MOB Subject 5 –Access methods Static and dynamic access techniques

MOB Subject 5 –Access methods Static and dynamic access techniques

MOB Subject 5 –Access methods Static and dynamic access techniques

1 Static access techniques

1.1 Space-division multiple-access (SDMA)

When one seeks to cover a broad territory by a whole of cells of the same dimension, one considers that the traffic is uniformly distributed, that the propagation law is the same and that the power of all the equipment is constant The same number of carrier will have thus to be allocated to each basic station

One will often use a hexagonal paving to cut out the zone has to cover, the hexagon being the polygon nearest to the circle which allows paving the plan One calls cluster, a set of adjacent cells which uses the unit of carriers one and only one time This cluster is repeated on all surface has to cover

Arithmetic and geometrical considerations allow showing that a cluster having a given number

of carrier is optimal if it is regular For a hexagonal paving, the size K of cluster verifies the relation:

with i and j entries positive natural or null.

The paving of the zone has to cover is realized of iterative manner in leaving of a cluster given

of K cells From one cells of this cluster, one moves vertically i cells to one of its sides, then j cells

in a direction forming an angle of +60 degrees with the initial direction The arrival point defines then the "homologous" cell of adjacent cluster By repeating this process for the set of cells of the cluster chosen at the origin while keeping these same directions, one defines an adjacent cluster One repeats then this process for the five other possible directions (other dimensions of the cell) to find the set of adjacent clusters In proceeding secondly little by little on the cluster, one then builds

a paving of the cover zone

Each cell of a cluster is seen allocated a subset of the carrier which is own for it The homologous cells of all the cluster use the same carrier The minimal distance between two cells (homologous)

using the same carrier is called reuse distance The size K of the cluster also constitutes the reuse

factor of the carrier

1 What is the principle of the reuse of frequencies in the domain of the cellular networks?

In a cellular network, the space is divided into cells, each served by a base station (i.e a cell is

a portion of the territory covered by a base station)

It was affecting to each cell (i.e at each base station), a number of carriers of the total bandwidth available based on estimated traffic in this cell

It is possible to reuse one even at the same carrier in different cells if they are sufficiently remote The reuse of frequencies thus allows an operator has to cover an unlimited geographic area by using a frequency band width limited

2 Discuss the advantages and disadvantages to use of cells of small size, in comparing with cells of big size in the cellular networks

Advantages:

- greater reuse of frequencies: optimization of bandwidth;

- increase battery life because the power of emission decreases;

- if a cell is malfunctioning or cluttered, possibility to switch to a neighboring cell.

Disadvantages:

- Roaming Management heavier if mobility;

- Increasing the level of interference between cells

3 List several manners to increase the capacity of a cellular network:

- Adding new carriers: usually when a system is implemented in a region, all carriers are not used Growth can then be managed using these carriers

- Borrowing frequency: in the simplest case, cells are over-used carrier will temporarily borrow the adjacent sub-cells used The carriers may also be dynamically allocated to cells

- The severing of cells: in practice, the traffic distribution and extent dansune cell are not uniform The cells can be very busy subdivided into smaller cells

- Sectorized cells: this is to divide the cell into sectors, each possessing its own set of carriers Typically we can consider three or six sectors per cell Each sector is assigned a distinct subset of carriers Directional antennas at the access points or base stations are used

to concentrate the signal in the region of the sector and reduce the interference and

"inter-sector"

- Microcells: for cellular networks with larger cells, the antennas must be placed on top of hills or tall buildings When cells become smaller, the antennae can be placed on buildings or even less high peaks of the streetlights The reduction of cell size accompanied by a reduction of the power radiated by both points of access by mobile devices The microcells may be used in congested areas, such as certain city streets, highways, the interior of large public buildings

4 Factor of reuse K.

a) Give the first entries which verify the equation (1) What do these values of K

correspond to?

The first K values are: 1, 3, 4, 7, 9, 12, 13, 16, 19, 21, 25, These values correspond

to the sizes of possible reasons

b) Show on the figure 1 the cluster for reuse for K = 19.

Figure 1: Hexagonal clusters

Figure 2: Hexagonal clusters with K = 19

Figure 3: Hexagonal clusters with K = 3

5 An analog cellular system has 33 MHz of band-width and uses two channels simplex of 25 kHz to provide a transfer service of full-duplex trace and the indication necessary

a) What is the number of full-duplex channels available by cell for a reuse factor K = 4, K =

7, K = 12?

