> From: zfs-discuss-boun...@opensolaris.org [mailto:zfs-discuss- > boun...@opensolaris.org] On Behalf Of Erik Trimble > > Here's how you calculate (average) how long a random IOPs takes: > > seek time + ((60 / RPMs) / 2))] > > 1 Random IOPs takes [8.5ms + 4.13ms] = 12.6ms, which translates to 78 > IOPS
While this is true, all drives nowadays use things like PIO command queueing, and other hardware optimization techniques. So even when you instruct the drive to do a bunch of random IO, the drive will make it less random in the controller before it instructs the arm to move about and so on. Generally speaking, these techniques will approx double the random IOPS, because with a random distribution of IO requests, on average it will be able to halve the randomness. Consider your nit picked. ;-) _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
