C     =================================================================
C     File: problem5.f
C     =================================================================

C     ******************************************************************
C     ******************************************************************

      subroutine inip(n,x,l,u,m,lambda,rho,equatn,linear,ml,meq,alin,
     +acol,aval,annz,b,fknown)

      implicit none

C     SCALAR ARGUMENTS
      integer annz,m,ml,meq,n
      double precision fknown

C     ARRAY ARGUMENTS
      logical equatn(*),linear(*)
      integer alin(*),acol(*)
      double precision aval(*),b(*),l(*),lambda(*),rho(*),u(*),x(*)

C     This subroutine must set some problem data. For achieving this 
C     objective YOU MUST MODIFY it according to your problem. See below 
C     where your modifications must be inserted.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     This subroutine has no input parameters.
C
C     On Return
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              initial point,
C
C     l        double precision l(n),
C              lower bounds on x,
C
C     u        double precision u(n),
C              upper bounds on x,
C
C     m        integer,
C              number of constraints (excluding the bounds),
C
C     lambda   double precision lambda(m),
C              initial estimation of the Lagrange multipliers,
C
C     rho      double precision rho(m),
C              initial penalty parameters.
C
C     equatn   logical equatn(m)
C              for each constraint j, set equatn(j) = .true. if it is an 
C              equality constraint of the form c_j(x) = 0, and set 
C              equatn(j) = .false. if it is an inequality constraint of 
C              the form c_j(x) <= 0,
C
C     linear   logical linear(m)
C              for each constraint j, set linear(j) = .true. if it is a 
C              linear constraint, and set linear(j) = .false. if it is a
C              nonlinear constraint.

C     LOCAL SCALARS
      integer i
      double precision seed

C     EXTERNAL FUNCTIONS
      double precision drand

C     Set problem data
C     write(*,*) 'Seed for the random initial point: '
C     read(*,*) seed
      seed = 123456.0d0

C     Number of variables
      n = 3

C     Lower and upper bounds
      l(1) = 1.0d-05
      l(2) = 1.0d-05
      l(3) = 0.0d0

      u(1) = 12.5d0
      u(2) = 37.5d0
      u(3) = 50.0d0

C     Initial point
      do i = 1,n
          x(i) = l(i) + ( u(i) - l(i) ) * drand(seed)
      end do

C     Number of constraints (equalities plus inequalities)

      m = 2

C     Lagrange multipliers approximation.

      do i = 1,m
          lambda(i) = 0.0d0
      end do

C     Initial penalty parameters

      do i = 1,m
          rho(i) = 10.0d0
      end do

C     For each constraint i, set equatn(i) = .true. if it is an equality
C     constraint of the form c_i(x) = 0, and set equatn(i) = .false. if 
C     it is an inequality constraint of the form c_i(x) <= 0.

      do i = 1,m
          equatn(i) = .true.
      end do

C     For each constraint i, set linear(i) = .true. if it is a linear
C     constraint, otherwise set linear(i) = .false.

      linear(1) = .false.
      linear(2) = .false.

C     Set the sparse matrix and the right hand side of the linear
C     constraints of the form ai^t x = bi for i = 1, ..., meq and
C     ai^t x >= bi for i = meq + 1,ml. These linear constraints will
C     be lower-level constraints. All the previous constraints will
C     be upper-lever constraints. The suggestions is to consider the
C     linear constraints as lower-level constraints, but it is not
C     mandatory.

      ml  = 1
      meq = 1

      annz = 3

C     x(1) + x(2) + x(3) = 50.0d0

      alin(1) = 1
      acol(1) = 1
      aval(1) = 1.0d0

      alin(2) = 1
      acol(2) = 2
      aval(2) = 1.0d0

      alin(3) = 1
      acol(3) = 3
      aval(3) = 1.0d0

      b(1) = 50.0d0

      fknown = 0.000000000000000d+00

      end

C     ******************************************************************
C     ******************************************************************

      subroutine evalf(n,x,f,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,n
      double precision f

C     ARRAY ARGUMENTS
      double precision x(n)

