Joint remote radio heads and baseband units pool resource scheduling for delay-aware traffic in cloud radio access networks

Abstract

Due to the centralized signal processing and powerful computational ability, cloud radio access networks (C-RAN) is considered as a promising technique to meet the increasing demand of high-data-rate services of the fifth generation wireless communications. This paper investigates the stochastic optimization problem of C-RAN by joint remote radio heads and baseband units pool resource scheduling to achieve the throughput utility maximization. We formulate a problem to maximize the average throughput utility while stabilizing all processing and transmission queues under the power, subcarrier and computational resources constraints. By utilizing Lyapunov optimization technique, the primal problem can be decomposed into four subproblems. Throughput utility maximization, admission control and computational resource allocation subproblems can be solved easily. For the joint power and subcarrier allocation subproblem, we utilize time-sharing and alternating approaches to obtain a feasible solution. By utilizing the results of four subproblems, a joint admission control and resource allocation (JACRA) algorithm is proposed. The simulation results are provided to show that the total average throughput of JACRA algorithm is increased by 4.02% compared with joint Hungarian and iterative waterfilling (JHIW) algorithm, but the total average backlogs of JACRA algorithm are decreased by 7.04% when the control parameter grows from 1 to 10.

Appendix

1.1 A

Proof (Proof of Lemma 1)

Let $U_{1}^{*}$ and $U_{2}^{*}$ be the optimal throughput utility value of problem (P1) and problem (P2), respectively. In order to prove problem (P1) and problem (P2) are equivalent. We prove $U_{1}^{*}\ge U_{2}^{*}$ firstly, and then we prove $U_{1}^{*}\le U_{2}^{*}$.

Firstly, we prove $U_{1}^{*}\ge U_{2}^{*}$. We assume $\gamma _{m}^{*}(t)$, $r_{m}^{*}(t)$, $\mu _{m}^{*}(t)$, $p_{lmi}^{*}(t)$ are the optimal solutions to achieve $U_{2}^{*}$ in problem (P2), and we denote them as $\rho _{2}^{*}(t)$. Because $U(\cdot )$ is a concave function, based on Jensen’s inequality, we can get

$$\begin{aligned} \sum _{n\in M} U(\overline{\gamma _{m}^{*}})\ge \sum _{n\in M}\overline{U(\gamma _{m}^{*})}=U_{2}^{*}. \end{aligned}$$

(37)

In addition, because $r_{m}^{*}\ge \gamma _{m}^{*}$ and $U(\cdot )$ is nondecreasing, we can get

$$\begin{aligned} \sum _{n\in M} U(\overline{r_{m}^{*}})\ge \sum _{n\in M}U(\overline{\gamma _{m}^{*}}). \end{aligned}$$

(38)

Because the constraints of problem (P2) include all the constraints of problem (P1), $\rho _{2}^{*}(t)$ is a feasible solution for problem (P1), and the utility value is not larger than $U_{1}^{*}$. Therefore, we can get

$$\begin{aligned} U_{1}^{*}\ge \sum _{n\in M} U(\overline{r_{m}^{*}})\ge U_{2}^{*}. \end{aligned}$$

(39)

Secondly, we prove $U_{1}^{*}\le U_{2}^{*}$. We assume $r_{m}^{*}(t)$, $\mu _{m}^{*}(t)$, $p_{lmi}^{*}(t)$ are the optimal solutions to achieve $U_{1}^{*}$ in problem (P1), and we denote them as $\rho _{1}^{*}(t)$. Because $\rho _{1}^{*}(t)$ satisfies the constraints (6b)–(6i) in problem (P1), it also satisfies the constraint (8d) in problem (P2). We set $\gamma _{m}(t)=\overline{r_{m}^{*}}$ which can satisfy the constraints (8b) and (8c) in problem (P2). Therefore, $\gamma _{m}(t)$ and $\rho _{1}^{*}(t)$ are the feasible solution of problem (P2). By the definition, we can get $\overline{U(\gamma _{m})}=\lim _{t\rightarrow \infty }(\frac{1}{t})\sum _{\tau =0}^{t-1}U(\gamma _{m}(\tau ))=\lim _{t\rightarrow \infty }(\frac{1}{t})\sum _{\tau =0}^{t-1}U(\overline{r_{m}^{*}})=U(\overline{r_{m}^{*}})$. Thus, we can get

$$\begin{aligned} U_{2}^{*}\ge \sum _{m\in \mathcal {M}}\overline{U(\gamma _{m})}=\sum _{m\in \mathcal {M}}U(\overline{r_{m}^{*}})=U_{1}^{*}. \end{aligned}$$

(40)

Based on Eqs. (39) and (40), we can get $U_{1}^{*}=U_{2}^{*}$. Therefore, problem (P1) and problem (P2) are equivalent.

