\usepackage{subfig}
\usepackage{listings}
\usepackage{colortbl}
+\usepackage{amsmath}
% \usepackage{sectsty}
% \usepackage{titlesec}
% \usepackage{secdot}
this case the fastest tasks have to wait at the synchronous barrier for the
slowest tasks to finish their job. In both two cases the overall execution time
of the program is the execution time of the slowest task as :
-\begin{equation} \label{eq:T1}
- Program Time=MAX_{i=1,2,..,N} (T_i) \hfill
+\begin{equation}
+ \label{eq:T1}
+ \textit{Program Time} = \max_{i=1,2,\dots,N} T_i
\end{equation}
where $T_i$ is the execution time of process $i$.
researchers see ~\cite{9,3,15,26}. The dynamic power of the CMOS processors
$P_{dyn}$ is related to the switching activity $\alpha$, load capacitance $C_L$,
the supply voltage $V$ and operational frequency $f$ respectively as follow :
-\begin{equation} \label{eq:pd}
- \displaystyle P_{dyn} = \alpha . C_L . V^2 . f
+\begin{equation}
+ \label{eq:pd}
+ P_{dyn} = \alpha \cdot C_L \cdot V^2 \cdot f
\end{equation}
The static power $P_{static}$ captures the leakage power consumption as well as
the power consumption of peripheral devices like the I/O subsystem.
-\begin{equation} \label{eq:ps}
- \displaystyle P_{static} = V . N . K_{design} . I_{leak}
+\begin{equation}
+ \label{eq:ps}
+ P_{static} = V \cdot N \cdot K_{design} \cdot I_{leak}
\end{equation}
where V is the supply voltage, N is the number of transistors, $K_{design}$ is a
design dependent parameter and $I_{leak}$ is a technology-dependent
parameter. Energy consumed by an individual processor $E_{ind}$ is the summation
of the dynamic and the static power multiply by the execution time for example
see~\cite{36,15} .
-\begin{equation} \label{eq:eind}
- \displaystyle E_{ind} = (P_{dyn} + P_{static} ) . T
+\begin{equation}
+ \label{eq:eind}
+ E_{ind} = ( P_{dyn} + P_{static} ) \cdot T
\end{equation}
The dynamic voltage and frequency scaling (DVFS) is a process that allowed in
modern processors to reduce the dynamic power by scaling down the voltage and
various frequency values in~\cite{3}. The reduction process of the frequency are
expressed by scaling factor \emph S. The scale \emph S is the ratio between the
maximum and the new frequency as in EQ~(\ref{eq:s}).
-\begin{equation} \label{eq:s}
- S=\:\frac{F_{max}}{F_{new}} \hfill \newline
+\begin{equation}
+ \label{eq:s}
+ S = \frac{F_{max}}{F_{new}}
\end{equation}
The value of the scale \emph S is grater than 1 when changing the frequency to
any new frequency value(\emph {P-state}) in governor. It is equal to 1 when the
consider the two powers metric for measuring the energy of the parallel tasks as
in EQ~(\ref{eq:energy}).
-\begin{equation} \label{eq:energy}
- E= \displaystyle \;P_{dyn}\,.\,S_1^{-2}\;.\,(T_1+\sum\limits_{i=2}^{N}\frac{T_i^3}{T_1^2})+\;P_{static}\,.\,T_1\,.\,S_1\;\,.\,N
+\begin{equation}
+ \label{eq:energy}
+ E = P_{dyn} \cdot S_1^{-2} \cdot
+ \left( T_1 + \sum_{i=2}^{N} \frac{T_i^3}{T_1^2} \right) +
+ P_{static} \cdot T_1 \cdot S_1 \cdot N
\hfill
\end{equation}
Where \emph N is the number of parallel nodes, $T_1 $ is the time of the slower
for the slower task. The scaling factor $S_1$, as in EQ~(\ref{eq:s1}), selects
from the set of scales values $S_i$. Each of these scales are proportional to
the time value $T_i$ depends on the new frequency value as in EQ~(\ref{eq:si}).
