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+++ b/main.tex
@@ -137,6 +137,13 @@ Thus the optimal solution must be $C=F=\text{mincut with capacity $c$}$.
 \end{frame}
 
 
+\begin{frame}{Plot $L(\mu)$}
+\begin{figure}
+\includegraphics[width=0.8\linewidth]{images/L.png}
+\end{figure}
+\end{frame}
+
+
 \begin{frame}{Breakpoints on $L(\mu)$}
 We have see that the first line segment is $L(\mu)=(\lambda_c-b)\mu$ where $\lambda_c$ is the value of mincut with capacity $c$. 
 
@@ -176,7 +183,8 @@ Again we first assume the $\mu$ is fixed. Then for each pair of constraints $\su
 IP\ref{IP} has an integrality gap of 2.
 \end{conjecture}
 
-Suppose that $\mu^*$ is the optimal solution to LP\ref{lp:dualcutint}. Let $\lambda^{fr}$ and $\lambda^{int}$ be the fractional and integral mincut with capacity $w_{\mu^*}$.
+Suppose that $\mu^*$ is the optimal solution to LP\ref{lp:dualcutint}.
+% Let $\lambda^{fr}$ and $\lambda^{int}$ be the fractional and integral mincut with capacity $w_{\mu^*}$.
 \newline
 
 Conjecture \ref{conj:gap2} implies $w_{\mu^*}(C^*)+b\mu^* \leq 2(\text{value of mincut with $w_{\mu^*}$})$, \newline which is stronger than Theorem \ref{thm:2approx}.