I would like to obtain the coefficients of a linear constraint c of a pyomo model m.
For instance, for
m= ConcreteModel()
m.x_1 = Var()
m.x_2 = Var()
m.x_3 = Var(within = Integers)
m.x_4 = Var(within = Integers)
m.c= Constraint(expr=2*m.x_1 + 5*m.x_2 + m.x_4 <= 2)
I would like to get the array c_coef = [2,5,0,1].
The answer to this question explains how to obtain all variables occurring in a linear constraint and I can easily use this to create the zero-coefficients for variables which don't occur in a constraint. However, I am struggling with the nonzero-coefficients. My current approach uses the private attribute _coef, that is c_nzcoef = m.c.body._coef which I probably should not use.
What would be the proper way to obtain the nonzero coefficients?
The easiest way to get the coefficients for a linear expression is to make use of the "Canonical Representation" data structure:
from pyomo.repn import generate_canonical_repn
# verify that the expression is linear
if m.c.body.polynominal_degree() == 1:
repn = generate_canonical_repn(m.c.body)
for i, coefficient in enumerate(repn.linear or []):
var = repn.variables[i]
This should be valid for any version of Pyomo from 4.0 through at least 5.3.
Related
This question is related to my previous question found here. I have managed to solve this problem (big thanks to #AirSquid!) My objective function is something like:
So the avgPrice_n variable is indexed by n. However, it is actually defined as
Meaning that it is indexed by n and i.
So at the moment my objective function is very messy as I have three sums. It looks something like (I expanded the brackets in the objective function and added each component separately, so the avgPrice_n*demand_n looks like):
expr += sum(sum(sum((1/12)*model.c[i]*model.allocation[i,n] for i in model.MP[t]) for t in model.M)*model.demand_n[n] for n in model.N)
And while this works, debugging was quite difficult because the terms are very long. So intead of using the actual definition of avgPrice_n, I was wondering if it would be possible to create a avgPrice_n variable, use this in the objective function and then create a constraint where I define avgPrice_n as I showed above.
The issue I am having is that I created my decision variable, x_{i,n}, as a variable but apparently I can't create a avgPrice_n as a variable where I index it by x_{i,n} as this results in a TypeError: Cannot apply a Set operator to an indexed Var component (allocation) error.
So as of now my decision variable looks like:
model.x = Var(model.NP_flat, domain = NonNegativeReals)
And I tried to create:
model.avg_Price = Var(model.x, domain = NonNegativeReals)
Which resulted in the above error. Any ideas or suggestions would be much appreciated!
You have a couple options. Realize you do not need the model.avg_price variable because you can construct it from other variables and you would have to make some constraints to constrain the value, etc. etc. and pollute your model.
The basic building blocks in the model are pyomo expressions, so you could put in a little "helper function" to build expressions (the cost function shown, which is dependent on n) which are not defined within the model, but just pop out an expression...totally legal). You can also "break up" large expressions into smaller expressions (like the other_stuff below) and then just kludge them all together in the objective (or where needed) this gives you the opportunity to evaluate them independently. I've made several models with an objective function that has a "cost" component and a "penalty" component by dividing it into 2 expressions.... Then when solved, you can inspect them independently.
My suggestion (if you don't like the triple sum in your current model) is to make an avg_cost(n) function to build the expression similar to what is done in the nonsensical function below, and use that as a substitute for a new variable.
Note: the initialization of the variables here is generally unnecessary. I just did it to "simulate solving" or they would be None...
Code:
import pyomo.environ as pyo
m = pyo.ConcreteModel()
m.N = pyo.Set(initialize=[0,1,2])
m.x = pyo.Var(m.N, initialize = 2.0)
def cost(n):
return m.x[n] + 2*m.x[n+1]
m.other_stuff = 3 * m.x[1] + 4 * m.x[2]
m.costs = sum(cost(n) for n in {0,1})
m.obj_expr = m.costs + m.other_stuff
m.obj = pyo.Objective(expr= m.obj_expr)
# inspect cost at a particular value of n...
print(cost(1))
print(pyo.value(cost(1)))
