如何修复 Python Gekko 最大方程长度错误
How to fix Python Gekko Max Equation Length error
我的代码适用于小变量。但是当我在 128*128 变量数组中执行时，出现以下错误：
APM model error: string > 15000 characters Consider breaking
up the line into multiple equations The may also be due to only
using newline character CR instead of CR LF (for Windows) or LF
(for MacOS/Linux) To fix this problem, save APM file with
appropriate newline characters
我不知道如何解决这个错误。
假设gekko不能运行那样的大数组变量。
我把代码放在下面。
from gekko import GEKKO
import numpy as np
m = GEKKO()
m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations max', \
# minlp iterations with integer solution
'minlp_max_iter_with_int_sol max', \
# treat minlp as nlp
'minlp_as_nlp min', \
# nlp sub-problem max iterations
'nlp_maximum_iterations 5000', \
# 1 = depth first, 2 = breadth first
'minlp_branch_method 1', \
# maximum deviation from whole number
'minlp_integer_tol 0.0005', \
# covergence tolerance
'minlp_gap_tol 0.0001' ,\
'max_cpu_time 10min']
n = 128
p = 8
q = 16
x = m.Array(m.Var,(n,n),integer=True)
for i in range(n):
for j in range(n):
x[i,j].value = 1
x[i,j].lower = 0
x[i,j].upper = 1
s = 60
a = m.Array(m.Param,(x.shape[0],))
a_ = [172,282,10, 264, 287, 442, 393, 428, 484, 444, 344, 250, 293, 34, 473, 110, 338, 62,
250, 205, 81, 336, 249, 199, 328, 447, 408, 82, 357, 334, 181, 133, 378, 79, 292, 103,
494, 382, 10, 477, 237, 267, 337, 313, 395, 110, 114, 381, 52, 232, 457, 69, 167, 344,
363, 284, 136 ,240, 258, 449, 119, 317, 370, 404, 197, 102, 428, 238, 321, 103, 282, 37,
41, 86, 496 ,31, 148, 245, 78, 219, 37, 115, 378, 129, 37, 448, 415, 368, 238, 406,
408, 100, 112 ,289 ,205, 329, 338, 396, 494, 145, 355, 45, 5, 220, 411, 167, 85, 369,
426, 442, 406 ,217, 57, 176, 12, 368, 444, 227, 337, 63, 267, 216, 227, 182, 408, 116,
385, 140]
for i in range(len(a)):
a[i] = a_[i]
O1 = np.array(range(n))
O_= np.transpose(np.nonzero(O1+1))
O = np.zeros((x.shape[0],x.shape[1]),dtype=int)
for i in range(n):
for j in range(n):
O[i,O_[i]] = 1
d = m.Array(m.Param,(p,q))
for i in range(p):
for j in range(q):
d[i,j] = (round(p/2)-int(i/2))*4-(j%4)
#condition
m.Equation(m.sum(abs(x[:,:]-O[:,:]))<=2*s)
for i in range(n):
m.Equation(m.sum(x[:,i])==1)
for i in range(n):
m.Equation(m.sum(x[i,:])==1)
#Objective
m.Obj(np.sum(a.T*np.sum(x*d.reshape(1,n),axis=1)))
#Set global options
m.options.IMODE = 3
#Solve simulation
m.solve(disp=False,debug=True)
下面是一个应该可以工作的版本，如果你让它解决足够长的时间，尽管解决方案不太可能，因为你有 128*128 = 16384 个二进制变量。有 2^16384 个解决方案，如果整数变量的数量大于几百，APOPT 等混合整数非线性规划求解器不太可能快速找到解决方案。它可以解决具有数千个变量的连续变量，但混合整数比较困难，因为它必须使用分支定界法来解决它们。如果您可以使问题线性化（用松弛变量代替 abs 函数），您可能想尝试像 CPLEX 或 Gurobi 这样的 MILP 求解器。
以下是关于您的模型的一些建议：
- 不要对模型表达式使用
