85 lines
2.6 KiB
Plaintext
85 lines
2.6 KiB
Plaintext
/* Number of people */
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param P_count, integer, > 0;
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/* Number of jobs */
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param J_count, integer, > 0;
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/* Number of days */
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param D_count, integer, > 0;
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param WL, integer, > 0;
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param WH, integer, > 0;
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param ML, integer, > 0;
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set P := 1..P_count;
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set J := 1..J_count;
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set D := 1..D_count;
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/* aanwezigheid x workload for that day */
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param Costs{p in P}, integer, >= 0;
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/* Person p likes to solve jobs j */
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param L{p in P, j in J} default 0, binary;
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/* Person p hates to solve jobs j */
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param H{p in P, j in J} default 0, binary;
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/* Person p is capable to perform job j */
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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{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] <= max_load[p,d];
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/* A person has at least 1 task */
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s.t. min_load_person{p in P}: sum{j in J, d in D} A[p,j,d] >= 1;
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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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s.t. max_load_person_total{p in P}: (sum{d in D, j in J} A[p,j,d] * Wl[j]) <= ML;
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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[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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s.t. error_lt{p in P}: error[p] >= ((sum{j in J, d in D} A[p,j,d] * Wl[j]) - Costs[p]);
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s.t. error_gt{p in P}: error[p] >= Costs[p] - (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", (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]) - Costs[p]);
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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]) - Costs[p])^2;
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end;
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