[英]Gekko returning incorrect successful solution
以下代码返回Successful Solution
Objective: 0.
。 但这不是最佳解决方案。 最佳解决方案是-6
。 通过阅读其他问题,我认为在目标 function 中使用非 Gekko 函数是一个问题,但我使用的唯一非 Gekko function 是np.matmul
。 np.matmul
是否适用于 gekko arrays? 注意sigma_post
是一个 nxn numpy 单位矩阵。
m = GEKKO(remote=False)
m.options.max_iter=1000
#m.options.ipopt_integer_tol=1
#m.solver_options = ['minlp_integer_tol 50']
#m.solver_options = ['minlp_max_iter_with_int_sol 1000',
# 'minlp_branch_method 1']
N = 2
b = m.Array(m.Var,(N,n), lb=0, ub=1, integer=True)
for i in range(N):
for j in range(n):
if j in [qb_index_range[0], rb_index_range[0], wr_index_range[0]]:
b[i][j].value = 1
else:
b[i][j].value = 0
print('b: ', b)
# CONSTRAINT: Each Lineup must be less than budget
z = np.array([None]*N)
for i in range(N):
z[i] = m.Intermediate(sum(b[i, :]*list(info_df['cost'])))
m.Equations([z[i] <= budget for i in range(N)])
# CONSTRAINT: Each Lineup has one QB
z_1 = np.array([None]*N)
for i in range(N):
z_1[i] = m.Intermediate(sum(b[i, qb_index_range[0]: qb_index_range[1]+1]))
m.Equations([z_1[i] == 1 for i in range(N)])
# CONSTRAINT: Each Lineup has one RB
z_2 = np.array([None]*N)
for i in range(N):
z_2[i] = m.Intermediate(sum(b[i, rb_index_range[0]: rb_index_range[1]+1]))
m.Equations([z_2[i] == 1 for i in range(N)])
# CONSTRAINT: Each Lineup has one WR
z_3 = np.array([None]*N)
for i in range(N):
z_3[i] = m.Intermediate(sum(b[i, wr_index_range[0]: wr_index_range[1]+1]))
m.Equations([z_3[i] == 1 for i in range(N)])
#OBJECTIVE: maximize with two lineups
sigma_1 = np.array([[None]*N for i in range(N)])
sig = np.matmul(np.matmul(b, sigma_post), b.T)
for i in range(N):
for j in range(N):
sigma_1[i][j] = m.Intermediate(sig[i][j])
m.Obj(-(sigma_1[0][0] + sigma_1[1][1]- 2*sigma_1[1][0]))
m.options.SOLVER = 1
m.solve(debug=0)
编辑:为了透明,理想情况下,我关心的目标 function 在下面,但上面详述的更简单的目标 function 会导致问题,所以我决定在那里开始故障排除。 以下目标抛出Warning: no more possible trial points and no integer solution Maximum iterations
mu
但mu
不存在于约束中。 非常感谢您的任何建议!
pi = 3.14159
eps = 1.0E-6
def normal_cdf(x, m):
return 1/(1+m.exp(-1.65451*x))
def normal_pdf(x, m):
return (1/((2*pi)**(.5)))*m.exp((x**2)/2)
def theta(s, m):
return m.sqrt(s[0][0]+s[1][1] - 2*s[0][1])
# OBJECTIVE: Maximize
mu_1 = np.array([None]*N)
for i in range(N):
mu_1[i] = m.Intermediate(np.matmul(b[i, :], mu))
inter = m.if2(theta(sigma_1, m)-eps, .5*mu_1[0]+.5*mu_1[1],
(mu_1[0]*normal_cdf((mu_1[0]-mu_1[1])/theta(sigma_1, m), m) + \
mu_1[1]*normal_cdf((mu_1[1]-mu_1[0])/theta(sigma_1, m), m) + \
theta(sigma_1, m)*normal_pdf((mu_1[0]-mu_1[1])/theta(sigma_1, m), m)))
m.Obj(-inter)
使用np.matmul
或任何其他允许对象而不是仅数值的 function 没有问题。 需要对象,因为b
是一个 Gekko 类型值的数组,需要这些值来计算具有自动微分的导数。 您还可以使用简化表达式的新@
运算符。 您最初的问题陈述不完整,缺少许多定义。 我添加了一些示例值,以便脚本可以在没有定义错误的情况下运行。 以下是帮助重现错误的指南。
N = 2
n = 3
qb_index_range = [0,2]
rb_index_range = [0,2]
wr_index_range = [0,2]
info_df = pd.DataFrame({'cost':np.ones(n)})
budget = 100
sigma_post = np.random.rand(n,n)
这是使用np.matmul()
的示例,它也可以是点积np.dot()
。
sigma_1 = np.matmul(np.matmul(b,sigma_post), b.T)
这也可以用矩阵乘法运算符来编写。
sigma_1 = b@sigma_post@b.T
这是完整的脚本。
from gekko import GEKKO
import numpy as np
import pandas as pd
m = GEKKO(remote=False)
m.options.max_iter=1000
N = 2
n = 3
b = m.Array(m.Var,(N,n), lb=0, ub=1, integer=True)
qb_index_range = [0,2]
rb_index_range = [0,2]
wr_index_range = [0,2]
info_df = pd.DataFrame({'cost':np.ones(n)})
budget = 100
sigma_post = np.eye(n)
for i in range(N):
for j in range(n):
if j in [qb_index_range[0], rb_index_range[0], wr_index_range[0]]:
b[i][j].value = 1
else:
b[i][j].value = 0
# CONSTRAINT: Each Lineup must be less than budget
z = [None]*N
for i in range(N):
z[i] = m.Intermediate(sum(b[i, :]*list(info_df['cost'])))
m.Equations([z[i] <= budget for i in range(N)])
# CONSTRAINT: Each Lineup has one QB
z_1 = [None]*N
for i in range(N):
z_1[i] = m.Intermediate(sum(b[i, qb_index_range[0]: qb_index_range[1]+1]))
m.Equations([z_1[i] == 1 for i in range(N)])
# CONSTRAINT: Each Lineup has one RB
z_2 = np.array([None]*N)
for i in range(N):
z_2[i] = m.Intermediate(sum(b[i, rb_index_range[0]: rb_index_range[1]+1]))
m.Equations([z_2[i] == 1 for i in range(N)])
# CONSTRAINT: Each Lineup has one WR
z_3 = np.array([None]*N)
for i in range(N):
z_3[i] = m.Intermediate(sum(b[i, wr_index_range[0]: wr_index_range[1]+1]))
m.Equations([z_3[i] == 1 for i in range(N)])
#OBJECTIVE: maximize with two lineups
#sigma_1 = np.matmul(np.matmul(b,sigma_post), b.T)
sigma_1 = b@sigma_post@b.T
m.Maximize(sigma_1[0][0] + sigma_1[1][1]- 2*sigma_1[1][0])
m.options.SOLVER = 1
m.solve(debug=0,disp=False)
print(b)
这产生了一个成功的解决方案。 由于原始问题陈述不完整,因此无法验证正确的解决方案。
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