A robust power control scheme for femtocell networks with probability constraint of channel gains

Abstract

In this paper, a robust power control algorithm is proposed in two-tier femtocell network system in order to address the uncertain channel gains of the interference links. In the algorithm, an outage probability is used to guarantee the communication qualities of users. As the deterministic form of the outage probability constraint is not convex of power, a novel and simple method is utilized to solve the power allocation problem. The allocation solution can be determined based on the average channel gains which are engaged with a dynamic factor with average signal-to-interference-plus-noise ratio. In the scheme, a dynamic power adjustment algorithm is developed. The algorithm attempts to guarantee the outage probability requirement of the macrocell user, and to achieve the optimal power allocation of femtocell users. Apart from the appropriate resource for the network, an admission control algorithm is adopted to remove the femtocell user whose communication quality can not be ensured. Numerical results demonstrate the effectiveness of the robust power control algorithm and also explain the admission control algorithm.

Appendices

Appendix

The detailed proofs of the proposition in Section 3 are given as follows

Appendix A

Proof

First introduces the Cauchy-Schwarz inequality

$$\begin{array}{@{}rcl@{}} |\langle X,Y \rangle|^{2}\leq\langle X,X\rangle\cdot\langle Y,Y \rangle \end{array} $$

(29)

we can elaborate the follow

$$\begin{array}{@{}rcl@{}} &&(X_{1}Y_{1}\,+\,X_{2}Y_{2}\,+\,\cdots\,+\,X_{N}Y_{N})\!\leq\!|X_{1}Y_{1}\,+\,X_{2}Y_{2}\,+\,\cdots\,+\,X_{N}Y_{N}| \\ &&\leq\left[\left( {X_{1}^{2}}+{X_{2}^{2}}+\cdots+{X_{N}^{2}}\right)\times\left( {Y_{1}^{2}}+{Y_{2}^{2}}+\cdots+{Y_{N}^{2}}\right)\right]^{1/2}\\ &&\leq \sqrt{\sum\limits_{i=1}^{N}{X_{i}^{2}}}\cdot\sum\limits_{i=1}^{N}|Y_{i}| \end{array} $$

then we can obtain

$$\begin{array}{@{}rcl@{}} &&\sum\limits_{j=0,j\neq i}^{N}p_{j}g_{i,j}=\sum\limits_{j=0,j\neq i}^{N}G_{i,j}\overline{g}_{i,j}p_{j}\\ &&\leq \sqrt{\sum\limits_{j=0,j\neq i}^{N}{p_{j}^{2}}\overline{g}_{i,j}^{2}}\cdot\sum\limits_{j=0,j\neq i}^{N}|G_{i,j}| \end{array} $$

(30)

formulation (11) can be rewritten as follows

$$\begin{array}{@{}rcl@{}} &&Pr\!\left[\sum\limits_{j=0,j\neq i}^{N}|G_{i,j}|\!\leq\!\frac{1}{\sqrt{{{\sum}_{j=0,j\neq i}^{N}{p_{j}^{2}}\overline{g}_{i,j}^{2}}}} \!\left( \frac{p_{i} g_{i,i}}{\widehat{\gamma}_{i}}\,-\,\sigma^{2}\right)\right]\\ &&\geq1-\varepsilon_{1} \end{array} $$

(31)

When G _{i, j} is assumed to be an exponential distribution with unit mean, we can obtain

$$\begin{array}{@{}rcl@{}} f_{G_{i,j}}(x)=\left\{ \begin{array}{ll} e^{-x}& x>0\\ 0 & \textit{else} \end{array} \right. \end{array} $$

(32)

We assume that Y = |x| = |G _{i, j} |. Based on Eq. 32, the distribution function of Y can be elaborated as

$$\begin{array}{@{}rcl@{}} F_{Y}(y)&=&P\{Y\leq y\}=P\{|x|\leq y\}\\ &=&{\int}_{-y}^{0}f_{G_{i,j}}(x)dx+{{\int}_{0}^{y}}f_{G_{i,j}}(x)dx\\ &=&{{\int}_{0}^{y}}e^{-x}dx=1-e^{-y} \end{array} $$

(33)

The probability density function of Y is $f_{Y}(y)=F_{Y}^{\prime }(y)$, hence,

$$\begin{array}{@{}rcl@{}} f_{|G_{i,j}|}(|x|)=\left\{ \begin{array}{ll} e^{-|x|}& |x|>0\\ 0 & \textit{else} \end{array} \right. \end{array} $$

(34)

