Stability of longest-queue-first scheduling in linear wireless networks with multihop traffic and one-hop interference

Abstract

We consider the stability of the longest-queue-first (LQF) scheduling policy in wireless networks with multihop traffic under the one-hop interference model. Although it is well known that the back-pressure algorithm achieves the maximal stability, its computational complexity is prohibitively high. In this paper, we consider LQF, a low-complexity scheduling algorithm, which has been shown to have near-optimal throughput performance in many networks with single-hop traffic flows. We are interested in the performance of LQF for multihop traffic flows. In this scenario, the coupling between queues due to multihop traffic flows makes the local-pooling-factor analysis difficult to perform. Using the fluid-limit techniques, we show that LQF achieves the maximal stability for linear networks with multihop traffic and a single destination on the boundary of the network under the one-hop interference model.

Appendix

1.1 Proof of Lemma 1

We first notice that if $\bar{Z}\in B$, then there are no adjacent dominating nodes; i.e., if $\bar{Z}\in B_i$ for some $i\in \{1, 2, \dots , N\}$, then $\bar{Z}\notin B_{i-1}$ and $\bar{Z}\notin B_{i+1}$. Let the dominating set at time $t$ be

$$\begin{aligned} I_{\text {dom}}(t) = \{i\in \{1, 2, \dots , N\}\mid \bar{Z}(t)\in B_i\}. \end{aligned}$$

Then we can easily check that $I_{\text {dom}}(t)\subseteq \bigcap \text {LQF}(\bar{Z}(t))$; i.e., all dominating links must be scheduled by LQF at time $t$. Due to the one-hop interference, there is no internal arrival to a scheduled link. Then for regular time $t$ and any $i\in I_{\text {dom}}(t)$,

$$\begin{aligned} \frac{\mathrm dA_i}{\mathrm dt}(t) = \frac{\mathrm dE_i}{\mathrm dt}(t) = \alpha _i \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm dD_i}{\mathrm dt}(t) = 1, \end{aligned}$$

while

$$\begin{aligned} \frac{\mathrm dD_{i-1}}{\mathrm dt}(t) = \frac{\mathrm dD_{i+1}}{\mathrm dt}(t) = 0. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\mathrm d \bar{Z}_i}{\mathrm dt}(t)-\frac{\mathrm d \bar{Z}_{i-1}}{\mathrm dt}(t) \le \alpha _i-1 \end{aligned}$$

(17)

and

$$\begin{aligned} \frac{\mathrm d \bar{Z}_i}{\mathrm dt}(t)-\frac{\mathrm d \bar{Z}_{i+1}}{\mathrm dt}(t) \le \alpha _i-1. \end{aligned}$$

(18)

Let $\bar{Z}_{\max } = \max _j\bar{Z}_j(0)$. We now present two propositions to complete the proof of Lemma 1.

Proposition 4

There exists $t_1\in (t_0, t_0+\bar{Z}_{\max }/(1-\alpha _i)]$ such that $\bar{Z}(t_1)\notin B_i$; i.e., link $l_i$ is not dominating any more at some time before $t_0+\bar{Z}_{\max }/(1-\alpha _i)$.

Proof

Indeed, if link $l_i$ remains dominating up to (and including) time $t_0+\bar{Z}_{\max }/(1-\alpha _i)$, then by (17), (18) and the absolute continuity of the fluids, we would have for any adjacent link $l_j$ of $l_i$,

$$\begin{aligned} \left[ \bar{Z}_i\left( t_0+\frac{\bar{Z}_{\max }}{1-\alpha _i}\right) -\bar{Z}_j\left( t_0+\frac{\bar{Z}_{\max }}{1-\alpha _i} \right) \right] -\left[ \bar{Z}_i(t_0)-\bar{Z}_j(t_0)\right] \le - \bar{Z}_{\max }. \end{aligned}$$

Hence

$$\begin{aligned} \bar{Z}_i\left( t_0+\frac{\bar{Z}_{\max }}{1-\alpha _i}\right) -\bar{Z}_j\left( t_0+\frac{\bar{Z}_{\max }}{1-\alpha _i}\right) \le \bar{Z}_i(t_0)-\bar{Z}_{\max } \le 0. \end{aligned}$$

