Chapter 1 Introduction Our goal q learn basic

Chapter 1: Introduction Our goal: q learn basic network terminologies q more depth, detail later in course q approach: v use Internet as example

Chapter 1: Introduction 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

What’s the Internet: a “service” view q communication infrastructure enables distributed apps: v v Enables apps to communicate Web, email, games, e-commerce, file sharing q communication services provided to apps: v Offers services

What’s the Internet: “nuts and bolts” view q millions of connected computing devices: called hosts or end systems v v e. g. , Laptops, workstations running network apps router server mobile local ISP q routers & switches: v forward packets (chunks of data) q communication links v e. g. , fiber, copper, radio, satellite workstation regional ISP company network

What’s the Internet: “nuts and bolts” view q Internet standards v IETF (Internet Eng. Task Force) • RFC: Request for comments v router server workstation mobile local ISP IEEE: for links/hardware E. g. , Ethernet regional ISP q network protocols v control sending/receiving of messages v e. g. , TCP, IP, HTTP, FTP, PPP company network

What’s a protocol? a computer network protocol: TCP connection request time TCP connection response Get http: //www. awl. com/kurose-ross <file>

What’s a protocol? human protocols: q “what’s the time? ” q “I have a question” q introductions … specific msgs sent … specific actions taken when msgs received, or other events network protocols: q machines rather than humans q all communication activity in Internet governed by protocols define (1) format, order of msgs sent and received among network entities, and (2) actions taken on msg transmission, receipt

Chapter 1: Introduction 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

A closer look at network structure: q network edge: applications and hosts q network core: routers v network of networks v q access networks, physical media: communication links

The network edge: service models q end systems (hosts): v v v run application programs e. g. Web, email at “edge of network” q client/server model v v client host requests, receives service from always-on server e. g. Web browser/server; email client/server q peer-to-peer model: v v minimal (or no) use of dedicated servers e. g. Skype, Bit. Torrent, Ka. Za. A

Chapter 1: Introduction 1 What is the Internet? 2 Network edge 3 Network core 4 Network access and physical media 5 Internet structure and ISPs 6 Protocol layers, service models 7 Delay & loss in packet-switched networks

The Network Core q mesh of interconnected routers q the fundamental question: how is data transferred through net? v circuit switching: dedicated circuit per call: telephone net v packet-switching: data sent thru net in discrete “chunks”

Network Core: Circuit Switching End-end resources reserved for “call” q dedicated resources: no sharing q call setup required q circuit-like (guaranteed) performance q same path for all chunks

Network Core: Circuit Switching network resources (e. g. , bandwidth) divided into “pieces” q allocated pieces per call q no sharing resource piece idle if not used by owning call

Network Core: Circuit Switching q Two ways of dividing bandwidth into “pieces” frequency division v time division v

Circuit Switching: FDM and TDM Example: Freq. Division Multiplx. (FDM) 4 users frequency time Time Division Multiplx. (TDM) frequency time

Network Core: Packet Switching 100 Mb/s Ethernet A B C 1. 5 Mb/s each end-to-end data stream is divided into packets q no dedication/reservation: all streams share resources q no setup is required q resources used as needed q each packet uses full link bandwidth q aggregate resource demand can exceed capacity q no guarantee

Network Core: statistical multiplexing 100 Mb/s Ethernet A B statistical multiplexing C 1. 5 Mb/s queue of packets waiting for output link D E Sequence of A & B packets does not have fixed pattern, shared on demand statistical multiplexing. Whereas in TDM, each host gets same slot (periodically)

Packet switching versus circuit switching A Circuit switching B B: has no packets to send A 2 Mb/s • 2 circuits (use TDM) • A reserves 1 circuit • B reserves 1 circuit Utilization = 50% only = 1 Mb/s Packet switching B 2 Mb/s Utilization = 100% = 2 Mb/s • statistical multiplex. • B uses full link since A is not using it

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

Numerical example q How long does it take to send a file of 640, 000 bits from host A to host B over a circuit-switched network? The link’s transmission rate = 0. 64 Mbps v Each link uses TDM with 10 slots/sec v 0. 5 sec to establish end-to-end circuit Let’s work it out! You have few minutes! v q Solution: v Bandwidth of circuit (in kbps)=. 64 x 1000/10 = 64 kbps v Time to send: 640 kbits/64 kbps + 0. 5 s = 10. 5 s

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

Packet switching versus circuit switching Packet switching allows more users to use network! q 3 Mb/s link q each user: v v 1 Mb/s when “active” active 1/3 of time q circuit-switching: v 3 users N users 3 Mbps link q packet switching: v 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? v With N=5 users, what are the chances that a user won’t get 1 Mb/s? v With N=6 users, what are the chances that a user won’t get 1 Mb/s?

Packet switching versus circuit switching Board …

Chapter 1: Introduction 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

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

Tier-1 ISP: e. g. , Sprint US backbone network DS 3 (45 Mbps) OC 3 (155 Mbps) OC 12 (622 Mbps) OC 48 (2. 4 Gbps) Seattle Tacoma Stockton San Jose Cheyenne Kansas City Chicago Roachdale New York Pennsauken Relay Wash. DC Anaheim Atlanta Fort Worth Orlando Chapter 1, slide:

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

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

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

Chapter 1: Introduction 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

Protocol “Layers” Networks are complex! q many “pieces”: v hosts v routers v links of various media v applications v protocols v hardware, software Question: Is there any hope of an organizing structure of network?

