Suppose we have several stations. In each station there are several operators. Each operator conducts several tasks during one cycle. Each of the tasks requires (several) materials. Materials are delivered in material containers with a certain size in m². The different tasks of the different operators, may require material from the same material container.
The question is now: What is the combination of tasks (and consequently the allocated material containers) that require the most floor space during one cycle, if one (or more generally some number k) task is (are) chosen for each operator?
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 $10^{20}$ possible combinations, which is too much.
But since I think that my description is confusing I will tell you the stupid solution principle for allowing k = 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 $10^{20}$ possible combinations. Now for each of those combinations, allocate the material containers to the tasks of the operators. 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.
Is there a smarter solution?
As an example was asked for. Imagine an assembly line for boats. The boat will go through several stations until it is assembled. At one of those stations, there are typically several employees (operators), which carry out tasks. For example there are 10 operators and each operator carries out 5 tasks in one station. The tasks are serial, which means, that he hast to start with task 1 and finish with task 5. Each operator carries out his own tasks, so the tasks are allocated to only one operator. However, to carry out the tasks, operators use material, which is delivered in material containers. For example, 10 operators carry out 5 tasks each and each requires material from 5 material containers. However, the material containers are not allocated to one specific operator, meaning, there are not $10 \times 5 \times 5$ material containers, but less, since the operators share those containers. This is due to the fact that there is not only one material in each container, but several, and the can, as described, serve different operators' tasks.
Now suppose that there is a material control principle (which is called a pull principle in manufacturing), which allows only the material containers that are allocated to one single task of each operator to be transported to the shop floor. This can be extended to allow not only the material allocated to only one task of each operator, but the material allocated to k-tasks of each operator. Example: if k=2, then there can be maximally as many material containers transported to the station, as are allocated to any two tasks of any operators, i.e., that are required by any of the operators for any of their two tasks.
Now further suppose we want to find out what the worst case scenario would be. Since it could be that each operator is getting stuck in his work at the worst possible time, meaning, that we want to find the situation in which those material containers are transported to the shop floor, that occupy the most space, given any k numbers of tasks that are allowed to be released to the shop floor: Example: if k = 2, we want to find the combination of k=2 tasks of each operator, which are connected to the largest total space requirement of the allocated material containers.
The general question is: how much material will be in the station in the worst case, when allowing the material containers allocated to k-tasks per operator into the station (meaning the rest of the material containers somewhere else).
Did this make it clearer?