C     This subroutine must compute the objective function. For achieving 
C     this objective YOU MUST MODIFY it according to your problem. See 
C     below where your modifications must be inserted.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     On Return
C
C     f        double precision,
C              objective function value at x,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the objective 
C              function. (For example, trying to compute the square root 
C              of a negative number, dividing by zero or a very small 
C              number, etc.) If everything was o.k. you must set it 
C              equal to zero.

      flag = 0

      f = 0.0d0

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalg(n,x,g,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,n

C     ARRAY ARGUMENTS
      double precision g(n),x(n)

C     This subroutine must compute the gradient vector of the objective 
C     function. For achieving these objective YOU MUST MODIFY it in the 
C     way specified below. However, if you decide to use numerical 
C     derivatives (we dont encourage this option at all!) you dont need
C     to modify evalg.
C
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     On Return
C
C     g        double precision g(n),
C              gradient vector of the objective function evaluated at x,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of any component 
C              of the gradient vector. (For example, trying to compute 
C              the square root of a negative number, dividing by zero or 
C              a very small number, etc.) If everything was o.k. you 
C              must set it equal to zero.

C     LOCAL SCALARS
      integer i

      flag = 0

      do i = 1,n
          g(i) = 0.0d0
      end do

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalh(n,x,hlin,hcol,hval,hnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,n,hnnz

C     ARRAY ARGUMENTS
      integer hcol(*),hlin(*)
      double precision hval(*),x(n)

C     This subroutine might compute the Hessian matrix of the objective 
C     function. For achieving this objective YOU MAY MODIFY it according 
C     to your problem. To modify this subroutine IS NOT MANDATORY. See 
C     below where your modifications must be inserted.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     On Return
C
C     hnnz     integer,
C              number of perhaps-non-null elements of the computed 
C              Hessian,
C
C     hlin     integer hlin(hnnz),
C              see below,
C
C     hcol     integer hcol(hnnz),
C              see below,
C
C     hval     double precision hval(hnnz),
C              the non-null value of the (hlin(k),hcol(k)) position 
C              of the Hessian matrix of the objective function must 
C              be saved at hval(k). Just the lower triangular part of
C              Hessian matrix must be computed,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the Hessian
C              matrix of the objective funtion. (For example, trying 
C              to compute the square root of a negative number, 
C              dividing by zero or a very small number, etc.) If 
C              everything was o.k. you must set it equal to zero.

      flag = 0

      hnnz = 0

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalc(n,x,ind,c,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer ind,flag,n
      double precision c

C     ARRAY ARGUMENTS
      double precision x(n)

C     This subroutine must compute the ind-th constraint of your 
C     problem. For achieving this objective YOU MUST MOFIFY it 
C     according to your problem. See below the places where your 
C     modifications must be inserted.
C
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     ind      integer,
C              index of the constraint to be computed,
C
C     On Return
C
C     c        double precision,
C              ind-th constraint evaluated at x,
C
C     flag     integer
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the 
C              constraint. (For example, trying to compute the square 
C              root of a negative number, dividing by zero or a very 
C              small number, etc.) If everything was o.k. you must set 
C              it equal to zero.

      flag = 0

      if ( ind .eq. 1 ) then
C         This seems to by a typo in the paper.
C         c = x(3) ** 2 / x(1) * x(2) ** 3 - 0.000169d0
          c = x(3) ** 2 / ( x(1) * x(2) ** 3 ) - 0.000169d0

      else ! if ( ind .eq. 2 ) then
          c = x(2) / x(1) - 3.0d0
      end if
 
      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evaljac(n,x,ind,indjac,valjac,jcnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,ind,n,jcnnz

C     ARRAY ARGUMENTS
      integer indjac(n)
      double precision x(n),valjac(n)

C     This subroutine must compute the gradient of the constraint ind. 
C     For achieving these objective YOU MUST MODIFY it in the way 
C     specified below.
C
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     ind      integer,
C              index of the constraint whose gradient will be computed,
C
C     On Return
C
C     jcnnz   integer,
C              number of perhaps-non-null elements of the computed 
C              gradient,
C
C     indjac   integer indjac(jcnnz),
C              see below,
C
C     valjac   double precision valjac(jcnnz),
C              the non-null value of the partial derivative of the 
C              ind-th constraint with respect to the indjac(k)-th 
C              variable must be saved at valjac(k).
C
C     flag     integer
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the 
C              constraint. (For example, trying to compute the square 
C              root of a negative number, dividing by zero or a very 
C              small number, etc.) If everything was o.k. you must set 
C              it equal to zero.