Therefore, Lemma 1 is proved. $\square$

1.2 B

Proof (Proof of Lemma 2)

By leveraging the fact that $(\max [a-b,0]+c)^{2}\le a^{2}+b^{2}+c^{2}-2a(b-c), \forall a, b, c\ge 0$ [33]. For Eq. (3), we have

$$\begin{aligned} \begin{aligned}&H_{m}(t+1)^{2}-H_{m}(t)^{2}=(\max [H_{m}(t)-\mu _{m}(t),0]+r_{m}(t))^{2}\\&-H_{m}(t)^{2}\le [\mu _{m}(t)^{2}+r_{m}(t)^{2}]+2H_{m}(t)[r_{m}(t)-\mu _{m}(t)]. \end{aligned} \end{aligned}$$

(41)

Similarly, for the queue $Q_{m}(t)$ and virtual queue $Z_{m}(t)$, we can get

$$\begin{aligned} Q_{m}(t+1)^{2}-Q_{m}(t)^{2}\le & {} [\mu _{m}(t)^{2}+c_{m}(t)^{2}]\nonumber \\&+\,\,2Q_{m}(t)[\mu _{m}(t)-c_{m}(t)]. \end{aligned}$$

(42)

$$\begin{aligned} Z_{m}(t+1)^{2}-Z_{m}(t)^{2}\le & {} [\gamma _{m}(t)^{2}+r_{m}(t)^{2}]\nonumber \\&+\,\,2Z_{m}(t)[\gamma _{m}(t)-r_{m}(t)]. \end{aligned}$$

(43)

Based on Eqs. (11), (41), (42) and (43), $\varDelta (\varvec{\Theta }(t))$ can be represented as

$$\begin{aligned} \begin{aligned}&\varDelta (\varvec{\Theta }(t))= \mathbf {E}\left[ \frac{1}{2}\sum _{m\in \mathcal {M}}[H_{m}(t+1)^{2}-H_{m}(t)^{2}\right. \\&\qquad \left. +\,\,Q_{m}(t+1)^{2}-Q_{m}(t)^{2}+Z_{m}(t+1)^{2}-Z_{m}(t)^{2}]\mid \varvec{\Theta }(t) \right] \\&\quad \le \mathbf {E}\left[ \frac{1}{2}\sum _{m\in \mathcal {M}}[\mu _{m}(t)^{2}+r_{m}(t)^{2}+\mu _{m}(t)^{2}+c_{m}(t)^{2}\right. \\&\qquad \left. +\,\,\gamma _{m}(t)^{2}+r_{m}(t)^{2}]\mid \varvec{\Theta }(t)\right] \\&\qquad +\,\,\mathbf {E}\left[ \sum _{m\in \mathcal {M}}[H_{m}(t)[r_{m}(t)-\mu _{m}(t)]+Q_{m}(t)[\mu _{m}(t)-c_{m}(t)] \right. \\&\qquad \left. +\,\,Z_{m}(t)[\gamma _{m}(t)-r_{m}(t)]]\mid \varvec{\Theta }(t)\right] \\&\quad \le D+\mathbf {E}[G(t)\mid \varvec{\Theta }(t)]. \end{aligned} \end{aligned}$$

(44)

Based on $r_{m}(t)\le B_{m}$, the constraints (6e), (6f), (6g) in problem (P1) and the constraint (8c) in problem (P2), we have

$$\begin{aligned} \begin{aligned}&\mathbf {E}\left[ \frac{1}{2}\sum _{m\in \mathcal {M}}[2r_{m}(t)^{2}+\gamma _{m}(t)^{2}+2\mu _{m}(t)^{2}+ c_{m}(t)^{2}]\mid \varvec{\Theta }(t)\right] \\&\quad \le \mathbf {E}\left[ \frac{1}{2}\sum _{m\in \mathcal {M}}[3B_{m}^{2}+2\min (\mu _{total}, C_{f})^{2}+C_{f}^{2}]\mid \varvec{\Theta }(t)\right] \\&\quad =\frac{1}{2}\sum _{m\in \mathcal {M}}[3B_{m}^{2}+2\min (\mu _{total}, C_{f})^{2}+C_{f}^{2}]=D. \end{aligned} \end{aligned}$$

