw materials in oil refineries

    13. determining optimal prices, in many retail and B2B settings, within the disciplines of

    pricing science

    Operations research is also used extensively in government where evidence-based policy is used.

    Q.No.2: What do you understand by Linear Programming Problem? What

    are the requirements of L.P.P.? What are the basic assumptions of

    L.P.P.?

    Answer: Linear programming problem (LPP):

    The standard form of the linear programming problem is used to develop the procedure for

    solving a general programming problem.

    A general LPP is of the form

    Max (or min) Z = c1x1 + c2x2 + +cnxn

    x1, x2, ... .xn are called decision variable. Subject to the constraints

    c1, c2,. Cn, a11, a12,. amn are all known constants

    Z is called the "objective function" of the LPP of n variables which is to be maximized or

    minimized.

  • 8/8/2019 MB0032 Operation Research Set1

    4/31

    Requirements of L.P.P :There are mainly four steps in the mathematical formulation of linear

    programming problem as a mathematical model. We will discuss formulation of those problems

    which involve only two

    variables.

    Identify the decision variables and assign symbols x and y to them. These decision

    variables are those quantities whose values we wish to determine.

    Identify the set of constraints and express them as linear equations/inequations in terms of the

    decision variables. These constraints are the given conditions.

    Identify the objective function and express it as a linear function of decision variables. It

    might take the form of maximizing profit or production or minimizing cost.

    Add the non-negativity restrictions on the decision variables, as in the physical problems,

    negative values of decision variables have no valid interpretation.

    There are many real life situations where an LPP may be formulated. The following examples

    will help to explain the mathematical formulation of an LPP.

    Example-1. A diet is to contain at least 4000 units of carbohydrates, 500 units of fat and 300

    units of protein. Two foods A and B are available. Food A costs 2 dollars per unit and food B

    costs 4 dollars per unit. A unit of food A contains 10 units of carbohydrates, 20 units of fat and 15

    units of protein. A unit of food B contains 25 units of carbohydrates, 10 units of fat and 20 units

    of protein. Formulate the problem as an LPP so as to find the minimum cost for a diet that

    consists of a mixture of these two foods and also meets the minimum requirements.

    Suggested answer:

    The above information can be represented as

  • 8/8/2019 MB0032 Operation Research Set1

    5/31

    Let the diet contain x units of A and y units of B.

    Total cost = 2x + 4y

    The LPP formulated for the given diet problem is

    Minimize Z = 2x + 4y subject to the constraints

    Basic Assumptions of L.P.P:

    Linear programming is applicable only to problems where the constraints and objective function

    are linear i.e., where they can be expressed as equations which represent straight lines. In real life

    situations, when constraints or objective functions are not linear, this technique cannot be used.

    Factors such as uncertainty, weather conditions etc. are not taken into consideration.

    There may not be an integer as the solution, e.g., the number of men required may be a

    fraction and the nearest integer may not be the optimal solution.i.e., Linear programming techniques may give practical valued answer which is not desirable.

    Only one single objective is dealt with while in real life situations, problems come with multi-

    objectives.

    Parameters are assumed to be constants but in reality they may not be so.

  • 8/8/2019 MB0032 Operation Research Set1

    6/31

    Q.No.3: Describe the different steps needed to solve a problem by simplex metho

    Answer: Simplex method

    The simplex method is a method for solving problems in linear programming. This method,

    invented by George Dantzig in 1947, tests adjacent vertices of the feasible set (which is apolytope) in sequence so that at each new vertex the objective function improves or is unchanged. The

    simplex method is very efficient in practice, generally taking 2m to 3m iterations at most (where m

    is the number of equality constraints), and converging in expected polynomial time for certain

    distributions of random inputs (Nocedal and Wright 1999, Forsgren 2002). However, its worst-case

    complexity is exponential, as can be demonstrated with carefully constructed examples (Klee

    and Minty 1972). Different types of methods for linear programming problems are interior point

    methods, whose complexity is polynomial for both average and worst case. These methods

    construct a sequence of strictly feasible points (i.e., lying in the interior of the polytope but never on

    its boundary) that converges to the solution. Research on interior point methods was spurred by a

    paper from Karmarkar (1984). In practice, one of the best interior-point methods is the predictor-

    corrector method of Mehrotra (1992), which is competitive with the simplex method, particularly

    for large-scale problems.

