4. A real-time system has three periodic events with periods 40, 80 and 120 ms E

Question

4. A real-time system has three periodic events with periods 40, 80 and 120 ms Explore and show the schedulability if each of the three kinds of events requires 15 ms of processing time. Assuming all events take equal time what is maximum time available for processing the events beyond which the system shall not be schedulable 5. Suppose we have a real-time system which works satisfactorily with events happening at 30 and 50 units of time which require processing time of 10 and 5 units of time. Suppose there is a need to expand the scope of operations and one more event to be service. What is the maximum time that can be assigned to the new event without jeopardizing schedulability 6. Consider the following set of tasks t, t2, t3 having priorities ti>t2 13. Consider the following schedule a. t3 starts at time Osec, executes for 1sec, requests resource R and then runs for 3 secs and completes b. starts at time 2sec, runs for sec, then requests resource R, and then runs for 4 secs and completes c. t2 starts at tie 3sec and runs for 5 secs Draw the schedule using rate monotonic scheduling and show where the priority inversion occurs. Also use the technique used to avoid priority inversion and draw the new schedule

Explanation / Answer

A real time system is different from different from other applications or processes in the sense it is mainly concerned with the objective of competing the task before their deadlines. The scheduling for real time can be static table driven or priority approaches and dynamica approaches. Overall the scheduling can be looked over to be composed of two parts, one is clock driven and the other is table driven.

4. Here, the periods are given as 40, 80, 120 ms. Using table driven schedule we create a schedule which will reapeat itself after every major cycle. The size of the major cycle is given by the LCM of the individual periods of the different events. Therefore the maximum possible time is equal to the size of the schedule which is:

LCM(40,80,120) = 240 ms

5. Here we have two processes where with periods p1=30, p2=50. Their processing time is respectively pr1=10, pr2=5. Now finding the cpu utilization for the existing tasks we get;

= pr1/p1 + pr2/p2

= 10/30 + 5/50

= 1/3 + 1/10

= 0.433

The value is less than 1 , which can be shown as

0.433 <= 1

Therefore a third event 'x' can be added as:

0.433 + x <= 1

=> x = 1-0.433

= 0.567

=> pr3/p3 = 0.567

=> pr3 = 0.567 X p3 , where p3 is the period of third event.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.