      flag = 0

      if ( ind .eq. 1 ) then

C         jcnnz = 3

C         indjac(1) = 1
C         valjac(1) = - x(2) ** 3 * x(3) ** 2 / x(1) ** 2

C         indjac(2) = 2
C         valjac(2) = 3.0d0 * x(2) ** 2 * x(3) ** 2 / x(1)

C         indjac(3) = 3
C         valjac(3) = 2.0d0 * x(2) ** 3 * x(3) / x(1)

          jcnnz = 3

          indjac(1) = 1
          valjac(1) = - x(3) ** 2 / ( x(1) ** 2 * x(2) ** 3 )

          indjac(2) = 2
          valjac(2) = - 3.0d0 * x(3) ** 2 / ( x(1) * x(2) ** 4 ) 

          indjac(3) = 3
          valjac(3) = 2.0d0 * x(3) / ( x(1) * x(2) ** 3 )

      else ! if ( ind .eq. 2 ) then

          jcnnz = 2

          indjac(1) = 1
          valjac(1) = - x(2) / x(1) ** 2

          indjac(2) = 2
          valjac(2) = 1.0d0 / x(1)

      end if

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalhc(n,x,ind,hclin,hccol,hcval,hcnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,ind,n,hcnnz

C     ARRAY ARGUMENTS
      integer hccol(*),hclin(*)
      double precision hcval(*),x(n)

C     This subroutine might compute the Hessian matrix of the ind-th
C     constraint. For achieving this objective YOU MAY MODIFY it 
C     according to your problem. To modify this subroutine IS NOT 
C     MANDATORY. See below where your modifications must be inserted.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     ind      integer,
C              index of the constraint whose Hessian will be computed,
C
C     On Return
C
C     hcnnz    integer,
C              number of perhaps-non-null elements of the computed 
C              Hessian,
C
C     hclin    integer hclin(hcnnz),
C              see below,
C
C     hccol    integer hccol(hcnnz),
C              see below,
C
C     hcval    double precision hcval(hcnnz),
C              the non-null value of the (hclin(k),hccol(k)) position 
C              of the Hessian matrix of the ind-th constraint must 
C              be saved at hcval(k). Just the lower triangular part of
C              Hessian matrix must be computed,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the Hessian
C              matrix of the ind-th constraint. (For example, trying 
C              to compute the square root of a negative number, 
C              dividing by zero or a very small number, etc.) If 
C              everything was o.k. you must set it equal to zero.

      flag = 0

      if ( ind .eq. 1 ) then

C         hcnnz = 6

C         hclin(1) = 1
C         hccol(1) = 1
C         hcval(1) = 2.0d0 * x(2) ** 3 * x(3) ** 2 / x(1) ** 3

C         hclin(2) = 2
C         hccol(2) = 1
C         hcval(2) = - 3.0d0 * x(2) ** 2 * x(3) ** 2 / x(1) ** 2

C         hclin(3) = 3
C         hccol(3) = 1
C         hcval(3) = - 2.0d0 * x(2) ** 3 * x(3) / x(1) ** 2

C         hclin(4) = 2
C         hccol(4) = 2
C         hcval(4) = 6.0d0 * x(2) * x(3) ** 2 / x(1)

C         hclin(5) = 3
C         hccol(5) = 2
C         hcval(5) = 6.0d0 * x(2) ** 2 * x(3) / x(1)

