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

If we add in Adam’s proposed constraint, we get

\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.

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