ccn nots final unit -3 (1)

Computer Communication Network: Unit 3- Multiple Access

Prof. Suresha V, Dept. Of E&C E. K V G C E, Sullia, D.K-574 327

UNIT 3 - MULTIPLE ACCESS

Learning Objectives:

Upon completion of this unit, the student should be able to:

Understand the need for multiple access in a shared channel.

Discuses many formal protocols have been devised to handle access to a shared link.

Explains the three categories of multiple access protocols.

Describe and analyze the different types of random access protocols.

Describe and analyze controlled access protocols.

Explains channelization protocols.

3.1. Introduction: Transmission technology can be categorized into two categories:

1. Point-to point networks

2. Broadcast networks

1. Point-to-point networks: Point-to-point networks are those in which when a message is sent

from one computer to another, it usually has to be sent via other computers in the network.

A point-to-point network consists of many connections between individual pairs of

computers.

2. Broadcast networks: Broadcast networks have a single communication channel that is

shared by all the machines on the network. A packet sent by one computer is received by all

other computers on the network. The packets that are sent contain the address of the

receiving computer; each computer checks this field to see if it matches its own address If it

does not then it is usually ignored; if it does then it is read. Broadcast channels are

sometimes known as multi-access channel.

The data link layer sublayers: The data link layer has two sublayers.This is shown in figure 3.1

Figure 3.1 Data link layer divided into two functionality-oriented sub-layers

The upper sublayer in the figure 3.1 is responsible for data link control. The lower sublayer is

responsible for resolving access to the shared media. If the channel is dedicated, do not need the

lower sub-layer.



IEEE has actually made this division for LANs. The upper sublayer that is responsible for flow

and error control is called the Logical Link Control (LLC) Layer. The lower sublayer that is mostly

responsible for multiple access resolution is called the Media Access Control (MAC) Layer.

When nodes or stations are connected and use a common link, called a multipoint or broadcast

link, we need a multiple-access protocol to coordinate access to the link.MAC coordinates

transmission between users sharing the spectrum. The main goals are to prevent collisions while

maximizing throughput and minimizing delay

Many formal protocols have been devised to handle access to a shared link.MAC protocol

categorize them into three groups, shown in figure 3.2

Figure 3.2 Taxonomy of multiple-access protocols

3.2 RANDOM ACCESS PROTOCOLS: In a random access method, each station has the right to the

medium without being controlled by any other station. However, if more than one station tries

to send, there is an access conflict leads to collision and the frames will be either destroyed or

modified. To avoid access conflict or to resolve it when it happens, each station follows a

procedure that answers the following questions:

1. When can the station access the medium?

2. What can the station do if the medium is busy?

3. How can the station determine the success or failure of the transmission?

4. What can the station do if there is an access conflict?

To address above conflicts by using following random Access protocols

1. ALOHA

2. Carrier Sense Multiple Access (CSMA)

3. Carrier Sense Multiple Access with Collision Detection (CSMA/CD)

4. Carrier Sense Multiple Access with Collision Avoidance (CSMD/CA)



1. ALOHA: It was the earliest random access method, developed at the University of Hawaii in

early 1970. It was designed for a radio (wireless) LAN, but it can be used on any shared medium.

Two types of ALOHA

a. Pure ALOHA.

b. Slotted ALOHA

a. Pure ALOHA: The Concept used in this protocol called “True Free for All”. The original

ALOHA protocol is called pure ALOHA.

Procedure for P-ALOHA:

Whenever a station has data, it transmits immediately

Receivers ACK all packets

No ACK = collision. Wait a random time and retransmit

Figure 3.3 shows an example of frame collisions in pure ALOHA

Figure 3.3 Frames in a pure ALOHA network

Figure 3.3 shows that only two frames survive: frame 1.1 from station 1 and frame 3.2 from

station 3. We need to mention that even if one bit of a frame coexists on the channel with one

bit from another frame, there is a collision and both will be destroyed. If all these stations try to

resend their frames after the time-out, the frames will collide again.

Algorithm for pure ALOHA protocol:

Pure ALOHA dictates that when the time-out period passes, each station waits a random amount

of time before resending its frame. The randomness will help avoid more collisions.



We call this time the “back-off time” TB.

1. Pure ALOHA has a second method to prevent congesting the channel with retransmitted

frames.

2. After a maximum number of retransmission attempts Kmax' a station must give up and

try later.

3. Figure 3.4 shows the procedure for pure ALOHA based on the above strategy.

Figure 3.4 Procedure for pure ALOHA protocol

4. The time-out period is equal to the maximum possible round-trip propagation delay,

which is twice the amount of time required to send a frame between the two most,

widely separated stations (2xTp).

5. The back-off time TB is a random value that normally depends on K (the number of

attempted unsuccessful transmissions).

6. The formula for TB depends on the implementation. One common formula is the binary

exponential back-off. In this method, for each retransmission, a multiplier in the range 0

to 2 K - 1 is randomly chosen and multiplied by Tp (maximum propagation time) or Tfr (the

average time required to send out a frame) to find TB. Note that in this procedure, the

range of the random numbers increases after each collision. The value of Kmax is usually

chosen as 15.



Advantages:

1. Trivially simple

2. No coordination between participants necessary, Hence no need for synchronization

3. No need for fixed length packets

4. Superior to fixed assignment when there is a large number of bursty stations.

5. Adapts to varying number of stations.

Disadvantages:

1. Collision factor is high, hence less efficiency

2. Theoretically proven throughput maximum of 18.4%.

3. Requires queuing buffers for retransmission of packets.

Example 3.1: The stations on a wireless ALOHA networks are a maximum of 600 km apart. If we

assume that signals propagate at 3 x 108 m/s, we find Tp = (600 x 105) / (3 x 108) = 2 ms. Now we

can find the value of TB for different values of K.

Solution:

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 °ms (0 x 2) or 2 ms (l x 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 to, 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 10.

