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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