3-01 What is the difference between data link (ie logical link) and link (ie physical link)? What is the difference between "circuit connected" and "data link connected"?
Answer: The difference between a data link and a link is that the data link must have some necessary procedures to control the transmission of data outside the link. Therefore, the data link has more hardware and required to implement the communication procedures than the link. software. "The circuit is connected" means that the node switches at both ends of the link have been turned on, and the physical connection has been able to transmit the bit stream, but the data transmission is not reliable, and the data link connection is established based on the physical connection. The data link is up. After that, because the data link connection has the functions of detection, confirmation and retransmission, the less reliable physical link becomes a reliable data link, and reliable data transmission is used as a data link. When disconnecting, the physical circuit connection does not necessarily follow the disconnection.
3-02 What functions does the link control in the data link layer include? Try to discuss the advantages and disadvantages of making the data link layer a reliable link layer.
Answer: Link management frame delimitation, flow control, error control, and data The advantages and disadvantages of distinguishing control information, transparent transmission, addressing, and reliable link layer depend on the application environment: for channels with serious interference, the reliable link layer can restrict the retransmission range to local links to prevent network-wide Transmission efficiency is impaired; for high-quality information
Therefore, the use of a reliable link layer will increase resource overhead and affect transmission efficiency.
3-03 What is the role of the network adapter? Which layer does the network adapter work?
Answer: The adapter (ie, the network card) implements the data link layer and the physical layer. The hardware and software network adapters work in TCP/IP The network interface layer in the protocol (the data link layer and the physical layer in OSI)
3-04 Why must the three basic issues of the data link layer (frame delimitation, transparent transmission and error detection) be resolved?
Answer: Frame delimitation is an inevitable requirement for packet switching. Transparent transmission avoids confusion between message symbols and frame delimitation symbols. Error detection prevents invalid data frames from combining errors and wasting transmission and processing resources on subsequent routes.
3-05 What will happen if frame delimitation is not performed at the data link layer?
Answer: Unable to distinguish between grouping and grouping Unable to determine the control domain and data domain of the grouping Unable to limit the scope of error correction to the exact part
What are the main features of 3-06 PPP agreement? Why doesn't PPP use frame numbers? What situation does PPP apply to? Why can't the PPP protocol enable reliable transmission at the data link layer?
Answer: Simple, provide unreliable datagram service, error detection, no error correction, no serial number and confirmation mechanism address field A is only set to 0xFF. The address field does not actually work. The control field C is usually set to 0x03. PPP is byte-oriented. When PPP is used in a synchronous transmission link, the protocol stipulates that hardware is used to complete bit filling (the same as HDLC). When PPP is used in asynchronous transmission, a special character filling method is used. Suitable for the case where the line quality is not too bad, PPP has no encoding and confirmation mechanism
3-07 The data to be sent is 1101011011. The generator polynomial using CRC is P(X)=X4+X+1. Try to find the remainder that should be added to the data. During the data transmission, the last 1 becomes 0. Can the receiver find out? If the last two 1s of the data become 0s during the data transmission, can the receiving end find out? After adopting CRC check, does the transmission of the data link layer become reliable transmission?
Answer: For binary division, 1101011011 0000 10011 is the remainder 1110, and the added check sequence is 1110. For binary division, both errors can be developed. Only CRC check is used, lack of retransmission mechanism, and the data link layer transmission is not reliable. Transmission.
3-08 The data to be sent is 101110. The CRCD generator polynomial is P(X)=X3+1. Try to find the remainder that should be added to the data.
Answer: For binary division, 101110 000 10011 The remainder added to the data is 011
3-09 The data part of a PPP frame (written in hexadecimal) is 7D 5E FE 27 7D 5D 7D 5D 65 7D 5E. What is the real data (written in hexadecimal)?
Answer: 7D 5E FE 27 7D 5D 7D 5D 65 7D 5E 7E FE 27 7D 7D 65 7D
The 3-10PPP protocol uses synchronous transmission technology to transmit the bit string 010111111111100. What kind of bit string becomes after zero bit stuffing? If the data part of the PPP frame received by the receiving end is 0001110111110111110110, what kind of bit string becomes after deleting the zero bits added by the sending end?
Answer: 01101111111 11111 00 011011111011111000
0001110111110111110110 000111011111 11111 110
3-11 Try to discuss under what conditions are transparent transmission and under what conditions are not transparent transmission. (Hint: Please figure out what "transparent transmission" is, and then consider whether it can meet its conditions.)
