*-----------------------------------------------------------
* Date created: February 21, 1996
* Example from A Ben-Israel, A. Ben-Tal, S. Zlobec, 
* Optimality in nonlinear programming: 
* A feasible directions approach.
* John Wiley & Sons, New York, 1981.
*
* Example from pp.75 (paragraph 7.2)
*
* Project: Test Problems with SPENBAR
*
* Dr. Neculai Andrei
* Research Institute for Informatics,
* 8-10, Averescu Avenue, Bucharest 1, Romania
* E-mail: nandrei@u3.ici.ro
* web: www.ici.ro/camo
*-----------------------------------------------------------
*
	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 A. Ben-Israel, A. Ben-Tal, S. Zlobec')
	write(1,11)
11	format(5x,'Example from pp.75.')
*
* Dimension of the problem:
*
	n=18
	m=13
	me=0
	mb=0
	nszc=30
	nsze=0
*
* Initial point:
*
	x(1)=1.d0
	x(2)=0.d0
	x(3)=2.d0
	x(4)=1.d0
	x(5)=0.d0
	x(6)=-1.5d0
	x(7)=2.d0
	x(8)=-8.d0
	x(9)=1.d0
	x(10)=0.d0
	x(11)=9.d0
	x(12)=0.d0
	x(13)=5.d0
	x(14)=1.d0
	x(15)=0.d0
	x(16)=6.16441d0
	x(17)=0.d0
	x(18)=32.6001d0
*
* Index vector for simple bounds: cb >= 0.
*
*
* Rows indices of the nonzeros of the Jacobian of the Inequalities
*
	icgc(1)=1
	icgc(2)=2
	icgc(3)=3
	icgc(4)=4
	icgc(5)=5
	icgc(6)=6
	icgc(7)=7
	icgc(8)=8
	icgc(9)=9
	icgc(10)=10
	icgc(11)=11
	icgc(12)=12
	icgc(13)=13
	icgc(14)=2
	icgc(15)=3
	icgc(16)=4
	icgc(17)=5
	icgc(18)=13
	icgc(19)=6
	icgc(20)=7
	icgc(21)=8
	icgc(22)=8
	icgc(23)=9
	icgc(24)=9
	icgc(25)=10
	icgc(26)=10
	icgc(27)=11
	icgc(28)=11
	icgc(29)=12
	icgc(30)=12
*
* Starting address of columns of the Jacobian of the Inequalities
*
	ipgc(1)=1
	ipgc(2)=14
	ipgc(3)=15
	ipgc(4)=16
	ipgc(5)=17
	ipgc(6)=18
	ipgc(7)=19
	ipgc(8)=20
	ipgc(9)=21
	ipgc(10)=22
	ipgc(11)=23
	ipgc(12)=24
	ipgc(13)=25
	ipgc(14)=26
	ipgc(15)=27
	ipgc(16)=28
	ipgc(17)=29
	ipgc(18)=30
	ipgc(19)=31
*
	return
	end
*

*--------------------------------------------------------------
* Date created: February 21, 1996
* Example from A.Ben-Israel, A. Ben-Tal, S. Zlobec.,   pp.75
*--------------------------------------------------------------
*
	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)-7.d0)**2 - x(2) + dexp(-x(3)) + x(3) + x(4)**2+
