# ECE/CS 372 introduction to computer networks. Lecture 2. Midterm scheduled for Tuesday, May 7 th

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1 ECE/CS 372 introduction to computer networks Lecture 2 Announcements: Please make sure to check the course s website on a regular basis Midterm scheduled for Tuesday, May 7 th Lab 1 is due on Tuesday, April 9 th ALL HW and LAB ASSIGNMENTS SHOULD BE HARD COPY, SOFT COPY IS NOT ACCEPTED Acknowledgement: slides drawn heavily from Kurose & Ross Chapter 1, slide: 1

2 Packet switching versus circuit switching Packet-switching Circuit-switching Resources sharing dedicated Congestion may lead to it admission control Overhead less overhead; more overhead; no connection setup reserve resources 1st Guarantee Best-effort provide guarantee no guarantee good for multimedia Chapter 1, slide: 2

3 Packet switching versus circuit switching Packet switching allows more users to use network! 3 Mb/s link each user: 1 Mb/s when active active 1/3 of time circuit-switching: 3 users Extra credit Due April 18 th packet switching: N users 3 Mbps link With N=4 users, what are the chances that a user won t get 1 Mb/s? I.e., what is the prob. that more than 3 (strictly) users are active? With N=5 users, what are the chances that a user won t get 1 Mb/s? With N=6 users, what are the chances that a user won t get 1 Mb/s? Chapter 1, slide: 3

4 Packet switching versus circuit switching Hint: answer basic probability questions and recall each user on average is active 1/3 of the time, so probability that a user is active is p=0.333 With N=4 users A, B, C and D what is the probability that exactly 1 user among the 4 is active? P 1 4 = Prob [ (A active & B, C, D are not) (B active & A, C, D are not) (C active & A, B, D are not) (D active & A, B, C are not) ] = p(1-p)(1-p)(1-p) + p(1-p)(1-p)(1-p) + p(1-p)(1-p)(1-p) + p(1-p)(1-p)(1-p) = 4p(1-p) 3 Chapter 1, slide: 4

5 Chapter 1: roadmap 1 What is the Internet? 2 Network edge 3 Network core 4 Internet structure and ISPs 5 Protocol layers, service models 6 Delay & loss in packet-switched networks Chapter 1, slide: 5

6 Internet structure: network of networks roughly hierarchical: tier 1, tier 2, and tier 3 at center: tier-1 ISPs e.g., MCI, Sprint, AT&T, Cable and Wireless, national/international coverage Tier-1 providers interconnect (peer) privately Tier 1 ISP NAP Tier 1 ISP Tier 1 ISP Tier-1 providers also interconnect at public network access points (NAPs) Chapter 1, slide: 6

7 Tier-1 ISP: e.g., Sprint Sprint US backbone network Seattle Tacoma DS3 (45 Mbps) OC3 (155 Mbps) OC12 (622 Mbps) OC48 (2.4 Gbps) Stockton San Jose Cheyenne Kansas City Chicago Roachdale New York Pennsauken Relay Wash. DC Anaheim Fort Worth Atlanta Orlando Chapter 1, slide: 7

8 Internet structure: network of networks Tier-2 ISPs: smaller (often regional) ISPs Connect to one or more tier-1 ISPs, possibly other tier-2 ISPs Tier-2 ISP is customer of tier-1 provider Tier-2 ISP Tier-2 ISP Tier 1 ISP NAP Tier-2 ISPs also peer privately with each other, interconnect at NAP Tier 1 ISP Tier-2 ISP Tier 1 ISP Tier-2 ISP Tier-2 ISP Chapter 1, slide: 8

9 Internet structure: network of networks Tier-3 ISPs and local ISPs last hop ( access ) network (closest to end systems) Local and tier- 3 ISPs are customers of higher tier ISPs connecting them to rest of Internet local ISP local ISP Tier 3 ISP Tier 1 ISP Tier-2 ISP Tier-2 ISP local ISP local ISP Tier 1 ISP local ISP Tier-2 ISP NAP Tier 1 ISP Tier-2 ISP local ISP local ISP Tier-2 ISP local ISP Chapter 1, slide: 9

10 Internet structure: network of networks a packet passes through many networks! local ISP Tier 3 ISP Tier-2 ISP local ISP Tier 1 ISP local ISP Tier-2 ISP NAP local ISP local ISP Tier 1 ISP Tier-2 ISP local ISP Tier 1 ISP Tier-2 ISP local ISP Tier-2 ISP local ISP Chapter 1, slide: 10

11 Chapter 1: roadmap 1 What is the Internet? 2 Network edge 3 Network core 4 Internet structure and ISPs 5 Protocol layers, service models 6 Delay & loss in packet-switched networks Chapter 1, slide: 11

12 Protocol Layers Networks are complex! many pieces : hosts routers links of various media applications protocols hardware, software Question: Is there any hope of an organizing structure of network? Chapter 1, slide: 12