numpy。 np.abs 和 np.sum 等函数不提供求解器所需的信息。您应该使用 m.abs2（或 m.abs3）和 m.sum。
- 数组
a 和 d 可以是常量的 numpy 数组，而不是 Gekko 参数。
- 您可以将至少一个表达式分解为 Intermediates 以帮助缩短求解时间。
- objective 函数是
a.T x[:,:] d 的标量结果。您可以使用来自 a[i] x[i,j] d[j] 的多个 m.Obj 函数，而不是矩阵乘法产生的一个大表达式。
- 我建议您从较小的问题开始，然后在您对模型和解决方案有信心时扩大到更大的问题。
from gekko import GEKKO
import numpy as np
m = GEKKO()
m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 1000', \
# minlp iterations with integer solution
'minlp_max_iter_with_int_sol 100', \
# treat minlp as nlp
'minlp_as_nlp 1', \
# nlp sub-problem max iterations
'nlp_maximum_iterations 5000', \
# 1 = depth first, 2 = breadth first
'minlp_branch_method 1', \
# maximum deviation from whole number
'minlp_integer_tol 0.0005', \
# covergence tolerance
'minlp_gap_tol 0.0001']
n = 128
p = 8
q = 16
x = m.Array(m.Var,(n,n),integer=True)
for i in range(n):
for j in range(n):
x[i,j].value = 1
x[i,j].lower = 0
x[i,j].upper = 1
s = 60
# don't need a gekko array here, constants are okay
#a = m.Array(m.Param,(x.shape[0],))
a = np.array([172,282,10, 264, 287, 442, 393, 428, 484, 444, 344, 250, 293, 34, 473, 110, 338, 62,
250, 205, 81, 336, 249, 199, 328, 447, 408, 82, 357, 334, 181, 133, 378, 79, 292, 103,
494, 382, 10, 477, 237, 267, 337, 313, 395, 110, 114, 381, 52, 232, 457, 69, 167, 344,
363, 284, 136 ,240, 258, 449, 119, 317, 370, 404, 197, 102, 428, 238, 321, 103, 282, 37,
41, 86, 496 ,31, 148, 245, 78, 219, 37, 115, 378, 129, 37, 448, 415, 368, 238, 406,
408, 100, 112 ,289 ,205, 329, 338, 396, 494, 145, 355, 45, 5, 220, 411, 167, 85, 369,
426, 442, 406 ,217, 57, 176, 12, 368, 444, 227, 337, 63, 267, 216, 227, 182, 408, 116,
385, 140])
# use .value to assign a value instead of over-writing the gekko variable
# with a floating point number
#for i in range(len(a)):
# a[i].value = a_[i]
O1 = np.array(range(n))
O_= np.transpose(np.nonzero(O1+1))
O = np.zeros((x.shape[0],x.shape[1]),dtype=int)
for i in range(n):
for j in range(n):
O[i,O_[i]] = 1
d = np.empty((p,q)) # doesn't need to be a gekko array
for i in range(p):
for j in range(q):
d[i,j] = (round(p/2)-int(i/2))*4-(j%4)
#condition - use m.abs2 or m.abs3 for versions of ABS that
# have continuous first and second derivatives
xabs = [[m.Intermediate(m.abs3(x[i,j]-O[i,j])) for j in range(n)] for i in range(n)]
# matrix summation
msum = m.sum([m.sum([xabs[i][j] for i in range(n)]) for j in range(n)])
m.Equation(msum<=2*s)
for i in range(n):
m.Equation(m.sum(x[:,i])==1)
for i in range(n):
m.Equation(m.sum(x[i,:])==1)
#Objective - can add multiple objectives
e = d.reshape(n)
for i in range(n):
for j in range(n):
m.Obj(a[i]*x[i,j]*e[j])
#m.Obj(m.sum(a.T*m.sum(x*d.reshape(1,n))))
#Set global options
m.options.IMODE = 3
