* To solve this problem, we also cap the util_avg of successive tasks to
* only 1/2 of the left utilization budget:
*
- * util_avg_cap = (1024 - cfs_rq->avg.util_avg) / 2^n
+ * util_avg_cap = (cpu_scale - cfs_rq->avg.util_avg) / 2^n
*
- * where n denotes the nth task.
+ * where n denotes the nth task and cpu_scale the CPU capacity.
*
- * For example, a simplest series from the beginning would be like:
+ * For example, for a CPU with 1024 of capacity, a simplest series from
+ * the beginning would be like:
*
* task util_avg: 512, 256, 128, 64, 32, 16, 8, ...
* cfs_rq util_avg: 512, 768, 896, 960, 992, 1008, 1016, ...
{
struct cfs_rq *cfs_rq = cfs_rq_of(se);
struct sched_avg *sa = &se->avg;
- long cap = (long)(SCHED_CAPACITY_SCALE - cfs_rq->avg.util_avg) / 2;
+ long cpu_scale = arch_scale_cpu_capacity(NULL, cpu_of(rq_of(cfs_rq)));
+ long cap = (long)(cpu_scale - cfs_rq->avg.util_avg) / 2;
if (cap > 0) {
if (cfs_rq->avg.util_avg != 0) {