Question

Use the dual simplex method to solve the following LP. Max z = -4xı - 6x2 - 18x3 Subject to 2x1 + 3x3 2 3 3x2 + 2x3 25 X1, X2, X3 20

Answer 1

① use the dual simplex method to solve the LP 0 Max z=-4x 6x g - 18x3 Subject to १५ + 3a 3 73 3xg 41 +223 715 221 4 3 710 Son The dual simplest method start with a better that optimal and infeasible basic sy". (1) sign to (2) by multiplying (1) one both Convert Maxm Z=-47,-6X2-18X3 Subto. -24 -34 x, 49, X3710 - 3X3 <-3 - QX3 <-5 Convert (s) sign by (=) sign by adding slack variables.

m Z=- 4x - 629 – 18xz Max Subto- ♡ -2x - 3x + x 4 =-3 --3Xg -223 tas - 5 X, X2, 131 tu, X5 710 Basic X 12 23 24 45 soin 4 6. o x4 10 -3 -2 O -3 X5 3 entring 0 learn 5 G -3 11) The tableau is optimal because all the reduced cost in Z- now are 70 because problem is maximize with G = 4, C2 = 6, C₂ =18, Cu=0, Cg=0. It is unfeasible because xy = -3, 85=-5. with the dual feasibility cough 45 = -5 is the learning Variable. The vratio shows de wis the euteuy variables, and the main { 1/3 [,1]

15 Basic ng 3 soln xy 24 23 14 0 le 4. Z -10 xy le entering -3 3- leaning 20 O تن wth 5 3 4 Ratio 14 -3 with the dual feasibility cond" x4 =-3 is the leaning variable. The vratio shows x, = Q = min { 1 ) 14 ) } is the entering variable.

15 Sol" xy Basic x 213 HQ Sol" ☺ -16 عع જ 8 O 0 elo ए 1 0 een 2 wfee No X 2 O 0 -1 3 cula 3 in this dual simplex methed in last tableau the optimality is maintained call wedured costs in the maximum problem) and feasibity also contained because all the att tastboltumt is 70 3. Here x = 2 ago culos Q and Z=-16 Also check in given objective fun z=-4x4 - 642-1843 z = -16