The number of available channels is N = 33,000 / 50 = 660

For a frequency reuse factor of K, each cell has Ncel = N / K full-duplex channels With K = 4, Ncel = N / K = 660 / 4 = 165 full-duplex channels

With K = 7, Ncel = N / K = 660 / 7 = 94 full-duplex channels With K = 12, Ncel = N / K = 660 / 12 = 55 full-duplex channels b) If one supposes that 1 MHz is dedicated to the control channels and that only a control channel is necessary by cluster, determine a reasonable distribution of the voice channels for each cell, for the 3 reuse factors given previously

32 MHz are dedicated to voice traffic, a total of 640 full-duplex channels

For K = 4, we can have 160 voice channels (4160 = 640)

For K = 7 then 640 / 7 = 91,4 which is not an integer It then looks for x number of cells with 91 voice channels and y the number of cells with 92 voice channels One solves the system of equations has two unknowns: x91 + y 92 = 640 and x + y = 7

We get x = 4 and y = 3

We can have 4 cells with 91 voice channels and 3 cells with 92 voice channels, remember (491 + 392 = 640)

For K = 12, we can have 8 cells with 53 voice channels and 4 cells with 54 voice channels (853 + 454 = 640)

1.2 Frequency-division multiple access (FDMA)

1 A cellular system uses the FDMA, with a spectrum allocation of 12,5 MHz in each direction,

a guard band of 10 kHz of each end of the allocation spectrum and a width of 30 kHz for each channel Which number of users can one have in maximum?

416 10

30

) 10 10 ( 2 10 5

,

12

3

3 6











2 The spectral efficacy of the FDMA for a cellular system is defined by: with

B c the band-width of the channel, N T the total number of trace channels in the covered zone,

and B w the band-width in a direction What is the maximum limit of η a?

The amount of bandwidth allocated to voice channels (BcNt) must be no greater than the

total bandwidth (Bw) Therefore η a 1

1.3 Time Division Multiple Access (TDMA)

In the traditional local networks, a physical support is sharing between several users It is thus important to define the access rules of the media for the frames emission Two families of solutions are possible: the static solutions and the dynamic solutions

1 Let a network be constituted of N connected stations on the same support Let T be the average stay time of a frame expressed in seconds, C the channel capacity in bits per second and M the

arrival rate of frames on the channel, expresses in frames per second It is supposed that the inter-arrival distribution of frames is exponential The size of a frame is distributed according

to an exponential law, of average L bits The digital values of these parameters are C = 64 kbits,

M = 10, L = 4 kbits.

a) What is the value of T?

To calculate the average time of stay, we consider a system has queue The frames of the stations are served one after the other in their order of arrival (FIFO

discipline) It has a single server.

Each frame with a size distributed according to an exponential distribution and the capacity of the channel being constant service time of a frame has an exponential distribution of average L/C or rate= C/L

Finally, the distribution is exponential interarrivals frames, rate  = M

It was therefore a M/M/1 queue The theory of the waiting can calculate the average time of stay:

T =



1

 T =

M L

C 1

6

1 10 4 64



b) The channel is divided in static way of N sub channels independent The traffic of the

stations is distributed in a uniform way on these various sub channels What becomes the average stay time of a frame?

Each sub-channel has a capacity of C / N bits per second The service time of a frame has an exponential distribution of average L / (C / N), or rate = C / (N L)

The distribution of inter-arrivals frames on each sub channel is exponential, rate = M/N

The average residence time is identical on each sub-channels Using the model has M/M/1 queue, we obtain:

s M

L

C N N

M LN C

T static

6

4 10 4

644













c) What do you deduce from?

For multiplexing static, the average time of stay T is N times greater than if all the frames had been ordered, as in a single queue using the single channel globally The static solutions are simple to introduce, but suffer from a very low yield You lose the multiplexing gain!

3 Dynamic access techniques

Pure Aloha: the principle of Aloha is to let talk to everyone when he wants it and without discrete

the moments when one can start to emit

Discrete Aloha: the principle is the same but with a discrete time i.e the moments when one can

start to emit are spaced out

CSMA (Carrier Sense Multiple Access): CSMA protocols used in the wire networks allow to

improve the output of the ALOHA protocol because they bring the certainty that the stations will be careful not to emit if they note that another station is emitting Another improvement

consists in introducing a mechanism allowing, in the event of collision, stopping immediately the continuation of transmission of a frame In others words, the transmitting stations seeing that the frame between collision ceases to emit They will start again the transmission (if the support is free) after a random duration This cannot nevertheless be set up in wireless networks

One distinguishes two types of persistent CSMA and non-persistent CSMA

Persistent CSMA: This is about the first CSMA protocol called also "1-persistent” because the

station which wants to emit maintains the listening of the channel and emits as soon as it notes that it became free (without carrier)

