Computer network training questions & self-made mind map
No one is reviewing Internet cafes just now? ? ? ? ? ? ? ? ! ! ! ! !
Except for the first file, the answers are all written by myself, which may not be accurate
Table of contents
Computer Network Review Questions---Special Training for Multiple Choice Questions
Questions and Answers Calculation Exercises
net plan
Part I Basic Knowledge
l The transport layer protocol of TCP/IP cannot provide connectionless services. ( × )
The TCP/IP protocol model includes a series of network protocols that form the basis of the Internet and is the core protocol of the Internet, among which UDP can provide connectionless services.
l ICMP packets are encapsulated in the data part of IP packets. ( √ )
The purpose of the network layer is the reliable transmission of IP datagrams between any two hosts. ( × )
The purpose of the network layer is to realize the transparent transmission of data between two end systems
l Clients can send and receive emails using the SMTP protocol. ( × )
No, the SMTP protocol is only used to send mail, and the POP3 protocol is used to receive mail.
l TELNET establishes a TCP connection between the client computer and the remote login server. ( √ )
l The firewall system cannot prevent damage to the system from the internal network. ( √ )
l The transport layer uses the process number (PID) to mark the application process of the communication between the hosts. ( × )
PID is the code name of each process, and each process has a unique PID number. It is randomly assigned by the system when the process is running, and does not represent a special process.
l The transport layer uses the port number to mark the application process of the communication between the hosts. ( √ )
l In the TCP protocol, the loss of an acknowledgment packet for a TCP segment does not necessarily cause the sender to retransmit the segment. ( √ )
LAN switches can isolate both collision domains and broadcast domains. ( × )
l The complexity of calculating the optimal path of the distance-vector routing algorithm is greater than that of the link-state routing algorithm. ( √ )
l The UDP protocol performs demultiplexing operations based on the destination port number in the UDP packet. ( √ )
l TCP only supports flow control, not congestion control. ( × )
TCP features: 1. Acknowledgment and receipt 2. Reasonable fragmentation and sorting of data 3 Flow control 4 Congestion control 5 Data verification
l The ARP protocol is used to solve the problem of mapping from IP addresses to MAC addresses of hosts or routers on the same LAN. ( √ )
l Obtain the MAC address of another host in the same LAN through the ARP protocol . ( √ )
l If the confirmation number of the TCP segment received by the sender is seq=1501, it means that the receiver expects to receive the segment with the sequence number 1501. ( × )
The sequence number (seq) is used to identify the data byte stream sent from the TCP sender to the TCP receiver, and it represents the first data byte in this segment.
ack: Indicates what is the next seq of the expected counterpart (receiver)
l According to the scope of coverage, computer networks can be divided into four types: local area network, metropolitan area network, wide area network and the Internet. ( √ )
l The main purpose of network interconnection is to connect multiple small networks to form a large network. ( × )
Network interconnection refers to connecting different networks to form a larger-scale network system and realize data communication, resource sharing and collaborative work between networks.
l The IP layer is the key for TCP/IP to realize network interconnection, but the IP layer does not provide reliability guarantee, so there is no reliability mechanism in the TCP/IP network. ( × )
The IP layer does not provide reliability guarantees , but the TCP protocol provides reliable connections,
l The main difference between IPv4 and IPv6 is the address encoding length. The length of each address in the former is 4 bytes, while the length of each address in the latter is 6 bytes. ( × )
An IPv6 address is 128 bits long, 8 bits per byte, 16 bytes
l TCP/IP can be used for communication between different processes on the same host. ( √ )
l The ICMP protocol is a part of the IP protocol. ( √ )
l The data transmission rate that the channel can support is directly proportional to the bandwidth of the channel. ( × )
l During the working process of the World Wide Web, there is no need to establish a TCP connection, but the HTTP protocol must be used. ( × )
The World Wide Web work process needs to establish a TCP connection
l In a CDMA system, the orthogonality between vector S and vector T is:
l In the CDMA system, it is known that the chip vector of station X is S, and station Y receives a chip vector T. If S·T=0 , then X did not send data to Y (T was sent by another station to Y); if S·T=+1 , then X sent bit 1 to Y; if S·T=-1 , then X sends bit 0 to Y.
l The data rate of 10BASE-T Ethernet is 10Mb/s, and the transmission distance is 100m. Using the CSMA/CD protocol, the contention period is 512 bit time (51.2 μ s ) , and the shortest frame length is 64 bytes (512 bits).
l The data rate of 100BASE -T Ethernet is 100Mb/s, and the transmission distance is 100m. CSMA/CD protocol is adopted in half-duplex, the contention period is 512 bit time (5.12 μ s ) , and the shortest frame length is 64 bytes (512 bits). The CSMA/CD protocol is not used in full duplex . [Note contention period changes]
l 100 0BASE-T Ethernet is also called Gigabit Ethernet or Gigabit Ethernet, with a data rate of 1000Mb/s and a transmission distance of 100m. The CSMA/CD protocol is adopted in half-duplex mode, the contention period is 512 bytes , and the shortest frame length is 64 bytes (512 bits). CSMA/CD protocol is not used in full duplex ,
l 10 Gigabit Ethernet is also called 10 Gigabit Ethernet, with a data rate of 10Gb/s. It only uses optical fiber to transmit media, and does not use CSMA/CD protocol. There are two physical layers: Local Area Network Physical Layer (LAN PHY) and Wide Area Network Physical Layer (WAN PHY)
lIn the TCP/IP protocol suite, TCP works at the transport layer, UDP works at the transport layer, and IP works at the network layer.
[Note that some textbooks refer to the transport layer as the transport layer and the network layer as the Internet layer]
l The URL indicates the location and access method of resources on the Internet . has the URL as
http://news.qq.com/38765/index.html, the access method is http ,
The resource location is news.qq.com/38765/index.html
l The general form of URL is protocol://host:port/path . If the URL is http://news.qq.com/38765/index.html, the protocol is http, the host is news.qq.com, the path is 38765/index.html, and the port is 80 .
[Note: If the port is not given in the URL, the default port of the protocol is used, and the default port of http is 80]
l The URL is http://news.qq.com/38765/index.html, the domain name is news.qq.com , the top-level domain name is com , the second-level domain name is qq , the third-level domain name is news , and the host name is news .
l DHCP uses the client-server mode. The host that needs an IP address broadcasts a discovery message (DHCPDISCOVER) to the DHCP server at startup . DHCP server reply offer message (DHCPOFFER)
l The IP address assigned by the DHCP server to the DHCP client is temporary, and this temporary period is called the lease period .
l Simple Mail Transfer Protocol (SMTP), which transmits the sender's mail from the sender's user agent to the receiver's mail server , but cannot receive mail. The post office protocol POP3 can be used to receive mail.
l The POP3 protocol is used by the recipient user agent to read mail from the mail server . POP3 uses the client server method, the mail server is the POP3 server, and the user agent is the POP3 client .
l The protocol is "horizontal", and the protocol is the rules governing the communication between peer entities .
l The service is "vertical". The service is provided by the lower layer to the upper layer through the interlayer interface. The lower layer entity is the service provider of the upper layer entity , and the upper layer entity is the service user of the lower layer entity .
l RED ( Random Early Detection) does not discard all packets at the end of the queue until the network congestion has already occurred, but randomly discards individual packets with probability p when early symptoms of network congestion are detected to avoid globalization congestion control.
l The advantage of RED is that when the average queue length exceeds the threshold TH min , a small number of packets will be discarded. As a result, the average length of the queue is reduced, thereby avoiding the occurrence of network congestion.
l When using the CSMA/CD protocol, because two stations send data at the same time, a collision will occur , so it is impossible for one station to send and receive at the same time.
