On Wed, Jun 01, 2011 at 06:17:08PM -0700, Erik Trimble wrote: > On Wed, 2011-06-01 at 12:54 -0400, Paul Kraus wrote: > Here's how you calculate (average) how long a random IOPs takes: > seek time + ((60 / RPMs) / 2))] > > A truly sequential IOPs is: > (60 / RPMs) / 2) > > For that series of drives, seek time averages 8.5ms (per Seagate). > So, you get > > 1 Random IOPs takes [8.5ms + 4.13ms] = 12.6ms, which translates to 78 > IOPS > 1 Sequential IOPs takes 4.13ms, which gives 120 IOPS. > > Note that due to averaging, the above numbers may be slightly higher or > lower for any actual workload.
Nahh, shouldn't it read "numbers may be _significant_ higher or lower" ...? ;-) Regards, jel. -- Otto-von-Guericke University http://www.cs.uni-magdeburg.de/ Department of Computer Science Geb. 29 R 027, Universitaetsplatz 2 39106 Magdeburg, Germany Tel: +49 391 67 12768 _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
