Queue-proportional rate allocation with per-link information in multihop wireless networks

Abstract

The backpressure scheduling algorithm for multihop wireless networks is known to be throughput optimal, but it requires each node to maintain per-destination queues. Recently, a clever generalization of processor sharing has been proposed which is also throughput optimal, but which only uses per-link queues. Here, we propose another algorithm, called Queue-Proportional Rate Allocation (QPRA), which also only uses per-link queues and allocates service rates to links in proportion to their queue lengths, and employs the Serve-In-Random-Order queueing discipline within each link. Through fluid limit techniques and using a novel Lyapunov function, we show that the QPRA algorithm achieves the maximum throughput. We demonstrate an advantage of QPRA by showing that, for the so-called primary interference model, it is able to develop a low-complexity scheduling scheme which approximates QPRA and achieves a constant fraction of the maximum throughput region, independent of network size.

Appendices

Appendix 1: Proof of Proposition 1

For any integer $t\ge 1$, let $R_l^{{\varSigma }}(t)\triangleq \sum _{\tau =0}^{t-1}R_{l}(\tau )$ be the total amount of service available to link l from time slot 0 to $t-1$. Let $H_l^{{\varSigma }}(t)\triangleq \sum _{\tau =0}^{t-1}H_{l}(\tau )$ be the cumulative number of packets departing from link l up to time slot $t-1$. Let $D_{l,r}^{{\varSigma }}(t)\triangleq \sum _{\tau =0}^{t-1}D_{l,r}(\tau )$ denote the cumulative number of route r packets departing from link l up to time slot $t-1$. Further, let $R_l^{{\varSigma }}(0)=H_l^{{\varSigma }}(0)=D_{l,r}^{{\varSigma }}(0)=0$. Then, the evolution of the queue length can be rewritten as

$$\begin{aligned} X_{l,r}(t)=X_{l,r}(0)+A_{l,r}^{{\varSigma }}(t)-D_{l_{-}^{(r)},r}^{{\varSigma }}(t)+D_{l,r}^{{\varSigma }}(t), \end{aligned}$$

(60)

for all pairs (l, r). Here, we note that $D_{l^{(r)}_{-},r}^{{\varSigma }}(t)=0$ if $l=l_1^{(r)}$.

For the purposes of our analysis, we interpolate the values of $A_{l,r}^{{\varSigma }}(t)$, $R_l^{{\varSigma }}(t)$, $H_l^{{\varSigma }}(t)$, and $D_{l,r}^{{\varSigma }}(t)$ to all real number $t\ge 0$ by linear interpolation between $\lfloor t\rfloor$ and $\lfloor t\rfloor +1$, where $\lfloor t\rfloor$ denotes the largest integer no greater than t. Then we have the following lemma.

Lemma 5

Under the QPRA algorithm, with probability 1, for any positive sequence $w_n\rightarrow \infty$ there exists a subsequence $w_{n_j}$ with $w_{n_j}\rightarrow \infty$ such that the following convergence holds uniformly over compact intervals of time t:

$$\begin{aligned}&\frac{1}{w_{n_j}}A_{l,r}^{{\varSigma }}(w_{n_j}t)\rightarrow \lambda _{l,r}t, \quad \forall (l,r), \end{aligned}$$

(61)

$$\begin{aligned}&\frac{1}{w_{n_j}}R_{l}^{{\varSigma }}(w_{n_j}t)\rightarrow \sigma _{l}^{{\varSigma }}(t), \quad \forall l\in \mathcal {L},\end{aligned}$$

(62)

$$\begin{aligned}&\frac{1}{w_{n_j}}H_{l}^{{\varSigma }}(w_{n_j}t)\rightarrow \phi _{l}^{{\varSigma }}(t), \quad \forall l\in \mathcal {L}, \end{aligned}$$

(63)

$$\begin{aligned}&\frac{1}{w_{n_j}}D_{l,r}^{{\varSigma }}(w_{n_j}t)\rightarrow \mu _{l,r}^{{\varSigma }}(t), \quad \forall (l,r),\end{aligned}$$

(64)

$$\begin{aligned}&\frac{1}{w_{n_j}}X_{l,r}(w_{n_j}t)\rightarrow x_{l,r}(t), \quad \forall (l,r),\end{aligned}$$

