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// SPDX-License-Identifier: EPL-2.0 OR GPL-2.0-or-later
// SPDX-FileCopyrightText: Bradley M. Bell <bradbell@seanet.com>
// SPDX-FileContributor: 2003-24 Bradley M. Bell
// ----------------------------------------------------------------------------
/*
{xrst_begin qp_box.cpp}
{xrst_spell
rl
}
abs_normal qp_box: Example and Test
###################################
Problem
*******
Our original problem is
.. math::
\begin{array}{rl}
\R{minimize} & x_0 - x_1 \; \R{w.r.t} \; x \in \B{R}^2 \\
\R{subject \; to} & -2 \leq x_0 \leq +2 \; \R{and} \; -2 \leq x_1 \leq +2
\end{array}
Source
******
{xrst_literal
// BEGIN C++
// END C++
}
{xrst_end qp_box.cpp}
*/
// BEGIN C++
# include <limits>
# include <cppad/utility/vector.hpp>
# include "qp_box.hpp"
bool qp_box(void)
{ bool ok = true;
typedef CppAD::vector<double> vector;
//
size_t n = 2;
size_t m = 0;
vector a(n), b(n), c(m), C(m), g(n), G(n*n), xin(n), xout(n);
a[0] = -2.0;
a[1] = -2.0;
b[0] = +2.0;
b[1] = +2.0;
g[0] = +1.0;
g[1] = -1.0;
for(size_t i = 0; i < n * n; i++)
G[i] = 0.0;
//
// (0, 0) is feasible.
xin[0] = 0.0;
xin[1] = 0.0;
//
size_t level = 0;
double epsilon = 99.0 * std::numeric_limits<double>::epsilon();
size_t maxitr = 20;
//
ok &= CppAD::qp_box(
level, a, b, c, C, g, G, epsilon, maxitr, xin, xout
);
//
// check optimal value for x
ok &= std::fabs( xout[0] + 2.0 ) < epsilon;
ok &= std::fabs( xout[1] - 2.0 ) < epsilon;
//
return ok;
}
// END C++
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