QUEUES , RANDOM GRAPHS AND BRANCHING PROCESSES "

In this paper it is shown that certain basic results of queueing theory can be used successfully in solving various problems of random graphs and branching processes.


I. INTRODUCTION
Some basic results of queueing theory turn out to be also useful in various other fields of mathematics.By way of illustration we will focus on two topics, namely, random graphs and branching processes.
In what follows we describe a general queueing model and clef'me a random graph and a branching process related to it.After deriving some specific results for the queueing model considered, we shall demonstrate their applications in the fields mentioned above.
The queueingmodel.Let us suppose that in the time interval (0,**) customers arrive at random at a counter and are served singly by one server in order of arrival.It is assumed that the server starts working at time t 0 and at that time (i 0, 1, 2,...) customers are already waiting for service.The initial customers are numbered 1, 2, ..., i and the customers arriving subsequently are numbered i+1, i + 2,... in the order of their arrivals.Denote by Vr the number of customers arriving during the service time of the r-th customer.This queueing model will be characterized by the initial queue size i and the sequence of random variables v 1, v 2, Vr, Throughout this paper we use the abbreviation (1) N r v + v 2 + +v r for r 1, 2, and write N O 0. We shall now consider the following random variables: p(i), the number of customers served in the initial busy period, for 1,2,... and p(0) 0. We write (2) "fn0= P{p(i) n } for I <in and f(o0) 1,1( )) =0 forn> I.
We define r(r 1, 2,...) as the queue size immediately after the r-th service ends, and write (3) Pi(n, k) P{r < k for I < r n and p(i) n} for I <i <n andk 1.
Following D. G. Kendall [10] we say that the initial customers (if i > 1) in the queue form the 0-th generation.The customers (ff any) arriving during the total service time of the initial i customers form the 1-st generation.Generally, the customers (ff any) arriving during the total service time of the customers in the (r-1)-th generation form the r-th generation for r 1, 2, Denote by Yr (r 0, 1, 2, ...) the number of customers in the r-th generation and def'me (4) x(i) sup{r: Tr > O, r > 0}.
An associated random graph.We associate a random graph with the queueing process considered.If i 1 and ff the initial busy period consists of n services, that is, ff p(i) n, then let us assume that the graph has vertex set 1, 2,.., n}.Furthermore, in the graph, two vertices r and s are joined by an edge ff and only ff the r-th customer arrives during the service time of the s-th customer.
If p(i) n in the queueing process and i I, then the graph is a roote tree with n labeled vertices, the root being labeled 1.If p(i) n and i > I, then the graph is a forest consisting of i distinct rooted trees whose roots arc labeled I, 2,.., respectively.The total number of vertices in the forest is n.
We define the height of the forest as the maximal distance between any vertex of the graph and the root of the tree which contains the given vertex.In the notation of (4), x(i) is the height of the random forest for i 1.If i 1, the forest becomes a tree.
An associated branching proeess.Now let us assume that in the queueing process vp v 2 yr,.., is a sequence of independent and identically distributed random variables for which (6) P{v r j} pj for j 0, I, 2,... and r I, 2, In this case the sequence of random variables (Y0' I,'", Yr'"" } forms a branching process.We can imagine that in a population initially we have i (i1) individuals and in each generation each individual reproduces independently of the others and has probability pj (j 0, 1, 2,...) of giving rise to j descendants in the following generation.The random variable Yr is the number of individuals in the r-th generation.Define p*(i) 0 + + + Yr + that is, p*(i) is the total number of individuals (total progeny) in the process.
The following discussion is based mostly on a simple combinatorial theorem which is a generalization of the classical ballot theorem.It can be formulated as follows: Theorem 1.Let us suppose that a box contains n cards marked with nonnegative integers k 1, k 2, k n respectively where k + k 2 + k n k n.
All the n cards are drawn without replacement from the box.Denote by v r the number drawn at the r-th drawing (r 1, 2,...,n).Then (9) P{V + + V < r for 1 < r < n} (n k)/n provided that all thepossible results are equally probable.
Proof.Denote by S(n, k) the number of favorable results, that is, the number of sequences (k 1, k2,..., k n) in which the sum of the first r numbers drawn is less than r for every r 1, 2, n.If we take into consideration that the last number drawn may be k 1, k 2, k n, then we can write down that (10) S(n, k) S(n-1, k-ki).
Theorem 2. Let v l, v2,... v n be interchangeable discrete random variables taking on nonnegative integers only, and define N r v + ...+ v r for r 1, 2,..., n.We have ( 13) Proof.If we apply Theorem I to every realization of the sequence of the random variables v 1, v2,..,vn, we get (13).
In a particular case Theorem 1 was formulated by J. Bertrand [2] in 1887 and was proved in the same year by D. Andr6 [1].The above generalizations have been given by L. Talcs [26].

