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Posted to issues@spark.apache.org by "SuYan (JIRA)" <ji...@apache.org> on 2014/10/31 10:26:33 UTC

[jira] [Closed] (SPARK-4167) Schedule task on Executor will be Imbalance while task run less than local-wait time

     [ https://issues.apache.org/jira/browse/SPARK-4167?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel ]

SuYan closed SPARK-4167.
------------------------
    Resolution: Not a Problem

> Schedule task on Executor will be Imbalance while task run less than local-wait time
> ------------------------------------------------------------------------------------
>
>                 Key: SPARK-4167
>                 URL: https://issues.apache.org/jira/browse/SPARK-4167
>             Project: Spark
>          Issue Type: Improvement
>          Components: Spark Core
>    Affects Versions: 1.1.0
>            Reporter: SuYan
>
> Recently, when run a spark on yarn job. it occurs executor schedules imbalance. 
> the procedure is that: 
> 1. because user's mistake, the spark on yarn job's input split contains 0 byte empty splits. 
> 1.1:
>  task0-99 , no-preference task(0 byte) 
> task100-800, node-local task 
> 1.2: user will run task 500 loops
> 1.3: 60 executor
>  2.
>  executor A only have 2 node-local task in the first loop, executor A first finished node-local-task, the it will run no-preference task, and the no-preference task in our situation have smaller input split than node-local task. So executor A finished all no-reference task, while others still run node-local job. 
> in the second loop, all task have process-local level, and all task finished in 3 seconds, so while executor A is still run process-local task while others are all finished process-local task. but all process-task run by executor A will finished in 3 seconds, so the local level will always be process-local. 
> it results other executors are all wait for executor A the same situation in the left loops. 
> To solve this situation, we let user to delete the empty input split.
>  but is still have implied imbalance, while in some loops, a executor got more process-local task than others in one loop, and this task all less-3 seconds task. and then in the left loops, the others executor will wait that executor to finished all process-local tasks.



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