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scheduler: simplify priority policy
Remove overdueness. User must submit calibration experiments with higher priority values for them to take precedence.
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b0f8141018
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@ -138,8 +138,8 @@ def _show_schedule(schedule):
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clear_screen()
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if schedule:
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l = sorted(schedule.items(),
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key=lambda x: (x[1]["due_date"] or 0,
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-x[1]["priority"],
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key=lambda x: (-x[1]["priority"],
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x[1]["due_date"] or 0,
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x[0]))
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table = PrettyTable(["RID", "Pipeline", " Status ", "Prio",
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"Due date", "File", "Experiment", "Arguments"])
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@ -17,8 +17,8 @@ class _ScheduleModel(DictSyncModel):
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parent, init)
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def sort_key(self, k, v):
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# order by due date, and then by priority and RID
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return (v["due_date"] or 0, -v["priority"], k)
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# order by priority, and then by due date and RID
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return (-v["priority"], v["due_date"] or 0, k)
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def convert(self, k, v, column):
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if column == 0:
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@ -81,14 +81,16 @@ class Run:
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self._notifier[self.rid]["status"] = self._status.name
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# The run with the largest priority_key is to be scheduled first
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def priority_key(self, now):
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def priority_key(self, now=None):
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if self.due_date is None:
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overdue = 0
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due_date_k = 0
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else:
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overdue = int(now > self.due_date)
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due_date_k = -self.due_date
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return (overdue, self.priority, due_date_k, -self.rid)
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if now is not None and self.due_date is not None:
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runnable = int(now > self.due_date)
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else:
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runnable = 1
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return (runnable, self.priority, due_date_k, -self.rid)
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@asyncio.coroutine
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def close(self):
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@ -240,10 +242,9 @@ class RunStage(TaskObject):
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next_irun = asyncio_queue_peek(self.inq)
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except asyncio.QueueEmpty:
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next_irun = None
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now = time()
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if not stack or (
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next_irun is not None and
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next_irun.priority_key(now) > stack[-1].priority_key(now)):
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next_irun.priority_key() > stack[-1].priority_key()):
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stack.append((yield from self.inq.get()))
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run = stack.pop()
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