***************************************************************************** * mckeown problem 2 ( math. prog. 9 (1975) 57-68 ) ***************************************************************************** subroutine getfun( x, n, f, m, ftf, fj, lfj, g, mode) implicit double precision (a-h,o-z) integer n, m, lfj, mode double precision x(n), f(m), ftf, fj(lfj,n), g(n) integer nprob, nprobs, nstart, nstrts common /PROBLM/ nprob, nprobs, nstart, nstrts integer nout common /IOUNIT/ nout logical lf, lj integer na, nb, nc, nd, nt, nh integer i, j double precision rhomin, rhomax double precision ddot double precision hmu, a0, g0, bx common /PARAM1/ hmu, a0(4), g0(4,3), bx(3) save /PARAM1/ double precision rmu, b0, d0 common /PARAM2/ rmu, b0(3,3), d0(4) save /PARAM2/ double precision rlamax, rlamin parameter (rlamin = 0.325172d0, rlamax = 13.7805d0) *======================================================================= if (mode .eq. 0) goto 20 if (mode .eq. -1) goto 10 if (mode .eq. -2) goto 30 na = mode / 1000 nt = mode - na*1000 nb = nt / 100 nh = nt - nb*100 nc = nh / 10 nd = nh - nc*10 lf = (na .ne. 0) .or. (nb .ne. 0) .or. (nd .ne. 0) lj = (nc .ne. 0) .or. (nd .ne. 0) if (lf .and. lj) goto 100 if (lf) goto 100 if (lj) goto 200 *----------------------------------------------------------------------- 10 continue numbr = 1 nprobs = 1 nstrts = 1 n = 3 m = 4 rmu = 1.d0 hmu = .5d0 * rmu rhomax = rmu / (rlamin * rlamin) rhomin = rmu / (rlamax * rlamax) a0(1) = 0.14272d0 a0(2) = -0.184918d0 a0(3) = -0.521869d0 a0(4) = -0.685306d0 d0(1) = 1.75168d0 d0(2) = -1.35195d0 d0(3) = -0.479048d0 d0(4) = -0.3648d0 g0(1,1) = -0.564255d0 g0(1,2) = 0.392417d0 g0(1,3) = -0.404979d0 g0(2,1) = 0.927589d0 g0(2,2) = -0.0735083d0 g0(2,3) = 0.535493d0 g0(3,1) = 0.658799d0 g0(3,2) = -0.636666d0 g0(3,3) = -0.681091d0 g0(4,1) = -0.869487d0 g0(4,2) = 0.586387d0 g0(4,3) = 0.289826d0 b0(1,1) = 2.95137d0 b0(1,2) = 4.87407d0 b0(1,3) = -2.0506d0 b0(2,1) = 4.87407d0 b0(2,2) = 9.39321d0 b0(2,3) = -3.93189d0 b0(3,1) = -2.0506d0 b0(3,2) = -3.93189d0 b0(3,3) = 2.64745d0 if (nout .gt. 0) write( nout, 9999) rmu,rhomin,rhomax,n,m return *----------------------------------------------------------------------- 20 continue x(1) = 0.1d0 x(2) = 0.1d0 x(3) = 0.1d0 return *----------------------------------------------------------------------- 30 continue ftf = 0.796555d0 return *----------------------------------------------------------------------- 100 continue call dcopy( m, a0, 1, f, 1) do 110 i = 1, n call daxpy( m, x(i), g0( 1, i), 1, f, 1) bx(i) = ddot( n, b0( 1, i), 1, x, 1) 110 continue call daxpy( m, (hmu*ddot( n, x, 1, bx, 1)), d0, 1, f, 1) if (nb .ne. 0) ftf = ddot( m, f, 1, f, 1) if (.not. lj) return 200 continue do 210 i = 1, m call dcopy( n, g0( i, 1), m, fj( i, 1), lfj) call daxpy( n, (rmu*d0(i)), bx, 1, fj( i, 1), lfj) 210 continue if (nd .eq. 0) return do 400 j = 1, n g(j) = ddot( m, fj( 1, j), 1, f, 1) 400 continue return 9999 format(/'1',70('=')//, * ' mckeown problem 2 : rmu = ', 1pe9.0/, * ' rhomin =', 1pe11.2, ' rhomax =', 1pe11.2//, * ' number of variables =', i4/, * ' number of functions =', i4// * ' ',70('=')/) end ************************************************************************ ************************************************************************ subroutine dfkdij( k, x, n, hess, lhess, linear) implicit double precision (a-h,o-z) logical linear integer k, n, lhess double precision x(n), hess(lhess,n) integer j double precision rmu, b0, d0 common /PARAM2/ rmu, b0(3,3), d0(4) save /PARAM2/ double precision zero parameter (zero = 0.d0) *======================================================================= linear = .false. do 100 j = 1, n call dcopy( n, zero, 0, hess( 1, j), 1) call daxpy( n, (rmu*d0(k)), b0( 1, j), 1, hess( 1, j), 1) 100 continue return end