INTRODUCTION TO OPERATIONS RESEARCH

Simplex Method Adapting to Other Forms

SIMPLEX METHOD  Until now, we have dealt with the standard form of the Simplex method  What if the model has a non-standard form?

 Equality Constraints x 1  Greater than Constraints + x 2 = 8 x 1 + x 2 ≥ 8  Minimizing  How do we get the initial BF solution?

EQUALITY CONSTRAINT

Original Form Augmented Form

Maximize Z = 3x 1 + 5x 2 Subject to: x 1 2x 2 ≤ 4 ≤ 12 3x 1 + 2x 2 = 18 x 1 ≥ 0, x 2 ≥ 0 Maximize Z = 3x 1 + 5x 2 Subject to: Z - 3x 1 x 1 + x 3 - 5x 2 = 4 2x 2 3x 1 + x 4 + 2x 2 = 12 = 0 = 18 x 1 , x 2 , x 3 , x 4 ≥ 0

ADAPTING EQUALITY CONSTRAINT

Original Form Artificial Form (Big M Method)

Maximize Z = 3x 1 + 5x 2 Maximize Z = 3x 1 + 5x 2 Subject to: x 1 2x 2 ≤ 4 ≤ 12 3x 1 + 2x 2 = 18 x 1 ≥ 0, x 2 ≥ 0 Subject to: Z - 3x 1 x 1 + x 3 - 5x 2 + Mx 5 = 4 2x 2 3x 1 + x 4 + 2x 2 = 12 + x 5 = 18 = 0 x 1 , x 2 , x 3 , x 4 , x 5 ≥ 0

SETTING UP THE SIMPLEX TABLEAU Z -3x 1 x 1 -5x 2 3x 1 x 2 2x 2 + x 3 + x 4 + Mx 5 + x 5 = 0 = 4 = 12 = 18

Basic Variable

Z x 3 x 4 x 5 Z 1 0 0 0 x 1 -3 1 0 3 Coefficient of: x 2 -5 x 0 3 0 2 2 1 0 0 x 4 0 0 1 0 x 5 M 0 0 1

Right Side

0 4 12 18

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 3 x 4 x 5 Z 1 0 0 0 x 1 -3 M -3 -2 M -5 1 0 3 Coefficient of: x 2 0 2 2 x 0 1 0 0 3 x 4 0 0 1 0 x 0 0 0 1 5

Right Side

-18 4 12 18 M To get to initial point, remove x 5 - 3x 1 - 5x 2 + Mx 5 = 0 (-M) ( 3x 1 (-3M-3)x 1 + 2x 2 + x 5 + (-2M-5)x 2 = 18) = -18M coefficient (M) from Z

Select Initial Point nonbasic variables: x 1 and x 2 (origin)

Initial BF solution: (x 1 , x 2 , x 3 , x 4 , x 5 ) = (0,0,4,12,18M)

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 3 x 4 x 5 Z 1 0 0 0 x 1 x 2 -3 M -3 -2 M -5 1 0 3 Coefficient of: 0 2 2 x 0 1 0 0 3 x 4 0 0 1 0 x 5 0 0 0 1

Right Side

-18 M 4 12 18

Optimality Test:

Are all coefficients in row (0) ≥ 0?

If yes , then

STOP

– optimal solution If no , then continue algorithm

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 3 x 4 x 5 Z 1 0 0 0 x 1 x 2 -3 M -3 -2 M -5 1 0 3 Coefficient of: 0 2 2 x 0 1 0 0 3 x 4 0 0 1 0 x 5 0 0 0 1

Right Side

-18 M 4 12 18 4/1 = 4 18/3 = 6

Select Entering Basic Variable

Choose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable

1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 3 x 4 x 5 Z x 1 x 4 x 5 1 0 0 0 Z 1 0 0 0 x 1 x 2 -3 M -3 -2 M -5 1 0 3 Coefficient of: 0 2 2 x 0 1 0 0 3 x 4 0 0 1 0 0 1 0 0 -2 M -5 3 M +3 0 2 2 1 0 -3 0 0 1 0 x 5 0 0 0 1 0 0 0 1

Right Side

-18 4 12 18 M -6 M +12 4 12 6

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 5 Z 1 0 0 0 x 1 0 1 0 0 Coefficient of: x 2 x 3 -2 M -5 3 M +3 0 2 2 1 0 -3 x 4 0 0 1 0 x 5 0 0 0 1

Right Side

-6 M +12 4 12 6

BF Solution: Optimality Test:

(x 1 , x 2 , x 3 , x 4 , x 5 ) = (4,0,0,12,6) Are all coefficients in row (0) ≥ 0?

