Optimal scheduling for dynamic channel allocation in wireless LANs

Abstract

Channel allocation schemes that have been used in cellular wireless ave limited applicability to Wireless LANs (WLANs) because of the small number of available channels and irregular cell geometries in WLAN environments. In this paper, we propose a dynamic, frame-based channel allocation architecture for WLANs. In this architecture, time is divided into a sequence of consecutive frames (in the order of milliseconds), and in each frame, only a non-interfering subset of access points (APs) is activated. Under broad traffic assumptions, we prove that the attainable system throughput can be optimized by scheduling APs and allocating channels in each frame such that a weighted sum of queue sizes at the activated APs is maximized. This optimality criterion for AP scheduling and channel allocation leads to a novel graph problem which is a variant of the well-known maximum independent set problem. We develop two heuristics for solving this problem. The first is a greedy heuristic that yields an approximation algorithm that has quadratic time complexity (in the number of APs) and, under certain conditions, yields a constant (6) factor approximation bound. The second heuristic is a graph decomposition heuristic. This heuristic, again under certain conditions, yields better approximation ratio, $1+\epsilon,$ but has complexity that grows exponentially with $1/\epsilon^{2}$ for arbitrarily small $\epsilon > 0.$ Using the ns2 simulator we conducted experiments to compare our frame-based approach to static channel allocation. Results of our simulation indicate that our approach is able to deliver system throughput improvements of more than 50%.

Appendix

We prove Theorem 1 through a sequence of three lemmas. To facilitate discussion, let us first introduce some new vector notation. Define Q(t), A(t), S(t), D(t), $\mathbf{\lambda},$ and $\mathbf{\mu}(\mathbf c),$ to be N dimensional vectors with entries Q _n (t), A _n (t), S _n (t), D _n (t), λ _n , and $\mu _n({\mathbf c}),$ respectively.

Lemma 5

Assume that there exists some channel allocation policy under which the system is stable. If the system follows policyP, then there exists some δ > 0, such that for all framest we have

$$ {\mathbf Q}^{\prime} (t) \cdot {\mathbf \lambda} \leq (1-\delta) {\mathbf Q^\prime} (t) \cdot {\mathbf \mu}({\mathbf c}^\ast(t)). $$

(14)

Proof

According to Lemma 1, since the system is assumed to be stable under some channel allocation policy, there exists a set of coefficients $\phi_n , n= 1, 2, \ldots, N,$ for which (4)–(6) are satisfied. From (6) it follows that there exists a δ > 0 such that

$$\lambda _n < (1-\delta)\sum _ {\forall {\mathbf c} \in {\mathcal C}} \phi _{\mathbf c} \cdot \mu_n ({\mathbf c}) , \quad n=1, 2, \ldots, N.$$

(15)

For example, validity of (15) can be easily verified for

$$\delta = \frac{1}{2} - \max \frac{\lambda _n}{2 \sum _ {\forall {\mathbf c} \in {\mathcal C}} \phi_{\mathbf c} \cdot \mu_n ({\mathbf c})} > 0.$$

We may express (15) in vector form as

$${\mathbf \lambda} < (1- \delta) \sum\limits_{\forall {\mathbf{c}} \in {{\mathcal{C}}}} \phi_{\mathbf c} \cdot {\mathbf \mu} ({\mathbf c} ). $$

(16)

Since Q(t) is composed of nonnegative entries, multiplying both sides of (16) by ${\mathbf Q^ \prime} (t),$ yields

$${\mathbf Q} ^{\prime} (t)\cdot {\mathbf \lambda} \leq (1-\delta) \sum \limits_{\forall {\mathbf c} \in {\mathcal C}} \phi_{\mathbf c} \cdot {\mathbf Q ^ \prime} (t)\cdot {\mathbf \mu}({\mathbf c} ).$$

(17)

Next, from the definition of policy P in (7) we have

$$\sum_{n=1}^N \mu _n ({\mathbf c}) \cdot Q_n(t) \leq \sum_{n=1}^N \mu _n ({\mathbf c}^\ast(t)) \cdot Q_n(t), \quad {\forall {\mathbf c} \in {\mathcal C}}.$$

Or, expressed in the vector form,

$${\mathbf Q}^{\prime} (t)\cdot {\mathbf \mu}({\mathbf c} ) \leq {\mathbf Q}^{\prime} (t)\cdot {\mathbf \mu}({\mathbf c}^\ast(t)).$$

