Figure 8.1 CSMA/CD worst-case collision detection.

Transcription

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

2 Figure 8.2 Hub configuration principles: (a) topology; (b) repeater schematic.

3 Figure 8.3 Ethernet/IEEE802.3 characteristics: (a) frame format; (b) operational parameters. (a) Bits /46 I/G L/C Bytes 7 1 2/6 2/6 2 46/ Preamble SFD DA SA Type/Length Data FCS 60/1512 bytes SFD = start-of-frame delimiter DA/SA = source/destination address I/G = individual (=0)/group (=1) address FCS = frame check sequence L/C = locally administered (=1)/ centrally administered (=0) (b) Bit rate Slot time Interframe gap Attempt limit Back off limit Jam size Maximum frame size (including FCS) Minimum frame size (including FCS) 10 Mbps (Manchester encoded) 512 bit times 9.6 microseconds bits 1518 bytes 512 bits

4 Figure 8.4 CSMA/CD MAC sublayer operation: (a) transmit; (b) receive.

5 Figure 8.5 Token ring network: principle of operation.

6 Figure 8.6 Token ring wiring configurations: (a) single hub; (b) station coupling unit; (c) multiple hubs/concentrators. (a) Station 1 Station 2 Station N MAC unit MAC unit MAC unit Hub/concentrator SCU SCU SCU Shielded twisted-pair (STP) drop cables 3 pairs per cable including power (b) to/from station to/from station Station coupling unit (SCU) Hub/concentrator Bypassed Inserted (c) to/from stations SCU SCU SCU SCU SCU SCU Hub/concentrator Trunk coupling unit (TCU) Bypassed Inserted Trunk cable STP or optical fiber

7 Figure 8.7 Token ring network frame formats and field descriptions: (a) token format; (b) frame format; (c) field descriptions.

8 Figure 8.8 Token ring MAC sublayer operation: (a) transmit; (b) receive.

9 Figure 8.9 Token generation and stack modifications: (a) token generation [Note: Sx = 0 if stack empty]; (b) stack modification. (a) Frame(s) queued and token received with P < Pm Transmit queued (waiting) frame(s) with P = Pr and R = 0 Pr > Rr/Pm Pr < Rr/Pm and Pr > Sx Pr < Rr/Pm and Pr = Sx Token (P = Pr, R = Rr/Pm) Token (P = Rr/Pm, R = 0) PUSH Pr to Sr PUSH P to Sx Token (P = Rr/Pm, R = 0) POP Sx PUSH P to Sx (b) Transmit token Token received with P = Sx Rr > Sr Token (P = Rr, R = 0) PUSH P to Sx Continue stacking Rr < Sr Token (P = Sr, R = Rr) POP Sx and Sr If Sx and Sr empty, cease stacking Transmit token

