constraints - 我如何管理每天的电池放电量？

Question

1.这个型号的电池充放电很疯狂，我怎么能让电池每天放电一两次。还是计算完全放电的次数并停在给定的次数？

2.是否可以在for循环中添加sum()

3.如何在给定的时间添加约束来强制停止、充电或放电电池。

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)

谢谢！

标签： constraintsbatterypyomocycle-sort