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  1. 1. Capacity of Multi-Channel Wireless Networks: Impact of Number of Channels and Interfaces∗ Technical Report March 2, 2005 Pradeep Kyasanur Nitin H. Vaidya Dept. of Computer Science, and Dept. of Electrical and Computer Engineering, and Coordinated Science Laboratory, Coordinated Science Laboratory, University of Illinois at Urbana-Champaign University of Illinois at Urbana-Champaign Email: Email: Abstract— This paper studies how the capacity of a Past research on wireless network capacity has typically static multi-channel network scales as the number of considered wireless networks with a single channel, nodes, n, increases. Gupta and Kumar have determined although the results are applicable to a wireless network the capacity of single-channel networks, and those bounds with multiple channels as well, provided that at each are applicable to multi-channel networks as well, provided node there is a dedicated interface per channel. With a each node in the network has a dedicated interface per channel. dedicated interface per channel, a node can use all the In this work, we establish the capacity of general multi- available channels simultaneously. However, the number channel networks wherein the number of interfaces, m, of available channels in a wireless network can be fairly may be smaller than the number of channels, c. We large (e.g., IEEE 802.11a [3] has provisioned for up to show that the capacity of multi-channel networks exhibits 12 non-overlapping channels), and it may not be feasible different bounds that are dependent on the ratio between c to have a dedicated interface per channel at each node. and m. When the number of interfaces per node is smaller When nodes are not equipped with a dedicated inter- than the number of channels, there is a degradation in face per channel, then capacity degradation may occur, the network capacity in many scenarios. However, one compared to using a dedicated interface per channel. important exception is a random network with up to O (log n) channels, wherein the network capacity remains In this paper, we characterize the impact of number of n at the Gupta and Kumar bound of Θ W log n bits/sec, channels and interfaces per node on the network capacity, and show that in certain scenarios, even with a single independent of the number of interfaces available at each node. Since in many practical networks, number of chan- interface per node, there is no capacity degradation. This nels available is small (e.g., IEEE 802.11 networks), this implies that it may be possible to build capacity-optimal bound is of practical interest. This implies that it may be multi-channel networks with as few as one interface per possible to build capacity-optimal multi-channel networks node. with as few as one interface per node. We also extend our When a dedicated interface per channel is not avail- model to consider the impact of interface switching delay, able, the available interfaces can potentially be switched and show that capacity losses due to switching delay can among different channels to use any of the available be avoided by using multiple interfaces. channels. Such an interface switching technique is often I. I NTRODUCTION used to improve channel utilization [4]–[6]. However, interface switching incurs a delay, which may reduce Previous research (e.g., [1], [2]) has characterized the the achievable network capacity. In this paper, we also capacity of wireless networks. One approach for enhanc- characterize the impact of interface switching delay on ing the network capacity is to use multiple channels. network capacity. We show that interface switching delay ∗ This research was supported in part by NSF grant ANI-0125859 has no impact on network capacity, even when there and a Vodafone Graduate Fellowship. are end-to-end delay constraints, provided that a few
  2. 2. 2 additional interfaces are provisioned for at each node. Kumar [1]). The network is said to transport one “bit- meter/sec” when one bit has been transported across a A. Modeling multi-channel multi-interface networks distance of one meter in one second. We consider a static wireless network containing n Random Networks: In the random network setting, nodes. We use the term “channel” to refer to a part of node locations are randomly chosen, and each node the frequency spectrum with some specified bandwidth. sets up one flow to a randomly chosen destination. There are c channels, and we assume that every node is The network capacity is defined to be the aggregate equipped with m interfaces, 1 ≤ m ≤ c. We assume that throughput over all the flows in the network, and is an interface is capable of transmitting or receiving data measured in terms of bits/sec. on any one channel at a given time. We use the notation We use the following notation to represent bounds: (m, c)-network to refer to a network with m interfaces 1) f (n) = O(g(n)) implies there exists some con- per node, and c channels. stant d and integer N such that f (n) ≤ dg(n) for We define two channel models to represent the data n > N. rate supported by each channel: 2) f (n) = o(g(n)) implies that limn→∞ f (n) = 0. g(n) Channel Model 1: In model 1, we assume that the 3) f (n) = Ω(g(n)) implies g(n) = O(f (n)). total data rate possible by using all channels is W . The 4) f (n) = ω(g(n)) implies g(n) = o(f (n)). total data rate is divided equally among the channels, 5) f (n) = Θ(g(n)) implies f (n) = O(g(n)) and and therefore the data rate supported by any one of the g(n) = O(f (n)). c channels is W/c. This was the channel model used 6) MINO (f (n), g(n)) is equal to f (n), if f (n) = by Gupta and Kumar [1], and we primarily use this O(g(n)), else, is equal to g(n). model in our analysis. In this model, as the number of The bounds for random networks hold with high channels increases, each channel supports a smaller data probability (whp). In this paper, whp implies with “prob- rate. This model is applicable to the scenario where the ability 1 when n → ∞.” total available bandwidth is fixed, and new channels are created by partitioning existing channels. C. Main Results Channel Model 2: In model 2, we assume that each channel can support a fixed data rate of W , independent Gupta and Kumar [1] have shown that in an arbitrary √ of the number of channels. Therefore, the aggregate data network, the network capacity scales as Θ (W n) bit- rate possible by using all c channels is W c. This model meters/sec, and in a random network, the network capac- is applicable to the scenario where new channels are n ity scales as Θ W log n bits/sec. Under the channel created by utilizing additional frequency spectrum. model 1, which was the model used by Gupta and Kumar The results presented in this paper are derived assum- [1], the capacity of a network with a single channel ing channel model 1. However, all the derivations are and one interface per node (that is, a (1, 1)-network applicable for channel model 2 as well, and the results in our notation) is equal to the capacity of a network for model 2 can be obtained by replacing W in the results with c channels and m = c interfaces per node (that of model 1 by W c. is, a (c, c)-network). Furthermore, under both channel models, the capacity of a (c, c)-network is at least as B. Definitions large as the capacity of a (m, c)-network, when m ≤ c We study the capacity of static multi-channel wireless (this is trivially true, by not using c − m interfaces in networks under the two settings introduced by Gupta and the (c, c)-network). In the results presented in this paper, Kumar [1]. we capture the impact of using fewer than c interfaces Arbitrary Networks: In the arbitrary network set- per node by establishing the loss in capacity, if any, of ting, the location of nodes, and traffic patterns can be a (m, c)-network in comparison to a (c, c)-network. controlled. Since any suitable traffic pattern and node The goal of this work is to study the impact placement can be used, the bounds for this scenario are of the number of channels c, and the number of applicable to any network. The arbitrary network bounds interfaces per node m, on the capacity of arbitrary and may be viewed as the best case bounds on network random networks. Our results show that the capacity is c capacity. The network capacity is measured in terms dependent on the ratio m , and not on the exact values of “bit-meters/sec” (originally introduced by Gupta and of either c or m (as shown in Lemma 2). We now state
