Reusable resources in a scheduling model

R.U.B wrote:
Hello,
>
I would like to ask for your help on a MIP model I have been working
on for a while. I am modeling reusable resources, teams of workers to
be exact, in a static model. In details, I have a scheduling model
which assigns jobs to machines where workers have to work for a
predefined time on each machine to complete the job. The issue is that
the # of workers is smaller than the # of machines and thus only a few
machines (as many as the number of workers) can be used at the same
time. The workers should be assigned to jobs, become busy, and then
become idle when the job is complete and get reassigned for another
job.
>
I tried to model the worker-job assignment using a binary variable.
It's easy to assign the worker for the first time (binary var =1) but
then I couldn't figure out any way to make the binary variable 0 after
some time and set it to 1 again later.
>
Could you help me with any idea of how to deal with this, preferably
without involving dynamic programming?
>

If workers are in some way unique (different workers have different
capabilities), you need binary assignment variables with three
subscripts, one each for worker, machine and time. For each worker-time
pair, the sum across all machines of those variables is <= 1. For each
job-machine-time combination, you need constraints that block that
assignment unless the requisite worker-machine-time binaries are equal to 1.

the other hand, if all workers look alike, rather than using binary
variables you are better off using general integer variables for the
number of workers assigned to each machine at each time. Use a
constraint that, in each time, the sum of those variables (across all
machines) is less than or equal to the number of workers in the pool.
You can do this for multiple worker pools (e.g., machine operators v.
material handlers).

Don't worry about forcing workers to become unassigned at the end of a
job. As long as the constraints prevent workers from being two places
at once, the solution will take care of that. Worst case, if the
solution has workers assigned to idle machines (which can only happen if
the workers are not needed elsewhere and there is no cost for have them
assigned), you can manually adjust the solution later. It will still be
optimal.

/Paul