Real-Time Architectures 2004/2005

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1 Real-Time Architectures 2004/2005 Scheduling Analysis I Introduction & Basic scheduling analysis Reinder J. Bril

2 Overview Algorithm and problem classes Simple, periodic taskset problem statement feasibility criteria (bounds) Basic analysis of example algorithms Rate Monotonic Scheduling Earliest Deadline First 2

3 Overview Algorithm and problem classes Simple, periodic taskset problem statement feasibility criteria (bounds) Basic analysis of example algorithms Rate Monotonic Scheduling Earliest Deadline First 3

4 Issues, questions of interest Relevant properties of algorithms cost functions, comparison, classification, optimality When to apply what algorithm criteria system types, parameters assumptions on execution environment OS & platform additional requirements (e.g. behaviour under overload) how to prove properties of a (method, taskset) combination static and dynamic tests to demonstrate feasibility 4

5 Properties of the task set System types periodic, sporadic tasks fixed/dynamic parameters deadline within period precedence relations & preemptability Criticality mix hard, firm and soft hard and firm: acceptance test overload possibilities Number and type of resources (e.g. processors) Modes: subsets of tasks statically defined / dynamically created 5

6 Priority static/dynamic Algorithm classes fixed priority: priority of job fixed dynamic task, fixed job priority dynamic Preemptive Online / offline online: e.g. admission/acceptance computation (guarantee), assignment of priorities offline precomputation of a table complex optimizations possible Cost functions e.g. maximum lateness,... 6

7 Overview Algorithm and problem classes Simple, periodic taskset problem statement feasibility criteria (bounds) Basic analysis of example algorithms Rate Monotonic Scheduling Earliest Deadline First 7

8 Problem System periodic, preemtable taskset, Z fixed parameters deadline equal to period non-blocking no precedence relations single processor Find one or more algorithms that work and for such an algorithm a criterion for feasibility an analysis method (and proof) a comparison with other algorithms 8

9 Feasibility criteria sufficient condition (easy to check) exact boundary feasible tasksets infeasible tasksets 9

10 Utilization criterion Recall U j = C j / T j utilization for task j U = U j total utilization Clearly, for U>1 the set is not schedulable by any algorithm (overload) proof: the amount of computation time in a hyperperiod H is the number of times each task releases a job times the computation time of that job, hence H C j *H/T j = H* C j / T j = H * U with U>1, the latter term exceeds H which is a contradiction for U=0 the set is schedulable (by any algorithm) A given utilization factor can be decreased by decreasing computation times increasing periods 10

11 Bounds Assume the scheduling algorithm is independent of computation times C i. These then can be varied to change the utilization factor. There is a utilization factor, dependent on algorithm and taskset such that computation times cannot be increased anymore without destroying feasibility U ub (Z, Alg) -- increase computation times to the limit Minimizing over tasksets gives the least utilization bound for the algorithm U lub (Alg) = (min Z:: U ub (Z, Alg)) Meaning: for tasksets with utilizations below U lub the algorithm will produce a feasible schedule U U lub : acceptance criterion ( sufficient condition ) 11

12 Feasibility criteria U = 1 U lub Z U U lub increasing computation times till U ub infeasible tasksets 12

13 Overview Algorithm and problem classes Simple, periodic taskset problem statement feasibility criteria (bounds) Basic analysis of example algorithms Rate Monotonic Scheduling Earliest Deadline First 13

14 Algorithm Rate Monotonic Scheduling fixed task priorities: higher priorities for shorter periods preemptive online algorithm Advantages of RMS simple, fixed priority assignment in-depth analysis available OS support deals reasonable with overload conditions NOTE: for now, assume tasks are sorted in order of increasing period 14

15 Questions U lub (RMS)? RMS (cnt d) U lub (RMS) maximal in some sense? what if U lub < U 1? what about the response time of RMS? Not in this slide set. 15

16 Example: utilization and schedulability Two tasks (written as (C,T)) (3,6) and (4,9) RMS yields deadline miss at t = 9 U = 51/54 = (2,4) and (4,8) is feasible with RMS U = 1 16

17 Utilization bound for RMS U lub (RMS) = n(2 1/n -1), n tasks result due to Liu & Layland (called LL-bound) converges to ln(2) ( 0.69) hence, in fact U lub (RMS) = ln(2) [independent of taskset] 17

18 Example: utilization bound Task set Z consisting of 3 tasks: Task Period T j Execution time C j Utilization U j τ τ τ Notes: U = , hence Z couldbe schedulable; U 1 +U 2 = 0.88 > LL(2) 0.83, therefore U > LL(3), hence another test required. 18

19 Optimality RMS is optimal among all fixed priority preemptive algorithms if a taskset can be scheduled feasibly with any fpp algorithm it can so with RMS 19

20 Algorithm Earliest Deadline First dynamic task priorities; job with nearest absolute deadline gets highest priority (fixed job priorities) preemptive minimize maximum lateness online algorithm Advantages optimal algorithm 20

21 Disadvantages Earliest Deadline First needs priority queue for storing deadlines logarithmic access needs dynamic priorities typically no OS support behaves badly under overload less predictable than RMA (which tasks will miss deadlines?) a jobs that missed its deadline is allowed to continue (causing a domino-effect of missed deadlines) difficult to handle relative importance 21

22 Schedulability EDF (cnt d) U lub (EDF) = 1, i.e. EDF can schedule a set Z of tasks if and only if U 1. Proof (see book Buttazzo): U > 1: obvious U 1: by means of a contradiction argument; see next slide 22

.....by instances with [a,d] entirely in [t 1, t 2 ] (EDF!

23 U lub (EDF) = 1 Suppose not task set with U 1 and not schedulable by EDF Choose [t 1, t 2 ] at t 2 overflow occurs t 1 such that: from t 1 onwards continuous utilization......by instances with [a,d] entirely in [t 1, t 2 ] (EDF!) hence t 1 is release time of some instance and job experiencing the overload is released in [t 1, t 2 ] Define C p (t 1, t 2 ) = ( i,k: t 1 a i,k, d i, k t 2 : C i ) the total amount of computation in [t 1, t 2 ] 23

24 U lub (EDF) = 1 C p (t 1, t 2 ) = i (t 2 t 1 ϕ i )/T i C i i (t 2 t 1 )/T i C i i ((t 2 t 1 )/T i ) C i (t 2 t 1 )U Overflow at t 2, hence: (t 2 t 1 ) < C p (t 1, t 2 ) Therefore (t 2 t 1 ) < (t 2 t 1 )U, and U > 1. 24

25 EDF (cnt d) Optimality (w.r.t. utilization): EDF is optimal among all dynamic priority preemptive algorithms; EDF is an optimal algorithm. 25

26 Example for RMA and EDF C 1 = 2, T 1 = 5, C 2 = 4, T 2 = 7; U = 2/5 + 4/7 = 34/ Schedulable under EDF, not under RMA. Number of preemptions of task 2: RMA: 5; EDF: 1!

Introduction to Embedded Systems

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