Joint subcarrier and power allocation for downlink cellular OFDMA systems by intercell interference limitation

Abstract

This paper addresses resource allocation for sum throughput maximization in a sectorized two-cell downlink orthogonal frequency-division multiple-access (OFDMA) system impaired by multicell interference. It is well known that the optimization problem for this scenario is NP-hard and combinational, which is here converted to a novel sum throughput maximization problem in cellular OFDMA systems based on the intercell interference limitation. Then, three subclasses of this new problem are solved. By the first subclass, on the assumption that subcarrier allocation parameters are fixed, an algorithm for optimal power allocation is obtained. However, the optimal resource allocation requires an exhaustive search, including the optimal power allocation which cannot be implemented in practice due to its high complexity. By the second subclass, the problem is reduced to a single cell case where the intercell interference in each subcarrier is limited to a certain threshold. Based on the solution of the single cell problem, a distributed resource allocation scheme with the aim of small information exchange between the coordinated base stations is proposed. By the third subclass, the centralized resource allocation for two adjacent cells as a general problem is solved. Here, the algorithm allocates simultaneously the subcarriers and the power of the considered two cells while the resource allocation parameters of both cells are coupled mutually. The present study shows that distributed and centralized resource allocation algorithms have much less complexity than the algorithm used in the exhaustive search and can be used in practice as efficient multicell resource allocation algorithms. Simulation results illustrate the performance improvements of the proposed schemes in comparison to the schemes with no intercell interference consideration.

Appendices

Appendix A: Optimum power allocation

After setting the derivative of $L(p_{k,n}^{A},p_{k,n}^{B})$ (from (14)) with regard to $p_{k,n}^{A}$ and $p_{k,n}^{B}$ to zero, respectively, the following equations are obtained:

$$ \rho_{{k,n}}^{A}.C_{k,n}^{\prime A} \bigl(p_{k,n}^{A}\bigr) - \lambda^{A} - \varOmega _{{k,n}}^{A} + v_{{k,n}}^{A} = 0 $$

(32)

$$ \rho_{{k,n}}^{B}.C_{k,n}^{\prime B} \bigl(p_{k,n}^{B}\bigr) - \lambda^{B} - \varOmega _{{k,n}}^{B} + v_{{k,n}}^{B} = 0 $$

(33)

where

$$ \begin{aligned} &C_{k,n}^{\prime A}\bigl(p_{k,n}^{A} \bigr) = \frac{\gamma_{{k,n}}^{A}}{\ln(2)(1 + \gamma _{{k,n}}^{A}p_{k,n}^{A})}, \\ &C_{k,n}^{\prime B}\bigl(p_{k,n}^{B} \bigr) = \frac{\gamma_{{k,n}}^{B}}{\ln(2)(1 + \gamma_{{k,n}}^{B}p_{k,n}^{B})} \end{aligned} $$

(34)

The KKT conditions impose that the following conditions must be satisfied for the optimal solution

$$ (1)\quad\lambda^{A}\Biggl(\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} - P_{{T}}^{A} \Biggr) = 0 $$

(35)

$$ (2)\quad\lambda^{B}\Biggl(\sum_{k = 1}^{K^{B}} \sum_{n = 1}^{N} p_{{k,n}}^{B} - P_{{T}}^{B} \Biggr) = 0 $$

(36)

$$ (3)\quad\varOmega_{{k,n}}^{A}\bigl(p_{{k,n}}^{A} - I_{n}^{A}\bigr) = 0 $$

(37)

$$ (4)\quad\varOmega_{{k,n}}^{B}\bigl(p_{{k,n}}^{B} - I_{n}^{B}\bigr) = 0 $$

(38)

$$ (5)\quad v_{{k,n}}^{A}p_{{k,n}}^{A} = 0 $$

(39)

$$ (6)\quad v_{{k,n}}^{B}p_{{k,n}}^{B} = 0 $$

(40)

Due to the fact that the proof of optimal $p_{{k,n}}^{A*}$ is the same as $p_{{k,n}}^{B*}$, we provide proof for $p_{{k,n}}^{A*}$ only.

