Lecture 5/6. Enrico Bini

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1 Advanced Lecture 5/6 November 8, 212

2 Outline 1 2 3

3 Theorem (Lemma 3 in [?]) Space of feasible C i N is EDF-schedulable if and only if i U i 1 and: t D n i=1 { max, t + Ti D i T i } C i t with D = {d i,k : d i,k = kt i + D i, i N, k N, d i,k D } and D = lcm(t 1,...,T n ) + max i {D i }. H = lcm(t 1,...,T n ) is often called the hyperperiod. Since we are investigating the space of C i, no smaller D can be considered. For given T i and D i, the space of EDF-schedulable C i is convex!

C i t with D = {d i,k : d i,k = kt i + D i, i N, k N, d i,k D } and D = lcm(t 1,...,T n ) + max i {D i }.

4 An example Let us assume T 1 = 4,D 1 = 5 and T 2 = 6,D 2 = 5. D = {5,9,11,13,17} Hence equations are [ U1 U 2 ]

5 An example: viewing the C i 1 1

6 Reducing the set of deadlines Hence only the equations [ U1 U 2 ] are needed to be checked, corresponding to D = {5,17} Finding a general method to reduce D without losing sufficiency would be an interesting contribution.

7 Optimal comp time Being the exact region convex the optimal assignment of computation times C i does not present big difficulties.

8 Outline 1 2 3

9 Deadline space: qualitative analysis Example: n = 2, T 1 = T 2, U 1 + U 2 1 We can reduce D 1 to C 1, but then D 2 = C 1 + C 2, or D 2 = C 2, but then D 1 = C 1 + C 2 convex comb of these two points are not EDF-sched Let us assume T 1 = T 2 =... = T n, and i U 1 1. For any permutation p : N N, a vertex has coordinates i D p(i) = j=1 C p(j)

convex comb of these two points are not EDF-sched Let us assume T 1 = T 2 =.

10 Another exact test Being the and deadlines into the set D and into the, dbf condition is unfit to show the space of feasible /deadline. Theorem The task set N is EDF-schedulable if and only if: n k N n \ {} i I k C j k j (k i 1)T i + D i j=1 where I k = {j : k j } is the set of non-zero indexes in k.

$Theorem The task set N is EDF-schedulable if and only if: n k N n \ {} i I$

11 EDF-schedulablity as covering problem k N n \ {} i : k i n C j k j (k i 1)T i + D i j=1 can be seen as the problem of finding a cover to N n \ {} with halfspaces of equations (C i T i )k i + j i C j k j T i + D i considering (k 1,...,k n ) as variables. Remember: i : k i, not just i N If U = 1 linearly dependent.

halfspaces of equations (C i T i )k i + j i C j k j T i + D i considering (k

12 Viewing the covering problem If T 1 = D 1 = 4, C 1 = 2, and T 2 = D 2 = 8, C 2 = 3, then have to cover N 2 \ {(,)} 2k 1 + 3k 2 2k 1 5k 2 5 k 2 i = i = 2 k 1 If U > 1 cannot cover, ever.

$2 \ {(,)} 2k 1 + 3k 2 2k 1 5k 2 5 k 2 i = 1 4 3 2 1$

13 Changing deadlines Deadline D i appears at the RHS of the i-th inequality only Changing D i means to translate boundary of the i-th halfspace 5 k 2 5 k D min 1 = 2 (= C 1 ) k D 2 = 5 k 1

boundary of the i-th halfspace 5 k 2 5 k 2 4 4 3 3 2 2 1 1

14 Space of feasible D i D D

15 Changing the The i-th period has an effect only on the i-th equation however T i appears in the coefficients and at the RHS changing T i is a rotation of the i-th halfspace the points in common between two equations with T i and T i are (C i T i )k i + C j k j = T i + D i j i (C i T i )k i + C j k j = T i + D i j i k i = 1 C j k j = D i C i j i

halfspace the points in common between two equations with T i and T i are (C i T i )k

16 Changing (n = 2) As T 1 changes, the 1st boundary rotates around k 1 = 1, k 2 = D 1 C 1 C 2 As T 2 changes, the 2nd boundary rotates around k 1 = D 2 C 2 C 1, k 2 = 1

17 Changing : example 5 k 2 C 1 = 2, D 1 = 4, and C 2 = 5, D 2 = 8 changing T 1 rotates around (1,.4) changing T 2 rotates around (1.5,1) 5 k 2 5 k k k k 1 T 1 = 27 T 1 = 8 T 1 = 6 T 2 = 27/5 T 2 = 2/3 T 2 = 15/2 Draw the f-space

5,1) 5 k 2 5 k 2 4 4 4 3 3 3 2 2 2 1 1 2 3 4 5 6 7 8 k 1 1 1 2 3 4 5 6 7 8

18 Space of deadlines: sufficient test By making some more restrictive hypothesis, we can derive the following sufficient EDF-test Theorem (Chantem et al. [?]) A task set is EDF-schedulable if: { i, D i T i + D min n i=1 U i D i n i=1 C i D min (1 n i=1 U i). with D min = min i D i. This condition has the advantage of being convex in the space of deadlines.

]) A task set is EDF-schedulable if: { i, D i T i + D min n i=1 U i D i n i=1 C i D min

19 Viewing sufficient cond If C 1 = 2, T 1 = 4 and C 2 = 4, T 2 = 7, we have D D 2

20 Outline 1 2 3

21 Optimal T i, D i assignment X = {T 1,D 1,...,T n,d n } Typical problem in control systems: given the execution times {C 1,..., C n } choose the sampling and deadlines (=delays) of all controllers such that a control cost (LQG?) is minimized min J(T 1,D 1,...,T n,d n ) T 1,D 1,...,T n,d n s.t N is EDF-schedulable

22 Difficulty of the problem We could try to find optimal T i and then D i 1 Suppose C 1 = C 2 = 1 2 Suppose that the optimal T i are T 1 = 2.1, T 2 = How much can we reduce the deadlines? very little. 4 However by choosing non-optimal T 1 = T 2 = 2, we can actually reduce one of the two deadlines by 1, possibly making the cost J smaller To best of my knowledge it is still unsolved (due to bizarre shape of the feasible region) solving this problem optimally would be a decent contribution

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