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Q uantitative E valuation of E mbedded S ystems.

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Presentation on theme: "Q uantitative E valuation of E mbedded S ystems."— Presentation transcript:

1 Q uantitative E valuation of E mbedded S ystems

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4 TDMA in a cyber physical system: preparation for the SDF3 assignment

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10 Period = P Slice = S Task = T P-S TP/S

11 Period = P Slice = S Task = T P-S q qT = rS S/q rr P-(S/q) 0 NEW!! Publication under submission... RTAS 2014 NEW!! Publication under submission... RTAS 2014

12 Comp. Inner control Physical World Comp. Emergency detection Comp. Image processing

13 Comp.1 Inner control Comp.2 Emergency detection Comp.3 Image processing

14 Comp.1 Inner control Comp.2 Emergency detection Comp.3 Image processing p2 p1 p3 p4 p5 p6 p7 p8 p9 p10 p11 p12 p13 p14 p15 p16 p17 p18

15 Comp. Inner control Comp. Emergency detection Comp. Image processing

16 Comp. Inner control Comp. Emergency detection Comp. Image processing

17 Comp.1 Inner control Comp.2 Emergency detection Comp.3 Image processing

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20 Latency 1 L2 L3

21 Period = P Slice = S Task = T P-S q qT = rS S/q rr P-(S/q) 0

22 Period = P Slice = S Task = T qT = rS One packet per slice in the network, therefore T = S = 0.01 ms One slice per node in the network But... each node schedules its routing in a TDMA fashion as well... So for each hop P = (C+1)*N*S where C is the number of connections and N is the total number of nodes in the network.

23 Period = P Slice = S Task = T qT = rS Three computations = three slices, so P = S1 + S2 + S3 Task times may be bigger than slice times! T1 = 0.5 ms T2 = 3 ms T3 = 7 ms It is part of the assignment to figure out how P should be chosen and divided over S1,S2 and S3.

24 Sensor 1 and 2 produce 1 packet every 2 ms Sensor 3 produces 50 packets every 100 ms Sensor 4 produces 10 packets every 20 ms Computation 1 needs 1 packet from sensor 1 and 2, and produces 1 packet for computation 2 and one for actor 1 Computation 2 takes 50 packets from computation 1 and 1 packet from computation 3 and produces 1 for actors 2 and 3 and for computation 3. Computation 3 takes 50 packets from sensor 3, 50 from sensor 4 and 1 packet from compation 2 and produces 1 for computation 2.

25 Sensor1 Sensor2 Sensor3 Sensor4 Actuator1 Actuator2 Actuator3 hop1hop2 hop3 hop4 hop5 hop6 hop7 hop8 hop9 hop10 hop11hop12 hop13 hop14 hop15 Comp1 Comp2 Comp ms 100 ms 20 ms


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