What is the lower bound of the following computation I came across this question was I was browsing online

An operation OP has the following characteristics

    OP Latency = 7 clocks
    OP cycles/issue =2


Derive the minimum number of cycles required for the following computation:

 X[0] OP X[1] OP X[2] OP ....... OP X[N-1] 


The answer was written as 7 + 2(n-2). But I was not able to achieve the result. Can someone help me to solve this.

My try: As suggested by D.W. to check the value for n=3,4,5
For n=3, I am getting lower bound as 7+2(2)

For n=4, I am getting lower bound as 7+3(2)

For n=5, I am getting lower bound as 2(7)+4(2)

For n=6, I am getting lower bound as 2(7)+5(2)

Am I going correct. But it deviates from the answer given
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.