34.10.3 Cumulative Scheduling

This example is a very small scheduling problem. We consider seven tasks where each task has a fixed duration and a fixed amount of used resource:

 Task Duration Resource `t1` 16 2 `t2` 6 9 `t3` 13 3 `t4` 7 7 `t5` 5 10 `t6` 18 1 `t7` 4 11

The goal is to find a schedule that minimizes the completion time for the schedule while not exceeding the capacity 13 of the resource. The resource constraint is succinctly captured by a `cumulative/4` constraint. Branch-and-bound search is used to find the minimal completion time.

This example was adapted from [Beldiceanu & Contejean 94].

```     :- use_module(library(clpfd)).
:- use_module(library(lists), [append/3]).

schedule(Ss, End) :-
length(Ss, 7),
Ds = [16, 6,13, 7, 5,18, 4],
Rs = [ 2, 9, 3, 7,10, 1,11],
domain(Ss, 1, 30),
domain([End], 1, 50),
after(Ss, Ds, End),
cumulative(Ss, Ds, Rs, 13),
append(Ss, [End], Vars),
labeling([minimize(End)], Vars). % label End last

after([], [], _).
after([S|Ss], [D|Ds], E) :- E #>= S+D, after(Ss, Ds, E).

%% End of file

| ?- schedule(Ss, End).
Ss = [1,17,10,10,5,5,1],
End = 23
```