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https://issues.apache.org/jira/browse/SOLR-8744?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=15297710#comment-15297710
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Noble Paul commented on SOLR-8744:
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bq. In OverseerTaskProcessor, handing off the lock object (up to and including 
the tpe.execute call) should probably move inside the try block; on 
successfully executing the task, null out the lock local, and then put an "if 
lock != null unlock" into a finally block.

The runner is always executed in a different thread. So the finally block will 
always evaluate {{lock != null}} to true. Actually, only in case of an 
exception, OverseerTaskProcessor should unlock it

bq.Do we need some kind of complete reset in the event stuff really blows up? 
Should there be a session-level clear that just unlocks everything?

Yes, we need it, but not at the session level. If I clear up everything for 
each session, it defeats the purpose. The session is valid for only one batch. 
The locks should survive for multiple batches.

A better solution is to periodically check if the running tasks are empty and 
if yes, just create a new {{LockTree}}. I was planning t do it anyway

bq. I think MigrateStateFormat could be collection level rather than cluster 
level?

I have set the lock levels more pessimistically. We should re-evaluate them

> Overseer operations need more fine grained mutual exclusion
> -----------------------------------------------------------
>
>                 Key: SOLR-8744
>                 URL: https://issues.apache.org/jira/browse/SOLR-8744
>             Project: Solr
>          Issue Type: Improvement
>          Components: SolrCloud
>    Affects Versions: 5.4.1
>            Reporter: Scott Blum
>            Assignee: Noble Paul
>              Labels: sharding, solrcloud
>         Attachments: SOLR-8744.patch
>
>
> SplitShard creates a mutex over the whole collection, but, in practice, this 
> is a big scaling problem.  Multiple split shard operations could happen at 
> the time time, as long as different shards are being split.  In practice, 
> those shards often reside on different machines, so there's no I/O bottleneck 
> in those cases, just the mutex in Overseer forcing the operations to be done 
> serially.
> Given that a single split can take many minutes on a large collection, this 
> is a bottleneck at scale.
> Here is the proposed new design
> There are various Collection operations performed at Overseer. They may need 
> exclusive access at various levels. Each operation must define the Access 
> level at which the access is required. Access level is an enum. 
> CLUSTER(0)
> COLLECTION(1)
> SHARD(2)
> REPLICA(3)
> The Overseer node maintains a tree of these locks. The lock tree would look 
> as follows. The tree can be created lazily as and when tasks come up.
> {code}
> Legend: 
> C1, C2 -> Collections
> S1, S2 -> Shards 
> R1,R2,R3,R4 -> Replicas
>                  Cluster
>                 /       \
>                /         \         
>               C1          C2
>              / \         /   \     
>             /   \       /     \      
>            S1   S2      S1     S2
>         R1, R2  R3.R4  R1,R2   R3,R4
> {code}
> When the overseer receives a message, it tries to acquire the appropriate 
> lock from the tree. For example, if an operation needs a lock at a Collection 
> level and it needs to operate on Collection C1, the node C1 and all child 
> nodes of C1 must be free. 
> h2.Lock acquiring logic
> Each operation would start from the root of the tree (Level 0 -> Cluster) and 
> start moving down depending upon the operation. After it reaches the right 
> node, it checks if all the children are free from a lock.  If it fails to 
> acquire a lock, it remains in the work queue. A scheduler thread waits for 
> notification from the current set of tasks . Every task would do a 
> {{notify()}} on the monitor of  the scheduler thread. The thread would start 
> from the head of the queue and check all tasks to see if that task is able to 
> acquire the right lock. If yes, it is executed, if not, the task is left in 
> the work queue.  
> When a new task arrives in the work queue, the schedulerthread wakes and just 
> try to schedule that task.



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