10 Figure 8.10 Token ring priority example. Rotation Station1 Pm = 2 F/T (P, R) Station7 Pm = 2 F/T (P, R) Station15 Pm = 4 F/T (P, R) Station17 Pm = 4 F/T (P, R) 0 T(0, 0) T(0, 0) T(0, 0) T(0, 0) 1 2 T(0, 0) Pr = 0, Rr = 0, Pm = 2 Pm > Pr F(0, 0) P = Pr = 0; Rr = 0 F(0, 4) Pr = 0, Rr = 4 Rr > Pr T(4, 0) P = Rr = 4; R = 0 Stacking St = 0, Sx = 4 F(0, 0) Pr = 0, Rr = 0, Pm = 2 Pm > Rr F(0, 2) P = Pr = 0; R = Pm = 2 T(4, 0) Pr = 4, Rr = 0, Pm = 2 Pr > Pm > Rr T(4, 2) P = Pr; R = Pm = 2 F(0, 2) Pr = 0, Rr = 2, Pm = 4 Pm > Pr F(0, 4) P = Pr = 0; R = Pm = 4 T(4, 2) Pr = 4, Rr = 2, Pm = 4 Pm = Pr F(4, 0) P = Pr = 4; R = 0 F(0, 4) Pr = 0, Rr = 4, Pm = 4 Pm = Rr F(0, 4) P = Pr = 0; R = Rr = 4 F(4, 0) Pr = 4, Rr = 0, Pm = 4 Pm > Rr F(4, 4) P = Pr; R = Pm = 4 3 F(4, 4) no frame to transmit F(4, 4) Sr = 0, Sx = 4 F(4, 4) Pr = 4, Rr = 4, Pm = 2 Pr < Rr F(4, 4) P = Pr; R = Rr F(4, 4) Pr = 4, Rr = 4 Rr = Pr T(4, 4) P = Pr = 4; R = Rr = 4 T(4, 4) Pr = 4, Rr = 4, Pm = 4 Pm = Pr F(4, 0) P = Pr; R = 0 4 F(4, 0) no frame to transmit F(4, 0) Sr = 0, Sx = 4 F(4, 0) Pr = 4, Rr = 0, Pm = 2 Pm > Rr F(4, 2) P = Pr; R = Pm F(4, 2) no frame to transmit F(4, 2) F(4, 2) Pr = 4, Rr = 2 Rr < Pr T(4, 2) P = Pr; R = Rr 5 T(4, 2) Pr = 4, Rr = 2, Sr = 0, Sx = 4 Pr = Sx, Rr > Sr T(2, 0) P = Pr = 2; R = 0 T(2, 0) Pr = 2, Rr = 0, Pm = 2 Pr = Pr F(2, 0) P = Pr; R = Rr F(2, 0) no frame to transmit F(2, 0) F(2, 0) no frame to transmit F(2, 0) Sr = 0, Sx = 2 6 F(2, 0) no frame to transmit F(2, 0) Sr = 0, Sx = 2 F(2, 0) Pr = 2, Rr = 0 Rr < Pr T(2, 0) P = Pr; R = Rr T(2, 0) no frame to transmit F(2, 0) F(2, 0) no frame to transmit F(2, 0) 7 T(2, 0) Pr = 2, Rr = 2, Sr = 0, Sx = 2 Pr = Sx, Rr < Sr T(0, 0) P = Sr; R = Rr T(0, 0) no frame to transmit T(0, 0) T(0, 0) no frame to transmit T(0, 0 T(0, 0) no frame to transmit T(0, 0 Cease stacking 8 T(0, 0) T(0, 0) T(0, 0) T(0, 0)

11 Figure 8.11 LAN interconnection: (a) repeaters; (b) bridges.

12 Figure 8.12 Transparent bridge schematic: (a) architecture; (b) application example.

13 Figure 8.13 Effect of dual paths on learning algorithm.

14 Figure 8.14 Active topology derivation example: (a) LAN topology; (b) root port selection; (c) designated port selection; (d) active topology.

15 Figure 8.14 Continued.

16 Figure 8.15 An example source routing bridged LAN: (a) topology; (b) routing table entries.

17 Figure 8.16 Token ring frame format: (a) position of routing information field; (b) structure of routing information field.

18 Figure 8.17 Source routing example: (a) topology; (b) spanning tree.

19 Figure 8.18 Typical establishmentwide LAN.

20 Figure 8.19 FDDI networking components.

21 Figure 8.20 Ring fault detection and isolation: (a) failure detection; (b) redundant ring configuration; (c) segment isolation; (d) station isolation. TCU = trunk coupling units

22 Figure 8.21 FDDI wiring schematic: (a) building; (b) establishment.

23 Figure 8.22 FDDI physical interface schematic.

24 Figure 8.23 FDDI line coding and framing detail: (a) 4B5B codes; (b) frame formats.

25 Figure 8.24 FDDI transmission example.

26 Figure 8.25 FDDI timed token rotation protocol example. Token rotation TRT Station 1 XMIT TRT Station 2 XMIT TRT Station 3 XMIT TRT Station 4 XMIT 0 T t + T l 0 T t + T l T t + T l T t + T l T t + T l 4 + T t + T l 4 + T t + T l 4 + T t + T l T t + T l 0 T t + T l T t + T l T t + T l T t + T l T t + T l 0 T t + T l T t + T l T t + T l T t + T l T t + T l 0 T t + T l 4 TRT = token rotation time XMIT = number of frames transmitted on this rotation of the token TTRT = target token rotation time T t = time to transmit the token T l = ring latency TTRT = 4 + T t + T l

27 Figure Base T: (a) use of wire pairs; (b) 8B6T encoding. (a) Station interface Hub interface Pair 1 Carrier sense/ collision detect Carrier sense/ collision detect Pair 2 Station Pair 3 Hub Pair 4 Station Hub transmissions Hub Station transmissions (b) 8-bit byte ternary symbol 0 40ns = 25 MTps (mega ternary (signals) per second) = 25 Mbaud Line idle

28 Figure BaseT transmission detail: (a) DC balance transmission rules; (b) 8B6T encoding sequence; (c) end of stream encoding.

