Performance Evaluation
4.1 Analysis for the Omnidirectional-antenna Net- Net-work
In this section, we analyze the Average Three-way Handshake Procedure Time (de-noted as ATHPT) in networks using fixed holdoff time schemes and those using our proposed schemes. We derive the ATHPT value based on the used holdoff time value and the number of 1-hop and two-hop neighboring nodes. Because ATHPT greatly affects when a node can transmit data, the analyses presented in this section greatly help us to obtain more insights into the performance of an IEEE 802.16 mesh CDS-mode network.
4.1.1 ATHPT in Networks using Identical Holdoff Time
Fig. 4.1 shows the transmission cycle of a node. Denote Hkas the holdoff time of node k and Ck as the contention time of node k. Denote the interval between two consecutive MSH-DSCH message transmissions for node k (i.e., the interval between two consecutive TxOpps won by node k) as τk, which is defined as follows:
τk= Hk+ Ck. (4.1)
Suppose that the scheduling information of nodes are known to each other. That is, all nodes are one-hop neighboring to each other. Further assume that each node uses the same holdoff time (i.e., Hk = H, ∀k ≥ 0, which is the default holdoff time setting in the
Figure 4.1: The transmission cycle of a node
IEEE 802.16 mesh CDS-mode network). In a network where nbr2(k) ≤ (2base+exp− 2exp), each node suspends itself for a sufficiently long interval; thus, it can win a TxOpp just after its holdoff time elapsed. (That is, Ck is only 1.) In this condition, τk is:
τk = H + 1. (4.2)
The proof is given below.
Theorem 4.1.1. Let G = (V, E) be an IEEE 802.16 mesh CDS-mode network, where each node vi ∈ V is 1-hop neighboring to each other and each vi uses the same holdoff time H = 2exp+base. Suppose that |V | ≤ (2exp+base− 2exp). Then, each node vi wins a TxOpp every (H + 1) TxOpps.
Proof. Let n be the number of nodes in the network (i.e., n = |v|) and m be the number of MSH-DSCH TxOpps in the interval of (2exp+base− 2exp). The MEA of the IEEE 802.16 mesh CDS-mode network should choose the winning node for each TxOpp from the eligible node list of that TxOpp. The chosen eligible node cannot voluntarily yield this TxOpp.
Thus, if n ≤ m and the network is stationary, the permutation of the winning nodes of m consecutive MSH-DSCH TxOpps is n!, followed by (m − n) idle MSH-DSCH TxOpps.
This proves that, in this condition, after node vi wins a TxOpp α, all other nodes will win their respective TxOpps in the interval [α + 1, α + H]. Because all nodes use the same holdoff time, the earliest TxOpp for which nodes other than vi can contend is (α + 1) + (H + 1) = α + H + 2. For this reason, the only eligible node on TxOpp (α + H + 1) (which is the earliest TxOpp for which node vi will contend again) is node vi
itself. Therefore, node vi wins a TxOpp every (H + 1) TxOpps.
Notice that, in the IEEE 802.16 mesh CDS mode, MSH-DSCH TxOpps are contained only in schedule control sub-frames. 4∗Nsf frames containing a schedule control sub-frame follows a frame containing a network control sub-frame, where Nsf denotes the number of frames containing schedule control sub-frames between two frames containing network
control sub-frames in multiples of 4. Thus, the interval between two MSH-DSCH message transmissions taking into account the frame usage of the network for node k is given as follows:
where Ndsch denotes the number of MSH-DSCH TxOpps per frame and Tf r denotes the length of a frame in milliseconds.
Combining Eq. 4.1 and Eq. 4.3, we obtain φk as follows:
φk = (H + 1) ∗Tf r(4Nsf + 1) 4NsfNdsch
. (4.4)
As can be seen, each node can win a TxOpp every (H + 1) TxOpp. Thus, in a THP, after the requesting node broadcasts its request IE out, the granting node must be able to broadcast its granting IE out before the next TxOpp won by the requesting node arrives.
This means that the requesting node can complete a THP within an interval of φk. Thus, we can obtain that ATHPT = φk, if nbr2(k) ≤ (2base+exp − 2exp). The formal proof is given below.
Theorem 4.1.2. Let G = (V, E) be an IEEE 802.16 mesh CDS-mode network, where each node vi ∈ V is 1-hop neighboring to each other and each vi uses the same holdoff time H = 2exp+base. Suppose that |V | ≤ (2exp+base− 2exp). Then, a Three-way Handshake Procedure (THP) is finished within (H + 1) TxOpps.
Proof. Suppose that node x intends to perform a THP with node y. Theorem 4.1.1 proves that, after node x wins a TxOpp α, all other nodes including node y will win their respective TxOpps in the interval [α + 1, α + H]. Thus, after node x transmitted its request IE on TxOpp α, node y can transmit its grant IE within the interval [α+1, α+H].
Finally, node x can transmit its confirm IE on TxOpp (α + H + 1). As a result, a THP can be completed using only (H + 1) TxOpps.
4.1.2 ATHPT in Networks using Non-identical Holdoff Times
Denote the requesting node and the granting node in a THP as NR and NG, respec-tively. The holdoff times of NR and NG are denoted as HR and HG, respectively, and the contention times of these two nodes are denoted as CR and CG, respectively. CR and CG
are random variables.
