suppose there is one workstation of interest at an assembly line in which a product stays for a few hours and is processed by the operators there before moving to the next station.
At the station there are several operators op = {op_1, op_2, ..., op_n}. Each operator has his own work instruction including several tasks t_op = {t_(op,1), t_(op,2), ..., t_(op,k)}. Tasks require material from material containers c = {c_1, c_2, ..., c_3}. One material container can be required by 1 or more tasks from 1 or more operators. Each container has a size in m².
What is the combination of tasks, if we choose one task from each task list t_op, which results in the largest (in terms of m²) combination of required material containers c.
In the next step, we could also ask what is the combination of 1, 2, ..., j tasks from each t_op that results in the largest combination.
I started with the idea of simply calculating the floor space for all possible combinations, however, if there are, e.g., 20 operators in one station and each operator can do 10 tasks, then this results in 20^10 possible combinations, which is too much. But since I was asked to edit with my own soluation, I will tell you the stupid solution principle for allowing j = 1 tasks on the shop floor: Combine each task of each operator with each task of each operator --> for 20 operators and 10 tasks we get 20^10 possible task combinations. Now for each of those combinations, allocate the material containers to the tasks. Now sum up the area of all material containers allocated to one combination. Then sort by largest --> This is the combination of tasks that leads to the largest space requirements.
Another option would be to formulate it as a discrete optimization problem, which would be relatively "easy".
Is there another solution?