With regard to condition $3 (\varOmega_{{k,n}}^{A}(p_{{k,n}}^{A} - I_{n}^{A}) = 0)$, we can consider two cases:

$$(\mbox{3-{I}}) \quad\mbox{if } p_{{k,n}}^{A} < I_{n}^{A} \quad\mbox{then } \varOmega_{{k,n}}^{A} = 0 $$

$$(\mbox{3-{II}}) \quad\mbox{if }\varOmega_{{k,n}}^{A} > 0 \quad\mbox{then } p_{{k,n}}^{A} = I_{n}^{A} $$

And with regard to condition 5 $(v_{{k,n}}^{A}p_{{k,n}}^{A} = 0)$, two cases are considered:

$$(\mbox{5-{I}}) \quad \mbox{if }p_{{k,n}}^{A} > 0 \quad \mbox{then } v_{{k,n}}^{A} = 0 $$

$$(\mbox{5-{II}}) \quad \mbox{if }v_{{k,n}}^{A} > 0 \quad \mbox{then } p_{{k,n}}^{A} = 0 $$

By combination of 3-I and 5-II we can obtain

If $\rho_{{k,n}}^{A}.C_{k,n}^{\prime A}(p_{k,n}^{A}) - \lambda^{A} < 0$ then $p_{{k,n}}^{A} = 0$, therefore

$$ \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} < 0 \quad\Rightarrow\quad p_{{k,n}}^{A} = 0 $$

(41)

By combination of 3-II and 5-I we can obtain

If $\rho_{{k,n}}^{A}.C_{k,n}^{\prime A}(p_{{k,n}}^{A} = I_{n}^{A}) - \lambda^{A} > 0$ then $p_{{k,n}}^{A} = I_{n}^{A}$, therefore

$$ \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} > I_{n}^{A} \quad\Rightarrow\quad p_{{k,n}}^{A} = I_{n}^{A} $$

(42)

By combination of 3-I and 5-I we can obtain

$$\begin{aligned} &{0 < \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma _{{k,n}}^{A}} < I_{n}^{A}} \\ &{\quad\Rightarrow\quad p_{{k,n}}^{A} = \frac{1}{\ln (2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} } \end{aligned}$$

(43)

3-II and 5-II do not occur jointly.

Due to Eqs. (41), (42) and (43) we can write

$$ p_{{k,n}}^{A} = \min\biggl(\biggl(\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}}\biggr)^{ +},I_{n}^{A}\biggr) $$

(44)

But if the condition $\lambda^{A}(\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} - P_{{T}}^{A} ) = 0$ is satisfied, the optimum value of λ ^A is obtained. Due to condition 1, two cases can be considered as follows:

$$(\mbox{1-{I}}) \quad\mbox{if }\lambda^{A} > 0\quad\mbox{then }\sum _{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} = P_{{T}}^{A} $$

$$(\mbox{1-{II}}) \quad\mbox{if }\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} < P_{{T}}^{A}\quad\mbox{then }\lambda^{A} = 0 $$

By combination of 1-II and (44) we can deduce that

$$\begin{aligned} &{\lambda^{A} = 0} \\ &{\quad\Rightarrow\quad p_{{k,n}}^{A} = \min \biggl(\biggl(\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma _{{k,n}}^{A}}\biggr)^{ +},I_{n}^{A} \biggr) = I_{n}^{A}} \end{aligned}$$

(45)

Thus if $\sum_{n = 1}^{N} I_{n}^{A} < P_{{T}}^{A}$, then $p_{{k,n}}^{A} = I_{n}^{A}$. And if $\sum_{n = 1}^{N} I_{n}^{A} \ge P_{{T}}^{A}$,we can conclude that the λ ^A>0 and for the optimal solution the condition $\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} = P_{{T}}^{A}$ must be satisfied. Therefore, the optimal value of λ ^A is chosen to fulfill the total power constraint with equality. We can summarize the optimal power allocation as follows

$$ \begin{aligned}[b] &p_{k,n}^{A*} = \left \{ \begin{array}{l@{\quad}l} I_{n}^{A}, & \sum_{n = 1}^{N} I_{n}^{A} < P_{{T}}^{A} \\ \min ((\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}})^{ +},I_{n}^{A}), & \sum_{n = 1}^{N} I_{n}^{A} \ge P_{{T}}^{A} \end{array} \right . \end{aligned} $$

