06/02/2006. Sharing the Medium. Peter Rounce Room Notes Courtesy of Graham Knight P.A.

Transcription

1 Sharing the Medium Peter Rounce Room 6.18 Notes Courtesy of Graham Knight P.A.Rounce

2 Sharing a link Introduction TDM, FDM - fixed allocations Statistical multiplexing - variable allocations Sharing a network Circuit switching Packet switching Queuing strategies in routers What about sharing the medium? 06/02/ We have looked at sharing links either by FDM or TDM. Both of these divide the capacity of a link into fixed capacity sub-channels each of which may be used by a separate activity. We have looked at statistical multiplexing where a link carries packets each of which belongs to a separate activity but which may be mixed on the link in an arbitrary order. We have looked at sharing a network. The networks we have considered have looked roughly like the telephone network with switches connected by link. We have seen that we can share the network by switching circuits or packets. Packet switching is the preferred choice as it is a form of statistical multiplexing which is favoured by the bursty nature of computer traffic, We now ask whether it is possible to eliminate the switches and share the medium directly. If we are to succeed we need a way of controlling access to the medium since it is dear that only one computer can be using it at any one time. Techniques for doing this are called "Medium Access Control" (MAC) algorithms P.A.Rounce

3 Bad ideas Approaches TDM - stations allocated time slots and take turns to transmit FDM - each station transmits on a different frequency (Rx need to listen on all frequencies) - Both waste resource when station not sending - hard to add stations need more slots or frequencies Good ideas? Contention we will examine this in detail Token passing Reserving timeslots Anything else? 06/02/ Using pure TDM and FDM would be a bad idea. These give fixed allocations to stations. Idle stations would waste their allocations. In fact there are other problems: For TDM we would nee done timeslot for each station. What happens when we add another station? For FDM we could allow each station to transmit on its own frequency. Receivers would now have to listen on all frequencies. Several ideas have been made to work: Contention. More-or-less like a conversation between a group of people. If one person is speaking, everyone else keeps quiet. We must decided what to do if two people start talking simultaneously. Token passing. There is one token which is passed around. Only the station with the token is allowed to transmit. Reserving timeslots. A bit like TDM but timeslots are not allocated to stations in rotation. Instead, there is a reservation protocol allowing stations to book timeslots as needed. There are several others including one interesting idea based on encryption P.A.Rounce

4 Aloha - simple contention Packet radio - University of Hawaii s "Contention-based" system - transmitters uncoordinated => collisions Collision recovery - wait a random time and try again How well will this work? host host 06/02/ host The ALOHA protocol was devised at the University of Hawaii in the 1970s. The protocol was first used on a packet radio network which was built to connect the various campuses of the University which were distributed amongst several islands, The transmitters were uncoordinated so it was quite possible for two or more to transmit simultaneously causing collisions. The ALOHA protocol attempts nothing to prevent this. It merely says "if a collision occurs wait a random time and try again". The random factor is necessary in order to prevent collisions occurring systematically on the re-transmissions. It is easy for a station to detect a collision; it just listens to the channel and detects whether the signal it transmitted has been altered in any way. Note that on a satellite network, as opposed to a packet radio network, there will be a considerable (propagation) delay before collisions can be detected. Geostationary satellites are 35,784 Km above the equator. At the speed of light the roundtrip delay is about 240 ms - more if you are not on the equator. Protocols which allow these sort of conflicts between stations in order to solve the channel allocation problem are called "contention" systems. There are many variants P.A.Rounce

