Today is the end of the morning summary of the software engineer exam (all from 1 to 4). After the summary today, the following will begin to test the afternoon case configuration. After reading it, remember that you can write more test papers from 2016 to 20, and practice more to see if there are any shortcomings. If you have any shortcomings, please leave a message.
Network planning and design
1. Example: Assuming that the production management network adopts the B/S model, the number of users who often work online is 300, and each user generates 2 transaction processing tasks per minute, and the average transaction size is 0.1MB, then this system What is the required information transmission rate?
The required transmission rate = the number of transactions generated by the number of users per unit of time. The size of the transaction volume. The
required transmission rate = 300 2/30 0.1*8 (conversion unit) = 8MB/s
1MB = 1Mb (bit)
2. Logic design: layered Network design (1. Access layer, low cost and high port density characteristics. User access and authentication 2. Convergence layer, (policy) access control list, VLAN routing selection, load balancing, speed up convergence. Core layer: (high speed) Forwarding) Link aggregation, IP routing configuration)
Computer hardware knowledge
1, CPU: (central processing unit = arithmetic unit + controller, instruction = operation code (function) + address code (address))
2, controller: 1, Program counter PC: points to the address of the next instruction, 2, instruction register IR, 3, address register AR, 4, data register DR, 5, instruction decoder.
3. Arithmetic unit: 1, arithmetic logic unit ALU, 2, general-purpose register, 3, data register, 4, program status sub-register.
Example: If the instruction pipeline divides an instruction into three parts: fetching, analyzing and executing, and The three backup times are 2ns, 2ns, and 1ns respectively. What is the total execution time for all 100 instructions?
Sampling pipeline: the longest time T=2ns, so the total time is (2+1+2) (total time) + (100-1) (number) 2 (the longest time) = 203
No pipeline: (2 +1+2) (total time) + (100-1) (number) 5 (total time) = 507
Speedup ratio: not adopted/adopted=507/203=2.463
memory type
1, RAM (random access memory: internal memory), ROM (read only memory: optical disk), SAM (sequential memory: tape), DAM (direct memory disk), CAM (connected memory: Content search)
Example: If the system memory is addressed by double bytes (each byte is 16 bits), how many double bytes are there in the address from B5000H to DCFFFH? If a memory chip with a storage capacity of 16K 8bit is used to form the memory, how many chips are required at least?
(Double byte 16bit, single byte 8bit, half byte 4bit)
DCFFF+1-B5000=28000 (hexadecimal)
memory capacity=highest byte+1-lowest byte
28000=0010 1000 0000 0000 0000 (binary) =2 17+2 15=2 10*(2 7+2^5)
2 10*(2 7+2 5)/1024 (bytes converted to kilobytes)=2 7+2^5=160KB
because 16K 8bit for storage so
(160 16)/(16 8)=20 (slices)
(kilobytes * double bytes)/single space = number of slices The
main control methods for data transmission between the host and peripherals are program control and interruption.
Black and white box test
1, white box test: the object can be regarded as a box that can be opened and the structure can be seen.
2. Black box testing; it can be regarded as a box that cannot be opened, so the internal operating principle of the software is not considered.
(PERT) plan review technology:
1. Critical path method: This longest activity time route is called the critical path.
Software intellectual property
1. Copyright holders and their powers: publishing rights, authorship rights, modification rights, distribution rights, etc.
2. Power protection period: protect the author's authorship right, modification right, and work integrity right. The protection period is the author's life and 50 years after his death. As of December 31 of the 50th year after death.
3. Limitation of power: not profitable.
4. Infringement: Any act that the infringer intentionally or negligently infringes the personal and property rights of the software protected by the Copyright Law and the Software Protection Regulations constitutes an infringement.