* Balbir Singh <balbir@linux.vnet.ibm.com> wrote:

quoted text
> I disagree. The cost is only adding a field to cfs_rq [...]

wrong. The cost is "only" of adding a field to cfs_rq and _updating it_, 
in the hottest paths of the scheduler:

@@ -256,6 +257,7 @@ static void __enqueue_entity(struct cfs_
                 */
                if (key < entity_key(cfs_rq, entry)) {
                        link = &parent->rb_left;
+                       rightmost = 0;
                } else {
                        link = &parent->rb_right;
                        leftmost = 0;
@@ -268,6 +270,8 @@ static void __enqueue_entity(struct cfs_
         */
        if (leftmost)
                cfs_rq->rb_leftmost = &se->run_node;
+       if (rightmost)
+               cfs_rq->rb_rightmost = &se->run_node;

quoted text
> [...] For a large number of tasks - say 10000, we need to walk 14 
> levels before we reach the node (each time). [...]

10,000 yield-ing tasks is not a common workload we care about. It's not 
even a rare workload we care about. _Especially_ we dont care about it 
if it slows down every other workload (a tiny bit).

quoted text
> [...] Doesn't matter if the data is cached, we are still spending CPU 
> time looking through pointers and walking to the right node. [...]

have you actually measured how much it takes to walk the tree that deep
on recent hardware? I have.

	Ingo
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