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
Related
I am trying to write a pyomo model in python3, but I am facing this error that I can't seem to figure out - 'list' object has no attribute 'is_expression_type'. Below is my pyomo model, any help would be appreciated.
R_avg_tolist = [[0.00043159478649482775,
0.00045388639592182584,
0.0006735271301199177,
0.00044026758948786,
0.0037176592984565836]]
Cov_list = [[5.884677519869241e-05,
5.756542207262417e-05,
6.017027849080026e-05,
6.180151597797322e-05,
-0.0005074353586106837],
[5.756542207262417e-05,
6.0380562653096757e-05,
6.613608499966434e-05,
6.737370769879904e-05,
-0.0005362752804115953],
[6.017027849080026e-05,
6.613608499966434e-05,
8.206495000024503e-05,
8.01694525889321e-05,
-0.0005958716888916681],
[6.180151597797322e-05,
6.737370769879904e-05,
8.01694525889321e-05,
0.00010129901491226823,
-0.000608829853150321],
[-0.0005074353586106837,
-0.0005362752804115953,
-0.0005958716888916681,
-0.000608829853150321,
0.007373689071617548]]
import pyomo.environ as pyo
# Optimization Problem
def create_model(rho,R_avg,Cov):
m = pyo.ConcreteModel()
init_x = {}
m.idx = pyo.Set(initialize=[0,1,2,3,4])
for i in m.idx:
init_x[i] = 0
m.x = pyo.Var(m.idx,initialize=init_x,bounds=(0,None))
def Obj_func(m):
b = []
mult_result = 0
for i in m.idx:
a = 0
for j in m.idx:
a+= m.x[j]*Cov[j][i]
b.append(a)
for i in m.idx:
mult_result += b[i]*m.x[i]
return mult_result
m.OBJ = pyo.Objective(rule=Obj_func)
def constraint1(m):
sum=0
for i in m.idx:
sum+=m.x[i]
return sum ==100
m.C1 = pyo.Constraint(rule=constraint1(m))
def constraint2(m):
sum=0
for i in m.idx:
sum += R_avg_tolist[i]*m.x[i]
return sum >=0.08
m.C2 = pyo.Constraint(rule=constraint2(m))
return m
When I run model using below code, I face the attribute error - 'list' object has no attribute 'is_expression_type'.
rho = 0.0008
model1 = create_model(rho,R_avg_tolist,Cov_list)
solver = SolverFactory('ipopt')
results = solver.solve(model1, tee = True)
Probably not what you want to hear, but your model has many syntax probs. It's obviously a course assignment... Do you have somebody (instructor/TA) to go over this with who can advise a bit?
You didn't include enough info about which line of code caused the issue, but there are several problem areas. I've posted many simple pyomo examples if you scan through some of them, you'll get some ideas, along with the documentation and whatever you have from your course notes....
A few pointers may help:
Do NOT overwrite keywords/functions by using them as variables. When you write:
sum = 0
sum ....
you are nuking the python function sum by making that name a variable and assigning it the value of 0. You should be using sum in several of your functions with verbiage like:
sum(m.X[i] for i in m.idx) # or similar
You seem to be confused on making valid pyomo expressions. That is the core job of pyomo ... to make expressions and fill the model. For example, in your constraint1, you can just make an expression (without a function) and add it to your model. You can do a 1-liner there because the constraint is not a "for each". You could:
m.C1 = pyo.Constraint(expr=sum(m.x[i] for x in m.idx) == 100)
In general, when you are starting:
Add 1 thing to your model, and then print the model:
model.pprint()
See if it looks right, if not, fix it. Then repeat!
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.
Hopefully someone can help me out. I am developing an optimization model where I minimize electricity costs over time (t) and over different transactions (s). (where: standard power(p)*costs of electricity (c) = electricity costs).
