LALP Assessment 3

LALP 3rd Continuous Assessment

Q1

Suppose it makes \(x\) product A and \(y\) product B.

Then the LP problem is:

\[\text{Maximize}\quad 2.4 x+2.6 y, \\ \left\{\ \begin{array} 5 x+4 y\leq 160\\ 2 x+3 y\leq 85\\ x+2 y\leq 50 \end{array} \right. \]

by drawing a graph, we can earn \(x^* = [20,15]\) and the objective value is \(87\).

Q2

Suppose it makes \(x\) chocolate cake and \(y\) cheese cake.

Then the LP problem is

\[\text{Maximize}\quad 1.2 x + 1.4 y, \\ \left\{\ \begin{array} 5 x+4 y\leq 160\\ 2 x+3 y\leq 85\\ x+2 y\leq 50 \end{array} \right. \]

by drawing a graph, we can earn \(x^* = [20,15]\) and the objective value is \(45\).

Q3

If the primal problem is given by (canonical form)

\[\begin{array}{cl} (L P) \quad \text { Minimize } & c^{T} x \\ \text { s.t. } & A x \geq b, \quad x \geq 0 \end{array} \]

then its dual problem is

\[\begin{array}{cl} (D P) \quad \text { Maximize } & b^{T} y \\ \text { s.t. } \quad & A^{T} y \leq c, y \geq 0 \end{array} \]

Hence the Dual of \(P\) is

\[\text { Maximise } 12 y_{1}+9 y_{2}+6 y_{3} \\ \begin{align} \text { subject to } 5 y_{1}+4 y_{2}+3 y_{3} &\leq 7,\\ 4 y_{1}+7 y_{2}+2 y_{3} &\leq 10,\\ y_{1} \geq 0,y_{2} \geq 0, y_{3} &\geq 0 \text {. } \end{align} \]

Q4

If the primal problem is given by

\[\begin{array}{cl} (L P) \quad \text { Minimize } & c^{T} x \\ \text { s.t. } & A x=b, \quad x \geq 0 \end{array} \]

then its dual problem is defined by

\[\begin{array}{cl} (D P) \quad \text { Maximize } & b^{T} y \\ \text { s.t. } \quad & A^{T} y \leq c \end{array} \]

where \(y \in R^{m}\).

Hence the Dual of \(P\) is

\[\text { Maximise } 12 y_{1}+9 y_{2}+6 y_{3} \\ \begin{align} \text { subject to } 5 y_{1}+4 y_{2}+3 y_{3} &\leq a,\\ 4 y_{1}+7 y_{2}+2 y_{3} &\leq 10, \\ y_{1} \in \mathbb{R}, y_{2} \in \mathbb{R}, y_{3} &\in \mathbb{R} \end{align} \]

Q5

Consider the following linear programming problem

\[\begin{align} (P): \text { Minimise } \qquad6 x_{1}+15 x_{2}+12 x_{3} \\ \text { subject to }\qquad \ \ \ \ x_{1}+2 x_{2}+4 x_{3} &\geq 3 \\ 3 x_{1}+5 x_{2}+3 x_{3} &\geq 4 \\ x_{1}, x_{2}, x_{3} &\geq 0 \end{align} \]

Given that \(\left(2, \frac{4}{3}\right)\) is an optimal solution to the dual problem.

Sol.

Introduce slack variables \(x_4\) and $ x_5 $ and transform the problem to the standard form:

\[\begin{align} (P): \text { Minimise } \qquad6 x_{1}+15 x_{2}+12 x_{3} \\ \text { subject to }\qquad \ \ \ \ \ x_{1}+2 x_{2}+4 x_{3}- x_4 &=3 \\ 3 x_{1}+5 x_{2}+3 x_{3} - x_5&= 4 \\ x_{1}, x_{2}, x_{3} &\geq 0 \end{align} \]

The optimality conditions for this LP are given as follows:

\[\begin{align} x_{1}+2 x_{2}+4 x_{3}- x_4 &=3 \\ 3 x_{1}+5 x_{2}+3 x_{3} - x_5&= 4 \\ x_{1}, x_{2}, x_{3},x_4,x_5 &\geq 0\\ y_1+3y_2 &\leq 6\\ 2y_1+5y_2&\leq 15\\ 4y_1+3y_2&\leq 12\\ y_1,y_2&\geq 0 \\ x_{1}+15 x_{2}+12 x_{3} &= 3y_1+4y_2 \end{align} \]

