I guess I'm reading it wrong, but doesn't "pi" mean "page-in" for
pages fetched from on-disk swap space? With so much available RAM,
why would it even be used?
Also, 4k*180561 = 722244Kb paged in, and 2048k*1207 = 2471936Kb
paged in, so in the second case the system actually does 3x more
IO (even if just some RAM ops) to page-in the data. Perhaps some
value in-between should give you better results in the particular
test run (smaller pages and requests amounting to 800mb-1gb, and
fewer requests having less overhead and RAM fragmentation).
This would also support the idea that an adaptive algorithm might
give certain benefits - like min/max block sizes in ZFS ;)
And you do skip the first few lines of vmstat output, right? (that
should show lifetime stats average during this OS uptime). Trust
it for dynamic data only after the second or third line.
Curiously,
//Jim
On 2013-02-17 16:46, Bob Friesenhahn wrote:
Notice that the real and system time were dramatically reduced by
using a larger page size. Additional testing shows system time being
cut in half and large reductions in reported user time as well.
'vmstat 10' while the application is running shows the huge effect of 4K
pagesize vs 2MB pagesize. With 4K pagesize, the 'pi' value reaches
180561 but with 2MB pagesize, the 'pi' value reaches only 1207:
kthr memory page disk faults cpu
r b w swap free re mf pi po fr de sr s1 s2 s3 s4 in sy cs
us sy id
0 0 0 127507048 110430156 0 180561 0 0 0 0 0 0 0 0 0 996 1120 460
2 1 97
kthr memory page disk faults cpu
r b w swap free re mf pi po fr de sr s1 s2 s3 s4 in sy cs
us sy id
0 0 0 113485140 96406824 284 1207 0 0 0 0 0 0 0 0 0 867 7630 497
2 1 97
-------------------------------------------
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