Re: [gentoo-user] OT: btrfs raid 5/6

[Rich Freeman <rich0@gentoo.org>]

On Sat, Dec 9, 2017 at 1:28 PM, Wols Lists <antlists@youngman.org.uk> wrote:
> On 09/12/17 16:58, J. Roeleveld wrote:
>> On Friday, December 8, 2017 12:48:45 AM CET Wols Lists wrote:
>>> On 07/12/17 22:35, Frank Steinmetzger wrote:
>>>>> (Oh - and md raid-5/6 also mix data and parity, so the same holds true
>>>>>
>>>>>> there.)
>>>>
>>>> Ok, wasn’t aware of that. I thought I read in a ZFS article that this were
>>>> a special thing.
>>>
>>> Say you've got a four-drive raid-6, it'll be something like
>>>
>>> data1   data2   parity1 parity2
>>> data3   parity3 parity4 data4
>>> parity5 parity6 data5   data6
>>>
>>> The only thing to watch out for (and zfs is likely the same) if a file
>>> fits inside a single chunk it will be recoverable from a single drive.
>>> And I think chunks can be anything up to 64MB.
>>
>> Except that ZFS doesn't have fixed on-disk-chunk-sizes. (especially if you use
>> compression)
>>
>> See:
>> https://www.delphix.com/blog/delphix-engineering/zfs-raidz-stripe-width-or-how-i-learned-stop-worrying-and-love-raidz
>>
> Which explains nothing, sorry ... :-(
>
> It goes on about 4K or 8K database blocks (and I'm talking about 64 MEG
> chunk sizes). And the OP was talking about files being recoverable from
> a disk that was removed from an array. Are you telling me that a *small*
> file has bits of it scattered across multiple drives? That would be *crazy*.

I'm not sure why it would be "crazy."  Granted, most parity RAID
systems seem to operate just as you describe, but I don't see why with
Reed Solomon you couldn't store ONLY parity data on all the drives.
All that matters is that you generate enough to recover the data - the
original data contains no more information than an equivalent number
of Reed-Solomon sets.  Of course, with the original data I imagine you
need to do less computation assuming you aren't bothering to check its
integrity against the parity data.

In case my point is clear a RAID would work perfectly fine if you had
5 drives with the capacity to store 4 drives wort of data, but instead
of storing the original data across 4 drives and having 1 of parity,
you instead compute 5 sets of parity so that now you have 9 sets of
data that can tolerate the loss of any 5, then throw away the sets
containing the original 4 sets of data and store the remaining 5 sets
of parity data across the 5 drives.  You can still tolerate the loss
of one more set, but all 4 of the original sets of data have been
tossed already.


-- 
Rich