12.2
Figure 12.1 Data link layer divided into two functionality-oriented sublayers
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Random Access-Data Communication Systems-Lecture Slides, Slides of Digital Systems Design
This lecture is part of lecture series on Data Communication Systems. It was delivered by Prof. Prajin Ahuja at Birla Institute of Technology and Science. Its main points are: Taxonomy, Random, FDMA, TDMA, CDMA, Contention, Aloha, Slotted, Vulnerable, CSMA, Collision, Access, Reservation, polling
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2011/2012
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Figure 12.
Data link layer divided into two functionality-oriented sublayers
Figure 12.
Taxonomy of multiple-access protocols discussed in this chapter
Figure 12.
Frames in a pure ALOHA network
Figure 12.
Procedure for pure ALOHA protocol
b. For K = 2, the range is {0, 1, 2, 3}. This means that T
B
can be 0, 2, 4, or 6 ms, based on the outcome of therandom variable. c. For K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. This
means that T
B^
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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Figure 12.
Vulnerable time for pure ALOHA protocol
The throughput for pure ALOHA is
S = G × e
− 2G
The maximum throughputS
max
= 0.184 when G= (1/2).
Note
A pure ALOHA network transmits 200-bit frames on ashared channel of 200 kbps. What is the throughput if thesystem (all stations together) producesa. 1000 frames per second
b. 500 frames per second
c. 250 frames per second.
Example 12.
SolutionThe 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 caseS = G× e
− 2 G
or S = 0.135 (13.5 percent). This means
that the throughput is 1000 × 0.135 = 135 frames. Only135 frames out of 1000 will probably survive.
Figure 12.
Frames in a slotted ALOHA network
The throughput for slotted ALOHA is
S = G × e
−G
The maximum throughputS
max
= 0.368 when G = 1.
Note
A slotted ALOHA network transmits 200-bit frames on ashared channel of 200 kbps. What is the throughput if thesystem (all stations together) producesa. 1000 frames per second
b. 500 frames per second
c. 250 frames per second.
Example 12.
SolutionThe 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 caseS = 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.
Example 12.4 (continued)
b. If the system creates 500 frames per second, this is
(1/2) frame per millisecond. The load is (1/2). In thiscase 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 caseS = G × e
− G or S = 0.195 (19.5 percent). This means
that the throughput is 250 × 0.195 = 49. Only 49frames out of 250 will probably survive.
Figure 12.
Vulnerable time in CSMA
Figure 12.
Behavior of three persistence methods