并行化与MPI循环?

时间:2015-08-09 11:38:53

标签: parallel-processing fortran mpi fortran90

我想将以下程序转换为MPI程序:

program pi

implicit none

integer, parameter :: DARTS = 50000, ROUNDS = 10, MASTER = 0

double precision :: pi_est
double precision :: homepi, avepi, pirecv, pisum
integer :: rank
integer :: i, n
integer, allocatable :: seed(:)

! we set it to zero in the sequential run
rank = 0

! initialize the random number generator
! we make sure the seed is different for each task
call random_seed()
call random_seed(size = n)
allocate(seed(n))
seed = 12 + rank*11
call random_seed(put=seed(1:n))
deallocate(seed)

avepi = 0
do i = 0, ROUNDS-1
   pi_est = dboard(DARTS)
   ! calculate the average value of pi over all iterations
   avepi = ((avepi*i) + pi_est)/(i + 1)
end do

   print *, "Pi is ", avepi

contains

   double precision function dboard(darts)

      integer, intent(in) :: darts

      double precision :: x_coord, y_coord
      integer :: score, n

      score = 0
      do n = 1, darts
         call random_number(x_coord)
         call random_number(y_coord)

         if ((x_coord**2 + y_coord**2) <= 1.0d0) then
            score = score + 1
         end if
      end do
      dboard = 4.0d0*score/darts

   end function

end program

我认为我必须将do循环分成n个部分,其中n是处理器的数量,将结果保存在向量上,然后计算向量的平均值。我不确定这是否正确,也不知道如何实施这一改变。

这就是我现在所知:

模块mpi_params.f90

module mpi_params
   USE MPI
   implicit none
   integer                              :: ierr, numprocs, proc_num, &
                                           points_per_proc, istart, iend
   integer, allocatable, dimension(:)   :: displs, recvcounts 
   doubleprecision, allocatable, dimension(:)   :: proc_contrib
contains
subroutine init_mpi_params(nn)
integer, intent(in)                     :: nn
integer                                 :: i
! Determine how many points to handle with each proc
 if ( mod(nn,numprocs)==0 ) then
    points_per_proc = nn/numprocs
 else
    points_per_proc = (nn-mod(nn,numprocs))/numprocs
    if (numprocs-1 == proc_num ) points_per_proc = nn - points_per_proc*(numprocs-1)
 end if
! Determine start and end index for this proc's points
istart = proc_num * points_per_proc + 1
if (numprocs-1 == proc_num ) istart = proc_num*(nn-mod(nn,numprocs))/numprocs +1
iend = istart + points_per_proc - 1
if (numprocs-1 == proc_num ) iend = nn
ALLOCATE(proc_contrib(points_per_proc))
!print *, 'about to allocate displs' 
allocate(displs(numprocs),source=(/(i*(nn-mod(nn,numprocs))/numprocs,i=0,numprocs-1)/))
!print *, 'about to allocate recvcounts'
allocate(recvcounts(numprocs),source=(nn-mod(nn,numprocs))/numprocs)
recvcounts(numprocs)=nn - points_per_proc*(numprocs-1)
if (numprocs-1 == proc_num ) recvcounts(numprocs) = iend-istart+1
end subroutine init_mpi_params

end module mpi_params

和程序piMPI.f90

program pi
    use mpi_params
    implicit none

    integer, parameter              :: DARTS = 50000, ROUNDS = 10, MASTER = 0
    double precision                :: pi_est
    double precision                :: homepi, avepi, pirecv, pisum
    integer                         :: rank
    integer                         :: i, n
    integer, allocatable            :: seed(:)
    double precision                :: y(ROUNDS)  


    call mpi_init(ierr)
    call mpi_comm_size(MPI_COMM_WORLD, numprocs, ierr)
    call mpi_comm_rank(MPI_COMM_WORLD, proc_num, ierr)
    CALL init_mpi_params(ROUNDS)

