Resource allocation for M2M-enabled cellular network using Nash bargaining game theory

Abstract

Machine-to-machine (M2M) communication is the form of communication among devices, which are the application of direct communication technology in particular case. For effectively improving the spectral efficiency and the performance of cellular networks by using M2M communication, this study investigates the resource allocation scheme to optimize the transmission performance of the M2M communication and cellular services from a Nash bargaining game theory point of view, which is a strong NP-hard problem. Firstly, we prove the existence of the Nash bargaining solution (NBS). To make the problem more tractable, we decompose it into two subproblems, namely, channel assignment subproblem and power allocation subproblem. The resource allocation scheme fully considers the quality requirements of cellular services and the performance of M2M service. Simulation results demonstrate that proposed resource allocation scheme results in improvement of system throughput and good adjustment effect on transmission performance of communication service.

Appendices

Appendix A: Proofs of Theorem 1

As described in Section 3, we set interference margin threshold 𝜖 for the power of MTCDs. According to SNR of UE i in 9, the transmission power of UE i satisfies

$$ {p}_{i}\geq \frac{\epsilon {\beta }_{i}N_{0}}{{g}_{ei}}={p}^{min}_{i} $$

(19)

When we adopt the minimum transmission power of UE i ${p}^{min}_{i}$, Eq. 8 is transformed into

$$ {p}^{max}_{ij}=\frac{\left( \epsilon -1 \right)N_{0}}{{g}^{I}_{ij}}\geq {p}^{max}_{ij},\forall i,j $$

(20)

For pairs of MTCDs, they will always increase their transmission power to improve their transmission rate until they attain the maximum power constraints. Hence, once the channel allocation identified as channel i for pairs of MTCD j, optimal power allocation of MTCDs is $p^{*}_{ij}=min\{p^{max}_{ij},P_{max}\}$.

This completes the proof of Theorem 1 .

Appendix B: Proofs of Theorem 2

Based on allocated channel and allocated power of MTCDs, the power allocation problem of UEs is transformed into

$$ \max\limits_{P} \sum\limits_{j=1}^{K} \ln \left( \sum\limits_{i=1}^{N_{c}}{c}^{*}_{ij}\log_{2} \left( 1+\frac {{p}_{i}{g}_{ei}}{N_{0}+{p}^{*}_{ij}{g}_{ij}^{I}}\right)-{\mu^{0}_{j}} \right) $$

(21)

subject to

$$\sum\limits_{j=1}^{K}\sum\limits_{i=1}^{N_{c}}{c}_{ij}{p}_{i}\leq {P}_{CU} \qquad\qquad\qquad\qquad\qquad\qquad\quad(27a)$$

$$p_{i}\geq 0,\forall i,j \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(27c) $$

The Lagrangian function of problem Eq. 19 is described as

$$\begin{array}{@{}rcl@{}} L\left( p_{i},\nu \right)&=&\sum\limits_{j=1}^{K} \ln \left( \sum\limits_{i=1}^{N_{c}}{c}^{*}_{ij}\log_{2} \left( 1+\frac {{p}_{i}{g}_{ei}}{N_{0}+{p}^{*}_{ij}{g}_{ij}^{I}}\right)-{\mu^{0}_{j}} \right)\\&&-\nu \left( \sum\limits_{j=1}^{K}\sum\limits_{i=1}^{N_{c}}{c}_{ij}{p}_{i}-{P}_{CU} \right) \end{array} $$

(22)

where μ≥0 are the Lagrange multipliers. Using the KKT conditions, we obtain

$$ \frac{\delta L}{\delta p_{i}} =\frac{{g}_{ei}}{\ln 2\times \left[{c}^{*}_{ij} {\log }_{2}\left( 1+\frac{{p}_{i}{g}_{ei}}{N_{0}+{p}^{*}_{ij}{g}^{I}_{ij}} \right) -{u}^{0}_{j}\right]\left( N_{0}+{p}^{*}_{ij}{g}^{I}_{ij}+{p}_{i}{g}_{ei}\right)}-\nu =0 $$

(23)

for identified ${c}^{*}_{ij}$ and ${p}^{*}_{ij}$. Hence,

$$ p^{*}_{i}={\Lambda} \left( c^{*}_{ij},{p}^{*}_{ij},{\nu }^{*}\right) $$

(24)

where Λ(∙) represents a function. Then,

$$ \frac{\delta L}{\delta \nu }\rightarrow \sum\limits_{j=1}^{K}\sum\limits_{i=1}^{N_{c}}{c}_{ij}{p}_{i}={P}_{CU} $$

(25)

Based on formula (24) and 25, we can solve $p^{*}_{i},\nu ^{*}$. Thus, we finally obtain the optimal power allocation strategy $Q^{*}_{N_{c}}=[p^{*}_{i}]$.

This completes the proof of Theorem 2.

