half-hardcoded work
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@@ -24,58 +24,187 @@ param H{p in P, j in J} default 0, binary;
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param C{p in P, j in J} default 1, binary;
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/* How many jobs need to be done on what day */
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param R{j in J, d in D}, integer, >= 0;
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param R{d in D, j in J}, integer, >= 0;
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/* hardcoded */
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param Q{p in P, j in J, d in D}, default 0, binary;
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/* workload */
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param Wl{j in J}, integer, >= 0;
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param max_load{p in P, d in D}, default 1, integer;
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/* Person p is allocated to do job j on day d */
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var A{p in P, j in J, d in D}, binary;
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var error{p in P}, integer, >= 0;
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s.t. hardcode{p in P, j in J, d in D}: A[p,j,d] >= Q[p,j,d];
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/* A person only has one task per day, at most */
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s.t. max_load_person{p in P, d in D}: sum{j in J} A[p,j,d] <= 1;
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s.t. max_load_person{p in P, d in D}: sum{j in J} A[p,j,d] <= max_load[p,d];
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/* A person has at least D-1 tasks */
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s.t. min_load_person{p in P}: sum{j in J, d in D} A[p,j,d] >= D_count-1;
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#s.t. min_load_person{p in P}: sum{j in J, d in D} A[p,j,d] >= min_load[p];
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/* A person does not perform the same job on all days */
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s.t. duplicate_jobs{p in P, j in J}: sum{d in D} A[p,j,d] <= D_count-1;
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/* Each task is allocated */
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s.t. all_allocated{j in J, d in D}: sum{p in P} A[p,j,d] == R[j, d];
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s.t. all_allocated{j in J, d in D}: sum{p in P} A[p,j,d] == R[d, j];
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/* A person only performs what (s)he is capable of */
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s.t. capability_person{p in P, j in J, d in D}: A[p,j,d] <= C[p,j];
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/* Maximize enjoyment */
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maximize enjoyment: sum{p in P, d in D, j in J} A[p,j,d] * (L[p, j] * WL - H[p, j] * WH);
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s.t. error_lt{p in P}: error[p] >= ((sum{j in J, d in D} A[p,j,d] * Wl[j]) - 4);
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s.t. error_gt{p in P}: error[p] >= 4 - (sum{j in J, d in D} A[p,j,d] * Wl[j]);
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/* Maximize enjoyment */
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# minimize error_diff: sum{p in P} error[p];
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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];
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solve;
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printf "Sum %d\n", enjoyment;
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printf{p in P, d in D, j in J : A[p,j,d] > 0} "%d %d %d %d\n", p, d, j, A[p,j,d] * (L[p, j] * WL - H[p, j] * WH);
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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));
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printf "p d j W l\n";
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printf ">>>>\n";
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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];
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printf "<<<<\n";
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printf "workloads\n";
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printf "p l\n";
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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));
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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;
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data;
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/* Test example */
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param P_count := 3;
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param J_count := 2;
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param D_count := 2;
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param WL := 3;
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param WH := 9;
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param P_count := 44;
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param J_count := 11;
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param D_count := 3;
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param WL := 1;
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param WH := 3;
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param Wl := 1 4, 2 4, 3 2, 4 2, 5 4, 6 3, 7 2, 8 1, 9 1, 10 2, 11 2;
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# de pol-parameter
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param max_load := 25,1,2; # 12,1,0 12,2,0 12,3,0;
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param R : 1 2 3 4 5 6 7 8 9 10 11 :=
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1 1 3 0 0 1 3 4 6 2 5 0
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2 1 0 6 4 1 3 4 6 2 5 5
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3 1 0 0 0 1 3 4 6 2 5 5;
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/* AUTOGEN STARTS HERE */
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param L : 5 6 7 8 9 10 :=
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1 0 1 0 1 0 0
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2 0 0 1 1 0 1
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3 0 0 1 0 0 1
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6 0 1 0 0 0 1
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7 0 0 0 1 0 1
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8 0 1 0 0 1 0
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9 0 1 1 0 0 1
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11 1 0 0 1 0 0
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13 0 1 0 0 1 0
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14 0 1 0 0 0 1
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16 0 1 0 0 0 0
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18 1 0 0 0 0 0
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19 0 0 1 0 1 1
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20 0 0 0 0 0 1
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21 0 0 0 1 1 0
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22 0 1 1 0 0 1
