On 6/2/2011 5:12 PM, Jens Elkner wrote:
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.
Nope. In terms of actual, obtainable IOPS, a 7200RPM drive isn't going
to be able to do more than 200 under ideal conditions, and should be
able to manage 50 under anything other than the pedantically worst-case
situation. That's only about a 50% deviation, not like an order of
magnitude or so.
--
Erik Trimble
Java System Support
Mailstop: usca22-123
Phone: x17195
Santa Clara, CA
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