.. _paintProduction: Paint production ================ The problem description in this section is taken from Section 7.5 *Paint production* of the book *Applications of optimization with Xpress-MP*. As a part of its weekly production a paint company produces five batches of paints, always the same, for some big clients who have a stable demand. Every paint batch is produced in a single production process, all in the same blender that needs to be cleaned between every two batches. The durations of blending paint batches 1 to 5 are respectively 40, 35, 45, 32, and 50 minutes. The cleaning times depend on the colors and the paint types. For example, a long cleaning period is required if an oil-based paint is produced after a water-based paint, or to produce white paint after a dark color. The times are given in minutes in the following table: +------------+----+----+----+----+----+ | CLEAN(i,j) | 1 | 2 | 3 | 4 | 5 | +------------+----+----+----+----+----+ | 1 | 0 | 11 | 7 | 13 | 11 | +------------+----+----+----+----+----+ | 2 | 5 | 0 | 13 | 15 | 15 | +------------+----+----+----+----+----+ | 3 | 13 | 15 | 0 | 23 | 11 | +------------+----+----+----+----+----+ | 4 | 9 | 13 | 5 | 0 | 3 | +------------+----+----+----+----+----+ | 5 | 3 | 7 | 7 | 7 | 0 | +------------+----+----+----+----+----+ Since the company also has other activities, it wishes to deal with this weekly production in the shortest possible time (blending and cleaning). Which is the corresponding order of paint batches? The order will be applied every week, so the cleaning time between the last batch of one week and the first of the following week needs to be counted for the total duration of cleaning. Formulation of model 1 ---------------------- As for the problem in :ref:section 11.3  we are going to present two alternative model formulations. The first one is closer to the Mathematical Programming formulation in *Applications of optimization with Xpress-MP*, the second uses a two-dimensional element constraint. Let :math:JOBS = \{1, ..., NJ\} be the set of batches to produce, :math:DUR_j the processing time for batch :math:j, and :math:CLEAN_{i,j} the cleaning time between the consecutive batches :math:i and :math:j. We introduce decision variables :math:succ_j taking their values in :math:JOBS, to indicate the successor of every job, and variables :math:clean_j for the duration of the cleaning after every job. The cleaning time after every job is obtained by indexing :math:CLEAN_{i,j} with the value of :math:succ_j. We thus have the following problem formulation: .. math:: &\text{minimize } \sum_{j \in JOBS}(DUR_j + clean_j)\\ &\forall j \in JOBS : succ_j \in {JOBS} \backslash \{j\}\\ &\forall j \in JOBS : clean_j = CLEAN_{j,succ_j}\\ &\text{all-different}(\bigcup\limits_{j \in JOBS}^{} succ_j)\\ The objective function sums up the processing and cleaning times of all batches. The last (*all-different*) constraint guarantees that every batch occurs exactly once in the production sequence. Unfortunately, this model does not guarantee that the solution forms a single cycle. Solving it indeed results in a total duration of 239 with an invalid solution that contains two sub-cycles 1 :math:\rightarrow 3 :math:\rightarrow 2 :math:\rightarrow 1 and 4 :math:\rightarrow 5 :math:\rightarrow 4. A first possibility is to add a disjunction excluding this solution to our model and re-solve it iteratively until we reach a solution without sub-cycles. .. math:: &succ_1 \ne 3 \vee succ_3 \ne 2 \vee succ_2 \ne 1 \vee succ_1 \ne 5 \vee succ_5 \ne 4\\ However, this procedure is likely to become impractical with larger data sets since it may potentially introduce an extremely large number of disjunctions. We therefore choose a different, a-priori formulation of the sub-cycle elimination constraints with a variable :math:y_j per batch