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Example

Example. Structured Cyclic Schedule. Question: The following system of periodic tasks is to be scheduled and executed according to a structured cyclic schedule with fixed frame size. T1 =(4, 1), T2=(5,1.8), T3=(20,1) and T4=( 20,2).

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Example

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  1. Example Structured Cyclic Schedule

  2. Question: The following system of periodic tasks is to be scheduled and executed according to a structured cyclic schedule with fixed frame size. T1=(4, 1), T2=(5,1.8), T3=(20,1) and T4=(20,2). • What is the appropriate frame size?( Show your work including LCM and GCD calculation ) • What do the resulting cyclic scheduler tables look like (show the first 4 frame only)?

  3. What is the appropriate frame size? Eq.1: Jobs to start and complete execution within a single frame: f ≥ max(e1, e2, …, en) f≥max(1,1.8,1,2) thus: f≥ 2

  4. Eq.2: The hyper-period should be an integer multiple of the frame size (f divides evenly into the period of at least one task): [pi/f]- pi/f =0

  5. How to approach equation 2: • Calculate the hyper-period = lcm (4,5,20,20) = 20 • Pick the smallest period and largest period of your tasks. (in this case 2 and 20) • Thus your possible value frame (f) values are 2, 20 and any number in between 2 and 20 which divides evenly with the hyper- period of the system. Dividing the hyper- period which is 20 with all the values in the range of 2 to 20, you shall found that 2,4,5,10 and 20 are the only number which divides evenly with 20 and the period of at least one task. Thus, your possible f value as per equation 2 is concern is: f {2, 4,5,10,20}

  6. Eq.3: At least one frame boundary between release time of a job and its deadline: 2*f – gcd(pi, f ) ≤ Di for i= 1, 2, …, n

  7. Let us try frame size 2 • 2 x 2 – gcd(4,2) ≤ 4 ≡ 4 – 2 ≤ 4TRUE • 2 x 2 – gcd(5,2) ≤ 5 ≡ 4 – 1 ≤ 5 TRUE • 2 x 2 – gcd (20,2) ≤ 20 ≡ 4 – 2 ≤ 20 TRUE • 2 x 2 – gcd (20,2) ≤ 20 ≡ 4 – 2 ≤ 20 TRUE Let us try frame size 4 • 2 x 4 – gcd (4,4) ≤ 4 ≡ 8 – 4 ≤ 4 TRUE • 2 x 4 – gcd (5,4) ≤ 5 ≡ 8 – 1 ≤ 5 FALSE Let us try frame size 5 • 2 x 5 – gcd (4,5) ≤ 4 ≡ 10 – 1 ≤ 4 FALSE Let us try frame size 10 • 2 x 10 – gcd (4,10) ≤ 4 ≡ 20 – 2 ≤ 4 FALSE Let us try frame size 20 • 2 x 20 – gcd (4,20) ≤ 4 ≡ 40 – 4 ≤ 4 FALSE

  8. The other possible values for f( as in eq 2) fails to fulfill equation 3. Thus, only the frame size 2 is suitable for scheduling.

  9. What do the resulting cyclic scheduler tables look like (show the first 4 frame only)

  10. Even though in this example we could successfully find a suitable frame size that satisfies all the three equations, it is quite probable that a suitable frame size may not exist for many problems. In such cases, to find a feasible frame size we have to split the task ( or a few tasks) that is (are) causing violation of the equations into smaller sub-tasks that can be scheduled in different frames.

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