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quadprog c++程序

一般编程问题

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  • 开发语言:Others
  • 实例大小:0.32M
  • 下载次数:7
  • 浏览次数:305
  • 发布时间:2019-10-29
  • 实例类别:一般编程问题
  • 发 布 人:robot666
  • 文件格式:.gz
  • 所需积分:2
 相关标签: c++ c++ UA 程序

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】

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【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;
}

标签: c++ c++ UA 程序

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