[GitHub] [bookkeeper] dlg99 commented on a change in pull request #2630: [doc] fix the no striping occurs condition to Qw=E

[GitBox <gi...@apache.org>]

dlg99 commented on a change in pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630#discussion_r585136102



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File path: site/docs/4.10.0/development/protocol.md
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@@ -64,7 +64,7 @@ Entry | Write quorum
 4 | B1, B2, B3
 5 | B2, B3, B4
 
-There are only **E** distinct write quorums in any ensemble. If **Q<sub>w</sub>** = **Q<sub>a</sub>**, then there is only one, as no striping occurs.
+There are only **E** distinct write quorums in any ensemble. If **Q<sub>w</sub>** = **E**, then there is only one, as no striping occurs.

Review comment:
       @michaeljmarshall To the best of my understanding:
   E >= Qw >= Qa. 
   There is no striping in case of (E, Qw, Qa) ==  (3, ,3 ,3) or (3, 3, 2). 
   There is striping in case of (5, 3, 3)
   
   if Qw > Qa the client will ack the write when Qa writes completed while (Qw-Qa) possibly failed or still in progress; it does not affect striping.
   
   If E > Qw, the write quorums will be spread across larger set of bookies (see RoundRobinDistributionSchedule)
   I.e. for E = 3 and Qw = 2, entries will go to
   e0: (b0, b1)
   e1: (b1, b2)
   e2: (b2, b0)
   e3: (b0, b1)
   etc.
   
   so the unique stripes in this case are (b0, b1), (b1, b2), (b2, b0)
   
   In case of Qw = 3 all writes go to (b0, b1, b2).
   IIRC, reads will still "rotate" between bookies to distribute reads more evenly, like e0 will be read from b0 (fallback to b1, b2), e1 from b1 (fallback to b2, b0) etc 
   
   Hope this helps




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