THE QUEUEING MODEL
In the queueing model defined in the Introduction, the probability (2), that is, the probability that the initial busy period consists of n services, is completely determined by the joint distribution of the random variables v, v2,..., v n.Since p(i) n if and only if N r > r i for < r < n and N n n i, we have (14) P{N r > r i for i r < n and N n n i} for 1 <in.
If v 1, v 2 v n is a sequence of independent and identically distributed random variables, then they are interchangeable and (16) holds unchangeably.In this case, as an alternative, we can use generating functions for the determination of f(n i).See L. Taktcs [24] and [27].Let us introduce the notations" P(Vr--j}--pj (j 0,1,2,...), ( 17) g(z) E Pj zi j=O for lzl (18) a= Ejpj, j=O fn f(n 1 for n : 1, and ( 19) f(w) E fawn for w g 1.Since the number of customers served in the initial busy period can be expressed as the sum of independent random variables each having the same distribution { fn }' we have (20 On the other hand, if we take into consideration that the number of customers arriving during the first service time may be j 0, 1, 2,..., then we can write that n-1 j=o for n > 1. Hence j=O for[ w < 1. Accordingly, z f(w), w 1, is a solution of the equation ( 23) z w g(z).
The function g(z) is regular in the unit disk z[ < I, and wg(z) [ w[ < z] if [z] = 1 and lw <1.Thus by Rouch's theorem, equation ( 23 Theorem 4. For an arbitrary set of random variables v 1, v2,..., v n the probability (3) can be expressed in thefollowing way: (25) Pi(n,k) P(i k < r N r < i for 1 < r < n and n N n i} ill <i<n and k 1. Proof.If p(i) =n, thenN r>r-i for lr<n, N n=n-i, and consequently r Nr" r + 1 for 1 r _< n.Now let us suppose again that v 1, v2,..., Vr is a sequence of independent and identically distributed random variables with the distribution P{Vr= j} =pj (j 0, 1, 2,...) and generating function g(z) defined by ( 17).
We are interested in determining the probabilities (5).Let us introduce the generating functions ( 26) hk(Z) Z P{x (I) k and p (I) n}z n for k 1, 2,... and zl_< 1.In analogy with (20), we have Z P{x(i) < k and p(i) _< n}z n [hk(Z)] for > 1, k 1 and z[ < 1.Thus the problem of finding (5) can be reduced to the determination of (26).

EXAMPLES FOR TI QUEUEING PROCESS
The following examples have some interest in their own, but they are also useful in studying random graphs.EXAMPLE 1.Let us suppose that in the queueing model defined in the Introduction v, v2,..., Vr,... is a sequence of independent random variables and the distribution of v r is given by ( 30) P{Vr j} e-Xpq r' (,A,pqr-1)J/j! for j 0, 1, 2,... and r 1, 2,... where , > 0, 0 < p < 1 and q 1 p.In this case the random variables v 1, v2,..., v n are not interchangeable, but by ( 14) we obtain (31) for 1 < < n, or equivalently, (32) f) and is a polynomial of degree (n i)(n + i 3)/2 in q.
We note that (35) ()(1) 11 n" i- for I i < n.This can be proved by ( 34).If we put q 1 in (34), we obtain that (36) and (35) follows from (36) by mathematical induction.EXAMPLE 2. Let us suppose that in the queueing process customers arrive in the time interval (0, **) in accordance with a Poisson process with parameter ., and the service firnes are independent random variables each having the same exponential distribution function (37) 1 e " for x 0, H(x) 0 for x < 0, and are independent of the arrival times.In this case v r v2,..., Vr,... is a sequence of independent and identically distributed random variables for which (38) P{v r j} pj q if j 0, 1, 2,... where p ./ (Z, + Ix) and q Ix / Now by ( 16) we obtain that (39) Pp(i)=n}=-pn qn for I < n, and (3) can also be expressed in the following way: (40) Pi(n, k) P{-i < r < k i for 1 < r < 2n and r/2 n -i} for 1 < i < n and k > 1 where the random variables r/0, r/l,..., Or describe a random walk on the real axis.A particle starts at x 0 and in each step it moves either a unit distance to the right with probability p or a unit distance to the left with probability q.The successive displacements are independent and r/r denotes the position of the particle after the r th step (r 1, 2,.. ); r/0 0. The probabilities (40) can easily be determined explicitly by using the reflection principle.See L.

