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