(45)

Therefore, Lemma 2 is proved. $\square$

1.3 C

Proof (Proof of Lemma 3)

For a fixed allocated time-share value, problem (P3-1) becomes a convex problem. We can use Lagrange dual method to solve this problem [34]. The Lagrange function of objective function can be written as

$$\begin{aligned} \begin{aligned} L(p_{mi},\lambda _{l_{m}})&= -\sum _{m\in \mathcal {M}}\sum _{i\in \mathcal {I}}Q_{m}W_{0}\log _{2}(1+p_{mi}g_{mi})\\&\quad +\lambda _{l_{m}} \left( \sum _{m\in \mathcal {M}_{l}}\sum _{i\in \mathcal {I}}p_{mi}-P_{l}\right) . \end{aligned} \end{aligned}$$

(46)

To maximize the Lagrange function, we only need to maximize the power allocation pointwise.

$$\begin{aligned} L(p_{mi},\lambda _{l_{m}}) = -\,\,Q_{m}W_{0}\log _{2}(1+p_{mi}g_{mi})+\lambda _{l_{m}} (p_{mi}-P_{l}). \end{aligned}$$

(47)

And then, use the Lagrange dual method differentiating $L(p_{mi},\lambda _{l_{m}})$ with respect to $p_{mi}$, and set the result to zero

$$\begin{aligned} \frac{\partial L(p_{mi},\lambda _{l_{m}})}{\partial p_{mi}} = 0, \end{aligned}$$

(48)

we can get $p_{mi}$

$$\begin{aligned} p_{mi} = \frac{Q_{m}W_{0}}{\lambda _{l_{m}}\ln 2}-\frac{1}{g_{mi}}. \end{aligned}$$

(49)

Because the value of power can not be negative, the power allocation is

$$\begin{aligned} p_{mi} = \left( \frac{Q_{m}W_{0}}{\lambda _{l_{m}}\ln 2}-\frac{1}{g_{mi}},0\right) ^{+},~~x^{+} =\max (x,0). \end{aligned}$$

(50)

Therefore, Lemma 3 is proved. $\square$

1.4 D

Proof (Proof of Lemma 4)

For a fixed allocated power, we can use Lagrange dual method to solve Eq. (30). The Lagrange function of objective function can be written as

$$\begin{aligned} \begin{aligned} L(\alpha _{mi},\beta _{i})&=-\sum _{m\in \mathcal {M}}\sum _{i\in \mathcal {I}}Q_{m}\alpha _{mi}W_{0}\log _{2}(1+p_{mi}g_{mi})\\&\quad +\beta _{i}\left( \sum _{m\in \mathcal {M}}\alpha _{mi}-1\right) . \end{aligned} \end{aligned}$$

(51)

To maximize the Lagrange function, we only need to maximize the power allocation pointwise.

$$\begin{aligned} \begin{aligned} L(\alpha _{mi},\beta _{i})&=-\,\,Q_{m}\alpha _{mi}W_{0}\log _{2}(1+p_{mi}g_{mi})\\&\quad +\beta _{i}(\alpha _{mi}-1). \end{aligned} \end{aligned}$$

(52)

And then, use the Lagrange dual method differentiating $L(\alpha _{mi},\beta _{i})$ with respect to $\alpha _{mi}$, and set the result to zero

$$\begin{aligned}&\frac{\partial (\alpha _{mi},\beta _{i})}{\partial \alpha _{mi}} = 0, \end{aligned}$$

(53)

$$\begin{aligned}&-\,\,Q_{m}W_{0}\left( \log _{2}(1+\frac{q_{mi}g_{mi}}{\alpha _{mi}})-\frac{\frac{q_{mi}g_{mi}}{\alpha _{mi}}}{ln2(1+\frac{q_{mi}g_{mi}}{\alpha _{mi}})}\right) \nonumber \\&\qquad +\,\,\beta _{i}=0. \end{aligned}$$

(54)

Therefore,

$$\begin{aligned} \begin{aligned}&u_{m}(y) = Q_{m}W_{0}\left( \log _{2}(1+y)-\frac{y}{\ln 2(1+y)}\right) = \beta _{i},\\&y = \frac{q_{mi}g_{mi}}{\alpha _{mi}}. \end{aligned} \end{aligned}$$

(55)

Therefore, Lemma 4 is proved. $\square$

1.5 E

Proof (Proof of Lemma 5)

