up

mhd/doc/review/answers.lyx

@@ -411,6 +411,548 @@ I would like the authors to give a more detailed explanation of how they
 \begin_inset Quotes erd
 \end_inset
 
+:
+\begin_inset Newline newline
+\end_inset
+
+We enriched the computational part of the part with a more precise evaluation
+ of the performance of the code: memory bandwidth and computational intensity.
+ This analysis confirms the excellent efficiency of the implementation.
+\begin_inset Formula 
+\[
+n_{\text{GB}}=\frac{\texttt{\texttt{Nx}\ensuremath{\times\texttt{Ny}}\ensuremath{\times prec}}\times4\times m}{1024^{3}},
+\]
+
+\end_inset
+
+where 
+\begin_inset Formula $prec$
+\end_inset
+
+ is the number of bytes for storing one floating point number (
+\begin_inset Formula $prec=4$
+\end_inset
+
+ for single precision and 
+\begin_inset Formula $prec=8$
+\end_inset
+
+ for double precision).
+ We then perform a given number of time iterations niter and measure the
+ elapsed time 
+\begin_inset Formula $t_{\text{elapsed}}$
+\end_inset
+
+ (with a specific features of the OpenCL library).
+ We perform two kind of experiments.
+ In the first experiment, we deactivate the numerical computations and only
+ perform the shift operations.
+ The memory bandwidth of the shift algorithm is then given by
+\begin_inset Formula 
+\[
+b=\frac{2\times n_{\text{GB}}\times niter}{t_{\text{elapsed}}}.
+\]
+
+\end_inset
+
+In the second experiment, we reactivate the computations and measure how
+ the bandwidth is reduced.
+ This allows to evaluating how the elapsed time is shared between memory
+ transfers and computations.
+ The results are given in Table xxx
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Tabular
+<lyxtabular version="3" rows="10" columns="5">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
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+\end_inset
+</cell>
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+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $b$
+\end_inset
+
+ (GB/s, only shift)
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
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+\begin_layout Plain Layout
+\begin_inset Formula $b$
+\end_inset
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+ (GB/s, full LBM)
+\end_layout
+
+\end_inset
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+\end_layout
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+\end_inset
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+\end_layout
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+
 
 \end_layout
 

mhd/doc/spring_mhd_lbm.bib

@@ -1714,7 +1714,7 @@ MRREVIEWER = {Philip D. Loewen},
 }
 
 @article{toth2000b,
-  title={The∇{\textperiodcentered} B= 0 constraint in shock-capturing magnetohydrodynamics codes},
+  title={The $\nabla \cdot b= 0$ constraint in shock-capturing magnetohydrodynamics codes},
   author={T{\'o}th, G{\'a}bor},
   journal={Journal of Computational Physics},
   volume={161},

mhd/doc/spring_mhd_lbm.tex

@@ -865,6 +865,15 @@ $
 	\texttt{nvtop}\footnote{\url{https://github.com/Syllo/nvtop}} indicates
 	that the GPU occupation is of $99\%$. This indicates a quasi-optimal
 	implementation.
+	In order to measure the efficiency of the implementation we perform a memory bandwidth test
+	on several grid sizes. One time step of the method implies the read access in the global memory  
+	to the set of fields of the previous time step. The local computations are done in registers. Then
+	there is another write access to global memory for storing the data of the next time step.
+	The size in memory in Gigabyte of one set of fields is
+	$$
+	n_{\text{GB}}=\frac{\texttt{\texttt{Nx}\ensuremath{\times}\texttt{Ny}\ensuremath{\times b}}\times4\times m}{1024^{3}}
+	,$$
+	where $b$ is the number of bytes for storing one floating point number ($b=4$ for single precision and $b=8$ for double precision), 
 	
 	\section{Numerical applications to MHD}
 	

mhd/vortex_lbm.py

@@ -166,6 +166,10 @@ def solve_ocl(m=_m, n=_n, nx=_nx, ny=_ny, Lx=_Lx, Ly=_Ly, Tmax=_Tmax,
                      levels=np.linspace(_minplot, _maxplot, 16))
         #fig = Figure(title=plot_title)
 
+    compute = False
+    print("Compute:")
+    print(compute)
+
     print("start OpenCL computations...")
     while t < Tmax + dt:
 
@@ -199,7 +203,10 @@ def solve_ocl(m=_m, n=_n, nx=_nx, ny=_ny, Lx=_Lx, Ly=_Ly, Tmax=_Tmax,
 
         t += dt
         #event = prg.time_step(queue, (nx * ny, ), None, wn_gpu, wnp1_gpu)
-        event = prg.time_step(queue, (nx * ny, ), None, fn_gpu, fnp1_gpu)
+        if compute :        
+            event = prg.time_step(queue, (nx * ny, ), None, fn_gpu, fnp1_gpu)
+        else :
+            event = prg.time_shift(queue, (nx * ny, ), None, fn_gpu, fnp1_gpu)
         #event = prg.time_step(queue, (nx * ny, ), None, wn_gpu, wnp1_gpu, wait_for = [event])
         event.wait()
         queue.finish()