Now I am trying to implement a cost component in the objective function that is based on the maximum power consumption that occurred (some alike: max(P[s,t])). However, np.max() returns an error because P[s,t] is an unsupported class for np.max(). Also the Gurobi function gp.max_(P[s,t]) also gives an unsupported class error. Is there someone who has a solution?
Code:
obj = gp.quicksum(p[s, t] * Cost_elect[t]e for t in range(T) for s in range(S)) + gp.max_(p_batt_ch[s,t]*fixed_cost for t in range(T) for s in range(S))
You need to assign the max constraint to a new auxiliary variable and the put this variable into the objective instead of the actual constraint.
maxobj = model.addVar()
max_constr = model.addConstr(maxobj == gp.max_(p_batt_ch[s,t] * fixed_cost
for t in range(T) for s in range(S)))
obj = gp.quicksum(p[s,t] * Cost_elect[t] for t in range(T) for s in range(S)) + maxobj)
Gurobi documentation
Using gurobi and python I am trying to solve a water balance(similar to the classic transportation problem) linear programming problem in the form of:
minimize c'x subject to:
Ax=b
lb<=x<=ub
A, L are sparse crs scipy matrices, c,b,lb,ub are vectors.
My problem should be updated for a number of steps and some elements are updated with new values. Specifically A is fixed, and all other elements get new values at each step. The following snippet works perfectly and is the basis I used so far (ignore the "self", as the model is embedded in a solver class, while "water_network is the graph object holding values and properties for each step):
### Snippet 1: Formulating/initializing the problem
# unitC is the c vector
# Bounds holds both lb and ub values for each x
self.model = gurobipy.Model()
rows, cols = len(self.water_network.node_list), len(self.water_network.edge_name_list)
self.x1 = []
for j in range(cols):
self.x1.append(self.model.addVar(lb=self.water_network.Bounds[j,0], ub=self.water_network.Bounds[j,1],obj=self.water_network.unitC[j]))
self.model.update()
self.EqualityConstraintA=[]
for i in range(rows):
start = self.water_network.A_sparse.indptr[i]
end = self.water_network.A_sparse.indptr[i+1]
variables = [self.x1[j] for j in self.water_network.A_sparse.indices[start:end]]
coeff = self.water_network.A_sparse.data[start:end]
expr = gurobipy.LinExpr(coeff, variables)
self.EqualityConstraintA.append(self.model.addConstr(lhs=expr, sense=gurobipy.GRB.EQUAL, rhs=self.water_network.b [i],name='A'+str(i)))
self.model.update()
self.model.ModelSense = 1
self.model.optimize()
The following simple snippet is used to update the problem at each step. Note i use the getConstrs function:
#### Snippet 2: Updating the constraints, working ok for every step.
self.model.setAttr("LB",self.model.getVars(), self.water_network.Bounds[:,0])
self.model.setAttr("UB", self.model.getVars(), self.water_network.Bounds[:,1])
self.model.setAttr("OBJ", self.model.getVars(), self.water_network.unitC)
self.model.setAttr("RHS", self.model.getConstrs(),self.water_network.b)
The problem arised when a new set of constraints should be added to the problem, in the form of:
Lx=0 where L is a sparse matrix that is updated every step! Now in the formulation I add the following just after the snippet 1:
self.EqualityConstraintL=[]
leakrows= len(self.water_network.ZeroVector)
for i in range(leakrows):
start = self.water_network.L_sparse.indptr[i]
end=self.water_network.L_sparse.indptr[i+1]
variables=[self.x1[j] for j in self.water_network.L_sparse.indices[start:end]]
coeff=self.water_network.L_sparse.data[start:end]
expr = gurobipy.LinExpr(coeff, variables)
self.EqualityConstraintL.append(self.model.addConstr(lhs=expr, sense=gurobipy.GRB.EQUAL, rhs=self.water_network.ZeroVector[i],name='L'+str(i)))
However, I can no longer use the getConstrs to update all constraints at once, as some need only the RHS changed and others need only the LHS changed. So I did the following for the update (Snippet 3):
self.model.setAttr("LB",self.model.getVars(), self.water_network.Bounds[:,0])
self.model.setAttr("UB", self.model.getVars(), self.water_network.Bounds[:,1])
self.model.setAttr("OBJ", self.model.getVars(), self.water_network.unitC)
