\newcommand{\JC}[2][inline]{%
\todo[color=red!10,#1]{\sffamily\textbf{JC:} #2}\xspace}
+\newcommand{\Xsub}[2]{\ensuremath{#1_\textit{#2}}}
+
+\newcommand{\Dist}{\textit{Dist}}
+\newcommand{\Eind}{\Xsub{E}{ind}}
+\newcommand{\Enorm}{\Xsub{E}{Norm}}
+\newcommand{\Eoriginal}{\Xsub{E}{Original}}
+\newcommand{\Ereduced}{\Xsub{E}{Reduced}}
+\newcommand{\Fdiff}{\Xsub{F}{diff}}
+\newcommand{\Fmax}{\Xsub{F}{max}}
+\newcommand{\Fnew}{\Xsub{F}{new}}
+\newcommand{\Ileak}{\Xsub{I}{leak}}
+\newcommand{\Kdesign}{\Xsub{K}{design}}
+\newcommand{\MaxDist}{\textit{Max Dist}}
+\newcommand{\Ntrans}{\Xsub{N}{trans}}
+\newcommand{\Pdyn}{\Xsub{P}{dyn}}
+\newcommand{\PnormInv}{\Xsub{P}{NormInv}}
+\newcommand{\Pnorm}{\Xsub{P}{Norm}}
+\newcommand{\Pstates}{\Xsub{P}{states}}
+\newcommand{\Pstatic}{\Xsub{P}{static}}
+\newcommand{\Sopt}{\Xsub{S}{opt}}
+\newcommand{\Tcomp}{\Xsub{T}{comp}}
+\newcommand{\TmaxCommOld}{\Xsub{T}{Max Comm Old}}
+\newcommand{\TmaxCompOld}{\Xsub{T}{Max Comp Old}}
+\newcommand{\Tmax}{\Xsub{T}{max}}
+\newcommand{\Tnew}{\Xsub{T}{New}}
+\newcommand{\Told}{\Xsub{T}{Old}}
+
\begin{document}
\title{Dynamic Frequency Scaling for Energy Consumption
Many researchers~\cite{9,3,15,26} divide the power consumed by a processor into
two power metrics: the static and the dynamic power. While the first one is
consumed as long as the computing unit is on, the latter is only consumed during
-computation times. The dynamic power $P_{dyn}$ is related to the switching
+computation times. The dynamic power $\Pdyn$ is related to the switching
activity $\alpha$, load capacitance $C_L$, the supply voltage $V$ and
operational frequency $f$, as shown in EQ~\eqref{eq:pd}.
\begin{equation}
\label{eq:pd}
- P_\textit{dyn} = \alpha \cdot C_L \cdot V^2 \cdot f
+ \Pdyn = \alpha \cdot C_L \cdot V^2 \cdot f
\end{equation}
-The static power $P_{static}$ captures the leakage power as follows:
+The static power $\Pstatic$ captures the leakage power as follows:
\begin{equation}
\label{eq:ps}
- P_\textit{static} = V \cdot N_{trans} \cdot K_{design} \cdot I_{leak}
+ \Pstatic = V \cdot \Ntrans \cdot \Kdesign \cdot \Ileak
\end{equation}
-where V is the supply voltage, $N_{trans}$ is the number of transistors,
-$K_{design}$ is a design dependent parameter and $I_{leak}$ is a
+where V is the supply voltage, $\Ntrans$ is the number of transistors,
+$\Kdesign$ is a design dependent parameter and $\Ileak$ is a
technology-dependent parameter. The energy consumed by an individual processor
to execute a given program can be computed as:
\begin{equation}
\label{eq:eind}
- E_\textit{ind} = P_\textit{dyn} \cdot T_{Comp} + P_\textit{static} \cdot T
+ \Eind = \Pdyn \cdot \Tcomp + \Pstatic \cdot T
\end{equation}
-where $T$ is the execution time of the program, $T_{Comp}$ is the computation
-time and $T_{Comp} \leq T$. $T_{Comp}$ may be equal to $T$ if there is no
+where $T$ is the execution time of the program, $\Tcomp$ is the computation
+time and $\Tcomp \leq T$. $\Tcomp$ may be equal to $T$ if there is no
communication, no slack time and no synchronization.
