ID
int64 1
1.96k
| Split
stringclasses 1
value | Domain
stringclasses 4
values | SubDomain
stringclasses 24
values | Format
stringclasses 1
value | Tag
stringclasses 2
values | Language
stringclasses 1
value | Question
stringlengths 15
717
| A
stringlengths 1
292
| B
stringlengths 1
232
| C
stringlengths 1
217
| D
stringlengths 1
192
| Answer
stringclasses 4
values | Explanation
stringlengths 21
1.43k
⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1,170
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
A channel samples once every 1/8s and transmits signals with 16 different states. The maximum data transmission rate is ( ).
|
16b/s
|
32b/s
|
48b/s
|
64b/s
|
B
|
From the problem statement, we know that the sampling frequency is 8 Hz. A signal with 16 different states can carry 4 bits of data, thus the maximum data transfer rate is 8 × 4 = 32 b/s.
|
1,171
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
After encoding 1 channel of analog signal into a digital signal, it is multiplexed with another 7 channels of digital signals onto a single communication line using synchronous TDM. The frequency range of the 1 channel of analog signal is 0~1kHz, and each sampling point is encoded into a 4-bit binary number using PCM. The data rate of the other 7 channels of digital signals is 7.2kb/s each. The minimum communication capacity required for the multiplexed line is ( ).
|
7.2kb/s
|
8kb/s
|
64kb/s
|
512kb/s
|
C
|
Channel 1 has a frequency of 1kHz, and the known frequency is 2kHz. Each code is a 4-bit binary number, so the data transmission rate is 8kb/s. The multiplexed rate for each branch must be equal, and the rates of the other 7 digital signals are all below 8kb/s. Therefore, they all need to use pulse stuffing to increase the data rate to 8kb/s. Then, by multiplexing these 8 signals, the required communication capacity is 8kb/s × 8 = 64kb/s.
|
1,172
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
To minimize data transmission delay in the network, the preferred switching method is ( ).
|
Circuit Switching
|
Message Switching
|
Packet Switching
|
ATM (Asynchronous Transfer Mode) switching
|
A
|
Circuit switching, although it has a larger delay in establishing connections, occupies the link continuously during data transmission, offering better real-time performance and smaller transmission delay.
|
1,173
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Regarding the characteristics of virtual circuit services and datagram services, the correct statement is ()
|
Virtual circuit service and datagram service are both connectionless services.
|
In datagram services, packets are transmitted along the same path in the network and arrive in the order they were sent.
|
After establishing a connection, packets must carry a virtual circuit identifier.
|
Packets in a virtual circuit may arrive in a different order than they were sent.
|
C
|
Virtual circuit service is connection-oriented, and packets belonging to the same virtual circuit are forwarded along the same route based on the same virtual circuit identifier, ensuring the orderly arrival of packets. In datagram service, the network independently selects routes for each packet, transmission does not guarantee reliability, nor does it guarantee the sequential arrival of packets.
|
1,174
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about datagrams and virtual circuits, the incorrect one is ( ).
|
In datagram mode, if a node loses a packet due to a fault, other packets can still be transmitted normally.
|
In datagram mode, each packet is independently routed and forwarded, with no inherent connection between different packets.
|
In virtual circuit mode, packets belonging to the same virtual circuit are forwarded along the same route.
|
Although virtual circuit mode is connection-oriented, it does not guarantee the orderly arrival of packets.
|
D
|
Datagram service is connectionless, does not provide reliability guarantees, and does not ensure the orderly arrival of packets. In virtual circuit service, packets belonging to the same virtual circuit are forwarded along the same route, ensuring the orderly arrival of packets.
|
1,175
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
The correct statement among the following is ( ).
|
Circuit switching is the actual exchange of physical lines, whereas virtual circuit switching is a logical connection, and a single physical line can only establish one logical connection.
|
The connection of a virtual circuit is temporary, and this connection is released when the session ends.
|
Datagram service does not provide reliable transmission, but it can ensure the orderly arrival of packets.
|
In datagram services, each packet must carry the source and destination addresses during transmission.
|
D
|
Circuit switching is the actual physical line switching, such as telephone lines; virtual circuit switching is a multiplexing technology, where each physical line can carry multiple logical connections. In virtual circuit services, packets only need to carry the virtual circuit identifier and do not need to include the source and destination addresses.
|
1,176
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
When describing the meaning of a physical layer interface pin being at a high level, this description pertains to ( ).
|
Mechanical properties
|
Electrical Characteristics
|
Functional Characteristics
|
Process Characteristics
|
C
|
The functional characteristics of the physical layer specify what significance a certain voltage level on a line represents.
|
1,177
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about physical layer devices, the incorrect one is ( ).
|
Repeaters only act on the electrical part of the signal.
|
The principle of using repeaters to extend the network transmission distance is that they amplify the attenuated signals.
