Messages in this thread | | | Date | Sun, 26 Oct 2003 17:43:48 +1100 | From | Nick Piggin <> | Subject | Re: Nick's scheduler v17 |
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Nick Piggin wrote:
> > > Andrew Theurer wrote: > >> On Friday 24 October 2003 13:10, Nick Piggin wrote: >> >>> Hi, >>> http://www.kerneltrap.org/~npiggin/v17/ >>> >>> Still working on SMP and NUMA. Some (maybe) interesting things I put >>> in are >>> - Sequential CPU balancing so you don't get a big storm of balances >>> every 1/4s. >>> - Balancing is trying to err more on the side of caution, I have to >>> start >>> analysing it more thoroughly though. >>> >> >> + >> + *imbalance /= 2; >> + *imbalance = (*imbalance + FPT - 1) / FPT; >> >> I think I see what is going on here, but would something like this >> work out better? >> > > Yeah, sorry its not well commented. Its still changing quite quickly. > >> >> *imbalance = min(this_load - load_avg, load_avg - max_load) >> >> That way you take just enough to either have busiest_queue or >> this_rq's length be the load_avg. I suppose you could take even >> less, but IMO, the /=2 is what I really don't like. Perhaps: >> > > That is _exactly_ what I had before! Thats probably the way to go. Thanks > for having a look at it. > >> >> >> *imbalance = min(this_load - load_avg, load_avg - max_load); >> *imbalance = (*imbalance + FPT - 1) / FPT; >> >> This should work well for intranode balances, internode balances may >> need a little optimization, since the load_avg really does not really >> represent the load avg of the two nodes in question, just one cpu >> from one of them and all the cpus from another. >>
Oh, actually, after my path, load_avg represents the load average of _all_ the nodes. Have a look at find_busiest_node. Which jogs my memory of why its not always a good idea to do your *imbalance min(...) thing (I actually saw this happening).
5 CPUs, 4 processes running on one cpu. load_avg would be 0.8 for all cpus. balancing doesn't happen. I have to think about this a bit more...
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