-\begin{equation} \label{eq:s1}
- S_1=MAX_{i=1,2,..,F} (S_i) \hfill
+\begin{equation}
+ \label{eq:s1}
+ S_1 = \max_{i=1,2,\dots,F} S_i
\end{equation}
-\begin{equation} \label{eq:si}
- S_i=\:S\: .\:(\frac{T_1}{T_i})=\: (\frac{F_{max}}{F_{new}}).(\frac{T_1}{T_i}) \hfill
+\begin{equation}
+ \label{eq:si}
+ S_i = S \cdot \frac{T_1}{T_i}
+ = \frac{F_{max}}{F_{new}} \cdot \frac{T_1}{T_i}
\end{equation}
Where $F$ is the number of available frequencies. In this paper we depend on
Rauber's energy model EQ~(\ref{eq:energy}) for two reasons : 1-this model used
optimal energy reduction derived from the EQ~(\ref{eq:energy}). He takes the
derivation for this equation (to be minimized) and set it to zero to produce the
scaling factor as in EQ~(\ref{eq:sopt}).
-\begin{equation} \label{eq:sopt}
- S_{opt}= {\sqrt [3~]{\frac{2}{n} \frac{P_{dyn}}{P_{static}} \Big(1+\sum\limits_{i=2}^{N}\frac{T_i^3}{T_1^3}\Big) }} \hfill
+\begin{equation}
+ \label{eq:sopt}
+ S_{opt} = \sqrt[3]{\frac{2}{n} \cdot \frac{P_{dyn}}{P_{static}} \cdot
+ \left( 1 + \sum_{i=2}^{N} \frac{T_i^3}{T_1^3} \right) }
\end{equation}
-%[\Big 3]
\section{Performance Evaluation of MPI Programs}
slower task. Secondly, we use these times for predicting the execution time for
any MPI program as a function of the new scaling factor as in the
EQ~(\ref{eq:tnew}).
-\begin{equation} \label{eq:tnew}
- \displaystyle T_{new}= T_{Max \:Comp \:Old} \; . \:S \;+ \;T_{Max\: Comm\: Old}
- \hfill
+\begin{equation}
+ \label{eq:tnew}
+ T_{new} = T_{\textit{Max Comp Old}} \cdot S + T_{\textit{Max Comm Old}}
\end{equation}
The above equation shows that the scaling factor \emph S has linear relation
with the computation time without affecting the communication time. The
For solving this problem, we normalize the energy by calculating the ratio
between the consumed energy with scaled frequency and the consumed energy
without scaled frequency :
-\begin{equation} \label{eq:enorm}
- E_{Norm}=\displaystyle\frac{E_{Reduced}}{E_{Orginal}}= \frac{\displaystyle \;P_{dyn}\,.\,S_i^{-2}\,.\,(T_1+\sum\limits_{i=2}^{N}\frac{T_i^3}{T_1^2})+\;P_{static}\,.\,T_1\,.\,S_i\;\,.\,N }{\displaystyle \;P_{dyn}\,.\,(T_1+\sum\limits_{i=2}^{N}\frac{T_i^3}{T_1^2})+\;P_{static}\,.\,T_1\,\,.\,N }
+\begin{equation}
+ \label{eq:enorm}
+ E_{Norm} = \frac{E_{Reduced}}{E_{Orginal}}
+ = \frac{ P_{dyn} \cdot S_i^{-2} \cdot
+ \left( T_1 + \sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
+ P_{static} \cdot T_1 \cdot S_i \cdot N }{
+ P_{dyn} \cdot \left(T_1+\sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
+ P_{static} \cdot T_1\, \cdot N }
\end{equation}
By the same way we can normalize the performance as follows :
-\begin{equation} \label{eq:pnorm}
- P_{Norm}=\displaystyle \frac{T_{New}}{T_{Old}}=\frac{T_{Max \:Comp \:Old} \;. \:S \;+ \;T_{Max\: Comm\: Old}}{T_{Old}} \;\;