13 Organization of air travel ticket (purchase) baggage (check) gates (load) runway takeoff ticket (complain) baggage (claim) gates (unload) runway landing airplane routing airplane routing airplane routing a series of steps Chapter 1, slide: 13

14 Layering of airline functionality ticket (purchase) ticket (complain) ticket baggage (check) baggage (claim baggage gates (load) gates (unload) gate runway (takeoff) runway (land) takeoff/landing airplane routing airplane routing airplane routing airplane routing airplane routing departure airport intermediate air-traffic control centers arrival airport Layers: each layer implements a service via its own internal-layer actions relying on services provided by layer below Chapter 1, slide: 14

15 Why layering? Dealing with complex systems: Easing assignment of tasks identify relationship among pieces of complex systems Easing maintenance, updating of system change of implementation of layer s service transparent to rest of system e.g., change in gate procedure doesn t affect rest of system Chapter 1, slide: 15

16 Internet protocol stack application: supporting network applications FTP, SMTP, HTTP transport: process-process data transfer TCP, UDP network: routing of datagrams from source to destination IP, routing protocols link: data transfer between neighboring network elements PPP, Ethernet physical: bits on the wire application transport network link physical 5-layer Internet Protocol Stack Chapter 1, slide: 16

17 source Encapsulation segment datagram frame message H l H t H n H t H n H t M M M M application transport network link physical link physical switch H l H n H n H t H t H t M M M M destination application transport network link physical H l H n H n H t H t M M network link physical H n H t M router Chapter 1, slide: 17

18 Chapter 1: roadmap 1 What is the Internet? 2 Network edge 3 Network core 4 Internet structure and ISPs 5 Protocol layers, service models 6 Delay & loss in packet-switched networks Chapter 1, slide: 18

19 Sources of packet delay 1. processing: check bit errors determine output link 2. queueing time waiting at output link for transmission depends on congestion level of router A B nodal processing queueing Chapter 1, slide: 19

20 Sources of packet delay 3. Transmission delay: R=link bandwidth (bps) L=packet length (bits) trans. delay = L/R 4. Propagation delay: d = length of physical link s = propagation speed in medium (~2x10 8 m/sec) propagation delay = d/s Note: s and R are very different quantities! A transmission propagation B nodal processing queueing Chapter 1, slide: 20

21 Caravan analogy 100 km 100 km ten-car caravan toll booth toll booth Cars run at 100 km/hr (speed of propagation) Booth takes 12 sec to service a car (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 = 12*10 = 120 sec = 2 mins Time for last car to propagate from 1st to 2nd toll both: =100km/(100km/hr)= 1 hr A: 1 hr 2 minutes Chapter 1, slide: 21

22 Caravan analogy (more) 100 km 100 km ten-car caravan toll booth Cars now propagate at 1000 km/hr Toll booth now takes 1 min to service a car Q: Will cars arrive to 2nd booth before all cars serviced at 1st booth? toll booth 6 mins to propagate 1 min to be serviced Yes! After 7 min, 1st car at 2nd booth and 8th car still at 1st booth. 1st bit of packet can arrive at 2nd router before packet is fully transmitted at 1st router! Chapter 1, slide: 22

23 Exercise 1 Packet length = L bits Host A trans. rate R = 1 Mbps Host B distance = 1 km, speed = 2x10 8 m/s Question: Which bit is being transmitted at the time the first bit arrives at Host B for Answer: First bit arrives after 1/R + d/s = 1/ /(2x10 8 ) = x10-6 = 6x10-6 = 6 µsec After 6 µsec 6 bits are already transmitted; so 7 th bit is being transmitted Chapter 1, slide: 23

24 Exercise 2 Transmission vs. propagation L=100Bytes Host A trans. rate R =? Host B distance = 2 km, speed = 2x10 8 m/s Question: At what rate (bandwidth) R would the propagation delay equal the transmission delay? Answer: Propagation delay = 2x10 3 (m)/2x10 8 (m/s) = 10-5 sec Transmission delay = 100x8 (bits)/r Prop. delay = trans. delay => R=10 5 x100x8 = 80 Mbps Chapter 1, slide: 24

25 Exercise 3 Voice over IP a=64kbps Host A L=48 Bytes Host A trans. rate R = 1Mbps delay_prop = 2msec Host B converts analog to digital at a=64kbps groups bits into L=48Byte packets sends packet to Host B as soon it gathers a packet Host B As soon as it receives the whole pkt, it converts it to analog Question: How much time elapses from the 1 st bit of 1 st packet is created until the last bit of the 1 st packet arrives at Host B? Chapter 1, slide: 25

26 Exercise 3 Voice over IP a=64kbps L=48 Bytes Host A trans. rate R = 1Mbps delay_prop = 2msec Host B Answer: Time to gather 1 st pkt: (48x8 (bits)) / (64x1000 (b/s)) = 6 msec Time to push 1 st pkt to link: (48x8 (bits)) / 1x10 6 (b/s) = msec Time to propagate: 2 msec Total delay = = msec Chapter 1, slide: 26

