1. Chapter 1: roadmap 1.1 what is the Internet? 1.2 network edge  end systems, access networks, links 1.3 network core  packet switching, circuit switching, network structure 1.4 delay, loss, throughput in networks 1.5 protocol layers, service models 1.6 networks under attack: security 1.7 history Delay Samples -1

2. How do loss and delay occur? packets queue in router buffers  packet arrival rate to link (temporarily) exceeds output link capacity  packets queue, wait for turn A B packet being transmitted (delay) packets queueing (delay) free (available) buffers: arriving packets dropped (loss) if no free buffers Delay Samples -2

3. Four sources of packet delay d proc : nodal processing  check bit errors  determine output link  typically < msec A B propagation transmission nodal processing queueing d queue : queueing delay  time waiting at output link for transmission  depends on congestion level of router d nodal = d proc + d queue + d trans + d prop Delay Samples -3

4. d trans : transmission delay:  L: packet length (bits)  R: link bandwidth (bps)  d trans = L/R d prop : propagation delay:  d: length of physical link  s: propagation speed in medium (~2x10 8 m/sec)  d prop = d/s d trans and d prop very different Four sources of packet delay propagation nodal processing queueing d nodal = d proc + d queue + d trans + d prop A B transmission * Check out the Java applet for an interactive animation on trans vs. prop delay Delay Samples -4

5. Caravan analogy  cars “propagate” at 100 km/hr  toll booth takes 12 sec to service car (bit transmission time)  car~bit; caravan ~ packet  Q: How long until caravan is lined up before 2nd toll booth?  time to “push” entire caravan through toll booth onto highway = 12*10 = 120 sec  time for last car to propagate from 1st to 2nd toll both: 100km/(100km/hr)= 1 hr  A: 62 minutes toll booth toll booth ten-car caravan 100 km Delay Samples -5

6. Caravan analogy (more)  suppose cars now “propagate” at 1000 km/hr  and suppose toll booth now takes one min to service a car  Q: Will cars arrive to 2nd booth before all cars serviced at first booth?  A: Yes! after 7 min, 1st car arrives at second booth; three cars still at 1st booth. toll booth toll booth ten-car caravan 100 km Delay Samples -6

7.  R: link bandwidth (bps)  L: packet length (bits)  a: average packet arrival rate traffic intensity = La/R  La/R ~ 0: avg. queueing delay small  La/R -> 1: avg. queueing delay large  La/R > 1: more “work” arriving than can be serviced, average delay infinite! average queueing delay La/R ~ 0 Queueing delay (revisited) La/R -> 1 * Check out the Java applet for an interactive animation on queuing and loss Delay Samples -7

8. “Real” Internet delays and routes  what do “real” Internet delay & loss look like?  traceroute program: provides delay measurement from source to router along end- end Internet path towards destination. For all i:  sends three packets that will reach router i on path towards destination  router i will return packets to sender  sender times interval between transmission and reply. 3 probes Delay Samples -8

9. “Real” Internet delays, routes 1 cs-gw (128.119.240.254) 1 ms 1 ms 2 ms 2 border1-rt-fa5-1-0.gw.umass.edu (128.119.3.145) 1 ms 1 ms 2 ms 3 cht-vbns.gw.umass.edu (128.119.3.130) 6 ms 5 ms 5 ms 4 jn1-at1-0-0-19.wor.vbns.net (204.147.132.129) 16 ms 11 ms 13 ms 5 jn1-so7-0-0-0.wae.vbns.net (204.147.136.136) 21 ms 18 ms 18 ms 6 abilene-vbns.abilene.ucaid.edu (198.32.11.9) 22 ms 18 ms 22 ms 7 nycm-wash.abilene.ucaid.edu (198.32.8.46) 22 ms 22 ms 22 ms 8 62.40.103.253 (62.40.103.253) 104 ms 109 ms 106 ms 9 de2-1.de1.de.geant.net (62.40.96.129) 109 ms 102 ms 104 ms 10 de.fr1.fr.geant.net (62.40.96.50) 113 ms 121 ms 114 ms 11 renater-gw.fr1.fr.geant.net (62.40.103.54) 112 ms 114 ms 112 ms 12 nio-n2.cssi.renater.fr (193.51.206.13) 111 ms 114 ms 116 ms 13 nice.cssi.renater.fr (195.220.98.102) 123 ms 125 ms 124 ms 14 r3t2-nice.cssi.renater.fr (195.220.98.110) 126 ms 126 ms 124 ms 15 eurecom-valbonne.r3t2.ft.net (193.48.50.54) 135 ms 128 ms 133 ms 16 194.214.211.25 (194.214.211.25) 126 ms 128 ms 126 ms 17 * * * 18 * * * 19 fantasia.eurecom.fr (193.55.113.142) 132 ms 128 ms 136 ms traceroute: gaia.cs.umass.edu to www.eurecom.fr 3 delay measurements from gaia.cs.umass.edu to cs-gw.cs.umass.edu * means no response (probe lost, router not replying) trans-oceanic link * Do some traceroutes from exotic countries at www.traceroute.org Delay Samples -9

10. Packet loss  queue (aka buffer) preceding link in buffer has finite capacity  packet arriving to full queue dropped (aka lost)  lost packet may be retransmitted by previous node, by source end system, or not at all A B packet being transmitted packet arriving to full buffer is lost buffer (waiting area) * Check out the Java applet for an interactive animation on queuing and loss Delay Samples -10