and :math:(NJ-1)^2 implication constraints. .. math:: &\forall j \in JOBS : rank_j \in \{1, ..., NJ\}\\ &\forall i \in JOBS, \forall j \in \{2, ..., NJ\}, i \ne j : succ_i = j \Rightarrow y_j = y_i + 1\\ &\forall j \in JOBS : rank_j \in \{1, ..., NJ\}\\ The variables :math:y_j correspond to the position of job :math:j in the production cycle. With these constraints, job 1 always takes the first position. Implementation of model 1 ------------------------- The implementation of the model formulated in the previous section is quite straightforward. The sub-cycle elimination constraints are implemented as logic relations with implies (a stronger formulation of these constraints is obtained by replacing the implications by equivalences, using KEquiv). .. tabs:: .. code-tab:: c++ // Number of paint batches (=jobs) int NJ = 5; // Durations of jobs int DUR[] = {40, 35, 45, 32, 50}; // Cleaning times between jobs int CLEAN[5][5] = { { 0, 11, 7, 13, 11 }, { 5, 0, 13, 15, 15 }, { 13, 15, 0, 23, 11 }, { 9, 13, 5, 0, 3 }, { 3, 7, 7, 7, 0 } }; // Cleaning times after a batch KIntArray CB; // Successor of a batch KIntVarArray succ; // Cleaning time after batches KIntVarArray clean; // Variables for excluding subtours KIntVarArray y; // Objective variable KIntVar * cycleTime; // Creation of the problem in this session KProblem problem(session,"B-5 Paint production"); // variables creation char name[80]; int j,i; for (j=0;jprintResume(); // Solution printing printf("Minimum cycle time: %i\n", sol->getValue(*cycleTime)); printf("Sequence of batches:\nBatch Duration Cleaning\n"); int first=0; do { printf("%i\t%i\t%i\n", first, DUR[first],sol->getValue(clean[first])); first = sol->getValue(succ[first]); } while (first!=0); .. code-tab:: py from kalis import * ### Data # Number of paint batches (=jobs) nb_jobs = 5 # Durations of jobs jobs_durations = [40, 35, 45, 32, 50] # Cleaning times between jobs jobs_cleaning_times = [[0, 11, 7, 13, 11], [5, 0, 13, 15, 15], [13, 15, 0, 23, 11], [9, 13, 5, 0, 3], [3, 7, 7, 7, 0]] max_cleaning_time = max(max(jobs_cleaning_times[j]) for j in range(nb_jobs)) ### Creation of the problem # Creation of the Kalis session session = KSession() # Creation of the optimization problem problem = KProblem(session, "B-5 Paint production") ### Variables creation # Successor of a batch batch_successor = KIntVarArray() for j in range(nb_jobs): batch_successor += KIntVar(problem, "succ(%d)" % j, 0, nb_jobs - 1) batch_successor[j].remVal(j) # Cleaning times after a batch batch_cleaning_time = KIntVarArray() for j in range(nb_jobs): batch_cleaning_time += KIntVar(problem, "clean(%d)" % j, 0, max_cleaning_time) # Variables for excluding subtours job_position_in_cycle = KIntVarArray() # i.e. y_j variables for j in range(nb_jobs): job_position_in_cycle += KIntVar(problem, "y(%d)" % j, 0, nb_jobs - 1) ### Constraints creation # Cleaning time after every batch: # i.e. "batch_cleaning_time[j] == jobs_cleaning_times[batch_succesor[j]]" for j in range(nb_jobs): K_cleaning_time = KIntArray() # convert Python list to KIntArray for i in range(nb_jobs): res = K_cleaning_time.add(jobs_cleaning_times[j][i]) kelt = KEltTerm(K_cleaning_time, batch_successor[j]) problem.post(kelt == batch_cleaning_time[j]) # One successor and on predecessor per batch problem.post(KAllDifferent("alldiff", batch_successor)) # Exclude subtours for i in range(nb_jobs): for j in range(1, nb_jobs): if i != j: problem.post( KGuard(batch_successor[i] == j, job_position_in_cycle[j] == (job_position_in_cycle[i] + 1))) # Set objective cycle_time = KIntVar(problem, "cycle_time", 0, 1000) batch_cleaning_times_sum = 0 for j in range(nb_jobs): batch_cleaning_times_sum += batch_cleaning_time[j] problem.post(cycle_time == (batch_cleaning_times_sum + sum(jobs_durations))) # First propagation to check inconsistency if