Control of wireless networks with flow level dynamics under constant time scheduling

Abstract

We consider a network control problem for wireless networks with flow level dynamics under the general k-hop interference model. In particular, we investigate the control problem in low load and high load regimes. In the low load regime, we show that the network can be stabilized by a regulated maximal scheduling policy considering flow level dynamics if the offered load satisfies a constraining bound condition. Because maximal scheduling is a general scheduling rule whose implementation is not specified, we propose a constant-time and distributed scheduling algorithm for a general k-hop interference model which can approximate the maximal scheduling policy within an arbitrarily small error. Under the stability condition, we show how to calculate transmission rates for different user classes such that the long-term (time average) network utility is maximized. This long-term network utility captures the real network performance due to the fact that under flow level dynamics, the number of users randomly change so instantaneous network utility maximization does not result in useful network performance. Our results imply that congestion control is unnecessary when the offered load is low and optimal user rates can be determined to maximize users’ long-term satisfaction. In the high load regime where the network can be unstable under the regulated maximal scheduling policy, we propose a cross-layer congestion control and scheduling algorithm which can stabilize the network under arbitrary network load. Through extensive numerical analysis for some typical networks, we show that the proposed scheduling algorithm has much lower overhead than other existing queue-length-based constant-time scheduling schemes in the literature, and it achieves performance much better than the guaranteed bound.

Appendices

Appendix 1

1.1 Proof of Proposition 1

Let $Q_l^s(t)$ and Q _l (t) be transmission queue lengths for class s and for all user classes at link l in time slot t, respectively. Similarly, let $P_l^s(t)$ and P _l (t) be regulator queue lengths for class s and for all user classes at link l in time slot t, respectively. Let us denote by $b_l^s$ and $a_l^s$ the previous link and next link of link l on the route of class-s users. Also, let $C_l^s(t)$ and $D_l^s(t)$ be the number of packets transmitted from regulator and transmission queues in time slot t, respectively. We have the following queue update equations

$$ Q_l^s(t+1)=Q_l^s(t)-D_l^s(t)+C_l^s(t) $$

(14)

$$ P_l^s(t+1)=P_l^s(t)-C_l^s(t)+D^s_{b_l^s}(t). $$

(15)

Thus, we have

$$ Q_l^s(t+1)+P_{a_l^s}^s(t+1)= Q_l^s(t) + P_{a_l^s}^s(t) + C_l^s(t) - C_{a_l^s}^s(t). $$

(16)

We will use the following Lyapunov function for the system

$$ V(\vec{P},\vec{Q})=V_1(\vec{Q}) + \xi V_2(\vec{P},\vec{Q}) $$

(17)

where

$$ \begin{aligned} V_1(\vec{Q}) =& \sum_{l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \frac{Q_k(t)}{R_k}\\ V_2(\vec{P},\vec{Q}) =& \sum_{l} \sum_{s} \left( P_{a_l^s}^s + Q_l^s\right)^2. \end{aligned} $$

In fact, this Lyapunov function was also used in [18]. Now, let us consider

$$ \begin{aligned} V_1 (\vec{Q}(t+1))-V_1(\vec{Q}(t))=& 2 \sum_{l} \frac{Q_l(t)}{R_l} \left( \sum_{k \in I_l} \frac{ ( Q_k(t+1) - Q_k(t) )}{R_k} \right) + \sum_{l} \left( \frac{ (Q_l(t+1)-Q_l(t))}{R_l} \right) \left( \sum_{k \in I_l} \sum_{s} \frac{ ( Q_k(t+1) - Q_k(t) )}{R_k} \right)\\ =& 2 \sum_{l} \frac{Q_l(t)} {R_l} \left( \sum_{k \in I_l} \sum_{s} \frac{( C_k^s(t)-D_k^s(t) )}{R_k} \right) + \sum_{l} \frac{ (C_l^s(t)-D_l^s(t))}{R_l} \left( \sum_{k \in I_l}\sum_{s} \frac{ ( C_k^s(t)-D_k^s(t) )} {R_k} \right). \end{aligned} $$

