This loop can easily be parallelized, since there are no loopcarried dependencies:
DO 10 I = 1,N A(I) = B(I) + 2 C(I) = A(I) + D(I) 10 CONTINUE
The next loop cannot be parallelized directly. An antidependence
on A exists from the second assignment statement to
the first. Addition of synchronization is needed to parallelize the
loop. (That, in turn, may slow execution.)
DO 10 I = 1,N A(I) = B(I) + 2 C(I) = A(I+1) + D(I) 10 CONTINUE