本文整理汇总了Python中pyomo.core.ConcreteModel.obj方法的典型用法代码示例。如果您正苦于以下问题：Python ConcreteModel.obj方法的具体用法？Python ConcreteModel.obj怎么用？Python ConcreteModel.obj使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类pyomo.core.ConcreteModel的用法示例。
在下文中一共展示了ConcreteModel.obj方法的8个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: test_indexedvar_noindextemplate
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def test_indexedvar_noindextemplate(self):
st_model = CreateConcreteTwoStageScenarioTreeModel(1)
st_model.StageVariables['Stage1'].add("x")
st_model.StageDerivedVariables['Stage1'].add("y")
st_model.NodeVariables['RootNode'].add("z")
st_model.NodeDerivedVariables['RootNode'].add("q")
st_model.StageCost['Stage1'] = "FirstStageCost"
st_model.StageCost['Stage2'] = "SecondStageCost"
scenario_tree = ScenarioTree(scenariotreeinstance=st_model)
self.assertEqual(len(scenario_tree.stages), 2)
self.assertEqual(len(scenario_tree.nodes), 2)
self.assertEqual(len(scenario_tree.scenarios), 1)
model = ConcreteModel()
model.s = Set(initialize=[1,2,3])
model.x = Var(model.s)
model.y = Var(model.s)
model.z = Var(model.s)
model.q = Var(model.s)
model.FirstStageCost = Expression(expr=0.0)
model.SecondStageCost = Expression(expr=0.0)
model.obj = Objective(expr=0.0)
scenario_tree.linkInInstances({'Scenario1': model})
root = scenario_tree.findRootNode()
self.assertEqual(len(root._variable_ids), 12)
self.assertEqual(len(root._standard_variable_ids), 6)
self.assertEqual(len(root._derived_variable_ids), 6)
for name in ("x", "y", "z", "q"):
for index in model.s:
self.assertEqual(
(name,index) in root._name_index_to_id, True)示例2: test_singletonvar_wildcardtemplate
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def test_singletonvar_wildcardtemplate(self):
st_model = CreateConcreteTwoStageScenarioTreeModel(1)
st_model.StageVariables['Stage1'].add("x[*]")
st_model.StageDerivedVariables['Stage1'].add("y[*]")
st_model.NodeVariables['RootNode'].add("z[*]")
st_model.NodeDerivedVariables['RootNode'].add("q[*]")
st_model.StageCost['Stage1'] = "FirstStageCost"
st_model.StageCost['Stage2'] = "SecondStageCost"
scenario_tree = ScenarioTree(scenariotreeinstance=st_model)
self.assertEqual(len(scenario_tree.stages), 2)
self.assertEqual(len(scenario_tree.nodes), 2)
self.assertEqual(len(scenario_tree.scenarios), 1)
model = ConcreteModel()
model.x = Var()
model.y = Var()
model.z = Var()
model.q = Var()
model.FirstStageCost = Expression(expr=0.0)
model.SecondStageCost = Expression(expr=0.0)
model.obj = Objective(expr=0.0)
scenario_tree.linkInInstances({'Scenario1': model})
root = scenario_tree.findRootNode()
self.assertEqual(len(root._variable_ids), 4)
self.assertEqual(len(root._standard_variable_ids), 2)
self.assertEqual(len(root._derived_variable_ids), 2)
for name in ("x", "y", "z", "q"):
for index in [None]:
self.assertEqual(
(name,index) in root._name_index_to_id, True)示例3: select_tear_mip_model
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def select_tear_mip_model(self, G):
