X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/mpi-energy.git/blobdiff_plain/4b6d80813dd8f58434a111f585d96547e5893a55..deaad9c91bd56f904fe3c5ff93664431a18d4f44:/paper.tex?ds=sidebyside

diff --git a/paper.tex b/paper.tex
index 42d2248..9c374ba 100644
--- a/paper.tex
+++ b/paper.tex
@@ -345,18 +345,18 @@ the consumed energy with scaled frequency and the consumed energy without scaled
 frequency:
 \begin{multline}
   \label{eq:enorm}
-  \Enorm = \frac{ \Ereduced}{\Eoriginal} \\
-      {} = \frac{\Pdyn \cdot S_1^{-2} \cdot
+  \Enorm(S) = \frac{ \Ereduced}{\Eoriginal} \\
+         {} = \frac{\Pdyn \cdot S^{-2} \cdot
              \left( T_1 + \sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
-               \Pstatic \cdot T_1 \cdot S_1 \cdot N}{
+               \Pstatic \cdot T_1 \cdot S \cdot N}{
              \Pdyn \cdot \left(T_1+\sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
                \Pstatic \cdot T_1 \cdot N }
 \end{multline}
 In the same way we can normalize the performance as follows:
 \begin{equation}
   \label{eq:pnorm}
-  \Pnorm = \frac{\Tnew}{\Told}
-         = \frac{\TmaxCompOld \cdot S + \TmaxCommOld}{
+  \Pnorm(S) = \frac{\Tnew}{\Told}
+            = \frac{\TmaxCompOld \cdot S + \TmaxCommOld}{
              \TmaxCompOld + \TmaxCommOld}
 \end{equation}
 The second problem is that the optimization operation for both energy and
@@ -377,7 +377,7 @@ direction.  Therefore, we inverse the equation of the normalized performance as
 follows:
 \begin{equation}
   \label{eq:pnorm_en}
-  \Pnorm^{-1} = \frac{ \Told}{ \Tnew}
+  \Pnorm^{-1}(S) = \frac{ \Told}{ \Tnew}
                = \frac{\TmaxCompOld +
                  \TmaxCommOld}{\TmaxCompOld \cdot S +
                  \TmaxCommOld}
@@ -417,29 +417,33 @@ the objective function described above.
 \begin{figure}[tp]
   \begin{algorithmic}[1]
     % \footnotesize
-    \State  Initialize the variable $\Dist=0$
-    \State Set dynamic and static power values.
-    \State Set $\Pstates$ to the number of available frequencies.
-    \State Set the variable $\Fnew$ to max. frequency,  $\Fnew = \Fmax $
-    \State Set the variable $\Fdiff$ to the difference between two successive
-           frequencies.
-    \For {$j := 1$ to $\Pstates $}
-      \State $\Fnew = \Fnew - \Fdiff $
-      \State $S = \frac{\Fmax}{\Fnew}$
-      \State $S_i = S \cdot \frac{T_1}{T_i}
-                  = \frac{\Fmax}{\Fnew} \cdot \frac{T_1}{T_i}$
-             for $i=1,\dots,N$
-      \State $\Enorm =
-          \frac{\Pdyn \cdot S_1^{-2} \cdot
+    \Require ~
+    \begin{description}
+    \item[$\Pstatic$] static power value
+    \item[$\Pdyn$] dynamic power value
+    \item[$\Pstates$] number of available frequencies
+    \item[$\Fmax$] maximum frequency
+    \item[$\Fdiff$] difference between two successive freq.
+    \end{description}
+    \Ensure $\Sopt$ is the optimal scaling factor
+
+    \State $\Sopt \gets 1$
+    \State $\Dist \gets 0$
+    \State $\Fnew \gets \Fmax$
+    \For {$j = 2$ to $\Pstates$}
+      \State $\Fnew \gets \Fnew - \Fdiff$
+      \State $S \gets \Fmax / \Fnew$
+      \State $\Enorm \gets
+          \frac{\Pdyn \cdot S^{-2} \cdot
                   \left( T_1 + \sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
-                  \Pstatic \cdot T_1 \cdot S_1 \cdot N }{
+                  \Pstatic \cdot T_1 \cdot S \cdot N }{
                 \Pdyn \cdot
                   \left(T_1+\sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
                   \Pstatic \cdot T_1 \cdot N }$
-      \State $\PnormInv = \Told / \Tnew$
+      \State $\PnormInv \gets \Told / \Tnew$
       \If{$(\PnormInv - \Enorm > \Dist)$}
-        \State $\Sopt = S$
-        \State $\Dist = \PnormInv - \Enorm$
+        \State $\Sopt \gets S$
+        \State $\Dist \gets \PnormInv - \Enorm$
       \EndIf
     \EndFor
     \State  Return $\Sopt$
@@ -479,7 +483,7 @@ program.
 \begin{figure}[tp]
   \begin{algorithmic}[1]
     % \footnotesize
-    \For {$k:=1$ to \textit{some iterations}}
+    \For {$k=1$ to \textit{some iterations}}
       \State Computations section.
       \State Communications section.
       \If {$(k=1)$}
