Operations Research - Linear Programming - Primal Simplex Tableaux Generator

by

Elmer G. Wiens

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The primal simplex method transforms an initial tableau into a final tableau containing the solutions to the primal and dual problems. Each stage of the algorithm generates an intermediate tableau as the algorithm gropes towards a solution.

This web page permits you to enter your own linear programming problem and generate these tableaux with a user friendly interface. You can specify up to 6 variables and 10 constraints in the primal problem, with any mixture of <=, >=, and = constraints.

To illustrate the procedure you need to follow, consider the problem:

To generate the form used to enter the l.p. data you must set the parameters that specify your l.p. problem as I have done below for the problem above.

After you fill in your form and click submit paramters, a web page will pop with a table, like the one below, where you can enter the data of your l.p. problem. Notice that you need neither multiply = constrainst by -1, nor convert >= constraints to <= constraints.

FormTable =

After you fill in your data and click submit, the program will automatically calculate a sequence of tableaux that solves the primal and dual l.p. problems. Examine the tableaux that follow to see how the primal simplex method proceeds to find the solution.

To perform a sensitivity analysis on your linear programming problem, change the data in the table above, and click Submit L.P. again.

Primal Simplex Algorithm.

The standard form for the initial primal simplex tableau is:

Initial Tableau

b

A

I_{3}

0

-c^{T}

O_{3}^{T}

The primal simplex algorithm calculates a sequence of tableaux. Tableau^{k} has the form:

Tableau^{k}

ß

U

B

µ

t^{T}

y^{T}

=

Tableau^{k}

ß

Û

µ

Ø

after consolidating the center 2 by 2 block of matrices.

The "three-phase method" of the primal simplex algorithm:

Phase 0 - drive all artificial variables (associated with = constraints) to zero, i.e. eliminate them from the basis;

Phase I - find a tableau with ß >= 0, i.e. a feasible primal program;

Phase II - generate tableaux that increase the value of µ, without dropping back into Phase 0 or I, until Ø >= 0 for all sign restricted variables, i.e. find a feasible primal program that maximizes the objective function.

Phase 0: Drive the artificial variables from the basis.

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Tableau^{0}

b^{0}

x^{0}_{1}

x^{0}_{2}

x^{0}_{3}

s^{0}_{1}

s^{0}_{2}

s*^{0}_{3}

row sum

b / Û_{k}

L^{0}_{1}

85

1

1

1

1

0

0

89

0

L^{0}_{2}

-90

-1.25

-0.5

-1

0

1

0

-91.75

0

L^{0}_{3}

55

0.6

1

0.5

0

0

1

58.1

0

P^{0}

0

-15

-10

-17

0

0

0

-42

Basis for Tableau^{0}: [s_{1}, s_{2}, s_{3}, ]. Value of Objective Function = 0.

Proceed to the next tableau as follows:

Phase 0: Drive the artificial variables from the basis.

A. In Tableau^{0}:

1. Select an artificial variable in the basis: s^{*}_{3}. The pivot row = 3.

2. Select a nonzero element in row L_{3} as pivot: Û_{3,2} = 1. The pivot column = 2.

B. To create Tableau^{1}:

3. Compute row L^{1}_{3} = L^{0}_{3} / (1).

4. Subtract multiples of row L^{1}_{3} from all other rows of Tableau^{0} so that x^{1}_{2} = e_{3} in Tableau^{1}.

Phase I: Goal: get ß >= 0.

Tableau^{1}

b^{1}

x^{1}_{1}

x^{1}_{2}

x^{1}_{3}

s^{1}_{1}

s^{1}_{2}

s*^{1}_{3}

row sum

b / Û_{k}

L^{1}_{1} = L^{0}_{1} - (1) * L^{1}_{3}

30

0.4

0

0.5

1

0

-1

30.9

75

L^{1}_{2} = L^{0}_{2} - (-0.5) * L^{1}_{3}

-62.5

-0.95

0

-0.75

0

1

0.5

-62.7

65.789

L^{1}_{3} = L^{0}_{3} / (1)

55

0.6

1

0.5

0

0

1

58.1

91.667

P^{1} = P^{0} - (-10) * L^{1}_{3}

550

-9

0

-12

0

0

10

539

Basis for Tableau^{1}: [s_{1}, s_{2}, x_{2}, ]. Value of Objective Function = 550.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Goal: get ß >= 0.

A. In Tableau^{1}:

1. Select a target row, r, with b_{r} < 0: b^{1}_{2} = -62.5, r = 2.

2. Select any column, col, with a negative entry in row = 2 as the pivot column: col = 1 associated with Û_{2,1} = -0.95 and constraint L_{2}.

3. Compute the ratios b_{i} / Û_{i,1} as per the last column.
Select the row with the least positive ratio as the pivot row: row = 2 associated with constraint L_{2}. Thus Û_{2,1} = -0.95 is the pivot; variable s_{2} will leave the basis; variable x^{2}_{1} will enter the basis.

