Steve, show your math on how you calculated 58MB/sec from two timed processes. On an array on one of my servers, I show that the time approximately doubles when running two processes instead of each by themselves. That still yields about the same throughput. Also, try running a benchmark like bonnie++: http://www.coker.com.au/bonnie++/ Steve Bergman wrote: > # Run two read operations, on different parts of /dev/md1 simultaneously > # This reads a total of 1GB of data > time dd if=/dev/md1 bs=4k count=131072 of=/dev/null & > time dd if=/dev/md1 skip=262144 bs=4k count=131072 of=/dev/null & > ===== > > The results show about 58MB/sec transferred, which is about the same as > hdparm is showing for each drive individually. > > Running the same thing, but reading the whole 1GB using one dd process > in the foreground gives identical results. > > Why am I not seeing higher numbers? > > Thanks, > Steve > > > _______________________________________________ > CentOS mailing list > CentOS at centos.org > http://lists.centos.org/mailman/listinfo/centos