Answer :
Absolutely, let's go through the steps one by one to pivot around the highlighted element in the simplex tableau.
We begin with the original simplex tableau:
[tex]\[ \left[ \begin{array}{rrrrrrrr} x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & z & \\ 2 & 2 & 1 & 1 & 0 & 0 & 0 & 10 \\ 1 & 2 & 3 & 0 & 1 & 0 & 0 & 40 \\ 3 & 1 & 1 & 0 & 0 & 1 & 0 & 20 \\ \hline -2 & -1 & -3 & 0 & 0 & 0 & 1 & 0 \end{array} \right] \][/tex]
### Step 1: Identify the pivot element
The pivot element is identified as [tex]\(x_3\)[/tex] in the first row. The cell with coordinates (row 1, column 3) possesses the element [tex]\(1\)[/tex].
### Step 2: Normalize the pivot row
To normalize the pivot row, we divide every element in the pivot row by the pivot element (which is 1 in this case).
[tex]\[ \text{First row:} \quad \left[ \frac{2}{1}, \frac{2}{1}, \frac{1}{1}, \frac{1}{1}, \frac{0}{1}, \frac{0}{1}, \frac{0}{1}, \frac{10}{1} \right] = [2.0, 2.0, 1.0, 1.0, 0.0, 0.0, 0.0, 10.0] \][/tex]
### Step 3: Perform row operations to zero out the rest of the pivot column
1. For the second row (row index 2): Subtract [tex]\(3\)[/tex] times the new pivot row from the second row to make the pivot column element zero.
[tex]\[ \begin{aligned} &\left[ 1, 2, 3, 0, 1, 0, 0, 40 \right] - 3 \times \left[ 2, 2, 1, 1, 0, 0, 0, 10 \right] \\ &=\left[ 1 - 6, 2 - 6, 3 - 3, 0 - 3, 1 - 0, 0 - 0, 0 - 0, 40 - 30 \right] = [-5.0, -4.0, 0.0, -3.0, 1.0, 0.0, 0.0, 10.0] \end{aligned} \][/tex]
2. For the third row (row index 3): Subtract [tex]\(1\)[/tex] times the new pivot row from the third row to make the pivot column element zero.
[tex]\[ \begin{aligned} &\left[3, 1, 1, 0, 0, 1, 0, 20\right] - 1 \times \left[2, 2, 1, 1, 0, 0, 0, 10 \right] \\ &= \left[3 - 2, 1 - 2, 1 - 1, 0 - 1, 0 - 0, 1 - 0, 0 - 0, 20 - 10 \right] = [1.0, -1.0, 0.0, -1.0, 0.0, 1.0, 0.0, 10.0] \end{aligned} \][/tex]
3. For the objective function row (row index 4): Add [tex]\(3\)[/tex] times the new pivot row to the fourth row to make the pivot column element zero.
[tex]\[ \begin{aligned} & \left[-2, -1, -3, 0, 0, 0, 1, 0\right] + 3 \times \left[2, 2, 1, 1, 0, 0, 0, 10\right] \\ &= \left[-2 + 6, -1 + 6, -3 + 3, 0 + 3, 0 + 0, 0 + 0, 1 + 0, 0 + 30 \right] = [4.0, 5.0, 0.0, 3.0, 0.0, 0.0, 1.0, 30.0] \end{aligned} \][/tex]
### Step 4: Write the updated tableau
The updated tableau after one pivot operation, with the element (1,3) being the pivot, is:
[tex]\[ \left[ \begin{array}{rrrrrrrr} x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & z & \\ 2.0 & 2.0 & 1.0 & 1.0 & 0.0 & 0.0 & 0.0 & 10.0 \\ -5.0 & -4.0 & 0.0 & -3.0 & 1.0 & 0.0 & 0.0 & 10.0 \\ 1.0 & -1.0 & 0.0 & -1.0 & 0.0 & 1.0 & 0.0 & 10.0 \\ \hline 4.0 & 5.0 & 0.0 & 3.0 & 0.0 & 0.0 & 1.0 & 30.0 \end{array} \right] \][/tex]
This is the simplified solution of the tableau after one pivot operation.
