Let's solve the given linear programming (LP) problem step by step using the **Big M method** and the **Two-Phase method**. --- ### **Problem Statement** Maximize: \[ z = x_1 + x_2 \] Subject to: \[ \begin{cases} 2x_1 + x_2 \geq 3 \quad (1) \\ 3x_1 + x_2 \leq 3 \quad (2) \\ 5x_1 + x_2 \leq 1 \quad (3) \\ x_1, x_2 \geq 0 \end{cases} \] --- ## **Step 1: Convert inequalities into equations** - Inequality (1): \(2x_1 + x_2 \geq 3\) To convert into an equation, subtract a surplus variable \(s_1\): \[ 2x_1 + x_2 - s_1 = 3, \quad s_1 \geq 0 \] Because the inequality is \(\geq\), we subtract surplus variable \(s_1 \geq 0\). - Inequality (2): \(3x_1 + x_2 \leq 3\) Add slack variable \(s_2\): \[ 3x_1 + x_2 + s_2 = 3, \quad s_2 \geq 0 \] - Inequality (3): \(5x_1 + x_2 \leq 1\) Add slack variable \(s_3\): \[ 5x_1 + x_2 + s_3 = 1, \quad s_3 \geq 0 \] --- ## **Step 2: Detect the feasibility issue** - For the first constraint: \[ 2x_1 + x_2 \geq 3 \] Since \(x_1, x_2 \geq 0\), the minimal value of \(2x_1 + x_2\) is 0; but to satisfy the constraint, the sum must be at least 3. - For the third constraint: \[ 5x_1 + x_2 \leq 1 \] Similarly, the minimal value of the left side is 0, but it must be less than or equal to 1. - The problem seems feasible at first glance, but the key is to check whether the constraints are compatible. --- ## **Step 3: Use the Big M method** Since the first constraint is a \(\geq\) inequality, to handle it in an LP maximization, introduce **artificial variables** to find an initial feasible solution. - For the \(\geq\) constraint (1): \[ 2x_1 + x_2 - s_1 + a_1 = 3 \] where \(a_1 \geq 0\) is the artificial variable. - For the \(\leq\) constraints (2) and (3): \[ 3x_1 + x_2 + s_2 = 3 \] \[ 5x_1 + x_2 + s_3 = 1 \] --- ## **Step 4: Set up the Big M objective function** Objective: \[ z = x_1 + x_2 - M a_1 \] where \(M\) is a very large positive number to penalize artificial variables. --- ## **Step 5: Write the initial tableau** Variables: \[ x_1, x_2, s_2, s_3, a_1 \] Constraints: \[ \begin{cases} 2x_1 + x_2 - s_1 + a_1 = 3 \quad (1) \\ 3x_1 + x_2 + s_2 = 3 \quad (2) \\ 5x_1 + x_2 + s_3 = 1 \quad (3) \end{cases} \] Objective function: \[ z = x_1 + x_2 - M a_1 \] --- ## **Step 6: Construct the initial simplex tableau** | Basic | \(x_1\) | \(x_2\) | \(s_2\) | \(s_3\) | \(a_1\) | RHS | |---------|---------|---------|---------|---------|---------|-------| | \(a_1\) | 2 | 1 | 0 | 0 | 1 | 3 | | \(s_2\) | 3 | 1 | 1 | 0 | 0 | 3 | | \(s_3\) | 5 | 1 | 0 | 1 | 0 | 1 | | **Z** | -1 | -1 | 0 | 0 | \(M\) | 0 | Note: The artificial variable \(a_1\) appears only in row 1 with coefficient 1. The objective function subtracts \(M a_1\) to penalize it. --- ## **Step 7: Iterative process to remove artificial variables** - Since \(a_1\) is artificial, we aim to eliminate it from the basis to find a feasible solution. - The initial solution has \(a_1 = 3\), which