1. CSC 335 Data Communications and Networking Lecture 7: Local Area Networking Dr. Cheer-Sun Yang Fall 2000

2. Motivation Up to this point, we’ve talked about point-to-point communication. We many need to connect many computers together. Local Area Network(LAN): if they are located in a relatively close geographic area. Metropolitan Area Network (MAN) : extends over entire city Wide Area Network (WAN) : extends across public switching network.

3. LAN Applications (1) Personal computer LANs –Low cost –Limited data rate Back end networks and storage area networks –Interconnecting large systems (mainframes and large storage devices) High data rate High speed interface Distributed access Limited distance Limited number of devices

4. LAN Applications (2) High speed office networks –Desktop image processing –High capacity local storage Backbone LANs –Interconnect low speed local LANs –Reliability –Capacity –Cost

5. LAN Architecture Topologies Protocol architecture –Physical Layer –Media access control –Logical Link Control

6. Topologies Bus: A single communication line, typically a twisted pair, coaxial cable, or optical fiber, represents the primary medium. Ring: packets can only be passed from one node to it’s neighbor. Star: A hub or a computer is used to connect to all other computers. Tree: no loop exists (logical connection).

7. LAN Topologies

8. Frame Transmission - Bus LAN

9. Ring Topology Repeaters joined by point to point links in closed loop –Receive data on one link and retransmit on another –Links unidirectional –Stations attach to repeaters Data in frames –Circulate past all stations –Destination recognizes address and copies frame –Frame circulates back to source where it is removed Media access control determines when station can insert frame

10. Frame Transmission Ring LAN

11. Star Topology Each station connected directly to central node –Usually via two point to point links Central node can broadcast –Physical star, logical bus –Only one station can transmit at a time Central node can act as frame switch

12. Bus and Tree Multipoint medium Transmission propagates throughout medium Heard by all stations –Need to identify target station Each station has unique address Full duplex connection between station and tap –Allows for transmission and reception Need to regulate transmission –To avoid collisions –To avoid hogging Data in small blocks - frames Terminator absorbs frames at end of medium

13. Protocol Architecture Protocol layering (IEEE 802.X) –Physical Layer –Media access control (MAC) Sublayer –Logical link control (LLC)

14. IEEE 802.X IEEE 802.3 : Ethernet LAN IEEE 802.4 : Token Bus LAN IEEE 802.5 : Token Ring LAN Other Ring Networks: FDDI, Slotted Rings. IEEE 802.6 : Distributed Queue Dual Bus (DQDB) MAN standard.

15. IEEE 802 vs. OSI Fig 6.4

16. LAN Protocols in Context

17. 802 Physical Layer Design Issues Encoding/decoding Preamble generation/removal Bit transmission/reception Transmission medium and topology

18. 802 Physical Layer Required hardware for connecting a PC to Ethernet directly: –Transceiver –Attachment Unit Interface (AUI) cable –Network Interface Card (NIC) also known as Network Adapter Required hardware for connecting a PC to a remote computer: modem (with the help of PPP)

19. Bus LANs Signal balancing –Signal must be strong enough to meet receiver’s minimum signal strength requirements –Give adequate signal to noise ration –Not so strong that it overloads transmitter –Must satisfy these for all combinations of sending and receiving station on bus –Usual to divide network into small segments –Link segments with amplifies or repeaters

20. Transmission Media Twisted pair –Not practical in shared bus at higher data rates Baseband coaxial cable –Used by Ethernet Broadband coaxial cable –Included in 802.3 specification but no longer made Optical fiber –Expensive –Difficulty with availability –Not used Few new installations –Replaced by star based twisted pair and optical fiber

21. Baseband Coaxial Cable Uses digital signaling Manchester or Differential Manchester encoding Entire frequency spectrum of cable used Single channel on cable Bi-directional Few kilometer range Ethernet (basis for 802.3) at 10Mbps 50 ohm cable

22. 10Base5 Ethernet and 802.3 originally used 0.4 inch diameter cable at 10Mbps Max cable length 500m Distance between taps a multiple of 2.5m –Ensures that reflections from taps do not add in phase Max 100 taps 10Base5

23. 10Base2 Cheapernet 0.25 inch cable –More flexible –Easier to bring to workstation –Cheaper electronics –Greater attenuation –Lower noise resistance –Fewer taps (30) –Shorter distance (200m)

24. Cable Specifications for 802.3 10BaseT: 10 Mbps, baseband, unshield twisted 10Base2: 10Mbps, Cat. 2 coaxial 10Base5: 10 Mbps, Cat. 5, Cat. 5e coaxial 100BaseTX: 100 Mbps, twisted cable (Fast Ethernet) 10Broad36: maximum segment length 3600 meters

25. Gigabit Ethernet 1000Base-SX –Short wavelength, multimode fiber 1000Base-LX –Long wavelength, Multi or single mode fiber 1000Base-CX –Copper jumpers <25m, shielded twisted pair 1000Base-T –4 pairs, cat 5 UTP Signaling - 8B/10B

26. Connectors T-connector: used to form a bus topology RJ-45 connectors: for connecting a PC to another PC, Ethernet, or hub. –Cross-over: a direct connection to another PC –Straight-through: connection with the Ethernet jack or hub.