We begin with the original simplex tableau:
[tex]\[ \left[ \begin{array}{rrrrrrrr} x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & z & \\ 2 & 2 & 1 & 1 & 0 & 0 & 0 & 10 \\ 1 & 2 & 3 & 0 & 1 & 0 & 0 & 40 \\ 3 & 1 & 1 & 0 & 0 & 1 & 0 & 20 \\ \hline -2 & -1 & -3 & 0 & 0 & 0 & 1 & 0 \end{array} \right] \][/tex]
### Step 1: Identify the pivot element
The pivot element is identified as [tex]\(x_3\)[/tex] in the first row. The cell with coordinates (row 1, column 3) possesses the element [tex]\(1\)[/tex].
### Step 2: Normalize the pivot row
To normalize the pivot row, we divide every element in the pivot row by the pivot element (which is 1 in this case).
[tex]\[ \text{First row:} \quad \left[ \frac{2}{1}, \frac{2}{1}, \frac{1}{1}, \frac{1}{1}, \frac{0}{1}, \frac{0}{1}, \frac{0}{1}, \frac{10}{1} \right] = [2.0, 2.0, 1.0, 1.0, 0.0, 0.0, 0.0, 10.0] \][/tex]
### Step 3: Perform row operations to zero out the rest of the pivot column
1. For the second row (row index 2): Subtract [tex]\(3\)[/tex] times the new pivot row from the second row to make the pivot column element zero.
[tex]\[ \begin{aligned} &\left[ 1, 2, 3, 0, 1, 0, 0, 40 \right] - 3 \times \left[ 2, 2, 1, 1, 0, 0, 0, 10 \right] \\ &=\left[ 1 - 6, 2 - 6, 3 - 3, 0 - 3, 1 - 0, 0 - 0, 0 - 0, 40 - 30 \right] = [-5.0, -4.0, 0.0, -3.0, 1.0, 0.0, 0.0, 10.0] \end{aligned} \][/tex]
2. For the third row (row index 3): Subtract [tex]\(1\)[/tex] times the new pivot row from the third row to make the pivot column element zero.
[tex]\[ \begin{aligned} &\left[3, 1, 1, 0, 0, 1, 0, 20\right] - 1 \times \left[2, 2, 1, 1, 0, 0, 0, 10 \right] \\ &= \left[3 - 2, 1 - 2, 1 - 1, 0 - 1, 0 - 0, 1 - 0, 0 - 0, 20 - 10 \right] = [1.0, -1.0, 0.0, -1.0, 0.0, 1.0, 0.0, 10.0] \end{aligned} \][/tex]
3. For the objective function row (row index 4): Add [tex]\(3\)[/tex] times the new pivot row to the fourth row to make the pivot column element zero.
[tex]\[ \begin{aligned} & \left[-2, -1, -3, 0, 0, 0, 1, 0\right] + 3 \times \left[2, 2, 1, 1, 0, 0, 0, 10\right] \\ &= \left[-2 + 6, -1 + 6, -3 + 3, 0 + 3, 0 + 0, 0 + 0, 1 + 0, 0 + 30 \right] = [4.0, 5.0, 0.0, 3.0, 0.0, 0.0, 1.0, 30.0] \end{aligned} \][/tex]
### Step 4: Write the updated tableau
The updated tableau after one pivot operation, with the element (1,3) being the pivot, is:
[tex]\[ \left[ \begin{array}{rrrrrrrr} x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & z & \\ 2.0 & 2.0 & 1.0 & 1.0 & 0.0 & 0.0 & 0.0 & 10.0 \\ -5.0 & -4.0 & 0.0 & -3.0 & 1.0 & 0.0 & 0.0 & 10.0 \\ 1.0 & -1.0 & 0.0 & -1.0 & 0.0 & 1.0 & 0.0 & 10.0 \\ \hline 4.0 & 5.0 & 0.0 & 3.0 & 0.0 & 0.0 & 1.0 & 30.0 \end{array} \right] \][/tex]
This is the simplified solution of the tableau after one pivot operation.
