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 ? rq->n
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 ? rq->n
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 + cf
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_uti
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
> ut
>> +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 possib
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 g
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