Operation ResearchCHAPTER 06 - THE BIG M METHOD

The Big M MethodI. 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.

Example Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 <= 11 -4x1+ x2 + 2x3 >= 3 2x1 – x3 = -1

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

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

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 x70 x4 1 -2 1 1 0 0 0 11M x6 -4 1 2 0 -1 1 0 3M x7 -2 0 1 0 0 0 1 1

CJ 0 0 0

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.

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Solution Iteration 2: Basic: x4=11-1=10, x6=3-2=1, x3=1. Z=1+M

-3 1 1 0 0 M Mx1 x2 x3 x4 x5 x6 x7

0 x4 3 -2 0 1 0 0 -1 10M x6 0 1 0 0 -1 1 -2 11 x3 -2 0 1 0 0 0 1 1

CJ 0 0 0 0

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.

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Solution Iteration 3: Basic: x4=12, x2=1, x3=1. Z=2

-3 1 1 0 0 M Mx1 x2 x3 x4 x5 x6 x7

0 x4 3 0 0 1 -2 2 -5 121 x2 0 1 1 0 -1 1 -2 11 x3 -2 0 1 0 0 0 1 1

CJ 0 0 0 0 0

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.

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Solution Iteration 4: Basic: x1=4, x2=1, x3=9. Z=-2

-3 1 1 0 0 M Mx1 x2 x3 x4 x5 x6 x7

-3 x1 1 0 0 1/3 -2/3 2/3 -5/3 41 x2 0 1 0 0 -1 1 -2 11 x3 0 0 1 2/3 -4/3 4/3 -7/3 9

CJ 0 0 0 0 0 0

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

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