  3. 3. 3 Capacity when c = m Capacity when c = m √ A W n W n D E Network capacity log n Network capacity W n Capacity loss F log n B log n W log log n Capacity loss W W C W log log n G n n log n 2 log log n 1 log n n log n n2 1 log n n n2 c c Ratio of channels to interfaces m Ratio of channels to interfaces m Fig. 1. Impact of number of channels on capacity scaling in arbitrary Fig. 2. Impact of number of channels on capacity scaling in random networks (figure is not to scale) networks (figure is not to scale) our main results under channel model 1. F in Figure 2 overlaps part of segment A-B in Figure 1), implying “randomness” does not incur 1. Results for arbitrary network: The network capacity a capacity penalty. of a (m, c)-network has two regions (see Figure 1) as c log log n 2 3) When m is Ω n log n , the network follows (from Theorem 1 and Theorem 2): log log c capacity is Θ W nmlog n n bits/sec (line F-G in c 1) When m is O(n), the network capacity is Figure 2). In this case, there is a larger capacity Θ W nm bit-meters/sec (segment A-B in Fig- c degradation than case 2. Furthermore, in this ure 1). Compared to a (c, c)-network, there is a region, the capacity of a (m, c)-random network capacity loss by a factor of 1 − m . c is smaller than that of a (m, c)-arbitrary network, c 2) When m is Ω(n), the network capacity is in contrast to case 2. Θ W nm bit-meters/sec (line B-C in Figure 1). c In this case, there is a larger capacity degradation 3. Other results: The results presented above are than case 1, as nm ≤ nm when m ≥ n. c c c derived under the assumption that there is no delay Therefore, there is always a capacity loss in arbitrary in switching an interface from one channel to another. networks whenever the number of interfaces per node However, we show that even if interface switching delay is fewer than the number of channels. is considered, the network capacity is not reduced, provided a few additional interfaces are provisioned for 2. Results for random network: The network capacity at each node. This implies that it is possible to hide of a (m, c)-network has three regions (see Figure 2) as the interface switching delay by using extra interfaces in follows (from Theorem 3 and Theorem 4): conjunction with carefully designed routing and trans- c 1) When m is O(log n), the network capacity is mission scheduling protocols. The rest of the paper is organized as follows. We n Θ W log n bits/sec (segment D-E in Figure 2). present related work in Section II. In Section III, we In this case, there is no loss compared to a (c, c)- establish the capacity of multi-channel networks under network. Hence, in many practical scenarios where arbitrary network setting. Section IV establishes the ca- c may be constant or small, a single interface per pacity of multi-channel networks under random network node suffices. setting. Section V characterizes the impact of interface c log log n 2 switching delay. We conclude in Section VI. 2) When m is Ω(log n) and also O n log n , the network capacity is Θ W nm bits/sec (seg- II. R ELATED W ORK c ment E-F in Figure 2). In this case, there is In their seminal work, Gupta and Kumar [1] derived some capacity loss. Furthermore, in this region, the the capacity of ad hoc wireless networks. The results capacity of a (m, c)-random network is the same are applicable to single channel wireless networks, or as that of a (m, c)-arbitrary network (segment E- multi-channel wireless networks where every node has a
  4. 4. 4 dedicated interface per channel. We extend the results Signal-to-Interference-Noise Ratio (SINR) (when path of Gupta and Kumar to those multi-channel wireless loss exponent is greater than 2). Therefore, the results networks where nodes may not have a dedicated interface in this paper are applicable under the physical model as per channel, and also consider the impact of interface well. We do not consider other physical layer character- switching delay on network capacity. istics such as channel fading in our analysis. Grossglauser and Tse [2] showed that mobility can We derive the capacity results for arbitrary and random improve network capacity, though at the cost of increased networks under the assumption that there is no switching end-to-end delay. Subsequently, other research [7], [8] delay. We extend our model to consider the impact of has analyzed the trade-off between delay and capacity switching delay in Section V. in mobile networks. Gamal et al. [9] characterize the In an arbitrary network, the location of nodes, and optimal throughput-delay trade-off for both static and traffic patterns can be controlled. Recall that the network mobile networks. In this paper, we adapt some of the is said to transport one “bit-meter/sec” when one bit has proof techniques presented by Gamal et al. [9] to the been transported across a distance of one meter in a multi-channel capacity problem. second. The network capacity of an arbitrary network is Recent results have shown that the capacity of wireless measured in terms of bit-meters per second, instead of networks can be enhanced by introducing infrastructure bits per second. The bit-meters/sec metric is a measure support [10]–[12]. Other approaches for improving net- of the “work” that is done by the network in transporting work capacity include the use of directional antennas bits. In the case of random networks, the average distance [13], and the use of unlimited bandwidth resources traveled by any bit is Θ(1), and therefore the “bit- (UWB) albeit with power constraints [14], [15]. meters/sec” and “bits/sec” capacity is of the same order. Li et al. [16] have used simulations to evaluate the ca- We assume that n nodes can be located anywhere pacity of multi-channel networks based on IEEE 802.11. on the surface of a torus of unit area, as in [9]. The Other research on capacity is based on considerations of assumption of a torus enables us to avoid technicalities alternate communication models [17]–[19]. arising out of edge effects, but the results are applicable Several researchers have proposed wireless protocols for nodes located on a unit square as well. We first for multi-channel networks (cf. [4]–[6], [20]–[22]). Some establish an upper bound on the network capacity of solutions are based on using a single interface at each arbitrary networks, and then construct a network to prove node [5], [21], [23], [24], while other solutions require the that bound is tight. a dedicated interface for each channel [20], [22]. More A. Upper bound on capacity recently, solutions have been proposed that require mul- tiple interfaces, but fewer interfaces than the number The capacity of multi-channel arbitrary networks of channels [4], [6], [25]. Although, there are several is limited by two constraints (described below), and proposals for multi-channel networks, it is not clear each of them is used to obtain a bound on the network how many interfaces are actually required to maximally capacity. The minimum of the two bounds (the bounds utilize the available channels. depend on ratio between the number of channels c and the number of interfaces m) is an upper bound on the III. C APACITY RESULTS FOR ARBITRARY NETWORKS network capacity. We derive the bounds under channel We model the impact of interference by using the model 1, and state the results under channel model 2 as protocol model proposed by Gupta and Kumar [1]. The well1 . transmission from a node i to a node j on some channel x is successful, if for every other node k simultaneously Constraint 1 – Interference constraint: The capacity transmitting on channel x, the following condition holds of arbitrary networks is constrained by interference. Using the proof techniques presented in [1] with d(k, j) ≥ (1 + ∆)d(i, j), ∆ > 0 some modifications to account for multiple interfaces where d(i, j) is the distance between nodes i and j , and and channels, a bound on the network capacity is ∆ is a “guard” parameter that ensures that concurrently O W nm bit-meters/sec. The detailed proof is in c transmitting nodes are sufficiently farther away from the Appendix I. receiver to prevent excessive interference. It is shown in [1] that the protocol model is equivalent 1 Recall that the results under channel model 2 can be obtained by to an alternate physical model that is based on received replacing W with W c in the results derived under channel model 1.
