Throughput improvement by mode selection in hybrid duplex wireless networks

Abstract

Hybrid duplex wireless networks, use half duplex (HD) as well as full duplex (FD) modes to utilize the advantages of both technologies. This paper tries to determine the proportion of the network nodes that should be in HD or FD modes in such networks, to maximize the overall throughput of all FD and HD nodes. Here, by assuming imperfect self-interference cancellation (SIC) and using ALOHA protocol, the local optimum densities of FD, HD and idle nodes are obtained in a given time slot, using Karush–Kuhn–Tucker (KKT) conditions as well as stochastic geometry tool. We also obtain the sub-optimal value of the signal-to-interference ratio (SIR) threshold constrained by fixed node densities, using the steepest descent method in order to maximize the network throughput. The results show that in such networks, the proposed hybrid duplex mode selection scheme improves the level of throughput. The results also indicate the effect of imperfect SIC on reducing the throughput. Moreover, it is demonstrated that by choosing an optimal SIR threshold for mode selection process, the achievable throughput in such networks can increase by around 5%.

Appendix

In [34] the standard form of an optimization problem with equality and inequality constraints is defined as follows:

$$\begin{aligned} \begin{aligned} \min&\quad f(x)&\\ s.t&\quad {g_j}(x) \leqslant 0 \quad&j = 1,2,\ldots ,m \\&\quad {h_k} = 0\quad \quad&j = 1,2,\ldots ,m \end{aligned} \end{aligned}$$

where f is the objective function, and g and h are the constraints of the problem. In this case, the KKT conditions can be written as follows:

$$\begin{aligned} \begin{array}{*{20}{l}} {\frac{{\partial f}}{{\partial {x_i}}} + \mathop {\sum }\limits _{j = 1}^m {\eta _j}\frac{{\partial {g_j}}}{{\partial {x_i}}} + \mathop {\sum }\limits _{j = 1}^p {\mu _k}\frac{{\partial {h_k}}}{{\partial {x_i}}} = 0,}&{}{i = 1,2, \ldots ,n} \\ {{\eta _j}{g_j} = 0,}&{}{j = 1,2, \ldots ,m} \\ {{g_j} \leqslant 0,\;}&{}{j = 1,2, \ldots ,m} \\ {{\eta _j} \geqslant 0,}&{}{j = 1,2, \ldots ,m} \\ {{h_k}\left( x \right) = 0,}&{}{k = 1,2, \ldots ,p} \end{array} \end{aligned}$$

where ${\mu _k}$ and ${\eta _k}$ are the Lagrange coefficients. The optimization problem in (10), which has an equality constraint and three inequality constraints, can be described in the following standard form:

$$\begin{aligned} \begin{array}{l} \begin{array}{*{20}{c}} {\min }&{}{f = - T} \end{array}\\ \begin{array}{*{20}{c}} {s.t.}&{}{{\lambda _1} + {\lambda _2} + {\lambda _3} = \lambda } \end{array}\\ \quad \quad - {\lambda _1} \le 0\\ \quad \quad - {\lambda _2} \le 0\\ \quad \quad - {\lambda _3} \le 0 \end{array} \end{aligned}$$

Applying the KKT conditions, leads to the following formulations:

$$\begin{aligned} \begin{aligned}&\left[ { \begin{array}{*{20}{c}} { - \log (1 + \theta )\exp ( - {\lambda _1}H - {\lambda _2}F)(1 - {\lambda _1}H - 2KH{\lambda _2})} \\ { - \log (1 + \theta )\exp ( - {\lambda _1}H - {\lambda _2}F)( - {\lambda _1}F + 2K - 2KF{\lambda _2})} \\ 0 \end{array}} \right] \\&\quad + \left[ { \begin{array}{*{20}{c}} { - 1}&{}0&{}0 \\ 0&{}{ - 1}&{}0 \\ 0&{}0&{}{ - 1} \end{array}} \right] \left[ { \begin{array}{*{20}{c}} {{\eta _1}} \\ {{\eta _2}} \\ {{\eta _3}} \end{array}} \right] + \mu \left[ { \begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right] = 0 \\ \end{aligned} \end{aligned}$$

where $\mu$ and ${\eta _i}$ with $i \in \{ 1,2,3\}$ are Lagrange coefficients and K, H and F are obtained from the existing equations in section III. Therefore, we have:

$$\begin{aligned} \begin{aligned}&A: - \log (1 + \theta )\exp ( - {\lambda _1}H - {\lambda _2}F)(1 - {\lambda _1}H - 2KH{\lambda _2}) \quad - {\eta _1} + \mu = 0 \\&B: - \log (1 + \theta )\exp ( - {\lambda _1}H - {\lambda _2}F)( - {\lambda _1}F + 2K - 2KF{\lambda _2}) \quad - {\eta _2} + \mu = 0 \\&C:0 - {\eta _3} + \mu = 0 \rightarrow {\eta _3} = \mu \\&D:{\eta _1}{\lambda _1} = 0 \\&E:{\eta _2}{\lambda _2} = 0 \\&F:{\eta _3}{\lambda _3} = 0 \\&G:{\lambda _i} \geqslant 0,i = \{ 1,2,3\} \\&H:\sum \limits _{i = 1}^3 {{\lambda _i} = \lambda } \\&I:{\eta _i} \geqslant 0,i = \{ 1,2,3\} \end{aligned} \end{aligned}$$

