10 B.Sufficient condition on r(n)of Theorem 5.1 The specific set of is and js are selected as follow.Let Suppose there are at most n sessions in a periodandu(n).Then we divide the unit square into let E:denote the event that s;is a failed session,where cells such that each cell is of size u2(n). i=1,2,...,n.Let each node have the transmission range Now,among these cells,pick ndo of them such that each of r=cr(n),where c 1.Using the existing approach in Section IV-B,we have them is at least away from others,as shown in Figure 8. selected cells =1 u(n) ≤ n(1-r2） ≤ ne-kndxr2 do 1 4nc2- For any c>1,the result follows. VI.THE CRITICAL TRANSMISSION RANGE FOR CONNECTIVITY OF STATIONARY K-HOP CLUSTERED NETWORKS In this section,our main result is the following theorem. Fig.8.Cell selection Theorem 6.1:For the stationary k-hop clustered networks, Pick one cluster member node from each of these selected the critical transmision rangeis(o）=t√，whee cells to form a set consisted of ndo nodes.Let Y denote the 0<d<1. set of such cluster member nodes.Note that any two nodes in Yare at a distance of at least away.Further,this distance is larger than kr(n)when n is sufficiently large. A.Necessary condition on r(n)of Theorem 6.1 Then,using the existing approach of computing the proba- Let gatat(n,r(n))denote the network where two nodes are bility that the network is disconnected,we have connected if their Euclidean distance is at most r(n)and we use the term disconnected to describe a cluster member whose Pd_stat(n,r(n)》 packets cannot reach a cluster head within k hops.Then,if 2 P(i is an disconnected cluster member there is at least one disconnected member node in the network. iEY we define that gstat is disconnected.Let P_stat(n,r(n))be in gstat(n,r(n))}) the probability that gstat is disconnected and we have the P(fi and j are disconnected cluster following proposition. i.jEY, ≠i Proposition 6.1:If wr2(n)=d where do<d. k2nd members in Gstat(n,r(n))}).(21) then liminf Pd_stat(n,r(n))>e-w(1-e), As to the first term in the right hand side of(21),it is 2+●如 bounded by where k limnoK(n),K>0. Proof:The proof applies similar techniques as that of -roj月产≥n,g: ‘ne=e-% Proposition 4.1,we will not considered all pairs of nodes i where 0<<1.To bound the second term,note that i,jY and j that are disconnected cluster members.The problem for this is that there are too many of these pairs.The intuitive are at least away,which will be larger than 2kr(n) explanation is depicted in the following figure. when n is sufficiently large.Thus,each term in the second summation is bounded by ,·cluster head more than kr(n)away P(and je discomcted))≤(1-2a(kr(m)）". 00 There are too many pairs ij Therefore, in this region such that both i,j are disconnected from the cluster head. Pa_stat((n,r(n)）≥0e-k-n2doe-2mr(r(n)2 ≥0eR-e-2r, Fig.7.An intuitive explanation of the problem in the above approach for all n Ne.x. Further,such i and j may be very close to each other,making The rest of steps are omitted and follows these in the earlier it difficult to bound the probability. proofs.10 B. Sufficient condition on r(n) of Theorem 5.1 Suppose there are at most n sessions in a period λb, and let Ei denote the event that si is a failed session, where i = 1, 2, . . . , n. Let each node have the transmission range r = cr(n), where c > 1. Using the existing approach in Section IV-B, we have P ( ∪n i=1 Ei ) ≤ ∑n i=1 P(Ei) ≤ n (( 1 − πr2 )n d )k ≤ ne−kndπr2 = 1 4nc 2−1 . For any c > 1, the result follows. VI. THE CRITICAL TRANSMISSION RANGE FOR CONNECTIVITY OF STATIONARY K-HOP CLUSTERED NETWORKS In this section, our main result is the following theorem. Theorem 6.1: For the stationary k-hop clustered networks, the critical transmission range is r(n) = 1 k √d log n πnd , where 0 < d < 1. A. Necessary condition on r(n) of Theorem 6.1 Let Gstat(n, r(n)) denote the network where two nodes are connected if their Euclidean distance is at most r(n) and we use the term disconnected to describe a cluster member whose packets cannot reach a cluster head within k hops. Then, if there is at least one disconnected member node in the network, we define that Gstat is disconnected. Let Pd stat(n, r(n)) be the probability that Gstat is disconnected and we have the following proposition. Proposition 6.1: If πr2 (n) = d0 log n+κ k2nd , where d0 < d, then lim inf n→∞ Pd stat(n, r(n)) ≥ e −κ (1 − e −κ ), where κ = limn→∞ κ(n), κ > 0. Proof: The proof applies similar techniques as that of Proposition 4.1, we will not considered all pairs of nodes i and j that are disconnected cluster members. The problem for this is that there are too many of these pairs. The intuitive explanation is depicted in the following figure. IR[YZKXNKGJ SUXKZNGTQXTG]G_ :NKXKGXKZUUSGT_VGOXYOP OTZNOYXKMOUTY[INZNGZHUZN OPGXKJOYIUTTKIZKJLXUSZNK IR[YZKXNKGJ Fig. 7. An intuitive explanation of the problem in the above approach Further, such i and j may be very close to each other, making it difficult to bound the probability. The specific set of is and js are selected as follow. Let u(n) = O (√ log n n ) . Then we divide the unit square into 1 u(n) × 1 u(n) cells such that each cell is of size u 2 (n). Now, among these cells, pick n d0 of them such that each of them is at least √ 1 nd0 away from others, as shown in Figure 8. YKRKIZKJIKRRY Fig. 8. Cell selection Pick one cluster member node from each of these selected cells to form a set consisted of n d0 nodes. Let Y denote the set of such cluster member nodes. Note that any two nodes in Y are at a distance of at least √ 1 nd0 away. Further, this distance is larger than kr(n) when n is sufficiently large. Then, using the existing approach of computing the proba￾bility that the network is disconnected, we have Pd stat(n, r(n)) ≥ ∑ i∈Y P({i is an disconnected cluster member in Gstat(n, r(n))}) − ∑ i,j∈Y, j̸=i P({i and j are disconnected cluster members in Gstat(n, r(n))}). (21) As to the first term in the right hand side of (21), it is bounded by n d0 ( 1 − π ( kr(n) )2 )n d ≥ n d0 · θ · 1 nd0 e −κ = θe−κ , where 0 < θ < 1. To bound the second term, note that i, j ∈ Y are at least √ 1 nd0 away, which will be larger than 2kr(n) when n is sufficiently large. Thus, each term in the second summation is bounded by P({i and j are disconnected}) ≤ ( 1 − 2π ( kr(n) )2 )n d . Therefore, Pd stat(n, r(n)) ≥ θe−κ − n 2d0 e −2n dπ(kr(n))2 ≥ θe−κ − e −2κ , for all n > Nθ,κ. The rest of steps are omitted and follows these in the earlier proofs.