l 以太网规定：以太网帧(即MAC帧)首部长为 14 字节，尾部(即FCS)长为 4 字节。以太网最短有效帧长 64 字节，凡是长度小于64字节的帧都是无效帧。以太网帧间最小间隙是 96 比特时间。以太网帧数据字段长度为 46～1500 字节。
l 在TCP/IP协议中，可以推导计算出：MAC帧首部加尾部长为 18字节 。MAC帧全长 64～1518 字节。
[注：为什么MAC帧数据字段最小长为46字节？ 因为64-18=46。为什么MAC帧最大长度为1518字节？ 因为1500+18=1518。]
l 如果上层IP数据报长度小于46字节，则MAC层会在IP数据报后面填充数据(通常填0xFF),使MAC帧数据字段长度达到 46 字节。此时MAC帧长度为 64 字节。
l MAC帧数据字段最大长度称为最大传送单元MTU。MTU的值为 1500 字节。
l TCP/IP标准规定：IP数据报首部固定部分长度 20 字节，可变部分长度0～40字节。IP数据报总长度最大 65535 字节。如果IP数据报分片，则IP数据报总长度指分片的总长度。所有主机和路由器必须有能力处理总长度为 576 字节的IP数据报。
[注： 教材上说576字节是IP数据报最小长度有误。576字节是对处理器缓冲能力提出的要求，早期的处理器能力低，所以定了这个指标]
l 若TCP连接初始化拥塞窗口cwnd=1个MSS,慢开始门限ssthresh=32个MSS,则4次成功发送后， cwnd值为 16 个MSS; 5次成功发送后, cwnd值为 32 个MSS; 6次成功发送后cwnd值为 33 个MSS。
l RIP路由协议是基于距离向量的路由选择协议。RIP的距离也叫 跳数 。每经过一个路由器，跳数就加一。距离等于 16 时相当于不可达。
l RIP协议的特点是：仅和 相邻 路由器交换信息，交换的信息是自己的 路由表 ，按固定时间间隔(通常为30s)交换信息，使用 距离向量 算法。RIP协议的缺点是： 坏消息 传播得慢。
l With regard to WWW documents, a static document refers to a WWW document whose content does not change during browsing. A dynamic document means that the content of the document is dynamically created by the application program when the browser accesses the World Wide Web server. The active document means that when the browser accesses the World Wide Web server, the server returns a copy of the active document program, and the program runs on the browser side to generate the corresponding document.
[Note: To understand the simple browsing process of static documents, dynamic documents and active documents:
The browser requests a static document (usually a static HTML file) from the server, the server sends the document to the browser, and the browser displays the document.
The browser requests a dynamic document from the server , the server runs the application, the application creates the document, the server sends the document to the browser, and the browser displays the document.
The browser requests the active document from the server , the server sends the active document program to the browser, the browser runs the program, the program creates the document, and the browser displays the document. ]
l Wireless local area network (WLAN) is divided into two categories, one with fixed infrastructure and the other without fixed infrastructure. 802.11WLAN is a wireless local area network with fixed infrastructure. A WLAN without a fixed infrastructure is called an ad hoc network, such as a wireless sensor network (WSN).
l The wireless local area network using the 802.11 series protocol is also called Wi-Fi, and its wireless fidelity technology is held by the Wi-Fi Alliance.
l 802.11 WLAN MAC layer adopts CSMA/CA protocol to solve the collision problem. The length of the 802.11 MAC frame header is 30 bytes, the length of the tail (FCS) is 4 bytes, and the data part does not exceed 2312 bytes.
l The minimum construction of 802.11 wireless LAN is BSS , a BSS consists of a base station and several mobile stations. The base station is the access point (AP), and the mobile station can connect to the BSS through the AP. Each AP has an SSID and a channel. In Windows, the network name of the wireless LAN is SSID .
l The standard of wireless personal area network (WPAN) is formulated by IEEE802.15 working group, currently widely used WPAN includes bluetooth system and ZigBee.
l Each IPv6 IP address occupies 16 bytes, that is, 128 bits, and uses colon hexadecimal notation. This notation allows zero compression . For example, the IP address FF08:0:0:0:0:0:0:A8 can be compressed into FF08::A8.
l The IPv6 loopback address is 0:0:0:0:0:0:0:1 , which can be written as ::1 .
l The computer network system of our network laboratory belongs to ( ② )
① Personal Area Network (PAN) ② Local Area Network (LAN) ③ Metropolitan Area Network (MAN) ④ Wide Area Network (WAN)
l The OSI/RM architecture proposed by the International Organization for Standardization (ISO) divides the protocol into ( 3 ) layers.
① 3 ② 5 ③ 7 ④ 9
l If the sending rate is 10Mb/s, the time required to send 576 bits is ( ② ).
① 576μs ② 57.6μs ③ 5.76μs ④ 0.576μs
l The following (①) protocol belongs to the application layer protocol of the TCP/IP architecture.
① FTP ② UDP ③ TCP ④ IP
l The real-time performance of the following exchange methods is the best ( ② ).
① Virtual circuit mode ② Circuit switching mode ③ Message switching mode ④ Packet switching mode.
l The line utilization rate is the highest among the following switching methods ( ④ ).
① Virtual circuit mode ② Circuit switching mode ③ Packet switching mode ④ Packet switching mode
l The medium access control method adopted by the MAC sublayer of the IEEE802.3 standard is ( ① ).
① CSMA/CD CSMA/CA ALOHA HDLC
l Divide the total bandwidth of the channel into several sub-frequency bands (or sub-channels), and each sub-channel transmits one signal. Such a channel multiplexing technique is ( ① ).
① Frequency Division Multiplexing (FDM) ② Time Division Multiplexing (TDM) ③ Wavelength Division Multiplexing (WDM) ④ Code Division Multiplexing (CDM)
l The description of the UDP datagram is incorrect ( ④ ).
① is connectionless ② is unreliable ③ does not provide confirmation ④ is connection-oriented and provides confirmation
l The IP address of a certain computer is 132.121.100.001, and the type of this address is ( ② ) type address.
① A ② B ③ C ④ D
l The subnet mask of a certain network is 255.255.255.224, and the number of hosts that can be connected to this network is ( ③ ).
① 14 ② 16 ③ 30 ④ 32
l Among the following addresses, ( ① ) matches 86.32/12.
① 86.33.224.123 ② 86.79.65.216 ③ 86.58.119.74 ④ 86.206.154
l If the 138.10.0.0 network is to be divided into 6 subnets, the subnet mask should be set to ( ④ ).
① 255.0.0.0 ② 255.255.0.0 ③ 255.255.128.0 ④ 255.255.224.0
l It is known that an IP address is 140.120.84.24/20, then the subnet mask of the network where the IP address is located is ( ③ ).
①255.255.0.0 ② 255.255.255.0 ③ 255.255.240.0 ④ 255.255.255.240
l The forward resolution of domain name service DNS is ( ④ ).
① Convert domain name to physical address ② Convert IP address to physical address
③ Convert IP address to domain name ④ Convert domain name to IP address
l When configuring the Ruijie switch, the following command ( ④ ) assigns the current access port to VLAN 8.
