Thank you for your reply Thomas,
Sorry, but I don't see how -m / --distribution could help in this case.
The problem is not how the tasks are distributed among all allocated nodes
but how to restrict the task distribution to an *arbitrary subset of
allocated nodes*.
I agree that manual solutions
I think I misunderstood what you were looking for.
You probably want to investigate the -m / --distribution option.
Default is "block", which means with a loop like
for i in {1..30}; do
srun -n1 job &
done
inside a job which was allocated nodes torus1, torus2, torus3,
the scheduler is
Thank you for your suggestion Thomas,
I have just tested it but no luck.
There is a shift in the node list to be used but only one of them is
accessed.
For example, with an allocation of three nodes ((torus6001, torus6002 and
torus6003), this code will confine the 30 tasks in torus6002 instead of
I've not used it, but when looking at srun man page (for something else)
notices a -r / --relative option which sounds like it might be what you
are looking for.
On Thu, 8 Dec 2016, Nigella Sanders wrote:
Hi Michael,
Actually, that was my first try but it didn't work. ?
srun finds it inconsis
Hi Michael,
Actually, that was my first try but it didn't work. 😑
srun finds it inconsistent with "-N -n1" and ends up using only the first
node provided in the -w list.
$ srun: Warning: can't run 1 processes on 2 nodes, setting nnodes to 1
Regards,
Nigella
2016-12-08 13:04 GMT+00:00 Michael
On Thu, Dec 8, 2016 at 5:48 AM, Nigella Sanders
wrote:
>
> All 30 tasks run always in the first two allocated nodes (torus6001 and
> torus6002).
>
> However, I would like to get these tasks using only the second and then
> third nodes (torus6002 and torus6003).
> Does anyone an idea about how to
