Messages in this thread | | | Date | Fri, 23 May 2008 13:14:31 +0530 | From | Srivatsa Vaddagiri <> | Subject | Re: fair group scheduler not so fair? |
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On Thu, May 22, 2008 at 06:17:37PM -0600, Chris Friesen wrote: > Peter Zijlstra wrote: > >> Given the following: >> root >> / | \ >> _A_ 1 2 >> /| |\ >> 3 4 5 B >> / \ >> 6 7 >> CPU0 CPU1 >> root root >> / \ / \ >> A 1 A 2 >> / \ / \ >> 4 B 3 5 >> / \ >> 6 7 > > How do you move specific groups to different cpus. Is this simply using > cpusets?
No. Moving groups to different cpus is just a group-aware extension to move_tasks() that is invoked as part of regular load balance operation. move_tasks()->sched_fair_class.load_balance() has been modified to understand how much various task-groups at various levels (ex: A at level 1, B at level 2 etc) contribute to cpu load. It moves tasks between cpus using this knowledge.
For ex: if we were to consider all tasks shown above to be in same cpu, CPU0, this is how it would look:
CPU0 CPU1 root root / | \ A 1 2 /| |\ 3 4 5 B / \ 6 7
Then cpu0 load = weight of A + weight of 1 + weight of 2 = 1024 + 1024 + 1024 = 3072
while cpu1 load = 0
load to be moved to cut down this imbalance = 3072/2 = 1536
move_tasks() running on CPU1 would try to pull iteratively tasks such that total weight moved is <= 1536.
Task moved Total Weight moved --------- ------------ 2 1024 3 1024 + 256 = 1280 5 1280 + 256 = 1536
resulting in:
CPU0 CPU1 root root / \ / \ A 1 A 2 / \ / \ 4 B 3 5 / \ 6 7
>> Numerical examples given the above scenario, assuming every body's >> weight is 1024: > >> s_(0,A) = s_(1,A) = 512 > > Just to make sure I understand what's going on...this is half of 1024 > because it shows up on both cpus?
not exactly ..as Peter put it:
s_(i,g) = W_g * rw_(i,g) / \Sum_j rw_(j,g)
In this case,
s_(0,A) = W_A * rw_(0, A) / \Sum_j rw_(j, A)
W_A = shares given to A by admin = 1024
rw_(0,A) = Weight of 4 + Weight of B = 1024 + 1024 = 2048 rw_(1,A) = Weight of 3 + Weight of 5 = 1024 + 1024 = 2048 \Sum_j rw_(j, A) = 4096
So,
s_(0,A) = 1024 *2048 / 4096 = 512
>> s_(0,B) = 1024, s_(1,B) = 0 > > This gets the full 1024 because it's only on one cpu.
Not exactly. rw_(0, B) = \Sum_j rw_(j, B) and that's why s_(0,B) = 1024
>> rw_(0,A) = rw(1,A) = 2048 >> rw_(0,B) = 2048, rw_(1,B) = 0 > > How do we get 2048? Shouldn't this be 1024?
Hope this is clarified from above.
>> h_load_(0,A) = h_load_(1,A) = 512 >> h_load_(0,B) = 256, h_load(1,B) = 0 > > At this point the numbers make sense, but I'm not sure how the formula for > h_load_ works given that I'm not sure what's going on for rw_.
-- Regards, vatsa
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