DVFS is a process that is allowed in modern processors to reduce the dynamic
ratio between the maximum and the new frequency as in EQ~\eqref{eq:s}.
\begin{equation}
\label{eq:s}
- S = \frac{F_\textit{max}}{F_\textit{new}}
+ S = \frac{\Fmax}{\Fnew}
\end{equation}
The value of the scaling factor $S$ is greater than 1 when changing the
frequency of the CPU to any new frequency value~(\emph{P-state}) in the
\begin{equation}
\label{eq:energy}
- E = P_\textit{dyn} \cdot S_1^{-2} \cdot
+ E = \Pdyn \cdot S_1^{-2} \cdot
\left( T_1 + \sum_{i=2}^{N} \frac{T_i^3}{T_1^2} \right) +
- P_\textit{static} \cdot T_1 \cdot S_1 \cdot N
- \hfill
+ \Pstatic \cdot T_1 \cdot S_1 \cdot N
\end{equation}
-where $N$ is the number of parallel nodes, $T_i$ for $i=1,\dots,N$ are
-the execution times and scaling factors of the sorted tasks. Therefore, $T1$ is
+where $N$ is the number of parallel nodes, $T_i$ and $S_i$ for $i=1,\dots,N$ are
+the execution times and scaling factors of the sorted tasks. Therefore, $T_1$ is
the time of the slowest task, and $S_1$ its scaling factor which should be the
highest because they are proportional to the time values $T_i$. The scaling
factors are computed as in EQ~\eqref{eq:si}.
\begin{equation}
\label{eq:si}
S_i = S \cdot \frac{T_1}{T_i}
- = \frac{F_\textit{max}}{F_\textit{new}} \cdot \frac{T_1}{T_i}
+ = \frac{\Fmax}{\Fnew} \cdot \frac{T_1}{T_i}
\end{equation}
In this paper we use Rauber and Rünger's energy model, EQ~\eqref{eq:energy}, because it can be applied to homogeneous clusters if the communication time is taken in consideration. Moreover, we compare our algorithm with Rauber and Rünger's scaling factor selection
method which uses the same energy model. In their method, the optimal scaling factor is
\begin{equation}
\label{eq:sopt}
- S_\textit{opt} = \sqrt[3]{\frac{2}{N} \cdot \frac{P_\textit{dyn}}{P_\textit{static}} \cdot
+ \Sopt = \sqrt[3]{\frac{2}{N} \cdot \frac{\Pdyn}{\Pstatic} \cdot
\left( 1 + \sum_{i=2}^{N} \frac{T_i^3}{T_1^3} \right) }
\end{equation}
of the new scaling factor as in EQ~\eqref{eq:tnew}.
\begin{equation}
\label{eq:tnew}
- \textit T_\textit{new} = T_\textit{Max Comp Old} \cdot S + T_{\textit{Max Comm Old}}
+ \Tnew = \TmaxCompOld \cdot S + \TmaxCommOld
\end{equation}
In this paper, this prediction method is used to select the best scaling factor
for each processor as presented in the next section.
