Allocating and scheduling tasks into rooms with conditions - optimization algorithm

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I need to find a suitable method to be based on, for developing an optimization algorithm which does the following:

Let's say we have N tasks to do, and we have M rooms that each one of them contains some specific number of infrastructure/conditions. Each task demands using room with a suitable conditions for the task.

For example, to get task A done we need to use water tap and gas piping, so we can only use rooms that contain those ones.

Also, for each task we have a predefined due date.

I hope I've explained it well enough.

So, I need to develop an algorithm which can allocate the tasks for each room in a proper scheduling, so I could do all of the tasks at the minimum total time and without exceeding deadline times (and if exceeding is inevitable, then getting the least worst answer).

What are an existing methods or algorithm I can get based on and learn from them? I though about 'Job Shop', but I wonder if there are other suitable algorithms that can handle problems like that.

linear-programming scheduling optimization algorithm
4 Answers

This is not an algorithm but a Mixed Integer Programming model. I am not sure if this is what you are looking for.

Assumptions: only one job can execute at the same time in a room. Jobs in different rooms can execute in parallel. Also, to keep things simple, I assume the problem is feasible (the model will detect infeasible problems but we don't return a solution if this is the case).

So we introduce a number of decision variables:

assign(i,j) = 1 if task i is assigned to room j
              0 otherwise

finish(i) = time job i is done processing

makespan = finishing time of the last job

With this we can formulate the MIP model:

enter image description here

The following data is used:

Length(i) = processing time of job i
M = a large enough constant (say the planning horizon)
DueDate(i) = time job i must be finished
Allowed(i,j) = Yes if job i can be executed in room j

Importantly, I assume jobs are ordered by due date. The first constraint says: if job i runs in room j then it finishes just after the previous jobs running in that room. The second constraint is a bound: a job must finish before its due date. The third constraint says: each job must be assigned to exactly one room where it is allowed to execute. Finally, the makespan is the last finish time.

To test this, I generated some random data:

----     37 SET use  resource usage

         resource1   resource2   resource3   resource4   resource5

task2                                              YES
task3                                                          YES
task5                                  YES
task7          YES
task9                      YES                     YES
task11                                                         YES
task12         YES                                 YES
task13         YES
task14         YES
task15                                 YES
task16                                 YES                     YES
task17                     YES
task20                     YES                     YES
task21         YES         YES
task23                     YES
task24                                             YES
task25         YES                                             YES
task26                                 YES
task28                                                         YES


----     37 SET avail  resource availability

        resource1   resource2   resource3   resource4   resource5

room1                     YES         YES         YES         YES
room2                                 YES         YES
room3                                             YES         YES
room4         YES         YES         YES                     YES
room5         YES                     YES         YES         YES

The set Allowed is calculated from use(i,r) and avail(j,r) data:

----     41 SET allowed  task is allowed to be executed in room

             room1       room2       room3       room4       room5

task1          YES         YES         YES         YES         YES
task2          YES         YES         YES                     YES
task3          YES                     YES         YES         YES
task4          YES         YES         YES         YES         YES
task5          YES         YES                     YES         YES
task6          YES         YES         YES         YES         YES
task7                                              YES         YES
task8          YES         YES         YES         YES         YES
task9          YES
task10         YES         YES         YES         YES         YES
task11         YES                     YES         YES         YES
task12                                                         YES
task13                                             YES         YES
task14                                             YES         YES
task15         YES         YES                     YES         YES
task16         YES                                 YES         YES
task17         YES                                 YES
task18         YES         YES         YES         YES         YES
task19         YES         YES         YES         YES         YES
task20         YES
task21                                             YES
task22         YES         YES         YES         YES         YES
task23         YES                                 YES
task24         YES         YES         YES                     YES
task25                                             YES         YES
task26         YES         YES                     YES         YES
task27         YES         YES         YES         YES         YES
task28         YES                     YES         YES         YES
task29         YES         YES         YES         YES         YES
task30         YES         YES         YES         YES         YES

We also have random due dates and processing times:

