Sparse sets in Pyomo Concrete Model-CodePudding

I want to create a sparse set in a ConcreteModel in pyomo. Here is a minimum example:

import pyomo
import pyomo.environ as pe

m = pe.ConcreteModel

m.P = pe.Set(initialize=['A', 'B', 'C'])
m.Q = pe.Set(initialize=[1, 2, 3, 4, 5, 6, 9, 10, 11])

D1 = {'A': [1, 2, 3], 'B': [9, 10, 11], 'C': [4, 5, 6]}
m.E = pe.Set(initialize=D1)

m.x = pe.Var(m.Q)

m.obj = pe.Objective(expr=1, sense=pe.minimize)


def constraint_rule(m: pe.ConcreteModel, p: str):
    return sum(m.x[i] for i in m.E[p]) <= 1


m.add_constraint = pe.Constraint(m.P, rule=constraint_rule)

opt = pe.SolverFactory('gurobi')
opt.solve(m)

When I run this model, I get the following message:

TypeError: valid_problem_types() missing 1 required positional argument: 'self'

Is this a problem in the construction of the set m.E?

CodePudding user response：

You're getting that error because you're not creating an instance of the pyomo.environ.ConcreteModel but using an alias to it. You need to use parenthesis in m = pe.ConcreteModel()

Now, I assume that you want to express your constraints something like this:

x[1]   x[2]   x[3] <= 1
x[9]   x[10]   x[11] <= 1
x[4]   x[5]   x[6] <= 1

using the actual sets created. Then you need to create model.E a a Subset of model.P, since the current values relays upong each value of model.P. Then you need to change it as follows:

m.E = pe.Set(m.P, initialize=D1)

The actual model will be something like this:

import pyomo
import pyomo.environ as pe

m = pe.ConcreteModel()

m.P = pe.Set(initialize=['A', 'B', 'C'])
m.Q = pe.Set(initialize=[1, 2, 3, 4, 5, 6, 9, 10, 11])

D1 = {'A': [1, 2, 3], 'B': [9, 10, 11], 'C': [4, 5, 6]}
m.E = pe.Set(m.P, initialize=D1)

m.x = pe.Var(m.Q)

m.obj = pe.Objective(expr=1, sense=pe.minimize)


def constraint_rule(m: pe.ConcreteModel, p: str):
    return sum(m.x[i] for i in m.E[p]) <= 1


m.add_constraint = pe.Constraint(m.P, rule=constraint_rule)

opt = pe.SolverFactory('gurobi')
opt.solve(m)

This will give you the following solution:

>>>m.x.display()
x : Size=9, Index=Q
    Key : Lower : Value : Upper : Fixed : Stale : Domain
      1 :  None :   1.0 :  None : False : False :  Reals
      2 :  None :   0.0 :  None : False : False :  Reals
      3 :  None :   0.0 :  None : False : False :  Reals
      4 :  None :   1.0 :  None : False : False :  Reals
      5 :  None :   0.0 :  None : False : False :  Reals
      6 :  None :   0.0 :  None : False : False :  Reals
      9 :  None :   1.0 :  None : False : False :  Reals
     10 :  None :   0.0 :  None : False : False :  Reals
     11 :  None :   0.0 :  None : False : False :  Reals

I assume that you understand that this only generate a feasible solution, since the OF is a constant