Pyomo Error: No value for uninitialized NumericValue object Variable
Pyomo Error: No value for uninitialized NumericValue object Variable
我已经建立了一个 multiobjective 模型，我正在使用 pyomo 的增强 epsilon 方法包来解决这个问题，它在 Github.
中可用
该模型为我提供了 objective 值，但是当我尝试打印出一些单独的变量值时，它为我提供了“未初始化的 NumericValue 对象无值”错误。我做了一些研究，发现这可能表明我的模型不可行，但奇怪的是，我有我想要的 objective 值。这是我没有数据的代码，因为我有很长的数据列表：
#Uncertain Demand Model
from pyomo.environ import *
import numpy as np
from pathlib import Path
from pyaugmecon import PyAugmecon
import pandas as pd
*Data entered here*
def model_sto():
model = ConcreteModel()
#Sets
E=RangeSet(6)
P=RangeSet(8)
L=RangeSet(2)
T=RangeSet(13)
S=RangeSet(4)
N=RangeSet(3)
A=RangeSet(0,53)
model.E = Set(initialize=E) #Raw material
model.P = Set(initialize=P) #Products
model.L = Set(initialize=L) #Production line
model.T = Set(initialize=T) #Planning periods in week
model.S = Set(initialize=S) #Seasons in months
model.N = Set(initialize=N) #Scenarios
model.A = Set(initialize=A) #Age of raw material inventory
#Production
model.f = Param(model.N, model.S, model.T, model.P, initialize=f) #Demand forecast
model.error = Param(model.N, model.S, model.T, model.P, initialize=error) #Forecast error
model.b = b #Lost sales penalty
model.prate= Param(model.P, model.L, initialize = prate) #Production rate of product p on line l
model.lcap=Param(model.L, initialize = lcap) #capacity of production line l in terms of time in week
model.rcon=Param(model.E, model.P, initialize =rcon) #raw material consumption rate
model.pa =Param(model.P, model.L, initialize = pa) #production assignment binary variable
model.oee =Param(model.L, initialize = oee) #overall equipment effificiency, not relevant for this model
model.i0 = 100 #initial inventory
model.u0 = 100 #initial inventory
#Inventory
model.hi = Param(model.E, initialize = hi)
model.hp = Param(model.P, initialize = hp)
#Environmental Impact - Not relevant for this case
model.EI = Param(model.E, initialize = EI)
model.EP = Param(initialize = EP)
model.ES = Param(initialize = ES)
#Decision variables
model.q = Var(model.P, model.L, model.T, model.S, within=NonNegativeReals) #production quantity
model.r = Var(model.P, model.L, model.T, model.S, model.N, within=NonNegativeReals) #recourse production quantity
model.w = Var(model.E, model.T, model.S, model.N, within=NonNegativeReals) #Raw material orders
model.i = Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Inventory level of final products
model.u = Var(model.A, model.E, model.T, model.S, model.N, within=NonNegativeReals) #Inventory level of raw material
model.y = Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Lost sales
model.sales=Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Sales
##Set of constraints
model.ct1productInventory=ConstraintList() #Final product inventory constraint
for n in model.N:
for s in model.S:
for t in model.T:
for p in model.P:
if t == model.T.first():
lhs = model.i[p,t,s,n]
rhs = model.i0 + sum(model.q[p,l,t,s] + model.r[p,l,t,s,n] for l in model.L) - model.sales[p,t,s,n]
if t!= model.T.first():
lhs = model.i[p,t,s,n]
rhs = model.i[p,t-1,s,n] + sum(model.q[p,l,t,s] + model.r[p,l,t,s,n] for l in model.L) - model.sales[p,t,s,n]
model.ct1productInventory.add (lhs == rhs)
model.ct2demand = ConstraintList() #Forecast + forecast error fulfilled from lost sales and sales
for n in model.N:
for s in model.S:
for t in model.T:
for p in model.P:
lhs = model.f[n,s,t,p] + model.error[n,s,t,p]*0.1
rhs = model.y[p,t,s,n] + model.sales[p,t,s,n]
