这里提供一种针对您的优化问题的方法(忽略您基于排列的方法)。
我将问题表述为混合整数问题,并使用专门的求解器计算出好的解决方案。
由于您的问题可能没有很好地形式化,因此可能需要进行一些修改。但总体的信息是:这个方法将很难被击败!
代码
import numpy as np
from cvxpy import *
""" Parameters """
N_POPULATION = 50
GROUPSIZES = [3, 6, 12, 12, 17]
assert sum(GROUPSIZES) == N_POPULATION
N_GROUPS = len(GROUPSIZES)
OBJ_FACTORS = [0.4, 0.1, 0.15, 0.35]
""" Create fake data """
age_vector = np.clip(np.random.normal(loc=35.0, scale=10.0, size=N_POPULATION).astype(int), 0, np.inf)
height_vector = np.clip(np.random.normal(loc=180.0, scale=15.0, size=N_POPULATION).astype(int), 0, np.inf)
weight_vector = np.clip(np.random.normal(loc=85, scale=20, size=N_POPULATION).astype(int), 0, np.inf)
skill_vector = np.random.randint(0, 100, N_POPULATION)
""" Calculate a-priori stats """
age_mean, height_mean, weight_mean, skill_mean = np.mean(age_vector), np.mean(height_vector), \
np.mean(weight_vector), np.mean(skill_vector)
""" Build optimization-model """
X = Bool(N_POPULATION, N_GROUPS)
D = Variable(4, N_GROUPS)
constraints = []
for p in range(N_POPULATION):
constraints.append(sum_entries(X[p, :]) == 1)
for g_ind, g_size in enumerate(GROUPSIZES):
constraints.append(sum_entries(X[:, g_ind]) == g_size)
group_deviations = [[], [], [], []]
for g_ind, g_size in enumerate(GROUPSIZES):
group_deviations[0].append((sum_entries(mul_elemwise(age_vector, X[:, g_ind])) / g_size) - age_mean)
group_deviations[1].append((sum_entries(mul_elemwise(height_vector, X[:, g_ind])) / g_size) - height_mean)
group_deviations[2].append((sum_entries(mul_elemwise(weight_vector, X[:, g_ind])) / g_size) - weight_mean)
group_deviations[3].append((sum_entries(mul_elemwise(skill_vector, X[:, g_ind])) / g_size) - skill_mean)
for i in range(4):
for g in range(N_GROUPS):
constraints.append(D[i,g] >= abs(group_deviations[i][g]))
obj_parts = [sum_entries(OBJ_FACTORS[i] * D[i, :]) for i in range(4)]
objective = Minimize(sum(obj_parts))
""" Build optimization-problem & solve """
problem = Problem(objective, constraints)
problem.solve(solver=GUROBI, verbose=True, TimeLimit=120)
print('Min-objective: ', problem.value)
""" Evaluate solution """
filled_groups = [[] for g in range(N_GROUPS)]
for g_ind, g_size in enumerate(GROUPSIZES):
for p in range(N_POPULATION):
if np.isclose(X[p, g_ind].value, 1.0):
filled_groups[g_ind].append(p)
for g_ind, g_size in enumerate(GROUPSIZES):
print('Group: ', g_ind, ' of size: ', g_size)
print(' ' + str(filled_groups[g_ind]))
group_stats = []
for g in range(N_GROUPS):
age_mean_in_group = age_vector[filled_groups[g]].mean()
height_mean_in_group = height_vector[filled_groups[g]].mean()