① vlan-menbership static ② vlan database
③ vlan8 ④ switchport access vlan 8
l 两台锐捷路由器R1和R2的连接与IP地址分配如图1-1所示，在R1中配置到达子网30.1.1.0的静态路由的命令是（ ① ）。
图1-1
① R1（config）# ip route 30.1.1.0 255.255.255.0 20.1.1.1
② R1（config）# ip route 30.1.1.0 255.255.255.0 30.1.1.2
③ R1（config）# ip route 30.1.1.1 255.255.255.0 10.1.1.1
④ R1（config）# ip route 30.1.1.2 255.255.255.0 10.1.1.1
l 图1-1中，主机PC1运行的Windows操作系统，本地连接属性配置为（ ① ）。
① IP地址：10.1.1.2 子网掩码: 255.255.255.0 网关: 10.1.1.1
② IP address: 10.1.1.2 Subnet mask: 255.255.255.0 Gateway: 20.1.1.2
③ IP address: 10.1.1.2 Subnet mask: 255.255.0.0 Gateway: 20.1.1.1
④ IP address: 10.1.1.2 Subnet mask: 255.255.0.0 Gateway: 10.1.1.1
Part II Ranking Terms
1. Point out the Chinese nouns corresponding to the following English nouns (see Appendix B of the textbook )
Internet ------ Internet
internet ------ Internet ( or Internet )
ISP ----------- Internet Service Provider
NGI ----------- Next Generation Internet
P2P ----------- peer-to-peer
LAN ----------- Local Area Network
WAN ----------- wide area network
PAN ----------- Personal Area Network
WLAN -----------Wireless Local Area Network
WMAN ------------Wireless Metropolitan Area Network
WPAN ------------ Wireless Personal Area Network
TCP/IP -------- Transmission Control Protocol/Internet Protocol
OSI/RM -------- Open Systems Interconnection Reference Model
SAP ----------- Service Access Point
SDU ----------- Service Data Unit
PDU ----------- Protocol Data Unit
FDM ------- frequency division multiplexing
TDM ------- time division multiplexing
STDM ------- statistical time division multiplexing
WDM -------- Wavelength Division Multiplexing
CDM -------- code division multiplexing
CDMA ------- code division multiple access
PCM -------- pulse code modulation (or pulse code modulation)
SONET ------ synchronous optical network
SDH -------- Synchronous Digital Hierarchy
DSL -------- Digital Subscriber Line
HFC -------- Fiber-optic coaxial hybrid network
MTU ---------- Maximum Transmission Unit
CRC --------- Cyclic Redundancy Check
PPP --------- Point-to-Point Protocol
PPPoE ------- Point-to-Point Protocol over Ethernet
MAC --------- Media Access Control
CSMA/CD ----- Carrier Sense Multipoint Access/Collision Detection
VLAN -------- virtual local area network
ARP ------------ Address Resolution Protocol
ICMP ----------- Internet Control Message Protocol
IGMP ----------- Internet Group Management Protocol
VPN ------------ Virtual Private Network
NAT ------------ Network Address Translation
TTL ------------ time to live
CIDR ----------- Classless Inter-Domain Routing
AS ------------- Autonomous System
IP ------------- Internet Protocol
ARQ -------- Automatic Repeat Request
MSS -------- maximum segment length
ACK -------- Acknowledgment
RTT -------- round trip time
RTO -------- timeout retransmission time
AIMD ------- addition increase multiplication decrease
RED -------- Random Early Detection
DNS ---------- Domain Name System
TLD ---------- top-level domain name
FTP ---------- File Transfer Protocol
TFTP --------- Trivial File Transfer Protocol
TELNET ------- Remote Terminal Protocol
DHCP ----------Dynamic Host Configuration Protocol
WWW ---------- World Wide Web
URL ---------- Uniform Resource Locator
HTTP ---------Hypertext Transfer Protocol
HTML ---------Hypertext Markup Language
SMTP --------- Simple Mail Transfer Protocol
POP ---------- Post Office Protocol
MIME --------- Common Internet Mail Extensions
Wi-Fi ---------Wireless Fidelity
SSID ---------- Service Set Identifier
2. Explain the following nouns
A network ---------- consists of a number of nodes and links connecting these nodes.
Client and server --- refer to two application processes of communication, the client is the requester of the service, and the server is the provider of the service.
Exchange ---------- Dynamically allocate transmission line resources in a certain way.
Bandwidth ---------- The highest data rate per unit time that can pass from one point of the network to another.
Latency ---------- The time it takes for data to travel from one end of the network to the other.
Protocol ---------- Rules, standards or conventions established for exchanging data in a network.
Baseband signal --- the signal from the source (that is, the basic frequency band signal)
Band-pass signal --- the signal after carrier modulation
Signal-to-noise ratio ----- The ratio of the average power of the signal to the average power of the noise.
Chip ------- In CDMA, each bit time is divided into m short intervals, and these m short intervals are called a chip.
Link -------- A section of physical line from a node to an adjacent node.
Bit error rate ------ The ratio of the transmitted erroneous bits to the total number of transmitted bits.
Contention period ------ Ethernet end-to-end round-trip time.
Fast Ethernet -- 100Base-T Ethernet. Can transmit 100Mb/s baseband signal on twisted pair.
Bridge ------------ An intermediate device used by the data link layer to extend an Ethernet.
Router ---------- An intermediate device (with the ability to select a path) used by the network layer to connect two networks.
longest prefix match --- select the route with the longest network prefix from the matching results
IP Multicast --------- Multicast on the Internet.
Local address ------- IP address assigned by this institution and valid only in this institution. Local address is also called private address
Well-known ports --- Ports with values ranging from 0 to 1023. Assigned by IANA to some services .
Reliable delivery --- The delivered data is error-free, not lost, not repeated, and arrives in order.
Socket ----- The IP address is spliced together with a colon and a port number.
Timeout retransmission --- After sending a packet, if the acknowledgment of the packet has not been received when the timeout timer expires, the packet will be resent.
Cumulative acknowledgment --- Sends an acknowledgment for the last packet that arrived in sequence.
Flow control --- Let the sender's sending rate not be too fast, so that the receiver can receive in time.
Congestion control --- Prevent excessive packets from being injected into the network, so that the routers or links of the network are not overloaded.
domain name ----------- the unique hierarchical name of a host or router on the Internet
Domain Name System --------- is a distributed network directory service system that provides mutual conversion services between domain names and IP addresses.
Authoritative domain name server -- the domain name server whose jurisdiction is the zone
A script ------------ is a program that is interpreted and executed by another program.
World Wide Web persistent connection -- The World Wide Web server maintains a client's TCP connection for a period of time so that the client can continue
HTTP request messages and response messages are transmitted on this connection.
Part Three Reflection Questions and Answers
l 网络适配器(又叫网卡)的作用是什么?
答: 网络适配器的重要作用是进行串行传输和并行传输的转换，将数据从计算机发送到局域网或从局域网接收数据上交给计算机。
适配器负责执行操作系统驱动程序的I/O指令，负责实现以太网协议。网络适配器发送或接收数据帧时不使用计算机CPU。当适配器收到有差错的帧时就丢弃。当网络适配器收到正确的帧时，就向计算机发送中断请求，将帧上交给网络层处理。当计算机发送IP数据报时，向下交给适配器，适配器将该IP数据报封装成帧发送到局域网。
[背景知识：网络适配器上装有网络控制器(处理器)]、存储器ROM和RAM，ROM内写有物理地址(即MAC地址)。计算机通过主板上的I/O并行总线与适配器通信，适配器通过电缆(双绞线)以串行方式与局域网通信。参见教材图3-15。]
l PPP协议的主要特点是什么？为什么PPP不使用帧编号？PPP适用于什么网络状况？
答：PPP协议的主要特点是：接收方每收到一个帧，就进行CRC检验。若CRC检验正确就收下该帧，否则就丢弃该帧。其他特点有：封装成帧、透明传输、支持多种网络协议和链路类型。
PPP不使用帧编号的原因是因为PPP不需要实现可靠传输。而编号是为了出错时可以使用编号重传。
PPP使用于线路质量好的网络。如果通信线路质量太差，传输出错就频繁。但PPP没有帧编号和确认机制，可靠传输依赖上层协议保证，这样就使数据传输效率降低。
l 用户在计算机上输入命令ping 127.0.0.1是，这个IP数据报将发送给谁？
答：127.0.0.1是环回地址，用于测试。主机将IP数据报发送给本主机的ICMP进行环回测试。而不是发送到因特网。
[Note: In Windows, you can also use the following command loopback test: ping localhost ]
l What are the similarities and differences between IP addresses 0.0.0.0/24 and 0.0.0.88/24?
