Energy-efficient power allocation in cognitive sensor networks: a coupled constraint game approach

Abstract

In this paper, we investigate power allocation in cognitive sensor networks, where cognitive users (cognitive enabled sensor nodes) opportunistically share the common spectrum with primary users (licensed devices). Consider that sensor nodes are self-interested to maximize their own utilities, we formulate the energy-efficient power allocation problem as a non-cooperative coupled constraint game, by taking the interference temperature into account. An energy efficiency-oriented utility function is defined as a new metric to evaluate the performance of power allocation. Firstly, we prove that there exist Nash equilibriums in the proposed game. Then, we prove that the power allocation game is a super-modular game under some conditions. Finally, we design centralized and distributed Game-based Efficiency-oriented Power Allocation algorithms (i.e., centralized GEPA and distributed GEPA) to obtain the Nash equilibriums. Extensive simulations are conducted to demonstrate that the proposed power allocation algorithms can achieve satisfactory performance in terms of energy efficiency, convergence speed and fairness in cognitive sensor networks.

Appendices

Appendices

Existence of Nash equilibriums

Proof

There are $n$ cognitive users involved in the energy-efficient power allocation game. Obviously, the first condition is satisfied. Note that the available transmission power of each user ranges from $p_{\min }$ to $p_{\max }$, which means that the strategy set of each cognitive user is an interval. Intervals and cartesian products of intervals are closed, bounded and convex. Therefore the second condition is satisfied.

As long as the third condition is satisfied, we can get the result that there exist Nash equilibriums. Now we analyze properties of the continuous utility function,

$$\begin{aligned} u_i (p_i ,p_{ - i} ) = \frac{{B\log (1 + S_i)}}{{p_i+\alpha }} \end{aligned}$$

(25)

Then we get the differential function of the utility function,

$$\begin{aligned} \frac{{\partial u_i (p)}}{{\partial p_i }} =B\log (e)\frac{{\frac{{h_{ii} (p_i + \alpha )}}{{I + \sum \nolimits _{j \in {\mathcal {N}}} {h_{ij} p_j } }} - \ln \left( 1 + \frac{{h_{ii} p_i }}{{I + \sum \nolimits _{j \ne i \in {\mathcal {N}}} {h_{ij} p_j}}}\right) }}{{(p_i + \alpha )^2 }} \end{aligned}$$

(26)

It is obvious that $u(0,p_{-i})=0$, $\left. {\frac{{\partial u_i (p)}}{{\partial p_i }}} \right| _{p_i = 0}>0$, and there exists only one constant $p_c$ such that $\left. {\frac{{\partial u_i (p)}}{{\partial p_i }}} \right| _{p_i = p_c }=0$. The utility function increases with $p_i$ within the interval $[p_{\min }, p_c]$ and decreases within the interval $[p_c, p_{\max }]$.

Definition 3

A function $f$: $S \rightarrow R$ defined on a convex subset $S$ to a real vector space is quasi-concave if for $\forall x,y \in S$, $\forall \lambda \in [0,1]$,

$$\begin{aligned} f(\lambda x + (1 - \lambda )y) \ge \min (f(x),f(y)) \end{aligned}$$

(27)

Given a random interval $[a, b] \in [p_{\min }, p_{\max }]$, if $a \ge p_c$, then the utility function is decreases within the interval $[a, b]$, and $u(b,p_{-i})=\min (u(a,p_{-i}),u(b,p_{-i}))$, so it is obvious that $\lambda a+(1-\lambda )b \le b$, and then $u_i(\lambda a+(1-\lambda )b,p_{-i})\ge u_i(b,p_{-i})$. Similarly, if $b \le p_c$, then the utility function is still quasi-concave due to the same reason. Otherwise, $\min u_i(p_i,p_{-i})=u_i(a,p_{-i})$ or $\min u_i(p_i,p_{-i})=u_i(b,p_{-i})$ over the internal $[a, b]$. We get the result that the utility function is a quasi-concave function over the available set $[p_{\min }, p_{\max }]$.

From the above, we conclude that the player set is finite, the strategy sets are closed, bounded, and convex, and the utility functions are continuous and quasi-concave in the strategy space. This completes the proof of Theorem 1, such that there exist Nash equilibriums in the energy-efficient power allocation game ${\mathcal {G}}$.$\square$

Convergence of lagrangian multiplier

Proof

According to Eq. (25), we can obtain the differential function,

$$\begin{aligned} \frac{{{\partial ^2}{u_i}}}{{\partial {p_i}^2}}&= B\log (e)\left[ \frac{{ - {A^2}{{({p_i} + \alpha )}^2}}}{{{{({p_i} + \alpha )}^3}{{(1 + A{p_i})}^2}}}\right. \nonumber \\&\left. - 2\frac{{A\frac{{({p_i} + \alpha )}}{{(1 + A{p_i})}} - \ln (1 + A{p_i})}}{{{{({p_i} + \alpha )}^3}}}\right] \end{aligned}$$

(28)

in which $A = \frac{{{h_{ii}}}}{{I + \sum \nolimits _{j\ne i, j\in {\mathcal {N}}} {{h_{ij}}{p_j}} }}$ and $A$ can be regarded as a constant since the transmission power of other cognitive users keeps unchanged. Notice the first part is always negative, i.e., $\frac{{ - {A^2}{{({p_i} + \alpha )}^2}}}{{{{({p_i} + \alpha )}^3}{{(1 + A{p_i})}^2}}}<0$. According to the analysis in Appendices, the second part satisfies

$$\begin{aligned} - 2\frac{{A\frac{{({p_i} + \alpha )}}{{(1 + A{p_i})}} - \ln (1 + A{p_i})}}{{{{({p_i} + \alpha )}^3}}}\left\{ \begin{array}{ll} < 0,&{} {p_i} < {p_c}\\ = 0, &{}{p_i} = {p_c}\\ > 0,&{} {p_i} > {p_c} \end{array} \right. \end{aligned}$$

(29)

There exists one point $p_t$ make $\frac{{{\partial ^2}{u_i}}}{{\partial {p_i}^2}}{\rm{{|}}_{{p_i}\rm{{ = }}{p_t}}} = 0$. Moreover, $p_t>p_c$.

Within the interval $[0, p_c]$, the utility function is increasing and concave. Since the relationship between the Lagrangian multiplier $\lambda$ and the transmission power $p_i$ is linear and the utility function is strictly concave, there exists a sufficiently small step size $\varepsilon$ that guarantee $p_i$ and $\lambda$ to converge to the optimal solution [35].

Within the interval $[p_c, p_t]$, the utility function is decreasing and concave. Within the interval $[p_t, \infty ]$, the utility function is deceasing and convex. With the similar reason, both the transmission power $p_i$ and $\lambda$ will converge to the optimal solution.$\square$