lec14_aper2_dy0 - Scheduling of Aperiodic and Sporadic Jobs...

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1 Scheduling of Aperiodic and Sporadic Jobs in Priority-Driven Systems - Chapter 7 – (Dynamic Priority Framework) Dynamic priority servers • We focused on fixed priority servers so far. Such servers can be used along with RM. • Now, we will focus on dynamic priority servers. They are used along with EDF
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2 The key idea: a server is given budget u s . Upon the aperiodic request of e , set the server budget e s = e and adjust the relative deadline D s such that e s /D s = u s , i.e., D s = e s /u s . Constant Utilization Server - 1 u s = 50% 3 06 5 16 Rule 1: initialization: e s = 0; and d s = 0; Rule 2: when an aperiodic request e arrives at time t , – A) If the queue is not empty, join the queue –B ) e l s e i f t < d s (d s is the current server deadline), do nothing (just join the queue) –e l s e d s = t + e/u s ; e s = e ; Rule 3: when t becomes equal to d s – A) If the queue is not empty, serve the head of the queue: d s = d s + e/u s ; e s = e . – B) else do nothing. 2 20 7 34 u s = 0.5 u s = 0.5 u s = 0.5 u s = 0.5 Constant Utilization Server - 2 •A t t = 3 , A 1 arrives, e s = ___ , d = _____ t t = 3 , A 1 arrives, e s = 1 , d = 3+1.0/0.25 = 7 t t = 6.9 , A 2 arrives, what should we do? t t = ___, e s = _____, d = ______ t t = 7, e s = 2.0, d = 7 + 2.0/0.25= 15 t t = 15, do nothing. Why? At t = 15.5, e s = 2.0, d = 15.5 + 2.0/0.25 = 23.5 0 3 6 9 12 15 18 21 24 0 7 15 23.5 1.0 2.0 2.0 3.0 6.9 15.5 u s =0.25 T 1 =(0.5, 3) T 2 =(1, 4) T 3 =(4.5, 19) Aperiodic arrivals
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3 Constant Utilization Server - 3 Giving a set of periodic tasks and a set of CUSs, if the total utilization is no more than 1, the system is schedulable. How can we prove the above? – What is the closest result that we know? How did we do it? Can we reuse the method directly? Can we convert this new problem into the old form by transformations? Constant Utilization Server Periodic Job T ij Which Part of this can be Reused? 1 12 1 1 tt ee t pp t t θ ⎢⎥ +> ⎣⎦ What needs to be modified in this figure?… Proof by contradiction : We assume that the total utilization is no more than 1 but J 22 misses its deadline. t θ 1 0 0 We sum up J 22 and all the jobs completed before J 22 . If J 22 misses its deadline t, the sum must be greater than t T 1 T 2 Computational demand of T 1 in [0, t] 2
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4 CUS: Counting the Demands p 0 0 d 1 d 2 d 3 Quiz: what is the assumption to be contradicted? can you explain what is going on here? θ 1 1 1 ) ( ) ( 3 2 1 3 2 1 3 2 1 > + > + + + > + × + + > + × + × + × p e u p e u p d d d p e u d d d p e u d u d u d s s s s s s TBS: Get it done earlier There is an idle time between 14 and 15. Suppose that A 3 arrives at 14 instead. CUS will not do anything until 15 and the CPU still idles, since the CUS already used its share within [7, 15]. One would like to use such idle time to execute earlier and thus achieving a better response time. 0 3 6 9 12 15 18 21 24 0 7 15 23.5 1.0 2..0 2..0 3.0 6.9 15.5 u s =0.25 T 1 =(0.5, 3) T 2 =(1, 4) T 3 =(4.5, 19) Aperiodic arrivals
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5 Total Bandwidth Server - 1 The key idea: same to CUS except it is possible to have an earlier replenishment of the budget. The sequence of deadlines is the same.
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lec14_aper2_dy0 - Scheduling of Aperiodic and Sporadic Jobs...

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