problem.propagate(): print("Problem is infeasible") sys.exit(1) ### Solve the problem # Set the solver solver = KSolver(problem) # Setting objective and sense of optimization problem.setSense(KProblem.Minimize) problem.setObjective(cycle_time) # Run optimization result = solver.optimize() # Solution printing if result: solution = problem.getSolution() solution.printResume() print("Minimum cycle time: %d" % solution.getValue(cycle_time)) print("Sequence of batches:") print("Batch Duration Cleaning") j = solution.getValue(batch_successor[0]) while j != 0: print("%d\t%d\t%d" % (j, jobs_durations[j], solution.getValue(batch_cleaning_time[j]))) j = solution.getValue(batch_successor[j]) .. code-tab:: java // Number of paint batches (=jobs) int NJ = 5; // Durations of jobs int[] DUR = {40, 35, 45 ,32 ,50}; // Cleaning times between jobs int[][] CLEAN = { { 0, 11, 7, 13, 11 }, { 5, 0, 13, 15, 15 }, { 13, 15, 0, 23, 11 }, { 9, 13, 5, 0, 3 }, { 3, 7, 7, 7, 0 } }; // Cleaning times after a batch KIntArray CB = new KIntArray(); // Successor of a batch KIntVarArray succ = new KIntVarArray(); // Cleaning time after batches KIntVarArray clean = new KIntVarArray(); // Variables for excluding subtours KIntVarArray y = new KIntVarArray(); // Objective variable KIntVar cycleTime = new KIntVar(); // Creation of the problem in this session KProblem problem = new KProblem(session,"B-5 Paint production"); // variables creation int j,i; for (j=0;j. As before, let :math:JOBS = \{1, ..., NJ\} be the set of batches to produce, :math:DUR_j the processing time for batch :math:j, and :math:CLEAN_{i,j} the cleaning time between the consecutive batches :math:i and :math:j. We introduce decision variables :math:rank_k taking their values in :math:JOBS, for the number of the job in position :math:k. Variables :math:clean_k with :math:k \in JOBS now denote the duration of the :math:k^{\text{th}} cleaning time. This duration is obtained by indexing :math:CLEAN_{i,j} with the values of two consecutive :math:rank_k variables. We thus have the following problem formulation. .. math:: &\text{minimize } \sum_{j \in JOBS}DUR_j + \sum_{k \in JOBS}clean_k\\ &\forall k \in JOBS : rank_k \in {JOBS} \backslash \{k\}\\ &\forall k \in \{1, ..., NJ-1\} : clean_k = CLEAN_{rank_k,rank_{k+1}}\\ & clean_{NJ} = CLEAN_{rank_{NJ},rank_{1}}\\ &\text{all-different}(\bigcup\limits_{k \in JOBS}^{} rank_k)\\ As in model 1, the objective function sums up the processing and cleaning times of all batches. Although not strictly necessary from the mathematical point of view, we use different sum indices for durations and cleaning times to show the difference between summing over jobs or job positions. We now have an all-different constraint over the rank variables to guarantee that every batch occurs exactly once in the production sequence. Implementation of model 2 ------------------------- The implementation of the second model uses the 2-dimensional version of the KElement` constraint in Artelys Kalis. .. tabs:: .. code-tab:: c++ // Number of paint batches (=jobs) int NJ = 5; // Durations of jobs int DUR[] = {40, 35, 45 ,32 ,50}; // Cleaning times between jobs KIntMatrix CLEAN(5,5,0,"CLEAN"); CLEAN[0][0] = 0;CLEAN[1][0] = 11;CLEAN[2][0] = 7;CLEAN[3][0] = 13;CLEAN[4][0] = 11; CLEAN[0][1] = 5;CLEAN[1][1] = 0;CLEAN[2][1] = 13;CLEAN[3][1] = 15;CLEAN[4][1] = 15; CLEAN[0][2] = 13;CLEAN[1][2] = 15;CLEAN[2][2] = 0;CLEAN[3][2] = 23;CLEAN[4][2] = 11; CLEAN[0][3] = 9;CLEAN[1][3] = 13;CLEAN[2][3] = 5;CLEAN[3][3] = 0;CLEAN[4][3] = 3; CLEAN[0][4] = 3;CLEAN[1][4] = 7;CLEAN[2][4] = 7;CLEAN[3][4] = 7;CLEAN[4][4] = 0; // Successor of a batch KIntVarArray rank; // Cleaning time after batches KIntVarArray clean; // Objective variable KIntVar * cycleTime; // Creation of the problem in this session KProblem problem(session,"B-5 Paint production"); // variables creation