  5. 5. 5 1 Constraint 2 – Interface constraint: The capacity of Group 1 m arbitrary networks is also constrained by the maximum Mapping number of bits that can be transmitted simultaneously (˜ − 1)m + 1 c over all interfaces in the network. Since each node has Group c ˜ c = cm ˜ m interfaces, there are a total of mn interfaces in the (m, c)-network. Each interface can transmit at a rate Individual Channels Channel groups of W bits/sec. Also, the maximum distance a bit can c ˜ Fig. 3. Lemma 1 construction: Forming c channel groups, with m travel in the network is O(1) meters. Hence, the total channels per group, in a (m, c)-network network capacity is at most O W nm bit-meters/sec. c c This bound is tight when m is Ω(n). channel group i, 1 ≤ i ≤ c, contains all channels j ˜ Combining the two constraints, the network capacity such that (i − 1)m + 1 ≤ j ≤ im. Assume that time on is O MINO W nm , W nmc c bit-meters/sec, under the channels is divided into slots of duration τ . Consider channel model 1. Therefore, we have the following any slot s. Suppose a node X in the (1, c)-network has ˜ theorem on the network capacity of arbitrary networks its interface on some channel i, 1 ≤ i ≤ c, in slot ˜ (Figure 1 has a pictorial representation). s. We simulate this behavior in the (m, c)-network by assigning the m interfaces of X in the slot s to the Theorem 1: The upper bound on the capacity of a m channels in the channel group i. In this fashion, in (m, c)-arbitrary network is as follows: any slot, the m interfaces of any node in the (m, c)- c 1) When m is O(n), network capacity is network are mapped to a channel group. The aggregate O W nm bit-meters/sec under channel data rate of each channel group is W m/c = W/˜ (since c c √ model 1 and O (W nmc) bit-meters/sec under c = m˜). Therefore, a channel group in the (m, c)- c channel model 2. network can support the same data rate as a channel 2) When m is Ω(n), network capacity is O W nm c in the (1, c)-network. This mapping allows the (m, c)- ˜ c bit-meters/sec under channel model 1 and network to mimic the behavior of (1, c)-network; the ˜ O (W nm) bit-meters/sec under channel model 2. W τ /˜ bits sent on some channel in any time slot s in c the (1, c)-network can be simulated by sending W τ /c ˜ √ bits (in the same slot s) on each of the m channels in The network capacity of a (c, c)-network is O (W n) bit-meters/sec under channel model 1, which was the the corresponding channel group of the (m, c)-network. result obtained by Gupta and Kumar [1]. When fewer Hence, a (m, c)-network can support the capacity of a interfaces are available, there is a capacity degradation (1, c) network, when c = m˜. ˜ c by at least a factor of 1 − m . Intuitively, the capacity c degradation arises because the total bits that can be Lemma 2: Suppose m and c are positive integers. 1 simultaneously transmitted decreases. Then, a (m, c)-network can support at least 2 the ca- c pacity supported by a 1, m -network. c c B. Constructive lower bound Proof: Suppose m = m . Then the result directly follows from the previous lemma. Otherwise, m < c, In this section, we construct a network to establish a c and we use c = m m of the channels in the (m, c)- lower bound on the network capacity. The lower bound network, and ignore the rest of the channels. This can matches the upper bound, implying that the bounds be viewed as a (m, c )-network, with a total data rate of are tight. We first establish two results that we use in W = W m m (as each channel supports W bits/sec). c c c the rest of the paper. The results are proved under the Using Lemma 1, a (m, c )-network with total data rate channel model 1, but hold for channel model 2 as well. c of W can support at least the capacity of a 1, m - network with total data rate of W . However, when Lemma 1: Suppose m, c, c are positive integers and ˜ c W < W , the (m, c )-network with total data rate W can c = m . Then, a (m, c)-network can support at least the ˜ achieve only a fraction W of the capacity of a 1, m - W c capacity supported by a (1, c)-network. ˜ network with total data rate W (instead of W ). Now, Proof: Consider a (m, c)-network. We group the c channels into c groups (numbered from 1 to c), with m ˜ ˜ W m c channels per group as shown in Figure 3. Specifically, = W c m
  6. 6. 6 c l = 2(1 + 2∆)r m = c m c m c c R4 ≥ c , since ≤ +1 m +1 m m 1 c S4 ≥ , since ≥1 2 m R1 S1 S2 R2 1 Hence, a (m, c)-network can support at least 2 the r∆ r r∆ r(1 + ∆) c capacity supported by a 1, m network. This implies S3 that asymptotically, a (m, c)-network has the same order c of capacity as a 1, m -network. R3 We now provide the following construction to establish that a capacity of Ω MINO W nm , W nm c c Fig. 4. The placement of nodes within a cell bit-meters/sec is achievable in a (1, c)-network under the channel model 1. The result is then extended to a (m, c)-network by using Lemma 2. senders are at least r(1 + ∆) distance from the cell boundary. Therefore, senders in adjacent cells of B are Step 1: We consider a torus of unit area. Let at least a distance of r(1 + ∆) away from B as well. k = min c, n . This implies that k ≤ c. Partition 8 Hence, under the protocol model of interference, the n the square area into 8k equal-sized square cells, and transmission between A and B is not interfered with by place 8k nodes in each cell. Since the total area is 1, any other transmission in the network, and this property each cell has an area of 8k , and sides of length l = 8k . n n holds for all communicating pairs. Step 2: The 8k nodes within each cell are distributed From the above construction, there are n pairs of 2 by placing k nodes at each of the eight positions shown nodes in the (1, c)-network, each transmitting at a rate in Figure 4. Nodes placed at locations S1, S2, S3, S4 act of W over a distance r = (1+2∆) 2k . Hence, the total c 1 n as senders, and nodes placed at remaining locations act capacity of the network (summing over all n nodes) is as receivers. The sender locations S1 through S4 are at nW W 1 nk 2 c r = c (1+2∆) 2 bit-meters/sec. Recall that k = a distance of r∆ from the center of the cell (recall that n min c, 8 . Substituting for k , we obtain the capacity ∆ is the “guard” parameter from the protocol model of of a (1, c)-network to be Ω MINO