Project ForgeComputer Organization & Architecture
Pipelining & Instruction Sets
GATE 2014
Hard
Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is
1.6(± 0.05 accepted)
Explanation
Time = InstructionCount × CPI / frequency, so frequency = IC×CPI/Time. With the same instruction count: f2 = IC×(1.2·CPI1)/(0.75·T1) = 1.6×(IC×CPI1/T1) = 1.6×f1 = 1.6 GHz.
Source: GATE CS 2014 (Session 1) (2014)
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