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Date:      Fri, 6 Jul 2012 07:52:00 -0700 (PDT)
From:      Jason Usher <jusher71@yahoo.com>
To:        Zaphod Beeblebrox <zbeeble@gmail.com>
Cc:        freebsd-fs@freebsd.org
Subject:   Re: vdev/pool math with combined raidzX vdevs...
Message-ID:  <1341586320.93244.YahooMailClassic@web122502.mail.ne1.yahoo.com>
In-Reply-To: <CACpH0MemwZDCXsh4USzeFHUO8fbW09TSOYyVPa2dWmKc8N%2B=_Q@mail.gmail.com>

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=0A=0A--- On Fri, 7/6/12, Zaphod Beeblebrox <zbeeble@gmail.com> wrote:=0A=
=0A> Is there some penalty for not=0A> googling some basic stats course?=A0=
 OK.=0A> This is from memory (hint: you probably should google).=0A=0A=0AAc=
tually I spent a few hours googling zfs probabilities, and I cannot find an=
y discussion of the changes in failure probability when multiple raidzX are=
 striped in a single pool.=0A=0AI did not try to google the math and do it =
myself, as I wouldn't trust my own results.=0A=0A=0A> Similarly,=0A> =0A> p=
(12drz2) =3D 12 * p(f) * 11 * p(f) * 10 * p(f)=0A> p(12drz3) =3D 12 * p(f) =
* 11 * p(f) * 10 * p(f) * 9 * p(f)=0A> =0A> ... again with those assumption=
s are more complex=0A> probabilities given=0A> your replacement strategy.=
=0A> =0A> ... so, again with simplistic assumptions,=0A> =0A> p(36drz3 --- =
12 drives, 3 groups) =3D p(12drz3) * 3=0A> =0A> A "vanilla" RAID-Z2 (if I m=
ake an assumption to what you're=0A> saying) is:=0A> =0A> p(36drz2) =3D 36 =
* p(f) * 35 * p(f)=0A> =0A> ... but I can't directly answer you question wi=
thout knowing=0A> a) the=0A> structure of the RAID-Z2 array and p(f).=A0 If=
 we use a=0A> 1% figure for=0A> p(f), then P(36drz3,12,3) =3D 0.035% and p(=
36drz2) =3D 4.3%=0A> =0A> ... that is the raid-Z2 case (one group of 36 dri=
ves, two=0A> redundant=0A> --- which is crazy) is 4.3% likely to fail where=
 the 3-group=0A> RAID-Z3=0A> is only 0.035% likely to fail.=A0 As a more sa=
ne=0A> comparison,=0A> p(36drz2,12,3) =3D 3.8%=0A=0A=0AOk, you're right - I=
 did not specify the structure of the raidz2 array.=0A=0AWhat I meant to co=
mpare was the failure probabilities of:=0A=0A- a single raidz2 vdev made up=
 of 12 disks (10 data, 2 parity)=0A=0Avs.=0A=0A- a single raidz3 vdev made =
up of 12 disks (9 data, 3 parity)=0A=0Avs.=0A=0A- a single raidz3 vdev made=
 up of 12 disks (9 data, 3 parity) which ALSO happens to be participating i=
n a stripe with two other identical raidz3 vdevs, all in one zpool.=0A=0A=
=0AI can see some probabilites for the first two examples here:=0A=0Ahttp:/=
/hardforum.com/showthread.php?t=3D1621123=0A=0A(search for first post on pa=
ge by "john4200")=0A=0Aas well as your own math, which appears to be the sa=
me.=0A=0A=0AAlso, I think we don't need to specify the failure rate, F, sin=
ce we are merely comparing three scenarios, and can compare results that st=
ill contain an 'F' variable in them ... that is, our answers can contain a =
yet undefined 'F', right ?=0A=0A=0AAs for myself, I have decided that raidz=
2 is "not enough" for me, and at the same time, would really, really like t=
o combine three raidz3 into a single zpool ... but I don't want to do that =
if that configuration brings me back to raidz2-ish failure probabilities...



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