MPI

1
MPI

• Message Passing Interface
• Beowulf and O2K
• Today we will discuss two simple MPI programs
• Important MPI makes fewer assumptions about the
  computing environment than pthreads and Open MP.

2
SPMD

• Single Program Multiple Data
• The same program is run on different processors
• Each processor has an identifier (rank)
• The rank of the processor is used to produce
  different results on the distinct processors.

3
Reference

• Parallel Programming with MPI
• By Peter S. Pacheco
• The website is http//fawlty.cs.usfca.edu/mpi.
• To get the user's guide, click on the "A
  User'sGuide to MPI" link in the first paragraph.
  Ghostscript and Adobe read it.

4
Compiling and Running MPI

• O2K
• cc ltcompile optionsgt filename(s).c lmpi
• mpirun np ofprocs a.out

5
Greetings

• include ltstdio.hgt
• include ltstring.hgt
• include "mpi.h"
•  
• main(int argc, char argv)
• int my_rank / rank
  of process /
• int p /
  number of processes /
• int source / rank
  of sender /
• int dest /
  rank of receiver /
• int tag 0 / tag
  for messages /
• char message100 / storage for
  message /
• MPI_Status status / return
  status for receive /

6
Greetings Continued

• / Start up MPI / MPI_Init(argc,
  argv)  / Find out process rank /
  MPI_Comm_rank(MPI_COMM_WORLD,
  my_rank)  / Find out number of
  processes / MPI_Comm_size(MPI_COMM_WORLD,
  p)

7
Greetings Continued

• if (my_rank ! 0)
• / Create message /
• sprintf(message, "Greetings from process
  d!",
• my_rank)
• dest 0
• / Use strlen1 so that '\0' gets
  transmitted /
• MPI_Send(message, strlen(message)1,
  MPI_CHAR,
• dest, tag, MPI_COMM_WORLD)
• else / my_rank 0 /
• for (source 1 source lt p source)
• MPI_Recv(message, 100, MPI_CHAR,
  source, tag,
• MPI_COMM_WORLD, status)
• printf("s\n", message)

8
Wrap up

• / Shut down MPI /
• MPI_Finalize()
• / main /

9
Using Matlab program for DD

• y f(t,y, y) y(a) c y(b) d
• h (b-a)/2n e guess1 f guess2
• Solve a, r (n1)h y(r) e ? Y1(0,,n1)
• Solve L (n-1)h, b y(L) f ? Y2(0,,n1)
• If (Y2(2) e Y1(n-1) f ) lt eps Exit
• Else e ? Y2(2) f ? Y1(n-1)
• Go to 1.

10
Trapezoid Rule

• Compute the integral of f using the trapezoid
  rule
• Single processor case
• MP case
• Communication

11
MPI Send

• int MPI_Send( void buffer,
• int count,
• MPI_Datatype dt,
• int
  destination,
• int tag,
• MPI_Comm communicator)

12
MPI Receive

• int MPI_Recv( void buffer,
• int count,
• MPI_Datatype dt,
• int source,
• int tag,
• MPI_Comm communicator
• MPI_Status status)

13
MPI Broadcast

• int MPI_Bcast( void message,
• int count,
• MPI_Datatype dt,
• int root,
• MPI_Comm communicator)

14
MPI Reduce

• int MPI_Reduce( void operand,
• void result,
• int count,
• MPI_Datatype dt,
• MPI_Op operator,
• int root,
• MPI_Comm communicator)

15
MPI_Op

• MPI_MAX MPI_LXOR
• MPI_MIN MPI_BXOR
• MPI_SUM MPI_MAXLOC
• MPI_PROD MPI_MINLOC
• MPI_LAND
• MPI_LOR
• MPI_BOR

16
MPI Matrix Multiply Design

• Where are the matrices at the beginning of the
  computation?
• Where will the result be at the end of the
  computation?
• What will each processor compute?

17
Example Lab 4

• Design and implement a Domain Decomposition
  program using matlab.
• Assume that there are 4 intervals.
• T, Y dd_ode(), where T and Y are the global
  grid and global solution respectively.
• dd_ode_4(a,b,c,f, left, right, bleft, bright,
  npts)
• Global mesh T left h right where h
  (right left)/(4npts)
• Thus npts is (nearly) the number of points per
  interval

18
Lab 4 (2)

• The 4 intervals become
• T(1), T(npts1) and
• T((k-1)npts), T( knpts1) for k 2, 3, 4
• Take a guess at the missing boundary values, use
  Solve_ode to solve the 4 bvps and then exchange
  information and iterate.
• Extra credit Write the routine for any number
  of intervals passing this number in the argument
  list.

19
Lab 4 (3)

• Comment on the efficiency of the program
• Compare the run time of the dd program to the
  runtime for solving the global problem using
  Solve_ode with n 4npts.
• Solve y y y cos(t3) y(0) 0, y(3) 0.
  For n 4500 2000.

20
Solving Tridiagonal Systems

• ai xi-1 di xi ci xi1 bi for i 1, ,n
• x0 and xn1 are both 0
• Use Gauss Elimination

21

• subroutine trid(sub, diag, sup, b, n, ans)
• integer n, i
• real(kind8), dimension(n) diag, sub, sup,
  ans, b
• if (n .le. 1) then
• ans(1) b(1)/diag(1)
• return
• end if
• do i 1, n
• ans(i) b(i)
• end do
• do i 2,n
• sub(i) sub(i)/diag(i-1)
• diag(i) diag(i) - sub(i) sup(i-1)
• ans(i) ans(i) - sub(i)ans(i-1)
• end do
• ans(n) ans(n)/diag(n)
• do i n-1, 1, -1
• ans(i) (ans(i) - sup(i)ans(i1))/diag(i)

22
integer n, i real(kind8),
dimension(), allocatable a, b, c, ans, d n
10 allocate( a(n), d(n), c(n), b(n), ans(n)) do
i1,n a(i) -1.d0 d(i) 2.d0 c(i)
-1.d0 b(i) 0.d0 enddo b(1) 1.d0 call
trid(a, d, c, b, n, ans) do i 1, n print,
i, ans(i) - (n1-i)/(n1.d0) end
do deallocate(a,d,c,b,ans) stop end