C         hclin(6) = 3
C         hccol(6) = 3
C         hcval(6) = 2.0d0 * x(2) ** 3 / x(1)

          hcnnz = 6

          hclin(1) = 1
          hccol(1) = 1
          hcval(1) = 2.0d0 * x(3) ** 2 / ( x(1) ** 3 * x(2) ** 3 )

          hclin(2) = 2
          hccol(2) = 1
          hcval(2) = 3.0d0 * x(3) ** 2 / ( x(1) ** 2 * x(2) ** 4 )

          hclin(3) = 3
          hccol(3) = 1
          hcval(3) = - 2.0d0 * x(3) / ( x(1) ** 2 * x(2) ** 3 )

          hclin(4) = 2
          hccol(4) = 2
          hcval(4) = 12.0d0 * x(3) ** 2 / ( x(1) * x(2) ** 5 )

          hclin(5) = 3
          hccol(5) = 2
          hcval(5) = - 6.0d0 * x(3) / ( x(1) * x(2) ** 4 )

          hclin(6) = 3
          hccol(6) = 3
          hcval(6) = 2.0d0 / ( x(1) * x(2) ** 3 )

      else ! if ( ind .eq. 2 ) then

          hcnnz = 2

          hclin(1) = 2
          hccol(1) = 1
          hcval(1) = - 1.0d0 / x(1) ** 2

          hclin(2) = 1
          hccol(2) = 1
          hcval(2) = 2.0d0 * x(2) / x(1) ** 3

      end if

      end

C     ******************************************************************
C     ******************************************************************

      subroutine evalhlp(n,x,m,lambda,p,hp,goth,flag)

      implicit none

C     SCALAR ARGUMENTS
      logical goth
      integer flag,m,n

C     ARRAY ARGUMENTS
      double precision hp(n),lambda(m),p(n),x(n)

C     This subroutine might compute the product of the Hessian of the
C     Lagrangian times vector p (just the Hessian of the objective 
C     function in the unconstrained or bound-constrained case). 
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     x        double precision x(n),
C              current point,
C
C     m        integer,
C              number of constraints,
C
C     lambda   double precision lambda(m),
C              vector of Lagrange multipliers,
C
C     p        double precision p(n),
C              vector of the matrix-vector product,
C
C     goth     logical,
C              can be used to indicate if the Hessian matrices were
C              computed at the current point. It is set to .false.
C              by the optimization method every time the current
C              point is modified. Sugestion: if its value is .false. 
C              then compute the Hessians, save them in a common 
C              structure and set goth to .true.. Otherwise, just use 
C              the Hessians saved in the common block structure,
C
C     On Return
C
C     hp       double precision hp(n),
C              Hessian-vector product,
C
C     goth     logical,
C              see above,
C              
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the 
C              Hessian-vector product. (For example, trying to compute 
C              the square root of a negative number, dividing by zero 
C              or a very small number, etc.) If everything was o.k. you 
C              must set it equal to zero.

      flag = - 1

      end

C     ******************************************************************
C     ******************************************************************

      subroutine endp(n,x,l,u,m,lambda,rho,equatn,linear)

      implicit none

C     SCALAR ARGUMENTS
      integer m,n

C     ARRAY ARGUMENTS
      logical equatn(m),linear(m)
      double precision l(n),lambda(m),rho(m),u(n),x(n)

C     This subroutine can be used to do some extra job after the solver
C     has found the solution,like some extra statistics, or to save the
C     solution in some special format or to draw some graphical
C     representation of the solution. If the information given by the
C     solver is enough for you then leave the body of this subroutine
C     empty.
C     
C     Parameters of the subroutine:
C
C     The paraemters of this subroutine are the same parameters of
C     subroutine inip. But in this subroutine there are not output
C     parameter. All the parameters are input parameters.

      end

C     ******************************************************************
C     ******************************************************************

      subroutine addc(n,l,u,f,ml,meq,alin,acol,aval,annz,b,fknown,mla,
     +annza,ind)

C     SCALAR ARGUMENTS
      integer annz,annza,meq,ml,mla,n
      double precision f,fknown

C     ARRAY ARGUMENTS
      integer alin(annz),acol(annz),ind(ml)
      double precision aval(annz),b(ml),l(n),u(n)

      mla   = ml
      annza = annz

      end

C     ******************************************************************
C     ******************************************************************

      subroutine setlv(n,linvar)

      implicit none

C     SCALAR ARGUMENTS
      integer n

C     ARRAY ARGUMENTS
      logical linvar(n)

