The GATE Grind

GATE 2020 CS – Question 53

Computer Organization and Architecture · Instruction Pipelining and Pipeline Hazards · 2 marks · Numerical answer

Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor. Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions, 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALU instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ______.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 2.15 to 2.18

Explanation

Non-pipelined time per instruction = 5/2.5 GHz = 2 ns. Pipelined CPI = 1 + 0.3×0.05×50 + 0.1×0.5×2 = 1.85, so time = 1.85/2 GHz = 0.925 ns. Speedup = 2/0.925 ≈ 2.16.