#################################################################################
# The Institute for the Design of Advanced Energy Systems Integrated Platform
# Framework (IDAES IP) was produced under the DOE Institute for the
# Design of Advanced Energy Systems (IDAES).
#
# Copyright (c) 2018-2024 by the software owners: The Regents of the
# University of California, through Lawrence Berkeley National Laboratory,
# National Technology & Engineering Solutions of Sandia, LLC, Carnegie Mellon
# University, West Virginia University Research Corporation, et al.
# All rights reserved. Please see the files COPYRIGHT.md and LICENSE.md
# for full copyright and license information.
#################################################################################
# TODO: Missing doc strings
# pylint: disable=missing-module-docstring
from functools import lru_cache
import pyomo.environ as pyo
import pyomo.dae as pyodae
from pyomo.common.errors import ApplicationError
[docs]def lp():
"""This provides a simple LP model for solver testing.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for m.x
"""
m = pyo.ConcreteModel()
m.x = pyo.Var(initialize=3)
m.y = pyo.Var(initialize=3)
m.c1 = pyo.Constraint(expr=m.x >= 1)
m.c2 = pyo.Constraint(expr=m.y >= 2)
m.c3 = pyo.Constraint(expr=m.x <= 5)
m.c4 = pyo.Constraint(expr=m.y <= 5)
m.obj = pyo.Objective(expr=m.x + m.y)
return m, 1
[docs]def milp():
"""This provides a simple MILP model for solver testing.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for m.x
"""
m = pyo.ConcreteModel()
m.x = pyo.Var(domain=pyo.Integers, initialize=3)
m.y = pyo.Var(domain=pyo.Integers, initialize=3)
m.c1 = pyo.Constraint(expr=m.x >= 0.5)
m.c2 = pyo.Constraint(expr=m.y >= 1.5)
m.c3 = pyo.Constraint(expr=m.x <= 5)
m.c4 = pyo.Constraint(expr=m.y <= 5)
m.obj = pyo.Objective(expr=m.x + m.y)
return m, 1
[docs]def nle():
"""This provides a simple system of nonlinear equations model for solver
testing.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for m.x
"""
m = pyo.ConcreteModel()
m.x = pyo.Var(initialize=-0.1)
m.eq1 = pyo.Constraint(expr=m.x**3 == 1)
return m, 1
[docs]def nlp():
"""This provides a simple NLP model for solver testing.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for m.x
"""
m = pyo.ConcreteModel()
m.x = pyo.Var(initialize=-0.1)
m.y = pyo.Var(initialize=1)
m.c = pyo.Constraint(expr=m.x >= 1)
m.obj = pyo.Objective(expr=m.x**2 + m.y**2)
return m, 1
[docs]def minlp():
"""This provides a simple MINLP model for solver testing.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for m.x and m.i
"""
m = pyo.ConcreteModel()
m.x = pyo.Var(initialize=-0.1)
m.y = pyo.Var(initialize=1)
m.i = pyo.Var(domain=pyo.Binary, initialize=0)
m.c = pyo.Constraint(expr=m.x >= 1)
m.obj = pyo.Objective(
expr=m.i * (m.x**2 + m.y**2) + (1 - m.i) * 4 * (m.x**2 + m.y**2)
)
return m, 1, 1
[docs]def dae(nfe=1):
"""This provides a DAE model for solver testing.
The problem and expected result are from the problem given here:
https://archimede.dm.uniba.it/~testset/report/chemakzo.pdf.
Args:
None
Returns:
(tuple): Pyomo ConcreteModel, correct solved value for y[1] to y[5] and y6
"""
model = pyo.ConcreteModel(name="chemakzo")
# Set problem parameter values
model.k = pyo.Param([1, 2, 3, 4], initialize={1: 18.7, 2: 0.58, 3: 0.09, 4: 0.42})
model.Ke = pyo.Param(initialize=34.4)
model.klA = pyo.Param(initialize=3.3)
model.Ks = pyo.Param(initialize=115.83)
model.pCO2 = pyo.Param(initialize=0.9)
model.H = pyo.Param(initialize=737)