"""
Generate a model for selecting tears from the given graph
Returns
-------
model
bin_list
A list of the binary variables representing each edge,
indexed by the edge index of the graph
"""
model = ConcreteModel()
bin_list = []
for i in range(G.number_of_edges()):
# add a binary "torn" variable for every edge
vname = "edge%s" % i
var = Var(domain=Binary)
bin_list.append(var)
model.add_component(vname, var)
# var containing the maximum number of times any cycle is torn
mct = model.max_cycle_tears = Var()
_, cycleEdges = self.all_cycles(G)
for i in range(len(cycleEdges)):
ecyc = cycleEdges[i]
# expression containing sum of tears for each cycle
ename = "cycle_sum%s" % i
expr = Expression(expr=sum(bin_list[i] for i in ecyc))
model.add_component(ename, expr)
# every cycle must have at least 1 tear
cname_min = "cycle_min%s" % i
con_min = Constraint(expr=expr >= 1)
model.add_component(cname_min, con_min)
# mct >= cycle_sum for all cycles, thus it becomes the max
cname_mct = mct.name + "_geq%s" % i
con_mct = Constraint(expr=mct >= expr)
model.add_component(cname_mct, con_mct)
# weigh the primary objective much greater than the secondary
obj_expr = 1000 * mct + sum(var for var in bin_list)
model.obj = Objective(expr=obj_expr, sense=minimize)
return model, bin_list示例4: makeNestedDisjunctions_NestedDisjuncts
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def makeNestedDisjunctions_NestedDisjuncts():
m = ConcreteModel()
m.x = Var(bounds=(0, 2))
m.obj = Objective(expr=m.x)
m.d1 = Disjunct()
m.d1.c = Constraint(expr=m.x >= 1)
m.d2 = Disjunct()
m.d2.c = Constraint(expr=m.x >= 1.1)
m.d1.d3 = Disjunct()
m.d1.d3.c = Constraint(expr=m.x >= 1.2)
m.d1.d4 = Disjunct()
m.d1.d4.c = Constraint(expr=m.x >= 1.3)
m.disj = Disjunction(expr=[m.d1, m.d2])
m.d1.disj2 = Disjunction(expr=[m.d1.d3, m.d1.d4])
return m示例5: _get_block_model
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def _get_block_model(self):
model = ConcreteModel()
model.s = Set(initialize=[1,2])
b = Block(concrete=True)
b.s = Set(initialize=[1,2])
b.x = Var()
b.X = Var(model.s)
model.b1 = b.clone()
model.b2 = b.clone()
model.b3 = b.clone()
model.b4 = b.clone()
model.B1 = Block(model.s, rule=lambda _,i: b.clone())
model.B2 = Block(model.s, rule=lambda _,i: b.clone())
model.B3 = Block(model.s, rule=lambda _,i: b.clone())
model.B4 = Block(model.s, rule=lambda _,i: b.clone())
model.FirstStageCost = Expression(expr=0.0)
model.SecondStageCost = Expression(expr=0.0)
model.obj = Objective(expr=0.0)
return model示例6: build_nonexclusive_model
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
#.........这里部分代码省略.........
# No Unit 1
m.x[2] == 0, m.x[3] == 0
]
])
m.unit2 = Disjunction(expr=[
[
# Unit 2
m.x[5] == log(m.x[4] + 1),
],
[
# No Unit 2
m.x[4] == 0, m.x[5] == 0
]
])
m.stage1_mix = Constraint(expr=m.x[3] + m.x[5] == m.x[6])
m.stage2_split = Constraint(expr=m.x[6] == sum(m.x[i] for i in (7, 9, 11, 13)))
m.unit3 = Disjunction(expr=[
[