(65)

$$\begin{aligned}&\frac{1}{w_{n_j}}Q_{l}(w_{n_j}t)\rightarrow q_{l}(t), \quad \forall l\in \mathcal {L}, \end{aligned}$$

(66)

where the limiting functions $\sigma _{l}^{{\varSigma }}(t), \phi _l^{{\varSigma }}(t), \mu _{l,r}^{{\varSigma }}(t), x_{l,r}(t),q_l(t)$ are Lipschitz continuous in $[0,\infty )$, which implies that these limiting functions are differentiable for almost all t. Let $\mathcal {T}$ be the set of time instants where these functions are differentiable. Then the following equations hold for all $t\in \mathcal {T}$:

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\sigma _{l}^{{\varSigma }}(t)=\sigma _l(t),\quad \forall l\in \mathcal {L}, \end{aligned}$$

(67)

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mu _{l,r}^{{\varSigma }}(t)=\mu _{l,r}(t),\quad \forall (l,r), \end{aligned}$$

(68)

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\phi _{l}^{{\varSigma }}(t)=\sigma _l(t),\quad \forall l\in \mathcal {L},\quad \text { whenever } q_l(t)>0, \end{aligned}$$

(69)

$$\begin{aligned}&\phi _{l}^{{\varSigma }}(t)=\sum _{r:l\in r}\mu _{l,r}^{{\varSigma }}(t), \quad \forall (l,r), \end{aligned}$$

(70)

$$\begin{aligned}&q_{l}(t)=\sum _{r:l\in r}x_{l,r}(t), \quad \forall (l,r), \end{aligned}$$

(71)

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}x_{l,r}(t)=\lambda _{l,r}+\mu _{l_{-}^{(r)},r}(t)-\mu _{l,r}(t),\quad \forall (l,r), \end{aligned}$$

(72)

where $\mu _{l_{-}^{(r)},r}(t)=0$ if $l=l_1^{(r)}$. Here, $\varvec{\sigma }(t)=(\sigma _l(t))_{l\in \mathcal {L}}$ is the longest vector within the capacity region $\varvec{\Lambda }$ that is in the same direction as $\mathbf {q}(t)$ and $\varvec{\mu }(t)=(\mu _{l,r})_{l\in \mathcal {L},r\in \mathcal {R}}$ satisfies

$$\begin{aligned}&\mu _{l_{-}^{(r)},r}(t)=0,\quad \text {whenever } l=l_1^{(r)},\end{aligned}$$

(73)

$$\begin{aligned}&\mu _{l,r}(t)=0, \quad \text {whenever } x_{l,r}(t)=0,\end{aligned}$$

(74)

$$\begin{aligned}&\frac{\mu _{l,r}(t)}{x_{l,r}(t)}=\frac{\mu _{l',r}(t)}{x_{l',r}(t)}, \end{aligned}$$

(75)

for any route $r\in l,r'\in l'$ with $x_{l,r}(t)>0$ and $x_{l',r'}(t)>0$.

Thus, Eqs. (9), (11), and (12) immediately follow from Lemma 5. By combining (68), (69), and (70), we have

$$\begin{aligned} \sigma _l(t)=\sum _{r:l\in r}\mu _{l,r}(t),\quad \text {if } q_l(t)>0, \end{aligned}$$

(76)

Since $\frac{\mu _{l,r}(t)}{x_{l,r}(t)}=\frac{\mu _{l,r'}(t)}{x_{l,r'}(t)}$ for all $r,r'\in l$ with $x_{l,r}(t)>0$ and $x_{l,r'}(t)>0$,

$$\begin{aligned} \frac{\mu _{l,r}(t)}{x_{l,r}(t)}=\frac{\sigma _l(t)}{q_l(t)}, \end{aligned}$$

(77)

for any route $r\in l$ with $x_{l,r}(t)>0$. Therefore, if $q_l(t)>0$, we have

$$\begin{aligned} \mu _{l,r}(t)=\frac{x_{l,r}(t)}{q_l(t)}\sigma _l(t), \quad \text {whenever } x_{l,r}(t)>0. \end{aligned}$$

(78)

Noting Eq. (74) and the convention $0/0=0$, we have Eq. (10).

Proof of Lemma 5

Equation (61) follows from the functional strong law of large numbers if $l=l_1^{(r)}$. If $l\ne l_1^{(r)}$, then $A_{l,r}^{{\varSigma }}(w_{n_j}t)=\lambda _{l,r}=0$ and thus Eq. (61) always holds.

