冯·诺依曼计算机

冯·诺依曼计算机由存储器、运算器、输出设施、输出设备和控制器五局部组成。

哈佛构造

哈佛构造是一种将程序指令存储和数据存储离开的存储器构造,它的次要特点是将程序和数据存储在不同的存储空间中,即程序存储器和数据存储器是两个独立的存储器,每个存储器独立编址、独立拜访,目标是为了加重程序运行时的访存瓶颈。哈佛架构的中央处理器典型代表ARM9/10及后续ARMv8的处理器,例如:华为鲲鹏920处理器。

组成计算机的根底硬件都须要与主板(Motherboard)连贯

计算机根底硬件 (2)

Opening the Box(Apple IPad2)

手机的内部结构 – 华为Mate30 Pro

主板(来自于 Tech Insights)

主板 反面

射频板

Inside the Processor (CPU)

  • Datapath(数据通路): performs operationson data
  • Control: sequences datapath, memory, ...
  • Register 寄存器
  • Cache memory 缓存

    • Small, fast: SRAM(动态随机拜访存储器)
      memory for immediate access to data

Intel Core i7-5960X

毅力号CPU曝光:250nm工艺、23年旧架构、主频仅233MHz

毅力号搭载的处理器是20多年前技术的产品。处理器型号为PowerPC 750处理器,与1998年苹果出品的iMac G3 电脑同款,PowerPC 750 处理器最高主频速度仅233MHz,且晶体管数量也只有600 万个,但单价仍高达20 万美元(约130万元)。抗辐射、耐凛冽-55~125℃

比照苹果最近推出的M1ARM 架构处理器领有最高主频3.2GHz,晶体管数量达160 亿个。

处理器发展趋势

支流CPU倒退门路

Through the Looking Glass

LCD screen: picture elements (pixels像素)

  • Mirrors content of frame buffer memory帧缓冲存储器

Touchscreen(触摸屏)

PostPC device

  • Supersedes(取代)keyboard and mouse
  • Resistive阻性 and Capacitive容性types

    • Most tablets, smart phones use capacitive
    • Capacitive allows multiple touches simultaneously(多点同时触控)

A Safe Place for Data

Volatile main memory(易失性主存)

  • Loses instructions and data when power off(断电)

Non-volatile secondary memory

  • Magnetic disk(磁盘)
  • Flash memory(闪存)
  • Optical disk (CDROM, DVD) 光盘

Networks 与其余计算机通信

  • Communication(通信), resource sharing(资源共享), nonlocal access(近程拜访)
  • Local area network (LAN): Ethernet,局域网/以太网
  • Wide area network (WAN): the Internet,广域网/互联网
  • Wireless network: WiFi, Bluetooth(蓝牙)

计算机根底硬件 (3)

Abstractions形象

The BIG Picture

  • Abstraction helps us deal with complexity

    • Hide lower-level detail
  • Instruction set architecture (ISA)指令集体系结构

    • The hardware/software (abstraction) interface
  • Application< ---- > binary interface利用二进制接口

    • The ISA plus system software interface
  • Implementation(区别于Architecture)

    • The details underlying the interface

半导体与集成电路

Technology Trends 处理器和存储器制作技术--趋势

Electronics technology continues to evolve

  • Increased capacity and performance
  • Reduced cost

Semiconductor Technology

  • Silicon硅: semiconductor 半导体
  • Add materials to transform properties属性:

    • Conductors
    • Insulators
    • Switch

设施列表

厂商在制作芯片的过程中,从前端工序、到晶圆制作工序,之后再到封装和测试工序,次要用到的设施顺次包含,单晶炉、气相内涵炉、氧化炉、低压化学气相沉积零碎、磁控溅射台、光刻机、刻蚀机、离子注入机、晶片减薄机、晶圆划片机、键合封装设施、测试机、分选机和探针台等

集成电路创造

  • 1952年,英国雷达研究所的科学家达默在一次会议上提出:能够把电子线路中的分立元器件,集中制作在一块半导体晶片上,一小块晶片就是一个残缺电路,这样一来,电子线路的体积就可大大放大,可靠性大幅提高。这就是初期集成电路的构想。
  • 1956年,美国材料科学专家富勒和赖斯创造了半导体生产的扩散工艺,这样就为创造集成电路提供了工艺技术根底。
  • 1958年9月,美国德州仪器公司的青年工程师杰克·基尔比(Jack Kilby),胜利地将包含锗晶体管在内的五个元器件集成在一起,基于锗资料制作了一个叫做相移振荡器的繁难集成电路,并于1959年2月申请了小型化的电子电路(Miniaturized Electronic Circuit)专利(专利号为No.31838743,批准工夫为1964年6月26日),这就是世界上第一块锗集成电路。