|
A hub is essentially equivalent to a multi-port repeater.
|
The network segments connected by physical layer devices still constitute a local area network (LAN), and segments with different data link layer protocols cannot be interconnected.
|
B
|
The principle of a repeater is to regenerate the attenuated signal rather than amplify it. The connected network segments still belong to the same local area network (LAN).
|
1,178
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
When a port of the hub receives data, it ( )
|
Broadcast from all ports
|
Broadcast out from all ports except the input port.
|
Forward out from the appropriate port based on the destination address.
|
Randomly select a port to forward.
|
B
|
A hub does not have addressing capabilities; it broadcasts data received from one port to all other ports except the input port.
|
1,179
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about repeaters and hubs, the incorrect one is ( )
|
Both operate at the Physical Layer of the OSI reference model.
|
Both can amplify and shape the signal.
|
The number of network segments interconnected via repeaters or hubs is not limited.
|
Repeaters typically have only 2 ports, while hubs usually have 4 or more ports.
|
C
|
The number of network segments interconnected by repeaters and hubs is limited, subject to the "5-4-3 rule," which allows for a maximum of 4 repeaters and 5 segments. All other statements are correct.
|
1,180
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the size of the send window in SR (Selective Repeat Protocol)? ()
|
2 ^ n
|
2^(n-1)
|
(2^n) - 1
|
n
|
B
| null |
1,181
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Which access control method is adopted by the IEEE 802.11 CSMA/CA wireless local area network? ()
|
CSMA
|
CSMA/CD
|
CSMA/CA
|
TDMA
|
C
| null |
1,182
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What functions are primarily involved by data link layer devices? ()
|
Framing, Error Control, Flow Control
|
Error Control
|
Flow Control
|
Transmission Management
|
A
| null |
1,183
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Which of the following is a primary function of the data link layer? ()
|
Source Address Verification
|
Mapping of source and destination addresses
|
Resolution of source and destination addresses
|
Translation of source and destination addresses
|
C
| null |
1,184
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What does the individual acknowledgment in CSMA/CA refer to? ()
|
The sender individually acknowledges the received ACK frames.
|
The sender individually acknowledges the received data frames.
|
The receiver individually acknowledges the received RTS frames.
|
The receiver individually acknowledges the received CTS frames.
|
B
| null |
1,185
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the limitation of parity check code when detecting errors in two positions? ()
|
The content can be detected.
|
Unable to detect
|
Can only detect a single-bit error.
|
Can only detect three-bit errors.
|
B
| null |
1,186
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the cyclic redundancy check code generated based on? ()
|
The length of the data
|
The content of the data
|
Data transmission rate
|
The storage location of data
|
B
| null |
1,187
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the MAC broadcast address?
|
00-00-00-00-00-00
|
FF-FF-FF-FF-FF-FF
|
11-11-11-11-11-11
|
AA-AA-AA-AA-AA-AA
|
B
| null |
1,188
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The most fundamental service provided by the Data Link Layer is to reliably transmit data to the network layer of the target machine at the adjacent node. What is the basic unit of data?
|
Data packet
|
frame
|
Data segment
|
Data Unit
|
B
| null |
1,189
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What does "transparent" refer to in the transparent transmission of the data link layer?
|
Data does not need to be encrypted.
|
Unlimited content, format and encoding of transmitted data.
|
Data can be transmitted to any node.
|
Data is not limited by physical media.
|
B
| null |
1,190
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What determines the Path MTU?
|
Data frame length of the data link layer
|
IP address of the network layer
|
Types of Physical Layer Media
|
Transport Layer Protocols
|
A
| null |
1,191
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the minimum length of an Ethernet frame in bytes?
|
46
|
54
|
64
|
76
|
C
| null |
1,192
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the broadcast address for Ethernet? ()
|
00-00-00-00-00-00
|
FF-FF-FF-FF-FF-FF
|
11-11-11-11-11-11
|
AA-AA-AA-AA-AA-AA
|
B
| null |
1,193
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the total length in bytes of the header and trailer of a data link layer frame? ()
|
14B
|
16B
|
18B
|
20B
|
C
| null |
1,194
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
How many bytes are occupied by the source address and the destination address of a frame, respectively? ()
|
4B, 4B
|
6B, 6B
|
8B, 8B
|
10B, 10B
|
B
| null |
1,195
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What do the first 24 bits of a MAC address represent? ()
|
Manufacturer's Mark
|
Network Identifier
|
Device Identification
|
Regional Identification
|
A
| null |
1,196
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
How does the data link layer implement flow control? ()
|
Stop-and-Wait Protocol
|
Stop-and-Wait Protocol and Sliding Window Protocol
|
ALOHA Protocol
|
CSMA/CD protocol
|
B
| null |
1,197
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
In the sliding window protocol, what does the size of the window refer to? ()
|
Sender's window size
|
Receiver's window size
|
Number of unacknowledged frames in the window
|
The total number of frames
|
A
| null |
1,198
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Which error detection method is most commonly used in the data link layer? ()
|
Parity Check
|
CRC Checksum
|
Checksum
|
LRC Check
|
B
| null |
1,199
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the function of ARP? ()
|
Parsing MAC Addresses
|
Parsing IP Addresses
|
Parsing Port Number
|
Resolve domain name
|
A
| null |
1,200
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What form is an ARP request broadcast in ()?
