微电网电池调度的约束优化

给定电力消耗、太阳能电池板发电、价格等输入（全部在给定时间 t），我们有一个电池，我们想要评估它在任何给定时间应该（放电）/充电多少。 该问题可以表述如下：

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)

目的是尽量减少购买的电量

具有以下约束：

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)

我遇到的问题是如何用 python (Scipy) 来表达这个问题，特别是更新电池电量。

我知道存在其他库（Pyomo、Pulp），欢迎提供解决方案。

你很幸运，我受到 Giorgio 的回答的激励来学习 pyomo （我主要使用 PULP），所以用你的问题作为一个机会来确保我理解所有的界面。我会将其发布在这里，以便将来我可以再次找到它：

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)