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#
# This file is part of CasADi.
#
# CasADi -- A symbolic framework for dynamic optimization.
# Copyright (C) 2010-2023 Joel Andersson, Joris Gillis, Moritz Diehl,
# KU Leuven. All rights reserved.
# Copyright (C) 2011-2014 Greg Horn
#
# CasADi is free software; you can redistribute it and/or
# modify it under the terms of the GNU Lesser General Public
# License as published by the Free Software Foundation; either
# version 3 of the License, or (at your option) any later version.
#
# CasADi is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
# Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public
# License along with CasADi; if not, write to the Free Software
# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
#
# nlpsol
# =====================
from casadi import *
from numpy import *
# In this example, we will solve a few optimization problems with increasing complexity
#
# Scalar unconstrained problem
# ============================
# $\text{min}_x \quad \quad (x-1)^2$ \\
# subject to $-10 \le x \le 10$ \\
#
# with x scalar
x=SX.sym('x')
nlp = {'x':x, 'f':(x-1)**2}
solver = nlpsol('solver', 'ipopt', nlp)
sol = solver(lbx=-10, ubx=10)
# The solution is obviously 1:
print(sol['x'])
assert(abs(sol['x']-1)<1e-9)
# Constrained problem
# ============================
# $\text{min}_x \quad \quad (x-1)^T.(x-1)$ \\
# subject to $-10 \le x \le 10$ \\
# subject to $0 \le x_1 + x_2 \le 1$ \\
# subject to $ x_0 = 2$ \\
#
# with $x \in \mathbf{R}^n$
n = 5
x=SX.sym('x',n)
# Note how we do not distinguish between equalities and inequalities here
nlp = {'x':x, 'f':mtimes((x-1).T,x-1), 'g':vertcat(x[1]+x[2],x[0])}
solver = nlpsol('solver', 'ipopt', nlp)
sol = solver(lbx=-10, ubx=10, lbg=[0,2], ubg=[1,2])
# $ 2 \le x_0 \le 2$ is not really as bad it looks.
# Ipopt will recognise this situation as an equality constraint.
# The solution is obviously [2,0.5,0.5,1,1]:
print(sol['x'])
for (i,e) in zip(list(range(n)),[2,0.5,0.5,1,1]):
assert(abs(sol['x'][i]-e)<1e-7)
# Problem with parameters
# ============================
# $\text{min}_x \quad \quad (x-a)^2$ \\
# subject to $-10 \le x \le 10$ \\
#
# with x scalar
x=SX.sym('x')
a=SX.sym('a')
a_ = 2
nlp={'x':x, 'p':a, 'f':(x-a)**2}
solver = nlpsol('solver', 'ipopt', nlp)
sol = solver(lbx=-10, ubx=10, p=a_)
# The solution is obviously a:
print(sol['x'])
assert(abs(sol['x']-a_)<1e-9)
# The parameter can change inbetween two solve calls:
sol = solver(lbx=-10, ubx=10, p=2*a_)
# The solution is obviously 2*a:
print(sol['x'])
assert(abs(sol['x']-2*a_)<1e-9)
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