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SMU (SIKKIM MANIPAL UNIVERSITY)-SOLVED ASSIGNMENT MB0032 SET-II (MBA SEM-II)

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SMU (SIKKIM MANIPAL UNIVERSITY)-SOLVED ASSIGNMENT MB0032 SET-II (MBA SEM-II) Powered By Docstoc
					MBA SEMESTER II                                                  REGISTRATION NO.:




                                          ASSIGNMENTS

                                        Subject code: MB0032

                                               Set 2

                       SUBJECT NAME: OPERATIONS RESEARCH



Q1. What are the important features of Operations Research? Describe in details the
different phases of Operations Research.

Ans.: Important features of OR are as follows :


   1. It is System oriented: OR studies the problem from overall point of view of organizations or
      situations since optimum result of one part of the system may not be optimum for some
      other part.

   2. 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.

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

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

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

   6. It gives Quantitative solution.

   7. It considers the human factors also.

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.

Operations Research (Subject Code: MB0032) Set 2                                                 1
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    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 reestablished
       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.

Q2. Describe a Linear Programming Problem in details in canonical form.

Ans.: 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:
Maximize Z = c1 x1+c2 x2 +------------- +cn xn
Subject to
a11 x1 + a12 x2 +------------- +a1n xn                      b1
a21 x1 + a22 x2 +------------- +a2n xn                      b2
--------------------------------------------------------------------------
---------------------------------------------------------------------------
am1 x1+am2 x2 + …… + amn xn bm
x1, x2, x3, … xn ³ 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 canonical 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 = c1 x1 + c2x2 + ……. + cn xn is equivalent to
Maximize – Z = – c1x1 – c2x2 – … – cnxn.

2. Any inequality in one direction ( or ) may be changed to an inequality in the opposite direction
( or ) by multiplying both sides of the inequality by –1.
For example 2x1+3x2 5 is equivalent to –2x1–3x2 –5


Operations Research (Subject Code: MB0032) Set 2                                                  2
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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+3x25 and 2x1+3x2 5 and
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.


Q3. What are the different steps needed to solve a system of equations by the simplex
method?

Ans.: Different Steps needed to solve a system of equations

   1) Introduce stack variables (Si’s) for “   ” type of constraint.

   2) Introduce surplus variables (Si’s) 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”
   5) Find Zj - Cj for each variable.
   6) 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.
   7) Zj = sum of [cost of variable x its coefficients in the constraints – Profit or cost coefficient
       of the variable].
   8) 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.
   9) 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.
   10) The element that lies both on key column and key row is called Pivotal element.
   11) Ratios with negative and “    ” value are not considered for determining key row.
   12) Once an artificial variable is removed as basic variable, its column will be deleted from next
       iteration.




Operations Research (Subject Code: MB0032) Set 2                                                        3
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    13) For maximization 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 functions.
    14) Values of artificial variables will always is – M for both maximization and minimization
        problems.
    15) The process is continued till all Zj - Cj   0.


Q4. 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.

Ans.: 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 objective is to develop an integral transportation schedule that meets all demands from the
inventory at a minimum total transportation cost.

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

              m n
Minimize z = ∑ ∑Cij Xij
             i=1 j=1

Subject to

 n
∑ Xij= ai; i=1, 2…... m
j=1
                                                            ( 2)
 m
∑Xij=bj=1, 2…………., n
i=1


With all xij ≥0 and integrals


Theorem: A necessary and sufficient condition for the existence of a feasible solution to the
transportation problem (2) is that



Operations Research (Subject Code: MB0032) Set 2                                                  4
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 m          n
∑ ai = ∑ bj
i=1     j=1

Basic Assumption behind Transportation Problem:

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.



MODI Method of Solving Transportation Problem:

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      specialized      to      the      format      of      table       it     involves:

a) Finding an integral basic feasible solution

b) Testing the solution for optimality

c) Improving the solution, when it is not optimal

d) Repeating steps (1) and (2) until the optimal solution is obtained


The solution of 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) Vogel’s approximation method. In the second stage we test the B.Fs for its optimality
either by MODI method or by stepping stone method.

Steps of MODI 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.



Operations Research (Subject Code: MB0032) Set 2                                                       5
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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 Δij = cij - ui - vj of all Δij ≥, then it is the optimal solution.

Step 7: If any Δij ≤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 Δij ≥ 0

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

Ans.: 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.




Operations Research (Subject Code: MB0032) Set 2                                                      6
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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.

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

Ans.: 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
                  n
Maximize      z = ∑ c j x j……………………………………………. (1)
                   j=1

Subject to the constraints

∑ aij xj≤bi                          i= 1, 2 , ...., m…………………(2)

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 ³ 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

Operations Research (Subject Code: MB0032) Set 2                                             7
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= 1.66 (for example). Then we formulate and solve two L.P.P’s 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 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 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.P’s (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).




Operations Research (Subject Code: MB0032) Set 2                                                  8

				
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