    Dantzig's simplex method should not be confused with the downhill simplex method (Spendley

    1962, Nelder and Mead 1965, Press et al. 1992). The latter method solves an unconstrained

    minimization problem in n dimensions by maintaining at each iteration n+1 points that define a

    simplex. At each iteration, this simplex is updated by applying certain transformations to it so that it

    "rolls downhill" until it finds a minimum.

    The Simplex Method is "a systematic procedure for generating and testing candidate vertex

    solutions to a linear program." (Gill, Murray, and Wright, p. 337) It begins at an arbitrary corner

    of the solution set. At each iteration, the Simplex Method selects the variable that will producethe largest change towards the minimum (or maximum) solution. That variable replaces one of its

    compatriots that is most severely restricting it, thus moving the Simplex Method to a different

    corner of the solution set and closer to the final solution. In addition, the Simplex Method can

    determine if no solution actually exists. Note that the algorithm is greedy since it selects the best

    choice at each iteration without needing information from previous or future iterations.

  • 8/8/2019 MB0032 Operation Research Set1

    7/31

    The Simplex Method solves a linear program of the form described in Figure 3 . Here, the

    coefficients cj represent the respective weights, or costs, of the variables xi. The minimized

    statement is similarly called the cost of the solution. The coefficients of the system of equations

    are represented by aij, and any constant values in the system of equations are combined on the

    right-hand side of the inequality in the variables bi.

    Combined, these statements represent a linear program, to which we seek a solution of minimum

    cost.

    Solving this linear program involves solutions of the set of equations. If no solution to the set of

    equations is yet known, slack variables x

    n+1, x n+2, ., x n+m, adding no cost to the solution,

    are introduced. The initial basic feasible solution (BFS) will be the solution of the linear program

    where the following holds:

    Once a solution to the linear program has been found, successive improvements are made to the

    solution. In particular, one of the non-basic variables (with a value of zero) is chosen to be

    increased so that the value of the cost function, , decreases. That variable is then

    increased, maintaining the equality of all the equations while keeping the other nonbasic variables

    at zero, until one of the basic (nonzero) variables is reduced to zero and thus removed from the

    basis. At this point, a new solution has been determined at a different corner of the solution set.

    The process is then repeated with a new variable becoming basic as another becomes nonbasic.

    Eventually, one of three things will happen. First, a solution may occur where no nonbasic

  • 8/8/2019 MB0032 Operation Research Set1

    8/31

    variable will decrease the cost, in which case the current solution is the optimal solution. Second,

    a non-basic variable might increase to infinity without causing a basic variable to become zero,

    resulting in an unbounded solution. Finally, no solution may actually exist and the Simplex

    Method must abort. As is common for research in linear programming, the possibility that the

    Simplex Method might return to a previously visited corner will not be considered here.

    The primary data structure used by the Simplex Method is "sometimes called a dictionary, since the

    values of the basic variables may be computed (looked up) by choosing values for the

    nonbasic variables." (Gill, Murray, and Wright, p. 337) Dictionaries contain a representation of the

    set of equations appropriately adjusted to the current basis. The use of dictionaries provide an intuitive

    understanding of why each variable enters and leaves the basis. The drawback to dictionaries,

    however, is the necessary step of updating them which can be time-consuming. Computer

    implementation is possible, but a version of the Simplex Method has evolved with a more efficient

    matrix-oriented approach to the same problem. This new implementation became known as the

    Revised Simplex Method.

    The steps of the Simplex Method also need to be expressed in the matrix format of the

    Revised

    Simplex Method. The basis matrix, B, consists of the column entries ofA corresponding to the

    coefficients of the variables currently in the basis. That is if x2 is the fourth entry of the basis,

    then [ a12 a22 am2]T is the fourth column of

    B. (Note that B is therefore an m X m matrix.)

    The non-basic columns ofA constitute a similar though likely not square, matrix referred to here

    as V.