C     LOCAL SCALARS
      integer i

      do i = 1,n
          linvar(i) = .false.
      end do

      end

C     ******************************************************************
C     ******************************************************************

      subroutine evalfi(n,y,f,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,n
      double precision f(2)

C     ARRAY ARGUMENTS
      double precision y(2*n)

C     This subroutine computes the intervalar objective function.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     y        double precision y(2*n),
C              bounds on x variables, y(2*i-1) <= x(i) <= y(2*i),
C
C     On Return
C
C     f        double precision,
C              objective function value at x,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the objective 
C              function. (For example, trying to compute the square root 
C              of a negative number, dividing by zero or a very small 
C              number, etc.) If everything was o.k. you must set it 
C              equal to zero.

      flag = 0

C     f = 0.0d0
      call ivl1(0.0d0,f)

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalhi(n,y,hlin,hcol,hval,hnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,n,hnnz

C     ARRAY ARGUMENTS
      integer hcol(*),hlin(*)
      double precision hval(*),y(2*n)

C     This subroutine computes the intervalar Hessian matrix of the 
C     objective function.
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     y        double precision y(2*n),
C              bounds on x variables, y(2*i-1) <= x(i) <= y(2*i),
C
C     On Return
C
C     hnnz     integer,
C              number of perhaps-non-null elements of the computed 
C              Hessian,
C
C     hlin     integer hlin(hnnz),
C              see below,
C
C     hcol     integer hcol(hnnz),
C              see below,
C
C     hval     double precision hval(hnnz),
C              the interval of the non-null value of the (hlin(k),hcol(k)) 
C              position of the Hessian matrix of the objective function must 
C              be saved at hval(P(k)). Just the lower triangular part of 
c              Hessian matrix must be computed,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the Hessian
C              matrix of the objective funtion. (For example, trying 
C              to compute the square root of a negative number, 
C              dividing by zero or a very small number, etc.) If 
C              everything was o.k. you must set it equal to zero.

      flag = 0

      hnnz = 0

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalci(n,y,ind,c,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer ind,flag,n
      double precision c(2)

C     ARRAY ARGUMENTS
      double precision y(2*n)

C     This subroutine computes the interval of the ind-th constraint of 
C     your problem. 
C
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     y        double precision y(2*n),
C              bounds on x variables, y(2*i-1) <= x(i) <= y(2*i),
C
C     ind      integer,
C              index of the constraint to be computed,
C
C     On Return
C
C     c        double precision,
C              interval of the ind-th constraint "evaluated at y",
C
C     flag     integer
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the 
C              constraint. (For example, trying to compute the square 
C              root of a negative number, dividing by zero or a very 
C              small number, etc.) If everything was o.k. you must set 
C              it equal to zero.

C     LOCAL ARRAYS
      double precision tmp(2)

C     EXTERNAL FUNCTION
      integer P

      flag = 0

      if ( ind .eq. 1 ) then
C         c = x(3) ** 2 / ( x(1) * x(2) ** 3 ) - 0.000169d0
          call power(y(P(3)),2,c)
          call power(y(P(2)),3,tmp)
          call mult(y(P(1)),tmp,tmp)
          call idiv(c,tmp,c)
          call scladd(-0.000169d0,c,c)

      else ! if ( ind .eq. 2 ) then
C         c = x(2) / x(1) - 3.0d0
          call idiv(y(P(2)),y(P(1)),c)
          call scladd(-3.0d0,c,c)
      end if

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evaljaci(n,y,ind,indjac,valjac,jcnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,ind,n,jcnnz

C     ARRAY ARGUMENTS
      integer indjac(n)
      double precision y(2*n),valjac(2*n)

C     This subroutine must compute the intervalar gradient of the 
C     constraint ind. 
C
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     y        double precision y(2*n),
C              bounds on x variables, y(2*i-1) <= x(i) <= y(2*i),
C
C     ind      integer,
C              index of the constraint whose intervalar gradient will be 
C              computed,
C
C     On Return
C
C     jcnnz   integer,
C              number of perhaps-non-null elements of the computed 
C              gradient,
C
C     indjac   integer indjac(jcnnz),
C              see below,
C
C     valjac   double precision valjac(jcnnz),
C              the interval of the non-null value of the partial 
C              derivative of the ind-th constraint with respect to the 
C              indjac(k)-th variable must be saved at valjac(P(k)).
C
C     flag     integer
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the 
C              constraint. (For example, trying to compute the square 
C              root of a negative number, dividing by zero or a very 
C              small number, etc.) If everything was o.k. you must set 
C              it equal to zero.

