2. The simplex tableau below is an intermediate step resulted from solving a linear programming problem using simplex method. (1) What is the current basic feasible solution? Is the current solution optimal, why? (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. 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? Iteration 1: Π x1 x2 x3 s1 s2 s3 s4 s5 constant 1 0 -200 -100 0 0 300 0 0 15000 0 0 10 2 1 0 -16 0 0 200 0 0 4 2 0 1 -8 0 0 100 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 0 0 0 1 0 0 0 0 1 150 Iteration 2: Π x1 x2 x3 s1 s2 s3 s4 $5 constant

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2. The simplex tableau below is an intermediate step resulted from solving a linear
programming problem using simplex method.
(1) What is the current basic feasible solution? Is the current solution optimal, why?
(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.
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?
Iteration 1:
Π
x1
x2
x3
s1
s2
s3
s4
s5
constant
1
0
-200
-100
0
0
300
0
0
15000
0
0
10
2
1
0
-16
0
0
200
0
0
4
2
0
1
-8
0
0
100
0
1
0
0
0
0
1
0
0
50
0
0
1
0
0
0
0
1
0
80
0
0
0
1
0
0
0
0
1
150
Iteration 2:
Π
x1
x2
x3
s1
s2
s3
s4
$5
constant
Transcribed Image Text:2. The simplex tableau below is an intermediate step resulted from solving a linear programming problem using simplex method. (1) What is the current basic feasible solution? Is the current solution optimal, why? (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. 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? Iteration 1: Π x1 x2 x3 s1 s2 s3 s4 s5 constant 1 0 -200 -100 0 0 300 0 0 15000 0 0 10 2 1 0 -16 0 0 200 0 0 4 2 0 1 -8 0 0 100 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 0 0 0 1 0 0 0 0 1 150 Iteration 2: Π x1 x2 x3 s1 s2 s3 s4 $5 constant
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