integer function isamax(n,sx,incx)
c
c     finds the index of element having max. absolute value.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),smax
      integer i,incx,ix,n
c
      isamax = 0
      if( n .lt. 1 ) return
      isamax = 1
      if(n.eq.1)return
      if(incx.eq.1)go to 20
c
c        code for increment not equal to 1
c
      ix = 1
      smax = abs(sx(1))
      ix = ix + incx
      do 10 i = 2,n
         if(abs(sx(ix)).le.smax) go to 5
         isamax = i
         smax = abs(sx(ix))
    5    ix = ix + incx
   10 continue
      return
c
c        code for increment equal to 1
c
   20 smax = abs(sx(1))
      do 30 i = 2,n
         if(abs(sx(i)).le.smax) go to 30
         isamax = i
         smax = abs(sx(i))
   30 continue
      return
      end
      real function sasum(n,sx,incx)
c
c     takes the sum of the absolute values.
c     uses unrolled loops for increment equal to one.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),stemp
      integer i,incx,m,mp1,n,nincx
c
      sasum = 0.0e0
      stemp = 0.0e0
      if(n.le.0)return
      if(incx.eq.1)go to 20
c
c        code for increment not equal to 1
c
      nincx = n*incx
      do 10 i = 1,nincx,incx
        stemp = stemp + abs(sx(i))
   10 continue
      sasum = stemp
      return
c
c        code for increment equal to 1
c
c
c        clean-up loop
c
   20 m = mod(n,6)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        stemp = stemp + abs(sx(i))
   30 continue
      if( n .lt. 6 ) go to 60
   40 mp1 = m + 1
      do 50 i = mp1,n,6
        stemp = stemp + abs(sx(i)) + abs(sx(i + 1)) + abs(sx(i + 2))
     *  + abs(sx(i + 3)) + abs(sx(i + 4)) + abs(sx(i + 5))
   50 continue
   60 sasum = stemp
      return
      end
      subroutine saxpy(n,sa,sx,incx,sy,incy)
c
c     constant times a vector plus a vector.
c     uses unrolled loop for increments equal to one.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),sy(1),sa
      integer i,incx,incy,ix,iy,m,mp1,n
c
      if(n.le.0)return
      if (sa .eq. 0.0) return
      if(incx.eq.1.and.incy.eq.1)go to 20
c
c        code for unequal increments or equal increments
c          not equal to 1
c
      ix = 1
      iy = 1
      if(incx.lt.0)ix = (-n+1)*incx + 1
      if(incy.lt.0)iy = (-n+1)*incy + 1
      do 10 i = 1,n
        sy(iy) = sy(iy) + sa*sx(ix)
        ix = ix + incx
        iy = iy + incy
   10 continue
      return
c
c        code for both increments equal to 1
c
c
c        clean-up loop
c
   20 m = mod(n,4)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        sy(i) = sy(i) + sa*sx(i)
   30 continue
      if( n .lt. 4 ) return
   40 mp1 = m + 1
      do 50 i = mp1,n,4
        sy(i) = sy(i) + sa*sx(i)
        sy(i + 1) = sy(i + 1) + sa*sx(i + 1)
        sy(i + 2) = sy(i + 2) + sa*sx(i + 2)
        sy(i + 3) = sy(i + 3) + sa*sx(i + 3)
   50 continue
      return
      end
      subroutine scopy(n,sx,incx,sy,incy)
c
c     copies a vector, x, to a vector, y.