5 Aloha Utilisation What proportion of the time is the network doing useful work? Let s build a mathematical model:- 1. Suppose all packets take T sec to transmit e.g. T=0.01 sec 2. Define Normalised Load G = number of packets generated in T sec 3. Define Normalised Throughput S = number of packets delivered in T sec N.B. S < G 4. Assume r collisions, hence r retransmissions in T sec We have S = G r or r =G-S (definition of r) 1 st Result: P(no collision) = S G e.g. 20 packets/sec => G = 0.2 e.g. if 20% collide, S = 80% of G, and so with G=0.2, S = /02/ In order to compare the various contention-based strategies we need to evaluate their performance. Ideally, when traffic is light all transmissions should succeed with minimal delay. When traffic is heavy the full capacity should be available and shared fairly between competing stations. Most schemes fall a long way short of this. Performance can be evaluated by direct measurement of real systems (not simple to arrange), by simulation or, sometimes by mathematical modelling. We will use the latter and will build some simple models of ALOHA and its derivatives. Generally the models have to be simple or the maths gets too hard! It is normal to assume that all packets are the same length (not usually the case in reality). We then take the time to transmit one packet (T sec) as our unit of time. We: then define: "Normalised Load" G = number of packets generated in T sec "Normalised Throughput" S = number of packets delivered in T sec G is the combined rate at which the stations launch packets into the network, S is the rate at which they arc successfully delivered. Collisions and so on mean that S < G. On a shared medium network we cannot do better than deliver one packet every packet time so S 1. For Aloha, we assume that a portion of the packets launched wilt be lost to collisions. The probability that an individual packet avoids this fate will be S/G. Our analysis of ALOHA is based on finding another estimate for this probability based on the likelihood of two transmissions colliding given that they are occurring at a particular rate. In order to do this we must make an assumption about the statistical properties of the transmission process. In fact we will assume that the intervals between transmissions have negative exponential distribution. This is a very common assumption despite the fact there is little experimental evidence to suggest that the intervals are not really distributed in that way. Basically it makes the maths easy! P.A.Rounce

6 12 Negative Exponential Distribution PDF: λe λt where λ is the mean arrival rate 10 8 Negative Exponential, mean 10 arrivals per sec Time (ms) Probability that next packet arrives between 50ms and 100ms from now is 0.239: is area under graph between 50ms and 100ms Negative Exponential Distribution is memory-less : how long waited doesn t matter! 06/02/ The graph shows the probability density function. Areas under the curve indicate the probability of events occurring at particular intervals. The numbers under the graph indicate the areas of the regions 0-50, , and so on. Thus, with a mean rate of 10 arrivals per second we would expect the probability of an inter-arrival gap of between 50ms and 100ms to be The negative exponential distribution has some interesting properties. For example, if each individual station generates packets with a negative exponential distribution then the combined distribution will be negative exponential with a mean the sum of the individual means. Negative exponential processes also have a "memoryless property - the probability of having to wait t sec for the next event is not affected by how long we have already been waiting P.A.Rounce

7 Negative Exponential Distribution memory-less An example: suppose there have been no arrivals for 150 ms, i.e. have waited this long. What is the probability that there will be an arrival during the next 50 ms? We want a conditional probability: P(arrival before 200ms no arrival before 150ms) By Bayes theorem: P( A B) = P( A B) P( B) P(arrival before 200ms and no arrival before 150ms) = (from graph) P(no arrival before 150ms) = 1 - ( ) = Thus P(arrival before 200ms no arrival before 150ms) = = This is (give or take a rounding error) the same as P(arrival in first 50ms). 06/02/ Thus the time that has elapsed since the last transmission does not affect the probability that a transmission will occur in the next 50ms, the next 100ms, etc. This makes it easier to develop the model P.A.Rounce

8 Poisson Distribution If inter-arrival times are Negative Exponential the probability of n arrivals in time t is given by the Poisson Distribution 0.70 n ( λt) e Pn ( t) = n! λt is mean arrival rate Probability Poisson probabilities, mean arrival rate 10 per second Probability of 1 arrival in the next 50ms Probability of 2 arrival in the next 100ms ms 100 ms 500 ms Number of arrivals 06/02/ If the inter-arrival times are negative exponential, it can be shown that the number of arrivals occurring during a given interval will have a Poisson distribution. The probability of n arrivals in time t is given by: n λt ( λt) e Pn ( t) = n! λ here is the mean arrival rate from the associated negative exponential distribution P.A.Rounce

9 Aloha Collisions: another formula for the probability of a collision t 0 -T t 0 t 0 +T Time A packet will collide with any transmission beginning in the interval (t 0 -T, t 0 +T) a contention time of 2T What is probability, P 0 (2T), of no transmissions in an interval 2T? Assume Poisson. G packets in T sec => λ = G packets/sec T P ( t) = n n ( λt) e n! λt, so P (2T ) = 0 (2G) e 0! 2G = e 2G This is our second form of the probability of no collision! 06/02/ We now return to the analysis of Aloha. The diagram shows that a packet transmitted at t 0 will be vulnerable to another packet arriving in the interval (t 0 -T, t 0 +T), i.e. a contention time of 2T sec. We can use the Poisson distribution to work out the possibility of this not happening. We assumed G packets in T sec which means: λ = G T packets/sec And from Poisson we calculate the probability of no arrivals in time 2T, i.e. P 0 (t=2t): n λt ( λt) e Pn ( t) = n!, so (2G) e P0 (2T ) = 0! 0 2G Of course, the probability of a collision is just 1-e 2G = e 2G P.A.Rounce

