Date: Fri, 6 Jul 2012 17:30:13 +0100 From: Martin Simmons <martin@lispworks.com> To: freebsd-fs@freebsd.org Subject: Re: vdev/pool math with combined raidzX vdevs... Message-ID: <201207061630.q66GUDPD019051@higson.cam.lispworks.com> In-Reply-To: <CACpH0MemwZDCXsh4USzeFHUO8fbW09TSOYyVPa2dWmKc8N%2B=_Q@mail.gmail.com> (message from Zaphod Beeblebrox on Fri, 6 Jul 2012 03:08:46 -0400) References: <1341537402.58301.YahooMailClassic@web122504.mail.ne1.yahoo.com> <CACpH0MemwZDCXsh4USzeFHUO8fbW09TSOYyVPa2dWmKc8N%2B=_Q@mail.gmail.com>
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>>>>> On Fri, 6 Jul 2012 03:08:46 -0400, Zaphod Beeblebrox said: > > Is there some penalty for not googling some basic stats course? OK. > This is from memory (hint: you probably should google). > > p(f) ... the probably of failure of one drive over some unit time (say > one year). A two drive RAID-0 array has probability p(2dr0) = 2 * > p(f) + p(f). That is (for the logic guys): the array fails if either > drive fails. This looks wrong to me because it can be > 1 if p(f) > 1/3 :-) I think it should be p(2dr0) = (2 * p(f) * (1 - p(f))) + (p(f) * p(f)) I.e. the probability of either drive failing and the other one not failing plus the probability of both drives failing simultaneously. Or another way of thinking about it: p(2dr0) = 1 - ((1 - p(f)) * (1 - p(f))) I.e. the inverse of the probability of neither drive failing __Martin
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