## 4 Model Construction
### Least Square Contribution of Trapezoidal Fuzzy Numbers Considering Contribution Excess of Players
\begin{array}{l}\sum\limits_{i\in N}\left[(e_{L}^{\rm C}(i,\overline{x})( \alpha)-\overline{k}_{L})^{2}+(e_{R}^{\rm C}(i,\overline{x})(\alpha)-\overline {k}_{R})^{2}\right]\\ =\sum\left[e_{L}^{\rm C}(i,\overline{x})(\alpha)^{2}+e_{R}^{\rm C}(i,\overline {x})(\alpha)^{2}\right]-2\sum\limits_{i\in N}\left[\overline{k}_{L}e_{L}^{\rm C }(i,\overline{x})(\alpha)+\overline{k}_{R}e_{R}^{\rm C}(i,\overline{x})( \alpha)\right]+n(\overline{k}_{L}^{2}+\overline{k}_{R}^{2})\end{array}  \tag{17}\]  
**Problem 2. ** Minimize \(\sum\limits_{i\in N}\left[e_{L}^{\rm C}(i,\overline{x})(\alpha)^{2}+e_{R}^{ \rm C}(i,\overline{x})(\alpha)^{2}\right]\),  
\[ \begin{array}{l}\mbox{s.t.}\left\{\begin{array}{l}\sum\limits_{i\in N} \overline{x}_{Li}(\alpha)=\widetilde{v}_{L}(N)(\alpha)\\ \sum\limits_{i\in N}\overline{x}_{Ri}(\alpha)=\widetilde{v}_{R}(N)(\alpha) \end{array}\right.\end{array}  \tag{18}\]  
This matter minimizes the sum of dissatisfaction of the players by minimizing the contribution excess of all the players so that the sum of satisfaction of the players is maximized. The Lagrangian function for Problem 2 is as follows:  
\[ \begin{array}{l}L(\overline{x}_{Li}(\alpha),\overline{x}_{Ri}(\alpha), \lambda,\gamma)=\sum\limits_{i\in N}\left[(e_{R}^{\rm C}(i,\overline{x})( \alpha))^{2}+(e_{R}^{\rm C}(i,\overline{x})(\alpha))^{2}\right]\\ +\lambda(\sum\limits_{i=1}^{n}\overline{x}_{Li}(\alpha)-\widetilde{v}_{L}(N)( \alpha))+\gamma(\sum\limits_{i=1}^{n}\overline{x}_{Ri}(\alpha)-\widetilde{v}_{ R}(N)(\alpha))\end{array}  \tag{19}\]  
Calculate the partial derivatives of the variables \(\overline{x}_{Li}(\alpha)\), \(\overline{x}_{Ri}(\alpha)\), \(\lambda\), and \(\gamma\), respectively, and let them be equal to zero:  
\[\left\{ \begin{array}{l}\dfrac{\partial L(\overline{x}_{Li}(\alpha), \overline{x}_{Ri}(\alpha),\lambda,\gamma)}{\partial\overline{x}_{Li}}=-2( \widetilde{v}_{Li}^{\rm C}(\alpha)-\overline{x}_{Li}(\alpha))+\lambda=0\\ \dfrac{\partial L(\overline{x}_{Li}(\alpha),\overline{x}_{Ri}(\alpha),\lambda,\gamma)}{\partial\lambda}=\sum\limits_{i=1}^{n}\overline{x}_{Li}(\alpha)- \widetilde{v}_{L}(N)(\alpha)=0\\ \dfrac{\partial L(\overline{x}_{Li}(\alpha),\overline{x}_{Ri}(\alpha),\lambda,\gamma)}{\partial\overline{x}_{Ri}}=-2(\widetilde{v}_{Ri}^{\rm C}(\alpha)- \overline{x}_{Ri}(\alpha))+\lambda=0\\ \dfrac{\partial L(\overline{x}_{Li}(\alpha),\overline{x}_{Ri}(\alpha),\lambda,\gamma)}{\partial\lambda}=\sum\limits_{i=1}^{n}\overline{x}_{Ri}(\alpha)- \widetilde{v}_{R}(N)(\alpha)=0\end{array} \right. \tag{20}\]  
The least square contribution of the trapezoidal fuzzy numbers considering the contribution excess of the players is calculated:  
\[\left\{ \begin{array}{l}\overline{x}_{Li}(\alpha)=\widetilde{v}_{Li}^{\rm C }(\alpha)+\dfrac{\widetilde{v}_{L}(N)(\alpha)-\sum\limits_{j=1}^{n}\widetilde {v}_{Lj}^{\rm C}(\alpha)}{n}\\ \overline{x}_{Ri}(\alpha)=\widetilde{v}_{Ri}^{\rm C}(\alpha)+\dfrac{ \widetilde{v}_{R}(N)(\alpha)-\sum\limits_{j=1}^{n}\widetilde{v}_{Rj}^{\rm C}( \alpha)}{n}\end{array} \right. \tag{21}\]  
At this point, the optimal solution of Problem 2 has been attained; the least square contribution of trapezoidal fuzzy numbers considering contribution excess of players can be presented as follows:  
\[\overline{x}_{i}^{\rm s}(\alpha)=[\widetilde{v}_{Li}^{\rm C}(\alpha)+( \widetilde{v}_{L}(N)(\alpha)-\sum\limits_{j=1}^{n}\widetilde{v}_{Lj}^{\rm C} (\alpha))/n,\widetilde{v}_{Ri}^{\rm C}(\alpha)+(\widetilde{v}_{R}(N)(\alpha)- \sum\limits_{j=1}^{n}\widetilde{v}_{Rj}^{\rm C}(\alpha))/n]\ (i\in N) \tag{22}\]Among them, \(\overline{x}_{i}^{*}(\alpha)\) is composed of two parts, namely the average of the marginal contribution and the residual contribution of player \(i\) to the alliance \(N\), which is in accordance with the principle of constructing the model.