In message <20050730221215.GA757_at_uk.tiscali.com>, Brian Candler writes: >On Sat, Jul 30, 2005 at 08:37:17PM +0200, Poul-Henning Kamp wrote: >> In message <20050730171536.GA740_at_uk.tiscali.com>, Brian Candler writes: >> >On Sat, Jul 30, 2005 at 03:29:27AM -0700, Julian Elischer wrote: >> >> >> >> The snapshot below is typical when doing tar from one drive to another.. >> >> (tar c -C /disk1 f- .|tar x -C /disk2 -f - ) >> >> >> >> dT: 1.052 flag_I 1000000us sizeof 240 i -1 >> >> L(q) ops/s r/s kBps ms/r w/s kBps ms/w d/s kBps ms/d %busy Name >> >> 0 405 405 1057 0.2 0 0 0.0 0 0 0.0 9.8| ad0 >> >> 0 405 405 1057 0.3 0 0 0.0 0 0 0.0 11.0| ad0s2 >> >> 0 866 3 46 0.4 863 8459 0.7 0 0 0.0 63.8| da0 >> >> 25 866 3 46 0.5 863 8459 0.8 0 0 0.0 66.1| da0s1 >> >> 0 405 405 1057 0.3 0 0 0.0 0 0 0.0 12.1| ad0s2f >> >> 195 866 3 46 0.5 863 8459 0.8 0 0 0.0 68.1| da0s1d >... >> >But if really is only 12.1% busy (which the 0.3 ms/r implies), >> >> "busy %" numbers is *NOT* a valid measure of disk throughput, please do >> not pay attention to such numbers! > >It seems to me that > reads/sec * milliseconds/read = milliseconds spent reading per second This is only true if there can be only one outstanding request at a time. That is not a valid assumption. Modern (SCSI) disks have tagged queueing for instance. You could have: Time Event 0.0 Issue request #0 0.1 Issue request #1 1.0 Request #0 comes back (service time = 1 second) 1.1 Request #1 comes back (service time = 1 second) The average service time is 1 second per request. The throughput is 2 requests per second. Using your formula, the drive has spent 2000 milliseconds per second handling requests. This is the good old bandwidth vs. delay discussion again... >and that the "busy %" is <milliseconds per second spent with one or more Maybe it would be more clear if you realize that, the definition is: %busy = 100 - %idle >I'm not saying geom is counting wrongly; I am 100% sure it counts correctly (well, actually, the %busy is not taking outstanding requests properly into account, it would cost a lot of work and math to do so.) >I am just agreeing with the OP that >the underlying reason for this poor utilisation is worth investigating. If the two processes are not deliberately sleeping (which I doubt), it has to be a scheduler bug. -- Poul-Henning Kamp | UNIX since Zilog Zeus 3.20 phk_at_FreeBSD.ORG | TCP/IP since RFC 956 FreeBSD committer | BSD since 4.3-tahoe Never attribute to malice what can adequately be explained by incompetence.Received on Sat Jul 30 2005 - 20:28:03 UTC
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