System Engineer Study Notes

Introduction: In the purpose of learning, little by little update notes, and later as a review, there is a need for reference. Given the limited capacity, there may be a problem Comments.

The first chapter basic computer

1.1 operating system features and classification

OS 5 major functions: process (processor) management, storage management, device management, file (information) management, user interface (job management)

The operating system can be divided into network operating system, time-sharing operating system, batch processing operating system, real-time operating systems, distributed operating system

1.2 file directory structure

'/'
D1
D2
W1
W2
f3
W3
f3
  • . "" Stands for the directory yourself, "..." represents the directory's parent directory. . "" For the root directory "..." represent themselves;
  • To determine the file in a directory tree structure in absolute and relative paths only one. Absolute path is constant, but the relative path just as the user changes the working directory constantly changing;
  • The current path is D1, the access to the file f2 is the absolute path / D1 / W2 / f2, the relative path is W2 / f2;
  • The current path is W1, the access to the file f2 absolute path is / D1 / W2 / f2, the relative path ... / W2 / f2;

---- file directory structure directory tree structure, windows, there are two disk formats are FAT32 (FAT16) and NTFS, NTFS disk fragmentation produces less than FAT32, and more secure; more likely to support a single file the maximum partition size supported is 2TB, while in window2000, FAT32 partition size supported is 32GB.

Pipe - "|" pipe symbol, UNIX the ">" symbol indicates output redirection, i.e., the input to the other contents of the file specified file if the file already exists, it is overwritten; and ">>", is then added way wrote the tail of another file.

1.3 virtual memory non-key

Virtual storage technology that is retained in memory or part of a program to place a copy of the entire address space in the external memory (hard disk). Role is: to solve the problem of insufficient capacity of the computer's main memory (in essence, the hardware cost savings).

Virtual storage technology can be divided into:

  1. Single continuous partition (not support multi-channel programming);
  2. A fixed partition (to support multi-channel programming, but prone to fragmentation);
  3. Variable partition (using a mosaic that is compact way to solve fragmentation problems, but more time-consuming);
  4. It can re-partition (to overcome the fragmentation problem fixed partition);
  5. Unsolicited paged (support multiprogramming, virtual storage is not supported);
  6. Tabbed request (i.e., support multiprogramming, virtual storage and support);
  7. Page segments of formula (segmented and binding of pagination).

Note: The theoretical basis of virtual storage management is the principle of locality program

Chapter II Computer Organization

2.1 Data encoding

1. original code - for summation; ✔

2. ---- inverted for subtraction, ✔

3. Complement - for subtraction operations; ✔

4. frameshift (complement of the first inversion) - for the exponent calculation.

Note: positive complement the original code, the same anti-code.
The original code, anti-code symbols, complement: 0 for positive, 1 for negative;
frameshift symbol: indicates a positive number, 0 negative;

Note: all binary storage

2.2 bus and interfaces

Bus is a set of signal lines for interconnecting and transmitting information (command, data and address).
1, with respect to the CPU bus by another chip position or point:
1) an internal bus (Internal Bus); ( Parallel Interface )
bus inside the CPU, data transfer between registers and between the arithmetic logic unit ALU and control of the components used.
2) an external bus (External Bus); ( serial interface )
the CPU and memory RAM (RamdomAccessMemory random access memory, such as a computer memory), ROM (Read Only Memory ROM, such as a hard disk, U disk, CD-ROM), and input and output passage for communication between the device interface.

Note: The bus speed is the biggest factor restricting the overall performance of the computer

2, the bus can be divided according to the functions:

1) the address bus (transmission address information);

2) a data bus (data transfer);

3) a control bus (transmitting various control signals).

3, the system interface is divided into

1) serial interface (transfer time only one): transmission speed is slow, complicated control, less communication connection, suitable for long distance transmission.
2) Parallel Interface (one can transmit multiple): transfer rate, control is simple, needs a lot of connection, suitable for long-distance transmission.

