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Subject: Re: Source code to measure it - results

Author: Robert Hyatt

Date: 09:08:58 07/16/03

Go up one level in this thread

On July 16, 2003 at 01:39:34, Andrew Dados wrote:

>On July 15, 2003 at 21:13:23, Jeremiah Penery wrote:
>>On July 15, 2003 at 20:19:34, Vincent Diepeveen wrote:
>>>On July 15, 2003 at 15:24:19, Gerd Isenberg wrote:
>>>Gerd use it with a bigger hashtable. Not such a small
>>>400MB is really the minimum to measure.
>>Measuring 90MB, something like 99.65% of the accesses should be to RAM and not
>>cache.  With 100MB, it's 99.8%.  Yet when I measure those two things, I get a
>>whole 6.1ns latency difference according to your test.  Even measuring only
>>20MB, 98.4% of the allocated memory can not be in cache. (All of this assumes
>>that the program takes up 100% of the cache, which it won't.)
>>There's something wrong that causes memory access time to be reported much
>>higher when testing larger 'hashtable' sizes.  Anything large enough to
>>overwhelm the cache should report similar, if not almost identical, results.
>>However, your program gives wildly different numbers.
>>Trying to allocate 12500000 entries. In total 100000000 bytes
>>  Average measured read read time at 1 processes = 183.935982 ns
>>Trying to allocate 11250000 entries. In total 90000000 bytes
>>  Average measured read read time at 1 processes = 177.806427 ns
>>Trying to allocate 43750000 entries. In total 350000000 bytes
>>  Average measured read read time at 1 processes = 253.592331 ns
>>In the last test, I can't be completely sure I wasn't paging at all.  I didn't
>>see the disk light flashing, but it's possible that this hit the disk more than
>>once, which would make the number look much higher than it should.
>>Still, relative to the other results people have given, this is not so bad,
>>since I have only PC2100 memory (133MHz DDR).
>Get rid of RNG.. do lookups every x+32 bytes (or so), after exhausting hashtable

That will fail on a PIV.  It's L2 cache line is 128 bytes.  You will see 1/4
the real latency, roughly, because 3 of every four probes will be a cache

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