On 6/30/2013 9:09 PM, Dave Chinner wrote: > On Sun, Jun 30, 2013 at 06:54:31PM -0700, aurfalien wrote: >> >> On Jun 30, 2013, at 6:38 PM, Dave Chinner wrote: >> >>> On Sun, Jun 30, 2013 at 04:42:06PM -0500, Stan Hoeppner wrote: >>>> 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. >>> >>> Right, that's how they are constructed, but not all RAID distributes >>> parity across different disks in the array. Some are symmetric, some >>> are asymmetric, some rotate right, some rotate left, and some use >>> statistical algorithms to give an overall distribution without being >>> able to predict where a specific parity block might lie within a >>> stripe... >>> >>> And at the other end of the scale, isochronous RAID arrays tend to >>> have dedicated parity disks so that data read and write behaviour is >>> deterministic and therefore predictable from a high level.... >>> >>> So, assuming that a RAID5/6 device has a specific data layout (be it >>> distributed or fixed) at the filesystem level is just a bad idea. We >>> simply don't know. Even if we did, the only thing we can optimise is >>> the thing that is common between all RAID5/6 devices - writing full >>> stripe widths is the most optimal method of writing to them.... >> >> Am I interpreting this to say; >> >> 16 disks is sw=16 regardless of parity? > > No. I'm just saying that parity layout is irrelevant to the > filesystem and that all we care about is sw does not include parity > disks. So, here's the formula aurfalien, where #disks is the total number of active disks (excluding spares) in the RAID array. In the case of RAID5 sw = (#disks - 1) RAID6 sw = (#disks - 2) RAID10 sw = (#disks / 2) [1] [1] If using the Linux md/RAID10 driver with one of the non-standard layouts such as n2 or f2, the formula may change. This is beyond the scope of this discussion. Visit the linux-raid mailing list for further details. -- Stan _______________________________________________ xfs mailing list xfs@xxxxxxxxxxx http://oss.sgi.com/mailman/listinfo/xfs
