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solve an equation (with max-statement) in python

i am seeking a solution for the following equation in Python.
345-0.25*t = 37.5 * x_a
'with'
t = max(0, 10-x_a)*(20-10) + max(0,25-5*x_a)*(3-4) + max(0,4-0.25*x_a)*(30-12.5)
'x_a = ??'
If there is more than one solution to the problem (I am not even sure, whether this can happen from a mathematical point of view?), I want my code to return a positive(!) value for x_a, that minimizes t.
With my previous knowledge in the Basics of Python, Pandas and NumPy I actually have no clue, how to tackle this problem. Can someone give me a hint?
For Clarification: I inserted some exemplary numbers in the equation to make it easier to gasp the problem. In my final code, there might of course be different numbers for different scenarios. However, in every scenario x_a is the only unknown variable.
Update
I thought about the problem again and came up with the following solution, which yields the same result as the calculations done by Michał Mazur:
import itertools
from sympy import Eq, Symbol, solve
import numpy as np
x_a = Symbol('x_a')
possible_elements = np.array([10-x_a, 25-5*x_a, 4-0.25*x_a])
assumptions = np.array(list(itertools.product([True, False], repeat=3)))
for assumption in assumptions:
x_a = Symbol('x_a')
elements = assumption.astype(int) * possible_elements
t = elements[0]*(20-10) + elements[1]*(3-4) + elements[2]*(30-12.5)
eqn = Eq(300-0.25*t, 40*x_a)
solution = solve(eqn)
if len(solution)>2:
print('Warning! the code may suppress possible solutions')
if len(solution)==1:
solution = solution[0]
if (((float(possible_elements[0].subs(x_a,solution))) > 0) == assumption[0]) &\
(((float(possible_elements[1].subs(x_a,solution))) > 0) == assumption[1]) &\
(((float(possible_elements[2].subs(x_a,solution)))> 0) == assumption[2]):
print('solution:', solution)
Compared to the already suggested approach this may have an little advantage as it does not rely on testing all possible values and therefore can be used for very small as well as very big solutions without taking a lot of time (?). However, it probably only is useful as long as you don't have more complex functions for t (even having for example 5 max(...) statements and therefore (2^5=)32 scenarios to test seems quite cumbersome).
As far as I'm concerned, I just realized that my problem is even more complex as i thought. For my project the calculations needed to derive the value of "t" are pretty entangled and can not be written in just one equation. However it still is a function, that only relies on x_a. So I am still hoping for a Python-Solution similar to the proposed Solver in Excel... or I will stick to the approach of simply testing out all possible numbers.

If you are interested in a solution a bit different than the Python one, then I will give you a hand. Open up your Excel, with Solver extention and plug in
the data you are interested in cheking, as the following:
Into the E2 you plug the command I just have writen, into E4 you plug in
=300-0,25*E2
Into the F4 you plug:
=40*F2
Then you open up your Solver menu
Into the Set Objective you put the variable t, which you want to minimize.
Into Changing Variables you put the a.
Into Constraint Menu you put the equality of E4 and F4 cells.
You check the "Make Unconstarained Variables be non-negative" which will prevent your a variable to go below 0. Your method of computing is strictly non-linear, so you leave this option there.
You click solve. The computed value is presented in the screen.
The python approach I can think of:
minimumval=10100
minxa=10000
eps=0.01
for i in range(100000):
k=i/10000
x_a=k
t = max(0, 10-x_a)*(20-10) + max(0,25-5*x_a)*(3-4) + max(0,4-0.25*x_a)*(30-12.5)
val=abs(300-0.25*t-40*x_a)
if (val<eps):
if t<minimumval:
minimumval=t
minxa=x_a
It isn't direct solution, as in it only controls the error that you make in the equality by eps value. However, it gives solution.

max_iters doesn't seem to work with GLPK_MI solver in Python

I'm debugging my code right now and since it's running with some datas and not with other ones, I wanted to set the 'max_iters' option to 1 to see if it works in only 1 iteration or if it needs more. I realised it doesn't seem to even use it. I tried putting a string "hello" instead of an int and it even worked. Do someone knows if it's a known problem?
self.prob.solve(solver="GLPK_MI", max_iters=1)
I'm using the CVXPY module with CVXOPT.
EDIT:
I want to do this because I don't get an error, it just continues to run forever. And with the project I'm working on it can take a lot of time to run so I wonder if it's really not working or if it's just a question of time.

