program count1
c count facilitates the determination of the execution
c time for specific operations, oepending on the value of ict.
c n is the number of outer loops.
c  For proper evaluation run the program with different compiler 
c      optimizations, e.g., -O1 -O2 etc
c   Note: Needs to be revisited to make certain all operations converge
c
      dimension e(10000),f(10000),g(10000)
      real a,a1,b,b1,c,c1,d
      integer ia,n,j,i
      a = 1.1
      b = 0.001
      b1 = 1.0000001
      c1 = 1.1
      c = 1./1.1
      d = 2.3
      ia = 3
      n = 10000
      do 21 j = 1,10000
         f(j) = 2.0
 21      g(j) = 1.0
      write(*,22)
c 22   format(5x,'value for ict and n',/)
 22   format(5x,'value for ict',/)
c      read(*,*) ict,n
      read(*,*) ict
c
      goto(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15),ict
 1    do 101 i = 1,n
         do 101 j = 1,10000
 101  continue
      goto 99
c
 2    do 102 i = 1,n
         do 102 j = 1,10000
           a = b+a
 102  continue
      goto 99
c
 3    do 103 i = 1,n
         do 103 j = 1,10000
           a = a-b
 103  continue
      goto 99
c
 4    do 104 i = 1,n
         a1 = a
         do 104 j = 1,10000
           a = a*b1
 104  continue
      goto 99
c
 5    do 105 i = 1,n
         a1 = a
         do 105 j = 1,10000
           a = a/b1
 105  continue
      goto 99
c
 6    do 106 i = 1,n
         do 106 j = 1,10000
            a = c+a
            a = 0.5*a
 106  continue
      goto 99
c
 7    do 107 i = 1,n
         do 107 j = 1,10000
           a = (a+d)*c
 107  continue
      goto 99
c
 8    do 108 i = 1,n
         do 108 j = 1,10000
           a = (a*c1+d)*c
 108  continue
      goto 99
c
 9    do 109 i = 1,n
         do 109 j = 1,10000
           a = (a*c1+d)/c1
 109  continue
      goto 99
c
 10   do 110 i = 1,n
         do 110 j = 1,10000
           a = i*c1+j
 110  continue
      goto 99
c
 11   do 111 i = 1,n
         do 111 j = 1,10000
           a = (i*c1+j)**c1
 111  continue
      goto 99
c
 12   do 112 i = 1,n
         do 112 j = 1,10000
           a = sqrt(i*c1+j)
 112  continue
      goto 99
c
 13   do 113 i = 1,n
         do 113 j = 1,10000
           a = sin(i*c1+j)
 113  continue
      goto 99
c
 14   do 114 i = 1,n
         do 114 j = 1,10000
           a = exp(b*(i*c1++j))
 114  continue
      goto 99
c
 15   do 115 i = 1,n
         do 115 j = 1,10000
           e(j) = f(j) + (i*c1+j)*g(j)
 115  continue
c

 99      continue
c
         write(*,35)ict,n
         write(*,*) a
 35      format(5x,' ict=',i7,5x, ' N =',i7)
         stop
         end