Relay selection for peer-to-peer cooperative OFDMA with channel distribution uncertainty

Abstract

Peer-to-peer relay technique provides the opportunity of cooperative communication between users to achieve the transmission diversity. In this paper, the problem of relay selection and subcarrier allocation in peer-to-peer cooperative OFDMA system with channel distribution uncertainty is studied. The quality of service (QoS) requirements are characterized with the probabilistic constraints. The Kullback-Leibler (KL) divergence is adopted to describe the uncertainty of the channel state information (CSI), which follows a distribution deviated within a certain distance from the reference distribution. Since the probabilistic constraint is intractable, we resort to a tractable upper bound instead. The problem is solved by the interior point cutting plane (IPCP) method, which proceeds in round whereby an iterative method is designed to combine both golden section method and interior point method to find the saddle point of the constraint. Simulation results show the effectiveness of our algorithm.

Appendix

1.1 Proof of Theorem 1

The case of k = 0 is obvious, we focus on the case of k > 0. The distribution functions of $\gamma _{s, m0}^{n}$ and $\gamma _{s, mk}^{n}$ are given as

$$\begin{aligned} f_{\gamma_{s, m0}^{n}}(\xi)=\frac{1}{\delta_{s, m0}^{n}}\exp\left(-\frac{\xi}{\delta_{s, m0}^{n}}\right), \\ f_{\gamma_{s, mk}^{n}}(\xi)=\frac{1}{\delta_{s, mk}^{n}}\exp\left(-\frac{\xi}{\delta_{s, mk}^{n}}\right), \end{aligned}$$

(34)

According to the theorem of total probability, the distribution function of $\sigma _{s, mk}^{n}\gamma _{r, mk}^{n}$ could be derived as follows:

$$\begin{aligned} f_{\sigma_{s, mk}^{n}\gamma_{s, mk}^{n}}(\xi)&=P_{s, mk}^{n} f_{\gamma_{s, mk}^{n}}(\xi)+(1-P_{s, mk}^{n})Dirac(0)\\ &= \frac{P_{s, mk}^{n} }{\delta_{s, mk}^{n}}\exp\left(-\frac{\xi}{\delta_{s, mk}^{n}}\right)+(1-P_{s, mk}^{n})Dirac(0). \end{aligned}$$

(35)

where Dirac(⋅) denotes the dirac delta function.

Denote $\lambda _{0}=\frac {1}{\delta _{s, m0}^{n}}$ and $\lambda _{1}=\frac {1}{\delta _{r, mk}^{n}}$, we have

$$M_{\gamma_{m0}^{n}}(s)=\frac{\lambda_{0}}{s+\lambda_{0}}, \ \ M_{\sigma_{s, mk}^{n}\gamma_{r, mk}^{n}}(s)=1-P_{s, mk}^{n}+P_{s, mk}^{n}\frac{\lambda_{1}}{s+\lambda_{1}}.$$

(36)

According to the properties of the moment generation function (MGF), the MGF of $\gamma _{mk}^{n}=\gamma _{m0}^{n}+\sigma _{s, mk}^{n}\gamma _{r, mk}^{n}$ can be expressed as

$$M_{\gamma_{mk}^{n}}(s)=\frac{\lambda_{0}}{s+\lambda_{0}}\left[(1-P_{s, mk}^{n})+P_{s, mk}^{n}\frac{\lambda_{1}}{s+\lambda_{1}}\right],$$

(37)

so the cumulative density function of $\gamma _{mk}^{n}$ can be given as

$$F_{\gamma_{mk}^{n}}(\xi)=\mathcal{L}^{-1}\left(\frac{M(s)}{s}\right) =1-e^{-\lambda_{0} \xi}+\frac{P_{s, mk}^{n}\lambda_{0}}{\lambda_{0}-\lambda_{1}}(e^{-\lambda_{0} \xi}-e^{-\lambda_{1} \xi}),$$

(38)

where $\mathcal {L}^{-1}(\cdot )$ denotes the inverse laplace transformation. Substituting λ ₀ and λ ₁ into the derivative of the cumulative density function, we can obtain the distribution function of $\gamma _{mk}^{n}$ with k > 0

$$f_{\gamma_{mk}^{n}}(\xi) =\frac{e^{-\frac{\xi }{\delta_{s, m0}^{n}}}}{\delta_{s, m0}^{n}}+\frac{e^{-\frac{\gamma_{th}}{\delta_{s, mk}^{n}}}{\delta_{r, mk}^{n}}}{{\delta_{r, mk}^{n}}-\delta_{s, m0}^{n}}\left( \frac{e^{-\frac{\xi }{\delta_{r, mk}^{n}} }}{\delta_{r, mk}^{n}}-\frac{e^{-\frac{\xi}{\delta_{s, m0}^{n}} }}{\delta_{s, m0}^{n}}\right).$$