Vulnerable time: The length of time in which there is a possibility of collision called “vulnerable

time”

Figure 3.5 Vulnerable time for pure ALOHA protocol



Assume that the stations send fixed-length frames with each frame taking Tfr to send. Figure 3.5

shows the vulnerable time for station A. Station A sends a frame at time t. Now imagine station

B has already sent a frame between t - Tfr and t. This leads to a collision between the frames

from station A and station B.

Vulnerable time in pure ALOHA is 2 times the frame transmission time.

Pure ALOHA vulnerable time = 2 x Tfr

Example 3.2: 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?

Solution

Average frame transmission time Tfr = Frame length/ transmission speed

= 200 bits / 200 kbps = 1 ms.

The vulnerable time is = 2* Tfr

= 2 × 1 ms = 2 ms.

This means no station should send later than 1ms before this station starts

transmission and no station should start sending during the 1ms period that station is

sending.

Throughput: The number of packets successfully transmitted through the channel per packet

time called throughput of Pure ALOHA. Let us call G the average number of frames generated by

the system during one frame transmission time. Then it can be proved that the average number

of successful transmissions for pure ALOHA is S = G x e-2G.

The maximum throughput Smax is 0.184, for G = 1/2. i.e., if one-half a frame is generated during

one frame transmission time (in other words, one frame during two frame transmission times),

then 18.4 percent of these frames reach their destination successfully.

Relation between G and S:



Example 1.3: 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 Tp = Frame length/ transmission speed

= 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. i.e. G = 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). i,e G= ½

• In this case S = G × e −2G or S = 0.184 (18.4 percent).

• This means that the throughput is 500 × 0.184 = 92

• 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



2. Slotted ALOHA: Slotted ALOHA was invented to improve the efficiency of pure ALOHA. Time is

divided into discrete time intervals(slot).A station can transmit only at the beginning of slot only.

Vulnerable time for slotted ALOHA is one-half that of pure ALOHA.

Slotted ALOHA vulnerable time = Tfr

Figure 1.6 Frames in a slotted ALOHA network

The throughput for slotted ALOHA is S = G × e−G.

The maximum throughput Smax = 0.368 when G = 1.

Vulnerable time for slotted ALOHA: Slotted ALOHA vulnerable time is Tfr, This is shown in fig 3.7

Figure 3.7 Vulnerable time for slotted ALOHA protocol



Example 1 : A slotted ALOHA network transmits 200-bit frames on a shared channel of 200kbps.

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:

a. The frame transmission time is 200/200 kbps or 1 ms.

If the system creates 1000 frames per second, this is 1 frame per millisecond.

The load is 1. i,e G=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).i,e G=1/2

• In this case S = G × e−G or S = 0.303 (30.3 percent).



c. If the system creates 250 frames per second, this is (1/4) frame per millisecond.

• The load is (1/4). I,e G=1/4

• In this case S = G × e −G or S = 0.195 (19.5 percent).



3. Carrier Sense Multiple Access (CSMA): CSMA is based on the principle "sense before

transmit" or "listen before talk. To minimize the chance of collision and, therefore, increase the

performance, the CSMA method was developed. The chance of collision can be reduced if a



station senses the medium before trying to use it. CSMA can reduce the possibility of collision,

but it cannot eliminate it. Figure 3.8 shows a Space/time model of the collision in CSMA

Figure 3.8 Space/time model of the collision in CSMA

The possibility of collision still exists because of propagation delay; when a station sends a

frame, it still takes time (although very short) for the first bit to reach every station and for every

station to sense it.

Vulnerable Time in CSMA: The vulnerable time for CSMA is the propagation time Tp. This is the

time needed for a signal to propagate from one end of the medium to the other. When a station

sends a frame, and any other station tries to send a frame during this time, a collision will result.

But if the first bit of the frame reaches the end of the medium, every station will already have

heard the bit and will refrain from sending.

Figure 3.9 Vulnerable time in CSMA

The vulnerable time for CSMA is the propagation time Tp



CSMA Protocols: There are several types of CSMA protocols:

a) 1 - Persistent CSMA.

b) Non - Persistent CSMA.

c) P - Persistent CSMA.

a) 1-Persistent CSMA: Procedure for this algorithm are as follows

Sense the channel.

If busy, keep listening to the channel and transmit immediately when the channel

becomes idle.

If idle, transmit a packet immediately. If collision occurs.

Wait a random amount of time and start over again.

The protocol is called 1-persistent because the host transmits with a probability of 1

whenever it finds the channel idle.

b) Non-Persistent CSMA: Procedure for this algorithm are as follows

Sense the channel.

If busy, wait a random amount of time and sense the channel again.

If idle, transmit a packet immediately.

If collision occurs.

Wait a random amount of time and start all over again.

Figure 3.10 Behavior of three persistence methods



c). P-persistent CSMA: Procedure for this algorithm is as follows

It is used if the channel has time slots with slot duration equal to or greater than the

maximum propagation time.

The p-persistent approach combines the advantages of the other two strategies. It

reduces the chance of collision and improves efficiency.

In this method, after the station finds the line idle it follows these steps:

1. With probability p, the station sends its frame.

2. With probability q = 1 - p, the station waits for the beginning of the next time slot

and checks the line again.

If the line is idle, it goes to step 1.

If the line is busy, it acts as though a collision has occurred and uses the back-off

procedure.

Figure 3.11 Flow diagrams for three persistence methods



3. Carrier Sense Multiple Access with Collision Detection (CSMA/CD): The CSMA method does

not specify the procedure following a collision. Carrier senses multiple access with collision

detection (CSMA/CD) augments the algorithm to handle the collision. CSMA/CD is used to

improve CSMA performance by terminating transmission as soon as a collision is detected.

IIIustration od CSMA/CD procedure:

Figure 3.13 Collision of the first bit in CSMA/CD

Let us look at the figure 3.13, first bits transmitted by the two stations involved in the collision.