11. Throughput  throughput: rate (bits/time unit) at which bits transferred between sender/receiver  instantaneous: rate at given point in time  average: rate over longer period of time server, with file of F bits to send to client link capacity R s bits/sec link capacity R c bits/sec server sends bits (fluid) into pipe pipe that can carry fluid at rate R s bits/sec) pipe that can carry fluid at rate R c bits/sec) Delay Samples -11

12. Throughput (more)  R s < R c What is average end-end throughput? R s bits/sec R c bits/sec  R s > R c What is average end-end throughput? link on end-end path that constrains end-end throughput bottleneck link R s bits/sec R c bits/sec Delay Samples -12

13. Throughput: Internet scenario 10 connections (fairly) share backbone bottleneck link R bits/sec RsRs RsRs RsRs RcRc RcRc RcRc R  per-connection end- end throughput: min(R c,R s,R/10)  in practice: R c or R s is often bottleneck Delay Samples -13

14. Sample#1  Takes L/R seconds to transmit (push out) packet of L bits on to link R bps  Entire packet must arrive at router before it can be transmitted on next link: store and forward  delay = 3L/R Example: L = 7.5 Mbits; message size R = 1.5 Mbps; link bandwidth message transmission time = L/R = 5 sec delay = 3L/R = 15 sec R R R L Delay Samples -14

15. Sample#2 Now break up the message into 5000 packets  Each packet 1,500 bits  1 msec to transmit packet on one link  pipelining: each link works in parallel  Delay reduced from 15 sec to 5.002 sec L Delay Samples -15

16. Calculate the total time required to transfer a 1,000-KB file in the following cases, assuming an RTT of 100 ms, a packet size of 1-KB data, and an initial 2×RTT of “handshaking” before data is sent. (a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously. Delay Samples -16 Sample#2

17.  Initial+ transmit time + Propagation Time=  2*RTT + Data/BW + 0.5*RTT  =2.5*RTT + 8*Mb/1.5Mb  =250ms+ 5.33sec= 0.25+5.33 = 5.58 Delay Samples -17 Sample#2 sol part a

18. (b) The bandwidth is 1.5 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. Delay Samples -18 Sample#2

19. part(a)+ 999*RTT = 5.58 + 999*RTT = 5.58 + 99.9 = 105.48 Delay Samples -19 Sample#2 sol part b

20. (c) The bandwidth is “infinite,” meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. Delay Samples -20 Sample#2

21.  2*RTT + 49*RTT + 0.5 * RTT  = 51.5 RTT  = 5.15 sec Delay Samples -21 Sample#2 sol part c

22. (d) The bandwidth is infinite, and during the first RTT we can send one packet (2^1 − 1), during the second RTT we can send two packets (2^2 − 1),during the third we can send four (2^3 − 1), and so on. Delay Samples -22 Sample#2

23. ??? Delay Samples -23 Sample#2 sol part d

24. Calculate the latency (from first bit sent to last bit received) for the following: (a) A 10-Mbps Ethernet with a single store-and- forward switch in the path, and a packet size of 5,000 bits. Assume that each link introduces a propagation delay of 10µs, and that the switch begins retransmitting immediately after it has finished receiving the packet. Delay Samples -24 Sample#3 10 Mbps

25.  2* 10 micro sec + 2* 500 micro sec  = 20+ 1000  == 1020 micro sec Delay Samples -25 Sample#3 sol part a

26. (b) Same as (a) but with three switches. Delay Samples -26 Sample#3

27.  4* 10 + 4*500  = 2040 mic s Delay Samples -27 Sample#3 sol part b

28. (c) Same as (a) but assume the switch implements “cut-through” switching: it is able to begin retransmitting the packet after the first 200 bits have been received. Delay Samples -28 Sample#3

29.  2*10 + (20 + 500)  =540 Delay Samples -29 Sample#3 sol part c

30. (d) Same as (c) but with three switches. Delay Samples -30 Sample#3

31.  4*10 + (500+60)  =600 Delay Samples -31 Sample#3 sol part d

32. Delay Samples -32 فریمی 1500 بایتی قرار است بر روی لینکی با پهنای باند 100 مگابیت‌برثانیه ارسال گردد. با فرض بر اینکه فاصله بین مبداء و مقصد برابر 250 کیلومتر باشد : الف - اگر بیت اول در لحظه صفر ارسال گردد، بیت آخر در چه زمانی ارسال خواهد شد؟ ب - اگر سرعت انتشار برابر 8^10×2 باشد، چه زمانی بیت اول به مقصد می‌رسد؟ ج - چه زمانی بیت آخر به مقصد می‌رسد؟ Sample#4

33.  a) Transmission Delay:  (1500*8)/(100*10^6)= 120 micro sec  b)Propagation Delay:  (25*10^3)/(2*10^8)= 1250 micro sec Delay Samples -33 Sample#4 sol parts a,b

34. All link have the same Bitrate: R All distances are the same: L Source Data Size: M bits Number of Packets: n Switches are Store and Forward. T p = L/(2*10^8) T t = (M/n)/R Sample#5

35. Phone R1 R2 Dest T= 3T p +2T t T= 3T p +3T t T= 3T p +4T t T= 3T p +5T t T= 3T p +(n+2)T t Sample#5

36.  Please calculate total delay, if number of intermediate switches reach to k? Delay Samples -36 Sample#5 T= (k+1)T p +(n+k)T t