+\begin{equation}
+ \label{eq:pnorm}
+ P_{Norm} = \frac{T_{New}}{T_{Old}}
+ = \frac{T_{\textit{Max Comp Old}} \cdot S +
+ T_{\textit{Max Comm Old}}}{T_{Old}}
\end{equation}
The second problem is the optimization operation for both energy and performance
is not in the same direction. In other words, the normalized energy and the
overheads. Our solution for this problem is to make the optimization process
have the same direction. Therefore, we inverse the equation of normalize
performance as follows :
-\begin{equation} \label{eq:pnorm_en}
- \displaystyle P^{-1}_{Norm}= \frac{T_{Old}}{T_{New}}=\frac{T_{Old}}{T_{Max \:Comp \:Old} \;. \:S \;+ \;T_{Max\: Comm\: Old}}
+\begin{equation}
+ \label{eq:pnorm_en}
+ P^{-1}_{Norm} = \frac{T_{Old}}{T_{New}}
+ = \frac{T_{Old}}{T_{\textit{Max Comp Old}} \cdot S +
+ T_{\textit{Max Comm Old}}}
\end{equation}
\begin{figure}
\centering
the minimum energy consumption with minimum execution time (better performance)
in the same time, see figure~(\ref{fig:r1}). Then our objective function has the
following form:
-\begin{equation} \label{eq:max}
- \displaystyle MaxDist = Max \;(\;\overbrace{P^{-1}_{Norm}}^{Maximize}\; -\; \overbrace{E_{Norm}}^{Minimize} \;)
+\begin{equation}
+ \label{eq:max}
+ \textit{MaxDist} = \max (\overbrace{P^{-1}_{Norm}}^{\text{Maximize}} -
+ \overbrace{E_{Norm}}^{\text{Minimize}} )
\end{equation}
Then we can select the optimal scaling factor that satisfy the
EQ~(\ref{eq:max}). Our objective function can works with any energy model or
\State - Calculate all available scales $S_i$ depend on $S$ as in EQ~(\ref{eq:si}).
\State - Select the maximum scale factor $S_1$ from the set of scales $S_i$.
\State - Calculate the normalize energy $E_{Norm}=E_{R}/E_{O}$ as in EQ~(\ref{eq:enorm}).
- \State - Calculate the normalize inverse of performance $P_{NormInv}=T_{old}/T_{new}$
-
- as in EQ~(\ref{eq:pnorm_en}).
- \If{ $(P_{NormInv}-E_{Norm}$ $>$ $Dist$) }
- \State $S_{optimal}=S$
+ \State - Calculate the normalize inverse of performance\par
+ $P_{NormInv}=T_{old}/T_{new}$ as in EQ~(\ref{eq:pnorm_en}).
+ \If{ $(P_{NormInv}-E_{Norm} > Dist$) }
+ \State $S_{optimal} = S$
\State $Dist = P_{NormInv} - E_{Norm}$
\EndIf
\EndFor
calculates the new frequency $F_i$ for each task proportionally to its time
value $T_i$. By substitution of the EQ~(\ref{eq:s}) in the EQ~(\ref{eq:si}), we
can calculate the new frequency $F_i$ as follows :
-\begin{equation} \label{eq:fi}
- F_i=\frac{F_{max} \; . \;T_i}{S_{optimal} \; . \;T_{max}} \hfill
+\begin{equation}
+ \label{eq:fi}
+ F_i = \frac{F_{max} \cdot T_i}{S_{optimal} \cdot T_{max}}
\end{equation}
According to this equation all the nodes may have the same frequency value if
they have balanced workloads. Otherwise, they take different frequencies when
