1. Multimedie- och kommunikationssystem, lektion 9 Kapitel 8: LAN. Multiple-access. CSMA/CD. The spanning tree algorithm.

2. Multiple Access

3. Figure 13.1 Multiple-access protocols

4. Evolution of random access protocols qAloha m“Try and error”. Developed in 1970 to be used on radio- LAN on Hawaiian islands. The access to the channel is random. qSlotted Aloha mImprovement to Aloha: Start transmission only at fixed time slots qCarrier Sense Multiple Access (CSMA) mStart transmission only if no transmission is ongoing qCSMA/CA=Collision Avoidance mUsed in today’s WLAN:s qCSMA/CD=Collision Detection mStop ongoing transmission if collision is detected mUsed in the Ethernet protocol

5. Figure 13.5 Collision in CSMA

6. Animeringar Animeringar som illustrerar tystnadsdetektering i CSMA: mwww.itm.mh.se/~mageri/animations/netbook/anim06_2- csma.movwww.itm.mh.se/~mageri/animations/netbook/anim06_2- csma.mov mwww.itm.mh.se/~mageri/animations/bjnil/anim1long.exewww.itm.mh.se/~mageri/animations/bjnil/anim1long.exe

7. Figure 8.1 CSMA/CD worst-case collision detection.

8. Figure 13.8 CSMA/CA procedure

9. CSMA/CD qSense for carrier. qIf carrier present, wait until carrier ends. qSend packet and sense for collision. qIf no collision detected, consider packet delivered. qOtherwise, abort immediately, perform “exponential back off” and send packet again. qCSMA/CD is used in traditional Ethernet LAN Animering som illustrerar kollisionshantering i CSMA/CD: mwww.itm.mh.se/~mageri/animations/bjnil/anim1.exewww.itm.mh.se/~mageri/animations/bjnil/anim1.exe

10. Figure 8.4 CSMA/CD MAC sublayer operation: (a) transmit;

11. Exponential Back-off qWhen a sender detects a collision, it sends a “jam signal”. mJam signal is necessary to make sure that all nodes are aware of the collision mLength of the jam signal 48 bits qWhen collision is detected, the sender resends the signal after a random time mThe random time is picked from an interval of 0 to 2 N x maximum propagation time mN is the number of attempted retransmission mLength of the interval increases with every retransmission

12. Figure 8.4 CSMA/CD MAC sublayer operation: (b) Receive.

13. Figure 8.31 LAN protocols: (a) protocol framework;

14. IEEE standards for LANs and similar technologies. IEEE 802.1 Station management 802.1d Transparent bridges 802.2 Logical link control (LLC) IEEE 802.3 CSMA/CD (Ethernet) bus IEEE 802.3u Fast Ethernet IEEE 802.3x Hop-by-hop switch flow control IEEE 802.3z Gigabit Ethernet IEEE 802.5 Token ring IEEE 802.11 Wireless LANs IEEE 802.15 Wireless Personal Area Networks (PANs) IEEE 802.16 Broadband Wireless Access (”WiMAX”) IEEE 802.20 Mobile Broadband Wireless Access

15. Traditional Ethernet qWork started back in 1973 by Bob Metcalfe and David Boggs from Xerox Palo Alto Research Center, as an improvement of the ALOHA qExperimental Ethernet implemented in 1975. qCooperative effort between Digital, Intel, and Xerox produced Ethernet Version 1.0 in 1980. qEthernet was adopted with modifications by the standards committees IEEE 802.3 and ANSI 8802/3. q Structure of Ethernet frame (Length)

16. Structure of Ethernet Frame qPreamble: m7 bytes with pattern 10101010 followed by one byte with pattern 10101011 mUsed to synchronize receiver, sender clock rates qAddresses: 6 bytes, the frame is received by all adapters on a LAN and dropped if address does not match qType: 2 bytes, is actually a length field in 802.3 qCRC: 4 bytes, checked at receiver, if error is detected, the frame is simply dropped qData payload: maximum 1500 bytes, minimum 46 bytes. If data is less than 46 bytes, pad with zeros to 46 bytes

