268 views

Is it 64?
Yes. Share your approach pls :)
At clock cycle=8, $I_{6}$'s execute instruction will end and hence at clock cycle=9, $I_{12}$'s Stage 1 will start and $I_{12}$ will complete at clock cycle=12. Now the remaining $I_{13}$ to $I_{16}$ instructions will complete at clock cycle=16. Since, cycle time of pipeline is 4 ns, hence Time Required =

4ns * 16 = 64 ns

After completion of S3 stage of I6, we can know that the next instruction is I12. So the S1 stage of I12 can start only after the S3 of I6.

edited by
thanks
soving a question is important but dont you think its too much lengthy
total instruction executed = 6 (1 to 6) + 5 (12 to 16) = 11

total cycles to execute instruction = 4(for first) + 10(for remaining) + 2 stall cycle(because branch address is known at third stage) = 16 cycle

time taken to execute 16 cycle = 16 x 4 = 64ns