Answer: The same point: both can be used as source address, but neither can be used as destination address. The masks are the same.
Differences: 0.0.0.0/24 means the local host of the network, and 0.0.0.88/24 means the 24th host of the network.
l What are the similarities and differences between IP addresses 255.255.255.255/16 and 128.1.255.255/16?
Answer: The same point: both are broadcast addresses, neither can be used as source address, but both can be used as destination address. The masks are the same.
The difference: 255.255.255.255/16 means that it only broadcasts to the local area network where the host is located, and all routers do not forward it. 128.1.255.255/16 means to broadcast to all hosts on the network 128.1.0.0.
l What is the function of the ARP protocol? When host A sends an IP datagram to host B in the local area network, how does ARP
Analytic address?
Answer: The role of ARP is to convert IP addresses into physical addresses, and RAPR converts physical addresses into IP addresses.
When host A sends an IP datagram to host B in the local area network, the ARP process of host A searches for the MAC address corresponding to the IP address of host B in the local ARP cache, and if found, fills it in the MAC frame and completes parse.
If the MAC address corresponding to the IP address of host B cannot be found in the local ARP cache, host A broadcasts and sends an ARP request packet in the local area network (this packet includes A's IP address and MAC address, B's IP address). After host B receives the ARP request packet, it sends an ARP response packet to host A (the packet contains A's IP address and MAC address, B's IP address and MAC address), and host A writes the obtained MAC address into its own ARP Cache, and fill in the MAC frame, complete the parsing process.
l A common feature of IP and UDP is that they are both connectionless. What is the main difference between them?
Answer: IP is a host-to-host communication protocol, while UDP is a process-to-process communication protocol.
l What are the similarities and differences between TCP communication service and UDP communication service?
Answer: The service provided by UDP is connectionless and does not guarantee reliable delivery; the service provided by TCP is connected and reliable delivery is guaranteed.
l Can a socket be connected to two remote sockets at the same time?
Answer: No. A socket can only be connected to one remote socket.
l Computer A initiates the establishment of a TCP connection with B. How does A and B establish a connection (that is, briefly explain the three-way handshake process for establishing a connection)?
Answer: (1) A sends a connection request TCP segment to B ( SYN=1 ; seq=x;). (2) B accepts the request and sends back a connection request response segment to A ( SYN=1;ACK=1 ;seq=y;ack=x+1;). (3) After receiving the response message segment, A sends a confirmation message segment for connection request confirmation to B ( ACK=1; seq=x+1; ack=y+1;).
l When two people meet in daily life, one greets and the other answers, and two handshakes are completed. Why is there a three-way handshake to establish a TCP connection?
Answer: It is to avoid errors caused by invalid connection requests (or in other words: to prevent the TCP server from mistaking delayed requests for new requests and causing errors).
An invalid connection request means that the first connection request sent by the TCP client has not arrived at the TCP server due to delay. The TCP client sends a second request, completes the transfer, and releases the connection. After that, the first connection request arrives at the TCP server (this is an invalid connection request), the TCP server mistakenly thinks it is a new connection, sends a confirmation message, agrees to establish a connection, and the TCP client ignores the TCP server because there is no request Confirmation, but the TCP server thinks that the connection is established, and has been waiting for the arrival of the data from the TCP client, wasting resources in vain.
l Why is there a four-way handshake to release a TCP connection?
Answer: After the TCP client sends a release connection message segment to the TCP server and gets confirmation, only the connection from the TCP client to the TCP server is closed, and the connection from the TCP server to the TCP client still exists, so the TCP server still needs to send a release connection message segment closes the connection. So a four-way handshake is required.
l If the acknowledgment of a certain TCP segment is lost, why not retransmit the segment?
Answer: Since the TCP protocol adopts a cumulative confirmation mechanism for byte streams; when the confirmation of a certain segment is lost, as long as the receiver still has data to send to the sender, the confirmation field will contain the information received earlier. The response information of the segment, so the sender does not need to resend the segment.
For example, suppose the message segment 3, 4, and 5 are in the sending port, and the sequence number 3 message segment is sent, but the confirmation of the sequence number 3 is not received, instead of resending the sequence number 3 message segment, but continuing to send the sequence number 4, 5 message segment . When the confirmation number is 6, it indicates that the message segments 3, 4, and 5 have been reliably received by the receiver.
l What are the main functions of the domain name system? Suppose the IP address of www.ccsu.cn is 202.101.208.10, and the default DNS address is 202.101.208.3. Briefly describe the domain name resolution process of DNS in the process of using a browser to access www.ccsu.cn.
Answer: The main function of the domain name system is to resolve domain names into corresponding IP addresses. The domain name resolution process of visiting www.ccsu.cn:
(1) Enter the address www.ccsu.cn in the browser address bar ;
(2) First search for the domain name in the domain name resolution cache of the main host, and if found, obtain the corresponding IP address immediately;
(3) If it is not in the local host cache, send a DNS request message to the local DNS server 202.101.208.3, requesting to resolve the domain name.
(4) After DNS receives the request, it searches its own cache and its mapping table. If it finds it, it sends a response message to the host that sent the request. If it does not find it, it makes a request to the upper-level DNS server until the resolution is successful or an error message is returned.
Part IV Computational Applications
[Calculation about CDMA]
l There are four CDMA communication stations A, B, C, and D communicating with each other. It is known that the chip vectors of A, B, C, and D are (-1-1-1+1+1-1+1+1 ),(-1-1+1-1+1+1+1-1),(-1+1-1+1+1+1-1-1), (-1+1-1-1- 1-1+1-1). Suppose C receives a chip sequence (-1+1-3+1-1-3+1+1). Question: Which station sent data, how much is the data?
Answer: T=(-1+1-3+1-1-3+1+1);
SA·T=(-1*(-1)+1*(-1)-3*(-1)+1*1-1*1-3*(-1)+1*1+1*1) /8=1;
∴ Station A sends bit 1 to C
SB·T=-1;
∴ Station B sends bit 0 to C
SD·T=1;
∴ Station D sends bit 1 to C
[Calculation related to Shannon's formula]
l It is known that the bandwidth of a certain channel is 2400Hz, and the maximum information transmission rate is 24kb/s. Find the signal-to-noise ratio S/N (expressed in decibels) of the channel.
Answer: It is known that C=24000b/s, W=2400Hz,
From Shannon's formula C=Wlog2 (1+S/N) to get S/N=2C/W-1
So S/N=224000/2400-1=1023
Converted to decibels: 10log10(S/N)=10log10(1023)=30dB
[Calculation about CRC]
l The FCS of a certain network adopts CRC, and its CRC generating polynomial is P(x)=x4+x+1. The data to be sent by a host on the network is 1101011011. Question: ① Find the FCS of this data. ②If the data frame received by the receiving end is 11010110101110, will the receiving end discard the frame?
answer:
- ∵P(x)=x4+x+1 ∴P=10011, n=4
Shift 1101011011 to the left by 4 bits to get 11010110110000
Divide 11010110110000 modulo 2 by 10011, the remainder=1110
∴FCS=1110
② Divide 11010110101110 modulo 2 by 10011 to get a remainder of 011. If the remainder is not 0, discard it.
[Calculation about MAC frames]
l The IEEE 802.3 protocol LAN transmits 11-byte data, and the data is packed into a MAC frame format. May I ask: ① What is the length of the frame? ② How many bytes is the data field length of the frame? ③ How many bytes need to be filled?
answer:
① ∵ 11 bytes is less than the minimum effective frame length of the MAC frame (64 bytes),
∴ The frame length is 64 bytes
② The data field length of the frame is 11 bytes
∵ The frame header length occupies 18 bytes
∴ Padding byte length = 64-18-11=35.
