GATE CS 2026 Set 2 Q.57: Instruction Pipelining and Hazards

A non-pipelined instruction execution unit that operates at 1.6 GHz clock takes an average of 5 clock cycles to complete the execution of an instruction. To improve the performance, the system was pipelined with a goal of achieving an average throughput of one instruction per clock cycle. However, it could operate only at 1.2 GHz due to pipeline overheads. While executing a program in the pipelined design, 30% of instructions encountered a stall of 2 cycles due to pipeline hazards.

The speed-up obtained by the pipelined design over the non-pipelined one for this program is ___________. (rounded off to two decimal places)

Note: $1\text{G}=10^9$

Your answer

Official answer and solution

Accepted answers: 2.3 to 2.4
From: Official GATE 2026 answer key

Solution

Compute average execution time per instruction $T = \frac{\text{CPI}}{f}$ for both designs and take their ratio.

Non-pipelined design:
- $\text{CPI}_{\text{non-pipe}} = 5$ , clock frequency $f_{\text{non-pipe}} = 1.6\text{ GHz}$ .
- Average time per instruction $= \frac{5}{1.6}\text{ ns} = 3.125\text{ ns}$ .
Pipelined design:
- $\text{CPI}_{\text{pipe}} = 1 + 0.30 \times 2 = 1.6$ , clock frequency $f_{\text{pipe}} = 1.2\text{ GHz}$ .
- Average time per instruction $= \frac{1.6}{1.2}\text{ ns} = \frac{4}{3}\text{ ns}$ .
Speed-up:

\text{Speed-up} = \frac{5 / 1.6}{1.6 / 1.2} = \frac{5 \times 1.2}{1.6 \times 1.6} = \frac{6}{2.56} = 2.34375 \approx 2.34

Final answer

2.34

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 57 of 65
Section: Computer Science and Information Technology

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