Optimization of Multi-Antenna Wireless Networks

Abstract

This chapter presents two applications of the general analytical framework on network design and optimization. The first application considers interference coordination. A tractable expression is first derived for the coverage probability of a user-centric interference nulling strategy, which then helps to effectively optimize the interference coordination range. The presented interference nulling strategy can achieve performance gains about 35–40% compared with the non-coordination case. The second application studies general multiuser MIMO heterogeneous networks (HetNets). Exact and asymptotic expressions of the coverage probabilities are presented, which reveal that the SIR invariance property of SISO HetNets does not hold for MIMO HetNets. Instead, the coverage probability may decrease as the network density increases. It is proved that the maximum coverage probability is achieved by activating only one tier of BSs, while the maximum ASE is achieved by activating all the BSs. This reveals a unique trade-off between the ASE and link reliability in multiuser MIMO HetNets. To achieve the maximum ASE while guaranteeing a certain link reliability, efficient algorithms are provided to find the optimal BS densities.

Appendix

Proof of Lemma 5.1

Since $\varPsi _u^{\left( 1\right) }$ is an inhomogeneous PPP with density $\lambda _{u}^{\left( 1\right) }\left( y\right) =$$\lambda _{b}p_{\mathrm{a}}\left( e^{-\pi \lambda _{b}\frac{\left\| y\right\| ^{2}}{\mu ^{2}}}-e^{-\pi \lambda _{b}\left\| y\right\| ^{2}}\right)$, the mean number of points in $\mathbb {R}^2$ is given by [37, Sect. 2.4.3]

$$\begin{aligned} \bar{K}=\int _{\mathbb {R}^{2}}\lambda _{u}^{\left( 1\right) }\left( y\right) dy. \end{aligned}$$

(5.73)

Applying the polar-Cartesian transformation, we have $\mathrm {d}y=r\mathrm {d}r\mathrm {d}\theta$, and $\bar{K}$ can be written as

$$\begin{aligned} \bar{K} = \!\!\int _{0}^{2\pi }\!\!\!\!\int _{0}^{\infty }\!\!\!\lambda _{b}p_{\mathrm{a}}\left( \!\!e^{-\pi \lambda _{b}\frac{r^{2}}{\mu ^{2}}}-e^{-\pi \lambda _{b}r^{2}}\!\!\right) rdrd\theta =p_{\mathrm{a}}\left( \mu ^{2}\!-\!1\right) , \end{aligned}$$

(5.74)

which is a finite value when $\mu$ is finite.

Based on the property of the PPP [37], the number of points of $\varPsi _u^{\left( 1\right) }$ in $\mathbb {R}^2$ is Poisson distributed with mean $\bar{K}$, which completes the proof.

Proof of Theorem 5.3

By substituting (5.55) into the power series $C\left( z\right)$, we obtain

$$\begin{aligned} C\left( z\right) =\sum _{i=0}^{\infty }c_{i}z^{i}\sim \sum _{j=1}^{K}\lambda _{j}P_{j}^{\delta }B_{j}^{\delta } \left( \frac{U_{k}B_{k}}{U_{j}B_{j}}\tau \right) ^{\delta } \frac{\varGamma \left( U_{j}+\delta \right) }{\varGamma \left( U_{j}\right) }\sum _{i=0}^{\infty } \frac{\delta }{\delta -i}\frac{\varGamma \left( i+1-\delta \right) }{\varGamma \left( i+1\right) }z^{i}, \end{aligned}$$

(5.75)

and it can be expressed as

$$\begin{aligned} C\left( z\right) \sim \sum _{j=1}^{K}\lambda _{j}P_{j}^{\delta }B_{j}^{\delta } \left( \frac{U_{k}B_{k}}{U_{j}B_{j}}\tau \right) ^{\delta }\frac{\varGamma \left( U_{j}+\delta \right) }{\varGamma \left( U_{j}\right) }\varGamma \left( 1-\delta \right) \left( 1-z\right) ^{\delta }. \end{aligned}$$

(5.76)

