Big-M Method Presentation

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A helpful presentation to understand Big-M method by animations by Nitesh Singh Patel, IMT, Nagpur

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Big-M Method Presentation

  1. 1. BIG-M METHOD A VARIANT OF SIMPLEX METHOD PRESENTED BY: NISHIDH VILAS LAD-2013176 NITESH BERIWAL-2013177 NITESH SINGH PATEL-2013178 NITIN BORATWAR-2013179 NITIN KUMAR SHUKLA-2013180 NOOPUR MANDHYAN-2013181
  2. 2. INTRODUCTION WHAT IS BIG-M METHOD?  The Big M method is a method of solving linear programming problems.  It is a variation of the simplex method designed for solving problems typically encompassing "greater-than" constraints as well as "less-than" constraints - where the zero vector is not a feasible solution.  The "Big M" refers to a large number associated with the artificial variables, represented by the letter M.
  3. 3. DRAWBACKS  If optimal solution has any artificial variable with non-zero value, original problem is infeasible  Four drawbacks of BIG-M method:  How large should M be?  If M is too large, serious numerical difficulties in a computer  Big-M method is inferior than 2 phase method  Here feasibility is not known until optimality
  4. 4. Steps In The Big-M Method  Add artificial variables in the model to obtain a feasible solution.  Added only to the ‘>’ type or the ‘=‘ constraints  A value M is assigned to each artificial variable  The transformed problem is then solved using simplex eliminating the artificial variables
  5. 5. Important Points To Remember Solve the modified LPP by simplex method, until any one of the three cases may arise.  If no artificial variable appears in the basis and the optimality conditions are satisfied  If at least one artificial variable in the basis at zero level and the optimality condition is satisfied  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.
  6. 6. Big M Method: Example 1  Minimize Z = 40x1 + 24x2  Subject to 20x1 + 50x2 >= 4800 80x1 + 50x2 >= 7200 x1 , x2 >= 0
  7. 7. Introducing Surplus Variable and Artificial Variable to obtain an Initial Solution  Minimize Z = 40x1 + 24x2+ 0s1 + 0s2 + MA1 + MA1  Subject to 20x1 + 50x2 –S1 + A1 = 4800 80x1 + 50x2 –S2 + A2 = 7200 x1 , x2 >= 0 S1 and S2 Surplus Variable A1 and A2 Artificial Variable
  8. 8. Basic Variable s Cb Xb X1 X2 S1 S2 A1 A2 Cj -40 -24 0 0 -M -M A1 -M 4800 A2 -M 7200 20 50 -1 0 1 0 80 50 0 -1 0 1 Z∑ (Cb-Xb) = 12000M -100M+40 -100M+24 M M 0 0 ∆j= ∑CbXj-Cj Min Ratio = Xb/Xk 4800/50 7200/50 Maximize Z = -40x1 – 24x2 – 0S1 – 0S2 - MA1 – MA2
  9. 9. Basic Variable s Cb Xb X1 X2 S1 S2 A1 A2 Cj -40 -24 0 0 -M -M X2 -24 96 A2 -M 2400 60 0 1 -1 -1 1 Z∑ (Cb-Xb) = -2400M - 2304 152/5- 60M 0 12/25- M M -12/25+2M 0 ∆j= ∑CbXj-Cj Min Ratio = Xb/Xk 480/2 2400/60 2/5 1 -1/50 0 1/50 0 BV Cb Xb X1 X2 S1 S2 A1 A2 A1 -M 4800 20 50 -1 0 1 0 A2 -M 7200 80 50 0 -1 0 1
  10. 10. Basic Variable s Cb Xb X1 X2 S1 S2 A1 A2 Cj -40 -24 0 0 -M -M X2 -24 80 X1 -40 40 1 0 1/60 -1/60 -1/60 1/60 Z∑ (Cb-Xb) = -3520 0 0 -2/75 78/150 2/75+M 38/75 +M ∆j= ∑CbXj-Cj Min Ratio = Xb/Xk -6000/2 2400/1 0 1 -2/75 1/150 2/75 -1/150 BV Cb Xb X1 X2 S1 S2 A1 A2 X2 -24 96 2/5 1 -1/50 0 1/50 0 A2 -M 24 60 0 1 -1 -1 1
