您好，我编写了一个 Pyomo 脚本来优化包含 3 个组件（蓄热器、锅炉和热泵）的调度模式。在定义有关热存储的约束之前，它似乎一直有效。谁能帮助我理解为什么没有最佳解决方案，以及哪个约束（不是排放约束）定义错误？谢谢您的帮助！
import pyomo.environ as pyo
from pyomo.environ import *
from pyomo.opt import SolverFactory
from pyomo.opt import TerminationCondition
# solver function
solver = SolverFactory("glpk")
model = ConcreteModel()
# creating a timeframe
number_of_period = 8760
model.time_set = pyo.Set(initialize=range(1, number_of_period+1), ordered=True)
# loading csv input files
input_assumptions = pd.read_csv('input_assumptions.csv')
input_price = pd.read_csv('prices.csv')
input_price_new = pd.DataFrame(input_price)
demand = input_assumptions['Total network demand (MWh)']
demand_new = pd.DataFrame(demand)
#parameters
model.Price = pyo.Param(model.time_set, initialize=dict(enumerate(input_price_new['Electricity prices'], 1)), within=pyo.Any)
model.Demand = pyo.Param(model.time_set, initialize=dict(enumerate(demand_new['Total network demand (MWh)'], 1)), within=pyo.Any)
model.HP_max = pyo.Param(model.time_set, initialize=dict(enumerate(input_assumptions['Max Heat pump'], 1)), within=pyo.Any)
#variables
model.b_min_cap = pyo.Param(initialize=0)
model.b_max_cap = pyo.Param(initialize=8.708)
model.b_soc = pyo.Var(model.time_set, domain = pyo.NonNegativeReals, bounds = (model.b_min_cap, model.b_max_cap))
model.b_discharge = pyo.Var(model.time_set, domain = pyo.NonNegativeReals, initialize = 0)
model.b_charge = pyo.Var(model.time_set, domain = pyo.NonNegativeReals, initialize = 0)
def HP_domain(model, i):
upper_HP = input_assumptions['Max Heat pump']
lower = input_assumptions['lower']
return(0, upper_HP[i-1])
model.HP_var = pyo.Var(model.time_set, domain = pyo.NonNegativeReals, bounds=HP_domain, initialize = 0)
model.Boiler_var = pyo.Var(model.time_set, domain = pyo.NonNegativeReals, bounds=(0, 24), initialize = 0)
#definition of the objective
def minimise(model):
return pyo.summation(model.Price, model.HP_var)
model.obj = pyo.Objective(rule=minimise, sense=pyo.minimize)
#constraints
def Soc_constraint(model, i):
if i == model.time_set.first():
return model.b_soc[i] == model.b_max_cap # starting at 100 % charge level
else:
return model.b_soc[i] == model.b_soc[i-1] - model.b_discharge[i-1] + model.b_charge[i-1]
model.soc_c = pyo.Constraint(model.time_set, rule=Soc_constraint)
def dispatch_rule(model, i):
return sum(model.b_discharge[i] for i in range(1, 8761)) + sum(model.b_charge[i] for i in range(1, 8761)) == 0
model.Summ = pyo.Constraint(rule=dispatch_rule)
def discharge_constraint_two(model, i):
return model.b_discharge[i] <= model.b_soc[i] # Maximum discharge rate within a single hour
model.discharge_two = pyo.Constraint(model.time_set, rule=discharge_constraint_two)
def charge_constraint_one(model, i):
return model.b_charge[i] <= model.HP_max[i] - model.Demand[i] # Maximum charge rate within a single hour
model.charge_one = pyo.Constraint(model.time_set, rule=charge_constraint_one)
def charge_constraint_two(model, i):