is positive; thus, the artificial variable is in the basis and needs to be removed. --- ## **Step 8: Pivot to remove \(a_1\)** - To eliminate \(a_1\), perform pivot operations. - Pivot on \(a_1\) in row 1, dividing row 1 by 2: \[ \text{Row 1} \rightarrow \text{Row 1} / 2 \] Resulting row: | \(x_1\) | \(x_2\) | \(s_2\) | \(s_3\) | \(a_1\) | RHS | |---------|---------|---------|---------|---------|-------| | 1 | 0.5 | 0 | 0 | 0.5 | 1.5 | - Now, eliminate \(a_1\) from other rows: - Row 2: subtract \(1 \times\) new row 1: \[ \text{Row 2} \rightarrow \text{Row 2} - 1 \times \text{Row 1} \] \[ (3 - 1 \times 1) = 2 \] \[ (1 - 1 \times 0.5) = 0.5 \] \[ (1 - 0) = 1 \] \[ (0 - 0) = 0 \] \[ (3 - 1.5) = 1.5 \] - Row 3: subtract \(2 \times\) new row 1: \[ \text{Row 3} \rightarrow \text{Row 3} - 2 \times \text{Row 1} \] \[ (5 - 2 \times 1) = 3 \] \[ (1 - 2 \times 0.5) = 0 \] \[ (0 - 0) = 0 \] \[ (1 - 2 \times 0) = 1 \] \[ (1 - 2 \times 1.5) = 1 - 3 = -2 \] - Objective function: subtract \(M \times\) row 1: \[ \text{Row Z} \rightarrow \text{Row Z} + M \times \text{Row 1} \] \[ (-1 + M \times 1) = M - 1 \] \[ (-1 + M \times 0.5) = M/2 - 1 \] \[ (0 + M \times 0) = 0 \] \[ (0 + M \times 0) = 0 \] \[ (M + M \times 0.5) = M + 0.5 M = 1.5 M \] --- ## **Step 9: Updated tableau** | Basic | \(x_1\) | \(x_2\) | \(s_2\) | \(s_3\) | \(a_1\) | RHS | |---------|---------|---------|---------|---------|---------|-------| | \(a_1\) | 1 | 0.5 | 0 | 0 | 0.5 | 1.5 | | \(s_2\) | 0 | 0.5 | 1 | 0 | -0.5 | 1.5 | | \(s_3\) | 0 | -0.5 | 0 | 1 | -1 | -2 | | **Z** | \(M - 1\) | \(M/2 - 1\) | 0 | 0 | \(1.5 M\) | 0 | - The artificial variable \(a_1\) is now replaced. - The goal is to eliminate the artificial variables entirely, which is achieved when the artificial variables are zero in the basis and their coefficients in the objective function are non-positive. --- ## **Step 10: Check the feasibility** - The RHS of the second and third rows: - Row 2: 1.5 (positive) - Row 3: -2 (negative) - The negative RHS in row 3 indicates infeasibility unless adjustments are made. --- ## **Step 11: Consider the feasibility** - Since the RHS of row 3 is negative, the LP is **infeasible** because the constraints cannot be satisfied simultaneously. - Hence, **the LP problem has no feasible solution**. --- ## **## **Conclusion:** ### **Final assessment:** - The LP problem is **infeasible** because the constraints are incompatible; specifically, the third constraint \(5x_1 + x_2 \leq 1\) conflicts with the demand of the first constraint \(2x_1 + x_2 \geq 3\), making it impossible to satisfy both simultaneously with \(x_1, x_2 \geq 0\). --- ## **Summary:** - Using the **Big M method**, we introduced artificial variables to find a feasible starting point. - The process revealed that the problem's constraints are incompatible. - Therefore, **the LP has no feasible solution**. --- **If you need to find the optimal solution for a feasible LP or modify the problem, please provide such data.**

