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    SubjectRe: [PATCH 02/18] sched: fix find_idlest_group mess logical
    On 12/11/2012 01:08 PM, Preeti U Murthy wrote:
    > Hi Alex,
    >
    > On 12/10/2012 01:52 PM, Alex Shi wrote:
    >> There is 4 situations in the function:
    >> 1, no task allowed group;
    >> so min_load = ULONG_MAX, this_load = 0, idlest = NULL
    >> 2, only local group task allowed;
    >> so min_load = ULONG_MAX, this_load assigned, idlest = NULL
    >> 3, only non-local task group allowed;
    >> so min_load assigned, this_load = 0, idlest != NULL
    >> 4, local group + another group are task allowed.
    >> so min_load assigned, this_load assigned, idlest != NULL
    >>
    >> Current logical will return NULL in first 3 kinds of scenarios.
    >> And still return NULL, if idlest group is heavier then the
    >> local group in the 4th situation.
    >>
    >> Actually, I thought groups in situation 2,3 are also eligible to host
    >> the task. And in 4th situation, agree to bias toward local group.
    >> So, has this patch.
    >>
    >> Signed-off-by: Alex Shi <alex.shi@intel.com>
    >> ---
    >> kernel/sched/fair.c | 12 +++++++++---
    >> 1 files changed, 9 insertions(+), 3 deletions(-)
    >>
    >> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
    >> index df99456..b40bc2b 100644
    >> --- a/kernel/sched/fair.c
    >> +++ b/kernel/sched/fair.c
    >> @@ -2953,6 +2953,7 @@ 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 *this_group = NULL;
    >> unsigned long min_load = ULONG_MAX, this_load = 0;
    >> int imbalance = 100 + (sd->imbalance_pct-100)/2;
    >>
    >> @@ -2987,14 +2988,19 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
    >>
    >> if (local_group) {
    >> this_load = avg_load;
    >> - } else if (avg_load < min_load) {
    >> + this_group = group;
    >> + }
    >> + if (avg_load < min_load) {
    >> min_load = avg_load;
    >> idlest = group;
    >> }
    >> } while (group = group->next, group != sd->groups);
    >>
    >> - if (!idlest || 100*this_load < imbalance*min_load)
    >> - return NULL;
    >> + if (this_group && idlest != this_group)
    >> + /* Bias toward our group again */
    >> + if (100*this_load < imbalance*min_load)
    >> + idlest = this_group;
    >
    > If the idlest group is heavier than this_group(or to put it better if
    > the difference in the loads of the local group and idlest group is less
    > than a threshold,it means there is no point moving the load from the
    > local group) you return NULL,that immediately means this_group is chosen
    > as the candidate group for the task to run,one does not have to
    > explicitly return that.

    In situation 4, this_group is not NULL.
    >
    > Let me explain:
    > find_idlest_group()-if it returns NULL to mark your case4,it means there
    > is no idler group than the group to which this_cpu belongs to, at that
    > level of sched domain.Which is fair enough.
    >
    > So now the question is under such a circumstance which is the idlest
    > group so far.It is the group containing this_cpu,i.e.this_group.After
    > this sd->child is chosen which is nothing but this_group(sd hierarchy
    > moves towards the cpu it belongs to). Again here the idlest group search
    > begins.
    >
    >> +
    >> return idlest;
    >> }
    >>
    >>
    > Regards
    > Preeti U Murthy
    >


    --
    Thanks
    Alex


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