return model.b_charge[i] <= 8.708 - model.b_soc[i] # Maximum charge rate within a single hour
model.charge_two = pyo.Constraint(model.time_set, rule=charge_constraint_two)
def Sum_Const(model, i):
return model.HP_var[i] + model.Boiler_var[i] + model.b_discharge[i] - model.b_charge[i] == model.Demand[i]
model.SumConst = pyo.Constraint(model.time_set, rule=Sum_Const)
def Summation_Const(model, i):
return sum(model.HP_var[i] for i in range(1, 8761)) + sum(model.Boiler_var[i] for i in range(1, 8761)) == sum(model.Demand[i] for i in range(1, 8761))
model.Summ = pyo.Constraint(rule=Summation_Const)
def Emission_Const(model, i):
total_gas = sum(model.Boiler_var[i] for i in range(1, 8761))
total_elec = sum(model.HP_var[i] for i in range(1, 8761))
total_demand = sum(model.Demand[i] for i in range(1, 8761))
return ((total_elec / 2.97 * 0.219) + (total_gas / 0.85 * 0.184)) / (total_demand * 0.984) == 0.1
model.Emission = pyo.Constraint(rule=Emission_Const)'''
GLPSOL--GLPK LP/MIP Solver 5.0
Parameter(s) specified in the command line:
--mipgap 0.015 --write C:\Users\MATE~1.TOT\AppData\Local\Temp\tmpyopw_8dk.glpk.raw
--wglp C:\Users\MATE~1.TOT\AppData\Local\Temp\tmp9aorspj7.glpk.glp --cpxlp
C:\Users\MATE~1.TOT\AppData\Local\Temp\tmpmlk9d5az.pyomo.lp
Reading problem data from 'C:\Users\MATE~1.TOT\AppData\Local\Temp\tmpmlk9d5az.pyomo.lp'...
17523 rows, 43800 columns, 96358 non-zeros
201490 lines were read
Writing problem data to 'C:\Users\MATE~1.TOT\AppData\Local\Temp\tmp9aorspj7.glpk.glp'...
201484 lines were written
GLPK Simplex Optimizer 5.0
17523 rows, 43800 columns, 96358 non-zeros
Preprocessing...
17521 rows, 35038 columns, 87595 non-zeros
Scaling...
A: min|aij| = 1.576e-06 max|aij| = 1.000e+00 ratio = 6.345e+05
GM: min|aij| = 7.640e-01 max|aij| = 1.309e+00 ratio = 1.713e+00
EQ: min|aij| = 5.836e-01 max|aij| = 1.000e+00 ratio = 1.713e+00
Constructing initial basis...
Size of triangular part is 17520
0: obj = 0.000000000e+00 inf = 6.458e+04 (4)
Perturbing LP to avoid stalling [11026]...
13932: obj = 2.831720898e+06 inf = 9.314e+02 (1) 84
17298: obj = 2.831720898e+06 inf = 9.314e+02 (1)
LP HAS NO PRIMAL FEASIBLE SOLUTION
glp_simplex: unable to recover undefined or non-optimal solution
If you need actual output for non-optimal solution, use --nopresol
Time used: 6.5 secs
Memory used: 37.8 Mb (39629941 bytes)
Writing basic solution to 'C:\Users\MATE~1.TOT\AppData\Local\Temp\tmpyopw_8dk.glpk.raw'...
61332 lines were written
Model unknown
欢迎来到该网站。有几件事您应该注意。
首先，您的模型存在错误，因此我不确定您正在运行的内容实际上会产生输出。具体来说，在您的几个约束中，您省略了需要传递给函数的索引。比如这个：
def Summation_Const(model, i):
return sum(model.HP_var[i] for i in range(1, 8761)) + sum(model.Boiler_var[i] for i in range(1, 8761)) == sum(model.Demand[i] for i in range(1, 8761))
model.Summ = pyo.Constraint(rule=Summation_Const)
你没有通过
i其次，我担心你们的排放限制。为什么它是一个等式约束而不是上限？ (
<=最重要的是，您需要关注结果语句，并看到它报告模型当前不可行。所以不仅仅是找不到最优解，而是你有一些逻辑错误。您可以单独注释掉这些约束，看看是否可以找到罪魁祸首或按照我在这篇文章中建议的方式执行某些操作：
我可以得到“输入假设.csv”和“价格.csv”吗？ 谢谢你