GATE CS 2017 Set 1 Q50 — Instruction Pipelining

The execution time of a pipeline with $k$ stages for $n$ instructions (with no hazards) is given by:

where $t_{\text{clk}}$ is the clock cycle time, determined by the maximum stage delay plus the buffer overhead:

1. Naive Pipeline (NP)

• Stages ($k_{\text{NP}}$): $5$ (IF, ID, OF, EX, WB)
• Stage delays: $5, 4, 20, 10, 3$ ns
• Buffer delay: $2$ ns
• Clock cycle time ($t_{\text{NP}}$): $$t_{\text{NP}} = \max(5, 4, 20, 10, 3) + 2 = 20 + 2 = 22 \text{ ns}$$
• Total execution time for $20$ instructions ($Time_{\text{NP}}$): $$Time_{\text{NP}} = (5 + 20 - 1) \times 22 = 24 \times 22 = 528 \text{ ns}$$

2. Efficient Pipeline (EP)

• Stages ($k_{\text{EP}}$): $6$ (IF, ID, OF1, OF2, EX, WB)
• Stage delays: $5, 4, 12, 8, 10, 3$ ns
• Buffer delay: $2$ ns
• Clock cycle time ($t_{\text{EP}}$): $$t_{\text{EP}} = \max(5, 4, 12, 8, 10, 3) + 2 = 12 + 2 = 14 \text{ ns}$$
• Total execution time for $20$ instructions ($Time_{\text{EP}}$): $$Time_{\text{EP}} = (6 + 20 - 1) \times 14 = 25 \times 14 = 350 \text{ ns}$$

3. Speedup Calculation

The speedup achieved by EP over NP is:

Rounding to two decimal places, the speedup is 1.51 (or 1.50 depending on the rounding convention used, both are accepted).