On Fri, 2012-04-06 at 09:14 +0200, Juri Lelli wrote:+/*It might be good to put a few words in as to why that is.. I know I
+ * Here we check if --at time t-- an entity (which is probably being
+ * [re]activated or, in general, enqueued) can use its remaining runtime
+ * and its current deadline _without_ exceeding the bandwidth it is
+ * assigned (function returns true if it can).
+ *
+ * For this to hold, we must check if:
+ * runtime / (deadline - t)< dl_runtime / dl_deadline .
always forget (but know where to find it by now), also might be good to
refer those papers Tommaso listed when Steven asked this a while back.
+ */
+static bool dl_entity_overflow(struct sched_dl_entity *dl_se, u64 t)
+{
+ u64 left, right;
+
+ /*
+ * left and right are the two sides of the equation above,
+ * after a bit of shuffling to use multiplications instead
+ * of divisions.
+ *
+ * Note that none of the time values involved in the two
+ * multiplications are absolute: dl_deadline and dl_runtime
+ * are the relative deadline and the maximum runtime of each
+ * instance, runtime is the runtime left for the last instance
+ * and (deadline - t), since t is rq->clock, is the time left
+ * to the (absolute) deadline. Therefore, overflowing the u64
+ * type is very unlikely to occur in both cases.
+ */
+ left = dl_se->dl_deadline * dl_se->runtime;
+ right = (dl_se->deadline - t) * dl_se->dl_runtime;
From what I can see there are no constraints on the values in
__setparam_dl() so the above left term can be constructed to be an
overflow.
Ideally we'd use u128 here, but I don't think people will let us :/