model.ct2demand.add (lhs == rhs)
model.ct3productionCap = ConstraintList() #Production capacity constraint
for n in model.N:
for s in model.S:
for t in model.T:
for l in model.L:
lhs = sum((model.q[p,l,t,s] + model.r[p,l,t,s,n])/(model.prate[p,l]*7*24) for p in model.P)
rhs = model.lcap[l]*model.pa[p,l]
model.ct3productionCap.add(lhs <= rhs)
model.ct4rawMaterInventory = ConstraintList()
for n in model.N:
for s in model.S:
for t in model.T:
for e in model.E:
if t==model.T.first(): #Initial inventory constraint
lhs = sum(model.u[a,e,t,s,n] for a in model.A)
rhs = model.u0 + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
if t!=model.T.first():
if e == model.E.first(): #Shelf-life constraint for Milk
lhs = sum(model.u[a,e,t,s,n] for a in A if a <=2)
rhs = sum(model.u[a,e,t-1,s,n] for a in A if a <=1) + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
if e != model.E.first(): #No shelf-life constraint for all other ingredients
lhs = sum(model.u[a,e,t,s,n] for a in A)
rhs = sum(model.u[a,e,t-1,s,n] for a in A) + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
model.ct4rawMaterInventory.add(lhs == rhs)
objective1 = sum((sum((model.hp[p]*model.i[p,t,s,n])*0.33333 for p in model.P) + sum((model.hi[e]*model.u[a,e,t,s,n])*0.33333 for a in A for e in model.E) + sum((model.b*model.y[p,t,s,n])*0.33333 for p in model.P)) for t in model.T for s in model.S for n in model.N)
objective2 = sum(sum((model.w[e,t,s,n]*model.EI[e])*0.33333 for e in model.E) + sum((model.q[p,l,t,s]*model.EP) +(model.r[p,l,t,s,n]*model.EP*0.33333) for p in model.P for l in model.L) + sum((model.i[p,t,s,n]*model.ES)*0.33333 for p in model.P) + sum(model.u[a,e,t,s,n]*model.EI[e] for e in model.E if e == 1 for a in A if a == 2) for t in model.T for s in model.S for n in N)
model.obj_list=ObjectiveList()
model.obj_list.add(expr=objective1, sense = minimize)
model.obj_list.add(expr=objective2, sense = minimize)
#By default, deactive all the objectives for PyAugmecon package
for o in range(len(model.obj_list)):
model.obj_list[o + 1].deactivate()
return model
model_type = 'model_sto'
# AUGMECON related options
options = {
'name': model_type,
'grid_points': 40,
}
solver_options = {}
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(model_sto(), options)
py_augmecon.solve()
# Options passed to Gurobi
print(py_augmecon.model.payoff) # this prints the payoff table
print(py_augmecon.unique_pareto_sols) # this prints the unique Pareto optimal solutions
for n in model.N:
print(value(model.w[e,t,s,n]))
在 运行 模型之后，我得到：
Solved 40 models for 40 unique Pareto solutions in 295.36 seconds
[[1.15809852e+08 5.72221560e+10]
[2.54428683e+08 5.31380259e+05]]
[[1.68191673e+08 3.22793965e+10]
[1.42790323e+08 4.40171657e+10]
[2.27040732e+08 7.33663710e+09]
[1.64729964e+08 3.37466177e+10]
[1.88961929e+08 2.34760697e+10]
[1.27744246e+08 5.13532714e+10]
[1.21725815e+08 5.42877137e+10]
[1.75115092e+08 2.93449542e+10]
[2.09732185e+08 1.46727428e+10]
[2.40887569e+08 1.46775252e+09]
[2.02808766e+08 1.76071851e+10]
[1.85500220e+08 2.49432908e+10]
[1.36771892e+08 4.69516080e+10]
[1.57845901e+08 3.66810600e+10]
[1.15809852e+08 5.72221560e+10]
[1.61268255e+08 3.52138388e+10]
[2.54428682e+08 5.31380259e+05]
[1.78576801e+08 2.78777331e+10]
[1.54827184e+08 3.81482811e+10]
[1.18734613e+08 5.57549348e+10]
[2.33964150e+08 4.40219481e+09]
[1.45799538e+08 4.25499445e+10]
[2.16655604e+08 1.17383005e+10]
[1.92423639e+08 2.20088485e+10]
[2.23579022e+08 8.80385824e+09]
