Multi-Objective Optimization for Secure Full-Duplex Wireless Communication Systems

Abstract

In traditional half-duplex (HD) communication systems, the HD base station (BS) can transmit artificial noise (AN) to jam the eavesdroppers for securing the downlink (DL) communication. However, guaranteeing uplink (UL) transmission is not possible with an HD BS because HD BSs cannot jam the eavesdroppers during UL transmission. In this chapter, we investigate the resource allocation algorithm design for secure multiuser systems employing a full-duplex (FD) BS for serving multiple HD DL and UL users simultaneously. In particular, the FD BS transmits AN to guarantee the concurrent DL and UL communication security. We propose a multi-objective optimization framework to study two conflicting yet desirable design objectives, namely total DL transmit power minimization and total UL transmit power minimization. To this end, the weighted Tchebycheff method is adopted to formulate the resource allocation algorithm design as a multi-objective optimization problem (MOOP). The considered MOOP takes into account the quality-of-service (QoS) requirements of all legitimate users for guaranteeing secure DL and UL transmission in the presence of potential eavesdroppers. Thereby, secure UL transmission is enabled by the FD BS, which would not be possible with an HD BS. Although the considered MOOP is non-convex, we solve it optimally by semidefinite programming (SDP) relaxation. Simulation results not only unveil the trade-off between the total DL transmit power and the total UL transmit power, but also confirm that the proposed secure FD system can guarantee concurrent secure DL and UL transmission and provide substantial power savings over a baseline system.

ⒸPortions of this chapter are reprinted from [19], with permission from IEEE.

Appendices

Appendix 1: Proof of Proposition 1

We start the proof by rewriting constraints C3 and C4 as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl} {}{\mbox{C3}}\mbox{: } &\displaystyle &\displaystyle \det({\mathbf{I}}_{N_{\mathrm{R}}} +{\mathbf{X}}^{-1}{\mathbf{L}}^H{\mathbf{W}}_k\mathbf{L}) \le 2^{R_{\mathrm{tol}_{k}}^{\mathrm{DL}}} \notag \\ {}&\displaystyle \overset{(a)}{\Longleftrightarrow}&\displaystyle \det({\mathbf{I}}_{N_{\mathrm{R}}} + {\mathbf{X}}^{-1/2}{\mathbf{L}}^H{\mathbf{W}}_k\mathbf{L}{\mathbf{X}}^{-1/2}) \le 2^{R_{\mathrm{tol}_{k}}^{\mathrm{DL}}}, \end{array} \end{aligned} $$

(10.23)

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} {\mbox{C4}}\mbox{: } &\displaystyle &\displaystyle \det({\mathbf{I}}_{N_{\mathrm{R}}} + P_j{\mathbf{X}}^{-1}{\mathbf{e}}_{j}{\mathbf{e}}_{j}^H) \le 2^{R_{\mathrm{tol}_{j}}^{\mathrm{UL}}} \notag \\ {} &\displaystyle \overset{(b)}{\Longleftrightarrow}&\displaystyle \det({\mathbf{I}}_{N_{\mathrm{R}}} + P_j{\mathbf{X}}^{-1/2}{\mathbf{e}}_{j}{\mathbf{e}}_{j}^H{\mathbf{X}}^{-1/2}) \le 2^{R_{\mathrm{tol}_{j}}^{\mathrm{UL}}}. \end{array} \end{aligned} $$

(10.24)

(a) and (b) hold due to a basic matrix equality, namely $\det (\mathbf {I}+\mathbf {AB})=\det (\mathbf {I}+\mathbf {BA})$. Then, we study a lower bound of (10.23) and (10.24) by applying the following Lemma.

Lemma 1 (Determinant Inequality [19])

For any semidefinite matrixA ≽0, the inequality holds where equality holds if and only if $\operatorname {{\mathrm {Rank}}}(\mathbf {A}) \le 1$.

