On 16 June 2016 at 00:46, Gilles Gouaillardet <gil...@rist.or.jp> wrote:
>
> Here is the idea on how to get the number of tasks per node
>
>
> MPI_Comm intranode_comm;
>
> int tasks_per_local_node;
>
> MPI_Comm_split_type(MPI_COMM_WORLD, MPI_COMM_TYPE_SHARED, 0, MPI_INFO_NULL,
> &intranode_comm);
>
> MPI_Comm_size(intranode_comm, &tasks_per_local_node)
>
> MPI_Comm_free(&intranode_comm);
>
>
>
> then you can get the available memory per node, for example
> grep ^MemAvailable: /proc/meminfo
> and then divide this by the number of tasks on the local node.
>
>
>
> now if distribution should be based on cpu speed, you can simply retrieve
> this value on each task, and then MPI_Gather() it to rank 0, and do the
> distribution.
>
>
> in any case, if you MPI_Gather() the task parameter you are interested in,
> you should be able to get rid of a static config file.
>
> non blocking collective are also available
> MPI_Igather[v] / MPI_Iscatter[v]
> if your algorithm can exploit this, that might be helpful
>
I now programmatically collect this info:
rank/core: cpufrequency maxmemavailable
=============================================
0 f0
m0
1 f1
m1
2 f2
m2
3 f3
m3
4 f4
m4
Let's say 1 task needs M bytes.
The maximum number of tasks I can do would be: sum(mi) / M
Now let's say I am given N tasks to do.
1. A solution that ignores the memory available is to spread N over
the 5 cores given the highest frequency cores the highest number of
tasks in a prorata way, ie f_i/ sum(f_i) => n_i for each core.
sum(n_i) = N
2. A 2nd stage could then be to ensure that no: n_i > m_i/M
which would then involve taking any excesses (n_i - m_i/M) and
spreading it over the cores.
Or
perhaps both cpufrequencies and maxmem could be considered in 1 go,
but I don't know how to do that?
Thanks
MM
