Stacking simple jobs with batch on Solaris 10 x86

Hey Guys,
I have a user that needs to load up a bunch of very low priority (nice 19) jobs on a 4 CPU machine so that once one completes, another one starts. I thought batch would be a good use for this. We don't have any mpi worries or anything, these are very monolithic standalone jobs.
I looked at the batch man page and it seems to imply that I can create a NEW queue in /etc/cron.d/queuedefs to accomplish this. However, I don't see how to format it such that many jobs can be "loaded" in a new queue.
Here is the default:
bash-3.00# cat /etc/cron.d/queuedefs
a.4j1n
b.2j2n90w
bash-3.00#Should I modify the b (batch) line to say:
b50j19n90wBut it seems like the above would simply allow 50 jobs to be started at a nice level of 19 - is that correct?? What I don't see is how to specify that only 4 jobs should be run at one time (4 CPUs) but please accept ~50 jobs into a queue and load them as sequentially as jobs finish.
Should a make a new queue to be listed after the "b" queue in queuedefs?
Can anyone comment on this? Is this out of scope for batch and queuedefs ?
Thanks,
Phil

One way to do this would be by using the shell construct of &&.
command_one && command_two && etc...
&& will tell command two to wait for command one to finish with an exit status of zero before continuing on. If command one fails for any reason then the rest of the commands would not finish.
So to get around that you could use:
command one
sleep <time>
command two
sleep <time>
etc...
or:
man wait
might be the best option.
alan

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Hi [email protected],
Q1) where to put Flash Recovery Area (FRA)?
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Michael Girdley
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has anyone been able to get Oracle's client 8.1.5 running on Solaris x86in conjunction with WLS to talk to back-end Oracle? Any help would be
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Stacking simple jobs with batch on Solaris 10 x86

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