A Robust Power Allocation Scheme for Relay Communication Networks Based on Seller–Buyer Game

Abstract

A robust power allocation algorithm is proposed in this paper based on robust game theory. Considering that the channel gains are usually changing with the dynamic environment in practice, the uncertainty will cause the estimation errors of channel gains and deteriorate the quality of services of users. A probability threshold method is used to deal with the uncertainty of the channel gains. To jointly consider the maximum utilities of source node and relay node, a seller–buyer game is introduced and it is proved that the game converges to the unique equilibrium point. Besides, a distributed iteration algorithm is proposed to reduce the information overhead. Numerical results show the effectiveness and robustness of the proposed power algorithm.

Appendices

Appendix 1

The detail proof of (6) in Sect. 3 is given as follows

Assuming that the channel gain $G={\overline{G}}+\Delta G$, the parameter ${\Delta {G}}$ is subject to standard normal distribution. In this, we define $X={\Delta {G}}$, that is

$$\begin{aligned} X\sim N(0,1) \end{aligned}$$

The density function of X is denoted as

$$\begin{aligned} \Psi (x)=\frac{1}{\sqrt{2\pi }}e^{-\frac{x^{2}}{2}} \end{aligned}$$

There is another random variable Y, it is described as

$$\begin{aligned} Y=uX \end{aligned}$$

where the parameter u is constant.

Then the distribution function is denoted as follows

$$\begin{aligned} F(Y\le y)=F(uX\le y)=F\left( X\le \frac{y}{u}\right) =\int _{-\infty }^{\frac{y}{u}}{\frac{1}{\sqrt{2\pi }} e^{-\frac{x^{2}}{2}}dx} \end{aligned}$$

There the density function of Y is formulated as

$$\begin{aligned} f(y)=[F(Y\le y)]^{\prime }=\frac{1}{u\sqrt{2\pi }}e^{\frac{y^{2}}{2u^{2}}} \end{aligned}$$

So the parameter Y is subject to the following relation.

$$\begin{aligned} Y\sim N(0,u^{2}) \end{aligned}$$

Appendix 2

The detailed proof process of (16) is given as follows

$$\begin{aligned} Pr[N\le L]= & {} \int _{-\infty }^{L}{\frac{1}{\sqrt{2\pi }\sqrt{a^{2}+b_i^{2}}}e^{-\frac{x^{2}}{2\left( a^{2}+b_i^{2}\right) }}dx}\\&=0.5+\int _{0}^{L}{\frac{1}{\sqrt{2\pi }\sqrt{a^{2}+b_i^{2}}}e^{-\frac{x^{2}}{2\left( a^{2}+b_i^{2}\right) }}dx} \end{aligned}$$

where define $A=\sqrt{a^{2}+b_i^{2}}$, then the expression is rewritten as follows

$$\begin{aligned} Pr[N\le L]= & {} 0.5+\int _{0}^{L}{\frac{1}{\sqrt{2\pi }A}e^{-\frac{x^{2}}{2A^{2}}}dx}\\ \end{aligned}$$

By Taylor series expansion, we know $e^{-\frac{x^{2}}{2A^{2}}}$ is expanded at $x=0$. So $e^{-\frac{x^{2}}{2A^{2}}}|_{x=0}=1-\frac{x^{2}}{2A^{2}}+\frac{x^{4}}{8A^{4}}$ is set up. Then the expression is formulated as

$$\begin{aligned} Pr[N\le L]= & {} 0.5+\frac{1}{\sqrt{2\pi }A}\int _{0}^{L}{\left( 1-\frac{x^{2}}{2A^{2}}+\frac{x^{4}}{8A^{4}}\right) dx}\\ \end{aligned}$$

By comparing the magnitude of the parameters, we know the other term excepting the first term is so small that can be ignored. So the expression is rewritten as follows

$$\begin{aligned} Pr[N\le L]= & {} {\frac{L}{\sqrt{2\pi }A}}\le \varepsilon ^{\prime } \end{aligned}$$

