MOB Subject 7 –IEEE 802.11 standards Wifi performances

MOB Subject 7 –IEEE 802.11 standards Wifi performances

Fladenmuller Baey - ”IEEE 802.11standards” 1/8

MOB Subject 7 –IEEE 802.11 standards

Wifi performances

1 Impact of the transmission quality

One is interested here in the impact of radio transmission errors on the performances of a wireless network and in the relation between transmission and fragmentation quality One takes here the standard IEEE 802.11b as an example

1 Let p be the binary error probability on a communication channel used by a local network and n be the length of a physical frame in bytes What is the error probability q on a frame?

It will be supposed that the errors on the bits are independent

A N : p = 10-6, n = 2370 bytes (maximum length of a MAC frame in conformity with the

802.11 standard (2346 bytes) + PHY header of 24 bytes)

A frame is error just if any bit has undergone a transmission error The corresponding probability can be written: (1 p)8n

The probability of error per frame can be written thus:

q = 1 – (1 – p)8n = 0,018781399

2 For an error probability q per frame, determine the average number N of transmissions has

to effect to succeed the correct transmission of a frame? For the digital application, one will

take p = 10-6

Reminder: 1 + 2q + 3q2 + … + iqi-1 + … =

) 1 (

1

2

q

 The probability that a frame need to be exactly i transmissions correctly transmitted can be written: p(i) = qi – 1(1 – q)

The average number of transmissions necessary to succeed a correct transmission is written so:

] ) 1 ( 1 [ 1

1 1

1 ) 1 (

1 )

1 ( ) 1 ( )

n i

i

p q

q

q iq

q q

iq i

ip N



























0191 ,1 ) 1

(

1



p

3 One wants to make a first coarse evaluation of the bit rate lost because of the retransmissions on error One thus does not take any other factor in account but the errors of transmission One supposes in first approximation that all bits of the physics frame are send with the same flow

a) For a error probability per frame q and a nominal capacity d of transmission, what is the

useful bit (goodput) and what is the lost flow at the time of the retransmissions for a channel which is used permanently (there is always a frame to transmit i.e is the channel is saturated)?

A D : d = 11 Mbit/s and p = 10-6

The useful throughput is written:

Mbps p

d p

d N

d

n

8 , 10 ) 1 ( )

1 (

8













The lost throughput during broadcasts can be written: (1 1)

N

d  or 0.2 Mbps

b) The same question that previously if one takes the header MAC + PHY of length l eequal

to 58 bytes into account (24 bytes of header PHY + 34 bytes of header MAC)

Length of the field data of a frame: ld

Length of the MAC + PHY header of a frame: le

Total length of a frame: n = ld + le

With these notations, useful throughput is written now (by reasoning on a frame):

Mbps p

d p

d N

d

n

53 , 10 ) 1 ( )

1 (

8













4 The wifi standard 802.11 envisages a functionality of fragmentation on the MAC level

Suppose that one splits up the contents of a physical frame of n bytes in k fragments (that

one will suppose all of identical size) before to send it, and that like previously one does not take any another factor into account but the transmission errors What becomes the useful bit?

A N.: d = 11 Mbit/s, length of the frame n = 2370 bytes, k = 4 fragments and p = 10-6 Length of the field data of a frame: ld

Length of the MAC + PHY header of a frame: le

Total length of a frame: n = ld + le

Data field length of a fragment: lf

Total length of a fragment: Lf = lf + le = e l e

k

l

n  With these notations, useful throughput is written now (by reasoning on a fragment):



  















  

















k l n

e e

e L

f

e

l k

l n k

l n p

d kL

l n p

d sent bits total nb

usefull bits nb

) 1 ( )

1 ( )

1 (

= 9,81Mbps

5 Split up the frames in small fragments increasingly to improve the useful bit is not possible What are the different factors which limit the use of the fragmentation in the wifi networks 802.11 to deal with the errors problem?

For a given BER, the transmission rate decreases as frame size decreases Nevertheless, the flow is useful in parallel DECREASE beyond a certain threshold as the proportion of bits used by the control (overhead of headers) will increase relative to the flow useful

2 Theoretical maximum flow of 802.11b

Fladenmuller Baey - ”IEEE 802.11standards” 3/8

Hypothesis:

- inter-frame interval to distributed access (DIFS) 802.11b: 50 µs

- hard copy of a back-off between 0 and 31 intervals of time

- duration of a time interval: 20 µs

- physical overhead: the transmission of each frame must be preceded of a "long" preamble of 192 bits transmitted to 1 Mbit/s or of a "short" preamble of 72 transmitted bits to 1 Mbit/s followed by 48 bits transmitted to 2 Mbit/s

All the equipment in front of must implement the long preamble; this is that which we retain here

- inter-frame short interval (SIFS) of 802.11b: 10 µs

- data volume to transfer: 1500 bytes

1 It is supposed that the data flow is fixed of 1 Mbit/s Determine the efficacy of channel with and without the RTS/CTS mechanism It is estimated that no frame is lost

See table below Result: 1880 µs.

