*-----------------------------------------------------------
* Date created: February 19, 1996
*
* Example from K. Schittkowski, 
* More Test Examples for Nonlinear Programming Codes.
* Springer Verlag, 1987, Problem 376, pp.195  
*
* Optimization of a multi-spindle automatic lathe.
*
* 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 K., Schittkowski.')
	write(1,11)
11	format(5x,'More Test Examples for Nonlinear Programming',
     1            ' Codes.')
	write(1,12)
12	format(5x,'Springer Verlag, 1987.')
	write(1,13)
13	format(5x,'Problem 376, pp.195.')
*
* Dimension of the problem:
*
	n=10
	m=14
	me=1
	mb=20
	nszc=48
	nsze=2
*
* Initial point:
*
	x(1)=9.9999d0
	x(2)=0.005d0
	x(3)=0.0080d0
	x(4)=100.d0
	x(5)=0.001699d0
	x(6)=0.001299d0
	x(7)=0.002699d0
	x(8)=0.00199d0
	x(9)=0.99d0
	x(10)=0.999d0
*
* 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
*1	
	icgc(1)=1
	icgc(2)=2
	icgc(3)=3
	icgc(4)=4
	icgc(5)=5
*2
	icgc(6)=6
	icgc(7)=7
	icgc(8)=8
	icgc(9)=9
	icgc(10)=10
	icgc(11)=11
	icgc(12)=12
	icgc(13)=13
*3
	icgc(14)=1
	icgc(15)=10
	icgc(16)=11
	icgc(17)=12
	icgc(18)=13
	icgc(19)=14
*4
	icgc(20)=1
	icgc(21)=2
	icgc(22)=3
	icgc(23)=4
	icgc(24)=5
	icgc(25)=6
	icgc(26)=7
	icgc(27)=8
	icgc(28)=9
	icgc(29)=10
	icgc(30)=11
	icgc(31)=12
	icgc(32)=13
	icgc(33)=14
*5
	icgc(34)=2
	icgc(35)=6
	icgc(36)=14
*6	
	icgc(37)=3
	icgc(38)=7
	icgc(39)=14
*7
	icgc(40)=4
	icgc(41)=8
*8
	icgc(42)=5
	icgc(43)=9
	icgc(44)=14
*9
	icgc(45)=2
	icgc(46)=6
*10
	icgc(47)=3
	icgc(48)=7
*
* Starting address of columns of the Jacobian of the Inequalities
*
	ipgc(1)=1
	ipgc(2)=6
	ipgc(3)=14
	ipgc(4)=20
	ipgc(5)=34
	ipgc(6)=37
	ipgc(7)=40
	ipgc(8)=42
	ipgc(9)=45
	ipgc(10)=47
	ipgc(11)=49
*
* Rows indices of the nonzeros of the Jacobian of the Equalities
*	
	icge(1)=1
	icge(2)=1
*
* Starting address of columns of the Jacobian of the Equalities
*
	ipge(1)=0
	ipge(2)=0
	ipge(3)=0
	ipge(4)=0
	ipge(5)=0
	ipge(6)=0
	ipge(7)=0
	ipge(8)=0
	ipge(9)=1
	ipge(10)=2
	ipge(11)=3
*
	return
	end
*

*--------------------------------------------------------------
* Date created: February 19, 1996
* Example from K., Schittkowski: Problem 376, pp.195
*--------------------------------------------------------------
*
	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)
	double precision t1,t2
*
* Objective function and its gradient:
*
	t1=0.15d0*x(1)+14.d0*x(2)-0.06
	t2=0.002d0+x(1)+60.d0*x(2)
*
	objf=-20000.d0*t1/t2
*
	gobj(1)=-20000.d0*(0.15d0*t2-t1)/(t2*t2)
	gobj(2)=-20000.d0*(14.d0*t2-60.d0*t1)/(t2*t2)
	gobj(3)=0.d0
	gobj(4)=0.d0
	gobj(5)=0.d0
	gobj(6)=0.d0
	gobj(7)=0.d0
	gobj(8)=0.d0
	gobj(9)=0.d0
	gobj(10)=0.d0
*
* Bounds on variables:
*
	cb(1)=x(1)
	cb(2)=10.d0-x(1)
	cb(3)=x(2)
	cb(4)=0.1-x(2)
	cb(5)=x(3)-0.5e-04
	cb(6)=0.0081-x(3)
	cb(7)=x(4)-10.d0
	cb(8)=1000.d0-x(4)
	cb(9)=x(5)-0.5e-04
	cb(10)=0.0017-x(5)
	cb(11)=x(6)-0.5e-04
	cb(12)=0.0013-x(6)
