Question #242479

Consider the following primal problem;


minimize x0=20x1+30x2+50x3+40x4

subject to:

4x1+6x2+x3+2x4≥12

2x1+x2+6x3+5x4≥14

x1+2x2+4x3+3x4≥8

xi≥0,i=1,2,3,4

Use simplex method to solve the dual of the above primal problem


1
Expert's answer
2021-10-18T12:36:31-0400

MIN Zx=20x1+30x2+50x3+40x4Subject to:4x1+6x2+x3+2x4122x1+x2+6x3+5x414x1+2x2+4x3+3x48and x1,x2,x3,x40;MIN \ Z_x = 20 x_1 + 30 x_2 + 50 x_3 + 40 x_4\\ Subject \ to:\\ 4 x_1 + 6 x_2 + x_3 + 2 x_4 ≥ 12\\ 2 x_1 + x_2 + 6 x_3 + 5 x_4 ≥ 14\\ x_1 + 2 x_2 + 4 x_3 + 3 x_4 ≥ 8\\ and \ x_1,x_2,x_3,x_4≥0;

Dual is (Solution steps of Dual by BigM method)MAX Zy=12y1+14y2+8y3Subject to:4y1+2y2+y3206y1+y2+2y330y1+6y2+4y3502y1+5y2+3y340and y1,y2,y30;Dual \ is \ (Solution \ steps \ of \ Dual\ by\ BigM \ method)\\ MAX\ Z_y = 12 y_1 + 14 y_2 + 8 y_3\\ Subject\ to:\\ 4 y_1 + 2 y_2 + y_3 ≤ 20\\ 6 y_1 + y_2 + 2 y_3 ≤ 30\\ y_1 + 6 y_2 + 4 y_3 ≤ 50\\ 2 y_1 + 5 y_2 + 3 y_3 ≤ 40\\ and\ y_1,y_2,y_3≥0;\\

Max:Z=12y1+14y2+8y3subject to:4y1+2y2+y3206y1+y2+2y330y1+6y2+4y3502y1+5y2+3y340and y1,y2,y30Max: Z=12 y_{1}+14 y_{2}+8 y_{3} \\subject \ to: 4 y_{1}+2 y_{2}+y_{3} \leq 20 \\ 6 y_{1}+y_{2}+2 y_{3} \leq 30 \\y_{1}+6 y_{2}+4 y_{3} \leq 50 \\ 2 y_{1}+5 y_{2}+3 y_{3} \leq 40 \\ and \ y_{1}, y_{2}, y_{3} \geq 0 \\

The problem is converted to canonical form by adding slack, surplus, and artificial variables as appropriate


1. As the constraint-1 is of type '≤' we should add slack variable S1S_1


2. As the constraint-2 is of type '≤' we should add slack variable S2S_2


3. As the constraint-3 is of type '≤' we should add slack variable S3S_3


4. As the constraint-4 is of type '≤' we should add slack variable S4S_4


After introducing slack variables


MaxZ=12y1+14y2+8y3+0S1+0S2+0S3+0S4subject to:4y1+2y2+y3+S1=206y1+y2+2y3+S2=30y1+6y2+4y3+S3=502y1+5y2+3y3+S4=40 and y1,y2,y3,S1,S2,S3,S40\operatorname{Max} Z=12 y_{1}+14 y_{2}+8 y_{3}+0 S_{1}+0 S_{2}+0 S_{3}+0 S_{4} \\ subject \ to:\\ \begin{aligned} 4 y_{1}+2 y_{2}+y_{3}+S_{1} =20 \\ 6 y_{1}+y_{2}+2 y_{3}+S_{2} =30 \\ y_{1}+6 y_{2}+4 y_{3} +S_3=50 \\ 2 y_{1}+5 y_{2}+3 y_{3} +S_{4} =40\\ \end{aligned}\\ \text { and } y_{1}, y_{2}, y_{3}, S_{1}, S_{2}, S_{3}, S_{4} \geq 0




The negative minimum Zj-Cj is -14 and its column index is 2. So, the entering variable is y2y_2 .


The minimum ratio is 8 and its row index is 4. So, the leaving basis variable is S4S_4 .


∴ The pivot element is 5.


Entering =y2y_2 , Departing =S4S_4 , Key Element =5





The negative minimum Zj-Cj is -6.4 and its column index is 1. So, the entering variable is y1.


The minimum ratio is 1.25 and its row index is 1. So, the leaving basis variable is S1.


 ∴ The pivot element is 3.2.


Entering =y1, Departing =S1, Key Element =3.2






Since all Zj-Cj≥0


Hence, the optimal solution arrives with value of variables as :

y1=1.25,y2=7.5,y3=0y_1=1.25,y_2=7.5,y_3=0


Max Z=120


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