*-----------------------------------------------------------
* Date created: July 22, 1996
* Copmbustion of Propane - Reduced Formulation.
*
* Averick, B.M., Carter, R.G., More, J.J.,
* The MINPACK-2 Test Problem Collection
* (Preliminary Version)
* Technical Memorandum No.150, May 1991.
*
* 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)
*
      double precision  p005, p05, p5
      parameter (p005=5.0d-3,p05=5.0d-2,p5=0.5d0)

*
* Information about the problem.
*
	write(1,10)
10	format(5x,'Example of nonlinear programming.')
	write(1,11)
11	format(5x,'Combustion of Propane - Reduced Formulation')
	write(1,12)
12	format(5x,'MINPACK-2 test problem 2.5, pp.9')
	write(1,13)
13	format(5x,'B.M. Averick, R.G. Carter, J.J. More,')
	write(1,14)
14	format(5x,'The MINPACK-2 test problem collection')
	write(1,15)
15	format(5x,'Technical Memorandum No.150, May 1991')
	write(1,16)
16	format(5x,'Argonne National Laboratory, Illinois 60439')
*
* Dimension of the problem:
*
	n=5		! No. of variables.
	m=0		! No. of inequality constraints.
	me=5		! No. of equality constraints.
	mb=10		! 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) = p005+10.d0
         x(2) = p005+10.d0
         x(3) = p05
         x(4) = p5+50.d0
         x(5) = p05
*
* 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)=26
*
	return
	end
*

*--------------------------------------------------------------
* Date created: July,22 1996
* Combustion of Propane - Reduced Formulation.
*--------------------------------------------------------------
*
	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 fjac(5,5)
      double precision k(10), r(10)
      double precision sqrtp

      double precision eight, forty, four, one, p, p005, p05, p5, rr,
     +                 ten, three, two, zero

      parameter (zero=0.0d0,p005=5.0d-3,p05=5.0d-2,p5=0.5d0,one=1.0d0,
     +          two=2.0d0,three=3.0d0,four=4.0d0,eight=8.0d0,ten=1.0d1,
     +          forty=4.0d1)
      parameter (p=forty,rr=ten)
*
      data k/0.0d0, 0.0d0, 0.0d0, 0.0d0, 1.930d-1, 2.597d-3, 3.448d-3,
     +     1.799d-5, 2.155d-4, 3.846d-5/
*
* Objective function and its gradient:
*
        objf=1.d0
*
	do i=1,n
	  gobj(1)=0.d0
	end do
*
* Bounds on variables:
*
	j=1
	do i=1,n
	  cb(j)=x(i)
	  cb(j+1)=10000.d0-x(i)
	  j=j+2
	end do
*
* Constraints. (Inequalities):
*
*
* Jacobian of the inequalities constraints:


c     Initialization.

      sqrtp = sqrt(p)
      r(5) = k(5)
      r(6) = k(6)/sqrtp
      r(7) = k(7)/sqrtp
      r(8) = k(8)/p
      r(9) = k(9)/sqrtp
      r(10) = k(10)/p

*
* Constraints. (Equalities):
*
         e(1) = x(1)*x(2) + x(1) - three*x(5)
         e(2) = two*x(1)*x(2) + x(1) + two*r(10)*x(2)**2 +
     +             x(2)*x(3)**2 + r(7)*x(2)*x(3) + r(9)*x(2)*x(4) +
     +             r(8)*x(2) - rr*x(5)
         e(3) = two*x(2)*x(3)**2 + r(7)*x(2)*x(3) +
     +             two*r(5)*x(3)**2 + r(6)*x(3) - eight*x(5)
         e(4) = r(9)*x(2)*x(4) + two*x(4)**2 - four*rr*x(5)
         e(5) = x(1)*x(2) + x(1) + r(10)*x(2)**2 + x(2)*x(3)**2 +
     +             r(7)*x(2)*x(3) + r(9)*x(2)*x(4) + r(8)*x(2) +
     +             r(5)*x(3)**2 + r(6)*x(3) + x(4)**2 - one
*
* Jacobian of the equalities constraints:
*
         fjac(1,1) = x(2) + one
         fjac(2,1) = two*x(2) + one
         fjac(3,1) = zero
         fjac(4,1) = zero
         fjac(5,1) = x(2) + one

         fjac(1,2) = x(1)
         fjac(2,2) = two*x(1) + four*r(10)*x(2) + x(3)**2 + r(7)*x(3) +
     +               r(9)*x(4) + r(8)
         fjac(3,2) = two*x(3)**2 + r(7)*x(3)
         fjac(4,2) = r(9)*x(4)
         fjac(5,2) = x(1) + two*r(10)*x(2) + x(3)**2 + r(7)*x(3) +
     +               r(9)*x(4) + r(8)

         fjac(1,3) = zero
         fjac(2,3) = two*x(2)*x(3) + r(7)*x(2)
         fjac(3,3) = four*x(2)*x(3) + r(7)*x(2) + four*r(5)*x(3) + r(6)
         fjac(4,3) = zero
         fjac(5,3) = two*x(2)*x(3) + r(7)*x(2) + two*r(5)*x(3) + r(6)

         fjac(1,4) = zero
         fjac(2,4) = r(9)*x(2)
         fjac(3,4) = zero
         fjac(4,4) = r(9)*x(2) + four*x(4)
         fjac(5,4) = r(9)*x(2) + two*x(4)

         fjac(1,5) = -three
         fjac(2,5) = -rr
         fjac(3,5) = -eight
         fjac(4,5) = -four*rr
         fjac(5,5) = zero
*
	do j=1,n
	   do i=1,n
	      ij=i+(j-1)*n
	      ge(ij)=fjac(i,j)
	   end do
	end do
*
	return
	end
*--------------------------------------------------------PROPR.for