[SOUND] We have defined three requirements for determining the largest possible frame size for a cyclic structured system. It might have looked a little bit complicated, so it's best that we now take an example so we can practice this skill a bit. So here we will show how to determine the largest possible frame size of a system with the tasks T1, T2, and T3 using the parameters shown. And remember that the largest frame size gives us a system with the minimal number of scheduling points, and can still guarantee the timeline for all the tasks. So we see that the longest task is task 3, right? This means that requirement 1 tells us that the frame must be larger or equal to 3. Requirement 2 tells us to list all candidates that divide the hyperperiod evenly. With the task of periods 8, 7, and 5 we have the candidates 8, 7, and 5, and also 4 because it divides 8 evenly. And also 2 because it also divides 8 evenly, and then 1. So what is left here to do is to check whether any of these candidates fulfill requirement 3. And this requires a little bit of more work. We start evaluating the largest frame, that is 8 with requirement 3 for all the tasks. We first check task 1 with period 5, and execution time 1 and deadline 5. We then check tasks 2 and 3 with their respective periods and execution times. With a frame equals 8, 2 times f equals 16, and the gcd of the period 5 and the frame 8 equals 1. If you don't want to calculate the greatest common divisor by hand you can use a calculator or MATLAB. We then have 16- 1 should be less or equal to 5, which is not true. f equals 8 is therefore too large. Now we move on to second largest frame size to see if this one would work. We check again 2f- gcd of 5 and 7. This turns out to be 13, which is not less or equal to 5.. So choosing 7 as the frame size does not work either. We then go down to f equals 5 and evaluate the same thing again. We check 2f- gcd of 5 and 5 which is 5. 5 is less or equals to 5, so a frame size of 5 would work for task 1. But when doing the same test again for task 2, we see that 10- 1 is not less or equal to 7. So a frame size of 5 does not work either. We now check a frame size of 4. So 2f- gcd of 5 and 4 is 7, which is not less or equal to 5. So even a frame size of 4 is too large. So we must evaluate f equals 2 in the same way. For task 1, 2f- gcd of 5 and 2 is 3, which is less than 5. For task 2, 2f- gcd of 7, which is a period of task 2, and 2 is 4. This is also less or equals to 7, so f equals 2 is okay for task 2. We finally check task 3 with 2f- gcd of 8 and 2, which is 2. And this is less than 8, so a frame size of 2 fulfills requirement 3. So wait a minute, didn't requirement 1 tell us that the frame must be equal or larger than the largest task? Well, yes it did, and f equals 2 does not fulfill this requirement. In this case, jobs from T3, which had an execution time of 3, must be split into two parts, J3 one and J3 two. Either J3 one has to have an execution time of 2, and J3 two an execution time of 1, or vice versa, or three jobs with an execution time of 1. And after this, a static schedule can be made in this frame. What we did in this lesson was to determine the largest possible frame size for a set of tasks, based on three requirements we learned in another lesson. The idea here was to have an as large frame size as possible, because it minimizes the context switches. But at the same time, we have to guarantee the feasibility for our tasks. This we showed with an example.