GATE CS 2022 Q51 — Pipeline Hazards

To find the speedup of processor $X_2$ over $X_1$, we first calculate the average Cycles Per Instruction ($\text{CPI}$) for both processors.

1. Calculation of $\text{CPI}$ for Processor $X_1$

• Base CPI: $1$
• Branch instruction frequency: $30\% = 0.3$
• Stall cycles per branch: $2$

$$\text{CPI}_1 = \text{Base CPI} + (\text{Branch Frequency} \times \text{Stall Cycles per Branch})$$ $$\text{CPI}_1 = 1 + (0.3 \times 2) = 1 + 0.6 = 1.6$$

2. Calculation of $\text{CPI}$ for Processor $X_2$

Processor $X_2$ uses a Branch Predictor Unit (BPU) with $80\%$ accuracy:

• Correct prediction rate: $80\% = 0.8$ (incurs $0$ stall cycles)
• Incorrect prediction rate: $20\% = 0.2$ (incurs the standard $2$ stall cycles)

The average stall cycles per branch instruction in $X_2$ is: $$\text{Average Stall Cycles} = (0.8 \times 0) + (0.2 \times 2) = 0.4$$

Now, we calculate the average $\text{CPI}$ for $X_2$: $$\text{CPI}_2 = \text{Base CPI} + (\text{Branch Frequency} \times \text{Average Stall Cycles})$$ $$\text{CPI}_2 = 1 + (0.3 \times 0.4) = 1 + 0.12 = 1.12$$

3. Speedup Calculation

Since both processors operate at the same clock frequency ($2\text{ GHz}$) and execute the same program $P$ (same instruction count), the speedup is given by the ratio of their CPIs:

$$\text{Speedup} = \frac{\text{Execution Time of } X_1}{\text{Execution Time of } X_2} = \frac{\text{CPI}_1}{\text{CPI}_2}$$ $$\text{Speedup} = \frac{1.6}{1.12} = \frac{10}{7} \approx 1.42857$$

Rounding off to two decimal places, the speedup obtained is 1.43.