On 29-Aug 10:15, Pavan Kondeti wrote:

On Fri, Aug 25, 2017 at 3:50 PM, Patrick Bellasi
[off-list ref] wrote:

quoted

When the scheduler looks at the CPU utlization, the current PELT value
for a CPU is returned straight away. In certain scenarios this can have
undesired side effects on task placement.

<snip>

quoted

+/**
+ * cpu_util_est: estimated utilization for the specified CPU
+ * @cpu: the CPU to get the estimated utilization for
+ *
+ * The estimated utilization of a CPU is defined to be the maximum between its
+ * PELT's utilization and the sum of the estimated utilization of the tasks
+ * currently RUNNABLE on that CPU.
+ *
+ * This allows to properly represent the expected utilization of a CPU which
+ * has just got a big task running since a long sleep period. At the same time
+ * however it preserves the benefits of the "blocked load" in describing the
+ * potential for other tasks waking up on the same CPU.
+ *
+ * Return: the estimated utlization for the specified CPU
+ */
+static inline unsigned long cpu_util_est(int cpu)
+{
+       struct sched_avg *sa = &cpu_rq(cpu)->cfs.avg;
+       unsigned long util = cpu_util(cpu);
+
+       if (!sched_feat(UTIL_EST))
+               return util;
+
+       return max(util, util_est(sa, UTIL_EST_LAST));
+}
+
 static inline int task_util(struct task_struct *p)
 {
        return p->se.avg.util_avg;

@@ -6007,11 +6033,19 @@ static int cpu_util_wake(int cpu, struct task_struct *p)

        /* Task has no contribution or is new */
        if (cpu != task_cpu(p) || !p->se.avg.last_update_time)
-               return cpu_util(cpu);
+               return cpu_util_est(cpu);

        capacity = capacity_orig_of(cpu);
        util = max_t(long, cpu_rq(cpu)->cfs.avg.util_avg - task_util(p), 0);

+       /*
+        * Estimated utilization tracks only tasks already enqueued, but still
+        * sometimes can return a bigger value than PELT, for example when the
+        * blocked load is negligible wrt the estimated utilization of the
+        * already enqueued tasks.
+        */
+       util = max_t(long, util, cpu_util_est(cpu));
+

We are supposed to discount the task's util from its CPU. But the
cpu_util_est() can potentially return cpu_util() which includes the
task's utilization.

You right, this instead should cover all the cases:

---8<---
 static int cpu_util_wake(int cpu, struct task_struct *p)
 {
-       unsigned long util, capacity;
+       unsigned long util_est = cpu_util_est(cpu);
+       unsigned long capacity;
 
        /* Task has no contribution or is new */
        if (cpu != task_cpu(p) || !p->se.avg.last_update_time)
-               return cpu_util(cpu);
+               return util_est;
 
        capacity = capacity_orig_of(cpu);
-       util = max_t(long, cpu_rq(cpu)->cfs.avg.util_avg - task_util(p), 0);
+       if (cpu_util(cpu) > util_est)
+               util = max_t(long, cpu_util(cpu) - task_util(p), 0);
+       else
+               util = util_est;
 
        return (util >= capacity) ? capacity : util;
 }
---8<---

Indeed:

- if *p is the only task sleeping on that CPU, then:
      (cpu_util == task_util) > (cpu_util_est == 0)
  and thus we return:
      (cpu_util - task_util) == 0

- if other tasks are SLEEPING on the same CPU, which however is IDLE, then:
      cpu_util > (cpu_util_est == 0)
  and thus we discount *p's blocked load by returning:
      (cpu_util - task_util) >= 0

- if other tasks are RUNNABLE on that CPU and
      (cpu_util_est > cpu_util)
  then we wanna use cpu_util_est since it returns a more restrictive
  estimation of the spare capacity on that CPU, by just considering
  the expected utilization of tasks already runnable on that CPU.

What do you think?

Thanks,
Pavan

Cheers Patrick

-- 
#include <best/regards.h>

Patrick Bellasi