  • 8/8/2019 MB0032 Operation Research Set1

    9/31

    ASSIGNMENTS

    Subject code: MB

    0032

    (2 credits)

    Set 2

    Marks 30

    SUBJECT NAME: OPERATIONS RESEARCH

    Note: Each Question carries 10 marks

    Q.No.1: What are the important features of Operations Research?

    Describe in details the different phases of Operations Research.

    Answer: Important features of OR are:i. It is System oriented: OR studies the problem from over all point of view of organizations or

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

    part.

    ii. It imbibes Inter - disciplinary team approach. Since no single individual can have a thorough

    knowledge of all fast developing scientific knowhow, personalities from different scientific and

    managerial cadre form a team to solve the problem.

    iii. It makes use of Scientific methods to solve problems.

    iv. OR increases the effectiveness of a management Decision making ability.

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

    vi. It gives Quantitative solution.

    vii. It considers the human factors also.

  • 8/8/2019 MB0032 Operation Research Set1

    10/31

    Phases of Operations Research

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

    i)Judgment Phase: This phase consists of

    a) Determination of the operation.

    b) Establishment of the objectives and values related to the operation.

    c) Determination of the suitable measures of effectiveness and

    d) Formulation of the problems relative to the objectives.

    ii) Research Phase: This phase utilizes

    a) Operations and data collection for a better understanding of the problems.

    b) Formulation of hypothesis and model.

    c) Observation and experimentation to test the hypothesis on the basis of additional data.

    d) Analysis of the available information and verification of the hypothesis using pre established

    measure of effectiveness.

    e) Prediction of various results and consideration of alternative methods.

    iii) Action Phase: It consists of making recommendations for the decision process by those who

    first posed the problem for consideration or by anyone in a position to make a decision,

    influencing the operation in which the problem is occurred.

    Q.No.2: Describe a Linear Programming Problem in details in canonical form.

    Answer: Linear Programming

    The Linear Programming Problem (LPP) is a class of mathematical programming in which the

    functions representing the objectives and the constraints are linear. Here, by optimization, we

    mean either to maximize or minimize the objective functions. The general linear programming

    model is usually defined as follows:

  • 8/8/2019 MB0032 Operation Research Set1

    11/31

    Maximize or Minimize

    Z = c1 x1 + c2 x 2 +.. +cn x n

    subject to the constraints,

    a11 x1 + a12 x2 + .+ a1n xn ~ b1

    a21 x1 + a22 x2 +..+ a2n xn ~ b2

    am1x1 + am2 x2 +. +amn xn ~ bm

    and x1 > 0, x2 >0, xn >0.

    Where cj, bi and aij (i = 1, 2, 3, .. m, j = 1, 2, 3.. n) are constants determined

    from the technology of the problem and xj (j = 1, 2, 3 n) are the decision variables. Here ~ is

    either (greater than) or = (equal). Note that, in terms of the above formulation the

    coefficient cj, aij, bj are interpreted physically as follows. Ifbi is the available amount of resources i,

    where aij is the amount of resource i that must be allocated to each unit of activityj, the worth per

    unit of activity is equal to cj.

    Canonical forms:

    The general Linear Programming Problem (LPP) defined above can always be put in the

    following form which is called as the canonical form:

    Maximise Z = c1 x1+c2 x2 + .+cn xn

    Subject to

    a11 x1 + a12 x2 +.. +a1n xn < b1

    a21 x1 + a22 x2 +.. +a2n xn < b2

    am1x1+am2 x2 + + amn xn < bm

  • 8/8/2019 MB0032 Operation Research Set1

    12/31

    x1, x2, x3, xn > 0.

    The characteristics of this form are:

    1) all decision variables are nonnegative.

    2) all constraints are of < type.

    3) the objective function is of the maximization type.

    Any LPP can be put in the cannonical form by the use of five elementary transformations:

    1. The minimization of a function is mathematically equivalent to the maximization of the

    negative expression of this function. That is, Minimize Z = c

    1 x1+ c

    2x

    2+ . + c

    nx

    n

    isequivalent to

    Maximize- Z = - c

    1x1 - c2x2 - - cnxn.