If yes , then

STOP

– optimal solution If no , then continue algorithm

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 5 Z 1 0 0 0 x 1 0 1 0 0 Coefficient of: x 2 x 3 -2 M -5 3 M +3 0 2 2 1 0 -3 x 4 0 0 1 0 x 5 0 0 0 1

Right Side

-6 M +12 4 12 6 12/2 = 6 6/2 = 3

Select Entering Basic Variable

Choose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable

1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 5 Z x 1 x 4 x 2 1 0 0 0 Z 1 0 0 0 0 1 0 0 x 1 0 1 0 0 Coefficient of: x 2 x 3 -2 M -5 3 M +3 0 2 2 1 0 -3 x 4 0 0 1 0 0 0 0 1 -9/2 1 3 -3/2 x 5 0 0 0 1

Right Side

-6 M +12 4 12 6 0 0 1 0 M +5/2 0 -1 1/2 27 4 6 3

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 2 Z 1 0 0 0 x 1 0 1 0 0 Coefficient of: x 2 0 0 x 1 3 -9/2 0 1 3 -3/2 x 4 0 0 1 0 x 5 M +5/2 0 -1 1/2

Right Side

27 4 6 3

BF Solution: Optimality Test:

(x 1 , x 2 , x 3 , x 4 , x 5 ) = (4,3,0,6,0) Are all coefficients in row (0) ≥ 0?

If yes , then

STOP

– optimal solution If no , then continue algorithm

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 2 Z 1 0 0 0 x 1 0 1 0 0 Coefficient of: x 2 0 0 x 1 3 -9/2 0 1 3 -3/2 x 4 0 0 1 0 x 5 M +5/2 0 -1 1/2

Right Side

27 4 6 3 4/1 = 4 6/3 = 2

Select Entering Basic Variable

Choose variable with negative coefficient having largest absolute value

Select Leaving Basic Variable

1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 4 x 2 Z x 1 x 3 x 2 1 0 0 0 Z 1 0 0 0 0 1 0 0 x 1 0 1 0 0 Coefficient of: x 2 0 0 x 1 3 -9/2 0 1 3 -3/2 x 4 0 0 1 0 x 5 M +5/2 0 -1 1/2

Right Side

27 4 6 3 0 0 0 1 0 0 1 0 3/2 -1/3 1/3 1/2 M +1 1/3 -1/3 0 36 2 2 6

ITERATIONS OF THE SIMPLEX METHOD

Basic Variable

Z x 1 x 3 x 2 Z 1 0 0 0 x 1 0 1 0 0 Coefficient of: x 2 0 0 x 0 0 3 x 4 3/2 -1/3 0 1 1 0 1/3 1/2 x 5 M +1 1/3 -1/3 0

Right Side

36 2 2 6

BF Solution: Optimality Test:

(x 1 , x 2 , x 3 , x 4 , x 5 ) = (2,6,2,0,0) Are all coefficients in row (0) ≥ 0?

If yes , then

STOP

– optimal solution If no , then continue algorithm

ADAPTING FOR MINIMIZATION Minimize Z = 3x 1 + 5x 2

Multiply by -1

Maximize -Z = -3x 1 - 5x 2

NEGATIVE RIGHT-HAND SIDE x 1 - x 2 ≤ -1

Multiply by -1

-x 1 + x 2 ≥ 1

ADAPTING FOR A ≥ CONSTRAINT x 1 - x 2 ≥ 1

Augmented Form

x 1 - x 2 x 5 ≥ 1

Change Inequality Big M

x 1 x 1 - x 2 - x 2 x 5 x + 5 x ≤ 1 6 ≤ -1

EXAMPLE OF ADAPTING FORM

Original Form Adaption Form

Minimize Z = 4x 1 + 5x 2 Subject to: 3x 1 5x 1 + x 2 + 5x ≤ 27 2 = 60 6x 1 + 4x 2 ≥ 60 x 1 ≥ 0, x 2 ≥ 0 Minimize Z = 4x 1 Maximize + 5x -Z = -4x 1 2 - 5x 2 Subject to: -Z + 4x 1 3x 1 + x 2 5x 1 6x 1 + 5x + 4x 2 2 + 5x 2 + Mx 4 + x 3 = 27 + x x 5 4 = 60 + x 6 + = 60 Mx 6 = 0