Then by continuity, there is some $t_1\in (t_0, t_0+\bar{Z}_{\max }/(1-\alpha _i)]$ such that $\bar{Z}_i(t_1)-\bar{Z}_j(t_1) = 0$, which contradicts our assumption that link $l_i$ remains dominating up to $t_0+\bar{Z}_{\max }/(1-\alpha _i)$. This completes the proof of the proposition. $\square$

Therefore, for any $i\in I_{\text {dom}}(t_0)$, there exists $t_1\in (t_0, t_0+\bar{Z}_{\max }/(1-\max _j\alpha _j)]$ such that $\bar{Z}(t_1)\notin B_i$.

Proposition 5

If $\bar{Z}(t_1)\notin B_i$, then $\bar{Z}(t)\notin B_i$ for any $t \ge t_1$.

Proof

Indeed, if $\bar{Z}(t_2)\in B_i$ for some $t_2 > t_1$, let $t_3 = \sup \{t < t_2\mid \bar{Z}(t)\notin B_i\}$. Then by Lipschitz continuity $\bar{Z}_i(t_3) = \bar{Z}_j(t_3)$ for some neighbor $l_j$ of link $l_i$ and $t_3 < t_2$. Since $\frac{\mathrm d}{\mathrm dt}(\bar{Z}_i(t)-\bar{Z}_j(t)) \le \max _k\alpha _k-1$ for almost all $t\in [t_3, t_2]$, we have

$$\begin{aligned} \bar{Z}_i(t_2)-\bar{Z}_j(t_2) \le \bar{Z}_i(t_3)-\bar{Z}_j(t_3)+(t_2-t_3)(\max _k\alpha _k-1)\le 0, \end{aligned}$$

which contradicts the assumption that $\bar{Z}(t_2)\in B_i$. Hence, $\bar{Z}(t)\notin B_i$ for any $t\ge t_1$. $\square$

Considering all $i$, we have $\bar{Z}(t)\notin B$ for any $t\ge t_0+\bar{Z}_{\max }/(1-\max _j\alpha _j)$.

1.2 Proof of Proposition 2

To show this proposition by contradiction, we suppose $J(t) = \emptyset$.

We first notice that for $j_1 = \min \{j\mid j\in J_0(t)\}$ we must have $\frac{\mathrm d}{\mathrm dt}\bar{W}_{j_1}(t) > 0$. If this is not the case, we would have $\frac{\mathrm d}{\mathrm dt}\bar{W}_{j_1}(t) < 0$ and $j_1 \ge 2$. Then it would follow that there exists some $\delta > 0$ such that for any $s\in (t, t+\delta )$ we have $\bar{Z}_1(s) > \bar{Z}_2(s)$ if $j_1 = 2$, and $\bar{Z}_{j_1-1}(s) > \max \{\bar{Z}_{j_1-2}(s), \bar{Z}_{j_1}(s)\}$ if $j_1 > 2$, which implies $\bar{Z}(s) \in B$, a contradiction.

We then conclude that if all $j\in \{1, 2, \dots , k\}\cap J_0(t)$ satisfy

$$\begin{aligned} \frac{\mathrm d}{\mathrm dt}\bar{W}_j(t) > 0, \end{aligned}$$

then either $k+1\notin J_0(t)$ or

$$\begin{aligned} \frac{\mathrm d}{\mathrm dt}\bar{W}_{k+1}(t) > 0. \end{aligned}$$

If this is not the case, there would exist $\delta > 0$ such that for any $s\in (t, t+\delta )$, we have $\bar{Z}_k(s) > \max \{\bar{Z}_{k-1}(s), \bar{Z}_{k+1}(s)\}$ if $\bar{W}_k(t) \ge 0$, or $\bar{Z}_j(s) > \max \{\bar{Z}_{j-1}(s), \bar{Z}_{j+1}(s)\}$ for some $j < k$ otherwise, either of which leads to a contradiction.

