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Note for Computer Network - CN By Sangu Mallikharjuna reddy

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UNIT – IV The Medium Access Control Sublayer-The Channel Allocation Problem-Static Channel Allocation-Assumptions for Dynamic Channel Allocation, Multiple Access Protocols-Aloha-Carrier Sense Multiple Access Protocols-Collision-Free Protocols-Limited Contention Protocols-Wireless LAN Protocols, Ethernet-Classic Ethernet Physical Layer-Classic Ethernet MAC Sublayer Protocol-Ethernet PerformanceFast Ethernet Gigabit Ethernet-10-Gigabit Ethernet-Retrospective on Ethernet, Wireless Lans-The 802.11 Architecture and Protocol Stack-The 802.11 Physical Layer-The802.11 MAC Sublayer Protocol-The 805.11 Frame Structure-Services THE MEDIUM ACCESS CONTROL SUBLAYER The protocols used to determine who goes next on a multi-access channel belong to a sublayer of the data link layer called the MAC (Medium Access Control) sublayer. • The MAC sublayer is especially important in LANs, particularly wireless ones because wireless is naturally a broadcast channel. • WANs, in contrast, use point-to-point links, except for satellite networks. THE CHANNEL ALLOCATION PROBLEM Static Channel Allocation • If there are N users, the bandwidth is divided into N equal-sized portions, with each user being assigned one portion. Since each user has a private frequency band, there is now no interference among users. • When there is only a small and constant number of users, each of which has a steady stream or a heavy load of traffic, this division is a simple and efficient allocation mechanism. • A wireless example is FM radio stations. Each station gets a portion of the FM band and uses it most of the time to broadcast its signal. • when the number of senders is large and varying or the traffic is suddenly changing(burst of data), FDM presents some problems. • If the spectrum is cut up into N regions and fewer than N users are currently interested in communicating, a large piece of valuable spectrum will be wasted. And if more than N users want to communicate, some of them will be denied permission for lack of bandwidth, even if some of the users who have been assigned a frequency band hardly ever transmit or receive anything. • A static allocation is a poor fit to most computer systems, in which data traffic is extremely bursty, often with peak traffic to mean traffic ratios of 1000:1. Consequently, most of the channels will be idle most of the time. • The poor performance of static FDM can easily be seen with a simple queuing theory calculation. Let us start by finding the mean time delay, T, to send a frame onto a channel of capacity C bps. • • • We assume that the frames arrive randomly with an average arrival rate of λ frames/sec, and that the frames vary in length with an average length of 1/μ bits. With these parameters, the service rate of the channel is μC frames/sec. A standard queueing theory result is T = 1/(μC − λ) Now let us divide the single channel into N independent subchannels, each with capacity C /N bps. The mean input rate on each of the subchannels will now be λ/N. Recomputing T, we get TN = 1/ (μ(C /N) − (λ/N)) = N/(μC − λ) = NT

