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IMPROVING RANDOM MULTIPLE ACCESS METHOD SCHEME FOR WIRE AND WIRELESS LOCAL NETWORKS
V. Zagursky, Dz. Zibin,
Key Words: Computer Networks, Rrandom multiple access method, Bernoulli scheme.
1. Introduction
Combinatorial and probability-theory methods are employed to analyze the random multiple access methods which have been proposed as the basis for a next generation of networks [3] . A cornerstone of the new method is a new algorithm for resolving conflicts (collisions in transmission), in which the parties involved in the collision "toss a coin." The technique for determining the length of each particular process is described; the average lengths of the processes are established for specified number of parties in conflict, as is the time loss involved in Implementing the given access method. Upper and lower bounds for the length of the processes are obtained.
Paper [1] proposed a random multiple access method (the ENET protocol) for LANs with a bus topology. The author of that paper assumed that the method can provide the basis for the development of a second generation of Ethernet networks. Below we offer an analysis and evaluation of this method. The proposed access method in question is a CSMA/CD method, but it differs 'substantially from the standard-access method proposed in ISO Recommendation 8802.3 [2]. It does not employ an algorithm for postponing repeat attempts at transmission by random time intervals. Instead, the parties involved in a collision "toss a coin," with a "transmit / no transmit" outcome. In addition, the active stations have four degrees of activity X = 0, 1, 2, 3, where X, the active station, monitors the base band (BB) and, if there is no carrier on it over X time segments, then it immediately initiates transmission. The duration of the time segment ts( 2(, where ( is the signal delay (transit time) between the two most distant stations in the network.
As is customary, the stations determine the state of the BB (free, successful transmission, collision). The access method can be clarified by a diagram (Fig. 1) which shows the states and transitions between them, and indicates the reasons for conditions for each transition. Transitions are instantaneous, while the functions that are executed in each state require a certain amount of time. The state diagram represents a partition of all the functions of the access method under consideration into subsets, which is maximally convenient for presenting and understanding them. Let us consider this diagram in greater detail.
Fig. 1. State and transition diagram for LAN with proposed protocol.
INIT is the initial state, in which the station waits for a transmitted frame to appear, or generates the frame itself. It also monitors the state of the BB, and, each time that the carrier disappears, it generates a pause of length 3tS, necessary to determine the instants at which the BB is freed up. We should recall that a busy BB becomes free not at the instant that the carrier disappears, but when the corresponding pause has elapsed. If carrier again appears during the time that the pause is being generated, this generation ceases and the BB is assumed to be continuously busy. As soon as a ready-to-transmit frame appears, the station goes into state A3 (transition 1). In state AX (X = 0, 1, 2, 3) the station is active and monitors the state of the BB. When the carrier disappears, it generates a pause of length XtS (for X = 0 it generates the minimum interval between frames, equal to 9.6 (sec), and, when this pause has elapsed, it goes into state TR, i.e., it begins to transmit (transitions 2, 5, 9, 10). Transition 1 does not interrupt generation of a 3tS pause, and therefore it can begin, e.g., in the initial state and terminate in A3. But the appearance of a new carrier, as in the initial state, terminates the generation of any pause. In the case of successful transmission, all stations go from Al to A0 (transition 7), while in the case of a collision they go to A2 (transition 8). State TR/B means that the station transmits its frame to the BB and monitors the transmission. If transmission is normal, then the station returns to the initial state after transmission is completed (transition 3). If a collision is detected, the station transmits another four bytes (i.e., it forces a collision) and then ceases to transmit. In addition, during the collision (forcing) the station "tosses the coin," and, if the coin comes up "transmit," then the station goes into state A0 (transition 4), while if it comes up "no transmit," the station goes to A1 (transition 6). Thus, when active stations are present, the pauses between transmissions do not exceed 3tS, while in the process of conflict resolution they do not exceed 2tS. New active stations are in A3 until the conflict-resolution process terminates, and then they begin their transmissions. No station is dropped from the list of active stations; sooner or later, they all gain access to the BB.
2. Preliminary comments.
Our proposed access method is simpler than the standard one, since pause generation and coin-tossing require less hardware than the generation of random delay times, whose range increases exponentially as the ordinal number of the repeat attempt at transmission of the frame increases. In addition, the standard access method also requires hardware for implementing p-persistent access.