(1) Ordinary telephone communication.
(2) Public telegraph communication provided by the telecommunications bureau.
(3) E-mail services provided by the Internet.
3-12 What are the working states of PPP protocol? When a user wants to use the PPP protocol to establish a connection with an ISP for communication, what kinds of connections need to be established? What problem does each connection solve?
What are the main features of 3-13 LAN? Why does the local area network adopt the broadcast communication method and the wide area network does not?
Answer: Local area network LAN refers to a computer communication network that interconnects limited communication devices in a small geographic area. From a functional point of view, the local area network has the following characteristics: (1) Shared transmission channel, in the local area network, Multiple systems are connected to a shared communication medium. (2) The geographical scope is limited and the number of users is limited. Usually a local area network serves only one unit, and is only connected in a relatively independent local area, such as a building or a concentrated building group. Generally speaking, the coverage area of the local area network is within 10m~10km or larger. From the perspective of network architecture and transmission detection reminders, the LAN also has its own characteristics: (1) The low-level protocol is simple (2) Without a separate network layer, the LAN architecture is only equivalent to the lowest two layers of OSI/RM ( 3) Two media access control technologies are used. Because the shared broadcast channel is used, and the channels can be used for different transmission media, the problem faced by the local area network is multi-source and multi-purpose connection management, which leads to multiple media access control technologies Stations in a local area network usually share communication media, and it is naturally appropriate to adopt a broadcast communication method, and a wide area network usually adopts a grid directly between stations.
3-14 What are the types of commonly used LAN network topologies? Which structure is the most popular now? Why did the early Ethernet choose the bus topology instead of the star topology, but now it uses the star topology instead?
Answer: Star network, bus network, ring network, tree network. The reliable star topology at the time was more expensive. People believed that the passive bus structure was more reliable, but practice proved that the bus Ethernet connected to a large number of sites It is prone to failure, and the use of dedicated ASIC chips can say that the hub of the star structure is very reliable. Therefore, the current Ethernet generally uses the topology of the star structure.
3-15 What is traditional Ethernet? What are the two main standards for Ethernet?
Answer: DIX Ethernet V2 standard LAN DIX Ethernet V2 standard and IEEE 802.3 standard
3-16 What is the symbol transmission rate of Ethernet with a data rate of 10Mb/s on physical media?
Answer: The symbol transmission rate is the baud rate. Ethernet uses Manchester encoding, which means that every bit sent has two signal periods. The data rate of standard Ethernet is 10MB/s, so the baud rate is twice the data rate, that is, 20M baud
3-17 Why is the LLC sub-layer standard already developed but rarely used now?
Answer: Because the local area network often used in the TCP/IP system is DIX Ethernet V2 instead of several types of local area networks in the 802.3 standard, the logical link control sublayer LLC (ie, the 802.2 standard) formulated by the 802 committee is no longer useful.
3-18 Try to explain the meaning of "10", "BASE" and "T" in 10BASE-T.
Answer: "10" in 10BASE-T means that the transmission rate of the signal on the cable is 10MB/s, "BASE" means that the signal on the cable is a baseband signal, and "T" stands for twisted pair star network, but 10BASE-T The communication distance is slightly shorter, and the distance from each station to the hub does not exceed 100m.
The CSMA/CD protocol used by 3-19 Ethernet is to access the shared channel in a contention manner. What are the advantages and disadvantages compared with the traditional time division multiplexing TDM?
Answer: The traditional time division multiplexing TDM is a static time slot allocation. The channel utilization rate is high when the load is uniform and high, and resources are wasted when the load is low or uneven. The CSMA/CD class dynamically uses idle new resources, and the channel when the load is low. The utilization rate is high, but the control is complicated, and the channel conflict is large when the load is high.
3-20 Assume that the data rate of a 1km long CSMA/CD network is 1Gb/s. Suppose the propagation rate of the signal on the network is 200,000km/s. Seek the shortest frame length that can use this protocol.
Answer: For a 1km cable, the one-way propagation time is 1/200000=5 as microseconds, and the round-trip propagation time is 10 microseconds. In order to work in accordance with CSMA/CD, the minimum frame transmission time cannot be less than 10 microseconds, in Gb/ Working at s rate, the number of bits that can be sent in 10 microseconds is equal to 10*10 -6/1*10 -9=10000, so the shortest frame is 10000 bits or 1250 bytes long
3-21 What is bit time? What are the benefits of using this time unit? How many microseconds is 100 bit time?