|
DataFrame
|
IP datagram
|
ARP frame
|
ICMP frame
|
A
| null |
1,201
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
How is data transmitted in the ring topology of the data link layer? ()
|
Unidirectional transmission
|
Bidirectional transmission
|
Clockwise transmission
|
Counterclockwise transmission
|
B
| null |
1,202
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Which transmission method of the data link layer is suitable for point-to-point communication? ()
|
Unicast
|
Broadcast
|
Multicast
|
Random Transmission
|
A
| null |
1,203
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the commonly used algorithm in the implementation of the retransmission mechanism? ()
|
Greedy algorithm
|
Sliding Window Algorithm
|
Dijkstra's algorithm
|
A* algorithm
|
B
| null |
1,204
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The following is not a function of the data link layer ( )
|
Frame Bounding function
|
Circuit management function
|
Error Control Function
|
Flow control function
|
B
| null |
1,205
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The services provided by the Data Link Layer to the Network Layer do not include ( ).
|
Unacknowledged connectionless service
|
Confirmed connectionless service
|
Unacknowledged connection-oriented service
|
Confirmed connection-oriented service
|
C
| null |
1,206
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Flow control is actually the control of ( ).
|
The sender's data traffic
|
Receiver's data traffic
|
Data traffic of the sender and receiver
|
Data traffic between any two nodes on the link
|
A
| null |
1,207
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Improving the signal-to-noise ratio can mitigate the impact of errors which are ( ).
|
Random errors
|
Transient Error
|
Data Loss Error
|
Interference Error
|
A
| null |
1,208
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
For a sliding window of size n, there can be at most () sent but unacknowledged.
|
0
|
n-1
|
n
|
n/2
|
B
| null |
1,209
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
In the Ethernet binary exponential backoff algorithm, after 11 collisions, a station will choose a random number between 0 and ().
|
255
|
511
|
1023
|
2047
|
C
| null |
1,210
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The correct statement about Ethernet is ( ).
|
The physical topology of Ethernet is a bus structure.
|
Ethernet provides connectionless service with acknowledgments.
|
The Ethernet reference model generally includes only the Physical Layer and the Data Link Layer.
|
Ethernet must use the CSMA/CD protocol.
|
C
| null |
1,211
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Among the following Ethernet types, the one that uses twisted pair as the transmission medium is ( ).
|
10BASE-2
|
10BASE-5
|
10BASE-T
|
10BASE-F
|
C
| null |
1,212
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Each Ethernet card has its own clock. In order for the network cards to communicate with each other and know when one bit ends and the next begins, i.e., to have the same frequency, they have adopted ()
|
Quantitative mechanism
|
Manchester encoding
|
Parity Check Mechanism
|
Fixed Interval Mechanism
|
B
| null |
1,213
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
In Ethernet, the reason why a large amount of broadcast information can degrade the overall network performance is ( ).
|
Every computer in the network must send an acknowledgment for each broadcast message.
|
Every computer in the network must process each broadcast message.
|
Broadcast messages are automatically routed to each network segment by the router.
|
Broadcast information cannot be automatically transmitted directly to the destination computer.
|
B
| null |
1,214
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The IEEE 802.3 standard specifies that if coaxial cable is used as the transmission medium, the maximum length of the transmission medium without repeaters must not exceed ( ).
|
500m
|
200m
|
100m
|
50m
|
A
| null |
1,215
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Among the following four types of Ethernet, the one that can only operate in full-duplex mode is ( ).
|
10BASE-T Ethernet
|
100BASE-T Ethernet
|
Gigabit Ethernet
|
10 Gigabit Ethernet
|
D
| null |
1,216
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
In the Ethernet defined by the 802.3 standard, the layer that implements the "frame sequencing" function is ( ).
|
Physical Layer
|
Medium Access Control sublayer (MAC)
|
Logical Link Control sublayer (LLC)
|
Network Layer
|
B
| null |
1,217
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Among the following statements about Wide Area Network (WAN) and Local Area Network (LAN), the correct one is ( ).
|
Wide Area Networks are similar to the internet, capable of connecting different types of networks.
|
In the OSI reference model hierarchy, both WAN and LAN involve the physical layer, data link layer, and network layer.
|
From the perspective of the Internet, WANs and LANs are equal.
|
A LAN is essentially an Ethernet network, and its logical topology is a bus structure.