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23 0 1 0 1 0 0
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24 0 0 0 0 1 0
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25 1 0 0 0 0 0
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26 0 1 1 0 0 1
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27 0 1 0 0 0 1
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28 0 1 0 1 0 0
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29 0 0 1 1 0 0
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30 0 1 1 0 0 0
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31 0 1 0 0 0 0
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32 0 0 1 0 1 1
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34 0 0 0 0 1 1
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35 0 1 0 1 1 0
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36 0 1 0 0 0 1
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38 0 0 1 0 1 1
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39 0 0 1 0 0 1
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40 0 1 0 0 0 1
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41 0 1 1 0 0 1
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43 0 1 0 0 0 0
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44 0 0 0 1 0 0;
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param H : 6 7 8 9 10 :=
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1 0 0 0 0 1
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2 1 0 0 1 0
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6 0 1 0 0 0
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7 0 1 0 0 0
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8 0 1 1 0 0
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9 0 0 1 0 0
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11 0 1 0 1 0
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13 0 1 1 0 0
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14 0 0 1 1 0
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16 0 0 0 0 1
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18 0 1 0 0 1
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19 1 0 1 0 0
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21 1 1 0 0 0
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22 0 0 1 1 0
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23 0 1 0 0 0
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24 0 0 1 0 0
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25 0 1 0 0 0
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26 0 0 1 0 0
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27 0 0 1 0 0
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28 0 1 0 1 0
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29 0 0 0 0 1
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30 0 0 0 1 0
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31 0 0 0 1 0
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32 1 0 1 0 0
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34 0 0 1 0 0
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35 0 1 0 0 1
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36 0 1 0 1 0
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38 1 0 1 0 0
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39 1 0 1 0 0
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40 0 0 0 1 0
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41 0 0 1 1 0;
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param C : 1 2 3 5 6 9 :=
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1 0 0 0 0 1 0
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2 0 1 0 0 0 0
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3 0 0 0 0 0 0
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4 1 0 0 0 0 0
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5 1 0 1 0 0 0
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6 0 1 0 0 1 0
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7 0 0 0 0 0 0
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8 0 0 0 0 1 1
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9 1 0 0 0 1 0
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10 0 0 0 0 0 0
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11 0 0 0 1 0 0
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12 1 0 0 0 0 0
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13 1 0 0 0 1 1
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14 0 0 0 0 1 0
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15 0 0 0 0 0 0
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16 0 0 0 0 1 0
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17 0 0 0 0 0 0
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18 0 0 0 1 0 0
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19 0 0 0 0 0 1
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20 0 0 0 0 0 0
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21 0 0 1 0 0 1
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22 0 0 0 0 1 0
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23 0 0 0 0 1 0
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24 0 0 1 0 0 1
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25 0 1 0 1 0 0
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26 0 0 0 0 1 0
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27 0 0 0 0 1 0
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28 0 0 0 0 1 0
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29 0 0 0 0 0 0
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30 0 0 0 0 1 0
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31 0 0 1 0 1 0
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32 0 0 0 0 0 1
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33 0 0 0 0 0 0
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34 0 0 0 0 0 1
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35 0 0 0 0 1 1
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36 0 0 0 0 1 0
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37 0 0 0 0 0 0
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38 0 0 0 0 0 1
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39 0 0 0 0 0 0
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40 0 0 0 0 1 0
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41 0 0 0 0 1 0
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42 0 0 0 0 0 0
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43 0 0 1 0 1 0
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44 0 0 1 0 0 0;
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param Q := 5,1,1,1 25,5,1,1 18,5,2,1 11,5,3,1;
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param L : 1 2 :=
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1 0 1
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2 1 0
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3 0 0;
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param H : 1 2 :=
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1 1 0
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2 0 1
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3 0 1;
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param R : 1 2 :=
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1 1 2
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2 2 1;
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param C[3,2] := 0;
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end;
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