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

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).

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.

Example: Minimize Z= 600X X 2 subject to constraints, 2X 1 + X 2 >or= 80 X 1 +2X 2 >or= 60 and X 1,X 2 >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=600X X 2 +0.S 1 +0.S 2 +MA 1 +MA 2 subject to constraints, 2X 1 + X 2 -S 1 +A 1 = 80 X 1 +2X 2 -S 2 +A 2 = 60 X 1,X 2,S 1,S 2,A 1,A 2 >or= 0

Step 2: Obtain a Basic Solution to the problem. We do this by putting the decision variables X 1 =X 2 =S 1 =S 2 =0, so that A 1 = 80 and A 2 =60. These are the initial values of artificial variables. Step 3: Form the Initial Tableau as shown.

It is clear from the tableau that X 2 will enter and A 2 will leave the basis. Hence 2 is the key element in pivotal column. Now,the new row operations are as follows: R 2 (New) = R 2 (Old)/2 R 1 (New) = R 1 (Old) - 1*R 2 (New)

It is clear from the tableau that X 1 will enter and A 1 will leave the basis. Hence 2 is the key element in pivotal column. Now,the new row operations are as follows: R 1 (New) = R 1 (Old)*2/3 R 2 (New) = R 2 (Old) – (1/2)*R 1 (New)

Since all the values of (C j -Z j ) are either zero or positive and also both the artificial variables have been removed, an optimum solution has been arrived at with X 1 =100/3, X 2 =40/3 and Z=80,000/3.