Although each station continues to send bits in the frame until it detects the collision. At time t1,

station A has executed its persistence procedure and starts sending the bits of its frame. At time

t2, station C has not yet sensed the first bit sent by A. Station C executes its persistence

procedure and starts sending the bits in its frame, which propagate both to the left and to the

right. The collision occurs sometime after time t2, Station C detects a collision at time t3 when it

receives the first bit of A's frame. Station C immediately aborts transmission. Station A detects

collision at time t4 when it receives the first bit of C's frame; it also immediately aborts

transmission. “A” transmits for the duration t4 – t1; “C” transmits for the duration t3 - t2 the

length of any frame divided by the bit rate. In this protocol must be more than either of these

durations. At time t4, the transmission of A:s frame, though incomplete, is aborted at time t3,

the transmission of B's frame, though incomplete, is aborted.

Minimum Frame Size: For CSMA/CD to work, It restriction on the frame size. Before sending the

last bit of the frame, the sending station must detect a collision, if any, and abort the

transmission.This is so because the station, once the entire frame is sent, does not keep a copy

of the frame and does not monitor the line for collision detection. Therefore,

Frame transmission time Tfr must be at least two times the maximum propagation

time Tp. i, e Tfr = 2Tp

The vulnerable time for CSMA/CD is twice the propagation time = 2Tp



Example 3.5

A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time

(including the delays in the devices and ignoring the time needed to send a jamming signal,) is

25.6 μs, what is the minimum size of the frame?

Solution

• The frame transmission time is Tfr= 2 × Tp = 51.2 μs.

• In the worst case, a station needs to transmit for a period of 51.2 μs to detect the

collision.

• The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes.

• This is actually the minimum size of the frame for Standard Ethernet.

Procedure for CSMA/CD: Algorithmic steps for CSMA/CD as shown in figure 3.15

Figure 3.15 Flow diagrams for the CSMA/CD



Energy Level: Level of energy in a channel can have three values:

1. Zero: At this level, the channel is idle

2. Normal: At this level, a station has successfully captured the channel and is sending its

frame

3. Abnormal: At this level, there is a collision and the level of the energy is twice the normal

level.

A station that has a frame to send or is sending a frame needs to monitor the energy level to

determine if the channel is idle, busy, or in collision mode. Figure 3.16 shows the situation.

Figure 3.16 Energy level during transmission, idleness, or collision

Throughput of CSMA/CD: The throughput of CSMA/CD is greater than that of pure or slotted

ALOHA. The maximum throughput occurs at a different value of G and is based on the

persistence method and the value of p in the p-persistent approach. For I-persistent method the

maximum throughput is around 50 percent when G =1.For non persistent method, the maximum

throughput can go up to 90 percent when G is between 3 and 8.

4. Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA): The basic idea behind

CSMA/CD is that a station needs to be able to receive while transmitting to detect a collision.

When there is no collision, the station receives one signal: its own signal. When there is a

collision, the station receives two signals: its own signal and the signal transmitted by a second

station. To distinguish between these two cases, the received signals in these two cases must be

significantly different. The signal from the second station needs to add a significant amount of

energy to the one created by the first station. In a wired network, the received signal has almost

the same energy as the sent signal because either the length of the cable is short or there are

repeaters that amplify the energy between the sender and the receiver. This means that in a

collision, the detected energy almost doubles.

However, in a wireless network, much of the sent energy is lost in transmission. The received

signal has very little energy. Therefore, a collision may add only 5 to 10 percent additional

energy. This is not useful for effective collision detection.CA is used to improve CSMA



performance by not allowing wireless transmission of a node if another node is transmitting,

thus reducing the probability of collision due to the use of a random time. Collisions are avoided

through the use of CSMA/CA's three strategies:

1. The inter frame space (IFS)

2. The contention window

3. Acknowledgments

Figure 3.17 Timing in CSMA/CA

1. Inter frame Space (IFS):

Collisions are avoided by deferring transmission even if the channel is found idle.

When an idle channel is found, the station does not send immediately.

It waits for a period of time called the inter-frame space or IFS.

Even though the channel may appear idle when it is sensed, a distant station may have

already started transmitting.

The distant station's signal has not yet reached this station. The IFS time allows the front

of the transmitted signal by the distant station to reach this station.

If after the IFS time the channel is still idle, the station can send, but it still needs to wait

a time equal to the contention time.

The IFS variable can also be used to prioritize stations or frame types. for example, a

station that is assigned shorter IFS has a higher priority.

2. Contention Window

The contention window is an amount of time divided into slots.

A station that is ready to send chooses a random number of slots as its wait time.

The number of slots in the window changes according to the binary exponential back- off

strategy.



This means that it is set to one slot the first time and then doubles each time the station

cannot detect an idle channel after the IFS time.

Contention window is that the station needs to sense the channel after each time slot.

If the station finds the channel busy, it does not restart the process; it just stops the

timer and restarts it when the channel is sensed as idle. This gives priority to the station

with the longest waiting time

3. Acknowledgment

With all these precautions, there still may be a collision resulting in destroyed data.

The positive acknowledgment and the time-out timer can help guarantee that the

receiver has received the frame.

Procedure for CSMA/CA: Figure 1.17 shows the procedure. Note that the channel needs to be

sensed before and after the IFS.

Figure 3.18 Flow diagram for CSMA/CA



3.3 CONTROLLED ACCESS: Here, every station consults one another to find which station has

the right to send. A station cannot send unless it has been authorized by other stations. There

are three popular controlled-access methods

1. Reservation

2. Polling

3. Token Passing

1. Reservation: In this method, a station needs to make a reservation before sending data. Time

is divided into intervals. In each interval, a reservation frame precedes the data frames sent in

that interval which is shown in figure 3.18

Figure 3.18 Reservation access methods

If there are N stations in the system, there are exactly N reservation minislots in the reservation

frame. Each minislot belongs to a station. When a station needs to send a data frame, it makes a

reservation in its own minislots. The stations that have made reservations can send their data

frames after the reservation frame.