Given that \(\left(2, \frac{4}{3}\right)\) is an optimal solution to the dual problem, quick check:

\[y_1+3y_2 = 6\\ 2y_1+5y_2< 15 \\ 4y_1+3y_2= 12\\ \]

Since the first constraint in the dual problem is an inequality, by the complementary slackness conditions, the first variable in the primal problem is zero, that is \(x^*_2 = 0\). Substituting this to the primal constraints, we obtain

\[\begin{align} \text { Minimise } \qquad6 x_{1}+12 x_{3} \\ x_{1}+4 x_{3} &\geq 3 \\ 3 x_{1}+3 x_{3} &\geq 4 \\ x_{1}, x_{3} &\geq 0\\ \end{align} \]

By drawing a graph, we can quickly obtain the reduced LP problem above, where the optimal solution is

\[x_1 = \frac{7}{9}\quad x_3 = \frac{5}{9} \]

Hence the minimun value of the primal problem is \(\frac{34}{3}\), and the optimal solution is \(\left[ \frac{7}{9},0, \frac{5}{9}\right]\)

Q6

Consider the following linear programming problem

\[\begin{array}{cl} \text { Minimise } & 2 x_{1}+5 x_{2}+4 x_{3} \\ \text { subject to } & x_{1}+2 x_{2}+4 x_{3} \geq 3 \\ & 3 x_{1}+5 x_{2}+3 x_{3} \geq 4 \\ & x_{1}, x_{2}, x_{3} \geq 0 \end{array} \]

Given that \(\left(\frac{2}{3}, \frac{4}{9}\right)\) is an optimal solution to the dual problem.

Sol.

Introduce slack variables \(x_4\) and $ x_5 $ and transform the problem to the standard form:

\[\begin{align} (P): \text { Minimise } \qquad2 x_{1}+5 x_{2}+4 x_{3} \\ \text { subject to }\qquad \ \ \ \ \ x_{1}+2 x_{2}+4 x_{3}- x_4 &=3 \\ 3 x_{1}+5 x_{2}+3 x_{3} - x_5&= 4 \\ x_{1}, x_{2}, x_{3} &\geq 0 \end{align} \]

The optimality conditions for this LP are given as follows:

\[\begin{align} x_{1}+2 x_{2}+4 x_{3}- x_4 &=3 \\ 3 x_{1}+5 x_{2}+3 x_{3} - x_5&= 4 \\ x_{1}, x_{2}, x_{3},x_4,x_5 &\geq 0\\ y_1+3y_2 &\leq 2\\ 2y_1+5y_2&\leq 5\\ 4y_1+3y_2&\leq 4\\ y_1,y_2&\geq 0 \\ 2x_{1}+ 5x_{2}+4 x_{3} &= 3y_1+4y_2 \end{align} \]

Given that \(\left(\frac{2}{3}, \frac{4}{9}\right)\) is an optimal solution to the dual problem, quick check:

\[y_1+3y_2 = 2\\ 2y_1+5y_2< 5 \\ 4y_1+3y_2= 4\\ \]

Since the first constraint in the dual problem is an inequality, by the complementary slackness conditions, the first variable in the primal problem is zero, that is \(x^*_2 = 0\). Substituting this to the primal constraints, we obtain

\[\begin{align} \text { Minimise } \qquad2 x_{1}+ 4x_{3} \\ x_{1}+4 x_{3} &\geq 3 \\ 3 x_{1}+3 x_{3} &\geq 4 \\ x_{1}, x_{3} &\geq 0\\ \end{align} \]

By drawing a graph, we can quickly obtain the reduced LP problem above, where the optimal solution is

\[x_1 = \frac{7}{9}\quad x_3 = \frac{5}{9} \]

Hence the minimun value of the primal problem is \(\frac{34}{9}\), and the optimal solution is \(\left[ \frac{7}{9},0, \frac{5}{9}\right]\)