Find Frame Size using Cyclic Real Time Scheduler Problem in Real Time System

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.