Wouldn't be better if you set the max iterations as a variable? (just a suggestion)
In any case, in CVXOPT you need to set the max number of iteration as
'maxiters' : 1
or you can set it as a variable and then call solver as per below
opts = {'maxiters' : 1}
self.prob.solve(solver="GLPK_MI", options = opts)

How does scipy.integrate.ode.integrate() work?

I have obviously read through the documentation, but I have not been able to find a more detailed description of what is happening under the covers. Specifically, there are a few behaviors that I am very confused about:
General setup
import numpy as np
from scipy.integrate import ode
#Constants in ODE
N = 30
K = 0.5
w = np.random.normal(np.pi, 0.1, N)
#Integration parameters
y0 = np.linspace(0, 2*np.pi, N, endpoint=False)
t0 = 0
#Set up the solver
solver = ode(lambda t,y: w + K/N*np.sum( np.sin( y - y.reshape(N,1) ), axis=1))
solver.set_integrator('vode', method='bdf')
solver.set_initial_value(y0, t0)
Problem 1: solver.integrate(t0) fails
Setting up the integrator, and asking for the value at t0 the first time returns a successful integration. Repeating this returns the correct number, but the solver.successful() method returns false:
solver.integrate(t0)
>>> array([ 0. , 0.20943951, 0.41887902, ..., 5.65486678,
5.86430629, 6.0737458 ])
solver.successful()
>>> True
solver.integrate(t0)
>>> array([ 0. , 0.20943951, 0.41887902, ..., 5.65486678,
5.86430629, 6.0737458 ])
solver.successful()
>>> False
My question is, what is happening in the solver.integrate(t) method that causes it to succeed the first time, and fail subsequently, and what does it mean to have an “unsuccessful” integration? Furthermore, why does the integrator fail silently, and continue to produce useful-looking outputs until I ask it explicitly whether it was successful?
Related, is there a way to reset the failed integration, or do I need to re-instantiate the solver from scratch?
Problem 2: solver.integrate(t) immediately returns an answer for almost any value of t
Even though my initial value of y0 is given at t0=0, I can request the value at t=10000 and get the answer immediately. I would expect that the numerical integration over such a large time span should take at least a few seconds (e.g. in Matlab, asking to integrate over 10000 time steps would take several minutes).
For example, re-run the setup from above and execute:
solver.integrate(10000)
>>> array([ 2153.90803383, 2153.63023706, 2153.60964064, ..., 2160.00982959,
2159.90446056, 2159.82900895])
Is Python really that fast, or is this output total nonsense?

Problem 0
Don’t ignore error messages. Yes, ode’s error messages can be cryptic at times, but you still want to avoid them.
Problem 1
As you already integrated up to t0 with the first call of solver.integrate(t0), you are integrating for a time step of 0 with the second call. This throws the cryptic error:
DVODE-- ISTATE (=I1) .gt. 1 but DVODE not initialized
In above message, I1 = 2
/usr/lib/python3/dist-packages/scipy/integrate/_ode.py:869: UserWarning: vode: Illegal input detected. (See printed message.)
'Unexpected istate=%s' % istate))
Problem 2.1
There is a maximum number of (internal) steps that a solver is going to take in one call without throwing an error. This can be set with the nsteps argument of set_integrator. If you integrate a large time at once, nsteps will be exceeded even if nothing is wrong, and the following error message is thrown:
/usr/lib/python3/dist-packages/scipy/integrate/_ode.py:869: UserWarning: vode: Excess work done on this call. (Perhaps wrong MF.)
'Unexpected istate=%s' % istate))
The integrator then stops at whenever this happens.
Problem 2.2
If you set nsteps=10**10, the integration runs without problems. It still is pretty fast though (roughly 1 s on my machine). The reason for this is as follows:
For a multi-dimensional system such as yours, there are two main runtime sinks when integrating:
Vector and matrix operations within the integrator. In scipy.ode, these are all realised with NumPy operations or ported Fortran or C code. Anyway, they are realised with compiled code without Python overhead and thus very efficient.
Evaluating the derivative (lambda t,y: w + K/N*np.sum( np.sin( y - y.reshape(N,1) ), axis=1) in your case). You realised this with NumPy operations, which again are realised with compiled code and very efficient. You may improve this a little bit with a purely compiled function, but that will grant you at most a small factor. If you used Python lists and loops instead, it would be horribly slow.
Therefore, for your problem, everything relevant is handled with compiled code under the hood and the integration is handled with an efficiency comparable to that of, e.g., a pure C program. I do not know how the two above aspects are handled in Matlab, but if either of the above challenges is handled with interpreted instead of compiled loops, this would explain the runtime discrepancy you observe.

