Gebali F. Analysis Of Computer And Communication Networks

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10.8 IEEE 802.11: PCF Function for Infrastructure Wireless LANs 369

For the case N = 5, the transition matrix is 5 ×5 and is given by

P =

⎡

⎢

⎣

yp(4, 0) 0 0 0

p(5, 2) p(4, 1) p(3, 0) 0 0

p(5, 3) p(4, 2) p(3, 1) p(2, 0) 0

p(5, 4) p(4, 3) p(3, 2) p(3, 1) p(1, 0)

p(5, 5) p(4, 4) p(3, 3) p(2, 2) p(1, 1)

⎤

⎥

⎦

(10.122)

10.8.5 1-Persistent IEEE 802.11/PCF Performance

Having obtained the transition matrix, we are able to find the performance of the

1-persistent IEEE 802.11/PCF channel.

The throughput of the system is given by

Th = s

[

1 − p(N, 0)

]

(

1 −s

)

(10.123)

= 1 −s

p(N, 0) (10.124)

The first term corresponds to the probability that the queue is empty and one

or more requests arrive and the second term corresponds to the probability that the

queue is not empty. Figure 10.32 shows the throughput of the 1-persistent IEEE

802.11 protocol when N = 10. The solid line is the throughput of 1-persistent IEEE

802.11/PCF, the dashed line represents the throughput of slotted ALOHA, and the

dotted line represents the throughput of pure ALOHA.

0 2.5 5 7.5 10

0.2

0.4

0.6

0.8

Input traffic

Throughput

Fig. 10.32 The throughput for 1-persistent IEEE 802.11/PCF versus the average input traffic when

N = 10. The solid line is the throughput of 1-persistent IEEE 802.11/PCF, the dashed line rep-

resents the throughput of slotted ALOHA, and the dotted line represents the throughput of pure

ALOHA

370 10 Modeling Medium Access Control Protocols

0 2.5

–2

–1

5 7.5 10

Input traffic

Number of Queued Users

Fig. 10.33 The average number of queued users with frames to send in 1-persistent IEEE

802.11/PCF versus the average input traffic when N = 10

Notice that the throughput reaches 100% as soon as the input traffic approaches

1 since at this level there is more than one user with frames to send and there are no

collisions. Further, the use of transmit buffers ensures that arriving requests are not

dropped as long as the buffer is not full.

The average number of queued users with frames to send is given by

N−1



i=0

(10.125)

Figure 10.33 shows the average number of queued users versus input traffic when

N = 10. Define p

as the access probability for one user which is given by

Th(user)

(in)

(10.126)

Figure 10.34 shows the access probability versus input traffic when N = 10. The

solid line is the access probability of 1-persistent IEEE 802.11/PCF, the dashed line

represents the access probability of slotted ALOHA, and the dotted line represents

the access probability of pure ALOHA.

Having found the acceptance probability, we are able to determine the average

number of attempts before a user gets access to the medium:

10.8 IEEE 802.11: PCF Function for Infrastructure Wireless LANs 371

0 2.5 5 7.5 10

0.2

0.4

0.6

0.8

Input traffic

Acceptance Probability

Fig. 10.34 The access probability in IEEE 802.11/PCF versus the average input traffic when

N = 10. The solid line is the access probability of 1-persistent IEEE 802.11/PCF, the dashed

line represents the access probability of slotted ALOHA, and the dotted line represents the access

probability of pure ALOHA

∞



k=0

(

1 − p

)

(10.127)

1 − p

(10.128)

Figure 10.35 shows the delay of IEEE 802.11 when N = 10. The solid line is the

delay of 1-persistent IEEE 802.11, the dashed line represents the delay of slotted

ALOHA, and the dotted line represents the delay of pure ALOHA.

10.8.6 1-Persistent IEEE 802.11/PCF User Performance

The previous section modeled the states of all the N users of the 1-persistent IEEE

802.11/PCF. In this section, we study an individual user, usually called the tagged

user. Specifically, we would like to study the access probability p

which represents

the probability that a request for transmitting data will be granted.

Because all users are the same, the throughput seen by the tagged user is given by

Th(user) =

(10.129)

Define p

as the access probability for our tagged user which is given by

372 10 Modeling Medium Access Control Protocols

0 2.5

–5

5 7.5 10

Input traffic

Delay

Fig. 10.35 Delay for the 1-persistent IEEE 802.11/PCF protocol versus the average input traf-

fic when N = 10. The solid line is the delay of 1-persistent IEEE 802.11/PCF, the dashed line

represents the delay of slotted ALOHA, and the dotted line represents the delay of pure ALOHA

Th(user)

(in)

(10.130)

Thus we obtain

1 −s

p(N, 0)

(10.131)

At light loading condition such that aN  1, the acceptance probability becomes

= 1, which makes sense since most of the time only one user has a request and is

immediately granted access to the medium. For heavy traffic a ≈ 1, the acceptance

probability p

≈ 1/N, which also makes sense since most of the time all users have

requests and only one is granted access with probability 1/N.