According to the mathematical relationship $\sqrt{X+Y}\le \sqrt{X}+\sqrt{Y}$, we get the following expression $A=\sqrt{a^{2}+b_i^{2}}\le \sqrt{a^{2}}+\sqrt{b_i^{2}}=a+b_i$. Because L is negative, the following expression is set up.

$$\begin{aligned} Pr[N\le L]= & {} {\frac{L}{\sqrt{2\pi }(a+b_i)}}\le \varepsilon ^{\prime }\\&\Rightarrow L-\varepsilon ^{\prime }\sqrt{2\pi }(a+b_i)\le 0 \end{aligned}$$

Appendix 3

The detailed proof process is given as follows

$$\begin{aligned} U_s(P_{r_i})= & {} g\ln \left( 1+\frac{P_sG_{s,d}+P_{r_i}G_{r_i,d}}{\sigma ^{2}}\right) -d_iP_{r_i}\\ \frac{\partial {U_s}}{\partial {P_{r_i}}}= & {} \frac{gG_{r_i,d}}{\sigma ^{2}+P_sG_{s,d}+P_{r_i}G_{r_i,d}}-d_i\\ \frac{\partial ^{2}U_s}{\partial ^{2}P_{r_i}}= & {} -\frac{gG_{r_i,d}^{2}}{\left( \sigma ^{2}+P_sG_{s,d}+P_{r_i}G_{r_i,d}\right) ^{2}}<0 \end{aligned}$$

So the function $U_{s}(P_{r_i})$ is concave, $U_{s}(P_{r_i})$ exists optimal solution.

Appendix 4

The detailed proof process is given as follows

$$\begin{aligned} \frac{\partial {U_{r_i}}}{\partial {d_i}}= & {} P_{r_i}+\frac{\partial {P_{r_i}}}{\partial {d_i}}\left( d_i-c_i\right) \\ \frac{\partial ^{2}U_{r_i}}{\partial {d_i^{2}}}= & {} 2\frac{\partial {P_{r_i}}}{\partial {d_i}}+\frac{\partial ^{2}P_{r_i}}{\partial {d_i^{2}}}(d_i-c_i)\\ \frac{\partial ^{2}P_{r_i}}{\partial {d_i^{2}}}= & {} \frac{2g\sigma ^{6}\left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] }{\left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] ^{4}} \end{aligned}$$

After list systematically, we regulate the following expression

$$\begin{aligned} \frac{\partial ^{2}{U_{r_i}}}{\partial {d_i^{2}}}= & {} \frac{2g\sigma ^{4}}{\left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] ^{2}}\\&\times \left[ \frac{-c_i\sigma ^{2}-u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) }{\left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] }\right] <0 \end{aligned}$$

So the function $U_{r_i}(P_{r_i})$ is concave, $U_{r_i}(P_{r_i})$ exists optimal solution.

Appendix 5

• (1) Positivity According to $\frac{\partial P_{r_i}^{*}}{\partial d_i}=-\frac{g\sigma ^{4}}{[d_i\sigma ^{2}+u_i({\overline{G}}_{r_i,d} +\varepsilon ^{\prime }\sqrt{2\pi })]^{2}}<0$. Moreover, if $c_i>0$ and $P_{r_i}\ge 0$, then by the following relation

  $$\begin{aligned} d_i=I(d_i)\triangleq c_i-\frac{P_{r_i}^{*}}{\partial P_{r_i}^{*}/\partial d_i} \end{aligned}$$

  So $I(d_i)>0$, each relay node starts improving its price from $c_i$.

• (2) Scalability Comparing $\alpha I(d_i)$ and $I(\alpha d_i)$ in an element-wise manner.

  $$\begin{aligned} \alpha I(d_i)-I(\alpha d_i)= & {} (\alpha -1)c_i+\alpha \left[ \frac{P_{r_i}(\alpha d_i)}{\partial P_{r_i}(\alpha d_i)/\partial d_i}-\frac{P_{r_i}(d_i)}{\partial P_{r_i}(d_i)/\partial d_i}\right] \end{aligned}$$

  where $\alpha >1$, $(\alpha -1)c_i>0$, so the problem reduces to proving that the second term is positive. Then, we get the following relationship.

