Operations Research - The Big M Method
- 1. Operations Research CHAPTER 06 - THE BIG M METHOD
- 2. The Big M Method I. Multiply the inequality constraints to ensure that the right hand side is positive. II. For any greater-than or equal constraints, introduce surplus and artificial variables. III. Choose a large positive M and introduce a term in the objective of the form M multiplying the artificial variables. IV. For less-than constraints, introduce slack variables so that all constraints are equalities. V. Solve the problem using the usual simplex method.
- 3. Example Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 <= 11 -4x1+ x2 + 2x3 >= 3 2x1 – x3 = -1
- 4. Solution Write the problem in standard form and let the right hand side positive. Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 + x4 = 11, Slack -4x1 + x2 +2x3 – x5 = 3, Surplus -2x1 + x3 = 1
- 5. Solution Write the problem in canonical form (add artificial variable to the 2nd and to the 3rd constraints). Add M to the artificial variables in the objective function In case the problem is a maximization problem, add –M as a coefficient to the artificial variables, Otherwise add M. Min Z= -3x1 + x2 + x3 + Mx6 + Mx7 S.T. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 – x5 + x6 = 3 -2x1 + x3 + x7 = 1
- 6. Solution Iteration 1: Basic: x4=11, x6=3, x7=1. Z=Mx6+Mx7=3M+M=4M -3 1 1 0 0 M M x1 x2 x3 x4 x5 x6 x7 0 x4 1 -2 1 1 0 0 0 11 M x6 -4 1 2 0 -1 1 0 3 M x7 -2 0 1 0 0 0 1 1 CJ 0 0 0
- 7. Solution C1= -3 – (0,M,M) = 6M-3 >0 C2= 1 – (0,M,M) = 1-M <0 C3= 1 – (0,M,M) = 1-3M <0 C5= 0 – (0,M,M) = M >0 x3 is entering, x3= min{11,3/2,1}=1, x7 is leaving. 1 -4 -2 -2 1 0 1 2 1 0 -1 0
- 8. Solution Iteration 2: Basic: x4=11-1=10, x6=3-2=1, x3=1. Z=1+M -3 1 1 0 0 M M x1 x2 x3 x4 x5 x6 x7 0 x4 3 -2 0 1 0 0 -1 10 M x6 0 1 0 0 -1 1 -2 1 1 x3 -2 0 1 0 0 0 1 1 CJ 0 0 0 0
- 9. Solution C1= -3 – (0,M,1) = -1 <0 C2= 1 – (0,M,1) = 1-M <0 C5= 0 – (0,M,1) = M >0 x2 is entering, x2= min{-5,1,inf.}=1, x6 is leaving. Basic: x4=12, x2=1, x3=1. Z=2. 3 0 -2 -2 1 0 0 -1 0
- 10. Solution Iteration 3: Basic: x4=12, x2=1, x3=1. Z=2 -3 1 1 0 0 M M x1 x2 x3 x4 x5 x6 x7 0 x4 3 0 0 1 -2 2 -5 12 1 x2 0 1 0 0 -1 1 -2 1 1 x3 -2 0 1 0 0 0 1 1 CJ 0 0 0 0 0
- 11. Solution C1= -3 – (0,1,1) = -1 <0 C5= 0 – (0,1,1) = 1 >0 x1 is entering, x1= min{4,inf.,-1/2}=4, x4 is leaving. Basic: x1=4, x2=1, x3=9. Z=-2. 3 0 -2 -2 -1 0
- 12. Solution Iteration 4: Basic: x1=4, x2=1, x3=9. Z=-2 -3 1 1 0 0 M M x1 x2 x3 x4 x5 x6 x7 -3 x1 1 0 0 1/3 -2/3 2/3 -5/3 4 1 x2 0 1 0 0 -1 1 -2 1 1 x3 0 0 1 2/3 -4/3 4/3 -7/3 9 CJ 0 0 0 0 0 0
- 13. Solution C5= 0 – (-3,1,1) = 1/3 >0 No entering variables. The current solution is optimal. x1=4, x2=1, x3=9, Z=-2 -2/3 -1 -4/3
