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Applications

Applications. CS 5270 Lecture 7. Outline. The Bounded Retransmission Protocol. The TTA model The verification issues Task arrival patterns and their schedulability. Periodic, aperiodic, sporadic tasks. More sophisticated patterns captured by timed automata.

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Applications

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  1. Applications CS 5270Lecture 7 Lecture 7

  2. Outline • The Bounded Retransmission Protocol. • The TTA model • The verification issues • Task arrival patterns and their schedulability. • Periodic, aperiodic, sporadic tasks. • More sophisticated patterns captured by timed automata. • Timed automata can also be used for schedulabilty analysis ! • Introduction to Regional Automata Lecture 7

  3. Case Studies • Available from the UPPAAL home page (“Examples”). • Bang & Olufsen Audio/Video Protocol. • Bang & Olufsen Power Down Protocol. • Commercial Field Bus Protocol. • Gear Box Controller. • Multimedia Stream. Lecture 7

  4. BRP • Bounded Retransmission Protocol (BRP). • Developed by Phillips Electronics Corporation. • A real-time bounded variant of the alternating-bit protocol. • Used to transfer in burst-mode a list of data (a file) • via an infra-red communication medium between AV equipment and a remote control unit. Lecture 7

  5. BRP • The medium is lossy! • The file is transmitted in chunks. • If an acknowledgment for a sent-chunk is not received “in time” the chunk is retransmitted. • If the number of retransmissions for the same chunk exceed a bound then the transmission is aborted. Lecture 7

  6. BRP • Timing aspects: • The sender has a timer to decide when to retransmit a chunk. • The receiver has a timer to detect when a transmission has been aborted by the sender. Lecture 7

  7. Sin Rout Sout Sender Receiver G F K B L A Lecture 7

  8. (d1, d2, ,,,,dn) ; a file consisting of n chunks of data. Sin Rout Sout Sender Receiver G F K B L A Lecture 7

  9. {IOK, INOK, IDK } Sin Rout Sout Sender Receiver G F K B L A Lecture 7

  10. The values of Sout • IOK • All the acknowledgments were received. • All the chunks were transmitted successfully and were received by the receiver. • INOK • Some ack. failed to arrive in time ; the MAX count of retransmissions for that chunk was exhausted without receiving an ack. • IDK • The ack. Were received for all the chunks except the last one. • Don’t know whether the transmission was successful or not. • This is due to asynchronous communication via a lossy channel. • Byzantine agreement is impossible! Lecture 7

  11. (e1, i1) (e2, i2) ….(ek, ik) Sin Rout Sout Sender Receiver G F K B L A Lecture 7

  12. (e1, i1) (e2, i2) ….(ek, ik) (d1, d2, ,,,,dn) Sin Rout Sout Sender Receiver G F K B L A Lecture 7

  13. Rout • (e1, i1) (e2, i2)……. (ek, ik) • 0 ≤ k ≤ n • ij2{IFST, IINC, IOK, INOK }, 0 < j ≤ k • IFST--- The first chunk of the file but not the last one. • IOK --- The last chunk of the file. • IINC --- For all other chunks. • INOK ---- Something has gone wrong. • In this case j = k and ek = * (no datum). Lecture 7

  14. The Specification • (ej, ij) • For every 0 < j ≤ k, if ij INOK then ej = dj • The datum delivered is the chunk that was sent. • If n > 1 then i1 = IFST • INOK is put out only if something at all was received. • If 1 < j < k then ij = IINC Lecture 7

  15. The Specification • ik = IOK OR ik = INOK • The last output must signal positive or negative termination. • ik = IOK implies k = n. • Successful transmission. • ik = INOK implies k > 1. • Unsuccessful only if something was received to start with. Lecture 7

  16. The Specification • If Sout = IOK then ik = IOK. • Should we demand the converse too? • If Sout = INOK then ik = INOK • If Sout = IDK then k = n. • ik = ? • If k = 0 then • Sout = IDK iff n = 1. • Sout = INOK iff n > 1. Lecture 7

  17. IOK a a b b c c (a, FST) (b, INC) (C, OK) Lecture 7

  18. ? a a b b c c (a, FST) (b, INC) (C, OK) Lecture 7

  19. IDK a a b b c c (a, FST) (b, INC) (C, OK) Lecture 7

  20. ? a a b b ? Lecture 7

  21. INOK a a b b (a, FST) (b, INC) (NOK, *) Lecture 7

  22. INOK a a b (a, FST) (NOK, *) Lecture 7

  23. INOK a Lecture 7

  24. The Sender Module • S reads the file (with n chunks d1, d2,…, dn) and sets the retry counter to 0. • It then starts sending over the chunks one by one: • Its sets a timer T1 and the first frame into the channel K. A frame is of the form (b1, b2, ab, d). • b1 (b2) indicates whether or not this chunk is the first (last) one. ab is the alternating bit. d is the chunk. • ab is used to distinguish between a retry and a fresh chunk. Lecture 7

