[jira] Commented: (HADOOP-3842) There is a window where the JobTracker is in the RUNNING state (i.e ready to accept jobs) and never executes them.

["Amar Kamat (JIRA)" <ji...@apache.org>]

    [ https://issues.apache.org/jira/browse/HADOOP-3842?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=12617373#action_12617373 ] 

Amar Kamat commented on HADOOP-3842:
------------------------------------

A simple approach would be to initialize the scheduler early in JobTracker's constructor before making the state {{RUNNING}}.

> There is a window where the JobTracker is in the RUNNING state (i.e ready to accept jobs) and never executes them.
> ------------------------------------------------------------------------------------------------------------------
>
>                 Key: HADOOP-3842
>                 URL: https://issues.apache.org/jira/browse/HADOOP-3842
>             Project: Hadoop Core
>          Issue Type: Bug
>          Components: mapred
>            Reporter: Amar Kamat
>            Priority: Blocker
>
> Prior to HADOOP-3412, job tracker could accept jobs without even offering service (i.e without {{JobTracker.offerService()}} being called). In such a case the job stays in JT's memory and job execution was guaranteed. With HADOOP-3412, {{JobTracker.submitJob()}} adds the job to JT's local structures and passes it to the scheduler. Scheduler gets initialized in {{JobTracker.offerService()}} and hence calling {{JobTracker.submitJob()}} before calling {{JobTracker.offerService()}} is actually a no-op. The job stays in JT's memory but never gets initialized. This is 
> - backward incompatible 
> - erroneous as there is a window where the jobtracker is ready to accept jobs, accepts them and never executes them.

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