I am working on some very large scale linear programming problems. (Matrices are currently roughly 1000x1000 and these are the 'mini' ones.)
I thought that I had the program running successfully, only I have realized that I am getting some very unintuitive answers. For example, let's say I were to maximize x+y+z subject to a set of constraints x+y<10 and y+z <5. I run this and get an optimal solution. Then, I run the same equation but with different constraints: x+y<20 and y+z<5. Yet in the second iteration, my maximization decreases!
I have painstakingly gone through and assured myself that the constraints are loading correctly.
Does anyone know what the problem might be?
I found something in the documentation about lpx_check_kkt which seems to tell you when your solution is likely to be correct or high confidence (or low confidence for that matter), but I don't know how to use it.
I made an attempt and got the error message lpx_check_kkt not defined.
I am adding some code as an addendum in hopes that someone can find an error.
The result of this is that it claims an optimal solution has been found. And yet every time I raise an upper bound, it gets less optimal.
I have confirmed that my bounds are going up and not down.
size = 10000000+1
ia = intArray(size)
ja = intArray(size)
ar = doubleArray(size)
prob = glp_create_prob()
glp_set_prob_name(prob, "sample")
glp_set_obj_dir(prob, GLP_MAX)
glp_add_rows(prob, Num_constraints)
for x in range(Num_constraints):
Variables.add_variables(Constraints_for_simplex)
glp_set_row_name(prob, x+1, Variables.variers[x])
glp_set_row_bnds(prob, x+1, GLP_UP, 0, Constraints_for_simplex[x][1])
print 'we set the row_bnd for', x+1,' to ',Constraints_for_simplex[x][1]
glp_add_cols(prob, len(All_Loops))
for x in range(len(All_Loops)):
glp_set_col_name(prob, x+1, "".join(["x",str(x)]))
glp_set_col_bnds(prob,x+1,GLP_LO,0,0)
glp_set_obj_coef(prob,x+1,1)
for x in range(1,len(All_Loops)+1):
z=Constraints_for_simplex[0][0][x-1]
ia[x] = 1; ja[x] = x; ar[x] = z
x=len(All_Loops)+1
while x<Num_constraints + len(All_Loops):
for y in range(2, Num_constraints+1):
z=Constraints_for_simplex[y-1][0][0]
ia[x] = y; ja[x] =1 ; ar[x] = z
x+=1
x=Num_constraints+len(All_Loops)
while x <len(All_Loops)*(Num_constraints-1):
for z in range(2,len(All_Loops)+1):
for y in range(2,Num_constraints+1):
if x<len(All_Loops)*Num_constraints+1:
q = Constraints_for_simplex[y-1][0][z-1]
ia[x] = y ; ja[x]=z; ar[x] = q
x+=1
glp_load_matrix(prob, len(All_Loops)*Num_constraints, ia, ja, ar)
glp_exact(prob,None)
Z = glp_get_obj_val(prob)
Start by solving your problematic instances with different solvers and checking the objective function value. If you can export your model to .mps format (I don't know how to do this with GLPK, sorry), you can upload the mps file to http://www.neos-server.org/neos/solvers/index.html and solve it with several different LP solvers.
Related
I have been trying to get into python optimization, and I have found that pyomo is probably the way to go; I had some experience with GUROBI as a student, but of course that is no longer possible, so I have to look into the open source options.
I basically want to perform an non-linear mixed integer problem in which I will minimized a certain ratio. The problem itself is setting up a power purchase agreement (PPA) in a renewable energy scenario. Depending on the electricity generated, you will have to either buy or sell electricity acording to the PPA.
The only starting data is the generation; the PPA is the main decision variable, but I will need others. "buy", "sell", "b1" and "b2" are unknown without the PPA value. These are the equations:
Equations that rule the problem (by hand).