Based on Eq. 34, we are aware that |G _{i, j} | is also an exponential distribution with parameter 1. Next, we introduce the Gamma (Γ) distribution. If X conforms a gamma distribution with the parameters α and β (i.e. X ∼ Γ(α, β)), the probability density function of X can be formulated as :

$$\begin{array}{@{}rcl@{}} f_{X}(x)=\left\{ \begin{array}{ll} \frac {\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}& x>0\\ 0 & \text{else} \end{array} \right. \end{array} $$

(35)

Assume that X and Y are independent to each other (i.e. X ∼ Γ(α ₁, β), Y ∼ Γ(α ₂, β) and Z = X + Y), the probability density function of Z can be formulated as,

$$\begin{array}{@{}rcl@{}} f_{Z}(z)={\int}_{-\infty}^{+\infty}f_{X}(x)f_{Y}(z-x)dx \end{array} $$

(36)

If and only if x > 0 and z − x > 0, the integrand is not zero. If z ≤ 0,f _Z (z) = 0. If z > 0, f _Z (z) is given as:

$$\begin{array}{@{}rcl@{}} f_{Z}(z)&=&{{\int}_{0}^{z}}\frac{\beta^{\alpha_{1}+\alpha_{2}}}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}x^{\alpha_{1}-1} (z-x)^{\alpha_{2}-1}e^{-\beta x-\beta(z-x)}dx \\ &=&\frac{\beta^{\alpha_{1}+\alpha_{2}}}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}e^{-\beta x}{{\int}_{0}^{z}} x^{\alpha_{1}-1}(z-x)^{\alpha_{2}-1}dx \\ &\overset{x=zt}{=}&\frac{\beta^{\alpha_{1}+\alpha_{2}}}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}e^{-\beta z}z^{\alpha_{1}+\alpha_{2}-1}{{\int}_{0}^{1}}t^{\alpha_{1}-1}(1-t)^{\alpha_{2}-1}dt\\ &=&\frac {\beta^{\alpha_{1}+\alpha_{2}}}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}e^{-\beta z}z^{\alpha_{1}+\alpha_{2}-1}\beta(\alpha_{1},\alpha_{2}) \end{array} $$

(37)

the β(α ₁, α ₂) is the function of β. Based on Appendix C, the relationship between Γ function and β function can be formulated as

$$\begin{array}{@{}rcl@{}} \frac{\beta(\alpha_{1},\alpha_{2})}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}=\frac{1}{\Gamma(\alpha_{1}+ \alpha_{2})} \end{array} $$

(38)

Hence,

$$\begin{array}{@{}rcl@{}} f_{Z}(z)=\left\{ \begin{array}{ll} \frac {\beta^{\alpha_{1}+\alpha_{2}}}{\Gamma(\alpha_{1}+\alpha_{2})}z^{\alpha_{1}+\alpha_{2}-1}e^{-\beta z}& z>0\\ 0 & \text{else} \end{array} \right. \end{array} $$

(39)

and Z ∼ Γ(α ₁ + α ₂, β). If X, Y and Z are independent to each other, X ∼ Γ(α ₁, β), Y ∼ Γ(α ₂, β), Z ∼ Γ(α ₃, β) and U = X + Y + Z. Based on the same token, we can obtain U ∼ Γ(α ₁ + α ₂ + α ₃, β). This conclusion can elaborate to multiple independent Γ distribution. If X _i ∼ Γ(α _i , β), then

$$\begin{array}{@{}rcl@{}} X_{1}+X_{2}+\cdots+X_{n}\sim{\Gamma}(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n},\beta) \end{array} $$

(40)

From the formulation (34), we know that |G _{i, j} |∼Γ(1,1). Based on Eq. 40, we can be obtained,

$$\begin{array}{@{}rcl@{}} \sum\nolimits_{j=0,j\neq i}^{N}|G_{i,j}|&\sim&{\Gamma}(|G_{i,0}|+|G_{i,1}|+|G_{i,2}|\\ &&+\cdots+|G_{i,i-1}|+0+|G_{i,i+1}|\\ &&+ \cdots+|G_{i,N}|,1)\\ &\sim&{\Gamma}(N,1) \end{array} $$

(41)

Hence, the probability density function of ${\sum }_{j=0,j\neq i}^{N}|G_{i,j}|$ can be elaborated as

$$\begin{array}{@{}rcl@{}} f\left( \sum\nolimits_{j=0,j\neq i}^{N}|G_{i,j}|\right)&=&\frac {1}{\Gamma(N)}\left( \sum\nolimits_{j=0,j\neq i}^{N}|G_{i,j}|\right)^{N-1}\\ &&e^{-{\sum}_{j=0,j\neq i}^{N}|G_{i,j}|} {}\\ \end{array} $$