29 Figure 8.28 Start-of-frame detail: (a) effect of NEXT; (b) preamble sequence. (a) Pair 1 Collision-detect line NEXT Pair 2 Station Pair 3 Hub Pair 4 NEXT = near-end crosstalk (b) SOS-1 SOS-1 SFD Data Pair 1 SOS-1 SOS-1 SFD Data Station Pair 3 SOS-1 SFD Data Time Pair 4 SOS = start of stream SFD = start frame delimiter

30 Figure 8.29 Fast Ethernet switch schematic. Fast Ethernet switch Control processor LIU 1 LIU N Switch memory LOU 1 LOU N Dual 100 Mbps drop cables Station 1 Station N LIU = line input unit LOU = line output unit

31 Figure 8.30 Example network configuration with a Fast Ethernet switch and 10/100BaseT hubs. Fast Ethernet switch Server Server 100 Base T Hub 100 Base T Hub 10 Base T Hub Stations Stations Stations 100Mbps duplex lines 10/100Mbps (CSMA/CD) half-duplex lines

32 Figure 8.31 LAN protocols: (a) protocol framework; (b) examples. (a) Network layer LLC sublayer MAC sublayer Link layer Convergence sublayer Physical medium-dependent sublayer Physical layer Transmission medium (b) IEEE Station management 802.1d Transparent bridges Logical link control (LLC) CSMA/CD (Ethernet) bus 802.3u Fast Ethernet 802.3x Hop-by-hop switch flow control 802.3z Gigabit Ethernet Token ring

33 Figure 8.32 Fast Ethernet media-independent interface. Transmit clock (TxClk) Transmit data Transmit data (TxD<0..3>) Transmit enable Transmit enable (TxEn) Carrier sense Carrier sense (CRS) MAC sublayer Collision detect Convergence sublayer Collision detect (CD) PMD Sublayer Receive clock (RxClk) Receive data Receive data (RxD<0..3>) Receive error Receive data available (RDv) Receive error (RErr) CSMA/CD MAC sublayer interface Media-independent interface

34 Figure 8.33 MAC user service primitives: (a) CSMA/CD; (b) token ring. (a) LLC layer MAC layer Correspondent s LLC layer MA_UNITDATA.request MA_UNITDATA.confirm MA_UNITDATA.indication (b) MA_UNITDATA.request MA_UNITDATA.indication MA_UNITDATA.confirm

35 Figure 8.34 LLC/MAC sublayer interactions. Source DTE Destination DTE Network LLC MAC LLC Network L_DATA.request (NPDU) MA_UNITDATA.req (UI) MA_UNITDATA.ind (UI) L_DATA.indication (NPDU) L_DATA.indication (NPDU) MA_UNITDATA.ind (UI) MA_UNITDATA.req (UI) L_DATA.request (NPDU) NPDU = network layer protocol data unit

36 Figure 8.35 Interlayer primitives and parameters. ECB = event control block

37 Figure 8.36 Example enterprise network architecture. Intersite communications facility Site LAN GW GW Site LAN GW Site LAN GW = intersite gateway (remote bridge or IP/IPX router)

38 Figure 8.37 Inverse multiplexing: (a) principle of operation; (b) reassembly schematic; (c) bonding protocol.

39 Figure 8.38 Public frame-relay network schematic.

40 Figure 8.39 Frame relay principles: (a) frame format; (b) frame routing; (c) frame relay schematic.

41 Figure 8.40 Schematic of large multisite enterprise network based on multiplexers and high bit rate leased circuits. Intersite enterprise backbone network Site A Site B Site telephony PBX MUX DS1/3 or E1/3 digital leased circuits MUX PBX Site telephony SM SM FRA FRA Site LAN RB RB Site LAN FRA SM MUX Site C RB Site LAN PBX Site telephony SM = subrate multiplexer FRA = frame relay adapter RB = remote bridge MUX = site multiplexer

