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Chapter 12Multiple Access

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

Figure 12.2 T axonomy of multiple-access protocols discussed in this chapter

1212--1 RANDOM ACCESS1 RANDOM ACCESS

I nI n randomrandom accessaccess or or contentioncontention methods,methods, nono stationstation isissuperior superior toto another another stationstation and and nonenone isis assigned assigned thethecontrol control over over another another. . NoNo stationstation permits,permits, or or doesdoes not not permit,permit, another another stationstation toto send send.. At At eacheach instance,instance, a a

stationstation that that hashas datadata toto send send usesuses aa procedureprocedure defined defined byby thethe protocol protocol toto makemake aa decisiondecision onon whether whether or or not not totosend send..

ALOHACarrier Sense Multiple AccessCarrier Sense Multiple Access with Collision DetectionCarrier Sense Multiple Access with Collision Avoidance

T opics discussed in this section:T opics discussed in this section:

Figure 12.3 F rames in a pure ALOHA network

Figure 12.4 P rocedure for pure ALOHA protocol

T he stations on a wireless ALOHA network are amaximum of 600 km apart. I f we assume that signalspropagate at 3 × 10 8 m/s, we find

T p = (600 × 10 5 ) / (3 × 10 8 ) = 2 ms.

Now we can find the value of T B for different values of K .

a . F or K = 1, the range is {0, 1}. T he station needs to|generate a random number with a value of 0 or 1. T hismeans that T B is either 0 ms (0 × 2) or 2 ms (1 × 2),based on the outcome of the random variable.

Example 12.1

b. F or K = 2, the range is {0, 1, 2, 3}. T his means that T B

can be 0, 2, 4, or 6 ms, based on the outcome of therandom variable.

c. F or K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. T hismeans that T B can be 0, 2, 4, . . . , 14 ms, based on theoutcome of the random variable.

d. We need to mention that if K > 10, it is normally set to10.

Example 12.1 (continued)

Figure 12.5 V ulnerable time for pure ALOHA protocol

A pure ALOHA network transmits 200-bit frames on ashared channel of 200 kbps. What is the requirement tomake this frame collision-free?

Example 12.2

SolutionAverage frame transmission time T fr is 200 bits/200 kbps or 1 ms. T he vulnerable time is 2 × 1 ms = 2 ms. T his meansno 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.

The throughput for pure ALOHA isS = G × e í2G .

The maximum throughputS max = 0.184 when G= (1/2).

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 12.3

SolutionT he frame transmission time is 200/200 kbps or 1 ms.a. I f the system creates 1000 frames per second, this is 1

frame per millisecond. T he load is 1. I n this caseS = G× e í2 G or S = 0.135 (13.5 percent). T his meansthat the throughput is 1000 × 0.135 = 135 frames. Only135 frames out of 1000 will probably survive.

b. I f the system creates 500 frames per second, this is(1/2) frame per millisecond. T he load is (1/2). I n thiscase S = G × e í2G or S = 0.184 (18.4 percent). T hismeans that the throughput is 500 × 0.184 = 92 and that only 92 frames out of 500 will probably survive. Notethat this is the maximum throughput case,percentagewise.

c. I f the system creates 250 frames per second, this is (1/4)frame per millisecond. T he load is (1/4). I n this caseS = G × e í 2G or S = 0.152 (15.2 percent). T his meansthat the throughput is 250 × 0.152 = 38. Only 38frames out of 250 will probably survive.

Figure 12.6 F rames in a slotted ALOHA network

The throughput for slotted ALOHA isS = G × e íG .

The maximum throughputS max = 0.368 when G = 1.

Figure 12.7 V ulnerable time for slotted ALOHA protocol

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 12.4

SolutionT he frame transmission time is 200/200 kbps or 1 ms.a. I f the system creates 1000 frames per second, this is 1

frame per millisecond. T he load is 1. I n this caseS = G× e íG or S = 0.368 (36.8 percent). T his meansthat the throughput is 1000 × 0.0368 = 368 frames.Only 386 frames out of 1000 will probably survive.

b. I f the system creates 500 frames per second, this is(1/2) frame per millisecond. T he load is (1/2). I n thiscase S = G × e íG or S = 0.303 (30.3 percent). T hismeans that the throughput is 500 × 0.0303 = 151.Only 151 frames out of 500 will probably survive.

c. I f the system creates 250 frames per second, this is (1/4)frame per millisecond. T he load is (1/4). I n this caseS = G × e íG or S = 0.195 (19.5 percent). T his meansthat the throughput is 250 × 0.195 = 49. Only 49frames out of 250 will probably survive.

