HPE Storage Users Group

Hello All

I have read the HP 3PAR StoreServ Storage Concepts Guide (http://h20565.www2.hp.com/hpsc/doc/public/display?docId=emr_na-c03606434)

When reading the discussion on page 37-38 about RAID Multi-Parity (essentially RAID 6) the diagram seems odd to me:



The document states:


V and X are taken from the same chunklet, therefore the same physical disk which seems to break the rules?
Also that is 7 data "steps" and only one parity step, which doesn't seem to be Multi-Parity.

Obviously I am misunderstanding something fundamental here, and my guess is that I am not properly grasping the 2 Sets in one Row.

Do we need to see the two sets as one RAID MP "set", which would see 14 data "steps" and 2 parity "steps".
Also that seems to imply that the parity steps may not be different parity syndromes (X-OR and Galois Field as in traditional RAID 6)?

Can anybody give a good explanation of how a RAID MP works, answer the specific question, or point me to some good resources on 3PAR raid concepts?

There does seem to be a dearth of resources unfortunately. All the resources I have been able to find do not go much further than discussing the basics (chunklets, LDs etc.), and leave the more complex considerations untouched.

Only resource I have come across so far is the following: http://h20564.www2.hp.com/hpsc/doc/public/display?docId=mmr_kc-0117156&sp4ts.oid=5074925



So what does this say?
Well two XOR parities don't provide the ability to recover from all two "step" loses.

The search for an answer continues ...

Note - Other Interesting Resources, not necessarily directly related:
http://www.techopsguys.com/2010/08/13/do-you-really-need-raid-6/
http://www.zdnet.com/article/why-raid-5-stops-working-in-2009/
http://igoro.com/archive/how-raid-6-dual-parity-calculation-works/

That illustration has completely confused and frustrated me.

So I think a key thing to focus on is:



So I believe the documentation may be in error.



We agree that V and X should never share a disk, ever. Thats a red flag.
The 3rd spindle that contains G,H,I,p2 is missing in the example set. Another red flag.
And lastly, we only have one parity p0 in the example and should have 2... 3 strikes?

That said, I suspect the document should say something like: "In the first RAID MP set in the following example, p0 and p1 are parity steps for data steps H, L, M, Q, U, and V."

Hopefully one of the HP-3Par engineers can confirm/correct our understanding of this picture and the explanation that goes with it.

_________________
Richard Siemers
The views and opinions expressed are my own and do not necessarily reflect those of my employer.

Well the parity for the first step is the data for the second step. I had to reread this at least 5 times (and delete my original post, i made before this one) before i finally understood.

I agree that either V or X must be the parity step.

Another weird issue is them saying that you need at least 8 drives.
A 3par can start off fine with 4 drives per node, and run raid 6+2 over those 4 drives.

Each drive failure would create a double parity failure, which is ugly and defies the usage of raid-6 imho, but it works.

Another thing is that in this example they work with 6+2, but the 3par supports 4+2 fine as well, that would again prove the need for 8 drives is nonsense.

_________________
The goal is to achieve the best results by following the clients wishes. If they want to have a house build upside down standing on its chimney, it's up to you to figure out how do it, while still making it usable.

Looks like a documentation mistake, probably a cut and paste job that no one looked into too deeply, to be honest I've never looked at the diagram either

"Another weird issue is them saying that you need at least 8 drives. A 3par can start off fine with 4 drives per node, and run raid 6+2 over those 4 drives."

Raid is built at the LD level and LD's are built per node, so for Raid 6 (min 4+2) means 2 nodes x 4+2 = 12 disks, for 6+2 that would be 16 disks (maybe they're talking per node). Minimum supported configuration is 8 drives to enable Raid 5 3+1, 2 x nodes x 3+1 = 8. But in theory you could get away with 6 drives and use raid 1 (metadata is triple mirrored) but that isn't a supported configuration.

John, point is that the documentation implies that 8 drives per raid-mp set are needed, not 8 drives per nodepair.

_________________
The goal is to achieve the best results by following the clients wishes. If they want to have a house build upside down standing on its chimney, it's up to you to figure out how do it, while still making it usable.

Yes appreciated, what I think they are attempting to describe is the layout of an LD rather than a raid set (multiple LD's) as this would vary based on the number or nodes so would be variable.

e.g for 4+2 and 6+2
2 Nodes = 12 disks -- 16 disks (minimum)
4 Nodes = 24 disks -- 32 disks
6 nodes = 36 disks -- 48 disks
8 nodes = 48 disks -- 64 disks

Once over the minimums you can get away without the exact numbers above but you'll have a number of chunklets left over per disk which are unable to satisfy the layout requirement of the CPG, but then again these may be able to be utilized by another CPG.

Even so the doc is inaccurate, with mag level each set (raidlet) (4+2, 6+2, 8+2, 10+2, 14+2) must exist on a separate disk, hence the above minimums. With cage the same is also true, but any 2 chunklets in a given set may share the same cage.

It should probably say something along the lines of The letters don't particularly matter, only that they are on separate disks.

There's a feedback address in the document for corrections, if anyone would like to request this