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Date: Tue, 29 Jan 2008 10:23:22 -0600
From: "Kevin Grittner" <Kevin.Grittner@wicourts.gov>
To: "Gregory Stark" <stark@enterprisedb.com>, "Matthew" <matthew@flymine.org>
Cc: "Mark Mielke" <mark@mark.mielke.cc>,
	<pgsql-performance@postgresql.org>
Subject: Re: RAID arrays and performance
References: <Pine.LNX.4.58.0712041213490.3731@aragorn.flymine.org>
	<87lk8aiqf3.fsf@oxford.xeocode.com>
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	<87bq96ii3b.fsf@oxford.xeocode.com>
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	<4755810E.4000707@mark.mielke.cc>
	<Pine.LNX.4.58.0712041637260.3731@aragorn.flymine.org>
	<87zlwq6yz9.fsf@oxford.xeocode.com>
	<Pine.LNX.4.64.0801291417330.4642@aragorn.flymine.org>
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	<Pine.LNX.4.64.0801291505180.4642@aragorn.flymine.org><Pine.LNX.4.64.0801291505180.4642@aragorn.flymine.org>
	(matthew@flymine.org's message of "Tue\,
	29 Jan 2008 15\:09\:03 +0000 \(GMT\)")
	<877ihsvdcb.fsf@oxford.xeocode.com>
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>>> On Tue, Jan 29, 2008 at  9:52 AM, in message
<877ihsvdcb.fsf@oxford.xeocode.com>, Gregory Stark <stark@enterprisedb.com>=

wrote:=20
=20
> 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.
=20
That seems right to me, based on the probabilities of any new
request hitting an already-busy drive.
=20
> 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
=20
What's the basis for that?
=20
-Kevin
=20