Having found the acceptance probability, we are able to determine the average

number of attempts before a user gets access to the medium:

∞



k=0

(

1 − p

)

(10.132)

1 − p

(10.133)

Example 10.4 A 1-persistent IEEE 802.11 employing the PCF function serves 8

customers and the channel bit rate is 1,920 kbps, which is used for a wireless micro-

cellular system. Each customer is assumed to issue 20 requests each second and the

average length of a frame is 5.12 kb. Obtain the performance of this system.

10.8 IEEE 802.11: PCF Function for Infrastructure Wireless LANs 373

First, we must find the time step value knowing the duration of an average frame

length:

time step value = 5.12/1, 920 = 2.7ms

To find the frame arrival probability a per time step, we need to estimate the

number of requests issued by the user over some observation time and we also

need to find the number of time steps during this same observation period. Take

the observation period to be 1 s. Thus the total user traffic during this period is

(in) = user request rate ×1 = 20 requests

The number of time steps during this period n is given by

n =

= 375

The number of requests from the binomial distribution is given from the relation

a × n = N

(in)

Thus we have the user request probability as

a = 0.0533

The transition matrix will be

P =

⎡

⎢

⎣

0.9357 0.6814 000000

0.0573 0.2687 0.7198 00000

0.0065 0.0454 0.2433 0.7603 0000

0.0005 0.0043 0.0343 0.2433 0.8031 0 0 0

00.0002 0.0026 0.0343 0.1810 0.8484 0 0

000.0001 0.0026 0.0153 0.1434 0.8962 0

0000.0001 0.0006 0.0081 0.1010 0.9467

000000.0002 0.0028 0.0533

⎤

⎥

⎦

The distribution vector for the request queue states is

s =



0.8988 0.0848 0.0145 0.0018 0.0002 0 0 0



We see that approximately 90% of the time the queue is empty, which is the value

of the first element of the vector. The performance figures are

374 10 Modeling Medium Access Control Protocols

Th = 0.4203 frames/time step

= 0.1198 requests

= 0.832

= 0.202 time steps

10.9 IEEE 802.11e: Quality of Service Support

The IEEE 802.11e provided enhancements over the legacy IEEE 802.11 to be able

to provide quality of service support such as voice over wireless and streaming

multimedia. True to our strategy throughout this book, we strive to provide sim-

plified analyses. The reader can then carry this approach further and develop more

sophisticated models.

Let us start by briefly explaining how quality of service is supported in this pro-

tocol. The distributed coordination function (DCF) of legacy IEEE 802.11 is now

replaced with enhanced distributed channel access (EDCA). High-priority traffic is

given better chance of accessing the channel compared to low-priority traffic. On

average, high-priority traffic waits less and experiences higher throughputs. We can

extend the analysis in Section 10.7 to describe the IEEE 802.11e using the following

assumptions:

1. There are two user priority classes: class 1 is the high-priority users and class 2

is the lower-priority users.

2. There are two contention windows: w

for class 1 and w

for class 2.

3. The probability that a user has a frame to send is a.

4. γ<1 is the probability that a transmitted frame belongs to class 1, and 1 − γ

the probability that the frame belongs to class 2 traffic.

The probability that a class 1 user with data to send reserves a particular reserva-

tion slot is given by

(10.134)

Similarly, α

= 1/w

is defined for class 2 users. Slot reservation requests are

representative of incoming traffic. The user population seen by each slot in the con-

tention windows of the two service classes is given by

(10.135)

(10.136)

The probability that k class 1 users attempt a transmission during a given reser-

vation slot is given by

10.9 IEEE 802.11e: Quality of Service Support 375





(1 −a

)

−k

;0≤ k ≤ N

(10.137)

Likewise, the probability that k class 2 users attempt a transmission during a

given reservation slot is given by





(1 −a

)

−k

;0≤ k ≤ N

(10.138)

Figure 10.36 shows the IEEE 802.11e channel state transition diagram. We notice

that as far as the channel is concerned, it could be in one of four states:

• idle when no frames are being transmitted

• collided when two or more frames are being transmitted simultaneously

• transmitting class 1 traffic when exactly one class 1 frame is being transmitted

• transmitting class 2 traffic when exactly one class 2 frame is being transmitted

The channel stays in the idle state with probability x given by

x = u

= (1 −a

)

(1 −a

)

(10.139)

The channel moves from idle to class 1 transmitting state with probability y

which is the probability that only one class 1 user in any of the w

reservation slots

requests a transmission and all users in the previous slots did not request to access

the channel. The probability y

for this event is given by

Idle

Class 2 Transmitting

Collided

...

