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Midterm Review

Midterm Review. CS460. Delay. Speed of light 3.0 x 10 8 meters/second in a vacuum 2.3 x 10 8 meters/second in a cable 2.0 x 10 8 meters/second in a fiber. Delay x Bandwidth Product. Delay. Bandwidth. Sequence Number.

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Midterm Review

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  1. Midterm Review CS460

  2. Delay • Speed of light • 3.0 x 108 meters/second in a vacuum • 2.3 x 108 meters/second in a cable • 2.0 x 108 meters/second in a fiber • Delay x Bandwidth Product Delay Bandwidth

  3. Sequence Number • Suppose you are designing a sliding window protocol for a 1-Mbps point-to-point link to the moon, which has a one-way latency of 1.25 second.  How many bits do you need for the sequence number? • 2.5 seconds * 1Mbps = 2.5Mbits = 312Kbytes. We need a sequence range that will cover double this value or 624K. So 20 bits are needed.

  4. Seven Layer Model Application Presentation Session Transport Application Presentation Session Transport Email, FTP, www integer size, big endian Synchronization(video/audio), name space reliability, congestion control (TCP) Routing Address framing Errors electrical signals Network Data Link Physical Network Data Link Physical IP Ether Network Data Link Physical Network Data Link Physical

  5. Process Model • Process per protocol • Context switch at each protocol boundary • No Semaphores • Incoming and outgoing queue • Slow • Process per message • Procedure call at each protocol boundary • Semaphores needed • Faster • Error prone because of multiple threads running in a single protocol

  6. Avoiding Copies • Assume Data rate of 600Mbps=73MBps • Assume 16MHz memory bus that is 16 bits wide, this results in 32MBps • For a MB message, one copy takes 1/32 sec • The resulting maximum data rate is 32MBps • Two copies take 1/32 sec + 1/32 sec = 1/16 sec • The resulting maximum data rate is 16MBps • Copies are necessary between user and system space if no special interface

  7. Links Sometimes you install your own Sometimes leased from the phone company STS=Synchronous Transport Signal=OC=Optical carrier

  8. Non-Return to Zero (NRZ) 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 Bits NRZ • Problem: Consecutive 1s or 0s • Low signal (0) may be interpreted as no signal • High signal (1) leads to baseline wander • Unable to recover clock • NRZI eliminates consecutive 1s

  9. 4B/5B • Problem: consecutive zeros • Idea: Every 4 bits of data is encoded in a 5-bit code, with the 5-bit codes selected to have no more than one leading 0 and no more than two trailing 0 (i.e., never get more than three consecutive 0s). • Resulting 5-bit codes are then transmitted using the NRZI encoding. Achieves 80% efficiency. • We already dealt with consecutive 1s with NRZI

  10. Bit stuffing Example • Original Data • 0011111011000011111100 • Bit Stuffed • 001111100110000111110100 • Receiver • 01111110 00111110110000111110100 01111110

  11. Sender: • multiply M(x) by xk; for our example, we get • x10 + x7 + x6+ x4(10011010000); • divide result by C(x) (1101); • Send 10011010000 - 101 = 10011010101, since this must be exactly divisible by C(x); 11111001 10011010 000 Message 1101 1001 1101 1000 1101 1011 1101 1100 1101 1 000 1 101 101 Remainder Generator 1101

  12. Class First 4 bits # Network Bits # Host Bits Network Number A 0xxx 7 24 1 to 127 B 10xx 14 16 128 to 191 C 110x 21 8 192 to 223 D 1110 28 Multicast 224 to 239 IP Addresses

  13. Class Maximum # of Networks Maximum # of Hosts Address Range Network Address Host Address A 128 16,777,214 1.*.*.* to 127.*.*.* a b.c.d B 16,384 65,534 128.*.*.* to 191.*.*.* a.b c.d C 2,097,152 254 192.*.*.* to 223.*.*.* a.b.c d IP Addresses

  14. Example A (3,3,0) B B3 (5,5,0) (7,7,0) (3,3,0) (5,5,0) C B5 (2,2,0) (5,5,0) D B7 K B2 (7,7,0) (2,2,0) (7,7,0) E F (1,1,0) (1,1,0) (1,1,0) B1 (1,1,0) G H (6,6,0) (4,4,0) (4,4,0) B6 B4 (6,6,0) (4,4,0) I J