  $$\begin{aligned} \frac{P_{r_i}(d_i)}{\partial P_{r_i}(d_i)/\partial d_i}= & {} -\frac{\left[ g\sigma ^{2}\left( {\overline{G}}_{r_i,d}+\Delta G_{r_i,d}\right) \right] \left[ d_i\sigma ^{2}+u_i({\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi })\right] }{g\sigma ^{4}\left( {\overline{G}}_{r_i,d}+\Delta G_{r_i,d}\right) }\\&\quad +\,\frac{\left[ \left( \sigma ^{2}+P_s{\overline{G}}_{s,d}+P_s\Delta G_{s,d}\right) \left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] \right] }{g\sigma ^{4}({\overline{G}}_{r_i,d}+\Delta G_{r_i,d})}\\&\quad \times \,\left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] \end{aligned}$$

  Similarly

  $$\begin{aligned} \frac{P_{r_i}(\alpha d_i)}{\partial P_{r_i}(\alpha d_i)/\partial d_i}= & {} -\frac{\left[ g\sigma ^{2}\left( {\overline{G}}_{r_i,d}+\Delta G_{r_i,d}\right) \right] \left[ \alpha d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] }{\alpha g\sigma ^{4}({\overline{G}}_{r_i,d}+\Delta G_{r_i,d})}\\&\quad +\,\frac{\left[ \left( \sigma ^{2}+P_s{\overline{G}}_{s,d}+P_s\Delta G_{s,d}\right) \left[ \alpha d_i\sigma ^{2}+u_i({\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi })\right] \right] }{\alpha g\sigma ^{4}({\overline{G}}_{r_i,d}+\Delta G_{r_i,d})}\\&\quad \times \,\left[ \alpha d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] \end{aligned}$$

  Because the signal of $\left( \frac{P_{r_i}(\alpha d_i)}{\partial P_{r_i}(\alpha d_i)/\partial d_i}-\frac{P_{r_i}(d_i)}{\partial P_{r_i}(d_i)/\partial d_i}\right)$ is difficult to determine. We define $m(t)=\frac{P_{r,i}(d_i)}{\frac{P_{r_i}(d_i)}{\partial P_{r_i}(d_i)/\partial d_i}}$. Where m(t) is a function about $d_i$. So the problem is transformed to judge the monotonicity of m(t).

  $$\begin{aligned} \frac{\partial m(t)}{\partial d_i}= & {} \frac{2\left( \sigma ^{2}+P_s{\overline{G}}_{s,d}+P_s\Delta G_{s,d}\right) \left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d}+\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] }{g\sigma ^{2}\left( {\overline{G}}_{r_i,d}+\Delta G_{r_i,d}\right) } -1 \end{aligned}$$

  After extensive numerical tests for a wide parameters, we observe the term $\frac{\partial m(t)}{\partial d_{i}}$ is positive. Since m(t) is negative, then $\left( \frac{P_{r_i}(\alpha d_{i})}{\partial P_{r_i}(\alpha d_{i})/\partial d_{i}}-\frac{P_{r_{i}}(d_{i})}{\partial P_{r_i}(d_{i})/\partial d_{i}}\right)$ is positive. Therefore, we claim that $\alpha I(d_{i})>I(\alpha d_i)$.

• (3) Monotonicity Suppose $d_{i}$ and $d_{i}^{\prime }$ are different price , and $d_{i}>d_{i}^{\prime }$.

  $$\begin{aligned} \frac{\partial I_i(d_i)}{\partial d_i}= & {} 1- \frac{2\left( \sigma ^{2}+P_s{\overline{G}}_{s,d}+P_s\Delta G_{s,d}\right) \left[ d_i\sigma ^{2}+u_i\left( {\overline{G}}_{r_i,d} +\varepsilon ^{\prime }\sqrt{2\pi }\right) \right] }{g\sigma ^{2}\left( {\overline{G}}_{r_i,d}+\Delta G_{r_i,d}\right) } \end{aligned}$$

  According to the analysis among the scalability, we know $\frac{\partial I_i(d_i)}{\partial d_i}<0$. Therefore, $I_{i}(d_{i})$ is monotonically decreasing.