Is there a way of changing the values of a constraint as the solver is running?
Basically, I have a constraint that depends on the value of a variable. The problem is that the constraint is evaluated based on the initial value of the variable, but isn't updated as the variable changes.
Here's a simple example:
from pyomo.environ import *
from pyomo.opt import SolverFactory
import numpy as np
# Setup
model = ConcreteModel()
model.A = Set(initialize = [0,1,2])
model.B = Set(initialize = [0,1,2])
model.x = Var(model.A, model.B, initialize=0)
# A constraint that I'd like to keep updating, based on the value of x
def changing_constraint_rule(model, a):
x_values = list((model.x[a, b].value for b in model.B))
if np.max(x_values) == 0:
return Constraint.Skip
else:
# Not really important what goes here, just as long as it updates the constraint list
if a == 1 : return sum(model.x[a,b] for b in model.B) == 0
else: return sum(model.x[a,b] for b in model.B) == 1
model.changing_constraint = Constraint(model.A, rule = changing_constraint_rule)
# Another constraint that changes the value of x
def bounding_constraint_rule(model, a):
return sum(model.x[a, b] for b in model.B) == 1
model.bounding_constraint = Constraint(
model.A,
rule = bounding_constraint_rule)
# Some objective function
def obj_rule(model):
return(sum(model.x[a,b] for a in model.A for b in model.B))
model.objective = Objective(rule=obj_rule)
# Results
opt = SolverFactory("glpk")
results = opt.solve(model)
results.write()
model.x.display()
If I run model.changing_constraint.pprint() I can see that no constraints have been made, since the initial value of the variable model.x was set to 0.
If it's not possible to change the constraint values while solving, how could I formulate this problem differently to achieve what I'm looking for? I've read this other post but couldn't figure it out from the instructions.
I am giving you the same answer in the other question by #Gabe:
Any if-logic you use inside of rules should not involve the values of
variables (unless it is based on the initial value of a variable, in
which case you would wrap the variable in value() wherever you use it
outside of the main expression that is returned).
for example:
model.x[a, b].value should be model.x[a, b].value()
But still this might not give you the solution what you are looking for.
Related
I am working on a small optimization model with some disjunctions. The way I did in a concrete model worked well:
from pyomo.environ import *
m = ConcreteModel()
m.d1 = Disjunct()
m.d2 = Disjunct()
m.d1.sub1 = Disjunct()
m.d1.sub2 = Disjunct()
m.d1.disj = Disjunction(expr=[m.d1.sub1, m.d1.sub2])
m.disj = Disjunction(expr=[m.d1, m.d2])
But now I tranfered the concrete model into an abstract formulation. I was able to fix everything instead of nesting the disjunctions. The way I did it was like:
#Disjunct 1
def _op_mode1(self, op_mode, t):
m = op_mode.model()
op_mode.c1 = po.Constraint(expr=m.x[t] == True)
#Disjunct 2
def _op_mode2(self, op_mode, t):
m = op_mode.model()
op_mode.c1 = po.Constraint(expr=m.x[t] == False)
#Disjunction 1
def _op_modes(self,m, t):
return [m.mode1[t], m.mode2[t]]
#Adding Components
self.model.del_component("mode1")
self.model.del_component("mode1_index")
self.model.add_component("mode1", pogdp.Disjunct(self.model.T, rule=self._op_mode1))
self.model.del_component("mode2")
self.model.del_component("mode2_index")
self.model.add_component("mode2", pogdp.Disjunct(self.model.T, rule=self._op_mode1))
self.model.del_component("modes")
self.model.del_component("modes_index")
self.model.add_component("modes", pogdp.Disjunction(self.model.T, rule=self._op_modes))`
As I previously mentioned, this works fine. But I haven`t found any way to nest the disjunctions. Pyomo alsways complains about the second layer of the disjuncts like "sub1".
Would anybody could give me a hint?
