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Posted to issues@bookkeeper.apache.org by GitBox <gi...@apache.org> on 2021/02/26 12:55:21 UTC
[GitHub] [bookkeeper] yangl opened a new pull request #2630: [doc] fix the no striping occurs condition to Qw=E
yangl opened a new pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630
Descriptions of the changes in this PR:
### Motivation
the no striping occurs condition should Qw=E, not the Qw=Qa
### Changes
update the condition to Qw=E
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[GitHub] [bookkeeper] michaeljmarshall commented on a change in pull request #2630: [doc] fix the no striping occurs condition to Qw=E
Posted by GitBox <gi...@apache.org>.
michaeljmarshall commented on a change in pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630#discussion_r584041212
##########
File path: site/docs/4.10.0/development/protocol.md
##########
@@ -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:
It seems to me that these two sentences should be combined to indicate that it is an either/or (exclusive) relationship between the first and the second. For example:
> There are only **E** distinct write quorums in any ensemble, unless **Q<sub>w</sub>** = **E**, then there is only one, as no striping occurs.
I'd be interested to know what one of the committers says though.
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[GitHub] [bookkeeper] michaeljmarshall commented on a change in pull request #2630: [doc] fix the no striping occurs condition to Qw=E
Posted by GitBox <gi...@apache.org>.
michaeljmarshall commented on a change in pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630#discussion_r585249009
##########
File path: site/docs/4.10.0/development/protocol.md
##########
@@ -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:
@dlg99 - thanks for your comment. I think what I wrote is consistent with your comment, please let me know if you think otherwise.
> There is no striping in case of (E, Qw, Qa) == (3, ,3 ,3) or (3, 3, 2).
In both cases above, E = Qw, and there is only 1 distinct write quorum.
> There is striping in case of (5, 3, 3)
In this case, E > Qw, and there are E distinct write quorums.
My only point is that I think the documentation would be clearer as a single sentence (with an obvious either/or relationship) instead of two.
No worries if we leave it.
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[GitHub] [bookkeeper] dlg99 merged pull request #2630: [doc] fix the no striping occurs condition to Qw=E
Posted by GitBox <gi...@apache.org>.
dlg99 merged pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630
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[GitHub] [bookkeeper] dlg99 commented on a change in pull request #2630: [doc] fix the no striping occurs condition to Qw=E
Posted by GitBox <gi...@apache.org>.
dlg99 commented on a change in pull request #2630:
URL: https://github.com/apache/bookkeeper/pull/2630#discussion_r585136102
##########
File path: site/docs/4.10.0/development/protocol.md
##########
@@ -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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