# Unit 3
m.x[8] == 2 * log(m.x[7]) + 3,
m.x[7] >= 0.2,
],
[
# No Unit 3
m.x[7] == 0, m.x[8] == 0
]
])
m.unit4 = Disjunction(expr=[
[
# Unit 4
m.x[10] == 1.8 * log(m.x[9] + 4),
],
[
# No Unit 4
m.x[9] == 0, m.x[10] == 0
]
])
m.unit5 = Disjunction(expr=[
[
# Unit 5
m.x[12] == 1.2 * log(m.x[11]) + 2,
m.x[11] >= 0.001,
],
[
# No Unit 5
m.x[11] == 0, m.x[12] == 0
]
])
m.unit6 = Disjunction(expr=[
[
# Unit 6
m.x[15] == sqrt(m.x[14] - 3) * m.x[23] + 1,
m.x[14] >= 5, m.x[14] <= 20,
],
[
# No Unit 6
m.x[14] == 0, m.x[15] == 0
]
])
m.stage2_special_mix = Constraint(expr=m.x[14] == m.x[13] + m.x[23])
m.stage2_mix = Constraint(expr=sum(m.x[i] for i in (8, 10, 12, 15)) == m.x[16])
m.stage3_split = Constraint(expr=m.x[16] == sum(m.x[i] for i in (17, 19, 21)))
m.unit7 = Disjunction(expr=[
[
# Unit 7
m.x[18] == m.x[17] * 0.9,
],
[
# No Unit 7
m.x[17] == 0, m.x[18] == 0
]
])
m.unit8 = Disjunction(expr=[
[
# Unit 8
m.x[20] == log(m.x[19] ** 1.5) + 2,
m.x[19] >= 1,
],
[
# No Unit 8
m.x[19] == 0, m.x[20] == 0
]
])
m.unit9 = Disjunction(expr=[
[
# Unit 9
m.x[22] == log(m.x[21] + sqrt(m.x[21])) + 1,
m.x[21] >= 4,
],
[
# No Unit 9
m.x[21] == 0, m.x[22] == 0
]
])
m.stage3_special_split = Constraint(expr=m.x[22] == m.x[23] + m.x[24])
m.stage3_mix = Constraint(expr=m.x[25] == sum(m.x[i] for i in (18, 20, 24)))
m.obj = Objective(expr=-10 * m.x[25] + m.x[1])
return m示例7: build_model
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def build_model():
"""
Base Model
Optimal solution:
Select units 1, 3, 8
Objective value -36.62
"""
m = ConcreteModel()
m.streams = RangeSet(25)
m.x = Var(m.streams, bounds=(0, 50), initialize=5)
m.stage1_split = Constraint(expr=m.x[1] == m.x[2] + m.x[4])
m.first_stage = Disjunction(expr=[
[
# Unit 1
m.x[2] == exp(m.x[3]) - 1,
m.x[4] == 0, m.x[5] == 0
],
[
# Unit 2
m.x[5] == log(m.x[4] + 1),
m.x[2] == 0, m.x[3] == 0
]
])
m.stage1_mix = Constraint(expr=m.x[3] + m.x[5] == m.x[6])
m.stage2_split = Constraint(expr=m.x[6] == sum(m.x[i] for i in (7, 9, 11, 13)))
m.second_stage = Disjunction(expr=[
[
# Unit 3
m.x[8] == 2 * log(m.x[7]) + 3,
m.x[7] >= 0.2,
] + [m.x[i] == 0 for i in (9, 10, 11, 12, 14, 15)],
[
# Unit 4
m.x[10] == 1.8 * log(m.x[9] + 4),
] + [m.x[i] == 0 for i in (7, 8, 11, 12, 14, 15)],
[
# Unit 5
m.x[12] == 1.2 * log(m.x[11]) + 2,
m.x[11] >= 0.001,
] + [m.x[i] == 0 for i in (7, 8, 9, 10, 14, 15)],
[
# Unit 6
m.x[15] == sqrt(m.x[14] - 3) * m.x[23] + 1,
m.x[14] >= 5, m.x[14] <= 20,
] + [m.x[i] == 0 for i in (7, 8, 9, 10, 11, 12)]
])
m.stage2_special_mix = Constraint(expr=m.x[14] == m.x[13] + m.x[23])
m.stage2_mix = Constraint(expr=sum(m.x[i] for i in (8, 10, 12, 15)) == m.x[16])
m.stage3_split = Constraint(expr=m.x[16] == sum(m.x[i] for i in (17, 19, 21)))
m.third_stage = Disjunction(expr=[
[
# Unit 7
m.x[18] == m.x[17] * 0.9,
] + [m.x[i] == 0 for i in (19, 20, 21, 22)],
[
# Unit 8
m.x[20] == log(m.x[19] ** 1.5) + 2,
m.x[19] >= 1,
] + [m.x[i] == 0 for i in (17, 18, 21, 22)],
[
# Unit 9
m.x[22] == log(m.x[21] + sqrt(m.x[21])) + 1,
