# Realizing Efficient Security and Privacy in IoT Networks
## 4 System Model
### Sequential Convex Approximation  
In attempt to realize a sequential convex approximation, the rate constraint and the objective function which constitute the optimization problem formulated in (13) is rewritten as (20) below. \[ \begin{array}{l}f(\mathrm{y},\mathrm{z})=\frac{1}{M}\sum\limits_{m\in\Theta _{2}}\left(\log(1+\frac{A_{m}y_{m}}{1+B_{m}z_{m}})-\log(1+\frac{C_{m}y_{m}}{1+ D_{m}z_{m}})\right)\\ =\frac{1}{M}\sum\limits_{m\in\Theta_{2}}\left(\log(1+B_{m}z_{m}+A_{m}y_{m})+ log(1+D_{m}z_{m})\right)\\ -\frac{1}{M}\sum\limits_{m\in\Theta_{2}}\left(\log(1+D_{m}z_{m}+C_{m}y_{m})+ log(1+B_{m}z_{m})\right)\\ =f_{1}(\mathrm{y},\mathrm{z})+f_{2}(\mathrm{y},\mathrm{z})\frac{1}{M}\sum \limits_{m}\left(\log(1+B_{m}z_{m}+A_{m}y_{m})\right)\\ -\frac{1}{M}\sum\limits_{m}\left(\log(1+B_{m}z_{m})\right)\geq C_{sd}\\ f_{3}(\mathrm{y})+f_{4}(\mathrm{y},\mathrm{z})\geq C_{sd}\end{array}  \tag{20}\]  
Evidently, \(f_{1}\) and \(f_{4}\) functions are concave while \(f_{2}\) and \(f_{3}\) functions are convex. Considering first-order Taylor series expansion of convex functions as the function's underestimator. By denoting the first-order Taylor series expansion of \(f_{2}\) and \(f_{3}\) around \((\mathrm{y}^{(h)},\mathrm{z}^{(h)})\) points and expressing them, respectively, as \(\tilde{f}_{2}^{(h)},\tilde{f}_{3}^{(h)}\), the following was realized,  
\[ \begin{array}{l}f_{1}(\mathrm{y},\mathrm{z})+f_{2}(\mathrm{y},\mathrm{z}) \geq f_{1}(\mathrm{y},\mathrm{z})+\tilde{f}_{2}^{(h)}(\mathrm{y},\mathrm{z}) \forall\mathrm{y},\mathrm{z}\in\mathit{M}\\ f_{4}(\mathrm{y},\mathrm{z})+f_{3}(z)\geq f_{4}(\mathrm{y},\mathrm{z})+\tilde{f }_{3}^{(h)}(z)\forall\mathrm{y},\mathrm{z}\in\mathit{M}\end{array}  \tag{21}\]  
where  
\[ \begin{array}{l}\tilde{f}_{2}^{(h)}(\mathrm{y},\mathrm{z})=f_{2}(\mathrm{y}^ {(h)},\mathrm{z}^{(h)})+\
abla f_{2}(\mathrm{y}^{(h)},\mathrm{z}^{(h)})^{R}( \left[\begin{array}{c}\mathrm{y}\\ \mathrm{z}\end{array}\right]\\ -\left[\begin{array}{c}\mathrm{y}^{(h)}\\ \mathrm{z}^{(h)}\end{array}\right])\\ \tilde{f}_{3}^{(h)}(\mathrm{z})=f_{3}(z^{(h)})+\
abla f_{3}(z^{(h)})^{R}( \mathrm{z}-\mathrm{z}^{(h)})\\ \
abla f_{2}(y,z)=-\frac{1}{M}\left[\begin{array}{c}\frac{C_{1}}{1+D_{1}z_{1 }+C_{1}y_{1}}\\ \vdots\\ \frac{D_{1}}{1+D_{M}z_{M}+C_{M}y_{M}^{4}}\\ \frac{D_{1}}{1+D_{1}z_{1}+C_{1}y_{1}}+\frac{B_{1}}{1+B_{1}z_{1}}\\ \vdots\\ \frac{D_{M}}{1+D_{M}z_{M}}+\frac{B_{M}}{1+B_{M}z_{M}}\end{array}\right]\\ \
abla f_{3}(z)=-\frac{1}{M}\left[\begin{array}{c}\frac{B_{1}}{1+B_{1}z_{1 }}\\ \vdots\\ \frac{B_{M}}{1+B_{M}z_{M}}\end{array}\right]\end{array}  \tag{22}\]Based on work done in [34] and assuming \(\frac{B_{m}}{D_{m}}<1\) we further simplify \(\Theta_{z}\) and obtain  
\[ \begin{array}{l}\Theta_{z}=\left\{m\left|z_{m}\geq 0,\frac{A_{m}}{1+B_{m}z_{m}}> \frac{C_{m}}{1+D_{m}z_{m}},\forall m\in\mathbb{N}\right.\right\}\\ =\left\{m\left|(B_{m}C_{m}-A_{m}D_{m})z_{m}>A_{m}-C_{m},\forall m\in\mathbb{N} \right.\right\}\\ =\left\{m\left|z_{m}\geq\frac{A_{m}-C_{m}}{B_{m}C_{m}-A_{m}D_{m}},1>\frac{A_{m }}{C_{m}}>\frac{B_{m}}{D_{m}},\forall m\in\mathbb{N}\right.\right\}\\ \cup\left\{m\left|z_{m}\geq 0,1<\frac{A_{m}}{C_{m}}>\frac{B_{m}}{D_{m}},\forall m \in\mathbb{N}\right.\right\}\end{array}  \tag{23}\]  
Thus, following iteration \(h\), we formulate the optimization problem as follows,  
\[ \begin{array}{l}\max\limits_{\begin{array}{c}y,\text{z}\\ \end{array}}\ f_{1}(\text{y},\text{z})+f_{2}^{(h)}(\text{y},\text{z})\\ s.t.f_{4}(\text{y},\text{z})+\tilde{f}_{3}^{(h)}(\text{z})\geq C_{sd}\\ \sum\limits_{m}z_{m}\leq\text{p}_{d},z_{m}\geq 0\forall m\in\mathbb{N}\\ z_{m}=0\forall m\in\left\{m\left|\frac{A_{m}}{C_{m}}<\frac{B_{m}}{D_{m}}<1, \forall m\in\mathbb{N}\right.\right\}\\ \cup\left\{m\left|\frac{B_{m}}{D_{m}}>1,\forall m\in\mathbb{N}\right.\right\} \\ z_{m}\geq\frac{A_{m}-C_{m}}{B_{m}C_{m}-A_{m}D_{m}}\forall m\in\left\{m\left|1> \frac{A_{m}}{C_{m}}>\frac{B_{m}}{D_{m}},\forall m\in\mathbb{N}\right.\right\} \\ \sum\limits_{m}y_{m}\leq\text{p}_{s},y_{m}\geq 0\forall m\in\mathbb{N}\end{array}