  • 2000年,集成电路问世42年当前,人们终于理解到他和他的创造的价值,他被授予了诺贝尔物理学奖。诺贝尔奖评审委员会已经这样评估基尔比:“为古代信息技术奠定了根底”。
  • 1959年7月,美国仙童半导体公司的诺伊斯,钻研出一种利用二氧化硅屏蔽的扩散技术和PN结隔离技术,基于硅平面工艺创造了世界上第一块硅集成电路,并申请了基于硅平面工艺的集成电路发明专利(专利号为No.2981877,批准工夫为1961年4月26日。尽管诺伊斯申请专利在基尔比之后,但批准在前)。
  • 基尔比和诺伊斯简直在同一时间别离创造了集成电路,两人均被认为是集成电路的发明者,而诺伊斯创造的硅集成电路更适于商业化生产,使集成电路从此进入商业规模化生产阶段。

Intel Core i7 Wafer

  • 300mm wafer, 280 chips, 32nm technology
  • Each chip is 20.7 x 10.5 mm

Integrated Circuit Cost

$Cost per die =\frac{\text { Cost per wafer }}{\text { Dies per wafer } \times \text { Yield }}$

$Dies per wafer \approx Wafer area/Die area$

$Yield =\frac{1}{(1+(\text { Defects per area } \times \text { Die area } / 2))^{2}}$

成品率

Defects per area:单位面积缺点

Die area:模具面积

Defining Performance

  • Which airplane has the best performance? 从不同的方面进行考查。

Response Time and Throughput

  • Response time响应工夫

    • How long it takes to do a task(the time between the start and completion of a task)
  • Throughput吞吐量

    • Total work done per unit time
    • e.g., tasks/transactions/… per hour
  • How are response time and throughput affected by

    • Replacing the processor with a faster version? 改善处理器
    • Adding more processors to do separate tasks? 增加更多的处理器
    • Queue ?采纳排队机制,改善吞吐量
  • We’ll focus on response time for now…

Relative Performance

Define Performance = 1/Execution Time

  • “X is n time faster than Y”
  • $Performance _{X} / Performance _{Y}
    = Execution time _{Y} / Execution time _{X}=n $

Example: time taken to run a program

  • 10s on A, 15s on B
  • Execution TimeB / Execution TimeA = 15s / 10s = 1.5
  • So A is 1.5 times faster than B

Measuring Execution Time

  • Elapsed time 消失工夫

    • Total response time, including all aspects

      • Processing, I/O, OS overhead, idle time
    • Determines system performance
  • CPU time(共享时,单独占用CPU工夫)

    • Time spent processing a given job

      • Discounts I/O time, other jobs’ shares
    • Comprises user CPU time and system CPU
      time
    • Different programs are affected differently by
      CPU and system performance

CPU Clocking

Operation of digital hardware governed(掌控) by a constant-rate clock (数字同步电路)

  • Clock period: duration of a clock cycle

    • e.g., 250ps = 0.25ns = 250×10^–12s
  • Clock frequency (rate): cycles per second

    • e.g., 4.0GHz = 4000MHz = 4.0×10^9Hz

CPU Time

CPU Time = CPU Clock Cycles x Clock Cycle Time =$\frac{\text { CPU Clock Cycles }}{\text { Clock Rate }}$

Performance improved by

  • Reducing number of clock cycles
  • Increasing clock rate
  • Hardware designer must often trade off(折中)clock rate against cycle count
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?

$Clock Cycles _{A}= CPU Time _{A} \times Clock Rate _{A}$

=$10 \mathrm{~s} \times 2 \mathrm{GHz}=20 \times 10^{9}$

=$\frac{1.2 \times 20 \times 10^{9}}{6 \mathrm{~s}}=\frac{24 \times 10^{9}}{6 \mathrm{~s}}=4 \mathrm{GHz}$

Instruction Count and CPI

$Clock Cycles = Instruction Count \times Cycles per Instruction$

$CPUTime = Instruction Count \times CPI \times Clock Cycle Time$

$=\frac{\text { Instruction Count } \times \mathrm{CPI}}{\text { Clock Rate }}$

  • 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 指令具备不同CPI

      • Average CPI affected by instruction mix

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?

CPI in More Detail

If different instruction classes take different numbers 每指令类CPI不同,且指令呈现频率不同

$\text { Clock Cycles }=\sum_{\mathrm{i}=1}^{n}\left(\mathrm{CPI}_{\mathrm{i}} \times \operatorname{Instruction~Count~}_{\mathrm{i}}\right)$

Weighted average CPI(均匀CPI)

$\mathrm{CPI}=\frac{\text { Clock Cycles }}{\text { Instruction Count }}=\sum_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{CPI}_{\mathrm{i}} \times \frac{\text { Instruction Count }_{\mathrm{i}}}{\text { Instruction Count }}\right)$

CPI Example

Alternative compiled code sequences using instructions in classes A, B, C (三类指令)

Which code sequence executes the most instructions? sequence2

Which will be faster?

What is the CPI for each sequence?