  25. The Sender Module • After sending the frame (b1, b2, ab, d), the sender module waits for an acknowledgment or a time-out. • If an ack. is received in time then T1 is reset. • The next frame (b1’, b2’, 1-ab, d’) is sent or (if b2 = 1 in the previous round), it signals Rout = IOK. • If it times out, the frame (b1, b2, ab, d) is resent after resetting the timer and incrementing the retry counter. • If MAX is exceeded in the process of incrementing the counter, the transmission is broken off; it signals Rout = INOK or Rout = IDK depending on n and how many ack messages were received. Lecture 7

  26. ab : = 0 idle Lecture 7

  27. ab : = 0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 Lecture 7

  28. ab : = 0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 ! (i = 1, i = n, ab, di) rc : = 0 Lecture 7

  29. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 (i = 1, i = n, ab, di) rc:=0 Lecture 7

  30. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 (x < T1)  (?ack) x := 0; ab := 1 -ab Lecture 7

  31. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 i < n i := i + 1 (x < T1)  (?ack) x := 0; ab := 1 -ab x ≤ 0 Lecture 7

  32. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 i < n i := i + 1 (x < T1)  (?ack) x := 0; ab := 1 -ab i = n ! IOK x ≤ 0 Lecture 7

  33. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 i < n i := i + 1 i = n ! IOK (x < T1)  (?ack) x := 0; ab := 1 -ab x ≤ 0 (rc = MAX), x = T1, i < n ! INOK ; x := 0 Lecture 7

  34. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 i < n i := i + 1 i = n ! IOK (x < T1)  (?ack) x := 0; ab := 1 -ab x ≤ 0 (rc = MAX), x = T1, i < n ! INOK ; x := 0 (rc = MAX), x = T1, i = n ! IDK ; x := 0 Lecture 7

  35. (rc < MAX )  (x = T1) ! (i = 1, i = n, ab, di) rc:= rc + 1; x : = 0 ab : = 0 (i = 1, i = n, ab, di) rc:=0 ? (d1, d2,..,dn) i := 1; x := 0 idle x ≤ 0 x ≤ T1 i < n i := i + 1 i = n ! IOK x = SYNCH ab := 0, (x < T1)  (?ack) x := 0; ab := 1 -ab x ≤ 0 (rc = MAX), x = T1, i < n ! INOK ; x := 0 (rc = MAX), x = T1, i = n ! IDK ; x := 0 x ≤ SYNCH Lecture 7

  36. The Receiver Module ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 exp_ab := rab Lecture 7

  37. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab =exp_ab , rb2 =1 !(IOK, d) exp_ab := rab w≤0 Lecture 7

  38. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 Lecture 7

  39. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) Lecture 7

  40. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 w≤0 rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) !ack; exp_ab = 1 – exp_ab z:=0 Lecture 7

  41. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 w≤0 z ≤ TR z < TR ?(rb1, rb2, rab, d) w :=0 rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) !ack; exp_ab = 1 – exp_ab z:=0 Lecture 7

  42. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) z = TR rb2 = 1 exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 w≤0 z ≤ TR z < TR ?(rb1, rb2, rab, d) w :=0 rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) !ack; exp_ab = 1 – exp_ab z:=0 Lecture 7

  43. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) z = TR rb2 = 1 exp_ab := rab rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 w≤0 z ≤ TR z < TR ?(rb1, rb2, rab, d) w :=0 rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) !ack; exp_ab = 1 – exp_ab z:=0 Lecture 7

  44. ?(rb1, rb2, rab, d) w≤0 z :=0; w:=0 rab = exp_ab, rb2 =1 !(IOK, d) z = TR rb2 = 0 !(INOK, *) z = TR rb2 = 1 exp_ab := rab z < TR ?(rb1, rb2, rab, d) w :=0 rab = exp_ab, rb1=0, rb2 = 0 !(IINC, d) w≤0 w≤0 z ≤ TR rab = exp_ab, rb1=1, rb2 = 0 !(IFST, d) exp-ab  rab !ack !ack; exp_ab = 1 – exp_ab z:=0 Lecture 7

  45. Verification • Premature time-outs do not occur. • In case of abortion, Sender waits sufficiently long so that the Receiver has reacted to the abortion before starting a new file. Lecture 7

  46. Verification • Using UPPAAL it was determined: • T1 > 2 £TD • TD the transmission delay of the channel. • SYNCH  TR  (2 £ MAX £ T1) + 3 £ TD • Both the verifier and the simulator had to be used! Lecture 7

  47. Task Scheduling Basic Idea: • Classical scheduling • Periodic • Aperiodic • Sporadic • Use timed automata to describe task arrivals. • Some of the control states have tasks associated with them. • Whenever a state is entered, its task is added to the ready queue. Lecture 7

  48. Periodic Task TSK x:=0 x = T Lecture 7

  49. Periodic Task Set TSK1 TSK2 TSK3 x:=0 x = T1 y:=0 y = T2 z:=0 z = T3 Lecture 7

  50. The Task Arrival Model TSK = (c, d) G : X TSK’ = (c’, d’) c computation time d relative deadline Whenever a task is released, it is added to the ready queue. Scheduling is done according to some policy (EDF); uniprocessor model. Lecture 7

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