我正在使用python中的Gekko库进行一些测试,但有一个小问题,我知道解决方案。完整的代码如下:
from gekko import GEKKO
P = [[3.0,3.55,5.18,7.9,5.98],
[1.56,1.56,2.48,3.15,2.38],
[1.49,4.96,6.4,9.4,6.5]]
M = [[1,2,3,4,5],
[6,7,8,9,10],
[11,12,13,14,15]]
mm = M
pp = P
c1 = [300,200,150,250,180]
qtde = [10,10,10]
flex = [0.2,0.2,0.2]
m = GEKKO(remote=False)
ni = 3
nj = 5
x = [[m.Var(lb=0,integer=True) for j in range(nj)] for i in range(ni)]
s = 0
expr = []
for i in range(ni):
for j in range(nj):
s += x[i][j]*pp[i][j]
expr.append(s)
s = 0
for i in range(ni):
for j in range(nj):
if mm[i][j] == 0:
m.Equation(x[i][j] == 0)
for i in range(len(flex)):
if flex[i] == 0:
m.Equation(sum([x[i][j] for j in range(nj)]) >= qtde[i])
else:
m.Equation(sum([x[i][j] for j in range(nj)]) >= qtde[i])
m.Equation(sum([x[i][j] for j in range(nj)]) <= (1+flex[i])*qtde[i])
b = m.Array(m.Var,nj,integer=True,lb=0,ub=1)
iv = [None]*nj
for j in range(nj):
iv[j] = m.sum([pp[i][j]*x[i][j] for i in range(ni)])
m.Equation(iv[j] >= b[j]*c1[j])
m.Obj(m.sum(expr))
m.options.SOLVER=1 # switch to APOPT
m.solver_options = ['minlp_gap_tol 1.0e-2',\
'minlp_maximum_iterations 50000',\
'minlp_max_iter_with_int_sol 50000',\
'minlp_branch_method 1',\
'minlp_integer_leaves 2']
m.solve()
for j in range(nj):
m.Equation((1 - b[j])*iv[j] == 0)
m.options.SOLVER=1
m.solve()
代码退出,错误为:Exception: @error: Solution Not Found
。这很奇怪,因为有一个明确的解决方案:
x = [[0,0,12,0,0],
[0,0,12,0,0],
[0,0,12,0,0]]
更奇怪的是,即使我大幅增加了变量qtde
(例如qtde = [40,40,40]
)的值,该算法也找不到解决方案。我编写约束的方式有误吗?
答案 0 :(得分:4)
有时候,求解者需要更好的初始猜测或选择范围的帮助,以远离有问题的解决方案。只需一个求解器调用,即可帮助解决问题。
lower = [0,0,4,0,0]
for i in range(ni):
for j in range(nj):
x[i][j].value = 5
x[i][j].lower = lower[j]
x[i][j].upper = 20
如果将所有发电单元的下限都设置为零,我总是会收到一条infeasible solution
消息。求解器似乎卡在全零的试验解中,或者当所有解都低于某个阈值时陷入困境。在这种情况下,我必须将中间单元的边界设置为4以上才能获得成功的解决方案,而其他单元为零。这是完整的代码:
from gekko import GEKKO
P = [[3.0,3.55,5.18,7.9,5.98],
[1.56,1.56,2.48,3.15,2.38],
[1.49,4.96,6.4,9.4,6.5]]
M = [[1,2,3,4,5],
[6,7,8,9,10],
[11,12,13,14,15]]
mm = M
pp = P
c1 = [300,200,150,250,180]
qtde = [10,10,10]
flex = [0.2,0.2,0.2]
m = GEKKO(remote=False)
ni = 3
nj = 5
x = [[m.Var(integer=True) for j in range(nj)] for i in range(ni)]
# Fix x at values to check the feasibility of the initial guess
#x = [[m.Param() for j in range(nj)] for i in range(ni)]
lower = [0,0,4,0,0]
for i in range(ni):
for j in range(nj):
x[i][j].value = 5
x[i][j].lower = lower[j]
x[i][j].upper = 20
s = 0
expr = []
for i in range(ni):
for j in range(nj):
s += x[i][j]*pp[i][j]
expr.append(s)
s = 0
for i in range(ni):
for j in range(nj):
if mm[i][j] == 0:
m.Equation(x[i][j] == 0)
for i in range(len(flex)):
if flex[i] == 0:
m.Equation(sum([x[i][j] for j in range(nj)]) >= qtde[i])
else:
m.Equation(sum([x[i][j] for j in range(nj)]) >= qtde[i])
m.Equation(sum([x[i][j] for j in range(nj)]) <= (1+flex[i])*qtde[i])
b = m.Array(m.Var,nj,value=0.5,integer=True,lb=0,ub=1)
iv = [None]*nj
for j in range(nj):
iv[j] = m.sum([pp[i][j]*x[i][j] for i in range(ni)])
m.Equation(iv[j] >= b[j]*c1[j])
m.Obj(m.sum(expr))
for j in range(nj):
m.Equation((1 - b[j])*iv[j] <= 1e-5)
print('Initial guess: ' + str(x))
# solve as NLP first to see iterations
#m.solver_options = ['minlp_as_nlp 1']
#m.options.SOLVER = 1
#m.solve(debug=0)
# solve as MINLP
m.options.SOLVER=1 # switch to APOPT
m.solver_options = ['minlp_gap_tol 1.0e-2',\
'minlp_maximum_iterations 50000',\
'minlp_max_iter_with_int_sol 50000',\
'minlp_branch_method 1',\
'minlp_integer_leaves 2']
m.options.SOLVER=1
m.solve(disp=False)
print('Final solution: ' + str(x))
使用完美的求解器,将不需要进行初步猜测,并且可以将范围从0
设置为infinity
。有些问题较难解决,例如混合整数变量的问题以及使用互补条件时的问题。您的问题同时存在,因此对于没有初始猜测或适当界限的求解器,我感到惊讶。