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+++ b/main.tex
@@ -178,7 +178,7 @@ A further reformulation (the new $x$ is $x-y$) gives us the following,
 \begin{aligned}
 \min&   &                   \sum_{e} x_e w(e)          &   & \\
 s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x+y$ is a cut)}\\
-&       &   \sum_{e} y_e c(e) &\leq B   &   &\text{(budget for $F$)}\\
+&       &   \sum_{e} y_e c(e) &\leq b   &   &\text{(budget for $F$)}\\
 % &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
 &       &   y_e,x_e&\in\{0,1\}      &   &\forall e
 \end{aligned}
@@ -248,13 +248,9 @@ We want to prove something like tree packing theorem for \autoref{lp:dualcutint}
     The optimum of \autoref{lp:dualcutint} is $\min \set{\frac{w(C\setminus F)}{B-c(F)}| \forall \text{cut $C$}, c(F)\leq b}$, where $B$ is the cost of mincut in $G$ and $b$ is the budget.
 \end{conjecture}
 
-I believe the previous conjecture is not likely to be true. This one seems better,
+I believe the previous conjecture is not likely to be true.
 
-\begin{conjecture}\label{conj:optimaldual}
-    $\lambda=\min \frac{w(C\setminus F)}{B-c(F)}$ is optimal for \autoref{lp:dualcutint}.
-\end{conjecture}
-
-Assuming \autoref{conj:optimaldual} is true, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$.
+\paragraph{Weight truncation} Assuming we know the optimal $\lambda$ to the LP dual, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$.
 
 \begin{conjecture}
     \autoref{lp:cutinterdict} has an integrality gap of 2.