On 10/23/2014 11:04 AM, George Smith wrote: > Uh oh, silence is never good :) Please let me know if I haven't > included key information, missed something obvious, etc. > > Here's maybe a clearer example of what I'm talking about. The output > file is from a read test using 5 threads: > > # grep 'read :' output.out > read : io=51200MB, bw=116673KB/s, iops=113 , runt=449367msec > read : io=51200MB, bw=189201KB/s, iops=184 , runt=277106msec > read : io=51200MB, bw=143385KB/s, iops=140 , runt=365650msec > read : io=51200MB, bw=114654KB/s, iops=111 , runt=457279msec > read : io=51200MB, bw=183110KB/s, iops=178 , runt=286324msec > > # grep READ output.out > READ: io=256000MB, aggrb=573269KB/s, minb=114653KB/s, > maxb=189201KB/s, mint=277106msec, maxt=457279msec > > > The sum of the threads is 747023, but aggrb is 573269. The bw= value > in each thread line is the amount of I/O (from io=) divided by the > time the I/O took. > > The aggrb= value is the total amount of I/O done (which is the sum of > each thread's io= value), divided by maxt, which seems to be the > maximum time seen during the run (which happens to be with my 4th > thread). > > So it appears that this is the discrepancy. I'm not sure if it's > correct to say the aggregate bandwidth is the total I/O divided by the > max time that one of the threads in the group took to complete. Seems > like taking the average time and dividing total I/O by that would be > more correct. > > Am I missing the spirit of what the READ line is supposed to be conveying to me? I think you deduced it correctly. And yes, it is a bit misleading, in that it differs from the sum of the reported bandwidths. Neither number really makes sense, however. -- Jens Axboe -- To unsubscribe from this list: send the line "unsubscribe fio" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