10 Aloha Performance We have from from our 1 st Result: P(no collision) = S/G We have from our 2 nd result: P(no collision) = e -2G Equating these gives P(no collision) So S = Ge ds dg = e 2G 2G 2Ge 2G S 2 G = e G 06/02/ = = 0 1 2G = 0 G = Smax = e = Maximum utilisation is 18.4% when G=0.5: less for other values of G We now equate the two measures of the probability of a collision and differentiate to obtain the maximum normalised throughput which is not very good! If Aloha were perfect we would deliver one packet every packet time, i.e. S=1. Thus we are using just 18.4% of the capacity a utilisation of 18.4% P.A.Rounce

11 Aloha Throughput 0.2 Normalised throughput S Remember G contains r re-transmissions, r = G - S Aloha normalised throughput Normalised load G 06/02/ The graph shows what happens when the offered load increases. Remember that the offered load includes retransmissions as a result of collisions. For example, when G=1, i.e. the offered load exactly matches the network capacity, the throughput is about Thus 86% of our transmissions are actually retransmissions. Trying to go faster than 18.4% utilisation just makes matters worse. Eventually very little is happening except retransmissions colliding with each other and throughput approaches zero. Does this mean that Aloha is hopeless? Not really but you have to ensure that the network has enough raw capacity so that G can be kept below 0.5. Remember G is the number of packets generated in the time to send one packet. A G of 3 means that there are on average 3 packets generated to be sent in the time it takes to send only one of them: lucky ones get through without collision and corruption when by chance there are any others generated in the contention period P.A.Rounce

12 Improvements for increased throughput: Slotted Aloha Packet ready for TX Packet transmitted t 0 -T t 0 t 0 +T Time contention time Packet transmissions are constrained to begin on slot boundaries - Must have a reliable time source to identify boundaries Packets now vulnerable for a contention time T Utilisation is now 36.8% but there is an extra ½T delay on average 06/02/ If we constrain transmissions to start on timeslot boundaries then any transmission becoming available during the interval (t 0, t 0 +T) will be deferred and will not collide. This reduces the vulnerable period to T. The analysis can be done as before and results in a doubling of the maximum utilisation. The down-side is that packets must wait for a timeslot an average of T/2 sec. additional delay. There must be some mechanism for ensuring that all stations are synchronised to the same slot boundaries. In a satellite network this can be done by having the satellite transmit regular pulses indicating timeslot boundaries P.A.Rounce

13 Alternative approach to increase throughput: Carrier Sense Multiple Access (CSMA) Listen before speaking Avoids collisions? A t 0 t 0 +T D Time B No! bit 1 Contention time ~ max p t bit 1 bit propagation time (but it helps) bit F time 06/02/ If a station continuously monitors the network it can discover when another station is transmitting and can defer its own transmission until the network is silent. This avoids many collisions. The buzz phrase for this is "carrier sense". Carrier sense does not avoid all collisions. The problem is that propagation delay ensures that another station's transmission may already have begun (legitimately) but has not yet reached us (see slide). The longer the propagation delay, the more likely this is to happen. In the extreme case of a satellite network, the propagation delay is about 240ms. Listening to the network may tell us that the network was silent 240ms ago but it does not help us to know what is happening now. Carrier sense cannot be used in such networks. However, provided the propagation delay is low, CSMA gains a lot over Aloha as we will see. This is the situation for most LANs (both wired and wireless) where the maximum distance between stations is nearly always less than a few Km. For example, the propagation delay on a 2Km optical fibre is x10 8 = 6.7µs. At 10Mbps only 67 bits can be transmitted in this time - a very small portion of the size of the average packet. However, note that at 1Gbps, a speed that is available commercially, 6700 bits or 837 bytes can be transmitted in 6.7µs - a sizeable portion of a packet P.A.Rounce

