## Lecture #3 notes

Just to clarify, today in lecture we only showed that for every non-negative cost function, there is an optimal solution to the LP that is an r-arborescence. It is not true if we allow negative costs, because nothing in our LP prevents us from picking multiple edges and reducing the costs further. Recall we were using the LP:

$\displaystyle \min \sum_a c_a x_a$
$\displaystyle \sum_{a \in \partial^+S} x_a \geq 1 ~~~~~~$ for all “interesting” sets $S$
$\displaystyle x \geq 0$

$\displaystyle \min \sum_a c_a x_a$
$\displaystyle \sum_{a \in \partial^+S} x_a \geq 1 ~~~~~~$ for all “interesting” sets $S$
$\displaystyle \sum_{a \in \partial^+v} x_a \leq 1$
$\displaystyle x \geq 0$

Then with this stronger LP we can indeed show that for all cost-functions, there is an optimal integer solution. Let’s sketch why. Now the dual looks like:

$\displaystyle \max \sum_S y_S - \sum_v z_v$
$\displaystyle \sum_{S: uv \in \partial^+S} y_S - z_u \leq c_{uv} ~~~~~~$ for all $uv \in A$
$\displaystyle y, z \geq 0$

The $y_S$ variables should be the same as before. What would you set the dual variables $z_u$ to? Easy: set it to the amount you’d have to raise the costs on edges going out of $u$ in order to make them all non-negative.