Lecture_3

# Lecture_3 - Chapter 1 Computer Abstractions and Technology...

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Chapter 1 Computer Abstractions and Technology

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CPU Clocking Operation of digital hardware governed by a constant-rate clock Clock (cycles) Data transfer and computation Update state Clock period Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250 × 10 –12 s Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0 × 10 9 Hz
CPU Time Performance improved by Reducing number of clock cycles Increasing clock rate Hardware designer must often trade off clock rate against cycle count

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CPU Time Example Computer A: 2GHz clock, 10s CPU time Designing Computer B Aim for 6s CPU time Can do faster clock, but causes 1.2 × clock cycles How fast must Computer B clock be?
Instruction Count and CPI Instruction Count for a program Determined by program, ISA and compiler Average cycles per instruction Determined by CPU hardware If different instructions have different CPI Average CPI affected by instruction mix

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CPI Example Computer A: Cycle Time = 250ps, CPI = 2.0 Computer B: Cycle Time = 500ps, CPI = 1.2 Same ISA Which is faster, and by how much? A is faster… …by this much
CPI in More Detail If different instruction classes take different numbers of cycles Weighted average CPI Relative frequency

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Alternative compiled code sequences using instructions in classes A, B, C Class A B C CPI for class 1 2 3 IC in sequence 1 2 1 2 IC in sequence 2 4 1 1 Sequence 1: IC = 5 Clock Cycles = 2 × 1 + 1
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## This note was uploaded on 03/07/2012 for the course CS 2506 taught by Professor Srinidhivaradarajan during the Spring '12 term at Virginia Tech.

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Lecture_3 - Chapter 1 Computer Abstractions and Technology...

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