diff --git a/latex/presentations/midterm/presentation.tex b/latex/presentations/midterm/presentation.tex
index 5767425..5b5ceac 100644
--- a/latex/presentations/midterm/presentation.tex
+++ b/latex/presentations/midterm/presentation.tex
@@ -15,6 +15,7 @@
 \usepackage{listings}
 \usepackage{graphicx}
 \usepackage{xcolor}
+\usepackage[binary-units]{siunitx}
 %\geometry{textheight=17.07cm,textwidth=6.9cm}
 %\usepackage{pgfpages}
 %\pgfpagesuselayout{resize to}[physical paper height=17.07cm,
diff --git a/latex/presentations/midterm/sections/decoding_algorithms.tex b/latex/presentations/midterm/sections/decoding_algorithms.tex
index a9082d2..48e44ad 100644
--- a/latex/presentations/midterm/sections/decoding_algorithms.tex
+++ b/latex/presentations/midterm/sections/decoding_algorithms.tex
@@ -19,7 +19,7 @@
                         _{\text{Parity constraint}},
             \hspace{5mm}\mathcal{A}\left( i \right) \equiv \left\{
                 j | j\in \mathcal{J},
-                \boldsymbol{H}_{i,j} = 1
+                \boldsymbol{H}_{j,i} = 1
             \right\},
             i \in \mathcal{I}
             \end{align*}
@@ -82,7 +82,10 @@
 
 \begin{frame}[t, fragile]
     \frametitle{Proximal Decoding: Algorithm}
-        \begin{algorithm}[caption={}, label={}]
+    \begin{itemize}
+        \item Resulting terative decoding algorithm:
+            \vspace{2mm}
+            \begin{algorithm}[caption={}, label={}]
 $\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
 for $k=0$ to $K-1$ do
     $\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{s}^{(k)}; \boldsymbol{y} \right) $
@@ -92,7 +95,8 @@ for $k=0$ to $K-1$ do
     If $\boldsymbol{\hat{x}}$ passes the parity check condition, break the loop.
 end for
 Output $\boldsymbol{\hat{x}}$
-        \end{algorithm}
+            \end{algorithm}
+    \end{itemize}
 \end{frame}
 
 
diff --git a/latex/presentations/midterm/sections/examination_results.tex b/latex/presentations/midterm/sections/examination_results.tex
index 91ac233..2b3148e 100644
--- a/latex/presentations/midterm/sections/examination_results.tex
+++ b/latex/presentations/midterm/sections/examination_results.tex
@@ -25,7 +25,7 @@
                             legend style={at={(0.05,0.05)},anchor=south west},
                             width=11.5cm,
                             height=8cm,
-                            ytick={0, 10e-1, 10e-2, 10e-3, 10e-4},
+                            ytick={0, 1e-1, 1e-2, 1e-3, 1e-4},
                             xtick={1, 2, 3, 4, 5},
                             ymax=1.2, ymin=0.8e-4,
                             xmin=0.9, xmax=5.6,
@@ -59,8 +59,13 @@
                 
             \end{figure}
 
-        \item Performance: $2800 \text{ transm.} / s$ - Intel Core i7-7700HQ @ 2.80GHz\\
-            ($\sim 10s$ for the shown plot)
+        \item $\mathcal{O}\left(n  \right) $ time complexity - same as BP;
+            Only multiplication and addition necessary \cite{proximal_paper}
+        \item Measured Performance: Between $\SI{0.5}{\mega\bit / \second}$ and
+            $\SI{2.5}{\mega\bit / \second}$ - Intel Core i7-7700HQ @ 2.80GHz\\
+            ($\sim \SI{10}{\second}$ for the shown plot)
+            \todo{Use the shown bitrate, or half?
+            ($n_{iterations} \cdot n$ or $n_{iterations} \cdot k$?)}
     \end{itemize}
 \end{frame}
 
diff --git a/latex/presentations/midterm/sections/theoretical_background.tex b/latex/presentations/midterm/sections/theoretical_background.tex
index 441f6ed..d95223d 100644
--- a/latex/presentations/midterm/sections/theoretical_background.tex
+++ b/latex/presentations/midterm/sections/theoretical_background.tex
@@ -14,7 +14,9 @@
         \item The standard message-passing algorithms used for decoding [LDPC and turbo codes]
             are often difficult to analyze. \cite{feldman_thesis}
         \item The iterative message–passing algorithms preffered in practice do not guarantee
-            optimality and may fail to decode correctly when the graph contains cycles \cite{ldpc_conv}
+            optimality and may fail to decode correctly when the graph contains cycles
+            \cite{ldpc_conv}
+
     \end{itemize}
 \end{frame}
 
@@ -62,7 +64,7 @@
                         \left(0,\frac{1}{2}\left(\frac{k}{n}\frac{E_b}{N_0}\right)^{-1}\right),
                     \hspace{2mm} \boldsymbol{y}, \boldsymbol{n} \in \mathbb{R}^n
             \end{align*}
-        \item All zeros assumption:
+        \item All-zeros assumption:
             \begin{align*}
                 \boldsymbol{c} = 0
             \end{align*}
@@ -172,12 +174,10 @@
                 \begin{align*}
                     N\left( j \right) \equiv \left\{
                         i | i\in \mathcal{I},
-                        \boldsymbol{H}_{i,j} = 1
+                        \boldsymbol{H}_{j,i} = 1
                     \right\},
                     j \in \mathcal{J}
                 \end{align*}
-                \todo{Is this correct? Shouldn't i and j be switched around?}
-            \item ``Illegal configurations''
                 \begin{align*}
                     S \subseteq N\left( j \right), \left| S \right| \text{odd}
                 \end{align*}