Common Interface Type:

  1. IDE (hard disk and CD-ROM)
  2. PCMCIA (laptop wireless Internet access in wireless network card)
  3. The P1394 (high-speed serial bus, a large number of external devices connected)
  4. USB (no security to speak)

2.3 Storage System

Note:

From fast to slow, the register memory Cache- ---- - secondary memory (hard disk, U disk)

In computer storage systems, Cache access the fastest, because the register is not for the storage system.

I.e. a random access memory mode called RAM, byte or word is addressed (one or more bytes), usually expressed in hexadecimal.

Note:

1T=1TB=1024GB

1G=1GB=1024MB

1M=1MB=1024KB

1K = 1KB = 1024B (bytes)

1B = 8b (one byte consists of 8 bits)

Six-sided pages effective recording disk is 6x2 -2 = 10

2.4 RAID (redundant array of disks) see information

RAID (redundant array of disks) is divided into 8 levels, a blocking technique is mainly used, and cross-technology techniques reunited;

1. RAID0 level ( no redundancy and no parity data block), only concerned for performance, capacity and price instead of the reliability of the application program;

2. RAID1 level ( disk mirroring array ), system files, data, and other important documents mainly used for storage, utilization ratio of 50% , from the backup role , it has no redundancy check

3. RAID2 level (error-correcting Hamming code disk array) for bulk data transfer , data transfer is not conducive to low-volume, less practical. Which has no redundancy check

4, RAID3 RAID4 grade and stage (using the parity disk array), quickly read data, write data due to computation of parity bits, very slowly.

5.RAID5 level (no parity disk array independent check disk) for high-volume and low-volume good read and write performance data. Note: RAID4, RAID5 block have adopted the technology. Which has redundant check

NOTE: RAID5 utilization stage (n-1) / n

6, RAID6 level (data having two separate and independent distributed hard check scheme), a separate data access path, expensive, performance improvement and rebate;

7, RAID7 stage (disk array having optimized high asynchronous I / O speed and high data rate), the highest level disk array, the upgrade of RAID6.

8, RAID10 (high reliability and high performance in combination): a plurality of stages and a stage RAID0 RAID1 composition, a RAID1 stage is redundant, is responsible for data reading and writing level RAID0 array, so the level referred RAID0 + 1. currently most widely used approach.

Chapter Computer Architecture

3.1 computer classification (abbreviated)

3.2 RISC and CISC compare

Reduced instruction set computer RISC (Reduced Instruction the Set Computer)
. 1) less the number of instructions;
2) less addressing mode instruction;
3) instruction length is fixed;
4) only a Load / Store instruction to access the memory;
5) to hardwired logic control based;
6) single-cycle instruction execution;
7) optimizing compiler.
Note: complex instruction set computer CISC (Complex Instruction Set Computer).

CISC RISK
Number of instructions many Select only the most common commands
Instruction complexity complex low
Instruction length Variety Short, fixed
Instruction execution cycle With big change command Mostly in a machine cycle is complete
Instruction format complex simple
Addressing many Minimum
It relates to a main memory access instruction many Very small, it has only two instructions
Decoding method Microprogram control Hardware circuit
System requirements for compilation low high

3.3 pipelined computer

✔N = instruction execution time required to complete the required instruction time + (n - 1) * a single instruction in the longest task time.
For example: If the instruction pipeline into the instruction fetch, analyze and execute three parts, and the time of three parts, respectively fetch 2ns, 2ns analysis, performed 1ns, then the longest subtask is 2ns, and therefore instructions 100 all the execution time required for completion of (2 + 2 + 1) + (100-1) × 2 = 203ns

The main factors that affect the pipeline ✔:

  1. Transfer instructions
  2. Access to shared resources conflicts;
  3. Respond to interrupts.

Chapter IV database system

4.1 ER model design

在数据库设计中被广泛用作数据建模的工具。(工具如:ROSE,PowerDesigner )

1, FIG ER ER model referred to, including the following three elements:
1), entity (type): is represented by a rectangular frame, box marked entity name;
2), Attribute: by ovals, and are connected by a connection with the entity ;
3), the link between the entities: represented by diamond, box marked contact name, and the diamond connected to the connection with the relevant entities, respectively, and indicate the type of contact in the connection.
4) Contact ER diagram attributed to three types:
① one contact (. 1:. 1);
② many relationship (. 1: n-);
③ contact-many (m: n);

4.2 Performance Evaluation

Performance evaluation (primarily from system reliability, fault tolerance, instruction cycle, to evaluate the response time and throughput)

  1. Computer system performance in response speed and throughput operations represented.

Response time: refers to a complete user makes a request to the system response time to complete the task given interval; job throughput: task refers to an amount per unit time to complete the system.

