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Real Time Systems




                    Notes
                                                 Figure 7.5: Non-optimality of the EDF and LST Algorithm



























                                   Source:  ansari.szabist-isb.edu.pk/RTS/RTS0508.ppt
                                   The system shown in the figure has three independent, non-preemptible jobs J1, J2 and J3. The
                                   explanation of the figure is as follows:
                                      The release times are 0, 2 and 4 and execution times are 10, 14 and 12.
                                      (a) shows the schedule produced by the EDF algorithm
                                      When J1 completes at time 3, J2 has already been released but not J3. Hence J2 is scheduled.

                                      When J3 is released at time 4, J2 is executing.
                                      Even though J3 has an earlier deadline and hence a higher priority, it must wait until J2
                                       completes s there is no preemption

                                      Thus J3 misses its deadline.
                                      (b) provides a feasible schedule.
                                      At time 3 when J1 completes, the processor is left idle, even though J2 is ready for execution.

                                      When J3 is released at 4, it can be scheduled ahead of J2, allowing both to meet deadline.
                                      (b) Can never be provided by a priority-driven scheduling algorithm as they never leave
                                       a processor idle when jobs are ready.

                                      This figure shows that the EDF algorithm is not optimal for scheduling preemptible jobs
                                       on more than one processor.
                                      J1, J2 and J3 have execution times 1, 1, 5 and deadlines 1, 2, 5 respectively and release
                                       time 0.
                                      On two processors, J1 and J2 are scheduled on the processors at time 0 because of higher
                                       priority so J3 misses deadline.

                                      An algorithm that could assign higher priority to J3 could do the job. (like LST)







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