How can I use bonnie++ to measure IOPS?

[dart]

The Hardware Tuning Factors page gives this command:
bonnie++ -d [/your volume mountpoint] -s [twice your system RAM in MB] -u root:root -qfb

How do I interpret the output:

Version  1.96       ------Sequential Output------ --Sequential Input- --Random-
Concurrency   1     -Per Chr- --Block-- -Rewrite- -Per Chr- --Block-- --Seeks--
Machine        Size K/sec %CP K/sec %CP K/sec %CP K/sec %CP K/sec %CP  /sec %CP
abc1234        192G           428063  75 208976  32           591613  36 280.3  67
Latency                        1374ms    1231ms              1593ms   97428us
Version  1.96       ------Sequential Create------ --------Random Create--------
abc1234             -Create-- --Read--- -Delete-- -Create-- --Read--- -Delete--
              files  /sec %CP  /sec %CP  /sec %CP  /sec %CP  /sec %CP  /sec %CP
                 16 11623  52 +++++ +++ +++++ +++ 22939  98 +++++ +++ +++++ +++
Latency               343us     615us     705us     140us      15us     303us
1.96,1.96,abc1234,1,1365178857,192G,,,,428063,75,208976,32,,,591613,36,280.3,67,16,,,,,11623,52,+++++,+++,+++++,+++,22939,98,+++++,+++,+++++,+++,,1374ms,1231ms,,1593ms,97428us,343us,615us,705us,140us,15us,303us

[sdvorak_splunk · Accepted solution]

The more memory you can use in the -s parameter, the more realistic the numbers (and longer the test takes to run). I used 6x memory in my testing, which was much more accurate than the 2x number I started with. Also, I recommend piping the output to bon_csv2html to get a nicely formatted, readable output. The number for IOPs is the Randon Seeks/sec number.

[jmantor]

Could you elaborate on how using 6* RAM is more accurate than using 2* RAM for the -s parameter, please?

[sdvorak_splunk]

it ensures you exhaust cache (since synthetic workloads tend to be more repetitive) in both the OS and the storage array.

[kristian_kolb]

hmm. linebreaking/tabs is acting up a bit in your output. screenshot?

[dart]

In the example given 280.3 is the number of measured IOPS