(39)

1.2 Proof of Proposition 1

$$\begin{aligned} &H_{m}\left(\theta, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ =&\bar{C}_{m} +\theta \log \mathbb{E}\left[\exp\left(-\frac{ \sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n}}{\theta}\right) \right]-\theta \log \epsilon\\ =&\theta \log \left\{\mathbb{E}\left[\exp{\left(\frac{\bar{C}_{m}-\sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n}}{\theta}\right)}\right]\right\}-\theta \log \epsilon, \end{aligned}$$

Equation (9) can be transformed as

$$\max_{f_{\gamma_{mk}^{n}}(\xi)\in Z_{mk}^{n}} \min\limits_{\theta\geq0} \mathbb{E}\left[\exp{\left(\frac{\bar{C}_{m}-\sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n}}{\theta}\right)}\right] \leq \epsilon.$$

(40)

Let us define function $F\left ({\sum }_{k=0}^{K} {\sum }_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar {C}_{m}\right )$ as:

$$F\left(\sum\limits_{k=0}^{K} \sum\limits_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m}\right)= \begin{cases} 1, \sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m} \leq 0, \\ 0, \sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m} > 0. \end{cases}$$

(41)

then $F\left ({\sum }_{k=0}^{K} {\sum }_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar {C}_{m}\right )$ can be bounded with an exponential function as

$$F\left(\sum\limits_{k=0}^{K} \sum\limits_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m}\right) \leq \exp{\left(-\frac{\sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m}}{\theta}\right)}.$$

(42)

By taking the expectation on both sides and minimizing 𝜃, we have

$$\begin{aligned} &Pr\left(\sum\limits_{k=0}^{K} \sum\limits_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m}\right)\\ \leq &\min\limits_{\theta\geq 0}\mathbb{E}\left[\exp{\left(\frac{\bar{C}_{m}-\sum\limits_{k=0}^{K} \sum\limits_{n=1}^{N} C_{mk}^{n} x_{mk}^{n}}{\theta}\right)}\right]. \end{aligned}$$

By the maximization over $f_{\gamma _{mk}^{n}}$ on both sides, the inequality also holds. Then we have

$$\begin{aligned} &\max_{f_{\gamma_{mk}^{n}}(\xi)\in Z_{mk}^{n}}Pr\left(\sum\limits_{k=0}^{K} \sum\limits_{n=1}^{N} C_{mk}^{n} x_{mk}^{n} - \bar{C}_{m}\right) \\ \leq &\max_{f_{\gamma_{mk}^{n}}(\xi)\in Z_{mk}^{n}}\min\limits_{\theta\geq 0}\mathbb{E}\left[\exp{\left(\frac{\bar{C}_{m}-\sum_{k=0}^{K} \sum_{n=1}^{N} C_{mk}^{n} x_{mk}^{n}}{\theta}\right)}\right]. \end{aligned}$$

(43)

Combine Eqs. (40) and (43), we have Eq. (14).

1.3 Proof of Theorem 2

It is obvious that the objection function, constraint (15a) and (15c) are all linear function of variable $x_{mk}^{n}$, so they are convex. Then, we demonstrate the constraint (15b) is convex. In fact, we only need to prove that the cumulate generating function ${\Phi }\left (\theta , x_{mk}^{n}, f_{\gamma _{mk}^{n}}(\xi )\right )$ is convex. Let a, b ∈ [0, 1] denote two variables satisfying a + b = 1. With k > 0, we have

$$\begin{aligned} &{\Phi}\left(\theta, a x_{mk}^{n1}+ b x_{mk}^{n2}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ =&\log\left[{\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W (a x_{mk}^{n1}+ b x_{mk}^{n2})}{2 \theta \log2}}f_{\gamma_{mk}^{n}}(\xi)d\xi\right]\\ =&\log\left\{\mathbb{E}\left[ e^{\frac{(a x_{mk}^{n1}+ b x_{mk}^{n2})W\log_{2}(1+\rho\xi)}{2\theta }}\right]\right\}\\ =&\log\left\{\mathbb{E}\left[ e^{\frac{a x_{mk}^{n1}W\log_{2}(1+\rho\xi)}{2\theta}}e^{\frac{b x_{mk}^{n2}W\log_{2}(1+\rho\xi)}{2\theta}} \right]\right\}, \\ \end{aligned}$$

(44)