|
B
| null |
1,218
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The functions provided by the PPP protocol include ( ).
|
A framing method
|
Link Control Protocol (LCP)
|
Network Control Protocol (NCP)
|
A, B, and C are all
|
D
| null |
1,219
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
According to the values of the first two bits of the control field in the HDLC protocol, HDLC can be divided into three categories, which do not include ( ).
|
Information Frame
|
Supervisory Frame
|
Acknowledgment Frame
|
Unnumbered Frame
|
C
| null |
1,220
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The following network connecting devices that operate at the data link layer are ( )
|
Repeaters and Hubs
|
Hubs and Bridges
|
Network bridges and local area network switches
|
Hubs and Local Area Network (LAN) Switches
|
C
| null |
1,221
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
Different network devices have different data transmission delays. Among the following devices, the one with the largest transmission delay is ( ).
|
Local Area Network (LAN) switch
|
Bridge
|
Router
|
Hub
|
C
| null |
1,222
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The main functions implemented by the local area network switch are ()
|
Physical Layer and Data Link Layer
|
Data Link Layer and Network Layer
|
Physical layer and network layer
|
Data Link Layer and Application Layer
|
A
| null |
1,223
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The number of collision domains and broadcast domains for a 16-port hub are ( ).
|
16.1
|
16,16
|
1,1
|
1,16
|
C
| null |
1,224
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Knowledge
|
English
|
The number of collision domains and broadcast domains for a 16-port Ethernet switch is ().
|
1,1
|
16,16
|
1,16
|
16,1
|
D
| null |
1,225
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
To avoid frame loss during transmission, the method adopted by the data link layer is ().
|
Numbering Mechanism
|
Cyclic Redundancy Check (CRC)
|
Hamming code
|
Timer Timeout Retransmission
|
D
|
To prevent data loss during transmission, in reliable data link layer protocols, the sender sets a timer for each transmitted data frame. If the timer expires and the acknowledgment for that frame has not yet been received, the sender will retransmit the frame. To ensure that the receiver does not receive duplicate frames, each transmitted frame must be numbered; both Hamming codes and Cyclic Redundancy Check (CRC) codes are used for error control.
|
1,226
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
For networks where the channel is relatively reliable and there is a high demand for real-time performance, it is appropriate for the data link layer to use ( ).
|
Unacknowledged Connectionless Service
|
Confirmed connectionless service
|
Unacknowledged connection-oriented service
|
Confirmed connection-oriented service
|
A
|
Unacknowledged connectionless service refers to the scenario where the source machine sends independent frames to the destination machine, which does not acknowledge these frames. No logical connection is established beforehand, nor is it necessary to release a logical connection afterwards. If a frame is lost due to noise on the line, the data link layer will not detect such an occurrence, nor will it respond. This type of service is very suitable when the error rate is very low. In such cases, the recovery task can be left to the higher layers. This service is also very appropriate for real-time communication, as in real-time communication, the lateness of data is worse than data corruption.
|
1,227
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The following protocol in the parentheses ( ) is not a standard of the data link layer.
|
ICMP
|
HDLC
|
PPP
|
SLIP
|
A
|
The Internet Control Message Protocol (ICMP) is a network layer protocol, while PPP is developed based on SLIP and both are data link layer protocols.
|
1,228
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Assuming the physical channel has a transmission success rate of 95%, and on average one network layer packet requires 10 data link layer frames to be sent, if the data link layer uses an unacknowledged connectionless service, then the success rate of sending a network layer packet is ( ).
|
0.4
|
0.6
|
0.8
|
0.95
|
B
|
To successfully send a network layer packet, 10 data link layer frames need to be successfully transmitted. The probability of successfully transmitting 10 data link layer frames is (0.95)^10 ≈ 0.598, which means there is only about a 60% success rate. This conclusion demonstrates that the efficiency of unacknowledged services over unreliable channels is very low. To improve reliability, acknowledged services should be introduced.
|
1,229
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about cyclic redundancy check, ( ) is incorrect.
|
A polynomial code with r check bits can detect all burst errors of length less than or equal to r.
|
The two parties in communication can directly use polynomial coding without prior agreement.
|
CRC checks can be performed using hardware.
|
There are some special polynomials that have become international standards due to their excellent properties.
|
B
|
When using polynomial coding, the sender and receiver must agree on a generator polynomial in advance. The sender obtains the checksum by modulo-2 division and appends it to the data when sending. Upon receiving the data, the receiver also needs to verify the correctness of the data according to the same generator polynomial.
|
1,230
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
From the perspective of sliding window, when the send window is 1 and the receive window is also 1, it is equivalent to the ARQ () mode.
