MiniZinc does not find a solution to the scheduling problem - minizinc

Question

I have a problem finding a solution using MiniZinc.
The task:
It is necessary to make a schedule of shifts for employees.
In one day there are three shifts: day (D), evening (E) and night (N).
It is necessary to draw up an optimal schedule, if possible avoiding undesirable situations:
Avoid single shifts (one shift between two breaks)
Avoid single breaks (shift, break, shift)
Avoid double breaks (shift, break, break, shift)
After a night shift should be a full day off (three breaks in a row)
To find a solution, I minimize the number of undesirable situations.
When I start the calculation, MiniZinc displays several intermediate variants, but does not find a final solution.
Is it possible to somehow optimize the calculations?
include "regular.mzn";
int: n = 21;
int: m = 6;
set of int: D = 1..n;
set of int: E = 1..m;
% Number of employees per shift
%|Sun |Mon |Tue |Wen |Thur |Fri |Sat |
array[D] of int: SHIFTS = [2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1];
/*2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1,
2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1,
2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2];*/
% The range of the number of shifts per employee for the period ([|from, to)
array[E, 1..2] of int: DC_SHIFTS = [|0, 10 %emp1
|0, 10 %emp2
|0, 10 %emp3
|0, 10 %emp4
|0, 10 %emp5
|0, 10 %emp6
|];
%-------------------------------------------------
% Variables
%-------------------------------------------------
array[E, D] of var 1..4: X;
% Counters of avoidable situations
var int: OS_PENALTY; % break, shift, break (single shift)
var int: NS_PENALTY; % night shift, not break, not break, not break (full day off after a night shift)
var int: DS_PENALTY; % shift, break, break, shift (two breaks between shifts)
var int: OO_PENALTY; % shift, break, shift (one break between shifts)
%-------------------------------------------------
% Constraints
%-------------------------------------------------
constraint
forall(d in D)(
sum(e in E)(bool2int(X[e, d] != 4)) = SHIFTS[d]
);
constraint
forall(e in E)(
sum(d in D)(bool2int(X[e, d] != 4)) >= DC_SHIFTS[e, 1]
/\
sum(d in D)(bool2int(X[e, d] != 4)) < DC_SHIFTS[e, 2]
);
constraint
forall(d in D)(
if d mod 3 = 1 then forall(e in E)(X[e, d] = 1 \/ X[e, d] = 4) else
if d mod 3 = 2 then forall(e in E)(X[e, d] = 2 \/ X[e, d] = 4) else
forall(e in E)(X[e, d] = 3 \/ X[e, d] = 4) endif endif
);
NS_PENALTY = sum(e in E, d in D where d < max(D) - 2)(bool2int(
X[e, d] = 3 \/ (X[e,d+1] != 4 /\ X[e,d + 2] != 4 /\ X[e,d + 3] != 4)
));
DS_PENALTY = sum(e in E, d in D where d < max(D) - 2)(bool2int(X[e, d] != 4 \/ X[e, d + 1] = 4 \/ X[e, d + 2] = 4 \/ X[e, d + 3] != 4));
OS_PENALTY = sum(e in E, d in D where d < max(D) - 1)(bool2int(X[e, d] = 4 /\ X[e, d + 1] != 4 /\ X[e, d + 2] = 4));
OO_PENALTY = sum(e in E, d in D where d < max(D) - 1)(bool2int(X[e, d] != 4 \/ X[e, d + 1] = 4 \/ X[e, d + 2] != 4));
%-------------------------------------------------
% Solve
%-------------------------------------------------
solve minimize OS_PENALTY + NS_PENALTY + DS_PENALTY + OO_PENALTY;
%-------------------------------------------------
% Output
%-------------------------------------------------
array[1..4] of string: rest_view = ["D", "E", "N", "-"];
output
[
rest_view[fix(X[e, d])] ++
if d = n then "\n" else "" endif
| e in E, d in D
];

Answer 1

I suggest the following changes to your model:
Change the declaration of X to array[E, D] of var 0..1: X; where 0 means break and 1 shift. Whether it's a day, evening or night shift is handled in the output section, where the results are transformed to show the shift type like if fix(X[e, d]) == 0 then "-" else rest_view[1 + (d-1) mod 3] endif.
Rewrite the constraints using globals like:
import "globals.mzn";
constraint
forall(d in D)(
exactly(SHIFTS[d], col(X, d), 1)
%sum(e in E)(bool2int(X[e, d] != 0)) = SHIFTS[d]
);
constraint
forall(e in E)(
global_cardinality_low_up(row(X, e), [1], [DC_SHIFTS[e, 1]], [DC_SHIFTS[e, 2] - 1])
%sum(d in D)(bool2int(X[e, d] != 0)) >= DC_SHIFTS[e, 1]
%/\
%sum(d in D)(bool2int(X[e, d] != 0)) < DC_SHIFTS[e, 2]
);
%constraint
% forall(d in D)(
% if d mod 3 = 1 then forall(e in E)(X[e, d] = 1 \/ X[e, d] = 4) else
% if d mod 3 = 2 then forall(e in E)(X[e, d] = 2 \/ X[e, d] = 4) else
% forall(e in E)(X[e, d] = 3 \/ X[e, d] = 4) endif endif
% );
Rewrite the penalties like:
NS_PENALTY = sum(e in E, d in 1..n - 3 where d mod 3 = 0)(bool2int(
X[e, d] = 1 /\ (sum(i in 1..3)(X[e,d+i]) > 0)
));
DS_PENALTY = sum(e in E, d in 1..n - 3)(bool2int(X[e, d] != 0 /\ X[e, d + 1] = 0 /\ X[e, d + 2] = 0 /\ X[e, d + 3] != 0));
OS_PENALTY = sum(e in E, d in 1..n - 2)(bool2int(X[e, d] = 0 /\ X[e, d + 1] != 0 /\ X[e, d + 2] = 0));
OO_PENALTY = sum(e in E, d in 1..n - 2)(bool2int(X[e, d] != 0 /\ X[e, d + 1] = 0 /\ X[e, d + 2] != 0));