parallelization problems with fpp under cf774.0.3 (long)

[Hemant Dandekar]

Hi, 

I am having trouble 'parallelizing' the outermost do 20 loop when I 
use the fpp preprocessor on a CRAY-2s running UNICOS 6.1 and cf77 4.0.3.
This loop does parallelize when I use the parallel fortran under IBM 
FORTVS 2.4 and 2.5.  I don't see an depedence in this loop at the
k level. Here is the the routine:

The subroutine calculates the convective term in a 3D cfd code. I
have removed the comment statements for sake of saving bandwidth.  

      subroutine convc1 (nlo,nhi,mlo,mhi,klo,khi,
     $                   f,indgeo,isimpl,confac)
      parameter          (nmax=65,mmax=45,kmax=45)
      dimension           f(0:nmax,0:mmax,0:kmax),
     1                    co1(0:nmax,0:mmax,0:kmax),
     1                    co2(0:nmax,0:mmax,0:kmax),
     1                    co3(0:nmax,0:mmax,0:kmax),
     1                    co4(0:nmax,0:mmax,0:kmax)
c
      common / admsh1 / x(0:nmax,0:mmax),y(0:nmax,0:mmax),
     1                  xne(0:nmax,0:mmax),yne(0:nmax,0:mmax)
      common / amsh1d / we(0:kmax),z(0:kmax),zne(0:kmax)
 
      common / conflu / cof(0:nmax,0:mmax,0:kmax)
      common / dotpro / dpl(0:nmax,0:mmax,0:kmax),
     1                  dpr(0:nmax,0:mmax,0:kmax),
     1                  dpb(0:nmax,0:mmax,0:kmax),
     1                  dpt(0:nmax,0:mmax,0:kmax)
      common / datpro / dw1(0:nmax,0:mmax,0:kmax),
     1                  dw2(0:nmax,0:mmax,0:kmax),
     1                  dw3(0:nmax,0:mmax,0:kmax),
     1                  dw4(0:nmax,0:mmax,0:kmax),
     1                  dpw1(0:nmax,0:mmax,0:kmax),
     1                  dpw2(0:nmax,0:mmax,0:kmax),
     1                  dpw3(0:nmax,0:mmax,0:kmax),
     1                  dpw4(0:nmax,0:mmax,0:kmax)
      do 10 k=klo,khi
      do 10 j=mlo,mhi
      do 10 i=nlo,nhi
        cof(i,j,k) = 0.0
   10 continue
      if(indgeo.eq.0) then
        confac = 2.0*2.0*8.0
        confc1 = confac * 0.5
      else
        confac = 2.0*4.0*4.0*4.0*6.0*6.0
        confc1 = confac * 0.5
      end if
      do 20 k=klo,khi
      do 20 j=mlo,mhi-1
      do 20 i=nlo,nhi-1
        f1  = f(i,j,k)
        f2  = f(i+1,j+1,k)
        f3  = f(i,j+1,k)
        f4  = f(i+1,j,k)
        f5  = f(i,j,k+1)
        f6  = f(i+1,j+1,k+1)
        f7  = f(i,j+1,k+1)
        f8  = f(i+1,j,k+1)
c
        zl  = ( (z(k+1) - z(k-1)) +  (zne(k+1) - zne(k-1))) / 4.0
c
        vnl = dpl(i,j,k)
        vnr = dpr(i,j,k)
        vnb = dpb(i,j,k)
        vnt = dpt(i,j,k)
c
        dwp1 = dpw1(i,j,k)
        dwp2 = dpw2(i,j,k)
        dwp3 = dpw3(i,j,k)
        dwp4 = dpw4(i,j,k)
c
        flb = f1
        flt = f3
        frb = f4
        frt = f2
        fbl = f1
        fbr = f4
        ftl = f3
        ftr = f2
c
        f1b = f1
        f1t = f5
        f2b = f2
        f2t = f6
        f3b = f3
        f3t = f7
        f4b = f4
        f4t = f8
c
        col = (vnl + abs(vnl))*flb + (vnl - abs(vnl))*flt
        cor = (vnr + abs(vnr))*frb + (vnr - abs(vnr))*frt
        cob = (vnb + abs(vnb))*fbl + (vnb - abs(vnb))*fbr
        cot = (vnt + abs(vnt))*ftl + (vnt - abs(vnt))*ftr
c
        co1(i,j,k) = (dwp1 + abs(dwp1))*f1b + (dwp1 - abs(dwp1))*f1t
        co2(i,j,k) = (dwp2 + abs(dwp2))*f2b + (dwp2 - abs(dwp2))*f2t
        co3(i,j,k) = (dwp3 + abs(dwp3))*f3b + (dwp3 - abs(dwp3))*f3t
        co4(i,j,k) = (dwp4 + abs(dwp4))*f4b + (dwp4 - abs(dwp4))*f4t
c
        cof(i,j,k)     = cof(i,j,k)     - (col + cob)*zl -
     $                 co1(i,j,k)*confc1
        cof(i+1,j+1,k) = cof(i+1,j+1,k) + (cor + cot)*zl -
     $                 co2(i,j,k)*confc1
        cof(i,j+1,k)   = cof(i,j+1,k)   + (col - cot)*zl -
     $                 co3(i,j,k)*confc1
        cof(i+1,j,k)   = cof(i+1,j,k)   - (cor - cob)*zl -
     $                 co4(i,j,k)*confc1
20    continue
      return
      end
Another section of the code which is exactly the same but with following
lines for terms 'flb to  ftr' does parallelize. 

        flb = f4 + 3.0*f1
        flt = f2 + 3.0*f3
        frb = f1 + 3.0*f4
        frt = f3 + 3.0*f2
        fbl = f3 + 3.0*f1
        fbr = f2 + 3.0*f4
        ftl = f1 + 3.0*f3
        ftr = f4 + 3.0*f2

Is it doing that because, it finds insufficient work to do in the previous
case and more work to do in the second case (where there are 8 additional
multiplications inside the loop)? Any comments/suggestions would be
appreciated.

thanks,
hemant

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Bitnet: v092qghg at ubvms.bitnet           | Hemant W. Dandekar
Internet: hemant at asterix.eng.buffalo.edu| 303, Furnas Hall, Chem. Eng.
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