(includes both Phases I & II)

Transcription

1 Minimize z=3x 5x 4x 7x 5x 4x subject to 2x x2 x4 3x6 0 x 3x3 x4 3x5 2x6 2 4x2 2x3 3x4 x5 5 and x 0 j, j ecause of the lack of a slack variable in each constraint, we must use Phase I to find an initial feasible basis. (includes both Phases I & II) Add variables X 9, X 0, X (artificial variables), and a Phase I objective of minimizing the sum of these three variables. Dennis ricker Dept of Mechanical & Industrial Engineering The University of Iowa Revised Simplex Method 09/23/04 page of 22 Revised Simplex Method 09/23/04 page 2 of 22 Phase One b phase one objective phase two objective Values of basic (artificial) variables are: i Xi Revised Simplex Method 09/23/04 page 3 of 22 Revised Simplex Method 09/23/04 page 4 of 22

2 Iteration Current partition: ( = basis, N = non-basis) = {9 0 }, N= { } asis inverse is Simplex multipliers (dual solution): i ca,, 0 0 [,,] 0 0 = [,, ] Compute each reduced cost: c j -A j, e.g. c = 0-[,,] [2 0]= -3 Reduced (Artificial) Costs j Cj set -3 N 2-3 N 3-5 N enter, since C j < 0 decrease in objective! 4-3 N 5-4 N 6-5 N 7 N 8 N Any nonbasic variables except x 7 & x 8 could be chosen to enter the basis. Revised Simplex Method 09/23/04 page 5 of 22 Revised Simplex Method 09/23/04 page 6 of 22 Entering variable is X[3] from set N Compute the substitution rates: A A = That is, one unit of x 3 will replace 3 units of the second basic variable (x 0 ) and 2 units of the third basic variable (x ). Current values of basic variables: , 0, x x x x A b L X d ratio inf minimum ratio = substitution rate, d = direction of change of basic variable, ratio = rhs/ for all Change of basis asic variable X[0] leaves basis New partition: ={9 3 }, N={ } Revised Simplex Method 09/23/04 page 7 of 22 Revised Simplex Method 09/23/04 page 8 of 22

3 Updating basis inverse matrix: affix column of substitution rates alongside the old inverse, and pivot: old inverse new inverse 0 0 A Current partition: Iteration = {9 3 }, N= { } asis inverse is Simplex multipliers (dual solution): 0 0 c, 0, 0 0 [, 2 A,] i Revised Simplex Method 09/23/04 page 9 of 22 Revised Simplex Method 09/23/04 page 0 of 22 Reduced cost of X, for example, is C = C A = Reduced (Artificial) Costs j Cj set any of variables, 4, 6, or 7 would improve N the phase I objective 2-3 N N enter 5 N N N 8 N N Entering variable is X[4] from set N Compute the substitution rates: A A 0 0 = L X d ratio min ratio Change of basis asic variable X[] leaves basis New partition: = {9 3 4}, N= { } Revised Simplex Method 09/23/04 page of 22 Revised Simplex Method 09/23/04 page 2 of 22

4 Updating basis inverse matrix Write the column of substitution rates alongside the old inverse matrix, and pivot: old inverse new inverse 2 3 A Current partition: Iteration = {9 3 4}, N= { } asis inverse is Simplex multipliers (dual solution): c A i [, 0, 0] , 2, 3 Revised Simplex Method 09/23/04 page 3 of 22 Revised Simplex Method 09/23/04 page 4 of 22 Reduced cost of X, for example, is c c A Reduced (Artificial) Costs j Cj set any of variables, 5, 6, & N would improve the phase I objective N N N enter N N N N Entering variable is X[6] from set N Compute the substitution rates A A = That is, one unit of x 6 will replace units of the first basic variable (x 9 ), one unit of the second (x 3 ), and require additional units of the third (x 4 ). L X d ratio min ratio Revised Simplex Method 09/23/04 page 5 of 22 Revised Simplex Method 09/23/04 page 6 of 22

5 Change of basis asic variable X[9] leaves basis New partition: = {6 3 4}, N= { } All artificial variables (9, 0, ) are now nonbasic. ***Feasibility has been achieved! End Phase One*** Optimal Phase One Solution i X[i] Cz Phase one objective = sum of artificial variables = 0 Revised Simplex Method 09/23/04 page 7 of 22 egin Phase II Iteration Current partition: = {6 3 4}, N= { } asis inverse is The costs of the basic variables are c c, c, c 4, 4, Compute the simplex multipliers: c A 4, 4, , ,.9892 Revised Simplex Method 09/23/04 page 8 of 22 Simplex multipliers (dual solution) i Reduced cost of x is, for example, 2 c A , , Reduced costs j Cj set only variable 2 would improve N the phase II objective N enter N N N Entering variable is X[2] from set N Compute the substitution rates: A 2 A L X d ratio min ratio Change of basis asic variable X[4] leaves basis New partition: = {6 3 2}, N= { } Revised Simplex Method 09/23/04 page 9 of 22 Revised Simplex Method 09/23/04 page 20 of 22

6 Update the basis inverse, by writing the substitution rates alongside the old basis inverse, and pivoting: ??? ??? 0 etc ??? When the simplex multipliers and the reduced costs have been computed, the resulting reduced costs are all nonnegative! Optimal Solution Objective function: Z= Primal Solution (reduced costs are all nonnegative) i L X[i] set reduced cost 0 0 N N N N N.5325 Dual Solution i Revised Simplex Method 09/23/04 page 2 of 22 Revised Simplex Method 09/23/04 page 22 of 22