CPU Pipeline: How to find average instruction execution time - pipeline

Question

In a CPU with a four (4)-stage pipeline composed of fetch, decode, execute, and write
back, each stage takes 10, 6, 8, and 8 ns, respectively. Which of the following is an
approximate average instruction execution time in nanoseconds (ns) in the CPU? Here, the
number of instructions to be executed is sufficiently large. In addition, the overhead for the
pipelining process is negligible, and the latency impact from all hazards is ignored.
a) 6
b) 8
c) 10
d) 32
Answer is 10ns.But i thought it might be 8ns since execute stage takes 8ns.please explain simply.thanks

Answer 1

Each instruction must go though the four stages. Once the pipeline is full, the flow of instructions in and out is determined by the duration of the longest stage:
Fetch|Decode|Exec|Write|
10ns | 6ns |8ns | 8ns |
-----+------+----+-----+
I7 I6 I5 --> I4 : I3 : I2 : I1 --> out
-----+------+----+-----+
I1..I7 are instructions. I1..I4 are in the pipeline, I5..I7 are
waiting to enter the pipeline.
After 6ns I3 is ready to move from Decode to Exec, but cannot because the stage Exec is still occupied by I2
After 2ns more (8ns total), I1 moves out of Write, I2 moves from Exec to Write, and I3 can finally move from Decode to Exec
I4 is still blocking Fetch, so I5 cannot enter
After 2ns more (10ns total) I4 moves from Fetch to Exec, and I5 can enter.
You see that the pipeline stalls until the longest stage is completed; one instruction enters the pipeline every 10ns. (The Decode stage will be idle 40% percent of the time, and the Exec and Write stages 20% of the time.)

Answer 2

In a pipelined situation, "the rate at which an output is produced" is determined by the slowest stage. It doesn't matter how fast the rest of the pipeline works, things are bound by the rate decoder operates. Therefore we could expect the pipeline to produce an output every 10 ns. "The rate at which an output is produced" can be interpreted as the average execution time. So its 10 ns.