(18)

Since the number of packets transmitted from regulator and transmission queues in each time slot are bounded, the second term in (18) can be bounded by a constant B ₁. Thus, we have

$$ \begin{aligned} E\left[ V_1(\vec{Q}(t+1))-V_1(\vec{Q}(t)) | \vec{P}(t),\,\vec{Q}(t) \right] \leq& 2 E\left[ \sum_{l} \frac{Q_l(t)}{R_l} \left( \sum_{k \in I_l} \sum_{s} \frac{ ( C_k^s(t)-D_k^s(t) )}{R_k} \right) | \vec{P}(t),\,\vec{Q}(t) \right] + B_1\\ \leq& 2 E\left[ \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \left( \sum_{k \in I_l} \sum_{s} \frac{C_k^s(t)} {R_k} - \sum_{k \in I_l} \frac{ D_k(t)}{R_k} \right) | \vec{P}(t),\,\vec{Q}(t) \right] + B_2. \end{aligned} $$

Let L be the largest number of hops traversed by any user class, we have

$$ E\left[ \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \sum_{s} \frac{C_k^s(t)}{R_k} | \vec{P}(t), \vec{Q}(t) \right] \leq \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \sum_{s} \frac{H_s^k \rho_s + L\epsilon}{R_k}. $$

(19)

Also, due to the definition of maximal scheduling policy, we have

$$ \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \frac{ D_k(t)}{R_k} \geq \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)} {R_l}. $$

(20)

For traffic load satisfying (5), we can find ε and δ small enough such that

$$ \sum_{k \in I_l} \sum_{s=1}^{S} \frac{H_s^k \rho_s + L\epsilon + \delta } {R_k} \leq 1, \quad \forall l \in E. $$

(21)

From (19), (20), (21), we have

$$ E\left[ V_1(\vec{Q}(t+1))-V_1(\vec{Q}(t)) | \vec{P}(t),\,\vec{Q}(t) \right] \leq -\delta \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} + B_2 \leq -\delta \sum_{l} \frac{Q_l(t)}{R_l} + B_3. $$

(22)

Now, using the procedure as in [18], we can obtain

$$ E\left[ V_2( \vec{P}(t+1), \vec{Q}(t+1))-V_2( \vec{P}(t),\,\vec{Q}(t)) | \vec{P}(t), \vec{Q}(t) \right] \leq \sum_{l} \sum_{s} R_l Q_l(t) - \eta \sum_{l} \sum_{s} P_{a_l^s}^s(t) + B_4. $$

(23)

Combining the results in (22), (23), we have

$$ E\left[ V( \vec{P}(t+1),\,\vec{Q}(t+1))-V( \vec{P}(t),\,\vec{Q}(t)) | \vec{P}(t),\,\vec{Q}(t) \right] \leq - \sum_{l} \sum_{s} \left[ \frac{\delta}{R_l} -\xi R_l \right] Q_l^s(t) - \xi \eta \sum_{l} \sum_{s} P_{a_l^s}^s(t) + B_5. $$

(24)

We can choose ξ small enough such that $\frac{\delta}{R_l} -\xi R_l > 0.$ Thus, the drift will be negative if the regulator and/or transmission queues become large enough. Therefore, the stability result follows by using theorem 2 of [39]. Note that the chosen Lyapunov function does not take regulator queue on the first hop of each user class into account. These regulator queues are, however, stable because their output rate is ρ _s +  ε which is larger than the average input load (i.e., ρ _s ).