     1		dexp(x(5)) + (x(6)+3.d0)**4 + (x(7)-4.d0)**2+
     1		dexp(-x(8)) + 12.d0*x(8) + dexp(x(9))+
     1		(x(9)+x(10))**2+
     1		(x(11)-x(12))**4 + dexp(x(13)) + dexp(-x(14))+
     1		(x(15)+x(16))**2 + dexp(x(17)) - 2.d0*x(17)+
     1		x(18)**2
*
	gobj(1)=2.d0*(x(1)-7.d0)
	gobj(2)=-1.d0
	gobj(3)=-dexp(-x(3)) + 1.d0
	gobj(4)=2.d0*x(4)
	gobj(5)=dexp(x(5))
	gobj(6)=4.d0*(x(6)+3.d0)**3
	gobj(7)=2.d0*(x(7)-4.d0)
	gobj(8)=-dexp(-x(8)) + 12.d0
	gobj(9)=dexp(x(9)) + 2.d0*(x(9)+x(10))
	gobj(10)=2.d0*(x(9)+x(10))
	gobj(11)=4.d0*(x(11)-x(12))**3
	gobj(12)=-4.d0*(x(11)-x(12))**3
	gobj(13)=dexp(x(13))
	gobj(14)=-dexp(-x(14))
	gobj(15)=2.d0*(x(15)+x(16))
	gobj(16)=2.d0*(x(15)+x(16))
	gobj(17)=dexp(x(17))-2.d0
	gobj(18)=2.d0*x(18)
*
* Bounds on variables:
*
*
* Constraints. (Inequalities):
*
	c(1)=   1.5d0 - x(1)*x(1)
	c(2)=   2.d0  - x(1)**2 - dexp(x(2))
	c(3)=   4.d0  + 2.d0*x(1) -(x(3)-1.d0)**2
	c(4)=   5.d0  - dexp(x(1)) - x(1) - x(4)
	c(5)=  11.d0  - (x(1)+2.d0)**2 - dexp(-x(5)) + x(5)
	c(6)=  15.d0  - dexp(-x(1)) - x(1)**2 - x(7)**2
	c(7)=  12.d0  - (x(1)-3.d0)**2 + x(8)
	c(8)=   2.d0  + 3.d0*x(1) - (x(9)-x(10))**2
	c(9)=  12.d0  - x(1)**4 - x(1) - x(11) - dexp(-x(12))
	c(10)= 20.d0  - dexp(x(1)) + 10.d0*x(1) - x(13)**2 + 5.d0*x(14)
	c(11)= 35.d0  - x(1)**4 + 5.d0*x(1) - dexp(-x(15)) - 
     1			(x(15)-x(16))**2
	c(12)=300.d0  - (x(1)+4.d0)**4 - dexp(x(17)) + 10.d0*x(18)
	c(13)= 10.d0  - 7.d0*x(1) + 2.d0*x(6)
*
* Jacobian of the inequalities constraints:
*
	gc(1)=-2.d0*x(1)
	gc(2)=-2.d0*x(1)
	gc(3)=2.d0
	gc(4)=-dexp(x(1))-1.d0
	gc(5)=-2.d0*(x(1)+2.d0)
	gc(6)=dexp(-x(1))-2.d0*x(1)
	gc(7)=-2.d0*(x(1)-3.d0)
	gc(8)=3.d0
	gc(9)=-4.d0*x(1)**3-1.d0
	gc(10)=-dexp(x(1))+10.d0
	gc(11)=-4.d0*x(1)**3+5.d0
	gc(12)=-4.d0*(x(1)+4.d0)**3
	gc(13)=-7.d0
*2
	gc(14)=-dexp(x(2))
	gc(15)=-2.d0*(x(3)-1.d0)
	gc(16)=-1.d0
	gc(17)=dexp(-x(5))+1.d0
	gc(18)=2.d0
	gc(19)=-2.d0*x(7)
	gc(20)=1.d0
	gc(21)=-2.d0*(x(9)-x(10))
	gc(22)=2.d0*(x(9)-x(10))
	gc(23)=-1.d0
	gc(24)=dexp(-x(12))
	gc(25)=-2.d0*x(13)
	gc(26)=5.d0
	gc(27)=dexp(-x(15))-2.d0*(x(15)-x(16))
	gc(28)=2.d0*(x(15)-x(16))
	gc(29)=-dexp(x(17))
	gc(30)=10.d0
*
	return
	end
*---------------------------------------------------BBZ72.for