Power optimization and subcarrier allocation for downlink MIMO-OFDMA based cognitive radio networks

Abstract

This paper presents a novel resource allocation framework for downlink transmissions in MIMO-OFDMA based cognitive radio (CR) networks. In this literature, due to the coexistence of primary and secondary users (SUs) in the same geographical area, the interference induced to the primary users’ (PUs) bands from SUs’ transmissions should be less than the tolerable interference threshold predefined by PU. Hence, the optimal resource allocation strategy is performed subject to the total power constraint in the cognitive base station and the interference power constraint for PUs so that the downlink sum throughput of the CR network can be maximize without causing unacceptable interference. To this aim, two sub-problems are presented and solved theoretically based on convex optimization framework that can be used in the main body of the proposed algorithm. Regarding to the first sub-problem, the resources are distributed among all SUs in a way that just the total power constraint is satisfied. Also, the sum throughput is maximized as the interference power onto the PUs’ bands keeps in a tolerable range in the second sub-problem. So, in order to find the optimum solution, our algorithm consists of three parts. The first two parts include these sub-problems while the last section includes a combination of them. For complexity reduction, two sub-optimal resource allocation algorithms are introduced based on the simplification of main problem constraints. The performance of different approaches is investigated via simulation. Conducted simulation results show that the optimal algorithm achieves better performance in comparison with benchmark algorithms and converges to sub-problem solutions in high and low regimes of interference thresholds.

Appendices

Appendix 1

We want to find the optimal solution for the optimization problem given by (7). Due to the convexity of the problem (7) with regard to $\left\{ {p_{{_{k,n,j} }}^{sub1} } \right\}$, the Lagrangian can be written as

$$\begin{aligned} & L\left( {\left\{ {p_{{_{k,n,j} }}^{sub1} } \right\} ,\left\{ {N_{k} } \right\} ,\left\{ {\mu_{k,n,j} } \right\} ,\lambda } \right) = \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {{ \log }_{2} \left( {1 + \frac{{\varLambda^{2}_{k,n,j} p_{{_{k,n,j} }}^{sub1} }}{{\sigma^{2} }}} \right)} } } \\ & - \lambda \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{sub1} - P_{T} } } } } \right) + \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\mu_{k,n,j} p_{{_{k,n,j} }}^{sub1} } } } \\ \end{aligned}$$

(27)

where $\left\{ {\mu_{k,n,j} } \right\}$ and $\lambda$ are non-negative Lagrangian multipliers for constraints or dual variables. Define the dual function

$$g\left( {\lambda ,\left\{ {\mu_{k,n,j} } \right\}} \right) = \mathop { \hbox{max} }\limits_{{\left\{ {p_{{_{k,n,j} }}^{sub1} } \right\} ,\left\{ {N_{k} } \right\}}} \left\{ {L\left( {\left\{ {p_{{_{k,n,j} }}^{sub1} } \right\}} \right)} \right\}$$

(28)

The Lagrangian dual problem can be expressed as

$$\begin{aligned} & \mathop { \hbox{min} }\limits_{{\lambda ,\left\{ {\mu_{k,n,j} } \right\}}} \left\{ {g\left( {\lambda ,\left\{ {\mu_{k,n,j} } \right\}} \right)} \right\} \\ & s.t.\;\lambda ,\left\{ {\mu_{k,n,j} } \right\} \ge 0 \\ \end{aligned}$$

(29)

The difference between the optimum solution and the solution obtained by dual problem is called the duality gap. Regardless of the convexity of the main problem, the dual problem is convex and the duality gap approaches zero when the number of subcarrier tends to infinity [41]. The KKT condition can be written as follows

$$\lambda \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }}^{{}} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{sub1\;*} - P_{T} } } } } \right) = 0$$

(30)

$$\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }}^{{}} {\sum\limits_{j = 1}^{L} {\mu_{k,n,j} p_{{_{k,n,j} }}^{sub1\;*} } } } = 0$$

