consumption. However, lowering the frequency of a CPU might increase the
execution time of an application running on that processor. Therefore, the
frequency that gives the best trade-off between the energy consumption and the
-performance of an application must be selected.\\
-In this paper, a new online frequencies selecting algorithm for heterogeneous
-platforms is presented. It selects the frequency which tries to give the best
+performance of an application must be selected.
+
+In this paper, a new online frequency selecting algorithm for heterogeneous
+platforms is presented. It selects the frequencies and tries to give the best
trade-off between energy saving and performance degradation, for each node
computing the message passing iterative application. The algorithm has a small
overhead and works without training or profiling. It uses a new energy model for
cluster composed of four different types of nodes having the characteristics
presented in Table~\ref{table:platform}, it takes on average \np[ms]{0.04} for 4
nodes and \np[ms]{0.15} on average for 144 nodes to compute the best scaling
-factors vector. The algorithm complexity is $O(F\cdot (N \cdot4) )$, where $F$
+factors vector. The algorithm complexity is $O(F\cdot N)$, where $F$
is the number of iterations and $N$ is the number of computing nodes. The
algorithm needs from 12 to 20 iterations to select the best vector of frequency
scaling factors that gives the results of the next sections.
\caption{Heterogeneous nodes characteristics}
% title of Table
\centering
- \begin{tabular}{|*{7}{l|}}
+ \begin{tabular}{|*{7}{r|}}
\hline
Node &Simulated & Max & Min & Diff. & Dynamic & Static \\
type &GFLOPS & Freq. & Freq. & Freq. & power & power \\
& & GHz & GHz &GHz & & \\
\hline
- 1 &40 & 2.5 & 1.2 & 0.1 & 20~W &4~W \\
+ 1 &40 & 2.50 & 1.20 & 0.100 & \np[W]{20} &\np[W]{4} \\
\hline
- 2 &50 & 2.66 & 1.6 & 0.133 & 25~W &5~W \\
+ 2 &50 & 2.66 & 1.60 & 0.133 & \np[W]{25} &\np[W]{5} \\
\hline
- 3 &60 & 2.9 & 1.2 & 0.1 & 30~W &6~W \\
+ 3 &60 & 2.90 & 1.20 & 0.100 & \np[W]{30} &\np[W]{6} \\
\hline
- 4 &70 & 3.4 & 1.6 & 0.133 & 35~W &7~W \\
+ 4 &70 & 3.40 & 1.60 & 0.133 & \np[W]{35} &\np[W]{7} \\
\hline
\end{tabular}