实例介绍
【实例简介】解决quadprog问题: 内点法求解线性规划问题; 通过内点法实现了线性规划问题的求解;
【实例截图】
【核心代码】
/* File main.cc This file contains just an example on how to set-up the matrices for using with the solve_quadprog() function. The test problem is the following: Given: G = 4 -2 g0^T = [6 0] -2 4 Solve: min f(x) = 1/2 x G x g0 x s.t. x_1 x_2 = 3 x_1 >= 0 x_2 >= 0 x_1 x_2 >= 2 The solution is x^T = [1 2] and f(x) = 12 Author: Luca Di Gaspero DIEGM - University of Udine, Italy l.digaspero@uniud.it http://www.diegm.uniud.it/digaspero/ LICENSE Copyright 2006-2009 Luca Di Gaspero This file is part of QuadProg . QuadProg is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. QuadProg 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 General Public License for more details. You should have received a copy of the GNU General Public License along with QuadProg ; if not, write to the Free Software Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA */ #include <iostream> #include <sstream> #include <string> #include "QuadProg .hh" int main (int argc, char *const argv[]) { Matrix<double> G, CE, CI; Vector<double> g0, ce0, ci0, x; int n, m, p; double sum = 0.0; char ch; n = 2; G.resize(n, n); { std::istringstream is("4, -2," "-2, 4 "); for (int i = 0; i < n; i ) for (int j = 0; j < n; j ) is >> G[i][j] >> ch; } g0.resize(n); { std::istringstream is("6.0, 0.0 "); for (int i = 0; i < n; i ) is >> g0[i] >> ch; } m = 1; CE.resize(n, m); { std::istringstream is("1.0, " "1.0 "); for (int i = 0; i < n; i ) for (int j = 0; j < m; j ) is >> CE[i][j] >> ch; } ce0.resize(m); { std::istringstream is("-3.0 "); for (int j = 0; j < m; j ) is >> ce0[j] >> ch; } p = 3; CI.resize(n, p); { std::istringstream is("1.0, 0.0, 1.0, " "0.0, 1.0, 1.0 "); for (int i = 0; i < n; i ) for (int j = 0; j < p; j ) is >> CI[i][j] >> ch; } ci0.resize(p); { std::istringstream is("0.0, 0.0, -2.0 "); for (int j = 0; j < p; j ) is >> ci0[j] >> ch; } x.resize(n); std::cout << "f: " << solve_quadprog(G, g0, CE, ce0, CI, ci0, x) << std::endl; std::cout << "x: " << x << std::endl; /* for (int i = 0; i < n; i ) std::cout << x[i] << ' '; std::cout << std::endl; */ /* FOR DOUBLE CHECKING COST since in the solve_quadprog routine the matrix G is modified */ { std::istringstream is("4, -2," "-2, 4 "); for (int i = 0; i < n; i ) for (int j = 0; j < n; j ) is >> G[i][j] >> ch; } std::cout << "Double checking cost: "; for (int i = 0; i < n; i ) for (int j = 0; j < n; j ) sum = x[i] * G[i][j] * x[j]; sum *= 0.5; for (int i = 0; i < n; i ) sum = g0[i] * x[i]; std::cout << sum << std::endl; }
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