(18)

Since $\phi_n , n= 1, 2, \ldots, N,$ are nonnegative, we get the following weighted sum of (18):

$$\begin{aligned}\sum\limits_{\forall {\mathbf c} \in {\mathcal C}}\phi_{\mathbf c} \cdot {\mathbf Q^\prime} (t)\cdot {\mathbf \mu}({\mathbf c} ) &\leq \sum\limits_{\forall {\mathbf c} \in {\mathcal C}} \phi_{\mathbf c} \cdot {\mathbf Q^\prime} (t)\cdot {\mathbf \mu}({\mathbf c} ^\ast(t))\\& = {\mathbf Q^\prime} (t)\cdot {\mathbf \mu}({\mathbf c} ^\ast(t) ),\end{aligned}$$

(19)

where the equality follows from (5). Finally, we establish the lemma by multiplying both sides of (19) by (1 − δ) and combining the result with (17). □

Lemma 6

Assume that there exists some channel allocation policy under which the system is stable. If the system follows policy P , then there exists a positive δ and a positive bounded σ, such that for all frames t we have

$$ E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] \leq \sigma - 2 \delta {\mathbf Q^\prime}(t) {\mathbf \mu}({\mathbf c} ^\ast(t)), $$

(20)

where $\|{\mathbf x}\|$ stands for the norm of vector ${\mathbf x}.$

Proof

Writing (6) in vector form, we get

$${\mathbf Q}(t+1) = {\mathbf Q}(t) - {\mathbf D}(t) + {\mathbf A} (t).$$

It follows that

$$\begin{aligned} E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] &= E [{\mathbf Q ^\prime} (t+1) {\mathbf Q} (t+1) - {\mathbf Q ^\prime} (t) {\mathbf Q} (t) | {\mathbf Q} (t) ] \\ &=E [{\mathbf D ^\prime} (t){\mathbf D} (t) | {\mathbf Q} (t) ] + E [{\mathbf A ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ]\\ &-2 E [{\mathbf D ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ] +2 E [{\mathbf Q ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ] \\ &- 2 E [{\mathbf Q ^\prime} (t){\mathbf D} (t) | {\mathbf Q} (t) ] . \end{aligned}$$

(21)

Since $D_n(t) \leq S_n(t), \forall n,$ we have

$$\begin{aligned} E [{\mathbf D ^\prime} (t){\mathbf D} (t) | {\mathbf Q} (t) ] &\leq E [{\mathbf S ^\prime} (t){\mathbf S} (t) | {\mathbf Q} (t) ]\\ &= E\left[\sum_ {n=1}^N S_n^2(t) |{\mathbf Q}(t)\right] \\ &= \sum _ {n=1}^N \overline{S_n^2(t)} , \end{aligned}$$

where the equality follows because S(t) and Q(t) are independent. Similarly, since A(t) and Q(t) are independent,

$$\begin{aligned} E [{\mathbf A ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ] &= E\left[\sum_ {n=1}^N A_n^2(t) |{\mathbf Q}(t)\right]\\ &=\sum _ {n=1}^N \overline{A_n^2} , \end{aligned}$$

and

$$E [{\mathbf Q ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ] = {\mathbf Q ^\prime} (t) E[{\mathbf A} (t)] = {\mathbf Q ^\prime} (t) \lambda .$$

Since D(t) and A(t) are nonnegative vectors,

$$E [{\mathbf D ^\prime} (t){\mathbf A} (t) | {\mathbf Q} (t) ] \geq 0.$$

Next, we define

$$H_n(t)\;\mathop=^{\hbox{def}}\;S_n(t) - D_n(t) = \max (0 , S_n(t) - Q_n(t)), $$

where the equality follows from (3). It follows that

$$\begin{aligned} Q_n(t) H_n(t) &= \left \{\begin{array}{ll} Q_n(t) [S_n(t)-Q_n(t)], &Q_n(t) \leq S_n(t) , \\ 0,&\hbox{otherwise.} \end{array}\right. \\ & \leq \sum_{n=1}^N S_n^2(t) .\end{aligned}$$

(22)