It was pointed out in [1] that, even when the BB is free for a prolonged period, all new active stations gain access to it only after a pause of duration 3tS. This is not a feature of the given access method, but is rather an omission by the author. Of course, this delay is in no way justified and can readily be eliminated. Such a station can initiate transmission immediately. For this, upon elapsing of the interval, the 3tS timer should issue an OK-to transmit ("BB free" signal) and maintain it until a new carrier appears. In other words, we need to refine the condition for transition 2 in the state diagram, as we have done. Yet another feature is mentioned only in passing in [l]. The description of the activity levels of the stations assumes that there is an absolute time scale that is the same for all stations, but this is impossible. Because of the fact that ((0.5tS, any pause in the BB at points of connection of different stations is of differing length , or, more specifically. Instead of a pause X2( (X = 0, 1, 2, 3) the stations "see" a pause of length X2(-(X + 1)2( in the BB, and this can lead to errors in determining the instants at which the BB becomes free. To eliminate this, the length of the time segment should contain a margin of time tZ, i.e., tS = 2( + tZ, that takes account of the accuracy of the timers and the requisite distance, and that permits adjacent values of the actual pauses in the BB to be reliably discriminated and permit method implement also for improving access mechanism at wireless networks [3].
3. Space of outcomes.
Paper [l] lacks an analysis of the access method or recommendations regarding the asymmetry of the coin. We will attempt to eliminate this shortcoming of the method .
As in all LANs with random multiple access, the first collision can occur either because the instants of appearance of active stations randomly coincide, or because of accumulation during transmission of-the next successive frame . Let us assume that m stations are involved in this collision (m = 2, 3, ..., N, where N is the number of subscribers in the network). Then these stations, independently of one another, toss a coin, and the results form the outcome of this stochastic process, which can be conveniently written as an m-bit binary number X0X1 - - - Xm-1. where Xe = 1 means that the coin of the e-th station has turned up "transmit," while Xe = 0 means that it has turned up "no transmit," e = 0, 1, ..., m - 1. The set of all m-bit binary numbers forms the space of outcomes M(m) of the random process comprising coin-tossing by all m parties involved in the collision. This space can be conveniently partitioned into m + 1 nonintersecting equivalence classes on the basis of the number of 1's in the outcomes, i.e., 00...0, (10...0), (110...0), - - - , (1...10), 11...1, where the parentheses denote all m-bit binary numbers with the indicated number of 1's. The number of outcomes in the 1-th class is Cmi, and their sum is Cm0+Cm1+...+Cmm-1+Cmm = 2m = |M(m)|. All outcomes of the same class lead to the same results. For example, in the case of the zero class, all m stations are brought to state Al and remain there a time tS. Then they all initiate repeat transmission, leading to repeat collision and a new coin-tossing, and so forth. Thus, the zero class lengthens the process by one pause and one collision, and yields no benefit.
In the case of the first class, one station goes to A0, while all the remaining ones go to A1. After a short pause, the first station initiates and successfully completes transmission, then all the remaining m - 1 station go from A1 to A0 and initiate transmission, thus leading to a new collision. In this case the useful result is one successful transmission and the absence of long pauses in the BB. As will be shown below, outcomes from the second through the (m-1)-th class also yield useful results, although somewhat more slowly. In the case of the m-th class, all m stations go to state A0 and, after the BB has become, free, the initiate transmission, leading to repeat collision of the same m transmissions. The process is prolonged by one collision without any positive result.
Assume that the probability of appearance of a 1 and 0 upon tossing the coin is p (0< p <1) and (1-p) respectively. The outcome at one station is entirely independent of the outcomes at the others, and therefore the probability of appearance of a next successive outcome with i 1's (i = 0, 1, ..., m) is equal to pi(l p)(m-i). All outcomes of the same class are equiprobable, and hence the probability that the next successive outcome will appear in the i-th class of cardinality Cmi is equal to P(m,i)=Cmipi(1-p)(m-i). Any outcome appears in one and only one class, and therefore EMBED Equation.3 . Thus, we have a classical Bernoulli scheme [4]. From dP(m,i)/dp=0 we can obtain the condition p=i/m such that the probability that the outcomes will appear in the i-th class is at a maximum and equal to
EMBED Equation.3
In the case of fixed finite i, this probability tends to the following limit as m increases:
EMBED Equation.3
In what follows, in cases in which m is fixed, we will write Pi instead of P(m,i) for simplicity. We should point out that P(m,i) for p = p0 and P(m, m-1) for p=1-p0 are equal, and hence the pair of probabilities P0 and (1 p0) yields mirror-symmetrical probability distributions over the classes, thus enabling us to simplify the calculations, since it is sufficient to consider p values in the range 0 < p ( 0.5.