|
Go-Back-N ARQ
|
Selective Repeat ARQ
|
Stop-and-Wait
|
Continuous ARQ
|
C
|
The working principle of the stop-and-wait protocol is: the sender must wait for an acknowledgment signal from the receiver after sending each frame before it can send the next frame; the receiver must send back an acknowledgment signal after receiving each frame to indicate readiness to receive the next frame. If the receiver does not send an acknowledgment signal, the sender must continue to wait.
|
1,231
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In the simple stop-and-wait protocol, when a loss occurs, the sender will wait indefinitely. The solution to this deadlock situation is ( ).
|
Error Checking
|
Frame Number
|
NAK mechanism
|
Timeout Mechanism
|
D
|
In the stop-and-wait protocol, the sender sets a timer after sending a frame and waits for an acknowledgment. If the acknowledgment is not received by the time the timer expires, the same frame is sent again to avoid falling into an indefinite wait.
|
1,232
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The data transmission rate of a channel is 4kb/s, and the one-way propagation delay is 30ms. To achieve a maximum channel utilization of 80% with the stop-and-wait protocol, the required minimum data frame length must be ( ).
|
160-bit
|
320-bit
|
560 bit
|
960-bit
|
D
|
The maximum channel utilization for the stop-and-wait protocol is (L/C)(L/C+2R) = L/(L+2RC) = L/(L+2x30msx4kb/s) = 80%, which yields L=960bit.
|
1,233
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The data link layer uses the Go-Back-N protocol for flow control and error control. The sender has already sent frames numbered 0 to 6. Upon timer expiration, only acknowledgments for frames 1, 3, and 5 have been received. The number of frames the sender needs to retransmit is ( ).
|
1
|
2
|
5
|
6
|
A
|
GBN typically employs cumulative acknowledgment, so upon receiving acknowledgment for packet 5, the receiver has acknowledged packets 1 to 5, hence the sender only needs to retransmit packet 6.
|
1,234
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
A data link layer protocol that uses the Selective Repeat protocol, with a 5-bit sequence number, can have a maximum receive window size of ( ).
|
15
|
16
|
31
|
32
|
B
|
In the Selective Repeat protocol, if n bits are used for numbering, to prevent the new window from overlapping with the old window after the receiver moves the window forward, the maximum size of the receive window should not exceed half of the sequence number range, that is, W_R <= 2^(n-1). Therefore, choose B.
|
1,235
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
For the sliding window protocol with unordered reception, if the number of bits for the sequence number is n, then the maximum size of the send window is ()
|
\(2^n-1\)
|
2n
|
2n-1
|
2^(n-1)
|
D
|
The question does not directly state that the Selective Repeat protocol is used, but it is implied indirectly. The mention of an out-of-order receiving sliding window protocol indicates that the receiving window is larger than 1, which leads to the conclusion that the Data Link Layer is using the Selective Repeat protocol. The maximum size of the sending window for the Selective Repeat protocol is 2^(n-1). Therefore, the correct answer is D.
|
1,236
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The channel multiplexing technique that divides the total bandwidth of the physical channel into several subchannels, with each subchannel transmitting one signal, is ( ).
|
Code Division Multiplexing
|
Frequency Division Multiplexing
|
Time Division Multiplexing
|
Space Division Multiplexing (SDM)
|
B
|
When the available bandwidth of a physical channel exceeds the bandwidth required for a single original signal, the total bandwidth of the physical channel can be divided into several subchannels with the same (or slightly wider) bandwidth as that required for transmitting a single signal. Each subchannel carries a different signal, and this is known as Frequency Division Multiplexing (FDM).
|
1,237
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following multiplexing techniques, the one with the capability of dynamically allocating time slots is ().
|
Synchronous Time Division Multiplexing
|
Statistical Time Division Multiplexing
|
Frequency Division Multiplexing (FDM)
|
Code Division Multiplexing
|
B
|
Asynchronous Time Division Multiplexing (also known as Statistical Time Division Multiplexing) is a dynamic time-division multiplexing technique that dynamically allocates time slots.
|
1,238
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In the following protocols, the one that will not experience collisions is ( )
|
TDM
|
ALOHA
|
CSMA
|
CSMA/CD
|
A
|
TDM is a method of static channel division, where nodes use the channel in a time-divided manner, avoiding collisions. In contrast, ALOHA, CSMA, and CSMA/CD are dynamic random access protocols that employ collision detection strategies to deal with collisions, and therefore, collisions may occur.
|
1,239
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following CSMA protocols, the ( ) protocol may still not transmit even when the medium is detected as idle.
|
1 - Adhere to CSMA
|
Non-persistent CSMA
|
p-persistent CSMA
|
None of the above
|
C
|
The p-persistent CSMA protocol is a compromise between the 1-persistent CSMA protocol and the non-persistent CSMA protocol. In the p-persistent CSMA protocol, after detecting that the channel is idle, data is sent with a probability p and not sent with a probability 1-p, with the aim of reducing the probability of collisions that occur when multiple nodes detect the channel as idle and transmit simultaneously in the 1-persistent CSMA.
