Using a cyclic real‐time scheduler, suggest a suitable frame size that can be used to schedule three periodic tasks T1, T2, and T3 with the following characteristics:
Find Frame Size using Cyclic Real Time Scheduler Problem in Real Time System
Answer:
For the given task set, an appropriate frame size is the one that satisfies all the required constraints. Let F be the appropriate frame size.
Constraint 1:
F>= max{execution time of a task}
F>= 30
Constraint 2:
The major cycle M for the given task set T is
M= LCM(100, 80, 150) = 1200
M should be an integral multiple of F. This consideration implies that F can take on the values 30, 40, 60, 100, 120, 150, 200, 300, 400, 600, 1200 and still satisfy Constraint 1.
Constraint 3:
To satisfy this constraint, we need to check whether a selected frame size F satisfies the inequality:
2F – GCD(F, pi) <= di for each pi.
Let us first try for F = 30.
T1: 2*30 – GCD(30, 100) <= 100
⇒ 60 – 10 <= 100
⇒ Satisfied
T2: 2*30 – GCD(30, 80) <= 80
⇒ 60 – 10 <= 80
⇒Satisfied
T3: 2*30 – GCD(30, 150) <= 150
⇒ 60 – 30 <= 150
⇒ Satisfied
Therefore, F=30 is a suitable frame size. We can carry out our computations for the other possible frame sizes, i.e. F = 40, 60, 100, 120, 150, 200, 300, 400, 600, 1200, to find out other possible solutions. However, it is better to choose the shortest frame size.
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