Figure 1: Format frame PHY layer of IEEE 802.11b (bits).

Figure 2: Format frame given sueh MAC layer of IEEE 802.11 (in bytes).

Table: Calculation of time sending and acquittal of 1500 bytes of data with 802.11b.

2 When the RTS/CTS mechanism is not implemented, it is estimated that only the date frames

can be lost When it is implemented, only RTS frames can be corrupted 20 µs are needed to

detect the frame absence (CTS or ACK), after which a DIFS delay is introduced Calculate

the probability p of frame loss from which the RTS/CTS mechanism is advantageous One will suppose p2 negligible

number of bits in length and µs Total and µs

overhead MAC 2 + 2 + 6 + 6 + 6 + 2 + 6 + 4 = 34 octets =

3 Environment heterogeneity

1 If two clients share the access point, one working in 11 Mbit/s and the other in 1 Mbit/s, what is the average flow of the access point, in considering that the supervision occupies the half time of the access point? The size of the frames and their quantity is the same for the two terminals

Sharing media is handed manner between the two terminals The overall flow of the access point is determined by the throughput of each client and the time during which they transmit He who emits a 1 Mbps generally require 11 times longer than that emits a 11 Mbit/s The issuance of a frame has 1 Mbit/s lasts 11 times longer than the issuance of a frame at 11 Mbit/s

Time frame for issuance of an x bits with 1 Mbit/s and with 11 Mbit/s (in s):

Rappel mathematical:

A discrete random variable is a random variable that takes a number countable assets The law of probability of a random variable X is given by { P (X = x), x value taken by X } Let X be a discrete random variable Called expectation of X and we denote by E (X) the quantity E (X) = xP(X x) The summation is over all values taken by X The expectation of the random variable is interpreted as its value average

Fixes:

 Time lost due to poor transmission of data:

TtotalData = TDIFS + TBO + TrData + TSIFS + 20 µs + TDIFS = 12904 µs

(it takes 20 µs to detect the absence of ACK)

 Time lost due to poor transmission of RTS:

TtotalRTS = TDIFS + TBO + TrRTS + TSIFS + 20 µs + TDIFS = 792 µs

The transmission time of a frame without error is given in the question precedente The average transmission time of one frame is written thus:

 Without RTS/CTS:

T(µs) = (1 – p) 13138 (no retransmission)

+ (1 – p) p (TtotalData+ 13138) (1 retransmission)

+ (1 – p) p2 (2TtotalData + 13138) (2 retransmission)

+

= 13138 + 12904 p by neglecting the terms p2

 With RTS/CTS:

T’(µs) = (1 – p) 13814 (no retransmission)

+ (1 – p) p (TtotalData+ 13814) (1 retransmission)

+ (1 – p) p2 (2TtotalData + 13814) (2 retransmission)

+

= 13814 + 792 p by neglecting the terms p2

T = T’ if p = 5,58%

Fladenmuller Baey - ”IEEE 802.11standards” 5/8

x

x

x

11

12

11



Average speed =

6

11 11

122x x  =1,83 Mbit/s

If the party oversight occupies 50% of the time the access point, the flow is more than 0,9 Mbit/s

In reality, if the party holds 50% of monitoring the bandwidth to 11 Mbps, it is proportionally much less than a 1 Mbit/s The real flow would be superior to 1 Mbps

2 Which solution would you recommend to maintain a high flow in the cell?

Two solutions:

- Disconnect customers who do not work at 11 Mbit / s This solution would oblige the operators of hotspots has increased the number of access points

- Limiting the access time of the slowest client Thus, if customers the slowest can take only half the time in emission,

3 If a Wifi card could automatically adapt its emission power to reach the access point, would that lengthen the lifespan of its batteries?

Yes, it would be more obligated to emit at full power all the time It is nevertheless the power is sufficient to avoid falling in the previous problem and be able to emit a maximum throughput There is no control of dynamic power provided for in the standard

4 Let a wifi network be working with the flow of 11 Mbit/s The access cards is the same that the access point can adapt their emission power

a) If one has decreased the transmission power of the access point, what would be the impact on the cell size?