W n , W n c c bit- interference), where r = 2(1+2∆) = (1+2∆) 2k . The l 1 n meters/sec under channel model 1, since ∆ is a constant. receiver locations R1 through R4 are at a distance of Using Lemma 2, the capacity of a (m, c)-network un- r(1 + ∆) from the center of the cell. Therefore, the n n distance between S1-R1, S2-R2, S3-R3, and S4-R4 is der channel model 1 is Ω MINO W c ,W c m m equal to r . Each receiver location is at a distance of r∆ 1 1 bit-meters/sec. Since c ≥ c , we have the capacity to m m from nearest edge of the cell, and each sender location is be Ω MINO W mn , W mnc c bit-meters/sec, which at a distance of r(1+∆) from the nearest edge of the cell. leads to the following theorem: Step 3: Label the k nodes in any location (S1 through Theorem 2: The achievable network capacity of a S4, R1 through R4) as 1 through k . The j th node (m, c)-arbitrary network is as follows: in each sender location, 1 ≤ j ≤ k , communicates c with the j th node in the nearest receiver location (at 1) When m is O(n), network capacity is a distance of r ) on channel j . Consider any pair of Ω W nm c bit-meters/sec under channel √ communicating nodes A and B that are located at, model 1 and Ω (W nmc) bit-meters/sec under say, S1 and R1 respectively. Then, the nearest senders channel model 2. within the cell, other than A (located at S1), which are 2) When m is Ω(n), network capacity is Ω W nm c c sending on the same channel as A are located at one bit-meters/sec under channel model 1 and of S2, S3, S4, and are at least a distance of r(1 + ∆) Ω (W nm) bit-meters/sec under channel model 2. away from B (located at R1). Similarly, in every cell,
  7. 7. 7 The upper bound (Theorem 1) and lower bound (The- Note that each node picks a destination node orem 2) on the order of the capacity of arbitrary networks randomly, and so a node may be the destination of match, indicating the bounds are tight. multiple flows. Let D(n) be the maximum number of flows for which a node in the network is a destination. C. Implications We use the following result to bound D(n). A common scenario of operation is when the number c of channels is not too large ( m = O(n)). Under this Lemma 3: The maximum number of flows for which scenario, the capacity of a (m, c)-network in the arbitrary a node in the network is a destination, D(n), is setting scales as Θ W nm under channel model 1. c log n Θ log log n , with high probability. Similarly, under channel model 2, the capacity of the Proof: The process of nodes selecting a random √ network scales as Θ (W nmc). Under either model, destination may be mapped to the well-known “Balls into the capacity of a (m, c)-network goes down by a factor Bins” problem [26]. Each source node may be viewed of 1 − m , when compared with a (c, c)-network. c as a “ball”, and each destination node may be viewed as Therefore, doubling the number of interfaces at each a “bin”. The process of selecting a destination node may node (as long as number of interfaces is smaller than be viewed as randomly dropping a “ball” into a “bin”. the number of channels) increases the channel capacity √ Based on this mapping, the proof of the lemma follows by a factor of only 2. Further, the ratio between m and from well-known results (cf. [26], Section 4). c decides the capacity, rather than the individual values of m and c. Increasing the number of interfaces may A. Upper bound result in a linear increase in the cost but only a sub-linear The capacity of multi-channel random networks (proportional to square-root of number of interfaces) is limited by three constraints, and each of them is increase in the capacity. Therefore, the optimal number used to obtain a bound on the network capacity. The of interfaces to use may be smaller than the number of minimum of the three bounds (the bounds depend on channels depending on the relationship between cost of ratio between the number of channels c and the number interfaces and utility obtained by higher capacity. of interfaces m) is an upper bound on the network Different network architectures have been proposed capacity. We derive the bounds under channel model 1, for utilizing multiple channels when the number of avail- and state the results under channel model 2. able interfaces is smaller than the number of available channels [4], [6], [22]. The construction used in proving Constraint 1 – Connectivity constraint: The capacity lower bound implies that maximal capacity is achieved of random networks is constrained by the need to when all channels are utilized. One architecture used in ensure the network is connected, so that every source- the past [22] is to use only m channels when m interfaces destination pair can successfully communicate. Gupta are available, leading to wastage of the remaining c − m and Kumar [1] have presented a bound on the network channels. That architecture results in a factor of 1 − m c capacity of O W n bits/sec based on this log n loss in capacity which can be significantly higher than requirement. This bound is applicable to multi-channel the optimal 1 − m loss (when m = O(n)). Hence, c c networks as well. in general, higher capacity may be achievable by archi- tectures that use all channels, possibly by dynamically Constraint 2 – Interference constraint: The capacity switching channels. of multi-channel random networks is also constrained IV. C APACITY RESULTS FOR RANDOM NETWORKS by interference (this is same as constraint 1 listed for We assume that n nodes are randomly located on arbitrary networks in Section III-A). This constraint the surface of a torus of unit area. Each node selects was already captured in the upper bound for arbitrary a destination randomly to which it sends λ(n) bits/sec. networks, and we had obtained a bound of O W nm c The highest value of λ(n) which can be supported by bit-meters/sec. In a random network, each of the n every source-destination pair with high probability is source-destination pairs are separated by an average defined as the per-node throughput of the network. The distance of Θ(1) meter. Consequently, the network traffic between a source-destination pair is referred to as capacity of random networks is at most O W nm c a “flow”. Since there are a total of n flows, the network bits/sec. capacity is defined to be nλ(n).
