Question

2. Solve the LPP by the dual simplex method Minimize: z = 3x1 + 2x2 Subject to:: x1 + x2 > 1 4x1 + x2 > 2 -X1 + 2x2 < 6 Xi > 0, i=1,2

Answer 1

Answer

Please find the solution below,

Problem is MIN Z = 321+ a22 subject to 21 +2221 4x1+ x2 za x + 2x256 and 2,xazo In order to apply the dual simplex method, convert min z to Max 2 and all > constraint to < consixount by multiplying 1. . Problem is Max 2 = -32,-222 subject to -x,-125-1 - 411-425-2 -xl taxa<6 and 21,222o The peoblem is conated 10 canonical forms by adding slack, sue plus and artificial variables as appropiate 1. As the constraint-1 is of type 's we should add slaek variables, Q. As the constraint-a is of type 's' we should add Black varübe sa s. As the constraint-3 $ f type ' we should add slack variable S3 Alte indkodlucing the sleek veomables max 2 = -32,-222 +os, + 05a + Osz subject to -21-22 +S - 4*, - 82 +52 -21+222 +53 = 6 and a,x2,5,,S2,8320 - 1 = -2

Hochon cy -3 O 0 ö B CB Хв 11 22 si S3 S2 S O -1 - 1 O O 92 O -2 - -1 0 S3 — 0 6 -1 2 O 0 O 2-0 29 0 O 0 3 2 20 zj-c R = 2j-2 Sa,j -0.751 -2. & sajao minimum negative XBS -2 and its ear incles is 2. 80 the leaving basis variable is sa maximum negative ratio is –0 45 and its cdumis index is 1 so entering variable is 821 : The pivot element is -4 Entering = 21, Depreeting = S2, key element = -4 + Ra (now) - Ralold)+(-4) + Ri (new) = Ricold )+Ra(new) + R3(news) = Rs (old)+ Rx (new)

Itection-2 Cj -3 - 0 153 B CB XB 21 22 sa si -0.25 SI -0.5 0 Goias) -0.25 21 -3 0.5 0:25 0 0:25 - 0 Sg 6.5 O O 0 075 2-15 zi -3 -0.75 DAS o zj-cj 0 1:25 O Rato= zj-ci -1.6667 1 -3 Si's &s, jco minimum negatine XB is 0.5 and its row indea is i, so the leaving basis variable is s Maximum negative raho is -1.6667 omel is dumb indese is 3 so the entering variable is az .: The pivot element is -0.75 Entering = 22, Deporting as, key element = -0.75 + Ri Chew ) = Ri Cold)(-0.75) +Ra (new) = Ra (old) -0.25R, (new) + R3 (new) = Rg (old)-2.25R, (new)

The Minimum value of Z = 2.3333

Ilaahon - 3 B СЕ XB 11 12 Si S2. 12. -2. o 3333 -1333 0.6667 O 0 11 -3 0.3333 0.3333 -0.3333 - 53 o o 3 2=-2 333 O zj -3 -2. 16667 0.3333 zjoj 0 16667 0 3333 Ratio - - since all 27-ci zo and all XBizo this the current solution as the optimal solutim. Hence, optimal solution is axxived with value of varvable as. 21 = 0.3333 212 = 0.6667 Max 2 = -2.3333 . min 2 = 2.3333