----     33 PARAMETER length  job length

task1  2.335,    task2  4.935,    task3  4.066,    task4  1.440,    task5  4.979,    task6  3.321,    task7  1.666
task8  3.573,    task9  2.377,    task10 4.649,    task11 4.600,    task12 1.065,    task13 2.475,    task14 3.658
task15 3.374,    task16 1.138,    task17 4.367,    task18 4.728,    task19 3.032,    task20 2.198,    task21 2.986
task22 1.180,    task23 4.095,    task24 3.132,    task25 3.987,    task26 3.880,    task27 3.526,    task28 1.460
task29 4.885,    task30 3.827


----     33 PARAMETER due  job due dates

task1   5.166,    task2   5.333,    task3   5.493,    task4   5.540,    task5   6.226,    task6   8.105
task7   8.271,    task8   8.556,    task9   8.677,    task10  8.922,    task11 10.184,    task12 11.711
task13 11.975,    task14 12.814,    task15 12.867,    task16 14.023,    task17 14.200,    task18 15.820
task19 15.877,    task20 16.156,    task21 16.438,    task22 16.885,    task23 17.033,    task24 17.813
task25 21.109,    task26 21.713,    task27 23.655,    task28 23.977,    task29 24.014,    task30 24.507

When I run this model, I get as results:

----    129 PARAMETER results  

                   start      length      finish     duedate

room1.task1                    2.335       2.335       5.166
room1.task9        2.335       2.377       4.712       8.677
room1.task11       4.712       4.600       9.312      10.184
room1.task20       9.312       2.198      11.510      16.156
room1.task23      11.510       4.095      15.605      17.033
room1.task30      15.605       3.827      19.432      24.507
room2.task6                    3.321       3.321       8.105
room2.task10       3.321       4.649       7.971       8.922
room2.task15       7.971       3.374      11.344      12.867
room2.task24      11.344       3.132      14.476      17.813
room2.task29      14.476       4.885      19.361      24.014
room3.task2                    4.935       4.935       5.333
room3.task8        4.935       3.573       8.508       8.556
room3.task18       8.508       4.728      13.237      15.820
room3.task22      13.237       1.180      14.416      16.885
room3.task27      14.416       3.526      17.943      23.655
room3.task28      17.943       1.460      19.403      23.977
room4.task3                    4.066       4.066       5.493
room4.task4        4.066       1.440       5.506       5.540
room4.task13       5.506       2.475       7.981      11.975
room4.task17       7.981       4.367      12.348      14.200
room4.task21      12.348       2.986      15.335      16.438
room4.task25      15.335       3.987      19.322      21.109
room5.task5                    4.979       4.979       6.226
room5.task7        4.979       1.666       6.645       8.271
room5.task12       6.645       1.065       7.710      11.711
room5.task14       7.710       3.658      11.367      12.814
room5.task16      11.367       1.138      12.506      14.023
room5.task19      12.506       3.032      15.538      15.877
room5.task26      15.538       3.880      19.418      21.713

Detail: based on the assignment I recalculated the start and finish times. The model can allow some slack here and there as long as it does not interfere with the objective and the due dates. To get rid of any possible slacks, I just execute all jobs as early as possible. Just back-to-back execution of jobs in the same room using the job ordering (remember I sorted jobs according to due date).

This model with 30 jobs and 10 rooms took 20 seconds using Cplex. Gurobi was about the same.

Augmenting the model to handle infeasible models is not very difficult. Allow jobs to violate the due date but at a price. A penalty term needs to be added to the objective. The due date constraint is in the above example a hard constraint, and with this technique, we make it a soft constraint.

user_56255340

I used a small variant of Alex's OPL CP Optimizer model on the data and it finds the optimal solution (makespan=19.432) within a couple of seconds and proves optimality in about 5s on my laptop. I think a big advantage of a CP Optimizer model is that it would scale to much larger instances and easily produce good quality solutions even if, for large instances, proving optimality may be challenging of course.

