* Date created: April 11, 1997
*
* Brown, A.A., Bartholomew-Biggs, M.C.,
* ODE versus SQP methods for constrained optimisation.
* Technical Report No. 179, June 1987.
* The Hatfield Polytechnic, College Lane, Hatfield
*                            
* Problem 16.                                     
*
* 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 of nonlinear programming.')
	write(1,11)
11	format(5x,'Problem 16, April 11, 1997')
	write(1,12)
12	format(5x,'Brown & Bartholomew-Biggs')
*
* Dimension of the problem:
*
	n=6		! No. of variables.
	m=0		! No. of inequality constraints.
	me=5		! No. of equality constraints.
	mb=2*n		! No. of simple bounds on variables.
	nszc=0		! No. of non-zeros in Jacobian of c(x)>=0.
	nsze=25		! No. of non-zeros in Jacobian of e(x) =0.

*
* Initial point:
*
	x(1)=6.d0
	x(2)=5.d0
	x(3)=4.d0
	x(4)=3.d0
	x(5)=2.d0
	x(6)=1.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 Inequalities
*
* Starting address of columns of the Jacobian of the Inequalities
*
* Rows indices of the nonzeros of the Jacobian of the Equalities
*
	icge(1)=1	
	icge(2)=2
	icge(3)=3	
	icge(4)=4	
	icge(5)=5	

	icge(6)=1	
	icge(7)=2	
	icge(8)=3
	icge(9)=4
	icge(10)=5

	icge(11)=1
	icge(12)=2	
	icge(13)=3	
	icge(14)=4	
	icge(15)=5
	
	icge(16)=1
	icge(17)=2	
	icge(18)=3
	icge(19)=4
	icge(20)=5
	
	icge(21)=1
	icge(22)=2	
	icge(23)=3	
	icge(24)=4	
	icge(25)=5	
*	
* Starting address of columns of the Jacobian of the Equalities
*
	ipge(1)=1
	ipge(2)=6
	ipge(3)=11
	ipge(4)=16
	ipge(5)=21
	ipge(6)=0
	ipge(7)=26
*
	return
	end


*--------------------------------------------------------------
* Date created: April 11, 1997
* Problem 16, Brown & Bartholomew-Biggs
*--------------------------------------------------------------
*
	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 =  (x(1)-x(2))**2 + (x(2)-x(3))**4 + 
     1		(x(3)-x(4))**6 + (x(4)-x(5))**8 +
     1		(x(5)-x(6))**2
*
	gobj(1)=  2.d0*(x(1)-x(2))
	gobj(2)= -2.d0*(x(1)-x(2))    + 4.d0*(x(2)-x(3))**3 
	gobj(3)= -4.d0*(x(2)-x(3))**3 + 6.d0*(x(3)-x(4))**5 
	gobj(4)= -6.d0*(x(3)-x(4))**5 + 8.d0*(x(4)-x(5))**7 
	gobj(5)= -8.d0*(x(4)-x(5))**7 + 2.d0*(x(5)-x(6)) 
	gobj(6)= -2.d0*(x(5)-x(6)) 
*
* Bounds on variables:
*
	j=1
	do i=1,n
	  cb(j)=x(i)+100.d0
	  cb(j+1)=100.d0-x(i)
	  j=j+2
	end do
*
* Constraints. (Inequalities):
*
*
* Jacobian of the inequalities constraints:
*
* Constraints. (Equalities):
*
	e(1) = x(1)+x(2)+x(3)+x(4)+x(5) - 15.d0

	e(2) =  x(1)*(x(2)+x(3)+x(4)+x(5)) + 
     1		x(2)*(x(3)+x(4)+x(5)) +
     1		x(3)*(x(4)+x(5)) +
     1		x(4)*x(5) - 85.d0

	e(3) =  x(1)*x(2)*(x(3)+x(4)+x(5)) +
     1		x(4)*x(5)*(x(1)+x(2)+x(3)) +
     1		(x(1)*x(3)+x(2)*x(3))*(x(4)+x(5)) - 225.d0

	e(4) =  x(1)*x(2)*(x(3)*x(4)+x(4)*x(5)+x(3)*x(5)) +
     1		x(3)*x(4)*x(5)*(x(1)+x(2)) - 274.d0

	e(5) = x(1)*x(2)*x(3)*x(4)*x(5) - 120.d0
*
* Jacobian of the equalities constraints:
*
c1
	ge(1)=1.d0
	ge(2)=x(2)+x(3)+x(4)+x(5)
	ge(3)=x(2)*(x(3)+x(4)+x(5)) + x(4)*x(5) +
     1		x(3)*(x(4)+x(5))
	ge(4)=x(2)*(x(3)*x(4)+x(4)*x(5)+x(3)*x(5)) +
     1		x(3)*x(4)*x(5)
	ge(5)=x(2)*x(3)*x(4)*x(5)
c2
	ge(6)=1.d0
	ge(7)=x(1) + (x(3)+x(4)+x(5))
	ge(8)=x(1)*(x(3)+x(4)+x(5)) + x(4)*x(5) +
     1		x(3)*(x(4)+x(5))
	ge(9)=x(1)*(x(3)*x(4)+x(4)*x(5)+x(3)*x(5)) +
     1		x(3)*x(4)*x(5)
	ge(10)=x(1)*x(3)*x(4)*x(5)
c3
	ge(11)=1.d0
	ge(12)=x(1) + x(2) + (x(4)+x(5))
	ge(13)=x(1)*x(2) + x(4)*x(5) + 
     1		(x(1)+x(2))*(x(4)+x(5))
	ge(14)=x(1)*x(2)*x(4) + x(1)*x(2)*x(5) +
     1		x(4)*x(5)*(x(1)+x(2))
	ge(15)=x(1)*x(2)*x(4)*x(5)
c4
	ge(16)=1.d0
	ge(17)=x(1) + x(2) + x(3) + x(5)
	ge(18)=x(1)*x(2) + x(5)*(x(1)+x(2)+x(3)) +
     1		x(1)*x(3)+x(2)*x(3)
	ge(19)=x(1)*x(2)*x(3) + x(1)*x(2)*x(5) +
     1		x(3)*x(5)*(x(1)+x(2))
	ge(20)=x(1)*x(2)*x(3)*x(5)
c5
	ge(21)=1.d0
	ge(22)=x(1) + x(2) + x(3) + x(4)
	ge(23)=x(1)*x(2) + x(4)*(x(1)+x(2)+x(3)) +
     1		x(1)*x(3)+x(2)*x(3)
	ge(24)=x(1)*x(2)*(x(4)+x(3)) +
     1		x(3)*x(4)*(x(1)+x(2))
	ge(25)=x(1)*x(2)*x(3)*x(4)
*
	return
	end
*
*---------------------------------------------BB16.for