Based on Eq. (12), we can get

$$\begin{aligned} \begin{aligned}&\varDelta (\varvec{\Theta }(t))-V\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] \\&\quad \le D- V\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] \\&\qquad +\mathbf {E}\left[ \sum _{m\in \mathcal {M}}[H_{m}(t)[r_{m}(t)-\mu _{m}(t)]+Q_{m}(t)[\mu _{m}(t)\right. \\&\qquad \left. -\,\,c_{m}(t)]+Z_{m}(t)[\gamma _{m}(t)-r_{m}(t)]\mid \varvec{\Theta }(t)\right] . \end{aligned} \end{aligned}$$

(56)

Equation (12) is to minimize the right hand of Eq. (56) when the $\varvec{\Theta }(t)$ is fixed. Therefore, for arbitrary feasible solution $r_{m}^{*}(t)$, $\gamma _{m}^{*}(t)$, $\mu _{m}^{*}(t)$, $p_{mi}^{*}(t)$ and $\alpha _{mi}^{*}(t)$ should satisfy the following inequality

$$\begin{aligned} \begin{aligned}&\varDelta (\varvec{\Theta }(t))-V\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] \\&\quad \le D- V\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}^{*}(t))\mid \varvec{\Theta }(t)\right] \\&\qquad +\sum _{m\in \mathcal {M}}H_{m}(t)\mathbf {E}[r_{m}^{*}(t)-\mu _{m}^{*}(t)\mid \varvec{\Theta }(t)]\\&\qquad +\sum _{m\in \mathcal {M}}Q_{m}(t)\mathbf {E}[\mu _{m}^{*}(t)-c_{m}^{*}(t)\mid \varvec{\Theta }(t)]\\&\qquad +\sum _{m\in \mathcal {M}}Z_{m}(t)\mathbf {E}[\gamma _{m}^{*}(t)-r_{m}^{*}(t)\mid \varvec{\Theta }(t)]. \end{aligned} \end{aligned}$$

(57)

We can get if problem (P1) can be solved, there must exist a feasible control strategy to satisfy the following inequality for arbitrary $\delta >0$ [25],

$$\begin{aligned} \begin{aligned}&\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}^{*}(t))\right] \ge U_{1}^{*}-\delta ,\\&\mathbf {E}[r_{m}^{*}(t)-\mu _{m}^{*}(t)]\le \delta ,\\&\mathbf {E}[\mu _{m}^{*}(t)-c_{m}^{*}(t)]\le \delta ,\\&\mathbf {E}[\gamma _{m}^{*}(t)-r_{m}^{*}(t)]\le \delta , \end{aligned} \end{aligned}$$

(58)

where $U_{1}^{*}$ is the optimal value of problem (P1). Let $\delta \rightarrow 0$, we substitute (57) into (56), we can get

$$\begin{aligned} \varDelta (\varvec{\Theta }(t))-V\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] \le D-VU_{1}^{*}. \end{aligned}$$

(59)

We take the expectation of Eq. (59), and based on the fact $L(\varvec{\Theta }(t))\ge 0$, we can get

$$\begin{aligned} U_{1}^{*}-\frac{1}{T}\sum _{t\in T}\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\right] \le \frac{D}{V}+\frac{\mathbf {E}[L(\varvec{\Theta }(0))]}{VT}. \end{aligned}$$

(60)

We take $T\rightarrow \infty$ for Eq. (60), we can get

$$\begin{aligned} U_{1}^{*}-\sum _{m=1}^{M}\overline{U_{m}(\gamma _{m})}\le \mathcal {O}(1/V). \end{aligned}$$

(61)

Therefore, Lemma 5 is proved. $\square$

1.6 F

Proof (Proof of Lemma 6)

Based on Eq. (59), we can get

$$\begin{aligned} \varDelta (\Theta (t))\le D+V\left[ \mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] -U_{1}^{*}\right] . \end{aligned}$$

(62)