[Calculation concerning IP address]
l The subnet mask of a class C network is 255.255.255.240, how many hosts can be connected to each subnet?
answer:
① The binary number corresponding to 240 is: 11110000;
② There are 4 consecutive 0s behind the subnet mask, so use 4 bits as the host code (0~15);
③ 0 and 15 cannot be assigned to the host; (because the host number is all 0 to represent the network, and 15 is the all 1 address, that is, the broadcast address)
④ Each subnet can connect up to 14 hosts.
l A Class C network 192.168.118.0, try to divide it evenly into 4 subnets, and calculate the effective host IP address range and corresponding network mask of each subnet.
Answer: Because the subnet mask of a Class C network is 255.255.255.0, it can be divided into 8 digits. 4 subnets occupy 2 digits. The mask is 11111111.11111111.11111111.11000000, which is 255.255.255.192
So the last 8 bits can be allocated like this:
xxx 00 000000 is assigned to subnet 1, the host number range is 00 00 0001-00 11 1110
xxx 01 000000 is assigned to subnet 2, the host number range is 01 00 0001-01 11 1110
xxx 10 000000 is assigned to subnet 3, the host number range is 10 00 0001-10 11 1110
xxx 11 000000 is assigned to subnet 4, the host number range is 11 00 0001-11 11 1110
The effective IP address range of subnet 1 is: 192.168.118.1～192.168.118.63
The effective IP address range of subnet 2 is: 192.168.118.65 ~ 192.168.118.126
The effective IP address range of subnet 3 is: 192.168.118.129 ~ 192.168.118.190
The effective IP address range of subnet 4 is: 192.168.118.193 ~ 192.168.118.254
The netmask for the 4 subnets is: 255.255.255.192
l What is the aggregate address block of the following 4 /24 address blocks?
212.56.132.0/24
212.56.133.0/24
212.56.134.0/24
212.56.135.0/24
Answer: [Aggregation method: find out the common prefix bits of these addresses]
Write the above address in binary:
11010100.00111000.10000100.00000000
11010100.00111000.10000101.00000000
11010100.00111000.10000110.00000000
11010100.00111000.10000111.00000000
Find the common prefix, there are 22 bits (the blue font part below)
11010100.00111000.10000100.00000000
11010100.00111000.10000101.00000000
11010100.00111000.10000110.00000000
11010100.00111000.10000111.00000000
These 22 bits are the aggregate address block 11010100.00111000.100001 00.00000000/22
Written in dotted decimal is 212.56.132.0/22
l Table 4-1 lists the routing table entries of router R. At time T, R receives an IP datagram D with a destination address of 176.192.71.132. According to the principle of longest prefix matching, which next-hop interface should datagram D be forwarded to?
Table 4-1 R routing table
| The router receives the network prefix |
Next hop |
| 176.17.164.0/30 |
Interface E0 |
| 176.192.71.128/26 |
Interface E1 |
| 176.192.71.128/28 |
Interface E2 |
| 176.192.0.0/16 |
Interface E3 |
A: Check that the address of datagram D matches the network prefix of the routing table entry
176.17.164.0/30 has a mask of 255.255.255.252 _ _
176.192.71.132 and 255.255.255.252 get 176.192.71.132 , which cannot match 176.17.164.0/30
176.192.71.128/26 has a mask of 255.255.255.192 _ _
176.192.71.132 and 255.255.255.192 get 176.192.71.128 , which can match 176.192.71.128/26
176.192.71.128/28 has a mask of 176.192.71.128 _ _
176.192.71.132 and 255.255.255.240 get 176.192.71.128 , which can match 176.192.71.128/28
176.192.0.0/16 has a mask of 255.255.0.0 _ _ _
176.192.71.132 and 255.255.0.0 get 176.192.0.0 , which can match 176.192.0.0/16
Among the above three matching network prefixes, the network prefix of 176.192.71.128/28 is the longest. Therefore, the corresponding next-hop interface should be selected for forwarding, and the forwarding interface is E2.
l The routing table of router A in the network is shown in Table 4.2, and the routing information received by Router B from router B is shown in Table 4.3. Please fill in the updated routing table of A in Table 4.4 according to the RIP protocol.
Table 4.2 Routing Table of Router A
| destination network |
next hop router |
distance |
| 192.168.1.0 |
B |
2 |
| 192.168.2.0. |
C |
2 |
| 192.168.3.0 |
F |
4 |
| 192.168.4.0 |
G |
5 |
Table 4.3 Routing Table of Router B
| destination network |
next hop router |
distance |
| 192.168.1.0 |
---- |
2 |
| 192.168.3.0 |
---- |
1 |
| 192.168.4.0 |
---- |
5 |
| 192.168.5.0 |
---- |
3 |
Table 4.4 Routing Table of Router B
| destination network |
next hop router |
distance |
Answer: In the first step, A modifies the received routing table (mantra: all next hops are changed to the sender, and all distances are increased by 1)
As shown in Table 4.5
Table 4.5 Router A's updated routing table
| destination network |
next hop router |
distance |
| 192.168.1.0 |
B |
3 |
| 192.168.3.0 |
B |
2 |
| 192.168.4.0 |
B |
6 |
| 192.168.5.0 |
B |
4 |
In the second step, A updates its own routing table (mantra: new network, add; same next hop, update; different next hop distance is shorter, update). As shown in Table 4.6
Table 4.6 Router A's updated routing table
| destination network |
next hop router |
distance |
|
| 192.168.1.0 |
B |
3 |
Same next hop, updated |
| 192.168.2.0. |
C |
2 |
----- |
| 192.168.3.0 |
B |
2 |
Different next hop, shorter distance, updated |
| 192.168.4.0 |
G |
5 |
Different next hops, larger distances, unchanged |
| 192.168.5.0 |
B |
4 |
new network, add |
Part V Experiment
To master the configuration of the router, read the experiment guide by yourself, and draw up the topic yourself:
Experiment 1 Static Routing
Experiment 4 RIP Routing
Experiment 5 OSPF Routing
==================================================== ==================================================== = SECOND FILE ======================================
Computer Network Review Questions---Special Training for Multiple Choice Questions
1. The English abbreviation "WWW" means ( B )
A) the Internet
b. World Wide Web
c. Wide area network
D. Metropolitan area network
2. Which of the following protocols belongs to the application layer protocol of the TCP/IP architecture ( A )
A. FTP
B. UDP
C. ARP
D. IP
3. Which of the following protocols is a LAN protocol? ( A )
A. CSMA / CD
B. PPP
C. ATM
D. Frame relay
4. Which of the following does not belong to the classification of computer networks according to the scope of function ( A )
A. Public Network (PAN)
B.Local Area Network (LAN)
C. Metropolitan Area Network (MAN)
D. Wide Area Network (WAN)
5. How many digits are there in the network prefix corresponding to 255.255.255.0 ( C )
A. 8
B. 16
C. 24
D. 32
6. Releasing the TCP connection requires 4 handshakes. How many handshakes must be done to establish a TCP connection? (C)
A. 1 time
B. 2 times
C. 3 times
D. 4 times
7. The ARP protocol is used for ( A )
A. Interpret an IP address as a MAC address
B. Interpret MAC address as IP address
C. Interpret the domain name as a MAC address
D. Interpret IP addresses as host domain names
8. The data units transmitted by the network layer, data link layer and physical layer are respectively ( D )
A. Messages, frames, packets
B. 比特、报文、帧
C. 报文、比特、帧
D. 分组、帧、比特
9.统一资源定位符 URL 的作用是 ( C ???? )