  8. 8. 8 Constraint 3 – Destination bottleneck constraint: The network capacity is O W nm bits/sec under √ c capacity of a multi-channel network is constrained by the channel model 1 and O (W nmc) bits/sec under data that can be received by a destination node. Consider channel model 2. 2 a node X which is the destination of the maximum 3) When m is Ω n log log n c , network capacity log n number (that is, D(n)) of flows. Recall that in a (m, c)- W mn log log n network, each channel supports a data rate of W bits/sec. is O c log n bits/sec under channel model c Therefore, the total data rate at which X can receive data W mn log log n 1 and O log n bits/sec under channel over m interfaces is Wcm bits/sec. Since X has D(n) model 2. incoming flows, the data rate of the minimum rate flow is Wm at most cD(n) bits/sec. Therefore, by definition of λ(n), The interesting observation from this theorem is that Wm c λ(n) ≤ cD(n) , implying that network capacity (which as long as m is O(log n), the number of interfaces has no W mn by definition is nλ(n)) is at most O cD(n) bits/sec. impact on channel capacity. This implies that when the Substituting for D(n) from Lemma 3, the network number of channels is O(log n) (which is the common log log case today), there is no loss in network capacity even if capacity is at most O W mnlog n n bits/sec. c each node has a single interface. The bound obtained from constraint 3 is applicable to any network, including mobile networks, as long B. Lower bound as the destination of every flow is randomly chosen The lower bound is established by constructing a among the nodes in the network. Even when m = c, routing scheme and a transmission schedule for any this bound implies that the per-flow throughput, λ(n), random network. The lower bound matches the upper is at most O W log log n bits/sec. Previous results on log n bound implying that the bounds are tight. We will capacity of mobile networks [2], [9], [27] have stated provide a construction for a (1, c)-network (a network a per-flow throughput of O(W ) bits/sec is possible. In wherein each node has a single interface) under channel our work, we choose the destination of a flow randomly model 1, and then invoke Lemma 2 to extend the from among n − 1 possible destinations, similar to result to a (m, c)-network. The steps involved in the Gupta and Kumar [1]. Considering our discussion construction are described next. above, the O(W ) bits/sec bound with mobility cannot apply when destination nodes are randomly chosen. The Cell construction previous results for mobile networks may hold under The surface of the unit torus is divided using a square other models of selecting destination nodes, wherein grid into square cells (see Figure 5), each of area a(n), each node is the destination of at most O(1) flows (for similar to the approach used in [9]. The size of the cell, example, such a constraint is satisfied when permutation a(n), has to be carefully chosen to meet multiple con- routing is used). straints (which are described later in the text). In particu- 100 log n c 2 1 lar, we set a(n) = min max n ,n , D(n) , Combining the three bounds, the network capacity is log n at most O MINO W n nm W mn log log n where D(n) = Θ log log n as described before. Intu- log n , W c , c log n bits/sec under channel model 1. From this, we itively, the three values that influence a(n) are based have the following theorem on the upper bound on on the three constraints that were described in the upper capacity of random networks (Figure 2 has a pictorial bound proof: cell size needed to ensure connectivity, cell representation). size needed when capacity is constrained by interference, and cell size needed when capacity is constrained by the maximum number of flows to any destination node, Theorem 3: The upper bound on the capacity of a respectively. (m, c)-random network is as follows: c We need to bound the number of nodes that are 1) When m is O(log n), network capacity is present in each cell. We state the bound here, and n O W log n bits/sec under channel model 1 and present a proof of the bound in Appendix II. n O Wc log n bits/sec under channel model 2. Lemma 4: If a(n) > 50 log n , then each cell has n c log log n 2 Θ (na(n)) nodes per cell, with high probability. 2) When m is Ω(log n) and also O n log n ,
  9. 9. 9 1/a(n) cells each of area a(n) A node in each cell through which the line passes is S used to relay traffic along that flow (we will describe the choice of the node later). Figure 5 shows an example of the cells used to route data for a flow between source S and destination D . In previously proposed constructions for proving lower bound on capacity [1], [9], it was immaterial D which node in a chosen cell forwarded packets for some flow. However, such an approach may “overload” certain Fig. 5. Routing through cells: Packets are routed through the cells nodes, leading to capacity degradation, when the number intersected by the line joining the source and the destination of interfaces per node is smaller than the number of channels. Consequently, it is important to ensure that the routing load is distributed among the nodes in a cell. This is a key extension to the routing procedure used in By construction, we ensure that a(n) ≥ 100 n n for log earlier capacity results [1]. large n (as max 100 n n , n is at least 100 n n , and log c log For each flow passing through a cell, one node in the 2 1 D(n) is asymptotically larger than 100 n n ). Thus, log cell is “assigned” to the flow. The assigned node of a with our choice of a(n), Lemma 4 holds for suitably flow in a cell is the only node in that cell which may large n, and each cell has Θ (na(n)) nodes per cell, whp. receive/transmit data along that flow. The assignment is The transmission range2 of each node, r(n), is set done using a flow distribution procedure as below: to be 8a(n). With this transmission range, a node in Step 1 – Assign source and destination nodes: For one cell can communicate with any node in its eight any flow that originates in a cell, the source node S is neighboring cells. Note that when the cell size a(n) assigned to the flow (S is necessarily in the originating increases, larger transmission range is required, as r(n) cell). Similarly, for any flow that terminates in a cell, the is dependent on a(n). destination node D is assigned to the flow. Since a single A transmission originating from a node S interferes node in each cell is allowed to receive or transmit data with another transmission from A to B, only if S is for a flow, it is required that the source and destination within a distance of (1 + ∆)r(n) of receiver B (using nodes be assigned to flows originating