*-----------------------------------------------------------
* Date created: February 26, 1996 
*
* Pibouleau, L., Floquet, P., Domenech, S.,
* Optimisation de procedes chimiques par une methode de
* gradient reduit. Partie II. Exemples d'illustration.
* Comparaison avec d'autres methodes.
* RAIRO Recherche Operationnelle, vol.19, no.4, 1985, pp.325
*
* Problem P1.
*
* Dr. Neculai Andrei,
* Research Institute for Informatics - Bucharest
* 8-10, Averescu Avenue,
* 71316 Bucharest - Romania
* E-mail: nandrei@u3.ici.ro
* web: http://www.ici.ro/camo/neculai/nandrei.htm
*-----------------------------------------------------------
*
	subroutine ini(n,m,mb,me,sb,x,icgc,ipgc,icge,ipge,
     1                 nszc,nsze,nb)
*
	double precision x(n)
	integer sb(mb)
	integer icgc(nszc),ipgc(n+1)
	integer icge(nsze),ipge(n+1)
*
* Information about the problem.
*
	write(1,10)
10	format(5x,'Example from Pibouleau et.al, Probleme P1')
	write(1,11)
11	format(5x,'RAIRO , 1985, pp.325-326.')
*
* Dimension of the problem:
*
	n=8
	m=0
	me=4
	mb=16
	nszc=0
	nsze=18
*
* Initial point:
*
	x(1)=1.d0
	x(2)=0.d0
	x(3)=1.d0
	x(4)=-0.5d0
	x(5)=1.d0
	x(6)=2.d0
	x(7)=3.d0
	x(8)=0.d0
*
* Index vector for simple bounds: cb >= 0.
*
	j=1
	do i=1,n
	  sb(j)=i
	  sb(j+1)=-i
	  j=j+2
	end do
*
* Rows indices of the nonzeros of the Jacobian of the Equalities
*
	icge(1)=1
	icge(2)=2
	icge(3)=4
*
	icge(4)=2
	icge(5)=4
*
	icge(6)=1
	icge(7)=3
*
	icge(8)=1
*
	icge(9)=1
	icge(10)=2
	icge(11)=3
	icge(12)=4
*
	icge(13)=4
*
	icge(14)=1
	icge(15)=2
	icge(16)=3
	icge(17)=4
*
	icge(18)=2
*
* Starting address of columns of the Jacobian of the Equalities
*
	ipge(1)=1
	ipge(2)=4
	ipge(3)=6
	ipge(4)=8
	ipge(5)=9
	ipge(6)=13
	ipge(7)=14
	ipge(8)=18
	ipge(9)=19
*
	return
	end
*

*--------------------------------------------------------------
* Date created: February 2, 1996
* Example from Pibouleau. Probleme P1.
*--------------------------------------------------------------
*
	subroutine prob(n,m,mb,me,sb,x,objf,gobj,c,gc,cb,e,ge,
     1                  nszc,nsze,nb)
*
* Calculate problem function at iterate x.
*
	double precision x(n),objf,gobj(n),c(m),gc(nszc)
	double precision cb(mb),e(me),ge(nsze)
*
* Objective function and its gradient:
*
*
*
	objf=1000.d0*x(1)*x(1)   +  x(2)*x(2)/1000.d0  +
     1		(x(3)-1.d0)**2   +  (x(4)-0.5d0)**2    +
     1		(x(5)-1.d0)**2   +  (x(6)-5.d0)**2     +
     1		(x(7)-3.d0)**2   +  110.d0*((x(8)-1.d0)**2)
*
	gobj(1)=2000.d0*x(1)
	gobj(2)=x(2)/500.d0
	gobj(3)=2.d0*(x(3)-1.d0)
	gobj(4)=2.d0*(x(4)-0.5)
	gobj(5)=2.d0*(x(5)-1.d0)
	gobj(6)=2.d0*(x(6)-5.d0)
	gobj(7)=2.d0*(x(7)-3.d0)
	gobj(8)=220.d0*(x(8)-1.d0)
*
*
* Bounds on variables:
*
	cb(1)=x(1)
	cb(2)=2.d0-x(1)
	cb(3)=x(2)+5.d0
	cb(4)=1.d0-x(2)
	cb(5)=x(3)
	cb(6)=2.d0-x(3)
	cb(7)=x(4)+1.d0
	cb(8)=2.d0-x(4)
	cb(9)=x(5)
	cb(10)=4.d0-x(5)
	cb(11)=x(6)+1.d0
	cb(12)=10.d0-x(6)
	cb(13)=x(7)
	cb(14)=6.d0-x(7)
	cb(15)=x(8)+1.d0
	cb(16)=1.d0-x(8)
*
* Constraints. (Equalities):
*
	e(1)=x(1)+2.d0*x(3)+x(4)+3.d0*x(5)-x(7)-2.5d0
	e(2)=x(1)+x(2)+3.d0*x(5)-4.d0*x(7)+x(8)+8.d0
	e(3)=-2.d0*x(3)+8.d0*x(5)-5.d0*x(7)+9.d0
	e(4)=3.d0*x(1)+x(2)+9.d0*x(5)+x(6)+17.d0*x(7)-65.d0
*
* Jacobian of the equalities constraints:
*
	ge(1)=1.d0
	ge(2)=1.d0
	ge(3)=3.d0
*
	ge(4)=1.d0
	ge(5)=1.d0
*
	ge(6)=2.d0
	ge(7)=-2.d0
*
	ge(8)=1.d0
*
	ge(9)=3.d0
	ge(10)=3.d0
	ge(11)=8.d0
	ge(12)=9.d0
*
	ge(13)=1.d0
*
	ge(14)=-1.d0
	ge(15)=-4.d0
	ge(16)=-5.d0
	ge(17)=17.d0
*
	ge(18)=1.d0
*
	return
	end
*
*-----------------------------------------PIB.for