Adaptive proportional fairness resource allocation for OFDM-based cognitive radio networks

Abstract

In this paper, we study the resource allocation problem in multiuser Orthogonal Frequency Division Multiplexing (OFDM)-based cognitive radio networks. The interference introduced to Primary Users (PUs) is fully considered, as well as a set of proportional rate constraints to ensure fairness among Secondary Users (SUs). Since it is extremely computationally complex to obtain the optimal solution because of integer constraints, we adopt a two-step method to address the formulated problem. Firstly, a heuristic subchannel assignment is developed based on the normalized capacity of each OFDM subchannel by jointly considering channel gain and the interference to PUs, which approaches a rough proportional fairness and removes the intractable integer constraints. Secondly, for a given subchannel assignment, we derive a fast optimal power distribution algorithm that has a complexity of O(L ² N) by exploiting the problem’s structure, which is much lower than standard convex optimization techniques that generally have a complexity of O((N + K)³), where N, L and K are the number of subchannels, PUs and SUs, respectively. We also develop a simple power distribution algorithm with complexity of only O(L + N), while achieving above 90 % sum capacity of the upper bound. Experiments show that our proposed algorithms work quite well in practical wireless scenarios. A significant capacity gain is obtained and the proportional fairness is satisfied perfectly.

Appendix: Proof of Theorem 1

We propose an L + 1 iterations procedure to solve (16). Define s + 1 intermediate variables, $u_1^s,u_2^s,\ldots,u_{s+1}^s \in {\mathbf R}^{N+K-1}$ at step $s, s=1,2,\ldots,L+1$, the procedure is illustrated as follows. Step 1: since H ₁ = H ₂ + G ₁ G ^T₁ , we have

$$ x = u_1^1 - \frac{G_1u_1^1}{1 + G_1^Tu_2^1}u_2^1, $$

(30)

and u ¹₁ , u ¹₂ can be calculated as

$$ \begin{aligned} H_{2}u_1^1 &= G_0\\ H_{2}u_2^1 &= G_1.\\ \end{aligned} $$

(31)

Step $s:\, s=2,\ldots,L$. As H _s  = H _s+1 + G _s G ^T _s , we have

$$ u_i^{s-1}= u_i^{s} - \frac{G_{s}^T u_i^{s}}{1 + G_{s}^T u_{s+1}^{s}}u_{s+1}^{s}, i=1,\ldots,s, $$

(32)

and we need to solve the following s + 1 equations

$$ H_{s+1}u_j^{s} = G_{j-1}, j=1,2,\ldots,s+1. $$

(33)

Step L + 1: we need to solve the following L + 2 equations at this step

$$ H u_j^{L+1} = G_{j-1}, j=1,2,\ldots,L+2. $$

(34)

Without loss of generality, each equation in (34) can be written as

$$ \left[ \begin{array}{ll} D & A^T\\ A & {\mathbf 0}_m\\ \end{array} \right ] \left[ \begin{array}{l} u\\ \nu\\ \end{array} \right ] = \left[ \begin{array}{l} G\\ {\mathbf 0}_v \\ \end{array} \right ], $$

(35)

where $x, G \in {\bf R}^{N\times1}$ and $\nu \in {\bf R}^{(K-1)\times 1}$. Recall that D is a diagonal matrix, denote $\theta_k = N - \sum_{i=k}^{N}N_i$, we have

$$ \begin{aligned} & \lambda_{\theta_{k}+i} u_{\theta_{k}+i}-\beta_1 \nu_k = h_{\theta_{k}+i}\\ & \beta_k\sum_{i=1}^{N_1}u_i -\beta_1 \sum_{i=1}^{N_k} u_{\theta_{k}+i}=0,\\ \end{aligned} $$

(36)

$i=1,2,\ldots,N_k$ for $k=1,2,\ldots,K$. And from (36), we have

$$ \begin{array}{ll} X_k = a_k +b_k \left (\beta_1\nu_k \right)\\ X_k = \frac{\beta_k}{\beta_1}X_1, \\ \end{array} $$

(37)

where $X_k = \sum\limits_{i=1}^{N_k} u_{\theta_{k}+i},\,a_k = \sum\limits_{i=1}^{N_k} \frac{h_{\theta_{k}+i}}{\lambda_{\theta_{k}+i}},\,b_k = \sum\limits_{i=1}^{N_k}\frac{1}{\lambda_{\theta_{k}+i}}$. Using the set of equations in (37), we can obtain

$$ \begin{aligned} X_1 &=\left( \sum\limits_{k=1}^{K}\frac{ a_k\beta_k}{b_k}\right)/\left(\sum\limits_{k=1}^{K}\frac{\beta_k^2}{b_k \beta_1}\right)\\ \nu_k &= \frac{\beta_k}{\beta_1^2 b_k}X_1-\frac{a_k}{b_k\beta_1}, k=2,\ldots,K.\\ \end{aligned} $$

(38)

As ν _k is worked out by (38), we can insert it to (36) to obtain u. The computation cost is O (K + N). Since K≪ N in practice, the complexity can be denoted as O (N).

Note that there are L + 2 equations in (34), each equation can be solved with complexity O (N). We carry out an inverse process by using the intermediate variables until Newton step x is worked out. In total, we should calculate L ² + 3L + 3 variables and the complexity of computing each variable is O (N). So the complexity of our proposed power allocation algorithm is O (L ² N).