CAYIY'S FORMULA
A forest is a simple graph that has no cycles.In other words, a forest is a simple graph, all of whose components are trees.
Denote by F(n, i), I -< i < n, the number of forests having vertex set { 1, 2,..., n} and i components which are trees such that vertices 1, 2,.., i all belong to different trees.We have (50) F(n, i) is the number of distinct trees with n labeled vertices.
Let us consider the queueing model defined in the Introduction.Suppose that at time t 0, i (1 i n) customers are waiting for service, and in the time interval (0, n] exactly n i customers arrive in such a way that independently of each other each customer may join the queue in any interval (r 1, r] (r I, 2,..., n) with probability 1 / n.The service times are assumed to be constant of unit length.In this case v I, v2,..., v n are interchangeable random variables with sum v 1 + v 2 +...+ v n n-i.If Pn(i) denotes the total number of customers served during the initial busy period, then by Theorem 3 we have ( 52) On the other hand, we have ( 53) for 1 < i n.To prove (53) let us consider the graph associated with the queueing process.The graph has vertex set { 1, 2,..., n}.There are n n i possible graphs, and they are equally probable.If Pn(i) n, then the graph consists of i tree components such that vertices 1, 2,..., i all belong to different trees.Conversely, to every such graph there corresponds a queueing process for which Pn(i) n.The number of favorable graphs (i.e.graphs corresponding to queueing processes for which pn(i) n) is F (n, i).This proves (53).A comparison of ( 52) and ( 53) implies (50).

RANDOM GRAPHS
Here we are concerned with a random graph cormeeted with the structure of polymers in chemistry.We define a random graph Fn(p) on the vertex set V n { 1, 2,..., n} in the following way.First, we form a complete graph K n on the vertex set V n.The complete graph is a simple graph which is undirected and has no loops or multiple edges.It contains all the possible () edges.If in K n, each edge, independently of the others, is either retained with probability p or removed with probability q where p + q 1 and 0<p<l, then we obtain Fn(P).Now let us f'LX a set of vertices (1 < < n) in V n, say { 1, 2,..., i}, and let us define Pn(i) as the number of vertices in the union of all those components of Fn(p) which contain at least one vertex of the set of vertices { 1, 2,..., i}.Thus Pn(i) is i plus the number of vertices in the set {i + 1,..., n} which are connected by a path (an uninterrupte sequence of edges) with one of the vertices in the set {1, 2,..., i}.
J.W. Kennedy [11] derived a recurrence formula for the determination of the distribution of pn(1) and calculated the probabilities P{Pn(1) k} for k 5.
In what follows we shall demonstrate that Pn(i) has the same distribution as the number of customers served in the initial busy period in an appropriate queueing process.
To describe the relevant queueing process let us suppose that n customers, numbered 1, 2,..., n, are served singly by one server.The service times are assumed to be constant of unit length.The server starts working at time t 0 and at that time i (1 < n) customers, I, 2,..., i, are already waiting for service.The server attends to these i customers and all the new arriving customers as long as they come.If there are no more customers in the system, the server leaves the system unattended for a time interval of unit length before resuming his duty again.
It is assumed that no customer joins the queue more than once, and that any customer who has not yet joined the queue until time t r 1 (r 1, 2,...) may join the queue in the time interval (r-1, r] with probability p (0 < p < 1) independently of the other customers.
The distribution of Pn(i) in the random graph Fn(p) is the same as the distribution of the number of customers served in the initial busy period in the queueing process defined above.By forming the product of ( 59), ( 60) and (61), we obtain the limit (58).
Theorem 7 confn'ms a conjecture of J.W. Kennedy [ 11] concerning the limit distribution of Pn(1)as np--a.Kennedy calculated the limit (58) for i I and k -< 5, and conjecmrexl that (58) is true for 1 and all k 1.
Let us suppose that all the possible C a trees are equally probable and choose a tree at random.Denote by o(n+l) the height of the U'e.We are interested in finding the distribution of o(n+1).To fred this distribution let us consider the queueing model discussezl in Example 2 in Section 3. If the initial queue size is i 1 and if the number of customers served in the initial busy period is p(1) n + 1, then the associated random graph can be described as an unlabeled, rooted, plane (ordered) tree with n + 1 vertices.The height of the tree o(n+1) < k if and only if in the associated queueing process (1) g k.Accordingly, ) has exactly one root z in the unit disk whenever [w[ < 1.Thus necessarily, f(w k ) P{z(i) < k and p(i) n} ai(n,k) pn-i qn for I < i < n and k I can be obtained by Theorem 5.In this case g(z) qz/(l-pz); and if we define k :> 1 and[ x[ < 1, then by (28) we obtain that qn-k c-aak-i/oc_i)!.