# inspect the pyomo expressions "other_stuff" and total costs...
print(m.other_stuff)
print(pyo.value(m.other_stuff))
print(m.costs)
print(pyo.value(m.costs))
# inspect the objective... which can be accessed by pprint() and display()
m.obj.pprint()
m.obj.display()
Output:
x[1] + 2*x[2]
6.0
3*x[1] + 4*x[2]
14.0
12.0
obj : Size=1, Index=None, Active=True
Key : Active : Sense : Expression
None : True : minimize : x[0] + 2*x[1] + x[1] + 2*x[2] + 3*x[1] + 4*x[2]
obj : Size=1, Index=None, Active=True
Key : Active : Value
None : True : 26.0
I have recently started in doing some OR, and have been trying to use Pyomo and NEOS to do some optimation problems. I have been following along with one of the UT Austin Pyomo lectures, and when my GLPT was being difficult to be installed, I moved on to NEOS. I am having some difficulty in now receiving a solved answer from NEOS.
What I have so far is this:
from pyomo import environ as pe
import os
os.environ['NEOS_EMAIL'] = 'my registered email'
model = pe.ConcreteModel()
model.x1 = pe.Var(domain=pe.Binary)
model.x2 = pe.Var(domain=pe.Binary)
model.x3 = pe.Var(domain=pe.Binary)
model.x4 = pe.Var(domain=pe.Binary)
model.x5 = pe.Var(domain=pe.Binary)
obj_expr = 3 * model.x1 + 4 * model.x2 + 5 * model.x3 + 8 * model.x4 + 9 * model.x5
model.obj = pe.Objective(sense=pe.maximize, expr=obj_expr)
con_expr = 2 * model.x1 + 3 * model.x2 + 4 * model.x3 + 5 * model.x4 + 9 * model.x5 <= 20
model.con = pe.Constraint(expr=con_expr)
solver_manager = pe.SolverManagerFactory('neos')
results = solver_manager.solve(model, solver = "minos")
print(results)
What I receive in return is number of solutions = 0, while I know for a fact that one exits. I also see that I don't have any bounds set, so how would I go about doing that? Once again, I am very new to this, and have not been able to find any sort of documentation regarding this elsewhere, or perhaps I just don't know how to look.
Thanks for any help!
This is a "problem" with the design of the current results object. For historical reasons, that field reports the number of solutions contained in the results object and is not the number of solutions generated by the solver. By default, Pyomo solvers directly load the solution returned by the solver into the original model (both for convenience and efficiency) and do not return it in the results object. You can change that behavior by providing load_solutions=False to the solve() call.
As for the bounds, what bounds are you referring to? Variable bounds are set using either the bounds= argument to the Var() declaration, or the domain= argument. For your example, because the variables are declared to be Binary, they all have bounds of [0..1]. Bounds on the objective are gathered by parsing the solver output. This is dependent on bother the solver that you are using (many do not report bounds information), and the interface used to parse the solver results.
As a final note, you are sending a MIP problem to a LP/NLP solver (minos). You will get fractional valies for your binary variables back from the solver.
To retrieve the solution from the model, you can use something like:
print(model.x1.value, model.x2.value, model.x3.value, model.x4.value, model.x5.value)
And using solver="cbc" you can avoid fractional values in this example.
I'm modeling a reoptimisation model and I would like to include a constraint in order to reduce the distance between the initial solution and the reoptimized solution. I'm doing a staff scheduling and to do so I wanna penalized each assignment in the reoptimized solution that is different from the initial solution.
Before I start, I'm new to optimisation model and the way I built the constraint may be wrong.
#1 Extract the data from the initial solution of my main variable
ModelX_DictExtVal = model.x.extract_values()
# 2 Create a new binary variable which activate when the main variable `ModelX_DictExtVal[x,s,d]` of the initial
#solution is =1 (an employee n works days d and sifht s) and the value of `model.x[n,s,d]` of the reoptimized solution are different.
model.alpha_distance = Var(model.N_S_D, within=Binary)
#3 Model a constraint to activate my variable.
def constraint_distance(model, n, s, d):
v = ModelX_DictExtVal[n,s,d]
if v == 1 and ModelX_DictExtVal[n,s,d] != model.x[n,s,d]:
return model.alpha_distance[n,s,d] == 1
elif v == 0:
return model.alpha_distance[n,s,d] == 0
model.constraint_distance = Constraint(model.N_S_D, rule = constraint_distance)
#4 Penalize in my objective function every time the varaible is equal to one
ObjFunction = Objective(expr = sum(model.alpha_distance[n,s,d] * WeightDistance
for n in model.N for s in model.S for d in model.D))
Issue: I'm not sure about what I'm doing in part 3 and I get an index error when v == 1.