14 CSMA come in 2 forms: Persistent and Non-persistent CSMA Persistent: stations transmit immediately the medium goes quiet : there will be many collisions. Non persistent if medium is busy wait a random time and retry Packet arrives Transmit here Persistent Packet arrives Random time Non-Persistent Transmit here 06/02/ As indicated in the previous slide, the naive CSMA strategy is for a station to wait until the medium is silent then begin transmission immediately. This is called "persistent CSMA". The problem with this strategy is that if two or more stations are waiting, they will almost certainly collide when they begin transmission. (Here we should note that the diagram above omits a potentially important detail; namely that it takes some time for the fact that the network is silent to propagate to all stations. Thus there must always be an interval between the end of one transmission and the beginning of the next. For the purpose of this discussion we assume a very small propagation delay so that this interval may be neglected). It should be clear that, when network load is heavy, persistent CSMA will generate very many collisions so that throughput will drop. Several solutions to this problem have been studied. One approach is "non-persistent CSMA". Here a station which finds the network busy "backs off' i.e. it waits a random time before listening again. The effectiveness of this strategy is examined in the next slide. Another approach is "p-persistent CSMA". Here a station continues to listen to a busy network waiting for it to become silent. When it does so, the station transmits with a probability p, defers (i.e. waits a random time) with a probability (1-p). If p is 1 we have persistent CSMA, If p is (say) 0.5, then two stations both waiting for silence will collide with probability The downside is that they will both defer with probability 0.25 leading to an unnecessary delay. Note. We are considering here the strategy to be adopted when a station finds that the network is busy and defers its transmission. Do not confuse this with the strategy to be adopted when, despite our efforts, a collision does occur. In particular, note that CSMA-CD (see later) handles collisions in a way which resembles that used by non-persistent CSMA to handle deferrals P.A.Rounce

15 Non-persistent CSMA throughput Successful deliveries occur at S per packet time (S<1) The arrows show instances of a station listening Short arrows show network free and packet transmission Long arrows show network busy that result in deferrals Arrows appear at the normalised load rate G Busy listens cause deferrals at a rate D. Clearly D = G-S time In unit time we deliver S packets on average Network is busy for a fraction S of the time: fixed length packets! A fraction S of all arrows, G, will be long arrows, so D = GS GS= G SS ( G+ 1) = GS = G G+ 1 06/02/ In this analysis we assume that the collision rate is negligible. This is reasonable provided we have a low propagation delay. (N.B. remember to distinguish between "network busy" and "collision". The former certainly occur, it is the latter we claim are negligible. The load is G per unit time with arrivals represented by the arrows in the top diagram. A portion of packets encounter a free network and are delivered. The normalised delivery rate is S. The remaining packets encounter a busy network and are deferred at a rate D. Clearly D=G-S. The fact that we deliver S packets in unit time means the network is busy for a fraction S of the total time. It follows that this fraction of the offered packets, G, will encounter a busy network. Thus D = GS. Combining these two results to eliminate D, we have: GS = G S S( G + 1) = G S = G G + 1 Actually, there is a possible flaw in this argument. If deferred transmissions were very likely to be retried after a very short interval we would get many more "busy" events than the proportion S would suggest since the retry would find the same transmission still in progress. However, it turns out that, if we assume that both the process generating packets and the process governing the random deferral times are both negative exponential, then their combined distribution will also be negative exponential. In this case, the proportionality argument is valid and the result stands P.A.Rounce

16 Persistent CSMA throughput (a) (b) (c) time If 0 or 1 packets arrive during a transmission (a or b) there will be no collision If >1 packet arrives during a transmission (c) there will be a collision P(More than one packet arrives in unit time) = 1 - e -G -Ge -G (assuming Poisson: 1- probability of 0 packets probability of 1 packet) In unit time we deliver S packets and suffer C collisions on average Network is busy for a fraction S+C of the time Fraction of transmissions which generate a collision is: C = 1 e S + C So, C( e + Ge Ge ) = S(1 e C = ( S + C)(1 e Ge C S + C 06/02/ ) Ge ) [1] [1a] In the persistent case we can ignore collisions occurring as a result of two stations just happening to transmit at about the same time. As in the non-persistent case, this is reasonable so long as the propagation delay is small. However, we cannot ignore the kinds of collisions illustrated in the slide. These occur when two or more stations want to transmit whilst the net is already busy. Both of these will wait until the transmission finishes and will then begin transmission, making a collision very likely - in fact, we will assume certainty. Notice that "transmissions" may be of two types: successful deliveries or collisions. Both of these last one packet time. Note also that not all arrivals result in transmissions (since two or more arrivals may result in one collision). We have a normalised delivery rate S and collision rate C. As in the non-persistent case, we argue that the net will be busy a fraction (S+C) of the total time. Therefore P(net is busy when a packet arrives) = (S+C) and P(A particular transmission is a collision) = C/(S+C) 1. What proportion of transmissions (of either type) are followed by a collision? This is just P(More than 1 packet arrives in unit time) which is 1 - e -G -Ge -G. It is also C/(S+C), so C = 1 e S + C So, C( e + Ge Ge ) = S(1 e C = ( S + C)(1 e Ge ) Ge ) [1] [1a] P.A.Rounce

