Yep, that sucks. Now what what does bonnie++ say? Steve Bergman wrote: > On Wed, 2006-09-13 at 08:42 -0700, Kirk Bocek wrote: >> 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. > > OK. > > 131072*4k = 524,288k per dd process. > > There are two of these processes running simultaneously, for a total > 1,048,576k of data. > > 1,048,576k / 1024k/M = 1024MB > > The dd's came back with wall clock times of 17.1 seconds and 18.85 > seconds. So the entire operation took a total of 18.85 seconds to read > 1024MB of data. > > 1024MB /18.85sec = 54.3MB/sec > > Thanks, > Steve > > > > _______________________________________________ > CentOS mailing list > CentOS at centos.org > http://lists.centos.org/mailman/listinfo/centos