Re: layout of far blocks in raid10

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On Tue, May 11, 2010 at 5:56 PM, Neil Brown <neilb@xxxxxxx> wrote:
> I'm not quite sure how to respond to this...  As a mathematician I would
> expect you to understand the important of precision in choosing words, yet
> you use the word "know" for something that is exactly wrong.  Either you mean
> "guess" or you have been seriously misinformed.  If it is the latter, then
> please let me know where this misinformation came from so I can see about
> getting it corrected.
>
> md/raid10 uses a simple cyclic layout in all cases.  It does so because this
> layout is completely general and works for all numbers of devices and copies.
>
> So you can only survive multiple device failures where are most N-1 are
> adjacent where N is the number of copies, and the first and last devices are
> treated as adjacent.

Mathematicians are sometimes wrong too, sadly.  :)  (And I'm only a
grad student!)  I believe this is where I got my info:

http://git.debian.org/?p=pkg-mdadm/mdadm.git;a=blob_plain;f=debian/FAQ;hb=HEAD

The answer to question 20 of that suggests that if you have four
disks, 0 1 2 3, then 0 and 1 form one pair and 2 and 3 form the other.
 If 2 fails, then 0 or 1 could still fail without data loss, but a
failure of 3 will cause data loss.  Obviously, you know what you're
talking about better than a Debian FAQ, so unless I'm misunderstanding
the FAQ or you or both, maybe you should talk to the author of that.

Testing with loopback files does seem to show that failing the second
and third drives in a four-drive RAID will cause the RAID to fail, as
I would predict from what you say and contrary to what I interpreted
that FAQ to mean, so hopefully now I understand correctly.

Thanks for the correction.  Next time I'll be more cautious.
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