Fundamentals of Computer Networks ECE 478578 Lecture 11

Fundamentals of Computer Networks ECE 478/578 Lecture #11: Ethernet Instructor: Loukas Lazos Dept of Electrical and Computer Engineering University of Arizona

Ethernet Components Cable segments (up to 2000 m) Transceivers : adaptor connected to the line Repeater: joins multiple segments Hubs: multi-way repeaters 2

Possible Topologies 3

Limitations of Ethernet Maximum hosts supported: 1, 024 Maximum # of repeaters between two hosts: 4 Minimum distance between transceivers: 2. 5 m Maximum distance covered by Ethernet: depends on the cable 4

Frame Format synchronization Layer of delivery Long enough for collision detection 46 size 1, 500 5

Duration of Contention Slots How long does it take for two stations to detect a collision? Α Α τ = d/c Β t 0 = 0 Β τ-ε Α Β e 2τ - ε A will detect collision after 2τ in the worst case. Minimum frame size 2τC (why? ) Ex. 1 Km cable, τ = 5μs, If C = 100 MBps, Frame size 100 bits C = 1 Gbps Frame size = 10, 000 bits 6

Ethernet Addressing Every host has a unique 48 -bit Ethernet address: Example 00001000 0000 00101011 11100100 10110001 00000010 0 8 : 0 0 : 2 b : e 4 : b 1 : 0 2 To ensure uniqueness, every Ethernet manufacturer has its own 24 bit prefix Regardless of addresses, each frame is received by EVERY adaptor An address with all 1 s is a broadcast address Address with first bit 1, is a multicast address 7

Exponential backoff Medium Access Protocol: CSMA/CD Each slot is 51. 2 μs long In case of collision First collision: wait for either 0 or 1 slot Second collision: wait for either 0, 1, 2, 3 slots … Kth collision: wait for [0. . 2 k - 1] slots Maximum # of slots: 1023 (k = 10) After 16 trials, report error 8

Problem 2. 44 Stations A, B attempt to access the medium. Time slot duration: 51. 2μs Collision on 1 st frame, A wins on second frame, Collision on 3 rd frame. Probability that A transmits before B KA = 0, 1, k. B = 0, 1, 2, 3 Via enumeration P(A) = 5/8 Suppose A wins backoff, and collision on next frame KA = 0, 1, k. B = 0… 7 Via enumeration P(A) = 13/16 Probability that A will win in all remaining backoffs: At each contention P(A wins in race i) (1 - 2 -i) P(A wins) prodi(1 - 2 -i) 3/4 Known as the capture effect 9