## 3 Materials and Methods: Workforce Capacity and Workload
### Computing the Required Workforce  
Now the necessary concepts have been defined, this section describes the following steps to find \(w\) and \(TP_{k}\). The process is illuminated with a numerical example. Let us consider an assembly process--although the example could be extended to other types of operations--where only two operations, X and Y, exist (and X precedes Y). Several units of two different models (A and B) are assembled each day. In Table 1, the amount of work (or workload) corresponding to the daily production (\(q_{ik}\)) is measured in man-hours per day for each operation and model. Notice that if the load \(q_{k}\) is measured in parts, its value is common across the assembly process, while if it is measured in man-hours, it may be different for each operation and \(q_{ik}\) must be used instead of \(q_{k}\). Let us assume that the available time is one eight-hour shift and that each operator's labor capacity is eight man-hours per shift. Equation (4) is used to compute unit workload per operation and model (\(Wu_{ik}\)). Next, the total unit workload per model (\(TWu_{k}\)) and the total multi-model unit workload of the system (\(TWu\)) are computed according to Equations (10) and (11). This way, the unit workloads of different processes are added to compute the unit workload of a multi-model system. Since the total multi-model unit workload (\(TWu\)) in Table 1 is twelve (twelve times what a person can perform in a day), the minimum number of employees to complete the task in the allowed time is twelve (Equation (12)). The ceiling function \(\cdot\) is used to return the smallest integer greater than or equal to \(TWu\) if \(TWu\) is not a whole number. \[TWu_{k}=\sum_{i}Wu_{ik} \tag{10}\]  
\[TWu=\sum_{k}TWu_{k} \tag{11}\]  
\[w=TWu=\sum_{k}\sum_{i}Wu_{ik} \tag{12}\]  
Now, our procedure redistributes the workforce capacity in order to complete the jobs in eight hours assuming a constant team of 12 cross-trained workers. Asterisks represent new values after w has been found. We assume that the system is in steady state so all operators can start working immediately with no wait. The plant just needs a fraction of the day to complete model A since its new labor capacity (\(CP^{*}{}_{k}=12\)) is greater than any model's workload (\(TWu_{k}\)). This way, capacity is adjusted to the desired workload. Equation (13) is derived taking into account that the time devoted to a model (\(TP_{k}\)) is inversely proportional to the available capacity (\(CP^{*}{}_{k}\)) and Equation (13) admits that the time devoted to each model has to be proportional to its relative workload. Since the job to perform has not changed, the real workload is not \(CP^{*}{}_{k}\) but it remains constant (\(TWu^{*}{}_{k}=TWu_{k}\)) within the scheduled time \(TP_{k}\). In our numerical example, according to Equation (14), a crew of twelve operators will be producing model A during 8/12 (\(TWu_{k}\)/\(TWu\)) of the available time or \(TP_{\text{A}}=5\) h and 20 min. Throughout the rest of the day, the plant will be producing model B.  
\[TWu^{*}_{k}\text{-}AT=CP^{*}_{k}\text{-}TP_{k} \tag{13}\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{3}{*}{**Model (\(k\))**} & \multicolumn{2}{c}{**Operation X**} & \multicolumn{2}{c}{**Operation Y**} & \multicolumn{2}{c}{**Total Unit**} \\  & **Load** & **Unit** & **Load** & **Unit** & **Workload** \\  & **Man-hours** & **Workload** & **Hours** & **Workload** & **Per Model** \\  & \(q_{ik}\) & \(Wu_{ik}=q_{ik}/CP_{ik}\) & \(q_{ik}\) & \(Wu_{ik}=q_{ik}/CP_{ik}\) & \(TWu_{ik}=\Sigma Wu_{ik}\) \\ \hline Model A (1) & 48 & 48/8 = 6 & 16 & 16/8 = 2 & 6 + 2 = 8 \\ Model B (2) & 8 & 8/8 = 1 & 24 & 24/8 = 3 & 1 + 3 = 4 \\ \hline \hline \multicolumn{4}{c}{Total multi-model unit workload \(TWu=\Sigma TWu_{ik}=\)} & 12 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Unit workload per operation (\(Wu_{ik}\)) and per model (\(TWu_{k}\)) and total load (\(TWu\)).