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Problems minimizing a function with 2 constraints - Python

I'm writing a program to minimize a function of several parameters subjected to constraints and bounds. Just in case you want to run the program, the function is given by:
def Fnret(mins):
Bj, Lj, a, b = mins.reshape(4,N)
S1 = 0; S2 = 0
Binf = np.zeros(N); Linf = np.zeros(N);
for i in range(N):
sbi=(Bi/2); sli=(Li/2)
for j in range(i+1):
sbi -= Bj[j]
sli -= Lj[j]
Binf[i]=sbi
Linf[i]=sli
for i in range(N):
S1 += (C*(1-np.sin(a[i]))+T*np.sin(a[i])) * ((2*Bj[i]*Binf[i]+Bj[i]**2)/(np.tan(b[i])*np.cos(a[i]))) +\
(C*(1-np.sin(b[i]))+T*np.sin(b[i])) * ((2*Bj[i]*Linf[i]+Lj[i]*Bj[i])/(np.sin(b[i])))
S2 += (gamma*Bj[0]/(6*np.tan(b[0])))*((Bi/2)*(Li/2) + 4*(Binf[0]+Bj[0])*(Linf[0]+Lj[0]) + Binf[0]*Linf[0])
j=1
while j<(N):
S2 += (gamma*Bj[j]/(6*np.tan(b[j])))*(Binf[j-1]*Linf[j-1] + 4*(Binf[j]+Bj[j])*(Linf[j]+Lj[j]) + Binf[j]*Linf[j])
j += 1
F = 2*(S1+S2)
return F
where Bj,Lj,a, and b are the minimization results given by N-sized arrays with N being an input of the program, I double-checked the function and it is working correctly. My constraints are given by:
def Brhs(mins): # Constraint over Bj
return np.sum(mins.reshape(4,N)[0]) - Bi
def Lrhs(mins): # Constraint over Lj
return np.sum(mins.reshape(4,N)[1]) - Li
cons = [{'type': 'eq', 'fun': lambda Bj: 1.0*Brhs(Bj)},
{'type': 'eq', 'fun': lambda Lj: 1.0*Lrhs(Lj)}]
In such a way that the sum of all components of Bj must be equal to Bi (same thing with Lj). The bounds of the variables are given by:
bounds = [(0,None)]*2*N + [(0,np.pi/2)]*2*N
For the problem to be reproducible, it's important to use the following inputs:
# Inputs:
gamma = 17.
C = 85.
T = C
Li = 1.
Bi = 0.5
N = 3
For the minimization, I'm using the cyipopt library (that is just similar to the scipy.optimize). Then, the minimization is given by:
from cyipopt import minimize_ipopt
x0 = np.ones(4*N) # Initial guess
res = minimize_ipopt(Fnret, x0, constraints=cons, bounds=bounds)
The problem is that the result is not obeying the conditions I imposed on the constraints (i.e. the sum of Bj or Lj values is different than Bi or Li on the results). But, for instance, if I only write one of the two constraints (over Lj or Bj) it works fine for that variable. Maybe I'm missing something when using 2 constraints and I can't find the error, it seems that it doesn't work with both constraints together. Any help would be truly appreciated. Thank you in advance!
P.S.: In addition, I would like that the function result F to be positive as well. How can I impose this condition? Thanks!

Not a complete answer, just some hints in arbitrary order:
Your initial point x0 is not feasible because it contradicts both of your constraints. This can easily be observed by evaluating your constraints at x0. Under the hood, Ipopt typically detects this and tries to find a feasible initial point. However, it's highly recommended to provide a feasible initial point whenever possible.
Your variable bounds are not well-defined. Evaluating your objective at your bounds yields multiple divisions by zero. For example, the denominator np.tan(b[i]) is zero if and only if b[i] = 0, so 0 isn't a valid value for all of your b[i]s. Proceeding similarly for the other terms, you should obtain 0 < b[i] < pi/2 and 0 ≤ a[i] < pi/2. Here, you can model the strict inequalities by 0 + eps ≤ b[i] ≤ pi/2 - eps and 0 ≤ a[i] ≤ pi/2 - eps, where eps is a sufficiently small positive number.
If you really want to impose that the objective function is always positive, you can simply add the inequality constraint Fnret(x) >= 0, i.e. {'type': 'ineq', 'fun': Fnret}.
In Code:
# bounds
eps = 1e-8
bounds = [(0, None)]*2*N + [(0, np.pi/2 - eps)]*N + [(0+eps, np.pi/2 - eps)]*N
# (feasible) initial guess
x0 = eps*np.ones(4*N)
x0[[0, N]] = [Bi-(N-1)*eps, Li-(N-1)*eps]
# constraints
cons = [{'type': 'eq', 'fun': Brhs},
{'type': 'eq', 'fun': Lrhs},
{'type': 'ineq', 'fun': Fnret}]
res = minimize_ipopt(Fnret, x0, constraints=cons, bounds=bounds, options={'disp': 5})
Last but not least, this still doesn't converge to a stationary point, so chances are that there's indeed no local minimum. From here, you can try experimenting with other (feasible!) initial points and double-check the math of your problem. It's also worth providing the exact gradient and constraint Jacobians.

