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Berkeley ELENG 122 - Midterm Review

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Slide 1AnnouncementsOverviewLayering: The ProblemLayering: SolutionLayeringLayering: InternetHourglassImplications of HourglassE2E Arguments: Where to Place Functionality?E2E Arguments: Moderate InterpretationOverviewLittle’s TheoremExampleExampleExampleExampleExampleExampleOverviewDefinitionsSending One PacketQueueingStore & ForwardStore & Forward: Various Capacities ExampleStore & Forward: Multiple Packet ExampleOverviewPacket ForwardingScalability ChallengeSolution: Hierarchical Addressing (IP Prefixes)Scalability ImprovedEasy to Add New HostsClassful AddressingClassful Addressing (cont’d)Classless Inter-Domain Routing (CIDR)Scalability: Address AggregationSlide 37OverviewAutomatic Repeat reQuest (ARQ)How Fast Can Stop-and-Wait Go?Allowing Multiple Packets in FlightSliding Window Example (This is NOT TCP !)Sliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleSliding Window ExampleWhat If Sending & Delivey Rates are 2 pkt/s?Performance with Sliding WindowOverviewEncodingNon-Return to Zero (NRZ)Clock CoordinationClock RecoveryNon-Return to Zero Inverted (NRZI)Manchester Encoding4-bit/5-bit (100Mb/s Ethernet)OverviewEthernet: CSMA/CD ProtocolMinimum Packet SizeMinimum Packet SizeMidterm Information1EE 122: Midterm ReviewIon StoicaTAs: Junda Liu, DK Moon, David Zatshttp://inst.eecs.berkeley.edu/~ee122/fa09 (Materials with thanks to Vern Paxson, Jennifer Rexford,and colleagues at UC Berkeley)2AnnouncementsMidterm InformationDate: 19 October 2008Time: 4:00 PM to 5:30 PMClosed book, open 8.5” x 11” crib sheet (both sides)No Blue Books; all answers on exam sheets we hand outNo calculators, PDAs, cell phones with cameras, etc.Please use PENCIL and bring ERASERIon, one additional office hour on Monday: 1-3pm3OverviewLayering and e2e ArgumentLittle TheoremPacket delaysIP Forwarding and AddressingStop-and-Wait and Sliding WindowBit encodingCSMA/CD & Ethernet & Ethernet4Layering: The ProblemRe-implement every application for every technology?No! But how does the Internet architecture avoid this?Telnet FTP NFSPacketradioCoaxial cableFiberopticApplicationTransmissionMediaHTTP5Layering: SolutionIntroduce an intermediate layer that provides a single abstraction for various network technologiesNew application just need to be written for intermediate layerNew transmission media just need to provide abstraction of intermediate layerSMTP SSH NFSPacketradioCoaxial cableFiberopticApplicationTransmissionMediaHTTPIntermediate layer6LayeringLayering is a particular form of modularizationSystem is broken into a vertical hierarchy of logically distinct entities (layers)Service provided by one layer is based solely on the service provided by layer belowRigid structure: easy reuse, performance suffers7Layering: InternetUniversal Internet layer:Internet has only IP at the Internet layerMany options for modules above IPMany options for modules below IPInternetNet access/PhysicalTransportApplicationIPLANPacketradioTCP UDPTelnet FTP DNS8Hourglass9Implications of HourglassSingle Internet layer module:Allows networks to interoperateAny network technology that supports IP can exchange packetsAllows applications to function on all networksApplications that can run on IP can use any networkSimultaneous developments above and below IP10E2E Arguments: Where to Place Functionality?Most influential paper about placing functionality is “End-to-End Arguments in System Design” by Saltzer, Reed, and Clark“Sacred Text” of the InternetEndless disputes about what it meansEveryone cites it as supporting their position11E2E Arguments: Moderate InterpretationThink twice before implementing functionality in the networkIf hosts can implement functionality correctly, implement it a lower layer only as a performance enhancementBut do so only if it does not impose burden on applications that do not require that functionality12OverviewLayering and e2e ArgumentLittle TheoremPacket delaysIP Forwarding and AddressingStop-and-Wait and Sliding WindowBit encodingCSMA/CD & Ethernet13Little’s TheoremAssume a system (e.g., router, network, checkout line in a supermarket) at which packets arrive at rate a(t)Let d(i) be the delay or service time of packet i , i.e., time packet i spends in the systemWhat is the average number of packets in the system? systema(t) – arrival rated(i) = delay of packet iIntuition:Assume arrival rate is a = 1 packet per second and the delay of each packet is s = 4 secondsWhat is the average number of packets in the system?14ExampleArrival rate = 1; delay = 4Time = 015ExampleArrival rate = 1; delay = 4Time = 1delay = 116ExampleArrival rate = 1; delay = 4Time = 2delay = 1delay = 217ExampleArrival rate = 1; delay = 4Time = 3delay = 2delay = 3delay = 118ExampleArrival rate = 1; delay = 4Time = 4delay = 3delay = 4delay = 2delay = 119ExampleArrival rate = 1; delay = 4Time = 4delay = 3delay = 2delay = 1Q: What is the average number of packets in system?A: number_of_packets_in_system = avg_arrival_rate x avg_delay20OverviewLayering and e2e ArgumentLittle TheoremPacket DelaysIP Forwarding and AddressingStop-and-Wait and Sliding WindowBit encodingCSMA/CD & Ethernet21DefinitionsLink bandwidth (capacity): maximum rate (in bps) at which the sender can send data along the link Propagation delay: time it takes the signal to travel from source to destinationPacket transmission time: time it takes the sender to transmit all bits of the packetQueuing delay: time the packet need to wait before being transmitted because the queue was not empty when it arrivedProcessing Time: time it takes a router/switch to process the packet header, manage memory, etc22Sending One PacketR bits per second (bps)T secondsP bitsBandwidth: R bpsPropagation delay: T sectimeTransmission time = P/RTPropagation delay =T = Length/speed1m/speed = 3.3 usec in free space 4 usec in copper 5 usec in fiber23QueueingThe queue has Q bits when packet arrives  packet has to wait for the queue to drain before being transmittedP bitstimeP/RTQ bitsQueueing delay = Q/RCapacity = R bpsPropagation delay = T sec24Packet 1Packet 1Store & ForwardPacket 1Queuing & processing delay of Packet 1 at Node 2Host 1 Host 2Node 1


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Berkeley ELENG 122 - Midterm Review

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