|
1,240
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In the non-persistent CSMA protocol, when the medium is busy, it ( ) until the medium is free.
|
Delay for a fixed time unit before listening.
|
Continue listening
|
Delay for a random time unit before listening again.
|
Stop listening
|
C
|
In the non-persistent CSMA protocol, a station listens to the channel before sending data. If the channel is busy, it stops listening and waits for a random period of time before listening again. If the channel is idle, it proceeds to send data.
|
1,241
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In the non-persistent CSMA protocol, when a station detects that the bus medium is idle, it ( ).
|
Transmit with a probability p
|
Immediate transmission
|
Transmit with a probability of 1-p
|
Transmit with a probability p after a delay of one time unit.
|
B
|
In the non-persistent CSMA protocol, when a station detects that the bus medium is idle, it will transmit data immediately.
|
1,242
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In Ethernet, when the data transmission rate increases, the slot time correspondingly decreases, which may affect collision detection. To effectively detect collisions, possible solutions include ( ).
|
Reduce the length of the cable medium or reduce the minimum frame size.
|
Reduce the length of the cable medium or increase the minimum frame size.
|
Increase the length of the cable medium or reduce the minimum frame size.
|
Increase the length of the cable medium or increase the minimum frame size.
|
B
|
The minimum frame length is equal to the number of bits sent during the contention period. Therefore, when the transmission rate is increased, the length of the cable medium can be reduced (to decrease the contention period time, i.e., to reduce the end-to-end delay of Ethernet), or the minimum frame length can be increased.
|
1,243
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
For a CSMA/CD Ethernet with a length of 10km and a data transmission rate of 10Mb/s, with a signal propagation rate of 200m/us, the minimum frame size is ( ).
|
20-bit
|
200 bit
|
100 bit
|
1000 bit
|
D
|
Round trip distance = 10000x2m, Round Trip Time RTT = 10000x2 / (200x10^6) = 10 μs, Minimum length = Bit width x RTT = 1000 bit
|
1,244
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In Ethernet, if a medium access collision occurs, the time for the next retransmission is determined by the binary exponential backoff algorithm. The reason for using the binary backoff algorithm is ( ).
|
This algorithm is simple.
|
This algorithm executes quickly.
|
This algorithm takes into account the impact of network load on collisions.
|
This algorithm is independent of the network size.
|
C
|
Ethernet uses CSMA/CD technology, and the greater the traffic on the network during load, the higher the probability of collisions. The binary exponential backoff algorithm can dynamically adapt to the number of sending stations, with the backoff delay value range forming a binary exponential relationship with the number of retransmissions n, thereby taking into account the impact of load on collisions.
|
1,245
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In Ethernet, the binary exponential backoff algorithm is used to handle collision issues. The data frame with the lowest probability of collision upon retransmission is ( ).
|
The frame retransmitted for the first time
|
Frames experiencing two collisions
|
Frame that has been retransmitted three times
|
Frame with four retransmissions
|
D
|
According to the IEEE 802.3 standard, Ethernet employs a binary exponential backoff algorithm in the event of a collision, where the wait time before retransmission is chosen randomly from an interval between 0 and 2^n-1. Consequently, a frame that has been retransmitted four times has the lowest probability of experiencing another collision.
|
1,246
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following descriptions of token ring networks, the incorrect one is ( )
|
Token ring networks are conflict-free.
|
At the same moment, only one piece of data is being transmitted on the ring.
|
All nodes on the internet share network bandwidth.
|
The time it takes for data to travel from one node to another can be calculated.
|
A
|
Token ring networks do not experience collisions because only the host that has obtained the token is allowed to transmit data.
|
1,247
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about token ring networks, the incorrect one is ( )
|
The utilization of media is relatively fair.
|
Under heavy load, the channel utilization is high.
|
Nodes can continuously hold the token until the data transmission is complete.
|
A token refers to a special type of control frame.
|
C
|
In a token ring network, nodes cannot hold the token indefinitely; instead, they obtain the right to send data within a certain period of time.
|
1,248
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In a token ring network, when all stations have data frames to send, the time a station waits to obtain the token and send data frames in the worst-case scenario is equal to ( ).
|
The total time for all sites to transmit tokens
|
The total time for all sites to transmit tokens and send frames.
|
Half of the total time for all sites to transmit tokens
|
Half of the total time for all sites to transmit tokens and send frames.
|
B
|
In the worst-case scenario, all stations need to send data, and the waiting time for a station to obtain the token is equal to the sum of the time taken for all stations to transmit the token and send frames.
|
1,249
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
On a broadcast channel with 3 stations A, B, and C, the medium access control uses channel division method, and the channel is divided using code division multiplexing technology. A and B want to send data to C. Assuming A's code sequence is +1, -1, -1, +1, +1, +1, -1, -1, the available code chip sequences for station B 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
|
D
|
The chip sequence of site B must be orthogonal to the chip sequence of site A, such that their inner product is 0. Option D is the only choice that meets this requirement.
