Distributed joint subcarrier and discrete power allocation for cognitive radio ad hoc networks

Abstract

One of the key problems in orthogonal frequency division multiple-access (OFDMA) cognitive radio (CR) ad hoc networks is to efficiently and fairly allocate subcarriers and powers in a distributed manner. However, two formidable shortcomings exist in most previous works. One is that the fairness issue has not been sufficiently taken into account so that different types of fairness among secondary users (SUs) may not be guaranteed. The other is that the transmission power of each SU is assumed to take any value in a continuous domain, whereas for practical CR ad hoc networks, the power level can only be quantized into discrete values. To overcome the above shortcomings, an optimization framework is first presented, where different types of fairness for resource allocation are considered and the transmission power of each SU is allowed to take only a finite number of discrete values. In particular, the fairness of resource allocation is guaranteed by associating each SU with a utility function for each subcarrier, where the utility function is allowed to be non-concave or non-differentiable so that our framework can deal with resource allocation for real-time applications. Furthermore, to solve the proposed non-convex integer optimization problem, a distributed algorithm with low complexity is proposed, according to which only limited cooperation among network entities is required. At last, simulation results verify that our algorithm has very good convergence and fairness performance, and then it may be applied to practical OFDMA-based CR ad hoc networks.

Appendix: A Proof of same dual function of problems (4) and (5)

In this part, we prove that problem (5) has the same dual function and dual problem with problem (4). First, by introducing the Lagrange multipliers $\mu _{m},$ $\tau _{n},$ the Lagrangian function of problem (5) can be given as follows:

$$\begin{aligned}&L({\mu },\tau ,\mathbf {prob})\nonumber \\&\quad =\sum \nolimits _{m=1}^{M}\sum \nolimits _{k=1}^{K} \sum \nolimits _{a_{m}^{k}} \sum \nolimits _{i=1}^{I_{a_{m}^{k}}}prob_{m}^{k}(i,a_{m}^{k})U_{m}^{k}(r_{m}^{k}(i,a_{m}^{k}))\nonumber \\&\qquad +\sum \nolimits _{m=1}^{M}\mu _{m}\Bigg \{P_{m}^{\max }-\sum \nolimits _{k=1}^{K}\sum \nolimits _{a_{m}^{k}} \sum \nolimits _{i=1}^{I_{a_{m}^{k}}}\nonumber \\&\qquad \times \,prob_{m}^{k}(i,a_{m}^{k})p_{m}^{k}(i,a_{m}^{k})\Bigg \} \nonumber \\&\qquad +\sum \nolimits _{n=1}^{N}\tau _{n}\Bigg \{ I_{n}^{th}-\sum \nolimits _{m=1}^{M}\sum \nolimits _{k=1}^{K}\sum \nolimits _{a_{m}^{k}} \sum \nolimits _{i=1}^{I_{a_{m}^{k}}}\nonumber \\&\qquad \times \,prob_{m}^{k}(i,a_{m}^{k})a_{m}^{k}b_{n}^{k}g_{mn}^{k}p_{m}^{k}(i,a_{m}^{k})\Bigg \}\nonumber \\&\quad =\sum \nolimits _{m=1}^{M}\sum \nolimits _{k=1}^{K}\sum \nolimits _{a_{m}^{k}} \sum \nolimits _{i=1}^{I_{a_{m}^{k}}}prob_{m}^{k}(i,a_{m}^{k})\nonumber \\&\qquad \times \,\Bigg \{ U_{m}^{k}(r_{m}^{k}(i,a_{m}^{k}))-\mu _{m}p_{m}^{k}(i,a_{m}^{k}) \nonumber \\&\qquad -\sum \nolimits _{n=1}^{N}\tau _{n}a_{m}^{k}b_{n}^{k}g_{mn}^{k}p_{m}^{k}(i,a_{m}^{k})\Bigg \}\nonumber \\&\qquad +\sum \nolimits _{m=1}^{M}\mu _{m}P_{m}^{\max }+\sum \nolimits _{n=1}^{N}\tau _{n}I_{n}^{th}. \end{aligned}$$

(23)

Then, the dual function can be given as

$$\begin{aligned} d({\mu },{\tau })= & {} {\max }_{\mathbf {prob}}L({\mu }, \tau ,\mathbf {prob}) \nonumber \\\le & {} \sum \nolimits _{m=1}^{M}\sum \nolimits _{k=1}^{K}\sum \nolimits _{a_{m}^{k}} \sum \nolimits _{i=1}^{I_{a_{m}^{k}}}prob_{m}^{k}\big (i,a_{m}^{k}\big )\nonumber \\&{\max }_{\begin{array}{c} a_{m}^{k},i=1,\ldots ,I_{a_{m}^{k}}, \\ p_{m}^{k}\big (i,a_{m}^{k}\big )\in X_{m}^{k}\big (a_{m}^{k}\big ) \end{array}}\nonumber \\&\times \Bigg \{ U_{m}^{k}\big (r_{m}^{k}\big (i,a_{m}^{k}\big )\big ) -\mu _{m}p_{m}^{k}\big (i,a_{m}^{k}\big )\nonumber \\&\quad -\sum \nolimits _{n=1}^{N}\tau _{n}a_{m}^{k}b_{n}^{k}g_{mn}^{k}p_{m}^{k}\big (i,a_{m}^{k}\big )\Bigg \}\nonumber \\&\quad +\sum \nolimits _{m=1}^{M}\mu _{m}P_{m}^{\max }+\sum \nolimits _{n=1}^{N}\tau _{n}I_{n}^{th}. \end{aligned}$$

(24)

Due to the equality $\sum _{a_{m}^{k}} \sum _{i=1}^{I_{a_{m}^{k}}}prob_{m}^{k}(i,a_{m}^{k})=1,\forall m,\forall k$, it follows that

$$\begin{aligned} d({\mu },{\tau })= & {} \sum \nolimits _{m=1}^{M}\sum \nolimits _{k=1}^{K}{\max } _{\begin{array}{c} a_{m}^{k},i=1,\ldots ,I_{a_{m}^{k}}, \\ p_{m}^{k}(i,a_{m}^{k}) \in X_{m}^{k}(a_{m}^{k}) \end{array}}\\&\times \Bigg \{U_{m}^{k}(r_{m}^{k}(i,a_{m}^{k}))-\mu _{m}p_{m}^{k}(i,a_{m}^{k})\\&-\sum \nolimits _{n}\tau _{n}a_{m}^{k}b_{n}^{k}g_{mn}^{k}p_{m}^{k}(i,a_{m}^{k})\Bigg \}\\&+\sum \nolimits _{m}\mu _{m}P_{m}^{\max }+\sum \nolimits _{n}\tau _{n}I_{n}^{th}. \end{aligned}$$

Similarly, it can be known that $d({\mu },{\tau })$ is also the dual function of (4), and then problem (4) has the same dual function and dual problem with problem (5). This completes the proof.