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Consider the following LP: Maximize z = 5x1 + 2x2 + 3x3; subject to x1 + 5x2 + 2x3 = 15; x1 - 5x2 - 6x3 <=20; x1, x2, x3 >= 0; The starting solution consists of artificial x4 for the first constraint and slack x5 for the second constraint. Using M = 100 for the artificial variables, the optimal tableau is given below. Use the optimal inverse matrix for the LP to find the optimal solution for the dual problem. Basic z x1 x5 xl 0 1 0 x2 23 5 -10 (y1,y2)=(3,2) x3 1 772 -8 x4 105 1 -1 x5 Solution 0 0 1 75 15 5

Consider the following LP: Maximize z = 5x1 + 2x2 + 3x3; subject to x1 + 5x2 + 2x3 = 15; x1 - 5x2 - 6x3 <=20; x1, x2, x3 >= 0; The starting solution consists of artificial x4 for the first constraint and slack x5 for the second constraint. Using M = 100 for the artificial variables, the optimal tableau is given below. Use the optimal inverse matrix for the LP to find the optimal solution for the dual problem. Basic z x1 x5 xl 0 1 0 x2 23 5 -10 (y1,y2)=(3,2) x3 1 772 -8 x4 105 1 -1 x5 Solution 0 0 1 75 15 5

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the following LP: Maximize z = 5x1 + 2x2 + 3x3; subject to x1 + 5x2 + 2x3 = 15; x15x2- 6x3 <=20; x1, x2, x3 >= 0; The starting solution consists of artificial x4 for the first constraint and slack x5 for the second constraint. Using M = 100 for the artificial variables, the optimal tableau is given below. Use the optimal inverse matrix for the LP to find the optimal solution for the dual problem. Basic Z x1 x5 xl 0 1 0 x2 23 5 -10 (y1,y2)=(3,2) (y1,y2)=(-1,3) (y1,y2)=(2,3) (y1,y2)=(5,0) (y1,y2)=(0,5) x3 7 2 -8 x4 105 1 -1 x5 0 0 1 Solution 75 15 5
Transcribed Image Text:Consider the following LP: Maximize z = 5x1 + 2x2 + 3x3; subject to x1 + 5x2 + 2x3 = 15; x15x2- 6x3 <=20; x1, x2, x3 >= 0; The starting solution consists of artificial x4 for the first constraint and slack x5 for the second constraint. Using M = 100 for the artificial variables, the optimal tableau is given below. Use the optimal inverse matrix for the LP to find the optimal solution for the dual problem. Basic Z x1 x5 xl 0 1 0 x2 23 5 -10 (y1,y2)=(3,2) (y1,y2)=(-1,3) (y1,y2)=(2,3) (y1,y2)=(5,0) (y1,y2)=(0,5) x3 7 2 -8 x4 105 1 -1 x5 0 0 1 Solution 75 15 5
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