Appendix 2

1.1 Proof of Proposition 2

Consider any link AB between node A and B. We will find the probability that at least one link in the interference set I _AB is scheduled. This event will be denoted as M _AB in the sequel. As mentioned before, RAMM scheduling algorithm includes BP-SIM scheduling algorithm for the one-hop (node exclusive) interference model proposed in [17] as a special case. In the following we prove Proposition 2 for k ≥ 2. For the special case of k = 1, we refer the readers to [17] for the proof and the corresponding analysis. Note that the proof for the case k ≥ 2 is very challenging and completely different from that for the case k = 1 in [17] due to the more complicated interference relationship.

We will illustrate some important definitions used in the proof in Fig. 15. Let I ₀ be the set of nodes which is at most k − 1 hops away from either A and B including A, B. For the grid network and link AB shown in Fig. 15 under the two-hop interference model, $I_0=\left\{ A, B, C_1,\ldots, C_6\right\}.$ Also, the interference set for node C ₆ (i.e., $I_{{C_6}}$) consists of all nodes in I ₀ and all “blank” nodes. Note that the notion of node interference set is different from that of link interference set provided in definition I. Here, all nodes in $I_{{C_6}}$ are at most 3 hops away from C ₆. We observe that all links incident to any nodes in I ₀ will belong to I _AB because they are within k hops from link AB. In the following, we will find the lower bound for the probability of M _AB by considering sub-cases in which there are i left nodes in the set I ₀. For convenience, we will use I ₀ to denote both the set itself and the corresponding number of nodes in I ₀. Let L _i be the event that there are i left nodes in the set I ₀. The probability that at least one link in the interference set of link AB is scheduled can be lower bounded as

$$ P_m \geq \sum_{i=1}^{I_0} {Pr} ( M_{AB} | L_i ) {Pr} ( L_i ). $$

(25)

Recall that for any particular node A, there are at most I nodes whose matching request can collide with that transmitted from node A. Note that these I nodes will be at most k +  1 hops away from mode A. To find the lower bound for the probability of M _AB , we will assume the worst-case scenario where each node has I interfering nodes. For convenience, we also use I _A to denote the set of these interfering nodes whose transmissions can collide with that from node A. We now consider the following cases.

Fig. 15

Interference set $I_{C_6}$ under two-hop interference model

1.2 There is only one left node in I ₀

We consider the following two sub-cases.

• If this left node is either A or B, then the left node will have at least one neighbor which is a right node. This case occurs with probability $2(1/2)^{I_0}.$ For ease of reference, we will refer to this left node as node C (i.e., C is either A or B). Also, there are at most I − I ₀ nodes whose matching requests can collide with that from node C. To find the lower bound for Pr (M _AB ), we assume that there are I − I ₀ such interfering nodes.

  Now, suppose there are i left nodes among these I − I ₀ interfering nodes. The matching request transmitted by node C will be successfully received if the backoff values of these i left nodes are larger than that of node C. Specifically, the matching request from node C will be successfully received and the corresponding link will be scheduled with a probability which is lower bounded by

  $$ F_1=\sum_{i=0}^{I-I_0} \left( \begin{array}{l} I-I_0 \\ i \end{array}\right) \left( \frac{1}{2}\right)^{I-I_0} \frac{1}{B} \sum_{m=1}^{B} \left(1- \frac{m}{B} \right)^i $$

  where we have broken the event into sub-cases where there are i left nodes among I − I ₀ interfering nodes and these i left nodes have backoff values larger than that of node C.

• If this left node is any node other than A and B then it can be any node among (I ₀ − 2) nodes. Again, we refer to this left node as node C. Note that all nodes which are within one hop from A or B including A and B are at most k +  1 hops away from C so they all belong to the interfering set I _C . Let x be the total number of nodes within one hop from A and B including A and B, then there are at most I − x nodes whose matching requests can collide with that from node C. This is because these x nodes belong to I ₀ and they are all right nodes except C if C is in I ₀. We will assume this worst-case scenario to calculate the lower bound of the matching probability in the following. Note that node C will have at least one neighbor which is a right node because it is the only left node in I ₀.

