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SubjectRe: fair clock use in CFS

* Daniel Hazelton <dhazelton@enter.net> wrote:

> [...] In a fair scheduler I'd expect all tasks to get the exact same
> amount of time on the processor. So if there are 10 tasks running at
> nice 0 and the current task has run for 20msecs before a new task is
> swapped onto the CPU, the new task and *all* other tasks waiting to
> get onto the CPU should get the same 20msecs. [...]

What happens in CFS is that in exchange for this task's 20 msecs the
other tasks get 2 msecs each. (and not only the one that gets on the CPU
next) So each task is handled equal. What i described was the first step
- for each task the same step happens (whenever it gets on the CPU, and
accounted/weighted for the precise period they spent waiting - so the
second task would get +4 msecs credited, the third task +6 msecs, etc.,
etc.).

but really - nothing beats first-hand experience: please just boot into
a CFS kernel and test its precision a bit. You can pick it up from the
usual place:

http://people.redhat.com/mingo/cfs-scheduler/

For example start 10 CPU hogs at once from a shell:

for (( N=0; N < 10; N++ )); do ( while :; do :; done ) & done

[ type 'killall bash' in the same shell to get rid of them. ]

then watch their CPU usage via 'top'. While the system is otherwise idle
you should get something like this after half a minute of runtime:

PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
2689 mingo 20 0 5968 560 276 R 10.0 0.1 0:03.45 bash
2692 mingo 20 0 5968 564 280 R 10.0 0.1 0:03.45 bash
2693 mingo 20 0 5968 564 280 R 10.0 0.1 0:03.45 bash
2694 mingo 20 0 5968 564 280 R 10.0 0.1 0:03.45 bash
2695 mingo 20 0 5968 564 280 R 10.0 0.1 0:03.45 bash
2698 mingo 20 0 5968 564 280 R 10.0 0.1 0:03.45 bash
2690 mingo 20 0 5968 564 280 R 9.9 0.1 0:03.45 bash
2691 mingo 20 0 5968 564 280 R 9.9 0.1 0:03.45 bash
2696 mingo 20 0 5968 564 280 R 9.9 0.1 0:03.45 bash
2697 mingo 20 0 5968 564 280 R 9.9 0.1 0:03.45 bash

with each task having exactly the same 'TIME+' field in top. (the more
equal those fields, the more precise/fair the scheduler is. In the above
output each got its precise share of 3.45 seconds of CPU time.)

then as a next phase of testing please run various things on the system
(without stopping these loops) and try to get CFS "out of balance" -
you'll succeed if you manage to get an unequal 'TIME+' field for them.
Please try _really_ hard to break it. You can run any workload.

Or try massive_intr.c from:

http://lkml.org/lkml/2007/3/26/319

which uses a much less trivial scheduling pattern to test a scheduler's
precision of scheduling)

$ ./massive_intr 9 10
002765 00000125
002767 00000125
002762 00000125
002769 00000125
002768 00000126
002761 00000126
002763 00000126
002766 00000126
002764 00000126
(the second column is runtime - the more equal, the more precise/fair
the scheduler.)

Ingo
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