17 Persistent CSMA Throughput [2] What fraction of arrivals is lost in collisions? This is P(net is busy & 1 or more further arrival occurs during the transmission) = (S+C)(1-e -G ) All arrivals not lost are delivered S = G - G(S+C)(1-e -G ) [2] Combining with [1a] gives: S G( e + Ge = G + e ) 06/02/ What proportion of arrivals are lost in collisions? This is P(net is busy & 1 or more further arrival occurs during the transmission). See case (c) in the picture in the previous slide. The analysis to find this probability is performed by taking a perspective that a packet has arrived for transmission. The probability that the network is busy at this time is S+C; the probability that another packet will arrive in the same time slot is (1-e -G ). Multiplying these 2 probabilities together gives the probability of a collision. This is (S+C)(1-e -G ). All packets not lost are delivered successfully. Therefore we have S = G - G(S+C)(1-e -G ) [2] Using [1] and [la] successively in [2] gives: GC(1 e ) GS(1 e ) S = G = 1 e Ge e + Ge G( e + Ge ) Which eventually gives S = G + e P.A.Rounce

18 Throughput Comparison Normalised throughput S Throughput comparison Aloha Non-persistent CSMA Persistent CSMA Normalised load G 06/02/ We can see that CSMA is always better than Aloha. When the normalised load is low, persistent CSMA wins. This is because non-persistent is wasting time by deferring packets for a random time when these could, most likely, be transmitted immediately. However, as the load increases, persistent CSMA performance collapses as the network becomes dominated by collisions. The non-persistent graph is actually misleading. At high loads the probability of collisions becomes significant thus invalidating one of the assumptions we made in our analysis. In reality the throughput would start to collapse as a result P.A.Rounce

19 CSMA with Collision detection (CSMA-CD) In CSMA, collisions last one packet time Why not stop once a collision is detected? How long does it take to detect a collision? A B t p p bit 1 bit 1 Answer: 2p bit F Example: 10Mbps Ethernet with 1500 byte packets: time t max = 1.2ms, but p = 10µs 06/02/ With both persistent and non-persistent CSMA a collision costs us one packet time. If we were to detect collisions as soon as we could and immediately stop transmission we would waste less time. Is this complication worthwhile? The diagram shows the worst case in which B starts to transmit just before the first bit of A's transmission arrives. This takes one propagation delay p. However, at this point A is unaware of the collision and will not become aware until the first bit of B's transmission reaches A. This takes another time p. Thus, in the worst case it takes 2p to detect a collision where p is the maximum propagation delay between two stations. So, if we add collision detection, the time wasted in a collision reduces from t to 2p. The original Ethernet (10Mbps) MAC protocol is CSMA with collision detection - CSMA- CD. The maximum size Ethernet packet is 1500 bytes and the transmission rate is 10Mbps. This means t max = 1.2ms. The maximum distance between two stations is 2Km which means that p is 10µs. So the saving certainly appears to be worthwhile P.A.Rounce

20 CSMA-CD Performance CSMA-CD is a persistent strategy with back-off on a collision When load is low, Negligible collisions Performs like persistent CSMA (Tx transmit as soon as channel notbusy) When load is high, many collisions Collisions handled by binary exponential back-off Effectively a deferral Performs like non-persistent CSMA Best of both worlds! But There is a cost (see next slide) CD is difficult in wireless networks 06/02/ CSMA-CD is an effective protocol and is used on the most widespread LAN - Ethernet Since it takes 2p (worst case) to detect a collision all transmissions must last at least 2p. CSMA-CD is a persistent strategy and, when load is low and there are few collisions, its performance is like that of persistent-csma. When the load increases, collision occur and are detected. Collisions are handled by binary exponential back-off. Stations involved defer transmission by a random time with mean t. If collisions recur the strategy is repeated but with means 2t, 4t, 8t etc. This means that CSMA-CD behaves much like non-persistent- CSMA at high loads. The down-side is the minimum frame-size requirement (see next slide) and the fact that collision detection is difficult in wireless systems if is difficult to construct equipment that can transmit and receive the same frequency simultaneously P.A.Rounce

