Subjects linear programming

Dual Simplex 8Ea5Ab

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Dual Simplex 8Ea5Ab


1. **State the problem:** We want to maximize $$Z = -4x_1 + 6x_2 - 18x_3$$ subject to $$2x_1 + 3x_3 \geq 4$$ $$-3x_2 + 2x_3 \geq 3$$ with bounds $$x_1 \geq 0, \quad x_2 \leq 0, \quad x_3 \geq -1.$$ 2. **Convert inequalities and variables for dual simplex:** Rewrite constraints as equalities by introducing surplus and artificial variables: $$2x_1 + 3x_3 - s_1 = 4$$ $$-3x_2 + 2x_3 - s_2 = 3$$ where $s_1, s_2 \geq 0$ are surplus variables. Since $x_2 \leq 0$, define $x_2' = -x_2 \geq 0$. Also, define $x_3' = x_3 + 1 \geq 0$ to handle $x_3 \geq -1$. Rewrite constraints: $$2x_1 + 3(x_3' - 1) - s_1 = 4 \Rightarrow 2x_1 + 3x_3' - s_1 = 7$$ $$3x_2' + 2(x_3' - 1) - s_2 = 3 \Rightarrow 3x_2' + 2x_3' - s_2 = 5$$ Objective function becomes: $$Z = -4x_1 - 6x_2' - 18(x_3' - 1) = -4x_1 - 6x_2' - 18x_3' + 18.$$ 3. **Set up initial tableau for dual simplex:** Basic variables: $s_1, s_2$ with initial values from constraints: $$s_1 = 2x_1 + 3x_3' - 7$$ $$s_2 = 3x_2' + 2x_3' - 5$$ Since $x_1, x_2', x_3' \geq 0$, initial $s_1 = -7 < 0$, $s_2 = -5 < 0$, so basic variables are negative, suitable for dual simplex. 4. **Dual simplex method steps:** - Identify leaving variable: basic variable with most negative value (here $s_1 = -7$). - Identify entering variable: among non-basic variables with negative coefficients in $s_1$ row, choose one that maintains feasibility. - Perform pivot to update tableau. - Repeat until all basic variables are nonnegative. 5. **Perform pivots:** - Pivot on $x_3'$ in $s_1$ row (coefficient 3). - Update $s_1$ row: $x_3'$ enters, $s_1$ leaves. - Update $s_2$ and objective function rows accordingly. 6. **Continue iterations:** - Next leaving variable is $s_2$ if negative. - Pivot on suitable entering variable. - Repeat until all $s_1, s_2 \geq 0$. 7. **Obtain optimal solution:** - After convergence, read values of $x_1, x_2', x_3'$. - Convert back: $x_2 = -x_2'$, $x_3 = x_3' - 1$. - Calculate maximum $Z$. 8. **Summary:** The dual simplex method starts with an infeasible but dual feasible solution and iteratively pivots to feasibility and optimality. **Note:** The full tableau and pivot calculations are extensive; this outlines the dual simplex approach to solve the problem.