REAL TIME BASICS

1. RTOS
r.k.tiwary

2. • At time 1, J2 completes and, hence, J3 becomes ready. J3
is placed in the priority queue
ahead 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 priority
among 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 ready
for execution. Consequently, processor P2 becomes idle.
• J7 finally completes at time 8. J6 and J8 can now be
scheduled and they are.

4. J1 J2 J3 J4
3 4 6 8
J7 J5 J6 J8
4 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).