(31)

$$\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{sub1} \le P_{T} } } }$$

(32)

$$p_{{_{k,n,j} }}^{sub1} \ge 0$$

(33)

$$\frac{\partial L}{{\partial p_{{_{k,n,j} }}^{sub1\;*} }} = \frac{{\varLambda^{2}_{k,n,j} }}{{{\text{ln(}}2 )\text{ }\left( {\sigma^{2} + \varLambda^{2}_{k,n,j} p_{{_{k,n,j} }}^{sub1\;*} } \right)}} - \lambda + \mu_{k,n,j} = 0$$

(34)

Rearranging the condition (34) we obtain

$$p_{{_{k,n,j} }}^{sub1\;*} = \left( {\frac{1}{{{\text{ln(}}2 )\text{ } (\lambda - \mu_{k,n,j} )}} - \frac{{\sigma^{2} }}{{\varLambda^{2}_{k,n,j} }}} \right)$$

(35)

Satisfying conditions (31) and (33), we obtain

$$p_{{_{k,n,j} }}^{sub1\;*} = \left( {\frac{1}{{\text{ln(2) (}\lambda \text{)}}} - \frac{{\sigma^{2} }}{{\varLambda^{2}_{k,n,j} }}} \right)^{ + }$$

(36)

where $( {\text{x)}}^{ + } = {\text{max(x,}}0 )$. For satisfaction of condition (30), if inserting $\lambda = 0$ in (36), the condition (32) will be violated. Thus, $\lambda > 0$ is chosen to fulfill the total power constraint with equality $\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{sub1\;*} = P_{T} } } }$. The optimum value of $\lambda$ can be found efficiently by iterative updating it utilizing bisection method until convergence of $\lambda$. Therefore, we can obtain the optimal solution based on (36).

Appendix 2

Similar to Appendix 1. we want to find the optimal solution for the optimization problem given by (10). So, the Lagrangian of problem (10) can be defined as

$$\begin{aligned} & L\left( {\left\{ {p_{{_{k,n,j} }}^{sub2} } \right\} ,\left\{ {N_{k} } \right\} ,\left\{ {\varOmega_{k,n,j} } \right\} ,\varOmega } \right) = \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {{ \log }_{2} \left( {1 + \frac{{\varLambda^{2}_{k,n,j} p_{{_{k,n,j} }}^{sub2} }}{{\sigma^{2} }}} \right)} } } \\ & \quad - \varOmega \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} p_{{_{k,n,j} }}^{sub2} I_{n,l} - I^{l} } } } } \right) + \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\varOmega_{k,n,j} p_{{_{k,n,j} }}^{sub2} } } } \\ \end{aligned}$$

(37)

where $\left\{ {\varOmega_{k,n,j} } \right\}$ and $\varOmega$ are the non-negative Lagrangian multipliers for constraints or dual variables. The dual function associated with (37) can be written as

$$g\left( {\varOmega ,\left\{ {\varOmega_{k,n,j} } \right\}} \right) = \mathop { \hbox{max} }\limits_{{\left\{ {p_{{_{k,n,j} }}^{sub2} } \right\} ,\left\{ {N_{k} } \right\}}} \left\{ {L\left( {\left\{ {p_{{_{k,n,j} }}^{sub2} } \right\}} \right)} \right\}$$

(38)

Therefore the dual problem is defined as

$$\begin{aligned} & \mathop { \hbox{min} }\limits_{{\varOmega ,\left\{ {\varOmega_{k,n,j} } \right\}}} \left\{ {g\left( {\varOmega ,\left\{ {\varOmega_{k,n,j} } \right\}} \right)} \right\} \\ & s.t.\;\varOmega ,\left\{ {\varOmega_{k,n,j} } \right\} \ge 0 \\ \end{aligned}$$

(39)

Regardless of the convexity of the main problem, the dual problem is convex and the duality gap approaches zero when the number of subcarrier tends to infinity [41]. Then, the KKT condition can be written as follows

$$\varOmega \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} p_{{_{k,n,j} }}^{sub2\;*} I_{n,l} - I^{l} } } } } \right) = 0$$

(40)

$$\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\varOmega_{k,n,j} p_{{_{k,n,j} }}^{sub2\;*} } } } = 0$$

