RTOS

r.k.tiwary

• At time 1, J2 completes and, hence, J3 becomes ready. J3 is placed in the priority queueahead of J7 and is scheduled on P2, the processor freed by J2.• At time 3, both J1 and J3 complete. J5 is still not released. J4 and J7 are scheduled.• At time 4, J5 is released. Now there are three ready jobs. J7 has the lowest priorityamong them. Consequently, it is preempted. J4 and J5 have the processors.• At time 5, J4 completes. J7 resumes on processor P1.• At time 6, J5 completes. Because J7 is not yet completed, both J6 and J8 are not readyfor execution. Consequently, processor P2 becomes idle.• J7 finally completes at time 8. J6 and J8 can now be scheduled and they are.

J1 J2 J3 J43 4 6 8

J7 J5 J6 J84 6 10 11

As an example, a partition and assignment of the jobs in Figure put J1, J2, J3,and J4 on P1 and the remaining jobs on P2. The priority list is segmented into two parts:(J1, J2, J3, J4) and (J5, J6, J7, J8).