Chapter 1: roadmap1.1 what is the Internet?1.2 network edge

end systems, access networks, links

1.3 network core packet switching, circuit switching, network

structure1.4 delay, loss, throughput in networks1.5 protocol layers, service models1.6 networks under attack: security1.7 history

Delay Samples -1

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

Four sources of packet delay

dproc: nodal processing check bit errors determine output link typically < msec

A

B

propagation

transmission

nodalprocessing queueing

dqueue: queueing delay

time waiting at output link for transmission

depends on congestion level of router

dnodal = dproc + dqueue + dtrans + dprop

Delay Samples -3

dtrans: transmission delay:

L: packet length (bits) R: link bandwidth (bps) dtrans = L/R

dprop: propagation delay: d: length of physical link s: propagation speed in

medium (~2x108 m/sec) dprop = d/sdtrans and dprop

very different

Four sources of packet delay

propagation

nodalprocessing queueing

dnodal = dproc + dqueue + dtrans + dprop

A

B

transmission

* Check out the Java applet for an interactive animation on trans vs. prop delay Delay Samples -4

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 100 km

Delay Samples -5

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 100 km

Delay Samples -6

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!

ave

rage

qu

eue

ing

d

ela

y

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

“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

3 probes

3 probes

Delay Samples -8

“Real” Internet delays, routes

1 cs-gw (128.119.240.254) 1 ms 1 ms 2 ms2 border1-rt-fa5-1-0.gw.umass.edu (128.119.3.145) 1 ms 1 ms 2 ms3 cht-vbns.gw.umass.edu (128.119.3.130) 6 ms 5 ms 5 ms4 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 ms7 nycm-wash.abilene.ucaid.edu (198.32.8.46) 22 ms 22 ms 22 ms8 62.40.103.253 (62.40.103.253) 104 ms 109 ms 106 ms9 de2-1.de1.de.geant.net (62.40.96.129) 109 ms 102 ms 104 ms10 de.fr1.fr.geant.net (62.40.96.50) 113 ms 121 ms 114 ms11 renater-gw.fr1.fr.geant.net (62.40.103.54) 112 ms 114 ms 112 ms12 nio-n2.cssi.renater.fr (193.51.206.13) 111 ms 114 ms 116 ms13 nice.cssi.renater.fr (195.220.98.102) 123 ms 125 ms 124 ms14 r3t2-nice.cssi.renater.fr (195.220.98.110) 126 ms 126 ms 124 ms15 eurecom-valbonne.r3t2.ft.net (193.48.50.54) 135 ms 128 ms 133 ms16 194.214.211.25 (194.214.211.25) 126 ms 128 ms 126 ms17 * * *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-oceaniclink

* Do some traceroutes from exotic countries at www.traceroute.orgDelay Samples -9

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 tofull buffer is lost

buffer (waiting area)

* Check out the Java applet for an interactive animation on queuing and loss Delay Samples -10

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, withfile of F bits

to send to client

link capacity

Rs bits/sec

link capacity

Rc bits/secserver sends

bits (fluid) into pipe

pipe that can carryfluid at rate

Rs bits/sec)

pipe that can carryfluid at rate

Rc bits/sec)

Delay Samples -11

Throughput (more) Rs < Rc What is average end-end throughput?

Rs bits/sec Rc bits/sec

Rs > Rc What is average end-end throughput?

link on end-end path that constrains end-end throughput

bottleneck link

Rs bits/sec Rc bits/sec

Delay Samples -12

Throughput: Internet scenario

10 connections (fairly) share backbone bottleneck link R bits/sec

Rs

Rs

Rs

Rc

Rc

Rc

R

per-connection end-end throughput: min(Rc,Rs,R/10)

in practice: Rc or Rs is often bottleneck

Delay Samples -13

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

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

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

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

(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

•part(a)+ 999*RTT•= 5.58 + 999*RTT• = 5.58 + 99.9•= 105.48

Delay Samples- 19

Sample#2 sol part b

(c) The bandwidth is “infinite,” meaning that we take transmit time to bezero, and up to 20 packets can be sent per RTT.

Delay Samples- 20

Sample#2

2*RTT + 49*RTT + 0.5 * RTT

= 51.5 RTT= 5.15 sec

Delay Samples- 21

Sample#2 sol part c

(d) The bandwidth is infinite, and during the first RTT we can send onepacket (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

???

Delay Samples- 23

Sample#2 sol part d

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

10 Mbps

2* 10 micro sec + 2* 500 micro sec

= 20+ 1000== 1020 micro sec

Delay Samples- 25

Sample#3 sol part a

(b) Same as (a) but with three switches.

Delay Samples- 26

Sample#3

4* 10 + 4*500= 2040 mic s

Delay Samples- 27

Sample#3 sol part b

(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

2*10 + (20 + 500)=540

Delay Samples- 29

Sample#3 sol part c

(d) Same as (c) but with three switches.

Delay Samples- 30

Sample#3

4*10 + (500+60)=600

Delay Samples- 31

Sample#3 sol part d

Delay Samples- 32

بZایتی قZرار اسZت بZر روی لینکی بZا پهنZای 1500فریمی بZر 100بانZد فZرض بZا گZردد. ارسZال مگابیت برثانیZه

کیلومZتر 250اینکZه فاصZله بین مبZداء و مقصZد برابZر باشد:

الZف- اگZر بیت اول در لحظZه صZفر ارسZال گZردد، بیت آخر در چه زمانی ارسال خواهد شد؟

برابZر انتشZار اگZر سZرعت باشZد، چZه 2×10^8ب- زمانی بیت اول به مقصد می رسد؟ 

ج- چه زمانی بیت آخر به مقصد می رسد؟

Sample#4

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

All link have the same Bitrate: All link have the same Bitrate: RRAll distances are the same: All distances are the same: LLSource Data Size: Source Data Size: M bitsM bitsNumber of Packets: Number of Packets: nnSwitches are Store and Forward.Switches are Store and Forward.TTpp= L/(2*10^8)= L/(2*10^8)TTtt= (M/n)/R= (M/n)/R

Sample#5

Phone

R1

R2

Dest

T= 3Tp+2Tt T= 3Tp+3TtT= 3Tp+4TtT= 3Tp+5Tt

T= 3TT= 3Tpp++(n+2)T(n+2)Ttt

Sample#5

Please calculate total delay, if number of intermediate switches reach to k?

Delay Samples- 36

Sample#5

T= T= (k+1)(k+1)TTpp+(+(n+kn+k)T)Ttt