Many greetings
Joerg
The issue with the latest model above is that you are declaring m.d1 and m.d2 for each element of m.T, but they overwrite each other each time since they have the same name. You should be seeing warning messages logged for this. So if you uncomment your pprint of the model, you'll see that you only have the last ones you declared (with constraints on x[10]). So the first 9 Disjunctions in m.disjunction_ are disjunctions of Disjuncts that do not exist. The simplest fix for this is to give the disjuncts unique names when you declare them:
import pyomo.environ as pyo
import pyomo.gdp as pogdp
model = pyo.ConcreteModel()
model.T = pyo.RangeSet(0, 10)
model.x=pyo.Var(model.T,bounds=(-2, 10))
model.y=pyo.Var(model.T,bounds=(20, 30))
# This was also a duplicate declaration:
#model.disjunction_ = pogdp.Disjunction(model.T)
def d1(m, t):
disj = pogdp.Disjunct()
disj.c1= pyo.Constraint(expr=m.x[t] <= 10)
m.add_component('d1_%s' % t, disj)
return disj
def d2(m, t):
disj = pogdp.Disjunct()
disj.c1= pyo.Constraint(expr=m.x[t] >= 10)
m.add_component('d2_%s' % t, disj)
return disj
# sum x,y
def obj_rule(m):
return pyo.quicksum(pyo.quicksum([m.x[t] + m.y[t]], linear=False) for t in
m.T)
model.obj = pyo.Objective(rule=obj_rule)
def _op_mode_test(m, t):
disj1 = d1(m, t)
disj2 = d2(m, t)
return [disj1, disj2]
model.disjunction_ = pogdp.Disjunction(model.T, rule=_op_mode_test)
However, it would be cleaner (and probably easier down the line) to index the Disjuncts by m.T as well, since that's basically what the unique names are doing.
Block (and hence Disjunct rules) are passed the block (or disjunct) to be populated as the first argument. So, an "abstract" equivalent too your concrete model might look something like this:
model = AbstractModel()
#model.Disjunct()
def d1(d):
# populate the `d` disjunct (i.e., `model.d1`) here
pass
#model.Disjunct()
def d2(d):
#d.Disjunct()
def sub1(sd):
# populate the 'sub1' disjunct here
pass
#d.Disjunct()
def sub2(sd):
# populate the 'sub2' disjunct here
pass
d.disj = Disjunction(expr=[d.sub1, d.sub2])
model.disj = Disjunction(expr=[model.d1, model.d2])
There is a more fundamental question as to why you are converting your model over to "abstract" form. Pyomo Abstract models were mostly devised to be familiar to people coming from modeling in AMPL. While they will work with block-structured models, as AMPL was never really designed with blocks in mind, similarly block-oriented Abstract models tend to be unnecessarily cumbersome.
Here ist our new model:
import pyomo.environ as pyo
import pyomo.gdp as pogdp
model = pyo.ConcreteModel()
model.T = pyo.RangeSet(0,10)
model.x=pyo.Var(model.T,bounds=(-2, 10))
model.y=pyo.Var(model.T,bounds=(20, 30))
model.disjunction_=pogdp.Disjunction(model.T)
def d1(m,t):
m.d1 = pogdp.Disjunct()
m.d1.c1= pyo.Constraint(expr=m.x[t] <=10)
def d2(m,t):
m.d2 = pogdp.Disjunct()
m.d2.c1= pyo.Constraint(expr=m.x[t] >=10)
# sum x,y
def obj_rule(m):
return pyo.quicksum(pyo.quicksum([m.x[t] + m.y[t]], linear=False) for t in m.T)
model.obj = pyo.Objective(rule=obj_rule)
def _op_mode_test(m,t):
d1(m,t)
d2(m,t)
return [m.d1,m.d2]
model.disjunction_=pogdp.Disjunction(model.T,rule=_op_mode_test)
#model.pprint()
pyo.TransformationFactory('gdp.bigm').apply_to(model)
solver = pyo.SolverFactory('baron')
solver.solve(model)
print(pyo.value(model.obj))
I think it has something to do with the RangeSet. For a single step it works, but with more than one steps it throws an error: AttributeError: 'NoneType' object has no attribute 'component'
It would be great if you could have a look on it.