Figure 12.8 Space/time model of the collision in CSMA

Figure 12.9 V ulnerable time in CSMA

Figure 12.10 B ehavior of three persistence methods

Figure 12.11 F low diagram for three persistence methods

Figure 12.12 Collision of the first bit in CSMA/CD

Figure 12.13 Collision and abortion in CSMA/CD

A network using CSMA/CD has a bandwidth of 10 Mbps.I f the maximum propagation time (including the delays inthe devices and ignoring the time needed to send ajamming signal, as we see later) is 25.6 s, what is the

minimum size of the frame?

Example 12.5

SolutionT he frame transmission time is T fr = 2 × T p = 51.2 s.T his means, in the worst case, a station needs to transmit

for a period of 51.2 s to detect the collision. T heminimum size of the frame is 10 Mbps × 51.2 s = 512bits or 64 bytes. T his is actually the minimum size of theframe for Standard Ethernet.

Figure 12.14 F low diagram for the CSMA/CD

Figure 12.15 Energy level during transmission, idleness, or collision

Figure 12.16 T iming in CSMA/CA

In CSMA/CA, the IFS can also be used todefine the priority of a station or a frame.

In CSMA/CA, if the station finds thechannel busy, it does not restart thetimer of the contention window;

it stops the timer and restarts it whenthe channel becomes idle.

Figure 12.17 F low diagram for CSMA/CA

1212--2 CONTROLLED ACCESS2 CONTROLLED ACCESS

I nI n controlled controlled accessaccess ,, thethe stationsstations consult consult oneone another another toto find find whichwhich stationstation hashas thethe right right toto send send.. AA stationstationcannot cannot send send unlessunless it it hashas beenbeen authorized authorized byby other other

stationsstations. . WeWe discussdiscuss threethree popular popular controlled controlled- -accessaccessmethodsmethods. .

ReservationPollingToken Passing

Figure 12.18 R eservation access method

Figure 12.19 Select and poll functions in polling access method

Figure 12.20 Logical ring and physical topology in token-passing access method

1212--3 CHANNELIZATION3 CHANNELIZATION

ChannelizationChannelization isis aa multiplemultiple--accessaccess method method inin whichwhichthethe availableavailable bandwidthbandwidth of of aa link link isis shared shared inin time,time,frequency,frequency, or or throughthrough code,code, betweenbetween different different stationsstations. .

I nI n thisthis section,section, wewe discussdiscuss threethree channelizationchannelizationprotocolsprotocols. .

Frequency-Division Multiple Access (FDMA)Time-Division Multiple Access (TDMA)Code-Division Multiple Access (CDMA)

We see the application of all thesemethods in Chapter 16 whenwe discuss cellular phone systems.

Figure 12.21 F requency-division multiple access ( F DMA)

In FDMA, the available bandwidthof the common channel is divided intobands that are separated by guard

bands.

Figure 12.22 T ime-division multiple access ( T DMA)

In TDMA, the bandwidth is just onechannel that is timeshared betweendifferent stations.

In CDMA, one channel carries alltransmissions simultaneously.

Figure 12.23 Simple idea of communication with code

Figure 12.24 Chip sequences

Figure 12.25 Data representation in CDMA

Figure 12.26 Sharing channel in CDMA

Figure 12.27 Digital signal created by four stations in CDMA

Figure 12.28 Decoding of the composite signal for one in CDMA

Figure 12.29 General rule and examples of creating Walsh tables

F ind the chips for a network witha. T wo stations b. F our stations

Example 12.6

Solution

We can use the rows of W 2 and W 4 in F igure 12.29:a. F or a two-station network, we have

[+ 1 +1] and [+ 1 í1].

b. F or a four-station network we have[+ 1 +1 +1 +1], [+ 1 í1 +1 í1],

[+ 1 +1 í1 í1], and [+ 1 í1 í1 +1].

What is the number of sequences if we have 90 stations inour network?

Example 12.7

Solution

T he number of sequences needs to be 2 m. We need tochoose m = 7 and N = 2 7 or 128. We can then use 90of the sequences as the chips.

P rove that a receiving station can get the data sent by aspecific sender if it multiplies the entire data on thechannel by the sender¶s chip code and then divides it bythe number of stations.

Example 12.8

SolutionLet us prove this for the first station, using our previousfour-station example. We can say that the data on the

channel D = ( d 1 c1 + d 2 c2 + d 3 c3 + d 4 c4).T he receiver which wants to get the data sent by station 1multiplies these data by c 1.

When we divide the result by N, we get d 1 .

aloha, csma

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