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-rw-r--r--kernel/sched/deadline.c10
1 files changed, 7 insertions, 3 deletions
diff --git a/kernel/sched/deadline.c b/kernel/sched/deadline.c
index 21f58d261134..3958bc576d67 100644
--- a/kernel/sched/deadline.c
+++ b/kernel/sched/deadline.c
@@ -289,7 +289,7 @@ static void replenish_dl_entity(struct sched_dl_entity *dl_se)
* arbitrary large.
*/
while (dl_se->runtime <= 0) {
- dl_se->deadline += dl_se->dl_deadline;
+ dl_se->deadline += dl_se->dl_period;
dl_se->runtime += dl_se->dl_runtime;
}
@@ -329,9 +329,13 @@ static void replenish_dl_entity(struct sched_dl_entity *dl_se)
*
* This function returns true if:
*
- * runtime / (deadline - t) > dl_runtime / dl_deadline ,
+ * runtime / (deadline - t) > dl_runtime / dl_period ,
*
* IOW we can't recycle current parameters.
+ *
+ * Notice that the bandwidth check is done against the period. For
+ * task with deadline equal to period this is the same of using
+ * dl_deadline instead of dl_period in the equation above.
*/
static bool dl_entity_overflow(struct sched_dl_entity *dl_se, u64 t)
{
@@ -355,7 +359,7 @@ static bool dl_entity_overflow(struct sched_dl_entity *dl_se, u64 t)
* of anything below microseconds resolution is actually fiction
* (but still we want to give the user that illusion >;).
*/
- left = (dl_se->dl_deadline >> 10) * (dl_se->runtime >> 10);
+ left = (dl_se->dl_period >> 10) * (dl_se->runtime >> 10);
right = ((dl_se->deadline - t) >> 10) * (dl_se->dl_runtime >> 10);
return dl_time_before(right, left);