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pardiso problem

littlewang
Beginner
397 Views

_CP8IUSO94HVLC5KIAIA(HC.png

littlewang_0-1624439432184.png

 

hello

As you can see in the picture, there is an error when I try to use pardiso to solve eqs. I don't know why it is. What should I do ?

SUBROUTINE SOL_PARDISO(n,lent,a,ia,ja,b,mtype)
C
C PARDISO : PARALLEL SPARSE DIRECT LINEAR SOLVER
C Based on '/sw/sdev/cmkl/10.0.2.18/examples/solver/source/
C pardiso_sym_f.f'
C
use omp_lib
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
INTEGER*8 PT(64)
INTEGER, PARAMETER :: dp = KIND(1.0D0)
C.. All other variables
INTEGER MAXFCT,MNUM,MTYPE,PHASE,N,NRHS,ERROR,error1,MSGLVL,lent
INTEGER iparm(64),idum(1)
REAL(KIND=DP) ddum(1)
dimension a(lent),ia(n+1),ja(lent),b(n),x(n)
DATA nrhs /1/, maxfct /1/, mnum /1/
C.. Set up PARDISO control parameter
do i = 1, 64
iparm(i) = 0
end do
iparm(1) = 3 ! no solver default
iparm(2) = 2 ! fill-in reordering from METIS
iparm(3) = 1 ! numbers of processors
iparm(4) = 0 ! no iterative-direct algorithm
iparm(5) = 0 ! no user fill-in reducing permutation
iparm(6) = 0 ! =0 solution on the first n compoments of x
iparm(7) = 0 ! not in use
iparm(8) = 9 ! numbers of iterative refinement steps
iparm(9) = 0 ! not in use
iparm(10) = 13 ! perturbe the pivot elements with 1E-13
iparm(11) = 1 ! use nonsymmetric permutation and scaling MPS
iparm(12) = 0 ! not in use
! if(mtype .eq. -2) then !Improved accuracy using (non-) symmetric weighted matching
iparm(13) = 0 ! Disable matching. Default for symmetric indefinite matrices
if(mtype .eq. 11) then
iparm(13) = 1 ! Enable matching. Default for nonsymmetric matrices.
endif
iparm(14) = 0 ! Output: number of perturbed pivots
iparm(15) = 0 ! not in use
iparm(16) = 0 ! not in use
iparm(17) = 0 ! not in use
iparm(18) = -1 ! Output: number of nonzeros in the factor LU
iparm(19) = -1 ! Output: Mflops for LU factorization
iparm(20) = 0 ! Output: Numbers of CG Iterations
iparm(24) =1
iparm(27) = 1 !turn on matrix checker on your matrix
error = 0 ! initialize error flag
msglvl = 0 !0/1 ! print statistical information
c mtype = -2 ! symmetric, indefinite
!! mtype = 11 ! real and nonsymmetric
C.. Initiliaze the internal solver memory pointer. This is only
C necessary for the FIRST call of the PARDISO solver.
do i = 1, 64
pt(i) = 0
end do


call mkl_set_dynamic(0)
 call mkl_set_num_threads(4)
call omp_set_num_threads(4)

c phase = -1
phase = 11 ! only reordering and symbolic factorization
CALL pardiso(pt,maxfct,mnum,mtype,phase,n,a,ia,ja,
1 idum,nrhs,iparm,msglvl,ddum,ddum,error)
! WRITE(*,*) 'Reordering completed ... '
IF (error .NE. 0) THEN
WRITE(*,*) 'The following ERROR was detected: ', error
STOP
END IF

c WRITE(*,*) 'Number of nonzeros in factors = ',iparm(18)
c WRITE(*,*) 'Number of factorization MFLOPS = ',iparm(19)

C.. Factorization.
phase=22 ! only factorization
CALL pardiso(pt,maxfct,mnum,mtype,phase,n,a,ia,ja,
1 idum,nrhs,iparm,msglvl,ddum,ddum,error)
! WRITE(*,*) 'Factorization completed ... '
IF (error .NE. 0)THEN
WRITE(*,*)'The following ERROR was detected: ',error
STOP
ENDIF

C.. Back substitution and iterative refinement
iparm(8) = 2 ! max numbers of iterative refinement steps
phase = 33 ! only factorization
do i=1,n
b(i)=1.d0
end do
CALL pardiso(pt, maxfct, mnum, mtype, phase, n, a, ia, ja,
1 idum, nrhs,iparm, msglvl, b, x, error)
! WRITE(*,*) 'Solve completed ... '
b=x
C.. Termiation and release of memory
c phase = -1 ! release internal memory
phase=11
CALL pardiso(pt,maxfct,mnum,mtype,phase,n,ddum,idum,idum,
1 idum,nrhs,iparm,msglvl,ddum,ddum,error)
IF (error .NE. 0)THEN
WRITE(*,*)'The following ERROR on release stage was detected: ',error
STOP
ENDIF

return
end
C

Thank you for all your reply.

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1 Solution
mecej4
Black Belt
379 Views

Why do you use PHASE=11 for releasing memory?

 

When you report a problem, please pay more attention to making the report complete and readable. In your report above, you did not show the error code, the screen-capture is so tiny as to be unreadable, and the program text is incomplete and poorly formatted. Please follow the guidelines in the "How to post ..." sticky post in this forum.

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2 Replies
mecej4
Black Belt
380 Views

Why do you use PHASE=11 for releasing memory?

 

When you report a problem, please pay more attention to making the report complete and readable. In your report above, you did not show the error code, the screen-capture is so tiny as to be unreadable, and the program text is incomplete and poorly formatted. Please follow the guidelines in the "How to post ..." sticky post in this forum.

MRajesh_intel
Moderator
324 Views

 Hi,

Thanks for accepting as a solution!

 

As this issue has been resolved, we will no longer respond to this thread. If you require any additional assistance from Intel, please start a new thread. Any further interaction in this thread will be considered community only.

 

Have a Good day.

 

Regards

Rajesh

 

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