Gurobi模型修改慢,我可以直接修改约束矩阵吗



我想更改现有模型中的系数。目前(使用Python API),我正在循环通过约束并调用model.chgCoeff,但速度相当慢。在Python和/或C API中,是否有更快速的方法可以直接访问约束矩阵?

下面的代码示例。它慢的原因似乎主要是因为循环本身;用任何其它操作替换CCD_ 2仍然是缓慢的。通常情况下,我会通过使用向量运算而不是循环来解决这个问题,但如果没有访问约束矩阵,我认为我无法做到这一点。

from __future__ import division
import gurobipy as gp
import numpy as np
import time
N = 300
M = 2000
m = gp.Model()
m.setParam('OutputFlag', False)
masks = [np.random.rand(N) for i in range(M)]
p = 1/np.random.rand(N)
rets = [p * masks[i] - 1 for i in range(M)]
v = np.random.rand(N)*10000 * np.round(np.random.rand(N))
t = m.addVar()
x = [m.addVar(vtype=gp.GRB.SEMICONT, lb=1000, ub=v[i]) for i in range(N)]
m.update()
cons = [m.addConstr(t <= gp.LinExpr(ret, x)) for ret in rets]
m.setObjective(t, gp.GRB.MAXIMIZE)
m.update()
start_time = time.time()
m.optimize()
solve_ms = int(((time.time() - start_time)*1000))
print('First solve took %s ms' % solve_ms)
p = 1/np.random.rand(N)
rets = [p * masks[i] - 1 for i in range(M)]
start_time = time.time()
for i in range(M):
    for j in range(N):
        if rets[i][j] != -1:
            m.chgCoeff(cons[i], x[j], -rets[i][j])
m.update()
update_ms = int(((time.time() - start_time)*1000))
print('Model update took %s ms' % update_ms)
start_time = time.time()
m.optimize()
solve_ms = int(((time.time() - start_time)*1000))
print('Second solve took %s ms' % solve_ms)
k = 2
start_time = time.time()
for i in range(M):
    for j in range(N):
        if rets[i][j] != -1:
            k *= rets[i][j]
solve_ms = int(((time.time() - start_time)*1000))
print('Plain loop took %s ms' % solve_ms)
R = np.array(rets)
start_time = time.time()
S = np.copy(R)
copy_ms = int(((time.time() - start_time)*1000))
print('np.copy() took %s ms' % copy_ms)

输出:

First solve took 1767 ms
Model update took 2051 ms
Second solve took 1872 ms
Plain loop took 1103 ms
np.copy() took 3 ms

在大小(2000300)约束矩阵上调用np.copy需要3ms。我错过的根本原因是整个模型更新不能那么快吗?

您不能使用Python接口在Gurobi中直接访问约束矩阵。即使可以,也不能执行np.copy操作,因为矩阵是CSR格式,而不是密集格式。要对约束进行大规模更改,最好通过删除约束并添加新的约束来修改约束矩阵。在您的情况下,对每个约束对象的更改都非常重要,因此您不会从一个温暖的开始中获得太多好处,因此不会因为不保留相同的约束对象而失去任何东西。

假设您在上面的代码中针对-1的特殊情况调整了rets数组,那么下面的代码将满足您的要求,并且速度会快得多。

for con in cons:               
    m.remove(con)
new_cons = [m.addConstr(t <= gp.LinExpr(ret, x)) for ret in rets]

最新更新