weight_mean_in_group = weight_vector[filled_groups[g]].mean()
skill_mean_in_group = skill_vector[filled_groups[g]].mean()
group_stats.append((age_mean_in_group, height_mean_in_group, weight_mean_in_group, skill_mean_in_group))
print('group-assignment solution means: ')
for g in range(N_GROUPS):
print(np.round(group_stats[g], 1))
""" Compare with input """
input_data = np.vstack((age_vector, height_vector, weight_vector, skill_vector))
print('input-means')
print(age_mean, height_mean, weight_mean, skill_mean)
print('input-data')
print(input_data)
输出(2分钟时间限制;商业求解器)
Time limit reached
Best objective 9.612058823514e-01, best bound 4.784117647059e-01, gap 50.2280
('Min-objective: ', 0.961205882351435)
('Group: ', 0, ' of size: ', 3)
[16, 20, 27]
('Group: ', 1, ' of size: ', 6)
[26, 32, 34, 45, 47, 49]
('Group: ', 2, ' of size: ', 12)
[0, 6, 10, 12, 15, 21, 24, 30, 38, 42, 43, 48]
('Group: ', 3, ' of size: ', 12)
[2, 3, 13, 17, 19, 22, 23, 25, 31, 36, 37, 40]
('Group: ', 4, ' of size: ', 17)
[1, 4, 5, 7, 8, 9, 11, 14, 18, 28, 29, 33, 35, 39, 41, 44, 46]
group-assignment solution means:
[ 33.3 179.3 83.7 49. ]
[ 33.8 178.2 84.3 49.2]
[ 33.9 178.7 83.8 49.1]
[ 33.8 179.1 84.1 49.2]
[ 34. 179.6 84.7 49. ]
input-means
(33.859999999999999, 179.06, 84.239999999999995, 49.100000000000001)
input-data
[[ 22. 35. 28. 32. 41. 26. 25. 37. 32. 26. 36. 36.
27. 34. 38. 38. 38. 47. 35. 35. 34. 30. 38. 34.
31. 21. 25. 28. 22. 40. 30. 18. 32. 46. 38. 38.
49. 20. 53. 32. 49. 44. 44. 42. 29. 39. 21. 36.
29. 33.]
[ 161. 158. 177. 195. 197. 206. 169. 182. 182. 198. 165. 185.
171. 175. 176. 176. 172. 196. 186. 172. 184. 198. 172. 162.
171. 175. 178. 182. 163. 176. 192. 182. 187. 161. 158. 191.
182. 164. 178. 174. 197. 156. 176. 196. 170. 197. 192. 171.
191. 178.]
[ 85. 103. 99. 93. 71. 109. 63. 87. 60. 94. 48. 122.
56. 84. 69. 162. 104. 71. 92. 97. 101. 66. 58. 69.
88. 69. 80. 46. 74. 61. 25. 74. 59. 69. 112. 82.
104. 62. 98. 84. 129. 71. 98. 107. 111. 117. 81. 74.
110. 64.]
[ 81. 67. 49. 74. 65. 93. 25. 7. 99. 34. 37. 1.
25. 1. 96. 36. 39. 41. 33. 28. 17. 95. 11. 80.
27. 78. 97. 91. 77. 88. 29. 54. 16. 67. 26. 13.
31. 57. 84. 3. 87. 7. 99. 35. 12. 44. 71. 43.
16. 69.]]
解决方案说明
- 这个解决方案看起来非常不错(关于平均偏差),而且只用了2分钟(我们事先决定了时间限制)
- 我们还得到了紧密的界限:
0.961是我们的解决方案;我们知道它不能低于4.784
可重复性
- 代码使用了numpy和cvxpy
- 使用了商业求解器
- 您可能需要使用非商业MIP求解器(支持提前终止的时间限制;获取当前最佳解决方案)
- 在cvxpy中支持的有效开源MIP求解器有:cbc(目前无法设置时间限制)和glpk(请查阅文档以获取时间限制支持)
模型决策
- 该代码使用L1范数惩罚,这会导致一个MIP问题
- 根据您的问题,使用L2范数惩罚可能是明智的(一个大偏差比许多小偏差更严重),这将导致一个更难的问题(MIQP / MISOCP)
{B,A},{C,D}和{C,D},{A,B}
都可以被抽样),但我怀疑这种方法的整体效率仍然很高(因为这些事件的概率很低)。该方法也非常简单。但是,每个有效的算法都取决于您的参数和成本函数,因此改进您的问题可能是明智的选择! - sascha