Note that for any $0\le t_1\le t_2$ we have

$$\begin{aligned} 0\le \frac{1}{w_{n}}R_{l}^{{\varSigma }}(w_{n}t_2)-\frac{1}{w_{n}}R_{l}^{{\varSigma }}(w_{n}t_1)\le t_2-t_1, \end{aligned}$$

(79)

where we use the fact that each link l can at most transfer one packet in one time slot. Thus, the sequence of functions $\{\frac{1}{w_{n}}R_{l}^{{\varSigma }}(w_{n}t)\}$ is uniformly equicontinuous, and since $R_{l}^{{\varSigma }}(0)=0$, the sequence is also uniformly bounded. Similarly, the sequence $\{\frac{1}{w_{n}}D_{l,r}^{{\varSigma }}(w_{n}t)\}$ is uniformly bounded and uniformly equicontinuous. Consequently, according to the Arzela–Ascoli Theorem, there must exist a subsequence of $\{w_n\}_{n\ge 1}$ for which both (62) and (64) hold. Since

$$\begin{aligned}&\frac{1}{w_{n}}H_{l}^{{\varSigma }}(w_{n}t)=\sum _{r:l\in r}\frac{1}{w_{n}}D_{l,r}^{{\varSigma }}(w_{n}t), \nonumber \\&\frac{1}{w_{n}}X_{l,r}(w_{n}t)=\frac{1}{w_{n}}A_{l,r}^{{\varSigma }}(w_{n}t)-\frac{1}{w_{n}}D_{l_{-}^{(r)},r}^{{\varSigma }}(w_{n}t)+\frac{1}{w_{n}}D_{l,r}^{{\varSigma }}(w_{n}t),\nonumber \\&\frac{1}{w_{n}}Q_{l}^{{\varSigma }}(w_{n}t)=\sum _{r:l\in r}\frac{1}{w_{n}}X_{l,r}^{{\varSigma }}(w_{n}t), \end{aligned}$$

we have (63), (65) and (66) by taking limits as $w_{n_j}\rightarrow \infty$.

Since the functions $R_l^{{\varSigma }}(t)$, $H_l^{{\varSigma }}(t)$, $D_{l,r}^{{\varSigma }}(t)$, $X_{l,r}(t)$, $Q_l(t)$ are Lipschitz continuous, the Lipschitz continuity of $\sigma _{l}^{{\varSigma }}(t)$, $\phi _l^{{\varSigma }}(t)$, $\mu _{l,r}^{{\varSigma }}(t)$, $x_{l,~r}(t)$, $q_l(t)$ also follows. Hence, these limiting functions are differentiable for almost all t. In the rest of proof we consider all $t\in \mathcal {T}$, where $\mathcal {T}$ is the set of time instants where the limiting functions are differentiable.

Next, we prove Eq. (67). $\sigma _l(\mathbf {q}(t))$ is continuous with respect to $\mathbf {q}$ when $q_l>0$. Therefore, for any $\epsilon >0$, there exists a $u>0$ such that for all $t'\in [t,t+u]$ we have

$$\begin{aligned} \left| \sigma _l(\mathbf {q}(t'))-\sigma _l(\mathbf {q}(t))\right| \le \epsilon . \end{aligned}$$

(80)

Since $\frac{1}{w_{n_j}}\mathbf {Q}(\lfloor w_{n_j}t'\rfloor )\rightarrow \mathbf {q}(t')$ uniformly over compact intervals of time, and $\sigma _l(a\mathbf {Q})=\sigma _l(\mathbf {Q})$ for any $a>0$, we have $\sigma _l(\mathbf {Q}(\lfloor w_{n_j}t'\rfloor ))\rightarrow \sigma _l(\mathbf {q}(t'))$ with probability 1. Thus, there exists an integer $J>0$ such that for all $j>J$ and $t'\in [t,t+u]$,

$$\begin{aligned} \sigma _l(\mathbf {q}(t'))-\epsilon \le \sigma _l(\mathbf {Q}(\lfloor w_{n_j}t'\rfloor ))\le \sigma _l(\mathbf {q}(t'))+\epsilon . \end{aligned}$$

(81)

