What is the lower bound of the following computation
The attempt in the edited question is almost there.  When you issue the instruction OP, you get the answer back seven cycles later. When you issue an OP instruction, you must wait two more cycles before issuing another. You need to issue $N-1$ OP instructions, so they're issued at times $0, 2, 4, \dots, 2(N-2)$, and the last one completes seven cycles after it's issued, so we get the required answer.



Having said that, the example in the question seems bad. Since OP is an instruction taking two arguments, it's unclear what X[0] OP X[1] OP X[2] even means. If it's supposed to mean S := X[0] OP X[1]; S := S OP X[2]; then there are dependencies between the instructions. The second OP needs to know the value returned by the first, so it can't be issued until the first one has finished.  If the instruction sequence is supposed to mean X[0] OP X[1]; X[1] OP X[2], then it should just be written that way, or even just OP; OP, since the arguments are irrelevant.