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									MBA SEMESTER II                                                     REGISTRATION NO.:


                                       Subject code: MB0032

                                               Set 1


Q1. Describe in details the different scopes of application of Operations Research.

Ans.: Scope of Operations Research (OR)

In general, whenever there is any problem simple or complicated, the OR techniques may be applied
to find the best solution. In this section we shall try to find the scope of OR by seeing its application
in various fields of everyday life.

i) In Defense Operations: In modern warfare the defense operations are carried out by a number
of independent components namely Air Force, Army and Navy. The activities in each of these
components can be further divided in four subcomponents

viz.: administration, intelligence, operations and training, and supply. The application of modern
warfare techniques in each of the components of military organizations requires expertise knowledge
in respective fields. Furthermore, each component works to drive maximum gains from its
operations and there is always a possibility that strategy beneficial to one component may have an
adverse effect on the other. Thus in defense operations there is a necessity to coordinate the
activities of various components which gives maximum benefit to the organization as a whole,
having maximum use of the individual components. The final strategy is formulated by a team of
scientists drawn from various disciplines who study the strategies of different components and after
appropriate analysis of the various courses of actions, the best course of action, known as optimum
strategy, is chosen.

ii) In Industry: the systems of modern industries are so complex that the optimum point of
operation in its various components cannot be intuitively judged by an individual. The business
environment is always changing and any decision useful at one time may not be so good some time
later. There is always a need to check the validity of decisions continually, against the situations. The
industrial revolution with increased division of labor and introduction of management
responsibilities has made each component an independent unit having their own goals. For example:
Production department minimize cost of production but maximizes output. Marketing department
maximizes output but minimizes cost of unit sales.

Finance department tries to optimize capital investment and personnel department appoints good
people at minimum cost. Thus each department plan their own objectives and all these objectives of
various department or components come to conflict with each other and may not conform to the
overall objectives of the organization. The application of OR techniques helps in overcoming this
difficulty by integrating the diversified activities of various components so as to serve the interest of

Operations Research (Subject Code: MB0032) Set 1                                                       1
MBA SEMESTER II                                                           REGISTRATION NO.:

the organization as a whole efficiently. OR methods in industry can be applied in the fields of
production, inventory controls and marketing, purchasing, transportation and competitive strategies

iii) Planning: In modern times it has become necessary for every government to have careful
planning, for economic development of the country. OR techniques can be fruitfully applied to
maximize the per capita income, with minimum sacrifice and time. A government can thus use OR
for framing future economic and social policies.

iv) Agriculture: With increase in population there is a need to increase agriculture output. But this
cannot be done arbitrarily. There are a number of restrictions under which agricultural production is
to be studied. Therefore there is a need to determine a course of action, which serves the best under
the given restrictions. The problem can be solved by the application of OR techniques.

v) In Hospitals: The OR methods can be used to solve waiting problems in outpatient department
of big hospitals. The administrative problems of hospital organization can also be solved by OR

vi) In Transport: Different OR methods can be applied to regulate the arrival of trains and
processing times, minimize the passengers waiting time and reduce congestion, formulate suitable
transportation policy, reducing the costs and time of transshipment.

vii) Research and Development: Control of R and D projects, product introduction planning etc.
and many more applications.

Q2. 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.?

Ans.: 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:

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 ³ 0, x2 ³ 0, -------------------- xn³ 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 £
(Less than), ³ (greater than) or = (equal). Note that, in terms of the above formulation the coefficient
cj, aij, bj are interpreted physically as follows. If bi is the available amount of resources

Operations Research (Subject Code: MB0032) Set 1                                                      2
MBA SEMESTER II                                                                        REGISTRATION NO.:

i, where aij is the amount of resource i, that must be allocated to each unit of activity j, 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:
Maximize 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
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 ³) may be changed to an inequality in the opposite direction (³
or £) by multiplying both sides of the inequality by –1.
For example 2x1+3x2 ³ 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 ³ 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 ³ – 5
or – 2x1 – 3x2 £ 5.

5. The variable which is unconstrained in sign (i.e., ³ 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 + ³ 0, x – £ 0.

Examples of a Linear Programming Problem:

Example 1: A firm engaged in producing 2 models, viz., Model A and Model B, performs only 3
operations painting, assembly and testing. The relevant data are as follows:

Operations Research (Subject Code: MB0032) Set 1                                                           3
MBA SEMESTER II                                                    REGISTRATION NO.:

Unit Sale Price                                      Hours required for each unit
                                                     Assembly         Painting             Testing
Model A Rs. 50.00                                      1.0 0.2                  0.0
Model B Rs. 80.00                                       1.5            0.2                   0.1

Total number of hours available each week is as under assembly 600, painting 100, and testing 30.
The firm wishes to determine the weekly product mix so as to maximize revenue.
Solution: Let us first write the notations as under:
Z: Total revenue
X1: Number of Units of Model A
X2: Number of Units of Model B
X1, X2: Are known as decision variables
b1: Weekly hours available for assembly
b2: Weekly hours available for painting
b3: Weekly hours available for testing.

