Date: Tue, 25 Apr 1995 14:54:57 +1000 From: Bruce Evans <bde@zeta.org.au> To: agl@mac.glas.apc.org, jgreco@brasil.moneng.mei.com Cc: freebsd-hackers@FreeBSD.org Subject: Re: Is P5 is faster than DX4 100 under FreeBSD? Message-ID: <199504250454.OAA15990@godzilla.zeta.org.au>
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>> Linux FAQ (or was it HOWTO?) states that bogomips are measured as >> DX2/DX4 Mhz * 0.5 >> P5 Mhz * 0.39 >... >Now, the 386/40 with Linux would be about 7.9 bogomips. The 386/40 with >FreeBSD would be about 8.6 bogomips (a little over the predicted 8?)... the >486dx2/80 would be 35.1 (a little under the predicted 40?), and the Pentium >would be 117.0, which does not seem reasonable, although the P90 does appear >to be at least 2x on most of my tests over a 486dx2/80... so perhaps the >P90 performs very well (and/or optimizes much better) on tight loops. >Interestingly enough, the Linux formula you gave would suggest that the P90 >would be rated at 35.1 bogomips. The bogimips test happened to have instructions that pipelined very poorly for Pentiums. This can't be fixed because then the test would be useless for historical comparisons. This doesn't matter because the test is bogus. On all x86's with an on-chip cache (i.e., for everything except i386's), the loop in the test takes a certain number of cycles that depends only on the cpu, so its results are given very accurately by the formula bogomips = speed_in_MHz / cycles_per_test / fudge_factor(= 2?) cycles_per_test is easy to calculate from the best-case instruction timings. >I then ftp'd a Linux boot disk and did some measurements on the machines I >had easy access to: >machine bogo >386dx/40 8.0 >486dx2/80 41.1 >pentium 90 36.2 For i386's the speed depends on the external cache and is difficult to predict. Bruce
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