  11. 11. Basic Variable s Cb Xb X1 X2 S1 S2 A1 A2 Cj -40 -24 0 0 -M -M X2 -24 144 S1 0 2400 60 0 1 -1 -1 1 Z∑ (Cb-Xb) = 3456 8/5 0 0 12/25 M M- 12/25 ∆j= ∑CbXj-Cj Min Ratio = Xb/Xk _ _ 8/5 1 0 -1/50 0 1/50 THE VALUE OF OBJECTIVE FUNCTION IS 3456 BV Cb Xb X1 X2 S1 S2 A1 A2 X2 -24 80 0 1 -2/5 1/150 2/75 -1/150 X1 -40 40 1 0 1/60 -1/60 -1/60 1/60 Hence Optimal Solution is ACHIEVED.
  12. 12. BIG-M Method : An Example  Maximize Z = 2x1 + 4x2  Subject to 2x1 + x2 <= 18 3x1 + 2x2 >= 30 x1 + 2x2 = 26 x1 , x2 >= 0
  13. 13. Introduce Surplus , Slack and Artificial variables  Maximize Z = 2x1 + 4x2 + 0S1 + 0S2 – MA1 – MA2  Subject to 2x1 + x2 + S1 = 18 3x1 + 2x2 – S2 + A1 = 30 x1 + 2x2 + A2 = 26 x1 , x2 , S1 , S2 , A1 , A2 >= 0 S1 -> Slack Variable S2 -> Surplus Variable A1 , A2 -> Artificial Variable M -> Large value
  14. 14. Max: Z= 2x1 + 4x2 + 0S1 + 0S2 – MA1 – MA2 2x1 + x2 + S1 = 18 3x1 + 2x2 – S2 + A1 = 30 x1 + 2x2 + A2 = 26 x1 , x2 , S1 , S2 , A1 , A2 >= 0 Z∑ (Cb*Xb) = -56M Basic Variables Cb Xb X1 X2 S1 S2 A1 A2 S1 0 18 A1 -M 30 A2 -M 26 2 1 1 0 0 0 3 2 0 -1 1 0 1 2 0 0 0 1 Min Ratio = Xb/Xk 18/1 30/2 26/2 Cj 2 4 0 0 -M -M Calculation: ∑(Cb*Xb) = 0*18 + (-M*30) + (-M*26) = -56M ∆j= ∑CbXj-Cj = (0*2)+(-M*3)+(-M*1)-2= -4M-2 ∆j= ∑CbXj-Cj = (0*1)+(-M*2)+(-M*2)-4 = -4M-4 ∆j= ∑CbXj-Cj = (0*1)+(-M*0)+(-M*0)-0 = 0 ∆j= ∑CbXj-Cj = (0*0)+(-M*(-1))+(-M*0)-0 = M ∆j= ∑CbXj-Cj = (0*0)+(-M*1)+(-M*0)-(-M) = 0 ∆j= ∑CbXj-Cj = (0*0)+(-M*0)+(-M*1)-(-M) = 0 ∆j= ∑CbXj-Cj0M0-4M-4-4M-2 0
  15. 15. BV Cb Xb X1 X2 S1 S2 A1 A2 S1 0 18 2 1 1 0 0 0 A1 -M 30 3 2 0 -1 1 0 A2 -M 26 1 2 0 0 0 1 Cj 2 4 0 0 -M -M Basic Variables Cb Xb X1 X2 S1 S2 A1 A2 S1 A1 X2 0 -M 4 Replace A2 by X2. Divide the key row X2 by key element 2. Now operate row X2 & S1. i.e. S1- X2 13 ½ 1 0 0 0 ½ 5 3/2 0 1 0 0 1 Now operate A1 & X2 i.e. A1-2 X2 4 2 0 0 -1 1 -1 Z∑ (Cb*Xb) = 52-4M Calculation: ∑(Cb*Xb) = 0*5 + (-M*4) + 4*13 = 52-4M ∆j= ∑CbXj-Cj = (0*3/2)+(-M*2)+(4*1/2)-2= -2M ∆j= ∑CbXj-Cj = (0*0)+(-M*0)+(4*1)-4 = 0 ∆j= ∑CbXj-Cj = (0*1)+(-M*0)+(4*0)-0 = 0 ∆j= ∑CbXj-Cj = (0*0)+(-M*(-1))+(4*0)-0 = M ∆j= ∑CbXj-Cj = (0*0)+(-M*1)+(4*0)-(-M) = 0 ∆j= ∑CbXj-Cj = (0*1)+(-M*(-1))+(4*1/2)-(-M) = 2+2M -2M 0 0 M 0 2+2M Select the least negative element i.e. -2M, this column will be taken as Xk . Now select the min. ratio which is 2, corresponding key element will be 2, & the key row would be the A1. Min. Ratio= Xb/Xk 10/3 2 26
  16. 16. BV Cb Xb X1 X2 S1 S2 A1 A2 S1 0 5 3/2 0 1 0 0 -.5 A1 -M 4 2 0 0 -1 1 -1 A2 4 13 1/2 1 0 0 0 .5 Cj 2 4 0 0 -M -M Basic Variables Cb Xb X1 X2 S1 S2 A1 A2 S1 X1 X2 0 2 4 Replace A1 by X1. Divide the key row X1 by key element 2 Now operate row X1 & S1. i.e. S1- 1.5 X1 2 0 0 1 ¾ -3/4 ¼ Now operate X1 & X2 i.e. X2-0.5 X1 Z∑ (Cb*Xb) = 52 Calculation: ∑(Cb*Xb) = 0*2 + (2*2) + 4*12 = 52 ∆j= ∑CbXj-Cj = (0*0)+(2*1)+(4*0)-2= 0 ∆j= ∑CbXj-Cj = (0*0)+(2*0)+(4*1)-4 = 0 ∆j= ∑CbXj-Cj = (0*1)+(2*0)+(4*0)-0 = 0 ∆j= ∑CbXj-Cj = (0*3/4)+(2*(-1/2))+(4*1/4)-0 = 0 ∆j= ∑CbXj-Cj = (0*-3/4)+(2*1/2)+(4*(-1/4))-(-M) = M ∆j= ∑CbXj-Cj = (0*1/4)+(2*(-1/2))+(4*3/4)-(-M) = 2+M 0 0 0 0 M 2+M 2 1 0 0 -1/2 ½ -1/2 12 0 1 0 ¼ -1/4 3/4 The optimum solution to the problem is X1=2, X2=12, S1=2 & other variable is 0. The objective function value is 52.

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