B. To create Tableau^{2}:

4. Compute row L^{2}_{2} = L^{1}_{2} / (-0.95).

5. Subtract multiples of row L^{2}_{2} from all other rows of Tableau^{1} so that x^{2}_{1} = e_{2} in Tableau^{2}.

Phase II: Goal: get Ø >= 0.

Tableau^{2}

b^{2}

x^{2}_{1}

x^{2}_{2}

x^{2}_{3}

s^{2}_{1}

s^{2}_{2}

s*^{2}_{3}

row sum

b / Û_{k}

L^{2}_{1} = L^{1}_{1} - (0.4) * L^{2}_{2}

3.684

0

0

0.184

1

0.421

-0.789

4.5

8.75

L^{2}_{2} = L^{1}_{2} / (-0.95)

65.789

1

-0

0.789

-0

-1.053

-0.526

66

0

L^{2}_{3} = L^{1}_{3} - (0.6) * L^{2}_{2}

15.526

0

1

0.026

0

0.632

1.316

18.5

24.583

P^{2} = P^{1} - (-9) * L^{2}_{2}

1142.105

0

0

-4.895

0

-9.474

5.263

1133

Basis for Tableau^{2}: [s_{1}, x_{1}, x_{2}, ]. Value of Objective Function = 1142.11.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Complete.

Phase II: Goal: get Ø >= 0.

A. In Tableau^{2}:

1. Select the pivot column, col, with the most negative value in Ø: col = 5, Ø_{5} = -9.474: s^{3}_{2} will enter the basis.

2. Compute the ratios b_{i} / Û_{i,5} as per the last column.
Select the row with the least positive ratio as the pivot row: row = 1 associated with constraint L_{1}. Thus Û_{1,5} = 0.421 is the pivot; variable s_{1} will leave the basis; variable s^{3}_{2} will enter the basis.

B. To create Tableau^{3}:

3. Compute row L^{3}_{1} = L^{2}_{1} / (0.421).

4. Subtract multiples of row L^{3}_{1} from all other rows of Tableau^{2} so that s^{3}_{2} = e_{1} in Tableau^{3}.

Tableau^{3}

b^{3}

x^{3}_{1}

x^{3}_{2}

x^{3}_{3}

s^{3}_{1}

s^{3}_{2}

s*^{3}_{3}

row sum

b / Û_{k}

L^{3}_{1} = L^{2}_{1} / (0.421)

8.75

0

0

0.438

2.375

1

-1.875

10.688

20

L^{3}_{2} = L^{2}_{2} - (-1.053) * L^{3}_{1}

75

1

0

1.25

2.5

0

-2.5

77.25

60

L^{3}_{3} = L^{2}_{3} - (0.632) * L^{3}_{1}

10

0

1

-0.25

-1.5

0

2.5

11.75

0

P^{3} = P^{2} - (-9.47) * L^{3}_{1}

1225

0

0

-0.75

22.5

0

-12.5

1234.25

Basis for Tableau^{3}: [s_{2}, x_{1}, x_{2}, ]. Value of Objective Function = 1225.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Complete.

Phase II: Goal: get Ø >= 0.

A. In Tableau^{3}:

1. Select the pivot column, col, with the most negative value in Ø: col = 3, Ø_{3} = -0.75: x^{4}_{3} will enter the basis.

2. Compute the ratios b_{i} / Û_{i,3} as per the last column.
Select the row with the least positive ratio as the pivot row: row = 1 associated with constraint L_{1}. Thus Û_{1,3} = 0.438 is the pivot; variable s_{2} will leave the basis; variable x^{4}_{3} will enter the basis.

B. To create Tableau^{4}:

3. Compute row L^{4}_{1} = L^{3}_{1} / (0.438).

4. Subtract multiples of row L^{4}_{1} from all other rows of Tableau^{3} so that x^{4}_{3} = e_{1} in Tableau^{4}.

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Tableau^{4}

b^{4}

x^{4}_{1}

x^{4}_{2}

x^{4}_{3}

s^{4}_{1}

s^{4}_{2}

s*^{4}_{3}

row sum

b / Û_{k}

L^{4}_{1} = L^{3}_{1} / (0.438)

20

0

0

1

5.429

2.286

-4.286

24.429

0

L^{4}_{2} = L^{3}_{2} - (1.25) * L^{4}_{1}

50

1

0

0

-4.286

-2.857

2.857

46.714

0

L^{4}_{3} = L^{3}_{3} - (-0.25) * L^{4}_{1}

15

0

1

0

-0.143

0.571

1.429

17.857

0

P^{4} = P^{3} - (-0.75) * L^{4}_{1}

1240

0

0

0

26.571

1.714

-15.714

1252.571

Basis for Tableau^{4}: [x_{3}, x_{1}, x_{2}, ]. Value of Objective Function = 1240.