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From: Frank Gruellich (frank.gruellich_at_[hidden])
Date: 2006-07-20 03:03:18


Hi,

George Bosilca wrote:
> On the all-to-all collective the send and receive buffers has to be able
> to contain all the information you try to send. On this particular case,
> as you initialize the envio variable to a double I suppose it is defined
> as a double. If it's the case then the error is that the send operation
> will send more data than the amount available on the envio variable.
>
> If you want to be able to do correctly the all-to-all in your example,
> make sure the envio variable has a size at least equal to:
> tam * sizeof(byte) * NPROCS, where NPROCS is the number of procs available
> on the mpi_comm_world communicator.

I'm unfortunately not that Fortran guy. Maybe the best would be to
submit the whole function at the beginning, it's neither secret nor big:

module alltoall
  use globales
  implicit none

contains
subroutine All_to_all

  integer,parameter :: npuntos = 24
  integer,parameter :: t_max = 2**(npuntos-1)
  integer siguiente,anterior,tam,rep,p_1,p_2,i,j,ndatos,rep2,o,k
  double precision time1,time2,time,ov,tmin,tavg
  double precision,dimension(t_max)::envio
  double precision,dimension(:),allocatable::recibe
  double precision,dimension(npuntos)::m_tmin,m_tavg
  double precision,dimension(npuntos)::tams

  rep2 = 10
  tag1 = 1
  tag2 = 2
  rep = 3

  allocate(recibe(t_max*nproc))
  siguiente = my_id + 1
  if (my_id == nproc -1) siguiente = 0
  anterior = my_id - 1
  if (my_id == 0) anterior = nproc- 1

  do i = 1,npuntos
    print *,'puntos',i
    tam = 2**(i-1)
    tmin = 1e5
    tavg = 0.0d0
    do j = 1,rep
      envio = 8.0d0*j
      call mpi_barrier(mpi_comm_world,ierr)
      time1 = mpi_wtime()
      do k = 1,rep2
        call mpi_alltoall(envio,tam,mpi_byte,recibe,tam,mpi_byte,mpi_comm_world,ierr)
      end do
      call mpi_barrier(mpi_comm_world,ierr)
      time2 = mpi_wtime()
      time = (time2 - time1)/(rep2)
      if (time < tmin) tmin = time
      tavg = tavg + time
    end do
    m_tmin(i) = tmin
    m_tavg(i) = tavg/rep
  end do
  call mpi_barrier(mpi_comm_world,ierr)
  print *,"acaba"

  if (my_id == 0) then
    open (1,file='Alltoall.dat')
    write (1,*) "#Prueba All to all entre todos los procesadores(",nproc,")"
    write (1,*) "#Precision del reloj:",mpi_wtick()*1.0d6,"(muS)"
    do i =1,npuntos
      write(1,900) 2*nproc*2**(i-1),m_tmin(i),m_tavg(i)!,ov
    end do
    close(1)
  end if
  900 FORMAT(I10,F14.8,F14.8)
  800 FORMAT(I10,F14.8,F14.8)
end subroutine
end module

Can you read this? (Sorry, I can't.) But the size_of envio seems to be
2**32 = 8388608 doubles, isn't it? I don't understand, why it should
depend on the number of MPI nodes, as you said.

Thanks for your help. Kind regards,

-- 
Frank Gruellich
HPC-Techniker
Tel.:   +49 3722 528 42
Fax:    +49 3722 528 15
E-Mail: frank.gruellich_at_[hidden]
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