Continue definitions

manuscrit/20_foundations/20_code_analyzers.tex

@@ -212,9 +212,37 @@ $\cyc{\kerK} = 1.5$.
 \end{example}
 
 \begin{remark}
-    Although we define $\cyc{\kerK}$ as the minimum over $\nat^*$, only so many
-    kernels may be aggregated until we find the minimum.
+    As $C(\kerK)$ depends on the microarchitecture of the processor considered,
+    the throughput $\cyc{\kerK}$ of a kernel $\kerK$ depends on the processor
+    considered.
+\end{remark}
 
+\medskip
+
+Although we define $\cyc{\kerK}$ as the minimum over $\nat^*$, only so many
+kernels may be aggregated until we find the minimum.
+
+\begin{lemma}\label{lem:cyc_k_conv}
+    Given a kernel $\kerK$,
+
+    \begin{enumerate}[(i)]
+
+        \item{}\label{lem:cyc_k_conv:low_n} the minimum considered in the
+            definition of $\cyc{\kerK}$ is reached for a small value of $n \leq
+            N_0$, $N_0$ being commensurate to the complexity of the
+            microarchitecture considered.
+
+        \item{}\label{lem:cyc_k_conv:conv} Furthermore, the sequence converges
+            towards $\cyc{\kerK}$:
+            \[
+                \lim_{n \to \infty} \dfrac{C(\kerK^n)}{n} = \cyc{\kerK}
+            \]
+
+    \end{enumerate}
+
+\end{lemma}
+
+\begin{proof}
     Indeed, as the number of resources that can be shared between instructions
     in a processor is finite (and relatively small, usually on the order of
     magnitude of 10), and their number of possible states is also finite (and
@@ -225,19 +253,40 @@ $\cyc{\kerK} = 1.5$.
 
     Thus, by the pigeon-hole principle, and as each state depends only on the
     previous one, the states visited by $\left(C(\kerK^n)\right)_{n \in
-    \nat^*}$ are periodic of period $p$. Take $r_0 \in \nat$ realizing
-    $\min_{0 < r \leq p}\left(\sfrac{C(\kerK^r)}{r}\right)$. As we are by hypothesis in
-    steady-state already, we have for any $n \in \nat^*$ such that $n = kp+r$,
-    $0 < r \leq p$, $k, r \in \nat$,
+    \nat^*}$ are periodic of period $p$. As such, and as we are by hypothesis
+    in steady-state already (and not only periodic from a certain rank), for
+    any $n \geq p$, we have
+    \[
+        C(\kerK^n) = C(\kerK^{n-p}) + C(\kerK^p)
+    \]
+
+    Take $r_0 \in \nat^*$ realizing
+    $\min_{0 < r \leq p}\left(\sfrac{C(\kerK^r)}{r}\right)$.
+
+    For any $n \in \nat^*$ such that $n = kp+r$, $0 < r \leq p$, $k, r \in \nat$,
     \begin{align*}
-        C(\kerK^n) &= k \cdot C(\kerK^p) + C(\kerK^r) \\
-            &\geq k \cdot C(\kerK^{r_0}) + C(\kerK^{r_0}) \\
-            &\geq (k+1) \cdot C(\kerK^{r_0}) \\
-        \implies \dfrac{C(\kerK^n)}{n} &\geq k \cdot \dfrac{C(\kerK^p)}{n} +
-        \dfrac{C(\kerK^r)}{n}
+        C(\kerK^n) &= k \cdot C(\kerK^p) + C(\kerK^r) & \textit{(by induction)} \\
+                   &= kp \dfrac{C(\kerK^p)}{p} + r \dfrac{C(\kerK^r)}{r} \\
+                   &\geq kp \cdot \dfrac{C(\kerK^{r_0})}{r_0} + r \dfrac{C(\kerK^{r_0})}{r_0} \\
+                   &\geq (kp+r) \dfrac{C(\kerK^{r_0})}{r_0} \\
+                   &\geq n \dfrac{C(\kerK^{r_0})}{r_0} \\
+        \implies \dfrac{C(\kerK^n)}{n} &\geq \dfrac{C(\kerK^{r_0})}{r_0} = \cyc{\kerK}
+    \end{align*}
+
+    Thus, $r_0$ realizes the minimum from the definition of $\cyc{\kerK}$, with
+    $r_0 \geq p$, commensurate with the complexity of the microarchitecture,
+    proving~(\ref{lem:cyc_k_conv:low_n}).
+
+    \medskip{}
+
+    For any $n > r_0$, we decompose $n = r_0 + m$ and $m = k'p + r'$, $0 < r'
+    \leq p$, $k', r' \in \nat$.
+
+    \begin{align*}
+        C(\kerK^n) = C(\kerK^{r_0}) + k'p \dfrac{C(\kerK^p)}{p} +
     \end{align*}
     \todo{}
-\end{remark}
+\end{proof}
 
