From: Nick Piggin <piggin@cyberone.com.au>
From: Frank Cornelis <frank.cornelis@elis.ugent.be>
In order to get the best possible resolution we need to use NR_CPUS instead
of the constant value 10. load is an int, so no need to worry about
overflows...
---
25-akpm/kernel/sched.c | 6 +++---
1 files changed, 3 insertions(+), 3 deletions(-)
diff -puN kernel/sched.c~sched-find_busiest_node-resolution-fix kernel/sched.c
--- 25/kernel/sched.c~sched-find_busiest_node-resolution-fix Fri Mar 19 15:28:33 2004
+++ 25-akpm/kernel/sched.c Fri Mar 19 15:28:33 2004
@@ -1097,7 +1097,7 @@ void sched_balance_exec(void)
* load_{t} = load_{t-1}/2 + nr_node_running_{t}
* This way sudden load peaks are flattened out a bit.
* Node load is divided by nr_cpus_node() in order to compare nodes
- * of different cpu count but also [first] multiplied by 10 to
+ * of different cpu count but also [first] multiplied by NR_CPUS to
* provide better resolution.
*/
static int find_busiest_node(int this_node)
@@ -1107,14 +1107,14 @@ static int find_busiest_node(int this_no
if (!nr_cpus_node(this_node))
return node;
this_load = maxload = (this_rq()->prev_node_load[this_node] >> 1)
- + (10 * atomic_read(&node_nr_running[this_node])
+ + (NR_CPUS * atomic_read(&node_nr_running[this_node])
/ nr_cpus_node(this_node));
this_rq()->prev_node_load[this_node] = this_load;
for_each_node_with_cpus(i) {
if (i == this_node)
continue;
load = (this_rq()->prev_node_load[i] >> 1)
- + (10 * atomic_read(&node_nr_running[i])
+ + (NR_CPUS * atomic_read(&node_nr_running[i])
/ nr_cpus_node(i));
this_rq()->prev_node_load[i] = load;
if (load > maxload && (100*load > NODE_THRESHOLD*this_load)) {
_