    2. Any inequality in one direction (< or >) may be changed to an inequality in the opposite

    direction (> or 5 is equivalent to -2x1-3x2 < -5.

    3. An equation can be replaced by two inequalities in opposite direction. For example, 2x1+3x2 =

    5 can be written as 2x1+3x2 < 5 and 2x1+3x2 > 5 or 2x1+3x2 < 5 and - 2x1 - 3x2 < - 5.

    4. An inequality constraint with its left hand side in the absolute form can be changed into two

    regular inequalities. For example: | 2x1+3x2 | < 5 is equivalent to 2x1+3x2 < 5 and

    2x1+3x2 > -

    5or - 2x1 - 3x2 < 5.

    5. The variable which is unconstrained in sign (i.e., > 0, < 0 or zero) is equivalent to the

    difference between 2 nonnegative variables. For example, if x is unconstrained in sign then x

    = (x + - x - ) where x + > 0, x - < 0.

  • 8/8/2019 MB0032 Operation Research Set1

    13/31

    Q.No.3: What are the different steps needed to solve a system of

    equations by the simplex method?

    Answer: To Solve problem by Simplex Method

    1. Introduce stack variables (Sis) for < type of constraint.

    2. Introduce surplus variables (Sis) and Artificial Variables (Ai) for > type of constraint.

    3. Introduce only Artificial variable for = type of constraint.

    4. Cost (Cj) of slack and surplus variables will be zero and that of Artificial variable will be M

    Find Zj Cj for each variable.

    5. Slack and Artificial variables will form Basic variable for the first simplex table. Surplus

    variable will never become Basic Variable for the first simplex table.

    6. Zj = sum of [cost of variable x its coefficients in the constraints - Profit or cost coefficient of the

    variable].

    7. Select the most negative value of Zj - Cj. That column is called key column. The variable

    corresponding to the column will become Basic variable for the next table.

    8. Divide the quantities by the corresponding values of the key column to get ratios select the

    minimum ratio. This becomes the key row. The Basic variable corresponding to this row will be

    replaced by the variable found in step 6.

    9. The element that lies both on key column and key row is called Pivotal element.

    10. Ratios with negative and a value are not considered for determining key row.

    11. Once an artificial variable is removed as basic variable, its column will be deleted from next

    iteration.

  • 8/8/2019 MB0032 Operation Research Set1

    14/31

    12. For maximisation problems decision variables coefficient will be same as in the objective

    function. For minimization problems decision variables coefficients will have opposite signs as

    compared to objective function.

    13. Values of artificial variables will always is - M for both maximisation and minimization

    problems.

    14. The process is continued till all Zj - Cj > 0.

    .

  • 8/8/2019 MB0032 Operation Research Set1

    15/31

    Primal

    Maximize

    From the above resource allocation model, the primal problem has n economic activities and m

    resources. The coefficient cj in the primal represents the profit per unit of activity j.

    Resource i, whose maximum availability is bi, is consumed at the rate aij units per unit of activity

    j.

    Interpretation of Duel Variables -

    For any pair of feasible primal and dual solutions,

  • 8/8/2019 MB0032 Operation Research Set1

    16/31

    (Objective value in the maximization problem) (Objective value in the minimization problem)

    At the optimum, the relationship holds as a strict equation. Note: Here the sense of optimization is

    very important. Hence clearly for any two primal and dual feasible solutions, the values of the

    objective functions, when finite, must satisfy the following inequality.

    The strict equality, z = w, holds when both the primal and dual solutions are optimal.

