> Von: linux-raid-owner@xxxxxxxxxxxxxxx [linux-raid-owner@xxxxxxxxxxxxxxx]" im Auftrag von "Roman Mamedov [rm@xxxxxxxxxxx] > Gesendet: Dienstag, 24. Februar 2015 00:58 > An: linux-raid@xxxxxxxxxxxxxxx > Betreff: RAID6 write I/O amplification? > > Hello, > > Got a bit of a "how does it actually work" question... > > Suppose I have an MD RAID6 of 8 drives, with 64KB chunk size. > > I am rewriting a 4KB filesystem sector somewhere on that RAID (not crossing > the stripe boundary). > > What's the amount of disk I/O in total this will result in? > > I assume the RAID will need to read data from all drives, recompute parity, > then write to the data stripe where the updated piece happened to be, and also > write to two parity stripes. > > Is this done at a stripe granularity, so 6x64KB reads, 3x64KB writes? > Or down to individual sectors (pages), i.e. 6x4KB reads, 3x4KB writes? > Or am I describing this algorithm correctly at all? Implementation will work on "internal" stripe granularity and that is 4K So your case will be 6x4KB read + 3x4KB write. That said, you can only reduce the I/O overhead by writing data that is larger than your configured stripe size (e.g. 64K). Looking at Neils development GIT you will find patches that allow read-modify-write cycles for RAID6. So we only need the old block, the old parities, recaluclate them and write the new block and the new parities. In your case that would reduce the I/Os to 3x4KB read + 3x4KB write. See http://git.neil.brown.name/?p=md.git;a=shortlog;h=refs/heads/devel I posted them 6 months ago but they did not made their way into the stable tree. Additionally it conatins patches to batch adjacent writes to be processed in less & larger I/Os. Currently Linux Raid will break each operation into 4K I/Os. Markus
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