Let's solve the LP step by step using **both the Big M method and the Two-Phase method**. --- ## 1. **Restate the Problem** \[ \begin{align*} \text{Maximize:} \quad & z = x_1 + x_2 \\ \text{Subject to:} \quad & 2x_1 + x_2 \geq 3 \\ & 3x_1 + x_2 \leq 3.5 \\ & 5x_1 + x_2 \leq 1 \\ & x_1, x_2 \geq 0 \end{align*} \] --- ## 2. **Standard Form (Introduce Slack, Surplus, Artificial variables)** Write each as an equation: - \(2x_1 + x_2 \geq 3\) ⇒ \(2x_1 + x_2 - s_1 + a_1 = 3\) (Surplus \(s_1\), Artificial \(a_1\)) - \(3x_1 + x_2 \leq 3.5\) ⇒ \(3x_1 + x_2 + s_2 = 3.5\) (Slack \(s_2\)) - \(5x_1 + x_2 \leq 1\) ⇒ \(5x_1 + x_2 + s_3 = 1\) (Slack \(s_3\)) - All variables \(\geq 0\). So, variables: \(x_1, x_2, s_1, a_1, s_2, s_3 \geq 0\) --- ## 3. **Big M Method** ### a) **Objective Function with Big M** \[ \text{Max } z = x_1 + x_2 - M a_1 \] --- ### b) **Initial Simplex Tableau** Let us order the variables as \(x_1, x_2, s_1, a_1, s_2, s_3\): | Basic | \(x_1\) | \(x_2\) | \(s_1\) | \(a_1\) | \(s_2\) | \(s_3\) | RHS | |---------|---------|---------|---------|---------|---------|---------|------| | \(a_1\) | 2 | 1 | -1 | 1 | 0 | 0 | 3 | | \(s_2\) | 3 | 1 | 0 | 0 | 1 | 0 | 3.5 | | \(s_3\) | 5 | 1 | 0 | 0 | 0 | 1 | 1 | | \(z\) | -1 | -1 | 0 | -M | 0 | 0 | 0 | --- ### c) **Check Feasibility** Let’s see if the constraints are compatible. From (1): \(2x_1 + x_2 \geq 3\) From (3): \(5x_1 + x_2 \leq 1\) Subtract (1) from (3): \[ [5x_1 + x_2] - [2x_1 + x_2] \leq 1 - 3 \implies 3x_1 \leq -2 \implies x_1 \leq -\frac{2}{3} \] But \(x_1 \geq 0\), so **no feasible solution exists!** --- ## 4. **Two-Phase Method** ### **Phase I: Minimize sum of artificial variables** Objective for Phase I: \(\min w = a_1\) The tableau is as above, but with the objective row: | Basic | \(x_1\) | \(x_2\) | \(s_1\) | \(a_1\) | \(s_2\) | \(s_3\) | RHS | |---------|---------|---------|---------|---------|---------|---------|------| | \(a_1\) | 2 | 1 | -1 | 1 | 0 | 0 | 3 | | \(s_2\) | 3 | 1 | 0 | 0 | 1 | 0 | 3.5 | | \(s_3\) | 5 | 1 | 0 | 0 | 0 | 1 | 1 | | \(w\) | 0 | 0 | 0 | 1 | 0 | 0 | 0 | We attempt to remove artificial variables from the basis via simplex, but as shown above, the constraints are inconsistent, so the minimum value of \(w\) is not zero. --- ## 5. **Conclusion** **Both methods show that the LP is infeasible.** The constraints: - \(2x_1 + x_2 \geq 3\) - \(5x_1 + x_2 \leq 1\) - \(x_1, x_2 \geq 0\) cannot be satisfied at the same time. Algebraically, the two inequalities together imply \(x_1 \leq -2/3\), which contradicts \(x_1 \geq 0\). --- ### **Final Answer** > **Using either the Big M method or the two-phase method, we see that the problem is infeasible: there is no solution that satisfies all constraints simultaneously.** > **Therefore, the optimal solution does not exist for this LP.**