Based on the Lyapunov theory, Eq. (62) means all the queues are rate stable and $\frac{1}{T}\sum _{t\in \mathcal {T}} \mathbf {E}[\gamma _{m}^{*}(t)-r_{m}^{*}(t)]\le 0$. On the Slater condition, if there is a feasible control strategy $\nu$, which can satisfy $\mathbf {E}[r_{m}^{\nu }(t)-\mu _{m}^{\nu }(t)]\le -\varepsilon$, $\mathbf {E}[\mu _{m}^{\nu }(t)-c_{m}^{\nu }(t)]\le -\varepsilon$ and $\mathbf {E}[\sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}^{\nu }(t))]=U_{\varepsilon }$.[35] Based on Eq. (57), we can get

$$\begin{aligned} \begin{aligned} \varDelta (\varvec{\Theta }(t))&\le D+V\left[ \mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\mid \varvec{\Theta }(t)\right] \right. \\&\quad \left. -U_{\varepsilon }\right] -\varepsilon \left( \sum _{m\in \mathcal {M}}H_{m}(t)+\sum _{m\in \mathcal {M}}Q_{m}(t)\right) . \end{aligned} \end{aligned}$$

(63)

We take the expectation of Eq. (63), we can get

$$\begin{aligned} \begin{aligned}&\frac{1}{T}\sum _{t\in \mathcal {T}}\sum _{m\in \mathcal {M}}(\mathbf {E}[H_{m}(t)]+\mathbf {E}[Q_{m}(t)])\le \frac{D}{\varepsilon }\\&\quad +\frac{V}{\varepsilon }\left[ \frac{1}{T}\sum _{t\in \mathcal {T}}\mathbf {E}\left[ \sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))\right] -U_{\varepsilon }\right] +\frac{\mathbf {E}[L(\varvec{\Theta }(0))]}{\varepsilon T}. \end{aligned} \end{aligned}$$

(64)

Due to $\frac{1}{T}\sum _{t\in \mathcal {T}}\mathbf {E}[\sum _{m\in \mathcal {M}}U_{m}(\gamma _{m}(t))]\le U_{1}^{*}$, we take $T\rightarrow \infty$ for Eq. (64), we can get

$$\begin{aligned} \begin{aligned}&\sum _{m\in \mathcal {M}}\overline{H_{m}}+\sum _{m\in \mathcal {M}}\overline{Q_{m}}\\&\quad =\lim _{T\rightarrow \infty }\sum _{t\in \mathcal {T}}\sum _{m\in \mathcal {M}}(\mathbf {E}[H_{m}(t)]+\mathbf {E}[Q_{m}(t)])\\&\quad \le \frac{D}{\varepsilon }+\frac{U_{1}^{*}-U_{\varepsilon }}{\varepsilon }V. \end{aligned} \end{aligned}$$

(65)

Therefore, Lemma 6 is proved. $\square$

1.7 G

Proof (Proof of Lemma 7)

Based on Eq. (7), we can get $Z_{m}(t)\ge 0$. And then, we utilize induction to prove $Z_{m}(t)\le V\omega _{m}+B_{m}$. We assume that $Z_{m}(t)\le V\omega _{m}+B_{m}$ at time slot t, then we prove it also holds at time slot $t+1$.

Firstly, for the case $Z_{m}(t)\le V\omega _{m}$, from Eq. (7), we can see that this queue can increase by at most $B_{m}$ at each time slot, therefore, we have $Z_{m}(t)\le V\omega _{m}+B_{m}$.

Secondly, we consider $V\omega _{m}<Z_{m}(t)\le V\omega _{m}+B_{m}$. At each time slot, BBU pool decides $\gamma _{m}(t)$ to maximize the following expression

$$\begin{aligned} VU(\gamma _{m}(t))-Z_{m}(t)\gamma _{m}(t). \end{aligned}$$

(66)

Based on the property of the maximum derivative, for any $\gamma _{m}(t)\ge 0$, we can get

$$\begin{aligned} \begin{aligned}&VU(\gamma _{m}(t))-Z_{m}(t)\gamma _{m}(t)\le VU(0)+V\omega _{m}\gamma _{m}(t)\\&\qquad -Z_{m}(t)\gamma _{m}(t)=VU(0)+\gamma _{m}(t)[V\omega _{m}-Z_{m}(t)]\\&\quad \le VU(0). \end{aligned} \end{aligned}$$

(67)

Equation (67) holds if and only if $\gamma _{m}(t)=0$ [36]. Then the algorithm will choose $\gamma _{m}(t)=0$ to maximize the expression (66), we can get

$$\begin{aligned} Z_{m}(t+1)=\max [Z_{m}(t)-r_{m}(t),0]\le Z_{m}(t)\le V\omega _{m}+B_{m}. \end{aligned}$$

(68)

$Z_{m}(t+1)\le V\omega _{m}+B_{m}$ is satisfied for these two cases. Therefore, Lemma 7 is proved. $\square$