    ! we set it to zero in the sequential run
    rank = 0

    ! initialize the random number generator
    ! we make sure the seed is different for each task
    call random_seed()
    call random_seed(size = n)
    allocate(seed(n))
    seed = 12 + rank*11
    call random_seed(put=seed(1:n))
    deallocate(seed)

    avepi = 0
    do i = istart, iend
       proc_contrib(i) = dboard(DARTS)
    end do

!!! MPI Reduce?
    call MPI_ALLGATHER(proc_contrib, points_per_proc, MPI_DOUBLE_PRECISION, &
                       y, points_per_proc, MPI_DOUBLE_PRECISION, &
                       MPI_COMM_WORLD, ierr)

     avepi = sum(y)/ROUNDS           
if (proc_num .eq. 0) then
    print *, "Pi is ", avepi
end if

    call mpi_finalize(ierr)

contains

   double precision function dboard(darts)

      integer, intent(in) :: darts

      double precision :: x_coord, y_coord
      integer :: score, n

      score = 0
      do n = 1, darts
         call random_number(x_coord)
         call random_number(y_coord)

         if ((x_coord**2 + y_coord**2) <= 1.0d0) then
            score = score + 1
         end if
      end do
      dboard = 4.0d0*score/darts

   end function

end program

我可以用以下代码编译此代码:

$ mpif90 mpi_params.f90 piMPI.f90

并使用1个或2个

处理器运行它
$ mpiexec -n 1 ./a.out  
Pi is    3.1369359999999999     
$ mpiexec -n 2 ./a.out 
Pi is    1.5679600000000000  

但结果似乎是错误的,n = 2。另外,如果我尝试使用3或更多版本运行它,我会收到以下错误:

$ mpiexec -n 3 ./a.out
Fatal error in PMPI_Allgather: Message truncated, error stack:
PMPI_Allgather(992)...............: MPI_Allgather(sbuf=0x213e9f0, scount=3, MPI_DOUBLE_PRECISION, rbuf=0x7ffc2638df80, rcount=3, MPI_DOUBLE_PRECISION, MPI_COMM_WORLD) failed
MPIR_Allgather_impl(838)..........: 
MPIR_Allgather(797)...............: 
MPIR_Allgather_intra(555).........: 
MPIDI_CH3U_Receive_data_found(131): Message from rank 2 and tag 7 truncated; 32 bytes received but buffer size is 24
Fatal error in PMPI_Allgather: Message truncated, error stack:
PMPI_Allgather(992)...............: MPI_Allgather(sbuf=0x24189f0, scount=3, MPI_DOUBLE_PRECISION, rbuf=0x7fff89575790, rcount=3, MPI_DOUBLE_PRECISION, MPI_COMM_WORLD) failed
MPIR_Allgather_impl(838)..........: 
MPIR_Allgather(797)...............: 
MPIR_Allgather_intra(532).........: 
MPIDI_CH3U_Receive_data_found(131): Message from rank 2 and tag 7 truncated; 32 bytes received but buffer size is 24

===================================================================================
=   BAD TERMINATION OF ONE OF YOUR APPLICATION PROCESSES
=   PID 5990 RUNNING AT UltraPro
=   EXIT CODE: 1
=   CLEANING UP REMAINING PROCESSES
=   YOU CAN IGNORE THE BELOW CLEANUP MESSAGES
===================================================================================

我做错了什么?

