On 9/25/2013 4:34 PM, Chris Murphy wrote:
> On Sep 25, 2013, at 3:18 PM, Stan Hoeppner <stan@xxxxxxxxxxxxxxxxx> wrote:
>> On 9/25/2013 7:56 AM, Stewart Webb wrote:
>>> Hi All,
>> Hi Stewart,
>>> I am trying to do the following:
>>> 3 x Hardware RAID Cards each with a raid 6 volume of 12 disks presented to
>>> the OS
>>> all raid units have a "stripe size" of 512 KB
>> Just for future reference so you're using correct terminology, a value
>> of 512KB is surely your XFS su value, also called a "strip" in LSI
>> terminology, or a "chunk" in Linux software md/RAID terminology. This
>> is the amount of data written to each data spindle (excluding parity) in
>> the array.
>> "Stripe size" is a synonym of XFS sw, which is su * #disks. This is the
>> amount of data written across the full RAID stripe (excluding parity).
>>> so given the info on the xfs.org wiki - I sould give each filesystem a
>>> sunit of 512 KB and a swidth of 10 (because RAID 6 has 2 parity disks)
>> Partially correct. If you format each /dev/[device] presented by the
>> RAID controller with an XFS filesystem, 3 filesystems total, then your
>> values above are correct. EXCEPT you must use the su/sw parameters in
>> mkfs.xfs if using BYTE values. See mkfs.xfs(8)
Small correction: su is a byte value. sw is an integer representing
the number of data spindles.
>>> all well and good
>>> But - I would like to use Linear LVM to bring all 3 cards into 1 logical
>>> volume -
>>> here is where my question crops up:
>>> Does this effect how I need to align the filesystem?
>> In the case of a concatenation, which is what LVM linear is, you should
>> use an XFS alignment identical to that for a single array as above.
> So keeping the example, 3 arrays x 10 data disks, would this be su=512k and
No. In this configuration, as far as XFS is concerned LVM doesn't exist
in the stack because it doesn't change the RAID geometry, so you ignore it.