Many thanks
Here is the code which works pretty fine with bigm, but not with mbigm or hull transformation:
import pyomo.environ as pyo
import pyomo.gdp as pogdp
model = pyo.ConcreteModel()
model.T = pyo.RangeSet(2)
model.x=pyo.Var(model.T,bounds=(1, 10))
model.y=pyo.Var(model.T,bounds=(1, 100))
def _op_mode_sub(m, t):
m.disj1[t].sub1 = pogdp.Disjunct()
m.disj1[t].sub1.c1= pyo.Constraint(expr=m.y[t] == 60)
m.disj1[t].sub2 = pogdp.Disjunct()
m.disj1[t].sub2.c1= pyo.Constraint(expr=m.y[t] == 100)
return [m.disj1[t].sub1, m.disj1[t].sub2]
def _op_mode(m, t):
m.disj2[t].c1= pyo.Constraint(expr=m.y[t] >= 3)
m.disj2[t].c2= pyo.Constraint(expr=m.y[t] <= 5)
return [m.disj1[t], m.disj2[t]]
model.disj1 = pogdp.Disjunct(model.T)
model.disj2 = pogdp.Disjunct(model.T)
model.disjunction1sub = pogdp.Disjunction(model.T, rule=_op_mode_sub)
model.disjunction1 = pogdp.Disjunction(model.T, rule=_op_mode)
def obj_rule(m, t):
return pyo.quicksum(pyo.quicksum([m.x[t] + m.y[t]], linear=False) for t in m.T)
model.obj = pyo.Objective(rule=obj_rule)
model.pprint()
gdp_relax=pyo.TransformationFactory('gdp.bigm')
gdp_relax.apply_to(model)
solver = pyo.SolverFactory('glpk')
solver.solve(model)
print(pyo.value(model.obj))
This paper describes Pyomo's Differential and Algebraic Equations framework. It also mentions multi-stage problems; however, it does not show a complete example of such a problem. Does such an example exist somewhere?
The following demonstrates a complete minimum working example of a multi-stage optimization problem using Pyomo's DAE system:
#!/usr/bin/env python3
#http://www.gpops2.com/Examples/OrbitRaising.html
from pyomo.environ import *
from pyomo.dae import *
from pyomo.opt import SolverStatus, TerminationCondition
import random
import matplotlib.pyplot as plt
T = 10 #Maximum time for each stage of the model
STAGES = 3 #Number of stages
m = ConcreteModel() #Model
m.t = ContinuousSet(bounds=(0,T)) #Time variable
m.stages = RangeSet(0, STAGES) #Stages in the range [0,STAGES]. Can be thought of as an integer-valued set
m.a = Var(m.stages, m.t) #State variable defined for all stages and times
m.da = DerivativeVar(m.a, wrt=m.t) #First derivative of state variable with respect to time
m.u = Var(m.stages, m.t, bounds=(0,1)) #Control variable defined for all stages and times. Bounded to range [0,1]
#Setting the value of the derivative.
def eq_da(m,stage,t): #m argument supplied when function is called. `stage` and `t` are given values from m.stages and m.t (see below)
return m.da[stage,t] == m.u[stage,t] #Derivative is proportional to the control variable
m.eq_da = Constraint(m.stages, m.t, rule=eq_da) #Call constraint function eq_da for each unique value of m.stages and m.t
#We need to connect the different stages together...