C     LOCAL ARRAYS
      double precision tmp(2)

C     EXTERNAL FUNCTION
      integer P

      flag = 0

      if ( ind .eq. 1 ) then

C         jcnnz = 3

C         indjac(1) = 1
C         valjac(1) = - x(2) ** 3 * x(3) ** 2 / x(1) ** 2
C         call power(y(P(2)),3,valjac(P(1)))
C         call sclmlt(-1.0d0,valjac(P(1)),valjac(P(1)))
C         call power(y(P(3)),2,tmp)
C         call mult(valjac(P(1)),tmp,valjac(P(1)))
C         call power(y(P(1)),2,tmp)
C         call idiv(valjac(P(1)),tmp,valjac(P(1)))

C         indjac(2) = 2
C         valjac(2) = 3.0d0 * x(2) ** 2 * x(3) ** 2 / x(1)
C         call power(y(P(2)),2,valjac(P(2)))
C         call sclmlt(3.0d0,valjac(P(2)),valjac(P(2)))
C         call power(y(P(3)),2,tmp)
C         call mult(valjac(P(2)),tmp,valjac(P(2)))
C         call idiv(valjac(P(2)),y(P(1)),valjac(P(2)))

C         indjac(3) = 3
C         valjac(3) = 2.0d0 * x(2) ** 3 * x(3) / x(1)
C         call power(y(P(2)),3,valjac(P(3)))
C         call sclmlt(2.0d0,valjac(P(3)),valjac(P(3)))
C         call mult(valjac(P(3)),y(P(3)),valjac(P(3)))
C         call idiv(valjac(P(3)),y(P(1)),valjac(P(3)))

          jcnnz = 3

          indjac(1) = 1
C         valjac(1) = - x(3) ** 2 / ( x(1) ** 2 * x(2) ** 3 )
          call power(y(P(3)),2,valjac(P(1)))
          call power(y(P(1)),2,tmp)
          call idiv(valjac(P(1)),tmp,valjac(P(1)))
          call power(y(P(2)),3,tmp)
          call idiv(valjac(P(1)),tmp,valjac(P(1)))
          call sclmlt(-1.0d0,valjac(P(1)),valjac(P(1)))

          indjac(2) = 2
C         valjac(2) = - 3.0d0 * x(3) ** 2 / ( x(1) * x(2) ** 4 ) 
          call power(y(P(3)),2,valjac(P(2)))
          call idiv(valjac(P(2)),y(P(1)),valjac(P(2)))
          call power(y(P(2)),4,tmp)
          call idiv(valjac(P(2)),tmp,valjac(P(2)))
          call sclmlt(-3.0d0,valjac(P(2)),valjac(P(2)))

          indjac(3) = 3
C         valjac(3) = 2.0d0 * x(3) / ( x(1) * x(2) ** 3 )
          call power(y(P(2)),3,tmp)
          call mult(y(P(1)),tmp,tmp)
          call idiv(y(P(3)),tmp,valjac(P(3)))
          call sclmlt(2.0d0,valjac(P(3)),valjac(P(3)))

      else ! if ( ind .eq. 2 ) then

          jcnnz = 2

          indjac(1) = 1
C         valjac(1) = - x(2) / x(1) ** 2
          call sclmlt(-1.0d0,y(P(2)),valjac(P(1)))
          call power(y(P(1)),2,tmp)
          call idiv(valjac(P(1)),tmp,valjac(P(1)))

          indjac(2) = 2
C         valjac(2) = 1.0d0 / x(1)
          call ivl1(1.0d0,valjac(P(2)))
          call idiv(valjac(P(2)),y(P(1)),valjac(P(2)))
      end if

      end

C     ******************************************************************
C     ******************************************************************
 
      subroutine evalhci(n,y,ind,hclin,hccol,hcval,hcnnz,flag)

      implicit none

C     SCALAR ARGUMENTS
      integer flag,ind,n,hcnnz

C     ARRAY ARGUMENTS
      integer hccol(*),hclin(*)
      double precision hcval(*),y(2*n)