c     uses unrolled loops for increments equal to 1.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),sy(1)
      integer i,incx,incy,ix,iy,m,mp1,n
c
      if(n.le.0)return
      if(incx.eq.1.and.incy.eq.1)go to 20
c
c        code for unequal increments or equal increments
c          not equal to 1
c
      ix = 1
      iy = 1
      if(incx.lt.0)ix = (-n+1)*incx + 1
      if(incy.lt.0)iy = (-n+1)*incy + 1
      do 10 i = 1,n
        sy(iy) = sx(ix)
        ix = ix + incx
        iy = iy + incy
   10 continue
      return
c
c        code for both increments equal to 1
c
c
c        clean-up loop
c
   20 m = mod(n,7)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        sy(i) = sx(i)
   30 continue
      if( n .lt. 7 ) return
   40 mp1 = m + 1
      do 50 i = mp1,n,7
        sy(i) = sx(i)
        sy(i + 1) = sx(i + 1)
        sy(i + 2) = sx(i + 2)
        sy(i + 3) = sx(i + 3)
        sy(i + 4) = sx(i + 4)
        sy(i + 5) = sx(i + 5)
        sy(i + 6) = sx(i + 6)
   50 continue
      return
      end
      real function sdot(n,sx,incx,sy,incy)
c
c     forms the dot product of two vectors.
c     uses unrolled loops for increments equal to one.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),sy(1),stemp
      integer i,incx,incy,ix,iy,m,mp1,n
c
      stemp = 0.0e0
      sdot = 0.0e0
      if(n.le.0)return
      if(incx.eq.1.and.incy.eq.1)go to 20
c
c        code for unequal increments or equal increments
c          not equal to 1
c
      ix = 1
      iy = 1
      if(incx.lt.0)ix = (-n+1)*incx + 1
      if(incy.lt.0)iy = (-n+1)*incy + 1
      do 10 i = 1,n
        stemp = stemp + sx(ix)*sy(iy)
        ix = ix + incx
        iy = iy + incy
   10 continue
      sdot = stemp
      return
c
c        code for both increments equal to 1
c
c
c        clean-up loop
c
   20 m = mod(n,5)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        stemp = stemp + sx(i)*sy(i)
   30 continue
      if( n .lt. 5 ) go to 60
   40 mp1 = m + 1
      do 50 i = mp1,n,5
        stemp = stemp + sx(i)*sy(i) + sx(i + 1)*sy(i + 1) +
     *   sx(i + 2)*sy(i + 2) + sx(i + 3)*sy(i + 3) + sx(i + 4)*sy(i + 4)
   50 continue
   60 sdot = stemp
      return
      end
      real function smach(job)
      integer job
c
c     smach computes machine parameters of floating point
c     arithmetic for use in testing only.  not required by
c     linpack proper.
c
c     if trouble with automatic computation of these quantities,
c     they can be set by direct assignment statements.
c     assume the computer has
c
c        b = base of arithmetic
c        t = number of base  b  digits
c        l = smallest possible exponent
c        u = largest possible exponent
c
c     then
c
c        eps = b**(1-t)
c        tiny = 100.0*b**(-l+t)
c        huge = 0.01*b**(u-t)
c
c     dmach same as smach except t, l, u apply to
c     double precision.
c
c     cmach same as smach except if complex division
c     is done by
c
c        1/(x+i*y) = (x-i*y)/(x**2+y**2)
c
c     then
c
c        tiny = sqrt(tiny)
c        huge = sqrt(huge)
c
c
c     job is 1, 2 or 3 for epsilon, tiny and huge, respectively.
c
c
      real eps,tiny,huge,s
c
      eps = 1.0
   10 eps = eps/2.0
      s = 1.0 + eps
      if (s .gt. 1.0) go to 10
      eps = 2.0*eps
c
      s = 1.0
   20 tiny = s
      s = s/16.0
      if (s*100. .ne. 0.0) go to 20
      tiny = (tiny/eps)*100.0
      huge = 1.0/tiny
c
      if (job .eq. 1) smach = eps
      if (job .eq. 2) smach = tiny
      if (job .eq. 3) smach = huge
      return
      end
      real function snrm2 ( n, sx, incx)
      integer          next
      real   sx(1),  cutlo, cuthi, hitest, sum, xmax, zero, one
      data   zero, one /0.0e0, 1.0e0/
c
c     euclidean norm of the n-vector stored in sx() with storage
c     increment incx .
c     if    n .le. 0 return with result = 0.
c     if n .ge. 1 then incx must be .ge. 1
c
c           c.l.lawson, 1978 jan 08
c
c     four phase method     using two built-in constants that are
c     hopefully applicable to all machines.
c         cutlo = maximum of  sqrt(u/eps)  over all known machines.