A. 用于万维网浏览器程序和服务器程序的信息交互.
B. 惟一地确定万维网上文档的地址
C. 便于多媒体文件的上传与下载
D. 上述三种都不准确
10. 拥塞控制的四种算法是 （ A ）
A. 慢开始、拥塞避免、快重传、快恢复
B. 快开始、拥塞避免、慢重传、快恢复
C. 快开始、拥塞避免、快重传、慢恢复
D. 慢开始、拥塞避免、慢重传、慢恢复
11.CSMA/CD协议中“碰撞检测”的含义是 ( B )
A. 发送前先检测
B．边发送边监听
C. 发送完再监听
D. 不发送也监听
12.滑动窗口协议主要用于进行（ C ）
A 差错控制
B 安全控制
C 流量控制
D 拥塞控制
13.以下哪个是MAC地址（ B ）
A OD-01-02-AA
B 00-01-22-OA-AD-01
C AO.O1.00
D 139.216.000.012.002
14.IP地址126.1.2.129/13的子网掩码可写为 C
A 255.224.0.0
B 255.240.0.0
C 255.248.0.0
D 255.255.0.0
15. The domain name resolution process usually uses two query algorithms ( D )
A. Spanning tree and shortest path
B. Carrier Sense and Collision Detection
C. Message Digest and Entity Authentication
D. Iteration and recursion
16. Which of the following agreements is not an e-mail agreement (A)
A、CMIP
B、IMAP
C、POP3
D、SMTP
17. Each end of the communication channel can be the sending end or the receiving end. Information can be transmitted from this end to the other end, or from that end to this end, but at the same time, there can only be one transmission of information. The way is ( B )
A simplex communication
B half duplex
C full duplex
D simulation
18. The following IPs belong to class D addresses ( C )
A 10.10.5.168 B 168.10.0.1
C 224.0.0.2 D 202.117.130.80
19. An IPv6 address is composed of a set of binary numbers, and its size is ( A )
A. 128
B. 64
C. 32
D. 256
20. When configuring a Ruijie switch, which of the following commands assigns the current access port to VLAN 8? (D)
A. vlan-menbership static
B. vlan database
C. vlan8
D. switchport access vlan 8
21. The network has had a positive impact on many aspects of social economy and life, "three networks" refers to (D)
A. Bus network, ring network, star network
B. IP network, NOVELL network, ETHERNET network
C. Telecommunications network, education network, Netcom network
D. Telecommunication networks, cable TV networks and computer networks
22. Which of the following is not part of time delay (A)
A. Delay-bandwidth product
B. Sending delay
C. Propagation delay
D. Processing Latency
23. The address length of the next generation Internet protocol IPv6 is ( C )
A．32 bit
B．64 bit
C．128 bit
D. 256 bit
24. The subnet mask of the IP address block 192.168.15.136/29 can be written as ( D )
A、255.255.255.192
B、255.255.255.224
C、255.255.255.240
D、255.255.255.248
25. In terms of network coverage, computer networks can be divided into local area networks, wide area networks and (B)
A physical network
B metropolitan area network
C packet switched network
D high speed network
26. In the following process of interaction between the DHCP server and the DHCP client, the error is (A)
A. The source IP address used by the DHCP client to broadcast the "DHCP Discovery" message is 127.0.0.1
B. After the DHCP server receives the "DHCP discovery" message, it broadcasts the "DHCP supply" message to the network
C. After the DHCP client receives the "DHCP supply" message, it requests the DHCP server to provide an IP address.
D. The DHCP server broadcasts the "DHCP Acknowledgment" message and assigns the IP address to the DHCP client
27. The area covered by the wide area network ranges from tens of kilometers to thousands of kilometers, and its communication subnet mainly uses ( B ).
A message switching technology
B packet switching technology
C file exchange technology
D circuit switching technology
28. If the IP address of a certain network is 192.12.69.248, then the network category of this address is ( C )
A. Class A address
B. Class B address
C. Class C address
D. Private address
29. Which of the following statements about domain names is wrong ( C )
A. The acronym for Domain Name System is DNS
B. The query from the host to the local domain name server is generally a recursive query
C. Use CHINA in the country's top-level domain name to represent China
D. The domain name is only a logical concept and cannot represent the physical location of the computer
30. The technical architecture of the Internet of Things (Internet of Things) can generally be divided into three layers: ( B )
A. Physical layer, link layer, network layer
B. Perception layer, network layer, application layer
C. Perception layer, application layer, management layer
D. Link layer, transport layer, network layer
31. The data transmission unit of the data link layer is ( D )
A. bit
B. Message
C. Segments
D. frame
32. The role of the DNS server is ( A )
A. Convert domain name to IP address
B. Convert IP address to domain name
C. Providing web page access services to users
D. Provision of virtual space services to users
33. Which of the following is not a national top-level domain name ( C )
A. US
B. UK
C. CHINA
D. CN
3 4. The IP used for the local software loopback test is ( B )
A. An IP address with all 0s
B. IP address with network number 127
C. IP address with all 1s
D. The IP address whose network number is 128.0.0.0
35. The basic means of packet switching network is to use ( A ) technology.
A. store and forward
B. No connection
C. Connection-oriented
D. Circuit switching
36. When the 100M Ethernet network card is in full-duplex mode, the actual total communication bandwidth is ( C ??? )
A. 400M bps
B. 10/100M bps
C. 200M bps
D. 50M bps
37. Divide the total bandwidth of the channel into several, each of which transmits 1 signal. The technology is ( A )
A. Frequency Division Multiplexing (FDM)
B. Time Division Multiplexing (TDM)
C. Wavelength Division Multiplexing (WDM)
D. Code Division Multiplexing (CDM)
38. The twisted-pair RJ-45 connector specifies the numbering of 8 pins ( A )
A 1, 2 send, 3, 6 receive
B. 1, 2, 3, 6 send, 4, 5, 7, 8 receive
C. 1, 2, 3, 4 send, 5, 6, 7, 8 receive
D. 1, 2 receive, 3, 4 receive
39. The most widely used local area network is (A) .
A Ethernet B Token Ring
C token bus network D peer-to-peer network
40. In wireless network technology literature, WiFi and WiMAX are respectively synonymous with ( B )
A. Wireless MAN and Wireless WAN
B. Wireless Local Area Network and Wireless Metropolitan Area Network
C. Wireless Ethernet and Wireless WAN
D. Wireless Ethernet and Wireless WAN
41. The following IPs that are not private addresses are ( D )
A. 10.1.1.1
B. 172.16.25.25
C. 192.168.69.248
D. 192.12.69.248
42. The following description of the UDP datagram is wrong ( D )
A. is connectionless
B. is unreliable
C. No Confirmation Provided
D. Ensure data transmission through sequence number and confirmation mechanism
43. In the network architecture, the data units transmitted by the network layer, data link layer and physical layer are ( D )