    Consider the optimal condition z = w first given that the primal problem represents a resource

    allocation model, we can think of z as representing profit in Rupees. Because bi represents the

    number of units available of resource i, the equation z = w can be expressed as profit (Rs) =

    (units of resource i) x (profit per unit of resource i) This means that the dual variables yi,

    represent the worth per unit of resource i [variables yi are also called as dual prices, shadow

    prices and simplex multipliers]. With the same logic, the inequality z < w associated with any two

    feasible primal and dual solutions is interpreted as (profit) < (worth of resources) This

    relationship implies that as long as the total return from all the activities is less than the worth of the

    resources, the corresponding primal and dual solutions are not optimal. Optimality is reached only

    when the resources have been exploited completely, which can happen only when the input equals the

    output (profit). Economically the system is said to remain unstable (non optimal) when the input (worth

    of the resources) exceeds the output (return). Stability occurs only when the two quantities are equal.

  • 8/8/2019 MB0032 Operation Research Set1

    17/31

    Q.No.5: How can use the Matrix Minimum method of finding the initial basic

    feasible solution in the transportation problem.

    Answer: The Initial basic feasible solution using Matrix Minimum Method

    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 always exists.

    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.

    Matrix Minimum Method

    Step 1:Determine the smallest cost in the cost matrix of the transportation table. Let it be cij ,

    Allocate xij = min ( ai, bj) in the cell ( i, j)

    Step 2: If xij = ai cross off the ith row of the transportation table and decrease bj by ai go to step

    3.

    if xij = bj cross off the ith column of the transportation table and decrease ai by bj go to step 3.

    if xij = ai= bj cross off either the ith row or the ith column but not both.

    Step 3: Repeat steps 1 and 2 for the resulting reduced transportation table until all the rim

    requirements are satisfied whenever the minimum cost is not unique make an arbitrary choice

    among the minima.

  • 8/8/2019 MB0032 Operation Research Set1

    18/31

    Q.No.6: Describe Integer Programming Problem? Describe the Gomorys All-I.P.P.method for solving the I.P.P. problem.

    Answer: Integer Programming Problem

    The Integer Programming Problem I P P is a special case of L P P where all or some variables are

    constrained to assume nonnegative integer values. This type of problem has lot of applications in

    business and industry where quite often discrete nature of the variables is involved in many

    decision making situations. Eg. In manufacturing the production is frequently scheduled in terms of

    batches, lots or runs In distribution, a shipment must involve a discrete number of trucks or

    aircrafts or freight cars .

    An integer programming problem can be described as follows:

    Determine the value of unknowns x1, x2, , xn so as to optimize z = c1x1 +c2x2 + . . .+

    cnxn subject to the constraints ai1 x1 + ai2 x2 + . . . + ain xn =bi , i = 1,2,,m and xj

    > 0 j = 1, 2, ,n where xjbeing an integral value forj = 1, 2, , k n.

    If all the variables are constrained to take only integral value i.e. k = n, it is called an all (or pure)

    integer programming problem. In case only some of the variables are restricted to take integral

    value and rest (n - k) variables are free to take any non negative values, then the problem is

    known as mixed integer programming problem.

    Gomorys All - IPP Method

    An optimum solution to an I. P. P. is first obtained by using simplex method ignoring the

    restriction of integral values. In the optimum solution if all the variables have integer values, the

    current solution will be the desired optimum integer solution. Otherwise the given IPP is

    modified by inserting a new constraint called Gomorys or secondary constraint which represents

    necessary condition for integrability and eliminates some non integer solution without losing any

    integral solution. After adding the secondary constraint, the problem is then solved by dual

    simplex method to get an optimum integral solution. If all the values of the variables in this

    solution are integers, an optimum inter-solution is obtained, otherwise another new constrained is

    added to the modified L P P and the procedure is repeated. An optimum integer solution will be

  • 8/8/2019 MB0032 Operation Research Set1

    19/31

    reached eventually after introducing enough new constraints to eliminate all the superior non

    integer solutions. The construction of additional constraints, called secondary or Gomorys

    constraints, is so very important that it needs special attention.

    The iterative procedure for the solution of an all integer programming problem is as follows:

    Step 1: Convert the minimization I.P.P. into that of maximization, if it is in the minimization

    form. The integrality condition should be ignored.

    Step 2: Introduce the slack or surplus variables, wherever necessary to convert the inequations

    into equations and obtain the optimum solution of the given L.P.P. by using simplex algorithm.