4. Flowchart of process.
The process of conflict resolution can be depicted as a flowchart or, more precisely, a locally finite connected oriented graph (Fig. 2) of tree type with a root. The graph vertices denote the following events: (C(m), after a time segments a collision of m transmissions occurs; bT, after b time segments a successful transmission occurs, where a, b ( {0, 1, 2}. The arc denotes transitions from one event (or vertex) to another. Each path (route from root to leaf) corresponds to one process, while all possible paths correspond to the set of all possible processes that pass through the first (root) vertex). Arcs that depart from the same vertex correspond to classes of outcomes, and therefore it is convenient to number them in the same order, while the probability Pi of appearing in the i-th class is the probability that a process will proceed further from vertex C(m) along the i-th arc.
We denote the length of the path (or of the path segment under consideration) by R, and we define it as the number of vertices that are encountered on this path (or segment). Then the list of all possible paths of length (R can be compiled either in the form of a string of events (sequence of graph vertices) or in numerical form (sequence of arc numbers). The probabilities of all the arcs making up a path are already known, and therefore we can use the multiplication rule to determine the probability PK that a process will follow the k-th path. For example, for m = 2 according to Fig. 2 we have for R ( 3. As we know, both forms of path notation uniquely designate a single process. Inspection of all paths of length (R means enumeration, with constraints, of all R-
Fig. 2. Flowchart of conflict-resolution process
00=C(2)-1C(2)-1C(2)...
01=C(2)-1C(2)-0T-0T.
02=C(2)-1C(2)-0C(2)....
1=C(2)-0T-0T.
20=C(2)-0C(2)-1C(2)...
21=C(2)-0C(2)-0T-0T.
22=C(2)-0C(2)-0C(2)...P00=P02
P01=P0P1
P02=P0P2
P1=P1
P20=P2P0
P21=P2P1
P22=P22
position numbers in the (m+1)-th number system; in the general case, different numbers have different bases. All possible paths of specified length, belonging to the initial vertex C(m), form a complete system of paths, and therefore EMBED Equation.3 . If all outcomes turn up either in the zero or m-th classes, then the length of this process R = (. Therefore the flowchart is also infinite, and is drawn only in part. At the same time, the most probable processes are short. The longer the process, the lower its probability, and the probability of an infinitely long process is zero.
It is more convenient to employ finite flowcharts in which the zero and m-th arcs are transformed into loops (Fig. 3), the zero loop yielding a delay tS. Flowcharts with loops are much simpler, since they do not explicitly show repeat collisions (number of steps around loops). In them, however, it is necessary to determine the average delay Dm of the process at the vertices C(m). For this we compile a list of all half-segments of length r = 1, 2, 3, ..., that close at vertex C(m), and we determine the probability p(m, x) of each of them. A segment of such a path begins by arriving at vertex C(m) and ends by departing from it along the j-th arc, where j = 1, 2, ..., m - 1. We also determine the instantaneous sum of the probabilities of all the paths already considered, and we terminate the length increase for r = rmax, when the instantaneous sum EMBED Equation.3 , with sufficient accuracy, becomes equal to 1. The length of each such segment of the process l(m,x)=z+cr, where z is the number of zero loops in the given segment, i.e., the number of delays of magnitude tS in A1-TR transitions; r is the number of repeat collisions C(m) on the given segment; and c is the average collision length in time segments. As we know, the collision length lies in the range (0-tS, and therefore, with a view to the worst case, we will assume that c = 1 in what follows.