Non-persistent CSMA: The station listens to the channel before emitting If the channel is free,

one emits directly its frame If the channel is already in use, the station does not remain in listening permanent the station, it waits a random time then repeats the procedure since the beginning

p-persistent CSMA: The time is divided into intervals, like "Discrete Aloha" When a station

wants to emit, it listens to the channel If the channel is free, it emits its frame with a probability

p and thus delays the emission at the next slot with a probability 1  p One repeats this procedure as long as the frame was not emitted correctly and that another station does not occupy the channel If another station emits, a random time is waited If at the beginning, the channel was occupied, the station awaits the next slot

Collision detection CSMA (CSMA/CD): One functions with 3 types of slots: idle slots, when no

station is emitting, busy slots, when a station is emitting its frame, contention slots when there is

a possible collision Once a station can emit, it sends its frame but it listens to the channel while

it emits, which returns the channel half-duplex considered that the controller uses already its ability of reception to control that there is no collision which occurs

MACA and MACAW:

- MACA (Multiple Access with Collision Avoidance) the idea consists in asking the authorization to the receiver to send the data This is done by the sending of a small frame to confirm that one can send the data frame One uses RTS (Request To Send) and CTS (Clear

to Send) frames

- MACAW (MACA for Wireless) One added to MACA the settlement to be able detect the lost frames

1 Pure Aloha

In order to determine performances of this access method of the simple Aloha protocol, one

will suppose that the frames are of length fixed t and one ignores the propagation time on the

support

Poisson process

A Poisson stochastic process of parameter  , is generally used to model the arrivals of clients

in a system;  is then the average rate of arrivals of clients in the system By definition, the probability that k clients arrive during x (between t and t + x) is independent of time t and the

"pass" process (it is the priority "without memory" of the Poisson process which makes that it

is largely used), and is given by:

One can then easily calculate the average number of clients who arrive in duration x:

(3)

a) If a machine starts to emit a frame at a moment t 0, what is the vulnerability period (time interval for which it should not be emission on behalf of other stations in order to avoid a collision)?

This frame is properly issued when the time t0 + t (and therefore properly received by the other stations are currently neglected or propagation time) if no other machine begins to issue during the time interval [t0, t , t0 + t] This "period of vulnerability" in length 2t b) If one suppose that the frames are generated in according to a Poisson process (of rate),

what is the average number G of frames emitted for duration of frame t?

It is therefore gives gives G = E[N(t)] =t

c) Calculate the probability q that no frame is emitted for the vulnerability period.

The probability that no frame is emitted during a period of vulnerability is given by:

q = P[N(2t) = 0] = e 2  t= e 2G

d) Deduce the average number S of frames emitted successfully for duration of frame t The latter relation makes it possible to connect the parameter S associated to the useful flow of support (an average number of frames transmitted successfully by time unit = S/t )

to the parameter G associated with traffic on the support (an average number of frames emitted in total by time unit = G/t ) The curve giving S according to G is represented on

the figure 4

It is given by : S = Gq = G e 2G

e) Comment this curve

Figure 4: Useful charge S according to the total charge G

f) Which value Gmax is the flow S maximum with?

Just calculate the derivative of S with respect to G and how to find, Gmax, it vanishes:

G G

G

G

e G Ge

e dG

dGe2   2 2  2 (12 )  2

This implies Gmax = 0.5

The maximum of S is attained when G = 0,5, that is to say when computers emit, on average, one frame every two units of time t

g) Deduce the efficacy of the pure Aloha protocol

The flow rate is maximum for G = Gmax = 0,5 and is Smax =

e

0,5 0,184.

Up to 36% of frames are successfully issued (as a length equal to 100t, a maximum of 18 frames on 50 will be issued with success), using only 18,4% of transmission time h) Which disadvantages do you see with the use of the pure ALOHA technique?

- Lack of efficiency: Just as the last bit of a frame coincides with the first bit of a frame so that there is collision and thus transmission of two frames

- The frame transmission is not interrupted by collisions

- The flow becomes very weak as the number of communications increases

i) A group of N stations share a pure ALOHA channel of 56 kbit/s Each station emits in

average a frame of 1000 bits for every 100 seconds; even if the preceding one was not sent (it is stored by the station to be retransmitted) By taking again the results obtained

for S max , determine the maximum value of N that one can have?