(42)

Based on Eqs. 31 and 42, the value of $\mathcal {P}_{f_{i}}$ can be expressed as

$$\begin{array}{@{}rcl@{}} &&\mathcal{P}_{f_{i}}=\frac {1}{\Gamma(N)}{\int}_{0}^{\frac{1}{\sqrt{{{\sum}_{j=0,j\neq i}^{N}{p_{j}^{2}}\overline{g}_{i,j}^{2}}}}\left( \frac{p_{i}g_{i,i}}{\widehat{\gamma}_{i}}-\sigma^{2}\right)}\\ &&\left( \!\sum\nolimits_{j=0,j\neq i}^{N}|G_{i,j}| \right)^{N-1} \!e^{-\left( {\sum}_{j=0,j\neq i}^{N}|G_{i,j}|\right)}d\! \left( \sum\nolimits_{j=0,j\neq i}^{N}\!|G_{i,j}|\right) {}\\ \end{array} $$

(43)

Based on the properties of the gamma function, Γ(N) = (N−1)!, the value of $\mathcal {P}_{f_{i}}$ can be obtained as follows

$$\begin{array}{@{}rcl@{}} \mathcal{P}_{f_{i}}&=&\frac {1}{\Gamma(N)}{\int}_{0}^{A_{i}}t^{N-1}e^{-t}dt \\ &=&\frac{1}{(N-1)!}\left[-(e^{-t}t^{N-1})\mid_{0}^{A_{i}}-{\int}_{0}^{A_{i}}e^{-t}(N-1)t^{N-2}dt\right]\\ &=&1-{\frac{e^{-{A_{i}}}}{(N-1)!}}\sum\nolimits_{k=0}^{N-1}\frac{(N-1)!}{(N-1-k)!}{A_{i}}^{N-1-k} \end{array} $$

(44)

where $A_{i}=\frac {1}{\sqrt {{{\sum }_{j=0,j\neq i}^{N}{p_{j}^{2}}\overline {g}_{i,j}^{2}}}}\left (\frac {p_{i}g_{i, i}}{\widehat {\gamma }_{i}}-\sigma ^{2}\right )$ and $t={\sum }_{j=0,j\neq i}^{N}|G_{i,j}|$. The value of $\mathcal {P}_{f_{i}}$ is obtained. □

Appendix B

Proof

Based on the Cauchy-Schwarz inequality, we can obtain:

$$\begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{N}p_{i}g_{i}=\sum\limits_{i=1}^{N}G_{i}\overline{g}_{i}p_{i}\leq \sqrt{\sum\limits_{i=1}^{N}{p_{i}^{2}}\overline{g}_{i}^{2}}\cdot\sum\limits_{i=1}^{N}|G_{i}| \end{array} $$

(45)

and (17) can be rewritten

$$\begin{array}{@{}rcl@{}} Pr\left[\sum\limits_{i=1}^{N}|G_{i}|\leq\frac{1}{\sqrt{{{\sum}_{i=1}^{N}{p_{i}^{2}}\overline{g}_{i}^{2}}}} \left( \frac{p_{0}g_{m}}{\widehat{\gamma}_{m}}-\sigma^{2}\right)\right]\geq1-\varepsilon_{2}{}\\ \end{array} $$

(46)

where $g_{i}=G_{i}{\bar {g_{i}}}$ and G _i is the exponential distribution with the parameter of 1. Similarly, we can obtain

$$\begin{array}{@{}rcl@{}} {\sum}_{i=1}^{N}|G_{i}|&\sim&{\Gamma}(|G_{1}|+|G_{2}|+\cdots+|G_{N}|,1)\\ &\sim&{\Gamma}(N,1) \end{array} $$

(47)

The probability density function of ${\sum }_{i=1}^{N}|G_{i}|$ is

$$\begin{array}{@{}rcl@{}} f\left( \sum\limits_{i=1}^{N}|G_{i}|\right)=\frac{1}{\Gamma(N)}\left( \sum\limits_{i=1}^{N}|G_{i}|\right)^{N-1}e^{-{\sum}_{i=1}^{N}| G_{i}|} \end{array} $$

(48)

