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, 6 2 3 4 5 6 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 2 3 4 5 6 7 8 9 0 b 0 0 0 0 0 0 0 0 0 phase one objective 3 5 4 7 5 4 0 0 0 0 0 0 phase two objective 2-0 0 3 0 0 0 0 0 0 3-3 2-0 0 0 2 0 4 2 3 0 0-0 0 5 Values of basic (artificial) variables are: i Xi 9 0 0 2 5 Revised Simplex Method 09/23/04 page 3 of 22 Revised Simplex Method 09/23/04 page 4 of 22
Iteration Current partition: ( = basis, N = non-basis) = {9 0 }, N= { 2 3 4 5 6 7 8} asis inverse is 0 0 0 0 0 0 Simplex multipliers (dual solution): i 2 3 0 0 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: 3 0 0 0 0 A A 0 0 3 = 3 0 0 2 2 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 00 9, 0, 0 0 2 x x x x A b 0 0 5 L X d ratio 9 0 0 0 - inf 0 0 2 3 4.0 minimum ratio 0 5 2 7.5 = 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={ 2 4 5 6 7 8 0} Revised Simplex Method 09/23/04 page 7 of 22 Revised Simplex Method 09/23/04 page 8 of 22
Updating basis inverse matrix: affix column of substitution rates alongside the old inverse, and pivot: 0 0 0 0 0 0 0 0 3 0 0 3 0 0 2 0 2 0 3 old inverse new inverse 0 0 A 0 0 3 0 2 3 Current partition: Iteration 2 ----------- = {9 3 }, N= { 2 4 5 6 7 8 0} asis inverse is 0 0 0 0.333333 0 0-0.666667 Simplex multipliers (dual solution): 0 0 c, 0, 0 0 [, 2 A,] 3 3 0 2 3 i 2-0.666667 3 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 -.33333 N the phase I objective 2-3 N 4-4.66667 N enter 5 N 6 -.66667 N 7-0.666667 N 8 N 0.66667 N Entering variable is X[4] from set N Compute the substitution rates: 0 0 4 A A 0 0 = 3 3 2 3 0 3 3 L X d ratio 9 0 0.000000 0.00000 3 0 4-0.333333 -- 0 7 3.666667.90909 min ratio Change of basis asic variable X[] leaves basis New partition: = {9 3 4}, N= { 2 5 6 7 8 0 } Revised Simplex Method 09/23/04 page of 22 Revised Simplex Method 09/23/04 page 2 of 22
Updating basis inverse matrix Write the column of substitution rates alongside the old inverse matrix, and pivot: 0 0 2 3 0 0 0 0 3 0 3 3 2 0 2 3 0 3 3 old inverse new inverse 2 3 A 0 3 0 2 3 Current partition: Iteration 3 ----------- = {9 3 4}, N= { 2 5 6 7 8 0 } asis inverse is 0.888-0.272727 0 0.272727 0.090909 0-0.888 0.272727 Simplex multipliers (dual solution): c A i 2 0.888 3-0.272727 3 2 [, 0, 0] 0 3 0 2 3, 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 0 0.888 0.272727 2 0 2.888 Reduced (Artificial) Costs j Cj set any of variables, 5, 6, &8-2.882 N would improve the phase I objective 2 2.0909 N 5-0.272727 N 6-3.36364 N enter 7 0.888 N 8-0.272727 N 0 0.8882 N.27273 N Entering variable is X[6] from set N Compute the substitution rates 2 3 37 3 6 A A 0 3 2 = 2 3 0 4 0 That is, one unit of x 6 will replace 3.36363 units of the first basic variable (x 9 ), one unit of the second (x 3 ), and require 0.36363 additional units of the third (x 4 ). L X d ratio 9 0 8.0909 3.36363 2.4054 min ratio 3 0 4.63636 0.54545 8.50000 4 0.90909-0.36363 -- Revised Simplex Method 09/23/04 page 5 of 22 Revised Simplex Method 09/23/04 page 6 of 22
Change of basis asic variable X[9] leaves basis New partition: = {6 3 4}, N= { 2 5 7 8 0 9} All artificial variables (9, 0, ) are now nonbasic. ***Feasibility has been achieved! End Phase One*** Optimal Phase One Solution i X[i] Cz 0 0 2 0 0 3 3.32432 0 4 2.78378 0 5 0 0 6 2.4054 0 7 0 0 8 0 0 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= { 2 5 7 8} asis inverse is 0.297297 0.054054-0.0808-0.6262 0.243243 0.3535 0.0808-0.6262 0.243243 The costs of the basic variables are c c, c, c 4, 4, 7 6 3 4 Compute the simplex multipliers: 0.297297 0.054054 0.80.8 c A 4, 4, 7 0.6262 0.243243 0.3535 0.0808 0.6262 0.243243.2973, 0.054054,.9892 Revised Simplex Method 09/23/04 page 8 of 22 Simplex multipliers (dual solution) i.2973 2 0.054054 3.9892 Reduced cost of x is, for example, 2 c A 3.2973, 0.054054,.9892 0.3535 0 Reduced costs j Cj set only variable 2 would improve 0.3535 N the phase II objective 2 -.37838 N enter 5 2.9892 N 7 0.054054 N 8.9892 N Entering variable is X[2] from set N Compute the substitution rates: 0.297297 0.054054 0.80.8 0.626 A 2 A 0.6262 0.243243 0.3535 0 0.7027 0.0808 0.6262 0.243243 4 0.8648 L X d ratio 6 0 2.4054-0.626 -- 3 0 3.32432 0.7027 4.730 4 0 2.78378 0.8648 3.28 min ratio Change of basis asic variable X[4] leaves basis New partition: = {6 3 2}, N= { 5 7 8 4} Revised Simplex Method 09/23/04 page 9 of 22 Revised Simplex Method 09/23/04 page 20 of 22
Update the basis inverse, by writing the substitution rates alongside the old basis inverse, and pivoting: 0.297297 0.054054 0.80.8 0.626??? 0 0.6262 0.243243 0.3535 0.7027??? 0 etc. 0.0808 0.6262 0.243243 0.8648??? When the simplex multipliers and the reduced costs have been computed, the resulting reduced costs are all nonnegative! Optimal Solution Objective function: Z= 37.9688 Primal Solution (reduced costs are all nonnegative) i L X[i] set reduced cost 0 0 N 0.4375 2 0 3.2875 0 3 0.0625 0 4 0 0 N.59375 5 0 0 N 2.5325 6 0 4.40625 0 7 0 0 N 0.325 8 0 0 N.5325 Dual Solution i.25 2 0.325 3.5325 Revised Simplex Method 09/23/04 page 2 of 22 Revised Simplex Method 09/23/04 page 22 of 22