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lys-0071 [83]
2 years ago
15

Consider two processes, p1 and p2, where p1 = 50, t1 = 25, p2 = 75, and t2 = 30.

Mathematics
1 answer:
12345 [234]2 years ago
8 0

Answer:

Given:

p1 = 50,

t1 = 25,

p2 = 75,

t2 = 30

Rate monotic scheduling is a priority scheduling where the shortest job time gets the highest priority. Priority is assigned based on process time. The higher priority will go to process with a short period while the process with a long period will be assigned a low priority because it requires CPU for more time.

Let us say that process P1, has a higher priority than process P2, with the rate monotic scheduler.

Let's take the initial conditions for both P1 & P2.

P1 at time, t = 0 (initial condition)

P2 at time, t = 25

P1 is later scheduled at t = 50,

P2 is scheduled at t = 75.

Let's determine scheduling would be possible.

CPU utilization of P1 = 25/50 = 0.5

CPU utilization of P2 = 30/75 =0.4

Total = 0.4 + 0.5 = 0.9 = 90%

Scheduling is possible as it uses 90% of CPU.

If P1 is assigned a higher priority, P2 will miss its deadline for completion of its CPU burst time at 75.

P1 will not be scheduled in time when it is assigned to a lower priority than P2, i.e P1 won't meet its deadline.

Although the rate monotic scheduler is guaranteed to to schedule the process to its deadline, there is no guarantee the two processes would be scheduled to meet their deadline.

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