Coalition Formation Game for Joint Power Control and Fair Channel Allocation in Device-to-Device Communications Underlaying Cellular Networks

Abstract

This paper jointly considers the channel allocation and power control problems for device-to-device (D2D) communications underlaying cellular networks, in order to mitigate the co-channel interference and improve the system performance. The problem is formulated as a coalition formation game in which each cellular user forms a coalition with multiple D2D pairs (D2DPs), then these users can share the same communication channel. We analyze a merge-and-split process that is usually used for finding the optimal coalition structure. Among the stability notions of a coalition formation game, e.g., ${\mathbb {D}}_{hp}$-stability and a stronger ${\mathbb {D}}_c$-stability, we present the conditions for which this merge-and-split process converges to the ${\mathbb {D}}_c$-stable structure, then that structure corresponds to the optimal solution. Since those conditions are shown not to hold in most practical scenarios of users’ position distribution, that idealized algorithm is not applicable. We propose a merge-and-split based algorithm that can be applied to practical systems. However, the simulation results show that this algorithm can not guarantee the fairness among the performance of D2DPs. We propose a scheduling mechanism that allows the game to control this unfairness. The analytical results are verified by a number of simulations with different users’ position scenarios. The simulation results also show that the proposed approach can control the unfairness while providing significant improvements in system performance.

Appendix: Proof of Proposition 1

From [15], the ${\mathbb {D}}_c$-stability exists if the two following conditions hold for the structure ${\mathbb {S}}$.

(i) For each coalition $S_i \in {\mathbb {S}}$ and each pair of disjoint D2DP sets $A_1, A_2$ such that $\{A_1 \cup A_2 \} \subseteq S_i$, the following condition must be satisfied

$$\begin{aligned} v(A_1 \cup A_2) > v(A_1) + v(A_2). \end{aligned}$$

(12)

For this condition to hold, we need to prove that

$$\begin{aligned} \begin{aligned} v(A_1 \cup A_2) =\,&\varepsilon \Bigg [v(A_1) + v(A_2) - \sum _{d \in A_1}{I(d, A_2)} \\&- \sum _{d \in A_2}{I(d, A_1)} - \sum _{d \in A_1}I(d, c, A_2) \Bigg ]. \end{aligned} \end{aligned}$$

(13)

First, consider a D2DP $d \in A$, its interference to other D2DPs is the difference between the sum SINR of A with and without d. That interference is

$$\begin{aligned} \begin{aligned}&\sum _{{\tilde{d}} \in \{A \backslash \{d\}\}}\Bigg [\frac{p_{{\tilde{d}}}g_{{\tilde{d}}}}{p_c g_{c{\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in \{A \backslash \{d\}\} \\ {\hat{d}} \ne {\tilde{d}} \end{array}}{p_{{\hat{d}}}g_{{\hat{d}}{\tilde{d}}}} + \sigma ^2} - \frac{p_{{\tilde{d}}}g_{{\tilde{d}}}}{p_c g_{c{\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in A \\ {\hat{d}} \ne {\tilde{d}} \end{array}} {p_{{\hat{d}}}g_{{\hat{d}}{\tilde{d}}} + \sigma ^2}}\Bigg ] \\&\quad = \sum _{{\tilde{d}} \in \{A \backslash \{d\} \}} \Bigg [\frac{p_{{\tilde{d}}}g_{{\tilde{d}}}}{p_c g_{c{\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in \{A \backslash \{d\} \} \\ {\hat{d}} \ne {\tilde{d}} \end{array}} {p_{{\hat{d}}}g_{{\hat{d}}{\tilde{d}}}} + \sigma ^2} \\&\qquad - \frac{p_{{\tilde{d}}}g_{{\tilde{d}}}}{p_c g_{c{\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in \{A \backslash \{d\} \} \\ {\hat{d}} \ne {\tilde{d}} \end{array}} {p_{{\hat{d}}}g_{{\hat{d}}{\tilde{d}}}} + p_d g_{d{\tilde{d}}} + \sigma ^2}\Bigg ] \\&\quad = \sum _{{\tilde{d}} \in \{A \backslash \{d\} \}}{\frac{p_{{\tilde{d}}}g_{{\tilde{d}}}p_d g_{d {\tilde{d}}}}{\left( p_cg_{c {\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in \{A \backslash \{d\} \} \\ {\hat{d}} \ne {\tilde{d}} \end{array}} {p_{{\hat{d}}}g_{{\hat{d}} {\tilde{d}}}} + \sigma ^2\right) \Bigg ( p_c g_{c {\tilde{d}}} + \sum \limits _{\begin{array}{c} {\hat{d}} \in A \\ {\hat{d}} \ne {\tilde{d}} \end{array}} {p_{{\hat{d}} } g_{{\hat{d}}{\tilde{d}}}} + \sigma ^2 \Bigg )}} \\&\quad = {I(d,A)}. \end{aligned} \end{aligned}$$

(14)

This means that the computation of the interference $\sum _{d \in A_1}{I(d, A_2)}$ and $\sum _{d \in A_2}{I(d, A_1)}$ in Eq. (13) is proved. Second, the interference generated by d to the CU c is the difference between that CU’s SINR with and without the existence of d. Similar to the computation in Eq. (14), this interference is

$$\begin{aligned} \begin{aligned}&{\frac{p_cg_{cB}p_dg_{dB}}{\left( \sum _{{\hat{d}} \in \{A \backslash \{d\} \}}{p_{{\hat{d}}}g_{{\hat{d}}B} + \sigma ^2} \right) \left( \sum _{{\hat{d}} \in A}{p_{{\hat{d}}}g_{{\hat{d}}B} + \sigma ^2} \right) }} \\&\quad = {I(d, c, A)}. \end{aligned} \end{aligned}$$

(15)

This means that the computation of the interference $\sum _{d \in A_1}I(d, c, A_2)$ in Eq. (13) is proved. From above analysis, when the condition (i) in Proposition 1 holds, we have $v(A_1 \cup A_2) > v(A_1) + v(A_2)$, which means that the condition (12) is satisfied.

(ii) For each D2DP set $A \subseteq N, A \not \subseteq S_i, S_i \in {\mathbb {S}}$, the following condition must be satisfied

$$\begin{aligned} \begin{aligned} v(A) < v(S_i \cap A) + v(S_j \cap A). \end{aligned} \end{aligned}$$

(16)

In fact, we have

$$\begin{aligned} \begin{aligned} v(A) =\,&\varepsilon \Bigg [v(A_1) + v(A_2) - \sum _{d_i \in A_1}{I(d_i, A_2)} \\&- \sum _{d_i \in A_2}{I(d_i, A_1)} - \sum _{d_i \in A_1}I(d_i, c, A_2) \Bigg ]; \end{aligned} \end{aligned}$$

(17)

and since $A_1 = \{A \cap S_i\}, A_2 = \{A \cap S_j\}, i \ne j$, we also have

$$\begin{aligned} v(A_1) + v(A_2) = v(A \cap S_i) + v(A \cap S_j). \end{aligned}$$

(18)

From Eqs. (8), (17) and (18), when the condition (ii) in Proposition 1 holds, we have $v(A) < v(S_i \cap A) + v(S_j \cap A)$, which means that the condition (16) is satisfied. $\square$