Appendix C: Proofs of Theorem 3

We define the optimal channel assignment solution, derived from the NBG algorithm, as $c_{ij^{\ast}}$ for UE i, namely pair j ^∗ of MTCDs is the optimal match for UE i, maximizing the payoff of the game players in formula (13). According to the NBG algorithm, there are maximum unit earnings ${\varphi }_{ij}=\frac {{g}_{ei}}{{g}_{ij}^{I}}$ for UE i. Based on the optimal channel and power allocation strategy, the payoff of game player i is $U_{j^{*}}$.

Let us suppose that there is global optimal solution $c_{ij^{\prime }}$ and $p_{i}^{\prime }$, which is different from $c_{ij^{*}}$ described above, namely j ^∗≠j ^′. The global optimal power solution $p_{i}^{\prime }$ is derived from the utility function in formula (??), relating to global optimal channel solution $c_{ij^{\prime }}$. Hence, the payoff of game player i is $U_{j}^{\prime }$. Therefore,

$$ U_{j^{*}}<U_{j^{\prime}} $$

(26)

Since optimal power allocation of MTCDs is derived from independent solution procedure, the global optimal transmission power of MTCDs is

$$ p_{ij}^{*}=min\{p_{ij}^{max},P_{max}\} $$

(27)

Hence, we discuss the global optimal solution of UEs for two different occasions, namely $p_{ij}^{*}=p_{ij}^{max}=\frac {\left (\epsilon -1 \right ) N_{0}}{g_{ij}^{I} }$ and $p_{ij}^{*}=P_{max}$.

When $p_{ij}^{*}=P_{max}$, according to global optimal solution $c_{ij^{\prime }}$ and $p_{i}^{\prime }$, the payoff of game player i is

$$ U_{j^{\prime}}=\log_{2} \left( 1+\frac{p_{i}^{\prime}g_{ei}}{N_{0}+P_{max}g_{ij^{\prime}}^{I}} \right) $$

(28)

while according to optimal solution $c_{ij^{*}}$ and $p_{i}^{*}$, the payoff of game player i is

$$ U_{j^{*}}=\log_{2} \left( 1+\frac{p_{i}^{*}g_{ei}}{N_{0}+P_{max}g_{ij*}^{I}} \right) $$

(29)

However, according to optimal solution $c_{ij^{*}}$ and $p_{i}^{\prime }$, the payoff of game player i is

$$ U_{j^{*}}^{\prime}=\log_{2} \left( 1+\frac{p_{i}^{\prime}g_{ei}}{N_{0}+P_{max}g_{ij^{*}}^{I}} \right) $$

(30)

It is obvious that $U_{j^{*} }\geq U_{j^{*}}^{\prime }$. Based on the NBG algorithm, the maximum unit earnings are ${\varphi }_{ij^{*}}=\frac {{g}_{ei}}{{g}_{ij^{*}}^{I}}$ for UE i. For the same UE i, $g_{ij^{*}}^{I}> g_{ij^{\prime }}^{I}$. Hence, $U_{j^{*}}^{\prime }>U_{j^{\prime }}$. Then, we know that

$$ U_{j^{*}}\geq U_{j^{*}}^{\prime}>U_{j^{\prime}} $$

(31)

which contradicts $U_{j^{*}}<U_{j^{\prime }}$.

Therefore, there is no other global optimal solution except for the solution derived from the NBG algorithm to achieve greater payoff of game player when $p_{ij}^{*}=P_{max}$. Namely, the global optimal solution is the solution of the NBG algorithm when $p_{ij}^{*}=P_{max}$.

When $p_{ij}^{*}=\frac {\left (\epsilon -1 \right ) N_{0}}{g_{ij}^{I} }$, according to global optimal solution $c_{ij^{\prime }}$ and $p_{i}^{\prime }$, the payoff of game player i is

$$ U_{j^{\prime}}=\log_{2} \left( 1+\frac{p_{i}^{\prime}g_{ei}}{\epsilon N_{0}} \right) $$

(32)

while according to optimal solution $c_{ij^{*}}$ and $p_{i}^{*}$, the payoff of game player i is

$$ U_{j^{*}}=\log_{2} \left( 1+\frac{p_{i}^{*}g_{ei}}{\epsilon N_{0}} \right) $$

(33)

However, according to optimal solution $c_{ij^{*}}$ and $p_{i}^{\prime }$, the payoff of game player i is

$$ U_{j^{*}}^{\prime}=\log_{2} \left( 1+\frac{p_{i}^{\prime}g_{ei}}{\epsilon N_{0}} \right) $$

(34)

Hence, $U_{j^{*}}\geq U_{j^{*}}^{\prime } = U_{j^{\prime }}$, which contradicts $U_{j^{*}}<U_{j^{\prime }}$.

Therefore, there is no other global optimal solution except for the solution derived from the NBG algorithm to achieve greater payoff of game player when $p_{ij}^{*}=\frac {\left (\epsilon -1 \right ) N_{0}}{g_{ij}^{I} }$. Namely, the global optimal solution is the solution of the NBG algorithm when $p_{ij}^{*}=\frac {\left (\epsilon -1 \right ) N_{0}}{g_{ij}^{I} }$.

This completes the proof of Theorem 3.