old_contents: 
  from collections import namedtuple
  
  job = namedtuple('job', ['start_time', 'finish_time', 'name']) # `name` in last position to use `<` on tuples for ordering.
  
  jobs = [ # some overlapping jobs...
      job(0, 3, 'A'),
      job(0, 7, 'B'),
      job(0, 3, 'C'),
      job(4, 7, 'D'),
      job(4, 10.5, 'E'),
      job(9, 12, 'F'),
      job(9, 12, 'G'),
      job(11.5, 16, 'H'),
      job(13, 16, 'I'),
      job(13, 16, 'J'),
  ] # everything ends at time 16.
  
  ordering = sorted(jobs)
  #print(ordering)
  
  label = {J: {'M₀', 'M₁', 'M₂', 'M₃', 'M₄'} for J in jobs} # each job can be assigned to any machine.
  
  machine = {} # answers.
  
  def overlaps(I, J):
      return I.finish_time > J.start_time # taking advantage of ordering
  
  for j, J in enumerate(ordering): # a greedy algorithm because the ordering is fixed and just one pass over it.
      for i, I in enumerate(ordering[:j]): # O(n^2) complexity because of the last job; btw, preprocess of overlappings may help to check only those ones, getting a linear time.
          if overlaps(I, J):
              label[J] -= {machine[I]} # remove the machine on which job `I` is allocated for possibilities about job `J`.
      machine[J] = label[J].pop() # let `pop` raises an exception if there is not an available label.
  
  print(machine) # show results.
  
  
new_contents: 
  from collections import namedtuple, defaultdict
  
  job = namedtuple('job', ['start_time', 'duration', 'deadline', 'name']) # `name` in last position to use `<` on tuples for ordering.
  
  def finish_time(jb):
      return jb.start_time + jb.duration
  
  jobs = [ # some overlapping jobs...
      job(0, 3, 3, 'A'),
      job(0, 7, 7, 'B'),
      job(0, 3, 3, 'C'),
      job(4, 3, 10, 'D'),
      job(4, 6.5, 13.5, 'E'),
      job(9, 3, 12, 'F'),
      job(9, 3, 12, 'G'),
      job(11.5, 4.5, 16, 'H'),
      job(13, 3, 16, 'I'),
      job(13, 3, 16, 'J'),
  ] # every job ends at time 16.
  
  
  label = {J.name: {'M₀', 'M₁', 'M₂', 'M₃', 'M₄'} for J in jobs} # each job can be assigned to any machine, initially.
  label['E'] = {'M₀'} # job 'E' can be performed on the first machine only.
  label['D'] = {'M₁', 'M₂'} # job 'E' can be performed on the first machine only.
  
  ordering = list(enumerate(sorted(jobs)))
  
  busy = defaultdict(list)
  busy.update({
      'M₀': [job(2, 3, None, 'cleaning'), job(14, 1, None, 'sunday')],
      'M₁': [job(5, 2, None, 'maintenance')],
  })
  
  def overlaps(I, J):
      return finish_time(I) > J.start_time # taking advantage of ordering
  
  def ontime(J):
      return finish_time(J) <= J.deadline
  
  def run(prefix, jobs, machine, label):
  
      if jobs: # still jobs to allocate.
  
          (j, J), *Js = jobs # unpacking.
          L = label[J.name].copy()
  
          for i, I in prefix: # O(n^2) complexity because of the last job; btw, preprocess of overlappings may help to check only those ones, getting a linear time.
              if overlaps(I, J):
                  label[J.name] -= {machine[I.name]} # remove the machine on which job `I` is allocated for possibilities about job `J`.
  
          for l in label[J.name]:
  
              J_delayed = J
              for B in busy[l]:
                  if overlaps(J_delayed, B):
                      J_delayed = J_delayed._replace(duration=J_delayed.duration + B.duration)
                  else:
                      break # assuming busy jobs are ordered too.
  
              if ontime(J_delayed):
                  machine[J.name] = l # an attempt to allocate job `J` on machine `l`.
                  yield from run([(j, J_delayed)] + prefix, Js, machine.copy(), label)
  
          label[J.name] = L
      else:
          assert len(machine) == len(ordering)
          yield machine
  
  sols = list(run([], ordering, {}, label.copy())) # execute for side effects on the list `sols`.
  
  """
  print(
      list(zip(sorted(map(lambda sol: (len(set(sol.values())), sol), sols),
                      key=lambda p: p[0]),
               range(10)))) # show results.
  """
  
  for sol in reversed(list(sorted(map(lambda sol: (len(set(sol.values())), sol), sols),
                      key=lambda p: p[0]))): # show results.
      print(sol)
  
  
subject: Update the INTERVAL-PARTITIONING solver in order to yield all feasible solutions.