1. Operations Research
(This is material is not full-fledged material. This covers only few topics)
Procedure of Simplex Method
The steps for the computation of an optimum solution are as follows:
Step-1: Check whether the objective function of the given L.P.P is to be maximized or
minimized. If it is to be minimized then we convert it into a problem of maximizing it by using
the result
Minimum Z = - Maximum(-z)
Step-2: Check whether all right hand side values of the constrains are non-negative. If any one
of values is negative then multiply the corresponding inequation of the constraints by -1,
so as to get all values are non-negative.
Step-3: Convert all the inequations of the constraints into equations by introducing
slack/surplus variables in the constraints. Put the costs of these variables equal to zero.
Step-4: Obtain an initial basic feasible solution to the problem and put it in the first column of
the simplex table.
Step-5: Compute the net evolutions Δj= Zj – Cj (j=1,2,…..n) by using the relation Zj – Cj = CB Xj
– Cj .
Examine the sign
(i) If all net evolutions are non negative, then the initial basic feasible
solution is an optimum solution.
(ii) If at least one net evolution is negative, proceed on to the next step.
Step-6: If there are more than one negative net evolutions, then choose the most
negative of them. The corresponding column is called entering column.
(i) If all values in this column are ≤ 0, then there is an unbounded solution to the
given problem.
(ii) If at least one value is > 0, then the corresponding variable enters the basis.
Step-7:Compute the ratio {XB / Entering column} and choose the minimum of these ratios.
The row which is corresponding to this minimum ratio is called leaving row. The common
element which is in both entering column and leaving row is known as the leading element
or key element or pivotal element of the table.
Step-8:Convert the key element to unity by dividing its row by the leading element itself
and all other elements in its column to zeros by using elementary row transformations.
Step-9: Go to step-5 and repeat the computational procedure until either an
optimum solution is obtained or there is an indication of an unbounded solution.

2. Artificial Variable Technique
– Big M-method
If in a starting simplex table, we don’t have an identity sub matrix (i.e. an obvious starting BFS), then we
introduce artificial variables to have a starting BFS. This is known as artificial variable technique. There is
one method to find the starting BFS and solve the problem i.e., Big M method.
Suppose a constraint equation i does not have a slack variable. i.e. there is no ith unit vector column in
the LHS of the constraint equations. (This happens for example when the ith constraint in the original
LPP is either ≥ or = .) Then we augment the equation with an artificial variable Ai to form the ith unit
vector column. However as the artificial variable is extraneous to the given LPP, we use a feedback
mechanism in which the optimization process automatically attempts to force these variables to zero
level. This is achieved by giving a large penalty to the coefficient of the artificial variable in the objective
function as follows:
Artificial variable objective coefficient
= - M in a maximization problem,
= M in a minimization problem
where M is a very large positive number.
Procedure of Big M-method
The following steps are involved in solving an LPP using the Big M method.
Step-1: Express the problem in the standard form.
Step-2:Add non-negative artificial variables to the left side of each of the equations corresponding to
constraints of the type ≥ or =. However, addition of these artificial variable causes violation of the
corresponding constraints. Therefore, we would like to get rid of these variables and would not allow
them to appear in the final solution. This is achieved by assigning a very large penalty (-M for
maximization and M for minimization) in the objective function.
Step-3:Solve the modified LPP by simplex method, until any one of the three cases may arise.
1. If no artificial variable appears in the basis and the optimality conditions are satisfied, then the
current solution is an optimal basic feasible solution.

3. 2. If at least one artificial variable in the basis at zero level and the optimality condition is satisfied
then the current solution is an optimal basic feasible solution.
3. If at least one artificial variable appears in the basis at positive level and the optimality condition
is satisfied, then the original problem has no feasible solution. The solution satisfies the contains
but does not optimize the objective function, since it contains a very large penalty M and is
called pseudo optimal solution.
Artificial Variable Technique
– Big M-method
Consider the LPP:
Minimize Z = 2 x1 + x2
Subject to the constraints
3 x 1 + x2 ≥ 9
x1 + x2 ≥ 6
x1, x2 ≥ 0
Putting this in the standard form, the LPP is:
Minimize Z = 2 x1 + x2
Subject to the constraints
3 x 1 + x 2 – s1 =9
x1 + x2 – s2 = 6
x1, x2 ,s1 , s2 ≥ 0
Here s1 , s2 are surplus variables.
Note that we do not have a 2x2 identity sub matrix in the LHS.
Introducing the artificial variables A1, A2 in the above LPP
The modified LPP is as follows:
Minimize Z = 2 x1 + x2 + 0. s1 + 0. s2 + M.A1 + M.A2
Subject to the constraints
3 x 1 + x 2 – s1 + A1 = 9

