/* Number of people */
param P_count, integer, > 0;
/* Number of jobs */
param J_count, integer, > 0;
/* Number of days */
param D_count, integer, > 0;
param WL, integer, > 0;
param WH, integer, > 0;
set P := 1..P_count;
set J := 1..J_count;
set D := 1..D_count;
/* Person p likes to solve jobs j */
param L{p in P, j in J} default 0, binary;
/* Person p hates to solve jobs j */
param H{p in P, j in J} default 0, binary;
/* Person p is capable to perform job j */
param C{p in P, j in J} default 1, binary;
/* How many jobs need to be done on what day */
param R{d in D, j in J}, integer, >= 0;
/* hardcoded */
param Q{p in P, j in J, d in D}, default 0, binary;
/* workload */
param Wl{j in J}, integer, >= 0;
param max_load{p in P, d in D}, default 1, integer;
/* Person p is allocated to do job j on day d */
var A{p in P, j in J, d in D}, binary;
var error{p in P}, integer, >= 0;
s.t. hardcode{p in P, j in J, d in D}: A[p,j,d] >= Q[p,j,d];
/* A person only has one task per day, at most */
s.t. max_load_person{p in P, d in D}: sum{j in J} A[p,j,d] <= max_load[p,d];
/* A person has at least D-1 tasks */
#s.t. min_load_person{p in P}: sum{j in J, d in D} A[p,j,d] >= min_load[p];
/* A person does not perform the same job on all days */
s.t. duplicate_jobs{p in P, j in J}: sum{d in D} A[p,j,d] <= D_count-1;
/* Each task is allocated */
s.t. all_allocated{j in J, d in D}: sum{p in P} A[p,j,d] == R[d, j];
/* A person only performs what (s)he is capable of */
s.t. capability_person{p in P, j in J, d in D}: A[p,j,d] <= C[p,j];
s.t. error_lt{p in P}: error[p] >= ((sum{j in J, d in D} A[p,j,d] * Wl[j]) - 4);
s.t. error_gt{p in P}: error[p] >= 4 - (sum{j in J, d in D} A[p,j,d] * Wl[j]);
/* Maximize enjoyment */
# minimize error_diff: sum{p in P} error[p];
maximize enjoyment: (sum{p in P, d in D, j in J} A[p,j,d] * (L[p, j] * WL - H[p, j] * WH)) - sum{p in P} error[p];
solve;
printf "Sum %d\n", (sum{p in P, d in D, j in J} A[p,j,d] * (L[p, j] * WL - H[p, j] * WH));
printf "p d j W l\n";
printf ">>>>\n";
printf{p in P, d in D, j in J : A[p,j,d] > 0} "%d %d %d %d %d\n", p, d, j, A[p,j,d] * (L[p, j] * WL - H[p, j] * WH), Wl[j];
printf "<<<<\n";
printf "workloads\n";
printf "p l\n";
printf{p in P} "%d %d\n", p, abs((sum{j in J, d in D : A[p,j,d] > 0} Wl[j]) - ((sum{d in D, j in J} Wl[j] * R[d,j]) / P_count));
