Messages in this thread | ![/](/images/icornerl.gif) | | From | Andrew Theurer <> | Subject | Re: Nick's scheduler v17 | Date | Mon, 27 Oct 2003 12:02:58 -0500 |
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> >> *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).
Oops, I meant avg_load, which you calculate in find_busiest_queue on the fly.
> 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...
Actually, if we use avg_load, I guess it would be 0, since this is an unsigned long. Maybe avg_load needs to have a min value of 1. Then if we apply:
*imbalance = min(max_load - avg_load, avg_load - this_load) min(4 - 1, 1 - 0)
And imbalance looks a lot better. Only concern would be an idle cpu stealing from another, leaving the other cpu idle. I guess a check could be put there.
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