I came across this question was I was browsing onlineAn operation OP has the following characteristics    OP Latency = 7 clocks    OP cycles/issue =2Derive 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=3,4,5For 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
What is the lower bound of the following computation
computer architecture
Your edit changes the question.  $7+2(n-2)$ is a lower bound on the number of cycles to get the answer, but it is not the minimum number of cycles.  For some values of $n$, the minimum number of cycles is strictly larger than $7+2(n-2)$.Example: if $n=3$, your goal is to compute $X[0] OP X[1] OP X[2]$.The optimal sequence is $Y := X[0] OP X[1]; Z := Y OP X[2]$.  This takes 14 cycles.It's not 9 cycles as you wrote: the value of $Y$ is not ready until 7 cycles have finished, and the computation of $Y OP X[2]$ cannot start until the value of $Y$ is ready; once you start the computation of $Y OP X[2]$, it will take another 7 cycles to get the final answer. Due to the dependencies, there is no way to do it in less than 14 cycles.  So, 14 cycles in the minimum number of cycles possible.What about the lower bound $7+2(n-2)$?  For $n=3$, we have $7+2(n-2)=7+2\times 1=9$.  So, the lower bound says 9.  That is a correct bound: the minimum is indeed $\ge 9$.  To say that something is a lower bound means that the true total time is always at least as large as the lower bound.So, $7+2(n-2)=9$ is a correct lower bound on the time taken for $n=3$.  It just isn't the best possible lower bound, but it is a lower bound.For general $n$, something similar happens.  $7+2(n-2)$ is a lower bound.  It's not a very good lower bound, and the actual time taken will usually be larger than $7+2(n-2)$.  But, it certainly won't be less -- and that's all that we mean when we say that $7+2(n-2)$ is a lower bound.How do we know that $7+2(n-2)$ is a lower bound?  Well, obviously, we have to perform at least $n-1$ OP operations.  After issuing each one, we have to wait 2 cycles before issuing the next one.  Also, there is a latency of 7 cycles to wait for the result from the last one.  Thus, we need at least $7+2(n-2)$ cycles in total, as explained in David Richerby's answer.That's why $7+2(n-2)$ is a valid lower bound, even though it isn't the minimum number of cycles.