With pure Aloha technique, the "usable bandwidth” is 0,184  56 Kbps = 10.3 Kbps Each station requires 1000bit / 100s = 10 bps The maximum number of stations that can share the channel is equal to N = 10300/10 = 1030 stations

2 Slotted Aloha Variant

An improvement of the pure ALOHA protocol consists in defining intervals of repetitive times (slots) and not in authorizing the transmitting stations that in beginning of each interval This protocol is known under the name of discrete ALOHA protocol

a) Calculate the vulnerability period of the discrete ALOHA protocol

It can reduce the period of vulnerability in half (compared to pure ALOHA, the latter from 2t to t) and thus increase the efficiency of access to the media transmissions can take place only moments precise: 0, 1, 2, , n, n +1, : A packet that arrives during the interval (n-1, n) begins its transmission at time (n) and ends at the moment (n + 1) a packet arriving in a slot given do not undergo collisions if no other package arrived on the same slot

b) Do you see a constraint imposed by this system?

It requires synchronization between the various stations which are connected

c) Calculate the probability p that a package is emitted successfully.

Always with the assumption of Poisson: p = e G

d) Explain S according to G and of p.

S = Gp = Ge G

e) Calculate the maximum efficacy of the Slotted Aloha protocol

The maximum efficient U is obtained by canceling the derivative of S with respect to G For G = 1, then U = 1/e = 36%

Note: node tries to transmit all its transmissions are successful and the efficiency is 100% This situation is not captured by our model! The Poisson assumption is true only when the traffic G is an aggregated traffic, generated by a large number of stations Therefore, the efficiency of 36% corresponds to the case or a large number of stations with comparable flow rates and sharing a common channel with Slotted Aloha protocol

3 Aloha Reservation

In the Aloha Reservation variant, the protocol alternates reservation phases with transmission phases During the reservation phase, the nodes use the Slotted Aloha protocol to try to reach the channel: a node which succeeds to reach the channel diffuses a reservation while emitting

a package containing its identification number At the end of the reservation phase, the transmission phase starts during which the nodes having made a reservation emit their package and this in the order of their reservation At the end of the transmission phase, a new

reservation phase starts, etc One notes T res the duration of a reservation slot, T transm the

transmission time of a package; it is supposed that T res = 0, 05 T transm

a) How many slots of reservation are necessary in average to succeed a reservation?

We wish to calculate the maximum efficiency of the protocol, that is to say, the maximum fraction of time during which packets can be issued successfully, or Smax

b) Calculate the efficacy U of the Slotted Aloha protocol.

Aloha Reservation Aloha Protocol A uses:

- phases of reservation, with 36% efficiency which also means that each slot has a probability of 0.36 to be used successfully! This also means that a reservation takes on average 1 / 0, 36 slots reservation

- phases with a transmission efficiency of 100% The efficiency is then given by Aloha:

U = Ttransmit / (Tres/0,36) + Ttransmit) = 1 / (1 + 2,8 Tres/ Ttransmit) = 0,871559633 Assuming that the length of a slot reservation is 5% of the length of the transmission of a packet, we obtain U = 1 / (1 + 2,8 5 / 95) = 0,87155963388%

4 CSMA and its derivatives

a) What are the protocols Carrier Sense sensitive in to the delays of propagation?

Carrier Sense Protocols are very sensitive to the propagation time In eect, if a station starts to issue, it is very unlikely that another station starts to issue just after but if the propagation time of the frame is long enough, it may be another station listens to the channel to send a frame, the channel is free (for her) and it emits its frame, which has the effect of creating a collision

b) Which conditions can a local network without wireless use a CSMA/CD protocol under?

If the issuer is able to detect a collision and that all radio transmitters are within reach of each other

c) Explain how a 1-persistent CSMA protocol functions

Prior to emit a frame, the station listens to whether the channel is free If so, it emits its frame, otherwise it will listen until the channel is free and it becomes free, it emits its frame If the frame emitted has created a collision, the station waits a random time before retry to emit its frame in a same way the We call 1-persistent because when the channel is free, the station emits with a probability of 1

d) Discus the performances of the non-persistent adjournment in comparing with the 1-persistent

The nonpersistent better use of bandwidth but increases latency compared to the 1-persistent Idea: If the two stations were willing to issue more patient there would be fewer collisions The non-persistent CSMA protocol agrees to be deliberately emit less press than its frame protocol 1-persistent

e) When a station wishes to emit, it listens to the trace If during a fixed delay baptized DIFS ("Distributed Inter Frame Spacing") no other station emits, the station starts to emit the exit

of DIFS delay If a station has already started to emit, the station in listening defers its transmission (1) and awaits the end of the transmission in progress At the exit of this waiting, it waits in more a random delay then transmits if no transmission takes place before the expiration of its delay, otherwise it waits defers its new transmission and passes

by again into (1) Is this about a persistent or non-persistent adjournment?

non-persistent