(46)

$$ \begin{aligned}[b] &p_{k,n}^{B*} = \left \{ \begin{array}{l@{\quad}l} I_{n}^{B}, & \sum_{n = 1}^{N} I_{n}^{B} < P_{{T}}^{B} \\ \min ((\frac{1}{\ln(2)\lambda^{B}} - \frac{1}{\gamma_{{k,n}}^{B}})^{ +},I_{n}^{B}), & \sum_{n = 1}^{N} I_{n}^{B} \ge P_{{T}}^{B} \end{array} \right . \end{aligned} $$

(47)

where (x)⁺=max(x,0). If $\sum_{n = 1}^{N} I_{n}^{A} \ge P_{T}$ or $\sum_{n = 1}^{N} I_{n}^{B} \ge P_{T}$, the optimal values of λ ^A or λ ^B is chosen to fulfill the total power constraint with equality.

Appendix B: Single cell problem

We differentiate $L(s_{k,n}^{A},\rho_{k,n}^{A})$ (from (20)) with regard to $s_{k,n}^{A},\rho _{k,n}^{A}$ and setting them to zero.

$$ \begin{aligned}[b] \frac{\partial L}{\partial s_{{k,n}}^{A}} &= \rho _{{k,n}}^{A}.C_{k,n}^{\prime A} \biggl(\frac{s_{{k,n}}^{A}}{\rho_{{k,n}}^{A}}\biggr) - \lambda^{A} - \varOmega _{{k,n}}^{A} + v_{{k,n}}^{A} \\ &= \frac{\gamma_{{k,n}}^{A}}{\ln(2)(1 + \gamma _{{k,n}}^{A}p_{k,n}^{A})} - \lambda^{A} - \varOmega_{{k,n}}^{A} + v_{{k,n}}^{A} = 0 \end{aligned} $$

(48)

$$ \begin{aligned}[b] \frac{\partial L}{\partial\rho_{{k,n}}^{A}} &= C_{k,n}^{A}\biggl(\frac{s_{{k,n}}^{A}}{\rho _{{k,n}}^{A}} \biggr) + \rho_{{k,n}}^{A}.\frac{\partial}{\partial\rho _{{k,n}}^{A}}C_{k,n}^{A} \biggl(\frac{s_{{k,n}}^{A}}{\rho_{{k,n}}^{A}}\biggr) - \varphi_{{n}}^{A} \\ &= \frac{1}{\ln(2)}\ln\biggl(1 + \frac{\gamma _{{k,n}}^{A}s_{k,n}^{A}}{\rho_{{k,n}}^{A}}\biggr) \\ &\quad{}- \frac{\frac{\gamma_{{k,n}}^{A}s_{k,n}^{A}}{\rho _{{k,n}}^{A}}}{\ln(2)(1 + \frac{\gamma _{{k,n}}^{A}s_{k,n}^{A}}{\rho_{{k,n}}^{A}})} - \varphi_{{n}}^{A} = 0 \end{aligned} $$

(49)

The KKT conditions that must be satisfied for obtaining the optimal solution are:

$$ (1)\quad\lambda^{A}\Biggl(\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} s_{{k,n}}^{A} - P_{{T}}^{A} \Biggr) = 0 $$

(50)

$$ (2)\quad\varOmega_{{k,n}}^{A}\bigl(s_{{k,n}}^{A} - I_{n}^{A}\bigr) = 0 $$

(51)

$$ (3)\quad v_{{k,n}}^{A}s_{{k,n}}^{A} = 0 $$

(52)

$$ (4)\quad\varphi_{n}^{A}\Biggl(\sum_{k^{A} = 1}^{K^{A}} \rho_{{k,n}}^{A} - 1\Biggr) = 0 $$