2. Polling: Polling works with topologies in which one device is designated as a primary station

and the other devices are secondary stations. All data exchanges must be made through the

primary device even when the ultimate destination is a secondary device shown in figure 3.19

Figure 3.19 Select and poll functions in polling access method



The primary device controls the link; the secondary devices follow its instructions. It is up to the

primary device to determine which device is allowed to use the channel at a given time. The

primary device, therefore, is always the initiator of the session, if the primary wants to receive

data, it asks the secondary’s if they have anything to send; this is called “Poll” function. If the

primary wants to send data, it tells the secondary to get ready to receive; this is called “select”

function. If the primary is neither sending nor receiving data, it knows the link is available. The

primary must alert the secondary to the upcoming transmission and wait for an

acknowledgment of the secondary's ready status. Before sending data, the primary creates and

ransmits a select (SEL) frame, one field of which includes the address of the intended secondary.

3. Token passing: In the token-passing method, the stations in a network are organized in a

logical ring. For each station, there is a predecessor and a successor. The predecessor is the

station which is logically before the station in the ring; the successor is the station which is after

the station in the ring. The current station is the one that is accessing the channel now. The right

to this access has been passed from the predecessor to the current station. The right will be

passed to the successor when the current station has no more data to send.

Figure 3.20 Logical ring and physical topology in token-passing access method

3.4 CHANNELIZATION: Channelization is a multiple-access method in which the available

bandwidth of a link is shared in time, frequency, or through code, between different stations.

Three channelization protocols.

Frequency-Division Multiple Access (FDMA)

Time-Division Multiple Access (TDMA)

Code-Division Multiple Access (CDMA)



1. Frequency-Division Multiple Access (FDMA): In FDMA, the available bandwidth of the

common channel is divided into bands that are separated by guard bands, see in figure 3.21.

Mainly used in Analog System or as a Voice channel. Main limitation is it is not efficiency

multiple access method.

Figure 3.21 Frequency-division multiple access (FDMA)

2. Time - Division Multiple Access (TDMA)

In TDMA, the bandwidth is just one channel that is timeshared between different stations.

Figure 3.22 Time-division multiple access (TDMA)

3. Code-Division Multiple Access (CDMA): In CDMA, one channel carries all transmissions

simultaneously. This is shown in figure 3.23

Each station assign with unique code



Assigned codes have two properties.

1. If we multiply each code by another, we get 0.

2. If we multiply each code by itself, we get 4 (the number of stations).

Figure 3.23 Simple idea of communication with code

Any station that wants to receive data from one of the other three multiplies the data on

the channel by the code of the sender.

For example, suppose stations 1 and 2 are talking to each other. Station 2 wants to hear

what station 1 is saying.

It multiplies the data on the channel by c1' the code of station 1. Because (c1 . c1) is 4,

but (c2 . c1), (c3 . c1), and (c4 . c1) are all 0s

Station 2 divides the result by 4 to get the data from station 1.

Data = (d1. C1 + d2 . C2 +d3 . C3 + d4 . c4) . C1

= d 1. C1. C1 + d2. C2 . C1 + d3 . C3 . C1 + d4 . C4. C1 =4 X d1

Chips

• CDMA is based on coding theory. Each station is assigned a code, which is a sequence of

numbers called chipsShown in Figure 3.24: The codes are for the previous example.

Figure 3.24 Chip sequences

We need to know that we did not choose the sequences randomly; they were carefully selected.



Orthogonal sequences and have the following properties:

1. Each sequence is made of N elements, where N is the number of stations.

2. If we multiply a sequence by a number, every element in the sequence is multiplied by that

element. This is called multiplication of a sequence by a scalar.

For example, 2. [+1 +1-1-1] = [+2+2-2-2]

3. If we multiply two equal sequences, element by element, and add the results, we get N,

Where N is the number of elements in the each sequence. This is called the inner product

of two equal sequences. For example, [+1 +1-1 -1] · [+1 +1 -1 - 1] = 1 + 1 + 1 + 1 = 4

4. If we multiply two different sequences, element by element, and add the results, we get 0.

This is called inner product of two different sequences.

For example, [+1 +1 -1 - 1] • [+1 +1 + 1 +1] = 1 + 1 - 1 - 1 = 0

5. Adding two sequences means adding the corresponding elements. The result is another

sequence. For example,[+1+1-1-1]+[+1+1+1+1]=[+2+2+0+0]

Data Representation in CDMA:

Figure 3.25 Data representation in CDMA

Rules for encoding:

1. If a station needs to send a 0 bit, it encodes it as -1

2. If it needs to send a 1 bit, it encodes it as +1.

3. When a station is idle, it sends no signal, which is interpreted as a 0

Encoding and Decoding in CDMA: Let a simple example show how four stations share the link

during a 1-bit interval. The procedure can easily be repeated for additional intervals. Assume

that stations 1 and 2 are sending a 0 bit and channel 4 is sending a 1 bit. Station 3 is silent. The

data at the sender site are translated to -1, -1, 0, and +1.Each station multiplies the

corresponding number by its chip (its orthogonal sequence) which is unique for each station. The

result is a new sequence which is sent to the channel. For simplicity, we assume that all stations

send the resulting sequences at the same time. The sequence on the channel is the sum of all

four sequences. Figure 3.26 shows the situation. Now imagine station 3, which we said is silent,



is listening to station 2. Station 3 multiplies the total data on the channel by the code for station

2, which is [ +1 -1 +1-1], to get [-1-1-3 +1]· [+1-1 +1-1] =-4/4 =-1 ...... bit 1

Figure 3.26 Sharing channel in CDMA

Illustration of Encoding and Decoding processes in CDMA

Figure 3.27 Encoding process in CDMA



Figure 3.28 Sharing channel in CDMA

Signal Level in Encoding and Decoding of CDMA signal:

The figure 3.29 and 3.30 shows the digital signal produced by each station and the data

recovered at the destination (using NRZ-L for simplicity) and the signal that is on the common

channel

Figure 3.29 Digital signal created by four stations in CDMA



Figure 3.30 Decoding of the composite signal for one in CDMA

Figure 3.28 shows how station 3 can detect the data sent by station 2 by using the code for

station 2. The total data on the channel are multiplied (inner product operation) by the signal

representing station 2 chip code to get a new signal. The station then integrates and adds the

area under the signal, to get the value -4, which is divided by 4 and interpreted as bit 0.

