release 2.16.196

Author: vlkong

examples/cp/basic/plant_location_with_cpo_callback.py

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+# --------------------------------------------------------------------------
+# Source file provided under Apache License, Version 2.0, January 2004,
+# http://www.apache.org/licenses/
+# (c) Copyright IBM Corp. 2015, 2016, 2018, 2020
+# --------------------------------------------------------------------------
+
+"""
+A ship-building company has a certain number of customers. Each customer is supplied
+by exactly one plant. In turn, a plant can supply several customers. The problem is
+to decide where to set up the plants in order to supply every customer while minimizing
+the cost of building each plant and the transportation cost of supplying the customers.
+
+For each possible plant location there is a fixed cost and a production capacity.
+Both take into account the country and the geographical conditions.
+
+For every customer, there is a demand and a transportation cost with respect to
+each plant location.
+
+While a first solution of this problem can be found easily by CP Optimizer, it can take
+quite some time to improve it to a very good one. We illustrate the warm start capabilities
+of CP Optimizer by giving a good starting point solution that CP Optimizer will try to improve.
+This solution could be one from an expert or the result of another optimization engine
+applied to the problem.
+
+In the solution we only give a value to the variables that determine which plant delivers
+a customer. This is sufficient to define a complete solution on all model variables.
+CP Optimizer first extends the solution to all variables and then starts to improve it.
+
+The solve is enriched with a CPO callback, available from version of COS greater or equal to 12.10.0.0.
+This callback displays various information generated during the solve, in particular intermediate
+solutions that are found before the end of the solve.
+"""
+
+from docplex.cp.model import CpoModel
+import docplex.cp.solver.solver as solver
+from docplex.cp.utils import compare_natural
+from collections import deque
+import os
+from docplex.cp.solver.cpo_callback import CpoCallback
+
+
+#-----------------------------------------------------------------------------
+# Initialize the problem data
+#-----------------------------------------------------------------------------
+
+# Read problem data from a file and convert it as a list of integers
+filename = os.path.dirname(os.path.abspath(__file__)) + "/data/plant_location.data"
+data = deque()
+with open(filename, "r") as file:
+    for val in file.read().split():
+        data.append(int(val))
+
+# Read number of customers and locations
+nbCustomer = data.popleft()
+nbLocation = data.popleft()
+
+# Initialize cost. cost[c][p] = cost to deliver customer c from plant p
+cost = list([list([data.popleft() for l in range(nbLocation)]) for c in range(nbCustomer)])
+
+# Initialize demand of each customer
+demand = list([data.popleft() for c in range(nbCustomer)])
+
+# Initialize fixed cost of each location
+fixedCost = list([data.popleft() for p in range(nbLocation)])
+
+# Initialize capacity of each location
+capacity = list([data.popleft() for p in range(nbLocation)])
+
+
+#-----------------------------------------------------------------------------
+# Build the model
+#-----------------------------------------------------------------------------
+
+mdl = CpoModel()
+
+# Create variables identifying which location serves each customer
+cust = mdl.integer_var_list(nbCustomer, 0, nbLocation - 1, "CustomerLocation")
+
+# Create variables indicating which plant location is open
+open = mdl.integer_var_list(nbLocation, 0, 1, "OpenLocation")
+
+# Create variables indicating load of each plant
+load = [mdl.integer_var(0, capacity[p], "PlantLoad_" + str(p)) for p in range(nbLocation)]
+
+# Associate plant openness to its load
+for p in range(nbLocation):
+      mdl.add(open[p] == (load[p] > 0))
+
+# Add constraints
+mdl.add(mdl.pack(load, cust, demand))
+
+# Add objective
+obj = mdl.scal_prod(fixedCost, open)
+for c in range(nbCustomer):
+    obj += mdl.element(cust[c], cost[c])
+mdl.add(mdl.minimize(obj))
+
+
+#-----------------------------------------------------------------------------
+# Solve the model, tracking objective with a callback
+#-----------------------------------------------------------------------------
+
+class MyCallback(CpoCallback):
+    def invoke(self, solver, event, jsol):
+        # Get important elements
+        obj_val = jsol.get_objective_values()
+        obj_bnds = jsol.get_objective_bounds()
+        obj_gaps = jsol.get_objective_gaps()
+        solvests = jsol.get_solve_status()
+        srchsts = jsol.get_search_status()
+        #allvars = jsol.get_solution().get_all_var_solutions() if jsol.is_solution() else None
+        solve_time = jsol.get_info('SolveTime')
+        memory = jsol.get_info('MemoryUsage')
+        print("CALLBACK: {}: {}, {}, objective: {} bounds: {}, gaps: {}, time: {}, memory: {}".format(event, solvests, srchsts, obj_val, obj_bnds, obj_gaps, solve_time, memory))
+
+if compare_natural(solver.get_solver_version(), '12.10') >= 0:
+    mdl.add_solver_callback(MyCallback())
+
+# Solve the model
+print("Solve the model")
+msol = mdl.solve(TimeLimit=10)
+msol.write()