# Problem variables ydot = dy/dt,
# (dy6/dt is not explicitly in the equations, so only 5 ydots i.e.
# y6 is an algebraic variable and y1 to y5 are differential variables)
model.t = pyodae.ContinuousSet(bounds=(0, 180))
model.y = pyo.Var(model.t, [1, 2, 3, 4, 5], initialize=1.0) #
model.y6 = pyo.Var(model.t, initialize=1.0) #
model.ydot = pyodae.DerivativeVar(model.y, wrt=model.t) # dy/dt
model.r = pyo.Var(model.t, [1, 2, 3, 4, 5], initialize=1.0)
model.Fin = pyo.Var(model.t, initialize=1.0)
# Equations
@model.Constraint(model.t)
def eq_ydot1(b, t):
return b.ydot[t, 1] == -2.0 * b.r[t, 1] + b.r[t, 2] - b.r[t, 3] - b.r[t, 4]
@model.Constraint(model.t)
def eq_ydot2(b, t):
return b.ydot[t, 2] == -0.5 * b.r[t, 1] - b.r[t, 4] - 0.5 * b.r[t, 5] + b.Fin[t]
@model.Constraint(model.t)
def eq_ydot3(b, t):
return b.ydot[t, 3] == b.r[t, 1] - b.r[t, 2] + b.r[t, 3]
@model.Constraint(model.t)
def eq_ydot4(b, t):
return b.ydot[t, 4] == -b.r[t, 2] + b.r[t, 3] - 2.0 * b.r[t, 4]
@model.Constraint(model.t)
def eq_ydot5(b, t):
return b.ydot[t, 5] == b.r[t, 2] - b.r[t, 3] + b.r[t, 5]
@model.Constraint(model.t)
def eq_y6(b, t):
return 0 == b.Ks * b.y[t, 1] * b.y[t, 4] - b.y6[t]
@model.Constraint(model.t)
def eq_r1(b, t):
return b.r[t, 1] == b.k[1] * b.y[t, 1] ** 4 * b.y[t, 2] ** 0.5
@model.Constraint(model.t)
def eq_r2(b, t):
return b.r[t, 2] == b.k[2] * b.y[t, 3] * b.y[t, 4]
@model.Constraint(model.t)
def eq_r3(b, t):
return b.r[t, 3] == b.k[2] / b.Ke * b.y[t, 1] * b.y[t, 5]
@model.Constraint(model.t)
def eq_r4(b, t):
return b.r[t, 4] == b.k[3] * b.y[t, 1] * b.y[t, 4] ** 2
@model.Constraint(model.t)
def eq_r5(b, t):
return b.r[t, 5] == b.k[4] * b.y6[t] ** 2 * b.y[t, 2] ** 0.5
@model.Constraint(model.t)
def eq_Fin(b, t):
return b.Fin[t] == b.klA * (b.pCO2 / b.H - b.y[t, 2])
# Set initial conditions and solve initial from the values of differential
# variables.
y0 = {1: 0.444, 2: 0.00123, 3: 0.0, 4: 0.007, 5: 0.0} # initial differential vars
for i, v in y0.items():
model.y[0, i].fix(v)
discretizer = pyo.TransformationFactory("dae.finite_difference")
discretizer.apply_to(model, nfe=nfe, scheme="BACKWARD")
return (
model,
0.1150794920661702,
0.1203831471567715e-2,
0.1611562887407974,
0.3656156421249283e-3,
0.1708010885264404e-1,
0.4873531310307455e-2,
)
[docs]@lru_cache(maxsize=10)
def ipopt_has_linear_solver(linear_solver):
"""Check if IPOPT can use the specified linear solver.
Args:
linear_solver (str): linear solver in {"ma27", "ma57", "ma77", "ma86",
"ma97", "pardiso", "pardisomkl", "spral", "wsmp", "mumps"} or other
custom solver.
Returns:
(bool): True if Ipopt is available with the specified linear solver or False
if either Ipopt or the linear solver is not available.
"""
m, x = nlp()
solver = pyo.SolverFactory("ipopt", options={"linear_solver": linear_solver})
try:
solver.solve(m)
except ApplicationError:
return False
try:
assert abs(x - pyo.value(m.x)) < 1e-8
except AssertionError:
return False # solver mysteriously doesn't work right
return True
