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how to show that y_i is the distance...?

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Yu Cong 2025-05-13 21:17:34 +08:00
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@ -120,7 +120,7 @@ s.t.& & \sum_i D_i d(s_i,t_i)&=1 & &\\
\begin{enumerate}
\item \ip{} $\geq$ \lp{}. Given any feasible solution to \ip{}, we can scale all $x_e$ and $y_i$ simultaneously with factor $1/\sum_i D_i y_i$. The scaled solution is feasible for \lp{} and gets the same objective value.
\item \lp{} $=$ \dual{}. by duality.
\item \metric{} $=$ \lp{}. It is easy to see \metric{} $\geq$ \lp{} since any feasible metric to \metric{} induces a feasible solution to \lp{}. In fact, any solution to \lp{} also induces a feasible metric.
\item \metric{} $=$ \lp{}. It is easy to see \metric{} $\geq$ \lp{} since any feasible metric to \metric{} induces a feasible solution to \lp{}. In fact, the optimal solution to \lp{} also induces a feasible metric. Consider a solution $x_e,y_i$ to \lp{}. Let $d$ be the shortest path metric on $V$ using edge length $x_e$. It suffices to show that $y_i$ is the shortest path distance fron $s_i$ to $t_i$. Suppose $y_i\leq d(s_i,t_i)$...
\end{enumerate}
\bibliographystyle{plainnat}