  15. Next Phase A (3,2,1) B B3 (5,1,1) (7,1,1) (5,1,1) C B5 (2,1,1) D B7 K B2 (7,1,1) E F (1,1,0) (1,1,0) (1,1,0) B1 (1,1,0) G H (4,1,1) B6 B4 (6,1,1) (4,1,1) I J

  16. Next Phase A B B3 (5,1,1) (5,1,1) C B5 (2,1,1) (7,1,1) D B7 K B2 E F (1,1,0) (1,1,0) (1,1,0) B1 (1,1,0) G H (4,1,1) B6 B4 (4,1,1) I J

  17. Routers? Hubs? Switches D 128.187.174.10 D 44.fe.34.56.32.d5 S 128.187.171.2 S fe.34.56.32.d5.29 D 128.187.174.10 D 44.fe.34.56.32.d5 S 128.187.171.2 S fe.34.56.32.d5.29 173 171 Hub H1 H2 H3 H7 H8 H9 172 174 H4 H5 H6 H10 H11 H12 D 128.187.174.10 D 44.fe.34.56.32.d5 S 128.187.171.2 S fe.34.56.32.d5.29 D 128.187.174.10 D 44.fe.34.56.32.d5 S 128.187.171.2 S fe.34.56.32.d5.29 H10= IP 128.187.174.10, Ethernet 44.fe.34.56.32.d5

  18. Forwarding • H1 LANE layer receives a packet for H2 • Since H2 is not in our cache, we send query to LES for H2 ATM address • Since the query and VC setup may take seconds, we start sending data through the BUS • We may wait until we receive several packets before setting up a separate virtual circuit • The LES replies with the ATM address of H2 • H1 sets up a direct VC to H2 and adds an entry in the cache for H2

  19. More Questions • What are the advantages and disadvantages of statistical multiplexing? • What are the three kinds of failures that network designers have to worry about? • Why is the delay x bandwidth product important? • What is jitter? Why would it be difficult to send video over a network with a high jitter value?

  20. More Questions • Given a source routing example, determine the destination that a message will go to. (Something like Figure 3.7) • Given a virtual circuit routing example, determine the destination that a message will go to. (Something like Figure 3.5) • Compare datagrams, virtual circuits and source routing. What are the advantages and disadvantages of each approach?

  21. More Questions • What is the spanning tree algorithm and which links would be eliminated in a problem similar to Figure 3.11? How do bridges learn host locations? • Why does an ATM cell have 48 bytes of payload? • What kind of routing does ATM use? • Given a banyan network similar to Figure 3.33, decide if there will be a contention free route for two packets.

  22. More Questions • What is the hidden node problem (page 138) and how does 802.11 deal with it? • Describe the process for selecting an AP with a mobile node (pages 140-141). • What is an internetwork? • What is the service model for IP? • How is fragmentation performed with IPv4? • Assume a host with an IP address of 222.123.250.3. To what class does this IP address belong? To what network is it connected?

  23. More Questions • What is an ARP packet and how is it used? How would ARP be implemented in a network such as ATM that doesn’t have broadcast capabilities? • What does an ARP cache contain? How does the ARP protocol work? • Is source routing practical? Explain. • What type of IP address class would be adequate for a corporation that needs to establish up to 100 sub-networks, each of which with up to 500 hosts? Justify your choice.

  24. More Questions • Assume the following routing table: What does the router do with the following packets? • a) 128.96.39.10 b) 128.96.40.12 c) 128.96.40.151 d) 192.4.153.17 e) 192.4.153.90 What are the problems with stop and wait, and how are they solved?

  25. Architecture LES Mac Mac Mac Mac Ethernet BUS Relay ATM Switch Mac LEC ATM Plug and Play Allows ATM hosts to connect Simplified Core Switch, complex edge switch

  26. Architecture Mac Mac Mac Mac Ethernet LEC Relay LES BUS ATM Switch Mac ATM PVC static configuration Simplified Edge switch, complex core switch

  27. Architecture POS Mac Mac Mac Mac Ethernet Relay Relay Ether Switch Mac POS Simplified Edge switch, point to point link, not scalable, time division multiplexing – leased line expense

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