Computer Abstractions & Technology

Abstraction

Abstraction in Computer Architecture

• hide lower-level details of computer systems (시스템의 자세한 사항들을 숨김으로)
• in order to facilitate design of sophisticated systems (복잡한 시스템 설계를 쉽게 함)

Instruction Set Architecture(ISA) (=Architecture)

• Interface between hardware and lower-level software
• programmer가 다루는 computer system의 속성(형태)
  • concepture structure(state) and functional behavior (operation)
  • as distinct from the organization of the data flows and controls, the logic design, and the physical implementation
• 명령어 집합 구조 : 하드웨어 최하위 계층 소프트웨어 사이의 인터페이스, 명령어, 레지스터, 메모리 접근, 입출력 등을 포함해서 정확히 작동하는 기계어 프로그램을 작성하기 위해 알아야 하는 모든 정보.

Computer System Inside

Classic 5 components

• Processor = Datapath + Control
• Memory
• Input & Output

Input/Output includes

• User-interface devices
  • display, keyboard, mouse
• Storage devices
  • HDD, ODD(Optical disc drive), Flash
• Network adpapters
  • for communicating with other computers

Inside the Processor(CPU)

• Datapath

  • performs operations on data
  • ALU, registers, internal buses

• Control

  • sequences(controls) the datapath, memory, and bus operations..

• cf.cache memory

  • called on-chip cache or level-1 cache
  • small fast SRAM memory for immediate access to data

CPU Clocking

• Digital hardware operations governed by a constant-rate clock
  

  • Clock period: duration of a clock cycle
    • e.g., 250ps = 0.25ns = 250 x 10^-12s
  • Clock frequecny(rate) : cycles per second (CPU 사이클)
    • e.g., 4.0GHz = 4000MHz = 4.0 x 10^9 Hz

CPU Performance and Its Factors

• CPU execution time

  • = CPU clock cycles x clock cycle time
  • = CPU clock cycles / clock rate(frequency)

• CPU 실행 시간 = 프로그램 CPU 클럭 사이클 수 X 클럭 사이클 시간

• CPU 실행 시간 = 프로그램 CPU 클럭 사이클 수 / 클럭 속도

• 클럭 속도와 클럭 사이클 시간은 역수 관계

• GHz, MHz : 사이클 / 초 (클럭 속도를 나타냄)

• Example:

  • Same instruction sets
  • Computer A : 4GHz, 10 seconds
  • Computer B : ?GHz, 6 seconds
  • B requires 1.2 times as many clock cycles as A.

• Answer:

  • CPU time(A) = CPU clock cycles(A) / clock rate(A)
  • 10 seconds = CPU clock cycles(A) / (4 x 10^9 cycles/sec)
  • CPU clock cycles(A) = 10 sec. X 4 X 10^9 cycles/sec
  • CPU time(B) = CPU clock cycles(B) / clock rate(B)
    • = 1.2 X CPU clock cycles(A) / clock rate(B)
  • 6 seconds = 1.2 X 40 X 10^9 cycles / clock rate(B)
  • clock rate(B) = 1.2 X 40 X 10^9 cycles / 6 seconds = 8GHz

CPI (Clock cycle Per Instruction) : 명령어 당 클럭 사이클 수

• CPU clock cylce 수 = 명령어의 수 X CPI
• CPU Time = 명령어 수 X CPI / 클럭 속도 = 명령어 수 X CPI X 클럭 사이클 시간

Multiprocessors

• Multicore microprocessors
  • More than one processor per chip
• Requires explicitly parallel programming
  • Compare with instruction level parallelism
    • Hardware executes multiple instructions at once
    • Hidden from the programmer
  • Hard to do
    • Programming for performance
    • Load balancing
    • Optimizing communication and synchronization

For Good Performance

• Algorithm
  • Determines number of operations executed
• Programming language, compiler, architecture
  • Determine number of machine instructions executed per operation
• Processor and memory system
  • Determine how fast instructions are executed
• I/O system (including OS)
  • Determines how fast I/O operations are executed

8 Great Ideas for Performance

• Design for Moore's Law
• Use abstraction to simplify design
• Make the common case fast
• Performance via parallelism
• Performance via pipelining
• Performance via prediction
• Hierarchy of memories
• Dependability via redundancy(중복을 통한 신뢰성)
  

..

• MIPS는 명령어 실행속도를 의미한다. (실행시간의 역수로 성능을 표시함)
• MIPS = 명령어 개수 / 실행시간 X 10^6
  • = 명령어 개수 / ((명령어 개수 X CPI X 10^6)/클럭속도) = 클럭 속도 / CPI X 10^6