# An Algorithm for Arranging Operators to Balance Assembly Lines and Reduce Operator Training Time
## 3 Methodology
### Cluster Algorithm
#### 3.3.1 Skills Sorting  
Whether two skills are grouped depends on their closeness. Closeness denotes how many operators have both of the skills. The more operators have the two skills, the closer the skills are. First, the skill with the highest closeness value is found and a new group is formed. Second, an unsorted skill follows a sorted skill, based on the highest value in its row in the closeness matrix. If there is no sorted skill to follow, the skill is placed in a new group. Finally, all skills are grouped.  
Thus, the skills can be sorted in three steps:  
Step 1. Let \(f\in G_{1}\), where \(f=\{x|b_{\text{rk}}=max.\{b_{\text{rs}}\},\ 1\leq k\leq v\}\). A tie is broken by selecting the smaller one.  
Step 2. Ignore all the grouped rows and form a new closeness matrix \(B^{\prime}u=\{b_{ij}\}\).  
Let  
\[\left\{\begin{array}{cc}f\in G_{k},&\quad\quad\text{if }=\{b_{ij}=max\{b_{ij} \}\ \wedge\ y\in G_{k}\\ h\in G_{new\ group},&\quad\quad\text{otherwise}\end{array}\right.\]  
Step 3. Repeat Step 2 until \(u\) skills are grouped.  
3.3.2. Operator Sorting  
Operator sorting aims to place operators in the group that will use the highest number of their skills.  
Thus, let \(v\in G_{y},\ y=\left\{x\middle|\sum\limits_{u\in G_{x}}so_{uv}=max.\left( \sum\limits_{u\in G_{i}}so_{uv}\right),\ 1\leq i\leq R\right\}\). A tie is broken by choosing the group with the lowest number of machines.  
3.4. _GT Efficiency_  
This work uses the GT efficiency measurement equation defined by [PERSON] and [PERSON] [65], while letting \(q\) be equal to 0.5, as follows:  
\[\eta=q\eta_{1}+(1-q)\eta_{2}=q\frac{e_{b}}{\sum\limits_{u\in G_{i}P_{i}}}+(1- q)\left(1-\frac{e_{0}}{uv-\sum\limits_{u\in G_{i}P_{i}}}\right) \tag{9}\]  
There are two concerns: the \"1\" in the group, and the \"1\" not in the group. The higher the GT efficiency, the more 1s are in the groups, and the fewer 1s are not in the groups. Thus, \(\sum\limits_{u\in G_{i}}^{c}\frac{e_{0}}{NS_{i}NO_{i}}\) is the ratio of the number of 1s in the groups to the number of 1s and 0s in all groups, while \(\frac{e_{0}}{uv-\sum\limits_{u\in G_{i}NO_{i}}}\) is the ratio of the number of 1s not in the groups to the number of 1s and 0s not in the groups.  
4. Numerical Examples  
4.1. _Example 1: Number of Operators at Each Workstation Decided_  
On an assembly line, 500 operators must be assigned to workstations (Figure 6).  
Let  
\[A=\left[\begin{array}{cccc}a_{11}&a_{12}&a_{13}\\ &a_{22}&a_{23}&a_{24}&a_{25}\end{array}\right]\]  
and  
\[\left\{t_{ij}\right\}=\left[\begin{array}{cccc}10&30&24&0&0\\ 0&5&40&20&15\end{array}\right]\]  
Let the work time per day be 8 h.  
Figure 6: Assembly line with seven workstations.  
\[\left\{t_{ij}\right\}=\left[\begin{array}{ccccc}10&30&&&\\ 20&40&15&\\ &&24&48\end{array}\right]\]  
and  
\[\left\{p_{ij}\right\}=\left[\begin{array}{ccccc}70&180&&&\\ 120&200&80&&\\ &&150&300\end{array}\right]\]  
Let the work hours per day be 8 h. Thus, the following matrix is obtained:  
\[\left\{\gamma_{ij}\right\}=\left[\begin{array}{ccccc}48&16&&&\\ 24&12&32&\\ &&20&10\end{array}\right]\]  
From Equation (2), the following matrix is obtained:  
\[\left\{\lambda_{ij}\right\}=\left[\begin{array}{ccccc}3360&2880&&\\ 2880&2400&2560&\\ &&3000&3000\end{array}\right]\]  
where \(\lambda_{line}\) = 2400. If the workers are rearranged, the output rate is increased without any increase in cost. Equation (6) is used in the method proposed in this study: \(\lambda_{line}\) = 528,000/187. Thus, the following matrix is obtained:  
\[\begin{array}{