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CprE 458/558: Real-Time Systems

CprE 458/558: Real-Time Systems. Mid-term Solutions. Problem 1 (precedence task graph). Do the following in each 1/80 second cycle Validate sensor data ( T v ) Do the 30-Hz avionics tasks, once every six seconds ( T a ) Keyboard input and mode selection ( T a1 )

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CprE 458/558: Real-Time Systems

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  1. CprE 458/558: Real-Time Systems Mid-term Solutions CprE 458/558: Real-Time Systems (G. Manimaran)

  2. Problem 1 (precedence task graph) • Do the following in each 1/80 second cycle • Validate sensor data (Tv) • Do the 30-Hz avionics tasks, once every six seconds (Ta) • Keyboard input and mode selection (Ta1) • Data normalization and coordinate transformation (Ta2) • Tracking reference update (Ta3) • Do the 30-Hz computations, once every six seconds (T30c) • Control laws of the outer pitch-control loop (T30c1) • Control laws of the outer roll-control loop (T30c2) • Control laws of the outer yaw- and collective-control loop (T30c3) • Do the 90-Hz computations once every two cycles, using outputs produced by 30-Hz computations and avionics tasks as inputs: (T90c) • Control laws of the inner pitch-control loop (T90c1) • Control laws of the inner roll- and collective-control loop (T90c2) • Compute the control laws of the inner yaw-control loop, using outputs produced by 90-Hz control-law computations as input. (Ty) • Output commands (To) • Carry out built-in-test (Tt) • Wait until the beginning of the next cycle (Tw) CprE 458/558: Real-Time Systems (G. Manimaran)

  3. Solution 1 Tv Every 6 cycles Every 6 cycles Ta1 Ta1 Ta1 T30c1 T30c2 T30c3 Every 2 cycles Every 2 cycles T90c1 T90c2 Ty To Tt Tw CprE 458/558: Real-Time Systems (G. Manimaran)

  4. Problem 2: Table driven Vs. Static priority CprE 458/558: Real-Time Systems (G. Manimaran)

  5. Problem 3: Task-set schedulable by EDF and not by RMS • One possible solution: • T1 = (30, 60) and T2 = (45, 100) • EDF schedules the task set as the utilization is 0.95 (< 1) • Here is the RMS schedule: T2 misses its deadline T1 T2 T1 T2 0 30 60 90 105 CprE 458/558: Real-Time Systems (G. Manimaran)

  6. Problem 4: Modify task parameters to reflect application specific priorities in RMS (without value) • RMS assigns priorities based on the periods, therefore an important task should have lesser period. In other words, split an important task into a set of sub tasks with each with smaller period and smaller computation time Split into 8 sub-tasks T21…T28 each equal to (0.5, 2) T2 = (2, 8) Split into 16 sub-tasks T31…T316 each equal to (1/16, 1) T3 = (4, 16) Now, according to RMS the priority order is: T1 < ( T21….T28 ) < ( T31……T316 ) Effectively: T1 < T2 < T3 CprE 458/558: Real-Time Systems (G. Manimaran)

  7. Problem 5: Response time order Background Increasing Response time Polling server Deferrable server PE server CprE 458/558: Real-Time Systems (G. Manimaran)

  8. Problem 6: Value Based Schedule EDF Schedule T1 T2 T1 T3 T1 T2 T1 T3 Value = V(T1) + V(T2) + V(T3) = 25 Clairvoyant Schedule T1 T2 T1 T5 T1 T2 T1 T5 Value = V(T1) + V(T2) + V(T5) = 70 Value ratio = 25 / 70 CprE 458/558: Real-Time Systems (G. Manimaran)

  9. Problem 7: Construct RMS schedule recognize blocking times RMS Schedule Direct blocking of T3 by T2 T3 T1 T2 T3 T2 T1 T2 T3 T3 T2 10 11 12 0 2 4 6 7 8 14 16 Priority inversion between T1 by T3 RMS Schedule with Priority Inheritance Protocol Direct blocking of T3 by T2 T3 T1 T2 T3 T2 T3 T1 T3 T2 0 Inheritance blocking of T1 by T2 CprE 458/558: Real-Time Systems (G. Manimaran)

  10. Problem 8: Best values of K and W • When U = E (case 1) • K = 1 and W = 0 • There is no flexibility, therefore having large K and W does not help much • When U = S and N (cases 2 and 3) • K = 1 and W = 0 • There is no possibility of hole creation anyways, therefore is no need of going for larger values • When there are two instances of the resources • Answers do not change for any of the three cases CprE 458/558: Real-Time Systems (G. Manimaran)

  11. Problem 9: Resource Reclaiming (1) • Two situations in which “no reclaiming” is better than all the four resource reclaiming algorithms studied in the class • When aw-ratio = 1, there is no hole to reclaim. The no-reclaiming scheme will avoid unnecessary overhead • When there are a lot of constraints. For example every task needs a single resource in exclusive mode. In this case there will be no holes and hence no-reclaiming will avoid unnecessary over-head CprE 458/558: Real-Time Systems (G. Manimaran)

  12. Problem 9: Resource Reclaiming (2) • Do Basic and Early start algorithms produce a valid post-run schedule? • Yes they do • Basic algorithm exploits only simultaneous holes. Therefore, it cannot result in any violations. • Early algorithm carefully maintains the relative ordering of the tasks and hence cannot result in any violations • Both the algorithms are very conservative in nature. RV algorithm is more exact and also results in a valid post-run schedule CprE 458/558: Real-Time Systems (G. Manimaran)

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