The simplex tableau below is an intermediate step resulted from solving a linear programming problem using simplex method. Iteration 1: Z x1 x2 x3 sl $2 $3 constant 1 0 -2 -1 0 0 3 15 0 0 1 2 0 1 -6 10 0 0 (2) 4 1 0 -8 8 0 1 0 0 0 0 1 4 (1) What is the current basic feasible solution? Is the current solution optimal, why? (5 points) X₁ = 4 S₁ = 8 S₂ = 10 No, it is not optimal, because there are ways to further improve, parameters in row 。 being negative. (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? (5 points) Iteration 2: ☑ x1 x2 x3 sl s2 s3 I 0 0 3 0-5 constant 23 0 0 2 1/2 0 -4 4 (X1, X2, S₂ I are in the basis at iteration 2.
The simplex tableau below is an intermediate step resulted from solving a linear programming problem using simplex method. Iteration 1: Z x1 x2 x3 sl $2 $3 constant 1 0 -2 -1 0 0 3 15 0 0 1 2 0 1 -6 10 0 0 (2) 4 1 0 -8 8 0 1 0 0 0 0 1 4 (1) What is the current basic feasible solution? Is the current solution optimal, why? (5 points) X₁ = 4 S₁ = 8 S₂ = 10 No, it is not optimal, because there are ways to further improve, parameters in row 。 being negative. (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? (5 points) Iteration 2: ☑ x1 x2 x3 sl s2 s3 I 0 0 3 0-5 constant 23 0 0 2 1/2 0 -4 4 (X1, X2, S₂ I are in the basis at iteration 2.
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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Can you show how to get to these answers.
Transcribed Image Text:The simplex tableau below is an intermediate step resulted from solving a linear
programming problem using simplex method.
Iteration 1:
Z x1
x2
x3
sl
$2
$3
constant
1
0
-2
-1
0
0
3
15
0
0
1
2
0
1
-6
10
0
0
(2)
4
1
0
-8
8
0
1
0
0
0
0
1
4
(1) What is the current basic feasible solution? Is the current solution optimal, why? (5
points)
X₁ = 4
S₁ = 8
S₂ = 10
No, it is not optimal, because there are
ways to further improve, parameters in
row 。 being negative.
(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? (5 points)
Iteration 2:
☑
x1
x2
x3
sl
s2
s3
I
0
0
3
0-5
constant
23
0
0
2
1/2
0 -4
4
(X1, X2, S₂ I are in the basis at iteration 2.
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