Class 1 Transmitting

Fig. 10.36 IEEE 802.11e channel state transition diagram

376 10 Modeling Medium Access Control Protocols

= u

+···+u

−1

= u

1 −

(

)

1 −u

(10.140)

In a similar fashion, the probability y

is given by

= u

1 −

(

)

1 −u

(

)

1 −v

−w

1 −v

(10.141)

We can now proceed to find the distribution vector at equilibrium using Fig. 10.36.

We can write the following equations:

= y

i (10.142)

= y

i (10.143)

c = zi (10.144)

It is easy to prove that the state probabilities are given by

i = 1/D (10.145)

= y

/D (10.146)

= y

/D (10.147)

c = z/D (10.148)

D = n(1 − x) +1 (10.149)

0 15 30

0.2

0.4

0.6

0.8

Input traffic

Throughput

Th1+Th2

CSMA/CA

Th1

Th2

Fig. 10.37 Throughput for the IEEE 802.11e protocol versus the average input traffic when

γ = 0.5, w1 = 4, w2 = 8, n = 10, and N = 32. Dashed line represents the throughput of

slotted ALOHA and the dotted line represents the throughput of pure ALOHA

10.9 IEEE 802.11e: Quality of Service Support 377

The throughputs for both priority classes are given by

= ny

/D (10.150)

= ny

/D (10.151)

Figure 10.37 shows the throughput of IEEE 802.11e when γ = 0.5, w1 = 4,

w2 = 8, n = 10, and N = 32. The average input traffic N

(in) is defined as

(in) = aN (10.152)

We notice, as expected, that class 1 throughput is higher than class 2 throughput.

What is really important for quality of service support is not the absolute value of

each throughput component since incoming traffic belonging to a certain service

class could be low to start with. In that case, the resulting throughput would be low

but this does not indicate low performance by any means. What is important here is

the packet acceptance probability p

for each class. We define the packet acceptance

probabilities for the two service classes as

γ N

(in)

(10.153)

(1 −γ )N

(in)

(10.154)

We would like to be assured that p

> p

for all levels of incoming traffic.

Figure 10.38 shows the frame access probability for the IEEE 802.11e protocol

–2

–1

16 32

Input traffic

Access Probability

CSMA/CA

80211e Class 1

80211e Class 2

Fig. 10.38 Frame access probability for the IEEE 802.11e protocol versus the average input traffic

when γ = 0.5, w1 = 4, w2 = 8, n = 10, and N = 32. Dashed line represents the access

probability of slotted ALOHA and the dotted line represents the probability of pure ALOHA

378 10 Modeling Medium Access Control Protocols

versus the average input traffic when γ = 0.5, w1 = 4, w2 = 8, n = 10, and

N = 32. Dashed line represents the throughput of slotted ALOHA and the dotted

line represents the throughput of pure ALOHA.

We note from the figure that the access probability for the class 1 traffic is higher

than that of class 2. This will lead to lesser delay for class 1 traffic, which is the

desired performance.

Problems

ALOHA Network

10.1 Use Equation (10.20) to find the maximum value for the throughput of an

ALOHA network and the value of a at the maximum.

10.2 Using the results of Section A.6 in Appendix A on page 599, prove that the

maximum throughput of the ALOHA network approaches the value 1/2e as

N →∞.

10.3 Assume an ALOHA network that is operating at its optimum conditions.

What is the average number of attempts for a user to be able to transmit a

frame under these conditions?

10.4 Assume an ALOHA network where the frame length is a multiple of some

unit of length with an upper limit on the maximum frame size. Draw a pos-

sible transition diagram for such system and write down the corresponding

state transition matrix.

10.5 Assume an ALOHA network where the propagation delay is bigger than the

frame time T . What would be a good choice for the time step of the Markov

chain? Draw a possible transition diagram for such system and write down

the corresponding state transition matrix.

10.6 Assume there are 25 users in an ALOHA network. What is the transmission

request probability a that corresponds to maximum throughput and what is

the value of the maximum throughput?

10.7 Assume there are 25 users in an ALOHA network and the probability that a

user request access is a = 0.06. What is the throughput of the channel and

what is the probability that a user will successfully transmit frame after three

unsuccessful attempts?

10.8 What is the average number of unsuccessful attempts before a user can trans-

mit a frame in the above problem?

Slotted ALOHA Network

10.9 What is the major difference between ALOHA and slotted ALOHA?

10.10 What are the major differences between the transition diagrams of ALOHA

and slotted ALOHA?

10.11 Write down the expressions for the steady-state distribution vectors for

ALOHA and slotted ALOHA and comment on their similarities and