Big M Method

A. R. Dani

Big M Method• In every ≥ (Greater Than) type Constraint – Add

a Surplus and Artificial Variable• In every = (Equality) type Constraint Add

Artificial Variable• Assign –M as Cost Coefficient for Artificial

variable in Objective function (where M is a large positive number)

• Use (Cj-Zj) to test optimality

• Objective Function is of Maximization Type

Big M Method• Maximize Z = 3x1-x2

• Subject to the constraints• 2x1 + X2 ≥ 2• 2x1 + X2 ≤ 3• X2 ≤ 4• X1, X2 ≥ 0• In the first step we introduce Surplus and

Artificial Variables in first constraints and slack variables in second and third constraint. The problem can be rewritten as

Big M Method

• Maximize Z = 3x1-x2

• Subject to the constraints

• 2x1 + X2 ≥ 2

• 2x1 + X2 ≤ 3

• X2 +≤ 4

• X1, X2 ≥ 0

• Maximize Z = 3x1-x2 + 0. S1 – M A1 + 0 . S3 + 0.S4

• Subject to the constraints

• 2x1 + X2 -S1 + A1 = 2

• 2x1 + X2 + S2 = 3

• X2 + S3 =4

• X1, X2 ≥ 0

• M is large Positive Number

• S1- Surplus variable

• S2,S3 – Slack Variables

• A1 – Artificial Variable

Big M Method

X1 X2 S1 A1 S2 S3 Bi

Cj 3 -1 0 –M 0 0

-M A1 2 1 -1 1 0 0 2

0 S2 2 1 0 0 1 0 3

0 S3 0 1 0 0 0 1 4

Zj -2M -M M -M 0 0

Cj-Zj 3+2M 1+M -M 0 0 0

Big M Method• It can be seen that some (Cj-Zj) are positive

• The maximum value is 3+2M, so X1 enters the basis

• The ratios are (2/2 = 1) and 3/2

• So A1 leaves the basis

• The element 2 is pivot element• So we divide first row by 2 and then make all

other elements in first column 0.

• The new table is obtained after R2-2R1

Big M Method

X1 X2 S1 A1 S2 S3 Bi

Cj 3 -1 0 –M 0 0

3 X1 1 1/2 -1/2 1/2 0 0 1

0 S2 0 0 1 -1 1 0 1

0 S3 0 1 0 0 0 1 4

Zj 3 3/2 -3/2 3/2 0 0

Cj-Zj 0 -5/2 3/2 -M-3/2 0 0

Big M Method• It can be seen that one (Zj-Cj) is positive

• The maximum value is 3/2, so S1 enters the basis

• S2 leaves the basis

• The element 1 is pivot element• So all other elements in the third column 0

(R2+1/2R2).

• The new table is

Big M Method

X1 X2 S1 A1 S2 S3 Bi

Cj 3 -1 0 –M 0 0

3 X1 1 1/2 0 0 1/2 0 3/2

0 S1 0 0 1 -1 1 0 1

0 S3 0 1 0 0 0 1 4

Zj 3 3/2 0 0 0 0

Cj-Zj 0 -5/2 0 -M 0 0

Big M Method

• So the optimum solution is obtained.

• X1 = 3/2 and X2 = 0