+We will describe here a new load-balancing strategy that we called
+\emph{best effort}. The general idea behind this strategy is, for a
+processor, to send some load to the most of its neighbors, doing its
+best to reach the equilibrium between those neighbors and himself.
+
+More precisely, when a processors $i$ is in its load-balancing phase,
+he proceeds as following.
+\begin{enumerate}
+\item First, the neighbors are sorted in non-decreasing order of their
+ known loads $x^i_j(t)$.
+
+\item Then, this sorted list is traversed in order to find its largest
+ prefix such as the load of each selected neighbor is lesser than:
+ \begin{itemize}
+ \item the processor's own load, and
+ \item the mean of the loads of the selected neighbors and of the
+ processor's load.
+ \end{itemize}
+ Let's call $S_i(t)$ the set of the selected neighbors, and
+ $\bar{x}(t)$ the mean of the loads of the selected neighbors and of
+ the processor load:
+ \begin{equation*}
+ \bar{x}(t) = \frac{1}{\abs{S_i(t)} + 1}
+ \left( x_i(t) + \sum_{j\in S_i(t)} x^i_j(t) \right)
+ \end{equation*}
+ The following properties hold:
+ \begin{equation*}
+ \begin{cases}
+ S_i(t) \subset V(i) \\
+ x^i_j(t) < x_i(t) & \forall j \in S_i(t) \\
+ x^i_j(t) < \bar{x} & \forall j \in S_i(t) \\
+ x^i_j(t) \leq x^i_k(t) & \forall j \in S_i(t), \forall k \in V(i) \setminus S_i(t) \\
+ \bar{x} \leq x_i(t)
+ \end{cases}
+ \end{equation*}
+
+\item Once this selection is completed, processor $i$ sends to each of
+ the selected neighbor $j\in S_i(t)$ an amount of load $s_{ij}(t) =
+ \bar{x} - x^i_j(t)$.
+
+ From the above equations, and notably from the definition of
+ $\bar{x}$, it can easily be verified that:
+ \begin{equation*}
+ \begin{cases}
+ x_i(t) - \sum_{j\in S_i(t)} s_{ij}(t) = \bar{x} \\
+ x^i_j(t) + s_{ij}(t) = \bar{x} & \forall j \in S_i(t)
+ \end{cases}
+ \end{equation*}
+\end{enumerate}
+
+\section{Other strategies}
+\label{Other}
+
+\textbf{Question} faut-il décrire les stratégies makhoul et simple ?
+
+\paragraph{simple} Tentative de respecter simplement les conditions de Bertsekas.
+Parmi les voisins moins chargés que soi, on sélectionne :
+\begin{itemize}
+\item un des moins chargés (vmin) ;
+\item un des plus chargés (vmax),
+\end{itemize}
+puis on équilibre avec vmin en s'assurant que notre charge reste
+toujours supérieure à celle de vmin et à celle de vmax.
+
+On envoie donc (avec "self" pour soi-même) :
+\[
+ \min\left(\frac{load(self) - load(vmin)}{2}, load(self) - load(vmax)\right)
+\]
+
+\paragraph{makhoul} Ordonne les voisins du moins chargé au plus chargé
+puis calcule les différences de charge entre soi-même et chacun des
+voisins.
+
+Ensuite, pour chaque voisin, dans l'ordre, et tant qu'on reste plus
+chargé que le voisin en question, on lui envoie 1/(N+1) de la
+différence calculée au départ, avec N le nombre de voisins.
+
+C'est l'algorithme~2 dans~\cite{bahi+giersch+makhoul.2008.scalable}.
