# On Optimal Cell Flashing for Reducing Delay and Saving Energy in Wireless Networks
## 3 Optimal Flashing Ratio  
Our problem is to find an optimal flashing ratio \(\beta\) so as to minimize the system cost function given below. Let the average delay be \(\phi(\beta)\), which is given by:  
\[\phi(\beta):=\sum_{i\in\mathcal{B}}\frac{1}{1-\rho_{i}(\beta)}. \tag{6}\]  
Let the average total power consumption be \(\psi(\beta)\). From [23], we know that power consumed by operating BS \(i\) is \(\rho_{i}(\beta)P_{i}+P_{o}\), where \(P_{i}\) denotes the transmit power, and \(P_{o}\) denotes static power when BS operates. Power consumed by sleeping BS is \(P_{s}\). Note that BS is on with the fraction of \(1-\frac{2}{3}\beta\), so \(\psi(\beta)\) is given by:  
\[\psi(\beta):=\sum_{i\in\mathcal{B}}\left\{\left(1-\frac{2}{3}\beta\right)( \rho_{i}(\beta)P_{i}+P_{o})+\frac{2}{3}\beta P_{s}\right\}. \tag{7}\]  
We then formulate our problem as optimization as follows. **Problem**:  
\[\min_{\beta} f(\beta)=\phi(\beta)+\eta\,\psi(\beta), \tag{8}\] \[s.t. 0\leq\beta\leq 1,\] (9) \[0\leq\rho_{i}(\beta)<1,\forall i\in\mathcal{B},\]  
where \(\eta\geq 0\) is the parameter that balances the tradeoff between delay and energy consumption. We analyze \(\phi(\beta)\) and \(\psi(\beta)\) first to investigate under which circumstance the optimal \(\beta^{*}\) can be found. Since all base stations are symmetric, we consider BS \(i\) in analysis hereafter. Lemma 1: \(\phi(\beta)\) _is a convex function of \(\beta\)._  
Proof: To check the convexity, we obtain the second order derivatives of \(\phi(\beta)\), which is given by:  
\[\phi^{\prime\prime}(\beta)=\sum_{i\in\mathcal{B}}\frac{\rho_{i}^{\prime\prime} (\beta)(1-\rho_{i}(\beta))+2\rho_{i}^{\prime}(\beta)^{2}}{(1-\rho_{i}(\beta) )^{3}}. \tag{10}\]  
We then obtain the second order derivatives of \(\rho_{i}(\beta)\), which is given by:  
\[\rho_{i}^{\prime\prime}(\beta)=\int_{\mathcal{L}}\frac{2\left(\frac{1}{3}c_{i }^{1}(x)-c_{i}^{0}(x)\right)^{2}\gamma(x)}{\left((1-\beta)c_{i}^{0}(x)+\frac{ \beta}{3}\ c_{i}^{1}(x)\right)^{3}}\ dx. \tag{11}\]  
Then, since \(\gamma(x)>0\) and \((1-\beta)c_{i}^{0}(x)+\frac{\beta}{3}\ c_{i}^{1}(x)>0\) for \(\forall\beta\), \(0\leq\beta\leq 1\), we have \(\rho_{i}^{\prime\prime}(\beta)>0\). Thus, it is obvious that \(\phi^{\prime\prime}(\beta)\geq 0\), which completes the proof. Lemma 2: \(\psi(\beta)\) _is a monotone decreasing function of \(\beta\), and we have:_  
\[\{\tilde{\beta}=1\ |\ \tilde{\beta}=\underset{\beta}{\mathrm{argmin}}\ \psi(\beta)\}. \tag{12}\]  
Proof: The derivative function of \(\psi(\beta)\) is:  
\[\psi^{\prime}(\beta)=\sum_{i\in\mathcal{B}}\frac{2}{3}\left(\left(-\rho_{i}( \beta)+\left(\frac{3}{2}-\beta\right)\rho_{i}^{\prime}(\beta)\right)P_{i}+P_{ s}-P_{o}\right). \tag{13}\]From Equation (5), we can obtain the coefficient of \(P_{t}\) in Equation (13):  
\[-\rho_{i}(\beta)+\left(\frac{3}{2}-\beta\right)\rho_{i}^{\prime}(\beta)=-\frac{1 }{2}\int_{\mathcal{L}}\frac{\gamma(x)\left(c_{i}^{1}(x)-c_{i}^{0}(x)\right)}{ \left((1-\beta)c_{i}^{0}(x)+\frac{\beta}{3}\ c_{i}^{1}(x)\right)^{2}}\ dx. \tag{14}\]  
Note that \(\gamma(x)>0\) and \(c_{i}^{0}(x)<c_{i}^{1}(x)\) because of the reduced interference power. Then, it is obvious that the coefficient of \(P_{t}\) in Equation (13) is negative and because operating power is larger than sleep power, the derivative function of \(\psi(\beta)\) is negative, which completes the proof. From Lemmas 1 and 2 we now know that \(\phi(\beta)\) is a convex function of \(\beta\) and \(\psi(\beta)\) is a monotone decreasing function of \(\beta\). Hereafter, we are interested in under which circumstance the optimal solution can be obtained. As one special case, we are interested in when the optimal \(\beta^{*}\) is \(1\), i.e., cell flashing all the time is optimal. The following proposition is about cell flashing in interference-limited regime. **Proposition 1** (Cell flashing in interference-limited regime). : Suppose that the traffic load density is uniform, i.e., \(\gamma(x)=\gamma\) for \(\forall x=\mathcal{L}\).