A. Messages, frames, packets
B. Bits, Messages, Frames
C. Messages, bits, frames
D. Packets, frames, bits
44. The MAC address is usually fixed on ( A ) of the network communication terminal.
A. Network card
B．Memory
C. hard disk
D. High speed buffer
45. Which of the following communication methods has the best real-time performance? ( B )
A. Message exchange
B. Circuit switching
C. Datagram Packet Switching
D. Virtual Circuit Switching
46. It is known that the polynomial used in cyclic redundancy detection is , then the divisor P should be ( A )
A. 1101
B. 11010
C. 0101
D. 1011
47. In the CSMA/CD agreement, the contention period is ( C )
A. 512 microseconds
B. 512 IFS
C. 512 bit times
D.512 SIFS
48. Intermediary devices are used to connect networks, among which (D)
A. Transponder
B. Bridge
c. Router
D. Gateway
49. Which of the following statements about the firewall is correct ( D ?? )
A. A firewall is a specially programmed switch
B. The network inside the firewall is generally called "untrusted network"
C. The network outside the firewall is generally called "trusted network"
D. Firewalls can be divided into network-level firewalls and application-level firewalls according to different technologies used
50. IPv6 address 0000:00AA:0000:0000:0000:0000:0000:0000 can be written with zero compression ( C )
A. 0：AA：0
B. 0：：AA： 0
C. 0：AA：：0
D. 0：：AA：： 0
51. Wavelength division multiplexing refers to frequency division multiplexing of light, and its English abbreviation is ( C )
A. FDM
B. TDM
C. WDM
D. CDM
52. In order to solve the problem of transparent transmission, the sender adds an escape character ( C ) to the data
A. EOT
B. SOH
C. ESC
D. CRC
53. When making a network cable, if the method of direct connection is used, the color order is ( C )
A. White orange, brown, white green, blue, white blue, green, white brown, brown
B. White orange, blue, white green, orange, white blue, green, white brown, brown
C. White orange, orange, white green, blue, white blue, green, white brown, brown
D. White green, green, white orange, blue, white blue, orange, white brown, brown
54. Using full-duplex work, the data transmission direction is: (A)
- two-way
- unidirectional
- Two-way, but not simultaneous transmission
- Neither is right
55. In the data link of the network, if the window size is 16, (B) bits are required to represent it.
A．3
B．4
C．5
D．6
56. In the OSI model, the relationship between the Nth layer and the N+l layer above it is: (A).
A. Layer N provides services for Layer N+1
B. N + 1 layer adds a header to the information received from N layer
C. Layer N utilizes the services provided by layer N+1
D. N layer has no effect on N+1 layer
57. In the OSI reference model, it is (B) that is responsible for implementing the routing function.
A．physical layer
B．Network layer
C．session layer
D．presentation layer
58. A campus network belongs to (A).
A．LAN
B．WAN
C．MAN
D．PAN
59. (B) in the IP address of the host or router on the same LAN must be the same.
A．host number
B．network number
C．last byte
D．none of the above is right
60. The following (B) address is an email address.
A．www.163.com
C．192.168.0.100
D．http://www.sohu.com
61. Which of the following communication methods has the best real-time performance? ( B )
A. Message exchange
B. Circuit switching
C. Datagram Packet Switching
D. Virtual Circuit Switching
62. When a host on network A sends a message to a host on network B, the router in between needs to check the ( C ) address.
A. Physics
b. port
C. IP
D. MAC
63. NAT technology is often used between the enterprise intranet and extranet, mainly for ( D )
A anti-virus
B error control
C flow control
D IP address translation
64. The RARP protocol is used for ( B )
A. Interpret an IP address as a MAC address
B. Interpret MAC address as IP address
C. Interpret the domain name as a MAC address
D. Interpret IP addresses as host domain names
65. In order to avoid the waste of IP addresses, it is necessary to divide the part of the host number in the IP address again, and divide it into two parts, namely (A).
A．Subnet number and host number
B．Subnet number and network number
C．host number and network number
D．Subnet number and extension number
66. Among the following Ethernet standards, the standard for wireless LAN is (D)
- IEEE802.3
- IEEE802.3i
- IEEE802.8
- IEEE802.11
6 7. Among the following protocols, the protocol that does not belong to the TCP/IP network layer is (C)
- ICMP
- ARP
- PPP
- RARP
68. The main function of a router is (A)
A path selection
B filter
C to increase speed
D flow control
69. Once the central node fails, the network topology of the entire network will be paralyzed (B)
A bus type
B star
C ring
D any type
70. Switches work in the OSI model (B)
A physical layer
B data link layer
C network layer
D application layer
71. The following IP addresses are legal IP host addresses (A)
A 1.255.255.2
B 127.2.3.5
C 255.23.200.9
D 192.157.78.254
72. The default port of the Web site is (A)
A 80
B 8000
C 8080
D 8008
73. The following (C) is not dependent on TCP
A SMTP
B POP3
C DHCP
D FTP
74. The following (B) protocol is the application layer protocol in the TCP/IP protocol stack
A ICMP
B HTTP
C TCP
D IP
75. Set a default route in the routing table, the destination address should be (D)
A 127.0.0.0
B 127.0.0.1
C 1.0.0.0
D 0.0.0.0
76.TCP is a connection-oriented protocol that provides (C) services
A simplex
B half duplex
C full duplex
D one-way
77. The unit to measure the data transmission rate on the network is bit/s, and its meaning is (C)
How many kilometers does a signal travel per second
How many thousand kilometers is the B signal transmitted per second
How many binary bits are transmitted per second in C
D How much data is transmitted per second
78.UDP is the (B) protocol of the TCP/IP protocol suite
A application layer
B transport layer
C network layer
D network interface layer
79. In PPP synchronous transmission, the data bit string 0111101111110111110 is framed and sent as ( D )
A 0111101111110111110
B 00111101111110111110
C 01111011111010111110
D 011110111110101111100
- What is a frame relay network? A
A. Wide area network B. Local area network C. ATM network D. Ethernet
==================================================== ========================== The third file ========================== ===================================================
Questions and Answers Calculation Exercises
- Describe the names and functions of each layer of the TCP/IP architecture and OSI architecture. P31
TCP/IP OSI
application layer physical layer
transport layer data link layer
internet layer network layer
network interface layer transport layer
session layer
presentation layer
application layer
TCP/IP Architecture
Application layer: Realize communication and interaction between application processes
Transport layer: Responsible for communication between processes in two hosts and providing general data transmission services. There are two protocols, TCP and UDP.
Network layer: Responsible for providing communication services for different hosts on the packet switching network.
Data link layer: 1. Combine data into data blocks and encapsulate them into frames; 2. Error control; 3. Flow control
OSI architecture
Physical layer: realize the transparent transmission of bit stream. Basic unit: bit.
Data link layer: physical address addressing, data framing, flow control, data error detection, retransmission, etc. At this layer, the unit of data is called a frame.
Network layer: path selection, data forwarding. Transparent transmission of packets from source host to destination host
Transport layer: port (port) to port (port) message
Session layer: establish, maintain and release user-oriented connections between two nodes, and manage and control sessions to ensure reliable transmission of session data.
Presentation layer: responsible for data format conversion, compression and decompression, encryption and decryption.
Application layer: The application layer provides an interface for the operating system or network applications to access network services.
- What are the classification methods of computer networks, and what are the categories according to different division methods? P20
According to the scope of the network: wide area network, metropolitan area network, local area network, personal area network
By network user: public network, private network
- Describe the main points of the network architecture of the five-layer protocol, including the main functions of each layer. P31
Physical layer: Consider how much voltage to use to represent 0 or 1, and identify the bits sent by the sender, and also determine how many pins the plug of the connecting cable has, and how to connect each pin.
Digital link layer: When transmitting data, the link layer encapsulates the IP datagrams passed down from the network layer into frames; when receiving data, after receiving a frame, it can extract the data part from it and hand it over to the network layer.
Network layer: responsible for providing communication services for different hosts on the packet switching network; selecting appropriate routes so that the packets transmitted by the host transport layer can find specific hosts through the routes in the network.
Transport layer: Responsible for providing general data transmission services for communication between processes in two hosts
Application layer: complete specific network applications through the interaction between application processes
4. The PPP protocol uses synchronous transmission technology to transmit the bit string 011011111111100. What kind of bit string does it become after zero bit padding? If the data part of the PPP frame received by the receiving end is 0001110111110111110110, what kind of bit string will become after deleting the zero bits added by the sending end?