So, based on #joni suggestions, I could find a stationary point respecting the constraints by adopting the trust-constr method of scipy.optimize.minimize library. My code is running as follows:
import numpy as np
from scipy.optimize import minimize
# Inputs:
gamma = 17
C = 85.
T = C
Li = 2.
Bi = 1.
N = 3 # for instance
# Constraints:
def Brhs(mins):
return np.sum(mins.reshape(4,N)[0]) - Bi/2
def Lrhs(mins):
return np.sum(mins.reshape(4,N)[1]) - Li/2
# Function to minimize:
def Fnret(mins):
Bj, Lj, a, b = mins.reshape(4,N)
S1 = 0; S2 = 0
Binf = np.zeros(N); Linf = np.zeros(N);
for i in range(N):
sbi=(Bi/2); sli=(Li/2)
for j in range(i+1):
sbi -= Bj[j]
sli -= Lj[j]
Binf[i]=sbi
Linf[i]=sli
for i in range(N):
S1 += (C*(1-np.sin(a[i]))+T*np.sin(a[i])) * ((2*Bj[i]*Binf[i]+Bj[i]**2)/(np.tan(b[i])*np.cos(a[i]))) +\
(C*(1-np.sin(b[i]))+T*np.sin(b[i])) * ((2*Bj[i]*Linf[i]+Lj[i]*Bj[i])/(np.sin(b[i])))
S2 += (gamma*Bj[0]/(6*np.tan(b[0])))*((Bi/2)*(Li/2) + 4*(Binf[0]+Bj[0])*(Linf[0]+Lj[0]) + Binf[0]*Linf[0])
j=1
while j<(N):
S2 += (gamma*Bj[j]/(6*np.tan(b[j])))*(Binf[j-1]*Linf[j-1] + 4*(Binf[j]+Bj[j])*(Linf[j]+Lj[j]) + Binf[j]*Linf[j])
j += 1
F = 2*(S1+S2)
return F
eps = 1e-3
bounds = [(0,None)]*2*N + [(0+eps,np.pi/2-eps)]*2*N # Bounds
cons = ({'type': 'ineq', 'fun': Fnret},
{'type': 'eq', 'fun': Lrhs},
{'type': 'eq', 'fun': Brhs})
x0 = np.ones(4*N) # Initial guess
res = minimize(Fnret, x0, method='trust-constr', bounds = bounds, constraints=cons, tol=1e-6)
F = res.fun
Bj = (res.x).reshape(4,N)[0]
Lj = (res.x).reshape(4,N)[1]
ai = (res.x).reshape(4,N)[2]
bi = (res.x).reshape(4,N)[3]
Which is essentially the same just changing the minimization technique. From np.sum(Bj) and np.sum(Lj) is easy to see that the results are in agreement with the constraints imposed, which were not working previously.

Penalty function method

I am trying to implement penalty function method for minimizing function. I need to find the minimum of Rosenbrok's function.
I am using this penalty function:
First of all, I have found the minimum using scipy.optimize.minimize:
from scipy.optimize import minimize, rosen
rz = lambda x: (1-x[0])**2 + 100*(x[1] - x[0]**2)**2;
h_1 = lambda x: (x[0] - 2 * x[1] + 2);
h_2 = lambda x: (-x[0] - 2 * x[1] + 6);
h_3 = lambda x: (-x[0] + 2 * x[1] + 2);
x0 = [2.3, 5];
cons = ({'type': 'ineq', 'fun': h_1},
{'type': 'ineq', 'fun': h_2},
{'type': 'ineq', 'fun': h_3})
minimize(rz, x0, constraints=cons)
The answer is x: array([ 0.99971613, 0.99942073])
Then I am trying to find the minimum using my implementation of penalty method:
x_c = [2.3, 3];
i = 1;
while i < 1000:
curr_func = lambda x: rz(x) + i*(h_1(x)**2 + h_2(x)**2 + h_3(x)**2)
x_c = minimize(curr_func, x_c).x;
i *= 1.2;
print(answer.x);
Which gives me [ 2.27402022 1.4157964 ] (if I increase the number of iterations, final values are even greater).
Where is the mistake in my implementation?
Thanks.
P.S. Function curr_func is specific for my constraints, of course, when they are all 'inequals' type.

