MILP: how to work a for loop into a constraint

I am trying to add a Constraint to my optimization (linear programming), that ensures, that a resource only works on one task at every given time (it's logical, that a resource/human cannot work on two screwing jobs at the same time).

My optimization variable is a boolean 3D array x[task, resource, time] which is 1 if at the time a task is started by the resource; 0 otherwise.

Time is considered in seconds of unit 1.

processing_time(task, ress) is a matrix with the indivudal times a ressource needs for every job there is.

I managed to build the for loops, that should be help up, but how do I combine it with/into a constraint for my optimization?

Explanation of the loops: I am looking at every processing time window for the jobs and and the sum over them are not allowed to be higher than 1 (so in the time window (processing time) of a task there is only the one (1) start of this one given task in the 3D array).

Mainly the "sum <= 1" is my Constraint.

So how do I get this to work in my optimization problem?

for ress = 1:nR
    for time = 1:nPT
        for task = 1:nT
            if (time-processing_time(task,ress) >0)
            sum (x(task,ress,(time-processing_time(task,ress)):time)<=1)
            end
        end
    end
end

Thanks ahead!

Answer 1

[Jaynik]

Hi Andra,

I understand that you are trying to add a constraint to optimize the assignment of the variable 'x'. I see that you are approaching the problem by getting the sum of 1s present in 'x'. Instead, you can create a boolean array of assigned resources assuming that the resources are numbered from 1 to nR. You can directly assign the index to "true" if the resource is available and "false" otherwise. Following is a psuedo code to perform the task.

if (time - processing_time(task, ress) > 0)
    % Check if the resource is available
    if resource_available(ress)
        time_starting = time - processing_time(task, ress) + 1; % Calculate the starting and ending time for the task
        time_ending = time;
        x(task, ress, time_starting:time_ending) = 1;   % Assign the task to the resource for the given time range
        resource_available(ress) = false;   % Update resource availability to false for the given time range
    else
        x(task, ress, time) = 0;
    end
end

Hope this helps!