We note that X ^−1∕2L ^HW _kLX ^−1∕2 ≽0 holds in (10.23). Thus, applying Lemma 1 to (10.23) yields

(10.25)

As a result, by combining (10.23) and (10.25), we have the following implications:

(10.26)

where λ _max(A) denotes the maximum eigenvalue of matrix A and (c) is due to the fact that holds for any A ≽0. Besides, if $\operatorname {{\mathrm {Rank}}}({\mathbf {W}}_k) \le 1$, we have

$$\displaystyle \begin{aligned} & \operatorname{{\mathrm{Rank}}}({\mathbf{X}}^{-1/2}{\mathbf{L}}^H{\mathbf{W}}_k\mathbf{L}{\mathbf{X}}^{-1/2}) \notag \\ &\quad \le \min\Big\{\operatorname{{\mathrm{Rank}}}({\mathbf{X}}^{-1/2}{\mathbf{L}}^H),\operatorname{{\mathrm{Rank}}}({\mathbf{W}}_k\mathbf{L}{\mathbf{X}}^{-1/2})\Big\} \notag \\ &\quad \le \operatorname{{\mathrm{Rank}}}({\mathbf{W}}_k\mathbf{L}{\mathbf{X}}^{-1/2}) \,\,\, \le 1.\\ {} \notag \end{aligned} $$

(10.27)

Then, equality holds in (10.25). Besides, in (10.26), is equivalent to $\lambda _{\mathrm {max}}({\mathbf {X}}^{-1/2}{\mathbf {L}}^H{\mathbf {W}}_k\mathbf {L}{\mathbf {X}}^{-1/2}) \le \xi _{k}^{\mathrm {DL}}$. Therefore, (10.23) and (10.26) are equivalent if $\operatorname {{\mathrm {Rank}}}({\mathbf {W}}_k) \le 1$.

As for constraint C4, we note that $\operatorname {{\mathrm {Rank}}}(P_j{\mathbf {X}}^{-1/2}{\mathbf {e}}_{j}{\mathbf {e}}_{j}^H{\mathbf {X}}^{-1/2}) \le 1$ always holds. Therefore, by applying Lemma 1 to (10.24), we have

(10.28)

Then, by combining (10.24) and (10.28), we have the following implications:

(10.29)

Appendix 2: Proof of Theorem 1

The SDP relaxed version of equivalent Problem 3 in (10.22) is jointly convex with respect to the optimization variables and satisfies Slater’s constraint qualification. Therefore, strong duality holds and solving the dual problem is equivalent to solving the primal problem [3]. For obtaining the dual problem, we first need the Lagrangian function of the primal problem in (10.22) which is given by

(10.30)

Here, Λ denotes the collection of terms that only involve variables that are independent of W _k. λ _k, μ _j, and π _i are the Lagrange multipliers associated with constraints C1, C2, and C9, respectively. Matrix ${\mathbf {D}}_k \in {\mathbb {C}^{N_{\mathrm {R}}\times N_{\mathrm {R}}}}$ is the Lagrange multiplier matrix for constraint $\widetilde {\text{C3}}$. Matrix ${\mathbf {Y}}_k \in {\mathbb {C}^{N_{\mathrm {T}}\times N_{\mathrm {T}}}}$ is the Lagrange multiplier matrix for the positive semidefinite constraint C7 on matrix W _k. For notational simplicity, we define Ψ as the set of scalar Lagrange multipliers for constraints C1, C2, C5, and C9 and Φ as the set of matrix Lagrange multipliers for constraints $\widetilde {\mbox{C3}}$, $\widetilde {\mbox{C4}}$, C6, and C7. Thus, the dual problem for the SDP relaxed problem in (10.22) is given by

$$\displaystyle \begin{aligned} & \underset{\Psi \ge 0, \boldsymbol{\Phi} \succeq \mathbf{0}}{\operatorname{\mathrm{maximize}}} \,\,\underset{{\mathbf{W}}_k,{\mathbf{Z}}_{\mathrm{AN}}\in\mathbb{H}^{N_{\mathrm{T}}},P_j,\tau}{\operatorname{\mathrm{minimize}}} \,\, {\mathcal{L}} \Big({\mathbf{W}}_k,{\mathbf{Z}}_{\mathrm{AN}},P_j,\Psi, \boldsymbol{\Phi}\Big)\notag\\ {} \hspace{-3mm} &\mbox{s.t.}\sum_{i=1}^2 \pi_i = 1.\end{aligned} $$

(10.31)

Constraint $\sum _{i=1}^2 \pi _i = 1$ is imposed to guarantee a bounded solution of the dual problem [3]. Then, we reveal the structure of the optimal W _k of (10.22) by studying the Karush–Kuhn–Tucker (KKT) conditions. The KKT conditions for the optimal ${\mathbf {W}}_k^*$ are given by

$$\displaystyle \begin{aligned}{\mathbf{Y}}_k^*,{\mathbf{D}}_k^* \succeq& \mathbf{0},\quad \lambda_k^*, \mu_j^*, \pi_i^* \ge 0, \end{aligned} $$