|
1,250
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
If a 10BASE-T Ethernet network with a coverage of 200m is designed using Category 5 UTP, the required equipment is ( ).
|
amplifier
|
Repeater
|
Network Bridge
|
Router
|
B
|
The transmission distance limit for 10BASE-T Ethernet over twisted pair cabling is 100 meters; to cover 200 meters, a repeater is required.
|
1,251
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The following description of Ethernet addresses is incorrect: ()
|
The Ethernet address is commonly referred to as the MAC address.
|
MAC address is also known as the local area network (LAN) hardware address.
|
The MAC address is obtained through domain name resolution.
|
Ethernet addresses are typically stored in the network card.
|
C
|
The MAC address is obtained through the ARP protocol and is unrelated to domain name resolution.
|
1,252
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
When two devices on the same local area network have the same static MAC address, a ( ) will occur.
|
The device that is first initiated exclusively uses this address; a second device cannot communicate.
|
The last guided device exclusively uses the address, and another device cannot communicate.
|
The two devices on the network are unable to communicate properly.
|
Both devices are capable of communication because they can read the entire content of the packets, knowing which packets are intended for them and not for other stations.
|
C
|
When two devices in the same local area network have the same static MAC address, they are both unable to communicate properly, as other devices cannot distinguish between the two.
|
1,253
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Fast Ethernet still uses the CSMA/CD protocol, which adopts ( ) to reduce the maximum electrical length to 100m, thereby increasing the data transmission rate of Ethernet to 100Mb/s.
|
Change the shortest inclination length
|
Change the Maximum Frame Length
|
Maintain the minimum frame length unchanged
|
Maintain the maximum frame length unchanged
|
C
|
Fast Ethernet increases the data transmission rate by reducing the maximum electrical length to 100m while maintaining the minimum frame length unchanged.
|
1,254
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The reason why wireless local area networks (WLANs) use CSMA/CA instead of CSMA/CD is that in wireless local area networks ( ).
|
Unable to transmit and receive simultaneously, cannot receive signals while transmitting.
|
No need for collision detection during transmission.
|
The broadcast nature of wireless signals ensures that there will be no conflicts.
|
The coverage is very small; omitting conflict detection does not affect correctness.
|
B
|
A wireless local area network (WLAN) cannot simply use the CSMA/CD protocol, particularly the collision detection part, for the following reasons: Firstly, on a WLAN adapter, the strength of the received signal is often much lower than the strength of the transmitted signal. Therefore, implementing collision detection would incur excessive hardware costs. Secondly, in a WLAN, not all stations can hear each other, leading to hidden and exposed station problems. The requirement that 'all stations can hear each other' is a fundamental basis for implementing the CSMA/CD protocol.
|
1,255
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
There are various methods to segment a virtual local area network (VLAN), () is not a correct method of segmentation.
|
Based on switch port segmentation
|
Based on NIC (Network Interface Card) Address Segmentation
|
Based on username partitioning
|
Based on network layer address division
|
C
|
There are generally three methods to segment VLANs. 1. Based on ports, several ports of a switch are grouped into a logical set; this method is the simplest and most effective. If a host leaves its original port, it may enter a new subnet. 2. Based on MAC addresses, hosts are divided into a subnet according to their MAC addresses; when a host physically moves from one switch to another, it still belongs to the original subnet. 3. Based on network layer addresses, VLANs are segmented according to network layer addresses or protocols, allowing VLANs to extend across routers and connect hosts from multiple local area networks.
|
1,256
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The following protocol that does not belong to the TCP/IP protocol suite is ().
|
ICMP
|
TCP
|
FTP
|
HDLC
|
D
|
The TCP/IP protocol suite mainly includes TCP, IP, ICMP, IGMP, ARP, RARP, UDP, DNS, FTP, HTTP, etc. HDLC is a bit-oriented data link layer protocol proposed by ISO, and it is not part of the TCP/IP protocol family.
|
1,257
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The function of LCP in the PPP protocol is ().
|
Negotiating Data Link Protocol options during the establishment phase
|
Configure network layer protocols
|
Check for errors in the data link layer and notify of error information.
|
Security control, ensuring the data security of both parties in communication.
|
A
|
In the default configuration of the PPP protocol, the address and control fields are always constants, so LCP provides the necessary mechanism to allow both parties to negotiate an option. During the establishment phase, LCP negotiates options within the data link protocol, but it is not concerned with the options themselves, only providing a mechanism for negotiation.
|
1,258
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about the PPP protocol and the HDLC protocol, the correct one is ( )
|
The PPP protocol is a network layer protocol, while the HDLC protocol is a data link layer protocol.
|
The PPP protocol supports half-duplex or full-duplex communication.