Sequence Generation:To generate chip sequences, we use a Walsh table, which is a two-

dimensional table with an equal number of rows and columns, as shown below. The number of

sequences in a Walsh table needs to be N = 2m.

Figure 3.31 General rule and examples of creating Walsh tables



Example 3.6: Find the chips for a network with

a. Two stations

b. Four stations

Solution

We can use the rows of W2 and W4 in Figure 3.31:

a. For a two-station network, we have

[+1 +1] and [+1 −1].

b. For a four-station network we have

[+1 +1 +1 +1], [+1 −1 +1 −1], [+1 +1 −1 −1], and [+1 −1 −1 +1].

Example 3.7

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

Solution

The number of sequences needs to be 2m. We need to choose m = 7 and N = 27 or 128. We can

then use 90 of the sequences as the chips.

Example 1.8

Prove that a receiving station can get the data sent by a specific sender if it multiplies the entire

data on the channel by the sender’s chip code and then divides it by the number of stations.

Solution

Let us prove this for the first station, using our previous four-station example. We can

say that the data on the channel D = (d1 ⋅ c1 + d2 ⋅ c2 + d3 ⋅ c3 + d4 ⋅ c4).

The receiver which wants to get the data sent by station 1 multiplies these data by c1. When we

divide the result by N, we get d1.



SUMMARY OF UNIT THREE: MULTIPLE ACCESS

Data link layer as two sublayers. The upper sublayer is responsible for data link control, and

the lower sublayer is responsible for resolving access to the shared media.

Many formal protocols have been devised to handle access to a shared link. We categorize

them into three groups: random access protocols, controlled access protocols, and

channelization protocols.

In random access or contention methods, no station is superior to another station and none

is assigned the control over another.

ALOHA allows multiple access (MA) to the shared medium. There are potential collisions in

this arrangement. When a station sends data, another station may attempt to do so at the

same time. The data from the two stations collide and become garbled.

To minimize the chance of collision and, therefore, increase the performance, the CSMA

method was developed. The chance of collision can be reduced if a station senses the

medium before trying to use it. Carrier sense multiple access (CSMA) requires that each

station first listen to the medium before sending. Three methods have been devised for

carrier sensing: I-persistent, nonpersistent, and P-persistent.

Carrier senses multiple access with collision detection (CSMA/CD) augments the CSMA

algorithm to handle collision. In this method, a station monitors the medium after it sends a

frame to see if the transmission was successful. If so, the station is finished. If, however,

there is a collision, the frame is sent again.

To avoid collisions on wireless networks, carrier sense multiple access with collision

avoidance (CSMA/CA) was invented. Collisions are avoided through the use three strategies:

the interframe space, the contention window, and acknowledgments.

In controlled access, the stations consult one another to find which station has the right to

send. A station cannot send unless it has been authorized by other stations. We discussed

three popular controlled-access methods: reservation, polling, and token passing.

In the reservation access method, a station needs to make a reservation before sending data.

Time is divided into intervals. In each interval, a reservation frame precedes the data frames

sent in that interval.

In the polling method, all data exchanges must be made through the primary device even

when the ultimate destination is a secondary device. The primary device controls the link;

the secondary devices follow its instructions.



In the token-passing method, the stations in a network are organized in a logical ring. Each

station has a predecessor and a successor. A special packet called a token circulates through

the ring.

Channelization is a multiple-access method in which the available bandwidth of a link is

shared in time, frequency, or through code, between different stations. We discussed three

channelization protocols: FDMA, TDMA, and CDMA.

In frequency-division multiple access (FDMA), the available bandwidth is divided into

frequency bands. Each station is allocated a band to send its data. In other words, each band

is reserved for a specific station, and it belongs to the station all the time.

In time-division multiple access (TDMA), the stations share the bandwidth of the channel in

time. Each station is allocated a time slot during which it can send data. Each station

transmits its data in its assigned time slot.

In code-division multiple access (CDMA), the stations use different codes to achieve multiple

access. CDMA is based on coding theory and uses sequences of numbers called chips. The

sequences are generated using orthogonal codes such the Walsh tables.

UNIT -3 MULTIPLE ACCESS QUESTION BANK

1. List three categories of multiple access protocols discussed in this chapter.

2. Define random access and list three protocols in this category.

3. Define controlled access and list three protocols in this category.

4. Define channelization and list three protocols in this category.

5. Explain why collision is an issue in a random access protocol but not in controlled access or

channelizing protocols.

6. Compare and contrast a random access protocol with a controlled access protocol.

7. Compare and contrast a random access protocol with a channelizing protocol.

8. Compare and contrast a controlled access protocol with a channelizing protocol.

9. Do we need a multiple access protocol when we use the 1ocalloop of the telephone company

to access the Internet? Why?

10. Do we need a multiple access protocol when we use one CATV channel to access the

Internet? Why?

11. We have a pure ALOHA network with 100 stations. If Tfr = 1, what is the number of frames/s

each station can send to achieve the maximum efficiency.

12. Repeat Exercise 11 for slotted ALOHA.



13. One hundred stations on a pure ALOHA network share a l-Mbps channel. If frames are 1000

bits long, find the throughput if each station is sending 10 frames per second.


15. In a CDMA/CD network with a data rate of 10 Mbps, the minimum frame size is found to be

512 bits for the correct operation of the collision detection process. What should be the

minimum frame size if we increase the data rate to 100 Mbps? To 1 Gbps? To 10 Gbps?

16. In a CDMA/CD network with a data rate of 10 Mbps, the maximum distance between any

station pair is found to be 2500 m for the correct operation of the collision detection process.