(10.32)

$$\displaystyle \begin{aligned} {\mathbf{Y}}_k^*{\mathbf{W}}_k^*=&\mathbf{0}, {} \end{aligned} $$

(10.33)

$$\displaystyle \begin{aligned} \nabla_{{\mathbf{W}}_k^*}{\mathcal{L}}=& \mathbf{0}, {}\\ {} \notag \end{aligned} $$

(10.34)

where ${\mathbf {Y}}_k^*$, ${\mathbf {D}}_k^*$, $\lambda _k^*$, $\mu _j^*,$ and $\pi _i^*$ are the optimal Lagrange multipliers for dual problem (10.31). $\nabla _{{\mathbf {W}}_k^*}{\mathcal {L}}$ denotes the gradient of Lagrangian function ${\mathcal {L}}$ with respect to matrix ${\mathbf {W}}_k^*$. The KKT condition in (10.34) can be expressed as

$$\displaystyle \begin{aligned} &{\mathbf{Y}}_k^*+ \frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}} \notag \\ &\quad = \varrho_1\pi_1{\mathbf{I}}_{N_{\mathrm{T}}}+ \sum_{j=1}^{J}\mu_j^*\rho{\mathbf{H}}_{\mathrm{SI}}^H\operatorname{{\mathrm{diag}}}({\mathbf{V}}_j){\mathbf{H}}_{\mathrm{SI}} + \mathbf{L}{\mathbf{D}}_k^*{\mathbf{L}}^H.\\ {} \notag \end{aligned} $$

(10.35)

Now, we divide the proof into two cases according to the value of ϱ ₁. First, for the case of 0 < ϱ ₁ ≤ 1, we define

$$\displaystyle \begin{aligned} {\mathbf{A}}_k^* =&\sum_{j=1}^{J}\mu_j^*\rho{\mathbf{H}}_{\mathrm{SI}}^H\operatorname{{\mathrm{diag}}}({\mathbf{V}}_j){\mathbf{H}}_{\mathrm{SI}} + \mathbf{L}{\mathbf{D}}_k^*{\mathbf{L}}^H, \end{aligned} $$

(10.36)

$$\displaystyle \begin{aligned} \boldsymbol{\Pi}_k^* =& \varrho_1\pi_1{\mathbf{I}}_{N_{\mathrm{T}}}+{\mathbf{A}}_k^*,\\ {} \notag \end{aligned} $$

(10.37)

for notational simplicity. Then, (10.35) implies

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {\mathbf{Y}}^*=\boldsymbol{\Pi}_k^*-\frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}. \end{array} \end{aligned} $$

(10.38)

Pre-multiplying both sides of (10.38) by ${\mathbf {W}}_k^*$, and utilizing (10.33), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {\mathbf{W}}_k^*\boldsymbol{\Pi}_k^* ={\mathbf{W}}_k^*\frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}. \end{array} \end{aligned} $$

(10.39)

By applying basic inequalities for the rank of matrices, the following relation holds:

$$\displaystyle \begin{aligned} \hspace{-4mm} \operatorname{{\mathrm{Rank}}}({\mathbf{W}}_k^*) \overset{(a)}{=}& \operatorname{{\mathrm{Rank}}}({\mathbf{W}}_k^*\boldsymbol{\Pi}_k^*) = \operatorname{{\mathrm{Rank}}}\Big({\mathbf{W}}_k^*\frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}\Big)\notag\\ \hspace{-4mm} \overset{(b)}{\le}& \min\Big\{\operatorname{{\mathrm{Rank}}}({\mathbf{W}}_k^*), \operatorname{{\mathrm{Rank}}}\Big(\frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}\Big)\Big\} \notag \\ \hspace{-4mm} \overset{(c)}{\le}& \operatorname{{\mathrm{Rank}}}\Big(\frac{\lambda_k^* {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}\Big) \le 1,\\ {} \notag \end{aligned} $$

(10.40)

where (a) is valid because $\boldsymbol {\Pi }_k^* \succ \mathbf {0}$, (b) is due to the basic result $\operatorname {{\mathrm {Rank}}}(\mathbf {A}\mathbf {B}) \le \min \big \{ \operatorname {{\mathrm {Rank}}}(\mathbf {A}), \operatorname {{\mathrm {Rank}}}(\mathbf {B})\big \}$, and (c) is due to the fact that $\min \{a,b\} \le a$. We note that ${\mathbf {W}}_k^* \neq \mathbf {0}$ for $\Gamma ^{\mathrm {DL}}_{\mathrm {req}_k} > 0$. Thus, $\operatorname {{\mathrm {Rank}}}({\mathbf {W}}_k^*)=1$.