Decrease in the size of the cell

b) Show that one would need much more access points to cover the same territory? Over the cell is smaller, it requires more access points

c) Does one thus increase the total capacity of the network?

Yes, the applications as a same frequency can be reused more often

d) Is the mobility reduced?

802.11 does not define how to implement the changes Intercel-lular, handovers in English The standard defines the functions of association and disassociation has an access point that can be used to establish the mechanisms of handovers, but it does not define the signaling between the access points, at least without the standard 802.11f The implementation of beneficial handover requires a system of signs that would be even more sophisticated than these handovers are many (many linked to the cell size) There are currently versions owners (CISCO) which set up a signal

We can therefore say that mobility is more limited with small cell sizes

5 Calculation of the maximum flow of 802.11g.

802.11g operates in the same frequency band that 802.11b and must be backwards compatible As the coding used by 802.11g is not recognized by the terminals 802.11b, the protection mechanisms were defined to limit the problems in the mixed environments 802.11b/802.11g These mechanisms consist mainly, of a terminal 802.11g operating in high flow, to persevere the radio medium in using slower reservation mechanisms, compatible with 802.11b

Hypothesis:

- SIFS: 10 µs

- short time interval (usable only when these is not terminal 802.11b): 9 µs

- long time interval (for the mixed environments): 20 µs

- DIFS: 2 time intervals + SIFS

- format of the physical frame 802.11g is given in Fig 1

- settlements are preceded and followed by 6 bits to which one adds 192 bits of stuffing all sent of 54 Mbit/s

Fig.1 – Frame format of PHY layer of IEEE 802.11g

Table: Calculation of time sending and acquittal of 1500 bytes of data with 802.11g.

a

v

number of bits in length and µs Total and µs

overhead MAC 2 + 2 + 6 + 6 + 6 + 2 + 6 + 4 = 34 octets =

ACK 14 octets + 6 bits before and after the

MAC frame + 92 bits of jam = 216 bits

and 54 Mbit/s

4

Fladenmuller Baey - ”IEEE 802.11standards” 7/8

Calculate the maximum flow when there are only terminals 802.11g (and thus no protection is necessary)

See Table 2 Result: 810,5 µs to transmit 12 000 bits data.

On or real throughput 12 000 bits /810,5 µs14.8 Mbps

6 What does the simultaneous use of equipment provided with 802.11b cards and g in the same network returns in to deal with a problem of hide terminal?

Normally, all the mobile radios that share a channel (including the access point) can agree However, this is not always the case There are cases or all mobile radios can hear and be heard by the access point but can not hear each other Under these conditions, the mechanism of listening before issuance does not work because the mobile radios are likely

to detect the channel as free and start to issue to the access point while it is already underway with reception terminal cover This is what we call the problem of hidden terminal

The coexistence of terminals using CCK and OFDM on the same channel (to 2.4 GHz) is very similar to the problem of hidden terminal since the terminal can not detect CCK and OFDM transmission However, with the implementation mechanism of RTS / CTS, OFDM terminals are able to operate without collision on the same channel as the wireless terminal

The mechanism RTS / CTS signaling induces more network but this overhead is relatively low The advantage is to allow migration to higher data rates for mobile radios operating in the 2.4 GHz band In the future, networks may no longer make use only of the OFDM in this band, which would abandon the use of the mechanism and RTS / CTS

The emergence of IEEE 802.11g is very beneque for the progress of wireless networks OFDM is the only authorized technology for high flow rates in the 2.4 GHz band Flow rates of up to 54 Mbps are now available in this band The compatibility with wireless equipment is also assured

7 What is the impact of the use of equipment mixed of 802.11b and 802.11g for the performances in the cell?

It is mandatory to use the mechanism RTS / CTS

8 One would like to develop a wifi network of future generation having better performances

as those of the current wifi network

a Show that a first disadvantage of current wifi network is not to have the possibility

to choose between the frequencies in the band of 2.4 GHz and the band of 5 GHz What do you deduce like improvement?

Should be able to choose the best signal based on interference in each band

b Show that a power control would allow increasing the global flow of a wifi network

If we adapt the power terminal and access point according to their distance, we limit the interference between cells The signal-to-noise ratio Ps / Pn being bigger, we limit the transmission errors (BER) and thus the probability of retransmission and fall of of actual flow The capacity increases

c A good part of the band-width is lost by the access point in cause of the supervision and starting temporization of terminals accesses towards the access point Can you propose improvements?

CSMA / CA is less effective when the flow increases because of the duration of timers It would therefore change the access technique