- 14. Example 2 Max Z= x1 + 2x2 + x3 S.T. x1 + x2 + x3 = 7 2x1 – 5x2 +x3 >= 10 In standard form: Max Z= x1 + 2x2 + x3 S.T. x1 + x2 + x3 = 7 2x1 – 5x2 +x3 – x4 = 10 In chanocial form: Max Z= x1 + 2x2 + x3 – Mx5 – Mx6 (Minus cuz its max S.T. x1 + x2 + x3 + x5 = 7 2x1 – 5x2 +x3 – x4 + x6 = 10
- 15. Solution Iteration 1: Basic: x5=7, x6=10. Z=-17M 1 2 1 0 -M -M x1 x2 x3 x4 x5 x6 -M x5 1 1 1 0 1 0 7 -M x6 2 -5 1 -1 0 1 10 CJ 0 0 - 17M
- 16. Solution C1= 1 – (-M,-M) = 1+3M C2= 2 – (-M,-M) = 2-4M C3= 1 – (-M,-M) = 1+2M C4= 0 – (-M,-M) = -M x1 is entering, x1= min{7,5}=5, x6 is leaving. 1 2 1 -5 1 1 0 -1
- 17. Solution Iteration 2: Basic: x5=2, x1=5. Z=5-2M 1 2 1 0 -M -M x1 x2 x3 x4 x5 x6 -M x5 0 7/2 1/2 1/2 1 -1/2 2 1 x1 1 -5/2 1/2 -1/2 0 1/2 5 CJ 0 0 0
- 18. Solution C2= 2 – (-M,1 ) = (9+7M)/2 C3= 1 – (-M,1 ) = (1+M)/2 C4= 0 – (-M,1 ) = (1+M)/2 x2 is entering, x2= min{4/7,-2}=4/7, x5 is leaving. 7/2 -5/2 1/2 1/2 1/2 -1/2
- 19. Solution Iteration 3: Basic: x2=4/7, x1=45/7. Z=53/7 1 2 1 0 -M -M x1 x2 x3 x4 x5 x6 2 x2 0 1 1/7 1/7 2/7 -1/7 4/7 1 x1 1 0 6/7 -1/7 5/7 1/7 45/7 CJ 0 0 0 0 53/7
- 20. Solution C3= 1 – ( 2,1 ) = -1/7 C4= 0 – ( 2,1 ) = -1/7 No entering variables, the current solution is optimal. x2=4/7, x1=45/7, z= 53/7 1/7 6/7 1/7 -1/7
- 21. Example of unrestricted variables Min Z= 3x1 + 2x2 + x3 S.T. 2x1 + 5x2 + x3 = 12 3x1+ 4x2 = 11 x2,x3 >=0, x1 is unrestricted x1= x1’ – x1’’
- 22. Example of unrestricted variables Min Z= 3x1’ – 3x1’’ + 2x2 + x3 + Mx4 + Mx5 S.T. 2x1’ – 2x1’’ + 5x2 + x3 + x4 = 12 3x1’ – 3x1’’ + 4x2 + x5 = 11 x1’, x1’’, x2, x3 >=0
- 23. Solution Iteration 1: Basic: x4=12, x5=11. Z=-Mx4-Mx5=23M X2 is entering and its equal to min{12/5,11/4}=12/5, so x4 is leaving ZJ is obtained by adding the products of the elements under that column with the corresponding CB values (the first column: 2M+3M) CJ 3 -3 2 1 M M CB Basis x1’ x1’’ x2 x3 x4 x5 M x4 2 -2 5 1 1 0 12 M x5 3 -3 4 0 0 1 11 ZJ 5M -5M 9M M M M 23M CJ-ZJ 3-5M -3+5M 2-9M 1-M 0 0
- 24. Solution Iteration 2: Basic: x2 and x5 X1’ is entering and its equal to min{6,1}=1, so x5 is leaving CJ 3 -3 2 1 M M C B Basi s x1’ x1’’ x2 x3 x4 x5 2 x2 2/5 -2/5 1 1/5 1/5 0 12/5 M x5 7/5 -7/5 0 -4/5 -4/5 1 7/5 ZJ 4/5-7/5M -4/5+7/5M 2 2/5+4/5M 2/5+4/5M -M 24/5-7/5M CJ- ZJ 11/5-7/5M -11/5+7/5M 0 3/5+4/5M 9/5M-2/5 0
- 25. Solution Iteration 1: Basic: x1’ and x2. X3 is entering and its equal to min{14/3,-7/4}=14/3, so x2 is leaving CJ 3 -3 2 1 M M CB Basis x1’ x1’’ x2 x3 x4 x5 2 x2 0 0 1 3/7 3/7 -2/7 2 3 x1’ 1 -1 0 -4/7 -4/7 5/7 1 ZJ 3 -3 2 -6/7 -6/7 11/7 7 CJ- ZJ 0 0 0 13/7 -M+6/7 -M-11/7
- 26. Solution Iteration 1: Basic: x4=12, x5=11. Z=-Mx4-Mx5=-23M No entering variables so the solution is optimal CJ 3 -3 2 1 -M -M CB Basis x1’ x1’’ x2 x3 x4 x5 1 x3 0 0 7/3 1 1 -2/3 14/3 3 x1’ 1 -1 4/3 0 0 1/3 11/3 ZJ 3 -3 19/3 1 1 -2/3 1 CJ- ZJ 0 0 -13/3 0 -M-1 -M-1/3