frequency:
\begin{multline}
\label{eq:enorm}
- E_\textit{Norm} = \frac{ E_\textit{Reduced}}{E_\textit{Original}} \\
- {} = \frac{P_\textit{dyn} \cdot S_1^{-2} \cdot
- \left( T_1 + \sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
- P_\textit{static} \cdot T_1 \cdot S_1 \cdot N }{
- P_\textit{dyn} \cdot \left(T_1+\sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
- P_\textit{static} \cdot T_1 \cdot N }
+ \Enorm = \frac{ \Ereduced}{\Eoriginal} \\
+ {} = \frac{\Pdyn \cdot S_1^{-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}{
+ \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}
- P_\textit{Norm} = \frac{T_\textit{New}}{T_\textit{Old}}
- = \frac{T_\textit{Max Comp Old} \cdot S +
- T_\textit{Max Comm Old}}{T_\textit{Max Comp Old} +
- T_\textit{Max Comm Old}}
+ \Pnorm = \frac{\Tnew}{\Told}
+ = \frac{\TmaxCompOld \cdot S + \TmaxCommOld}{
+ \TmaxCompOld + \TmaxCommOld}
\end{equation}
The second problem is that the optimization operation for both energy and
performance is not in the same direction. In other words, the normalized energy
follows:
\begin{equation}
\label{eq:pnorm_en}
- P^{-1}_\textit{Norm} = \frac{ T_\textit{Old}}{ T_\textit{New}}
- = \frac{T_\textit{Max Comp Old} +
- T_\textit{Max Comm Old}}{T_\textit{Max Comp Old} \cdot S +
- T_\textit{Max Comm Old}}
+ \Pnorm^{-1} = \frac{ \Told}{ \Tnew}
+ = \frac{\TmaxCompOld +
+ \TmaxCommOld}{\TmaxCompOld \cdot S +
+ \TmaxCommOld}
\end{equation}
\begin{figure}
\centering
our objective function has the following form:
\begin{equation}
\label{eq:max}
- \textit{Max Dist} = \max_{j=1,2,\dots,F}
- (\overbrace{P^{-1}_\textit{Norm}(S_j)}^{\text{Maximize}} -
- \overbrace{E_\textit{Norm}(S_j)}^{\text{Minimize}} )
+ \MaxDist = \max_{j=1,2,\dots,F}
+ (\overbrace{\Pnorm^{-1}(S_j)}^{\text{Maximize}} -
+ \overbrace{\Enorm(S_j)}^{\text{Minimize}} )
\end{equation}
where $F$ is the number of available frequencies. Then we can select the optimal
scaling factor that satisfies EQ~\eqref{eq:max}. Our objective function can
\begin{figure}[tp]
\begin{algorithmic}[1]
% \footnotesize
- \State Initialize the variable $Dist=0$
- \State Set dynamic and static power values.
- \State Set $P_{states}$ to the number of available frequencies.
- \State Set the variable $F_{new}$ to max. frequency, $F_{new} = F_{max} $
- \State Set the variable $F_{diff}$ to the difference between two successive
- frequencies.
- \For {$j:=1$ to $P_{states} $}
- \State $F_{new}=F_{new} - F_{diff} $
- \State $S = \frac{F_\textit{max}}{F_\textit{new}}$
- \State $S_i = S \cdot \frac{T_1}{T_i}
- = \frac{F_\textit{max}}{F_\textit{new}} \cdot \frac{T_1}{T_i}$
+ \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 $S_i \gets S \cdot \frac{T_1}{T_i}
+ = \frac{\Fmax}{\Fnew} \cdot \frac{T_1}{T_i}$
for $i=1,\dots,N$
- \State $E_\textit{Norm} =
- \frac{P_\textit{dyn} \cdot S_1^{-2} \cdot
+ \State $\Enorm \gets
+ \frac{\Pdyn \cdot S_1^{-2} \cdot
\left( T_1 + \sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
- P_\textit{static} \cdot T_1 \cdot S_1 \cdot N }{
- P_\textit{dyn} \cdot
+ \Pstatic \cdot T_1 \cdot S_1 \cdot N }{
+ \Pdyn \cdot
\left(T_1+\sum_{i=2}^{N}\frac{T_i^3}{T_1^2}\right) +
- P_\textit{static} \cdot T_1 \cdot N }$
- \State $P_{NormInv}=T_{old}/T_{new}$
- \If{$(P_{NormInv}-E_{Norm} > Dist)$}
- \State $S_{opt} = S$
- \State $Dist = P_{NormInv} - E_{Norm}$
+ \Pstatic \cdot T_1 \cdot N }$
+ \State $\PnormInv \gets \Told / \Tnew$
+ \If{$(\PnormInv - \Enorm > \Dist)$}
+ \State $\Sopt \gets S$
+ \State $\Dist \gets \PnormInv - \Enorm$
\EndIf
\EndFor
- \State Return $S_{opt}$
+ \State Return $\Sopt$
\end{algorithmic}
\caption{Scaling factor selection algorithm}
\label{EPSA}
\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 {$(k=1)$}
frequency $F_i$ as follows:
\begin{equation}
\label{eq:fi}
- F_i = \frac{F_\textit{max} \cdot T_i}{S_\textit{optimal} \cdot T_\textit{max}}
+ F_i = \frac{\Fmax \cdot T_i}{\Sopt \cdot \Tmax}
\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
