tex: add NFBP graphs

latex/content.tex

@@ -184,7 +184,7 @@ results in the form of histograms.
 
 \footnotetext{The code is available in Annex \ref{annex:probabilistic}}
 
-We will try to approximate $ \mathbb{E}[X] $ and $ \mathbb{E}[V] $ with $
+We will try to approximate $ \mathbb{E}[R] $ and $ \mathbb{E}[V] $ with $
 	\overline{X_N} $ using $ {S_n}^2 $. This operation will be done for both $ R =
 	2 $ and $ R = 10^6 $ simulations.
 
@@ -206,25 +206,46 @@ variance and further determine the Confidence Interval (95 \% certainty).
 \paragraph{2 simulations} We first ran $ R = 2 $ simulations to observe the
 behavior of the algorithm and the low precision of the results.
 
-% TODO graph T_i 2 sim
+\begin{figure}[h]
+	\centering
+	\includegraphics[width=0.8\textwidth]{graphics/graphic-NFBP-Ti-2-sim}
+	\caption{Histogram of $ T_i $ for $ R = 2 $ simulations and $ N = 50 $ items (number of items per bin)}
+	\label{fig:graphic-NFBP-Ti-2-sim}
+\end{figure}
 
-On this graph, we can see each value of $ T_i $. Our calculations have yielded
+On this graph (figure \ref{fig:graphic-NFBP-Ti-2-sim}), we can see each value
-that $ \overline{T_1} = 1.0 $ and $ {S_N}^2 = 2.7 $. Our student coefficient is
+of $ T_i $. Our calculations have yielded that $ \overline{T_1} = 1.0 $ and $
-$ t_{0.95, 2} = 4.303 $.
+	{S_N}^2 = 2.7 $. Our Student coefficient is $ t_{0.95, 2} = 4.303 $.
+
+We can now calculate the Confidence Interval for $ T_1 $ for $ R = 2 $ simulations :
 
 \begin{align*}
-	\overline{T_1} = \sum_{k=1}^{2} {T_1}_k & = 1.0                                                                 \\
+	IC_{95\%}(T_1) & = \left[ 1.0 \pm 1.96 \frac{\sqrt{2.7}}{\sqrt{2}} \cdot 4.303 \right] \\
-	IC_{95\%}(T_1)                          & = \left[ 1.0 \pm 1.96 \frac{\sqrt{2.7}}{\sqrt{2}} \cdot 4.303 \right] \\
+	               & = \left[ 1 \pm 9.8 \right]                                            \\
-	                                        & = \left[ 1 \pm 9.8 \right]                                            \\
 \end{align*}
 
-With two simulations, we obtain $ \overline{T_1} = 1.0 $.
+We can see that the Confidence Interval is very large, which is due to the low
+number of simulations. Looking at figure \ref{fig:graphic-NFBP-Ti-2-sim}, we
+easily notice the high variance.
+
+\begin{figure}[h]
+	\centering
+	\includegraphics[width=0.8\textwidth]{graphics/graphic-NFBP-Vi-2-sim}
+	\caption{Histogram of $ V_i $ for $ R = 2 $ simulations and $ N = 50 $ items (size of the first item in a bin)}
+	\label{fig:graphic-NFBP-Vi-2-sim}
+\end{figure}
+
+On the graph of $ V_i $ (figure \ref{fig:graphic-NFBP-Vi-2-sim}), we can see
+that the sizes are scattered pseudo-randomly between $ 0 $ and $ 1 $, which is
+unsuprising given the low number of simulations. The process determinig the statistics
+is the same as for $ T_i $, yielding $ \overline{V_1} = 0.897 $, $ {S_N}^2 =
+	0.2 $ and $ IC_{95\%}(V_1) = \left[ 0.897 \pm 1.3 \right] $. In this particular run,
+the two values for $ V_1 $ are high (being bouded between $ 0 $ and $ 1 $).
 
 
+\paragraph{1 000 000 simulations} In order to ensure better precision, we then
+ran $ R = 10^6 $ simulations with $ N	= 50 $ different items each.
 
-IC observed
-
-We then ran $ R = 10^6 $ simulations with $ N = 50 $ different items each.
 With 10 6 simulations, we obtain Xn barre = cf graphe
 Calcul Sn carre
 IC observed
@@ -261,8 +282,6 @@ were needed to store $ n $ items.
 
 
 
-\cite{hofri:1987}
-% TODO mettre de l'Histoire
 
 \section{Next Fit Dual Bin Packing algorithm (NFDBP)}
 
@@ -505,7 +524,10 @@ then calculate the statistics (which iterates multiple times over the array).
 
 \subsection{Optimal algorithm}
 
-\cite{bin-packing-approximation:2022}
 
 \sectionnn{Conclusion}
 
+
+\nocite{bin-packing-approximation:2022}
+\nocite{hofri:1987}
+