To the second question, yes, the output might be nonsense. Local errors, be they from discretization or floating point operations, accumulate with a compounding factor which is about the Lipschitz constant of the ODE function. In a first estimate, the Lipschitz constant here is K=0.5. The magnification rate of early errors, that is, their coefficient as part of the global error, can thus be as large as exp(0.5*10000), which is a huge number.
On the other hand it is not surprising that the integration is fast. Most of the provided methods use step size adaptation, and with the standard error tolerances this might result in only some tens of internal steps. Reducing the error tolerances will increase the number of internal steps and may change the numerical result drastically.

Making a Qobj out of a picos variable

I need to write a semidefinite program that minimizes the trace of an operator, say R, subject to the constraint that tr_A(R)^{Tb} >>0 . That means that R represents a 3 qubit quantum system and the trace over the first system gives you an operator that represents the remaining 2 qubit systems. Taking the partial transpose with respect to one of the qubits, you get the partially transposed quantum state of the restricted 2 qubit system. It is this state that I want to make positive semidefinite.
I am using PICOS (to write the SDP) and qutip (to do the operations).
P = pic.Problem()
Rho = P.add_variable('Rho',(n,n),'hermitian')
P.add_constraint(pic.trace(Rho)==1)
P.add_constraint(Rho>>0)
RhoQOBJ = Qobj(Rho)
RhoABtr = ptrace(RhoQOBJ, [0,1])
RhoABqbj = partial_transpose(RhoABtr, [0], method='dense')
RhoAB = RhoABqbj.full()
Problem: I need to make Rho a Qobj, for qutip to be able to understand it, but Rho above is only an instance of the Variable class. Anyone has any idea on how to do this?
Also I looked here, http://picos.zib.de/tuto.html#variables , it became even more confusing as this function puts the instance in a dictionary and only gives you back a key.

You need to be able to output a numpy array or sparse matrix to convert to a Qobj. I could not find anything in the picos docs that discusses this option.

I am seeing this post very late, but maybe I can help... I am not sure what the function Qobj() is doing, can you please tell me more about it.
Otherwise, there is now a new partial_transpose() function in PICOS (version released today), which hopefully does what you need.
Best,
Guillaume.

Iterations in python consume whole memory - faulty code?

I have written some very simple trial-and-error code in Sage (a computer algebra system written in python where you can use regular python syntax in scripting). The little code snippet creates a polynomial and does some calculations with the coefficients, especially it determines the Groebner basis for the ideal generated by three expressions in the coefficients.
The problem is: This program goes on and eats up all my memory until it's killed by the kernel. Every iteration consumes only like 200kB, but this memory is never freed again.
Here is the code. The details are not that important and very bulky, therefore left out:
R = PolynomialRing(QQ, 2, 'bc', order='lex')
expr1, expr2, expr3 = ...
for i in range (0,50):
for j in range(i+1,50):
for k in range(j+1,50):
for l in range(k+1,50):
for m in range(l+1,50):
for n in range(m+1,50):
poly = (x-i)*(x-j)*(x-k)*(x-l)*(x-m)*(x-n)
r = poly.coeffs()
p1 = expr1.substitute(r...)
p2 = expr2.substitute(r...)
p3 = expr3.substitute(r...)
I = (p1, p2, p3)*R
B = I.groebner_basis()
As far as I understood python's memory management, the variables in the loop body are freed every so often. Now, it may be a programming problem, an internal python problem or some problem in the Sage routines. I don't know. Can you spot a problem with my code or is it something else?

The problem doesn't appear to be your loops (in python2.7, OS-X 10.5.8):
a = 0
for i in range (0,50):
for j in range(i+1,50):
for k in range(j+1,50):
for l in range(k+1,50):
for m in range(l+1,50):
for n in range(m+1,50):
a += 1
print( a )
Which takes very little additional memory on both python2.x and python3.x.
And it really doesn't take all that long to run either:
time python test.py
15890700
real 0m6.015s
user 0m5.940s
sys 0m0.032s
Perhaps something is funky when running with sage? Or maybe it's something else in your loops that is causing the problem...

The origin of the bug might be the call method on multivariate polynomials. Something as innocent as:
` for i in xrange(really_big_number):
polynomial(1,0,0,0)==0 `
will explode.
This might either happen in p1 = expr1.substitute(r...) or well inside the algorithm for the Groebner Basis.