By induction we have $\frac{\mathrm d}{\mathrm dt}\bar{W}_j(t) > 0$ for all $j\in J_0(t)$, which also leads to contradiction since by letting $j_2 = \max \{j\mid j\in J_0(t)\}$ there exists $\delta > 0$ such that for any $s\in (t, t+\delta )$ we have $\bar{Z}_N(s) > \bar{Z}_{N-1}(s)$ if $j_2 = N$, and $\bar{Z}_{j_2}(s) > \max \{\bar{Z}_{j_2-1}(s), \bar{Z}_{j_2+1}(s)\}$ if $j_2 \ne N$. Then $\bar{Z}(s)\in B$, which is a contradiction. This completes the proof of Proposition 2.

1.3 Proof of Proposition 3

Note that $u \ge 2$. By the definition of $u, \bar{W}_u(t_1) = \frac{\mathrm d}{\mathrm dt}\bar{W}_u(t_1) = 0$, i.e., $\bar{Z}_u(t_1) = \bar{Z}_{u-1}(t_1)$ and $\frac{\mathrm d}{\mathrm dt}\bar{Z}_u(t_1) = \frac{\mathrm d}{\mathrm dt}\bar{Z}_{u-1}(t_1)$.

We first claim that there exists $\delta > 0$ such that for any $t\in (t_1, t_1+\delta )$,

$$\begin{aligned} 0 < \bar{Z}_1(t) < \bar{Z}_2(t) < \dots < \bar{Z}_{u-1}(t). \end{aligned}$$

Indeed, if $J_0(t_1)\cap \{1, 2, \dots , u-1\} = \emptyset$, then $\bar{W}_i(t_1)\ne 0$ for $i = 1, 2, \dots , u-1$. Otherwise for any $j\in J_0(t_1)\cap \{1, 2, \dots , u-1\}$, by the definitions of $u$ and $J_0(\cdot )$, we have $\bar{W}_j(t_1) = 0$ and $\frac{\mathrm d}{\mathrm dt}\bar{W}_j(t_1) \ne 0$, so there exists $\delta _j > 0$ such that $\bar{W}_j(t) \ne 0$ for any $t\in (t_1, t_1+\delta _j)$. In either case, there exists $\delta > 0$ such that $\bar{W}_i(t)\ne 0$ for any $t\in (t_1, t_1+\delta )$ and any $i\in \{1, 2, \dots , u-1\}$. If $\bar{W}_i(t) < 0$ for some $i\in \{1, 2, \dots , u-1\}$ and some $t\in (t_1, t_1+\delta )$, then $\bar{Z}_j(t) > \max \{\bar{Z}_{j-1}(t), \bar{Z}_{j+1}(t)\}$ for some $j\le i$, which contradicts Lemma 1. Hence $\bar{W}_i(t) > 0$ for $i = 1, 2, \dots , u-1$; i.e., $0 < \bar{Z}_1(t) < \bar{Z}_2(t) < \dots < \bar{Z}_{u-1}(t)$. The claim then follows.

In the actual system with this strict order of queues, either all odd links up to $l_u$ get scheduled at a time slot, or all even links up to $l_u$ get scheduled. Then in our fluid limits we would have

$$\begin{aligned} \mu _i(t) = \mu _{i+2}(t)\quad \forall i\in \{1, 2, \dots , u-2\} \end{aligned}$$

for any regular time $t\in (t_1, t_1+\delta )$. Then by the absolute continuity,

$$\begin{aligned} \bar{D}_3(t)-\bar{D}_1(t) = \bar{D}_3(t_1)-\bar{D}_1(t_1)+\int _{t_1}^t(\mu _3(s) -\mu _1(s))\mathrm ds = \bar{D}_3(t_1)-\bar{D}_1(t_1). \end{aligned}$$

By the definition of derivatives, we have

$$\begin{aligned} \frac{\mathrm d}{\mathrm dt}(\bar{D}_3(t_1)-\bar{D}_1(t_1)) = \lim _{t\rightarrow t_1^+}\frac{(\bar{D}_3(t)-\bar{D}_1(t)) -(\bar{D}_3(t_1)-\bar{D}_1(t_1))}{t-t_1} = 0. \end{aligned}$$

So $\mu _1(t_1) = \mu _3(t_1)$. Similarly, we have $\mu _i(t_1) = \mu _{i+2}(t_1)$ for $i = 1, 2, \dots , u-2$.