Data Block size on EVAs

[k20pattu]

Gurus,

I believe the data is written within the EVA in the form of stripes on each disks. Does anyone know what is the stripe size, whether it is configurable, if yes - how to configure it?

Thanks
Ketan

[Uwe Zessin · Accepted solution]

The internal "chunk size" is 128KB and not configurable (this is driven by special hardware chips).
The "stripe size" depends on the VRAID-level:
D=1 data chunk
- 0 = 4D
- 1 = 4D and each D mirrored on two disks
- 5 = 4D+P
- 6 = 4D+PQ

[k20pattu]

Thanks.

How are chunk size/stripe size/data chunks related? Is there a document with these details?

Ketan

[Uwe Zessin]

The chunk size defines how many 512-byte blocks are stored sequentially on a single disk drive. The 'next' chunk is very likely stored on another disk drive.

The stripe size defines how many chunks are included in that 'stripe'.

A 'data chunk' stores user data. P+Q chunks are used to store redundany information to be able to recover in case a disk holding a particular data chunk fails.

There is no single, official document (from HP) that describes this. A lot of this information can only obtained by training or discussion with people 'who know'.

[k20pattu]

Chunk size is fixed at 128KB (256 x 512 Bytes blocks). How many chunks in a single stripe - is it dependent on RAID?

RAID0 - 1 Stripe (size 128KB) = 1 Chunk of 128KB
RAID1 - 1 Stripe (size 256KB) = 2 Chunks of 128KB chunks
RAID5 - 1 Stripe (size 640KB) = 5 Chunks of 128KB
RAID6 - 1 Stripe (size 768KB) = 6 Chunks of 128KB

Thanks
Ketan

[Uwe Zessin]

Yes, the stripe size depends on the VRAID-level:

VRAID-0 - stripe size = 4 chunks

VRAID-1 - stripe size = 4 chunks
........ but chunks are mirrored, so physically there are 8 chunks in a stripe


VRAID-5 - stripe size = 5 chunks
........ 4 chunks store data and 1(one) stores redundancy information (P)

VRAID-6 - stripe size = 6 chunks
........ 4 chunks store data and 2(two) store redundancy information (P+Q)


You can mix different VRAID-levels in a disk group. A disk group need not have an amount of disk drives that is a multiple of the stripe size of a particular VRAID-level (e.e. VRAID-5: diskgroup need not have N*5 disk drives). Due to the way the EVA virtualization works, all VRAID-data is pretty equally striped across all disk drives in that disk group.

There is one exception due to the way VRAID-1 is implemented:
In a disk group with an odd (un-even) number of disk drives (e.g. 9, 11, 17, 111), one of the disk drives cannot store VRAID-1 data, because it is missing a 'partner'.


You may have seen documentation that suggests a disk group be N*8 disk drives - well, that is an optimization for VRAID-1. All 8 chunks of a stripe can be stored on 8 different disk drives.

[k20pattu]

I thought stripe size will remain same as 512KB but apparantly that is not the case.

Anyway, many thanks for this information.

Ketan