Fakultät für informatik informatik 12 technische universität dortmund Lab 3: Scheduling Solution - Session 10 - Heiko Falk TU Dortmund Informatik 12 Germany.

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Fakultät für informatik informatik 12 technische universität dortmund Lab 3: Scheduling - Session 10 - Peter Marwedel Heiko Falk TU Dortmund Informatik.

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fakultät für informatik informatik 12 technische universität dortmund Lab 3: Scheduling Solution - Session 10 - Heiko Falk TU Dortmund Informatik 12 Germany

- 2 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 1: Rate Monotonic Scheduling Parameters: p 1 = 5, c 1 = 3p 2 = 8, c 2 = 3 Processor Utilization: μ = Σ ( c i / p i )  n( 2 (1/n) – 1) μ = 3/5 + 3/8 = 39/40 = > 2( 2 (1/2) - 1) =  A process may miss its deadline. i =1 n

- 3 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 1: Rate Monotonic Scheduling Schedule:

- 4 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 2: Aperiodic Scheduling Least Laxity Schedule: Laxity l i = d i - c i

- 5 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 2: Aperiodic Scheduling Earliest Deadline First Schedule:

- 6 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

- 7 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

- 8 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

- 9 - technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2 T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2 T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2 T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 3: Latest Deadline First Scheduling T2 T3 T1 T4 T5 c 1 = 2d 1 = 15 c 2 = 5d 2 = 20 c 3 = 4d 3 = 12 c 4 = 3d 4 = 15 c 5 = 3d 5 = 20 Stack: Execution order: In opposite order! T2T3T5T4T1 Resulting Schedule:

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 4: Resource Constrained Scheduling

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 4: Resource Constrained Scheduling Problems of this Schedule  Task T4 of highest Priority thwarted by T1, T2 and T3:  At time 5, T4 has to wait until T1 frees the printer.  T1 can only free the printer after T2 and T3 finished execution  Despite highest priority, T4 terminates very late at time 34  Priority Inversion!  Ways out: Scheduling with Priority Inheritance

technische universität dortmund fakultät für informatik  h. falk, informatik 12, 2008 TU Dortmund Exercise 4: Resource Constrained Scheduling