ECE 556 Linear Programming Ting-Yuan Wang Electrical and Computer Engineering University of Wisconsin-Madison March

Outline: Related Courses: CS525 Linear Programming CS726 Nonlinear Programming Theory and Applications CS730 Nonlinear Programming Algorithms Jordan Exchange Linear Programming (Simplex Method) Phase II Phase I

Jordan Exchange ( pivot operation) Consider a linear system of m equations:

Tableau Form

A Jordan exchange with pivot A rs is the process of interchanging the dependent variable y r and the independent variables x s. Process: 1.Solve the rth equation for x s in terms of x 1, x 2, …, x s-1, y r, x s+1,…,x n. Note A rs ≠ 0 2.Substitute for x s in the remaining equations 3.Write the new system in a new tableau form

Tableau Form

The Simplex Method A linear program (or a linear programming problem) is the problem of minimizing (or maximizing) a linear function subject to linear inequalities and linear equalities. The Simplex Method: First find a feasible vertex of the standard linear program. If none exists, the problem is infeasible. Starting at this feasible vertex, move to the adjacent vertex such that the objective function z strictly decreases. If no such adjacent vertex exists, then stop, the current vertex is a solution of the problem or the objective is unbounded.

LP standard form: Objective function Constraints Bounds LP canonical form:

Tableau form Basic variables (slack variables) Non-basic variables feasible: b  0

Example 1 Minimize: Subject to: X1X21 X3 =121 X4 =210 X5 =11 X6 =1-413 X7 =-4123 Z3-60 Tableau Form

Pivot Selection Rules 1.Pricing (Pivot Column s Selection ): The pivot column is any column s with a negative element in the bottom row. We choose the most negative element as pivot column, which gives the most steepest local descent in the objective function z. 2.Ratio Test (Pivot Row r Selection ): The pivot row is any row r such that

X1X51 X3 =3-23 X4 =31 X2 =11 X6 =-349 X7 =-324 Z-36-6 X6X51 X3 =212 X4 =310 X2 =-1/31/34 X1 =-1/34/33 X7 =1-515 Z12-15 Pivot (X 1,X 5 ) Pivot (X 1,X 6 )

Geometric Illustration Vertex 1: N{1,2} Vertex 3: N{5,6} Vertex 2: N{1,5} X1=0 X2=0 Feasible region X7=0 X5=0 X3=0 X4=0 X6=0 Z=-6 Z=0

Phase II Procedure 1.Formulate the problem into standard form. 2.Create an initial feasible tableau. 3.Determine the pivot column s by pricing rule. If none exists, then tableau is optimal. 4.Determine the pivot row r by ratio test. If none exists, then tableau is unbounded. 5.Exchange X B and X N using Jordan exchange on H rs. 6.Go to step (3).

Example 2 X1X5X7X4 X2= X6= X3= Z= X1X2X3X4 X5=-304 X6=-2003 X7=0-43 Z= X1X5X3X4 X2= X6= X7= Z= X1X5X7X4 X2= X6= X3= Z= Pivot (X 7,X 3 ) Pivot (X 6,X 1 ) Pivot (X 5,X 2 )

Example 3 X1X2 X3=12 X4= 6 Z=10 Pivot (X 4,X 2 ) X1X4 X3=-28 X2= 6 Z=21-6 X2=0 X4=0 X3=0 X1=0 Vertex 1: N{1,2} Vertex 2: N{1,4} Z=-6 Z=0

Example 4 X1X2 X3=21 X4=11 Z= 0 Pivot (X 3,X 2 ) X1X3 X2=21 X4=12 Z=-31 X2=0 X4=0 X3=0 X1=0 Vertex 2: N{1,3} Vertex 1: N{1,2} Unbounded !! Z=0

Phase I Procedure 1.If b>0, introduce the artificial variable x 0 ≥ 0 in all the constraints that are violated and set z 0 = x 0. 2.The first pivot is chosen in the x 0 column and the row with worst infeasibility. Then do Jordan exchange. 3.Apply the standard simplex pivot rules until an optimal tableau is obtained. If the optimal value is positive, the original problem has no feasible point. 4.Strike out the column corresponding to x 0 and the row corresponding to z 0. 5.Go to Phase II.

Infeasible ?? X0=0 X3=0 X1=0 X5=0 X2=0 X4=0 Infeasible vertex 1 Plane{(x1,x2,x0)|x0=0}

X0X0 X 3 =0,X 0 =2 X 1 =0,X 0 =2X 5 =0,X 0 =2 X 2 =0,X 0 =2 X 4 =0,X 0 =2 feasible vertex 2 (0,0,2) X1X2X01 X3=111 X4=211-2 X5=004 Z0=0010 X1X2X41 X3=011 X0=-212 X5=004 Z0=-212 X0X2X41 X3=0.5 0 X1= X5= Z0=1000 Plane{(x1,x2,x0)|x0=2}

Example 5 X1X2X3X41 X5=-304 X6=-2003 X7=0-43 X8=1120 X9=140 Z= Pivot (X 8,X 0 ) X1X2X3X4X01 X5=-3004 X6= X7=0-403 X8=11201 X9=1401 Z= Z0= Phase I Add column Add row

X1X2X3X4X81 X5=-3004 X6= X7=0-403 X0= X9= Z= Z0= X1X2X0X4X81 X5=-3004 X6= X7= X3= X9= Z= Z0= Pivot (X 0,X 3 ) X1X2X4X81 X5=-304 X6=-2003 X7=21-21 X3= X9=-3021 Z= Delete row X 0 & column Z 0 Pivot (X 3,X 2 ) Go to Phase II

X1X3X4X81 X5=26-31 X6=20 2 X7=1-2 2 X2=-2011 X9= Z=27-4 X1X3X4X51 X8= X6= X7= X2= X9= Z= X1X7X4X51 X8= X6= X3= X2= X9= 02 Z= X6X7X4X51 X8= X1= X3= X2= X9= Z= Pivot (X 5,X 8 ) Pivot (X 7,X 3 ) Pivot (X 6,X 1 )