Again, suppose there are i left nodes among potential interfering nodes in I _C . The matching request transmitted by node C will be successfully received if the backoff values of these i left nodes are larger than that from node C and the matching request from node C is transmitted to a right node. Now, we consider the following two sub-cases. For the first case, if C is one of x nodes (i.e., x one-hop neighbors of A or B) but not A and B. Then, this case occurs with probability $(x-2)(1/2)^{I_0}.$ In this case, the matching request from node C will be successfully received and the corresponding link will be scheduled with a probability which is lower bounded by

$$ F_2=\sum_{i=0}^{I-x-1} \left( \begin{array}{l}I-x-1 \\ i \end{array}\right) \left( \frac{1}{2}\right)^{I-x} \frac{1}{B} \sum_{m=1}^{B} \left(1- \frac{m}{B} \right)^i $$

where i is the number of left node. Also, in calculating the lower bound for P _m we assume that C transmits its matching request to a right node which is not an one-hop neighbor of A or B. Hence, there are at most I − x − 1 left nodes which can collide with the matching request from C. For the second case, if C is not one of x nodes (i.e., x one-hop neighbors of A or B). Then, this case occurs with probability $(I_0-x)(1/2)^{I_0}.$ In this case, the matching request from node C will be successfully received and the corresponding link will be scheduled with a probability which is lower bounded by

$$ F_3=\sum_{i=0}^{I-x-2} \left( \begin{array}{l} I-x-2 \\ i \end{array}\right) \left( \frac{1}{2}\right)^{I-x} \frac{1}{B} \sum_{m=1}^{B} \left(1- \frac{m}{B} \right)^i $$

where in calculating the lower bound for P _m we assume that C transmits its matching request to a right node which is not an one-hop neighbor of A or B. Hence, there are at most I − x − 2 left nodes which can collide with the matching request from C.

1.3 There are two or more left nodes in I ₀

Suppose node C in the set I ₀ becomes left and wins the contention. Then node C should have the smallest backoff value among all the nodes whose matching requests can collide with the matching request from C. Also, node C should send the matching request to a node which is a right one. For ease of reference, we will refer to this right node as node D in the sequel. In general, D can belong to set I ₀ or not. However, to find the lower bound of P _m , we assume that D belongs to I ₀; therefore, there are at most I ₀ − 2 other left nodes besides C and D in I ₀.

As before, we assume the worst-case scenario where there are I nodes whose transmissions can collide with that of node C. Recall that all x nodes which are one-hop neighbors of A or B belong to the set I. Similar to the previous case, we consider the following two sub-cases. For the first case, C is one of x nodes (i.e., x one-hop neighbors of A or B). In this case the matching probability can be lower bounded as

$$ F_4= x \sum_{i=1}^{I_0-2} \sum_{j=0}^{I-x-1} \left( \begin{array}{l} I_0-2\\ i \end{array}\right) \left( \frac{1}{2} \right)^{I_0} \left( \begin{array}{l}I-x-1 \\ j \end{array}\right) \left( \frac{1}{2}\right)^{I-x} \frac{1}{B} \sum_{m=1}^B \left( 1 - \frac{m}{B} \right)^{i+j} $$

where i is the number of left nodes besides C and D in the set I ₀. And j is the number of left nodes which belong to I but are not one-hop neighbors of A or B (i.e., there is no link between these nodes and A or B). We will denote this set as $I \setminus x$ in the sequel. In general, D can belong to $I \setminus x$ or not; however, to find the lower bound for P _m , we only allow j takes values from 0 to I − x − 1. In addition, C can be any node among x nodes so we have a factor of x before the sum. Also, we require that all left nodes (i left nodes belonging I ₀ and j left nodes belonging to $I \setminus x$) achieve larger backoff values than that of node C.