21 CSMA-CD Protocol used on the Ethernet Must have t > 2p or collisions may be missed For a given p there is a minimum frame-size 10Mbps Ethernet, min. frame is 64bytes t p p A bit 1 bit 1 bit F B time 06/02/ The original commercial Ethernet ran at 10Mbps and had a maximum cable length of 2.5Km hence p = x10 8 sec = 12.5µs. Thus t > 25µs, if collisions are to be detected. At 10Mbps we can transmit 10 bits per µs so the minimum frame size should be 250 bits - about 32 bytes, In fact we have to allow a bit for the electronics to react and then add on a little bit more just to be safe. Hence the 10Mbps Ethernet standard specifies a minimum frame size of 64 bytes. This is unfortunate if we only want to send one byte but maybe not too bad. However the situation changes as the bit-rate increases. Fast Ethernet uses 100Mbps and reduced the maximum cable length to 200m to compensate. Gigabit Ethernet (1000Mbps) increases the minimum frame size to 512 bytes. However, gigabit Ethernet is often used as a duplex pointto-point link in which case there is no contention and the restriction can be dropped P.A.Rounce

22 Ethernet Originally on coax cable, now mainly on twisted pair (next slide) Faster versions often use optical fibre Binary exponential back-off Coax Twisted Pair Optical/ twisted pair Frequency, f 10Mbps 100Mbps 1000Mbps Span, s 2500m 200m 200m Min Frame 64 bytes 64 bytes 512 bytes Speed of light, s m/s m/s m/s Bit Tx time, p (=s/c) 12.6 s 1.01 s 1.01 s Pkt Tx size, (>2pf/8) 31.5 bytes 25.3 bytes 253 bytes 06/02/ Ethernet comes in a host of varieties using various different media at three bit rates. 10Mbps is widespread. The original topology was a bus using coax cable. "Spurs" could be added so one might have a main cable running round a building with spurs into each office. Now 10Mbps Ethernet is mainly on twisted pair using a hub - effectively the bus is put in a box. 100Mbps ("Fast Ethernet") is a fairly straightforward speed-up. The network span (and hence the propagation delay) has had to be reduced in order to keep t > 2p. As increase frequency, the speed of light stays the same, so p stays the same unless span of network is decreased, but number of bits transmitted in time p increases as frequency increases. Fortunately the trend is towards small LANs connected by devices called Ethernet switches so the 200m span is not too much of a problem. Each mini LAN has only a few hosts which reduces contention and hence collisions. Note that a switch is not the same as a hub: hosts connected to hubs by twisted pair which allow high data rates. 1000Mbps ("Gigabit Ethernet") is now available. Making transmission work economically and reliably at such frequencies is a challenge. Most versions use optical fibre though a standard for copper wire does exist. Reducing the span still further was not an option. Therefore the minimum frame size has been increased to 512 bytes (frame is just the LAN word for a packet). This could be very wasteful if many small frames are to be transmitted as these must be padded. A variety of bit encodings are used. The original version used Manchester encoding but some variants use NRZ P.A.Rounce

23 Ethernet Originally on coax cable, now mainly on twisted pair (next slide) Faster versions often use optical fibre Original 10Mbps single coax cable Modern 10/100/1000Mbs Bus Hub Bus with spurs 06/02/ Ethernet comes in a host of varieties using various different media at three bit rates. 10Mbps is widespread. The original topology was a bus using coax cable. "Spurs" could be added so one might have a main cable running round a building with spurs into each office. Now 10Mbps Ethernet is mainly on twisted pair using a hub - effectively the bus is put in a box. 100Mbps ("Fast Ethernet") is a fairly straightforward speed-up. Each host to hub cable has 2 twisted pairs: one pair taking signals from host to hub, and one pair taking signals from hub to host: signals arriving at hub are transmitted out on all outputs, so hosts receive a merged signal so still a contention network P.A.Rounce

