Hello,
I'm planning to buy a new desktop machine, and since my computations
utilize NaN values heavily, I'd like to know whether Intel Core 2
processors suffer from the same slowdown with NaN values as Pentium. For
details, see http://www.cygnus-software.com/papers/x86andinfinity.html
If someone has Octave on a machine with a Core 2 processor, please run
the following commands and tell the results. (Or a more detailed test
from ftp://ftp.cygnus-software.com/pub/specialnumbers.zip)
a=zeros(300,300);tic;b=(1.0+a)*a;toc
a=zeros(300,300);tic;b=(NaN+a)*a;toc
Below are results from my current machines. Since SSE2 in Pentium isn't
affected with the slowdown, it seems that for some reason the commands
above don't utilize SSE2. AMD processors aren't affected.
Pentium M, Octave 3.0.1 MSVC2005 SSE2
octave-3.0.1.exe:8> a=zeros(300,300);tic;b=(1.0+a)*a;toc
Elapsed time is 0.0430298 seconds.
octave-3.0.1.exe:9> a=zeros(300,300);tic;b=(NaN+a)*a;toc
Elapsed time is 28.8307 seconds.
Pentium 4 (Family 15, Model 2), Octave 3.0.1 MSVC2005 SSE2
octave-3.0.1.exe:2> a=zeros(300,300);tic;b=(1.0+a)*a;toc
Elapsed time is 0.0257161 seconds.
octave-3.0.1.exe:3> a=zeros(300,300);tic;b=(NaN+a)*a;toc
Elapsed time is 15.7125 seconds.
AMD Turion 64 X2, Octave 3.0.1 MSVC2008 SSE3
octave-3.0.1.exe:3> a=zeros(300,300);tic;b=(1.0+a)*a;toc
Elapsed time is 0.0244939 seconds.
octave-3.0.1.exe:4> a=zeros(300,300);tic;b=(NaN+a)*a;toc
Elapsed time is 0.0251131 seconds.
Thank you,
Olli
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