+
\documentclass[smallextended]{svjour3}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
the amount of load of processor $i$ at time $t+1$ is given by:
\begin{equation}
x_i(t+1)=x_i(t)-\sum_{j\in V(i)} s_{ij}(t) + \sum_{j\in V(i)} r_{ji}(t)
+\label{eq:ping-pong}
+\end{equation}
+
+
+Some conditions are required to ensure the convergence. One of them can be
+called the \texttt{ping-pong} condition which specifies that:
+\begin{equation}
+x_i(t)-\sum _{k\in V(i)} s_{ik}(t) \geq x_j^i(t)+s_{ij}(t)
\end{equation}
+for any processor $i$ and any $j \in V(i)$ such that $x_i(t)>x_j^i(t)$. This
+condition aims at avoiding a processor to send a part of its load and beeing
+less loaded after that.
+
+Nevertheless, we think that this condition may lead to deadlocks in some
+cases. For example, if we consider only three processors and that processor $1$
+is linked to processor $2$ which is also linked to processor $3$ (i.e. a simple
+chain wich 3 processors). Now consider we have the following values at time $t$:
+\begin{eqnarray*}
+x_1(t)=10 \\
+x_2(t)=100 \\
+x_3(t)=99.99\\
+ x_3^2(t)=99.99\\
+\end{eqnarray*}
+In this case, processor $2$ can either sends load to processor $1$ or processor
+$3$. If it sends load to processor $1$ it will not satisfy condition
+(\ref{eq:ping-pong}) because after the sending it will be less loaded that
+$x_3^2(t)$. So we consider that the \texttt{ping-pong} condition is probably to
+strong. Currently, we did not try to make another convergence proof without this
+condition or with a weaker condition.
\section{Best effort strategy}
