# Cooperative Game-Based Energy Efficiency Management over Ultra-Dense Wireless Cellular Networks
## 4 Bargaining Problem Solutions for Ultradense Wireless Cellular Networks
### Solution of Cooperative Bargaining Game
 \tag{37}\]  
Finally, the closed-form solution of MeNB\({}_{\text{m}}\) can be obtained, as follows:  
\[p_{m}^{*}=\frac{1-\alpha_{m}-\beta_{m}}{\sum\limits_{n=1}^{N}\sum\limits_{j=1 }^{I}\frac{S_{m,(n,j)}}{S_{n,j}}\frac{h_{n,j}^{2}}{h_{n}}+\lambda_{m}} \tag{38}\]  
\(\qed\)  
**Remark 3**: _According to Equation (38), we conclude the following:_  
1. _When_ \(\alpha_{m}\) _is large, MeNB_\({}_{\text{m}}\) _focuses more on the EE (e.g.,_ \(\alpha_{m}>1/3\)_); thus, less transmission power is needed for MeNB_\({}_{\text{m}}\)_._
2. _When_ \(\beta_{m}\) _is large, MeNB_\({}_{\text{m}}\) _focuses more on the DE (e.g.,_ \(\beta_{m}>1/3\)_); thus, less transmission power is needed for MeNB_\({}_{\text{m}}\)_._
3. _When_ \(N\) _and_ \(J\) _are large, that is, many SeNBs with many SUEs are deployed in the macro-cell controlled by MeNB_\({}_{\text{m}}\)_, if these SeNBs are deployed near MeNB_\({}_{\text{m}}\), as a result, MeNB_\({}_{\text{m}}\) provides a lower transmission power. In this sense, the strategy takes full account of the benefits to the opponents._
4. _When_ \(h_{n}\) _is large, the aggregate CINRs of all SUEs are good enough; thus, MeNB_\({}_{\text{m}}\) _can provide a large transmission power to enhance its capacity._  
**Corollary 4**: _The closed-form power solution of P-SeNB in Equation (31) is given by:_  
\[p_{n}^{*}=\frac{1-\alpha_{n}-\beta_{n}}{\sum\limits_{m=1}^{M}\sum\limits_{i=1} ^{I}\frac{S_{n,(n,j)}}{S_{n,j}}\frac{h_{n,j}^{2}}{h_{n}}+\lambda_{n}} \tag{39}\]  
_where \(\lambda_{n}\) is the Lagrange multiplier corresponding to the SeNBn power constraint._  
**Proof:** Similarly, for the problem in Equation (31), introducing the Lagrangian multipliers \(\lambda_{n}\) corresponding to the SeNB\({}_{n}\) power constraint \(0\leq p_{n}\leq p_{n}^{MAX}\), which yields:  
\[L_{n}=\sum\limits_{n=1}^{N}\left\{\ln p_{n}-(\alpha_{n}+\beta_{n})\ln(a_{sistem} *p_{n}+b_{sistem})\right\}+\sum\limits_{m=1}^{M}\ln h_{m}-\sum\limits_{n=1}^{N} \lambda_{n}(p_{n}-p_{n}^{MAX}) \tag{40}\]  
According to the KKT conditions, the power allocation for SeNB\({}_{n}\) is determined by obtaining the first-order derivative in Equation (40) with respect to \(p_{n}\), which is given as:  
\[\frac{\partial L_{n}}{\partial p_{n}}=\frac{1}{p_{n}}-\frac{(\alpha_{n}+\beta_ {n})\cdot a_{sistem}}{a_{sistem}*p_{n}+b_{sistem}}+\sum\limits_{m=1}^{M}\frac{ 1}{h_{m}}\frac{\partial h_{m}}{\partial p_{n}}-\lambda_{n} \tag{41}\]  
According to the definition of \(h_{m}=\sum_{i=1}^{I}h_{m,i}\) and the definition of \(h_{m,i}\) in Equation (2), we obtain:  
\[\frac{\partial h_{m}}{\partial p_{n}}=\frac{\partial}{\partial p_{n}}\left\{ \sum\limits_{i=1}^{I}h_{m,i}\right\}=\sum\limits_{i=1}^{I}\frac{\partial h_{ m,i}}{\partial p_{n}}=-\sum\limits_{i=1}^{I}\frac{g_{n,\{m,i\}}}{g_{m,i}}h_{m,i}^{2} \tag{42}\]  
By substituting Equation (42) into Equation (41), we obtain:  
\[\frac{\partial L_{n}}{\partial p_{n}}=\frac{1}{p_{n}}-\frac{(\alpha_{n}+\beta _{n})\cdot a_{sistem}}{a_{sistem}*p_{n}+b_{sistem}}-\sum\limits_{m=1}^{M}\sum \limits_{i=1}^{I}\frac{g_{n,\{m,i\}}}{g_{m,i}}\frac{h_{m,i}^{2}}{h_{m}}- \lambda_{n} \tag{43}\]  
Similarly, to obtain the closed-form solution of P-SeNB, we assume that \(b_{sistem}\) = 0. Then:  
\[\frac{\partial L_{n}}{\partial p_{n}}=\frac{1}{p_{n}}-\frac{(\alpha_{n}+\beta _{n})}{p_{n}}-\sum\limits_{m=1}^{M}\sum\limits_{i=1}^{I}\frac{g_{n,\{m,i\}}}{ g_{m,i}}\frac{h_{m,i}^{2}}{h_{m}}-\lambda_{n} \tag{44}\]  
Finally, we obtain the closed-form solution of SeNB\({}_{n}\), as follows:  
\[p_{n}^{*}=\frac{1-\alpha_{n}-\beta_{n}}{\sum\limits_{m=1}^{M}\sum\limits_{i= 1}^{I}\frac{g_{n,\{m,i\}}}{g_{m,i}}\frac{h_{m,i}^{2}}{h_{m}}+\lambda_{n}} \tag{45}\]  
\(\square\)  
**Remark 4:**_According to Equation (45), we conclude the following:_  
1. _When_ \(\alpha_{n}\) _is large, SeNB_\({}_{n}\) _focuses more on the EE (e.g.,_ \(\alpha_{n}\) _>_ 1/3_); thus, less transmission power is needed for SeNB_\({}_{n}\)_._
2.