# Utility Distribution Strategy of the Task Agents in Coalition Skill Games
## 3 Definition of the Problem Model
### Utility Distribution Strategy of Task Agent  
As for the utility distribution scheme of task agent \(t_{k}\in T\), it is assumed that \(u_{k}=\sum_{s_{j}\in S\wedge ST_{j,k}=1}TS_{k,j}(\tau)\), and if \(ST_{j,k}=1\), \(TS_{k,j}(\tau)\geq 0\). At initialization, the utility of the task agent is averagely distributed among the required skills, and then the task agent adjusts its utility distribution scheme according to the inadequacy degree of the skills. If the task agent only needs one skill,the adjustment is not needed. To understand how \(t_{k}\in T\) adjusts its utility distribution scheme, it is only need to answer the following 4 questions:  
1. Among the skills needed by \(t_{k}\in T\), whose share of utility needs to be increased? Let \({S_{k}}^{ncat}(\tau)\) denote the set of skills of this type: \({s_{j}}\in{S_{k}}^{nced}(\tau)\) must satisfy the following three conditions: (1) \(ST_{j,k}=1\), (2) \(R^{1}(k,j,\tau)=\Phi\) and (3) \(R^{2}(k,j,\tau)=\Phi\).
2. For \({s_{j}}\in{S_{k}}^{nced}(\tau)\), what is the minimum increase? It is denoted with \({f^{min}(k,j,\tau)}\), whose computing method is given in Sub-procedure 2, where \(\xi_{1}=1\), if \(u_{RTS_{i,0}(\tau)}\geq u_{k}\), otherwise \(\xi_{1}=0\)).
3. From which skills can \(t_{k}\) adjust shares of utility to \({s_{j}}\in{S_{k}}^{nced}(\tau)\)? Let \(S^{lend}(k,j,\tau)\) denote the set of skills of this type: \[S^{lend}(k,j,\tau)\leftarrow\{{s_{j^{\prime}}}\in S\,{\downarrow}\,{j^{\prime }}\
eq j\wedge ST_{j^{\prime},k}=1\wedge(R^{1}(k,j,\tau)\cup R^{2}(k,j,\tau)) \
eq\Phi\}.\]
4. For \({s_{j^{\prime}}}\in S^{lend}(k,j,\tau)\), what is the maximum decrease? It is denoted with \({f^{max}(k,j,j^{\prime})}\), whose computing method is shown in Sub-procedure 3. Where \(k^{sc}\) denotes the serial number of the task agents which will be selected by \(r_{i}\) if \(t_{k}\) was out of consideration. The method to compute the values of \(\xi_{1}\) and \(\xi_{2}\) is: if \(u_{k}\leq u_{k^{sec}}\), \(\xi_{1}\gets 1\), otherwise, \(\xi_{1}\gets 0\). If \(u_{k}\leq u_{RTS_{i,0}(\tau)}\), \(\xi_{2}\gets 1\), otherwise, \(\xi_{2}\gets 0\). If \(k^{sec}=0\), \(\xi_{1}\gets 0\), \(\xi_{2}\gets 0\).  
**Sub-Procedure 3** Compute the value of \({f^{max}(k,j,\tau)}\)  
```
1:IF\(R^{1}(k,j,\tau)\)\(\
eq\)\(\Phi\)\(R^{2}(k,j,\tau)\)\(\
eq\)\(\Phi\)THEN
2:\({f^{max}(k,j,j^{\prime})}\leftarrow\)\(\max\limits_{r_{i}\in R^{1}(k,j,\tau)}(f(i,k,\tau)-f(i,k^{sec},\tau)-\xi_{1})\), \(\max\limits_{r_{i}\in R^{2}(k,j,\tau)}(f(i,k,\tau)-f(i,RTS_{i,0}(\tau),\tau)- \xi_{2}))\);
3:ELSEIF\(R^{1}(k,j^{\prime},\tau)\)\(\
eq\)\(\Phi\)THEN
4:\({f^{max}(k,j,j^{\prime})}\leftarrow\)\(\max\limits_{r_{i}\in R^{2}(k,j,\tau)}(f(i,k,\tau)-f(i,k^{sec},\tau)-\xi_{1})\);
5:ELSEIF\(R^{2}(k,j^{\prime},\tau)\)\(\
eq\)\(\Phi\)THEN
6:\({f^{max}(k,j,j^{\prime})}\leftarrow\)\(\max\limits_{r_{i}\in R^{2}(k,j,\tau)}(f(i,k,\tau)-f(i,RTS_{i,0}(\tau),\tau)- \xi_{2})\);
7:ELSE
8:\({f^{max}(k,j,j^{\prime})}\)\(\leftarrow\)\(0\);
9:ENDIF
10:\({f^{max}(k,j,j^{\prime})}\)\(\leftarrow\)\(\min\)\(({f^{max}(k,j,j^{\prime})},TS_{k,j^{\prime}}(\tau))\).
```  
**Sub-Procedure 2** Compute the value of \({f^{min}(k,j,\tau)}\)