LSF Version 7.3 - Using Platform LSF HPC

SUMMARY: ( time unit: second )
Total number of done jobs: 0 Total number of exited jobs: 1
Total CPU time consumed: 0.4 Average CPU time consumed: 0.4
Maximum CPU time of a job: 0.4 Minimum CPU time of a job: 0.4
Total wait time in queues: 8.0
Average wait time in queue: 8.0
Maximum wait time in queue: 8.0 Minimum wait time in queue: 8.0
Average turnaround time: 43 (seconds/job)
Maximum turnaround time: 43 Minimum turnaround time: 43
Average hog factor of a job: 0.01 ( cpu time / turnaround time )
Maximum hog factor of a job: 0.01 Minimum hog factor of a job: 0.01
Example job submissions
The following script defines a job requiring 128 CPUs:
#!/bin/sh
#myscript
./sequential_pre_processor
prun -n 64 parallel1 &
prun -n 64 parallel2 &
wait
./sequential_prog
prun -n 128 parallel3
./write_results
Submit the job with the following command:
$ bsub -n 128 ./myscript
$ bsub -n 128 prun parallel_app
The following job assumes that prun is specified in the JOB_STARTER parameter of
the default queue definition in
lsb.queues.
$ bsub -n 128 parallel_app
prun with no arguments uses all CPUs allocated to it.
To enforce a multi-threaded application to run within a node, use
-extsched "nodes=1". For example:
$ bsub -n 3 -extsched "RMS[RMS_SLOAD;nodes=1]" prun mt_app
To enforce that a job only takes nodes where there is no other job running on, use
RMS[
ptile= cpus_per_node ]. For example, on an ES40 machine:
$ bsub -n 20 -extsched "RMS[RMS_SLOAD;ptile=4]" prun
parallel_app
For more information
◆ About MANDATORY_EXTSCHED and DEFAULT_EXTSCHED, see
“Configuring default and mandatory topology scheduling options” on page 112
◆ About job operations, see Administering Platform LSF