2 个答案:

答案 0 :(得分:2)

如果我已经理解了你的代码,并且我总是可以理解你的代码,那么它就是一个直接的蒙特卡罗计算pi的值,具有很好的特性,初学并行程序员,简单地计算更多(随机)数字将提高总估计的准确性。要进行M计算,您可以让一个流程计算所有流程,或者P流程计算它们M/P,然后取平均值以获得相同的准确度。在这种方法中,在程序结束时将本地值最终减少到全局值之前,不需要传递任何消息。

首先让每个进程计算它要运行的迭代次数,让每个进程通过使用程序参数和调用mpi例程找出num_procs等来自行解决这个问题。

我认为您的代码大纲应该是这样的:

program main
    ! all processes make same declarations, including variables to be used
    ! to calculate pi, and parameters

    call mpi_init(...)
    ...
    ! calculate pi independently on each process, no MPI calls necessary
    ! each process uses program parameters to calculate own contribution
    call mpi_reduce(local_pi, master_pi, 1, mpi_double_precision, mpi_sum, 0, &
               mpi_comm_world, ierr)
    if (proc_num==0) write(*,*) 'pi = ', master_pi/num_procs
    call mpi_finalize

那是关于它的。

答案 1 :(得分:0)

如果有人正在寻找可以编译的代码,这是我的工作解决方案:

program pi
    use mpi_params
    implicit none

    integer, parameter              :: DARTS = 500000, ROUNDS = 100, MASTER = 0
    double precision                :: pi_est
    double precision                :: homepi, avepi, pirecv, pisum
    integer                         :: rank
    integer                         :: i, n
    integer, allocatable            :: seed(:)
    double precision                :: y 
    double precision                :: sumpi


    call mpi_init(ierr)
    call mpi_comm_size(MPI_COMM_WORLD, numprocs, ierr)
    call mpi_comm_rank(MPI_COMM_WORLD, proc_num, ierr)
    CALL init_mpi_params(ROUNDS)

    ! we set it to zero in the sequential run
    rank = 0

    ! initialize the random number generator
    ! we make sure the seed is different for each task
    call random_seed()
    call random_seed(size = n)
    allocate(seed(n))
    seed = 12 + rank*11
    call random_seed(put=seed(1:n))
    deallocate(seed)

    y=0.0d0
    do i = istart, iend
       y = y + dboard(DARTS)
    end do

    call mpi_reduce(y, sumpi, 1, mpi_double_precision, mpi_sum, 0, &
                    mpi_comm_world, ierr)


if (proc_num==0) write(*,*) 'pi = ', sumpi/ROUNDS

    call mpi_finalize(ierr)

contains

   double precision function dboard(darts)

      integer, intent(in) :: darts

      double precision :: x_coord, y_coord
      integer :: score, n

      score = 0
      do n = 1, darts
         call random_number(x_coord)
         call random_number(y_coord)

         if ((x_coord**2 + y_coord**2) <= 1.0d0) then
            score = score + 1
         end if
      end do
      dboard = 4.0d0*score/darts

   end function

end program

和额外的模块

module mpi_params
   USE MPI
   implicit none
   integer                              :: ierr, numprocs, proc_num, &
                                           points_per_proc, istart, iend
   doubleprecision, allocatable, dimension(:)   :: proc_contrib
contains
subroutine init_mpi_params(nn)
integer, intent(in)                     :: nn
integer                                 :: i
! Determine how many points to handle with each proc
 if ( mod(nn,numprocs)==0 ) then
    points_per_proc = nn/numprocs
 else
    points_per_proc = (nn-mod(nn,numprocs))/numprocs
    if (numprocs-1 == proc_num ) points_per_proc = nn - points_per_proc*(numprocs-1)
 end if
! Determine start and end index for this proc's points
istart = proc_num * points_per_proc + 1
if (numprocs-1 == proc_num ) istart = proc_num*(nn-mod(nn,numprocs))/numprocs +1
iend = istart + points_per_proc - 1
if (numprocs-1 == proc_num ) iend = nn
ALLOCATE(proc_contrib(points_per_proc))
end subroutine init_mpi_params

end module mpi_params

此代码可以使用

进行编译
mpif90 mpi_params.f90 piMPI.f90

并以

运行
time mpiexec -n 10 ./a.out

比@HighPerformanceMark提出的解决方案更复杂,因为我想保留拆分do循环的想法(对我正在处理的其他代码很有用)