(41)

$$\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} p_{{_{k,n,j} }}^{sub2} I_{n,l} \le I^{l} } } }$$

(42)

$$p_{{_{k,n,j} }}^{sub2} \ge 0$$

(43)

$$\frac{\partial L}{{\partial p_{{_{k,n,j} }}^{sub2\;*} }} = \frac{{\varLambda^{2}_{k,n,j} }}{{{\text{ln(}}2 )\text{ }\left( {\sigma^{2} + \varLambda^{2}_{k,n,j} p_{{_{k,n,j} }}^{sub2\;*} } \right)}} - \varOmega \left| {g_{n,j}^{l} } \right|^{2} I_{n,l} + \varOmega_{k,n,j} = 0$$

(44)

Rearranging the condition (44) we obtain

$$p_{{_{k,n,j} }}^{sub2\;*} = \left( {\frac{1}{{{\text{ln(}}2 )\text{ }\left( {\varOmega \left| {g_{n,j}^{l} } \right|^{2} I_{n,l} - \varOmega_{k,n,j} } \right)}} - \frac{{\sigma^{2} }}{{\varLambda^{2}_{k,n,j} }}} \right)$$

(45)

Satisfying conditions (41) and (43), we obtain

$$p_{{_{k,n,j} }}^{sub2\;*} = \left( {\frac{1}{{{\text{ln(}}2 )\left( {\varOmega \left| {g_{n,j}^{\,l} } \right|^{2} I_{n,l} } \right)}} - \frac{{\sigma^{2} }}{{\varLambda^{2}_{k,n,j} }}} \right)^{ + }$$

(46)

where $( {\text{x)}}^{ + } = {\text{max(x,}}0 )$. For satisfaction of condition (40), if inserting $\varOmega = 0$ in (46), the condition (42) will be violated. Thus, $\varOmega > 0$ is chosen to fulfill the interference power constraint with equality as $\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} I_{n,l} p_{{_{k,n,j} }}^{sub2\;*} = I^{l} } } }$. The optimum value of $\varOmega$ can be found efficiently by iterative updating it utilizing bisection method until convergence of $\varOmega$. Therefore, we can obtain the optimal solution based on (46).

Appendix 3

Proof:

Based on the theorem 2, the problem (6) should be solved with jointly satisfaction of two constraints. Therefore, the Lagrangian will be derived as

$$\begin{aligned} & L\left( {\left\{ {p_{k,n,j} } \right\} ,\left\{ {N_{k} } \right\} ,\left\{ {\mu_{k,n,j} } \right\} ,\lambda_{1} ,\lambda_{2} } \right) = \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {{ \log }_{2} \left( {1 + \frac{{\varLambda^{2}_{k,n,j} p_{k,n,j} }}{{\sigma^{2} }}} \right)} } } \\ & \quad - \lambda_{\,1} \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{k,n,j} - P_{T} } } } } \right) - \lambda_{\,2} \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} p_{{_{k,n,j} }} I_{n,l} - I^{l} } } } } \right) \\ & \quad + \sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\mu_{k,n,j} p_{k,n,j} } } } \\ \end{aligned}$$

(47)

where $\left\{ {\mu_{k,n,j} } \right\}$, $\lambda_{1}$ and $\lambda_{2}$ are the non-negative Lagrangian multipliers for constraints. The dual function and dual problem can be defined as

$$g\left( {\lambda_{1} ,\lambda_{2} ,\left\{ {\mu_{k,n,j} } \right\}} \right) = \mathop { \hbox{max} }\limits_{{\left\{ {p_{k,n,j} } \right\} ,\left\{ {N_{k} } \right\}}} \left\{ {L\left( {\left\{ {p_{k,n,j} } \right\}} \right)} \right\}$$

(48)

$$\begin{aligned} \mathop { \hbox{min} }\limits_{{\lambda_{1} ,\lambda_{2} ,\left\{ {\mu_{k,n,j} } \right\}}} \left\{ {g (\lambda_{1} ,\lambda_{2} ,\left\{ {\mu_{k,n,j} } \right\}} \right\} )\hfill \\ s.t.\;\lambda_{1} ,\lambda_{2} ,\left\{ {\mu_{k,n,j} } \right\} \ge 0 \hfill \\ \end{aligned}$$