Performance Summary

$\text { CPU Time }=\frac{\text { Instructions }}{\text { Program }} \times \frac{\text { Clock cycles }}{\text { Instruction }} \times \frac{\text { Seconds }}{\text { Clock cycle }}$

Performance depends on

  • Algorithm: affects IC(指令数), possibly CPI
  • Programming language: affects IC, CPI
  • Compiler: affects IC, CPI
  • Instruction set architecture: affects IC, CPI, Tc

Power Trends

In CMOS IC technology

$\text { Power }=\frac{1}{2} \text { Capacitive load } \times \text { Voltage }^{2} \times \text { Frequency }$

Capacitive load:负载电容。

Reducing Power

Suppose a new CPU has

  • 85% of capacitive load of old CPU
  • 15% voltage and 15% frequency reduction

$\frac{P_{\text {new }}}{P_{\text {old }}}=\frac{C_{\text {old }} \times 0.85 \times\left(V_{\text {old }} \times 0.85\right)^{2} \times F_{\text {old }} \times 0.85}{C_{\text {old }} \times V_{\text {old }}^{2} \times F_{\text {old }}}=0.85^{4}=0.52$

  • The power wall (功率墙)

    • We can’t reduce voltage further 可能低压泄露
    • We can’t remove more heat 可能sleep
  • How else can we improve performance?

Constrained by power, instruction-level parallelism, memory latency(受到功率、指令级并行性、内存提早的制约)

Multiprocessors(多核)
  • Multicore microprocessors

    • More than one processor per chip
  • Requires explicitly parallel programming

    • Compare with instruction level parallelism(e.g.流水线)

      • Hardware executes multiple instructions at once
      • Hidden from the programmer (程序员不可见)
  • Hard to do

    • Programming for performance 编程难度减少
    • Load balancing 负载平衡
    • Optimizing communication and synchronization

A R M提供更多计算外围

多核架构单位芯片面积提供更强算力,更合乎分布式业务的需要

A R M多核高并发劣势,匹配互联网分布式架构

随着多核A R M CPU的性能一直加强,应用领域一直扩大

A R M服务器级别处理器一览

SPEC CPU Benchmark

  • Programs used to measure performance

    • Supposedly typical of actual workload
  • Standard Performance Evaluation Corp (SPEC)

    • Develops benchmarks for CPU, I/O, Web, …
  • SPEC CPU2006

    • Elapsed time to execute a selection of programs

      • Negligible I/O, so focuses on CPU performance
    • Normalize relative to reference machine(参考机器)
    • Summarize as geometric mean of performance ratios

      • CINT2006 (integer) and CFP2006 (floating-point)

$\sqrt[n]{\prod_{\mathrm{i}=1}^{n} \text { Execution time ratio }_{i}}$

CINT2006 for Intel Core i7 920

SPEC Power Benchmark

Power consumption of server at different workload levels

  • Performance: ssj_ops/sec
  • Power: Watts (Joules/sec)

SPECpower_ssj2008 for Xeon X5650

Pitfall(陷阱): Amdahl’s Law

Improving an aspect of a computer and expecting a proportional improvement in overall performance

$T_{\text {improved }}=\frac{T_{\text {affected }}}{\text { improvement factor }}+T_{\text {unaffected }}$

Example: multiply accounts for 80s/100s

Speedup(E)=1/{(1-P)+P/S}

Amdahl's law次要的用`途是指出了在计算机体系结构设计过程中,某个部件的优化对整个构造的优化帮忙是有下限的,这个极限就是当S->时, speedup(E)= 1/(1-P);也从另外一个方面阐明了在体系结构的优化设计过程中,应该筛选对整体有重大影响的部件来进行优化,以失去更好的后果。

Fallacy舛误: Low Power at Idle

Look back at i7 power benchmark

  • At 100% load: 258W
  • At 50% load: 170W (66%)
  • At 10% load: 121W (47%)

Google data center

  • Mostly operates at 10% – 50% load
  • At 100% load less than 1% of the time

Consider designing processors to make power proportional to load

Pitfall: MIPS as a Performance Metric

MIPS: Millions of Instructions Per Second

  • Doesn’t account for 思考

    • Differences in ISAs between computers
    • Differences in complexity between instructions
  • $\begin{aligned}
    \text { MIPS } &=\frac{\text { Instruction count }}{\text { Execution time } \times 10^{6}} \
    &=\frac{\text { Instruction count }}{\frac{\text { Instruction count } \times \mathrm{CPI}}{\text { Clock rate }} \times 10^{6}}=\frac{\text { Clock rate }}{\mathrm{CPI} \times 10^{6}}
    \end{aligned}$
  • CPI varies between programs on a given CPU

Concluding Remarks

Cost/performance is improving

  • Due to underlying technology development

Hierarchical layers of abstraction

  • In both hardware and software

Instruction set architecture

  • The hardware/software interface

Execution time: the best performance measure

Power is a limiting factor

  • Use parallelism to improve performance