(53)

As seen in Appendix A, by combination of conditions 1, 2, 3 and (48), we can obtain the optimal power allocation as follows:

$$ s_{k,n}^{A*} = \left \{ \begin{array}{l@{\quad}l} I_{n}^{A}, & \sum_{n = 1}^{N} I_{n}^{A} < P_{{T}}^{A} \\ \min ((\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}})^{ +},I_{n}^{A}), & \sum_{n = 1}^{N} I_{n}^{A} \ge P_{{T}}^{A} \end{array} \right . $$

(54)

Let $x = 1 + \gamma_{{k,n}}^{A}p_{k,n}^{A}$ and $c = \ln(2).\varphi _{{n}}^{A}$. Because $\varphi_{{n}}^{A}$ is the same for different users in cell A, it is constant over k. Therefore, (49) can be expressed as

$$ \begin{aligned}[b] \ln(x) = 1 - x^{ - 1} + c &\quad\Rightarrow\quad x = e^{ - x^{ - 1}}e^{1 + c} \\ &\quad\Rightarrow\quad- x^{ - 1}e^{ - x^{ - 1}} = - e^{ - (1 + c)} \end{aligned} $$

(55)

Observe that this is in the form of Lembert-W function W(c) [28] which is the solution to W(c)e ^W(c)=c. Thus, we can write

$$ \begin{aligned}[b] &- x^{ - 1} = W\bigl( - e^{ - (1 + c)}\bigr) \\ &\quad\Rightarrow\quad x = 1 + \frac{\gamma_{{k,n}}^{A}s_{k,n}^{A}}{\rho _{k,n}^{A}} = \frac{ - 1}{W( - e^{ - (1 + c)})} \\ &\quad\Rightarrow\quad\frac{\gamma _{{k,n}}^{A}s_{k,n}^{A}}{\rho_{k,n}^{A}} = \frac{ - 1}{W( - e^{ - (1 + c)})} - 1 \end{aligned} $$

(56)

Therefore $\frac{\gamma_{{k,n}}^{A}s_{k,n}^{A}}{\rho_{k,n}^{A}}$ is constant for all users and to maximize $\rho_{k,n}^{A}$, the subcarrier n should be allocated to user $k_{n}^{*}$ selected by

$$ k_{n}^{*} = \arg\max_{k}\bigl(\gamma _{{k,n}}^{A}s_{k,n}^{A}\bigr)\quad\mbox{for }\forall n $$

(57)

By substituting $s_{k,n}^{A}$ from Eq. (54) and since λ ^A is the same for different users, we can obtain

$$ k_{n}^{*} = \arg\max_{k}\bigl(\gamma _{{k,n}}^{A}\bigr)\quad\mbox{for }\forall n $$

(58)

where $k_{^{n}}^{*}$ is the user which subcarrier n is exclusively assigned to. By this scheme, the condition 4 is satisfied, too.

Appendix C: Centralized resource allocation problem

For optimal power allocation, we differentiate $L(s_{k,n}^{A},s_{k,n}^{B}, \rho_{{k,n}}^{A},\rho _{{k,n}}^{B})$ (from (25)) with regard to $s_{k,n}^{A},s_{k,n}^{B}$ and setting them to zero.

$$ \begin{aligned}[b] \frac{\partial L}{\partial s_{{k,n}}^{A}} & = \rho_{{k,n}}^{A}. C_{k,n}^{\prime A} \biggl(\frac{s_{{k,n}}^{A}}{\rho_{{k,n}}^{A}}\biggr) - \lambda^{A} \\ &\quad{}- \varOmega _{{k,n}}^{A}\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\bigl \vert \bar{H}_{{l,n}}^{B} \bigr \vert ^{2} + v_{{k,n}}^{A} \\ & = \frac{\gamma_{{k,n}}^{A}}{\ln(2)(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A})} - \lambda^{A} \\ &\quad{}- \varOmega_{{k,n}}^{A}\sum _{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\bigl \vert \bar{H}_{{l,n}}^{B} \bigr \vert ^{2} + v_{{k,n}}^{A} = 0 \end{aligned} $$

