Multiple Access (Chapter 12)-Data Communication and Computer Networks-Lecture Slides, Slides of Data Communication Systems and Computer Networks

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Chapter 12

Multiple Access

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Figure 12.1 Data link layer divided into two functionality-oriented sublayers

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12-1 RANDOM ACCESS

In random access or contention methods, no station is superior to another station and none is assigned the control over another. No station permits, or does not permit, another station to send. At each instance, a station that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send.

ALOHA Carrier Sense Multiple Access Carrier Sense Multiple Access with Collision Detection Carrier Sense Multiple Access with Collision Avoidance

Topics discussed in this section:

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Figure 12.3 Frames in a pure ALOHA network

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The stations on a wireless ALOHA network are a maximum of 600 km apart. If we assume that signals propagate at 3 × 10^8 m/s, we find Tp = (600 × 10^5 ) / (3 × 10^8 ) = 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.

Example 12.

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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

Example 12.1 (continued)

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A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free?

Example 12.

Solution Average frame transmission time Tfr is 200 bits/200 kbps or 1 ms. The vulnerable time is 2 × 1 ms = 2 ms. This means no station should send later than 1 ms before this station starts transmission and no station should start sending during the one 1-ms period that this station is sending.

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The throughput for pure ALOHA is

S = G × e −2G^.

The maximum throughput

Smax = 0.184 when G= (1/2).

Note

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Example 12.3 (continued)

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 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. Docsity.com

Figure 12.6 Frames in a slotted ALOHA network

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Figure 12.7 Vulnerable time for slotted ALOHA protocol

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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.

Example 12.

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. Docsity.com

Figure 12.9 Vulnerable time in CSMA

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Figure 12.10 Behavior of three persistence methods

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