24 Twisted Pair Cables - two wires twisted around each other Tx V + V - V + + e + V - + e - Rx 2 wires driven to opposite voltages at Tx, e.g. +0.6V, -0.6V Direction of voltage difference carries binary value. Thus, Tx sets wires to V + and V - Rx measures voltage difference received. Rx receives V + + e + and V - + e - : e +, e - are noise signals Voltage difference at Rx = (V + - V - ) + (e + - e - ) = applied voltage For twisted pair, e + == e - to first order so they cancel on subtraction 06/02/ In a twisted pair cable, there are 2 wires twisted around each other, so that the 2 wires are near enough the same length, and follow almost exactly the same path from Rx to TX, suffering the same noise effects. Signalling is done by placing a differential signal across the 2 wires with voltages of the same magnitude but different sign. Thus if the wires are labelled A and B, then the following binary values might be sent:- Logic 0 A = + 0.6V B= -0.6V voltage diff = V Logic 1 A = - 0.6V B= +0.6V voltage diff = V There are several benefits to this approach of which 2 are:- 1) Noise elimination: the noise voltage picked up by each wire is the same to the first order, so that the voltages seen by the Rx are (V A + V noise ) and (V B + V noise ). The Rx measures the voltage difference, which elimiates the noise so that the differential voltage (V A V B ) is measured, I.e. the original Tx value. 2) Reduction of electro-magnetic radiation there are legal limits on the EM radiation from electronic equipment. Because the voltages applied to the wires have the same magnitude but opposite sign, the currents produced in the 2 wires are in opposite directions, these produce opposite magnetic fields. The changes in the currents (and their magnetic field) that produce EM radiation are in opposite directions in the 2 wires so that they cancel each other too a large extent reducing the EM radiation. 3) Twisted pair cables are cheaper, thinner and more flexible than co-axial cables: so preferred technology P.A.Rounce

25 Token passing [1] Token: is passed around a ring token holder transmits token is then released Topology: physical ring logical ring Bits stored on ring: 1Km ring holds 50 bits some bits at receiver, some at transmitter, some on ring token ring (physical ring) token bus (logical ring) 06/02/ The idea of token passing is very simple. A small, special bit pattern called the token is passed around all stations connected to the media and only a station holding the token can transmit. After transmission it releases the token, passing it to the next station in the ring. All messages transmitted on a ring circulate through all stations in a known order and can potentially return to their source. Note that the number of bits on a ring at any one time can be quite small. For example, at 10Mb/s one bit is transmitted every 0.1µs. If the propagation speed is 2x10 8 m/s (= 200m/µs) then each bit occupies 20m. If a ring has a circumference of 1Km it will only have 50 bits "stored" on the cable. In addition to these bits there will probably be at least one bit stored at each station as these act as repeaters (regenerating the input signal on the output). Thus, one can expect the total number of bits on the ring to be of the order of so an average size data frame (say 512 bytes) does not fit on the ring. During the transmission of such a frame, part of it will have been received, part will be on the medium and part will still be waiting at the transmitter. The token may pass along a physical ring topology, or a logical ring topology. On an idle ring, the token circulates continuously and each of the participating stations merely relays the bit-stream (and hence the token) around the ring. Once it has the token, a station can use the full bandwidth of the medium for its transmission. On the IBM token ring, the transmitted bits travel around the ring and back to the transmitter - which then removes them. The receiving station may set some bits at the end of a data frame to indicate correct reception. Once transmission is complete the transmitting station re-generates the token followed by "idle fill' data until the token (or a data frame) returns. When the ring is idle, with the token circulating, each station acts as a simple bit-repeater. For this to work the token must be small enough to fit on the ring in its entirety. The token contains some bits which specify a priority, and stations may then grab a high-priority token only if they themselves are configured as high priority stations P.A.Rounce

