CS 640 Lecture 4: 09/11/2014

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1 CS 640 Lecture 4: 09/11/2014 A) Bandwidth-delay product B) Link layer intro C) Encoding, Framing, Error Detection D) Multiple access Ethernet A. Bandwidth-delay product This in the above example is C * D (the definitions of C and D change similarly for multi-link paths as above). When an application has C * D bits outstanding that is, these bytes have left the sender and are yet to reach the receiver, then it is keeping pipe full, or fully utilizing the available capacity. If the application keeps any fewer bytes outstanding, then it isn t fully utilizing the network. Examples C = 1 Gbps, D = 50 ms, BW-delay = 50 Mb or 6.25 MB C = 10 Gbps, D = 1 µs, BW-delay = 10 Kb or 1.25 KB (This is usually seen in data centers) C = 1 Mbps, D = 500 ms, BW-delay = 0.5 Mb or 62.5 KB (This is usually seen in satellites) B. Link layer intro Problem statement: getting packets across the physical network For now, we will focus on one segment of a network i.e., there are no switches Key issues that arise here: links operate on signal electromagnetic. We are dealing with digital info encoding we are sending packets over the links need other end to know when packet was received framing there may be errors in transmission or reception what to do? error detection there may be multiple senders contending on a shared medium. How to get people to take turns and share fairly multiple access

2 The rest of the lecture is about how the link layer supports these. C.1 Encoding Problem statement: get the underlying link to send 0s and 1s. step 1: represent zero and one using physical link level signals. step 2: transmit the bits Often you need to do more. Some of the approaches are NRZ encoding(non Return to Zero): zero - low; 1 high. This has two issues: baseline wander and timing. Both arising with long sequence of 0s or 1s. NRZI (NRZ Inverted) : transition from current so signal 1; no transition for 0. Solves the long sequence of 1s, but not the 0s. Manchester encoding: Every bit has a transition in the middle. Zero low to high; 1 high to low. Downside receiver has half the time to detect signal change. Therefore capacity offered is low. 4B/5B. Match each 4Bits string to 5Bit string. Ensure that the encoded string has no more than one leading 0 or two trailing 0s. The transmit using NRZI. NRZI - solves the many 1s problem. The property of not too many leading trailing 0s in the encoding strings solves the many 0s problems. Manchester Early Ethernet 4B/5B; 10B/12B modern high speed Ethernet. C.2 Framing Problem statement: need the other end of the link to know when a packet transmission started and ended. What makes this challenging? Packets can be of different length and knowing start is hard are you just hearing noise? Approach: add delimiters Byte oriented add special bytes at the beginning and end. If the special byte appears in payload then you would need to escape it. Another approach have a starting special byte, and then include a byte counter.

3 Bit oriented add special bit sequences at beginning and end. You would need to escape individual bits in case they appear in the payload. Enough to understand this idea at this high level. C.3 Error detection Classic problem a lot of this is math, so we will gloss over. In networks there are many choices: don t detect errors. Let higher layer deal with it end to end. detect errors and retroactively correct it. proactively correct errors. Generally later two are used in many network technologies even though end to end sanity checks take place. This is for cost and efficiency reasons. Error detection is common. Some possible approaches: Add parity bits. Add parity bytes. Compute some function over payload and add that to end of message receiver checks. Common ideas here: check sum IP layer uses this. Simple to implement but not robust. cyclic redundancy check Ethernet uses this. A little more complex, but more robust to more different error patterns. D. Multiple access Ethernet One of the most popular LAN technology. Designed for multiple access over a shared medium Problem: how to arbitrate to ensure fair and efficient access? Ethernet presents one way of doing this. We will consider another way in the context of WiFi. Ethernet - also called CSMA-CD: Carrier sense multiple access with collision detection CS sender knows when medium is busy/idle CD sender also listens while transmitting, so can identify when its frame has interfered with someone else.

4 D.1 Frame format 64 bit preamble Dest Addr Src Addr Type Body CRC Addresses are the 6 byte LL addresses we talked about in last class. Preamble is something two nodes use to identify start of frame. Ethernet is bit oriented. D.2 Receiver Simple each frame is received by all connected nodes. Nodes recognize frames: addressed to them sent to a special broadcast address sent to a multicast group address that they are part of. That s all a receiver does. D.3 Transmitter Has all the smarts. When sender has frame to send and line is idle, sends immediately. If busy, then wait for idle and send immediately What can go wrong? Because there is no centralized control two senders can send at the same time. Sender detects a collision; transmits a jamming sequence of 32 bits. Thus a sender will minimally transmits 96bits preamble + jamming sequence. What is the worst case here? Senders A and B are at either end of an Ethernet segment and so it takes them a while to detect collisions. Say maximum propagation delay across the segment is D. Say one sender A starts sending a frame at t. First byte of this frame reaches the other end at t + D. Say just prior to this, a different sender B at the other end transmits a signal, detects a collision and transmits a jamming frame

5 Unfortunately the first sender A will not know of this until t+2d. A needs to be transmitting at t+2d to know this is a collision. Hence we need Ethernet frames to be long enough for transmissions to span 2D so that collisions can be detected. In Ethernet segments that are at most 10 KM long, 2D = 51.2 µs. At 10 Mbps, therefore, the frame needs to be at least 512 bits long. When collision is detected, sender does binary exponential back off. At the ith try, the send waits for a period chosen from {0, 1 * 51.2microsec,, 2^(i-1) * 51.2} and tries again.

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