Now we can determine the average length of the process segment, in other words, the average delay of the process at vertex C(m) in time segments, via the formula EMBED Equation.3 . Finally, it is necessary to determine the new probability distribution Qj that the process will proceed further from C(m) along the j-th arc, where j = 1, 2, ..., m - 1, since entries into loops are already taken into account in Dm. The probabilities Qj can be determined in the form of a sum of probabilities p(m, x) of those segments that terminate by exiting from C(m) along the j-th arc, i.e.,
EMBED Equation.3
or by the formula
EMBED Equation.3 where a = rmax-1. The calculations are made only for modest m values, since, as m increases, the delay Dm (1 and Qj ( Pi, j=i, while P0, Pm ( 0.
Now we can determine the length of the entire specified process that begins at vertex C(m):
Tk=(Dm+(d+(T.
where =(Dm is the sum of the average delays at those vertices that are encountered in the k-th process; (d is the sum of the pauses, equal to 2t S, which are encountered in the k-th process in A2-TR transitions; (T is the sum of the time intervals that are occupied by all m successful transmissions (in time segments). Here (T is the useful part of the process, while (Dm+(d = Yk is the time loss required to implement the access method.
Knowing the losses Yk and the probabilities Pk for each process, we can determine the average value 7 of the time loss with respect to all processes belonging to C(m), using the formula
EMBED Equation.3
where k = 1, 2, ..., is the ordinal number of the process.
Example. Let us determine the time losses for elementary processes in accordance with Fig. 3 for p = 0.5:
m=2 1=C(2)-0T-0T Y2=D2=2.5
m=3 11=C(3)-0T-0C(2)-0T-0T Y11=D3+D2=4
21= C(3)- 0C(2)-0T-0T-2T Y21=D3+D2+2=6
Y3=Q1Y11+Q2Y21=0.5(4+0.5(6=5.
m=4 111=C(4)-0T-0C(3)-0T-0C(2)-0T-0T
121=C(4)-0T-0C(3)-0C(2)-0T-0T-2T
211=C(4) -0C(2) -0T-0T-2C(2)-0T-0T
311=C(4)-0C(3)-0T-0C(2)-0T-0T-2T
321=C(4)-0C(3)-0C(2)-0T-0T-2C(2)-0T-0T.
Y111=D4+D3+D2=1,22+1,5+2,5=5,22
Y121=D4+D3+D2+2=5,22+2=7,22
Y211=D4+D2+2+D2=1,22+2,5+2+2,5=8,22
Y311=D4+D3+D2+2=5,22+2=7,22
Y321=D4+D3+D2+2+D3=7,22+2,5=9,72
Y4=0,2855(6,22+0,429(8,22+0,2855(8,47=7,22
Fig. 3. Flowcharts with loops for m = 2, 3, 4
5. Calculation of process length with respect to sub diagrams.
The amount of labor involved in the above calculations increases rapidly with m, since it is necessary to make repeated .calculations for the same sub diagrams that are common to many processes. For example, for m = 4 in Fig. 3 we have three different sub diagrams with initial vertex C(2), and two with C(3). Sub diagrams with identical initial vertices differ by virtue of dissimilar initial conditions, i.e., the different number s of stations in state A2, e.g.,
m=2 s=0 C(2)-0T-0T; s=1 C(2)-0T-0T-2T;
s=2 C(2)-0T-0T-C(2)-0T-0T;
m=3 s=0 C(3)-0T-0C(2)-0T-0T; C(3)-0C(2)-0T-0T-2T;
s=1 C(3)-0T-0C(2)-0T-0T-2T;
C(3)-0C(2)-0T-0T-2C(2)-0T-0T.
In accordance with Fig. 1, after all stations transmit their frames from A0 and A1, a 2tS pause will appear on the BB, and then all s > 0 stations go from A2 to the TR state. Here, if s = 1, successful transmission follows; if s ( 2, a collision C(s) follows.
Let us determine the average time losses Y(m,s) for elementary processes for m = 2, 3, ... and s = 0, 1, 2, ..., e.g.,
Y(2,0)=D2=2,5; Y(2,1)=D2+2=4,5; Y(2,2)=D2+2+D2=7,0.