Figure 4.2: Example case where HR ≫ HG
If HR ≫ HG (as shown in Fig. 4.2), then ATHPT is dominated by the transmission cycle of the requesting node and can be defined as follows:
ATHPT = HR+ E[CR]. (4.5)
Deriving E[CR] is difficult because it is involved with the holdoff times used by neigh-boring nodes. In [2], however, Cao et al. show that the distribution of a node’s contention time can be approximately modeled by a geometric distribution. Following this assump-tion, E[CR] = 1p, where p is the probability for a node to win a TxOpp. Since, for node x, the maximum number of nodes contending for a TxOpp is |nbr2(x)|, p is given as follows:
p(x) ≥ 1
|nbr2(x)|. (4.6)
We take the lower-bound value |nbr1
2(x)| for p(x) to estimate the lower-bound performance of a network using static non-identical holdoff times. The value of ATHPT in such a network is thus derived as follows:
ATHPT = HR+ nbr2(NR). (4.7)
As shown in Fig. 4.3, if HR ≪ HG, then ATHPT = eG+ eR, where eG denotes the residual time of τG and eR denotes the residual time of τR, where τG = HG + CG and τR = HR+ CR. Thus, the ATHPT value in this condition can be defined as follows:
ATHPT = E[eG] + E[eR]. (4.8)
To simplify the analysis, we adopt the assumptions used in [2], which is defined here.
Let Zk(t) be the number of transmission times of node k up to slot t. That is, Zk(t) is a counting process with inter-arrival time τk. Suppose that {τk = Hk+ Ck, ∀k ≥ 0} is independent and identically distributed (i.i.d.). Then, Zk(t) is a renewal process and τk
Figure 4.3: Example case where HR ≪ HG
is the renewal interval. With this assumption, we know that the expected value of the residual time of next arrival is one half of the expected value of the observed inter-arrival time (denoted as τk0) [18]. The formal expression is given as follows:
E[ek] = τk0
2 , (4.9)
where τk0 ≥ Hk+ E[Ck] = Hk+ nbr2(k).
Following this, ATHPT can be defined as:
ATHPT ≥ HG+ nbr2(NG)
2 + HR+ nbr2(NR)
2 . (4.10)
If HR ≃ HG, the value of ATHPT depends on whether (HR+ CR) is larger than eG. If this condition is satisfied, a THP can be completed within a transmission cycle of the requesting node. Otherwise, a THP should be completed using the time amount of (eG+ eR). The results derived from this case (Eq. 4.11) perfectly match the results derived from the cases where HR≫ HG and HR ≪ HG. Thus, by combining these equations, we can
4.1.3 ATHPT in Networks using Dynamic Holdoff Times
Our proposed dynamic holdoff time scheme makes the requesting node win a TxOpp later than and closest to the next TxOpp won by the granting node. Thus, as shown in Fig. 4.4, using this design ATHPT can be modeled as follows:
ATHPT = E[e′G] + E[e′R], (4.13)
Figure 4.4: Example case where HR ≫ HG when using the proposed dynamic holdoff time scheme
where E[e′G] denotes the expected value of the residual time of the granting node’s next TxOpp and E[e′R] denotes the expected value of the residual time of the requesting node’s next TxOpp. Similar to E[eG] in Eq. 4.9, E[e′G] basically can be modelled as follows:
E[e′G] ≥ HG+ nbr2(NG)
2 . (4.14)
However, our proposed dynamic holdoff time scheme makes the holdoff time of a node depend on those of its neighboring nodes. Thus, assuming that all nodes in the network have data to send at all time, we can know that the HG in Eq. 4.14 is not fixed and ap-proaches to the minimum value of nodes’ holdoff times (denoted as Hmin) in the network, if the network is in the stationary and steady state. Thus, E[e′G] is corrected as follows:
E[e′G] ≥ Hmin+ nbr2(NG)
2 . (4.15)
On the other hand, E[e′R] can be defined as follows:
E[e′R] = E[HR′ ] − E[eG] + nbr2(NR). (4.16) Because in the dynamic holdoff time scheme the requesting node can find a TxOpp later than and closest to the next TxOpp won by the granting node, HR′ is either 2⌊log2(eG)⌋
or 2⌈log2(eG)⌉. Thus, |HR′ − eG| ≤ 2⌈log2(eG)⌉− 2⌊log2(eG)⌋. Combining these results, ATHPT can be written as follows:
0 10 20 30 40 50 60 70
Dynamic Static HT-16 HT-32 HT-64
Tthp (TxOpp)
Analysis Simulation
Figure 4.5: Comparison between ATHPT values derived from the theoretical model and those obtained by simulations
ATHPT ≤ Hmin+ nbr2(NG) + nbr2(NR) + 2⌈log2(eG)⌉− 2⌊log2(eG)⌋
2 + ǫ (4.17)
≤ Hmin+ nbr2(NG) + nbr2(NR) + 2⌈log2(Hmin+nbr2(NG)
2 )⌉− 2⌊log2(Hmin +nbr2(NG)
2 )⌋
2 + ǫ.
(4.18) Because 2⌈log2(Hmin+nbr2(NG))⌉− 2⌊log2(Hmin+nbr2(NG))⌋ ≤ 2⌊log2(Hmin+nbr2(NG))⌋, we can de-rive ATHPT as follows:
ATHPT ≤ Hmin+ nbr2(NG) + nbr2(NR) + 2⌊log2(Hmin+nbr2(NG))⌋−1
2 + ǫ. (4.19)
Eq. 4.19 can be used to evaluate ATHPT. We compare the ATHPT values derived from this analytical model and those obtained from simulations in a 10-node chain network.
The results are plotted in Fig. 4.5. As can be seen, the results shown in Fig. 4.5 shows that the ATHPT values derived from our analytical model greatly matches those obtained by the simulation experiments.