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ECEN4533 Data Communications Lecture #18 18 February 2013 Dr. George Scheets

ECEN4533 Data Communications Lecture #18 18 February 2013 Dr. George Scheets. Problems: 2011 Exam #1 Corrected Design #1 Due 18 February (Live) 1 week after you get them back (DL) Exam #1: 22 February (Live), < 1 March (DL).

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ECEN4533 Data Communications Lecture #18 18 February 2013 Dr. George Scheets

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  1. ECEN4533 Data CommunicationsLecture #18 18 February 2013Dr. George Scheets Problems: 2011 Exam #1 Corrected Design #1 Due 18 February (Live) 1 week after you get them back (DL) Exam #1: 22 February (Live), < 1 March (DL)

  2. ECEN4533 Data CommunicationsLecture #19 20 February 2013Dr. George Scheets • Read 11.1 - 11.3 • Problems: 2012 Exam #1 • Exam #122 February (Live), < 1 March (DL)

  3. Possible Test Topics S Reading HW Lectures Anything in the circles is fair game.

  4. Exponentially Distributed Packet Length(Somewhat decent fit to real world traffic) Bin Count Bytes

  5. Little's Rule Under steady-state conditionsE[K(t)] = λ E[T]E[Kq(t)] = λ E[Tq]E[# in server] = λ E[Ts]regardless of PDF's involved.

  6. M/G/1 • Exponentially distributed IAT • Arbitrary packet size distribution • Single Server • E[Tq] = E[Ts2]/[2(1-ρ)] • E[Ts2] = σTs2 + E[Ts]2

  7. M/M/1 Queues • Exponentially Distributed IAT's • Exponentially Distributed Packet Sizes • E[Ts] = σTs if Exponential • Single Server • E[Tq] = ρE[Ts]/(1-ρ) • Multiple Input, Multiple Output Switch? • Repeat analysis once per output trunk • Base on input traffic exiting that trunk

  8. Classical M/M/1 Queuing Theory Average Queue Size Dropped Packet Probability 0% 100% Offered Load

  9. Finite Buffer Queuing Performance Probability of dropped packets Average Delay for delivered packets 0% 100% Trunk Offered Load

  10. M/M/a Queues Exponentially Distributed IAT's Exponentially Distrubuted Packets Multiple Servers (# = a) Queue servicing "a" output trunks Trunks have identical loads M/M/1 versus equal speed M/M/a EX) M/M/1 @ 100 Mbps had E[T] = 172.5 μsec M/M/2 @ 2x50 Mbps had E[T] = 185.4 μsec Want a big trunk to minimize delay thru switch

  11. M/M/1 Queues with Priorities Exponentially Distributed IAT's Exponentially Distrubuted Packets Single Server Hi priority traffic to head of queue Gets output more speedily Packet is server is not prempted Low priority traffic slower to exit Overall average ≈ same as M/M/1

  12. Queuing with Priorities Overall Average Stays ≈ the Same M/M/1 Low Priority Average Delay High Priority 0% 100% Offered Load

  13. M/D/1 Queues Exponentially Distributed IAT's Fixed Packet Size (i.e. Cells) Single Server E[Tq] = ρ[Ts]/[2(1-ρ)] Given equivalent loads and same average sizes, fixed size cells are moved faster.

  14. Classical Queuing Theory Average # in System M/M/1 ρ/(1-ρ) M/D/1 ρ2/(1-ρ) 0% 100% Offered Load

  15. Armed with • The average service time E[Ts] • An equation for E[Time in Queue] or E[Time in System] • Little's Rule • Average # packets = E[Time] E[Packet Arrival Rate]where E[Packet Arrival Rate] = λ packets/second • You can find a large number of parameters • E[T], E[Tq], E[K(t)], E[Kq(t)]

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