[1.39781107e+08 4.54843868e+10]
[1.51817969e+08 3.96155023e+10]
[1.33762677e+08 4.84188291e+10]
[1.71653383e+08 3.08121754e+10]
[2.37425860e+08 2.93497367e+09]
[1.48808754e+08 4.10827234e+10]
[1.24735030e+08 5.28204925e+10]
[2.06270476e+08 1.61399640e+10]
[2.30502441e+08 5.86941595e+09]
[1.82038511e+08 2.64105120e+10]
[2.20117313e+08 1.02710794e+10]
[1.95885348e+08 2.05416274e+10]
[1.99347057e+08 1.90744062e+10]
[1.30753461e+08 4.98860503e+10]
[2.13193894e+08 1.32055217e+10]]
ERROR: evaluating object as numeric value: w[6,13,4,1]
(object: <class 'pyomo.core.base.var._GeneralVarData'>)
No value for uninitialized NumericValue object w[6,13,4,1]
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_20860/2659247016.py in <module>
2737
2738 for n in model.N:
-> 2739 print(value(model.w[e,t,s,n]))
pyomo\core\expr\numvalue.pyx in pyomo.core.expr.numvalue.value()
pyomo\core\expr\numvalue.pyx in pyomo.core.expr.numvalue.value()
ValueError: No value for uninitialized NumericValue object w[6,13,4,1]
所以给出了结果，它们是合理的，因为它们给了我预期的结果，但是当我尝试打印单个变量时，我得到了一个错误。我尝试打印其他变量并得到相同的错误。任何指导或帮助将不胜感激！
非常感谢！
这只是一种预感，但我认为您的命名空间中发生了一些奇怪的事情。 model 未在您的函数之外定义，并且您在调用该段中的函数时没有“捕获”您创建的实例：
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(model_sto(), options)
py_augmecon.solve()
所以，当您稍后检查变量时，我不确定 model 是如何成为已知变量的。
试试这个（或类似的）：
my_model = model_sto() # <--note I'm catching the returned model
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(my_model, options)
py_augmecon.solve()
然后在后面的代码中，迭代 my_model 中的值而不是 model
for n in my_model.N:
print(value(my_model.w[e,t,s,n]))
增强...
啊啊，我现在在你的错误中看到你有一个 ipykernel 正在运行，可能来自笔记本。这解释了我上面的部分困惑。你有一个旧的 model 实例漂浮在周围，它与你解决的问题无关。这可能发生在您重复 运行 不同代码段的笔记本中。如果你是 re-running 东西以避免这些类型的名称冲突，请经常重置内核，我认为如果你这样做，代码将 barf 并且根本无法识别 model。或者，您可以将其从笔记本中剥离出来，然后单独 运行。
我已经建立了一个 multiobjective 模型，我正在使用 pyomo 的增强 epsilon 方法包来解决这个问题，它在 Github.
中可用该模型为我提供了 objective 值，但是当我尝试打印出一些单独的变量值时，它为我提供了“未初始化的 NumericValue 对象无值”错误。我做了一些研究，发现这可能表明我的模型不可行，但奇怪的是，我有我想要的 objective 值。这是我没有数据的代码，因为我有很长的数据列表：
#Uncertain Demand Model
from pyomo.environ import *
import numpy as np
from pathlib import Path
from pyaugmecon import PyAugmecon
import pandas as pd
*Data entered here*
def model_sto():
model = ConcreteModel()
#Sets
E=RangeSet(6)
P=RangeSet(8)
L=RangeSet(2)
T=RangeSet(13)
S=RangeSet(4)
N=RangeSet(3)
A=RangeSet(0,53)
model.E = Set(initialize=E) #Raw material
model.P = Set(initialize=P) #Products
model.L = Set(initialize=L) #Production line
model.T = Set(initialize=T) #Planning periods in week
model.S = Set(initialize=S) #Seasons in months
model.N = Set(initialize=N) #Scenarios
model.A = Set(initialize=A) #Age of raw material inventory
#Production
model.f = Param(model.N, model.S, model.T, model.P, initialize=f) #Demand forecast
model.error = Param(model.N, model.S, model.T, model.P, initialize=error) #Forecast error
model.b = b #Lost sales penalty
model.prate= Param(model.P, model.L, initialize = prate) #Production rate of product p on line l
model.lcap=Param(model.L, initialize = lcap) #capacity of production line l in terms of time in week
model.rcon=Param(model.E, model.P, initialize =rcon) #raw material consumption rate
model.pa =Param(model.P, model.L, initialize = pa) #production assignment binary variable
model.oee =Param(model.L, initialize = oee) #overall equipment effificiency, not relevant for this model
model.i0 = 100 #initial inventory
model.u0 = 100 #initial inventory
#Inventory
model.hi = Param(model.E, initialize = hi)
model.hp = Param(model.P, initialize = hp)
