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

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.

Recommendations