Problem:
As the comment above the calculation of max_pull in the function states,
there is a need to ensure that negative results of the subtractions do
not wrap around to large numbers. This has not been implemented for the
(max_load - busiest_load_per_task) expression and the possible
consequences are for undesirable movement of tasks from one group to
another group. E.g. consider a numa system with two nodes, each
node containing four processors. If there are two processes in node-0
and with node-1 being completely idle, one of those processes will be
moved to node-1 whereas the desired behavior is to retain those two
processes in node-0.
Fix:
Make sure that max_load is greater than busiest_load_per_task before
making the calculation. If it isn't max_pull will be zero and we skip
directly to out_balanced.
Signed-off-by: Peter Williams <[email protected]>
--
Peter Williams [email protected]
"Learning, n. The kind of ignorance distinguishing the studious."
-- Ambrose Bierce
Index: MM-2.6.X/kernel/sched.c
===================================================================
--- MM-2.6.X.orig/kernel/sched.c 2006-03-27 11:07:52.000000000 +1100
+++ MM-2.6.X/kernel/sched.c 2006-03-27 12:08:47.000000000 +1100
@@ -2179,6 +2179,8 @@ find_busiest_group(struct sched_domain *
* by pulling tasks to us. Be careful of negative numbers as they'll
* appear as very large values with unsigned longs.
*/
+ if (max_load <= busiest_load_per_task)
+ goto out_balanced;
/* Don't want to pull so many tasks that a group would go idle */
max_pull = min(max_load - avg_load, max_load - busiest_load_per_task);
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