Questions: Consider the following intermediate tableau (not the Initial Simplex tableau): [ P x1 x2 x3 s1 s2 RHS 0 0 2 -1 1 -1 4 0 1 0 1 0 1 8 1 0 -3 -2 0 2 16 ] a) Determine the pivot column and the pivot element, and perform the necessary row operations to convert the pivot element into 1 and the other numbers in the pivot column to 0. Once you obtain the new tableau this way, look at the numbers you have in the objective row (the bottom row) and enter each one as requested in each box below. Note: Where applicable, fractions must be entered as 2 / 5, -1 / 3, and so on. Under column x1 in the objective row, you have: Under column x2 in the objective row, you have: Under column x3 in the objective row, you have: Under column s1 in the objective row, you have: Under column s2 in the objective row, you have: Under column RHS in the objective row, you have:

Solution Steps

To solve this problem, we need to follow these steps:

1. Identify the pivot column by selecting the column with the most negative value in the objective row.
2. Determine the pivot element by selecting the smallest positive ratio of RHS to the corresponding element in the pivot column.
3. Perform row operations to make the pivot element 1 and all other elements in the pivot column 0.
4. Update the tableau and extract the values from the objective row.

The pivot column is determined by selecting the column with the most negative value in the objective row. In this case, the pivot column is \( x_2 \) (column index 2).

Next, we calculate the ratios of the right-hand side (RHS) to the corresponding elements in the pivot column. The ratios are: \[ \text{ratios} = \left[ \frac{2}{1}, \infty \right] = [2, \infty] \] The smallest positive ratio is \( 2 \), which corresponds to the pivot row \( 0 \).

The pivot element is \( 2 \). We convert it to \( 1 \) by dividing the entire pivot row by \( 2 \): \[ \text{New Row 0} = \left[ 0, 0, 1, 0, 0, 0, 2 \right] \] Next, we perform row operations to make all other entries in the pivot column \( 0 \). The updated tableau becomes: \[ \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 & 0 & 1 & 8 \\ 1 & 0 & 0 & -2 & 0 & 2 & 22 \end{bmatrix} \]

The objective row is now: \[ \text{Objective Row} = \left[ 1, 0, 0, -2, 0, 2, 22 \right] \] From this row, we extract the following values:

• Under column \( x_1 \): \( 0 \)
• Under column \( x_2 \): \( 0 \)
• Under column \( x_3 \): \( -2 \)
• Under column \( s_1 \): \( 0 \)
• Under column \( s_2 \): \( 2 \)
• Under column RHS: \( 22 \)

Final Answer