Answer :
To set up the initial simplex tableau for the given problem, we need to understand the structure of the tableau based on the objective function and constraints. We deal with a linear optimization problem in which we seek to maximize or minimize an objective function subject to certain constraints.
### Objective Function and Constraints
We have an objective function:
[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 \][/tex]
We will rewrite the objective function with negative coefficients because the simplex method handles maximization by converting it to a minimization form:
[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 + 0s_1 + 0s_2 + 0s_3 \][/tex]
### Construct the Simplex Tableau
Based on the standard form and the constraints provided in the problem, we can create the initial simplex tableau.
1. First Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 1s_1 + 0s_2 + 0s_3 \leq 600
\][/tex]
2. Second Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0.5x_2 & + & 0.125x_3 & + & 0.125x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 1s_2 + 0s_3 \leq 300
\][/tex]
3. Third Constraint Row:
[tex]\[
\begin{array}{ccccccc}
1x_1 & + & 0.5x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 0s_2 + 1s_3 \leq 400
\][/tex]
4. Objective Row: We take the negative of each coefficient from the objective function:
[tex]\[
\begin{array}{cccccccc}
-110x_1 & - & 90x_2 & - & 80x_3 & - & 70x_4 & \\
\end{array}
\][/tex]
+
[tex]\[
0s_1 + 0s_2 + 0s_3 + 0 \text{(initial profit)}
\][/tex]
### Assembling the Simplex Tableau
Putting it all together, the initial simplex tableau is:
[tex]\[
\begin{array}{cccccccc}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \text{RHS} \\
\hline
0 & 0 & 0.25 & 0 & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.125 & 0.125 & 0 & 1 & 0 & 300 \\
1 & 0.5 & 0.25 & 0 & 0 & 0 & 1 & 400 \\
\hline
-110 & -90 & -80 & -70 & 0 & 0 & 0 & 0 \\
\end{array}
\][/tex]
This tableau will be used in the simplex algorithm to iteratively find the optimal solution to the problem.
### Objective Function and Constraints
We have an objective function:
[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 \][/tex]
We will rewrite the objective function with negative coefficients because the simplex method handles maximization by converting it to a minimization form:
[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 + 0s_1 + 0s_2 + 0s_3 \][/tex]
### Construct the Simplex Tableau
Based on the standard form and the constraints provided in the problem, we can create the initial simplex tableau.
1. First Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 1s_1 + 0s_2 + 0s_3 \leq 600
\][/tex]
2. Second Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0.5x_2 & + & 0.125x_3 & + & 0.125x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 1s_2 + 0s_3 \leq 300
\][/tex]
3. Third Constraint Row:
[tex]\[
\begin{array}{ccccccc}
1x_1 & + & 0.5x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 0s_2 + 1s_3 \leq 400
\][/tex]
4. Objective Row: We take the negative of each coefficient from the objective function:
[tex]\[
\begin{array}{cccccccc}
-110x_1 & - & 90x_2 & - & 80x_3 & - & 70x_4 & \\
\end{array}
\][/tex]
+
[tex]\[
0s_1 + 0s_2 + 0s_3 + 0 \text{(initial profit)}
\][/tex]
### Assembling the Simplex Tableau
Putting it all together, the initial simplex tableau is:
[tex]\[
\begin{array}{cccccccc}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \text{RHS} \\
\hline
0 & 0 & 0.25 & 0 & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.125 & 0.125 & 0 & 1 & 0 & 300 \\
1 & 0.5 & 0.25 & 0 & 0 & 0 & 1 & 400 \\
\hline
-110 & -90 & -80 & -70 & 0 & 0 & 0 & 0 \\
\end{array}
\][/tex]
This tableau will be used in the simplex algorithm to iteratively find the optimal solution to the problem.