Given the equation G, each ${\lambda _i}$, $i \in \{ 1,2,3\}$ has two states: one is equal to zero and one to be positive. Therefore, there are eight modes to set them up (the whole-zero state is eliminated with respect to the relation H, and we have seven states):

Case 1: $\lambda _1=0, \lambda _2=0, \lambda _3>0 \Rightarrow \eta _3=\mu =0$

from A: $\eta _1= - \log (1 + \theta ) \geqslant 0 \rightarrow \times$

from B: $\eta _2= - 2K\log (1 + \theta ) \geqslant 0 \rightarrow \times$

Case 2: $\lambda _1= \lambda , \lambda _2=0, \lambda _3=0 \Rightarrow \eta _1=0$

from A: $\eta _3 = \mu = \log (1 + \theta )\exp ( - \lambda H)(1 - \lambda H) \geqslant 0$$\rightarrow \lambda H < 1$

from B: $\eta _2 = \mu - \log (1 + \theta )\exp ( - \lambda H)( - \lambda F + 2K) \geqslant 0$$\rightarrow 1 - \lambda H + \lambda F - 2K \geqslant 0$

Case 3: $\lambda _1 = 0,\lambda _2 = \lambda ,\lambda _3 = 0 \Rightarrow {\eta _2} = 0$

from B: $\eta _3 = \mu = \log (1 + \theta )exp( - \lambda F)(2K - 2K\lambda F) \geqslant 0$$\rightarrow \lambda F \leqslant 1$

from A: $\eta _1 = \log (1 + \theta )exp( - \lambda F)(2K - 2K\lambda (F - H) - 1) \geqslant 0$

$\rightarrow 2K - 2K\lambda (F - H) \geqslant 1$

Case 4: $\lambda _1 = 0,{\lambda _2}> 0,{\lambda _3} > 0 \Rightarrow {\eta _2} = 0,{\eta _3} = \mu = 0$

from B: $2K - 2KF\lambda _2 = 0 \rightarrow \lambda _2 = \frac{1}{F} > 0$

from H:$\lambda _3 = \lambda - \frac{1}{F} > 0 \rightarrow \frac{1}{F} < \lambda$

from A:$\eta _1 = 0 - \log (1 + \theta )exp( - 1)(1 - 2K\frac{H}{F}) \geqslant 0$$\rightarrow F \leqslant 2KH$

Case 5: $\lambda _1> 0,{\lambda _2} = 0,{\lambda _3} > 0 \Rightarrow {\eta _1} = 0,{\eta _3} = \mu = 0$

from A: ${1 -\lambda _1}H = 0 \rightarrow {\lambda _1} = \frac{1}{H} > 0$

from H: ${\lambda _3} = \lambda - \frac{1}{H} > 0 \rightarrow \frac{1}{H} < \lambda$

from B: ${\eta _2} = 0 - \log (1 + \theta )exp( - 1)( - \frac{F}{H} + 2K) \geqslant 0$$\rightarrow F \geqslant 2KH$

Case 6: ${\lambda _1}> 0,{\lambda _2} > 0,{\lambda _3} = 0 \Rightarrow {\eta _1} = 0,{\eta _2} = 0$

from A,B,H: $\left\{ { \begin{array}{*{20}{l}} {{\lambda _1} = \lambda - \frac{{2K - \lambda F + \lambda H - 1}}{{H - F - 2KH + 2KF}}> 0} \\ {{\lambda _2} = \frac{{2K - \lambda F + \lambda H - 1}}{{H - F - 2KH + 2KF}} > 0} \end{array}} \right.$

from A:${\eta _3} = \mu \geqslant 0 \rightarrow {\lambda _1}H + 2KH{\lambda _2} \leqslant 1$

Case 7: ${\lambda _1}> 0,{\lambda _2}> 0,{\lambda _3} > 0 \Rightarrow {\eta _1} = 0,{\eta _2} = 0,{\eta _3} = \mu = 0$

from A,B: $\left\{ \begin{array}{*{20}{l}} {1 - {\lambda _1}H - 2KH{\lambda _2} = 0} \\ { - {\lambda _1}F + 2K - 2KF{\lambda _2} = 0} \end{array}\right.$

$\rightarrow F = 2KH{\text { for }}\forall \mathop {{\lambda _i}}\limits _{i = 1,2,3} > 0 \, \& \, {\lambda _1} + {\lambda _2} + {\lambda _3} = \lambda$