Computer Architecture Notes

2 分钟读完

This note is taken while reading book: Computer Architecture: A Quantitative Approach, 6th Edition.

Chapter 1: Fundamentals of Quantitative Design and Analysis

Concepts

ISA

  1. Class: GPR, Stack, Accumulator
  2. Memory addressing: byte/others, aligned or not
  3. Addressing modes: imm, base+displacement, base+scaled index + displacement
  4. Types & size of operands: supported FPs and ints
  5. Operations: allowed operations of data transfer, arithmetic, logical, control
  6. Control flow ins: conditional/unconditional branch, procedure call/return
  7. Encoding: compressed or not, etc.

Ideas

  • Embedded processors (especially ARM) compose the biggest part of CPU market.
  • Even a very small portion of downtime can lead to huge financial loss for service provider.
  • Design goals: application area, SW compatibility, OS requirement, standards
  • Processors and compilers can be tuned for benchmarks, thus real-world apps will be the best gauge for performance.
  • Enhancing common cases would achieve better performance.
  • IC: Moore’s law is no longer applicable. $Die size +$, $transistor +$
  • Efforts in making processor faster: ILP (pipeline, frequency) -> DLP/TLP/RLP (SIMD, multicore) -> application-specific
  • Power consumption: $static power ++$, $voltage -$ -> $dynamic energy –$

Formulas

Cost of IC (C for Cost)

\(C = \frac{C_{Die} + C_{packaging\&test}}{Final\ test\ yield}\) \(C_{die} = \frac{C_{Wafer}}{Die\ per\ wafer * Die\ yield}\) \(Dies\ per\ wafer = \frac{\pi * (Wafer\ diameter/2)^2 }{Die\ area} - \frac{\pi * Wafer\ diameter}{\sqrt{2 * Die\ area}}\)

Dependability

Reliability: measure of the continuous service accomplishment from a reference instant. MTTF, FIT, MTTR.
Availability: measure of the service accomplishment w.r.t. the alteration between the two states of accomplishment and interruption.
\(Module availability = \frac{MTTF}{MTTF + MTTR}\)

Performance

\[n = \frac{Exec\ time_Y}{Exec\ time_X} = \frac{Perf_X}{Perf_Y}\]

Amdahl’s law

\(Speedup = \frac{Perf.\ for \ entire\ task\ with\ enhancement}{Perf.\ for \ entire\ task\ without\ enhancement}\)
Thus, $Exec\ time_{new} = Exec\ time_{old} * ((1-Fraction_{enhanced}) + \frac{Fraction_{enhanced}}{Speedup_{enhanced}})$

CPU time

\(CPU\ time = \frac{Instructions}{Program} * \frac{Clock\ cycles}{Instructions} * \frac{Seconds}{Clock cycle} = \frac{Seconds}{Program}\)

  • Clock cycle time: HW tech. & org.
  • CPI: Org. & ISA
  • #Instruction: ISA & compiler

留下评论