The Big-M method is used to handle artificial variables in linear programming problems. It assigns very large coefficients to the artificial variables in the objective function, making them undesirable to include in optimal solutions. This removes the artificial variables from the basis. As an example, the document presents a linear programming problem to minimize an objective function subject to constraints, and shows the steps of converting it to an equivalent problem using artificial variables and applying the Big-M method to arrive at an optimal solution without artificial variables.

1. Artificial Variable Technique
(The Big-M Method)
ATISH KHADSE

2. Big-M Method of solving LPP
The Big-M method of handling instances with artificial
variables is the “commonsense approach”. Essentially, the
notion is to make the artificial variables, through their
coefficients in the objective function, so costly or unprofitable
that any feasible solution to the real problem would be
preferred....unless the original instance possessed no feasible
solutions at all. But this means that we need to assign, in the
objective function, coefficients to the artificial variables that are
either very small (maximization problem) or very large
(minimization problem); whatever this value,let us call it Big M.
In fact, this notion is an old trick in optimization in general; we
simply associate a penalty value with variables that we do not
want to be part of an ultimate solution(unless such an outcome
Is unavoidable).

3. Indeed, the penalty is so costly that unless any of the
respective variables' inclusion is warranted algorithmically,
such variables will never be part of any feasible solution.
This method removes artificial variables from the basis. Here,
we assign a large undesirable (unacceptable penalty)
coefficients to artificial variables from the objective function
point of view. If the objective function (Z) is to be minimized,
then a very large positive price (penalty, M) is assigned to
each artificial variable and if Z is to be minimized, then a very
large negative price is to be assigned. The penalty will be
designated by +M for minimization problem and by –M for a
maximization problem and also M>0.

4. Example: Minimize Z= 600X1+500X2
subject to constraints,
2X1+ X2 >or= 80
X1+2X2 >or= 60 and X1,X2 >or= 0
Step1: Convert the LP problem into a system of
linear equations.
We do this by rewriting the constraint inequalities as
equations by subtracting new “surplus & artificial variables"
and assigning them zero & +M coefficientsrespectively in the
objective function as shown below.
So the Objective Function would be:
Z=600X1+500X2+0.S1+0.S2+MA1+MA2
subject to constraints,
2X1+ X2-S1+A1 = 80
X1+2X2-S2+A2 = 60
X1,X2,S1,S2,A1,A2 >or= 0

5. Step 2: Obtain a Basic Solution to the problem.
We do this by putting the decision variables X1=X2=S1=S2=0,
so that A1= 80 and A2=60.
These are the initial values of artificial variables.
Step 3: Form the Initial Tableau as shown.
Cj 600 500 0 0 M M
Min.Ratio
Basic
Basic (XB/Pivotal
CB Variab X1 X2 S1 S2 A1 A2 Col.)
Soln(XB)
le (B)
M A1 80 2 1 -1 0 1 0 80
M A2 60 1 2 0 -1 0 1 60
Zj 3M 3M M M M M
Cj - Zj 600-3M 500-3M M M 0 0

6. It is clear from the tableau that X2 will enter and A2 will
leave the basis. Hence 2 is the key element in pivotal
column. Now,the new row operations are as follows:
R2(New) = R2(Old)/2
R1(New) = R1(Old) - 1*R2(New)
Cj 600 500 0 0 M
Min.Ratio
Basic
Basic (XB/Pivota
CB Variab X1 X2 S1 S2 A1 l Col.)
Soln(XB)
le (B)
M A1 50 3 2 0 -1 1 2 1 100/3
500 X2 30 1 2 1 0 - 1/2 0 60
Zj 3M/2+250 500 M M/2-250 M
Cj - Zj 350-3M/2 0 M 250-M/2 0

7. It is clear from the tableau that X1 will enter and A1 will
leave the basis. Hence 2 is the key element in pivotal
column. Now,the new row operations are as follows:
R1(New) = R1(Old)*2/3
R2(New) = R2(Old) – (1/2)*R1(New)
Cj 600 500 0 0 Min.
Basic Ratio
Varia Basic (XB/P
CB X1 X2 S1 S2 ivotal
ble Soln(XB)
(B) Col.)
600 X1 100/3 1 0 2 3 1 3
500 X2 40/3 0 1 1 3 2 3
Zj 600 500 700 3 400 3
Cj - Zj 0 0 700 3 400 3

8. Since all the values of (Cj-Zj) are either zero or positive
and also both the artificial variables have been removed,
an optimum solution has been arrived at with X1=100/3 ,
X2=40/3 and Z=80,000/3.