ERROR: Rule failed when generating expression for constraint
constraint_distance with index (0, 'E', 6): ValueError: Constraint
'constraint_distance[0,E,6]': rule returned None
I am wondering since I am reusing the same model for re-optimization if the model keeps the value of the initial solution of model.x [n, s, d] to do the comparison ModelX_DictExtVal [n, s, d]! = model.x [n, s, d] during the re-optimization phase instead of the new assignments...
You are right to suspect part 3. :)
So you have some "initial values" that could be either the original schedule (before optimizing) or some other preliminary optimization. And your decision variable is binary, indexed by [n,s,d] if I understand your question.
In your constraint you cannot employ an if-else structure based on a comparison test of your decision variable. The value of that variable is unknown at the time the constraint is built, right?
You are on the right track, though. So, what you really want to do is to have your alpha_distance (or penalty) variable capture any changes, indicating 1 where there is a change. That is an absolute value operation, but can be captured with 2 constraints. Consider (in pseudocode):
penalty = |x.new - x.old| # is what you want
So introduce 2 constraints, (indexed fully by [n,s,d]):
penalty >= x.new - x.old
penalty >= x.old - x.new
Then, as you are doing now, include the penalty in your objective, optionally multiplied by a weight.
Comment back if that doesn't make sense...
I'm trying to model a multi-component reaction diffusion system using FiPy. Each component, \phi_i, has a diffusive term and multiple bi-molecular reaction terms (\phi_m, \phi_n). Each component's time evolution is given by the following equation:
\frac{d\phi_i}{dt} = D_i\nabla \phi_i + \sum_{m,n} k_{m,n}\phi_m \phi_n ,
which I represent in FiPy as
eq = TransientTerm() == DiffusionTerm(D_i) + ReactionTerm
where
if i!=m and i!=n: ReactionTerm = k_mn * \phi_m * phi_n
elif i==m and i!=n: ReactionTerm = ImplicitSourceTerm(k_mn * \phi_n)
elif i!=m and i==n: ReactionTerm = ImplicitSourceTerm(k_mn * \phi_m)
How do I represent the ReactionTerm in the case that i==m==n? In other words, how does one represent a term quadratically dependent on the variable being solved for, as shown in the following equation:
\frac{d\phi_i}{dt} = \phi_i \cdot \phi_i
Until now, I have been representing it explicitly, but I wonder if there is a way to represent it implicitly like I do for the other terms which include the component being solved for.
Implicitness is linear in all cases. ImplicitSourceTerm(coeff=k_mn * phi_i) will do.
In all cases, you need to sweep the nonlinearities, whether they come from quadratic terms in the solution variable or dependencies on other solution variables.
I'm trying to solve a nonlinear problem in fenics. Instead of knowing the nonlinear function as an expression, I only know the values in the form of data.
In other words, if i am solving nabla(k(u).nabla(u)) = C, I do not know k(u) or cannot write it in terms of x[0] etc, I only know corresponding values of k for all values of u.
At the moment i have written some interpolating function and called it into 'k' which can project it to a function space, but it does not seem to work when i put it into the variational form and solve.
Below is some code to clarify the problem: I was using method of manufactured solutions where u=x^2 with k=1+u^2 as a test, where k_data = list of data for u=[0,(1+xmax^2)].
I've been told i can look into subclasses in Expression but i have no idea how to do this without knowing it in terms of x[0]. please help!
u = Function(V)
v = TestFunction(V)
u.interpolate(Expression("x[0]"))
k = interpolate_table(u,kappa_data)
plot(project(k,V))
C = Expression("-2.0-10.0*x[0]*x[0]*x[0]*x[0]")
diff_part = k*inner(grad(u),grad(v))*dx
source_part = C*v*dx
Res = diff_part - source_part
J = derivative(Res,u)
problem = NonlinearVariationalProblem(Res, u, bcs, J)
solver = NonlinearVariationalSolver(problem)
solver.solve()
prm = solver.parameters['newton_solver']
prm['relative_tolerance'] = 1E-6
prm['absolute_tolerance'] = 1E-6
prm['maximum_iterations'] = 100