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Multiple access protocols are essential in enabling multiple users to access a shared communication channel efficiently. They play a crucial role in various communication systems, including wireless networks, satellite communication, and local area networks. These notes cover topics such as TDMA, CSMA, CDMA, FDMA, and ALOHA along with solved examples and apt diagrams.
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Detection
Avoidance Computer Networks
3 The data link layer is the second layer in the OSI (Open Systems Interconnection) model and the layer responsible for the reliable transfer of data between two devices connected on the same network. It is concerned with addressing, framing, and error detection and correction. One of the important aspects of the data link layer is handling multiple access to the shared communication medium, especially in shared network topologies. Multiple access refers to the ability of multiple devices to communicate over the same network medium simultaneously. In shared network topologies, such as Ethernet and Wi-Fi, multiple devices contend for access to the shared communication channel. The data link layer must manage this contention and ensure that data packets are transmitted efficiently and without collisions.
Aloha is a multiple access protocol used in data communication networks to manage access to a shared communication channel. It was one of the first multiple access methods developed for early computer networks and played a significant role in the development of modern networking protocols. The Aloha protocol was originally developed at the University of Hawaii in the early 1970s. There are two variants of the Aloha protocol:
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10 The stations on a wireless ALOHA network are a maximum of 600 km apart. If we assume that signals propagate at 3 × 108 m/s, we find Tp = (600 × 103 ) / (3 × 108 ) = 2 ms. Now we can find the value of TB for different values of K. a. For K = 1, the range is {0, 1}. The station needs to| generate a random number with a value of 0 or 1. This means that TB is either 0 ms (0 × 2) or 2 ms (1 × 2), based on the outcome of the random variable. b. For K = 2, the range is {0, 1, 2, 3}. This means that TB can be 0, 2, 4, or 6 ms, based on the outcome of the random variable. c. For K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. This means that TB can be 0, 2, 4,... , 14 ms, based on the outcome of the random variable. d. We need to mention that if K > 10, it is normally set to
11 Figure 5 Vulnerable time for pure ALOHA protocol
13 A pure ALOHA network transmits 200 - bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a. 1000 frames per second b. 500 frames per second c. 250 frames per second.
Solution The frame transmission time is 200 / 200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e −2 G or S = 0.135 (13.5 percent). This means that the throughput is 1000 × 0.135 = 135 frames. Only 135 frames out of 1000 will probably survive. b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this case S = G × e −2G^ or S = 0.184 (18.4 percent). This means that the throughput is 500 × 0.184 = 92 and that means only 92 frames out of 500 will probably survive. Note that this is the maximum throughput case, percentagewise. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G × e −2G^ or S = 0.152 (15.2 percent). This means that the throughput is 250 × 0.152 = 38. Only 38 frames out of 250 will probably survive.
14 Figure 6 Frames in a slotted ALOHA network
−G
16 A slotted ALOHA network transmits 200 - bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a. 1000 frames per second b. 500 frames per second c. 250 frames per second.
Solution The frame transmission time is 200/200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e−G^ or S = 0.368 (36.8 percent). This means that the throughput is 1000 × 0.0368 = 368 frames. Only 386 frames out of 1000 will probably survive. b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this case S = G × e−G^ or S = 0.303 (30.3 percent). This means that the throughput is 500 × 0.0303 = 151. Only 151 frames out of 500 will probably survive. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G × e −G or S = 0.195 (19.5 percent). This means that the throughput is 250 × 0.195 = 49. Only 49 frames out of 250 will probably survive.
17 Figure 8 Space/time model of the collision in CSMA
19 Figure 10 Behavior of three persistence methods
20 Figure 11 Flow diagram for three persistence methods
