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CS 5565 Network Architecture and Protocols

CS 5565 Network Architecture and Protocols. Lecture 26 Godmar Back. Announcements. Project 2B Updated simulator available, see website Public test cases have been posted Part/1 due tonight 11:59pm, Part/2 due Wed May 4 11:59pm 2 nd Research Paper assignment

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CS 5565 Network Architecture and Protocols

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  1. CS 5565Network Architecture and Protocols Lecture 26 Godmar Back

  2. Announcements • Project 2B • Updated simulator available, see website • Public test cases have been posted • Part/1 due tonight 11:59pm, Part/2 due Wed May 4 11:59pm • 2nd Research Paper assignment • Maltz HotNets Position Paper • Current reading assignment: • Chapter 5, 6.1 – 6.3 (including CDMA & 802.11) • Final May 10, 3:25pm, MCB 216 • Please budget 3 hours. • Course evaluations May 4 • Must attend class to fill out evaluations CS 5565 Spring 2005

  3. A R B Recap: ARP Send datagram from A to B via R • Q. How would you implement ARP? • Q. What about security? CS 5565 Spring 2005

  4. Ethernet “dominant” wired LAN technology: • cheap <$20 for 100Mbps! • first widely used LAN technology • standardized in 802.3 series of IEEE standards • simpler, cheaper than token LANs and ATM • kept up with speed race: 10 Mbps – 10 Gbps Metcalfe’s Ethernet sketch CS 5565 Spring 2005

  5. Ethernet Model • Connectionless: • No handshaking between sending and receiving adapter • Unreliable: • No acks (or nacks) – higher-level protocols must compensate, but low BER (bit error rate) • CSMA/CD: • no slots • random access protocol CS 5565 Spring 2005

  6. Star Topology • Bus topology popular through mid 90s • Now star topology prevails • Connection choices: switch or hub • Hubs rarely used nowadays hub or switch CS 5565 Spring 2005

  7. Ethernet Frame Structure Sending adapter encapsulates IP datagram (or other network layer protocol packet) in Ethernet frame Preamble: • 7 bytes with pattern 10101010 followed by one byte with pattern 10101011 • used to synchronize receiver, sender clock rates  7   6   6  2  -----------  4 <05DC 802.3 Length 0800 IP 0806 ARP 809B Appletalk … CS 5565 Spring 2005

  8. 1. Adaptor receives datagram from net layer & creates frame 2. If adapter senses channel idle, it starts to transmit frame. If it senses channel busy, waits until channel idle and then transmits 3. If adapter transmits entire frame without detecting another transmission, the adapter is done with frame ! 4. If adapter detects another transmission while transmitting, aborts and sends jam signal 5. After aborting, adapter enters exponential backoff: after the mth collision, adapter chooses a K at random from {0,1,2,…,2m-1}. Adapter waits K·512 bit times and returns to Step 2 Ethernet CSMA/CD algorithm CS 5565 Spring 2005

  9. Jam Signal: make sure all other transmitters are aware of collision; 48 bits Bit time: .1 microsec for 10 Mbps Ethernet ;for K=1023, wait time is about 50 msec Exponential Backoff: Goal: adapt retransmission attempts to estimated current load heavy load: random wait will be longer first collision: choose K from {0,1}; delay is K· 512 bit transmission times after second collision: choose K from {0,1,2,3}… after ten collisions, choose K from {0,1,2,3,4,…,1023} Ethernet’s CSMA/CD (more) See/interact with Java applet on AWL Web site: highly recommended ! CS 5565 Spring 2005

  10. CSMA/CD efficiency • tprop = max prop between 2 nodes in LAN • ttrans = time to transmit max-size frame • Efficiency goes to 1 as tprop goes to 0 • Goes to 1 as ttrans goes to infinity • Much better than ALOHA, but still decentralized, simple, and cheap CS 5565 Spring 2005

  11. Efficiency of 10Mbps Ethernet • Backoff slot times is 512-bit • NB: does not imply CSMA/CD Ethernet is slotted CS 5565 Spring 2005