According to the $H\ddot {o}lder's$ inequality, we have:

$$\begin{aligned} &{\Phi}\left(\theta, a x_{mk}^{n1}+ b x_{mk}^{n2}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ \leq & \log\left\{\left(\mathbb{E}\left[ e^{\frac{ x_{mk}^{n1}W\log_{2}(1+\rho\xi)}{2\theta }}\right]\right)^{a} \left(\mathbb{E}\left[ e^{\frac{x_{mk}^{n2}W\log_{2}(1+\rho\xi)}{2\theta }} \right]\right)^{b}\right\}\\ =&a{\Phi}\left(\theta, x_{mk}^{n1}, f_{\gamma_{mk}^{n}}(\xi)\right)+ b {\Phi}\left(\theta, x_{mk}^{n2}, f_{\gamma_{mk}^{n}}(\xi)\right). \end{aligned}$$

(45)

The case of k = 0 can be proved similarly. Therefore, the constraint (15b) is convex and the problem (15) is a convex optimization problem.

1.4 Proof of Theorem 3

To prove that the problem (21) is convex over 𝜃, we only need to prove that the cumulate generating function ${\Phi }\left (\theta , x_{mk}^{n}, f_{\gamma _{mk}^{n}}(\xi )\right )$ is convex over 𝜃. Assume there exists 𝜃 ₁ and 𝜃 ₂, $a, b\in \left [0, 1\right ]$ which satisfying a + b = 1, when k > 0, we have

$$\begin{aligned} &{\Phi}\left(a\theta_{1}+b\theta_{2}, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ =&\log\left[{\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2(a\theta_{1}+b\theta_{2}) \log2}}f_{\gamma_{mk}^{n}}(\xi)d\xi\right]\\ =&\log\left\{\mathbb{E}\left[ e^{\frac{x_{mk}^{n}W\log_{2}(1+\rho\xi)}{2(a \theta_{1}+b\theta_{2})}}\right]\right\} \end{aligned}$$

(46)

Because

$$\begin{aligned} &\frac{1}{a\theta_{1}+b\theta_{2}}\\ =&\frac{a\theta_{2}+b\theta_{1}}{(a\theta_{2}+b\theta_{1})(a\theta_{1}+b\theta_{2})}\\ =&\frac{a\theta_{2}+b\theta_{1}}{(a^{2}+b^{2})\theta_{1}\theta_{2}+ab({\theta_{1}^{2}}+{\theta_{2}^{2}})}\\ \leq&\frac{a\theta_{2}+b\theta_{1}}{(a^{2}+b^{2})\theta_{1}\theta_{2}+2ab\theta_{1}\theta_{2}}\\ =&\frac{a\theta_{2}+b\theta_{1}}{\theta_{1}\theta_{2}}=\frac{a}{\theta1}+\frac{b}{\theta2}, \end{aligned}$$

(47)

we have

$$\begin{aligned} &{\Phi}\left(a\theta_{1}+b\theta_{2}, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ \leq&\log\left\{\mathbb{E}\left[e^{x_{mk}^{n}\frac{W}{2}\log_{2}(1+\rho\xi)(\frac{a}{\theta_{1}}+\frac{b}{\theta_{2}})}\right]\right\}\\ =&\log\left\{\mathbb{E}\left[e^{\left(\frac{x_{mk}^{n}\frac{W}{2}\log_{2}(1+\rho\xi)}{\theta_{1}}\right)a}e^{\left(\frac{x_{mk}^{n}\frac{W}{2}\log_{2}(1+\rho\xi)}{\theta_{2}}\right)b}\right]\right\}. \end{aligned}$$

(48)

With $H\ddot {o}lder's$ inequality, we have

$$\begin{aligned} &{\Phi}\left(a\theta_{1}+b\theta_{2}, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right)\\ \leq&\log\left\{\left(\mathbb{E}\left[e^{\frac{x_{mk}^{n}\frac{W}{2}\log_{2}(1+\rho\xi)}{\theta_{1}}}\right]\right)^{a} \left(\mathbb{E}\left[e^{\frac{x_{mk}^{n}\frac{W}{2}\log_{2}(1+\rho\xi)}{\theta_{2}}}\right]\right)^{b}\right\}\\ =&a{\Phi}\left(\theta_{1}, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right)+b{\Phi}\left(\theta_{2}, x_{mk}^{n}, f_{\gamma_{mk}^{n}}(\xi)\right). \end{aligned}$$

(49)

The proof is similar for the case k = 0, which is omitted. In conclusion, ${\Phi }\left (\theta , x_{mk}^{n}, f_{\gamma _{mk}^{n}}(\xi )\right )$ is convex over 𝜃, the problem (21) is convex over 𝜃.