What should be the maximum distance if we increase the data rate to 100 Mbps? To 1 Gbps?

To 10 Gbps?

17. In Figure 3.13, the data rate is 10 Mbps, the distance between station A and C is 2000 m,

and the propagation speed is 2 x 108 m/s. Station A starts sending a long frame at time t1 =0;

station C starts sending a long frame at time t2 =3, if The size of the frame is long enough to

guarantee the detection of collision by both stations. Find:

a. The time when station C hears the collision (t3)'

b. The time when station A hears the collision (t4)'

c. The number of bits station A has sent before detecting the collision.

d. The number of bits station C has sent before detecting the collision.

18. Repeat Exercise 17 if the data rate is 100 Mbps.

19. Calculate the Walsh table Ws from W4 in Figure 3.31.

20. Recreate the W2 and W4 tables in Figure 3.29 using WI = [-1]. Compare the recreated

tables with the ones in Figure 3.31.

21. Prove the third and fourth orthogonal properties of Walsh chips for W4 in Figure 3.31.

22. Prove the third and fourth orthogonal properties of Walsh chips for W4 recreated in Q20

23. Repeat the scenario depicted in Figures 3.29 to 3.30 if both stations 1 and 3 are silent.

24. A network with one primary and four secondary stations uses polling. The size of a data

frame is 1000 bytes. The size of the poll, ACK, and NAK frames are 32 bytes each. Each station

has 5 frames to send. How many total bytes are exchanged if there is no limitation on the

number of frames a station? can send in response to a poll?

25. Repeat question 24 if each station can send only one frame in response to a poll.



UNIT 3- MULTIPLE ACCESS - SOLVED EXAMPLES

1. Determine the maximum throughput that can be achieved using ALOHA and slotted ALOHA

protocols.

Solution:

We know that throughput of the ALOHA protocol S = Ge-2G

To find the maximum throughputs that can be achieved using pure ALOHA protocol

we have to calculate the derivative of the above equation and equating it to zero.

dS/dG = e-2G

– 2Ge-2G

= 0

Smax = 18.39, when G = 0.5

Thus the best taffic utilization one can hope for using pure ALOHA is 0.5/e=0.1839

Erlangs.

Similar for slotted ALOHA:

Throughput of the ALOHA protocol S = Ge

-G

To find the maximum throughputs that can be achieved using Slotted ALOHA

protocol we have to calculate the derivative of the above equation and equating it to

zero.

dS/dG = e-G

– Ge-G

= 0

Smax = 36.78, when G = 1.

2. In a pure ALOHA system the packet arrival times forms a Poisson process having rate of

103 packets/sec. The channel bit rate equals 10 Mbps. Each packet comprises of 1000 bits. What

is the normalized throughput of the system? Find the number of bits per packet that will maximize

the throughput.

Solution:

Normalized throughput, S, expresses throughput as a fraction of channel capacity and

equals S= L λs /R. Hence the normalized throughput of the system is

S=1000 [bits/packet] • 1000 [packets/sec] / 107 [bits/sec] = 0.1.

Maximal normalized throughput equals 0.18.

Hence the number of bits per packet that will maximize the throughput = 0.18 • 107/ 10

3

= 1800 bits/packet.

3. Three stations share a 1 Mbps pure ALOHA channel. The average bit rate transmitted from

each of the three stations is R1=150 kbits/s, R2=200 kbits/s and R3=400 kbits/s. The size of each

packet is 2000 bits/packet. Assume that the arrival process is Poisson.



1. What is the normalized total traffic on the channel?

2. What is the normalized throughput?

3. What is the probability of successful transmission?

4. Find the arrival rate of successful packets.

Solution:

1. λ1=R1/L is the first station average packet arrival rate.

λ1=75 packets/s, λ2=100 packets/s, λ3=200 packets/s.

Total average packet arrival rate equals λ = λ1 + λ2 + λ3 = 375 packets/s.

The transmission time of each packet equals T = 2000 [bits/packet] /1000000 [bits/s]=2

ms/packet.

The normalized total traffic G= λT = 375 • 0.002 = 0.75

2. The normalized throughput S = G e-2G

= 0.17

3. The probability of successful transmission Ps = P(K=0) = e-2G

= 0.22

4. The arrival rate of successful packets λs = Ps• λ = 83.7 packets/s.

4. We have a pure ALOHA network with 100 stations. If Tfr = 1μs, what is the number of

Frames/s each station can send to achieve the maximum efficiency.

Solution:

Given, The number of stations in a network ‘n’=100

The frame transmission time Tfr = 1μs

To achieve the maximum efficiency in pure ALOHA, G = 1/2.

If we let ns to be the number of stations

let nfs to be the number of frames a station can send per second.

Then load G = ns × nfs × Tfr

1/2 = 100 × nfs × 1 μs

nfs = 5000 frames/s

Thefore, total number of frames each station can send to achieve the maximum efficiency

is 5000 frames/s


Solution:

Given, The number of stations in a network ‘n’=100

The frame transmission time Tfr = 1μs

To achieve the maximum efficiency in Slotted ALOHA, G = 1.



If we let ns to be the number of stations

Let nfs to be the number of frames a station can send per second.

Then load G = ns × nfs × Tfr

1 = 100 × nfs × 1 μs

nfs = 10,000 frames/s

Thefore, total number of frames each station can send to achieve the maximum efficiency

is 10,000 frames/s

6. One hundred stations on a pure ALOHA network share a l-Mbps channel. If frames are 1000

bits long, find the throughput if each station is sending 10 frames per second.

Solution:

We can first calculate Tfr and G, and then the throughput S.

Frame transmission time Tfr = frame size/ data rate = (1000 bits) / 1 Mbps = 1 ms

G = ns × nfs × Tfr = 100 × 10 × 1 ms = 1

For pure ALOHA, through put S = G × e−2G

= 1× e−2

≈ 13.53 percent.

This means that each station can successfully send only 1.35 frames per second.


Solution:

We can first calculate Tfr and G, and then the throughput S.