Using pyomo, I was trying to set up the problem as:
# Dataframe with my Generation information:
January = Data['Full_Data'][(Data['Full_Data']['Month'] == 1) & (Data['Full_Data']['Year'] == 2011)]
Gen = January['Producible (MWh)']
Time = len(Generacion)
M=100
# Model variables and definition:
m = ConcreteModel()
m.IDX = range(time)
m.PPA = Var(initialize = 2.0, bounds =(1,7))
m.compra = Var(m.IDX, bounds = (0, None))
m.venta = Var(m.IDX, bounds = (0, None))
m.b1 = Var(m.IDX, within = Binary)
m.b2 = Var(m.IDX, within = Binary)
And then, the constraint; only the first one, as I was already getting errors:
m.b1_rule = Constraint(
expr = (((Gen[i] - PPA)/M for i in m.IDX) <= m.b1[i])
)
which gives me the error:
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-5-5d5f5584ebca> in <module>
1 m.b1_rule = Constraint(
----> 2 expr = (((Generacion[i] - PPA)/M for i in m.IDX) <= m.b1[i])
3 )
pyomo\core\expr\numvalue.pyx in pyomo.core.expr.numvalue.NumericValue.__ge__()
pyomo\core\expr\logical_expr.pyx in pyomo.core.expr.logical_expr._generate_relational_expression()
AttributeError: 'generator' object has no attribute 'is_expression_type'
I honestly have no idea what this means. I feel like this should be a simple problem, but I am strugling with the syntax. I basically have to apply a constraint to each individual data from "Generation", there is no sum involved; all constraints are 1-to-1 contraints set so that the physical energy requirements make sense.
How do I set up the constraints like this?
Thank you very much
You have a couple things to fix. First, the error you are getting is because you have "extra parenthesis" around an expression that python is trying to convert to a generator. So, step 1 is to remove the outer parenthesis, but that will not solve your issue.
You said you want to generate this constraint "for each" value of your index. Any time you want to generate copies of a constraint "for each" you will need to either do that by making a constraint list and adding to it with some kind of loop, or use a function-rule combination. There are examples of each in the pyomo documentation and plenty on this site (I have posted a ton if you look at some of my posts.) I would suggest the function-rule combo and you should end up with something like:
def my_constr(m, i):
return m.Gen[i] - m.PPA <= m.b1[i] * M
m.C1 = Constraint(m.IDX, rule=my_constr)
I was trying to solve certain set of constraints using z3 in python. My code:
import math
from z3 import *
### declaration
n_co2 = []
c_co2 = []
alpha = []
beta = []
m_dot_air = []
n_pir = []
pir_sensor = []
for i in range(2):
c_co2.append(Real('c_co2_'+str(i)))
n_pir.append(Real('n_pir_'+str(i)))
n_co2.append(Real('n_co2_'+str(0)))
alpha.append(Real('alpha_'+str(0)))
beta.append(Real('beta_'+str(0)))
m_dot_air.append(Real('m_dot_air_'+str(0)))
pir_sensor.append(Real('pir_sensor_'+str(0)))
s = Solver()
s.add(n_co2[0]>0)
s.add(c_co2[0]>0)
s.add(c_co2[1]>=0.95*c_co2[0])
s.add(c_co2[1]<=1.05*c_co2[0])
s.add(n_co2[0]>=0.95*n_pir[1])
s.add(n_co2[0]<=1.05*n_pir[1])
s.add(c_co2[1]>0)
s.add(alpha[0]<=-1)
s.add(beta[0]>0)
s.add(m_dot_air[0]>0)
s.add(alpha[0]==-1*(1+ m_dot_air[0] + (m_dot_air[0]**2)/2.0 + (m_dot_air[0]**3)/6.0 ))
s.add(beta[0]== (1-alpha[0])/m_dot_air[0])
s.add(n_co2[0]== (c_co2[1]-alpha[0]*c_co2[0])/(beta[0]*19.6)-(m_dot_air[0]*339)/19.6)
s.add(n_pir[1]>=0)
s.add(pir_sensor[0]>=-1)
s.add(pir_sensor[0]<=1)
s.add(Not(pir_sensor[0]==0))
s.add(n_pir[1]==(n_pir[0]+pir_sensor[0]))
#### testing
s.add(pir_sensor[0]==1)
s.add(n_pir[1]==1)
s.add(n_co2[0]==1)
print(s.check())
print(s.reason_unknown())
print(s.model())
The output of the code:
sat
[c_co2_0 = 355,
c_co2_1 = 1841/5,
m_dot_air_0 = 1,
n_co2_0 = 1,
n_pir_1 = 1,
pir_sensor_0 = 1,
n_pir_0 = 0,
beta_0 = 11/3,
alpha_0 = -8/3,
/0 = [(19723/15, 1078/15) -> 1793/98,
(11/3, 1) -> 11/3,
else -> 0]]
What is the significance "/0 = ..." part of the output model.