 \medskip
 
@@ -258,6 +307,47 @@ stead.
 In the literature or in analyzers' reports, the throughput of a kernel is often
 referred to as its \emph{IPC} (its unit).
 
-\todo{Measure of $\cyc{\kerK}$}
+\begin{notation}[Experimental measure of $\cyc{\kerK}$]
+    We note $\cycmes{\kerK}{n}$ the experimental measure of $\kerK$, realized
+    by:
+    \begin{itemize}
+        \item sampling the hardware counter of total number of instructions
+            retired and the counter of total number of cycles elapsed,
+        \item executing $\kerK^n$,
+        \item sampling again the same counters, and noting respectively
+            $\Delta_n\text{ret}$ and $\Delta_{n}C$ their differences,
+        \item noting $\cycmes{\kerK}{n} = \dfrac{\Delta_{n}C\cdot
+            \card{\kerK}}{\Delta_n\text{ret}}$, where $\card{\kerK}$ is the
+            number of instructions in $\kerK$.
+    \end{itemize}
+\end{notation}
+
+\begin{lemma}
+    For any kernel $\kerK$,
+    $\cycmes{\kerK}{n} \xrightarrow[n \to \infty]{} \cyc{\kerK}$.
+\end{lemma}
+\begin{proof}
+    For an integer number of kernel iterations $n$,
+    $\sfrac{\Delta_n\text{ret}}{\card{\kerK}} = n$. While measurement
+    errors may make $\Delta_{n}\text{ret}$ fluctuate slightly, this
+    fluctuation will be below a constant threshold:
+    \[
+        \abs{\dfrac{\Delta_n\text{ret}}{\card{\kerK}} - n}
+        \leq E_\text{ret}
+    \]
+
+    The same way, and due to the pipelining effects we noted below
+    the definition of $\cyc{\kerK}$,
+    \[
+        \abs{\Delta_{n}C - C(\kerK^n)} \leq E_C
+    \]
+    with $E_C$ a constant.
+
+    As such, for a given $n$,  \todo{}
+\end{proof}
+
+Given this property, we will use $\cyc{\kerK}$ to refer to $\cycmes{\kerK}{n}$
+for large values of $n$ in this manuscript whenever it is clear that this value
+is a measure.
 
 \subsubsection{Basic block of an assembly-level program}

manuscrit/include/macros.tex

@@ -10,9 +10,12 @@
 \newcommand{\calI}{\mathcal{I}}
 \newcommand{\calB}{\mathcal{B}}
 
+\newcommand{\abs}[1]{\left| #1 \right|}
+
 \newcommand{\cyc}[1]{\overline{#1}}
 \newcommand{\cycF}[1]{\overline{#1}^\textbf{F}}
 \newcommand{\cycB}[1]{\overline{#1}^\textbf{B}}
+\newcommand{\cycmes}[2]{\overline{#1}^{\textit{M(}#2\textit{)}}}
 
 \newcommand{\nat}{\mathbb{N}}