def eq_stage_continuity(m,stage):
if stage==m.stages.last(): #The last stage doesn't connect to anything
return Constraint.Skip #So skip this constraint
else:
return m.a[stage,T]==m.a[stage+1,0] #Final time of each stage connects with the initial time of the following stage
m.eq_stage_continuity = Constraint(m.stages, rule=eq_stage_continuity)
#Boundary conditions
def _init(m):
yield m.a[0,0] == 0 #Initial value (at zeroth stage and zeroth time) of `a` is 0
yield ConstraintList.End
m.con_boundary = ConstraintList(rule=_init) #Repeatedly call `_init` until `ConstraintList.End` is returned
#Objective function: maximize `a` at the end of the final stage
m.obj = Objective(expr=m.a[STAGES,T], sense=maximize)
#Get a discretizer
discretizer = TransformationFactory('dae.collocation')
#Disrectize the model
#nfe (number of finite elements)
#ncp (number of collocation points within finite element)
discretizer.apply_to(m,nfe=30,ncp=6,scheme='LAGRANGE-RADAU')
#Get a solver
solver = SolverFactory('ipopt', keepfiles=True, log_file='/z/log', soln_file='/z/sol')
solver.options['max_iter'] = 100000
solver.options['print_level'] = 1
solver.options['linear_solver'] = 'ma27'
solver.options['halt_on_ampl_error'] = 'yes'
#Solve the model
results = solver.solve(m, tee=True)
print(results.solver.status)
print(results.solver.termination_condition)
#Retrieve the results in a pleasant format
r_t = [t for s in sorted(m.stages) for t in sorted(m.t)]
r_a = [value(m.a[s,t]) for s in sorted(m.stages) for t in sorted(m.t)]
r_u = [value(m.u[s,t]) for s in sorted(m.stages) for t in sorted(m.t)]
plt.plot(r_t, r_a, label="r_a")
plt.plot(r_t, r_u, label="r_u")
plt.legend()
plt.show()
I'm looking to create a simple indicator variable in Pyomo. Assuming I have a variable x, this indicator function would take the value 1 if x > 0, and 0 otherwise.
Here's how I've tried to do it:
model = ConcreteModel()
model.A = Set(initialize=[1,2,3])
model.B = Set(initialize=['J', 'K'])
model.x = Var(model.A, model.B, domain = NonNegativeIntegers)
model.ix = Var(model.A, model.B, domain = Binary)
def ix_indicator_rule(model, a, b):
return model.ix[a, b] == int(model.x[a, b] > 0)
model.ix_constraint = Constraint(model.A, model.B,
rule = ix_indicator_rule)
The error message I get is along the lines of Avoid this error by using Pyomo-provided math functions, which according to this link are found at pyomo.environ...but I'm not sure how to do this. I've tried using validate_PositiveValues(), like this:
def ix_indicator_rule(model, a, b):
return model.ix[a, b] == validate_PositiveValues(model.x[a, b])
model.ix_constraint = Constraint(model.A, model.B,
rule = ix_indicator_rule)
with no luck. Any help is appreciated!
You can achieve this with a "big-M" constraint, like this:
model = ConcreteModel()
model.A = Set(initialize=[1, 2, 3])
model.B = Set(initialize=['J', 'K'])
# m must be larger than largest allowed value of x, but it should
# also be as small as possible to improve numerical stability
model.m = Param(initialize=1e9)
model.x = Var(model.A, model.B, domain=NonNegativeIntegers)
model.ix = Var(model.A, model.B, domain=Binary)
# force model.ix to be 1 if model.x > 0
def ix_indicator_rule(model, a, b):
return model.x <= model.ix[a, b] * model.m
model.ix_constraint = Constraint(
model.A, model.B, rule=ix_indicator_rule
)
But note that the big-M constraint is one-sided. In this example it forces model.ix on when model.x > 0, but doesn't force it off when model.x == 0. You can achieve the latter (but not the former) by flipping the inequality to model.x >= model.ix[a, b] * model.m. But you can't do both in the same model. Generally you just pick the version that suits your model, e.g., if setting model.ix to 1 worsens your objective function, then you would pick the version shown above, and the solver will take care of setting model.ix to 0 whenever it can.
Pyomo also offers disjunctive programming features (see here and here) which may suit your needs. And the cplex solver offers indicator constraints, but I don't know whether Pyomo supports them.
I ended up using the Piecewise function and doing something like this:
DOMAIN_PTS = [0,0,1,1000000000]
RANGE_PTS = [0,0,1,1]
model.ix_constraint = Piecewise(
model.A, model.B,
model.ix, model.x,
pw_pts=DOMAIN_PTS,
pw_repn='INC',
pw_constr_type = 'EQ',
f_rule = RANGE_PTS,
unbounded_domain_var = True)
def objective_rule(model):
return sum(model.ix[a,b] for a in model.A for b in model.B)
model.objective = Objective(rule = objective_rule, sense=minimize)
It seems to work okay.