|
The network layer at both ends of the PPP protocol must run the same network layer protocol.
|
The PPP protocol is a byte-oriented protocol, while the HDLC protocol is a bit-oriented protocol.
|
D
|
PPP and HDLC protocols are both data link layer protocols, so option A is incorrect. HDLC protocol is a bit-oriented data link layer protocol. Based on the characteristics of PPP, options B and C are incorrect, while option D is correct.
|
1,259
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
Among the following descriptions of data link layer devices, the incorrect one is ( ).
|
A network bridge divides the network into multiple segments, and a failure in one segment will not affect the operation of another segment.
|
Bridges can interconnect different physical layers, different MAC sublayers, and Ethernet networks with varying speeds.
|
The bandwidth occupied by each port node of the switch will not decrease with the increase in the number of port nodes, and the total bandwidth of the entire switch will increase as the number of port nodes increases.
|
Using a switch, Virtual Local Area Networks (VLANs) can be implemented, which can isolate collision domains but cannot isolate broadcast domains.
|
D
|
The advantage of a switch is that the bandwidth occupied by each port node will not decrease with the increase in the number of port nodes, and the total bandwidth of the switch will increase as the number of port nodes increases. Additionally, switches can be used to implement Virtual Local Area Networks (VLANs), which can isolate not only collision domains but also broadcast domains. Therefore, option C is correct, and option D is incorrect.
|
1,260
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The following () is not a benefit of segmenting networks using bridges.
|
Reduce the scope of the collision domain.
|
Under certain conditions, the bandwidth of the network was increased.
|
Filtering data between network segments
|
Reduced the scope of the broadcast domain.
|
D
|
A network bridge can isolate information, dividing the network into multiple segments, creating secure segments to prevent unauthorized access from users in other segments. Due to the segmentation of the network, each segment operates relatively independently, and a failure in one segment will not affect the operation of another segment. Therefore, options B and C are correct. Based on the characteristics of network bridges, option A is correct, and option D is incorrect.
|
1,261
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
The reason why switches provide better network performance than hubs is ( ).
|
The switch supports multiple pairs of users communicating simultaneously.
|
Switches use error control to reduce the error rate.
|
Switches expand the network coverage area.
|
The switch requires no configuration and is more convenient to use.
|
A
|
Switches can isolate collision domains and operate in full-duplex mode, allowing multiple pairs of nodes to communicate simultaneously in the network, thus enhancing network utilization. This is an advantage of switches.
|
1,262
|
Test
|
Computer Network
|
Data Link Layer
|
Multiple-choice
|
Reasoning
|
English
|
In Ethernet A, 80% of the traffic is within the local area network, and the remaining 20% is between the local area network and the Internet. Conversely, in local area network B, one uses a hub, and the other uses a switch. The switch should be placed in the local area network ().
|
Ethernet A
|
Ethernet B
|
Any Ethernet
|
None are suitable.
|
A
|
Switches can isolate collision domains, while the network segments connected by a hub belong to a single collision domain. If a hub is used in Ethernet A, which has a high volume of traffic, the collision domain will be enlarged, reducing the overall network efficiency. Therefore, it is necessary to use switches to isolate the communication between different network segments.
|
1,263
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the function of a subnet mask? ()
|
Distinguish between network and host parts
|
Encrypted network traffic
|
Access Control
|
Limited Bandwidth
|
A
| null |
1,264
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
What are the characteristics of the OSPF protocol?
|
Link State Protocol based
|
Distance Vector-based Protocol
|
UDP-based protocol
|
Protocol Based on MAC Address
|
A
| null |
1,265
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the function of the ICMP protocol? ()
|
Provide reliable data transmission
|
Sending control information in the network
|
Routing Selection
|
Perform data encryption
|
B
| null |
1,266
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
What type of service does the IP protocol provide? ()
|
Reliable Service
|
Connectionless Service
|
Connection-oriented service
|
Unreliable Service
|
B
| null |
1,267
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the function of TTL in the IP protocol? ()
|
Limit the size of data packets
|
Limit the Time to Live (TTL) of data packets
|
Limit the transmission rate of data packets
|
Limiting the priority of data packets
|
B
| null |
1,268
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
Why is it necessary to subnet a network? ()
|
Improve network performance
|
Reduces network traffic, improves network performance, and facilitates network management.
|
Enhancing Network Reliability
|
Increase network bandwidth
|
B
| null |
1,269
|
Test
|
Computer Network
|
Network Layer
|
Multiple-choice
|
Knowledge
|
English
|
Given an IP address of 192.168.2.50 with a subnet mask of 255.255.255.192, what is the range of its subnet? ()
|
192.168.2.0 - 192.168.2.63
|
192.168.2.0 - 192.168.2.127
|
192.168.2.0 - 192.168.2.255
|
192.168.2.64 - 192.168.2.127
|
B
| null |
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