Frame transmission time Tfr = frame size/ Channel B.W = (1000 bits) / 1 Mbps = 1 ms

G = ns × nfs × Tfr = 100 × 10 × 1 ms = 1

For Slotted ALOHA, through put S = G × e−G

= 1× e−1

≈ 27.06 percent.

This means that each station can successfully send only 2.706 frames per second.

7. 10,000 stations are competing for the use of a single slotted ALOHA channel. The average

station makes 18 requests/ hour. A slot is 125 µsec. What is the approximate total

channel load?

Solution:

Average requests for 10000 stations = 104 x 18 / (60 x 60) = 50 requests/sec

Average slots number = 1 / (125 x 10-6

) = 8000 slots/sec.

Total channel load = Average requests / average slots number

= 50 / 8000 = 0.0625

Hence, the total channel load is 0.0625 request/slot.



8. In a CDMA system the four chip sequences are:

A = (-1 -1 -1 +1 +1 -1 +1 +1)

B = (-1 -1 +1 -1 +1 +1 +1 -1)

C = (-1 +1 -1 +1 +1 +1 -1 -1)

D = (-1 +1 -1 -1 -1 -1 +1 -1) in bipolar form.

If the received sequence is (-1 +1 -3 +3 +1 -1 -1 +1) .What is the data transmitted by the

four stations.

Solution:

Station A = (-1 -1 -1 +1 +1 -1 +1 +1) . (-1 +1 -3 +3 +1 -1 -1 +1) = +8/8= +1 = bit 1

Station B = (-1 -1 +1 -1 +1 +1 +1 -1) . (-1 +1 -3 +3 +1 -1 -1 +1) = - 8/8= -1 = bit 0

Station C = (-1 +1 -1 +1 +1 +1 -1 -1) . (-1 +1 -3 +3 +1 -1 -1 +1) = + 8/8= +1 = bit 1

Station D = (-1 +1 -1 -1 -1 -1 +1 -1) . (-1 +1 -3 +3 +1 -1 -1 +1) = 0 = Silent

Data transmitted by the four stations are 101 and station D is silent

10 . In a CDMA/CD network with a data rate of 10 Mbps, the minimum frame size is found to be

512 bits for the correct operation of the collision detection process. What should be the

minimum frame size if we increase the data rate to 100 Mbps? To 1 Gbps? To 10 Gbps?

Solution:

Given, Data rate =10Mbps

The frame size = 512 bits

Let us find the relationship between the minimum frame size and the data rate. We

know that Tfr = (frame size) / (data rate)

For CSMA/CD Tfr = 2 × Tp = 2 × Distance / (propagation speed)

9. frame size = [2 × (distance) / (propagation speed)] × (data rate)]

10. (frame size) = K × (data rate)

11. This means that minimum frame size is proportional to the data rate (K is a constant).

12. When the data rate is increased, the frame size must be increased in a network with a fixed

length to continue the proper operation of the CSMA/CD. we mentioned that the minimum

frame size for a data rate of 10 Mbps is 512 bits. We calculate the minimum frame size

based on the above proportionality relationship

o Data rate = 10 Mbps → minimum frame size = 512 bits

o Data rate = 100 Mbps → minimum frame size = 5120 bits

o Data rate = 1 Gbps → minimum frame size = 51,200 bits

o Data rate = 10 Gbps → minimum frame size = 512,000 bits



11. In a CDMA/CD network with a data rate of 10 Mbps, the maximum distance between any

station pair is found to be 2500 m for the correct operation of the collision detection process.

What should be the maximum distance if we increase the data rate to 100 Mbps? To 1 Gbps?

To 10 Gbps?

Solution:

Given, Data rate =10Mbps

Maximum distance between any station pair = 2500 mts

Let us find the relationship between the maximum distance and the data rate. We

know that Tfr = (frame size) / (data rate)

For CSMA/CD Tfr = 2 × Tp = 2 × Distance / (propagation speed)

Tp = Distance / Data rate = 2500/ 10×106

= 250 μs

We calculate the maximum distance based on the above proportionality relationship

o Data rate = 10 Mbps → maximum distance = 2500 mts

o Data rate = 100 Mbps → 250 μs = maximum distance/ data rate

250 μs = distance / 100 ×106

Maximum Distance = 250 μs × 100 ×106

= 25000 mts or 25 Km.

Similarly for:

o Data rate = 1 Gbps → maximum distance = 250 Km.

o Data rate = 10 Gbps → maximum distance = 2500 Km.

12. Figure 3.33, the data rate is 10 Mbps, the distance between station A and C is 2000 m, and

the propagation speed is 2 x 108 m/s. Station A starts sending a long frame at time t1 =0μs;

station C starts sending a long frame at time t2 b =3μs, if the size of the frame is long enough to

guarantee the detection of collision by both stations. Find:

a. The time when station C hears the collision (t3)

b. The time when station A hears the collision (t4)

c. The number of bits station A has sent before detecting the collision.

d. The number of bits station C has sent before detecting the collision.

Figure 3.33 Collision of the first bit in CSMA/CD



Solution:

We have t1 = 0 and t2 = 3 μs

a. t3 − t1= (2000 m) / (2 × 108 m/s) =10 μs → t3 = 10 μs + t1 = 10 μs

b. t4 − t2 = (2000 m) / (2 × 108 m/s) =10 μs → t4 = 10 μs + t2 = 13 μs

c. Tfr(A) = t4 − t1 = 13 − 0 = 13 μs → BitsA = 10 Mbps × 13 μs = 130 bits

d. Tfr(C) = t3 − t2 = 10 − 3 = 07μs → BitsC = 10 Mbps × 07 μs = 70 bits

13. Repeat Exercise 7 if the data rate is 100 Mbps

Solution:

We have t1 = 0 and t2 = 3 μs

a. t3 − t1= (2000 m) / (2 × 108 m/s) =10 μs → t3 = 10 μs + t1 = 10 μs

b. t4 − t2 = (2000 m) / (2 × 108 m/s) =10 μs → t4 = 10 μs + t2 = 13 μs

c. Tfr(A) = t4 − t1 = 13 − 0 = 13 μs → BitsA = 100 Mbps × 13 μs = 1300 bits

d. Tfr(C) = t3 − t2 = 10 − 3 = 07μs → BitsC = 100 Mbps × 07 μs = 700 bits

14. Calculate the Walsh table W8 from W4 in figure below.

Solution:

W8 = [

]



15. Prove the third and fourth orthogonal properties of Walsh chips for W4

Solution:

16. A network with one primary and four secondary stations uses polling. The size of a data

frame is 1000 bytes. The size of the poll, ACK, and NAK frames are 32 bytes each. Each station

has 5 frames to send. How many total bytes are exchanged if there is no limitation on the number

of frames a station? Can send in response to a poll?