    Step 3: Test the integrality of the optimum solution

    a) If the optimum solution contains all integer values, an optimum basic feasible integer solution has

    been obtained.

    b) If the optimum solution does not include all integer values then proceed onto next step.

    Step 4: Examine the constraint equations corresponding to the current optimum solution. Let

    these equations be represented by

    Where n denotes the number of variables and m the number of equations.

    Choose the largest fraction ofbis ie to find {bi}i

    Let it be [bk 1]

    or write is as

    f ko

  • 8/8/2019 MB0032 Operation Research Set1

    20/31

    Step 5: Express each of the negative fractions if any, in the k th row of the optimum simplex

    table as the sum of a negative integer and a nonnegative fraction.

    Step 6: Find the Gomorian constraint

    and add the equation

    to the current set of equation constraints.

    Step 7: Starting with this new set of equation constraints, find the new optimum solution by dual

    simplex algorithm. (So that Gsla (1) is the initial leaving basic variable).

    Step 8: If this new optimum solution for the modified L.P.P. is an integer solution. It is also

    feasible and optimum for the given I.P.P. otherwise return to step 4 and repeat the process until an

    optimum feasible integer solution is obtained.

  • 8/8/2019 MB0032 Operation Research Set1

    21/31

    ASSIGNMENTS

    Subject code: MB

    0032

    (3 or 4 credits)

    Set 2

    Marks 60

    SUBJECT NAME : OPERATIONS RESEARCH

    Note: Each Question carries 10 marks

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

    limitations of the Operations Research?

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

    1. Definition of the problem.

    2. Construction of the model.

    3. Solution of the model.

    4. Validation of the model.

    5. Implementation of the final result.

    1. 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, and (3) a recognition of the limitations, restrictions and requirements of the

    system.

    2. 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 problem and helps tackling it in a well

    organized manner. If the resulting model fits into one of the common mathematical models, a

    convenient solution may be obtained by using mathematical techniques. If the mathematical

    relationships of the model are too complex to allow analytic solutions, a simulation model may

    be

  • 8/8/2019 MB0032 Operation Research Set1

    22/31

    more appropriate. There are various types of models which can be constructed under different

    conditions.

    3. Solution of the model

    Once an appropriate model has been formulated, the next stage in the analysis calls for its

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

    model implies determination of a specific set of decision variables that would yield an Optimum

    solution. An Optimum solution is one which maximize or minimize the performance of any

    measure in a model subject to the conditions and constraints imposed on the model.

    4. Validation the model

    A model is a good representative of a system, then the Optimal solution must improve the

    systems performance. A common method for testing the 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. The problem here

    is that there is no assurance that future performance will continue to duplicate past behavior. Also,

    since the model is based on careful examination of past data, the comparison should always reveal

    favorable results. In some instances this problem may be overcome by using data from trial 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.

    5. Implementation of the final result

    The optimal solution obtained from a model should be applied practice 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 will reach its

    peak in this phase.

    Limitations of OR

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

    i) Magnitude of computation: Modern problem involve large number of variables and hence to find

    interrelationship, among makes it difficult.

    ii) Non - quantitative factors and Human emotional factor cannot be taken into account.

  • 8/8/2019 MB0032 Operation Research Set1

    23/31

    iii) There is a wide gap between the managers and the operation researches

    iv) Time and Money factors when the basic data is subjected to frequent changes then

    incorporation of them into OR models is a costly affair.

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

    Answer: The Standard Form Of LPP

    The characteristics of the standard form are:

    1. All constraints are equations except for the non negativity condition which remain inequalities (>,

    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 inequality constraints can be changed to equations by adding or subtracting the left hand side of

    each such constraints by a nonnegative variable. The nonnegative 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 substracted from a constraint inequality of the form to change it

    to an equation 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 by (-1) wherever necessary.

    The remaining characteristics are achieved by using the elementary transformations introduced with

    the canonical form.

    The Standard Form Of The LPP

    Any standard form of the L.P.P. is given by

  • 8/8/2019 MB0032 Operation Research Set1

    24/31

    Fundamental Theorem Of L.P.P.