or terminating the interference definition of protocol model). Since the from them. distance between A to B is at most r(n), the distance Step 2 – Balance distribution of remaining flows: After between the two transmitters, S and A, must be less step 1 is complete, we are left with only those flows that than (2 + ∆)r(n) if the transmissions were to interfere. pass through a cell. Each such remaining flow passing Thus, any transmission can possibly interfere with through a cell is assigned to the node in the cell that only those transmissions from transmitters within a has the least number of flows assigned to it so far. This distance of (2 + ∆)r(n). Therefore, nodes in a cell step balances the assignment of flows to ensure that all can be interfered with by only nodes in cells within nodes are assigned (nearly) the same number of flows. a distance of (2 + ∆)r(n), and this interfering area The node assigned to a flow will receive packets from can be completely enclosed in a larger square of side some node in the previous cell and send the packet to a 3(2 + ∆)r(n) (this is a loose bound). Consequently, node in the next cell. 2 there are at most (3(2+∆)r(n)) = 72(2 + ∆)2 interfering Each node is the originator of one flow. Each node a(n) is the destination of at most D(n) flows, which by cells (recall r(n) = 8a(n)). Hence, the number of log n Lemma 3 is Θ log log n . Therefore, step 1 of the flow interfering cells, kinter ≤ 72(2 + ∆)2 , is a constant that only depends on ∆ (and is independent of a(n) and n). distribution procedure assigns to each node at most 1 + D(n) flows. We use the following result to bound the number of Routing Scheme source-destination lines that pass through any cell; we Packets are routed through the cells that lie along the omit the proof as it has already been presented earlier straight line joining the source and the destination node. in [9]. 2 Transmission range is defined to be the maximum distance over which any node can communicate. Lemma 5: The maximum number of source-
  10. 10. 10 destination lines that intersect any cell (including lines such that any scheduled transmission is interference- originating and terminating in the cell) is O n a(n) , free. By using the two-step process, each transmission with high probability. in a mini-slot satisfies both constraint 1 and constraint 2. Step 2 of the flow distribution procedure carefully Step 1 – Build a routing graph: We build a graph, assigns the remaining flows among the nodes in the called the “routing graph”, whose vertices are the nodes cell to ensure all nodes end up with nearly same in the network. One edge is inserted between all node number of flows. By Lemma 4, each cell has Θ (na(n)) pairs, say A and B , for every flow on which A and B nodes, and by Lemma 5 at most O n a(n) flows are consecutive nodes (the routing scheme for selecting pass through a cell. Therefore, step 2 will assign to nodes along a flow was described earlier). Therefore, by this construction, every hop3 in the network along any any node in the network at most O √ 1 flows. a(n) flow is associated with one edge in the routing graph. Therefore the total flows assigned to any node is at The resulting routing graph is a multi-graph4 in which most O 1 + D(n) + √ 1 . When choosing the size each node has at most O √ 1 edges, since each a(n) a(n) of a(n) earlier, the maximum value of a(n) was at flow through a node can result in at most two edges, 2 most D(n) , which implies √ 1 1 is at least D(n). one incoming and one outgoing, and we have already a(n) Hence, the total flows assigned to any node is always shown that each node is assigned to at most O √ 1 a(n) asymptotically dominated by √ 1 , and is therefore flows. It is a well-known result [28] that a multi-graph a(n) with at most e edges per vertex can be edge-colored 5 equal to O √1 flows. a(n) with at most 3e colors. Therefore, the routing graph can 2 Scheduling transmissions be edge colored with at most O √ 1 colors. a(n) The transmission scheduling scheme is responsible for We use edge coloring to ensure that when a generating a transmission schedule for each node in the transmission is scheduled along a edge, the interfaces (1, c)-network that satisfies the following constraints: on the nodes at either end of the edge are free, thereby Constraint 1: When a node X transmits a packet to satisfying constraint 1. We divide every 1 second period a node Y over a channel j for some flow, X and Y into O √ 1 “edge-color” slots each of length a(n) should not be scheduled to transmit/receive at the same time for any other flow (since each node is assumed to Ω a(n) seconds, and each of these edge-color slots have a single interface). is associated with an unique edge color. An edge is Constraint 2: Any two simultaneous transmissions on scheduled for transmission in the slot associated with its any channel should not interfere. edge color. Since edge coloring ensures that at a vertex, The multi-channel construction differs from the mech- all edges connected to the vertex use different colors, anisms used in earlier constructions [1], [9] in two ways. each node will have at most one transmission/reception First, the scheduling is on a per-node basis since flows scheduled in any edge-color slot. By construction, each are distributed among nodes, whereas in the past work it edge corresponds to a hop in the network. Therefore was sufficient to schedule on a per-cell basis. Second, this scheme ensures that during every 1 second interval, since there is a single interface, but c channels are along any flow in the network, one transmission is available (recall that we are assuming a (1, c)-network scheduled on each hop of a flow. for now), the schedule has to additionally ensure that a single transmission/reception is scheduled for a node at Step 2 – Build an interference graph: In step 2, each any time (constraint 1 above). edge-color slot is further sub-divided into “mini-slots” We build a suitable schedule using a two-step process. as explained below, and every node has an opportunity In the first step, we satisfy constraint 1 by scheduling to transmit in some mini-slot. We develop a schedule transmissions in “edge-color” slots so that at every node for using mini-slots, which satisfies constraint 2. The during any edge-color slot, at most one transmission 3 A hop is a pair of consecutive nodes on a flow. or reception is scheduled. In the second step, we 4 A graph with possibly multiple edges between a pair of nodes. satisfy constraint 2 by dividing each edge-color slot 5 Edge-coloring requires any two edges incident on a common into “mini-slots”, and assigning mini-slots to channels vertex to use different colors.