Combining with (80), we have

$$\begin{aligned} \sigma _l(\mathbf {q}(t))-2\epsilon \le \sigma _l(\mathbf {Q}(\lfloor w_{n_j}t'\rfloor ))\le \sigma _l(\mathbf {q}(t))+2\epsilon . \end{aligned}$$

(82)

By the definition of the limit in (62), for any $t'\in [t,t+u]$ we have

$$\begin{aligned} \sigma _l^{{\varSigma }}(t')-\sigma _l^{{\varSigma }}(t)=\lim _{j\rightarrow \infty }\frac{1}{w_{n_j}}\sum _{k=\lfloor w_{n_j}t\rfloor }^{\lfloor w_{n_j}t'\rfloor -1}R_l(k). \end{aligned}$$

(83)

Define the filtration $\mathcal {F}_k, k=1,2,\ldots$, where $\mathcal {F}_k$ is the $\sigma -$algebra generated by the random variables $A_{l,r}^{{\varSigma }}(\lfloor w_{n_j}t\rfloor +k')$, $R_{l}^{{\varSigma }}(\lfloor w_{n_j}t\rfloor +k')$, $D_{l,r}^{{\varSigma }}(\lfloor w_{n_j}t\rfloor +k')$, $X_{l,r}(\lfloor w_{n_j}t\rfloor +k')$ for all pairs (l, r) and for $k'=0,1,2,\ldots , k-1$. Let

Therefore, $\sum _{m=0}^{k-1}Y_m, k=1,2,\ldots$, is a martingale with respect to the filtration $\mathcal {F}_k, k=1,2,\ldots$. Further, $\mathbb {E}\left[ Y_k^2\right]$ is always bounded for all k. Hence, using a strong law of large numbers for martingales [5], we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{1}{k}\sum _{m=0}^{k-1}Y_m=0, \quad \text {with probability } 1, \end{aligned}$$

(84)

which implies that

(85)

By using (82), we have

$$\begin{aligned} (t'-t)(\sigma _l(\mathbf {q}(t))-2\epsilon )\le \sigma _l^{{\varSigma }}(t')-\sigma _l^{{\varSigma }}(t)\le (t'-t)(\sigma _l(\mathbf {q}(t))+2\epsilon ), \end{aligned}$$

for $t'\in [t,t+u]$. Since we assume that $\sigma _l^{{\varSigma }}(t)$ is differentiable at t (i.e., $t\in \mathcal {T}$), we have

$$\begin{aligned} \sigma _l(\mathbf {q}(t))-2\epsilon \le \frac{\mathrm{d}}{\mathrm{d}t}\sigma _l^{{\varSigma }}(t)\le \sigma _l(\mathbf {q}(t))+2\epsilon . \end{aligned}$$

(86)

Finally, since this is true for any $\epsilon >0$, Eq. (67) follows. The proof of Eq. (68) follows a similar technique and is omitted here for brevity.

Next, we consider Eq. (69). If $q_l(t)>0$, then there exists a positive u such that for all $t'\in [t,t+u]$, $q_l(t')>0$. This implies that for all sufficiently large j, the backlog $Q_l(\lfloor w_{n_j}t'\rfloor )$ at link l is larger than 1 for all $t'\in [t,t+u]$. Therefore, the available service to link l will be fully utilized between $\lfloor w_{n_j}t\rfloor$ and $\lfloor w_{n_j}(t+u)\rfloor$. We thus have

$$\begin{aligned} R_l^{{\varSigma }}(\lfloor w_{n_j}t'\rfloor )-R_l^{{\varSigma }}(\lfloor w_{n_j}t\rfloor )=H_l^{{\varSigma }}(\lfloor w_{n_j}t'\rfloor )-H_l^{{\varSigma }}(\lfloor w_{n_j}t\rfloor ), \end{aligned}$$

for all $t\le t'\le t+u$. Dividing both sides by $w_{n_j}$ and taking limits as $w_{n_j}\rightarrow \infty$, we have

$$\begin{aligned} \sigma _{l}^{{\varSigma }}(t')-\sigma _{l}^{{\varSigma }}(t)=\phi _{l}^{{\varSigma }}(t')-\phi _{l}^{{\varSigma }}(t), \end{aligned}$$

(87)

for all $t\le t'\le t+u$, which implies $\frac{\mathrm{d}}{\mathrm{d}t}\phi _l^{{\varSigma }}(t)=\frac{\mathrm{d}}{\mathrm{d}t}\sigma _l^{{\varSigma }}(t)$. By combining with (67), we have Eq. (69).