printf "workload_dev: %d\n", sum{p in P} abs((sum{j in J, d in D : A[p,j,d] > 0} Wl[j]) - ((sum{d in D, j in J} Wl[j] * R[d,j]) / P_count))^2;
data;
/* Test example */
param P_count := 44;
param J_count := 11;
param D_count := 3;
param WL := 1;
param WH := 3;
param Wl := 1 4, 2 4, 3 2, 4 2, 5 4, 6 3, 7 2, 8 1, 9 1, 10 2, 11 2;
# de pol-parameter
param max_load := 25,1,2; # 12,1,0 12,2,0 12,3,0;
param R : 1 2 3 4 5 6 7 8 9 10 11 :=
1 1 3 0 0 1 3 4 6 2 5 0
2 1 0 6 4 1 3 4 6 2 5 5
3 1 0 0 0 1 3 4 6 2 5 5;
/* AUTOGEN STARTS HERE */
param L : 5 6 7 8 9 10 :=
1 0 1 0 1 0 0
2 0 0 1 1 0 1
3 0 0 1 0 0 1
6 0 1 0 0 0 1
7 0 0 0 1 0 1
8 0 1 0 0 1 0
9 0 1 1 0 0 1
11 1 0 0 1 0 0
13 0 1 0 0 1 0
14 0 1 0 0 0 1
16 0 1 0 0 0 0
18 1 0 0 0 0 0
19 0 0 1 0 1 1
20 0 0 0 0 0 1
21 0 0 0 1 1 0
22 0 1 1 0 0 1
23 0 1 0 1 0 0
24 0 0 0 0 1 0
25 1 0 0 0 0 0
26 0 1 1 0 0 1
27 0 1 0 0 0 1
28 0 1 0 1 0 0
29 0 0 1 1 0 0
30 0 1 1 0 0 0
31 0 1 0 0 0 0
32 0 0 1 0 1 1
34 0 0 0 0 1 1
35 0 1 0 1 1 0
36 0 1 0 0 0 1
38 0 0 1 0 1 1
39 0 0 1 0 0 1
40 0 1 0 0 0 1
41 0 1 1 0 0 1
43 0 1 0 0 0 0
44 0 0 0 1 0 0;
param H : 6 7 8 9 10 :=
1 0 0 0 0 1
2 1 0 0 1 0
6 0 1 0 0 0
7 0 1 0 0 0
8 0 1 1 0 0
9 0 0 1 0 0
11 0 1 0 1 0
13 0 1 1 0 0
14 0 0 1 1 0
16 0 0 0 0 1
18 0 1 0 0 1
19 1 0 1 0 0
21 1 1 0 0 0
22 0 0 1 1 0
23 0 1 0 0 0
24 0 0 1 0 0
25 0 1 0 0 0
26 0 0 1 0 0
27 0 0 1 0 0
28 0 1 0 1 0
29 0 0 0 0 1
30 0 0 0 1 0
31 0 0 0 1 0
32 1 0 1 0 0
34 0 0 1 0 0
35 0 1 0 0 1
36 0 1 0 1 0
38 1 0 1 0 0
39 1 0 1 0 0
40 0 0 0 1 0
41 0 0 1 1 0;
param C : 1 2 3 5 6 9 :=
1 0 0 0 0 1 0
2 0 1 0 0 0 0
3 0 0 0 0 0 0
4 1 0 0 0 0 0
5 1 0 1 0 0 0
6 0 1 0 0 1 0
7 0 0 0 0 0 0
8 0 0 0 0 1 1
9 1 0 0 0 1 0
10 0 0 0 0 0 0
11 0 0 0 1 0 0
12 1 0 0 0 0 0
13 1 0 0 0 1 1
14 0 0 0 0 1 0
15 0 0 0 0 0 0
16 0 0 0 0 1 0
17 0 0 0 0 0 0
18 0 0 0 1 0 0
19 0 0 0 0 0 1
20 0 0 0 0 0 0
21 0 0 1 0 0 1
22 0 0 0 0 1 0
23 0 0 0 0 1 0
24 0 0 1 0 0 1
25 0 1 0 1 0 0
26 0 0 0 0 1 0
27 0 0 0 0 1 0
28 0 0 0 0 1 0
29 0 0 0 0 0 0
30 0 0 0 0 1 0
31 0 0 1 0 1 0
32 0 0 0 0 0 1
33 0 0 0 0 0 0
34 0 0 0 0 0 1
35 0 0 0 0 1 1
36 0 0 0 0 1 0
37 0 0 0 0 0 0
38 0 0 0 0 0 1
39 0 0 0 0 0 0
40 0 0 0 0 1 0
41 0 0 0 0 1 0
42 0 0 0 0 0 0
43 0 0 1 0 1 0
44 0 0 1 0 0 0;
param Q := 5,1,1,1 25,5,1,1 18,5,2,1 11,5,3,1;
end;