But when I change the type of n_pir from Real to Int, z3 cannot solve it. Although we saw that we have an Int solution for n_pir. Reason of unknown:
smt tactic failed to show goal to be sat/unsat (incomplete (theory arithmetic))
How this problem can be solved? Could anyone please provide reasoning about this problem?
For the "/0" part: It's an internally generated constraint from converting real to int solutions. You can totally ignore that. In fact, you shouldn't really even look at the value of that, it's an artifact of the z3py bindings and should probably be hidden from the user.
For your question regarding why you cannot make 'Real' to 'Int'. That's because you have a non-linear set of equations (where you multiply or divide two variables), and non-linear integer arithmetic is undecidable in general. (Whereas non-linear real arithmetic is decidable.) So, when you use 'Int', solver simply uses some heuristics, and in this case fails and says unknown. This is totally expected. Read this answer for more details: How does Z3 handle non-linear integer arithmetic?
Z3 does come with an NRA solver, you can give that a try. Declare your solver as:
s = SolverFor("NRA")
But again you're at the mercy of the heuristics and you may or may not get a solution. Also, watch out for z3py bindings coercing constants to when you mix and match arithmetic like that. A good way is to write:
print s.sexpr()
before you call s.check() and take a look at the output and convince yourself that the translation has been done correctly. For details on that, see this question: Python and Z3: integers and floating, how to manage them in the correct way?
I dont know if this question has been asked before in SO, I will go ahead and post it here, I am attempting to solve a simple system with a PID controller, my system of differential equations are given below. I am basically attempting to code very basic PID algorithm. The structure of my control u depends on both derivative and integral of error term. I dont have any problem with the derivative term, it is the integral term that is creating problem in my code. The problem crops up when I assign s=0 in the beginning
and use it in my function as described in my code below. Is there a way to bypass it? I tried assigning s and told as global variables, but it didnt solve my problem. In a nutshell what I am doing is- I am adding state x1 every time and multiplying by dt(which is denoted by t-told).
Kindly help me iron out this issue, PFA my code attached below.
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
plt.style.use('bmh')
t0=0
y0=[0.1,0.2]
kp,kd,ki=2,0.5,0.8
s,told=0,0
def pid(t,Y):
x1,x2=Y[0],Y[1]
e=x1-1
de=x2
s=(x1+s)
integral=s*(t-told)
told=t
#ie=
u=kp*e+kd*de+ki*integral
x1dot=x2
x2dot=u-5*x1-2*x2
return[x1dot,x2dot]
solver=ode(pid).set_integrator('dopri5',rtol=1e-6,method='bdf',nsteps=1e5,max_step=1e-3)
solver.set_initial_value(y0,t0)
t1=10
dt=5e-3
sol = [ [yy] for yy in y0 ]
t=[t0]
while solver.successful() and solver.t<t1:
solver.integrate(solver.t+dt)
for k in range(2): sol[k].append(solver.y[k]);
t.append(solver.t)
print(len(sol[0]))
print(len(t))
x1=np.array(sol[0])
x2=np.array(sol[1])
e=x1-1
de=x2
u=kp*e+kd*de
for k in range(2):
if k==0:
plt.subplot(2,1,k+1)
plt.plot(t,sol[k],label='x1')
plt.plot(t,sol[k+1],label='x2')
plt.legend(loc='lower right')
else:
plt.subplot(2,1,k+1)
plt.plot(t,u)
plt.show()
You are making assumptions on the solver and the time steps that it visits that are not justified. With your hacking of the integral, even if it were mathematically sound (it should look like integral = integral + e*(t-told), which gives an order 1 integration method), you reduce the order of any integration method, probably down to 1, if you are lucky only to order 2.
