On 04/03/2013 10:22 AM, Paul Turner wrote:
> On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi wrote:
>> On 04/02/2013 05:02 PM, Namhyung Kim wrote:
> + cfs_util = (FULL_UTIL - rt_util) > rq->util ? rq->util
> + : (FULL_UTIL - rt_util);
> + nr_running = rq->nr_running ?
On 04/03/2013 10:22 AM, Paul Turner wrote:
On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi alex@intel.com wrote:
On 04/02/2013 05:02 PM, Namhyung Kim wrote:
+ cfs_util = (FULL_UTIL - rt_util) rq-util ? rq-util
+ : (FULL_UTIL - rt_util);
+ nr_running = rq-nr_running ?
On 04/03/2013 10:22 AM, Paul Turner wrote:
> On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi wrote:
>> On 04/02/2013 05:02 PM, Namhyung Kim wrote:
> + cfs_util = (FULL_UTIL - rt_util) > rq->util ? rq->util
> + : (FULL_UTIL - rt_util);
> + nr_running = rq->nr_running ?
On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi wrote:
> On 04/02/2013 05:02 PM, Namhyung Kim wrote:
>>> > + cfs_util = (FULL_UTIL - rt_util) > rq->util ? rq->util
>>> > + : (FULL_UTIL - rt_util);
>>> > + nr_running = rq->nr_running ? rq->nr_running : 1;
>> This can be cleaned up with
On 04/02/2013 05:02 PM, Namhyung Kim wrote:
>> > + cfs_util = (FULL_UTIL - rt_util) > rq->util ? rq->util
>> > + : (FULL_UTIL - rt_util);
>> > + nr_running = rq->nr_running ? rq->nr_running : 1;
> This can be cleaned up with proper min/max().
>
>> > +
>> > + return rt_util +
On 04/02/2013 10:38 PM, Vincent Guittot wrote:
>> +static unsigned int max_rq_util(int cpu)
>> > +{
>> > + struct rq *rq = cpu_rq(cpu);
>> > + unsigned int rt_util = scale_rt_util(cpu);
>> > + unsigned int cfs_util;
>> > + unsigned int nr_running;
>> > +
>> > +
On 30 March 2013 15:34, Alex Shi wrote:
> Since the rt task priority is higher than fair tasks, cfs_rq utilization
> is just the left of rt utilization.
>
> When there are some cfs tasks in queue, the potential utilization may
> be yielded, so mulitiplying cfs task number to get max potential
>
>> +nr_running = rq->nr_running ? rq->nr_running : 1;
>
> This can be cleaned up with proper min/max().
yes, thanks
>
>> +
>> +return rt_util + cfs_util * nr_running;
>
> Should this nr_running consider tasks in cfs_rq only? Also it seems
> there's no upper bound so that it can
Hi Alex,
On Sat, 30 Mar 2013 22:34:57 +0800, Alex Shi wrote:
> Since the rt task priority is higher than fair tasks, cfs_rq utilization
> is just the left of rt utilization.
>
> When there are some cfs tasks in queue, the potential utilization may
> be yielded, so mulitiplying cfs task number to
Hi Alex,
On Sat, 30 Mar 2013 22:34:57 +0800, Alex Shi wrote:
Since the rt task priority is higher than fair tasks, cfs_rq utilization
is just the left of rt utilization.
When there are some cfs tasks in queue, the potential utilization may
be yielded, so mulitiplying cfs task number to get
+nr_running = rq-nr_running ? rq-nr_running : 1;
This can be cleaned up with proper min/max().
yes, thanks
+
+return rt_util + cfs_util * nr_running;
Should this nr_running consider tasks in cfs_rq only? Also it seems
there's no upper bound so that it can possibly exceed
On 30 March 2013 15:34, Alex Shi alex@intel.com wrote:
Since the rt task priority is higher than fair tasks, cfs_rq utilization
is just the left of rt utilization.
When there are some cfs tasks in queue, the potential utilization may
be yielded, so mulitiplying cfs task number to get max
On 04/02/2013 10:38 PM, Vincent Guittot wrote:
+static unsigned int max_rq_util(int cpu)
+{
+ struct rq *rq = cpu_rq(cpu);
+ unsigned int rt_util = scale_rt_util(cpu);
+ unsigned int cfs_util;
+ unsigned int nr_running;
+
+ cfs_util = (FULL_UTIL -
On 04/02/2013 05:02 PM, Namhyung Kim wrote:
+ cfs_util = (FULL_UTIL - rt_util) rq-util ? rq-util
+ : (FULL_UTIL - rt_util);
+ nr_running = rq-nr_running ? rq-nr_running : 1;
This can be cleaned up with proper min/max().
+
+ return rt_util + cfs_util * nr_running;
On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi alex@intel.com wrote:
On 04/02/2013 05:02 PM, Namhyung Kim wrote:
+ cfs_util = (FULL_UTIL - rt_util) rq-util ? rq-util
+ : (FULL_UTIL - rt_util);
+ nr_running = rq-nr_running ? rq-nr_running : 1;
This can be cleaned up with
On 04/03/2013 10:22 AM, Paul Turner wrote:
On Tue, Apr 2, 2013 at 7:15 PM, Alex Shi alex@intel.com wrote:
On 04/02/2013 05:02 PM, Namhyung Kim wrote:
+ cfs_util = (FULL_UTIL - rt_util) rq-util ? rq-util
+ : (FULL_UTIL - rt_util);
+ nr_running = rq-nr_running ?
Since the rt task priority is higher than fair tasks, cfs_rq utilization
is just the left of rt utilization.
When there are some cfs tasks in queue, the potential utilization may
be yielded, so mulitiplying cfs task number to get max potential
utilization of cfs. Then the rq utilization is sum of
Since the rt task priority is higher than fair tasks, cfs_rq utilization
is just the left of rt utilization.
When there are some cfs tasks in queue, the potential utilization may
be yielded, so mulitiplying cfs task number to get max potential
utilization of cfs. Then the rq utilization is sum of
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