On 19/05/2015 11:11, Miyamae, Takeshi wrote: > Hi Loic, > >> to recover D6 which is in rack 2 it may be necessary to use P2 from rack 1 > > Because D6 can be recovered from P3 as well, P2 is not necessarily used to recover D6. > However, when D6 is recovered from P3, D5 which is in rack 1 must be read. > Therefore, I believe Ceph-LRC is more efficient when network resources is poor such as > in case of Geo-Replication. SHEC should be used in a single data center that has enough > network resources. Thanks for the detailed explanation, it is clear now :-) Cheers > > Best regards, > Takeshi Miyamae > > -----Original Message----- > From: Loic Dachary [mailto:loic@xxxxxxxxxxx] > Sent: Tuesday, May 19, 2015 3:39 PM > To: Miyamae, Takeshi/宮前 剛 > Cc: Ceph Development > Subject: What crush ruleset for a given SHEC configuration ? > > Hi Takeshi, > > In the context of http://ceph.com/docs/master/rados/operations/erasure-code-shec/ it would be useful to have a more detailed explanation of why SHEC is more efficient during recovery (in the introduction). > > Am I correct to assume that SHEC does not provide a way to control the locality of the chunks ? For instance in the following scenario: > > rack 1 has 10 OSDs > rack 2 has 10 OSDs > > a crush ruleset is made to provide 15 OSDs with 7 in the first rack, 8 in the last rack: the first 7 are in rack 1, the last 8 in rack 2. When SHEC is used with such a crush ruleset, it cannot guarantee that the loss of one chunk in rack 2 can always be recovered with chunks from rack 2. When reading at figure 3 of > > https://wiki.ceph.com/Planning/Blueprints/Hammer/Shingled_Erasure_Code_%28SHEC%29 > > with D1 to D5, P1 and P2 in rack 1 and D6 to D10, P3, P4, P5 in rack 2, my understanding is that to recover D6 which is in rack 2 it may be necessary to use P2 from rack 1. And to recover D5 which is in rack 1 it may be necessary to use P3 from rack 2. > > Maybe I'm missing something ? Thanks in advance for your explanations :-) > > Cheers > -- Loïc Dachary, Artisan Logiciel Libre
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