char name[80]; int k,j,i; for (j=0;jprintResume(); // Solution printing printf("Minimum cycle time: %i\n", sol->getValue(*cycleTime)); printf("Sequence of batches:\nBatch Duration Cleaning\n"); for (k=0;kgetValue(rank[k]), DUR[sol->getValue(rank[k])], sol->getValue(clean[k])); } .. code-tab:: py ### Data # Number of paint batches (=jobs) nb_jobs = 5 # Durations of jobs jobs_durations = [40, 35, 45, 32, 50] # Cleaning times between jobs jobs_cleaning_times = [[0, 11, 7, 13, 11], [5, 0, 13, 15, 15], [13, 15, 0, 23, 11], [9, 13, 5, 0, 3], [3, 7, 7, 7, 0]] max_cleaning_time = max(max(jobs_cleaning_times[j]) for j in range(nb_jobs)) # Setting data as a KIntMatrix K_cleaning_times_matrix = KIntMatrix(5, 5, 0, "CLEAN") for i in range(nb_jobs): for j in range(nb_jobs): K_cleaning_times_matrix.setMatrix(i, j, jobs_cleaning_times[i][j]) ### Creation of the problem # Creation of the Kalis session session = KSession() # Creation of the problem in this session problem = KProblem(session, "B-5 Paint production") ### Variable creation # Successor of a batch batch_rank = KIntVarArray() for j in range(nb_jobs): batch_rank += KIntVar(problem, "rank(%d)" % j, 0, nb_jobs - 1) # Cleaning time after batches batch_cleaning_time = KIntVarArray() for j in range(nb_jobs): batch_cleaning_time += KIntVar(problem, "clean(%d)" % j, 0, max_cleaning_time) ### Constraints creation # Set cleaning times after every batch for k in range(nb_jobs): if k < nb_jobs - 1: kelt = KEltTerm2D(K_cleaning_times_matrix, batch_rank[k], batch_rank[k + 1]) problem.post(KElement2D(kelt, batch_cleaning_time[k], "element")) else: kelt = KEltTerm2D(K_cleaning_times_matrix, batch_rank[k], batch_rank[0]) problem.post(KElement2D(kelt, batch_cleaning_time[k], "element")) # One position for every job problem.post(KAllDifferent("alldiff", batch_rank)) # Set objective cycle_time = KIntVar(problem, "cycle_time", 0, 1000) batch_cleaning_times_sum = 0 for j in range(nb_jobs): batch_cleaning_times_sum += batch_cleaning_time[j] problem.post(cycle_time == (batch_cleaning_times_sum + sum(jobs_durations))) # First propagation to check inconsistency if problem.propagate(): print("Problem is infeasible") sys.exit(1) ### Solve the problem # Set the solver solver = KSolver(problem) # Setting objective and sense of optimization problem.setSense(KProblem.Minimize) problem.setObjective(cycle_time) # Run optimization result = solver.optimize() # Solution printing if result: solution = problem.getSolution() solution.printResume() print("Minimum cycle time: %d" % solution.getValue(cycle_time)) print("Sequence of batches:") print("Batch Duration Cleaning") for k in range(nb_jobs): job_id = solution.getValue(batch_rank[k]) print("%d\t%d\t%d" % (job_id, jobs_durations[job_id], solution.getValue(batch_cleaning_time[k]))) .. code-tab:: java // Cleaning times between jobs KIntMatrix CLEAN = new KIntMatrix(5, 5, 0, "CLEAN"); CLEAN.setMatrix(0, 0, 0); CLEAN.setMatrix(1,0,11); CLEAN.setMatrix(2,0,7); CLEAN.setMatrix(3,0,13); CLEAN.setMatrix(4,0,11); CLEAN.setMatrix(0,1,5); CLEAN.setMatrix(1,1,0); CLEAN.setMatrix(2,1,13); CLEAN.setMatrix(3,1,15); CLEAN.setMatrix(4,1,15); CLEAN.setMatrix(0,2,13); CLEAN.setMatrix(1,2,15); CLEAN.setMatrix(2,2,0); CLEAN.setMatrix(3,2,23); CLEAN.setMatrix(4,2,11); CLEAN.setMatrix(0,3,9); CLEAN.setMatrix(1,3,13); CLEAN.setMatrix(2,3,5); CLEAN.setMatrix(3,3,0); CLEAN.setMatrix(4,3,3); CLEAN.setMatrix(0,4,3); CLEAN.setMatrix(1,4,7); CLEAN.setMatrix(2,4,7); CLEAN.setMatrix(3,4,7); CLEAN.setMatrix(4,4,0); // Successor of a batch KIntVarArray rank = new KIntVarArray(); // Cleaning time after batches KIntVarArray clean = new KIntVarArray(); // Objective variable KIntVar cycleTime = new KIntVar(); // Creation of the problem in this session KProblem problem = new KProblem(session, "B-5 Paint production"); // variables creation int k,j; for (j=0; j