27. Repeaters Transmits in both directions Joins two segments of cable No buffering No logical isolation of segments If two stations on different segments send at the same time, packets will collide Only one path of segments and repeaters between any two stations

28. Media Access Control Sublayer Assembly of data into frame with address and error detection fields Disassembly of frame –Address recognition –Error detection Govern access to transmission medium –Not found in traditional layer 2 data link control –Also known as Contention protocols (section 3.4)

29. Collision vs. Contention When the communication link is used by one station to transmit a frame, another station connecting to the same link tries to send a packet– collision Contention: accessing the medium with the consideration that a collision may occur. Contention Protocols: the protocol is designed to deal with collision using contention. Collision-free Protocols: the protocol is designed so that collision will not occur.

30. Contention Protocols Pure ALOHA Slotted ALOHA Carrier Sense Multiple Access (CSMA) Persistent and non-persistent CSMA CSMA with Collision Detection (CSMA/CD)

31. Collision-Free Protocols A Bit-Map Protocol: reservation protocol Binary Countdown

32. Pure Aloha Packet Radio When station has frame, it sends Station listens (for max round trip time)plus small increment If ACK, fine. If not, retransmit If no ACK after repeated transmissions, give up Frame check sequence (as in HDLC)

33. Pure Aloha(cont’d) If frame OK and address matches receiver, send ACK Frame may be damaged by noise or by another station transmitting at the same time (collision) Any overlap of frames causes collision Max utilization 18% (WHY?)

34. The Efficiency of Pure Aloha G = the traffic measured as the average number of frames generated per slot S = the success rate, success frame / slot = G e –2G (pure Aloha) S = G e –G (slotted Aloha) S = Pr[no frame is generated]= e -G Pr[k frames are generated] = G k e –G / k ! This is called a probability distribution function(pdf) for Poisson distribution. (e = 2.7818…)

35. The Efficiency of Pure Aloha = G e –2G (pure Aloha) S = G * P 0 If there is no negative acknowledgement frame received after sending out one frame, the transmission is successful. So P 0 = Pr[no frames are generated in 2 time slots] = e -G * e –G = e –2G

36. The Efficiency of Pure Aloha S = G * P 0 = G e –2G (pure Aloha) We need to find the value of G such that S is maximized. S’ = G (-2) e –2G + e –2G = (1 – 2G) * e –2G Let S’ = 0 => G = ½ When G = ½, S = 1/ 2e = 0.184 = 18%

37. Slotted ALOHA A computer is not allowed to send until the beginning of the next slot. Time in uniform slots equal to frame transmission time When a frame is allowed to be transmitted, there is no collision. Need central clock (or other sync mechanism) Transmission begins at slot boundary Max utilization 37% (WHY?)

38. The Efficiency of Slotted Aloha = G e –G (slotted Aloha) S = G * P 0 If there is no other frame received after sending out one frame, the transmission is successful. So P 0 = Pr[no frames are generated in one time slots] = e -G

39. The Efficiency of Slotted Aloha S = G * P 0 = G e –G (slotted Aloha) We need to find the value of G such that S is maximized. S’ = G (-1) e –G + e –G = (1 –G) * e –G Let S’ = 0 => G = 1 When G = 1, S = 1/ e = 0.368 = 37%

40. Carrier Sense Multiple Access (CSMA) Protocols Protocols in which stations listen for a carrier (i.e., a transmission) and act accordingly are called carrier sense protocols. –1-persistent CSMA –Non-persistent CSMA –p-persistent CSMA

41. CSMA Propagation time is much less than transmission time All stations know that a transmission has started almost immediately First listen for clear medium (carrier sense)

42. If Busy? If medium is idle, transmit If busy, listen for idle then transmit immediately No ACK then retransmit If two stations are waiting, it is called a collision.

43. 1-persistent CSMA When a station has data to send, it first listens to the channel to see if anyone else is transmitting at that moment. If the channel is busy, the station waits until it becomes idle. The station retransmits with a probability of 1 when it finds that the channel is idle.