Thus, the power series $\bar{P}\left( z\right)$ is given by

$$\begin{aligned} \bar{P}\left( z\right) =\frac{A}{C\left( z\right) }\sim \frac{A\frac{1}{\varGamma \left( 1-\delta \right) } \left( 1-z\right) ^{-\delta }}{\sum _{j=1}^{K}\lambda _{j}P_{j}^{\delta }B_{j}^{\delta } \left( \frac{U_{k}B_{k}}{U_{j}B_{j}}\tau \right) ^{\delta } \frac{\varGamma \left( U_{j}+\delta \right) }{\varGamma \left( U_{j}\right) }}. \end{aligned}$$

(5.77)

Based on the above expression, the coefficient $t_n$ is given by

$$\begin{aligned} \bar{p}_{n}=\frac{1}{n!}\bar{P}(z)^{\left( n\right) }\left( z\right) \mid _{z=0}\sim \frac{A\frac{1}{n!}\left( \delta \right) _{n}\frac{1}{\varGamma \left( 1-\delta \right) }}{\sum _{j=1}^{K}\lambda _{j}P_{j}^{\delta }B_{j}^{\delta }\left( \frac{U_{k}B_{k}}{U_{j}B_{j}} \tau \right) ^{\delta }\frac{\varGamma \left( U_{j}+\delta \right) }{\varGamma \left( U_{j}\right) }}. \end{aligned}$$

(5.78)

As $p_{\mathrm{s}}\left( k\right) =\sum _{n=0}^{M_k-1}\bar{p}_{n}$, and the sum $\sum _{n=0}^{M_k-1}\frac{1}{n!}\left( \delta \right) _{n}= \frac{\varGamma \left( M_k+\delta \right) }{\varGamma \left( 1+\delta \right) \varGamma \left( M_k\right) }$, the coverage probability is given by

$$\begin{aligned} p_{\mathrm{s}}\left( k\right) \sim \frac{A\frac{\varGamma \left( M_k+\delta \right) }{\varGamma \left( 1-\delta \right) \varGamma \left( 1+\delta \right) \varGamma \left( M_k\right) }}{\sum _{j=1}^{K}\lambda _{j}P_{j}^{\delta }B_{j}^{\delta }\left( \frac{U_{k}B_{k}}{U_{j}B_{j}} \tau \right) ^{\delta } \frac{\varGamma \left( U_{j}+\delta \right) }{\varGamma \left( U_{j}\right) }}. \end{aligned}$$

(5.79)

Using the equality $\varGamma \left( 1+\delta \right) \varGamma \left( 1-\delta \right) =\frac{\pi \delta }{\sin \left( \pi \delta \right) } =\frac{1}{\mathrm{sinc}\left( \delta \right) }$, we obtain (5.56).

Proof of Lemma 5.3

We consider the function $y=\frac{\mathbf {c}^{T}\varvec{\lambda }}{\mathbf {d}^{T}\varvec{\lambda }}$, where $\varvec{\lambda }=\left[ \lambda _{1},\lambda _{2},\ldots \lambda _{K}\right] ^{T}$, $\mathbf {c}=\left[ c_1,c_2,\ldots ,c_K\right]$ and $\mathbf {d}=\left[ d_1,d_2,\ldots ,d_K\right]$. The partial derivative of $f\left( \varvec{\lambda }\right)$ with respect to $\lambda _i$ is then given by

$$\begin{aligned} \frac{\partial y}{\partial \lambda _{i}} = \frac{\sum _{j=1}^{K}d_{i}d_{j}\lambda _{j}\left( \frac{c_{i}}{d_{i}}-\frac{c_{j}}{d_{j}}\right) }{\left( \mathbf {d}^{T}\varvec{\lambda }\right) ^{2}}. \end{aligned}$$

(5.80)

It shows that changing $\lambda _i$ will not change the sign of $\frac{\partial y}{\partial \lambda _{i}}$, i.e., y is monotonic with respect to $\lambda _i$. Moreover, y is independent of $\varvec{\lambda }$ if $\frac{c_i}{d_i}=\frac{c_j}{d_j}$ for all $i,j\in \left\{ 1,2,\ldots ,K\right\}$.

Proof of Lemma 5.4

In this proof, we consider a more general case where the optimization problem is given by

$$\begin{aligned} \mathscr {P}_{w/\,cons} : \underset{\varvec{\lambda }}{\text {maximize}}&\frac{\mathbf {c}^{T}\varvec{\lambda }}{\mathbf {d}^{T}\varvec{\lambda }} \\ \text {subject to}&\mathbf {A}\varvec{\lambda }\le \mathbf {b}, \nonumber \\&\varvec{\lambda }\ge \mathbf {0}, \nonumber \end{aligned}$$

(5.81)

where $\mathbf {A}\varvec{\lambda }\le \mathbf {b}$ represents an arbitrary linear constraint, $\mathbf {A}$ is a $n\times K$ matrix, and $\mathbf {b}$ is a $n\times 1$ vector with positive elements. We will find the optimal solution $\varvec{\lambda }^\star$ in the following derivation.