A mathematically correct method to implement this system is to introduce a third variable x3 for the integral of e, that is, the derivative of x3 is e. That the correct order 1 system has to be of dimension 3 can be read of the fact that (eliminating e) your system has 3 differentiation/integration operations. With that your system becomes
def pid(t,Y):
x1, x2, x3 =Y
e=x1-1
x1dot = x2
edot = x1dot
x3dot = e
u=kp*e+kd*edot+ki*x3
x2dot=u-5*x1-2*x2
return[x1dot, x2dot, x3dot]
Note that there are no global dynamic variables necessary, only the constants (which could also be passed as parameters, whatever seems more efficient or readable).
Now you will also need an initial value for x3, it was not visible from the system what the integration variable would have to be, your code seems to suggest 0.
First of all you need to include "s" Variable into the pid function.
'
def pid(s, t, Y): ...
'
Easiest solution I can see right now is to create a class with s and told as properties of this class:
class PIDSolver:
def __init__(self)
self.t0=0
self.y0=[0.1,0.2]
self.kp,self.kd,self.ki=2,0.5,0.8
self.s,self.told=0,0
def pid(t,Y):
x1,x2=Y[0],Y[1]
e=x1-1
de=x2
self.s=(x1+self.s)
integral=self.s*(t-self.told)
self.told=t
#ie=
u=self.kp*e+self.kd*de+self.ki*integral
x1dot=x2
x2dot=u-5*x1-2*x2
return[x1dot,x2dot]
For the first part of your problem. Use pidsolver = PIDSolver() in the next part of your solution.
I solved this problem myself by using set_f_params() method and passing a list in itz argument. Also I passed a 3rd argument in pid() i.e pid(t,Y,arg). And lastly I assigned s,told=arg[0],arg[1].
As I'm really struggleing to get from R-code, to Python code, I would like to ask some help. The code I want to use has been provided to my from withing the mathematics forum of stackexchange.
https://math.stackexchange.com/questions/2205573/curve-fitting-on-dataset
I do understand what is going on. But I'm really having a hard time trying to solve the R-code, as I have never seen anything of it. I have written the function to return the sum of squares. But I'm stuck at how I could use a function similar to the optim function. And also I don't really like the guesswork at the initial values. I would like it better to run and re-run a type of optim function untill I get the wanted result, because my needs for a nearly perfect curve fit are really high.
def model (par,x):
n = len(x)
res = []
for i in range(1,n):
A0 = par[3] + (par[4]-par[1])*par[6] + (par[5]-par[2])*par[6]**2
if(x[i] == par[6]):
res[i] = A0 + par[1]*x[i] + par[2]*x[i]**2
else:
res[i] = par[3] + par[4]*x[i] + par[5]*x[i]**2
return res
This is my model function...
def sum_squares (par, x, y):
ss = sum((y-model(par,x))^2)
return ss
And this is the sum of squares
But I have no idea on how to convert this:
#I found these initial values with a few minutes of guess and check.
par0 <- c(7,-1,-395,70,-2.3,10)
sol <- optim(par= par0, fn=sqerror, x=x, y=y)$par
To Python code...
I wrote an open source Python package (BSD license) that has a genetic algorithm (Differential Evolution) front end to the scipy Levenberg-Marquardt solver, it functions similarly to what you describe in your question. The github URL is:
https://github.com/zunzun/pyeq3
It comes with a "user-defined function" example that's fairly easy to use:
https://github.com/zunzun/pyeq3/blob/master/Examples/Simple/FitUserDefinedFunction_2D.py
along with command-line, GUI, cluster, parallel, and web-based examples. You can install the package with "pip3 install pyeq3" to see if it might suit your needs.
