Simplex Method

Adapting to Other Forms

Until now, we have dealt with the standard form of the Simplex method

What if the model has a non-standard form? Equality Constraints x1 + x2 = 8

Greater than Constraints x1 + x2 ≥ 8

Minimizing How do we get the initial BF solution?

Original Form

Maximize Z = 3x1 + 5x2

Subject to:x1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 = 18

x1 ≥ 0, x2 ≥ 0

Augmented Form

Maximize Z = 3x1 + 5x2

Subject to:Z - 3x1 - 5x2 = 0

x1 + x3 = 4

2x2 + x4 = 12

3x1 + 2x2 = 18

x1 , x2 , x3 , x4 ≥ 0

Original Form

Maximize Z = 3x1 + 5x2

Subject to:x1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 = 18

x1 ≥ 0, x2 ≥ 0

Artificial Form (Big M Method)

Maximize Z = 3x1 + 5x2

Subject to:Z - 3x1 - 5x2 + Mx5 = 0x1 + x3 = 4

2x2 + x4 = 12

3x1 + 2x2 + x5 = 18

x1 , x2 , x3 , x4 , x5 ≥ 0

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 -3 -5 0 0 M 0x3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18

Z -3x1 -5x2 + Mx5 = 0x1 + x3 = 4

x2 + x4 = 123x1 2x2 + x5 = 18

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18

Select Initial Pointnonbasic variables: x1 and x2 (origin)Initial BF solution: (x1, x2, x3, x4, x5) = (0,0,4,12,18M)

To get to initial point, remove x5 coefficient (M) from Z - 3x1 - 5x2 + Mx5 = 0

(-M) ( 3x1 + 2x2 + x5 = 18) (-3M-3)x1 + (-2M-5)x2 = -18M

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18

Optimality Test: Are all coefficients in row (0) ≥ 0?

If yes, then STOP – optimal solutionIf no, then continue algorithm

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18

Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable1. Select coefficient in pivot column > 0

2. Divide Right Side value by this coefficient3. Select row with smallest ratio

4/1 = 4

18/3 = 6

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18Z 1 0 -2M-5 3M+3 0 0

-6M+12

x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 -2M-5 3M+3 0 0-

6M+12

x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6

BF Solution: (x1, x2, x3, x4, x5) = (4,0,0,12,6)Optimality Test: Are all coefficients in row (0) ≥ 0?

If yes, then STOP – optimal solutionIf no, then continue algorithm

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 -2M-5 3M+3 0 0-

6M+12

x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6

Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable1. Select coefficient in pivot column > 0

2. Divide Right Side value by this coefficient3. Select row with smallest ratio

12/2 = 6

6/2 = 3

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 -2M-5 3M+3 0 0-

6M+12

x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3

BF Solution: (x1, x2, x3, x4, x5) = (4,3,0,6,0)Optimality Test: Are all coefficients in row (0) ≥ 0?

If yes, then STOP – optimal solutionIf no, then continue algorithm

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3

Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable1. Select coefficient in pivot column > 0

2. Divide Right Side value by this coefficient3. Select row with smallest ratio

4/1 = 4

6/3 = 2

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3Z 1 0 0 0 3/2 M+1 36 x1 0 1 0 0 -1/3 1/3 2 x3 0 0 0 1 1/3 -1/3 2 x2 0 0 1 0 1/2 0 6

Basic Variable

Coefficient of:Right Side

Z x1 x2 x3 x4 x5

Z 1 0 0 0 3/2 M+1 36 x1 0 1 0 0 -1/3 1/3 2 x3 0 0 0 1 1/3 -1/3 2 x2 0 0 1 0 1/2 0 6

BF Solution: (x1, x2, x3, x4, x5) = (2,6,2,0,0)Optimality Test: Are all coefficients in row (0) ≥ 0?

If yes, then STOP – optimal solutionIf no, then continue algorithm

Minimize Z = 3x1 + 5x2

Multiply by -1

Maximize -Z = -3x1 - 5x2

x1 - x2 ≤ -1

Multiply by -1

-x1 + x2 ≥ 1

x1 - x2 ≥ 1

x1 - x2 - x5 ≥ 1

Change Inequality

x1 - x2 - x5 ≤ 1

x1 - x2 - x5 + x6 ≤ -1Big M

AugmentedForm

Original Form

Minimize Z = 4x1 + 5x2

Subject to:3x1 + x2 ≤ 27

5x1 + 5x2 = 60

6x1 + 4x2 ≥ 60

x1 ≥ 0, x2 ≥ 0

Adaption Form

Minimize Z = 4x1 + 5x2

Maximize -Z = -4x1 - 5x2

Subject to:-Z + 4x1 + 5x2 + Mx4 + Mx6 = 03x1 + x2 + x3 = 27

5x1 + 5x2 + x4 = 60

6x1 + 4x2 - x5 + x6 = 60