*---------------------------------------------------------
* Date created:  December 23, 1997
* 
* Hock-Schittkowski, 
* Test Examples for Nonlinear Programming.
* Springer-Verlag, 1981.
*
* Problem 108, pp,117
*
* 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)
*
	write(1,10)
10	format(5x,'Example of a Nonlinear Programming Problem.')
	write(1,11)
11	format(5x,'Hock-Schittkowski, Prob 108, pp.117')
*
      n=9
      m=13
      me=0
      mb=2*n
      nszc=39
      nsze=0
*
* Initial Point
*
	do i=1,n
	  x(i)=1.d0
	end do
*
* Index vector for simple bounds:
*
      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.
*
*
* Starting address of the columns of the Jacobian of the Equalities.
*
*
* Rows indices of the nonzeros of the Jacobian of the INEqualities.
*
	icgc(1)=3
	icgc(2)=4
	icgc(3)=11
	icgc(4)=12
c2
	icgc(5)=3
	icgc(6)=4
	icgc(7)=11
	icgc(8)=12
c3
	icgc(9)=1
	icgc(10)=5
	icgc(11)=6
	icgc(12)=8
	icgc(13)=12
c4
	icgc(14)=1
	icgc(15)=5
	icgc(16)=6
	icgc(17)=12
c5
	icgc(18)=2
	icgc(19)=3
	icgc(20)=5
	icgc(21)=9
	icgc(22)=13
c6
	icgc(23)=2
	icgc(24)=3
	icgc(25)=5
	icgc(26)=9
c7
	icgc(27)=4
	icgc(28)=6
	icgc(29)=7
	icgc(30)=9
c8
	icgc(31)=4
	icgc(32)=6
	icgc(33)=7
	icgc(34)=9
c9
	icgc(35)=7
	icgc(36)=8
	icgc(37)=10
	icgc(38)=11
	icgc(39)=13
*
* Starting address of the columns of the Jacobian of the INEqualities.
*
	ipgc(1)=1
	ipgc(2)=5
	ipgc(3)=9
	ipgc(4)=14
	ipgc(5)=18
	ipgc(6)=23
	ipgc(7)=27
	ipgc(8)=31
	ipgc(9)=35
	ipgc(10)=40
*
      return
      end
*

*--------------------------------------------------------------
* Date created:  December 23, 1997
* Hock-Schittkowski, Problem 108, pp.117.
*--------------------------------------------------------------
*
      subroutine prob(n,m,mb,me,sb,x,objf,gobj,c,gc,cb,e,ge,
     1                nszc,nsze,nb)
*
* Calculate problem functions at iterate x.
*
      double precision x(n),objf,gobj(n),c(m),gc(nszc),
     *   cb(mb),e(me),ge(nsze)
*
* Objective function, and its gradient.
*
        objf=  -(x(1)*x(4) - x(2)*x(3) + x(3)*x(9) -
     1		 x(5)*x(9) + x(5)*x(8) - x(6)*x(7))/2.d0
*
	gobj(1)=-x(4)/2.d0
	gobj(2)= x(3)/2.d0
	gobj(3)= x(2)/2.d0 - x(9)/2.d0
	gobj(4)=-x(1)/2.d0
	gobj(5)= x(9)/2.d0 - x(8)/2.d0
	gobj(6)= x(7)/2.d0
	gobj(7)= x(6)/2.d0
	gobj(8)=-x(5)/2.d0
	gobj(9)=-x(3)/2.d0 + x(5)/2.d0
*
* Bounds on variables.
*
	j=1
	do i=1,n
	  cb(j)=x(i)
	  cb(j+1)=1.d38-x(i)
	  j=j+2
	end do
*
*
* Equality Constraints.
*
*
* Jacobian of the equality constraints.
*
*
* INEquality Constraints.
*
*
	c(1)=1.d0 - x(3)**2 - x(4)**2
	c(2)=1.d0 - x(5)**2 - x(6)**2
	c(3)=1.d0 - (x(1)-x(5))**2 - (x(2)-x(6))**2
	c(4)=1.d0 - (x(1)-x(7))**2 - (x(2)-x(8))**2
	c(5)=1.d0 - (x(3)-x(5))**2 - (x(4)-x(6))**2
	c(6)=1.d0 - (x(3)-x(7))**2 - (x(4)-x(8))**2
	c(7)=1.d0 - x(7)**2        - (x(8)-x(9))**2
	c(8)=x(3)*x(9)
	c(9)=x(5)*x(8) - x(6)*x(7)
	c(10)=1.d0 - x(9)**2
	c(11)=1.d0 - x(1)**2       - (x(2)-x(9))**2
	c(12)=x(1)*x(4) - x(2)*x(3)
	c(13)=-x(5)*x(9)
*
* Jacobian of the INEquality constraints.
c1
	gc(1)=-2.d0*(x(1)-x(5))
	gc(2)=-2.d0*(x(1)-x(7))
	gc(3)=-2.d0*x(1)
	gc(4)=x(4)
c2
	gc(5)=-2.d0*(x(2)-x(6))
	gc(6)=-2.d0*(x(2)-x(8))
	gc(7)=-2.d0*(x(2)-x(9))
	gc(8)=-x(3)
c3
	gc(9)=-2.d0*x(3)
	gc(10)=-2.d0*(x(3)-x(5))
	gc(11)=-2.d0*(x(3)-x(7))
	gc(12)=x(9)
	gc(13)=-x(2)
c4
	gc(14)=-2.d0*x(4)
	gc(15)=-2.d0*(x(4)-x(6))
	gc(16)=-2.d0*(x(4)-x(8))
	gc(17)=x(1)
c5
	gc(18)=-2.d0*x(5)
	gc(19)=2.d0*(x(1)-x(5))
	gc(20)=2.d0*(x(3)-x(5))
	gc(21)=x(8)
	gc(22)=-x(9)
c6
	gc(23)=-2.d0*x(6)
	gc(24)=2.d0*(x(2)-x(6))
	gc(25)=2.d0*(x(4)-x(6))
	gc(26)=-x(7)
c7
	gc(27)=2.d0*(x(1)-x(7))
	gc(28)=2.d0*(x(3)-x(7))
	gc(29)=-2.d0*x(7)
	gc(30)=-x(6)
c8
	gc(31)=2.d0*(x(2)-x(8))
	gc(32)=2.d0*(x(4)-x(8))
	gc(33)=-2.d0*(x(8)-x(9))
	gc(34)=x(5)
c9
	gc(35)=2.d0*(x(8)-x(9))
	gc(36)=x(3)
	gc(37)=-2.d0*x(9)
	gc(38)=2.d0*(x(2)-x(9))
	gc(39)=-x(5)
*
      return
      end
*---------------------------------------------- HS108.for