Pfthb

I have a following problem: I would like to schedule some meetings (tasks) using pyschedule library and allow brakes between them when cumulatively they take too long (more than 4 time slots). In the same time I would like to allow tasks taking more than the maximum 4 time slots. Let's say that I have a 1 person and 3 meetings:

person = scenario.Resource('person')
meeting1 = scenario.Task('meeting1', 1)
meeting2 = scenario.Task('meeting2', 2)
meeting3 = scenario.Task('meeting3', 5)

Then the desired solution would be for example [meeting1, meeting2, break, meeting3].
I've tried to make a restriction:

MAX_CONSECUTIVE_SLOTS = 4
for slot in range(HORIZON):
 scenario += person[slot:slot + MAX_CONSECUTIVE_SLOTS + 1] <= MAX_CONSECUTIVE_SLOTS

but this works only when all meetings are no longer than MAX_CONSECUTIVE_SLOTS. I've also try to combine this condition with the number of tasks per time slice:

meeting1.count = 1
meeting2.count = 1
meeting3.count = 1

for slot in range(HORIZON):
 scenario += (person[slot:slot + MAX_CONSECUTIVE_SLOTS + 1] <= MAX_CONSECUTIVE_SLOTS) or 
 (person['count'][slot:slot + MAX_CONSECUTIVE_SLOTS + 1] <= 1)

But person['count'][n:m] apparently means the number of tasks finished in given time slice, when I need the number of tasks overlapping this slice.

I am using mip.solve solver. Any help would be much appreciated.

Here is a solution based on timnon's answer on Github. It is based on stress parameter, which has to be limited for any person. Each meeting increases person's stress by the amount equal to the meetings length in case of shorter meetings, or by the amount equal to the stress limit for longer meetings. Each break reduces person stress.

from pyschedule import Scenario, solvers


horizon = 20
stress_limit = 4

S = Scenario('test', horizon=horizon)

meeting1 = S.Task('meeting1', 1, stress=1)
meeting2 = S.Task('meeting2', 2, stress=2)
meeting3 = S.Task('meeting3', 5, stress=stress_limit)
breaks = S.Tasks('break', schedule_cost=0, num=3, stress=-stress_limit)

person = S.Resource('person')
meeting1 += person
meeting2 += person
meeting3 += person
breaks += person

for t in range(horizon + 1):
 S += person['stress'][:t] <= stress_limit
 S += person['stress'][:t] >= 0

S.clear_solution()
S.use_flowtime_objective()

if solvers.mip.solve(S, msg=0, kind='CBC'):
 print(S.solution())
else:
 print('no solution found')

Edit: this solution works well, but starts to be really slow with greater number of users and meetings.