27 Nodal delay d = d + d + d + nodal proc queue trans d prop d proc = processing delay typically a few microsecs or less d queue = queuing delay depends on congestion d trans = transmission delay = L/R, significant for low-speed links d prop = propagation delay a few microsecs to hundreds of msecs Chapter 1, slide: 27

28 Queueing delay (more insight) Packet arrival rate = a packets/sec queue Packet length = L bits Link bandwidth = R bits/sec Every second: al bits arrive to queue Every second: R bits leave the router Question: what happens if al > R? Answer: queue will fill up, and packets will get dropped!! al/r is called traffic intensity Chapter 1, slide: 28

29 Queueing delay: illustration 1 packet arrives every L/R seconds queue Link bandwidth = R bits/sec Packet length L bits Arrival rate: a = 1/(L/R) = R/L (packet/second) Traffic intensity = al/r = (R/L) (L/R) = 1 Average queueing delay = 0 (queue is initially empty) Chapter 3, slide: 29

30 Queueing delay: illustration N packet arrive simultaneously every LN/R seconds queue Link bandwidth = R bits/sec Packet length L bits Arrival rate: a = N/(LN/R) = R/L packet/second Traffic intensity = al/r = (R/L) (L/R) = 1 Average queueing delay (queue is time 0)? {0 + L/R + 2L/R + + (N-1)L/R}/N = L/(RN){1+2+ +(N-1)} =L(N-1)/(2R) Note: traffic intensity is same as previous scenario, but queueing delay is different Chapter 3, slide: 30

31 Queueing delay: behavior Packet arrival rate = a packets/sec queue Packet length = L bits Link bandwidth = R bits/sec La/R ~ 0: avg. queuing delay small La/R -> 1: delays become large La/R > 1: more work than can be serviced, average delay infinite! (this is when a is random!) Chapter 1, slide: 31

32 Packet-switching: store-and-forward L R R R Entire packet must arrive at router before it can be transmitted on next link: store and forward Takes L/R seconds to transmit (push out) packet of L bits on to link of R bps delay = 3L/R (assuming zero propagation delay) more on this next Chapter 1, slide: 32

33 Store-and-forward: illustration distance = d meters; speed of propagation = s m/sec transmission rate of link = R bits/s L d R L d/2 d/2 R R delay (one packet only) = L/R + d/s Example: d/s = 0.5 sec L = 10 Mbits R = 1 Mbps delay = 10.5 sec delay (one packet only) = L/R + ½d/s + L/R + ½d/s = 2L/R + d/s Example: d/s = 0.5 sec L = 10 Mbits R = 1 Mbps delay = 20.5 sec Chapter 1, slide: 33

34 Store-and-forward & queuing delay distance = d meters; speed of propagation = s m/sec transmission rate of link = R1 and R2 bits/s Consider sending two packets A and B back to back Case 1: Assume R1 < R2 L R1 d R2 Case 2: Assume R1 > R2 Q: is there a queuing delay? how much is this delay? Answer (queue is empty initially): Time for last bit of 2 nd pkt to arrive at router: d1= L/R1 + L/R1 + d/(2s) Time for last bit of 1 st pkt to leave router: d2= L/R1 + d/(2s) + L/R2 Queuing delay = d2 d1 = L/R2 L/R1 if positive, otherwise 0. Hence: when R1 < R2, queuing delay = d2 d1 = 0 when R1 > R2, queuing delay = d2 d1 = L/R2 L/R1 Chapter 1, slide: 34

35 Throughput analysis Host A L R R R Host B Suppose: Host A has huge file of size F bits to send to Host B File is split into N packets, each of length L bits (i.e., N=F/L) Ignore propagation delay for now Question 1: how long it takes to send the file? A: (N+2)L/R = (F+2L)/R Question 2: what is the average throughput achieved when sending the file? A: NL/[(N+2)L/R]=NR/(N+2) = FR/(F+2L) = R/(1+2L/F) Note: throughput = number of total bits sent / total time taken Chapter 1, slide: 35

36 Throughput analysis Host A L d/3 d/3 d/3 R R R Host B Suppose: Host A has huge file of size F bits to send to Host B File is split into N packets, each of length L bits (i.e., N=F/L) Do NOT ignore propagation delay (assume prop. speed = s m/s) Question 1: how long it takes to send the file? A: (N+2)L/R + d/s = (F+2L)/R + d/s Question 2: what is the average throughput achieved when sending the file? A: NL/[(N+2)L/R +d/s]=fr/[(n+2)l + dr/s] = FR/[F+2L+dR/s] Chapter 1, slide: 36

37 Introduction: Summary Covered a ton of material! Internet overview Network protocol Network edge, core, access network Packet-switching versus circuit-switching Internet/ISP structure layering and service models performance: delay and throughput analysis Chapter 1, slide: 37

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