MZN 2.4.1: std definition of cumulative

[matsc-at-sics-se]

This should fail, but succeeds:

$ cat /tmp/bug.mzn
include "globals.mzn";
constraint cumulative([1,2],[1,1],[2,2],1);
solve satisfy;
$ minizinc /tmp/bug.mzn
----------

Here, the definition optimizes cumulative to alldifferent/disjunctive, but forgets to check a necessary condition.

[schutta]

This constraint needs to be added in the "then" part of the first if-then-else statement from the standard cumulative definition in cumulative.mzn.

forall(i in index_set(r))(r[i] <= b)

Furthermore, the test whether all tasks are disjunctive seems to be too restrictive and should be replaced by testing whether the two tasks with the lowest lower bound on the resource variable are disjunctive, e.g.,

if card(index_set(s)) = 0 then
  true
elseif card(index_set(s)) = 1 then
  r[min(index_set(r))] <= b
else
  let { array[int] of int: sort_r = [lb(r[i]) | i in index_set(r)]; } in
  if sort_r[1] + sort_r[2] > ub(b) then
    % disjunctive case
    [...]
  else
    % cumulative case
    [...]
  endif
endif

Note that the similar optimisation from cumulative to alldifferent/disjunctive should be done for cumulative_opt.

[matsc-at-sics-se]

This part doesn't look quite right:

elseif card(index_set(s)) = 1 then
  r[min(index_set(r))] <= b

I would suggest:

elseif card(index_set(s)) = 1 then
  r[min(index_set(r))] <= b \/ d[min(index_set(d))] = 0

[matsc-at-sics-se]

Also, your suggestion:

forall(i in index_set(r))(r[i] <= b)

can remove solutions if the corresponding d[i] = 0.

[schutta]

You are right. I forgot to consider the zero-duration case. Thus, it should be

forall(i in index_set(r))(r[i] <= b \/ d[i] = 0)

and for the case of one task as you suggested.