给定输入,如电力消耗,太阳能电池板发电,价格(所有在给定的时间t),我们有一个电池,我们想要评估它应该(dis)/charge在任何特定的时间。问题可以表述如下:
Pt = price of electricity at time t
Lt = consumption of electricity at time t
Zt = charge of battery at time t (how much is in the battery)
St = Electricity generated from solar generator at time t
Qt = amount the battery (dis)/charges at time t
我们试图优化的函数是Ct = Pt *(Lt - St - Qt)。
这样做的目的是尽量减少用电量。
有以下限制:
Lt - St - Qt >= 0 (our demand has to be non-negative)
Qmin <= Qt <= Qmax ( the battery can only (dis)/charge between certain values at any given time)
Zmin <= Zt <= Zmax. (the battery has to be within its capacity, i.e. you can't discharge more than the battery holders, and you can charge more than the battery can hold)
Zt+1 = Zt + Qt+1 ( this means that the battery level at the next time step is equal to the battery level at the previous time step plus the amount that was (dis)/charged from the battery)
我所面临的问题是如何在python中制定这个问题,特别是更新电池的级别。
我知道其他图书馆(Pyomo,Pulp)也存在,欢迎加入解决方案。
发布于 2019-07-13 22:18:00
幸运的是,我被Giorgio的回答激发了学习pyomo (我主要是用户)的动机,所以利用你的问题来确保我理解了所有的界面。我会把它贴在这里,这样以后我可以自己再找到它:
import pyomo.environ as pyomo
import numpy as np
# create model
m = pyomo.ConcreteModel()
# Problem DATA
T = 24
Zmin = 0.0
Zmax = 2.0
Qmin = -1.0
Qmax = 1.0
# Generate prices, solar output and load signals
np.random.seed(42)
P = np.random.rand(T)*5.0
S = np.random.rand(T)
L = np.random.rand(T)*2.0
# Indexes
times = range(T)
times_plus_1 = range(T+1)
# Decisions variables
m.Q = pyomo.Var(times, domain=pyomo.Reals)
m.Z = pyomo.Var(times_plus_1, domain=pyomo.NonNegativeReals)
# objective
cost = sum(P[t]*(L[t] - S[t] - m.Q[t]) for t in times)
m.cost = pyomo.Objective(expr = cost, sense=pyomo.minimize)
# constraints
m.cons = pyomo.ConstraintList()
m.cons.add(m.Z[0] == 0.5*(Zmin + Zmax))
for t in times:
m.cons.add(pyomo.inequality(Qmin, m.Q[t], Qmax))
m.cons.add(pyomo.inequality(Zmin, m.Z[t], Zmax))
m.cons.add(m.Z[t+1] == m.Z[t] - m.Q[t])
m.cons.add(L[t] - S[t] - m.Q[t] >= 0)
# solve
solver = pyomo.SolverFactory('cbc')
solver.solve(m)
# display results
print("Total cost =", m.cost(), ".")
for v in m.component_objects(pyomo.Var, active=True):
print ("Variable component object",v)
print ("Type of component object: ", str(type(v))[1:-1]) # Stripping <> for nbconvert
varobject = getattr(m, str(v))
print ("Type of object accessed via getattr: ", str(type(varobject))[1:-1])
for index in varobject:
print (" ", index, varobject[index].value)发布于 2019-07-11 22:50:51
根据我的经验(线性/ MIP)优化是这类应用程序的一种有效方法。在我看来(是的),Pyomo是一个很好的工具:
文档相当广泛,托管在这里:https://pyomo.readthedocs.io/en/latest/index.html
你可以在这里找到更多的资料:examples.html
另外,这是对Pyomo的一个相当广泛的介绍的链接,它深入到相当高级的主题,如随机优化和双级问题。
最后,唯一具体的问题,你的情况是,你可能想把损失的充放电电池。首先,定义两个充放电的自变量(它们都是非负的),这样你就可以把电池的能量平衡写成一个约束,把时间t的能量状态( SOE )和时间上的SOE联系起来。
祝好运!
https://stackoverflow.com/questions/56968971
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