Hei all,
I am trying to set up an abstract model for a very simple QP of the form
min (x-x0)^2
s.t.
A x = b
C x <= d
I would like to use an abstract model, as I need to resolve with changing parameters (mainly x0, but potentially also A, b, C, d). I am right now struggeling with simply setting the parameters in the model instance. I do not want to use an external data file, but rather internal python variables. All examples I find online use AMPL formatted data files.
This is the code I have right now
import pyomo.environ as pe
model = pe.AbstractModel()
# the sets
model.n = pe.Param(within=pe.NonNegativeIntegers)
model.m = pe.Param(initialize = 1)
model.ss = pe.RangeSet(1, model.n)
model.os = pe.RangeSet(1, model.m)
# the starting point and the constraint parameters
model.x_hat = pe.Param(model.ss)
model.A = pe.Param(model.os, model.ss)
model.b = pe.Param(model.os)
model.C = pe.Param(model.os, model.os)
model.d = pe.Param(model.ss, model.os)
# the decision variables
model.x_projected = pe.Var(model.ss)
# the cosntraints
# A x = b
def sum_of_elements_rule(model):
value = model.A * model.x_projected
return value == model.d
model.sumelem = pe.Constraint(model.os, rule=sum_of_elements_rule)
# C x <= d
def positivity_constraint(model):
return model.C*model.x_projected <= model.d
model.bounds = pe.Constraint(model.ss, rule=positivity_constraint)
# the cost
def cost_rule(model):
return sum((model.x_projected[i] - model.x[i])**2 for i in model.ss)
model.cost = pe.Objective(rule=cost_rule)
instance = model.create_instance()
And somehow here I am stuck. How do I set the parameters now?
Thanks and best, Theo
I know this is an old post but a solution to this could have helped me so here is the solution to this problem:
## TEST
data_init= {None: dict(
n = {None : 3},
d = {0:0, 1:1, 2:2},
x_hat = {0:10, 1:-1, 2:-100},
b = {None: 10}
)}
# create instance
instance = model.create_instance(data_init)
This creates the instance in an equivalent way than what you did but in a more formal way.
Ok, I seemed to have figured out what the problem is. If I want to set a parameter after I create an instance, I need the
mutable=True
flag. Then, I can set the parameter with something like
for i in range(model_dimension):
getattr(instance, 'd')[i] = i
The model dimension I need to choose before i create an instance (which is ok for my case). The instance can be reused with different parameters for the constraints.
The code below should work for the problem
min (x-x_hat)' * (x-x_hat)
s.t.
sum(x) = b
x[i] >= d[i]
with x_hat, b, d as parameters.
import pyomo.environ as pe
model = pe.AbstractModel()
# model dimension
model.n = pe.Param(default=2)
# state space set
model.ss = pe.RangeSet(0, model.n-1)
# equality
model.b = pe.Param(default=5, mutable=True)
# inequality
model.d = pe.Param(model.ss, default=0.0, mutable=True)
# decision var
model.x = pe.Var(model.ss)
model.x_hat = pe.Param(model.ss, default=0.0, mutable=True)
# the cost
def cost_rule(model):
return sum((model.x[i] - model.x_hat[i])**2 for i in model.ss)
model.cost = pe.Objective(rule=cost_rule)
# CONSTRAINTS
# each x_i bigger than d_i
def lb_rule(model, i):
return (model.x[i] >= model.d[i])
model.state_bound = pe.Constraint(model.ss, rule=lb_rule)
# sum of x == P_tot
def sum_rule(model):
return (sum(model.x[i] for i in model.ss) == model.b)
model.state_sum = pe.Constraint(rule=sum_rule)
## TEST
# define model dimension
model_dimension = 3
model.n = model_dimension
# create instance
instance = model.create_instance()
# set d
for i in range(model_dimension):
getattr(instance, 'd')[i] = i
# set x_hat
xh = (10,1,-100)
for i in range(model_dimension):
getattr(instance, 'x_hat')[i] = xh[i]
# set b
instance.b = 10
# solve
solver = pe.SolverFactory('ipopt')
result = solver.solve(instance)
instance.display()
I am just starting with Pyomo and I have a big problem.