  11. 11. 11 One Second schedule decides on which mini-slot within an edge- Edge−color slot color slot and on what channel a node may transmit, and the same schedule is used in every edge-color slot. 1 We build another graph, called the “interference 2 graph”, wherein, vertices are nodes in the network, and c−1 there is an edge between two nodes if they may interfere with each other. Since every cell has at most some c constant kinter number of cells that may interfere with each other, and each cell has Θ (na(n)) nodes, each Mini−slot node has at most O (na(n)) edges in the interference graph. It is well-known that a graph with maximum Fig. 6. Transmission schedule degree e can be vertex-colored6 with at most e + 1 colors [28]. Therefore, the graph can be vertex-colored with O (na(n)) colors, i.e., at most k1 na(n) colors for can be transported. Since k1 na(n) ≤ k1 na(n) + 1, √ c c some constant k1 . Transmissions of two nodes assigned W a(n) the same vertex-color do not interfere with each other. we have, λ(n) = Ω k1 na(n)+c bits/sec. Depend- Hence, they can be scheduled to transmit on the same ing on the asymptotic order of c, either na(n) or channel at the same time. On the other hand, nodes c will dominate the denominator of λ(n). Hence, √ colored with different colors may interfere with each √W W a(n) λ(n) = Ω MINO , c bits/sec. Since other, and need to be scheduled either on different n a(n) channels, or at different time slots on the same channel. each flow is scheduled to receive one mini-slot on We divide each edge-color slot into k1 na(n) each hop during every 1 second interval, every source- c destination flow can support a per-node throughput of mini-slots on every channel, and number the slots on λ(n) bits/sec, and therefore, the network capacity is √ each channel from 1 to k1 na(n) . There is a total of c W n a(n) nλ(n) = Ω MINO √W , bits/sec. c k1 na(n) mini-slots across the c channels. Channels c c a(n) are numbered from 1 to c. A node which is allocated Recall that we choose a(n) to be 100 log n c 2 a color p, 1 ≤ p ≤ k1 na(n) is allowed to transmit in 1 min max n ,n , D(n) , where mini-slot p on channel (p mod c) + 1. The node c log n may actually transmit if the edge-coloring has allocated D(n) = Θ log log n . Substituting for the three an outgoing edge from the node to the corresponding values, and then applying Lemma 2 to extend the results edge-color slot. to an (m, c)-network, we have the following theorem. Figure 6 depicts a schedule of transmissions on the Theorem 4: The achievable capacity of a (m, c)- network developed after the two-step scheduling process. random network is as follows: The first step allocates one edge-color slot for each c log n 1) When m is O(log n), a(n) = Θ n , and the hop of every flow. The second step decides within each n edge-color slot when the transmitter node on a hop may network capacity is Ω W log n bits/sec under actually transmit a packet. n channel model 1 and Ω W c log n bits/sec un- As seen in step 1, each edge-color slot is of length der channel model 2. Ω a(n) seconds. As seen in step 2, each edge-color c log log n 2 k1 na(n) 2) When m is Ω(log n) and also O n log n , slot is sub-divided into c mini-slots. Therefore, √ a(n) = Θ c , and the network capacity is a(n) mn each mini-slot is of length Ω k1 na(n) seconds. Each Ω W nm bits/sec under channel model 1 and c W √ c channel can transmit at the rate of c √bits/second. Ω (W nmc) bits/sec under channel model 2. W a(n) c log log n 2 Hence, in each mini-slot, λ(n) = Ω k1 na(n) bits 3) When m is Ω n log n , a(n) = c c log log n 2 Θ log n , and the network capacity 6 Vertex-coloring requires any two vertices sharing a common edge W mn log log n to use different colors. is Ω c log n bits/sec under channel model
  12. 12. 12 W mn log log n 1 and Ω log n bits/sec under channel number of nodes in the neighborhood7 of a node. model 2. When the number of channels is large (specifically, ω(log n)) and each node has a single interface, there The lower bound matches the upper bound (Theorem is a capacity loss when compared to a single channel 3) implying that the bounds are tight. Recall that the network. This capacity loss arises because the number transmission range r(n) has been set to 8a(n). Hence, of channels is more than the number of interfaces in a the transmission range increases in case 2 and case “neighborhood”. The lower bound construction suggests 3 of Theorem 4 as compared to case 1 (since a(n) that the cell size should be chosen such that the number increases). This implies that in multi-channel networks of nodes in each neighborhood is equal to the number with large number of channels, higher transmission of channels. Thus, an optimal strategy for maximiz- power is necessary for meeting capacity bounds. ing capacity when number of channels is large is to sufficiently increase the cell size a(n), which implies a larger transmission range r(n) is needed to allow communication with neighboring cells. However, there C. Implications is still some capacity loss because larger transmission range (than that is needed for connectivity alone) lowers The above result implies that the capacity of multi- capacity by “consuming” more area. channel random networks with total channel data rate of W is the same as that of a single channel network with D. Optimal routing and transmission scheduling ap- c data rate W as long as the ratio m is O(log n). When proaches the number of nodes n in the network increases, we can The construction used in demonstrating that the lower also scale the number of channels (for example, by using bound is achievable can be used to develop optimal rout- additional bandwidth, or by dividing available bandwidth ing and transmission scheduling approaches. The lower into multiple sub-channels). Even then, as long as the bound construction suggests that load balancing (i.e., channels are scaled at a rate not more than log n, there is distributing flows) among nodes in a given neighborhood no loss in capacity even if a single interface is available is essential for full utilization of multiple channels. In at each node. In particular, if the number of channels a single channel network, load balancing is sometimes c is a fixed constant, independent of the node density, used to balance energy consumption across nodes, or to then as the node density increases beyond some threshold improve resilience of the network. However, load bal- density (at which point c ≤ log n), there is no loss in ancing in the same neighborhood is not always required capacity even if just a single interface is available per in single channel networks for maximizing capacity. node. Thus, this result may be used to roughly estimate For example, consider a simple scenario with two the number of interfaces each node has to be equipped flows from A to B and C to D as shown in Figure 7. with for a given node density and a given number of The flows pass through a cell with two nodes E and F. channels. Assume that node E is being used to forward data for In a single channel random network, i.e., a (1, 1)- flow A-B. In a single channel network with channel rate network, the capacity bottleneck arises out of the channel W , the per-flow throughput is the same whether node E becoming fully utilized, and not because interface at any or node F is chosen to forward data along flow C-D. In node is fully utilized. On an average, the interface of a particular, the per-flow throughput is W as the channel 1 4 node in a single channel network is busy only for X rate is split between receiving and sending data at the in- fraction of the time, where X is the average number termediate nodes. This is because E and F interfere with of nodes that interfere with a given node. In a (1, 1)- each other, and therefore cannot simultaneously transmit. random network with n nodes, each node on an average Now, consider the scenario wherein two channels of data has Θ(log n) neighbors to maintain connectivity [1]. This rate W are available, but each node has a single interface. 2 implies that in a single channel network, each interface is In this scenario, if node E is chosen to forward data along 1 busy for only Θ log n time. Intuitively, our construction both flows A-B and C-D, the interface on node E can above utilizes this slack time of interfaces to support up transmit/receive at most W bits/sec, leading to a lower 2 to O(log n) channels without loss in capacity. In general, there is no loss in capacity in a random network as long 7 The neighborhood of a node consists of all other nodes that may as the number of channels is smaller than the average interfere with it.
