1. **Problem Statement:** We have a linear programming problem (P) with objective function $$w = 5x_1 + \alpha x_2 - x_3$$ and constraints: $$3x_1 + 2x_2 = \beta$$ $$2x_2 + 4x_3 \leq 6$$ $$-x_1 + 2x_3 \geq -1$$ with variable bounds $$x_1 \text{ free}, x_2 \geq 0, x_3 \leq 0$$. Given the final tableau for $$\alpha=2$$ and $$\beta=4$$, we analyze: (a) For which $$\alpha$$ values does the basic solution remain optimal when $$\beta=4$$? (b) For which $$\beta$$ values does the basic solution remain feasible when $$\alpha=2$$? (c) How does the objective value $$w$$ depend on $$\alpha, \beta$$ if the solution is feasible and optimal? (d) For $$\alpha=2, \beta=4$$, adding variable $$x_4 \geq 0$$ with column $$a_{3\times4} = (-1,-1,0)^T$$ and objective coefficient $$\gamma$$, for which $$\gamma$$ values is the optimum objective value 7? --- 2. **Key Concepts and Formulas:** - Optimality depends on the reduced costs in the tableau; nonnegative reduced costs for maximization imply optimality. - Feasibility depends on the right-hand side values (basic variables) being nonnegative (or satisfying bounds). - Objective value $$w$$ can be expressed from the tableau as a function of parameters. - Adding a variable affects optimality depending on its reduced cost. --- 3. **Step-by-step Solutions:** **(a) Optimality with respect to $$\alpha$$ when $$\beta=4$$:** - From the tableau, the reduced costs depend on $$\alpha$$. - The condition for optimality is that all reduced costs $$\geq 0$$. - Given the tableau and problem data, the critical reduced cost leads to $$\alpha \leq \frac{11}{3}$$. **Answer:** $$\alpha \leq \frac{11}{3}$$. **(b) Feasibility with respect to $$\beta$$ when $$\alpha=2$$:** - Feasibility requires the basic variables' values (right-hand side) to satisfy bounds. - From the tableau, the basic solution is feasible if $$1 \leq \beta \leq 5$$. **Answer:** $$1 \leq \beta \leq 5$$. **(c) Objective value $$w$$ as a function of $$\alpha, \beta$$:** - Using the tableau and parameters, the objective value is: $$w = \alpha - 0.25\beta + 6$$ **Answer:** $$w = \alpha - 0.25\beta + 6$$. **(d) Effect of adding variable $$x_4 \geq 0$$ with column $$(-1,-1,0)^T$$ and coefficient $$\gamma$$:** - The reduced cost for $$x_4$$ is $$\gamma - 1$$. - For the current solution to remain optimal with objective value 7, reduced cost must be $$\geq 0$$. - Thus, $$\gamma \geq -1$$. **Answer:** $$\gamma \geq -1$$. --- **Summary:** (a) $$\alpha \leq \frac{11}{3}$$ (b) $$1 \leq \beta \leq 5$$ (c) $$w = \alpha - 0.25\beta + 6$$ (d) $$\gamma \geq -1$$