(59)

$$ \begin{aligned}[b] \frac{\partial L}{\partial s_{{k,n}}^{B}} &= \rho_{{k,n}}^{B}.C_{k,n}^{\prime B} \biggl(\frac{s_{{k,n}}^{B}}{\rho_{{k,n}}^{B}}\biggr) - \lambda^{B} \\ &\quad{}- \varOmega _{{k,n}}^{B}\sum_{l = 1}^{K^{A}} \rho_{{l,n}}^{A}\bigl \vert \bar{H}_{{l,n}}^{A} \bigr \vert ^{2} + v_{{k,n}}^{B} \\ &= \frac{\gamma_{{k,n}}^{B}}{\ln(2)(1 + \gamma_{{k,n}}^{B}p_{k,n}^{B})} - \lambda^{B} \\ &\quad{}- \varOmega_{{k,n}}^{B}\sum _{l = 1}^{K^{A}} \rho_{{l,n}}^{A}\bigl \vert \bar{H}_{{l,n}}^{A} \bigr \vert ^{2} + v_{{k,n}}^{B} = 0 \end{aligned} $$

(60)

The KKT conditions that must be satisfied for the optimal solution is

$$ (1)\quad\lambda^{A}\Biggl(\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} s_{{k,n}}^{A} - P_{{T}}^{A} \Biggr) = 0 $$

(61)

$$ (2)\quad\lambda^{B}\Biggl(\sum_{k = 1}^{K^{B}} \sum_{n = 1}^{N} s_{{k,n}}^{B} - P_{{T}}^{B} \Biggr) = 0 $$

(62)

$$ (3)\quad\varOmega_{{k,n}}^{A}\Biggl(s_{{k,n}}^{A}. \sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B} \bigl \vert \bar{H}_{{l,n}}^{B} \bigr \vert ^{2} - I \Biggr) = 0 $$

(63)

$$ (4)\quad\varOmega_{{k,n}}^{B}\Biggl(s_{{k,n}}^{B}. \sum_{l = 1}^{K^{A}} \rho_{{l,n}}^{A} \bigl \vert \bar{H}_{{l,n}}^{A} \bigr \vert ^{2} - I \Biggr) = 0 $$

(64)

$$ (5)\quad v_{{k,n}}^{A}s_{{k,n}}^{A}=0 $$

(65)

$$ (6)\quad v_{{k,n}}^{B}s_{{k,n}}^{B}=0 $$

(66)

$$ (7)\quad\varphi_{n}^{A}\Biggl(\sum_{k = 1}^{K^{A}} \rho_{{k,n}}^{A} - 1\Biggr) = 0 $$

(67)

$$ (8)\quad\varphi_{n}^{B}\Biggl(\sum_{k = 1}^{K^{B}} \rho_{{k,n}}^{B} - 1\Biggr) = 0 $$

(68)

The conditions $v_{{k,n}}^{A}s_{{k,n}}^{A} = 0$ and $\varOmega_{{k,n}}^{A}(s_{{k,n}}^{A}.\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B} \vert \bar{H}_{{l,n}}^{B} \vert^{2} - I) = 0$ show that

$$(\mbox{5-{I}}) \quad\mbox{if }s_{{k,n}}^{A} > 0 \quad\mbox{then } v_{{k,n}}^{A} = 0 $$

$$(\mbox{5-{II}}) \quad\mbox{if }v_{{k,n}}^{A} > 0 \quad\mbox{then } s_{{k,n}}^{A} = 0 $$

$$(\mbox{3-{I}}) \quad\mbox{if }s_{{k,n}}^{A}.\sum _{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\bigl \vert \bar{H}_{{l,n}}^{B} \bigr \vert ^{2} < I\quad\mbox{then } \varOmega_{{k,n}}^{A} = 0 $$

$$(\mbox{3-{II}}) \quad\mbox{if }\varOmega_{{k,n}}^{A} > 0 \quad\mbox {then } s_{{k,n}}^{A} = \frac{I}{\sum_{l = 1}^{K^{B}} \rho _{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} $$

