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

Question. Max (Z) = -40,-64-18.33 Subject to 20,+303 > tu 3027203 >5 42) oli, da, zo Solution: We have both the inequalities (I) and (2) have the sign of >, therefore Convert the sign of > into a by multikdying both the sides of inequalities ) and (3) by - I, we feve Max (Z) = -40- 68-18 03 subject to -291-3033.-3 (3) -306-20 <-5 all, l2, 9370 Now Convert the inequalities (3) end (4) into equations by adding black variables ay and its to the left hand sides of inequalities (3) and (4) lesbectively. +4)

(e) as we have Max(Z) = -498,-64-1893 to Oly to go subject to -21, too -30% tout outs = -3 oli-34-2012 toalut Is = -5 ملا جلا رہاادرلاردافراد Simbelex table - I G -4 -6 -18 Basic GB XB G Ola Ol Dy I valiable -3 lu 0 als -5 0 -2 0 6 Idebastute 18 Zg- ↑ entering Note: G: Coefficient of of in the objective function z bivot for j = 1,2,3,4,5 XB : value of basic variable CB : coefficient of basic baliable in the objective function Zi element

O Zg C873 - Gj, fol j = 1,33,4,55. Note : Initially the lack variables de and as the basic variables, since they are providing standard basis Vectors and Now in the simplex table -I, all the Zj-G , fol j = 1,2,3,4,5 ale non-negative, therefore the solution given by 044 = -3 and as =-5 is obtinury but not feasible. y=-3 and as are negative we have in the column of XB as =-5 is most negative, therefore the black viable car is the departure variable from the basiso Now for entering variable, find the maximum of the lations of Z;- 9 to the Contesponding negative elements in the low of the departuse valiable als rii Max 1 2 3 = Max {-2,-9. which is cofresbonding to the element -3 6

(4) in the column vector of la Hence to is the entering valiable in to the basis and the element as in the Column vector of Ia is the pivot elemente Simhlex terbale -II: -6 -18 Basic O Variable Cå a4 ola é -65/30 23 -Y3 O 14 2. t departure Zg- bivatelement Pentering we have geht the ZgCg, for j = 1,2,3,4,5 are non-negative, the sefore the solution la = √ and Oly=-3 is optimung but not feasible 04 =-3 is negative? Now we have in the column vectos of XB only oly=-3 is negative, therefore the baliable ay is the departure variable

's) 4 2 from the basis. Now for entering variable, find the maximum of the lattose of Zj-Cj, fol j = 1,3,3145 to the corresponding negative elements in the low of the departure valiable au loco, in the first low. 14 Max ? Max ( -2, -143 = -2 which is Colfebording to the edenent a in the column vecto-l of 311, therefore the variable di is the entering Variable in to the basis and the element aa in the columy vector of all is the bivot elemente Sinplex table-th; -18 Basic GXBOX Ivariable 3/2 - |-615/3 -13 2 Z-C ہوا۔ ت o ola Of 94 ols -4 32 1 o 0 all dla 213 8 2

We have all the Zg-Cj, fof 9 = 1,2,3,4,5 (6) in the simplex table - Il are nou-negative and both the basic variables oli 2 / 2 and Ola = I are non-negative, therefore the solution di = 2/3 la = √ given by simplex Eable-I is optimum basic feasible solution Hence the optimum basic feasible solution is given by all Ye= 013- f is non-basic Variable, the reforte 9330 and Meexitum (Z) 6x = -18x0 -2 X3-25-O -6_o -16 تمام ادا х