python混合整数线性规划_Python 实现整数线性规划：分枝定界法（Branch and Bound）...
今天做作业，要实现整数线性规划的分枝定界法算法。找了一些网上的博客，发现都很屎，感觉自己写的这个比较清楚、规范，所以在此记录。如有错误，请指正。
from scipy.optimize import linprog
import numpy as np
import math
import sys
from queue import Queue
class ILP():
def __init__(self, c, A_ub, b_ub, A_eq, b_eq, bounds):
# 全局参数
self.LOWER_BOUND = -sys.maxsize
self.UPPER_BOUND = sys.maxsize
self.opt_val = None
self.opt_x = None
self.Q = Queue()
# 这些参数在每轮计算中都不会改变
self.c = -c
self.A_eq = A_eq
self.b_eq = b_eq
self.bounds = bounds
# 首先计算一下初始问题
r = linprog(-c, A_ub, b_ub, A_eq, b_eq, bounds)
# 若最初问题线性不可解
if not r.success:
raise ValueError('Not a feasible problem!')
# 将解和约束参数放入队列
self.Q.put((r, A_ub, b_ub))
def solve(self):
while not self.Q.empty():
# 取出当前问题
res, A_ub, b_ub = self.Q.get(block=False)
# 当前最优值小于总下界，则排除此区域
if -res.fun < self.LOWER_BOUND:
continue
# 若结果 x 中全为整数，则尝试更新全局下界、全局最优值和最优解
if all(list(map(lambda f: f.is_integer(), res.x))):
if self.LOWER_BOUND < -res.fun:
self.LOWER_BOUND = -res.fun
if self.opt_val is None or self.opt_val < -res.fun:
self.opt_val = -res.fun
self.opt_x = res.x
continue
# 进行分枝
else:
# 寻找 x 中第一个不是整数的，取其下标 idx
idx = 0
for i, x in enumerate(res.x):
if not x.is_integer():
break
idx += 1
# 构建新的约束条件(分割
new_con1 = np.zeros(A_ub.shape[1])
new_con1[idx] = -1
new_con2 = np.zeros(A_ub.shape[1])
new_con2[idx] = 1
new_A_ub1 = np.insert(A_ub, A_ub.shape[0], new_con1, axis=0)
new_A_ub2 = np.insert(A_ub, A_ub.shape[0], new_con2, axis=0)
new_b_ub1 = np.insert(
b_ub, b_ub.shape[0], -math.ceil(res.x[idx]), axis=0)
new_b_ub2 = np.insert(
b_ub, b_ub.shape[0], math.floor(res.x[idx]), axis=0)
# 将新约束条件加入队列，先加最优值大的那一支
r1 = linprog(self.c, new_A_ub1, new_b_ub1, self.A_eq,
self.b_eq, self.bounds)
r2 = linprog(self.c, new_A_ub2, new_b_ub2, self.A_eq,
self.b_eq, self.bounds)
if not r1.success and r2.success:
self.Q.put((r2, new_A_ub2, new_b_ub2))
elif not r2.success and r1.success:
self.Q.put((r1, new_A_ub1, new_b_ub1))
elif r1.success and r2.success:
if -r1.fun > -r2.fun:
self.Q.put((r1, new_A_ub1, new_b_ub1))
self.Q.put((r2, new_A_ub2, new_b_ub2))
else:
self.Q.put((r2, new_A_ub2, new_b_ub2))
self.Q.put((r1, new_A_ub1, new_b_ub1))
def test1():
""" 此测试的真实最优解为 [4, 2] """
c = np.array([40, 90])
A = np.array([[9, 7], [7, 20]])
b = np.array([56, 70])
Aeq = None
beq = None
bounds = [(0, None), (0, None)]
solver = ILP(c, A, b, Aeq, beq, bounds)
solver.solve()
print("Test 1's result:", solver.opt_val, solver.opt_x)
print("Test 1's true optimal x: [4, 2]\n")
def test2():
""" 此测试的真实最优解为 [2, 4] """
c = np.array([3, 13])
A = np.array([[2, 9], [11, -8]])
b = np.array([40, 82])
Aeq = None
beq = None
bounds = [(0, None), (0, None)]
solver = ILP(c, A, b, Aeq, beq, bounds)
solver.solve()
print("Test 2's result:", solver.opt_val, solver.opt_x)
print("Test 2's true optimal x: [2, 4]\n")
if __name__ == '__main__':
test1()
test2()
运行结果截图：