By combination of 3-I and 5-II we can obtain

$$\mbox{If} \quad\rho_{{k,n}}^{A}.C_{k,n}^{\prime A}(p_{k,n}^{A}) - \lambda^{A} < 0 \quad \mbox{then }p_{{k,n}}^{A} = 0 $$

$$ \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} < 0 \quad\Rightarrow\quad p_{{k,n}}^{A} = 0 $$

(69)

By combination of 3-II and 5-I we can obtain

$$\begin{aligned} &\mbox{If} \quad\rho_{{k,n}}^{A}.C_{k,n}^{\prime A}\biggl(p_{{k,n}}^{A} = \frac{I}{\sum_{l = 1}^{K^{B}} \rho _{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \biggr) - \lambda^{A} > 0 \quad\mbox{then } \\ &\quad s_{{k,n}}^{A} = \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar {H}_{{l,n}}^{B} \vert ^{2}} \end{aligned} $$

$$ \begin{aligned}[b] &\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} > \frac {I}{\sum_{l = 1}^{K^{B}} \rho _{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \\ &\quad\Rightarrow\quad p_{{k,n}}^{A} = \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \end{aligned} $$

(70)

By combination of 3-I and 5-I we can obtain

$$ \begin{aligned}[b] &0 < \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} < \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \\ &\quad\Rightarrow\quad p_{{k,n}}^{A} = \frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}} \end{aligned} $$

(71)

3-II and 5-II do not occur jointly.

We can summarize Eqs. (69), (70) and (71) as follows

$$ p_{{k,n}}^{A} = \min\biggl(\biggl(\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{{k,n}}^{A}}\biggr)^{ +},\frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar {H}_{{l,n}}^{B} \vert ^{2}} \biggr) $$

(72)

But if the condition $\lambda^{A}(\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} - P_{{T}}^{A} ) = 0$ is satisfied the optimum value of λ ^A is obtained. Due to this condition, two cases can be considered:

$$(\mbox{1-{I}}) \quad\mbox{if }\lambda^{A} > 0\quad\mbox{then }\sum _{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} = P_{{T}}^{A} $$

$$(\mbox{1-{II}}) \quad\mbox{if }\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} p_{{k,n}}^{A} < P_{{T}}^{A}\quad\mbox{then }\lambda^{A} = 0 $$

By combination of 1-II and (72) we can deduce that

$$ \begin{aligned}[b] &\lambda^{A} = 0 \quad\Rightarrow\quad \\ &p_{{k,n}}^{A} = \min \biggl(\biggl(\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma _{{k,n}}^{A}}\biggr)^{ +}, \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \biggr) \\ &\hphantom{p_{{k,n}}^{A}}= \frac{I}{\sum_{l = 1}^{K^{B}} \rho _{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} \end{aligned} $$

(73)

Thus if

$$\sum_{n = 1}^{N} \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} < P_{{T}}^{A} $$

then

$$p_{{k,n}}^{A} = \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar{H}_{{l,n}}^{B} \vert ^{2}} $$

and if

$$\sum_{n = 1}^{N} \frac{I}{\sum_{l = 1}^{K^{B}} \rho_{{l,n}}^{B}\vert \bar {H}_{{l,n}}^{B} \vert ^{2}} \ge P_{{T}}^{A}, $$

we can conclude that the λ ^A>0 and for the optimal solution, the condition $\sum_{k = 1}^{K^{A}} \sum_{n = 1}^{N} s_{{k,n}}^{A} = P_{{T}}^{A}$ must be satisfied. Therefore, the optimal value of λ ^A is chosen to fulfill the total power constraint with equality. We can write the optimal power allocation in cell A as follows:

