Setting Up the Simplex Tableau
+ x4
+ x5 =
=
= =
18 + 2x2
3x1
12 2x2
4 + x3
x1
0 - 5x2
-3x1 Z
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Iterations of the Simplex Method in Tabular Form
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Iterations of the Simplex Method in Tabular Form
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Optimality Test: Are all coefficients in row (0) 0?
Iterations of the Simplex Method in Tabular Form
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Select Entering Basic Variable
Iterations of the Simplex Method in Tabular Form
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Pivot Column
Select Leaving Basic Variable
1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient
Iterations of the Simplex Method in Tabular Form
18/2 = 9 1
0 0
2 3
0 (3)
x5
12/2 = 6 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Pivot Column
Select Leaving Basic Variable
1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient
Iterations of the Simplex Method in Tabular Form
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Pivot Column Pivot Row
Pivot Number
Use Gaussian Elimination
Iterations of the Simplex Method in Tabular Form
Coefficient of:
(1)
Right Side x5
x4 x3
Basic Var
0
Iterations of the Simplex Method in Tabular Form
Coefficient of:
(1)
Right Side x5
x4 x3
Basic Var
0
Iterations of the Simplex Method in Tabular Form
6 1
-1 0
0 3
0 (3)
x5
6 0
1/2 0
1 0
0 (2)
x2
4 0
0 1
0 1
0 (1)
x3
30 0
5/2 0
0 -3
1 (0)
Z
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Select 2(degrees of freedom) nonbasic variables: x1and x4
Iterations of the Simplex Method in Tabular Form
6 1
-1 0
0 3
0 (3)
x5
6 0
1/2 0
1 0
0 (2)
x2
4 0
0 1
0 1
0 (1)
x3
30 0
5/2 0
0 -3
1 (0)
Z
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Optimality Test: Are all coefficients in row (0) 0?
Iterations of the Simplex Method in Tabular Form
6 1
-1 0
0 3
0 (3)
x5
6 0
1/2 0
1 0
0 (2)
x2
4 0
0 1
0 1
0 (1)
x3
30 0
5/2 0
0 -3
1 (0)
Z
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Select Leaving Basic Variable
1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient
Iterations of the Simplex Method in Tabular Form
6/3 = 2 1
-1 0
0 3
0 (3)
x5
6 0
1/2 0
1 0
0 (2)
x2
4/1 = 4 0
0 1
0 1
0 (1)
x3
30 0
5/2 0
0 -3
1 (0)
Z
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Select Leaving Basic Variable
1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient
Iterations of the Simplex Method in Tabular Form
6 1
-1 0
0 3
0 (3)
x5
6 0
1/2 0
1 0
0 (2)
x2
4 0
0 1
0 1
0 (1)
x3
30 0
5/2 0
0 -3
1 (0)
Z
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Use Gaussian Elimination
Iterations of the Simplex Method in Tabular Form
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Iterations of the Simplex Method in Tabular Form
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
1
Iterations of the Simplex Method in Tabular Form
36 1
3/2 0
0 0
1 (0)
Z
2 -1/3
1/3 1
0 0
0 (1)
x3
6 0
1/2 0
1 0
0 (2)
x2
2 1/3
-1/3 0
0 1
0 (3)
x1
x2 x1
Z
Coefficient of:
Eq.
Right Side x5
x4 x3
Basic Var
2
Optimality Test: Are all coefficients in row (0) 0?
Tie-Breaking for Entering Basic Variable
18 1
0 0
2 3
0 (3)
x5
12 0
1 0
2 0
0 (2)
x4
0 -3
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Choice of entering basic variable is arbitrary–
Tie-Breaking for Leaving Basic Variable (Degeneracy)
18/3 = 6 1
0 0
3 3
0 (3)
x5
12/2 = 6 0
1 0
2 0
0 (2)
x4
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x3
0 0
0 0
Z
Right Side x5
x4 x3
Basic Var
Might encounter degeneracy(one or more BF solutions equal to zero) if there are redundant constraints (a BF solution with more than nconstraints passing through it)
No Leaving Basic Variable – Unbounded Z
0 -5
x2
1 -3
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 1
x3
0 0
Z
Right Side x3
Basic Var
No candidates for leaving basic variable –
Entering basic variable could be increased indefinitelywithout giving negative values to any current basic variables
Multiple Optimal Solutions
3 1/2
0 -3/2
1 0
0 (3)
x2
6 -1
1 3
0 0
0 (2)
x4
0 0
x2
1 0
x1
0 1
Z
Coefficient of:
(1) (0)
Eq.
4 0
0 1
x1
18 1
0 0
Z
Right Side x5
x4 x3
Basic Var
Final tableau has one or more nonbasic variable with a coefficient of zero in row (0)
Simplex Method Definitions
Slack Variables
– Added to convert functional inequality constraints to
equivalent equality constraints
Augmented Solution
– Solution for the original variables that is
augmented by the slack variables
(3,2) ĺ (3,2,1,8,5)
Basic Solution
– Augmented corner-point solution
Basic Feasible (BF) Solution
– Augmented CPF solution
Degrees of Freedom
– Number of variables – number of equations
5 – 3 = 2
Number of variables that are set equal to zero