C     This subroutine might compute the intervalar Hessian matrix of 
C     the ind-th constraint. 
C     
C     Parameters of the subroutine:
C
C     On Entry:
C
C     n        integer,
C              number of variables,
C
C     y        double precision y(2*n),
C              bounds on x variables, y(2*i-1) <= x(i) <= y(2*i),
C
C     ind      integer,
C              index of the constraint whose Hessian will be computed,
C
C     On Return
C
C     hcnnz    integer,
C              number of perhaps-non-null elements of the computed 
C              Hessian,
C
C     hclin    integer hclin(hcnnz),
C              see below,
C
C     hccol    integer hccol(hcnnz),
C              see below,
C
C     hcval    double precision hcval(hcnnz),
C              the interval of the non-null value of the (hclin(k),hccol(k)) 
C              position of the Hessian matrix of the ind-th constraint must 
C              be saved at hcval(P(k)). Just the lower triangular part of 
C              Hessian matrix must be computed,
C
C     flag     integer,
C              You must set it to any number different of 0 (zero) if 
C              some error ocurred during the evaluation of the Hessian
C              matrix of the ind-th constraint. (For example, trying 
C              to compute the square root of a negative number, 
C              dividing by zero or a very small number, etc.) If 
C              everything was o.k. you must set it equal to zero.

C     LOCAL ARRAYS
      double precision tmp(2)

C     EXTERNAL FUNCTIONS
      integer P

      flag = 0

      if ( ind .eq. 1 ) then

C         hcnnz = 6

C         hclin(1) = 1
C         hccol(1) = 1
C         hcval(1) = 2.0d0 * x(2) ** 3 * x(3) ** 2 / x(1) ** 3
C         call power(y(P(2)),3,hcval(P(1)))
C         call sclmlt(2.0d0,hcval(P(1)),hcval(P(1)))
C         call power(y(P(3)),2,tmp)
C         call mult(hcval(P(1)),tmp,hcval(P(1)))
C         call power(y(P(1)),3,tmp)
C         call idiv(hcval(P(1)),tmp,hcval(P(1)))

C         hclin(2) = 2
C         hccol(2) = 1
C         hcval(2) = - 3.0d0 * x(2) ** 2 * x(3) ** 2 / x(1) ** 2
C         call power(y(P(2)),2,hcval(P(2)))
C         call sclmlt(-3.0d0,hcval(P(2)),hcval(P(2)))
C         call power(y(P(3)),2,tmp)
C         call mult(hcval(P(2)),tmp,hcval(P(2)))
C         call power(y(P(1)),2,tmp)
C         call idiv(hcval(P(2)),tmp,hcval(P(2)))

C         hclin(3) = 3
C         hccol(3) = 1
C         hcval(3) = - 2.0d0 * x(2) ** 3 * x(3) / x(1) ** 2
C         call power(y(P(2)),3,hcval(P(3)))
C         call sclmlt(-2.0d0,hcval(P(3)),hcval(P(3)))
C         call mult(hcval(P(3)),y(P(3)),hcval(P(3)))
C         call power(y(P(1)),2,tmp)
C         call idiv(hcval(P(3)),tmp,hcval(P(3)))

C         hclin(4) = 2
C         hccol(4) = 2
C         hcval(4) = 6.0d0 * x(2) * x(3) ** 2 / x(1)
C         call sclmlt(6.0d0,y(P(2)),hcval(P(4)))
C         call power(y(P(3)),2,tmp)
C         call mult(hcval(P(4)),tmp,hcval(P(4)))
C         call idiv(hcval(P(4)),y(P(1)),hcval(P(4)))

C         hclin(5) = 3
C         hccol(5) = 2
C         hcval(5) = 6.0d0 * x(2) ** 2 * x(3) / x(1)
C         call power(y(P(2)),2,hcval(P(5)))
C         call sclmlt(6.0d0,hcval(P(5)),hcval(P(5)))
C         call mult(hcval(P(5)),y(P(3)),hcval(P(5)))
C         call idiv(hcval(P(5)),y(P(1)),hcval(P(5)))