First, we consider the optimization problem without the constraint $\mathbf {A}\varvec{\lambda }\le \mathbf {b}$, i.e., the problem is given by

$$\begin{aligned} \mathscr {P}_{w/o\,cons} : \underset{\varvec{\lambda }}{\text {maximize}}&\frac{\mathbf {c}^{T}\varvec{\lambda }}{\mathbf {d}^{T}\varvec{\lambda }} \\ \text {subject to}&\varvec{\lambda }\ge \mathbf {0}. \nonumber \end{aligned}$$

(5.82)

Without loss of generality, we assume $\frac{c_{1}}{d_{1}}\ge \frac{c_{2}}{d_{2}}\ge \cdots \ge \frac{c_{K}}{M_k}$, and then from (5.80), we find that to maximize $p_\mathrm{s}$, $\lambda _K$ should be 0 since $\frac{\partial p_{\mathrm{s}}}{\partial \lambda _{K}}\le 0$.

Then, repeating the same procedure, we find $\lambda _{K-1}=0$, $\lambda _{K-2}=0$, and $\lambda _{2}=0$ successively. Finally, the objective function $p_\mathrm{s}$ is equal to $\frac{c_1}{d_1}$ with any $\lambda _1>0$, which is the solution of $\mathscr {P}_{w/o\,cons}$.

Second, we consider the optimization Problem $\mathscr {P}_{w/\,cons}$, and we want to prove that the optimal $\varvec{\lambda }^\star$ of Problem $\mathscr {P}_{w/o\,cons}$ is also the optimal solution of problem $\mathscr {P}_{w/\,cons}$. To do so, we need to prove (1) $\varvec{\lambda }^\star$ is feasible for $\mathscr {P}_{w/\,cons}$, and (2) the optimal solution of $\mathscr {P}_{w/\,cons}$ is $\varvec{\lambda }^\star$.

To prove the feasibility, we assume $\varvec{\lambda }_0^\star$ and $\varvec{\lambda }_1^\star$ are optimal solutions of $\mathscr {P}_{w/o\,cons}$, but $\mathbf {A}\varvec{\lambda }_1^\star > \mathbf {b}$. Since $\mathbf {b}>0$, and $\varvec{\lambda }_0^\star =k\varvec{\lambda }_1^\star$ for any $k>0$, we have $\exists k>0$, $\mathbf {A}\varvec{\lambda }_0^\star =k\mathbf {A}\varvec{\lambda }_1^\star \le \mathbf {b}$, i.e., there exists an optimal solution of $\mathscr {P}_{w/o\,cons}$, which is feasible to $\mathscr {P}_{w/\,cons}$.

Finally, since $\mathscr {P}_{w/\,cons}$ has one more constraint than $\mathscr {P}_{w/o\,cons}$, the solution of $\mathscr {P}_{w/\,cons}$ should be the subset of the solution of $\mathscr {P}_{w/o\,cons}$ and the maximum value of $\mathscr {P}_{w/\,cons}$ will no greater than the maximum value of $\mathscr {P}_{w/o\,cons}$. As we have proved that the optimal solution of $\mathscr {P}_{w/o\,cons}$ is feasible in $\mathscr {P}_{w/\,cons}$, we obtain that the maximum value of $\mathscr {P}_{w/\,cons}$ is equal to $\frac{c_1}{d_1}$.

Now we come back to the general MIMO HetNet case. The original problem (5.59) indicates $\mathbf {A}=\mathbf {I}_K$ and $\frac{c_{i}}{d_{i}}= \frac{\varGamma \left( M_{i}+\delta \right) /\varGamma \left( M_{i}\right) }{\varGamma \left( U_{i}+\delta \right) /\varGamma \left( U_{i}\right) }$. Therefore, the maximum $p_\mathrm{s}$ is obtained by only deploying one tier which has the maximal value of $\frac{c_{i}}{d_{i}}$.