c         cuthi = minimum of  sqrt(v)      over all known machines.
c     where
c         eps = smallest no. such that eps + 1. .gt. 1.
c         u   = smallest positive no.   (underflow limit)
c         v   = largest  no.            (overflow  limit)
c
c     brief outline of algorithm..
c
c     phase 1    scans zero components.
c     move to phase 2 when a component is nonzero and .le. cutlo
c     move to phase 3 when a component is .gt. cutlo
c     move to phase 4 when a component is .ge. cuthi/m
c     where m = n for x() real and m = 2*n for complex.
c
c     values for cutlo and cuthi..
c     from the environmental parameters listed in the imsl converter
c     document the limiting values are as follows..
c     cutlo, s.p.   u/eps = 2**(-102) for  honeywell.  close seconds are
c                   univac and dec at 2**(-103)
c                   thus cutlo = 2**(-51) = 4.44089e-16
c     cuthi, s.p.   v = 2**127 for univac, honeywell, and dec.
c                   thus cuthi = 2**(63.5) = 1.30438e19
c     cutlo, d.p.   u/eps = 2**(-67) for honeywell and dec.
c                   thus cutlo = 2**(-33.5) = 8.23181d-11
c     cuthi, d.p.   same as s.p.  cuthi = 1.30438d19
c     data cutlo, cuthi / 8.232d-11,  1.304d19 /
c     data cutlo, cuthi / 4.441e-16,  1.304e19 /
      data cutlo, cuthi / 4.441e-16,  1.304e19 /
c
      if(n .gt. 0) go to 10
         snrm2  = zero
         go to 300
c
   10 assign 30 to next
      sum = zero
      nn = n * incx
c                                                 begin main loop
      i = 1
   20    go to next,(30, 50, 70, 110)
   30 if( abs(sx(i)) .gt. cutlo) go to 85
      assign 50 to next
      xmax = zero
c
c                        phase 1.  sum is zero
c
   50 if( sx(i) .eq. zero) go to 200
      if( abs(sx(i)) .gt. cutlo) go to 85
c
c                                prepare for phase 2.
      assign 70 to next
      go to 105
c
c                                prepare for phase 4.
c
  100 i = j
      assign 110 to next
      sum = (sum / sx(i)) / sx(i)
  105 xmax = abs(sx(i))
      go to 115
c
c                   phase 2.  sum is small.
c                             scale to avoid destructive underflow.
c
   70 if( abs(sx(i)) .gt. cutlo ) go to 75
c
c                     common code for phases 2 and 4.
c                     in phase 4 sum is large.  scale to avoid overflow.
c
  110 if( abs(sx(i)) .le. xmax ) go to 115
         sum = one + sum * (xmax / sx(i))**2
         xmax = abs(sx(i))
         go to 200
c
  115 sum = sum + (sx(i)/xmax)**2
      go to 200
c
c
c                  prepare for phase 3.
c
   75 sum = (sum * xmax) * xmax
c
c
c     for real or d.p. set hitest = cuthi/n
c     for complex      set hitest = cuthi/(2*n)
c
   85 hitest = cuthi/float( n )
c
c                   phase 3.  sum is mid-range.  no scaling.
c
      do 95 j =i,nn,incx
      if(abs(sx(j)) .ge. hitest) go to 100
   95    sum = sum + sx(j)**2
      snrm2 = sqrt( sum )
      go to 300
c
  200 continue
      i = i + incx
      if ( i .le. nn ) go to 20
c
c              end of main loop.
c
c              compute square root and adjust for scaling.
