We begin by adding slack variables and converting inequalities into equalities.

The initial tableau is then as follows:

The correct second tableau and third tableau and some of their calculations follow. The optimal solution, given in the third tableau, is $X_1=18, X_2=4, S_1=0, S_2=0,$ and profit $=\$190.$

1. Stpng 1 and 2 To go from the first to the second tableau, we note that the pivot column (in the first tableau) is $X_1,$ which has the highest $C_j-Z_j$ value, $9. The pivot row is $S_1$, since 40/2 is less than 30/1, and the pivot number is 2.

2. Step 3 The new $X_1$ row is found by dividing each number in the old $S_1$ row by the pivot number—that is, $2/2=1, 1/2=0.5, 1/2=0.5, 0/2=0,$ and $40/2=20.$

3. Step 4 The new values for the $S_2$ row are computed as follows:

   $\left(\begin{array}{l}\text{Number in}\\ {\text{new S}}_{\text{2}}\text{ row}\end{array}\right)\times \left(\begin{array}{l}\text{Number in}\\ {\text{old S}}_{\text{2}}\text{ row}\end{array}\right)-\left[\left(\begin{array}{l}\text{Number below}\\ \text{pivot number}\end{array}\right)\times \left(\begin{array}{l}\text{Corresponding}\\ \text{number in}\\ \text{new }{X}_{\text{1}}\text{ row}\end{array}\right)\right]$
   0	$=$	1	$-[\left(1\right)$	$\times$	$\left(1\right)]$
   2.5	$=$	3	$-[\left(1\right)$	$\times$	$\left(0.5\right)]$
   $20.5$	$=$	0	$-[\left(1\right)$	$\times$	$\left(0.5\right)]$
   1	$=$	1	$-[\left(1\right)$	$\times$	$\left(0\right)]$
   10	$=$	30	$-[\left(1\right)$	$\times$	$\left(20\right)]$

4. Step 5 The following new $Z_j$ and $C_j-Z_j$ rows are formed:

   $$\begin{array}{ll}{Z}_j\left(\text{for }{X}_{\text{1}}\right)\text{ = \$9}\left(\text{1}\right)\text{ + 0}\left(\text{0}\right)\text{ = \$9}\hfill & C_j\text{− }{Z}_j\text{ = \$9 − \$9 = 0}\hfill \\ Z_j\left(\text{for }{X}_{\text{2}}\right)\text{ = \$9}\left(\text{0}\text{.5}\right)\text{ + 0}\left(\text{2}\text{.5}\right)\text{ = \$4}\text{.5}\hfill & C_j\text{ − }{Z}_j\text{ = \$7 − 4}\text{.5 = \$2}\text{.5}\hfill \\ Z_j\left(\text{for }{S}_{\text{1}}\right)\text{ = \$9}\left(\text{0}\text{.5}\right)\text{ + 0}\left(\text{−0}\text{.5}\right)\text{ = \$4}\text{.5}\hfill & C_j\text{ − }{Z}_j\text{ = 0 − 4}\text{.5 = \$4}\text{.5}\hfill \\ Z_j\left(\text{for }{S}_{\text{2}}\right)\text{ = \$9}\left(\text{0}\right)\text{ + 0}\left(\text{1}\right)\text{ = \$0}\hfill & C_j\text{ − }{Z}_j\text{ = 0 − 0 = 0}\hfill \\ Z_j\left(\text{profit}\right)\text{ = \$9}\left(\text{20}\right)\text{ + 0}\left(\text{10}\right)\text{ = \$180}\hfill & \hfill \end{array}$$

   

   7.1-29 Full Alternative Text

   This solution is not optimal, and you must perform stpng 1 to 5 again. The new pivot column is $X_2,$ the new pivot row is $S_2,$ and 2.5 (circled in the second tableau) is the new pivot number.

   

   7.1-30 Full Alternative Text

   The final solution is $X_1=18, X_2=4,$ and profit $=\$190.$

1. Shadow price $=\mathrm{-}\left(C_j-Z_j\right)$

   For constraint 1, shadow price $=\mathrm{-}\left(\mathrm{-}4\right)=4.$

   For constraint 2, shadow price $=\mathrm{-}\left(\mathrm{-}1\right)=1.$

2. For constraint 1, we use the $S_1$ column.

   Quantity	S1	Ratio
   18	0.6	$18/\left(0.6\right)=30$
   4	$\mathrm{-}0.2$	$4/\left(\mathrm{-}0.2\right)=\mathrm{-}20$

   The smallest positive ratio is 30, so we may reduce the right-hand side of constraint 1 by 30 units (for a lower bound of $40-30=10$). Similarly, the negative ratio of $\mathrm{-}20$ tells us that we may increase the right-hand side of constraint 1 by 20 units (for an upper bound of $40+20=60$).

3. $$\begin{array}{ll}\text{The maximum possible profit}\hfill & \text{= Original profit + 10}\left(\text{shadow price}\right)\hfill \\ \hfill & \text{= 190 + 10}\left(\text{4}\right)\text{ = 230}\hfill \end{array}$$

   The values for the basic variables are found using the original quantities and the substitution rates:

   Original Quantity	$S_1$	New Quantity
   18	0.6	$18+\left(0.6\right)\left(10\right)=24$
   4	$\mathrm{-}0.2$	$4+\left(\mathrm{-}0.2\right)\left(10\right)=2$

   $$\begin{array}{l}{X}_{\text{1}}\text{ = 24, }{X}_{\text{2}}\text{ = 2, }{S}_{\text{1}}\text{ = 0, }{S}_{\text{2}}\text{ = 0 }\left(\text{both slack variables remain nonbasic variables}\right)\\ \text{profit = 9}\left(\text{24}\right)\text{ + 7}\left(\text{2}\right)\text{ = 230 }\left(\text{which was also found using the shadow price}\right)\end{array}$$

4. Let $\Delta =$ change in profit for $X_1.$

   

   7.1-31 Full Alternative Text

   For this solution to remain optimal, the $C_j-Z_j$ values must remain negative or zero.

   $$\begin{array}{rrr}\hfill -\text{4 }-\left(\text{0}\text{.6}\right)\Delta & \hfill \le & 0\hfill \\ \hfill -\text{4}& \hfill \le & \left(0.6\right)\Delta \hfill \\ \hfill -\text{20/3}& \hfill \le & \Delta \hfill \end{array}$$

   and

   $$\begin{array}{rrr}\hfill -1\text{ + }\left(\text{0}\text{.2}\right)\Delta & \hfill \le & 0\hfill \\ \hfill \left(0.2\right)\Delta & \hfill \le & 1\hfill \\ \hfill \Delta & \hfill \le & 5\hfill \end{array}$$

   So the change in profit $\left(\Delta \right)$ must be between $\mathrm{-}20/3$ and 5. The original profit was 9, so this solution remains optimal as long as the profit on $X_1$ is between $2.33=9-20/3$ and $14=9+5.$