Simplex Minimization 7F0Be7

1. **State the problem:** We want to minimize the objective function $$f = x_1 + x_2 + 3x_3 + 2x_5$$ subject to the constraints: $$\begin{cases} x_1 + x_2 + x_3 = 4, \\ x_1 - 3x_2 + x_4 = 2, \\ 4x_1 + 2x_2 + x_5 = 7, \end{cases}$$ with non-negativity conditions $$x_j \geq 0, \quad j=1,5.$$ 2. **Rewrite constraints for simplex method:** The system is already in equality form with slack variables where needed. Variables $x_4$ and $x_5$ appear as slack or surplus variables. 3. **Set up the initial simplex tableau:** Variables: $x_1, x_2, x_3, x_4, x_5$ Objective: minimize $f = x_1 + x_2 + 3x_3 + 2x_5$ 4. **Convert minimization to maximization:** Maximize $$-f = -x_1 - x_2 - 3x_3 - 2x_5$$ 5. **Initial basic variables:** From constraints, choose basic variables $x_3, x_4, x_5$ and express $x_1, x_2$ as non-basic. 6. **Express basic variables from constraints:** From first constraint: $$x_3 = 4 - x_1 - x_2$$ From second constraint: $$x_4 = 2 - x_1 + 3x_2$$ From third constraint: $$x_5 = 7 - 4x_1 - 2x_2$$ 7. **Substitute into objective:** $$f = x_1 + x_2 + 3(4 - x_1 - x_2) + 2(7 - 4x_1 - 2x_2)$$ Simplify: $$= x_1 + x_2 + 12 - 3x_1 - 3x_2 + 14 - 8x_1 - 4x_2$$ $$= (x_1 - 3x_1 - 8x_1) + (x_2 - 3x_2 - 4x_2) + (12 + 14)$$ $$= (-10x_1) + (-6x_2) + 26$$ 8. **Since $x_1, x_2 \\geq 0$, to minimize $f$, maximize $x_1$ and $x_2$ subject to $x_3, x_4, x_5 \\geq 0$:** From $x_3 \\geq 0$: $$4 - x_1 - x_2 \\geq 0 \\Rightarrow x_1 + x_2 \\leq 4$$ From $x_4 \\geq 0$: $$2 - x_1 + 3x_2 \\geq 0 \\Rightarrow x_1 \\leq 2 + 3x_2$$ From $x_5 \\geq 0$: $$7 - 4x_1 - 2x_2 \\geq 0 \\Rightarrow 4x_1 + 2x_2 \\leq 7$$ 9. **Graphical solution:** The feasible region is bounded by these inequalities and $x_1, x_2 \\geq 0$. 10. **Check vertices of feasible region:** - At $(x_1, x_2) = (1.5, 0)$: $$f = -10(1.5) - 6(0) + 26 = -15 + 26 = 11$$ - At $(x_1, x_2) = (0, 3)$: $$f = -10(0) - 6(3) + 26 = -18 + 26 = 8$$ 11. **Check intersection of constraints:** Solve system: $$x_1 + x_2 = 4$$ $$4x_1 + 2x_2 = 7$$ Multiply first by 2: $$2x_1 + 2x_2 = 8$$ Subtract second: $$(2x_1 + 2x_2) - (4x_1 + 2x_2) = 8 - 7$$ $$-2x_1 = 1 \\Rightarrow x_1 = -0.5$$ (not feasible since $x_1 \\geq 0$) 12. **Check other vertices:** At $(x_1, x_2) = (0, 0)$: $$f = 26$$ At $(x_1, x_2) = (1.75, 2.25)$ (intersection of $x_1 \\leq 2 + 3x_2$ and $4x_1 + 2x_2 \\leq 7$) is outside feasible region due to $x_1 + x_2 \\leq 4$. 13. **Optimal solution:** Minimum $f = 8$ at $(x_1, x_2) = (0, 3)$. 14. **Find other variables:** $$x_3 = 4 - 0 - 3 = 1$$ $$x_4 = 2 - 0 + 3(3) = 2 + 9 = 11$$ $$x_5 = 7 - 4(0) - 2(3) = 7 - 6 = 1$$ 15. **Check non-negativity:** All variables $x_j \\geq 0$ for $j=1,5$ and others. **Final answer:** $$x_1 = 0, x_2 = 3, x_3 = 1, x_4 = 11, x_5 = 1$$ Minimum value of $$f = 8$$. **Note:** To implement in Excel, set up the tableau with variables and constraints, then use Excel Solver to minimize $f$ subject to constraints.