A reliability
1, MTBF = 1 / failure rate ([lambda])
① failure rate ([lambda]): refers to the device or system failure occurs in the expected number of times per unit of time;
② MTBF: from t = 0, the event of a failure the system continues to run expectations of time.
For example: Suppose the same model 1000 computer 1,000 hours under the conditions specified, including 10 fails, then one thousand hours of such computer reliability (1000-10) /1000=0.99, the failure rate is 10 / ( 1000 1000) = 1 10 (-5), as MTBF relationship with the failure rate of 1 / failure rate, the mean time to failure is 10 (5) hours.
2, MTTR = 1 / repair rate ([mu])
repair rate ([mu]) is the average number of times per unit time repairable system.
3, MTBF = mean time between failures + MTTR

4.3 Reliability Calculation

  1. Series system

    Reliability multiplying R 1) = N sub-system reliability = R1xR2x ...... xRn;

    2) the total failure rate failure rate λ = N subsystems = λ1 + λ2 + ...... + λn;

  2. Parallel System

    1) Reliability R = 1- (1-R1) × (1-R2) × (1-R3) ...... × (1-Rn);

    2) failure rate λ = 1 / ((1 / λ) × 1 + (1 / λ) × (1/2) + ...... + (1 / λ) × (1 / n)).

    电路接通
    R
    R
    R
    R
    电路接通

    FIG: Assuming that each member R = 0.9, the failure rate λ = 0.1

    After two series reliability R component = 0.9 × 0.9 = 0.81
    failure rate of the two components in series = 0.1 + 0.1 = [lambda] 0.2
    R l- Total = (1-0.81) × (1-081) = .9639
    [lambda] total = 1 / ((1 / 0.2 ) × 1 + (1 / 0.2) × 1/2)

4.4 Fault Tolerance

提高可靠性的技术可分为避错技术和容错技术

避错:指预防和避免系统在运行中出错;
容错:指系统在其某一组件故障存在的情况下不失效,仍然能够正常工作的特性。注:容错功能一般通过冗余组件设计来实现;计算机系统的容错性通常可以从系统的可靠性、可用性和可测性等方面衡量;
冗余技术就是在容错技术的基础上实现,分为:

  1. 硬件冗余(以检测或屏蔽故障为目的而增加一定硬件设备的方法);
  2. 软件冗余(增加一些在正常运行时所不需要的软件);
  3. 信息冗余( 添加一些信息,以保证运行结果正确性的方法。纠错码就是信息冗余的例子)。
  4. 时间冗余(使用附加一定时间的方法完成系统功能。这些附加的时间主要用在故障检测 、复执或故障屏蔽上)。

4.5 指令周期

1. 时钟频率(主频、时钟脉冲)

反映机器速度的,时钟频率越高,计算机的工作速度就越快。

2. 时钟周期=1/时钟频率

是计算机中最基本、最小的时间单位。在一个时钟周期内,CPU仅完成一个最基本动作;

3. 机器周期=n个时钟周期

即完成一个基本操作如取指令、存储器读/写所需要的时间;一个机器周期由若干个时钟周期组成;

4. 指令周期=n个机器周期

即执行一条指令所需要的时间;

注:
频率的单位是赫兹,简称赫,以符号Hz表示,频率单位还有千赫(KHz)、兆赫(MHz)、吉赫(GHz);1兆赫相当于1000千赫(KHz),也就是10^6赫兹;1s=10的9次方ns。
其中MIPS 的含义是“百万条指令/每秒

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