$$ p_{k,n}^{A*} = \left \{ \begin{array}{l} \frac{I}{\sum_{k = 1}^{K^{B}} \rho_{k,n}^{B*}\vert \bar{H}_{{k,n}}^{B} \vert ^{2}}, \quad\sum_{n = 1}^{N} I_{n}^{A}(\rho_{k,n}^{B*}) < P_{{T}}^{A} \\ \min((\frac{1}{\ln(2)\lambda^{A}} - \frac{1}{\gamma_{k,n}})^{ +},I_{n}^{A}(\rho _{k,n}^{B*})), \\ \quad\sum_{n = 1}^{N} I_{n}^{A}(\rho_{k,n}^{B*}) \ge P_{{T}}^{A} \end{array} \right . $$

(74)

and similarly, by combination of the conditions 2, 4, and 6 in Eq. (60), we can write the optimal power allocation in cell B as follows:

$$ p_{k,n}^{B*} = \left \{ \begin{array}{l} \frac{I}{\sum_{k = 1}^{K^{A}} \rho_{k,n}^{A*}\vert \bar{H}_{{k,n}}^{A} \vert ^{2}}, \quad\sum_{n = 1}^{N} I_{n}^{B}(\rho_{k,n}^{A*}) < P_{{T}}^{B} \\ \min((\frac{1}{\ln(2)\lambda^{B}} - \frac{1}{\gamma_{k,n}})^{ +},I_{n}^{B}(\rho _{k,n}^{A*})), \\ \quad\sum_{n = 1}^{N} I_{n}^{B}(\rho_{k,n}^{A*}) \ge P_{{T}}^{B} \end{array} \right . $$

(75)

Therefore, the optimal power allocation is obtained through (74) and (75). It can be observed that the optimal power allocation is a function of the subcarrier allocation in the adjacent cell. Therefore, the subcarrier allocation and power distribution must be performed jointly.

Now we must determine the subcarrier allocation scheme. We differentiate $L(s_{k,n}^{A},s_{k,n}^{B},\rho_{{k,n}}^{A},\rho_{{k,n}}^{B})$ with regard to $\rho _{{k,n}}^{A}$, $\rho_{{k,n}}^{B}$ and setting them to zero.

$$\begin{aligned} \frac{\partial L}{\partial\rho_{{k,n}}^{A}} =& C_{k,n}^{A}\biggl (\frac{s_{{k,n}}^{A}}{\rho_{{k,n}}^{A}} \biggr) + \rho_{{k,n}}^{A}.\frac{\partial}{ \partial\rho_{{k,n}}^{A}}C_{k,n}^{A} \biggl(\frac{s_{{k,n}}^{A}}{\rho_{{k,n}}^{A}}\biggr) \\ &{}- \bigl \vert \bar{H}_{{k,n}}^{A} \bigr \vert ^{2}\sum_{m = 1}^{K^{B}} \varOmega_{{m,n}}^{B}s_{{m,n}}^{B} - \varphi _{{n}}^{A} \\ =& \frac{1}{\ln(2)}\ln\bigl(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A} \bigr) - \frac{\gamma _{{k,n}}^{A}p_{k,n}^{A}}{\ln(2)(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A})} \\ &{}- \bigl \vert \bar{H}_{{k,n}}^{A} \bigr \vert ^{2}\sum_{m = 1}^{K^{B}} \varOmega_{{m,n}}^{B}s_{{m,n}}^{B} - \varphi _{{n}}^{A} = 0 \end{aligned}$$