011011111011111000
00011101111111111110
5. What are the characteristics of a routing protocol in an ideal state ? P151
Algorithms must be correct and complete
Algorithms should be computationally simple
Algorithms adapt to changes in traffic and network topology
Algorithms should be stable
algorithm is fair
algorithm is optimal
6. What are the network topologies of commonly used LANs? Which one is the most commonly used at the moment? P82
Star network, ring network, bus network
star network
- How many types of IP addresses are there? What are the two fields for each type of address? What are the lengths of the fields? P119
A Network number 1 Host number 3
B Network number 2 Host number 2
C Network No. 3 Host No. 1
8. What is the function of ping command, what is the function of ping 127.0.0.1? P149
Used to test connectivity between two hosts
The purpose is to detect whether the circuit of the machine is normal, to test the connectivity between the machine and the machine,
- Please explain the working process of a browser using the HTTP protocol to access a web page. P267
Each World Wide Web (WWW) site has a server process that constantly listens to port 80 of TCP (Transmission Control Protocol) in order to find out whether a browser initiates a connection establishment request to it. Once the connection establishment request is monitored and the TCP connection is established , the browser sends a request to the World Wide Web server to browse a certain page, and the World Wide Web server returns the requested page as a response, and finally the TCP connection is released.
9. Please explain in detail the three-way handshake process of TCP connection establishment. P238
The TCP protocol provides reliable connection services and uses three-way handshake to establish connections.
The first handshake: When the connection is established, the client sends a SYN packet (SYN=1) to the server, randomly generates a data packet of seq=X to the server, and waits for the server to confirm;
The second handshake: After the server receives the request to confirm the online information, it must confirm the client's SYN ( ack=X+1 ), and at the same time, it also sends a SYN package, that is, SYN+ACK package, that is, SYN=1, ACK=1 , and randomly generate packets with seq=Y;
The third handshake: the client receives the SYN + ACK packet from the server and sends an acknowledgment packet ACK (ack=Y+1) to the server. After the packet is sent, the client and server enter the ESTABLISHED state and complete the three-way handshake. The server starts sending data.
- Please explain the domain name resolution process, and what are the two ways to query the domain name? Try to explain their difference. P258
The two domain name query methods are recursive query and iterative query:
Recursive query, the query from the host to the local domain name server generally uses recursive query, if the local domain name server does not know the IP address of the host query domain name, the local domain name server will continue to send queries to other root domain name servers as a DNS (computer domain name) client Request message, instead of letting the host to perform the next query by itself.
Iterative query, the query from the local domain name server to the root domain name server generally uses iterative query. When the root domain name server receives the iterative query request message sent by the local domain name server, it either gives the query IP address, or tells the local domain name server the next step Which domain name server do you want to query, and then let the local domain name server perform the next query by itself.
- What is the difference between TCP and UDP protocols? P205
13. Please briefly describe the working process of the ARP protocol. P124
1. The ARP process broadcasts an ARP request packet on the local area network
2. All ARP processes running on all hosts on the local area network receive this ARP request packet
3. The IP address of host B is the same as the IP address to be queried in the ARP request packet, so it accepts the ARP request packet, sends an ARP response packet to host A, and writes its own hardware address in the ARP response packet
4. After host A receives the ARP response packet from host B, it writes the mapping from the IP address of host B to the hardware address in its ARP cache
- Please briefly describe the CSMA/CD protocol data transmission process. P85
For a station to send data, it must first listen to the channel. If the channel is idle, send data immediately and perform collision detection; if the channel is busy, continue to listen to the channel until the channel becomes idle, then continue to send data and perform collision detection. If a station detects a collision while sending data, it will immediately stop sending data and wait for a random long time, repeating the above process.
Mantra: listen first and then send, listen and send at the same time, stop the conflict, delay and resend
Calculation problems:
- The data to send is 1101011011 . The generator polynomial using CRC is . Find the remainder that should be added after the data .
According to the title: M = 1101011011 P = 10011 then n = 4, then 2 n M = 11010110110000 According to binary division, R = 1110
2，试辨认以下IP地址的网络类别。
(1) 128.36.199.3 B
(2) 21.12.240.17 A
(3) 183.194.76.253 B
(4) 192.12.69.248 C
- 一个主机的IP地址是136.160.160.134，子网掩码是255.255.255.224，试求该主机所在的网络地址。
136.160.160.10000110
255.255.255.11100000
136.160.160.10000000
136.160.160.128
4。某单位分配到一个地址块136.23.12.64/26。现在需要进一步划分两个一样大的子网。试问：1、 每个子网的前缀有多长？2、 每一个子网中有多少个地址？3、 每一个子网的地址块是什么？4、 每一个子网可分配给主机使用的最小地址和最大地址是什么？
27
136.23.12.64/27 136.23.12.010 00001-136.23.12.01011110
136.23.12.96/27 136.23.12.011 00001-136.23.12.01111110
- 已知地址块中的一个地址是140.120.84.24/20。试求这个地址块中的最小地址和最大地址。地址掩码是什么？地址块中共有多少个地址？相当于多少个C类地址？
140.120.80.0-140.120.95.255
255.255.255.240
16
6. Assume that the routing table of router A in the network has the following items (these three columns represent "destination network", "distance" and "next hop router")
N1 4 B
N2 2 C
N3 1 F
N4 5 G
Now A receives the routing information from C (these two columns represent "destination network" and "distance" respectively):
N1 2
N2 1
N3 3
N4 7
Find the updated routing table of router A.
N1 3 C
N2 2 C
N 3 1 F
N4 5 G
7. If the PPP data frame 1011110 is encoded using the Manchester encoding method during synchronous transmission, please draw its waveform and explain the reason.
Rationale: An upward transition in the center of a bit period represents a 0, and a downward bar in the center of a bit period represents a 1.
- A certain department is assigned an IP address of 190.85.145.64 with a subnet mask of 255.255.255.192. There are currently 5 computers in each of the three offices in the department. What is the minimum address that each office can assign to the host computer? (Known: 190=10111110, 85=1010101, 145=10010001, 64=1000000, 192=11000000, 224=11100000, 248=11111000, 252=11111100, 240=11110000)
190.85 .145.01000000
255.255.255.11000000
190.85 .145.01000000(64)/26
190.85 .145.01000 001/28
190.85 .145.01010 001/28
190.85 .145.01100 001/28
9. The Internet construction standard published in 1999 stipulates four algorithms of congestion control, slow start, congestion avoidance, fast retransmission, and fast recovery. The known relationship between the congestion window of the first 12 rounds and the transmission round is shown in the figure. If the sender receives three repeated acknowledgments in the 12th round and starts to execute the fast retransmission algorithm , please draw the following 13 to 22 rounds in the figure and explain the reason.
The congestion window changes to 12 13 14 15 16 17 18 19 20 21 22. According to the fast retransmission algorithm, if three repeated confirmations are received in a row, the threshold value and the congestion window should be adjusted to half, and then the congestion window will be increased by one every round
10. The relationship between the size of the TCP congestion window cwnd and the transmission round n is as follows:
| n cwnd |
1 1 |
2 2 |
3 4 |
4 8 |
5 16 |
6 32 |
7 33 |
8 34 |
9 35 |
10 36 |
11 37 |
12 38 |
13 39 |
| n cwnd |
14 40 |
15 41 |
16 42 |
17 21 |
18 22 |
19 23 |
20 24 |
21 25 |
22 26 |
23 1 |
24 2 |
25 4 |
26 8 |
( 1 ) Try to draw the relationship curve between the congestion window and the transmission round.
( 2 ) Indicates the time interval during which TCP works in the slow start phase. Round 6
( 3 ) Indicate the time interval during which TCP works in the congestion avoidance phase. Round 16
( 4 ) When sending in the 1st round, the 6th round and the 22nd round, what is the threshold ssthresh set to respectively?
32 32 21