17. Figure 14.2 802.3 MAC frame

18. Figure 14.3 Minimum and maximum length

19. Figure 14.10 Categories of traditional Ethernet

20. Figure 14.12 Connection of stations to the medium using 10Base2

21. Reflektioner Animering: Se www.itm.mh.se/~mageri/animations/ledningsreflex/www.itm.mh.se/~mageri/animations/ledningsreflex/

22. Classic 10Mbps Ethernet qFour different implementation at the physical layer for the baseband 10Mbps Ethernet mThick Ethernet (10base5) – obsolete Thick coaxial cable (0.5” diameter) 500meter max length, bus physical topology mThin Ethernet (10base2 802.3a) - obsolete RG58 coaxial cable 185 meter max length, bus physical topology mTwisted Pair Ethernet (10baseT 802.3i) 4 pair UTP (unshielded twisted pair) cable 100 meter max length, star physical topology mFiber-link Ethernet (10Base-FL) Fiber cable connected to external transceiver Star topology is used

23. Fast Ethernet qGo from 10mbit/s to 100mbit/s q3 competing standards: m100Base-TX m100Base-T4 m100VG-Anylan q100Base-T4 and 100VG-Anylan are the losers (were not very well accepted). q100Base TX is the winner. It is almost a standard everywhere.

24. 100Base - TX q100 Mbps over 2 pairs of wire (just like 10base-T) qRequires Category 5 UTP wiring or STP qDe facto standard today qVery small price difference with 10Mbps-only equipment qHas clearly won over 100baseT4 and 100VG-Anylan by now

25. 100Base-FX qFast Ethernet with fiber optic cables qUses two optical fibers, one for transmission and one for reception

26. Gigabit Ethernet qProvides speeds of 1000 Mbps (i.e., one billion bits per second capacity) for half-duplex and full-duplex operation. qUses Ethernet frame format and MAC technology mCSMA/CD access method mBackward compatible with 10Base-T,100Base-T and 100BaseTX qCan be shared (hub) or switched

27. Gigabit Ethernet Implementations qFiber m1000 Base – SX Short wavelengths, two fiber-optic cables m1000 Base – LX Long wavelengths, two fiber-optic cables qCopper m1000 Base – CX Uses shielded twisted pair copper jumpers m1000 Base – TX Uses category 5 twisted pair copper cable

28. 1000Base - T qFour pairs of Category 5 UTP qIEEE 802.3ab ratified in June 1999. qCategory 5, 6 and 7 copper up to 100 meters qUses encoding scheme 4D-PAM5 qFive level of pulse amplitude modulation are used qComplicated technique

29. Limitations of Ethernet Technologies qDistance (the length of the cable) m200 m in Thin Ethernet (10Base2) m100 m in twisted pair Ethernet (10BaseT or 100BaseT or Fast Ethernet) qNumber of collisions when too many stations are connected to the same segment qThe situation is similar in other LAN technologies

30. Devices that Extend Local Networks qPhysical layer devices ( Repeaters and hubs) qMAC layer devices ( Bridges and two-layer switches ) qNetwork layer devices ( Routers and three layers switches)

31. Figure 16.2 Repeater A repeater connects segments of a LAN. A repeater forwards every frame bit-by-bit; it has no packet queues, no filtering capability and no collision detection.

32. Figure 16.3 Function of a repeater A repeater is a regenerator

33. Hubs A hub is a multiport repeater used in 10BaseT and Fast Ethernet Hubs give a possibility to have a physical star topology but logical bus topology.

