EDF priority tie-breaks.wip-robust-tie-break

Instead of tie-breaking by PID (which is a static
priority tie-break), we can tie-break by other
job-level-unique parameters. This is desirable
because tasks are equaly affected by tardiness
since static priority tie-breaks cause tasks
with greater PID values to experience the most
tardiness.

There are four tie-break methods:

1) Lateness.  If two jobs, J_{1,i} and J_{2,j} of
tasks T_1 and T_2, respectively, have equal
deadlines, we favor the job of the task that
had the worst lateness for jobs J_{1,i-1} and
J_{2,j-1}.

Note: Unlike tardiness, lateness may be less than
zero. This occurs when a job finishes before its
deadline.

2) Normalized Lateness.  The same as #1, except
lateness is first normalized by each task's
relative deadline.  This prevents tasks with short
relative deadlines and small execution requirements
from always losing tie-breaks.

3) Hash. The job tuple (PID, Job#) is used to
generate a hash. Hash values are then compared.
A job has ~50% chance of winning a tie-break
with respect to another job.

Note: Emperical testing shows that some jobs
can have +/- ~1.5% advantage in tie-breaks.
Linux's built-in hash function is not totally
a uniform hash.

4) PIDs. PID-based tie-break used in prior
versions of Litmus.

Conflicts:

	litmus/edf_common.c

1 files changed, 91 insertions, 19 deletions

diff --git a/litmus/edf_common.c b/litmus/edf_common.c
index 668737f0fbf9..52205df3ea8b 100644
--- a/litmus/edf_common.c
+++ b/litmus/edf_common.c
@@ -14,6 +14,32 @@
 
 #include <litmus/edf_common.h>
 
+#ifdef CONFIG_EDF_TIE_BREAK_LATENESS_NORM
+#include <litmus/fpmath.h>
+#endif
+
+#ifdef CONFIG_EDF_TIE_BREAK_HASH
+#include <linux/hash.h>
+static inline long edf_hash(struct task_struct *t)
+{
+        /* pid is 32 bits, so normally we would shove that into the
+         * upper 32-bits and and put the job number in the bottom
+         * and hash the 64-bit number with hash_64(). Sadly,
+         * in testing, hash_64() doesn't distribute keys were the
+         * upper bits are close together (as would be the case with
+         * pids) and job numbers are equal (as would be the case with
+         * synchronous task sets with all relative deadlines equal).
+         *
+         * A 2006 Linux patch proposed the following solution
+         * (but for some reason it wasn't accepted...).
+         *
+         * At least this workaround works for 32-bit systems as well.
+         */
+        return hash_32(hash_32((u32)tsk_rt(t)->job_params.job_no, 32) ^ t->pid, 32);
+}
+#endif
+
+
 /* edf_higher_prio -  returns true if first has a higher EDF priority
  *                    than second. Deadline ties are broken by PID.
  *
@@ -63,32 +89,78 @@ int edf_higher_prio(struct task_struct* first,
 
 #endif
 
-        /* Determine the task with earliest deadline, with
-         * tie-break logic.
-         */
-        if (unlikely(!is_realtime(second_task))) {
-                return 1;
-        }
-        else if (earlier_deadline(first_task, second_task)) {
-                /* Is the deadline of the first task earlier?
-                 * Then it has higher priority.
-                 */
+        if (earlier_deadline(first_task, second_task)) {
                 return 1;
         }
         else if (get_deadline(first_task) == get_deadline(second_task)) {
-                /* Need to tie break */
-
-                /* Tie break by pid */
-                if (first_task->pid < second_task->pid) {
+                /* Need to tie break. All methods must set pid_break to 0/1 if
+                 * first_task does not have priority over second_task.
+                 */
+                int pid_break;
+                
+                
+#if defined(CONFIG_EDF_TIE_BREAK_LATENESS)
+                /* Tie break by lateness. Jobs with greater lateness get
+                 * priority. This should spread tardiness across all tasks,
+                 * especially in task sets where all tasks have the same
+                 * period and relative deadlines.
+                 */
+                if (get_lateness(first_task) > get_lateness(second_task)) {
                         return 1;
                 }
-                else if (first_task->pid == second_task->pid) {
-                        /* If the PIDs are the same then the task with the
-                         * inherited priority wins.
-                         */
-                        if (!second_task->rt_param.inh_task) {
+                pid_break = (get_lateness(first_task) == get_lateness(second_task));
+                
+                
+#elif defined(CONFIG_EDF_TIE_BREAK_LATENESS_NORM)
+                /* Tie break by lateness, normalized by relative deadline. Jobs with
+                 * greater normalized lateness get priority.
+                 *
+                 * Note: Considered using the algebraically equivalent
+                 *      lateness(first)*relative_deadline(second) >
+                                        lateness(second)*relative_deadline(first)
+                 * to avoid fixed-point math, but values are prone to overflow if inputs
+                 * are on the order of several seconds, even in 64-bit.
+                 */
+                fp_t fnorm = _frac(get_lateness(first_task),
+                                                   get_rt_relative_deadline(first_task));
+                fp_t snorm = _frac(get_lateness(second_task),
+                                                   get_rt_relative_deadline(second_task));
+                if (_gt(fnorm, snorm)) {
+                        return 1;
+                }
+                pid_break = _eq(fnorm, snorm);
+                
+                
+#elif defined(CONFIG_EDF_TIE_BREAK_HASH)
+                /* Tie break by comparing hashs of (pid, job#) tuple.  There should be
+                 * a 50% chance that first_task has a higher priority than second_task.
+                 */
+                long fhash = edf_hash(first_task);
+                long shash = edf_hash(second_task);
+                if (fhash < shash) {
+                        return 1;
+                }
+                pid_break = (fhash == shash);
+#else
+                
+                
+                /* CONFIG_EDF_PID_TIE_BREAK */
+                pid_break = 1; // fall through to tie-break by pid;
+#endif
+
+                /* Tie break by pid */
+                if(pid_break) {
+                        if (first_task->pid < second_task->pid) {
                                 return 1;
                         }
+                        else if (first_task->pid == second_task->pid) {
+                                /* If the PIDs are the same then the task with the
+                                 * inherited priority wins.
+                                 */
+                                if (!second_task->rt_param.inh_task) {
+                                        return 1;
+                                }
+                        }
                 }
         }
         return 0; /* fall-through. prio(second_task) > prio(first_task) */