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(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points) Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,

(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points) Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,

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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(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points)
Transcribed Image Text:(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points)
Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,
Transcribed Image Text:Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,
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