Here is my version of the CP Optimizer model:

using CP;

int N = 30; // Number of tasks
int M = 5;  // Number of rooms
int Length [1..N] = ...; // Task length
int DueDate[1..N] = ...; // Task due date
{int} Rooms[1..N] = ...; // Possible rooms for task

tuple Alloc { int job; int room; }
{Alloc} Allocs = {<i,r> | i in 1..N, r in Rooms[i]};

dvar interval task[i in 1..N] in 0..DueDate[i] size Length[i];
dvar interval alloc[a in Allocs] optional;

minimize max(i in 1..N) endOf(task[i]);
subject to {
  forall(i in 1..N) { alternative(task[i], all(r in Rooms[i]) alloc[<i,r>]); }
  forall(r in 1..M) { noOverlap(all(a in Allocs: a.room==r) alloc[a]); }
}

Note also that the MIP model exploits a problem specific dominance rule that the tasks allocated to a particular room can be ordered by increasing due-date. While this is perfectly true for this simple version of the problem, this assumption may not hold anymore in the presence of additional constraints (as for instance a minimal start time for the tasks). The CP Optimizer formulation does not make this assumption.

user_95895080

Within CPLEX you can rely on MIP but you could also use CPOptimizer scheduling.

In OPL your model would look like

using CP;

int N = 30; // nbTasks
int M = 10; // rooms

range Tasks = 1..N;
range Rooms = 1..M; 

int taskDuration[i in Tasks]=rand(20);
int dueDate[i in Tasks]=20+rand(20);
int possible[j in Tasks][m in Rooms] = (rand(10)>=8);

dvar interval itvs[j in Tasks][o in Rooms] optional in 0..100 size taskDuration[j] ;
dvar interval itvs_task[Tasks];
dvar sequence rooms[m in Rooms] in all(j in Tasks) itvs[j][m];


execute {
        cp.param.FailLimit = 10000;
}

minimize max(j in Tasks) endOf(itvs_task[j]);

subject to {
  // alternative
  forall(t in Tasks) alternative(itvs_task[t],all(m in Rooms)itvs[t][m]);  

  // one room is for one task at most at the same time
  forall (m in Rooms)
    noOverlap(rooms[m]);

  // due dates  
  forall(j in Tasks) endOf(itvs_task[j]) <=dueDate[j]; 

}

and give

Gantt view of room allocation

user_37255960

In Pyomo Erwin's MIP can be implemented like:

        ################################################################################
        # Sets
        ################################################################################

        model.I = Set(initialize=self.resource_usage.keys(), doc='jobs to run')
        model.J = Set(initialize=self.resource_availability.keys(), doc='rooms')
        model.ok = Set(initialize=self.ok.keys())

        ################################################################################
        # Params put at model
        ################################################################################

        model.length = Param(model.I, initialize=self.length)
        model.due_date = Param(model.I, initialize=self.due_date)

        ################################################################################
        # Var
        ################################################################################

        model.x = Var(model.I, model.J, domain=Boolean, initialize=0, doc='job i is assigned to room j')
        model.finish = Var(model.I, domain=NonNegativeReals, initialize=0, doc='finish time of job i')
        model.makespan = Var(domain=NonNegativeReals, initialize=0)

        ################################################################################
        # Constraints
        ################################################################################

        M = 100

        def all_jobs_assigned_c(model, i):
            return sum(model.x[ii, jj] for (ii, jj) in model.ok if ii == i) == 1

        model.all_jobs_assigned_c = Constraint(model.I, rule=all_jobs_assigned_c)

        def finish1_c(model, i, j):
            return sum(
                model.length[ii] * model.x[ii, jj] for (ii, jj) in model.ok if jj == j and ii <= i
            ) - M * (1 - model.x[i, j]) <= model.finish[i]

        model.finish1_c = Constraint(model.I, model.J, rule=finish1_c)

        model.finish2_c = Constraint(
            model.I, rule=lambda model, i: model.finish[i] <= model.due_date[i]
        )

        model.makespan_c = Constraint(
            model.I, rule=lambda model, i: model.makespan >= model.finish[i]
        )

        ################################################################################
        # Objective
        ################################################################################
        def obj_profit(model):
            return model.makespan

        model.objective = Objective(rule=obj_profit, sense=minimize)

Solving with CBC took with 4 cores about 2min and results in:

schedule_result

user_82121290
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