Does it stay in the queue indefinitely? Can you post an example of one queue item that isn't being processed.

Is the SLA agent turned on, on all nodes?

Yes, I am guessing it remains indefinitely. It is a Dev environment and if I want to clear entire SLA queue, what is the standard way? Restarting system with clearing cache won't remove SLA queue, will it?

We have two app nodes. How can I check both SLA agent is turned on?

Thanks,

Could you please provide the logs if you can trace the sla through SMA.

Hi

This link may help you

https://community.pega.com/video-library/support-guide-how-troubleshoot-sla-broken-queue-entries

Thanks

Hi,

I think your email server is not responding within defined time due that send email is getting failed.

Please try to end why email server is not responding, it may have an authentication issue.  

Hi,

Please find below information to begin the investigation.

SMA - Agent Administration - Queue Management. See if you can find the any queue entry for SLA agent task in Broker-Process queue.

Click on individual item in Broker queue and click on detail to get the detail of failure. Try to fix the cause.

If there is no entry in Broken process queue then find entries in Scheduled task. This is were all up coming execution will be stored. If you see there huge pile up then probably agent is unable to process them due to any issue in agent or agent is not able to pick them due to huge load.

Check how many nodes agent is running and any of them are facing any issue? You can check this in Rule log.

-Rajesh

Hi

Below link's might be helpful for you to debug the issue.

https://community.pega.com/video-library/support-guide-how-troubleshoot-sla-broken-queue-entries

https://community.pega.com/video-library/support-guide-how-trace-batch-requestors-through-system-management-application-sma

Thanks. Another question - if we have 100 scheduled SLA tasks, are 100 tasks processed in parallel by two nodes or sequential? We want faster processing like 50 by app #1 and the rest of 50 by app #2.