(49)

Similar to previous appendixes, by deriving the KKT conditions for dual problem, the following terms should be satisfied

$$\begin{aligned} & (1 )\text{ }\lambda_{1} \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{*} - P_{T} } } } } \right) = 0 \\ & (2 )\text{ }\lambda_{2} \left( {\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{\,l} } \right|^{2} p_{{_{k,n,j} }}^{*} I_{n,l} - I^{l} } } } } \right) = 0 \\ & (3 )\text{ }\mu_{k,n,j} p_{{_{k,n,j} }}^{*} = 0 \\ & (4 )\text{ }\frac{\partial L}{{\partial p_{{_{k,n,j} }}^{*} }} = 0 \\ \end{aligned}$$

(50)

Now, by satisfying the two last conditions (3) and (4) in (50), the allocated power to user k on the subcarrier n and jth spatial subchannel can be specified as

$$\text{ }p_{{_{k,n,j} }}^{*} = \left( {\frac{1}{{{\text{ln(}}2 )\text{ } (\lambda_{1} + \lambda_{2} \left| {g_{n,j}^{l} } \right|^{2} I_{n,l} )}} - \frac{{\sigma^{2} }}{{\varLambda^{2}_{k,n,j} }}} \right)^{ + }$$

(51)

However, there are still two unaddressed conditions (1) and (2) in (50). Hence, the following cases are considered to solve these conditions.

$$\text{ }\lambda_{1} = 0,\;\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{*} < P_{T} } } }$$

(52)

$$\text{ }\lambda_{1} > 0,\;\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{*} = P_{T} } } }$$

(53)

$$\text{ }\lambda_{2} = 0,\;\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{l} } \right|^{2} p_{{_{k,n,j} }}^{*} I_{n,l} < I^{l} } } }$$

(54)

$$\text{ }\lambda_{2} > 0,\;\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{l} } \right|^{2} p_{{_{k,n,j} }}^{*} I_{n,l} = I^{l} } } }$$

(55)

As the power values depend on the values of $\lambda_{1}$ and $\lambda_{2}$, the optimal solution needs a search over these values which must be updated jointly. If $\lambda_{1} = 0$ and $\lambda_{2} > 0$, the optimal solution is determined based on sub-problem 2. Similarly, if $\lambda_{2} = 0$ and $\lambda_{1} > 0$, the optimal solution is defined based on the sub-problem 1 which has been expressed in the previous section. On the other hand, according to (50) the conditions (52) and (54) cannot be satisfied jointly. Therefore, in order to satisfy (53) and (54), the optimal values of $\lambda_{\,1}$ and $\lambda_{\,2}$ are chosen to meet jointly the total power and interference constraints with equalities $\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {\left| {g_{n,j}^{l} } \right|^{2} p_{{_{k,n,j} }}^{*} I_{n,l} = I^{l} } } }$ and $\sum\limits_{k = 1}^{K} {\sum\limits_{{n \in N_{k} }} {\sum\limits_{j = 1}^{L} {p_{{_{k,n,j} }}^{*} = P_{T} } } }$. This can be efficiently done by iterative updating these values using an ellipsoid or subgradient-based method [16]. To the best of our knowledge, the ellipsoid search is one of the best methods which can be applied while its complexity and convergence rate is desired [17]. Here, the power is allocated based on (51) and the subcarrier allocation can be performed based on per subcarrier optimization problem as follows

$$\kappa_{n} = \mathop {{ \arg }\;{ \hbox{max} }}\limits_{k} \;\left\{ {\sum\limits_{j = 1}^{L} {{ \log }_{2} \left( {1 + \frac{{\varLambda_{{_{k,n,j} }}^{2} p_{{_{k,n,j} }}^{*} }}{{\sigma^{2} }}} \right) - \sum\limits_{j = 1}^{L} {\left( {\lambda_{1} + \lambda_{2} \left| {g_{n,j}^{\,l} } \right|^{2} I_{n,l} } \right)p_{{_{k,n,j} }}^{*} } } } \right\}$$

(56)