在Pyomo存储库中看到了“magn汽车示例”的精彩实现之后,我想继续用新功能和约束扩展问题,但我发现了开发过程中的下一个问题。有人能解决它们吗?
1)增加了新的约束“电动汽车”:现在加速度受到遵守的限制,直到使用确定的速度然后使用恒定功率模型。我无法像我想的那样实现这个约束。它被评论,但Pyomo抱怨约束与变量有关。 (现在Umax取决于车速)。
2)增加了新的舒适加速度和加加速度约束。看起来他们工作正常,但如果Pyomo大师监督他们并告诉我他们是否真的以正确的方式实施,那应该是好的。
3)关于最后一个,为了减少冗长。有没有什么方法可以将accelerationL和accelerationU结合在一个独特的约束中?对于jerkL和jerkU也一样。
4)最后一个特征是速度限制约束分为两个步骤。同样,我无法使其工作,所以它在代码中被注释。有人敢解决吗?
# Ampl Car Example (Extended)
#
# Shows how to convert a minimize final time optimal control problem
# to a format pyomo.dae can handle by removing the time scaling from
# the ContinuousSet.
#
# min tf
# dx/dt = v
# dv/dt = u - R*v^2
# x(0)=0; x(tf)=L
# v(0)=0; v(tf)=0
# -3 <= u <= 1 (engine constraint)
#
# {v <= 7m/s ===> u < 1
# u <= { (electric car constraint)
# {v > 7m/s ===> u < 1*7/v
#
# -1.5 <= dv/dt <= 0.8 (comfort constraint -> smooth driving)
# -0.5 <= d2v/dt2 <= 0.5 (comfort constraint -> jerk)
# v <= Vmax (40 kmh[0-500m] + 25 kmh(500-1000m])
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
m.R = Param(initialize=0.001) # Friction factor
m.L = Param(initialize=1000.0) # Final position
m.T = Param(initialize=50.0) # Estimated time
m.aU = Param(initialize=0.8) # Acceleration upper bound
m.aL = Param(initialize=-1.5) # Acceleration lower bound
m.jU = Param(initialize=0.5) # Jerk upper bound
m.jL = Param(initialize=-0.5) # Jerk lower bound
m.NFE = Param(initialize=100) # Number of finite elements
'''
def _initX(m, i):
return m.x[i] == i*m.L/m.NFE
def _initV(m):
return m.v[i] == m.L/50
'''
m.tf = Var()
m.tau = ContinuousSet(bounds=(0,1)) # Unscaled time
m.t = Var(m.tau) # Scaled time
m.x = Var(m.tau, bounds=(0,m.L))
m.v = Var(m.tau, bounds=(0,None))
m.u = Var(m.tau, bounds=(-3,1), initialize=0)
m.dt = DerivativeVar(m.t)
m.dx = DerivativeVar(m.x)
m.dv = DerivativeVar(m.v)
m.da = DerivativeVar(m.v, wrt=(m.tau, m.tau))
m.obj = Objective(expr=m.tf)
def _ode1(m, i):
if i==0:
return Constraint.Skip
return m.dt[i] == m.tf
m.ode1 = Constraint(m.tau, rule=_ode1)
def _ode2(m, i):
if i==0:
return Constraint.Skip
return m.dx[i] == m.tf * m.v[i]
m.ode2 = Constraint(m.tau, rule=_ode2)
def _ode3(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] == m.tf*(m.u[i] - m.R*m.v[i]**2)
m.ode3 = Constraint(m.tau, rule=_ode3)
def _accelerationL(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] >= m.aL*m.tf
m.accelerationL = Constraint(m.tau, rule=_accelerationL)
def _accelerationU(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] <= m.aU*m.tf
m.accelerationU = Constraint(m.tau, rule=_accelerationU)
def _jerkL(m, i):
if i==0:
return Constraint.Skip
return m.da[i] >= m.jL*m.tf**2
m.jerkL = Constraint(m.tau, rule=_jerkL)
def _jerkU(m, i):
if i==0:
return Constraint.Skip
return m.da[i] <= m.jU*m.tf**2
m.jerkU = Constraint(m.tau, rule=_jerkU)
'''
def _electric(m, i):
if i==0:
return Constraint.Skip
elif value(m.v[i])<=7:
return m.a[i] <= 1
else:
return m.v[i] <= 1*7/m.v[i]
m.electric = Constraint(m.tau, rule=_electric)
'''
'''
def _speed(m, i):
if i==0:
return Constraint.Skip
elif value(m.x[i])<=500:
return m.v[i] <= 40/3.6
else:
return m.v[i] <= 25/3.6
m.speed = Constraint(m.tau, rule=_speed)
'''
def _initial(m):
yield m.x[0] == 0
yield m.x[1] == m.L
yield m.v[0] == 0
yield m.v[1] == 0
yield m.t[0] == 0
m.initial = ConstraintList(rule=_initial)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=value(m.NFE), wrt=m.tau, scheme='BACKWARD')
#discretizer = TransformationFactory('dae.collocation')
#discretizer.apply_to(m, nfe=value(m.NFE), ncp=4, wrt=m.tau, scheme='LAGRANGE-RADAU')
solver = SolverFactory('ipopt')
solver.solve(m,tee=True)
print("final time = %6.2f" %(value(m.tf)))
t = []
x = []
v = []
a = []
u = []
for i in m.tau:
t.append(value(m.t[i]))
x.append(value(m.x[i]))
v.append(3.6*value(m.v[i]))
a.append(10*value(m.u[i] - m.R*m.v[i]**2))
u.append(10*value(m.u[i]))
import matplotlib.pyplot as plt
plt.plot(x, v, label='v (km/h)')
plt.plot(x, a, label='a (dm/s2)')
plt.plot(x, u, label='u (dm/s2)')
plt.xlabel('distance')
plt.grid('on')
plt.legend()
plt.show()
提前多多感谢, 巴勃罗
答案 0 :(得分:0)
(1)您不应该将Pyomo约束规则视为解算器使用的回调。您应该将它们更多地视为生成约束对象容器的函数,在构造模型时,每个索引都会调用一次约束对象。意味着在if
语句中使用变量是无效的,除非您实际上只使用其初始值来定义约束表达式。有办法表达我认为你想要做的事情,但它们涉及在问题中引入二进制变量,在这种情况下你不能再使用Ipopt。
(2)无法提供任何帮助。语法看起来很好。
(3)Pyomo允许您从约束规则返回双边不等式表达式(例如,L&lt; = f(x)&lt; = U),但它们不能涉及L和U位置中的变量表达式。它看起来并不像您所指的约束可以组合成这种形式。
(4)见(1)