#Solve simulation
m.solve(disp=True,debug=True)
我的代码适用于小变量。但是当我在 128*128 变量数组中执行时，出现以下错误：
APM model error: string > 15000 characters Consider breaking up the line into multiple equations The may also be due to only using newline character CR instead of CR LF (for Windows) or LF (for MacOS/Linux) To fix this problem, save APM file with appropriate newline characters
我不知道如何解决这个错误。
假设gekko不能运行那样的大数组变量。
我把代码放在下面。
from gekko import GEKKO
import numpy as np
m = GEKKO()
m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations max', \
# minlp iterations with integer solution
'minlp_max_iter_with_int_sol max', \
# treat minlp as nlp
'minlp_as_nlp min', \
# nlp sub-problem max iterations
'nlp_maximum_iterations 5000', \
# 1 = depth first, 2 = breadth first
'minlp_branch_method 1', \
# maximum deviation from whole number
'minlp_integer_tol 0.0005', \
# covergence tolerance
'minlp_gap_tol 0.0001' ,\
'max_cpu_time 10min']
n = 128
p = 8
q = 16
x = m.Array(m.Var,(n,n),integer=True)
for i in range(n):
for j in range(n):
x[i,j].value = 1
x[i,j].lower = 0
x[i,j].upper = 1
s = 60
a = m.Array(m.Param,(x.shape[0],))
a_ = [172,282,10, 264, 287, 442, 393, 428, 484, 444, 344, 250, 293, 34, 473, 110, 338, 62,
250, 205, 81, 336, 249, 199, 328, 447, 408, 82, 357, 334, 181, 133, 378, 79, 292, 103,
494, 382, 10, 477, 237, 267, 337, 313, 395, 110, 114, 381, 52, 232, 457, 69, 167, 344,
363, 284, 136 ,240, 258, 449, 119, 317, 370, 404, 197, 102, 428, 238, 321, 103, 282, 37,
41, 86, 496 ,31, 148, 245, 78, 219, 37, 115, 378, 129, 37, 448, 415, 368, 238, 406,
408, 100, 112 ,289 ,205, 329, 338, 396, 494, 145, 355, 45, 5, 220, 411, 167, 85, 369,
426, 442, 406 ,217, 57, 176, 12, 368, 444, 227, 337, 63, 267, 216, 227, 182, 408, 116,
385, 140]
for i in range(len(a)):
a[i] = a_[i]
O1 = np.array(range(n))
O_= np.transpose(np.nonzero(O1+1))
O = np.zeros((x.shape[0],x.shape[1]),dtype=int)
for i in range(n):
for j in range(n):
O[i,O_[i]] = 1
d = m.Array(m.Param,(p,q))
for i in range(p):
for j in range(q):
d[i,j] = (round(p/2)-int(i/2))*4-(j%4)
#condition
m.Equation(m.sum(abs(x[:,:]-O[:,:]))<=2*s)
for i in range(n):
m.Equation(m.sum(x[:,i])==1)
for i in range(n):
m.Equation(m.sum(x[i,:])==1)
#Objective
m.Obj(np.sum(a.T*np.sum(x*d.reshape(1,n),axis=1)))
#Set global options
m.options.IMODE = 3
#Solve simulation
m.solve(disp=False,debug=True)
下面是一个应该可以工作的版本，如果你让它解决足够长的时间，尽管解决方案不太可能，因为你有 128*128 = 16384 个二进制变量。有 2^16384 个解决方案，如果整数变量的数量大于几百，APOPT 等混合整数非线性规划求解器不太可能快速找到解决方案。它可以解决具有数千个变量的连续变量，但混合整数比较困难，因为它必须使用分支定界法来解决它们。如果您可以使问题线性化（用松弛变量代替 abs 函数），您可能想尝试像 CPLEX 或 Gurobi 这样的 MILP 求解器。
以下是关于您的模型的一些建议：
- 不要对模型表达式使用
numpy。np.abs和np.sum等函数不提供求解器所需的信息。您应该使用m.abs2（或m.abs3）和m.sum。 - 数组
