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- 1. Artificial Variable Technique (The Big-M Method) ATISH KHADSE
- 2. Big-M Method of solving LPPThe Big-M method of handling instances with artificialvariables is the “commonsense approach”. Essentially, thenotion is to make the artificial variables, through theircoefficients in the objective function, so costly or unprofitablethat any feasible solution to the real problem would bepreferred....unless the original instance possessed no feasiblesolutions at all. But this means that we need to assign, in theobjective function, coefficients to the artificial variables that areeither 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; wesimply associate a penalty value with variables that we do notwant to be part of an ultimate solution(unless such an outcomeIs unavoidable).
- 3. Indeed, the penalty is so costly that unless any of therespective 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 functionpoint of view. If the objective function (Z) is to be minimized,then a very large positive price (penalty, M) is assigned toeach artificial variable and if Z is to be minimized, then a verylarge negative price is to be assigned. The penalty will bedesignated by +M for minimization problem and by –M for amaximization problem and also M>0.
- 4. Example: Minimize Z= 600X1+500X2subject to constraints,2X1+ X2 >or= 80 X1+2X2 >or= 60 and X1,X2 >or= 0Step1: Convert the LP problem into a system oflinear equations.We do this by rewriting the constraint inequalities asequations by subtracting new “surplus & artificial variables"and assigning them zero & +M coefficientsrespectively in theobjective function as shown below.So the Objective Function would be:Z=600X1+500X2+0.S1+0.S2+MA1+MA2subject 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/PivotalCB Variab X1 X2 S1 S2 A1 A2 Col.) Soln(XB) le (B)M A1 80 2 1 -1 0 1 0 80M 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 willleave the basis. Hence 2 is the key element in pivotalcolumn. Now,the new row operations are as follows:R2(New) = R2(Old)/2R1(New) = R1(Old) - 1*R2(New) Cj 600 500 0 0 M Min.Ratio Basic Basic (XB/PivotaCB Variab X1 X2 S1 S2 A1 l Col.) Soln(XB) le (B) M A1 50 3 2 0 -1 1 2 1 100/3500 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/PCB X1 X2 S1 S2 ivotal ble Soln(XB) (B) Col.)600 X1 100/3 1 0 2 3 1 3500 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 positiveand 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.

its nice! and easy to understand