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DateThu, 30 Jan 1997 16:01:55 -0800
FromDaniel Quinlan <>
Subjectstrange `time' results on dual-processor P6 systems
We are attempting to use time(1) to gather some basic performance
statistics on both single and dual-processor P6-200 systems.  The
kernel version is 2.0.27.  The results appear to be quite wrong.

On a single-processor system, the time values for a test program are:

  9.63user 0.18system 0:09.80elapsed 100%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (49major+3912minor)pagefaults 0swaps

  9.67user 0.16system 0:09.84elapsed 99%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (49major+3912minor)pagefaults 0swaps

  9.62user 0.19system 0:09.81elapsed 99%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (49major+3912minor)pagefaults 0swaps

On the dual-processor system, the time values are:

  0.02user 10.47system 0:10.50elapsed 99%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (0major+0minor)pagefaults 0swaps

  0.43user 10.06system 0:10.50elapsed 99%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (0major+0minor)pagefaults 0swaps

  1.73user 8.76system 0:10.50elapsed 99%CPU (0avgtext+0avgdata 0maxresident)k
  0inputs+0outputs (0major+0minor)pagefaults 0swaps

The elapsed time seems correct on both systems.  However, the
proportion of user to system time is more or less reversed and quite
wrong for the dual-processor system.

Any ideas?

This is the test program:

------- start of cut text --------------
#include <stdio.h>
int a[4000000];
main()
{
        int i, j;
        for( i=0; i<30; ++i )
        {
                for( j=0; j<4000000; ++j )
                        ++a[j];
        }
        return( 0 );
}
------- end ----------------------------
-- 
Daniel Quinlan (at work)        Daniel Quinlan (otherwise)
quinlan@transmeta.com           http://www.pathname.com/~quinlan/

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