By elaborating (46) and (48), the probability of $\mathcal {P}_{m}$ can be obtained as

$$\begin{array}{@{}rcl@{}} \mathcal{P}_{m}&=&\frac {1}{(N-1)!}{{\int}_{0}^{B}}t^{N-1}e^{-u}du \\ &=&\frac{1}{(N-1)!}\left[-(e^{-u}u^{N-1}){\mid_{0}^{B}}-{{\int}_{0}^{B}}e^{-u}(N-1)u^{N-2}du\right]\\ &=&1-\frac{e^{-B}}{(N-1)!}\sum\nolimits_{l=0}^{N-1}\frac{(N-1)!}{(N-1-l)!}B^{N-1-l} \end{array} $$

(49)

where we define $B=\frac {1}{\sqrt {{{\sum }_{i=1}^{N}{p_{i}^{2}}\overline {g}_{i}^{2}}}}\left (\frac {p_{0} g_{m}}{\widehat {\gamma }_{m}}-\sigma ^{2}\right )$, $u={\sum }_{i=1}^{N}|G_{i}|$. The value of $\mathcal {P}_{m}$ is obtained. □

Appendix C

Proof

First, the gamma function is defined, ${\Gamma }(\alpha )={\int }_{0}^{+\infty }x^{\alpha -1}e^{-x}dx$. Then, the beta function is defined as $\beta (p,q)={{\int }_{0}^{1}}x^{p-1}(1-x)^{q-1}dx$. Then we prove the relationship between gamma function and beta function. Given that

$$\begin{array}{@{}rcl@{}} \beta(p,q)&=&{{\int}_{0}^{1}}x^{p-1}(1-x)^{q-1}dx \\ &\overset{x={1-y}}{=}&{{\int}_{1}^{0}}(1-y)^{p-1}y^{q-1}d(1-y)\\ &=&{{\int}_{0}^{1}}y^{q-1}(1-y)^{p-1}dy=\beta(q,p) \end{array} $$

(50)

Hence, we can obtain

$$\begin{array}{@{}rcl@{}} \beta(p+1,q+1)&=&{{\int}_{0}^{1}}x^{p}(1-x)^{q}dx\\ &=&\frac{q}{p+1}{{\int}_{0}^{1}}x^{p+1}(1-x)^{q-1}dx\\ &=&\frac{q}{p+1}\{ {{\int}_{0}^{1}}[x^{p}\!-x^{p}(1\,-\,x)](1-x)^{q-1}dx\}\\ &=&\frac{q}{p+1}[\beta(p+1,q)-\beta(p+1,q+1)] \\ \end{array} $$

(51)

and

$$\begin{array}{@{}rcl@{}} \beta(p+1,q+1)=\frac{q}{p+q+1}\beta(p+1,q) \end{array} $$

(52)

Based on Eq. 52, we can obtain the following

$$\begin{array}{@{}rcl@{}} \beta(m+1,n+1)&=&\frac{n}{m+n+1}\beta(m+1,n)\\ &=&\frac{n}{m+n+1}\cdot\frac{n-1}{m+n}\beta({m+1,n-1})\\ &=&\frac{n}{m+n+1}\cdot\frac{n-1}{m+n}\cdots\frac{1}{m+2}\beta({m+1,1}) \\ \end{array} $$

(53)

As β(p, q) = β(q, p), we can elaborate the following

$$\begin{array}{@{}rcl@{}} \beta({m+1,1})=\beta(1,m+1) \end{array} $$

(54)

Hence,

$$\begin{array}{@{}rcl@{}} \beta(1,m+1)&=&\frac{m}{m+1}\beta(1,m)=\frac{m}{m+1}\cdot\frac{m-1}{m}\beta(1,m-1)\\ &=&\frac{m}{m+1}\cdot\frac{m-1}{m}\cdots\frac{1}{2}\beta(1,1) \end{array} $$

(55)

Since β(1, 1) = 1, we can obtain

$$\begin{array}{@{}rcl@{}} \beta(m+1,n+1)&=&\frac{n}{m+n+1}\cdot\frac{n-1}{m+n}\cdots\frac{1}{m+2}\\ &&\cdot\frac{m}{m+1}\cdot\frac{m-1}{m}{\cdots} \frac{1}{2}\cdot1\\ &=&\frac{\Gamma(m+1){\Gamma}({n+1})}{\Gamma(m+n+2)} \end{array} $$

(56)

So we obtain

$$\begin{array}{@{}rcl@{}} \frac{\beta(\alpha_{1},\alpha_{2})}{\Gamma(\alpha_{1}){\Gamma}(\alpha_{2})}=\frac{1}{\Gamma(\alpha_{1}+ \alpha_{2})} \end{array} $$

(57)

□