On Thu, Jun 2, 2011 at 11:49 PM, Erik Trimble <erik.trim...@oracle.com> wrote:
> 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.
So is there a way to read these real I/Ops numbers ?
iostat is reporting 600-800 I/Ops peak (1 second sample) for these
7200 RPM SATA drives. If the drives are doing aggregation, then how to
tell what is really going on ?
--
{--------1---------2---------3---------4---------5---------6---------7---------}
Paul Kraus
-> Senior Systems Architect, Garnet River ( http://www.garnetriver.com/ )
-> Sound Coordinator, Schenectady Light Opera Company (
http://www.sloctheater.org/ )
-> Technical Advisor, RPI Players
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