MOB Subject 7 –IEEE 802.11 standards Wifi performances

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MOB Subject 7 –IEEE 802.11 standardsWifi performances Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 1/8MOB Subject 7 –IEEE 802.11 standardsWifi performances1. Impact of the transmission qualityOne is interested here in the impact of radio transmission errors on the performances of awireless network and in the relation between transmission and fragmentation quality. One takeshere the standard IEEE 802.11b as an example.1. Let p be the binary error probability on a communication channel used by a local networkand 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 the802.11 standard (2346 bytes) + PHY header of 24 bytes).A frame is error just if any bit has undergone a transmission error. The correspondingprobability can be written: (1 p)8nThe probability of error per frame can be written thus:q = 1 – (1 – p)8n = 0,0187813992. For an error probability q per frame, determine the average number N of transmissions hasto effect to succeed the correct transmission of a frame? For the digital application, one willtake p = 10-6.Reminder: 1 + 2q + 3q2 + … + iqi-1 + … =)1(12qThe probability that a frame need to be exactly i transmissions correctly transmitted can bewritten: p(i) = qi – 1(1 – q)The average number of transmissions necessary to succeed a correct transmission is writtenso:])1(1[1111)1(1)1()1()(8211niipqqqiqqqiqiipN  0191,1)1(18np3. One wants to make a first coarse evaluation of the bit rate lost because of theretransmissions on error. One thus does not take any other factor in account but the errors oftransmission. One supposes in first approximation that all bits of the physics frame are sendwith the same flow.a) For a error probability per frame q and a nominal capacity d of transmission, what is theuseful bit (goodput) and what is the lost flow at the time of the retransmissions for achannel which is used permanently (there is always a frame to transmit i.e. is thechannel is saturated)?A. D. : d = 11 Mbit/s and p = 10-6.Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 2/8The useful throughput is written:MbpspdpdNdGnn8,10)1()1(188The lost throughput during broadcasts can be written:)11(Nd or 0.2 Mbpsb) The same question that previously if one takes the header MAC + PHY of length leequalto 58 bytes into account (24 bytes of header PHY + 34 bytes of header MAC).Length of the field data of a frame: ldLength of the MAC + PHY header of a frame: leTotal length of a frame: n = ld + leWith these notations, useful throughput is written now (by reasoning on a frame):MbpspdpdNdGnn53,10)1()1(188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 (thatone will suppose all of identical size) before to send it, and that like previously one does nottake any another factor into account but the transmission errors. What becomes the usefulbit?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: ldLength of the MAC + PHY header of a frame: leTotal length of a frame: n = ld + leData field length of a fragment: lfTotal length of a fragment: Lf = lf + le =eelklnWith these notations, useful throughput is written now (by reasoning on a fragment):eefflklneeeLfeLpdlklnklnpdkLlnpdsentbitstotalnbusefullbitsnbG888)1()1()1(= 9,81Mbps5. 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 networks802.11 to deal with the errors problem?For a given BER, the transmission rate decreases as frame size decreases. Nevertheless, theflow is useful in parallel DECREASE beyond a certain threshold as the proportion of bitsused by the control (overhead of headers) will increase relative to the flow useful .2. Theoretical maximum flow of 802.11bMaster of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 3/8Hypothesis:- 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 transmittedbits 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 weretain here.- inter-frame short interval (SIFS) of 802.11b: 10 µs- data volume to transfer: 1500 bytes1. It is supposed that the data flow is fixed of 1 Mbit/s. Determine the efficacy of channel withand 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 framescan be lost. When it is implemented, only RTS frames can be corrupted. 20 µs are needed todetect the frame absence (CTS or ACK), after which a DIFS delay is introduced. Calculatethe probability p of frame loss from which the RTS/CTS mechanism is advantageous. Onewill suppose p2 negligible.number of bits in length and µs Total and µsDIFS 50back-offAverage = 15,5  20 µs310overhead PHY 192 bits and 1 Mbit/s 192overhead MAC 2 + 2 + 6 + 6 + 6 + 2 + 6 + 4 = 34 octets =272 bits and 11 Mbit/s25data 1500 octets = 12000 bits 1091SIFS 10overhead PHY 192ACK 14 octets = 112 10Total 1880Master of Computer Science 1 - MOB Mobile Internet and Surrounding43. Environment heterogeneity1. 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 occupiesthe half time of the access point? The size of the frames and their quantity is the same forthe two terminals.Sharing media is handed manner between the two terminals. The overall flow of the accesspoint is determined by the throughput of each client and the time during which theytransmit. He who emits a 1 Mbps generally require 11 times longer than that emits a 11Mbit/s. The issuance of a frame has 1 Mbit/s lasts 11 times longer than the issuance of aframe 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. Thelaw 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) thequantity E (X) = x)(XxP. The summation is over all values taken by X. Theexpectation 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 µsThe transmission time of a frame without error is given in the question precedente. The averagetransmission 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%.Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 5/8xxx111211Average speed =61111122xx=1,83 Mbit/sIf the party oversight occupies 50% of the time the access point, the flow is more than 0,9Mbit/s.In reality, if the party holds 50% of monitoring the bandwidth to 11 Mbps, it isproportionally 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 theoperators of hotspots has increased the number of access points.- Limiting the access time of the slowest client. Thus, if customers the slowest can takeonly 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 thepower is sufficient to avoid falling in the previous problem and be able to emit a maximumthroughput. 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 thatthe access point can adapt their emission power.a) If one has decreased the transmission power of the access point, what would be theimpact 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 inEnglish. The standard defines the functions of association and disassociation has anaccess point that can be used to establish the mechanisms of handovers, but it doesnot define the signaling between the access points, at least without the standard802.11f . The implementation of beneficial handover requires a system of signs thatwould be even more sophisticated than these handovers are many (many linked tothe 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.Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 6/85. Calculation of the maximum flow of 802.11g.802.11g operates in the same frequency band that 802.11b and must be backwardscompatible. As the coding used by 802.11g is not recognized by the terminals 802.11b, theprotection mechanisms were defined to limit the problems in the mixed environments802.11b/802.11g. These mechanisms consist mainly, of a terminal 802.11g operating inhigh 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 ofstuffing 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.avnumber of bits in length and µs Total and µsDIFS2  9 + 1028back-offAverage = 15,5  9 µs139,5overhead PHY 192 bits and 1 Mbit/s 192overhead MAC 2 + 2 + 6 + 6 + 6 + 2 + 6 + 4 = 34 octets =272 bits and 54 Mbit/s5data 1500 octets = 12 000 bits 222Preamble 12signal extension 6SIFS 10overhead PHY 192ACK 14 octets + 6 bits before and after theMAC frame + 92 bits of jam = 216 bitsand 54 Mbit/s4Total 810,5Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 7/8Calculate the maximum flow when there are only terminals 802.11g (and thus noprotection 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 Mbps6. What does the simultaneous use of equipment provided with 802.11b cards and g in thesame 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 beheard by the access point but can not hear each other. Under these conditions, themechanism of listening before issuance does not work because the mobile radios are likelyto detect the channel as free and start to issue to the access point while it is alreadyunderway with reception terminal cover. This is what we call the problem of hiddenterminal.The coexistence of terminals using CCK and OFDM on the same channel (to 2.4 GHz) isvery similar to the problem of hidden terminal since the terminal can not detect CCK andOFDM transmission. However, with the implementation mechanism of RTS / CTS,OFDM terminals are able to operate without collision on the same channel as the wirelessterminal.The mechanism RTS / CTS signaling induces more network but this overhead is relativelylow. The advantage is to allow migration to higher data rates for mobile radios operating inthe 2.4 GHz band. In the future, networks may no longer make use only of the OFDM inthis 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. Flowrates of up to 54 Mbps are now available in this band. The compatibility with wirelessequipment is also assured.7. What is the impact of the use of equipment mixed of 802.11b and 802.11g for theperformances 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 performancesas those of the current wifi network.a. Show that a first disadvantage of current wifi network is not to have the possibilityto 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 networkIf we adapt the power terminal and access point according to their distance, we limitthe interference between cells. The signal-to-noise ratio Ps / Pn being bigger, welimit the transmission errors (BER) and thus the probability of retransmission andfall of of actual flow. The capacity increases.Master of Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802.11standards” 8/8c. A good part of the band-width is lost by the access point in cause of the supervisionand starting temporization of terminals accesses towards the access point. Can youpropose improvements?CSMA / CA is less effective when the flow increases because of the duration oftimers. It would therefore change the access technique. . Internet and SurroundingFladenmuller Baey - ”IEEE 802. 1 1standards 5/8xxx 1112 11Average speed = 6111 1122xx=1,83 Mbit/sIf the party oversight occupies. 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 802. 1 1standards 1/ 8MOB Subject 7 –IEEE
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