Since the objective (goal) of the firm is to maximize its revenue, the model can be stated as follows:

The objective function, Z = 50x1 + 80x2 is to be maximized subject to the constraints

1.0 x1+1.5x2 £ 600, (Assembly constraints)
0.2 x1+0.2x2 £ 100, (Painting constants)
0.0 x1+0.1x2 £ 30, (Testing constraints) and
x1 ³ 0, x2 ³ 0, The Non negativity conditions.

Requirements of L.P.P

i. Decisions variables and their relationship
ii. Well defined objective function
iii. Existence of alternative courses of action
iv. Nonnegative conditions on decision variables.

Basic assumptions of L.P.P

1. Linearity: Both objective function and constraints must be expressed as linear inequalities.
2. Deterministic: All coefficients of decision variables in the objective and constraints expressions
should be known and finite.
3. Additivity: The value of objective function for the given values of decision variables and the total
sum of resources used must be equal to sum of the contributions earned from each decision variable
and the sum of resources used by decision variables respectively.
4. Divisibility: The solution of decision variables and resources can be any nonnegative values
including fractions.

Operations Research (Subject Code: MB0032) Set 1                                                         4
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Q3. Describe the different steps needed to solve a problem by simplex method.

Ans.: Simplex Method

Consider a LPP given in the standard form, to optimize z = c1 x1 + c2 x2 + ---+ cn xn Subject to
a11 x1 + a12 x2 + -- + an x n          S1 = b1 a21 x1 + a22 x2 + ----+ a2n xn≥ S2 = b2
am1 x1 + am2 x2 + -- + amn xn Sm = bm x1, x2, --- xn, S1, S2 ---, Sm ≥ 0.

To each of the constraint equations add a new variable called an artificial variable on the left hand
side of every equation which does not contain a slack variable. Then every constraint equation
contains either a slack variable or an artificial variable. The introduction of slack and surplus
variables does not alter either the constraints or the objective function. So such variables can be
incorporated in the objective function with zero coefficients. However, the artificial variables do
change the constraints, since these are added only to one side i.e., to the left hand side of the
equations. The new constraint equations so obtained is equivalent to the original equations if and
only if all artificial variables have value zero. To guarantee such assignments in the optimal solutions,
artificial variables are incorporated into the objective function with very large positive coefficient M
in the minimization program and very large negative coefficient – M in the maximization program.
These coefficients represent the penalty incurred in making a unit assignment to the artificial
variable. Thus the standard form of LPP can be given as follows: Optimize Z = CT X Subject to
AX = B, and X 0 Where X is a column vector with decision, slack, surplus and artificial variables,
C is the vector corresponding to the costs, A is the coefficient matrix of the constraint equations
and B is the column vector of the right hand side of the constraint equations.

To solve problem by Simplex Method

    1. Introduce stack variables (Si’s) for “ ” type of constraint.
    2. Introduce surplus variables (Si’s) and Artificial Variables (Ai) for “ ” type of constraint.
    1. Introduce only Artificial variable for “=” type of constraint.
    2. Cost (Cj) of slack and surplus variables will be zero and that of Artificial variable will be “M”
    3. Find Zj - Cj for each variable.
    4. 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.
    5. Zj = sum of [cost of variable x its coefficients in the constraints – Profit or cost coefficient
        of the variable].
    6. 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.

Operations Research (Subject Code: MB0032) Set 1                                                       5
MBA SEMESTER II                                                   REGISTRATION NO.:

   7. 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.
   8. The element that lies both on key column and key row is called Pivotal element.
   9. Ratios with negative and “ ” value are not considered for determining key row.
   10. Once an artificial variable is removed as basic variable, its column will be deleted from next
   11. 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.
   12. Values of artificial variables will always is – M for both maximization and minimization
   13. The process is continued till all Zj - Cj ≤ 0.

Q4. Describe the economic importance of the Duality concept.

Ans.: Economic importance of duality

The linear programming problem can be thought of as a resource allocation model in which the
objective is to maximize revenue or profit subject to limited resources. Looking at the problem from
this point of view, the associated dual problem offers interesting economic interpretations of the L.P
resource allocation model. We consider here a representation of the general primal and dual
problems in which the primal takes the role of a resource allocation model.

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.

Economic importance of Dual variables:

For any pair of feasible primal and dual solutions, (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

Operations Research (Subject Code: MB0032) Set 1                                                    6
MBA SEMESTER II                                                     REGISTRATION NO.:

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.

Q5. How can you use the Matrix Minimum method to find the initial basic feasible solution
in the transportation problem?

Ans.: 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 goes to step

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.

