GATE 2022 CS Q.61: solution and answer

Official answer: 1.42 to 1.45Source: Official answer key, GATE 2022 CS (IIT Kharagpur)

1. Compute the CPI of $X_1$. Without a BPU, every branch (30% of instructions) stalls for 2 cycles, so:
   $$\text{CPI}(X_1)=1+0.30\times 2=1.6$$

2. Model the branch outcomes on $X_2$. The BPU gives 0 stall cycles for a correctly predicted branch, and the problem states there is "neither any savings nor any additional stalls for wrong predictions" — i.e. a mispredicted branch stalls exactly as it did on $X_1$, for 2 cycles. With 80% prediction accuracy, the expected stall per branch is:
   $$0.8\times 0+0.2\times 2=0.4\text{ cycles}$$

3. Compute the CPI of $X_2$, applying that expected stall only to the 30% branch fraction:
   $$\text{CPI}(X_2)=1+0.30\times 0.4=1.12$$

4. Both processors run at the same clock frequency and execute the same program $P$ (same instruction count $N$), so execution time is proportional to CPI alone:
   $$T=N\times \text{CPI}\times \text{Clock Period}$$
   Hence:
   $$\text{Speedup}=\frac{T(X_1)}{T(X_2)}=\frac{\text{CPI}(X_1)}{\text{CPI}(X_2)}=\frac{1.6}{1.12}$$

5. Evaluate and round to two decimal places:
   $$\frac{1.6}{1.12}=\frac{160}{112}=\frac{10}{7}=1.42857\dots \approx 1.43$$