#Environmental Impact - Not relevant for this case
model.EI = Param(model.E, initialize = EI)
model.EP = Param(initialize = EP)
model.ES = Param(initialize = ES)
#Decision variables
model.q = Var(model.P, model.L, model.T, model.S, within=NonNegativeReals) #production quantity
model.r = Var(model.P, model.L, model.T, model.S, model.N, within=NonNegativeReals) #recourse production quantity
model.w = Var(model.E, model.T, model.S, model.N, within=NonNegativeReals) #Raw material orders
model.i = Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Inventory level of final products
model.u = Var(model.A, model.E, model.T, model.S, model.N, within=NonNegativeReals) #Inventory level of raw material
model.y = Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Lost sales
model.sales=Var(model.P, model.T, model.S, model.N, within=NonNegativeReals) #Sales
##Set of constraints
model.ct1productInventory=ConstraintList() #Final product inventory constraint
for n in model.N:
for s in model.S:
for t in model.T:
for p in model.P:
if t == model.T.first():
lhs = model.i[p,t,s,n]
rhs = model.i0 + sum(model.q[p,l,t,s] + model.r[p,l,t,s,n] for l in model.L) - model.sales[p,t,s,n]
if t!= model.T.first():
lhs = model.i[p,t,s,n]
rhs = model.i[p,t-1,s,n] + sum(model.q[p,l,t,s] + model.r[p,l,t,s,n] for l in model.L) - model.sales[p,t,s,n]
model.ct1productInventory.add (lhs == rhs)
model.ct2demand = ConstraintList() #Forecast + forecast error fulfilled from lost sales and sales
for n in model.N:
for s in model.S:
for t in model.T:
for p in model.P:
lhs = model.f[n,s,t,p] + model.error[n,s,t,p]*0.1
rhs = model.y[p,t,s,n] + model.sales[p,t,s,n]
model.ct2demand.add (lhs == rhs)
model.ct3productionCap = ConstraintList() #Production capacity constraint
for n in model.N:
for s in model.S:
for t in model.T:
for l in model.L:
lhs = sum((model.q[p,l,t,s] + model.r[p,l,t,s,n])/(model.prate[p,l]*7*24) for p in model.P)
rhs = model.lcap[l]*model.pa[p,l]
model.ct3productionCap.add(lhs <= rhs)
model.ct4rawMaterInventory = ConstraintList()
for n in model.N:
for s in model.S:
for t in model.T:
for e in model.E:
if t==model.T.first(): #Initial inventory constraint
lhs = sum(model.u[a,e,t,s,n] for a in model.A)
rhs = model.u0 + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
if t!=model.T.first():
if e == model.E.first(): #Shelf-life constraint for Milk
lhs = sum(model.u[a,e,t,s,n] for a in A if a <=2)
rhs = sum(model.u[a,e,t-1,s,n] for a in A if a <=1) + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
if e != model.E.first(): #No shelf-life constraint for all other ingredients
lhs = sum(model.u[a,e,t,s,n] for a in A)
rhs = sum(model.u[a,e,t-1,s,n] for a in A) + model.w[e,t,s,n] - sum((model.q[p,l,t,s]+model.r[p,l,t,s,n])*model.rcon[e,p] for p in model.P for l in model.L)
model.ct4rawMaterInventory.add(lhs == rhs)
objective1 = sum((sum((model.hp[p]*model.i[p,t,s,n])*0.33333 for p in model.P) + sum((model.hi[e]*model.u[a,e,t,s,n])*0.33333 for a in A for e in model.E) + sum((model.b*model.y[p,t,s,n])*0.33333 for p in model.P)) for t in model.T for s in model.S for n in model.N)
objective2 = sum(sum((model.w[e,t,s,n]*model.EI[e])*0.33333 for e in model.E) + sum((model.q[p,l,t,s]*model.EP) +(model.r[p,l,t,s,n]*model.EP*0.33333) for p in model.P for l in model.L) + sum((model.i[p,t,s,n]*model.ES)*0.33333 for p in model.P) + sum(model.u[a,e,t,s,n]*model.EI[e] for e in model.E if e == 1 for a in A if a == 2) for t in model.T for s in model.S for n in N)