Solution:

Polling and Data Transfer

Frame 1 for all four stations: 4 × [poll + frame + ACK)]





Polling and Sending NAKs

Station 1: [poll + NAK]




Total Activity:

24 polls + 20 frames + 20 ACKs + 4 NAKs

Total Bytes:

24×32 + 20×1000 + 20×32 + 4×32 = 21536 bytes



UNIVERSITY QUESTIONS AND ANSWER

1. a). Explain CSMA and show the behavior of the three persistence methods of CSMA. Compare

the vulnerable times in CSMA and CSMA/CD.

(Dec 09/ Jan 10 – 10marks –Answer refer page no. 9 to 13)

b). 10,000 stations are competing for the use of a single slotted ALOHA channel. The average

station makes 18 requests/ hour. A slot is 125 µsec. What is the approximate total

channel load? (Dec 09/ Jan 10 – 05 marks –Answer refer page no. 32)

c). In a CDMA system the four chip sequences are:

A=(-1 -1 -1 +1 +1 -1 +1 +1)

B=(-1 -1 +1 -1 +1 +1 +1 -1)

C=(-1 +1 -1 +1 +1 +1 -1 -1)

D=(-1 +1 -1 -1 -1 -1 +1 -1) in bipolar form.

If the received sequence is (-1 +1 -3 +3 +1 -1 -1 +1) what is the data transmitted by the four

stations. (Dec 09/ Jan 10 – 05 marks –Answer refer page no. 33)

2. a). A slotted ALOHA network transmits 200 bit frames using a shared channel with 200 kbps

bandwidth. Find the throughput if the system produces 500 frames

(May/ June 10 – 03 marks –Ans refer page no.9)

b). A network using CSMA/CD has a bandwidth of 10 mbps. If the maximum propagation

time is 25.6 µsec, what is the minimum size of the frame?


c). Explain token passing method of controlled access of the channel.


d). What is channelization in the context of multiple access? What are the various available

channelization techniques? List the properties of orthogonal sequences used in CDMA.

(May/ June 10 – 08 marks –Ans refer page no.20 to 23)

3 a). Explain Pure ALOHA protocol. (December 10 – 06 marks –Answer refer page no.3 to 5)

b). Pure ALOHA networks transmits 200-bit frames on shared channel of 200Kbps. What is the

throughput if the system produces: i)1000 frames/sec ii)500 frames/sec iii)250 frames/sec

(December 10 – 04marks –Answer refer page no.07)

c). Discuss the three controlled access methods.

(December 10 – 10marks –Answer refer page no.18 to 19)



4 a). Define random access method explain three different protocols in this category.

(June/July 2011 – 10marks –Answer refer page no.2 to 3 and 8 to 11)

b). Explain reservation, pooling and token passing in controlled access method.

((June/July 2011 – 10marks –Answer refer page no.18 to 19)

5.a) Write the different physical topologies used in the logical ring method and explain briefly.


b). In CSMA/CD, the data rate is 10Mbps,the distance between the stations ‘A’ and ‘C’ is

2000m and propagation is 2*10^8m.Station A starts sending a long frame at time t1=0;station C

starts sending along frame at t2=3µs. The size of the frame is long enough to guarantee the

detection of collision by the stations. Find:

i. The time when station C hears the collision (t3).

ii. The time when station A hears the collision (t4).

iii. The number of bits station A has sent before detection the collision.

iv. The number of bits station C has sent before detection the collision.


6. a). Compare ALOHA with slotted ALOHA. What are the reasons for poor channel utilization

in ALOHA systems? How the same is improved in CSMA.

(December 2012 – 08 marks –Answer refer page no.5 and 7)

b). Discuss the concept of

i) 1-persistent CSMA ii) Non-persistent CSMA.

(December 2012 – 06 marks –Answer refer page no.11 and 12)

c). Explain the working of CSMA/CD. Suppose a point to point link is set up between earth and

MARS. The distance from earth to mars is approximately 55 Gm and data travels over the

link at a speed of light 3×108 m/s. Calculate the minimum round trip propagation time.

(December 2012 – 06 marks –Answer refer page no.13)

Solution: Minimum round trip propagation time = 2 * Tp = 2*183.33 = 366.66 seconds

Where Tp = Distance earth and MARS / Speed of light = 55 *109

/ 3*108

= 183.33 sec

7 a). Explain how collision are avoided through use of “IFS’, contention window and

acknowledgments in CSMA/CA. with help of the flow chart show the proceduer for

CSMA/CA. ( June/July 2013 – 10marks –Ans refer page no. 15 to 17)



b).Explain the ‘Token Passing’ method of controlled access of the channel

(June/July 2013 – 06marks –Answer refer page no. 19)

c). Slotted ALOHA networks transmits 200-bit frames on shared channel of 200Kbps. What is

the throughput if the system produces: i) 1000 frames/sec ii) 500 frames/sec iii) 250 frames/se

(June /July 2013 – 04marks –Answer refer page no. 9)

ccn nots final unit -3 (1)

Documents

access conflict

multipleaccess protocol

controlled access protocols

multiple access resolution

random access method

broadcast link

shared link

media access control