  13. 13. 13 C V. I MPACT OF SWITCHING DELAY The previous discussion on multi-channel capacity has E not considered the impact of interface switching delay. When the number of interfaces at each node is smaller A F B than the number of channels, interfaces may have to be switched between channels. Switching an interface D from one channel to another may incur a switching delay, say S , as the interface hardware has to change Fig. 7. Need for balancing load among nodes in a neighborhood the frequency of operation. For example, existing IEEE 802.11-based wireless interfaces require [4] between few tens to hundreds of microseconds to switch from one channel to another. Switching delay is, however, per-flow throughput of W . Instead, if node F is chosen 8 independent of the number of nodes in the network. to forward data along C-D, and links A-E and E-B use In the case of arbitrary networks, capacity bounds one channel, while C-F and F-D use the other channel, are met without requiring interface switching at all (as a higher per-flow throughput of W can be achieved. 4 was shown in the construction used for lower bound). This example highlights the need for distributing flows Hence, switching delay will not impact the capacity of (“load”). The routing protocol should therefore explicitly arbitrary networks. Even in the case of random networks, try to balance load among nodes in every neighborhood, the upper bound proofs do not mandate interfaces to be and select routes with lower load. switched, and the bounds may be tight with switching In the transmission scheduling scheme used for lower delay as well8 . Hence, it appears that the tight capacity bound construction, it suffices for a node to always upper bound in random networks is independent of the transmit on a specific channel without requiring to switch interface switching delay. channels for different packets (recall that the same mini- In the construction used to prove lower bound in slot on a specific channel is used by a node in all random networks, interfaces may have to be switched “edge-color” slots). However, a node may have to switch between channels (when receiving data). In the worst channels for receiving data. An alternate construction case, an interface may have to be switched between is to use a scheduling scheme which ensures that a channels for every packet transmission. Each packet of node receives all data on a specific channel, but may size L bits requires a transmission time of T = Lc W have to switch channels when sending data. It can be seconds (since channel data rate is W bits/sec). One way c shown that the alternate construction is equivalent to of hiding the interface switching delay S is to insert a the lower bound construction by modifying the mini-slot “guard” slot of duration S between two “edge-color” assignment to be done on a per-receiver basis instead of slots (we call this the “guard slot” approach) to ensure a per-sender basis. This intuition can be used to develop there is sufficient time for interface switching. However, a practical scheme that uses two interfaces per node. this approach degrades the network capacity by a factor One interface can be used for receiving data and is S of T +S . The capacity reduction can be minimized by always fixed to a single channel. The second interface sending extremely large packets (L λ) resulting in can be used for sending data and is switched between T S . However, this significantly increases end-to-end channels, as necessary. Existing multi-channel protocols latency. Hence, although switching delay will not in the have often required tight synchronization among nodes. asymptotic sense affect the scaling properties of network The use of two interfaces, with a dedicated interface on capacity, if there is an end-to-end delay constraint, then a fixed channel obviates the need for tight synchroniza- switching delay may reduce the achievable network tion as a node receives data on a well-known channel. capacity by a factor proportional to the magnitude of the Furthermore, using a fixed channel for reception does switching delay. We will next define the end-to-end delay not degrade capacity since it is based on the (optimal) constraint, and describe an alternate approach that does alternate construction. not lose capacity while meeting the end-to-end delay We have already used some of the insights gained from this work to develop routing and channel assignment 8 This is a conjecture that we have not yet proved. However, there protocols [6], [25] that are well-suited for multi-channel is no change in the lower bound with extra interfaces, and therefore, networks. we speculate that the conjecture may be true.
  14. 14. 14 (v−1) slots constraint. We use the end-to-end delay definition from [9]. Each 1 T S T packet is assumed to have a size L, and L is scaled with 2 respect to the throughput obtained for each end-to-end (v−1) slots flow. If each flow can transport λ bits/sec, then each flow is assumed to send packets of size L = λ. By the v−1 v lower bound construction provided before, if packet sizes are set to λ bits, each packet traverses one hop in one v slots second. Therefore, the end-to-end delay of a flow will be equal to the number of hops on the flow, when there Fig. 8. Constructing a virtual interface with zero switching delay is no interface switching latency. Let us assume that the minimum end-to-end delay in the absence of interface switching latency is Dopt . A reasonable delay constraint switching delay is provisioned for. By this construction, in the presence of switching latency is to require that the the simulated virtual interface can continuously trans- end-to-end delay is at most a small constant multiple of mit/receive, with 0 switching delay. Therefore, a network Dopt ; otherwise applications may see a large increase in the end-to-end delay. using v interfaces having switching delay S , can mimic We now describe an alternate approach that can the behavior of a (1, c)-network built with interfaces completely hide the impact of switching delay by having switching delay 0. deploying multiple interfaces. This approach assumes that each of the given interfaces has a switching delay From the previous lemma, by increasing the number S , and uses the interfaces to simulate a virtual interface of interfaces at each node by a factor of v , switching having 0 switching delay. By this construction, the use delay is completely hidden. Suppose, each node cannot S of additional interfaces per node can hide the switching be provisioned with v = T +1 interfaces. Then, if each delay, leading to the same capacity and end-to-end node has at least 2 interfaces, by using larger packets, delay bounds derived earlier assuming no interface we can still achieve the same capacity as a network switching delay. built with interfaces having no switching delay. Suppose, p interfaces are actually available, with 2 ≤ p < v . Lemma 6: Suppose that the time required for packet Then we increase the packet size by a factor of v p transmission in a (1, c)-network is T = Lc . Then to increase packet transmission time T . This hides the W a S + 1, c -network built with interfaces having interface switching delay, and results in no capacity loss. T switching delay S , can achieve the same capacity and However, larger packet size increases end-to-end delay end-to-end delay as a (1, c)-network built with interfaces by a constant factor of v . Note that the previously p having 0 switching delay. discussed “guard slot” approach may increase the end-to- Proof: Let us assume that each node has v = T + S end delay by an arbitrarily large factor, or lose capacity 1 interfaces, each having a switching delay S . We build when at most a constant increase in the end-to-end delay a virtual interface with zero switching delay by using the can be tolerated. Therefore, by our construction, as long v physical interfaces, as shown in Figure 8. We consider as each node has at least two interfaces, there is no loss any time interval of length vT . We divide this time into in network capacity, although there may be a increase in v slots of length T , and only allow the ith interface, the end-to-end latency by a constant factor independent 1 ≤ i ≤ v , to transmit/receive in slot i. Thus, each of n. It is an open question if switching delay can be physical interface is used for transmission/reception in successfully hidden with only one interface per node. one slot, and is idle for the next (v − 1) slots of total S VI. C ONCLUSIONS duration (v − 1)T seconds. Since v = T + 1, we have: In this paper, we have derived the lower and upper S bounds on the capacity of static multi-channel wireless (v − 1)T = T T networks. We have shown that in an arbitrary network, ≥ S there is a loss in network capacity when the number Hence, between two successive operations of a physical of interfaces per node is smaller than the number of interface there is at least a gap of S , which ensures that channels. However, we have shown that surprisingly, in a