Question

Consider the following tableau: 21 81 82 RHS P 0 1 1 0 3 3 2 7 2 0 0 3 1 12 -7 –12 0 0 0 a) Determine the pivot element and perform all the pivot operations for the entire pivot column to obtain the next tableau. In this next tableau that you obtained in the objective row, enter in each box below the number you have under each column. Note: Where applicable, fractions must be entered as 2/5, -1/3, and so on. Under column t1 in the objective row, you have: Under column 22 in the objective row, you have: Under column si in the objective row, you have: Under columns, in the objective row, you have: Under column RHS in the objective row, you have: b) Given the tableau that you obtained above, three interpretations are possible. In the box below, type or copy and paste whichever answer shown in boldface letters below that you think is the correct choice. 1. Enter solved if you think you have obtained the final tableau. 2. Enter ready for another set of pivot operations if you think more pivot operations are possible. 3. Enter no solution if you think the tableau has no solution.

Answer 1

We have,

	Basic variables	P	$x_1$	12	$S_1$	S2	RHS
RI	$S_1$	0	1	$\frac{3}{2}$	$\frac{1}{2}$	0	3
$R_2$	S2	0	3	7	0	1	12
R3		1	-7	-12	0	0	0

First of all we find the pivot column, for this we look for the most negative entry in R₃, which in our case is -12(highlighted in red colour). So, x₂ is our pivot column and the incoming variable.

Now, we look for our pivot row. To find the pivot row, we divide each entry in the RHS column by the entry in the corresponding in the pivot column. In this case, we’ll get $\small \frac{3}{\frac{3}{2}}=2$ as the ratio for the first row and $\small \frac{12}{7}=1.71$ for the ratio in the second row. The pivot row is the row corresponding to the smallest ratio, in this case 1.71. So, our pivot row is R₂ and thus S₂ is the outgoing variable

Pivot element is the entry at the intersection of pivot row and pivot column, which in this case is 7(highlighted by green colour).

We now perform $\small R_2\rightarrow \frac{R_2}{7}$ to make the pivot element 1, we get:

		P	x1	x2	S1	S2	RHS
R1	S1	0	1	$\frac{3}{2}$	$\frac{1}{2}$	0	3
R2	S2	0	$\small \frac{3}{7}$	1	0	$\small \frac{1}{7}$	$\small \frac{12}{7}$
R3		1	-7	-12	0	0	0

Now we perform the following row operations to get convert the pivot column to a unit column:

$\small R_1\rightarrow R_1-\frac{3}{2}R_2$

$\small R_3\rightarrow R_3+12R_2$

		P	x1	x2	S1	S2	RHS
R1	S1	0	$\small \frac{5}{14}$	0	$\frac{1}{2}$	$\small \frac{-3}{14}$	$\small \frac{3}{7}$
R2	x2	0	$\small \frac{3}{7}$	1	0	$\small \frac{1}{7}$	$\small \frac{12}{7}$
R3		1	$\small \frac{-13}{7}$	0	0	$\small \frac{12}{7}$	$\small \frac{144}{7}$

• Under column $x_1$ in the objective row we have  $\small \frac{-13}{7}$
• Under column
  12
  in the objective row we have 0
• Under column $S_1$ in the objective row we have 0
• Under column
  S2
  in the objective row we have $\small \frac{12}{7}$
• Under column RHS in the objective row we have   $\small \frac{144}{7}$

Since, we still have one negative entry left in the R₃ row so more pivot operations are required.

The correct choice is 2. ready for another set of pivot operations.

Hope this helps!