Equations (70) and (71) follow from the equation $H_l^{{\varSigma }}(t)=\sum _{r:l\in r}D_{l,r}^{{\varSigma }}(t)$ and $Q_{l}(t)=\sum _{r:l\in r}X_{l,r}(t)$ by taking limits as $w_{n_j}\rightarrow \infty$, respectively.

Finally, by using the queue-evolution Eq. (60) and taking limits as $w_{n_j}\rightarrow \infty$, we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}x_{l,r}(t)=\lambda _{l,r}+\frac{\mathrm{d}}{\mathrm{d}t}\mu _{l_{-}^{(r)},r}^{{\varSigma }}(t)-\frac{\mathrm{d}}{\mathrm{d}t}\mu _{l,r}^{{\varSigma }}(t). \end{aligned}$$

(88)

By utilizing Eq. (68), we have Eq. (72). $\square$

Appendix 2: Proof of Lemma 1

We prove it by contradiction. Assume

$$\begin{aligned} \frac{\mathrm{D}^+}{\mathrm{d}x^+}f(x)>\max _{i\in \mathcal {K}}\left\{ \frac{\mathrm{D}^+}{\mathrm{d}x^+}f_i(x)\right\} . \end{aligned}$$

(89)

Then, for a sufficiently small $\epsilon >0$, there exists a decreasing sequence $\{u_k,k=1,2,\ldots \}$ with $\lim _{k\rightarrow \infty }u_k=0$ such that

$$\begin{aligned} \frac{f(x+u_k)-f(x)}{u_k} \ge \max _{i\in \mathcal {K}}\left\{ \frac{\mathrm{D}^+}{\mathrm{d}x^+}f_i(x)\right\} +\epsilon , \quad \forall k=1,2,\ldots . \end{aligned}$$

Note that $f(x)=f_i(x),\forall i\in \mathcal {K}$. Since there are a finite number of locally Lipschitz continuous functions $f_i(x), i=1,2,\ldots ,K$, there must exist a $j\in \mathcal {K}$ and a decreasing subsequence $\{u_{t_k},k=1,2,\ldots \}$ of $\{u_{k},k=1,2,\ldots \}$ such that $f_j(x+u_{t_k})=f(x+u_{t_k})=\max _{i=1,2,\ldots ,K}f_i(x+u_{t_k}), \forall k=1,2,\ldots$, which implies that

$$\begin{aligned} \frac{f_j(x+u_{t_k})-f_j(x)}{u_{t_k}} \ge \max _{i\in \mathcal {K}}\left\{ \frac{\mathrm{D}^+}{\mathrm{d}x^+}f_i(x)\right\} +\epsilon , \quad \forall k=1,2,\ldots . \end{aligned}$$

Therefore, we obtain the contradiction

$$\begin{aligned} \frac{\mathrm{D}^+}{\mathrm{d}x^+}f_j(x)\ge \max _{i\in \mathcal {K}}\left\{ \frac{\mathrm{D}^+}{\mathrm{d}x^+}f_i(x)\right\} +\epsilon . \end{aligned}$$

(90)

Hence, we have the desired result.

Appendix 3: Proof of Lemma 2

(i) Assume that there exists a $t_1>0$ and a $\zeta >0$ such that $g(t_1)=\zeta$. Since $g(0)=0$, according to the continuity property of the function $g(\cdot )$ there exists a $t_2\in (0,t_1)$ such that $g(t_2)=\zeta /2$ and $g(t)\ge \zeta /2>0$ for any $t\in (t_2,t_1]$. Since $g(t)>0$ for any $t\in [t_2,t_1]$ and $\frac{\mathrm{D}^+}{\mathrm{d}t^+}g(t)\le 0$ whenever $g(t)>0$, we have $g(t_1)\le g(t_2)$, which contradicts that $g(t_1)=\zeta >g(t_2)=\zeta /2$. Therefore, $g(t)=0$ for all $t\ge 0$.

(ii) Since $\frac{\mathrm{D}^+}{\mathrm{d}t^+}g(t)\le -\gamma$ whenever $g(t)>0$, the function $g(\cdot )$ will first hit zero at time $T=g(0)/\gamma$. Then, by using the technique in (i), we can show that $g(t)=0$ for all $t\ge T$.