C-Optを使ってGurobiモデルを解くためのwrapper

C-Optは中国製のGurobiもどきだ．ベンチマークでは２位が多い．マニュアルはここにあり，学生は無料で１年使える．日本からでもOK．

https://guide.coap.online/copt/en-doc/index.html#

Cardinal Optimizer (COPT) User Guide — Cardinal Optimizer (COPT) User~Guide, Ver6.5

以下のコードをまずまわしておく．

#c-opt wrapper
import coptpy as cp
from coptpy import COPT, quicksum, tuplelist, multidict, tupledict

GRB = COPT

class Model():
    def __init__(self, name=""):
        env = cp.Envr()
        self.ObjVal = None
        self.model = env.createModel(name)
        
    def addVar(self, *args, **kwargs):
        return self.model.addVar(*args, **kwargs)
    
    def addConstr(self, *args, **kwargs):
        return self.model.addConstr(*args, **kwargs)

    def setObjective(self, *args, **kwargs):
        return self.model.setObjective(*args, **kwargs)
    
    def getVars(self, *args, **kwargs):
        return self.model.getVars(*args, **kwargs)
    
    def getConstrs(self, *args, **kwargs):
        return self.model.getConstrs(*args, **kwargs)
    
    def getSOS(self, *args, **kwargs):
        return self.model.SOS(*args, **kwargs)
    
    def setParam(self, *args, **kwargs):
        return self.model.setParam(*args, **kwargs)
    
    def optimize(self):
        ret = self.model.solve()
        self.ObjVal = self.model.objval 
        self.status = self.model.status
        return ret  

例として多制約ナップサックの難しい問題例を作って，Gurobiモデルを書いて解く．

a = {}
b, v = [], []
m = 5
n = 100
for i in range(m):
    sumA = 0.0
    for j in range(n):
        a[i, j] = 1.0 - math.log(random.random(), 2)
        sumA += a[i, j]
    b.append(0.25 * sumA)
for j in range(n):
    sumA = 0.0
    for i in range(m):
        sumA += a[i, j]
    v.append(10.0 * sumA / m + 10.0 * random.random())

J = list(range(n))
I = list(range(m))
model = Model("mkp")
x = {}
for j in J:
    x[j] = model.addVar(vtype="B", name=f"x({j})")
#model.update()

constr = {}
for i in I:
    for j in J:
        constr[i,j] = model.addConstr(quicksum(a[i, j] * x[j] for j in J) <= b[i], f"Capacity({i},{j})")

model.setParam(GRB.Param.TimeLimit, 10.0)

model.setObjective(quicksum(v[j] * x[j] for j in J), GRB.MAXIMIZE)

model.optimize()

すると C-Optで解ける．

解の確認はこうする．

if model.status == GRB.OPTIMAL:
    print("Objective value =", model.ObjVal)
    print("Solution:")
    for j in x:
        print(f"x[{j}]={x[j].X}")