I want to run an Abstract Model without using the terminal. I can do it with a concrete model but I have serious problems to do it in with the abstract one.
I just want to use F5 and run the code.
This ismy program:
import pyomo
from pyomo.environ import *
#
# Model
#
model = AbstractModel()
#Set: Indices
model.Unit = Set()
model.Block = Set()
model.DemBlock = Set()
#Parameters
model.EnergyBid = Param(model.Unit, model.Block)
model.PriceBid = Param(model.Unit, model.Block)
model.EnergyDem = Param(model.DemBlock)
model.PriceDem = Param(model.DemBlock)
model.Pmin = Param(model.Unit)
model.Pmax = Param(model.Unit)
#Variables definition
model.PD = Var(model.DemBlock, within=NonNegativeReals)
model.PG = Var(model.Unit,model.Block, within=NonNegativeReals)
#Binary variable
model.U = Var(model.Unit, within = Binary)
#Objective
def SocialWellfare(model):
SocialWellfare = sum([model.PriceDem[i]*model.PD[i] for i in model.DemBlock]) - sum([model.PriceBid[j,k]*model.PG[j,k] for j in model.Unit for k in model.Block ])
return SocialWellfare
model.SocialWellfare = Objective(rule=SocialWellfare, sense=maximize)
#Constraints
#Max and min Power generated
def PDmax_constraint(model,p):
return ((model.PD[p] - model.EnergyDem[p])) <= 0
model.PDmax = Constraint(model.DemBlock, rule=PDmax_constraint)
def PGmax_constraint(model,n,m):
return ((model.PG[n,m] - model.EnergyBid[n,m])) <= 0
model.PGmax = Constraint(model.Unit, model.Block,rule = PGmax_constraint)
def Power_constraintDW(model,i):
return ((sum(model.PG[i,k] for k in model.Block))-(model.Pmin[i] * model.U[i]) ) >= 0
model.LimDemandDw = Constraint(model.Unit, rule=Power_constraintDW)
def Power_constraintUP(model,i):
return ((sum(model.PG[i,k] for k in model.Block) - (model.Pmax[i])*model.U[i])) <= 0
model.LimDemandaUp = Constraint(model.Unit, rule=Power_constraintUP)
def PowerBalance_constraint(model):
return (sum(model.PD[i] for i in model.DemBlock) - sum(model.PG[j,k] for j in model.Unit for k in model.Block)) == 0
model.PowBalance = Constraint(rule = PowerBalance_constraint)
model.pprint()
instance = model.create('datos_transporte.dat')
## Create the ipopt solver plugin using the ASL interface
solver = 'ipopt'
solver_io = 'nl'
opt = SolverFactory(solver,solver_io=solver_io)
results = opt.solve(instance)
results.write()
Any help with the last part??
Thanks anyway,
I think your example is actually working. Starting in Pyomo 4.1, the solution returned from the solver is stored directly into the instance that was solved, and is not returned in the solver results object. This change was made because generating the representation of the solution in the results object was rather expensive and was not easily parsed by people. Working with the model instance directly is more natural.
It is unfortunate that the results object reports the number of solutions: 0, although this is technically correct: the results object holds no solutions ... but the Solver section should indicate that a solution was returned and stored into the model instance.
If you want to see the result returned by the solver, you can print out the current status of the model using:
instance.display()
after the call to solve(). That will report the current Var values that were returned from the solver. You will want to pay attention to the stale column:
False indicates the value is not "stale" ... that is, it was either set by the user (before the call to solve()), or it was returned from the solver (after the call to solve()).
True indicates the solver did not return a value for that variable. This is usually because the variable was not referenced by the objective or any enabled constraints, so Pyomo never sent the variable to the solver.
[Note: display() serves a slightly different role than pprint(): pprint() outputs the model structure, whereas display() outputs the model state. So, for example, where pprint() will output the constraint expressions, display() will output the numerical value of the expression (using the current variable/param values).]
Edited to expand the discussion of display() & pprint()