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• • • • • Ex: if C is 100 Mbps, the mean frame length, 1/μ, is 10,000 bits, and the frame arrival rate, λ, is 5000 frames/sec, then T = 1/(μC − λ) T = 200 μsec. Now if 100-Mbps network with 10 networks of 10 Mbps each and statically allocate each user to one of them, the mean delay would jump from 200 μsec to 2 msec. TN = 1/ (μ(C /N) − (λ/N)) = N/(μC − λ) =NT=200*10=2000 μsec =2msec. The mean delay for the divided channel is N times worse than that of without dividing. (TN = NT) • This same result says that a bank lobby full of ATM machines is better off having a single queue feeding all the machines than a separate queue in front of each machine. Assumptions for Dynamic Channel Allocation • There are the following five key assumptions: • Independent Traffic • Single Channel. • Observable Collisions. • Continuous or Slotted Time. • 1. Carrier Sense or No Carrier Sense The model consists of N independent stations each with a program or user that generates frames for transmission. Once a frame has been generated, the station is blocked and does nothing until the frame has been successfully transmitted. 2. A single channel is available for all communication. All stations can transmit on it and all can receive from it. 3. All stations can detect that a collision has occurred. A collided frame must be transmitted again later. 4. Time may be assumed continuous, frame transmission can begin at any instant. Alternatively, time may be slotted or divided into discrete intervals (called slots). Frame transmissions must then begin at the start of a slot. 5. With the carrier sense assumption, stations can tell if the channel is in use before trying to use it. Station will transmit only when channel is free. MULTIPLE ACCESS PROTOCOLS ALOHA • Abramson’s work, called the ALOHA system, used ground based radio broadcasting. The basic idea is applicable to any system in which uncoordinated users are competing for the use of a single shared channel. • Two versions of ALOHA: pure and slotted. They differ w r t whether time is continuous, as in the pure version, or divided into discrete slots into which all frames must fit.(slotted aloha) Pure ALOHA • In the ALOHA system, after each station has sent its frame to the central computer, this computer rebroadcasts the frame to all of the stations. A sending station can thus listen for the broadcast from the hub to see if its frame has received properly.

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• In other systems, such as wired LANs, the sender might be able to listen for collisions while transmitting. • If the frame was destroyed, the sender just waits a random amount of time and sends it again. The waiting time must be random or the same frames will collide over and over, in lockstep. Systems in which multiple users share a common channel in a way that can lead to conflicts are known as contention systems. • In pure ALOHA, frames are transmitted at completely arbitrary times • We have made the frames all the same length because the throughput of ALOHA systems is maximized by having a uniform frame size rather than by allowing variablelength frames. • Whenever two frames try to occupy the channel at the same time, there will be a collision and both will be garbled. If the first bit of a new frame overlaps with just the last bit of a frame that has almost finished, both frames will be totally destroyed (i.e., have incorrect checksums) and both will have to be retransmitted later. • The checksum does not (and should not) distinguish between a total loss and a near miss. Let the ‘‘frame time’’ denote the amount of time needed to transmit the standard, fixedlength frame (i.e., the frame length divided by the bit rate). At this point, we assume that the new frames generated by the stations are well modeled by a Poisson distribution with a mean of N frames per frame time. (The infinite population assumption is needed to ensure that N does not decrease as users become blocked.) • • If N > 1, the user community is generating frames at a higher rate than the channel can handle, and nearly every frame will suffer a collision. For reasonable throughput, we would expect 0 < N < 1. • In addition to the new frames, the stations also generate retransmissions of frames that previously suffered collisions. Let us further assume that the old and new frames combined are well modeled by a Poisson distribution, with mean of G frames per frame time. Clearly, G ≥ N. At low load (i.e., N ∼ 0), there will be few collisions, hence few retransmissions, so G ∼N. • • At high load, there will be many collisions, so G > N. Under all loads, the throughput, S = GP0, where P0 is the probability that a frame does not suffer a collision

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• • The probability that k frames are generated during a given frame time, in which G frames are expected, is given by the Poisson distribution Pr[k ] =Gk e−G /k! so the probability of zero frames is just e−G. In an interval two frame times long, the mean number of frames generated is 2G. • The probability of no frames being initiated during the entire vulnerable period is thus given by P0 = e−2G. Using S = G P0, we get S = Ge−2G The maximum throughput occurs at G = 0.5, with S = 1/2e, which is about 0.184. In other words, the best we can hope for is a channel utilization of 18%. This result is not very encouraging, but with everyone transmitting at will, we could hardly have expected a 100% success rate. Slotted ALOHA • Divide time into discrete intervals called slots, each interval corresponding to one frame. • A station is not permitted to send whenever the user types a line. Instead, it is required to wait for the beginning of the next slot. Thus, the continuous time ALOHA is turned into a discrete time one. • This halves the vulnerable period.(t instead of 2t) • The probability of no other traffic during the same slot as our test frame is then P0 = e−G

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