Seems like I have been able to fix the problem.
def model (par,x):
n = len(x)
res = np.array([])
for i in range(0,n):
A0 = par[2] + (par[3]-par[0])*par[5] + (par[4]-par[1])*par[5]**2
if(x[i] <= par[5]):
res = np.append(res, A0 + par[0]*x[i] + par[1]*x[i]**2)
else:
res = np.append(res,par[2] + par[3]*x[i] + par[4]*x[i]**2)
return res
def sum_squares (par, x, y):
ss = sum((y-model(par,x))**2)
print('Sum of squares = {0}'.format(ss))
return ss
And then I used the functions as follow:
parameter = sy.array([0.0,-8.0,0.0018,0.0018,0,200])
res = least_squares(sum_squares, parameter, bounds=(-360,360), args=(x1,y1),verbose = 1)
The only problem is that it doesn't produce the results I'm looking for... And that is mainly because my x values are [0,360] and the Y values only vary by about 0.2, so it's a hard nut to crack for this function, and it produces this (poor) result:
Result
I think that the range of x values [0, 360] and y values (which you say is ~0.2) is probably not the problem. Getting good initial values for the parameters is probably much more important.
In Python with numpy / scipy, you would definitely want to not loop over values of x but do something more like
def model(par,x):
res = par[2] + par[3]*x + par[4]*x**2
A0 = par[2] + (par[3]-par[0])*par[5] + (par[4]-par[1])*par[5]**2
res[np.where(x <= par[5])] = A0 + par[0]*x + par[1]*x**2
return res
It's not clear to me that that form is really what you want: why should A0 (a value independent of x added to a portion of the model) be so complicated and interdependent on the other parameters?
More importantly, your sum_of_squares() function is actually not what least_squares() wants: you should return the residual array, you should not do the sum of squares yourself. So, that should be
def sum_of_squares(par, x, y):
return (y - model(par, x))
But most importantly, there is a conceptual problem that is probably going to plague this model: Your par[5] is meant to represent a breakpoint where the model changes form. This is going to be very hard for these optimization routines to find. These routines generally make a very small change to each parameter value to estimate to derivative of the residual array with respect to that variable in order to figure out how to change that variable. With a parameter that is essentially used as an integer, the small change in the initial value will have no effect at all, and the algorithm will not be able to determine the value for this parameter. With some of the scipy.optimize algorithms (notably, leastsq) you can specify a scale for the relative change to make. With leastsq that is called epsfcn. You may need to set this as high as 0.3 or 1.0 for fitting the breakpoint to work. Unfortunately, this cannot be set per variable, only per fit. You might need to experiment with this and other options to least_squares or leastsq.
I have an optimization problem and I write a python program to solve it. I used Pulp with the CPLEX solver:
import pulp
prob = LpProblem("myProblem", LpMinimize)
x = pulp.LpVariable.dicts("p", range( K ), 0, 1, pulp.LpContinuous)
prob += pulp.lpSum( x[k] for k in range( K ) )
...
# Rest of the constraints
status = prob.solve( pulp.CPLEX( msg = 0 ) )
I get the error:
File "C:\Anaconda\lib\site-packages\pulp\solvers.py", line 468, in readsol
raise PulpSolverError, "Unknown status returned by CPLEX: "+statusString
pulp.solvers.PulpSolverError: Unknown status returned by CPLEX: infeasible
My question is : How can I test if the problem is infeasible or not? I want to prevent this event like if problem is infeasible then return 0.
I tried :
if prob.status == 'infeasible':
...
and I tried
if pulp.LpStatusInfeasible == 'infeasible':
...
Is your 'problem' finding whether a given problem instance is feasible or not, or are you actually interested in the solution if it is feasible. Rather than just trap the error when the model is infeasible, I would examine your problem and try to add some slack variables and penalty costs to give you some more information when the problem would otherwise be infeasible.
So rather than add a hard constraint like
sum(x) <= K
you could try something like
sum(x) <= K + penaltyVar
and add a term into your objective like 1000000 * penaltyVar so that the solver really doesn't want to use that penalty variable as non-zero.
Adding these slack/penalty variables in various places in your model can help identify where the constraints are too tight and making your model infeasible.
Don't just ignore the answer above though, as it is still valuable to trap the error.
I think you can solve this by caging the statement inside a try-exception clause.
for example:
# ...
try:
status = prob.solve(pulp.CPLEX(msg = 0))
except PulpSolverError:
# infeasible
return 0
return status