	cb(13)=x(7)-0.5e-04
	cb(14)=0.0027-x(7)
	cb(15)=x(8)-0.5e-04
	cb(16)=0.002-x(8)
	cb(17)=x(9)-0.5e-04
	cb(18)=1.d0-x(9)
	cb(19)=x(10)-0.5e-04
	cb(20)=1.d0-x(10)
*
* Constraints. (Inequalities):
*
	c(1)=x(1)-0.75d0/(x(3)*x(4))
	c(2)=x(1)-x(9)/(x(4)*x(5))
	c(3)=x(1)-x(10)/(x(4)*x(6))-10.d0/x(4)
	c(4)=x(1)-0.19d0/(x(4)*x(7))-10.d0/x(4)
	c(5)=x(1)-0.125d0/(x(4)*x(8))
	c(6)=10000.d0*x(2)-0.00131d0*x(9)*(x(5)**0.666)*(x(4)**1.5)
	c(7)=10000.d0*x(2)-0.001038d0*x(10)*(x(6)**1.6)*(x(4)**3)
	c(8)=10000.d0*x(2)-0.000223d0*(x(7)**0.666)*(x(4)**1.5)
	c(9)=10000.d0*x(2)-0.000076*(x(8)**3.55)*(x(4)**5.66)
	c(10)=10000.d0*x(2)-0.000698*(x(3)**1.2)*x(4)*x(4)
	c(11)=10000.d0*x(2)-0.00005*(x(3)**1.6)*(x(4)**3)
	c(12)=10000.d0*x(2)-0.00000654*(x(3)**2.42)*(x(4)**4.17)
	c(13)=10000.d0*x(2)-0.000257*(x(3)**0.666)*(x(4)**1.5)
	c(14)=30.d0-2.003d0*x(4)*x(5)-1.885d0*x(4)*x(6)-
     1		    0.184d0*x(4)*x(8)-2.d0*x(4)*(x(3)**0.803)
*
* Jacobian of the inequalities constraints:
*1
	gc(1)=1.d0
	gc(2)=1.d0
	gc(3)=1.d0
	gc(4)=1.d0
	gc(5)=1.d0
*2
	gc(6)=10000.d0
	gc(7)=10000.d0
	gc(8)=10000.d0
	gc(9)=10000.d0
	gc(10)=10000.d0
	gc(11)=10000.d0
	gc(12)=10000.d0
	gc(13)=10000.d0
*3
	gc(14)=0.75d0/(x(3)*x(3)*x(4))
	gc(15)=-0.000698d0*1.2d0*(x(4)**2)*(x(3)**0.2)
	gc(16)=-0.00005d0*1.6d0*(x(4)**3)*(x(3)**0.6)
	gc(17)=-0.00000654d0*2.42d0*(x(3)**1.42)*(x(4)**4.17)
	gc(18)=-0.000257*0.666*(x(4)**1.5)/(x(3)**0.334)
	gc(19)=-2.d0*0.803d0*x(4)/(x(3)**0.197)
*4
	gc(20)=0.75d0/(x(3)*x(4)*x(4))
	gc(21)=x(9)/(x(4)*x(4)*x(5))
	gc(22)=x(10)/(x(6)*x(4)*x(4))+10.d0/(x(4)**2)
	gc(23)=0.19/(x(4)*x(4)*x(7))+10.d0/(x(4)**2)
	gc(24)=0.125d0/(x(4)*x(4)*x(8))
	gc(25)=-0.00131d0*1.5d0*x(9)*(x(5)**0.666)*(x(4)**0.5)
	gc(26)=-0.001038d0*3.d0*x(10)*(x(6)**1.6)*(x(4)**2)
	gc(27)=-0.000223*1.5d0*(x(7)**0.666)*(x(4)**0.5)
	gc(28)=-0.000076*5.66d0*(x(8)**3.55)*(x(4)**4.66)
	gc(29)=-0.000698*2.d0*(x(3)**1.2)*x(4)
	gc(30)=-0.00005d0*3.d0*(x(3)**1.6)*(x(4)**2)
	gc(31)=-0.00000654*4.17d0*(x(3)**2.42)*(x(4)**3.17)
	gc(32)=-0.000257*1.5d0*(x(3)**0.666)*(x(4)**0.5)
	gc(33)=-2.003d0*x(5)-1.885d0*x(6)-0.184d0*x(8)-2.d0*(x(3)**0.803)
*5
	gc(34)=x(9)/(x(5)*x(5)*x(4))
	gc(35)=-0.00131*x(9)*(x(4)**1.5)/(x(5)**0.334)
	gc(36)=-2.003d0*x(4)
*6
	gc(37)=x(10)/(x(4)*x(6)*x(6))
	gc(38)=-0.001038d0*1.6d0*(x(4)**3)*(x(6)**0.6)*x(10)
	gc(39)=-1.885d0*x(4)
*7
	gc(40)=0.19d0/(x(4)*x(7)*x(7))
	gc(41)=-0.000223d0*0.666*(x(4)**1.5)/(x(7)**0.334)
*8
	gc(42)=0.125d0/(x(4)*x(8)*x(8))
	gc(43)=-0.000076d0*3.55d0*(x(8)**2.55)*(x(4)**5.66)
	gc(44)=-0.184d0*x(4)
*9
	gc(45)=-1.d0/(x(4)*x(5))
	gc(46)=-0.00131d0*(x(5)**0.666)*(x(4)**1.5)
*10
	gc(47)=-1.d0/(x(4)*x(6))
	gc(48)=-0.001038d0*(x(6)**1.6)*(x(4)**3)
*
* Constraints. (Equality):
*
	e(1)=x(9)+x(10)-0.225d0
*
* Jacobian of the equalities constraints:
*
	ge(1)=1.d0
	ge(2)=1.d0
*
	return
	end
*-------------------------------------------------S376.for