Since D _n (t) = S _n (t) − H _n (t), we have

$$\begin{aligned} E [{\mathbf Q} ^\prime (t){\mathbf D} (t) | {\mathbf Q} (t) ]&= E [{\mathbf Q ^\prime} (t){\mathbf S} (t) | {\mathbf Q} (t) ] - E \left[\sum_{n=1}^N Q_n(t)H_n(t)| {\mathbf Q} (t) \right] \\ & \geq {\mathbf Q ^\prime} (t) {\mathbf\mu}({\mathbf c} ^\ast(t)) - \sum_{n=1}^N \overline{S_n^2(t)} ,\end{aligned} $$

(23)

where the inequality follows from the independence of S(t) and Q(t) and from (22). Finally, by combining (21) with the subsequent results, we get

$$\begin{aligned} E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] &\leq \sum_{n=1}^N \overline{A_n^2} + 3 \sum_{n=1}^N \overline{S_n^2(t)}\\ &+ 2 {\mathbf Q ^\prime} (t) \lambda- 2 {\mathbf Q ^\prime} (t) {\mathbf \mu}({\mathbf c} ^\ast(t)).\end{aligned} $$

(24)

Now let us define

$$\sigma\; \mathop=^{\hbox{def}}\; \sum_{n=1}^N \overline{A_n^2} + 3 \sum_{n=1}^N \overline{S_n^2(t)}. $$

Note that σ is bounded since the arrival and service rate processes A _n (t) and S _n (t) are assumed to have bounded second moments. The lemma may now be established by multiplying (14) by 2 and adding it to (24), which leads to (20).□

Lemma 7

Let there be some channel allocation policy under which the system is stable, and assume that the system follows policyP. Then for any frame t and for all nonnegative vectorsQ(t),

$$E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] < \infty ].$$

(25)

Furthermore, for any$\epsilon > 0,$ there exists a positive bounded number K, such that for all nonnegative vectorsQ(t) satisfying$\| {\mathbf Q}(t)\| \geq K,$ we have

$$E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] < -\epsilon .$$

(26)

Proof

Inequality (25) is a direct result of Lemma 6. In order to show (26), we must first specify K for any given $\epsilon > 0.$ For this purpose, define the channel allocation vector $\hat{\mathbf c}_n, n=1,2, \ldots, N$ as one for which only cell n is activated and all other cells are inactive. Clearly, $\hat{\mathbf c}_n$ is feasible, and $\mu_n(\hat{\mathbf c}_n) > 0.$ Define $\hat{\mu}\; \mathop=^{\hbox{def}}\;\min_{\forall n} \mu_n(\hat{\mathbf c}_n) > 0.$ It follows that $\hat{\mu} > 0.$ We can now choose K as follows

$$K = \frac{(\sigma + \epsilon)N}{2 \delta \hat{\mu}},$$

(27)

where δ and σ are given in Lemma 6.

Now, let queue m be the longest queue at the beginning of frame t, i.e., let $Q_m(t) \geq Q_n(t), \forall n.$ It follows that

$$Q_m(t) \geq \frac{1}{N} \| {\mathbf Q}(t) \|.$$

(28)

Also, according to the definition of policy P,

$$\begin{aligned} {\mathbf Q^\prime} (t) {\mathbf\mu}({\mathbf c} ^\ast(t)) & \geq {\mathbf Q^\prime} (t) {\mathbf \mu}(\hat{\mathbf c}_m) \geq Q_m(t) {\mathbf \mu} _m (\hat{\mathbf c}_m)\\ & \geq \frac{1}{N} \| {\mathbf Q}(t) \| \hat{\mu},\end{aligned} $$

(29)

where the last inequality follows from (28) and from the definition of $\hat{\mu}.$

We can now establish the second part of the lemma. Let $\| {\mathbf Q}(t)\| \geq K.$ From Lemma 6 and (29) we get

$$\begin{aligned} E [ \| {\mathbf Q}(t+1) \|^2 - \| {\mathbf Q}(t) \|^2 | {\mathbf Q}(t)] &\leq \sigma - 2 \delta \frac{1}{N} \| {\mathbf Q}(t) \| \hat{\mu}\\ &\leq\sigma - 2 \frac{\delta}{N}K \hat{\mu} = -\epsilon , \end{aligned}$$

(30)

where the equality follows from (27).□

Proof of Theorem 1

Consider the Markov chain associated with the whole queuing system. Define a quadratic Lyapunov function L(Q) as $L ({\mathbf Q}) \;\mathop=^{\hbox{def}}\;\|{\mathbf Q}\|^2.$ The theorem follows in view of Lemma 7 and Foster’s criterion for irreducible chains.□