GATE CSE 2020 Q53: Speedup of a 5-stage pipeline with memory and branch stalls
- Computer Organization and Architecture
- 2 marks
- NAT
- No negative marking
Question
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 _______.
Try it: numerical answer: type the value
Show answer and explanation
Answer: 2.15 to 2.18 (official GATE 2020 final answer key)
Concept used: Speedup = (non-pipelined time per instruction) / (pipelined CPI × pipelined cycle time).
First step: Non-pipelined: ns. Pipelined CPI , at 0.5 ns per cycle.
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Source: official GATE 2020 CS question paper and final answer key, organised by IIT Delhi. Question paper · Answer key
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