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

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

3. Possible Topologies 3

4. Limitations of Ethernet Maximum hosts supported:1,024 Maximum # of repeaters between two hosts: 4 Minimum distance between transceivers: 2.5m Maximum distance covered by Ethernet: depends on the cable 4

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

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

7. Ethernet Addressing Every host has a unique 48-bit Ethernet address: Example 00001000 00000000 00101011 11100100 10110001 00000010 7 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 1s is a broadcast address Address with first bit 1, is a multicast address

8. 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 … K th collision: wait for [0..2 k - 1] slots Maximum # of slots: 1023 (k = 10) After 16 trials, report error 8

9. 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 K A = 0, 1, k B = 0, 1, 2, 3 Via enumeration P(A) = 5/8 Suppose A wins backoff, and collision on next frame K A = 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)  prod i (1- 2 -i )  3/4 Known as the capture effect 9