34. Hub’s Limitations qHubs and repeaters resolve the problem with the distance, but does not resolve the problem with collisions. qA hub network can have lower throughput than several separate networks.  The maximum througput of the three separate networks = 3x10Mbps  The throughput of the connected network = 10Mbps

35. Bridges – A Simple Example B1 P1 P2 LAN segment 1 LAN segment 2 H1 H4 H2 H3 H6 H5  A frame from H1 to H4 is forwarded by the bridge  A frame from H1 to H3 is dropped by the bridge

36. Figure 8.12 Bridge filtering A bridge has a table used for filtering

37. Figure 16.5 Bridge A bridge has a table used in filtering decisions

38. How Does a Bridge Know Where the Nodes are Located? qOld bridges: Static tables mAdministrator typed them in and maintained them mToo much trouble mLack of flexibility qToday: Dynamic tables mBridges learn themselves mAdministrators don’t have to maintain them! mTransparent operation. Plug and Play!

39. Learning (transparent) bridge

40. Figure 8.13 Effect of dual paths on learning algorithm

41. Figure 16.10 Forwarding ports and blocking ports Dotted lines = blocking (non-active redundant) ports. May be used if one of the other bridges or links fails. Continuous black lines = forwarding (active) ports. These constitute a spanning tree (ett spännande träd) without loops.

42. The Spanning Tree Algorithm 1.Assign costs to each port, based on for example delay, distance, bandwith, or number of hops (1 per port). 2.Elect a root bridge. (The bridge with lowest ID number.) 3.Calculate the Root Path Cost for each bridge, i.e. the cost of the min- cost path (the “nearest” path) to the root. 4.Choose a root port for every bridge for minimum root path cost. 5.Chose a designated bridge for each LAN, for minimum cost between the LAN and the root bridge. Mark the corresponding port as a designated port. 6.Mark the root ports and designated ports as forwarding (active) ports, and the others as blocking (non-active) ports.

43. Figure 16.9 Applying spanning tree Root ports: Minimum one star. Designated ports: Two stars. The other ports are blocking ports.

44. Another example B3 B5 B7 B2 B1 B6 B4 B8 Cost for each port is 1 (hop-count)

45. The Root Bridge and the Spanning Tree B3 B5 B7 B2 B1 B6 B4 Root B8 B2B4B5B7 B8 B1 Spanning Tree: * * * * * * ** * A spanning tree is a connected graph which has no loops (cycles) **

46. Figure 8.14 Active topology derivation example: (a) LAN topology.

47. (b) Root port selection. PC = Port cost. RPC = Root Path Cost. RP = Root Port. Dashed lines are non-root ports.

48. (c) Designated port selection. DPC = Designated Port Cost.

49. (d) Active topology. DP = Designated ports. RP = Root ports. The rest (dashed lines) are non-active

50. Example 8.2: Spanning Tree To illustrate how the various elements of the spanning tree algorithm work, consider the bridged LAN shown in Figure 8.14(a). The unique identifier of each bridge is shown inside the box representing the bridge together with the port numbers in the inner boxes connecting the bridge to each segment. Typically, the additional bridges on each segment are added to improve reliability in the event of a bridge failure. Also, assume that the LAN is just being brought into service, all bridges have equal priority, and all segments have the same designated cost (bit rate) associated with them. Determine the active (spanning tree) topology.

52. Figure 8.17 Source routing example: (a) topology;

53. (b) Spanning tree.

54. Example 8.3: Source routing Assume the bridged LAN shown in Figure 8.17(a) is to operate using source routing. Also assume that all bridges have equal priority and all rings have the same designated cost (bit rate). Derive the following when station A wishes to send a frame to station B : (i) the active spanning tree for the LAN, (ii) all the paths followed by the single-root broadcast frame(s), (iii) all the paths followed by the all-routes broadcast frame(s), (iv) the route (path) selected by A.

56. Figure 14.17 Collision domains in a nonbridged and bridged network

57. Figure 14.18 Switched Ethernet

58. Figure 8.30 Example network configuration with a Fast Ethernet switch and 10/100BaseT hubs.

59. Figure 16.15 Virtual LANs (VLANs)

60. VLANs create broadcast domains. Note:

61. Figure 16.16 Two switches in a backbone using VLAN software