Answer: Bit time is the time required to send more than one bit. It is the reciprocal of the transmission rate. It is convenient to establish the relationship between the length of the message and the transmission delay. "Bit time" is converted to "microseconds". You must first know the data rate, such as data The rate is 10Mb/s, then 100 bit time is equal to 10 microseconds.
3-22 Assume that a station in a 10Mb/s Ethernet using the CSMA/CD protocol detects a collision when sending data, and selects a random number r=100 when executing the backoff algorithm. How long does the station need to wait before sending data again? What if it is 100Mb/s Ethernet?
Answer: For a 10mb/s Ethernet, the Ethernet sets the contention period as 51.2 microseconds, and 100 contention periods need to be backed up. The waiting time is 51.2 (microseconds)*100=5.12ms for 100mb/s Ethernet Internet, Ethernet set the contention period to 5.12 microseconds, 100 contention period to back up, the waiting time is 5.12 (microseconds) * 100 = 512 microseconds
Formula 3-23 (3-3) indicates that the limit channel utilization of Ethernet has nothing to do with the number of stations connected to the Ethernet. Can it be deduced from this that the utilization rate of Ethernet is also independent of the number of sites connected to the Ethernet? Please explain your reasons.
Answer: The actual Ethernet sends data at random moments, and the maximum channel utilization of
Ethernet is based on the assumption that the Ethernet uses a special scheduling method (it is no longer CSMA/CD), so that each The sending of the node does not collide.
3-24 Assume that sites A and B are on the same 10Mb/s Ethernet network segment. The propagation delay between these two stations is 225 bit time. Now suppose that A starts to send a frame, and B also sends a frame before the end of A's transmission. If A sends the shortest frame allowed by the Ethernet, can A send its own data before detecting a collision with B? In other words, if A does not detect a collision before the transmission is completed, can you be sure that the frame sent by A will not collide with the frame sent by B? (Hint: In the calculation, it should be considered that when each Ethernet frame is sent to the channel, a few bytes of preamble and frame delimiter should be added before the MAC frame)
Answer: Set at t=0 A Start sending, at t=(64+8)*8=576 bit time, A should finish sending. t=225 bit time, B detects the signal of A. As long as B sends data before t=224 bit time, A must detect a collision before the transmission is completed, and it can be sure that it will not send collisions in the future. If A does not detect a collision before the transmission is completed, then it can be sure The frame sent by A will not collide with the frame sent by B (of course, it will not collide with other stations).
3-25 Stations A and B in the previous question sent data frames at the same time at t=0. When t=255 bit time, A and B have detected a collision at the same time, and the interference signal transmission is completed in t=255+48=273 bit time. A and B choose different r values to back off in the CSMA/CD algorithm. Assume that the random numbers chosen by A and B are rA=0 and rB=1, respectively. When does A and B start to retransmit their data frames? When does the data frame retransmitted by A arrive at B? Will the data retransmitted by A collide with the data retransmitted by B again? Will B stop sending data at the scheduled retransmission time?
Answer: When t=0, A and B start to send data T1=225 bit time, A and B both detect collision (tau) T2=273 bit time, A and B end the transmission of interference signals (T1+48) T3= 594 bit time, A starts sending (T2+Tau+rA Tau+96) T4=785 bit time, B detects the channel again. (T4+T2+Tau+Rb Tau) If it is idle, B will send data at T5=881 bit time, otherwise it will back off. (T5=T4+96) A retransmitted data arrives at B at 819 bit time, B first detects that the channel is busy, so B stops sending at the predetermined 881 bit time
There are only two stations on the 3-26 Ethernet, and they send data at the same time, causing a collision. Therefore, the retransmission is performed according to the truncated binary exponential backoff algorithm. The number of retransmissions is denoted as i, i=1, 2, 3,.... Try to calculate the probability of the first retransmission failure, the probability of the second retransmission, the probability of the third retransmission failure, and the average number of retransmissions I before a station successfully sends data.
Answer: Record the probability of the i-th retransmission as pi. Obviously, the probability of failure in the first retransmission is 0.5, the probability of failure in the second retransmission is 0.25, and the probability of failure in the third retransmission is 0.125. The average number of retransmissions I=1.637
3-27 Assume that 80% of the communication volume on an Ethernet is carried out on the local local area network, and the remaining 20% of the communication volume is carried out between the local area network and the Internet. The other Ethernet situation is reversed. One of these two Ethernets uses an Ethernet hub, and the other uses an Ethernet switch. Which network do you think the Ethernet switch should be used for?