(76)

$$\begin{aligned} \frac{\partial L}{\partial\rho_{{k,n}}^{B}} =& C_{k,n}^{B}\biggl (\frac{s_{{k,n}}^{B}}{\rho_{{k,n}}^{B}} \biggr) + \rho_{{k,n}}^{B}.\frac{\partial}{ \partial\rho_{{k,n}}^{B}}C_{k,n}^{B} \biggl(\frac{s_{{k,n}}^{B}}{\rho_{{k,n}}^{B}}\biggr) \\ &{}- \bigl \vert \bar{H}_{{k,n}}^{B} \bigr \vert ^{2}\sum_{k = 1}^{K^{A}} \varOmega_{{k,n}}^{A}s_{{k,n}}^{A} - \varphi _{{n}}^{B} \\ =& \frac{1}{\ln(2)}\ln\biggl(1 + \frac{\gamma _{{k,n}}^{B}s_{k,n}^{B}}{\rho _{{k,n}}^{B}}\biggr) - \frac{\frac{\gamma_{{k,n}}^{B}s_{k,n}^{B}}{\rho_{{k,n}}^{B}}}{\ln (2)(1 + \frac{\gamma_{{k,n}}^{B}s_{k,n}^{B}}{\rho_{{k,n}}^{B}})} \\ &{}- \bigl \vert \bar{H}_{{k,n}}^{B} \bigr \vert ^{2}\sum_{k = 1}^{K^{A}} \varOmega _{{k,n}}^{A}s_{{k,n}}^{A} - \varphi _{{n}}^{B} = 0 \end{aligned}$$

(77)

If we assume user m is assigned to subcarrier n in cell B, due to (59) we have

$$ \varOmega_{{m,n}}^{B}\bigl \vert \bar{H}_{{k,n}}^{A} \bigr \vert ^{2} = \frac{\gamma_{{m,n}}^{B}}{\ln (2)(1 + \gamma_{{m,n}}^{B}p_{m,n}^{B})} - \lambda^{B} + v_{{m,n}}^{B} $$

(78)

By substituting (78) in (76), we have

$$\begin{aligned} \frac{\partial L}{\partial\rho_{{k,n}}^{A}} =& \log\bigl(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A} \bigr) - \frac{\gamma_{{k,n}}^{A}p_{k,n}^{A}}{\ln(2)(1 + \gamma _{{k,n}}^{A}p_{k,n}^{A})} \\ &{}- \frac{\gamma _{{m,n}}^{B}p_{m,n}^{B}}{\ln(2)(1 + \gamma_{{m,n}}^{B}p_{m,n}^{B})} + \lambda^{B}p_{m,n}^{B} \\ &{}- v_{{m,n}}^{B}p_{m,n}^{B} - \varphi _{{n}}^{A} \end{aligned}$$

(79)

Due to condition 6, $v_{{m,n}}^{B}p_{m,n}^{B} = 0$ and $\varphi _{{n}}^{A}$ is the same for different users. Therefore, the optimal subcarrier allocation in cell A is

$$\begin{aligned} k_{{n}}^{A*} =& \arg\max_{k,m}\biggl\{ \log \bigl(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A}\bigr) - \frac{\gamma _{{k,n}}^{A}p_{k,n}^{A}}{\ln(2)(1 + \gamma_{{k,n}}^{A}p_{k,n}^{A})} \\ &{}+ \lambda^{B}p_{m,n}^{B} - \frac{\gamma_{{m,n}}^{B}p_{m,n}^{B}}{\ln(2)(1 + \gamma _{{m,n}}^{B}p_{m,n}^{B})}\biggr\} \end{aligned}$$

(80)

where $k_{{n}}^{A*}$ is the user assigned to subcarrier n in cell A.

By the same analysis based on the combination of (60) and (77), we can write for cell B

$$\begin{aligned} k_{{n}}^{B*} =& \arg\max_{k,m}\biggl\{ \log \bigl(1 + \gamma_{{k,n}}^{B}p_{k,n}^{B}\bigr) - \frac{\gamma _{{k,n}}^{B}p_{k,n}^{B}}{\ln(2)(1 + \gamma_{{k,n}}^{B}p_{k,n}^{B})} \\ &{}+ \lambda^{A}p_{m,n}^{A} - \frac{\gamma_{{m,n}}^{A}p_{m,n}^{A}}{\ln(2)(1 + \gamma _{{m,n}}^{A}p_{m,n}^{A})}\biggr\} \end{aligned}$$

(81)

where $k_{{n}}^{B*}$ is the user assigned to subcarrier n in cell B.