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From: Gregory Stark <gsstark@mit.edu>
To: Kevin Grittner <Kevin.Grittner@wicourts.gov>
Cc: Matthew <matthew@flymine.org>
Cc: Mark Mielke <mark@mark.mielke.cc>
Cc: pgsql-performance@postgresql.org
Subject: Re: RAID arrays and performance
Date: 29 Jan 2008 11:45:20 -0500
Message-ID: <873asginrz.fsf@stark.xeocode.com> (raw)
In-Reply-To: <479EFE9A.EE98.0025.0@wicourts.gov>
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	<877ihsvdcb.fsf@oxford.xeocode.com>
	<479EFE9A.EE98.0025.0@wicourts.gov>

"Kevin Grittner" <Kevin.Grittner@wicourts.gov> writes:

> >>> On Tue, Jan 29, 2008 at  9:52 AM, in message
> <877ihsvdcb.fsf@oxford.xeocode.com>, Gregory Stark <stark@enterprisedb.com>
> wrote: 
>  
> > I got this from a back-of-the-envelope calculation which now that I'm trying
> > to reproduce it seems to be wrong. Previously I thought it was n(n+1)/2 or
> > about n^2/2. So at 16 I would have expected about 128 pending i/o requests
> > before all the drives could be expected to be busy.
>  
> That seems right to me, based on the probabilities of any new
> request hitting an already-busy drive.
>  
> > Now that I'm working it out more carefully I'm getting that the expected
> > number of pending i/o requests before all drives are busy is
> >  n + n/2 + n/3 + ... + n/n
>  
> What's the basis for that?

Well consider when you've reached n-1 drives; the expected number of requests
before you hit the 1 idle drive remaining out of n would be n requests. When
you're at n-2 the expected number of requests before you hit either of the two
idle drives would be n/2. And so on. The last term of n/n would be the first
i/o when all the drives are idle and you obviously only need one i/o to hit an
idle drive.

-- 
  Gregory Stark
  EnterpriseDB          http://www.enterprisedb.com




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