Question

Consider the Linear Programming Problem P:
\[ \text{Maximize } 3x_1 + 2x_2 + 5x_3 \]
subject to
\[ \begin{aligned} x_1 + 2x_2 + x_3 &\le 44, \\ x_1 + 2x_3 &\le 48, \\ x_1 + 4x_2 &\le 52, \\ x_1 \ge 0, \ x_2 \ge 0, \ x_3 &\ge 0 \end{aligned} \]
The optimal value of the problem P is equal to ..................

Hint:

When performing the simplex method, be meticulous with the row operations as arithmetic errors are common. Always double-check your pivot selection and calculations before moving to the next iteration. A final check of the solution against the constraints is a good practice to ensure feasibility.

Answer 1

Step 1: Understanding the Concept:
This is a standard Linear Programming Problem (LPP). We need to find the maximum value of a linear objective function subject to a set of linear inequality constraints. The Simplex Method is an efficient algorithm for solving such problems.
Step 2: Key Formula or Approach:
We will use the Simplex Method.
1. Convert the inequalities into equalities by introducing slack variables \(s_1, s_2, s_3 \ge 0\).
2. Set up the initial simplex tableau. The initial basic feasible solution is \(x_1 = x_2 = x_3 = 0\).
3. Select the pivot column: the column with the most negative indicator in the objective function row (Z-row).
4. Select the pivot row: perform the minimum ratio test. For each positive entry in the pivot column, divide the RHS value by the entry. The row with the smallest ratio is the pivot row.
5. Perform pivot operations (row operations) to make the pivot element 1 and all other elements in the pivot column 0.
6. Repeat steps 3–5 until there are no negative indicators in the Z-row. The solution is then optimal.
Step 3: Detailed Explanation or Calculation:
The problem is:
Maximize \(Z = 3x_1 + 2x_2 + 5x_3\) or \(Z - 3x_1 - 2x_2 - 5x_3 = 0\).
Constraints:
\[ \begin{aligned} x_1 + 2x_2 + x_3 + s_1 &= 44, \\ x_1 + 2x_3 + s_2 &= 48, \\ x_1 + 4x_2 + s_3 &= 52 \end{aligned} \] Initial Tableau: \[ \begin{array}{c|cccccc|c} \text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & 1 & 2 & 1 & 1 & 0 & 0 & 44 \\ s_2 & 1 & 0 & 2 & 0 & 1 & 0 & 48 \\ s_3 & 1 & 4 & 0 & 0 & 0 & 1 & 52 \\ \hline Z & -3 & -2 & -5 & 0 & 0 & 0 & 0 \end{array} \] Iteration 1:
Pivot column: \(x_3\) (most negative is -5).
Ratio test: \(44/1 = 44\), \(48/2 = 24\). Minimum is 24.
Pivot row: Row 2 (\(s_2\)). Pivot element is 2.
Entering variable: \(x_3\). Leaving variable: \(s_2\).
Row operations: \(R_2 \to R_2/2\), \(R_1 \to R_1 - R_{2,new}\), \(R_Z \to R_Z + 5R_{2,new}\).
Tableau 1: \[ \begin{array}{c|cccccc|c} \text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & 1/2 & 2 & 0 & 1 & -1/2 & 0 & 20 \\ x_3 & 1/2 & 0 & 1 & 0 & 1/2 & 0 & 24 \\ s_3 & 1 & 4 & 0 & 0 & 0 & 1 & 52 \\ \hline Z & -1/2 & -2 & 0 & 0 & 5/2 & 0 & 120 \end{array} \] Iteration 2:
Pivot column: \(x_2\) (most negative is -2).
Ratio test: \(20/2 = 10\), \(52/4 = 13\). Minimum is 10.
Pivot row: Row 1 (\(s_1\)). Pivot element is 2.
Entering variable: \(x_2\). Leaving variable: \(s_1\).
Row operations: \(R_1 \to R_1/2\), \(R_3 \to R_3 - 4R_{1,new}\), \(R_Z \to R_Z + 2R_{1,new}\).
Tableau 2 (Optimal): \[ \begin{array}{c|cccccc|c} \text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline x_2 & 1/4 & 1 & 0 & 1/2 & -1/4 & 0 & 10 \\ x_3 & 1/2 & 0 & 1 & 0 & 1/2 & 0 & 24 \\ s_3 & 0 & 0 & 0 & -2 & 1 & 1 & 12 \\ \hline Z & 0 & 0 & 0 & 1 & 2 & 0 & 140 \end{array} \] Step 4: Final Answer:
Since all indicators in the Z-row are non-negative, the tableau is optimal.
The optimal value is \(Z = 140\).
The solution is \(x_1 = 0\), \(x_2 = 10\), \(x_3 = 24\).
Step 5: Why This is Correct:
The simplex algorithm was applied correctly. Each pivoting step improved the objective function value. The final tableau has no negative indicators in the objective row, confirming optimality. The value of Z from this final tableau is 140.
Verification:
\(Z = 3(0) + 2(10) + 5(24) = 0 + 20 + 120 = 140\).
Constraints:
1. \(0 + 2(10) + 24 = 44 \le 44\) (OK)
2. \(0 + 2(24) = 48 \le 48\) (OK)
3. \(0 + 4(10) = 40 \le 52\) (OK)
Solution is feasible and optimal.