model.obj_list=ObjectiveList()
model.obj_list.add(expr=objective1, sense = minimize)
model.obj_list.add(expr=objective2, sense = minimize)
#By default, deactive all the objectives for PyAugmecon package
for o in range(len(model.obj_list)):
model.obj_list[o + 1].deactivate()
return model
model_type = 'model_sto'
# AUGMECON related options
options = {
'name': model_type,
'grid_points': 40,
}
solver_options = {}
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(model_sto(), options)
py_augmecon.solve()
# Options passed to Gurobi
print(py_augmecon.model.payoff) # this prints the payoff table
print(py_augmecon.unique_pareto_sols) # this prints the unique Pareto optimal solutions
for n in model.N:
print(value(model.w[e,t,s,n]))
在 运行 模型之后，我得到：
Solved 40 models for 40 unique Pareto solutions in 295.36 seconds
[[1.15809852e+08 5.72221560e+10]
[2.54428683e+08 5.31380259e+05]]
[[1.68191673e+08 3.22793965e+10]
[1.42790323e+08 4.40171657e+10]
[2.27040732e+08 7.33663710e+09]
[1.64729964e+08 3.37466177e+10]
[1.88961929e+08 2.34760697e+10]
[1.27744246e+08 5.13532714e+10]
[1.21725815e+08 5.42877137e+10]
[1.75115092e+08 2.93449542e+10]
[2.09732185e+08 1.46727428e+10]
[2.40887569e+08 1.46775252e+09]
[2.02808766e+08 1.76071851e+10]
[1.85500220e+08 2.49432908e+10]
[1.36771892e+08 4.69516080e+10]
[1.57845901e+08 3.66810600e+10]
[1.15809852e+08 5.72221560e+10]
[1.61268255e+08 3.52138388e+10]
[2.54428682e+08 5.31380259e+05]
[1.78576801e+08 2.78777331e+10]
[1.54827184e+08 3.81482811e+10]
[1.18734613e+08 5.57549348e+10]
[2.33964150e+08 4.40219481e+09]
[1.45799538e+08 4.25499445e+10]
[2.16655604e+08 1.17383005e+10]
[1.92423639e+08 2.20088485e+10]
[2.23579022e+08 8.80385824e+09]
[1.39781107e+08 4.54843868e+10]
[1.51817969e+08 3.96155023e+10]
[1.33762677e+08 4.84188291e+10]
[1.71653383e+08 3.08121754e+10]
[2.37425860e+08 2.93497367e+09]
[1.48808754e+08 4.10827234e+10]
[1.24735030e+08 5.28204925e+10]
[2.06270476e+08 1.61399640e+10]
[2.30502441e+08 5.86941595e+09]
[1.82038511e+08 2.64105120e+10]
[2.20117313e+08 1.02710794e+10]
[1.95885348e+08 2.05416274e+10]
[1.99347057e+08 1.90744062e+10]
[1.30753461e+08 4.98860503e+10]
[2.13193894e+08 1.32055217e+10]]
ERROR: evaluating object as numeric value: w[6,13,4,1]
(object: <class 'pyomo.core.base.var._GeneralVarData'>)
No value for uninitialized NumericValue object w[6,13,4,1]
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_20860/2659247016.py in <module>
2737
2738 for n in model.N:
-> 2739 print(value(model.w[e,t,s,n]))
pyomo\core\expr\numvalue.pyx in pyomo.core.expr.numvalue.value()
pyomo\core\expr\numvalue.pyx in pyomo.core.expr.numvalue.value()
ValueError: No value for uninitialized NumericValue object w[6,13,4,1]
所以给出了结果，它们是合理的，因为它们给了我预期的结果，但是当我尝试打印单个变量时，我得到了一个错误。我尝试打印其他变量并得到相同的错误。任何指导或帮助将不胜感激！
非常感谢！
这只是一种预感，但我认为您的命名空间中发生了一些奇怪的事情。 model 未在您的函数之外定义，并且您在调用该段中的函数时没有“捕获”您创建的实例：
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(model_sto(), options)
py_augmecon.solve()
所以，当您稍后检查变量时，我不确定 model 是如何成为已知变量的。
试试这个（或类似的）：
my_model = model_sto() # <--note I'm catching the returned model
#Solve using PyAugmecon Package
py_augmecon=PyAugmecon(my_model, options)
py_augmecon.solve()
然后在后面的代码中，迭代 my_model 中的值而不是 model
for n in my_model.N:
print(value(my_model.w[e,t,s,n]))
增强...
啊啊，我现在在你的错误中看到你有一个 ipykernel 正在运行，可能来自笔记本。这解释了我上面的部分困惑。你有一个旧的 model 实例漂浮在周围，它与你解决的问题无关。这可能发生在您重复 运行 不同代码段的笔记本中。如果你是 re-running 东西以避免这些类型的名称冲突，请经常重置内核，我认为如果你这样做，代码将 barf 并且根本无法识别 model。或者，您可以将其从笔记本中剥离出来，然后单独 运行。