    Given a set ofm 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 Method

    The drawback of the penalty cost method is the possible computational error 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

    the 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

    simplex method.

    Examples:

    7) Use the two phase method to

    Maximise z = 3x1 - x2

    Subject to 2x1 + x2 > 2

  • 8/8/2019 MB0032 Operation Research Set1

    25/31

    x1 + 3x2 < 2

    x2 < 4,

    x1, x2 > 0

    Rewriting in the standard form,

    Maximize z = 3x1 - x2 + 0S1 - MA1 + 0.S2 + 0.S3

    Subject to 2x1 + x2 - S1 + A1 = 2

    x1 + 3x2 + S2 = 2

    x2 + S3 = 4,

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

    Phase I :Consider the new objective,

    Maximize Z* = - A1

    Subject to 2x1 + x2 - S1 + A1 = 2

    x1 + 3x2 + S2 = 2

    x2 + S3 = 4,

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

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

    x1 enters the basic set replacing A1.

    The first iteration gives the following table:

  • 8/8/2019 MB0032 Operation Research Set1

    26/31

    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 + 0S3

    Subject to

    x2 + S3 = 4

    x1, x2, S1, S2, S3 > 0

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

    Proceeding to the next iteration, we get the following table:

  • 8/8/2019 MB0032 Operation Research Set1

    27/31

    Q.No.4: What do you understand by the transportation problem? What is thebasic assumption behind the transportation problem? Describe the

    MODI method of solving transportation problem.

    Answer: 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 the fictitious destination

    to all sources and from all destinations to the fictitious sources are assumed to be zero so that total cost

    of transportation will remain the same.

    The 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

  • 8/8/2019 MB0032 Operation Research Set1

    28/31

    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 rale 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.

    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 value to find ij = cij - ui - vj 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 are -ve and +ve. Examine the quantities

    allocated at -ve places. Select the minimum. Add it at +ve places and subtract from -ve place. Step 8:

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

  • 8/8/2019 MB0032 Operation Research Set1

    29/31

    Q.No.5: Describe the North-West Corner rule for finding the initial basic feasible

    solution in the transportation problem.

    Answer: North West Corner Rule

    Step1: The first assignment is made in the cell occupying the upper left hand (north west) corner

    of the transportation table. The maximum feasible amount is allocated there, that is x11 = min

    (a1,b1)

    So that either the capacity of origin O1 is used up or the requirement at destination D1 is satisfied or

    both. This value of x11 is entered in the upper left hand corner (small square) of cell (1, 1) in the

    transportation table

    Step 2: If b1 > a1 the capacity of origin O, is exhausted 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. Move down vertically to the second row and make the second allocation of

    magnitude

    x21 = min (a2, b1 - x21) in the cell (2,1). This either exhausts the capacity of origin O2 or

    satisfies the remaining demand at destination D1.

    If a1 > b1 the requirement at destination D1 is satisfied but the capacity of origin O1 is not

    completely exhausted. Move to the right 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 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. 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: 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 lower

    right corner of the transportation table until all the rim requirements are satisfied.

  • 8/8/2019 MB0032 Operation Research Set1

    30/31

    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

    Subject to the constraints

    xj is integer valued ,j = 1, 2, .., r (< n) ---- (3)

    xj > 0 . j = r + 1, .., n ----------(4)

    Further let us suppose that for each integer valued xj, we can assign lower and upper bounds for the

    optimum values of the variable by

    Lj xj Uj j = 1, 2, . r ------------- (5)

    The following idea is behind the branch and bound technique

    Consider any variable xj, and let I be some integer value satisfying Lj I Uj - 1. Then clearly an

    optimum solution to (1) through (5) shall also satisfy either the linear constraint.

    x j > I + 1 ------------- ( 6)

    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) forj = 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 optimal solution satisfying (3), and

    thus some more computations are necessary. We now discuss step wise the algorithm that

  • 8/8/2019 MB0032 Operation Research Set1

    31/31

    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 forz, say z (0) at 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 z (0) , we also have a list of L.P.Ps (to be called master list)

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