44. Non-persistent CSMA When a station has data to send, it first listens to the channel to see if anyone else is transmitting at that moment. If the channel is busy, the station waits until it becomes idle. The station does not keep trying. It waits for a random number of time and retries.

45. Non-persistent CSMA This applies to slotted channels. When a station has data to send, it first listens to the channel to see if anyone else is transmitting at that moment. If the channel is idle, it transmits with a probability p. With a probability of 1-p, it defers until the next slot. If the next slot is also idle, it transmits or defers again with probability p and q.

46. CSMA Max utilization depends on propagation time (medium length) and frame length. Longer frame and shorter propagation gives better utilization. Collisions still can be a problem, especially with p-persistent CSMA. One way to reduce the frequency of collision with CSMA is to lower the probability that a station will send when a previous is done. (Fig. 3.26) Smaller values of p => fewer collision.

47. Any Other Way? Is there another way to improve the successful rate? Yes if there is a way to detect collision prior to transmission. Why is this faster?

48. Collisions with and without Detection Without collision detection, a station must send and then wait for 2 time slots before another attempt to send. With collision detection, a station can stop transmission if collision detection requires less time than sending a frame.

49. Collision Detection On baseband bus, collision produces much higher signal voltage than signal Collision detected if cable signal greater than single station signal Signal attenuated over distance Limit distance to 500m (10Base5) or 200m (10Base2) For twisted pair (star-topology) activity on more than one port is collision Special collision presence signal

50. CSMA/CD With CSMA, collision occupies medium for duration of transmission Stations listen while transmitting If medium idle, transmit If busy, listen for idle, then transmit If collision detected, jam then ease transmission After jam, wait random time then start again –Binary exponential back off

51. CSMA/CD Operation

52. Binary Exponential Back Off If a station’s frame collides for the first time, wait 0 or 1 time slot (chosen randomly) before trying again. If it collides a second time, wait 0, 1, 2, or 3 slots (again, chosen randomly). After a third collision, wait anywhere from 0 to 2 n –1 slots if n 10, wait between 0 to 1024 (2 10 ) slots. After 16 collisions, give up. Further recovery is up to the upper layer, such as a user.

53. IEEE 802.3 Frame Format

54. Frame Format (802.3) Start of frame delimiter: 10101011 Destination address Source address Data length field Data field Pad field: the data field must be at least 46 octets. Frame check sequence: using 32-bit CRC.

55. Efficiency of 802.3(p.375) P: the probability that a frame is sent without a collision P s: the probability that a station sends The probability of a collision = 1 – P. The probability of a transmission requiring exactly N attempts = = N-1 collisions followed by a success = N * P s * ( 1- P s ) N - 1

56. Efficiency of 802.3 (p.376-377) We would like to know under what conditions the largest number of frames are sent successfully.

57. Efficiency of 802.3(p.375) The probability of a transmission requiring exactly N attempts =P = N * P s * ( 1- P s ) N – 1 dP/dP s = N(1-P s ) N-1 + N P s (N-1)(1-P s ) N-2 = N (1-P s ) N-2 [1- P s – P s (N –1) ] Let dP/dP s = 0 => P s = 1/N

58. Efficiency of 802.3 How many time slots has passed before a frame is sent successfully?

59. The Efficiency of 802.3 The contention period = the number of time slots passed before a successful transmission

60. The Efficiency of 802.3 Assume that the probability of a success in each attempt = p. The probability of a collision = 1 – p. The probability of a transmission requiring exactly i+1 attempts = P (i+1) = i collisions followed by a success = p (1- p) i

61. The Efficiency of 802.3 The contention period = = (1-p)/p

62. The Efficiency of 802.3 The contention period (C) = (1/p) –1 0 <= p <= 1 (p: the successful rate) If p -> 1 C -> 0 If p -> 0, C = large We’ve found that when P s =1/N, p is maximized. So, C = (1-1/N) 1-N -1 when p is maximized. If N->large, C = 2.718 – 1 = 1.718 (close to 2).

63. The Utilization Rate of Ethernet The percent utilization (U) does not depend on the number of stations in practice. A station will try to send regardless of how many other stations there are. The previous result often is used as benchmarks against which measures are made to estimate efficiency.

64. The Utilization Rate of Ethernet The percent utilization (U) is defined as the amount of time spent on transmitting a frame as a percentage of the total time spent on contending and transmitting. Assume: R = transmission rate F = number of bits in a frame T = slot time So U =

65. Required Reading Section 6.1, 3.4, 6.2