[rfc][patch] kernel/sched.c oddness?

[Matthew Dobson]

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This snippet of code appears wrong...  Either that, or the accompanying 
comment is wrong?

from kernel/sched.c::find_busiest_queue():

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*imbalance = (max_load - nr_running) / 2;
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/* It needs an at least ~25% imbalance to trigger balancing. */
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if (!idle && (*imbalance < (max_load + 3)/4)) {
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	busiest = NULL;
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	goto out;
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}

The comment says 25% imbalance, but the code is really checking for a 
~50% imbalance.  The attatched patch moves the division by two to the 
pull_task function where the imbalance number is actually used.  This 
patch makes the code match the comment, and divides the imbalance by two 
where it is needed.

Please let me know if I've misinterpreted what this code is supposed to 
be doing, -or- if we really want the comment to match the current code.

Cheers!

-Matt

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diff -Nur --exclude-from=/usr/src/.dontdiff linux-2.5.40-vanilla/kernel/sched.c linux-2.5.40-sched_cleanup/kernel/sched.c
--- linux-2.5.40-vanilla/kernel/sched.c	Tue Oct  1 00:07:35 2002
+++ linux-2.5.40-sched_cleanup/kernel/sched.c	Wed Oct  2 11:22:41 2002
@@ -689,7 +689,7 @@
 	if (likely(!busiest))
 		goto out;
 
-	*imbalance = (max_load - nr_running) / 2;
+	*imbalance = (max_load - nr_running);
 
 	/* It needs an at least ~25% imbalance to trigger balancing. */
 	if (!idle && (*imbalance < (max_load + 3)/4)) {
@@ -746,6 +746,11 @@
 	task_t *tmp;
 
 	busiest = find_busiest_queue(this_rq, this_cpu, idle, &imbalance);
+	/*
+	 * We only want to steal a number of tasks equal to 1/2 the imbalance,
+ 	 * otherwise, we'll just shift the imbalance to the new queue.
+	 */
+	imbalance >>= 1;
 	if (!busiest)
 		goto out;
 

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