a和d可以是常量的 numpy 数组，而不是 Gekko 参数。 - 您可以将至少一个表达式分解为 Intermediates 以帮助缩短求解时间。
- objective 函数是
a.T x[:,:] d的标量结果。您可以使用来自 a[i] x[i,j] d[j] 的多个m.Obj函数，而不是矩阵乘法产生的一个大表达式。 - 我建议您从较小的问题开始，然后在您对模型和解决方案有信心时扩大到更大的问题。
from gekko import GEKKO
import numpy as np
m = GEKKO()
m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 1000', \
# minlp iterations with integer solution
'minlp_max_iter_with_int_sol 100', \
# treat minlp as nlp
'minlp_as_nlp 1', \
# nlp sub-problem max iterations
'nlp_maximum_iterations 5000', \
# 1 = depth first, 2 = breadth first
'minlp_branch_method 1', \
# maximum deviation from whole number
'minlp_integer_tol 0.0005', \
# covergence tolerance
'minlp_gap_tol 0.0001']
n = 128
p = 8
q = 16
x = m.Array(m.Var,(n,n),integer=True)
for i in range(n):
for j in range(n):
x[i,j].value = 1
x[i,j].lower = 0
x[i,j].upper = 1
s = 60
# don't need a gekko array here, constants are okay
#a = m.Array(m.Param,(x.shape[0],))
a = np.array([172,282,10, 264, 287, 442, 393, 428, 484, 444, 344, 250, 293, 34, 473, 110, 338, 62,
250, 205, 81, 336, 249, 199, 328, 447, 408, 82, 357, 334, 181, 133, 378, 79, 292, 103,
494, 382, 10, 477, 237, 267, 337, 313, 395, 110, 114, 381, 52, 232, 457, 69, 167, 344,
363, 284, 136 ,240, 258, 449, 119, 317, 370, 404, 197, 102, 428, 238, 321, 103, 282, 37,
41, 86, 496 ,31, 148, 245, 78, 219, 37, 115, 378, 129, 37, 448, 415, 368, 238, 406,
408, 100, 112 ,289 ,205, 329, 338, 396, 494, 145, 355, 45, 5, 220, 411, 167, 85, 369,
426, 442, 406 ,217, 57, 176, 12, 368, 444, 227, 337, 63, 267, 216, 227, 182, 408, 116,
385, 140])
# use .value to assign a value instead of over-writing the gekko variable
# with a floating point number
#for i in range(len(a)):
# a[i].value = a_[i]
O1 = np.array(range(n))
O_= np.transpose(np.nonzero(O1+1))
O = np.zeros((x.shape[0],x.shape[1]),dtype=int)
for i in range(n):
for j in range(n):
O[i,O_[i]] = 1
d = np.empty((p,q)) # doesn't need to be a gekko array
for i in range(p):
for j in range(q):
d[i,j] = (round(p/2)-int(i/2))*4-(j%4)
#condition - use m.abs2 or m.abs3 for versions of ABS that
# have continuous first and second derivatives
xabs = [[m.Intermediate(m.abs3(x[i,j]-O[i,j])) for j in range(n)] for i in range(n)]
# matrix summation
msum = m.sum([m.sum([xabs[i][j] for i in range(n)]) for j in range(n)])
m.Equation(msum<=2*s)
for i in range(n):
m.Equation(m.sum(x[:,i])==1)
for i in range(n):
m.Equation(m.sum(x[i,:])==1)
#Objective - can add multiple objectives
e = d.reshape(n)
for i in range(n):
for j in range(n):
m.Obj(a[i]*x[i,j]*e[j])
#m.Obj(m.sum(a.T*m.sum(x*d.reshape(1,n))))
#Set global options
m.options.IMODE = 3
#Solve simulation
m.solve(disp=True,debug=True)