4. x1 + x2 – s2 + A2 = 6
x1, x2 , s1 , s2 , A1 , A2 ≥ 0
Note that we now have a 2x2 identity sub matrix in the coefficient matrix of the constraint equations.
Now we can solve the above LPP by the Simplex method.
But the above objective function is not in maximization form. Convert it into maximization form.
Max Z = -2 x1 – x2 + 0. s1 + 0. s2 – M A1 – M A2
Cj: -2 -2 0 0 -M -M
B.V CB XB X1 X2 S1 S2 A1 A2 MR
XB/X1
A1 -M 9 1 -1 0 1 0 3
-M -1
A2 6 1 1 0 0 1 6
Zj -15M -4M -2M M M -M -M
Δj -4M+2 -2M+1 M M 0 0
B.V CB XB X1 X2 S1 S2 A1 A2 MR
XB/X1
A1 -M 9 1 -1 0 1 0 3
-M -1
A2 6 1 1 0 0 1 6
Zj -15M -4M -2M M M -M -M
Δj -4M+2 -2M+1 M M 0 0

5. TRANSPORTATION PROBLEM
Transportation problem is one of the subclasses of LPP’s in which the objective is to
transport various quantities of a single homogeneous commodity, that are initially stored at
various origins to different destinations in such a way that the total transportation cost is
minimum. To achieve this objective we must know the amount and location of available
supplies and the quantities demanded. In addition, we must know the costs that result from
transporting one unit of commodity from various origins to various destinations.
Mathematical Formulation of the Transportation Problem
A transportation problem can be stated mathematically as a Linear Programming Problem as
below:
Minimize Z =
subject to the constraints
= ai , i = 1, 2,…..,m
= bj , j = 1, 2,…..,m
xij ≥ 0 for all i and j
Where, ai = quantity of commodity available at origin i
bj= quantity of commodity demanded at destination j
cij= cost of transporting one unit of commodity from ith origin to jth
destination
xij = quantity transported from ith origin to jth destination
Tabular form of the Transportation Problem
To D1 D2 ……. Dn Supply
From
O1 c11 c12 ……. c1n a1
O2 c21 c22 ……. c2n a2
. . . ……. . .
. . . . .
. . . . .
Om cm1 cm2 ……. cmn am
Demand b1 b2 ……. bn

6. NORTH - WEST CORNER RULE
Step1:Identify the cell at North-West corner of the transportation matrix.
Step2:Allocate as many units as possible to that cell without exceeding supply or
demand; then cross out the row or column that is exhausted by this assignment
Step3:Reduce the amount of corresponding supply or demand which is more by allocated
amount.
Step4:Again identify the North-West corner cell of reduced transportation matrix.
Step5:Repeat Step2 and Step3 until all the rim requirements are satisfied.
Vogel’s Approximation Method (VAM)
Step-I: Compute the penalty values for each row and each column. The penalty will be
equal to the difference between the two smallest shipping costs in the row or column.
Step-II: Identify the row or column with the largest penalty. Find the first basic variable which
has the smallest shipping cost in that row or column. Then assign the highest
possible value to that variable, and cross-out the row or column which is
exhausted.
Step-III: Compute new penalties and repeat the same procedure until all the rim
requirements are satisfied.
An example for Vogel’s Method
Find the IBFS of the following transportation problem by using Penalty Method.
D1 D2 D3
Supply
10
6 7 8
15
15 80 78
15 5 5
Step 1: Compute the penalties in each row and each column .
Supply Row Penalty
10 7-6=1
6 7 8
15 78-15=63
15 80 78
Demand 15 5 5
Column Penalty 15-6=9 80-7=73 78-8=70

7. Step 2: Identify the largest penalty and choose least cost cell to corresponding this penalty
Supply Row Penalty
10 7-6=1
6 7 8
15 78-15=63
15 80 78
Demand 15 5 5
Column Penalty 15-6=9 80-7=73 78-8=70
Step-3: Allocate the amount 5 which is minimum of corresponding row supply and column
demand and then cross out column2
Supply Row Penalty
5
10 7-6=1
6 7 8
15 78-15=63
15 80 78
Demand 15 5 5
Column Penalty 15-6=9 80-7=73 78-8=70
Step-4: Recalculate the penalties
Supply Row Penalty
5
5 8-6=2
6 7 8
15 78-15=63
15 80 78
Demand 15 X 5
Column Penalty 15-6=9 78-8=70

8. Step-5: Identify the largest penalty and choose least cost cell to corresponding this penalty
Supply Row Penalty
5
5 8-6=2
6 7 8
15 78-15=63
15 80 78
Demand 15 X 5
Column Penalty 15-6=9 78-8=70
Step-6: Allocate the amount 5 which is minimum of corresponding row supply and column
demand, then cross out column3
Supply Row Penalty
5 5
5 8-6=2
6 7 8
15 78-15=63
15 80 78
Demand 15 X X
Column Penalty 15-6=9
Step-7: Finally allocate the values 0 and 15 to corresponding cells and cross out column 1
D1 D2 D3 Supply
0 5 5
X
O1 6 7 8
15
X
O2 15 80 78
Demand X X X
Solution of the problem
Now the Initial Basic Feasible Solution of the transportation problem is
X11=0, X12=5, X13=5, and X21=15 and
Total transportation cost = (0x6)+(5x7)+(5x8)+(15x15)
= 0+35+40+225
= 300.