c
      snrm2 = xmax * sqrt(sum)
  300 continue
      return
      end
      subroutine srot (n,sx,incx,sy,incy,c,s)
c
c     applies a plane rotation.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),sy(1),stemp,c,s
      integer i,incx,incy,ix,iy,n
c
      if(n.le.0)return
      if(incx.eq.1.and.incy.eq.1)go to 20
c
c       code for unequal increments or equal increments not equal
c         to 1
c
      ix = 1
      iy = 1
      if(incx.lt.0)ix = (-n+1)*incx + 1
      if(incy.lt.0)iy = (-n+1)*incy + 1
      do 10 i = 1,n
        stemp = c*sx(ix) + s*sy(iy)
        sy(iy) = c*sy(iy) - s*sx(ix)
        sx(ix) = stemp
        ix = ix + incx
        iy = iy + incy
   10 continue
      return
c
c       code for both increments equal to 1
c
   20 do 30 i = 1,n
        stemp = c*sx(i) + s*sy(i)
        sy(i) = c*sy(i) - s*sx(i)
        sx(i) = stemp
   30 continue
      return
      end
      subroutine srotg(sa,sb,c,s)
c
c     construct givens plane rotation.
c     jack dongarra, linpack, 3/11/78.
c
      real sa,sb,c,s,roe,scale,r,z
c
      roe = sb
      if( abs(sa) .gt. abs(sb) ) roe = sa
      scale = abs(sa) + abs(sb)
      if( scale .ne. 0.0 ) go to 10
         c = 1.0
         s = 0.0
         r = 0.0
         go to 20
   10 r = scale*sqrt((sa/scale)**2 + (sb/scale)**2)
      r = sign(1.0,roe)*r
      c = sa/r
      s = sb/r
   20 z = 1.0
      if( abs(sa) .gt. abs(sb) ) z = s
      if( abs(sb) .ge. abs(sa) .and. c .ne. 0.0 ) z = 1.0/c
      sa = r
      sb = z
      return
      end
      subroutine sscal(n,sa,sx,incx)
c
c     scales a vector by a constant.
c     uses unrolled loops for increment equal to 1.
c     jack dongarra, linpack, 3/11/78.
c
      real sa,sx(1)
      integer i,incx,m,mp1,n,nincx
c
      if(n.le.0)return
      if(incx.eq.1)go to 20
c
c        code for increment not equal to 1
c
      nincx = n*incx
      do 10 i = 1,nincx,incx
        sx(i) = sa*sx(i)
   10 continue
      return
c
c        code for increment equal to 1
c
c
c        clean-up loop
c
   20 m = mod(n,5)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        sx(i) = sa*sx(i)
   30 continue
      if( n .lt. 5 ) return
   40 mp1 = m + 1
      do 50 i = mp1,n,5
        sx(i) = sa*sx(i)
        sx(i + 1) = sa*sx(i + 1)
        sx(i + 2) = sa*sx(i + 2)
        sx(i + 3) = sa*sx(i + 3)
        sx(i + 4) = sa*sx(i + 4)
   50 continue
      return
      end
      subroutine sswap (n,sx,incx,sy,incy)
c
c     interchanges two vectors.
c     uses unrolled loops for increments equal to 1.
c     jack dongarra, linpack, 3/11/78.
c
      real sx(1),sy(1),stemp
      integer i,incx,incy,ix,iy,m,mp1,n
c
      if(n.le.0)return
      if(incx.eq.1.and.incy.eq.1)go to 20
c
c       code for unequal increments or equal increments not equal
c         to 1
c
      ix = 1
      iy = 1
      if(incx.lt.0)ix = (-n+1)*incx + 1
      if(incy.lt.0)iy = (-n+1)*incy + 1
      do 10 i = 1,n
        stemp = sx(ix)
        sx(ix) = sy(iy)
        sy(iy) = stemp
        ix = ix + incx
        iy = iy + incy
   10 continue
      return
c
c       code for both increments equal to 1
c
c
c       clean-up loop
c
   20 m = mod(n,3)
      if( m .eq. 0 ) go to 40
      do 30 i = 1,m
        stemp = sx(i)
        sx(i) = sy(i)
        sy(i) = stemp
   30 continue
      if( n .lt. 3 ) return
   40 mp1 = m + 1
      do 50 i = mp1,n,3
        stemp = sx(i)
        sx(i) = sy(i)
        sy(i) = stemp
        stemp = sx(i + 1)
        sx(i + 1) = sy(i + 1)
        sy(i + 1) = stemp
        stemp = sx(i + 2)
        sx(i + 2) = sy(i + 2)
        sy(i + 2) = stemp
   50 continue
      return
      end