Answer: The hub is a physical layer device, which simulates the common contention of the shared medium of the bus, and becomes the bottleneck of the communication capacity of the LAN. The switch is a link-layer device, which can realize transparent switching. The local area network is connected to the Internet through a router. When the communication between the local area network and the Internet is the main component, a centralized data flow facing the router is formed. The use of a hub has a greater conflict. Improved. When the traffic in the local area network is the main component, the use of switches to improve external traffic is not obvious
3-28 There are 10 stations connected to the Ethernet. Try to calculate the bandwidth that each station can get in the three cases. (1) 10 stations are connected to a 10Mb/s Ethernet hub; (2) 10 stations are connected to a 100Mb/s Ethernet hub; (3) 10 stations are connected to a 10Mb/s Ethernet switch .
Answer: (1) 10 stations are connected to a 10Mb/s Ethernet hub: 10mbs (2) 10 stations are connected to a 100mb/s Ethernet hub: 100mbs (3) 10 stations are connected to a 10mb/s s Ethernet switch: 10mbs
3-29 When upgrading from 10Mb/s Ethernet to 100Mb/s, 1Gb/S and 10Gb/s, what technical problems need to be solved? Why can Ethernet eliminate its competitors in the process of development and expand its application range from local area networks to metropolitan area networks and wide area networks?
Answer: Technical question: To keep parameter a small, you can reduce the maximum cable length or increase the minimum frame length. The method used in 100mb/s Ethernet is to keep the minimum frame length unchanged, but The maximum cable length of a network segment is reduced to 100m, and the time interval between frames is changed from the original 9.6 microseconds to the current 0.96 microseconds. Gigabit Ethernet still maintains the maximum length of a network segment of 100m, but adopts "carrier extension" "The shortest frame length is still 64 bytes (so that compatibility can be maintained), and the contention time is increased to 512 bytes. And use "packet burst" to reduce overhead. The frame format of 10 Gigabit Ethernet is exactly the same as that of 10mb/s, 100mb/s and 1Gb/s Ethernet. Gigabit Ethernet also retains the minimum Ethernet sum specified by the standard. Maximum frame length, which enables users to easily communicate with lower-speed Ethernet when upgrading their existing Ethernet. Due to the high data rate, Gigabit Ethernet no longer uses copper wire but only uses optical fiber as the transmission medium. It uses long-distance (more than km) optical transceivers and single-mode optical fiber interfaces so that it can work in wide areas.
3-30 What are the characteristics of Ethernet switches? How to form a virtual local area network with it?
Answer: Ethernet switches are link-layer devices that can realize transparent switching. Virtual local area network VLANs are logical groups that are composed of some local area network segments and have nothing to do with physical locations. These network segments have certain common requirements. The virtual local area network protocol allows a 4-byte identifier to be inserted into the Ethernet frame format, called a VLAN tag, to indicate which virtual local area network the workstation sending the frame belongs to.
What are the working principles and characteristics of 3-31 bridges? What are the similarities and differences between bridges, repeaters and Ethernet switches?
Answer: The bridge works at the data link layer. It forwards the received frames according to the destination address of the MAC frame. The bridge has the function of filtering frames. When the bridge receives a frame, it does not forward the frame to all interfaces, but first checks the destination MAC address of the frame, and then determines which interface the frame is forwarded to. The forwarder works at the physical layer. Simply forward the signal, and the Ethernet switch without filtering capability is a link-layer device, which can be regarded as a multi-port bridge
3-32 Figure 3-35 shows that there are five sites connected to three LANs respectively, and they are connected by bridges B1 and B2. Each bridge has two interfaces (1 and 2). At the beginning, the forwarding tables in both bridges were empty. Later, the following stations sent data frames to other stations: A sent to E, C sent to B, D sent to C, and B sent to A. Try to fill in the relevant data in Table 3-2.
The forwarding table in the 3-33 bridge is established with a self-learning algorithm. If some sites always do not send data but only receive data, is there no item corresponding to such a site in the forwarding table? If you want to send a data frame to this site, can the bridge forward the data frame to the destination address correctly?
Answer: There is no project corresponding to such a site; the bridge can use broadcast to correctly forward the data frame to the destination address