42 Summary Figure 8.41 Summary of the topics discussed in this chapter relating to enterprise networks. Enterprise networks PBX (telephony) (Chapter 7) LANs (data) Legacy LANs Ethernet Token ring LAN interconnection Transparent bridges Source routing bridges FDDI backbone networks High-speed LANs Fast Ethernet hubs/repeaters Fast Ethernet switches Gigabit Ethernet Enterprise LAN interconnection technologies Remote bridges/routers (Chapter 9) ISDN connections Frame relay High bit rate leased circuits Multisite enterprise network example

43 Example 8.1 A token ring network has been configured to operate with four priority classes: 0, 2, 4, and 8, with 8 the highest priority. After a period of inactivity when no transmissions occur on successive rotations of the token, four stations generate frames to send as follows: Station 1 1 frame of priority 2 Station 7 1 frame of priority 2 Station 15 1 frame of priority 4 Station 17 1 frame of priority 4 Assuming the order of stations on the ring is in increasing numerical order and that station 1 receives the token first with a zero priority and reservation field, derive and show in table form the transmissions made by each station for the next eight rotations of the token. Include in your table the values in the priority and reservation fields both as each new token is generated and as each frame circulates around the ring. Also include the actions taken by the stacking station. Answer: The transmissions made by each station for the next eight rotations of the token are shown in Figure On the first rotation of the token, station 1 seizes the token and initiates the transmission of its waiting frame. Also on this rotation the reser vation field in the frame is raised by station 7 to 2 and then by station 15 to 4.

44 8.1 Continued On the second rotation, station 1 reads the reservation field from the frame and determines it must release the token with a priority of 4. Since it is raising the ring priority, it must become a stacking station and saves the current ring priority (0) on stack St and the new priority (4) on Sx. The token then rotates and is seized by station 15. Also on this rotation station 17 raises the reservation field from 0 to 4. On the third rotation, station 15 releases the token with a priority and reservation field of 4. Station 17 therefore seizes the token and initiates the transmission of its waiting frame. On the fourth rotation, station 7 updates the reservation field from 0 to 2 and this causes the token to be released by station 17 with the same priority (4) but a reservation value of 2. On the fifth rotation, since station 1 is a stacking station, it detects Rr is greater than Sr and hence lowers the priority of the token/ring from 4 to 2 and saves the lower priority on the stack. Station 7 is therefore able to transmit its waiting frame. On the sixth rotation, station 7 releases the token with the same priority since no reservations have been made. On the seventh rotation, station 1 detects the reservation field in the token is less than the priority field and hence reduces the priority to 0 and thereby ceases to be a stacking station. The token has thus returned to its initial state and continues rotating until further frames are generated.

45 Example 8.2 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. Answer: (i) First the exchange of configuration BPDUs will establish bridge B1 as the root bridge since this has the lowest identifier. (ii) After the exchange of configuration BPDUs, the root path cost (RPC) of each port will have been computed. These are shown in Figure 8.14(b). (iii) The root port (RP) for each bridge is then chosen as the port with the lowest RPC. For example, in the case of bridge B3, port 1 has an RPC of 1 and port 2 an RPC of 2, so port 1 is chosen. In the case of B2, both ports have the same RPC and hence port 1 is chosen since this has the smaller identifier. The selected RPs are also shown in the figure. (iv) B1 is the root bridge so all its ports have a designated port cost (DPC) of 0. Hence they are the designated ports for segments S1, S2, and S3. (v) For S4, port 1 of B5 is an RP and hence is not involved in selecting the designated port. The two other ports connected to S4 both have a DPC of 1. Hence port 2 of B3 is selected as the designated port because it has a lower identifier than B4. (vi) For S5, the only port connected to it is port 2 of B5 and hence this is selected. (vii)finally, for S6 both ports have a DPC of 1, so port 2 of B4 is selected rather than port 2 of B6. The DPCs are shown in Figure 8.14(c) and the resulting active topology is thus as shown in Figure 8.14(d). Note that the DPC of a port is always equal to the RPC of the root port of the bridge.

