On Jun 30, 2013, at 7:47 PM, Stan Hoeppner wrote: > 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 I totally got your original post with the cup o M&Ms. Just wanted his take on it is all. And I'm on too many mailing lists as it is :) - aurf _______________________________________________ xfs mailing list xfs@xxxxxxxxxxx http://oss.sgi.com/mailman/listinfo/xfs
