Lecture-08 - Lecture #08 Todays Lecture Objectives Review...

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Lecture #08
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Today’s Lecture Objectives • Review • Ethernet • Repeaters, Bridges, Switches, Routers, and Gateways • MAC Addressing
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IEEE 802 Family of Standards • 802.1: Introduction • 802.2: Logical Link Control – Interface to Network Layer • 802.3: Ethernet • 802.4: Token Ring • 802.5: Token Bus • 802.6: Distributed Queue Dual Bus • 802.7: Broadband LANs (not active) • … • 802.11: Wireless LANs
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Cabling Types 10 Base 5: Thicknet (up to 500m) 10 Base 2: Thinnet (up to 200m) 10 Base T: Cat-3 or Cat-5 100 Base T4: 4 Cat-3 (up to 100m) 00 Base TX: 2 Cat (up to 100m) 100 Base TX: 2 Cat-5 (up to 100m) 100 Base FX: full duplex multi-mode fiber (up to 2 km) 1000 Base T: Cat-5 or Cat-6 1000 Base SX: full duplex multi-mode 1000 Base LX: full duplex single-mode
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Manchester Encoding • Each bit has a transition • Allows clocks in sending and receiving nodes to synchronize to each other – no need for a centralized, global clock among nodes!
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Ethernet Frame Structure • Sending adapter encapsulates IP datagram (or other network layer protocol packet) in Ethernet frame Preamble: • 7 bytes with pattern 10101010 followed by one byte with pattern 10101011 • used to provide rough synchronization for sender and receiver
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Ethernet Frame Structure • Addresses: 6 bytes – if adapter receives frame with matching destination address, or with broadcast address (eg ARP packet), it passes data in frame to net-layer protocol – otherwise, adapter discards frame • Type: indicates the higher layer protocol (mostly IP but others may be supported such as Novell IPX and AppleTalk) • CRC: checked at receiver, if error is detected, the frame is simply dropped
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Minimum Packet Size • Why put a minimum frame size? – Give a host enough time to detect collisions • In Ethernet, minimum frame size = 64 bytes o 6 yte addresses, 2 yte type, 4 yte CRC, – two 6-byte addresses, 2-byte type, 4-byte CRC, and 46 bytes of pad • What is the relationship between minimum packet size and the length of the LAN?
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Minimum Packet Size • Why is min needed? – To detect collisions – Provide motivation on why needed
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Minimum Packet Size (more) propagation delay (d) a) Time = t; Host 1 starts to send frame Host 1 Host 2 propagation delay (d) Host 1 Host 2 b) Time = t + d; Host 2 starts to send a frame just before it hears from host 1’s frame propagation delay (d) Host 1 Host 2 c) Time = t + 2*d; Host 1 hears Host 2’s frame b detects collision LAN length = (min_frame_size)*(light_speed)/(2*bandwidth) = = (8*64b)*(2.5*10 8 mps)/(2*10 7 bps) = 6400m approx
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Exponential Backoff Algorithm • To recover from a collision… • Ethernet uses the exponential backoff algorithms to determine when a station can retransmit after a collision Algorithm: Set “slot time” equal to 512bit time After first collision wait 0 or 1 slot times After i-th collision, wait a random number between 0 and 2 i -1 time slots Do not increase random number range, if i=10 Give up after 16 collisions
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Beyond Ethernet “Plain/Slow” Ethernet: 802.3 Switched Ethernet (1995) Use buffering and scheduling to avoid collisions
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Lecture-08 - Lecture #08 Todays Lecture Objectives Review...

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