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From: Erich Focht <efocht@ess.nec.de>
To: "Andrew Theurer" <habanero@us.ibm.com>,
       "Michael Hohnbaum" <michael@hbaum.com>
Subject: Re: [Lse-tech] Re: NUMA scheduler 2nd approach
Date: Tue, 14 Jan 2003 16:13:27 +0100
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Cc: "Martin J. Bligh" <mbligh@aracnet.com>, "Robert Love" <rml@tech9.net>,
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On Tuesday 14 January 2003 17:52, Andrew Theurer wrote:
> > I suppose I should not have been so dang lazy and cut-n-pasted
> > the line I changed.  The change was (((5*4*this_load)/4) + 4)
> > which should be the same as your second choice.
> >
> > > We def need some constant to avoid low load ping pong, right?
> >
> > Yep.  Without the constant, one could have 6 processes on node
> > A and 4 on node B, and node B would end up stealing.  While making
> > a perfect balance, the expense of the off-node traffic does not
> > justify it.  At least on the NUMAQ box.  It might be justified
> > for a different NUMA architecture, which is why I propose putting
> > this check in a macro that can be defined in topology.h for each
> > architecture.
>
> Yes, I was also concerned about one task in one node and none in the
> others. Without some sort of constant we will ping pong the task on every
> node endlessly, since there is no % threshold that could make any
> difference when the original load value is 0..  Your +4 gets rid of the 1
> task case.

That won't happen because:
 - find_busiest_queue() won't detect a sufficient imbalance
 (imbalance == 0),
 - load_balance() won't be able to steal the task, as it will be
 running (the rq->curr task)

So the worst thing that happens is that load_balance() finds out that
it cannot steal the only task running on the runqueue and
returns. But we won't get there because find_busiest_queue() returns
NULL. It happens even in the bad case:
   node#0 : 0 tasks
   node#1 : 4 tasks (each on its own CPU)
where we would desire to distribute the tasks equally among the
nodes. At least on our IA64 platform...

Regards,
Erich



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