  12. Ethernet Physical Layer CS 5565 Spring 2005

  13. Manchester Encoding • Used in 10BaseT • Each bit has a transition • Allows clocks in sending and receiving nodes to synchronize to each other • no need for a centralized, global clock among nodes! CS 5565 Spring 2005

  14. Fast Ethernet: 100BaseT 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 • 4B5B Bit Encoding • MLT-3 (Multi-Level Threshold) Source: http://de.wikipedia.org CS 5565 Spring 2005

  15. Gigabit Ethernet • uses standard Ethernet frame format • allows for point-to-point links and shared broadcast channels • in shared mode, CSMA/CD is used; short distances between nodes required for efficiency • supports hubs, called here “Buffered Distributors” • Though you wouldn’t use them • full-duplex at 1 Gbps for point-to-point links • 1000BaseT: • 8B10B coding with 5 signal levels • See IOL page for more information CS 5565 Spring 2005

  16. Hubs • Hubs are essentially physical-layer repeaters: • bits coming from one link go out all other links • at the same rate • no frame buffering • no CSMA/CD at hub: adapters detect collisions • provides net management functionality • Single collision domain CS 5565 Spring 2005

  17. hub hub hub hub Interconnecting LANs • Hypothetical design: • backbone hub interconnects LAN segments • Disadvantage: • individual segment collision domains become one large collision domain • can’t interconnect 10BaseT & 100BaseT CS 5565 Spring 2005

  18. Switches • Link layer device • stores and forwards Ethernet frames • examines frame header and selectively forwards frame based on MAC dest address • when frame is to be forwarded on segment, uses CSMA/CD to access segment • transparent • hosts are unaware of presence of switches • plug-and-play, self-learning • switches do not need to be configured CS 5565 Spring 2005

  19. switch hub hub hub Forwarding 1 3 2 • How do determine onto which LAN segment to forward frame? • Looks like a routing problem... • NB: Switches are called bridges if #interfaces == 2 CS 5565 Spring 2005

  20. Self Learning • A switch has a switch table • entry in switch table: • (MAC Address, Interface, Time Stamp) • stale entries in table dropped (TTL can be 60 min) • switchlearns which hosts can be reached through which interface • when frame received, switch “learns” location of sender: incoming LAN segment • records sender/location pair in switch table CS 5565 Spring 2005

  21. Switch Example Suppose C sends frame to D address interface switch 1 1 1 2 3 A B E G 3 2 hub hub hub A I F D G B C H E • Switch receives frame from C • notes in bridge table that C is on interface 1 • because D is not in table, switch forwards frame into interfaces 2 and 3 • frame received by D CS 5565 Spring 2005

  22. Switch Example (2) Suppose D replies back with frame to C. address interface switch 1 1 2 3 1 A B E G C hub hub hub A I F D G B C H E • Switch receives frame from from D • notes in bridge table that D is on interface 2 • because C is in table, switch forwards frame only to interface 1 • frame received by C CS 5565 Spring 2005

  23. switch hub hub hub Switch: Traffic Isolation • switch installation breaks subnet into LAN segments • switch filters packets: • same-LAN-segment frames not usually forwarded onto other LAN segments • segments become separate collision domains collision domain collision domain collision domain CS 5565 Spring 2005

  24. Switches: Dedicated Access A • Switch with many interfaces • Hosts have direct connection to switch • No collisions; full duplex • A-to-E and B-to-F simultaneously • cut-through switching: frame forwarded from input to output port without first collecting entire frame • slight reduction in latency C’ B switch D F E This is the dominant technology today CS 5565 Spring 2005

  25. Switches: Flow Control • What if A=100Mbps, E=10Mbps? • host sends frames faster than switch can send them out of outgoing port • undetected losses due to overflow could occur • IEEE 802.3 provides LLC layer that has flow control facilities • “pause” frames to get sender to stop CS 5565 Spring 2005

  26. Summary • Ethernet • Interconnecting LANs • Hubs vs. Switches CS 5565 Spring 2005

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