The problem you have is that the h_i in your formula are for equality constraints, whereas the problem you are solving is for inequality constraints, which correspond to the g_i in your formula. Hence, your penalty function should be using terms like min(0, h_1(x))**2 instead of h_1(x)**2. To see why this is the case, just think about what happens if i = 1000 and x is the desired solution (1, 1). Then, the penalty will include a term i * h_1(x)**2 = 1000, which is huge.
Note that I used min instead of max because it seems like the inequality you want to enforce is h_1(x) >= 0. That means as long as h_1(x) >= 0, the penalty should be zero, but as soon as h_1(x) goes negative, you start penalizing. If it's actually h_1(x) <= 0 you want, then you use max (then you'll have to switch h_1 with -h_1 when you use scipy.optimize.minimize).
BTW, since i is usually an index variable, it's probably better to name the penalty weight something else, like a.

algebraic constraint to terminate ODE integration with scipy

I'm using Scipy 14.0 to solve a system of ordinary differential equations describing the dynamics of a gas bubble rising vertically (in the z direction) in a standing still fluid because of buoyancy forces. In particular, I have an equation expressing the rising velocity U as a function of bubble radius R, i.e. U=dz/dt=f(R), and one expressing the radius variation as a function of R and U, i.e. dR/dT=f(R,U). All the rest appearing in the code below are material properties.
I'd like to implement something to account for the physical constraint on z which, obviously, is limited by the liquid height H. I consequently implemented a sort of z<=H constraint in order to stop integration in advance if needed: I used set_solout in order to do so. The situation is that the code runs and gives good results, but set_solout is not working at all (it seems like z_constraint is never called actually...). Do you know why?
Is there somebody with a more clever idea, may be also in order to interrupt exactly when z=H (i.e. a final value problem) ? is this the right way/tool or should I reformulate the problem?
thanks in advance
Emi
from scipy.integrate import ode
Db0 = 0.001 # init bubble radius
y0, t0 = [ Db0/2 , 0. ], 0. #init conditions
H = 1
def y_(t,y,g,p0,rho_g,mi_g,sig_g,H):
R = y[0]
z = y[1]
z_ = ( R**2 * g * rho_g ) / ( 3*mi_g ) #velocity
R_ = ( R/3 * g * rho_g * z_ ) / ( p0 + rho_g*g*(H-z) + 4/3*sig_g/R ) #R dynamics
return [R_, z_]
def z_constraint(t,y):
H = 1 #should rather be a variable..
z = y[1]
if z >= H:
flag = -1
else:
flag = 0
return flag
r = ode( y_ )
r.set_integrator('dopri5')
r.set_initial_value(y0, t0)
r.set_f_params(g, 5*1e5, 2000, 40, 0.31, H)
r.set_solout(z_constraint)
t1 = 6
dt = 0.1
while r.successful() and r.t < t1:
r.integrate(r.t+dt)

You're running into this issue. For set_solout to work correctly, it must be called right after set_integrator, before set_initial_value. If you introduce this modification into your code (and set a value for g), integration will terminate when z >= H, as you want.
To find the exact time when the bubble reached the surface, you can make a change of variables after the integration is terminated by solout and integrate back with respect to z (rather than t) to z = H. A paper that describes the technique is M. Henon, Physica 5D, 412 (1982); you may also find this discussion helpful. Here's a very simple example in which the time t such that y(t) = 0.5 is found, given dy/dt = -y:
import numpy as np
from scipy.integrate import ode
def f(t, y):
"""Exponential decay: dy/dt = -y."""
return -y
def solout(t, y):
if y[0] < 0.5:
return -1
else:
return 0
y_initial = 1
t_initial = 0
r = ode(f).set_integrator('dopri5')
r.set_solout(solout)
r.set_initial_value(y_initial, t_initial)
# Integrate until solout constraint violated
r.integrate(2)
# New system with t as independent variable: see Henon's paper for details.
def g(y, t):
return -1.0/y
r2 = ode(g).set_integrator('dopri5')
r2.set_initial_value(r.t, r.y)
r2.integrate(0.5)
y_final = r2.t
t_final = r2.y
# Error: difference between found and analytical solution
print t_final - np.log(2)

Why is odeint time evolving these two differently in python?