26 Token passing [2] Maximum latency when each station has a maximum size frame to transmit T = N ( T + T + T ) + T MAX S F T R R T length of a bit 8 V 2 10 = = R L B T round trip delay on ring F R = R N = R S F R 6 R=16Mbps = 12.5m = = s maximum frame transmission time = = s number of bits on ring L R =1000m LR 1000 NR = + NS = + 20 = 100 LB 12.5 N s =no of stations =20 token transmission time ST 24 6 TT = = = s R maximum latency TMAX = NS ( TF + TT + TR ) + TR = 20 ( ) = s TMAX 6ms 06/02/ Some mechanism must be introduced to ensure that the station holding the token does not hog the network. There is a maximum transmission length and a requirement that a station always releases the token after transmission. At start-up, or if the token is corrupted and lost, no station can transmit, a protocol is required to generate a new token. The possible solution of having a specialised "monitor" station to perform this task is not favoured, as it would introduce a single point of failure. All stations monitor the ring to check that a token is circulating. If a station finds no token it will generate a new token, if two stations do so simultaneously they will back-off and re-transmit the token in a fashion similar to that used for CSMA/CD. One important characteristic of token passing MAC algorithms is that it is possible to determine the maximum delay before a transmission succeeds (the maximum latency). This deterministic property of the token ring distinguishes it from the CSMA/CD LAN and is important in some time-critical applications such as process control. What is the longest time a station must wait before grabbing a free token? In the worst case, all stations have maximum length frames ready to transmit. Suppose a station has the token, it transmits a frame, followed by a free token. As soon as the last bit of the free token reaches the next station this next station can begin its own transmission. Typical token ring values are shown below. N S : number of stations (20) T T : token transmission time L R : length of ring (1000m) T F : maximum frame transmission time L B : length of bit T R : round trip delay on ring S T : token size (24 bits) S F : frame size (4096 bits) R: data rate (16Mb/s) N R : number of bits on ring V: propagation speed (2x 10Sm/s) T MAX : maximum latency/delay Note that N is chosen here as 20, but token ring/bus can handle 100s of stations per ring/bus P.A.Rounce

27 Token Passing - Performance If only one station is trying to transmit then the time for each transmission is T MIN =T F + T T + 2xT R E.g. T MIN = 2.56x x x6.25x10-6 = 2.70x10-4 sec Utilisation is 2.56x x10-4 = 95% but diminishes when ring length increases If all stations are trying to transmit the system behaves like TDM At light loads, CSMA-CD tends to be more efficient At heavy loads, token passing wins 06/02/ At light loads the time for the token to circulate back to a transmitting station becomes significant; it imposes an unnecessary gap between transmissions. At heavy loads, when every station has something to transmit, token passing allocates the network to each station in turn and so behaves like TDM. Arguably, this is fair. It also allows the utilisation to remain high and there is no throughput collapse as seen with CSMA techniques when the network becomes dominated by collisions P.A.Rounce

28 Code Division Multiple Access Division of capacity through use of codes Complex, difficult to engineer Good performance efficient use of bandwidth Used in wireless networks Good noise immunity Bits divided into chips 64 or 128 chips per bit Each station has a unique (carefully allocated) chip sequence 06/02/ In the examples here we will assume there are 8 bits per chip. It turns out that the demonstration is simpler if we represent a 1 chip as + 1 and a zero chip as 1. Note that we are now sending 8 signals per bit so we need 8 times the bandwidth we would need with a more conventional encoding. However, things turn out not to be as bad as they seem. Suppose we have a 2Mbps channel and divide this between 20 stations using TDM. Each station will get 100Kbps. Using CDMA the channel delivers 2 Megachips per sec. However, the coding scheme allows each station to transmit at that rate. Assuming 8 chips per bit this gives 250Kbps to each station P.A.Rounce

29 CDMA Example Assume 8 chips per bit, 4 stations A-D A ( ) B ( ) C ( ) D ( ) Chip sets are orthogonal e.g. A.C = ( )/8 = ( )/8 = 0 To transmit a 1 send the chip set. To transmit a 0 send the complement. 06/02/ Chip sets are carefully chosen to be orthogonal. I.e. the normalised inner product of any pair is zero. This means that when we compare two chip sets we find as many pairs the same as we find different. To send a 1 a station sends its chip set; so A sends ( ) To send a 0 a station sends the complement; so A sends ( ) Note that, if A B = 0 then A B = 0. Also A A = P.A.Rounce

30 CDMA Example [2] Suppose A sends 1, C 0 and D 1 Combined signal will be: A 1: C 0: D 1: S : To retrieve D s transmission, compute S.D Note that S : D : ( )/8 = 1 D ( A + C + D) = D A + D C + D D = = 1 06/02/ We make two simplifying assumptions; the stations are in bit synchrony, i.e.bits are sent at exactly the same time, and the signals are of equal power so they add in a simple arithmetic way. The slide shows three bits being added. We can then extract whichever station we want simply by forming the inner product with its chip sequence. In practice neither of our assumptions would be true. We would also have to contend with noise. However, it turns out that, provided the chip sequences are long enough, it is possible for the receiver to synchronise with the bits from an individual station. We can also cope with unequal power provided this is constrained to some limit P.A.Rounce