For the second case, C belong to the set $I \setminus x.$ In this case, the matching probability can be lower bounded as

$$ F_5= \left( I_0-x \right) \sum_{i=1}^{I_0-2} \sum_{j=0}^{I-x-2} \left( \begin{array}{l} I_0-2 \\ i \end{array}\right) \left( \frac{1}{2} \right)^{I_0} \left( \begin{array}{l} I-x-2 \\ j \end{array}\right) \left( \frac{1}{2}\right)^{I-x} \frac{1}{B} \sum_{m=1}^B \left( 1 - \frac{m}{B} \right)^{i+j} $$

where C can be one of I ₀ − x nodes so we have the factor $\left( I_0 - x \right)$ before the sum. Also, to calculate the lower bound for the matching probability, we assume D always belong to $I \setminus x,$ so j can be at most I − x − 2.

Substitute results of all considered cases into (25), the matching probability is lower bounded by

$$ P_m \geq P_m^{0} = 2 (1/2)^{I_0} F_1 + (x-2)(1/2)^{I_0}F_2 + (I_0-x)(1/2)^{I_0}F_3 + F_4 + F_5 $$

(26)

where F ₁, F ₂, F ₃, F ₄, and F ₅ are defined above.

From the lower bound of the matching probability $P_m^{0}$ derived above, we can calculate the lower bound of P _m for any link AB as

$$ p^{*} = \min_{x, I_0} P_m^{0} $$

where we find the minimum of $P_m^{0}$ over all possible x and I ₀. Note that possible values of x and I ₀ will be in the range of [3, 2d ^*] and $[3, I_0^{*}],$ respectively. Here, x and I ₀ are at least three for the network to be connected (i.e., A and B should have at least one one-hop neighbor). It is observed that p ^* is independent of the network size and depends only on B, d ^*, I, $I_0^{*}.$ Now, we can choose K such that at least one link in I _l of a backlogged link l is scheduled with probability greater than μ after K rounds as follows:

$$ K=\min_{k \geq 1} \left\{ k: (1-p^{*})^k \leq 1- \mu \right\}. $$

(27)

Thus, we can choose K which is independent of network size and only depends on B, d ^*, I, $I_0^{*},$ μ such that performance guarantee arbitrarily close to the constraining bound can be achieved.

Example

For the grid network and two-hop interference model, we have I = 22, I ₀ = x can take values of 5, 6, 8. With maximum backoff value B = 10, by using the analysis presented above, the minimum number of scheduling rounds to achieve μ = 0.9 is K = 60. In fact, this calculation is quite conservative because it considers the worst case scenario. In practice, the size of interference sets decreases quickly over scheduling rounds, so the required value of K is much smaller.

Appendix 3

1.1 Proof of Proposition 4

The proof is similar to that of Proposition 1 (i.e., we use the same Lyapunov function and proof procedure). In particular, we have a similar bound as in (19) as follows:

$$ E\left[ \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \sum_{s} \frac{C_k^s(t)}{R_k} | \vec{P}(t),\,\vec{Q}(t) \right] \leq \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} \sum_{k \in I_l} \sum_{s} \frac{H_s^k x_s^{*} + L\epsilon}{R_k}. $$

(28)

Due to the constraints of the optimization problem in (10), we can find ε and δ small enough such that

$$ \sum_{k \in I_l} \sum_{s=1}^{S} \frac{H_s^k x_s^{*} + L\epsilon + \delta } {R_k} \leq 1, \quad \forall l \in E. $$

(29)

From (28) and (29), we have

$$ E\left[ V_1(\vec{Q}(t+1))-V_1(\vec{Q}(t)) | \vec{P}(t),\,\vec{Q}(t) \right] \leq -\delta \sum_{l:Q_l(t) \geq R_l} \frac{Q_l(t)}{R_l} + B_6 \leq -\delta \sum_{l} \frac{Q_l(t)}{R_l} + B_7. $$

(30)

The remaining steps to obtain negative drift when regulator and/or transmission queues become large enough are the same as in the proof of Proposition 1.