Question

Given a pipeline with 5 stages, the delay for each stage is as follows:

\[ \begin{array}{|c|c|} \hline \textbf{Stage} & \textbf{Delay (ns)} \\ \hline 1 & 250 \\ 2 & 150 \\ 3 & 100 \\ 4 & 200 \\ 5 & 50 \\ \hline \end{array} \]

The buffer delay is 10 ns. Find the time for \( n = 1000 \) instructions.

• A. 261.040 microseconds
• B. 200.050 microseconds
• C. 150.030 microseconds
• D. 100.020 microseconds

Hint:

For pipeline performance, the total time for \( n \) instructions is the product of the total delay per instruction and \( n-1 \), factoring in the delays for each pipeline stage and any buffer delays.

Answer 1

The Correct Option is A

Step 1: Calculate the total delay for each stage.

The total delay for the pipeline is the sum of the delays for all the stages plus the buffer delay.

Total delay per instruction is: \[ \text{Total Delay} = (250 + 150 + 100 + 200 + 50) \, \text{ns} + 10 \, \text{ns (buffer)} \] \[ \text{Total Delay} = 750 \, \text{ns} + 10 \, \text{ns} = 760 \, \text{ns} \]

Step 2: Calculate the time for \( n = 1000 \) instructions.

For \( n \) instructions, the time required is: \[ \text{Time} = (n - 1) \times \text{Total Delay} \] \[ \text{Time} = (1000 - 1) \times 760 \, \text{ns} = 999 \times 760 \, \text{ns} = 759240 \, \text{ns} \] Convert to microseconds: \[ \text{Time} = \frac{759240}{10^3} \, \mu\text{s} = 261.040 \, \mu\text{s} \]

Thus, the correct answer is \( 261.040 \, \mu\text{s} \).