Construct the dual from the following problem:
Maximize: $$ z(x_1, x_2) = 2x_1 + 7x_2 $$
Subject to: $$ \begin{cases} -2x_1 + 3x_2 \leq 14, \ x_1 + x_2 \leq 8, \ x_1 \geq 0, \ x_2 \geq 0. \end{cases} $$
Using the graphical method find the primal and dual solutions.
The dual problem is constructed as follows:
Minimize: $$ w(y_1, y_2) = 14y_1 + 8y_2 $$
Subject to: $$ \begin{cases} -2y_1 + y_2 \geq 2, \ 3y_1 + y_2 \geq 7, \ y_1 \geq 0, \ y_2 \geq 0. \end{cases} $$
Primal Solution Space: The feasible region for the primal problem is plotted with constraints:
$-2x_1 + 3x_2 = 14$ $x_1 + x_2 = 8$
Vertices (A, B, C):
- A:
$(0, 0)$ - B:
$(6, 2)$ from intersection of$x_1=0$ and$x_1 + x_2 = 8$ - C:
$(2, 6)$ from intersection of$-2x_1 + 3x_2 = 14$ and$x_1 + x_2 = 8$
Optimal solution: $$ (x_1, x_2) = (2, 6), \quad F(2, 6) = 46 $$
Dual Solution Space: The feasible region for the dual problem is plotted with constraints:
$-2y_1 + y_2 = 2$ $3y_1 + y_2 = 7$
Vertices (A, B):
- A:
$(0, 0)$ - B:
$(1, 4)$ from intersection of both lines
Optimal solution: $$ (y_1, y_2) = (1, 4), \quad w(1, 4) = 46 $$
Construct the dual from the following problem:
Minimize: $$ z(x_1, x_2) = -2x_1 - 3x_2 $$
Subject to: $$ \begin{cases} -4x_1 + 2x_2 \geq 4, \ x_1 + x_2 \geq 6, \ x_1 \geq 0, \ x_2 \geq 0. \end{cases} $$
Using the graphical method find the primal and dual solutions.
The dual problem is constructed as follows:
Maximize: $$ w(y_1, y_2) = 4y_1 + 6y_2 $$
Subject to: $$ \begin{cases} -4y_1 + y_2 \leq -2, \ 2y_1 + y_2 \leq -3, \ y_1 \geq 0, \ y_2 \geq 0. \end{cases} $$
Primal Solution Space: The feasible region for the primal problem is unbounded (no finite solution exists).
Dual Solution Space: Constraints:
$-4y_1 + y_2 = -2$ $2y_1 + y_2 = -3$
No intersection point within non-negative constraints, so the feasible set is empty.
Construct the dual from the following problem:
Maximize: $$ z(x_1, x_2, x_3) = x_1 + 2x_2 - x_3 $$
Subject to: $$ \begin{cases} -x_1 + 4x_2 - 2x_3 \leq 12, \ x_1 + x_2 + 2x_3 \leq 17, \ 2x_1 - x_2 + 2x_3 = 4, \ x_1 \geq 0, \ x_2 \geq 0, \ x_3 \geq 0. \end{cases} $$
Using the M-method find the primal and dual solutions.
The dual problem is constructed as follows:
Minimize: $$ w(y_1, y_2, y_3) = 12y_1 + 17y_2 + 4y_3 $$
Subject to: $$ \begin{cases} -y_1 + y_2 + 2y_3 \geq 1, \ 4y_1 + y_2 - y_3 \geq 2, \ -2y_1 + 2y_2 + 2y_3 \geq -1, \ y_1 \geq 0, \ y_2 \geq 0. \end{cases} $$
Simplex Method Iterations (M-method): (Note: Full tableau details omitted for brevity; key results provided.)
Optimal solution from iterations: $$ (x_1, x_2, x_3) = (14, 15, 2), \quad z_{\text{max}} = -400. $$
Using the "Advanced Simplex Method" solve:
Maximize: $$ z(x_1, x_2, x_3, x_4, x_5, x_6) = 9x_1 + 10x_2 + 16x_3 $$
Subject to: $$ \begin{cases} 18x_1 + 15x_2 + 12x_3 + x_4 = 360, \ 6x_1 + 4x_2 + 8x_3 + x_5 = 192, \ 5x_1 + 3x_2 + 3x_3 + x_6 = 180, \ x_i \geq 0 \quad (i=1,\ldots,6). \end{cases} $$
Iteration Steps:
- Initial tableau: Basic variables
$X_B^0 = (x_4, x_5, x_6)$ - Entering variable:
$P_3$ (due to$\min(-9, -10, -16) = -16$ ) - Leaving variable:
$x_5$
Iteration 1:
Basic variables:
Optimal solution: $$ (x_1, x_2, x_3, x_4, x_5, x_6) = (0, 8, 20, 0, 0, 96), \quad z_{\text{max}} = 400. $$
Using the "Advanced Simplex Method" solve:
Maximize: $$ z(x_1, x_2, x_3, x_4, x_5, x_6) = 8x_2 + 7x_4 + x_6 $$
Subject to: $$ \begin{cases} x_1 - 2x_2 - 3x_4 - 2x_6 = 12, \ 4x_2 + x_3 - 4x_4 - 3x_6 = 12, \ 5x_2 + 5x_4 + x_5 + x_6 = 25, \ x_i \geq 0 \quad (i=1,\ldots,6). \end{cases} $$
Iteration Steps:
- Initial tableau: Basic variables
$X_B^0 = (x_1, x_3, x_5)$ - Entering variable:
$P_2$ (due to$\min(-8, -7, -1) = -8$ ) - Leaving variable:
$x_3$
Iteration 1:
Basic variables:
Optimal solution: $$ (x_1, x_2, x_3, x_4, x_5, x_6) = (23, 4, 0, 1, 0, 0), \quad z_{\text{max}} = 39. $$