*--------------------------------------------------------------
* Date created: November 22, 1996
* Bayreuth University, Mathematics Department
*  
* Hansen Problem.
* Chemical Mixing Problem. 
*--------------------------------------------------------------
*
	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)
*
	real*8 a(18)

*
* Coefficients of the objective function:
*
	a(1)=100.d0
	a(2)=89.73d0
	a(3)=10.27d0
	a(4)=0.00037d0
	a(5)=0.0147d0
	a(6)=0.982d0
	a(7)=0.d0
	a(8)=0.0001d0
	a(9)=0.22d0
	a(10)=0.66d0
	a(11)=0.114d0
	a(12)=0.002d0
	a(13)=0.004d0
	a(14)=0.245d0
	a(15)=0.734d0
	a(16)=0.0147d0
	a(17)=0.0022d0
	a(18)=0.0044d0

* Objective function and its gradient:
*
	objf = 0.d0
	do i=1,n
	  objf = objf + (x(i)-a(i))**2
	end do  
	
	do i=1,n
	  gobj(i) = 2.d0*(x(i)-a(i))
	end do  
		
* Bounds on variables:
*
*	j=1
*	do i=1,n
*	  cb(j) = x(i)
*	  cb(j+1) = 110.d0 - x(i)
*	  j=j+2
*	end do
*
* Constraints. (Inequalities):
*
*
* Jacobian of the inequalities constraints:
*
*
* Constraints. (Equalities):
*
	e(1)=  x(1)*x(9)  - x(2)*x(14) - x(3)*x(4)
	e(2)=  x(1)*x(10) - x(2)*x(15) - x(3)*x(5)
	e(3)=  x(1)*x(11) - x(2)*x(16) - x(3)*x(6)
	e(4)=  x(1)*x(12) - x(2)*x(17) - x(3)*x(7)
	e(5)=  x(1)*x(13) - x(2)*x(18) - x(3)*x(8)
	
	e(6)= x(4) + x(5) + x(6) + x(7) + x(8)      - 1.d0
	e(7)= x(9) + x(10) + x(11) + x(12) + x(13)  - 1.d0
	e(8)= x(14) + x(15) + x(16) + x(17) + x(18) - 1.d0
	
	e(9)=    x(14) - 66.67d0  * x(4)
	e(10)=   x(15) - 50.d0    * x(5)
	e(11)=   x(16) - 0.01d0   * x(6)
	e(12)=   x(17) - 100.d0   * x(7)
	e(13)=   x(18) - 33.33d0  * x(8)

* Jacobian of the equalities constraints:
*
C1
	ge(1)=x(9)
	ge(2)=x(10)
	ge(3)=x(11)
	ge(4)=x(12)
	ge(5)=x(13)
C2
	ge(6)=-x(14)
	ge(7)=-x(15)
	ge(8)=-x(16)
	ge(9)=-x(17)
	ge(10)=-x(18)
C3
	ge(11)=-x(4)
	ge(12)=-x(5)
	ge(13)=-x(6)
	ge(14)=-x(7)
	ge(15)=-x(8)
C4
	ge(16)=-x(3)
	ge(17)=1.d0
	ge(18)=-66.67d0
C5
	ge(19)=-x(3)
	ge(20)=1.d0
	ge(21)=-50.d0
C6
	ge(22)=-x(3)
	ge(23)=1.d0
	ge(24)=-0.015d0
C7
	ge(25)=-x(3)
	ge(26)=1.d0
	ge(27)=-100.d0
C8
	ge(28)=-x(3)
	ge(29)=1.d0
	ge(30)=-33.33d0
C9
	ge(31)=x(1)
	ge(32)=1.d0
C10
	ge(33)=x(1)
	ge(34)=1.d0
C11
	ge(35)=x(1)
	ge(36)=1.d0
C12
	ge(37)=x(1)
	ge(38)=1.d0
C13
	ge(39)=x(1)
	ge(40)=1.d0
C14
	ge(41)=-x(2)
	ge(42)=1.d0
	ge(43)=1.d0
C15
	ge(44)=-x(2)
	ge(45)=1.d0
	ge(46)=1.d0
C16
	ge(47)=-x(2)
	ge(48)=1.d0
	ge(49)=1.d0
C17
	ge(50)=-x(2)
	ge(51)=1.d0
	ge(52)=1.d0
C18
	ge(53)=-x(2)
	ge(54)=1.d0
	ge(55)=1.d0

*
	return
	end