Pipelining and Pipeline Hazards Speedup (Variation 9)

Question

A non-pipelined processor takes 10 ns to execute an instruction. The same processor is designed as a 5-stage pipeline. The stages take 1.2 ns, 1.5 ns, 1.8 ns, 1.4 ns, and 1.1 ns, respectively. The pipeline registers introduce a delay of 0.2 ns.

What is the ideal speedup achieved by the pipelined processor over the non-pipelined processor for a very large number of instructions?

Kiran_Kumar · Answer

[Variation 9 Explanation]

1. **Non-pipelined time ($T_{non-pipe}$):** $10$ ns.
2. **Pipelined cycle time ($T_{pipe}$):** determined by slowest stage + register overhead: $\max(1.2, 1.5, 1.8, 1.4, 1.1) + 0.2 = 1.8 + 0.2 = 2.0$ ns.
3. **Ideal Speedup ($S$):** $T_{non-pipe} / T_{pipe} = 10 / 2.0 = 5.0$.

So the speedup is **5.0**.

Pradyumna_Rao · Answer

[Variation 9 Explanation]

### Alternative Approach / Shortcut Method

We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy.

Let's apply the standard boundary cases:
- Let's check with small values of $N$ (e.g. $N=1, 2, 3$).
- By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed.

This alternative proof validates our selected consensus solution!

Kiran_Kumar · Answer

[Variation 9 Explanation]

### Critical Warnings & Common Student Pitfalls

Many students make simple mistakes when solving this type of problem in the exam pressure:
1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly.
2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations.
3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds.

Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!

Community Explanations (3)