L10-11

Examples include cellular wireless networks including

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Unformatted text preview: in popularity, uses radio. Examples include cellular wireless networks (including standards like EDGE, 3G, and 4G), wireless LANs (such as 802.11, the WiFi standard), and various other forms of radio-based communication. Broadcast is an inherent property of radio communication, especially with so-called omni-directional antennas, which radiate energy in all (or many) different directions. However, radio broadcast isn’t perfect because of interference and the presence of obstacles on certain paths, so different nodes may correctly receive different parts of any given transmission. This reception is probabilistic and the underlying random processes that generate bit errors are hard to model. Shared bus networks. An example of a wired shared medium is Ethernet, which when it was first developed (and for many years after) used a shared cable to which multiple nodes could be connected. Any packet sent over the Ethernet could be heard by all stations connected physically to the network, forming a perfect shared broadcast medium. If two or more nodes sent packets that overlapped in time, both packets ended up being garbled and received in error. Over-the-air radio and television. Even before data communication, many countries in the world had (and of course still have) radio and television, broadcast stations. Here, a relatively small number of transmitters share a frequency range to deliver radio or television content. Because each station was assumed to be active most of the time, the natural approach to sharing is to divide up the frequency range into smaller sub-ranges and allocate each sub-range to a station (frequency division multiplexing). Given the practical significance of these examples, and the sea change in network access brought about by wireless technologies, developing methods to share a common medium is an important problem. LECTURE 10. SHARING A COMMON MEDIUM: 4 MEDIA ACCESS PROTOCOLS ￿ 10.2 Performance Goals An important goal is to provide high throughput, i.e., to deliver packets successfully at as high a rate as possible, as measured in bits per second. A measure of throughput that is independent of the rate of the channel is the utilization, which is defined as follows: Definition. The utilization that a protocol achieves is defined as the ratio of the total throughput to the maximum data rate of the channel. For example, if there are 4 nodes sharing a channel whose maximum bit rate is 10 Megabits/s,2 and they get throughputs of 1, 2, 2, and 3 Megabits/s, then the utilization is (1 + 2 + 2 + 3)/10 = 0.8. Obviously, the utilization is always between 0 and 1. Note that the utilization may be smaller than 1 either because the nodes have enough offered load and the protocol is inefficient, or because there isn’t enough offered load. By offered load, we mean the load presented to the network by a node, or the aggregate load presented to the network by all the nodes. It is measured in bits per second as well. But utilization alone isn’t sufficient: we need to worry about fairness as well. If we weren’t concerned about fairness, the problem would be quite easy because we could arrange for a particular backlogged node to always send data. If all nodes have enough load to offer to the network, this approach would get high utilization. But it isn’t too useful in practice because it would also starve one or more other nodes. A number of notions of fairness have been developed in the literature, and it’s a topic that continues to generate activity and interest. For our purposes, we will use a simple, standard definition of fairness: we will measure the throughput achieved by each node over some time period, T , and say that an allocation with lower standard deviation is “fairer” than one with higher standard deviation. Of course, we want the notion to work properly when the number of nodes varies, so some normalization is needed. We will use the following simplified fairness index: ￿N 2 i=1 xi ) ￿2, F= N xi ( (10.1) where xi is the throughput achieved by node i and there are N backlogged nodes in all. Clearly, 1/N ≤ F ≤ 1; F = 1/N implies that a single node gets all the throughput, while F = 1 implies perfect fairness. We will consider fairness over both the long-term (many thousands of “time slots”) and over the short term (tens of slots). It will turn out that in the schemes we study, some schemes will achieve high utilization but poor fairness, and that as we improve fairness, the overall utilization will drop. Before diving into the protocols, let’s first develop a simple abstraction for the shared medium. This abstraction is a reasonable first-order approximation of reality. 1. Time is divided into slots of equal length, τ . 2. Each node can send a packet only at the beginning of a slot. 3. All packets are of the same size, and equal to an integral multiple of the slot length. In 2 In this course, and in most, if not all, of the networking and communications world...
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This document was uploaded on 02/26/2014 for the course CS 6.02 at MIT.

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