# Update A rhs...
for i in range(len(self.water_network.edge_name_list)):
self.model.setAttr("RHS", self.model.getConstrs()[i],self.water_network.b[i])
# Update L expr...
x1=self.model.getVars()
n=len(self.water_network.node_list) # because there are n rows in the A constrains, and L constraints are added after
# Now i rebuild the LHS expressions
for i in range(len(self.water_network.ZeroVector)):
start = self.water_network.L_sparse.indptr[i]
end=self.water_network.L_sparse.indptr[i+1]
variables=[x1[j] for j in self.water_network.L_sparse.indices[start:end]]
coeff=self.water_network.L_sparse.data[start:end]
expr = gurobipy.LinExpr(coeff, variables)
self.model.setAttr("LHS",self.model.getConstrs()[n+i],expr)
self.model.update()
self.model.optimize()
When I run the problem, it initializes fine, but at the second step it returns this error:
File "model.pxi", line 1709, in gurobipy.Model.setAttr
TypeError: object of type 'Constr' has no len()
and the offending line is:
self.model.setAttr("RHS", self.model.getConstrs()[i],self.water_network.b[i])
Two questions: 1) why is that happening? replacing getConstrs()[i] with getConstrByName('A'+str(i)) also fails with the exact same error. How to update the RHS/LHS of a specific constraint?
2) Is there a way to more efficiently update the RHS on the constraints contained in the self.EqualityConstraintA list and then the LHS on the other constraints contained in the self.EqualityConstraintL list ?
Many thanks in advance!
Di
The setAttr function on the model object is for
setting attributes globally on the model
setting attributes for a list of variables
setting attributes for a list of constraints
The individual constraint and variable objects have their own setAttr functions to set attributes on single variables and constraints. In your case,
for i in range(len(self.water_network.edge_name_list)):
self.model.getConstrs()[i].setAttr('RHS', self.water_network.b[i])
Which could be replaced by the more pythonic (and likely more efficient)
m = self.model
constrs = m.getConstrs()[:len(self.water_network.edge_name_list)]
m.setAttr('RHS', constrs, self.water_network.b)
My question is if there's a way to take some values in a function that are not
integrated in odeint.
Exemple: if I have a derivative dy(x)/dt = A*x+ln(x) and before to get this equation I computed A throught of a intermediate equation like A = B*D . I would like to take the A's value during the process.
More detailed (only exemple):
def func(y,t)
K = y[0]
B = 3
A = cos(t**2) + B
dy/dt = A*t+ln(t)
return [dy/dt]
Can I take A's values of function?
The answer for Josh Karpel
The code is like that:
def Reaction(state,t):
# Integrate Results
p = state[0]
T = state[1]
# function determine enthalpy of system
f1(T,p) = enthalpy
# function determine specific volume of system
f2(T,p) = specific volume
# function determine heat release by reactions
f3(T,p,t) = heat release by reactions
# Derivatives
dp/dt = f(T,p,enthalpy,specific volume,heat release by reactions)
dT/dt = f(T,p,enthalpy,specific volume,heat release by reactions)
The real code is bigger than that. But, I would like to know if there is a way to store the values of f1 (enthalpy), f2 (specific volume), f3 (heat release) as a vector or tuple during the process of solution of odeint with the same size of p and T.
It's not entirely clear what you want, but it sounds like you need to pass another value to the function you're integrating over. There are two options I can think of:
scipy.integrate.odeint takes an args argument which contains extra arguments to be passed to the integrand function, which could then have signature y(t, A).
You could use functools.partial to construct a new function which has the argument A for the integrand function y(t, A) already set.