examples/cp/basic/trimloss.py

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+# --------------------------------------------------------------------------
+# Source file provided under Apache License, Version 2.0, January 2004,
+# http://www.apache.org/licenses/
+# (c) Copyright IBM Corp. 2020
+# --------------------------------------------------------------------------
+
+"""
+The trim loss problems arises in the paper industry.
+
+The problem is to cut wide papers rolls into sub rolls (orders).
+The wide roll are cut into pieces with a cutting pattern.
+
+A cutting pattern defines the blades positions for cutting the roll.
+A maximum number of orders is allowed in a cutting pattern (here it is 6).
+When cutting a wide roll, we can have a loss of paper that is wasted.
+This loss is contrained to be not more than a given value (here it is 100)
+
+An order is characterised by a demand, a roll width, and a maximum number of
+time it can appear in a cutting pattern.
+
+The goal is to meet the demand while minimizing the roll used and the number
+of different cutting patterns used for production.
+
+In this example we also use:
+- extra constraints to avoid assigning orders to unused patterns,
+- lexicographic constraints to break symmetries between cutting patterns
+- strong constraints to have a better domain reduction by enumerating possible
+  patterns configurations
+All this makes the proof of optimality rather fast.
+"""
+
+from docplex.cp.model import *
+from sys import stdout
+
+#-----------------------------------------------------------------------------
+# Initialize the problem data
+#-----------------------------------------------------------------------------
+
+# Data
+ROLL_WIDTH    = 2200    # Width of roll to be cutted into pieces
+MAX_WASTE     = 100     # Maximum waste per roll
+MAX_ORDER_PER_CUT = 5   # Maximum number of order per cutting pattern 
+
+# Orders demand, width and max occurence in a cutting pattern
+ORDER_DEMAND     = (  8,  16,  12,   7,  14,  16)
+ORDER_WIDTH      = (330, 360, 380, 430, 490, 530)
+ORDER_MAX_REPEAT = (  2,   3,   3,   5,   3,   4)
+# Number of different order types 
+NUM_ORDER_TYPE = len(ORDER_DEMAND)
+# Maximum number of cutting pattern
+NUM_PATTERN_TYPE = 6
+# Maximum of time a cutting pattern is used
+MAX_PATTERN_USAGE = 16
+# Cost of using a pattern
+PATTERN_COST  = 0.1
+# Cost of a roll
+ROLL_COST = 1
+
+PATTERNS = range(NUM_PATTERN_TYPE)
+ORDERS   = range(NUM_ORDER_TYPE)
+
+
+#-----------------------------------------------------------------------------
+# Build the model
+#-----------------------------------------------------------------------------
+
+model = CpoModel()
+
+# Decision variables : pattern usage
+patternUsage = [model.integer_var(0, MAX_PATTERN_USAGE, "PatternUsage_"+str(p)) for p in PATTERNS]
+
+# Decision variables : order quantity per pattern
+x = [[model.integer_var(0, max, "x["+str(o)+","+str(p)+"]")
+        for (o, max) in enumerate(ORDER_MAX_REPEAT)]
+        for p in PATTERNS]
+
+# Maximum number of orders per cutting pattern
+for p in PATTERNS :
+    model.add(sum(x[p]) <= MAX_ORDER_PER_CUT)
+
+# Roll capacity
+usage = [0] + [v for v in range(ROLL_WIDTH - MAX_WASTE, ROLL_WIDTH+1)]   # usage is [0, 2100..2200]
+rollUsage = [model.integer_var(domain = usage, name = "RollUsage_"+str(p)) for p in PATTERNS]
+
+for p in PATTERNS :
+    model.add(sum(ORDER_WIDTH[o] * x[p][o] for o in ORDERS) == rollUsage[p])
+
+# Production requirement
+for o in ORDERS :
+    model.add(model.sum(x[p][o] * patternUsage[p] for p in PATTERNS) >= ORDER_DEMAND[o])
+
+# Objective
+model.add(minimize(model.sum((patternUsage[p] > 0) * PATTERN_COST + patternUsage[p] * ROLL_COST
+                             for p in PATTERNS)))
+
+# Extra constraint to avoid assigning orders to an unused pattern
+for p in PATTERNS :
+    model.add((patternUsage[p] == 0) == (rollUsage[p] == 0))
+
+# Extra lexicographic constraint to break symmetries
+for p in range(NUM_PATTERN_TYPE - 1) :
+    model.add(model.lexicographic([patternUsage[p]] + x[p], [patternUsage[p+1]] + x[p+1]))
+
+# Strong constraints to improve the time to prove optimality
+for p in PATTERNS :
+    model.add(model.strong(x[p]))
+
+# KPIs : Number of rolls, of pattern used and total loss of paper
+model.add_kpi(model.sum([patternUsage[p] for p in PATTERNS]), "Rolls")
+model.add_kpi(model.sum([(patternUsage[p] > 0) for p in PATTERNS]), "Patterns")
+model.add_kpi(model.sum([patternUsage[p] * (ROLL_WIDTH - rollUsage[p]) for p in PATTERNS]), "Loss")
+
+
+#-----------------------------------------------------------------------------
+# Solve the model and display the result
+#-----------------------------------------------------------------------------
+
+print("Solve the model...")
+msol = model.solve(LogPeriod=1000000, TimeLimit=300)
+if msol:
+    print("patternUsage = ")
+    for p in PATTERNS:
+        l = ROLL_WIDTH - msol[rollUsage[p]]
+        stdout.write("Pattern {} , usage = {}, roll usage = {}, loss = {}, orders =".format(p, msol[patternUsage[p]], msol[rollUsage[p]], l))
+        for o in ORDERS:
+            stdout.write(" {}".format(msol[x[p][o]]))
+        stdout.write('\n')
+else:
+    print("No solution found")