Then, for the case of ϱ ₁ = 0, we show that we can always construct a rank-one optimal solution ${\mathbf {W}}_k^{**}$. We note that the problem in (10.22) with ϱ ₁ = 0 is equivalent to a total UL transmit power minimization problem which is given by

$$\displaystyle \begin{aligned}&\hspace{0mm}\underset{{\mathbf{W}}_k,{\mathbf{Z}}_{\mathrm{AN}}\in\mathbb{H}^{N_{\mathrm{T}}},P_j} {\operatorname{\mathrm{minimize}}}\,\, \sum_{j=1}^{J}P_j \notag\\ & \hspace{-10mm} \mbox{s.t.} \hspace{5mm} \mbox{C1},\mbox{C2},\widetilde{\mbox{C3}},\widetilde{\mbox{C4}},\mbox{C5},\mbox{C6},\mbox{C7},{\mbox{C9}}.\end{aligned} $$

(10.41)

We first solve the above convex optimization problem and obtain the UL transmit power $P_j^{**}$, the DL beamforming matrix ${\mathbf {W}}_k^*$, and the AN covariance matrix ${\mathbf {Z}}_{\mathrm {AN}}^*$. If $\operatorname {{\mathrm {Rank}}}({\mathbf {W}}_k^*)=1,\forall k$, then the globally optimal solution of problem (10.19) for ϱ ₁ = 0 is achieved. Otherwise, we substitute $P_j^{**}$ and ${\mathbf {Z}}_{\mathrm {AN}}^*$ into the following auxiliary problem:

(10.42)

Since the problem in (10.42) shares the feasible set of problem (10.41), problem (10.42) is also feasible. Now, we claim that for a given $P_j^{**}$ and ${\mathbf {Z}}_{\mathrm {AN}}^*$ in (10.42), the solution ${\mathbf {W}}_k^{**}$ of (10.42) is a rank-one matrix. First, the gradient of the Lagrangian function for (10.42) with respect to ${\mathbf {W}}_k^{**}$ can be expressed as

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {\mathbf{Y}}^{**}=\boldsymbol{\Pi}_k^{**}-\frac{\lambda_k^{**} {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}, \end{array} \end{aligned} $$

(10.43)

where

$$\displaystyle \begin{aligned} \boldsymbol{\Pi}_k^{**}=&{\mathbf{I}}_{N_{\mathrm{T}}}+{\mathbf{A}}_k^{**} \,\,\,\,\text{ and} \end{aligned} $$

(10.44)

$$\displaystyle \begin{aligned} {\mathbf{A}}_k^{**}=& \sum_{j=1}^{J}\mu_j^{**}\rho{\mathbf{H}}_{\mathrm{SI}}^H\operatorname{{\mathrm{diag}}}({\mathbf{V}}_j){\mathbf{H}}_{\mathrm{SI}} + \mathbf{L}{\mathbf{D}}_k^{**}{\mathbf{L}}^H.\\ {} \notag \end{aligned} $$

(10.45)

${\mathbf {Y}}_k^{**}$, ${\mathbf {D}}_k^{**}$, $\lambda _k^{**}$, and $\mu _j^{**}$ are the optimal Lagrange multipliers for the dual problem of (10.42). Pre-multiplying both sides of (10.43) by the optimal solution ${\mathbf {W}}_k^{**}$, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {\mathbf{W}}_k^{**}\boldsymbol{\Pi}_k^{**} ={\mathbf{W}}_k^{**}\frac{\lambda_k^{**} {\mathbf{H}}_k}{\Gamma^{\mathrm{DL}}_{\mathrm{req}_k}}. \end{array} \end{aligned} $$

(10.46)

We note that $\boldsymbol {\Pi }_k^{**}$ is a full-rank matrix, i.e., $\boldsymbol {\Pi }_k^{**} \succ \mathbf {0}$, and (10.46) has the same form as (10.39). Thus, we can follow the same approach as for the case of 0 < ϱ _i ≤ 1 for showing that ${\mathbf {W}}_k^{**}$ is a rank-one matrix. Also, since ${\mathbf {W}}_k^{**}$ is a feasible solution of (10.41) for $P_j^{**}$, an optimal rank-one matrix ${\mathbf {W}}_k^{**}$ for the case of ϱ ₁ = 0 is constructed.