Then for m = 3, s = 0 both sub diagrams are already known, and therefore Y(m,i)=Y(3,1)= D3+Y(2,0)=1,5+2,5=4 for the first of them, while Y(3,2)=D3+Y(2,1)=1,5+4,5=6, for the second, and the average value Y(3,0)=Q1Y(3,1)+ Q2Y(3,2)=0,5(4+0,5(6=5. Similarly for s=1 we obtain D3+Y(2,1)=6; D3+Y(2,2)=1,5+7,0= 8,5; Y(3,1)=0,5(6+0,5(8,5 =7,25.
Fig. 4. Diagram with sub diagrams.
Thus, upon going to the next m value, instead of. the entire flowchart it is sufficient to consider a very simple diagram (Fig. 4) whose sub diagrams are already known, and to determine from it the average time losses in the processes that go through C(m) and then through the j-th sub diagram:
Y(m)j=Dm+Y(m,s) ,
where m1 = m - 1 and s1 = 0, if j = 1; m1 = j and s1 = m - 1, if 2 ( j if m - 1. The average time loss with respect to all processes belonging to C(m) is
EMBED Equation.3
Fig. 5. Distribution of time losses over sub circuits for p = 0.5, c = 1 (a) and 0.5 (b).
The distribution of the time loss over the sub diagram (over classes of outcomes in C(m)) is shown in Fig. 5 for moderate m and for c = 1 and c = 0.5. As we can see, this calculation technique greatly reduces the amount of labour involved.
6. Limits of process length.
Let us determine the limits of the processes that go through C(m). The minimum length for specified Dm is displayed by the first process (uppermost with respect to the flowchart with loops) C(m) 0T0C(m-l)0T...0C(2)-0T0T, which departs from each vertex along the first arc. The time losses in it are Y(m)i=Dm+Dm-1+ +D2. The maximum length is displayed by the last process (lowest according to the flowchart with loops), .which departs from each successive vertex via the last arc, i.e.,
C(m)-0C(m-l)-0C(m-2)- ... -0C(2)-0T-0T-
-2C(m-2)-0C(m-3)- ... -0C(2)-0T-0T-
-- -- --
-2 | C(2)-0T-0T;
| T.
In it there are h = m/2 (m even) or h = (m+1)/2 (m odd) stages (rows in the notation) and h - 1 long pauses, and therefore
EMBED Equation.3
In the limiting case, when all outcomes always end up in the first class upon the first attempt (when Dm=1 for all m), we obtain the lower bound for the time losses for processes that go through C(m), equal to Y(m)min= m-1. As, already noted, the upper bound may be infinite, and therefore m-l(Y(m)<(. At least for modest m, the lower bound is quite realistically achievable, while the upper bound is asymptotic, i.e., in effect unachievable (Fig. 6).
Fig. 6. Time losses plotted against m for c = 1 (a) and 0.5 (b).
We can similarly make calculations for other p values and obtain different relationships (Figs. 7 and 8). Calculations show that, as m increases, the minimum of Dm shifts to .the left within the range 0.58 (p0(0,5, and becomes consistently flatter and broader. It does not coincide with the minimum of Ym, which also shifts leftward and, for m(5, lies below p = 0.5. The time loss Involved in one successful transmission (see Fig. 8) increases slowly as m ( N, and may have an asymptotic limit. As m increases, therefore, the method under consideration deteriorates slightly.
Fig. 7. Average delay at vertex plotted against p for c - 1 (a) and 0.5 (b).
Fig. 8. Time losses involved in one successful transmission.
7. References
[1] K.Molloy, "Collision resolution on the CSMA/CD bus, Computer Networks and ISDN Systems,no. 9. pp. 209-214, 1985.
[2] Standard for Local Area Networks. CSMA/CD. Access Method and Physical LayerSpecifications. IS 8802/3, 1986-09.
[3] Raja Jurdak, Cristina Videira Lopes, and Pierre Baldi, A Survey, Classsification and Comparative
Analysis of Medium Access Control Protocols for AD HOC etworks, IEEE COMMUNICATIONS The Electronic Magazine of Original Peer-Reviewed Survey Articles, pp 1-16, 2004.
[4] C..W. Yi, P.J. Wan, X.Y. Li, and O. Frieder, Asymptotic distribution of the number of isolated nodes in wireless ad hoc networks with Bernoulli nodes, IEEE Wireless Communications and Networking Conference (WCNC), New Orleans, Louisiana, Procreedings pp.131-135, March 2003.
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