# Model of Multi-Branch Trees for Efficient Resource Allocation
## 2 Methods
### Notation  
Let us consider that _an organization_ is considering distributing its budget to _projects_ where the number of projects is set to \(N\). Each project belongs to _a department_. A department has at least one project. A manager of a department is called _a director_. The top of the organization is called _an executive_. The executive manages all departments either directly or indirectly. If the executive manages them indirectly, there are mid-level departments managed by mid-level executives managing either departments or mid-level departments. This organizational structure is drawn as a tree structure using graph theory.  
Figure 1 shows an example of an organization. There are 10 projects (\(P_{1},P_{2},\ldots,P_{10}\)) that require money, 5 departments (whose directors are \(E_{3},E_{4},E_{5},E_{7}\), and \(E_{8}\)), and 3 mid-level executives (\(E_{1},E_{2}\), and \(E_{6}\)) and 1 executive (\(E_{0}\)) in the organization. The departments own projects are \(\{P_{1},P_{2}\}\) for \(E_{3}\), \(\{P_{3},P_{4}\}\) for \(E_{4}\), \(\{P_{5}\}\) for \(E_{5}\), \(\{P_{6},P_{7}\}\) for \(E_{7}\), and \(\{P_{8},P_{9},P_{10}\}\) for \(E_{8}\).  
We define a multi-branch tree structure \(T_{(n,m)}\), where \(n\) is the number of leaf nodes representing projects, and \(m\) is that of internal nodes representing _evaluators_ consisting of directors, mid-executives, and an executive. \(P\) denotes a set of leaf nodes \(\{P_{i}\}\) (\(i\in\{1,\ldots,n\}\)), and \(E\) denotes a set of non-leaf nodes \(\{E_{j}\}\) (\(j\in\{0,\ldots,m\}\)). We call \(E_{0}\) the root node (the executive) and the others internal nodes. The example tree of Figure 1 is defined as \(T_{(10,8)}\) with 10 projects and 9 evaluators.  
For any \(j\in\{0,\ldots,m\}\), we denote by \(T_{j}\) a tree consisting of \(E_{j}\) and all of its descendent nodes. Note that \(T_{0}=T_{(n,m)}\). Let \(E^{j}\) be the set of all child nodes of \(E_{j}\) and \(c(j)\) the set of the index numbers of evaluators in \(E^{j}\). A tree \(T_{k}\) (\(k\in c(j)\)) is called a _child tree of \(E_{j}\)_. Here, we call \(S_{j}\) the set of child trees of \(E_{j}\). For example, \(S_{2}=\{T_{5},T_{6}\}\) is obtained from \(E_{2}\) in the example tree \(T_{(10,8)}\) of Figure 1.  
Next, we define a map \(l:\{1,\ldots,n\}\times\{0,\ldots,m\}\longrightarrow\mathbb{Z}\cup\{\infty\}\) as  
\[l(i,j)=h,\]  
where \(h\) means the length of the path from \(E_{j}\) to \(P_{i}\). In contrast, we define a map \(e:\{1,\ldots,n\}\times\mathbb{Z}\longrightarrow\{0,\ldots,m\}\) as  
\[e(i,h)=j,\]  
which outputs an index number of \(E_{j}\), where \(h\) means the length of the path from \(E_{j}\) to \(P_{i}\). Note that if \(P_{i}\) is not a descendant leaf node of \(E_{j}\), we have \(l(i,j)=\infty\), and \(e(i,h)\) is not defined.  
For any \(i\in\{1,\ldots,n\}\), we obtain only one shortest path from \(E_{0}\) to \(P_{i}\) and call it _path_\(i\). For example, we consider \(P_{7}\) in Figure 1. The parent node of \(P_{7}\) is \(E_{7}\), and the parent node of \(E_{7}\) is \(E_{6}\). Repeating this procedure, we finally reach the root node, \(E_{0}\). We then solve _path_\(7\) as follows.  
Note that every _path_\(i\) for \(i\in\{1,\ldots,n\}\) is a sequence from \(E_{0}\) to \(P_{i}\).