On 6/30/2013 1:43 PM, aurfalien wrote:
> I understand swidth should = #data disks.
No. "swidth" is a byte value specifying the number of 512 byte blocks
in the data stripe.
"sw" is #data disks.
> And the docs say for RAID 6 of 8 disks, that means 6.
>
> But parity is distributed and you actually have 8 disks/spindles working for
> you and a bit of parity on each.
>
> So shouldn't swidth equal disks in raid when its concerning distributed
> parity raid?
No. Lets try visual aids.
Set 8 coffee cups (disk drives) on a table. Grab a bag of m&m's.
Separate 24 blues (data) and 8 reds (parity).
Drop a blue m&m in cups 1-6 and a red into 7-8. You just wrote one RAID
stripe. Now drop a blue into cups 3-8 and a red in 1-2. Your second
write, this time rotating two cups (drives) to the right. Now drop
blues into 5-2 and reds into 3-4. You've written your third stripe,
rotating by two cups (disks) again.
This is pretty much how RAID6 works. Each time we wrote we dropped 8
m&m's into 8 cups, 6 blue (data chunks) and 2 red (parity chunks).
Every RAID stripe you write will be constructed of 6 blues and 2 reds.
XFS, or EXT4, or any filesystem, can only drop blues into the first 6
cups of a stripe. The RAID adds the two reds to every stripe.
Maybe now you understand why sw=6 for an 8 drive RAID6. And now maybe
you understand what "distributed parity" actually means--every stripe is
shifted, not just the parity chunks.
--
Stan
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