I have two programs, one that can take in N coupled ODEs and one that uses 2 coupled ODEs. In the case that I input the 2 same ODEs into both codes, with the same time span, I get different answers. I know the correct answer, so I can deduce that my N many program is wrong.
Here is the code for the 2 equation dedicated one:
# solve the coupled system dy/dt = f(y, t)
def f(y, t):
"""Returns the collections of first-order
coupled differential equations"""
#v11i = y[0]
#v22i = y[1]
#v12i = y[2]
print y[0]
# the model equations
f0 = dHs(tRel,vij)[0].subs(v12,y[2])
f1 = dHs(tRel,vij)[3].subs(v12,y[2])
f2 = dHs(tRel,vij)[1].expand().subs([(v11,y[0]),(v22,y[1]),(v12,y[2])])
return [f0, f1, f2]
# Initial conditions for graphing
v110 = 6
v220 = 6
v120 = 4
y0 = [v110, v220, v120] # initial condition vector
sMesh = np.linspace(0, 1, 10e3) # time grid
# Solve the DE's
soln = odeint(f, y0, sMesh)
and here is the N equation dedicated one:
def f(y, t):
"""Returns the derivative of H_s with initial
values plugged in"""
# the model equations
print y[0]
for i in range (0,len(dh)):
for j in range (0,len(y)):
dh[i] = dh[i].subs(v[j],y[j])
dhArray = []
for i in range(0,len(dh)):
dhArray.append(dh[i])
return dhArray
sMesh = np.linspace(0, 1, 10e3) # time grid
dh = dHsFunction(t, V_s).expand()
soln = odeint(f, v0, sMesh)
where dHs(tRel,vij) = dHsFunction(t,V_s) i.e. the exact same ODEs. Similarly y0 and v0 are the exact same. But when I print y[0] in the N many case, I get an output of:
6.0
5.99999765602
5.99999531204
5.97655553477
5.95311575749
5.92967598021
5.69527820744
5.46088043467
5.2264826619
2.88250493418
0.53852720647
-1.80545052124
-25.2452277984
-48.6850050755
-72.1247823527
-306.522555124
as opposed to the 2 dedicated case of:
6.0
5.99999765602
5.99999765602
5.99999531205
5.99999531205
5.98848712729
5.98848712125
5.97702879748
5.97702878476
5.96562028875
5.96562027486
5.91961750442
5.91961733611
5.93039037809
5.93039029335
5.89564277275
5.89564273736
5.86137647436
5.86137638807
5.82758984835
etc.
where the second result is the correct one and graphs the proper graphs.
Please let me know if more code is needed or anything else. Thanks.

Your second version for f modifies the value of the global variable dh.
On the first call, you substitute in values in it, and these same values are then used in all subsequent calls.
Avoid that by using e.g. dh_tmp = list(dh) inside the function.

ODEs with infinite initlal condition in python

I have a second order differential equation that I want to solve it in python. The problem is that for one of the variables I don't have the initial condition in 0 but only the value at infinity. Can one tell me what parameters I should provide for scipy.integrate.odeint ? Can it be solved?
Equation:
Theta needs to be found in terms of time. Its first derivative is equal to zero at t=0. theta is not known at t=0 but it goes to zero at sufficiently large time. all the rest is known. As an approximate I can be set to zero, thus removing the second order derivative which should make the problem easier.

This is far from being a full answer, but is posted here on the OP's request.
The method I described in the comment is what is known as a shooting method, that allows converting a boundary value problem into an initial value problem. For convenience, I am going to rename your function theta as y. To solve your equation numerically, you would first turn it into a first order system, using two auxiliary function, z1 = y and z2 = y', and so your current equation
I y'' + g y' + k y = f(y, t)
would be rewitten as the system
z1' = z2
z2' = f(z1, t) - g z2 - k z1
and your boundary conditions are
z1(inf) = 0
z2(0) = 0
So first we set up the function to compute the derivative of your new vectorial function:
def deriv(z, t) :
return np.array([z[1],
f(z[0], t) - g * z[1] - k * z[0]])
If we had a condition z1[0] = a we could solve this numerically between t = 0 and t = 1000, and get the value of y at the last time as something like
def y_at_inf(a) :
return scipy.integrate.odeint(deriv, np.array([a, 0]),
np.linspace(0, 1000, 10000))[0][-1, 0]
So now all we need to know is what value of a makes y = 0 at t = 1000, our poor man's infinity, with
a = scipy.optimize.root(y_at_inf, [1])