Downlink resource allocation for on-demand multimedia services in high-speed railway communication systems

Abstract

With the rapid development of high-speed railway (HSR) system, there is an increasing demand on providing high throughput and continuous multimedia (CM) services for HSR passengers. In this paper, we investigate the downlink resource allocation problem for on-demand CM services in HSR OFDMA systems with a cellular/infostation integrated network architecture. Considering both the integrity and continuity of service transmission, the resource allocation problem is formulated as a two-stage optimization programming. The aim of this study is to maximize the total reward of delivered services then to minimize the weighted total number of cumulative discontinuity packets over the trip of the train. To resolve the difficulty of multi-stage optimization, an equivalent one-stage programming is proposed. Since the resultant mixed integer programming is NP-hard in general, we reformulate it as a sparse $\ell _{0}$-minimization problem and then relax it to a linear programming. Furthermore, a reweighted $\ell _1$-minimization technique is applied to improve the system performance. Guided by the proposed optimization approaches, we next develop two efficient online resource allocation algorithms for practical systems,where a-priori knowledge of future service arrivals and channel gains is not available. Finally, simulation results are provided to validate the feasibility and effectiveness of the proposed algorithms.

Appendix: proof of Lemma 1

Proof

Notice that problem P2 is equivalent to problem (8), and the problem (8) is equivalent to

$$\begin{aligned} (\mathbf {P5})~~\max _{\{x_{hks}\}_{s\in \mathcal {S}^{*}}}&~ \alpha \mathrm {tr}(\mathbf {W}^\mathrm {T}\mathbf {B}\mathbf {X}) \end{aligned}$$

(21a)

$$\begin{aligned} \hbox {s.t}.~&~\mathrm {({7\hbox {b}})-({7\hbox {e}})}, \end{aligned}$$

(21b)

then we only need to prove that the optimization problem P1 and P5 is equivalent to the problem P3.

Let M, N and Q be the optimal values of problem P1, P5 and P3, respectively. First, to prove $Q \ge M+N$. Let $x_{hks}^{*}$ be an optimal solution achieving M, N in problem P1 and P5, respectively. Since the constraints in problem P3 include all of the constraints of the original problem P1 and P5, $x_{hks}^{*}$ is a feasible solution for problem P3 which gives an objective value $M+N$ that is not larger than the optimal value Q. Thus we conclude that $Q \ge M+N$.

Next, to prove $M+N \ge Q$. Let $x_{hks}^{**}$ and $\mathcal {S}^{**}$ be the optimal solution and optimal service set of problem P3, respectively. Note that the objective function (9a) of the problem P3 is composed of two items, where the first term $\sum _{s\in \mathcal {S}}w_s\psi _{s}$ is discontinuous (combination of a number of $w_s$) with minimal improvement is $w^{\star }$. Recall that $0 < \alpha < \frac{w^{\star }}{{\mathrm{tr}}(\mathbf {W}^\mathrm {T}\mathbf {B}\mathbf {X}^{*})}$, which implies $\alpha {\mathrm{tr}}(\mathbf {W}^{{\mathrm {T}}}\mathbf {B}\mathbf {X})< w^{\star }$. Hence, the total contribution from the second term in (9a) cannot exceed $w^{\star }$, regardless of the resource allocation $\mathbf {X}$. Thus, let $Q = \hat{M}+\hat{N}$, where $\hat{M}$ (combination of a number of $w_s$) and $\hat{N}$ $(0<\hat{N}<w^{\star })$ be the optimal value of the first and second item in (9a), respectively. Since $x_{hks}^{**}$ satisfies the constraints (5b)–(5e), it is also a feasible solution for the problem P1 which gives an objective value $\hat{M}$ that is not larger than M (i.e. $\hat{M}\le M$). Assume that $\hat{M}< M$, and we will derive a contradiction. If $\hat{M}< M$, then we will have $\hat{M}+w^{\star }\le M$, and the optimal value of problem P3 should be $Q > M \ge \hat{M}+w^{\star }>\hat{M}+\hat{N}$. This contradicts the optimality of $Q = \hat{M}+\hat{N}$. Thus,we get that $\hat{M}= M$. As we know, once $x_{hks}^{**}$ is determined, the optimal service set $\mathcal {S}^{**}$ is also determined by the problem P1. Thus $x_{hks}^{**}$ is a feasible solution of problem P5 in $\mathcal {S}^{**}$ with an objective value $\hat{N}$ that is not larger than N. Therefore, we obtain $Q = \hat{M}+\hat{N}= M+\hat{N}\le M+N$.

From the above analysis, we can conclude $Q = M+N$, and that an optimal solution for the problem P3 can be directly turned into an optimal solution for the problem P1 and P5. The proof is complete. $\square$