91Ó°ÊÓ

Every linear maximization problem has a dual minimization problem, and they relate to each other. Any feasible value of the dual LP is an upper bound on the original primal LP. For any feasible solution of the dual is an upper bound on any feasible solution of the primal.

Consider the given Linear programming,

$$\begin{gathered} \max x+y \\ 2x+yâ\copyright ½3 \\ x+3yâ\copyright ½5 \\ x,yâ\copyright ¾0 \end{gathered}$$

To find the dual LP, minimize the maximum value with multipliers as follows,$\min 3a+5b$

Subject to

$$\begin{gathered} 2a+bâ\copyright ¾1 \\ a+3bâ\copyright ¾1 \\ a,bâ\copyright ¾0 \end{gathered}$$

Therefore, the above bound is the dual LP.

Consider the primal LP, and with the multipliers and Inequality,

For the value of X

$$\begin{gathered} 2x+yâ\copyright ½3 \\ x+3yâ\copyright ½5 \\ 5xâ\copyright ¾4 \end{gathered}$$

For the value of Y

$$\begin{gathered} 2x+yâ\copyright ½3 \\ x+3yâ\copyright ¾5 \\ 5yâ\copyright ¾7 \end{gathered}$$

Therefore, the optimal solution of primal LP is $x=\frac{4}{5}$, $y=\frac{7}{5}$,

Consider the dual LP,

$\min 3a+5b$

$$\begin{gathered} 2a+bâ\copyright ¾1 \\ a+3bâ\copyright ¾1 \\ a,bâ\copyright ¾0 \end{gathered}$$

Compare, inequality as follows,

$$\begin{gathered} 2a+bâ\copyright ¾1 \\ a+3bâ\copyright ¾1 \\ 2a+5bâ\copyright ½1 \\ a,bâ\copyright ¾0 \end{gathered}$$

Therefore, the optimal solution of dual LP is $a=\frac{2}{5}$ , $b=\frac{1}{5}$.