From: Ramiro Willmersdorf Newsgroups: comp.parallel Subject: Unrolling murders performance Date: 3 Jun 1999 16:06:43 GMT Organization: Deja.com - Share what you know. Learn what you don't. Approved: bigrigg@cs.cmu.edu Message-Id: <7j696j$mno$1@goldenapple.srv.cs.cmu.edu> Originator: bigrigg@ux6.sp.cs.cmu.edu Xref: ukc comp.parallel:15644 Hi, Is there a reason for this an unrolled loop such as : *************************************************************** do 100 iter = 1, niter *---!MIC$ DOALL PRIVATE(i, istep), SHARED(a,b,c,d) do 20 i = 1, (nstep-1)*NROLL+1, nroll * 0 c(i) = c(i) + a(i)*a(i) + b(i)*b(i) d(i) = c(i) + a(i)*b(i) * 1 c(i+1) = c(i+1) + a(i+1)*a(i+1) + b(i+1)*b(i+1) d(i+1) = c(i+1) + a(i+1)*b(i+1) * 2 c(i+2) = c(i+2) + a(i+2)*a(i+2) + b(i+2)*b(i+2) d(i+2) = c(i+2) + a(i+2)*b(i+2) * 3 c(i+3) = c(i+3) + a(i+3)*a(i+3) + b(i+3)*b(i+3) d(i+3) = c(i+3) + a(i+3)*b(i+3) * 4 c(i+4) = c(i+4) + a(i+4)*a(i+4) + b(i+4)*b(i+4) d(i+4) = c(i+4) + a(i+4)*b(i+4) 20 continue * Remaining iterations do 30 i = nstep*nroll+1, BIG c(i) = c(i) + a(i)*a(i) + b(i)*b(i) d(i) = c(i) + a(i)*b(i) 30 continue 100 continue *************************************************************** to run *four* times as slow as the original loop: *************************************************************** do 100 iter = 1, niter *--- !MIC$ DOALL PRIVATE(i), SHARED(a,b,c,d) do 20 i = 1, BIG c(i) = c(i) + a(i)*a(i) + b(i)*b(i) d(i) = c(i) + a(i)*b(i) 20 continue 100 continue *************************************************************** The loop just above is representative of the loop that uses most time in a fortran program I'm trying to paralellize with a Sun Enterprise 450 with 4 processors. ``BIG'' is very big :) It runs Solaris 2.6 with Workshop Fortran 4.2. Obviously, with such light loops and long vectors, when I try to force paralellism, the performace just sucks, it takes twice as long to run. That's actually what I expected. ``Hey, no problem, I'll just unroll the loop, and eventually it's *gotta* get enough load tomake the startup costs low enough. The problem is that when I run the loop as unrolled above, it takes about *four* times as long to run! 22 seconds against 5 seconds, give or take. I hardly expected that. The unrolled loop even parallelizes very well, however, it still takes longer than the sequential loop since it's baseline is so much slower. The loops in the original program are actually somewhat more complex, but I don't want to tackle them before I understand what's going on here. I'd be very grateful for any insight, Ramiro. Sent via Deja.com http://www.deja.com/ Share what you know. Learn what you don't. -- Articles to bigrigg+parallel@cs.cmu.edu (Admin: bigrigg@cs.cmu.edu) Archive: http://www.hensa.ac.uk/parallel/internet/usenet/comp.parallel