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SubjectRe: [RFC 2/2] sched/fair: prefer a CPU in the "lowest" idle state
On 01/31/2013 05:19 AM, Sebastian Andrzej Siewior wrote:
> If a new CPU has to be choosen for a task, then the scheduler first selects
> the group with the least load. This group is returned if its load is lower
> compared to the group to which the task is currently assigned.
> If there are several groups with completely idle CPU(s) (the CPU is in
> an idle state like C1) then the first group is returned.
> This patch extends this decision by considering the idle state of CPU(s)
> in the group and the first group with a CPU in the lowest idle state
> wins (C1 is prefered over C2). If there is a CPU which is not in an idle
> state (C0) but has no tasks assigned then it is consider as a valid target.
> Should there be no CPU in an idle state at disposal then the loadavg is
> used as a fallback.
>
> Signed-off-by: Sebastian Andrzej Siewior <bigeasy@linutronix.de>
> ---
> include/linux/sched.h | 1 +
> kernel/sched/core.c | 6 ++++--
> kernel/sched/fair.c | 24 ++++++++++++++++++++++++
> 3 files changed, 29 insertions(+), 2 deletions(-)
>
> diff --git a/include/linux/sched.h b/include/linux/sched.h
> index d211247..c2f6a44 100644
> --- a/include/linux/sched.h
> +++ b/include/linux/sched.h
> @@ -934,6 +934,7 @@ struct sched_domain {
> unsigned int wake_idx;
> unsigned int forkexec_idx;
> unsigned int smt_gain;
> + unsigned int prefer_lp;
> int flags; /* See SD_* */
> int level;
>
> diff --git a/kernel/sched/core.c b/kernel/sched/core.c
> index 26058d0..fad16e6 100644
> --- a/kernel/sched/core.c
> +++ b/kernel/sched/core.c
> @@ -4971,7 +4971,7 @@ set_table_entry(struct ctl_table *entry,
> static struct ctl_table *
> sd_alloc_ctl_domain_table(struct sched_domain *sd)
> {
> - struct ctl_table *table = sd_alloc_ctl_entry(13);
> + struct ctl_table *table = sd_alloc_ctl_entry(14);
>
> if (table == NULL)
> return NULL;
> @@ -5001,7 +5001,9 @@ sd_alloc_ctl_domain_table(struct sched_domain *sd)
> sizeof(int), 0644, proc_dointvec_minmax, false);
> set_table_entry(&table[11], "name", sd->name,
> CORENAME_MAX_SIZE, 0444, proc_dostring, false);
> - /* &table[12] is terminator */
> + set_table_entry(&table[12], "prefer_lp", &sd->prefer_lp,
> + sizeof(int), 0644, proc_dointvec_minmax, false);
> + /* &table[13] is terminator */
>
> return table;
> }
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index 5eea870..bff9800 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -23,6 +23,7 @@
> #include <linux/latencytop.h>
> #include <linux/sched.h>
> #include <linux/cpumask.h>
> +#include <linux/cpuidle.h>
> #include <linux/slab.h>
> #include <linux/profile.h>
> #include <linux/interrupt.h>
> @@ -3181,8 +3182,10 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
> int this_cpu, int load_idx)
> {
> struct sched_group *idlest = NULL, *group = sd->groups;
> + struct sched_group *idle_group = NULL;
> unsigned long min_load = ULONG_MAX, this_load = 0;
> int imbalance = 100 + (sd->imbalance_pct-100)/2;
> + int least_idle_cpu = INT_MAX;
>
> do {
> unsigned long load, avg_load;
> @@ -3208,6 +3211,25 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
> load = target_load(i, load_idx);
>
> avg_load += load;
> + if (!local_group && sd->prefer_lp && least_idle_cpu) {
> + int idle_level;
> +
> + idle_level = cpuidle_get_state(i);
> + /*
> + * Select the CPU which is in the lowest
> + * possible power state. Take the active
> + * CPU only if its run queue is empty.
> + */
> + if (!idle_level) {
> + if (idle_cpu(i)) {
> + least_idle_cpu = idle_level;
> + idle_group = group;
> + }
> + } else if (least_idle_cpu > idle_level) {
> + least_idle_cpu = idle_level;
> + idle_group = group;
> + }
> + }
> }
>
> /* Adjust by relative CPU power of the group */
> @@ -3221,6 +3243,8 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
> }
> } while (group = group->next, group != sd->groups);
>
> + if (idle_group)
> + return idle_group;

Hi, Sebastian

I'm not sure, but just concern about this case:

group 0 cpu 0 cpu 1
least idle 4 task

group 1 cpu 2 cpu 3
1 task 1 task

The previous logical will pick group 1 and now it will take group 0, and
that cause more imbalance, doesn't it?

May be check that state in find_idlest_cpu() will be better?

Regards,
Michael Wang

> if (!idlest || 100*this_load < imbalance*min_load)
> return NULL;
> return idlest;
>



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