C         hclin(6) = 3
C         hccol(6) = 3
C         hcval(6) = 2.0d0 * x(2) ** 3 / x(1)
C         call power(y(P(2)),3,hcval(P(6)))
C         call sclmlt(2.0d0,hcval(P(6)),hcval(P(6)))
C         call idiv(hcval(P(6)),y(P(1)),hcval(P(6)))

          hcnnz = 6

          hclin(1) = 1
          hccol(1) = 1
C         hcval(1) = 2.0d0 * x(3) ** 2 / ( x(1) ** 3 * x(2) ** 3 )
          call power(y(P(3)),2,hcval(P(1)))
          call power(y(P(1)),3,tmp)
          call idiv(hcval(P(1)),tmp,hcval(P(1)))
          call power(y(P(2)),3,tmp)
          call idiv(hcval(P(1)),tmp,hcval(P(1)))
          call sclmlt(2.0d0,hcval(P(1)),hcval(P(1)))

          hclin(2) = 2
          hccol(2) = 1
C         hcval(2) = 3.0d0 * x(3) ** 2 / ( x(1) ** 2 * x(2) ** 4 )
          call power(y(P(3)),2,hcval(P(2)))
          call power(y(P(1)),2,tmp)
          call idiv(hcval(P(2)),tmp,hcval(P(2)))
          call power(y(P(2)),4,tmp)
          call idiv(hcval(P(2)),tmp,hcval(P(2)))
          call sclmlt(3.0d0,hcval(P(2)),hcval(P(2)))

          hclin(3) = 3
          hccol(3) = 1
C         hcval(3) = - 2.0d0 * x(3) / ( x(1) ** 2 * x(2) ** 3 )
          call power(y(P(1)),2,tmp)
          call idiv(y(P(3)),tmp,hcval(P(3)))
          call power(y(P(2)),3,tmp)
          call idiv(hcval(P(3)),tmp,hcval(P(3)))
          call sclmlt(-2.0d0,hcval(P(3)),hcval(P(3)))

          hclin(4) = 2
          hccol(4) = 2
C         hcval(4) = 12.0d0 * x(3) ** 2 / ( x(1) * x(2) ** 5 )
          call power(y(P(3)),2,hcval(P(4)))
          call idiv(hcval(P(4)),y(P(1)),hcval(P(4)))
          call power(y(P(2)),5,tmp)
          call idiv(hcval(P(4)),tmp,hcval(P(4)))
          call sclmlt(12.0d0,hcval(P(4)),hcval(P(4)))

          hclin(5) = 3
          hccol(5) = 2
C         hcval(5) = - 6.0d0 * x(3) / ( x(1) * x(2) ** 4 )
          call idiv(y(P(3)),y(P(1)),hcval(P(5)))
          call power(y(P(2)),4,tmp)
          call idiv(hcval(P(5)),tmp,hcval(P(5)))
          call sclmlt(-6.0d0,hcval(P(5)),hcval(P(5)))

          hclin(6) = 3
          hccol(6) = 3
C         hcval(6) = 2.0d0 / ( x(1) * x(2) ** 3 )
          call power(y(P(2)),3,tmp)
          call mult(y(P(1)),tmp,tmp)
          call ivl1(2.0d0,hcval(P(6)))
          call idiv(hcval(P(6)),tmp,hcval(P(6)))

      else ! if ( ind .eq. 2 ) then

          hcnnz = 2

          hclin(1) = 2
          hccol(1) = 1
C         hcval(1) = - 1.0d0 / x(1) ** 2
          call ivl1(-1.0d0,hcval(P(1)))
          call power(y(P(1)),2,tmp)
          call idiv(hcval(P(1)),tmp,hcval(P(1)))

          hclin(2) = 1
          hccol(2) = 1
C         hcval(2) = 2.0d0 * x(2) / x(1) ** 3
          call sclmlt(2.0d0,y(P(2)),hcval(P(2)))
          call power(y(P(1)),3,tmp)
          call idiv(hcval(P(2)),tmp,hcval(P(2)))

      end if
 
      end