/*
	minimize 4*x1^2 + x2^2
	subject to -x1 -x2 +1		<= 0
		   -(1/4)x1 - x2 +(1/2) <= 0

	inequality conditions are converted into
	equational conditions shown below by introducing
	slack variables l1 and l2
	
	minimize 4*x1^2 + x2^2
	subject to -x1 -x2 +1		+l1^2 = 0
		   -(1/4)x1 - x2 +(1/2) +l2^2 = 0
		   
	this converted problem will be solved by use of
	the program shown below
								*/

#include <stdio.h>
#include <math.h>
#include <values.h>
#include "opt.h"

#define	Debug		0
#define	Static		static

Static dbl obj(int n, dbl *x)
{
	dbl	x1, x2, l1, l2;
	
	x1 = x[0];
	x2 = x[1];
	l1 = x[2];
	l2 = x[3];
	return 4.00*x1*x1 + x2*x2;
}

Static dbl *gradobj(int n, dbl *x, dbl *d)
{
	dbl	x1, x2, l1, l2;
	
	x1 = x[0];
	x2 = x[1];
	l1 = x[2];
	l2 = x[3];
	d[0] = 8.00*x1;
	d[1] = 2.00*x2;
	d[2] = 0.00;
	d[3] = 0.00;
	return d;
}

Static dbl *cond(int n, dbl *x, int l, dbl *d)
{
	dbl	x1, x2, l1, l2;
	
	x1 = x[0];
	x2 = x[1];
	l1 = x[2];
	l2 = x[3];
	d[0] = - x1 - x2 + l1*l1 + 1.00;
	d[1] = - 0.25*x1 - x2 + l2*l2 + 0.50;
	return d;
}

Static dbl **gradcond(int n, dbl *x, int l, dbl **d)
{
	dbl	x1, x2, l1, l2;
	
	x1 = x[0];
	x2 = x[1];
	l1 = x[2];
	l2 = x[3];
	(d[0])[0] = -1;
	(d[0])[1] = -1;
	(d[0])[2] = 2*l1;
	(d[0])[3] = 0;
	(d[1])[0] = -0.25;
	(d[1])[1] = -1;
	(d[1])[2] = 0;
	(d[1])[3] = 2*l2;
	return d;
}

#define	NVAR	4
#define	NCND	2

main()
{
	dbl	x[NVAR];
	enum status	stat;
	
	x[0] = 0.1;
	x[1] = 0.1;
	x[2] = 0.0001;
	x[3] = 0.0001;
	matrixprintformat(" %l16.10f ");
	MultiplierMethodSetConsts(1000, 1.0e-6);
	stat = MultiplierMethodEqConds(NVAR, obj, gradobj,
				       NCND, cond, gradcond, x, 10, 1.0e-4);
	if (stat == success) {
		fprintf(stderr, "----- solution -----\n");
		vectorfprint(stderr, NVAR, x);
	} else {
		fprintf(stderr, "status = %d\n", stat);
	}
}