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 s2 $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 S₁₁ = 8 S₂ = 10 No, it is not optimal, because there are ways to further improve, parameters in row 。 being negative.
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 s2 $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 S₁₁ = 8 S₂ = 10 No, it is not optimal, because there are ways to further improve, parameters in row 。 being negative.
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter12: Algebra Of Matrices
Section12.CR: Review Problem Set
Problem 37CR
Question
Can you show step by step on how to calculate the feasible solution
![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
s2
$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
S₁₁ = 8
S₂ = 10
No, it is not optimal, because there are
ways to further improve, parameters in
row 。 being negative.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeefd442-7c64-4a51-8ed1-c0196ac6a13e%2Fb9355ca5-8a99-4d14-ae6b-6e62868fa366%2F4un438o_processed.png&w=3840&q=75)
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
s2
$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
S₁₁ = 8
S₂ = 10
No, it is not optimal, because there are
ways to further improve, parameters in
row 。 being negative.
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