From mboxrd@z Thu Jan  1 00:00:00 1970
From: Michael Evans <mjevans1983@gmail.com>
Subject: Re: What RAID type and why?
Date: Sun, 7 Mar 2010 12:40:24 -0800
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To: Goswin von Brederlow <goswin-v-b@web.de>
Cc: Neil Brown <neilb@suse.de>, Guy Watkins <linux-raid@watkins-home.com>, Greg Freemyer <greg.freemyer@gmail.com>, Mark Knecht <markknecht@gmail.com>, Linux-RAID <linux-raid@vger.kernel.org>
List-Id: linux-raid.ids

On Sun, Mar 7, 2010 at 4:52 AM, Goswin von Brederlow <goswin-v-b@web.de=
> wrote:
> Neil Brown <neilb@suse.de> writes:
>
>> On Sat, 06 Mar 2010 18:17:44 -0500
>> "Guy Watkins" <linux-raid@watkins-home.com> wrote:
>>
>>> }
>>> } At a minimum I would build a 3-disk raid 6. =A0raid 6 does a lot =
of i/o
>>> } which may be a problem.
>>>
>>> If he only needs 3 drives I would recommend RAID1. =A0Can still loo=
se 2 drives
>>> and you don't have the RAID6 I/O overhead.
>>>
>>
>> and as md/raid6 requires at least 4 drives, RAID1 is not just the be=
st
>> solution to survive two failures on a 3-device array, it is the only=
 solution.
>>
>> NeilBrown
>
> Except that there also is raid10 with 3 mirrors. :)
>
> MfG
> =A0 =A0 =A0 =A0Goswin
>
> PS: Why doesn't raid6 still not allow 3 drives for the special case o=
f
> converting raid1 -> raid6?
>
>
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That should be obvious:

Possible stripes:

Start:
1, 1, 1;
2, 2, 2;

'raid6' overtake...
1, q, Q;
2, q, Q;

'raid6' overtake with missing;
1, (missing 2), q, Q;
3, (missing 4), q, Q;

In the first overtake case you have the requirement of generating 200%
parity, which probably won't work for the algorithm and is a silly
idea in general since it's computationally far less expensive to store
another copy of either form of data instead.

In the second you're gaining the space of a second disk at the cost of
being already degraded; why not just go for raid 5 instead?

You can overtake raid5 later with raid6 if you add more devices.
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