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A72: minor changes

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Théophile Bastian 2023-09-23 16:45:00 +02:00
parent eef1889478
commit ca430288b8
1 changed files with 9 additions and 8 deletions
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17
manuscrit/40_A72-frontend/30_manual_frontend.tex
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@ -78,14 +78,15 @@ throughput is a linear function of the number of \uops{} in the kernel~---, we
then know that either $\cyc{\kerK} = \cycF{\kerK}$ or $\cyc{\kerK} =
\cycB{\kerK}$.
For a given instruction $i$, we then construct a sequence $\kerK_k$ of kernels
For a given instruction $i$ and for a certain $k \in \nat$, we then construct a
kernel $\kerK_k$
such that:
\begin{enumerate}[(i)]
\item\label{cnd:kerKk:compo} for all $k \in \nat$, $\kerK_k$ is composed of the instruction $i$,
\item\label{cnd:kerKk:compo} $\kerK_k$ is composed of the instruction $i$,
followed by $k$ basic instructions;
\item\label{cnd:kerKk:linear} the kernels $\kerK_k$ are simple enough to exhibit this purely linear
\item\label{cnd:kerKk:linear} the kernel $\kerK_k$ is simple enough to exhibit this purely linear
frontend behaviour;
\item\label{cnd:kerKk:fbound} after a certain rank, $\cycB{\kerK_k} \leq \cycF{\kerK_k}$.
\item\label{cnd:kerKk:fbound} $\cycB{\kerK_k} \leq \cycF{\kerK_k}$.
\end{enumerate}
We denote by $\mucount{}\kerK$ the number of \uops{} in kernel $\kerK$. Under
@ -123,7 +124,7 @@ incorrect for this instruction, we should measure
$\ceil{\cyc{\imath}} \leq \cyc{\kerK_{k_0}}$ and $\cyc{\kerK_{k_0}} +
\sfrac{1}{3} = \cyc{\kerK_{k_0+1}}$. For instructions $i$ where it is not the
case, increasing $k_0$ by 3 or using other basic instructions eventually
yielded satisfying measures. Finally, we then obtain
yielded satisfying measures. Finally, we obtain
\[
\mucount{}i = 3 \cyc{\kerK_{k_0}} - k_0
@ -332,7 +333,7 @@ kernel, frontend-wise.
\bigskip{}
This model, however, is not satisfactory in many cases. For instance, the
kernel $\kerK' = \texttt{ADDV} + 2x\basic{Int01}$ is predicted to run in $1.5$
cycles, as depicted in \autoref{fig:frontend_nocross_addv_2add}; however, a
\pipedream{} measure yields $\cyc{\kerK'} = 1.35 \simeq 1\,\sfrac{1}{3}$
kernel $\kerK' = \texttt{ADDV} + 2\times\basic{Int01}$ is predicted to run in
$1.5$ cycles, as depicted in \autoref{fig:frontend_nocross_addv_2add}; however,
a \pipedream{} measure yields $\cyc{\kerK'} = 1.35 \simeq 1\,\sfrac{1}{3}$
cycles.