Why layering? Dealing with complex systems: q Easing assignment of tasks v identify relationship among pieces of complex systems q Easing maintenance, updating of system change of implementation of layer’s service transparent to rest of system v e. g. , change in gate procedure doesn’t affect rest of system v

Internet protocol stack q application: supporting network applications v FTP, SMTP, HTTP q transport: process-process data transfer v TCP, UDP q network: routing of datagrams from source to destination v IP, routing protocols q link: data transfer between neighboring network elements v PPP, Ethernet q physical: bits “on the wire” application transport network link physical

Encapsulation source message segment Ht M datagram Hn Ht M frame Hl Hn Ht M M application transport network link physical switch destination M Ht M Hn Ht Hl Hn Ht M M application transport network link physical Hn Ht Hl Hn Ht M M network link physical Hn Ht M router

ISO/OSI Model: late 70’s application presentation session transport application transport network link data link physical 7 -layer ISO/OSI model (OSI: open system interconnections) 5 -layer Internet Protocol Stack

Chapter 1: Introduction 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

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

Sources of packet delay 3. Transmission delay: 4. Propagation delay: q R=link bandwidth (bps) q d = length of physical link q L=packet length (bits) q s = propagation speed in medium (~2 x 108 m/sec) q propagation delay = d/s q trans. delay = L/R Note: s and R are very different quantities! transmission A propagation B nodal processing queueing

Caravan analogy 100 km ten-car caravan toll booth q Cars run at 100 km/hr (speed of propagation) q Booth takes 12 sec to service a car (transmission time) q Car ~ bit; caravan ~ packet q Q: How long until caravan is lined up before 2 nd toll booth? 100 km toll booth q Time to “push” entire caravan through toll booth = 12*10 = 120 sec = 2 mns q Time for last car to propagate from 1 st to 2 nd toll both: =100 km/(100 km/hr)= 1 hr q A: 1 hr 2 minutes

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

Exercise 1 Packet length = L bits Host A trans. rate R = 1 Mbps Host B distance = 1 km, speed = 2 x 108 m/s Question: r 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 = 1/106 + 103/(2 x 108) = 10 -6 + 5 x 10 -6 = 6 µsec After 6 µsec 6 bits are already transmitted; so 7 th bit is being transmitted

Exercise 2 Transmission vs. propagation L=100 Bytes Host A Question: trans. rate R = ? Host B distance = 2 km, speed = 2 x 108 m/s r At what rate (bandwidth) R would the propagation delay equal the transmission delay? Answer: r Propagation delay = 2 x 103 (m)/2 x 108 (m/s) = 10 -5 sec r Transmission delay = 100 x 8 (bits)/R r Prop. delay = trans. delay => R=105 x 100 x 8 = 80 Mbps

Exercise 3 Voice over IP L=48 Bytes a=64 Kbps Host A trans. rate R = 1 Mbps delay_prop = 2 msec Host B r Host A v converts analog to digital at a=64 Kbps groups bits into L=48 Byte packets sends packet to Host B as soon it gathers a packet v As soon as it receives the whole pckt, it converts it to analog v v r Host B r Question: v 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?

Exercise 3 Voice over IP L=48 Bytes a=64 Kbps Host A trans. rate R = 1 Mbps delay_prop = 2 msec Host B Answer: r Time to gather 1 st pkt: 48 x 8 (bits)/64 x 1000 (b/s) = 6 msec r Time to push 1 st pkt to link: 48 x 8 (bits)/1 x 106 (b/s) = 0. 384 msec r Time to propagate: 2 msec r Total delay = 6 + 0. 384 + 2 = 8. 384 msec

Nodal delay q dproc = processing delay v typically a few microsecs or less q dqueue = queuing delay v depends on congestion q dtrans = transmission delay v = L/R, significant for low-speed links q dprop = propagation delay v a few microsecs to hundreds of msecs

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

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 = a. L/R = (R/L) (L/R) = 1 Average queueing delay = 0 (queue is initially empty)

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 = a. L/R = (R/L) (L/R) = 1 Average queueing delay (queue is empty @ time 0) ? {0 + L/R + 2 L/R + … + (N-1)L/R}/N = L/(RN){1+2+…+(N-1)} =L(N-1)/(2 R) Note: traffic intensity is same as previous scenario, but queueing delay is different

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

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 q Takes L/R seconds to transmit (push out) packet of L bits on to link of R bps q delay = 3 L/R (assuming zero propagation delay) more on this next…

Store-and-forward: illustration q distance = d meters; speed of propagation = s m/sec q transmission rate of link = R bits/s L d R q delay (one packet only) = L/R + d/s Example: q d/s = 0. 5 sec q L = 10 Mbits q R = 1 Mbps q delay = 10. 5 sec L d/2 R q delay (one packet only) = L/R + ½d/s + L/R + ½d/s = 2 L/R + d/s Example: q d/s = 0. 5 sec q L = 10 Mbits q R = 1 Mbps q delay = 20. 5 sec

Store-and-forward & queuing delay q distance = d meters; speed of propagation = s m/sec q transmission rate of link = R 1 and R 2 bits/s q Consider sending two packets A and B back to back d L R 1 q Case 1: Assume R 1 < R 2 q Case 2: Assume R 1 > R 2 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: d 1= L/R 1 + d/(2 s) Time for last bit of 1 st pkt to leave router: d 2= L/R 1 + d/(2 s) + L/R 2 Queueing delay = d 2 – d 1 = L/R 2 – L/R 1 if positive, otherwise 0. Hence: when R 1 < R 2, queueing delay = d 2 – d 1 = 0 when R 1 > R 2, queueing delay = d 2 – d 1 = L/R 2 – L/R 1

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

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

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