Example 2: Obtain an initial basic feasible solution to the following T.P. using the matrix minima

             D1        D2        D3               D4
      O1      1         2            3            4
    8 8             Capacity
      O2      4         3            2            0
                                                      1 10
              0         2            2            1

              4         6            8        6         24


Operations Research (Subject Code: MB0032) Set 1                                                       7
MBA SEMESTER II                                                      REGISTRATION NO.:

Where 0i and Di denote ith origin and jth destination respectively.

Solution: The transportation table of the given T.P. has 12 cells. Following the matrix minima
method. The first allocation is made in the cells (3, 1) the magnitude being x31 = 4. This satisfies the
requirement at destination D1 and thus we cross off the first column from the table. The second
allocation is made in the cell (2, 4) magnitude x24 = min (6, 8) =6. Cross off the fourth column of
the table. This yields the table (i)

We choose arbitrarily again to make the next allocation in cell (2,3) of magnitude x23 = min (2,8) =
2 cross off the second row this gives table (iii). The last allocation of magnitude x23 = min (6 , 6) =
6 is made in the cell (3,3). Now all the rim requirements have been satisfied and hence an initial
feasible solution has been determined. This solution is shown in table ( iv) Since the cells do not
form a loop, the solution is basic one. Moreover the solution is degenerate also. The transportation
cost according to the above route is given by


Q6. Describe the Integer Programming Problem. Describe the Gomory’s All-I.P.P. method
for solving the I.P.P. problem.

Ans.: All And Mixed I P P

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 ³ 0 j = 1, 2, … ,n where xj
being an integral value for j = 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.

Gomory’s all – IPP Method

Operations Research (Subject Code: MB0032) Set 1                                                        8
MBA SEMESTER II                                                      REGISTRATION NO.:

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 Gomory’s or secondary constraint which represents necessary
condition for inerrability 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 reached eventually after
introducing enough new constraints to eliminate all the superior non integer solutions. The
construction of additional constraints, called secondary or Gomory’s constraints is so very important
that it needs special attention.

Construction of Gomory’s Constraints
Consider an L P P for which an optimum non – integer basic feasible solution has been attained.
With usual notations, let this solution be displayed in the following simplex table.

YB        XB                 Y1                 Y2                      Y3                     Y4
Y2        Y10                Y11                Y12                     Y13                    Y14
Y3        Y20                Y21                Y22                     Y23                    Y24
          Y00                Y01                Y02                     Y03                    Y04

Clearly the optimum basic feasible solution is given by xB = [ ] 2 3 x, x=[ , ] 10 20 y y ; Max z = y00
since xB is a non integer solution. We assume that y10 is fractional.
The constraint equation is

y10 = y11 x1 + y12 x2 + y13 x3+ y14 x4 ( 1 4 0 ) x = x = reduces to
y10 = y11 x1 + x2 + y14 x4 (0) 1 4 x = x = (

     1) because x2 and x3 are basic variables ( which implies that y12 = 1 and y13 = 0 )
The above equation can be rewritten as
x2 = y10 y11
x1 y14
x4 Which is a linear combination of non basic variables.
Now, since y10 ³ 0 the fractional part of y10 must also be non negative. We split over each of yij in
(1) Into an integral part Iij , and a non negative fractional part, f1j for j = 0,1,2,3,4. After this break
up (1) may be written as
I10 + f10 = (I11 + f11) x2 + (I14 + f14) x4
f10 f11
x2 f14
x4 = x2 + I11 x1 + I14x4 I10
Comparing (1) and (2) we come to know that if we add an additional constraint in such a way that
the L.H.S. of (2) is an integer, then we shall be forcing the non integer
y10 towards an integer.
This is what is needed.

Operations Research (Subject Code: MB0032) Set 1                                                         9
MBA SEMESTER II                                                   REGISTRATION NO.:

The desired Gomory’s constraint is f10 – f11 x1 – f11x4 ≤ 0.
Let it be possible to have f10 – f11 x1 – f11 x4 = h where h > 0 is an integer. Then f10 = h + f11 x1
f14 x4 is greater than one. This contradicts that 0 < fij < 1 for j = 0, 1, 2, 3, 4.
Thus Gomory’s constraint is.

When Gsla (1) is slack variable in the above first Gomory constraint.

This additional constraint is to be included in the given L.P.P. in order to move further towards
Obtaining an optimum all integer solution. After the addition of this constraint, the optimum
Simplex table looks like as given below

    yB                      xB                         y1    y2     y3      y4     Gsla(1)

y1                         y10                        y11   y12     y13 y14       0
y2                        y20                          y21 y22       y23  y24     0
Gsla(1)                   – f10                       – f11 0         0    – f14   1
                          y00                           y01 y02       y03 y04    y05

Since – f10 is negative. The optimal solution is infeasible and thus the dual simplex method is to be
applied for obtaining an optimum feasible solution. After obtaining this solution, the above referred
procedure is applied for constructing second Gomory’s constraint. The process is to be continued
so long as an all – integer solution has not been obtained.

Operations Research (Subject Code: MB0032) Set 1                                                    10

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