In the following, we prove that ${\Phi }\left (\theta , x_{mk}^{n}, f_{\gamma _{mk}^{n}}(\xi )\right )$ is concave over $f_{\gamma _{mk}^{n}}(\xi )$, when k > 0, for a, b ∈ [0, 1] satisfying a + b = 1, we have

$$\begin{aligned} &{\Phi}\left(\theta, x_{mk}^{n}, af_{\gamma_{mk}^{n}}^{1}(\xi)+bf_{\gamma_{mk}^{n}}^{2}(\xi)\right)\\ =&\log\left[{\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2\theta \log2}}\left(af_{\gamma_{mk}^{n}}^{1}(\xi)+b f_{\gamma_{mk}^{n}}^{2}(\xi)\right)d\xi\right]\\ =&\log\left[a{\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2\theta \log2}}f_{\gamma_{mk}^{n}}^{1}(\xi)d\xi+b\!\! {\int}_{0}^{\infty}\!(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2\theta \log2}}f_{\gamma_{mk}^{n}}^{2}(\xi)d\xi\right]\\ \geq&a\log\left[{\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2\theta \log2}}f_{\gamma_{mk}^{n}}^{1}(\xi)d\xi\right]\\ &+b\log\left[ {\int}_{0}^{\infty}(1+\rho\xi)^{-\frac{W x_{mk}^{n}}{2\theta \log2}}f_{\gamma_{mk}^{n}}^{2}(\xi)d\xi\right]\\ =&a{\Phi}\left(\theta, x_{mk}^{n}, f_{\gamma_{mk}^{n}}^{1}(\xi)\right)+b{\Phi}\left(\theta, x_{mk}^{n}, f_{\gamma_{mk}^{n}}^{2}(\xi)\right). \end{aligned}$$

(50)

The proof of k = 0 can be obtained similarly. Thus, ${\Phi }\left (\theta , x_{mk}^{n}, f_{\gamma _{mk}^{n}}(\xi )\right )$ is concave over $f_{\gamma _{mk}^{n}}(\xi )$.

1.5 Proof of Proposition 2

Since problem (22) is convex, so we can adopt the Lagrangian dual method to solve this problem, whereby the Lagrangian function is given as:

$$\begin{aligned} &L(f_{\gamma_{mk}^{n}}(\xi), \lambda_{mk}^{n}, \mu_{mk}^{n})\\ =&{\int}_{0}^{+\infty} l(\xi)f_{\gamma_{mk}^{n}}(\xi)d\xi-\lambda_{mk}^{n}\left({\int}_{0}^{+\infty}f_{\gamma_{mk}^{n}} (\xi) \log \frac{f_{\gamma_{mk}^{n}} (\xi)}{f_{\gamma_{mk}^{n}}^{0} (\xi)} d\xi - \chi_{mk}^{n} \right)\\ &- \mu_{mk}^{n} \left({\int}_{0}^{+\infty} f_{\gamma_{mk}^{n}} (\xi) d \xi -1 \right). \end{aligned}$$

(51)

The KKT conditions can be obtained as:

$$\begin{array}{*{20}l} l(\xi)-\lambda_{mk}^{n}\left(\log \frac{f_{\gamma_{mk}^{n}}(\xi)}{f_{\gamma_{mk}^{n}}^{0}(\xi)} + 1\right)-\mu_{mk}^{n}=0, \end{array}$$

(52)

$$\begin{array}{*{20}l} {\int}_{0}^{+\infty}f_{\gamma_{mk}^{n}}(\xi)d\xi - 1=0, \end{array}$$

(53)

$$\begin{array}{*{20}l} \lambda_{mk}^{n}\left({\int}_{0}^{+\infty}f_{\gamma_{mk}^{n}} (\xi) \log \frac{f_{\gamma_{mk}^{n}} (\xi)}{f_{\gamma_{mk}^{n}}^{0} (\xi)} d\xi - \chi_{mk}^{n} \right)=0, \end{array}$$

(54)

$$\begin{array}{*{20}l} \lambda_{mk}^{n}\geq0. \end{array}$$

(55)

With Eq. (52), we have

$$f_{\gamma_{mk}^{n}}^{*}(\xi)=f_{\gamma_{mk}^{n}}^{0}(\xi)e^{\frac{l(\xi)-\mu_{mk}^{n*}}{\lambda_{mk}^{n*}}-1}$$

(56)

where $\lambda _{mk}^{n*}$ and $\mu _{mk}^{n*}$ are the solutions given by Eqs. (53–55). By substituting Eq. (56) into these equations, we can obtain the Eqs. in (25) .