m.x[21] >= 4,
] + [m.x[i] == 0 for i in (17, 18, 19, 20)]
])
m.stage3_special_split = Constraint(expr=m.x[22] == m.x[23] + m.x[24])
m.stage3_mix = Constraint(expr=m.x[25] == sum(m.x[i] for i in (18, 20, 24)))
m.obj = Objective(expr=-10 * m.x[25] + m.x[1])
return m示例8: criticalityCheck
# 需要导入模块: from pyomo.core import ConcreteModel [as 别名]
# 或者: from pyomo.core.ConcreteModel import obj [as 别名]
def criticalityCheck(self, x, y, z, rom_params, worstcase=False, M=[0.0]):
model = self.model
self.setVarValue(x=x,y=y,z=z)
self.setBound(x, y, z, 1e10)
self.deactiveExtraConObj()
self.activateRomCons(x, rom_params)
optGJH = SolverFactory('contrib.gjh')
optGJH.solve(model, tee=False, symbolic_solver_labels=True)
g, J, varlist, conlist = model._gjh_info
l = ConcreteModel()
l.v = Var(varlist, domain=Reals)
for i in varlist:
#dummy = model.find_component(i)
l.v[i] = 0.0
l.v[i].setlb(-1.0)
l.v[i].setub(1.0)
if worstcase:
if M.all() == 0.0:
print('WARNING: worstcase criticality was requested but Jacobian error bound is zero')
l.t = Var(range(0, self.ly), domain=Reals)
for i in range(0, self.ly):
l.t[i].setlb(-M[i])
l.t[i].setub(M[i])
def linConMaker(l, i):
# i should be range(len(conlist) - 1)
# because last element of conlist is the objective
con_i = model.find_component(conlist[i])
isEquality = con_i.equality
isROM = False
if conlist[i][:7] == '.' + self.TRF.name + '.rom':
isROM = True
romIndex = int(filter(str.isdigit, conlist[i]))
# This is very inefficient
# Fix this later if these problems get big
# This is the ith row of J times v
Jv = sum(x[2] * l.v[varlist[x[1]]] for x in J if x[0] == i)
if isEquality:
if worstcase and isROM:
return Jv + l.t[romIndex] == 0
else:
return Jv == 0
else:
lo = con_i.lower
up = con_i.upper
if lo is not None:
level = lo.value - con_i.lslack()
if up is not None:
return (lo.value <= level + Jv <= up.value)
else:
return (lo.value <= level + Jv)
elif up is not None:
level = up.value - con_i.uslack()
return (level + Jv <= up.value)
else:
raise Exception(
"This constraint seems to be neither equality or inequality: " + conlist(i))
l.lincons = Constraint(range(len(conlist)-1), rule=linConMaker)
l.obj = Objective(expr=sum(x[1] * l.v[varlist[x[0]]] for x in g))
# Calculate gradient norm for scaling purposes
gfnorm = sqrt(sum(x[1]**2 for x in g))
opt = SolverFactory(self.config.solver)
opt.options.update(self.config.solver_options)
#opt.options['halt_on_ampl_error'] = 'yes'
#opt.options['max_iter'] = 5000
results = opt.solve(
l, keepfiles=self.keepfiles, tee=self.stream_solver)
if ((results.solver.status == SolverStatus.ok)
and (results.solver.termination_condition == TerminationCondition.optimal)):
l.solutions.load_from(results)
if gfnorm > 1:
return True, abs(l.obj())/gfnorm
else:
return True, abs(l.obj())
else:
print("Waring: Crticality check fails with solver Status: " + str(results.solver.status))
print("And Termination Conditions: " + str(results.solver.termination_condition))
return False, infinity