From owner-freebsd-fs@FreeBSD.ORG Fri Jul 6 16:40:46 2012 Return-Path: Delivered-To: freebsd-fs@freebsd.org Received: from mx1.freebsd.org (mx1.freebsd.org [69.147.83.52]) by hub.freebsd.org (Postfix) with ESMTP id 4FFE81065672 for ; Fri, 6 Jul 2012 16:40:46 +0000 (UTC) (envelope-from martin@lispworks.com) Received: from lwfs1-cam.cam.lispworks.com (mail.lispworks.com [193.34.186.230]) by mx1.freebsd.org (Postfix) with ESMTP id DA5508FC19 for ; Fri, 6 Jul 2012 16:40:45 +0000 (UTC) Received: from higson.cam.lispworks.com (higson.cam.lispworks.com [192.168.1.7]) by lwfs1-cam.cam.lispworks.com (8.14.5/8.14.5) with ESMTP id q66GUDx3049521; Fri, 6 Jul 2012 17:30:13 +0100 (BST) (envelope-from martin@lispworks.com) Received: from higson.cam.lispworks.com (localhost.localdomain [127.0.0.1]) by higson.cam.lispworks.com (8.14.4) id q66GUD2e019055; Fri, 6 Jul 2012 17:30:13 +0100 Received: (from martin@localhost) by higson.cam.lispworks.com (8.14.4/8.14.4/Submit) id q66GUDPD019051; Fri, 6 Jul 2012 17:30:13 +0100 Date: Fri, 6 Jul 2012 17:30:13 +0100 Message-Id: <201207061630.q66GUDPD019051@higson.cam.lispworks.com> From: Martin Simmons To: freebsd-fs@freebsd.org In-reply-to: (message from Zaphod Beeblebrox on Fri, 6 Jul 2012 03:08:46 -0400) References: <1341537402.58301.YahooMailClassic@web122504.mail.ne1.yahoo.com> Subject: Re: vdev/pool math with combined raidzX vdevs... X-BeenThere: freebsd-fs@freebsd.org X-Mailman-Version: 2.1.5 Precedence: list List-Id: Filesystems List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Fri, 06 Jul 2012 16:40:46 -0000 >>>>> 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