46 Example 8.3 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. Answer: (i) (a) Bridge B1 has the lowest identifier and is selected as the route. (b) The root ports for each bridge are then derived as shown. (c) The designated ports for each segment can now be derived and these are as shown. (d) The active topology is as shown in Figure 8.17(b). (ii) Paths of single-route broadcast frames: R1 B1 R2 B2 R3 R2 B3 R5 B6 R6 R1 B4 R4 B5 (iii)paths of all-routes broadcast frames: R6 B6 R5 B3 R2 B2 R3 B2 B1 R1 B1 B4 R4 B5 R5 B3 (iv) Since each ring has the same bit rate, the route (path) selected is either: R1 B1 R2 B3 R5 B6 R6 or R1 B4 R4 B5 R5 B6 R6

47 Example 8.4 Assuming a signal propagation delay in the fiber of 5 µs per 1 km, derive the latency of the following FDDI ring configurations in both time and bits assuming a usable bit rate of 100Mbps. (i) 2 km ring with 20 stations, (ii) 20 km ring with 200 stations, (iii) 100 km ring with 500 stations. Answer: Ring latency, T l = Signal propagation delay, T p + N station latency, T s where N is the number of stations. (i) T l = = 30 µs or 3000 bits (ii) T l = = 300 µs or bits (iii) T l = = 1000 µs or bits Note that the above values assume that the primary ring only is in use. If a fault occurred, the three signal propagation delay values would each be doubled. Also, for dual attach stations, the station latency would be doubled.

48 Example 8.5 The timed token rotation protocol is to be used to control access to a four-station FDDI ring network. All frames to be transmitted are of the same length and the TTRT to be used is equivalent to the transmission of four frames plus any ring latency. After an idle period when no frames are ready to send, all four stations receive a block of frames to send. Assuming the time for the token to rotate around the ring in the idle state is equal to the time to transmit the token, T t, plus the ring latency T 1, derive and show in table form the number of frames each station can transmit for the next four rotations of the token. Answer: The number of frames that each station can transmit on the first four rotations of the token are shown in diagrammatic form in Figure They are derived as follows. After the idle period, the TRT of all stations will be T t + T 1 since no frames have been transmitted. Once frames become ready to send, on receipt of the token station 1 computes a THT of 4 and hence transmits (XMIT) four waiting frames before passing on the token. However, since station 1 has sent four frames, none of the other stations is able to send any frames for this rotation of the token. This is so since their TRT is now greater than 4 in each case and hence their corresponding computed THT is negative. On the second rotation of the token, the TRT of station 1 has incremented to 4 plus the ring latency, hence it is not able to send any waiting frames. This means that the TRT of station 2 is less than the TTRT and the computed THT is again 4 hence it can send four waiting frames. Again this will block stations 3 and 4 from sending any waiting frames on this rotation of the token. On the third rotation of the token, the THT of stations 1 and 2 are both 4 plus the ring latency and hence neither can send frames. The TRT of station 3, however, is this time less than the TTRT and the computed THT is 4 hence it can transmit four waiting frames. Again station four is blocked. Finally, on the fourth rotation of the token, stations 1, 2, and 3 are still all blocked and station 4 can transmit four frames. This simple example shows that the available transmission capacity is shared in an equitable way between all four stations.

49 Example 8.6 Derive the maximum obtainable throughput and the maximum access delay for the following three ring configurations. Assume a TTRT of 4 ms has been chosen for each configuration. (i) 2 km ring with 20 stations, (ii) 20 km ring with 200 stations, (iii) 100 km ring with 500 stations. Answer: The three ring configurations are the same as those used in Example 8.4 and hence the same computed ring latency times will be used here. Now: Maximum available thoughput, U max = n(ttrt T l ) nttrt + T l Maximum access delay, A max = (n 1) TTRT + 2T l (i) From Example 8.4, T l = 0.03 ms. Hence: 20(4 0.03) U max = = and A max = = ms

50 8.6 Continued (ii) From Example 8.4, T l = 0.3 ms. Hence: 200 (4 0.3) U max = = and A max = = ms (iii) From Example 8.4, T l = 1 ms. Hence: 500 (4 1) U max = = and A max = = s